McQuiston HVAC Analysis Design 6th Solutions

SOLUTION MANUAL

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Requests for permission or further information should be addressed to the Permission Department, John Wiley & Sons, Inc, 111 River Street, Hoboken, NJ 07030.

Chapter 1

1-1

(a) 98 Btu/(hr-ft-F) x1.7307 = 170 W/(m-K) (b) 0.24 Btu/(Ibm-F) x4186.8=1.0 kJ/kg-K (c)

0.04 Ibm/(ft-hr) µNs x1.488 = 16.5 3600 sec/hr m2

(d) 1050

Btu J 1 2.20462 Ibm MJ x x = 2.44 Ibm 9.48x10−4 Btu kg kg

(e) 12,000

(f) 14.7

1-2

Ibf in2

Btu 1 x = 3.52 kW Ibm 3.412 x 6894.76 = 101 kPa

(a) 120 kPa x

(b) 100

(c) 0.8

lbf / in2 = 17.4 lbf/in2 6.89476kPa

W x 0.5778 = 57.8 Btu/hr-ft-F m −K

W 2

m −K

x 0.1761 = 0.14 Btu/hr-ft2-F

(d) 10-6 N-s/m2 x

1 lbm = 6.7 x 10-7 1.488 ft − sec

(e) 1200 kW x 3412 = 4.1 x 10-6 Btu/hr

2

(f) 1000

1-3

1-4

1 Btu 1 kg Btu kJ x x = 430 1.055 kJ 2.2046 lbm lbm kg

Hp = 50 (ft) x 0.3048 (

m ) = 15.2 m ft

∆P =

15.2 m 9.807 N x ( ) x 1000 (kg/m3) = 149 kPa 1000 Pa/kPa 1 kg

∆P =

m kg 4 9.807 N (ft) x 0.3048 ( ) x ( ) x 1000 ( 3 ) ft 12 1 kg m

∆ P = 996 Pa ≈ 1.0 kPa 1-5 TOTAL BILL = ENERGY CHARGE + DEMAND CHARGE + METER CHARGE

( 96,000 )

kw - hrs ( 0.045 ) $ / kw − hr + ( 624 ) kw (11 − 50 ) $ / kw

+ $68 = $4,320 + $7,176 + $68 = $11,564

1-6

7 AM to 6 PM

(11)

11 hrs/day, 5 days/wk

hrs days (22) = 242 hrs / month day months

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

3

ratio =

1-7

( 624 ) kw = 1.57 ⎛ ( 96,000 ) kw − hr ⎞ ⎜ ⎟ 242 hr ( ) ⎝ ⎠

This is a trial and error solution since eq. 1-1 cannot be solved explicitly for i. Answer converges at just over 4.2% using eq. 1-1

1-8 Determine present worth of savings using eq. 1-1

⎡ $1000 ( ) ⎢1⎢⎣ P=

0.012 ⎞ ⎛ ⎜ 1+ ⎟ 12 ⎠ ⎝

−(12 )(12 ) ⎤

⎛ 0.012 ⎞ ⎜ ⎟ ⎝ 12 ⎠

⎥ ⎥⎦

P = $134,000

1-9

= VA = 2 x 3.08 x 10-3 = 6.16 x 10-3m3/s (a) Q = 6.16 x 10-3 x 998 = 6.15 kg/s = ρQ m (b) A=

π 4

(0.3)2 = 7.07 x 10-2 m2

= 7.07x10-2 x 4 = 0.283 m3 / s; ρ = 1.255 kq/m3 Q = 1.225 x 0.283 = 0.347 kg/s m

1-10

V = 3x10x20 = 600m3


4

= 600 x 1 x 1 = 4.17 x 10-2 m3/s Q i 4 3600

1-11

p ∆T q = mc

c p = 4.183 kJ/(kg-K)

ρ = 983.2 kg/m3

1-11 (cont’d) 3 q = (1) m ( 983.2 ) kg ( 4.183 ) kJ ( 5 )c = 20,564 kJ s kg − K s m3

q = 20,564 kw

1-12

q wat = −q air 11,200(1)(10) =

=

5000x60x14.7x144x0.24(t 2 − 50) (53.35x510)

11,200 = 5601.5 (t2-50); t2 = (11,200/5601.5) + 50 = 70 F 1-13 Diagram as in 1-12 above.

q wat = - q air 1.5 (4186)(90-t2) = 2.4 (1.225)(1.0)(30-20)(1000) 6279(90-t2) = 29,400


5 t2 = 90 -

29,400 = 85.3 C 6279

1-14 q = hA(ts- t ∞ ) A= π (1/12) x 10 = 2.618 ft2

t s = t ≈ 212 F sur q = 10x2.618x(212-50) = 4241 Btu/hr 1-15

A= π x 0.25x4 = 3.14 16 m2

q = hA(ts- t ∞ ) q 1250 = ; h = 4.42 W/(m2 – C) h= A(t s -t ∞ ) 3.1416(100 − 10)

xρ =Q p (t2-t1) ; m 1-16 q = mc

ρ = P/RT = 14.7x144/53.35(76+460) ρ = 0.074 lbm/ft3 = 5000x0.074x60 = 22,208 lbm/hr m

c p = 0.24 Btu/lbm-F q = 22,208x0.24(58-76) = -95,939 Btu/hr Negative sign indicates cooling

1cp (t3-t1) + 1-17 m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

6

2cp2 (t3-t2) = 0 m c p1 = c p2 t3 =

1t1 + (m 1+ (m

2t 2 ) m 2) m

ρ = 1000x 1 = Q m 2 1

14.7x144 = 73.5 lbm/min 53.35(460 + 50)

1-17 (cont’d)

ρ = 600x 2 =Q m 2 2 t3 =

14.7x144 = 46.7 lbm/min 53.35(460 + 50)

(73.5x80) + (46.7 x 50) = 68.3 F (73.5 + 46.7)


7 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Requests for permission or further information should be addressed to the Permission Department, John Wiley & Sons, Inc, 111 River Street, Hoboken, NJ 07030.

Chapter 2 2-1 through 2-20 Solutions are not furnished since many acceptable responses exist for each problem. It is not expected that the beginning student can handle these questions easily. The objective is to make the student think about the complete design problem and the various functions of the system. These problems are also intended for use in class discussions to enlarge the text material.


Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Requests for permission or further information should be addressed to the Permission Department, John Wiley & Sons, Inc, 111 River Street, Hoboken, NJ 07030.

Chapter 3 3-1

(a) Pv = φ r Ps = 0.45(3.17)kPa = 1.43 kPa or 0.45(0.435) = 0.196 psia Pa = 101 – 1.43 = 99.57 kPa or 14.696-0.196 = 14.5 psia (b)

P 1430 Pv = 0.0104 kg/m3 = RvT or ρv = v ; ρv = Rv T 462.5(297) ρv 0.196(144) = 0.00062 lbv/ft3 85.78(535)

or

(c) W =

or

3-2

0.6219 (1.43) = 0.00893 kgv/kga (99.57)

0.6219(0.196) = 0.00854 lbv/lba 14.5

(a) English Units – t = 80F; P = 14.696 psia; Pv = 0.507 psia Table A-1a W = 0.6219

Pv 0.6219 (0.507) = = 0.0222 lbv/lba Pa (14.696 − 0.507)

i = 0.24t + W(1062.2 + 0.444t) i = 0.24 (80) + 0.0222[1061.2 + 0.444(80)] = 43.55 Btu/lbm

8

Ra T 53.35(460 + 80) = = 13.61 ft3/lbm Pa (14.696 − 0.507)144

v=

(b) English Units – 32F, 14.696 psia Pv = 0.089 psia (Table A-1) 3-2 (cont’d) W=

0.6219(0.089) lbmv = 0.00379 (14.696 − 0.089) lbma

i = 0.24(32) + 0.00379 [1061.2 + 0.444(32)] = 11.76 Btu/lbma

53.35(492) = 12.48 ft3/lbma (14.696 − 0.089)144

v= 3-2

(a) SI Units – 27C; 101.325 kPa Pv = 3.60 kPa, Table A-1b W = 0.6219

Pv 0.6219(3.6) kgv = = 0.0229 Pa (101.325 − 3.6) kga

i = 1.0t + W(2501.3 + 1.86t) kJ/kga i = 27 + 0.0229(2501.3 + 1.86 x 27) = 85.43 kJ / kga v=

Ra T 0.287(300) = =0.88 m3 /kga Pa (101.325 - 3.6)

(b) SI Units 0.0C; 101.325 kPa Pv = 0.61 kPa, Table A-1b W=

0.6219(0.61) =0.00377 kgv/kga (101.325 - 0.61)


9

i = 0.0 + 0.00377 (2501.3 – 1.86 x 0.0) = 9.42 kJ/kga

0.287(273) = 0.778 m3 /kga (101.325 - 0.61)

v=

3-3

(a) English Units – 5000 ft elevation, P = 12.24 psia = 24.92 in.Hg t = 80 F; Pv = 0.507 psia (Table A-1a) W = 0.6219

Pv 0.6219(0.507) = = 0.0269 lbv/lba Pa (12.24 - 0.507)

i = 0.24(80) + 0.0269 [1061.2 + 0.444(80)] = 48.7 Btu/lbma v=

RaT 53.35(540) = = 17.05 ft3 / lbma Pa (12.24 - 0.507) 144

(b) English Units – t = 32 F, Pv = 0.089 psia ( Table A-1a) W=

0.6219(0.089) = 0.00456 lbmv/lbma (12.24 − 0.089)

i = 0.24(32) + 0.00456 [1061.2 + 0.444(32)] =12.58 Btu/lbma v= 3-3

53.35(492) = 15.00 ft3/lbma (12.24 − 0.089)144

(a) SI Units -27 C, 1500 m elevation P = 99.436 + 1500(-0.01) = 84.436 kPa Pv = 3.60 kPa, Table A-1b


10 W=

0.6219x3.60 = 0.0277 kgv/kga (84.436 − 3.60)

i = 27 + 0.0277 (2501.3 + 1.86 x 27) = 97.68 kJ/kga

3-3 (cont’d) v=

0.287x300 (84.436 - 3.60)

= 1.065 m3 / kga

(b) SI Units – 0.0C; 1500m or 84.436 kPa Pv = 0.61 kPa; Table A-1b W=

0.6219 x 0.61 = 0.00453 kgv / kga (84.436 - 0.61)

i = 0.0 + 0.00453 (2501.3 – 0.0 x 1.86) = 11.33 kJ / kga v=

3-4 (a)

0.287 x 273 = 0.935 m3 / kga (84.436 - 0.61)

English Units – 70F, Pv = 0.363 psia

Pv = φ Pg = 0.75(0.363) = 0.272 psia W=

0.6219 (0.272) (14.696 - 0.272)

= 0.0117 lbmv / lbma

i = 0.24 (70) + 0.0117 [1061.2 + 0.444 (70)] = 29.58 Btu / lbma (b) Pv = 0.75 (0.363) = 0.272 psia; P = 12.24 psia


11 W=

0.6219 (0.272) = 0.0141 lbmv / lbma (12.24 - 0.272)

i = 0.24(70) + 0.0141 [1061.2 + 0.444 (70)] = 32.20 Btu/ lbma 3-4

SI Units – (a) 20C, 75% RH, Sea Level

3-4 (cont’d) Ps = 2.34 kPa; Pv = 0.75 x 2.34 = 1.755 kPa

W=

0.6219 x 1.755 = 0.0110 kgv / kga (101.325 - 1.755)

i = 1.0 t + W(2501.3 + 1.86t) i = 20 + 0.0110(2501.3 + 1.86 x 20) = 47.92 kJ / kga (b)

20C, 75% RH, 1525m P = 99.436 – 0.01 x 1525 = 84.186 kPa Ps = 2.34 KPa; Pv = 0.75 x 2.34 = 1.755 kPa W=

0.6219 x 1.755 = 0.0132 kgv / kga (84.186 - 1.755)

i = 20 + 0.0132(2501.3 + 1.86 x 20) = 53.51 kJ / kga 3-5 English Units – t = 72 Fdb; φ = 50 %; P = 14.696 psia

φ=

Pv or Pv = φPs ; Pv = 0.5(0.3918) = 0.196 psia Ps


12 Air dewpoint = saturated temp. at 0.196 psia = 52.6 F Moisture will condense because the glass temp. 40 F is below the dew point temp. 3-5

SI Units – t = 22C ; 50% ; P = 100 kPa Pv = φ Ps ; Pv = 0.5(2.34) = 1.17 kPa

3-5 (cont’d) Air dewpoint = sat.temp. at 1.17 kPa = 9.17 C Glass temp. of 4 C is below the dewpoint of 9.17 C, therefore, moisture will ccondense on the glass

3-6 English Units (a) At 55F, 80% RH, va = 13.12 ft3 / lba and ρ a = 0.0752 lbma / ft3

a = 5000 (0.0762) = 381 lbma / min = 22,860 lbma / hr m (b) Using PSYCH ρ a = 0.0610 lbma / ft3 or va = 16.4 ft3 / lba

a = 5000 (0.061) = 305 lbma / min m

= 18,300 lbma / hr

3-6 SI Units – (a) t = 13 C and relative humidity 80%

a = 2.36 / 0.82 = 2.88 kga / s then va ≈ 0.820 m3 / kga; m (b) Assuming same conditions

a = 2.36 / 0.985 = 2.40 kga / s v a = 0.985 m3 / kga ; m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

13

3-7 English Units – t = 80F, 60% RH (a) Pv = φ Ps = 0.6 (0.507) = 0.304 psia

t dp = (t sat @ Pv ) = 64.5 F (b) Same as (a) above 3-7 SI Units – (a) 27 C, 60% RH, Sea Level Ps = 3.57 kPa; Pv = 0.6 x 3.57 = 2.14 kPa

t dp =(t sat at Pv ) ≈ 18.4 C (b) Same as (a) above 3-8 t dp ≤ 9C (48F)

φ ≤ 42% ;

W ≤ 0.0071 kgv / kga (lbv / lba)

Chart 1a & 1b


14

ASHRAE PSYCHROMETRIC CHART NO.1 R

90

AMERICAN S OCIETY OF HEATING, REFRIGERATING AND AIR-CONDITIONING ENGINEERS, INC.

SEA LEVEL

1.0

60

55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

R

50

.028

60 1 .0

85 15

-2 00 0 -1 00 0

2. 0

0. 6

SENSIB LE HEAT TOTAL HEAT

5000 3000

0 .5

4 .0 8 .0

Qs Qt -2.

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

-8 -4 .0.0 0

45

85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

.026

.0

0 .8

-

50 0

80

.024

BU

LB

TE

MP E

55

RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENTHALPY HU MIDITY RATIO

'W

75

35

50

14 .5

.018

R

D

A

N

ER

U

M

P

PO

TE N TI O

.014

HUMIDITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

A R TU SA

25

E

% 90

ME

% 60

-C

60

.5

50

% 50

45

A IR

40 35

W=0.0071

13

45

15

Room

%

4 0%

40 1 3.

30 %

0

35

20 % 1 2. 10

3-9

48 (9)

15

72 (22)

20

110

105

100

95

90

85

80

75

70

65

60

55

50

45

40

35

5

Y I VE H UMIDIT 10% RE LAT

45

.012

.010

40

.008

.006

35 .004

.002

115

55

RY .D LB ER

50

T. P

dp

U.F

42 %

% 70

DR Y BUL B T EMPERAT UR E - °F

80

55

LU VO

20

1 4 .0

65

60

12 0

R PE U -B T Y AL P

70

65

N

TH

.016

TU

F

30

O

Problem 3-8

E

D

-°

R

F

Y

A

IR

75

70

25

ENTHALPY - BT U PER POUND OF DRY AIR

(a,b,d) Using the Properties option of PSYCH: Relative Humidity = 0.59 or 59% Enthalpy = 30.4 Btu/lbma Humidity Ratio = 0.0114 lbu/lba (c) Again using the Properties option At W=0.0114 lbv/lba; RH = 1.00 or 100% The dew point = tdb or twb = 59.9 F


30

15

3-9 (cont’d) (e) Using the Density of Dry Air option: Mass Density = 0.070 lba/ft3

3-10

Using program PSYCH (a) tdb = 102.6; twb = 81.1F 75 Fdb; 65 fwb; 14.2 psia

ν = 58.7 lbm/hr (b) m

2 = 1027 cfm Q 3-11

t1 = 80 / 67 F; t2 = 55 F and sat.; assume std. barometer (a) W1 – W2 = 0.0112 – 0.0092 = 0.002 lbv / lba (b) ql = 31.5 - 29.3 = 2.2 Btu / lba (c) qs = 29.3 – 23.2 = 6.1 Btu / lba (d) q = ql + qs = 8.3 Btu / lba

3-12

(a) W2* = W1 =

0.6219 (0.3095) = 0.0134 kgv / kga (14.696 − 0.3095) 0.24 (65 - 80) + ( 0.0134 x 1056.5) = 0.00993 lbv / lba (1096 - 33)

also W1 = 0.6219 Pv1 / (P – Pv1) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

16 Pv1 = (0.00993 x 14.696) / ( 0.6219 + 0.00993) = 0.231 psia 3-12 (cont’d)

0.231 = 0.46 or 46% 0.507

φ1 =

(b) P = 29.42 – (0.0009 x 5000) = 24.92 in.Hg. or P = 12.24 psia

W2* =

0.6219 x (0.3095) = 0.01613 lbv/lba (12.24 - 0.3095)

0.24(65 − 80) + (0.01613 x 1056.5) = 0.01265 lbv / lba ( 1096 - 33) or kgv / kga

W1 =

Pv1 = 0.01265 x 12.24 / ( 0.6219 + 0.01265) = 0.244 psia

φ1 = 3-13

0.244 = 0.48 or 48% 0.507

(a) Sea Level

Dry Bulb, F

Wet Bulb, F

85 75 74.6 88.6 100

60 59.6 65.1 70 85.8

Dew point F 40.6 49.2 60.1 60.9 81.7

Humid. Ratio, lba/lbv 0.0053 0.0074 0.0111 0.01143 0.0235

Mass Enthalpy Rel. Density Btu/lba Humid., % lba/ft3 26.6 21 0.072 26.1 40 0.073 30 60 0.073 33.8 40 0.071 50 56 0.068


17 (a) 5000 ft. Dry Bulb, F

Wet Bulb, F

85 75 71.2 102.7 100

60 58.6 61.6 70 81.3

Dew point F 45.1 49.2 56.7 55.8 76.1

Humid. Ratio, lba/lbv 0.0076 0.0089 0.0118 0.01143 0.0235

Mass Enthalpy Rel. Density Btu/lba Humid., % lba/ft3 28.7 25 0.060 27.7 40 0.061 30 60 0.061 37.3 22 0.058 50 47 0.057

(c) Note effect of barometric pressure.


18 3-14 ASHRAE PSYCHROMETRIC CHART NO.1 55


R

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.

SEA LEVEL

1.0

1 .0

60

R

50

.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6

SENSIB LE HEAT T OTAL HEAT

50 00

300 0

0 .5

4 .0 8 .0

Qs Qt

-2.

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

-8 -4 .0.0 0

-0 .5 - 0.4 -0.3

0. 3

-1 .0

.026

45 85

0

50 0

WE T

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENTHAL PY HU MIDITY RAT IO

'W

75

35

50 14 .5

.018

F

R ER

U

M

P

PO

TE N TI O

.014

HUMIDITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

A R TU SA

25

E

.0 V

% 90

Max RH=49.6 % ME

% 60

-C

U.F

60

13

50

40

% 50

45

35

IR YA

.5

45

15

W=0.0083

Room

%

4 0%

40 30%

1 3. 0

35

20 %

3-15

52 (11)

15

72 (22)

20

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

.010

40

.008

.006

35

.004


Use Chart 1b, SI

(i1 − i2 ) = (b) q = m

.012

25

(a) td = 10 C; SHF = 0.62 2.4 (57.1 - 34) = 63.95 kJ / s = 63.95 k W 0.867

q s = 63.95 (0.62) = 39.65 kW 3-15

45

.002

115

55

dp

R B. D RL PE

50

T.

% 70

DR Y BUL B T EMPERATUR E - °F

80

55

U OL

20

14

65

60

12 0

R PE U -B T Y AL P

70

65

N

TH

.016

A

N

D

O

30

TU

Problem 3-14

E

D

-°

R

F

Y

A

IR

75

70

Use Chart 1a, IP (a) td = 52 F; SHF = 0.63


30

19 3-15 (cont’d)

5000(60) (32 - 22.6)= 203,317. Btu/hr 13.87

q =

(b)

ASHRAE PSYCHROMETRIC CHART NO.1 55


90


SEA LEVEL

1.0

1 .0

60

R

50

.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6


50 00

300 0

0 .5

4 .0 8 .0

Qs Qt

-2.

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

-8 -4 .0.0 0

85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

.026

45

50 0

80

.024

BU

LB

TE MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENT HAL PY HU MIDIT Y RAT IO

'W

75

35

50 14 .5

.018

A

IR

75

R TU TI O A R


U -B T

TU

Y

25

1

SA

AL P

.014

E

%

L B. ER

55 %

DR

1 3 .5

4 0%

40 1 3.

30%

0

35

20 %

52 (10) 55 (13)

15

20

80 (27)

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

.012

.010

40

.008

.006

35

.004

.002

115

% 50


IR YA

50 45

35

T. P

2

45 40

U.F

% 70

60

15

60

-C

ADP

ME

80

LU VO

% 90

55

50

1 4 .0

65

60

20

45

65

N

TH

70

N

PE

TE

R

M

P

PO

ER

U

A

N

D

O

30

.016

70

E

F

D

-°

R

F

Y

Problem 3-15

12 0

R

30

25


q s = 203,317 (0.63) = 128,089. Btu/hr 3-16

(a) i1 = 30 Btu / lba; v1 = 13.78 ft3 / lba; W = 0.0103

lbv ; φ1 = 50% lba

(b) i1 = 51.6 kJ / kga v1 = 0.86 m3 / kga Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

20

3-16 (cont’d) W1 = 0.0103

kgv kga

φ1 = 50% 3-17

Use the Heat Transfer option of program PSYCH:

q = 148,239 Btu/hr q s = 102,235 Btu/hr SHF = 0.69

3-18 Use the Heat Transfer option of program PSYCH for sensible heat transfer only:

q s = − 178,911 Btu/hr Negative sign indicates heating.

3-19 Use the program PSYC to compute the various properties at 85/68 F; sea level and 6000 ft elevation. Elevation ft 0 6000

Enthalpy Btu/lbm 32.2 36.3

Rel. Hum percent 42 45

Hum. Ratio lbv/lba 0.0107 0.0144

Density lba/ft3 0.072 0.058

a = 5000 x 0.072 x 60 = 21,600 lba/hr At sea level: m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

21 3-19 (cont’d)

a = 5000 x 0.057 x 60 = 17,100 lba/hr At 6000 feet:: m Percent Decrease at 6000 ft:

PD = 3-20

(21,600 − 17,100)100 = 20.8% 21,600

Use the program PSYC to compute the heat transfer rates at 1000 and 6000 feet elevation: (a) at 1000 ft, q = 200,534 Btu/hr

= 190,224 Btu/hr (b) at 6000 ft, q (c) PD =

(200,534 − 190,224)100 = 5.1 % 200,543

3-21 (a) English Units – PB = 29.92 in.Hg. ; q = 0

∆i = iw = 180.2 + 0.8 (970.2) ∆W iw = 956.4 Btu / lbv From chart 1a; t2 = 91.5 F


22



90


SEA LEVEL

1.0

1 .0

60

R

50

.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6


50 00 300 0

0 .5

4 .0 8 .0

Qs Qt

-8 -4 .0.0 -2.0

0.4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

.026

45

50 0

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENT HAL PY HU MIDITY RATIO

'W

2

75

35

50 14 .5

F

R

O D

A

N

ER

U

M

P

PO

TE N

70

TI O

.014

A R TU SA

25

14

E

65

.0 V ME

% 60

-C U.F

55 60

.5

50

% 50

45

A IR

40 35

% 13

45

15

RY .D LB ER

50

T. P

% 70

1

4 0%

40 30%

1 3. 0

35

20 %

15

20

110

105

100

95

90

85

80

65

60

55

50

45

40

35

75

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

.012

.010

.006

35

.004

.002

25

91.5 (32) 98 (38)


3-21 (a)

SI Units – PB = 101.325 kPa

∆i = iw = 419.04 + (0.8 x 2257) ∆W iW = 2224.6 kJ / kg From chart 1b; t2 = 32 C (b) Use Humidification (adiabatic) option to obtain 91.5 F db.

3-22

40

.008

115

80

55

U OL

% 90

DR Y BULB T EMPERATUR E - °F

60

20

45

65 HUMIDITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

R PE U -B T Y AL P

.016

N

TH

60 %

TU

Problem 3-21

70

E

D

-°

R

F

Y

A

IR

75

30

.018

12 0

R

PB = 29.92 in.Hg.; q = 0 (a) Using chart 1a


30

23

3-22 (cont’d)

∆i = iw = 1090 Btu / lbm ∆W

From table A-1

i-if 1090 - 196.1 = ifg 960.1

x=

x = 0.931 or about 93 % (b)

x will be the same


90

AMERICAN S OCIETY OF HEATING , REFRIGERATING AND AIR-CONDITIO NING ENGINEERS, INC.

SEA LEVEL

1.0

1 .0

60

55


R

50

.028

60

85 15

-2 00 0 -1 00 0

2.0

0. 6


50 00 300 0

0 .5

0. 4

0 200

0 .1

15 00

-0 .1

-0 .2

0.2

.026

45

85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

4 .0 8 .0 -8 -4 .0.0 -2. 0

Qs Qt

.0

0 .8

-

50 0

80

.024

BU

LB

TE

MP E

a

55

RA TU

40

1090

RE

- °F

.022

1 000

80

.020

'h

ENT HAL PY HU MIDIT Y RATIO

'W

75

35

50 14 .5

.018

R A

N

ER

U

M

P

PO

TE N

70

TI O

.014

HUMI DITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

A R TU SA

25

E

% 90

ME

%

60

-C U.F

55

IR

% 50

YA

1

.5

50

45

DR

13

45 40

35

%

4 0%

40 1 3.

30%

0

35

20 %

15

80

20

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5 10

Y I VE H UM IDIT 70

1 2.

10% RE LAT

.012

.010

40

.008

.006

35 .004

.002

115

60 15

L B. ER

50

T. P

% 70


80

55

LU VO

20

65

60

1 4 .0

60

45

65

12 0

R PE U -B T Y AL P

b

N

TH

.016

TU

O

30

D

Problem 3-22

E

F

D

-°

R

F

Y

A

IR

75

70

25



30

24

3-23 Assume PB = 101.325 kPa; q = 0

∆i 272.1 = iw = kJ / kg ∆W 1000 iw = 0.272 (on scale) t2 = 22.6 C ASHRAE PSYCHROMETRIC CHART NO.1 11 0

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 101.325 kPa Copyright 1992

R

30

SEA LEVEL

10 0

0.7

0. 6


10 .0

0.5

1 .0

1 .5 2. 0

-5. 0

4 .0

Qs Qt

0.0

80

1 .0

EM

PE R

24

AT

UR

E-

°C

11 0 22

0 .9

0

5.0

4. 0

0 .1

WE TB UL BT

-0.2

0.2

30

0.272

-4.0 -2 .0

-0 .5

0. 3

-1 .0

12 0 26

90

-2.0

0. 4

4

0 .8

-

28

30 0 .9

1.0

0

R


12

2

2 .0

3 .0

2.5

70

25 20

'h

ENTHALPY HU MIDITY RAT IO

'W

10 0 25

18

C

R

-°

D

16

R

E

O F

TU

20

A EM

P

ER

20

A R TU SA

HUMI DITY RATIO - GRAM S M OISTURE PER KI LOGR AM D RY AIR

TI O

P J -K LP Y A E

N

20

8

80 %

0 .8

40

TH

14

N

E

T

R

90

2

90

15

%

% 80 0 .8 6V

15

%

U OL -C UB IC M

5

0%

4 0. 8

10

ET

5

ER

0 .8

5

2

3 0%

YA IR

20

22.6

30

38

8

70

6

4

60

40

ENTHALPY - KJ PER KILOGRAM OF DRY AIR

For adia. humidification (a)

80 10

2

40

20

15

10

5

8

10

HU MID ITY 25

0

0 .7

IVE 10% RE LAT

35

0 .8

20 %

30

10

R gD Rk PE

40 %

3-24

1

ME

% 60

20

45

70

10

D RY BU LB TEMPERA TU RE - °C

30

12

50

50

K

IL O G

R

AM

Problem 3-23

0

Y

0 .9

AI R

60

∆i = iw = 1131 Btu / lbw ∆W


50

25

3-24 (cont’d)

a (i2 - i1) q c = m

a = 2000 x 60 / 13.14 m a = 9132 lba / hr m

i1 = 18.1 Btu / lba ; i2 = 29.7 Btu / hr q c = 9132 (29.7 - 18.1) = 105,931 Btu / hr w = m a (W3 - W2 ) ; W3 = 0.0167; W2 = 0.0032 lbv/lba m

w = 9132 (0.01 67 - 0.0032) = 123.3 lbw / hr m


26



R

90


SEA LEVEL

1.0

1 .0

60

R

50

.028

60

85 1 5 .0

0 .8 -200 0 -1 00 0

2.0

-

0. 6


50 00 300 0

0 .5

4 .0 8 .0

Qs Qt

-4 .0.0 -2.0

0.4

0 200

0 .1

0.2

- 0.4 -0.3

-0 .1

-0 .2

15 00

0

1131

85

0

WE T

0

-0 .5

0. 3

-1.

.026

45

-8

50 0

.024

BU

80

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENTHALPY HU MIDIT Y RAT IO

'W

75

35

50 14 .5

.018

3

E

F

D

-°

R

F

Y

A

IR

75

R TU ER M

P

PO

TE N TI O

.014


A R TU SA

25

E

% 90

ME

% 60

U.F

4 0%

40

1

1 3.

30%

0

35

20 %

15

60 (16)

20

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

Y VE H UM IDIT 70

1 2.

10% REL ATI

10

45

25

.012

.010

40

.008

.006

35

.004

2 .002

115

% 50

DR Y BULB TEMPERATUR E - °F

IR YA

50

45

35

% 1 3 .5

45 40

DR L B. ER

55 60

15

T. P

% 70

50

30 %

-C

80

55

LU VO

20

1 4 .0

65

60

12 0

R PE U -B T Y AL P

70

65

N

TH

.016

A

N

D

O

30

U

Problem 3-24

70

110 (43)


(b) Solution similar to (a)

3-25 English Units – See diagram for construction on chart 1a.

32 2000 Q 2 = = 1= 12 3000 Q3 3 Layout 2L/3 on the chart and read: W3 = 0.007 lbv/lba Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

30

27 I3 = 22.2 Btu/lba ASHRAE PSYCHROMETRIC CHART NO.1 55


R

90


SEA LEVEL

1.0

1 .0

60

R

50

.028

60

85 1 5 .0

0 .8 -200 0 -1 00 0

2.0

-

0. 6


50 00 300 0

0 .5

4 .0 8 .0

Qs Qt

-4 .0.0 -2.0

0.4

0 200

0 .1

0.2

- 0.4 -0.3

-0 .1

-0 .2

15 00

0

85

0

WE T

0

-0 .5

0. 3

-1.

.026

45

-8

50 0

80

.024

BU

LB

TE

MP E

55 RA TU

40

1 000

77

RE

- °F

.022

80

.020

'h


'W

75

35

50 14 .5

.018

R A ER

U

M

P

PO

TE N

70

TI O

.014


A R TU SA

25

E

% 90

U.F T. P DR Y BULB TEMPERATUR E - °F

IR YA

4 0%

40 30%

1 3. 0

35

20 %

58.4 (15)

20

110

105

100

95

90

85

80

75

65

15

40 (4)

SI Units –

60

55

50

45

40

35

5

10

Y VE H UM IDIT 70

1 2.

10% REL ATI

.012

.010

.008

.006

35

.004

25

100 (38)


Same procedure as above, read:

i3 = 34 kJ / kga W3 = 0.007 kgv / kga 3-26

40

.002

115

%

% 50

1

3-25

DR L B. ER

50

45

35

-C

60

1 3 .5

45

35

60

55

3

40

%

% 70

50

15

ME

80

55

LU VO

52

1 4 .0

65

60

20

45

65

12 0

R PE U -B T Y AL P

2

N

TH

.016

TU

O N

D

Problem 3-25

70

E

F

D

-°

R

F

Y

A

IR

75

30

English Units – Layout the given data on Chart 1a as shown for problem 3-25.

a1 = 2000(60) 12.66 = 9,479lba hr m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

30

28 3-26 (cont’d)

a2 = 1000(60) 14.44 = 4,155lba hr m a1 m 32 9479 = = = 0.695 a1+m a2 9479 + 4155 12 m

Layout distance 32 on line from 1 to 2 to locate point 3 for the mixture. Read: i3 = 21.5 Btu/lbm W3 = 0.0067 lbu/lba For W, % Error =

For I, % Error =

3-27 SHF =

(0.007 − 0.0067)100 = 4.5 0.0067

(22.2 − 21.5)100 = 3.3 21.5

250,000 = 0.8 200,000

or SHF =

59 = .81 73


29



R

R

90


SEA LEVEL

1.0

50

.028

60

85 1 5 .0

0.8

1 .0

60

0 .8 -2 00 0 -1 00 0

2. 0

-

0. 6


50 00

300 0

0 .5

4 .0 8 .0

Qs Qt

-2.

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

-8 -4 .0.0 0

-0 .5 4 - 0. -0.3

0. 3

-1. 0

.026

45 85

0

50 0

WE T

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h


'W

75

35

50 14 .5

.018

E

F

D

-°

R

F

Y

A

IR

75

R TU TE

M

P

28.2

N


R TU SA

25

E

ME

% 60

-C U .F

1

60

50

% 50

45

35

IR YA

1 3 .5

45 40

%

4 0%

40 30%

1 3. 0

35

20 %

3-28

53 (12)

15

75 (24)

20

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

45

.012

.010

.008

.006

35

.004

25

ENTHALPY - BT U PER POUND O F DRY AIR

Refer to diagram for 3-27

a (i1 - i2 ); i1 = 28.2; i2 = 21.5 (a) q = m

a = 250,000 / (28.2 - 21.5) = 37,313 lba / hr m =m a v 2 = 37,313 x 13.09 / 60 = 8,140 ft 3 / min Q

= 3.85 m3 / s (b) similar procedure; Q


40

.002

115

55

2 15

DR L B. ER

50

T. P

% 70

DR Y BULB T EMPERATUR E - °F

80

55

LU VO

20

% 90

1 4 .0

21.5

50 %

65

60

12 0

A

-B T Y AL P

.014

65

N

TH

70

TI O

U

PE

R

PO

ER

U

Problem 3-27

A

N

D

O

30

.016

70

30

30 3-29 (a) Use the AirQuantity option of program PSYCH, iterating on the relative humidity and setting the minimum outdoor Air Quantity to 0.01, NOT ZERO. Use the properties option to find the entering wet bulb temperature of 62.6F. Then

φ = 0.852

(iterated)

ts = 56F

= 9,360 cfm Q s (b) Proceed as above

φ = 0.882 ts = 56F

= 10,014 cfm Q s 3-30

Proceed as in 3-29 above.

φ = 0.92 ts = 56.1 ≈ 56 F

≤ = 11,303 cfm Q s 3-31

(a) SHF =

500,000 = 0.91 550,000


31

3-31 (cont’d) (b)

a (i2 -i1) q = m

a = q/(i 2 -i1) or m a = m

550,000 (34.3 − 22.8)

a =47,826lba hr m a v 2 47,826 =m Q = x 14.62=11,654 cfm or 5.5 m3/s 2 60 60


32


R

90

AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIR-CONDITIONING ENGINEERS, INC.

SEA LEVEL

1.0

50

.028

60

85 1 5 .0

0.91

1.0

60

55


0.8 -2000

2. 0

-

0. 6

SENSIBLE HEAT TOTAL HEAT

5000

300 0

0 .5 0 200

0 .1

15 00

-0 .1

-0 .2

0.2

-8 -4 .0.0 0

85

0

WE T

0

-0 .5 4 - 0. -0.3

0. 3

-1.

.026

45

-2.

0.4

0

-1 00 0

4 .0 8 .0

Qs Qt

50 0

.024

BU

LB

TE MP E

80

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h


'W

75

34.3 35

50 14 .5

.018

D

-°

R

F

Y

A

IR

75

E R TU

D

O

F

30

TI O

E

% 90

ME

% 60

U.F

2

1 4 0%

40 1 3.

30 %

0

35

20 % 1 2.

10

15

72 (22)

20

110

105

100

95

90

85

80

75

70

65

60

55

50

45

40

35

5

Y IVE H UMIDIT 10% RE LAT

25

1400 x 60 = 5,915.5 14.2

i2 =

-5 x 12,000 +38.5 5,915.5

.010

40

.008

.006

35

.004

30

115 (46)

q +i a 1 m

a m

.012


=m a (i2 -i1) 3-32 q

i2 =

45

.002

115

50 %


IR YA

50

45

35

% 1 3 .5

45 40

DR L B. ER

55 60

15

T. P

% 70

50

30 %

-C

80

55

LU VO

20

1 4 .0

65

60


A R TU SA

25

22.8

12 0

U -B T Y AL P

.014

65

N

TH

70

N

PE

TE

R

M

P

PO

ER

U

A

N

Problem 3-31

.016

70

i 2 = 2 8 .3 6 Btu/lba Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

33 Then from Chart 1a, t2= 67F ASHRAE PSYCHROMETRIC CHART NO.1 55


R

R

90


SEA LEVEL

1.0

1 .0

60

50

.028

60

85 1 5 .0

0 .8

-2 00 0 -1 00 0

2. 0

-

0. 6


50 00

300 0

0 .5

4 .0 8 .0

Qs Qt

0 200

0 .1

15 00

-0 .1

-0 .2

0.2

-8 -4 .0.0

85

0

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

45

-2.

0. 4

.026

50 0

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80 'h


'W

75

35

.020

75 50 14 .5

.018

F

R ER

U

P M TE N TI O

.014

A R


U -B T

TU

Y

25

SA

AL P

70

14

E

65

.0 V U OL

% 90

2

ME

% 60

-C U .F

55 60

IR YA

.5

50

40

% 50

45

35

% 13

45

15

R .D LB ER

50

T. P

% 70

4 0%

40 1 3.

30%

0

35

20 %

55

15

67

20

90

110

105

100

95

90

85

80

70

65

60

55

50

45

40

35

5

10

Y I VE H UM IDIT 75

1 2.

10% RE LAT

.012

.010

.008

.006

35

.004

25

ENTHALPY - BT U PER PO UND O F DRY AIR

3-33 Use Adiabatic Mixing option of PSYCH with the Properties option to enter requested data. Assume volume flow rates of 3 to 1 to obtain. Tmix,db = 84.2 F Tmix,wb = 71.3 F

3-34 Use Program PSYCH at Sea Level elevation Iteration on the supply volume flow rate is required. This is the same as the leaving air quantity for the coil. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

40

.002

115

80

ADP55

DR Y BULB T EMPERAT UR E - °F

60

20

45

65

12 0

PO R PE

28.4

N

TH

.016

1

A

N

D

O

30

TU

Problem 3-32

E

D

-°

R

F

Y

A

IR

75

70

30

34 3-34 (cont’d)

(a) Supply air quantity is 9,384 cfm. (b) The outdoor air quantity is 938 cfm. (c) Air enters the coil at 74.6 F db, 60.5 F wb at a rate of 9,740 cfm (d) The coil capacity is 248,256 Btu/hr. The amount of air returned is: (9,740 – 939) = 8,802 cfm.

3-35 Use Program PSYCH at 5,000 ft elevation Iteration on the supply volume flow rate is required. This is the same as the leaving air quantity for the coil. (a) Supply air quantity is 11,267 cfm. (b) The outdoor air quantity is 1,127 cfm. (c) Air enters the coil at 74.6 F db, 62.1 F wb at a rate of 11,697 cfm (d) The coil capacity is 334,143 Btu/hr. The amount of air returned is: (11,697 – 1,127) = 10,570 cfm.

3-36

= 1000 cfm Q 0 (a) From Chart 1a

t s =120 / 74 F s= m

q 200,000 = (is -ir ) (37.2 − 22.8)


35

1 = 13,889 lb/hr = m

=m sv s = m s (14.78)/60 = 3,421 ft 3 / min Q s /v = 1000 x 60 / 12.61 = 4758 lb/hr o= Q (b) m o o r 13,889 − 4758 m = = 0.66; From Chart 1a t1 = 61/ 47 F 1 m 13,889

t 3 - t1 = (119 − 61)

w= m s (Ws -W2 ) = 13,889 (0.0075 - 0.0036) = 54.2 lbm/hr (c) m

1(i3 -i1) =13,889 (32.8 − 18.6) = 197,224 Btu/hr (d) q f = m


36



R

90


SEA LEVEL

1.0

50

.028

60

85 1 5 .0

50 00

0.8

1 .0

60

R

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6


300 0

0 .5

4 .0 8 .0

Qs Qt

-8 -4 .0.0 -2.0

0.4

85

0

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

50 0

WE T

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

80

1 000

1150

-0 .5 - 0.4 -0.3

0. 3

-1 .0

0

.026

45

.020

'h


'W

75

35

50 14 .5

.018

E

F

D

-°

R

F

Y

A

IR

75

R TU ER

U

P M TE N TI O


A R TU SA

25

E

14

65

.0 V ME

% 60

U.F

55

IR YA

.5

DR

13

50

40

% 50

r

45

35

%

4 0%

1

40

30%

1 3.

1

0

35

20 %

(a) t s = 120 / 71.4 F

61

72

20

110

105

100

95

90

85

80

65

60

55

50

45

40

35

3-37

15

75

5

40

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

0

25

.012

.010

.008

40

s

.006

35

.004

3 .002

115

60

45

15

B. RL PE

50

T.

% 70

47

30 %

-C

80

55

U OL

% 90


60

20

45

.014

12 0

R PE U -B T Y AL P

70

65

N

TH

.016

A

N

D

O

30

PO

Problem 3-36

70

30

120


Use Chart 1Ha

s = 200,000 /(38.7 − 24.0) = 13,605 lba/hr = m 1 m = 13,605 x 17.85 / 60 = 4048 cfm Q s

0 = (1000 / 15 .2) x 60 = 3947 lba/hr (b) m r 13,605 − 3947 m = = 0.71; t1 = 62.8 / 47 F 1 m 13,605

t 3 -t1 = (119.5 − 62.8) w =m s (w s -W1) = 13,605 (0.0088 - 0.0046) = 57.14 lbw/hr (c) m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

37 (d)

3-38

q f = 13,605 (33.8 - 20.2) = 185,028 Btu/hr

Assume fan power and heat gain are load on the space s = m

9384 x 60 = 42,915 lbm/hr; Prob 3-34 13.12

W fan + qduct = ms (is − ic ) = (4 x 2545) + 1000 = 11,180 Btu / hr ic = 20.8 −

11,180 = 20.54 Btu/lbm 42,915

State c is required condition leaving coil Part a, b, and c are same as prob. 3-34;

1(i1-ic ) = 42,915 (26.8 - 20.54) = 268,648 Btu/hr (d) q coil =m


38



R

90


SEA LEVEL

1.0

1 .0

60

R

50

.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6


50 00

300 0

0 .5

4 .0 8 .0

Qs Qt

-2.

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

-8 -4 .0.0 0

-0 .5 - 0.4 -0.3

0. 3

-1 .0

.026

45 85

0

50 0

WE T

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

0 1 000

80

.020

'h


'W

75

35

50 14 .5

.018

R TU

O D

A

N

ER

U


M TE N TI O R

A

.014

TU SA

25

E

% 90

1

% 60

-C U.F

13

IR

% 50

45

35

YA

.5

s 50

40

%

4 0%

40 30%

1 3. 0

35

20 %

3-39

55

15

72

20

100

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

45

.012

.010

40

.008

.006

35

.004

.002

115

60

45

15

r

55

c

DR B. RL PE

50

T.

% 70


80

55

ME

20

U OL .0 V

20.54

14

65

60

12 0

P

PO R PE U -B T Y AL P

50 %

70

65

N

TH

.016

70

E

F

D

Problem 3-38

-°

R

F

Y

A

IR

75

30

25


s (i r − i s ); W s (i s − i c ) q r = m fan = m

(a) ic = 28 Btu/lbm; ir = 33.7 Btu/lbm Using Chart 1Ha

q r = 1,320,000 Btu/hr

W fan = 30 x 2545 = 76350 Btu/hr

W fan = 30 x 2545 = 76,350 = ma (is -ic )


30

39

a(ir -is ) q s = 1,320,000 = m ASHRAE PSYCHROMETRIC CHART NO.4 55

NORMAL TEMPERATURE

60 85

BAROMETRIC PRESSURE: 24.896 INCHES OF MERCURY Copyright 1992

R

R

50


0.8

1.0

.028

80 1 .0

60

1 8 .0

5000 FEET

0 .8

-

SENSIBL E HEAT TOTAL HEAT

50 00 300 0

0 .5

0. 4

0 200

0.2

0 .1

ET

0

BU

LB

TE MP

.024

ER

75 50 0

AT U

40

RE

55 -° F .022

17 .5

15 00

-0 .1

0

80 W

.0

-0 .5 -0 .4 - 0.3 -0 .2

0. 3

-1

45

-1 000

4 .0 8 .0 -8 -4.0.0 -2. 0

Qs Qt

.026

-2 000

2. 0

0 .6

75 10 00

.020 'h

ENTHALPY HU MIDIT Y RATIO

35

'W

70 50

UR AT

D F O

ER

.016

M

U

TE

PO

50 %

TI O

N RA TU

65

.014

r

s

HUMIDITY RATIO - POUNDS MOISTURE PER PO UND DRY AIR

SA

59 (15) c

60

% 90

55

16

% 80

.5 V

20

-C

% 60

U .F

50

45

%

ME

70

U OL

55

50

T. P

30 %

20%

10

15

62.5 (17)

20

80 (27)

90 (32)

110

105

100

95

90

85

80

75

70

65

60

55

50

45

40

35

I TY TIV E HUMID 10% RELA

.010

.006

35

.004

25

Solve simultaneous:

W fa n + q s = m a (ir -i c ) 1 ,3 2 0 ,0 0 + 7 6 ,3 5 0 (3 3 .7 -2 8 )

a = 2 4 4 ,9 7 4 lb a /h r m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

40

.008


Two unknowns & two equations

a= m

.012

.002

115

0

IR YA

4 0%

1 5 .5

35

%

R .D LB

50


45

40

ER

40 35

16 .0

15

45

60

12 0

PE R TU -B

LP Y A TH

25

EN

65

P

ND

30

17 .0

Problem 3-39

70

E

RY

-°

F

AI R

.018

30

40

a) is = ir - ( q s m is = 33.7 -

1,320,000 =28.3 Btu/lba 244,974

Locate points on the condition line on Chart 1 Ha and point c is on cooler process line horz. to left of points. Read ts = 62.5 F, tc = 61.6F.

= 244,974 x16.2 = 66,143cfm (a) Q s 60 = 31.2 m 3 s (b) Q s

3-40

English Units –Tucson, Arizona, Elevation 2,556 ft.

imin =i0 =31.1 Btu/lba and sat. air ; t min =64.5 F; PSYCH Shreveport, Louisiana, Elevation 259 ft. imin =i0 = 42.5 Btu/lba and sat. air ; tmin = 76.8 F; PSYCH

SI Units – Tucson, Arizona

imin =i0 = 51.5 kJ/kga ; tmin =18.1 C; Chart 1b Shreveport, Louisiana imin =i0 =75.5 kJ/kga ; tmin =24.8 C; Chart 1b


41



R

R

90


259 FEET

1.0

60

50

.028

60 1 .0

85

15 .0

0 .8

-

SENSIBL E HEAT T OTAL HEAT

50 00 300 0

0 .5

-8 -4.0.0 -2. 0

-0 .4 - 0.3

-0 .1

-0 .2

0 200

15 00

0 .1

85 0

.0

-0 .5

0. 3

-1

0.2

45

0. 4

0

-1 000

4 .0 8 .0

Qs Qt

WE TB UL

80

50 0

.024

BT EM

PE R

40

55 AT

UR

E°F

.022

10 00

80 .020 'h


.026

-2 000

2.0 0 .6

75

'W

TLO

35

50 .018

-° F E AT U R

U

TE M

PE

PO

O N

70

.014


AT I R U AT S

25

E

65

60

% 90 UM OL 0V

%

60

%

E-

70

CU

55

.FT

50

%

1 3.

50

5

40 35

IR YA

45

4 0%

40 30% 1 3 .0

35

20 %

10

15

76.8

20

95

110

105

100

95

90

85

80

75

HUMIDITY

70

65

60

55

50

45

40

35

IVE 10% RE LAT

.012

.010

40

.008

.006

35

.004

.002

115

45

15

DR LB . ER .P

% 60


50

1 4.

80

55

20

45

65

12 0

ER P TU

-B Y LP A

.016

N

TH

SL

R

30

N

D

O

F

D

R

Problem 3-40 Shreveport, LA

14 .5

Y

AI R

75

70

25



30

42



R

R

85 50


2556 FEET

1.0

1 .0

.028

85

60

5 16 .

60

0 .8

-

SENSIBL E HEAT T OTAL HEAT

50 00 300 0

0 .5

0. 4

0 200

0.2

0 .1

15 00

-0 .1

-0 .2

0

80 0

.024

.0

55

-0 .5 -0 .4 - 0.3

0. 3

-1

45

-1 000

4 .0 8 .0 -8 -4.0.0 -2. 0

Qs Qt

.026

-2 000

2. 0

0 .6

50 0

80

40

WE

TB

UL

75

.022

BT

EM P

TU

10 00

RE

- °F

1 6 .0

ER A

.020

'h


'W

35

75

50 .018

70

R TU

N

EM

P

O U P

65

T

R

A

TLO

R

%

EN

60

55

% 80

% 60

CU

15

1 4.

. LB ER .P .FT

% 50

45

5

4 0%

40

DR

1 4 .0

IR YA

30 %

35

TA

20%

10

15

64.6

20

102

110

105

100

95

90

85

80

75

70

65

60

55

50

45

40

35

ITY TIV E HU MID 10% RELA

45

.012

.010

40

.00 8

.00 6

35

.00 4

.00 2

115

50

E-

45

UM OL 0V

50

40

1 5.

% 7 055


20

3-41


TU SA

65

60 90

35

.014

12 0

TI O

N

PE TU -B LP Y A

70

1 5 .5

25

TH

.016

A

30

ER

D

O F

D

E

R

Y

-°

F

A

IR

Problem 3-40 Tucson, Arizona

25


s (ir − i s ) q = m

s = 12,000 /(28.2 − 19.1) = 1,319 lba/hr ton (a) m = 1319 x 15.6 = 343 cfm/ton Q s 60

o r1 m 13 = = = 0.55 or 55% s r0 23.5 m ≈ 0.046 m3 / s - kW (b) Q s

0 /m s ≈ 55% m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

30

43



R

60 85

R

50


5000 FEET

1.0

.028

80 1 .0

60

1 8 .0

0 .8

50 00

-

300 0

0.7


0 .5

-0.1

BU

LB

TE MP

.024

ER

75 50 0

40

AT U

RE

55 - °F .022

17 .5

0 200

15 00

0 .1

ET

0

.0

.5 -0 -0 .4 - 0.3 -0 .2

0. 3

-1

0.2

80 W

-8

0. 4

0

45

-1 000

4 .0 8 .0 -4.0.0 -2. 0

Qs Qt

.026

-2 000

2.0

0 .6

75 10 00

.020 'h

ENT HALPY HU MIDIT Y RAT IO

35

'W

70 50

PE RA

O

F

TU

DR

RE

Y

-°

F

AI R

.018

N IO

% 90

40 %

60

U.F

% 60

T. P

10 %

1

IR YA

4 0%

0

1 5 .5

30 %

20%

10

50 (10)

2 (i2 − i3 ); m 2 = 3-42 q = m

15

75 (24)

20

110

105

100

95

90

85

80

75

70

HUMIDI TY

65

60

55

50

45

40

35

IVE 10% REL AT

45

.012

.010

40

.008

.006

35

.004

.002

115

DR


%

L B.

50

ER

45

40

16 .0

40

-C

s

ME

50

45

r

%

LU VO

55 70

35

1 6 .5

% 80

50

35

HUMI DITY RATIO - POUNDS MOISTURE PER POUND DRY AIR

55

.014

12 0

AT SA T

UR

65

60

25

20

15

.016

TE

PO PE R TU -B AL PY

TH EN

65

M

UN

D

30

17 .0

Problem 3-41

70

25

100 (38)


500,000 (41.1 − 21.9)

2 = 26,042 lba/hr m

= 26042 x 14.55/60 = 6315 cfm Q 2 0 = 0.25 x 26,042 = 6511 lba/hr m 0 /m 3 = 0.25; t mix = 67.5 / 49.5 F m


30

44 3-42 (cont’d) Preheat Coil:

0c p (t 4 -t 0 ) = 6511 x 0.24 (60-6) = 84,383 Btu/hr q ph = m

Heat Coil:

2 (i5 -i1) = 26,042 (28.4 - 20) = 218,753 Btu/hr q h = m Humidifier:

w= m 2 (W2 -W5 ) = 26,042 (0.0144 - 0.0035) m = 283.9 lbw/hr = 2.98 m3 / s; q = 24.7 kW; q = 64.1 kW; (b) Q 2 ph h

w = 0.036 kg/s m


45 ASHRAE PSYCHROMETRIC CHART NO.1 55


R

90


SEA LEVEL

1.0

1 .0

60

R

50

.028

60

85 1 5 .0

0 .8 -2 000

2.0

-

0. 6


50 00 300 0

0 .5 0 200

0 .1

0.2

-0 .1

-0 .2

15 00

50 0

WE T

.024

BU

80

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

80

1 000

1153

85

0

0

-0 .5 - 0.4 -0.3

0. 3

-1.

.026

45

-8 -4 .0.0 -2.0

0.4

0

-1 00 0

4 .0 8 .0

Qs Qt

.020

'h

ENT HAL PY HU MIDITY RATIO

'W

75

35

50 14 .5

.018

R TU ER

U

M

P

PO

TE N

70

TI O

.014


A R TU SA

25

1 4 .0

E

65

ME

% 60

-C U.F

55 60

50 %

45

35

3

4 0%

40 1 3.

1

30%

0

35

IR YA

50

40

% 1 3 .5

45

15

30 %

DR L B. ER

50

T. P

% 70

5 20 %

3-43

15

60 (16)

70 (21)

20

110

105

100

95

90

85

80

75

65

4

60

55

50

45

40

35

5

10

70

1 2.

10%

H UM IDITY REL ATI VE

.012

.010

40

.008

.006

35

.004

.002

115

80

55

LU VO

% 90

DR Y BUL B TEMPERATUR E - °F

60

20

45

65

12 0

R PE U -B T Y AL P

2

N

TH

.016

A

N

D

O

Problem 3-42

70

E

F

D

-°

R

F

Y

A

IR

75

30

30

25

105 (40)


a (ir − is ) Use Chart 1a; q d = m a = q d /(ir − i s ) or m a = 150 x 12,000 / (28.4-22) = 28,125 lbm/hr (a) m

= 28,125 x 13.25/60 = 61,211 cfm Q d = 0.20 Q m

= 1,242 cfm Q d

m = 1,242 x 60/13.5 = 5,521 lbm/hr [vm assumed] m im =ir − 1.8 x 12,000/5,521 = 24.5 Btu/lbm; tm = 62 / 57 F


46

= 2.93 m3 /s; Q = .59 m3 /s; t = 17/14 C (b) Q d m m ASHRAE PSYCHROMETRIC CHART NO.1 55


R

90


1 .0

.028

60

85 1 5 .0

0.8

0 .8 -2 00 0 -1 00 0

2. 0

-

0. 6

50 00 300 0


0 .5

0 .1

15 00

-0 .1

-0 .2

0.2

.026

45 85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

4 .0 8 .0 -8 -4 .0.0 -2. 0

Qs Qt

0. 4

0 200

0.6

SEA LEVEL

1.0

50

50 0

.024

BU

LB

80

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h


'W

75

35

50 14 .5

R TU

O

A

N

ER

U

M

P

PO

TE N TI O

.014

A R


R PE U -B T

TU

Y

25

SA

AL P

70

E

U OL ME

60

-C U.F

% 70

T. P L B. ER

55

IR YA

50

DR

1 3 .5

45

%

% 50

45

4 0%

40 1 3.

30%

0

35

62 (17) 20 %

3-44

15

60 (16)

75 (24)

20

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

.012

.010

40

.008

.006

35

.004

.002

115

60


50

35

r

m 80 %

55

40

.0 V

% 90

s

15

14

65

60

20

45

65

N

TH

.016

70

E

F

D

-°

R

F

Y

A

IR

75

30

D

Problem 3-43

.018

12 0

R

60

25


15.0 x 12,000 a = (a) m = 29,508 lba/hr (31.2 - 25.1)

= 29,508 x 16.0/60 = 7,869 cfm; Q = 0.2 x Q Q d m s

= 1,574 cfm

m =1,574 x 60/16.2 = 5,829 lba/hr (v massumed) m im = 35.7 − 1.8 x 12,000/5,829 = 27.5 Btu/lba; Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

30

47

tm = 62.5 / 58 F =3.7 m3 / s; Q = 0.74 m3 /s; t = 17 /14.4 C (b) Q s m m

3-45

Use Chart 1a;

r m 10 = 0.8 = 1 m 0r

[Both design and min. load condition]

s is = ir - q m / m s = m

Q 50 x 12,000 d = ir − i s (29.35 - 22.3)

s = 85,106 lba/hr; m s is constant for all conditions m

i s' = 29.35 − 25 x 12,000/85, 106 = 25.83 Btu/lba (a) From Chart 1a; t s' = 64 F

si s + m b i1' = (m c +m b ) i s' (b) m b (i s − i s' ) 24.2 − 25.8 m = = 0.271 = c (i s' − i1' ) 25.8 − 31.7 m

(b) From chart 1a; t d = 49 F for both cases


48



90


SEA LEVEL

1.0

.028

60 85 1 5 .0

0.9

1 .0

50

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6


50 00 300 0

0 .5

0 200

0 .1

15 00

-0 .1

-0 .2

0.2

85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

50 0

80

.024

BU

LB

TE

MP E

55 RA TU

0

40

RE

- °F

.022

1 000

80

.020

'h


'W

75

35

50 14 .5

0'

.018

75

R TU ER

U

M

P

PO

TE

R

N

PE

TI O

U

.014

1'

R

A

-B T

TU

Y

25

SA

AL P

70

65

E

N

TH

50 %

A

N

D

O

30

.016

70

E

F

D

-°

R

F

Y

A

IR

Problem 3-45

.026

45

-8

0. 4

0

4 .0 8 .0 -4 .0.0 -2. 0

Qs Qt


60

R

LU VO

1

% 90

1 4 .0

65

% 60

-C U.F

r

55

s

60

IR YA

% 50

45

35

DR

50

40

% 1 3 .5

45

15

L B. ER

50

T. P

% 70

4 0%

40 30%

1 3. 0

35

20 %

3-46

15

55 (13) 64 (18)

20

77 (25) 85 (29)

95 (35)

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y I VE H UM IDIT 70

1 2.

10% RE LAT

45

.012

.010

40

.008

.006

35

.004

.002

115

80

55

ME

s' 20


60

12 0

R

25 ENTHALPY - BT U PER POUND OF DRY AIR

Refer to problem 3-45. Results are similar.

3-47 (a) It is probably impossible to cool the air from 1 to 2 in one process. The extension of line 12 does not intersect the saturation curve. (b) Cool the air to state 1' and then heat to state 2.


30

49



R

90


SEA LEVEL

1.0

1 .0

60

R

50

.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6


50 00 300 0

0 .5

0. 4

0 200

0 .1

15 00

-0 .1

-0 .2

0.2

.026

45 85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

4 .0 8 .0 -8 -4 .0.0 -2. 0

Qs Qt

50 0

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENTHAL PY HU MIDIT Y RAT IO

'W

75

35

50 14 .5

.018

R A

N

ER

U

M

P

PO

TE N

70

TI O

.014


A R TU SA

25

E

U OL .0 V

% 90

% 60

-C U .F

55

50

% 50

45

35

A IR

40

% .5

45

15

60

13

2

4 0%

40 1 3.

30%

0

35

20 %

3-48

15

52 (11) 60 (16)

20

80 (27)

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

.010

40

.008

.006

35

.004

25


c sh m = =.837 (a) s ch m

h cs m = = 0.163 s ch m c 0.837 m = = 5.14 h 0.163 m s (ir − is ) q = m s = m

.012

.002

115

1'

RY B. D RL PE

50

T.

% 70


80

55

ME

54

14

1

65

60

20

45

65

12 0

R PE U -B T Y AL P

67

N

TH

.016

90 %

TU

O

30

D

Problem 3-47

E

F

D

-°

R

F

Y

A

IR

75

70

50 x 12,000 = 93,750 lba/hr (28.2-21.8)


30

50

= 93,750 x 13.2/60 = 20,625 cfm Q s = 9.7 m3 /s (b) Q s ASHRAE PSYCHROMETRIC CHART NO.1 55


R

90


SEA LEVEL

1.0

1 .0

60

R

50

.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6


0 .5

50 00 300 0

0.65

4 .0 8 .0

Qs Qt

0. 4

0 200

0. 1

0.2

-0 .1

-0 .2

15 00

0

-8 -4 .0.0 -2. 0

85

0

WE T

0

.5 -0 - 0.4 -0.3

0. 3

-1.

.026

45

50 0

.024

BU

80

LB

TE

MP E

55 RA TU

RE

40

- °F

.022

1 000

80

.020

'h


'W

75

35

50 14 .5

.018

R TU

O D

A

N

ER

U

M

P

PO

N

70


TI O A R TU SA

25

.014

90 %

65

E

.0 V

% 90

%

ME

r

60

-C U.F

s % 50

45

35

20 % h

IR YA

50

.5

45 40

% 13

c

4 0%

40 30%

1 3. 0

35

20 %

3-49

52 (11)

15

75 (24)

20

90 (32)

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y VE H UM IDIT 70

1 2.

10% REL ATI

.012

.010

c h 10.1 m m 36 = = 0.9; = = 0.10 ; s 46.3 s 46.3 m m

s = m

40

.008

.006

35

.004

25

ENTHALPY - BT U PER POUND O F DRY AIR

See diagram of problem 3-48 (a)

45

.002

115

55 60

15

R .D LB ER

50

T. P

% 70


80

55

U OL

20

14

65

60

12 0

TE

R PE U -B T Y AL P

.016

N

TH

70

E

F

D

Problem 3-48

-°

R

F

Y

A

IR

75

30

c m 0.9 = = 9.0 h 0.10 m

50 x 12,000 = 83,333 lba/hr (30.1 - 22.9)

=83,333 x 15.67/60 = 21,763 cfm Q s Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

30

51

=10.3 m3 /s (b) Q s

3-50

(a) See diagram for problem 3-48 c m c (ir -ic ); m c = 0.714 x m s = 0.837 x 93,750 = 0.837; q c = m s m

= 78,469 x 13.04/60 = 17,054 cfm c = 78,469 lba/hr ; Q m c

q c = 78,469 (28.2-20.6) = 596,364 Btu/hr =8.1 m3 /s; q = 175 kW (b) Q c c

3-51 SI Units (a) On the basis of volume flow rate using Chart 1b:

= 13 Q = 0.69 x 1.18 = 0.815 m3/s Q 2 3 12 =Q -Q = 1.18 − 0.815 = 0.365 m3/s and Q 1 3 2

(b)

a3 (i4 -i3 ) = q 34 = m q 34 =

Q 3 (i -i ) 4 3 v3

1.18 (47.8-41.0) = 9.6 kW 0.835


52 ASHRAE PSYCHROMETRIC CHART NO.1 11 0


R

SEA LEVEL

10 0

0 .8 0. 7 0. 6

Problem 3-51 SENSIB LE HEAT T OTAL HEAT

10 .0

0.5

1 .0

1 .5 2. 0

-5.0

4 .0

Qs Qt

-4.0

80

1 .0

MP ER

24

AT

UR

E-

°C

11 0 22

0 .9

0

5.0

4. 0

0 .1

WE TB UL BT E

-0.2

0.2

30

0.0

.0

-0 .5

0. 3

-1 .0

12 0 26

90

-2.0

-2

0. 4

4

-

28

30 0 .9

1.0

0 30


12

R

2

2 .0

3 .0

2.5

70

25 20

'h


'W

10 0 25

C

R

-°

D

R TU A

50

90 14

E

N

8

SA

1

HUMIDITY RATIO - G RAM S M OISTURE PER KILOGR AM D RY AIR

R

20

TU

LP Y

40

%

%

U OL ME IC M

%

UB

50

4 0.8

10

-C

% 60

20

ET

5

ER

0 .8

g Rk PE

40 % 2

DR

3 0%

IR YA

20

15

10

5

8

10

12

HUM ID ITY 25

0

0 .7

IVE 10% RE LAT

20

17.2

24

29

35

0 .8

20 %

30

5

10

30

12

80 10

8

70

6

4

60

2

45

10

4

15

6V

11

70

2

0 .8

% 80 3

30


90

40

14.7 15

50

A

-K

J

TI O

P

N

E

T

R

K

EM

P

ER

20

0 .8

A

16

50 %

E

O F AM R IL O G

0

Y

0 .9

AI R

Problem 3-51

TH

18

60

40

ENTHALPY - KJ PER KILO GRAM O F DRY AIR

English Units

= 640 cfm; q = 33,684 Btu/hr (a) Q 1 34 3-52 (a),(b) From Chart 1b, states 1.4 and ADP are known. Based on approx. 11.8 C db, 11.2 C wb, and 90% RH locate state 2. Then for full load design condition air is cooled from 1 to 2 and the room process proceeds from 2 to 4. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

50

53 For the high latent load condition, the air at 2 is reheated to state 3 where it enters the space and the process proceeds to state 4. Q a (i4 -i2 ) = 2 (i4 -i2 ) (c) q 24 = m v2

=35 x 0.817 (47.7-32) ; Q 2

a (i1-i2 ) = q 12 = m

= 1.82 m3/s Q 2

1.82 (60.6-32) 0.817

q 12 = 63.7 kW

a (i4 -i3 )= q 34 = m

1.82 (47.7-39.4) 0.817

q 34 = 18.5 kW q 23 = q 24 - q 34 = 35-18.5=16.5 kW


54 ASHRAE PSYCHROMETRIC CHART NO.1 11 0

NORMAL TEMPERATURE

SEA LEVEL

10 0

0. 6


10 .0

0.5

-4.0

30

0.0

.0

0

WE TB UL BT E

MP ER

80

1 .0

-0 .5

5.0

0. 3

-1.

24

AT

UR

E-

°C

11 0

-0.2

22

0 .9

0

4. 0

0 .1

26

90

-2.0

-2

0. 4

0.2

12 0

-5. 0

4 .0

Qs Qt

4

1 .5 2.0

0.7

0 .9

1 .0

0 .8

-

28

30

Problem 3-52

1.0

0

R


12

30

BAROMETRIC PRESSURE: 101.325 kPa Copyright 1992

R

2

2 .0

3 .0

2.5

70

25 20

'h


'W

10 0 25

18

C

R

-°

D

R

E

O F

TU

AM R

A

16

ER

20

90

EM T N R

%

ME -C

3

UB IC M ET

5

ER

0 .8

R gD Rk PE

40 % 2

3 0%

19

20

23

30

20

15

5

10

8

11.8

10

HU MID ITY 25

0

0 .7

IVE 10% RE LAT

9

3-52

IR YA

0 .8

20 %

27

35

5

10

30

12

80 10

8

70

6

4

60

2

45

4 0 .8

% 50

U OL

% 60

10

6V

%

10

2

0 .8

4

15

70


SA

% 80

40

TU

LP Y A TH N E

90

11 ADP

8

17

14

30

HUMIDITY RATIO - GRAM S M OISTURE PER KILOGR AM D RY AIR

A

-K

J

TI O

P

1 20

50

K R E

14

0 .8

40

15

20

21

P

IL O G

50

0

Y

Problem 3-52

0 .9

AI R

60

40

ENTHALPY - KJ PER KILO GRAM O F DRY AIR

English Units (a),(b) See above

= 4103cfm ; q 12 =221,243 Btu/hr (c) Q 2

q 34 = 67,498 Btu/hr; q 23 = 52,502 Btu/hr 3-53 English Units (a)

s (ir -is ); m s = 5000 x 60/13.2 = 22,727 lba/hr q=m (specific volume value of 13.2 ft3/lbm is assumed.)


50

55

s =28.2 − 10 x 12,000 / 22,727 = 22.9 Btu/lba is = ir - q /m

t s = to = 57.5 F; Ws =Wo = 0.0083 lbv/lba (b)

r 0m m m = = 0.462 r m 0r s m m

r =0.462 x 22,727 = 10,500 lba/hr m o = 22,727 − 10,500 = 12,227 lba/hr m = 10,500 x 13.68/60 = 2,394 cfm Q r = 12,227 x 12.11/60 = 2,468 cfm Q o

r 0'm' m = =0.578 (c) m' m 0'r

r =0.578 x 22,727 = 13,131 lba/hr; m o' = 9,596 lba/hr m =13,131 x 13.68/60 = 2,994 cfm; Q = 9,596 x 13.48/60 Q r o' = 2,156 cfm

s (im' -is ) = 22,727 (28.4 - 22.8) = 127,271 Btu/hr (d) q c = m


56



90


1.0

1 .0

50

.028

60

85 1 5 .0

50 00

0.8

SEA LEVEL

60

R

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6


300 0

0 .5

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

85

0

50 0

WE T

.024

BU

LB

80

TE

MP E

55 RA TU

40

RE

- °F

.022

80

1 000

1150

.026

45

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

4 .0 8 .0 -8 -4 .0.0 -2. 0

Qs Qt

.020

'h


'W

75

35

50 14 .5

A

IR

75

R TU M TE

R

N

PE

TI O

U

.014

R TU SA

0'

E

ME

% 60

-C U.F

r

60

% 13

50

YA

.5

45

IR

% 50

45

4 0%

m

40

43 (6)

1 3.

30%

0

35

55

s

ADP

35

DR L B. ER

50

T. P

% 70

20 %

10

40 (4)

15

70 (21)

57.5 (14) 65 (18)

75 (24)

20

110

105

100

95

90

85

80

75

Y IVE H UM IDIT 70

65

60

55

50

40

35

5

10% RE LAT 45

1 2.

0

.012

.010

40

.008

.006

35

.004

25


3-53 SI Units (a) ts = 14.2C; Ws = 0.0083 kgv/kga

=1.17m3 s =1.13m3 s ; Q (b) Q r o

=1.41m3 s ; Q =1.02m3 s (c) Q r o' (d) q c = 37.3 kW 3-54

45

.002

115

80

55

U OL .0 V

20

50 %

m'

14

65 % 90


25

60

40


A

-B T Y AL P

70

65

N

TH

90 %

P

PO

ER

U

A

N

D

O

30

.016

70

E

F

D

-°

R

F

Y

Problem 3-53

15

.018

12 0

R

(a) Any combination that will yield an enthalpy less than 57.0 kJ/kga or 33 Btu/lba


30

57

s = 5 / 0.84 = 5.95 kga/s = m r (b) m

o mr m = =0.36 r 0r m

o = 0.36 x 5.95 = 2.14 kga/s m = 2.14 x 0.852 = 1.82 m3 /s = 3,857cfm Q o (c) tad = 15.4 C or 60F

o /q n = (im -is )/(ir -is ) = 1.0 (Essentially, no difference) (d) q ASHRAE PSYCHROMETRIC CHART NO.1 11 0


R

SEA LEVEL

10 0

0.7

0. 6


10 .0

0.5

1 .5 2. 0

-5.0

4 .0

Qs Qt

5.0

0. 3

-1. 0

-4.0 -2 .0

WE TB UL BT

80

1 .0

EM

PE R

24

AT

UR

E-

°C

11 0 22

0 .9

0

4. 0

0 .1

30

0.0

-0.2

0.2

12 0 26

90

-2.0

0. 4 -0 .5

0.6

1 .0

4

0 .8

-

28

30

0 .9

1.0

0 30


12

R

2

2 .0

3 .0

2.5

70

25 20

'h


'W

10 0

25

18

60

-°

C

R D

16

R

E

O F

TU

AM

A

50

20

90

r

SA

E

N

8

TH

0

HUMIDITY RATIO - G RAM S M OISTURE PER KI LO GR AM D RY AIR

TU

A

R

LP Y

A

m2

0 .8

40

s%

90

15

% 80 0 .8 6V

15

%

U OL ME

10

UB IC M ET

5

ER

0 .8 2

3 0%

18 (64)

20

25 (77)

35

20

15

10

8

HUM ID ITY 25

0 .7 5

A IR

0

20 (68)

IVE 10% RE LAT

10

RY

0 .8

20 %

30

5

10

gD Rk PE

40 %

30

80 10

8

70

6

4

60

2

45

5

0%

4 0. 8

10

-C

% 60

20


70

40

30

12

50

J -K

14

0

TI O

P

N

E

T

R

K

EM

P

ER

R IL O G

0

Y

AI R

57

0 .9

Problem 3-54

40

ENTHALPY - KJ PER KI LO GRAM O F DRY AIR


50

58

SHF =

3-55

−424,000 = -4 530,000 − 424,000

Construct condition line on Chart 1a with preheat and mixing processes.

sen = -424,000 = m sc p (t r -t s ) (a) q s = m

−424,000 = 88,333 lba/hr 0.24 (75 − 95)

=88,333 x 14.07/60 = 20,714 cfm or 9.8 m3 /s Q s (b)

r hm m r = 0.33 x 88,333 lba/hr = =0.33; m mm hr

= 29,150 x 13.68/60 = 6,646 cfm or 3.14 m3 /s r =29,150 lba/hr; Q m r h m h = 0.67 x 88,333 =1 − 0.33 = 0.67; m m m

= 59,183 x 13.1/60 h = 59,183 lba/hr; Q m h = 12,922 cfm or 6.1 m3 /s (at heated condition) Q h

hc p (th -t o ) = 59,183 x 0.24 (60-35) (c) q ph =m

355,098 Btu/hr or 104 kW q= (d) q m =88,333 x 0.24 (95 - 65) = 635,998 Btu/hr or 186 kW Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

59



R

R

90


SEA LEVEL

1.0

1 .0

60

50

.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6


50 00 300 0

0 .5

0. 4

0 200

0 .1

15 00

-0 .1

-0 .2

0.2

.026

45 85

0

-4

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

4 .0 8 .0 -8 -4 .0.0 -2. 0

Qs Qt

50 0

WE T

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h


'W

75

35

50 14 .5

.018

F

R ER

U

M

P

PO

TE N TI O

.014


A R TU SA

25

-C

r

T. P U.F

% 70

55 60

IR

50 %

45

35

YA

50

40

%

1 3 .5

45

15

20 %

DR L B. ER

50

4 0%

40

m 1 3.

30%

0

35

s 20 %

15

60 (16)

75 (24)

20

95 (35)

110

105

100

95

90

85

80

75

Y IVE H UM IDIT 70

60

55

50

45

40

35

5

35 (2)

10

10% RE LAT 65

1 2.

h

0

.012

.010

40

.008

.006

35

.004

25


3-56 Refer to chart 1a. (a)

x 60 (i -i ) a3 (i4 -i3 ) = Q q 34 = m 3 4 3 v3 x Q 3

45

.002

115

60


E

%

ME

80

55

LU VO

20

50 %

% 90

1 4 .0

65

60

12 0

R PE U -B T Y AL P

70

65

N

TH

.016

A

N

D

O

30

TU

Problem 3-55

E

D

-°

R

F

Y

A

IR

75

70

q 34v 3 (1750 x 13.23) = 60(i4 -i3 ) 60(28.1-23)

= 75.7 or 76 cfm = 0.040 m3 /s Q 3


30

60 (b) t3db = 58.5 F and 80% RH or 15 C

31 ; Q3 = 0.754 x 75.7 = 57 cfm or 0.028 m3 /s 12

= (c) Q 2

= 76 - 57 = 19 cfm or 0.012 m3 /s Q 1 ASHRAE PSYCHROMETRIC CHART NO.1 55


90


1 .0

.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

0. 6


300 0

0 .5

4 .0 8 .0

Qs Qt

-8 -4 .0.0 0

-2.

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

85

0

50 0

80

1 000

RE

- °F

.022

.020

50

90 %

.018

70

R TU

O

.016

50 % 70

.014

R

1

TU

62

E

N

SA

25

65

ME

% 60

-C

80

LU VO

% 90

55

T. P L B. ER

55

3

60

%

DR

1 3 .5

IR YA

50

2

U.F

4

% 70

50

1 4 .0

20

% 50

45

4 0%

40 30%

1 3. 0

35

20 %

58.5

75

84

20

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

3-57 (a)

50

15

Y I VE H UM IDIT 70

1 2.

10% RE LAT

10

45

65

A

-B T

TI O

U

N

PE

TE

R

M

P

PO

ER

U

A

N

D

70

E

F

D

-°

R

F

Y

A

IR

75

30

Y AL P

55 RA TU

14 .5

TH

MP E

75

60

35

TE

80 'W

Problem 3-56

45

LB

40

35

40

.024

BU

'h


15

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

.026

45

HUMI DITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

-

.012

.010

40

.008

.006

35

.004

.002

12 0

50 00

0.8

SEA LEVEL

1.0

50

115

60

R


R

25


Refer to Chart 1

A reheat system is required. Process 1-2 is for the coil. Process 3-4 is defined by the SHF = 0.5 Process 2-3 represents the required heat. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

30

61 State 3 is defined by the intersection of the reheat and space condition lines.

(b)

a3 (i4 -i3 ) = q 34 = m

x 60 Q 3 (i4 -i3 ) v3

= q 34 v 3 = 100,000 x 13.4 Q 3 60(i4 -i3 ) 60(28.2-23.9) = 5,194 cfm or 2.5 m3 /s Q 3 (c)

a (i1-i2 ) = q 12 = m

5,194 x 60 (34.2-20.2) 13.4

q 12 = 325,594 Btu/hr or 95.4 kW q 23 =

5,194 x 60 (23.9-20.2) 13.4

q 23 =86,050 Btu/hr or 25.2 kW


62



90


SEA LEVEL

1.0

1 .0

60

R

50

.028

60 85 15

-2 00 0 -1 00 0

2.0 0. 6


50 00

300 0

0 .5

0. 4

0 200

0 .1

15 00

-0 .1

-0 .2

0.2

.026

45 85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

4 .0 8 .0 -8 -4 .0.0 -2. 0

Qs Qt

.0

0 .8

-

50 0

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENT HALPY HU MIDIT Y RAT IO

'W

75

35

50 14 .5

75

R TU A

.016

ER

U

50 %

M

P

PO

TE N

70

TI O

1 4 .0 ME

% 60

-C U.F

4

E

55 60

IR

% 50

45

35

3

YA

50

40

% .5

2

13

15

R B. D RL PE

50

T.

% 70

ADP45

4 0%

40 30%

1 3. 0

35

20 %

66

15

75

20

85

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

45 51

Y IVE H UM IDIT 70

1 2.

10% RE LAT

10

.010

40

.008

.006

35

.004

.002

25

ENTHALPY - BT U PER PO UND OF DRY AIR

3-58 Assume room temperature humidity of 50% and layout the state & processes on required from point c to s.

Supply Air:

sc p (t s -tr ) q sen = 120,000 x 0.5 = 60,000 Btu/hr = m s = m

45

.012

115

80

55

LU VO

% 90

56

20

1 65

60

HUMI DITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

A R TU

62

SA

25

.014

65


R PE U -B T Y AL P

70

E

D F N

D

O

30

N

TH

70

-°

R

F

Y

A

IR

Problem 3-57

.018

12 0

R

60,000 = 53,192 lba/hr 0.24 (75-70.3)


30

63

=53,192 x 16.33/60 = 14,477 cfm or 6.8 m3 /s Q s Mixed Air:

o = 53,192 x 0.333 = 17,703 lba/hr m = 17,713 x 17.2/60 = 5,078 cfm or 2.4 m3 /s Q o

r = 53,192 − 17,713 = 35,479 lba/hr m =35,479 x 16.5/60 = 9,757 cfm or 4.6 m3 /s Q r Reheat:

c c p (t s -t c ) = 53,192 x 0.24 (70.3-55.2) q rh = m

= 192,768 Btu/hr or 56.5 kW Coil:

m (im -ic ) = 53,192 (34.4 - 24.2) = 542,558 Btu/hr or 159 kW q c =m

=

( 200,412 − 190,109 )100 = 5 .1 % 200,412


64


NORMAL TEMPERATURE R

60 85

BAROMETRIC PRESSURE: 24.896 INCHES OF MERCURY Copyright 1992

R

50


5000 FEET

1.0

.028

80 1 .0

60

1 8 .0

0 .8 0 .6


50 00 300 0

0 .5

0. 4 0. 3

0 200

0.2

0 .1

75

50 0

40

BU

TE MP

.024

ER

AT U

RE

55 -° F .022

50 %

75 10 00

.020 'h

ENTHALPY HU MIDIT Y RATIO

LB

75

17 .5

15 00

-0 .1

0

80 WE T

0

.0

-0 .5 -0 .4 - 0.3 -0 .2

0.6 0.5

-1

45

-1 000

4 .0 8 .0 -8 -4.0.0 -2. 0

Qs Qt

.026

-2 000

2. 0

-

35

'W

70

0 50

UR ER P M

U

TE

PO

.016

TI O

N RA TU

65

SA

.014

m

60

HUMIDITY RATIO - POUNDS MOISTURE PER PO UND DRY AIR

PE R TU -B LP Y A

90 %

EN

TH

25 60

% 90

55

%

ME -C

% 60

U .F

50

45

s

U OL

70

.5 V

55

c

50

16

r

% 80

20

T. P IR YA

30 %

10

55 (13)

15

75 (24)

20

90 (32)

110

105

100

95

90

85

70 (21) 80

65

60

55

50

45

40

35

I TY TIV E HUMID 10% RELA

75

20%

70

1 5 .5

35

4 0%

.010

.008

.006

35

.004

25



40

.002

115

40

%

R .D LB

50


45

ER

40 35

16 .0

15

45

.012

12 0

AT

D F O ND

30

65

17 .0

Problem 3-58

70

E

RY

-°

F

AI R

.018

30

Exοerpts from this wοrk may be reproduced by instructors for distribution on a not-for-proΓrt basis for testing or instructional purposes only tο students enrolled in courses for which the textboοk has been adopted. Αny other reproduction or trαnsΙαtiοn ofthis work beyond thαt permitted by Sections ]07 or ]08 of the 1976 United Stαtes Copyright Αct withοut the permissiοn of the copyright owner is unΙαwful. Requests for permission or further informαtion should be αddressed to the Permission Depαrtment, John Wiley & Sons, Ιnc, Ι ] ] Riνer Street' Hoboken, NJ 07030'

Chapter 4

4-1

(a) comfortable (b) too warm

(c) comfortabΙe (d) too dry

4-2

(a) comfortable (b) too warm

(c) comfortable (d) too dry

4-3

(a) Assume sedentary dry using equation 4-4a,

bulb of 78

to,act =

75

-

F,

clo

=

o.5, met. = 1 .8,

5.4(1 + 0.5)(1.8

-

1

Relative humidity should be less than 50% (b) Should wear a S\Ι/eater or light jacket and slacks.

(clo = 0.8)

4-4

Use fig 4-1

(a) Summer,

to =

76 F or 24 C; Winter,

(b) Use equation 4-4a aS a guide, met =

3.0,

tdb

:76

F

\Λ/ith

to =

72 F or 22 C

clo = 0.2,

.2) = 71 F

o/

t

to =76

4-5

-5.4 (1+0.2)(3-1.2) = 64 F [winterorSummer]

From fig 4-3 temperature can rise about3.2 F.(j.g C)

t=68 +3.2=71.2 Fort=20+ 1.8= 21.8C

4-6 4-7

From fig 4-3 @200 fpm, temp rise ρ 5.3 F (2.9 C) with t,,"-t _ 9 F (5 c), temp rise ε 6.5 F (3.6 c) to = (t,

T,fn

+t^r)|2, then using Eq.

='6* Cν\l2

4-1

σg _Tr) = (53s)4 +(O.103 x 109) (4o)1Ι2(78_74)

tmft:82For27'8C to=(74+82)Ι2 = 78F or25.6C

4-8

Compute the operative temperature, Τ,xn

= φ4q4

to = (84

+

to

(O.103 x 1o911eo11/'(εo _76)= 83.5 F or 28.6 C

+76)12= 79.8 F or 26.5 C

From Fig 4-1, to

=

79'8 F and 50 % R.Η. is out of the comfort

zone. Recommend lowering to to about 77 F or 25 C. tu

4-9

x72 F

Use Eq. 4-4 to estimate a value of the operative temperature

to, active, assuming to for sedentary activities is 78 F (25.6 C) with met = 2.0. to, active = 78

-

5.4 (1 + 0.5) (2

-

1.2) = 71.5 F, (22C)

Exοerpts fiom this work may be reprοduοed by instruοtors for distribution on a not-1br-proΓit basis for testing or instruοtionaΙ puφoses only to students enrolled in οοurses for whiοh the textbook has been adopted' Αny οther reproduction οr trαnsιαtion of this wοrk beyοnd ιhaι permiιted by Secιions 107 οr ]08 ofthe Ι976 Uniιed Stqιes CopνriPhι Αcιwithouι ιhe oermission οfthe cοpyrighι οwner is unlωυful.

Αs an approximation Tmrt = 2To

_Τ,

and

Tflx =

Tno

* ci1Ι21Tg

_ Τ,

)

Eq '

(4-1)

eliminating Tmrt between the 2 equations

2(Τo_T3)4

= Tno

*CV1/21τn _Tr)

where all temperatures are absolute

Solve by trial and error with T, =72+ 460

=

532 R

and Te =(71.5+460)=531.5 R, C=0.103 x 1Oe, V=30

ta=85F(30C) Cold surroundings require high ambient air temperature for comfort, even with high activity level.

4-10

(a) Most occupants will be uncomfortable because the relative humidity is more than 60%, even with

trx

=

t,

(b) The lightest weight possible. Short sleeves, shorts, open neck, etc.

(c) Lower relative humidity

if possible by adjusting the cooling

system to remove more moisture. CouΙd also increase the relative air motion to highest values, perhaps use fans.

4-11

(a) Even

if the suit

was heavy weight, many executives would be

ΕXceφts from this work may be reproduοed by instτuctors for distribution on a not-for-pro1'it basis for testing or instructional purpοses onΙy to students enrollοd in οourses for which thΘ teΧtbook has been adοpted. Αny other reproducιion or trαnsιαtion οf ιhis νοrk beyond ιhαt permitted by Secιions ] 07 or 108 o{ the ] 97 6 United Stαtes Copyrighι Αcι ινithout the permissiοn of the coρyright oινner is unΙωνfuΙ.

cool if sedentary.

(b) Would definitely be cold, especiaΙly hands and feet.

(c) Probably would be comfortable

in typical work cΙothes

(d) Probably would be comfortable since they would keep their coats on and would be walking around.

(e) Cold to very cold

4-12

Determine relative temperatures difference between inside and outside.

68 - 45 23 = 7 4 _ 45 29 Costs 74 68

4-13

- 45 29 - 45 23

are79o/o of that for increased setting, or

Costs are increased by

26o/o if

thermostat is raised.

Too much air motion in the cold winter months tends to cause drafts and make people uncomfortabΙe. Air velocity just sufficient to prevent large temperature gradients from floor to ceiling is best for winter. Τhe opposite is true for hot summer months. Higher air velocity tends to compensate for high temperature and humidity.

4-14

(a) Raising the chiΙled water temperature will cause the cooling coil to operate with a higher surface temperature and the relative humidity in the space will tend to rise if the latent heat gain is signifΙcant such as would be the case with many occupants, this could lead to u

ncomfortable cond itions.

(b) Yes, during the unoccupied hours the space load may be almost totalΙy ΕΧοerpts from thrs work may be reproduced by instructors for distribution on a not-fοr-profit basis for testing or instruοtional puφoses only to students enrolled in οourses for whiοh the textbοok has been adopled' Αny other reproducιion or ιrαnsιαιion of this νοrk beyοnd ιhαι permiιιed by Sections ] 07 or 108 οf the Ι 97 6 United Stαtes Copyright Αcι τν ithοuι ιhe permissiοn οf the cοpyright oτνner is unlατνful.

SensibΙe heat gain and the load is much less than the design value. ln this case the chiΙled water temperature may be increased.

4-15

Τhese fans may bring air down in the Summer, increasing the velocity of air in the occupied zone and providing improved comfort. ln the winter, air may be drawn upward, pushing the warm air at the

ceiling downward where it can increase the temperature in the

occupied zone without increasing significantly the air motion below the fan.

4'16

(a) Τable 4-2 gives a minimum required amount of ventilation air of 15 ft3 /min per occupant. this is the minimum amount of outdoor air that should be used under any circumstances.

Therefore, (Qo)rin = 15(30)

=

450

ft3/min

(b) on the basis of floor area, the occupancy wouΙd be 25 and the minimum ventilation requirement would be

Q,

=

15 (25) = 375 ft3 /min. lt would be better to design for

floor area if lowest air flow is desired. With 30 actuaΙ student air flow is such a case wouΙd be insufficient.

4-17

Use Eq. 4-5, Solving for C, Cs

=

(QtC"

+

N)/Qt

=

C"

+

(N/at)

= (2001196 + (O.25l9oο)

Ξ

478 x 1o-6 = 478 ppm

ΕXcΘφtS from this work may be reprοduced by instruοtors for distribution on a not_tbr-profit basis for tosting or instruοtional purposes only to students enΙoιled in courses for which the textbook has been adopted. Αny οιher reprοduction or ιrαnsιαιiοn of ιhis νork beyond ιhaι permiιιed by Secιions ] 07 or 1 08 of ιhe Ι 97 6 Uniιed Stαιes Copyrighι Αcι w ithοul the permission of the cοpyrighι oιυner is unlαινful.

or using Sl Units

c, =

4-18

=

(2oo / 106)+ (0. 118 t 0.472x 9OO)

(2OO /

106) + (278 t 106):478 ppm

n = number of people to occupy a room

N=n(5.Oml/s) Solving Eq. 4-5 for Ν N = Qt (C, n

- C") = n (5.0) ml/s-Person

: Qt (C, - C")

:

/ (5.0)

2.8 (1000-280) / 5.ο

n = 403 persons or 0.0069 m3 /s

- person

For English Units: n=

6000 (1oOO - 28Ox 10-6) / O.O107

= 404 persons

4-19

or 14.8 cfm/person

Use the M-100 media of fig. 4-8. From table 4-3, select a

12x24 x 8

unit; 650 cfm, ΔP = 0.4 in. wg

At ΔP = 0.25 in. wg. each unit will handle Q

=

Ql |o'25 Ι o.40]1l2

=

650 [O.25 t o'4oJ1l2

:514 cfm/unit. Then the number

of units

EΧcerpts fiom this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructiοnal puφoses only to

StudΘnts enroιιed in οourses for which the teΧtbook has been adopted. Αny oιher reproduction or ιrαnsιαtιon οf ιhis νork beyond ιhαt by Secιions 1 07 οr 1 08 of ιhe Ι 976 Uniιed Stαtes Copyrighι Αcι lυithοuι ιhe permission οf the copyright oνner is unΙανful.

permiιιed

;g μ=(2000

4-20

l

514) = 3.89 or

4. This

is a satisfactory number.

Use the M-100 media from TabΙe 4-3 select a O.3 x O.6 x-O.2 unit. This is rated at 0.3 m'/s with 1oo pa pressure drop. 60 pa the alΙowabte flow rate for each unit would be Q = (0.3) (60/1 OOf tz = 0.23 m3/s

Αt ΔP

=

require 1'Oolo'23 = 4'34 units. This requires at Ιeast 5 filter units, but since this is an odd number, recommend 1.OO m3/s wouΙd

using six units. Trying the 0.6 x 0.6 x 0.2 filter the allowable flow per unit would be

Q

= (0.62) (60/1}q1t2 =

0.48 This would require more than two units of this size. Εconomies would determine the best choice.

4-21

Solving Εq' 4-1O for Q Q

=

Qr [ΔP / ΔP,]1'2

N = ss00/235 = VeΙ = Q/Α =

4-22

= 9OO

[o'1 l0'15]1|2 =735 cfm/module

T.4g [must be integer] Use g modules

ΨΨ (2)(8)= 344fpm '|-

=

5.7 fps

Solving Εq' 4-1O for Q Q = Qr [ΔP

/

ΔP,J1l2 = (o'42) |24

l 37

'4]1t2 =

0.336

m=(2.8)/0.336=8.3 Use 9 modules, a 3 x 3 arrangement.

Exceφts from this work may be τeproduοed by instructors for'distribution on a not_1br-proΓit basis for testing or instructionaΙ purposes only tο students enrolled in courses for which the textbook has been adopted. Αny other ,rpr:orl'u"Ιιon or trαnsιαιion of this work beyond thαι permilιecΙ by Sectiοns 1 07 or 108 οf the ] 976 t]niιed StαιeS Copyrι?ht Αcι w-ithοut thi permis'ιδ" oj ιn, copyrighι οwner ιi unlανful.

/3

VelocitY

4-23

=

a

(2.8)m3

FAcΕ AREA

M-200; 0.6x0.6 xO.2;

/s

(0.3)(0.6)(e)m2

O.4Om3

=1.73m1s

/s/module

Use Eq. 4-10 ΔP

=

Δη ta / Qι.]' :1OO

Velocity

4-24

=

a

A

lO,4OtO.42l2 =90.7 Pa

0.4 =2.22m1s (0,3)(o 6)

No solution exists due to the fixed air quantity for the unit. This part of the problem is intended to show the student that typical direct expansion equipment cannot be used in this \May. lt also shows that the load due to outdoor air is very large.

4-25

exhaust

sHF= 0.7

Γho

:

o'25 rh"; Locate point 1 on psychrometric Chart at82'4 F db and

66.8 F wb it = 31.4 Btu / lbm and v1 = 13.9 ft3 /lbm fbr tΘSting oΙ instructional puφosΘS only to Exοerpts from this wοrk ιnay be reproduced by instructors foΙ distribution on a nοt-for-proftt basis or ιrαnsιαtion οf ιhis wοrk beyοnd ιhαι permiιted students errrolled in courses fbr whiοh the textbook has been adopted. Αny other reprodicrion is unlανful' by Sectiοns 107 οr ]08 ofthe t976 Llnited Stαtes Copyrilht Αctv,ithοut ιhe peιmissiοn ofιhe copyright oνner

74

Q1, =

Φl

rhi (ii - is) = at

=

/V

35ο Ι 12,ooo Qls is = 31

.-

,

(60)

(i1

-

is)

= (350 l 12'000) (6ο / v1)

''''J8?rε;

=

''

(i1

_i.)

23'46 Btu / lbm

Locate on psychrometric chart' ts = 65'6 F db' 55'5 F wb Q.r =

lil,

tr,

= Γil1

O,

=

(ir.

=

- is) = 36'000;

^ -ΨΨ- 23.46)

(27 .6

rh, (vr,

=

=

:8695'7

Ψ(13.4)

Qt" = 8695.7 (31.4 -

Qr

ir =27

=

'6 Btu / lbm lb / hr

1940 cfm

23.46) = 69,000 Btu / hr = 5.75 tons

5.75 (350) = 2014 cfm

(ο) Design filters for 2014 cfm, use M-200 media

of fig 4-8.

Try the 24x24x8 units of table 4-3. 920 cfm @0.4 in. wg.

For max. ΔP of 0'125 in.wat. Q = 920 [0. 125 tO.4O]

1t2

=514 cfm / module;

n = 2014 I 514 = 3'92, use 4 modules

4-26

Use the M-15 media, η

=

93 % from fig' 4-3'

From table 4-2,60 cfm / person is required, outdoor air. purpοses only to on a not-for-profit basis for testing or instruοtional Excerpts tiom this work ιnay be reproduced by instructors fοr distribution thαι permiιιed beyond work thιs or trαnslαιionλf Αny oιher reprοduction students enrolled in οourses for which tl-ιe textbook ι-,u, υ..n uαopt"J ' copyright oινner is unlαwfuΙ' ofιhe permission thi \,'ithout Αcι Copyrighι Stαtes tJnited ιhe-|97|6 108 o7 by Sectiοns ] 07 οr

75

-A fresh air balance on the filter gives QrEt

+

Qo

=

Qs

where Q,. is recirculated air, Qs is outdoor air and

Q. is supply

air.

8, = (60 - 20) 10.93 = 43.0; Q,

=

43.0 +20

=

63.0 cfm / person

or the total amount of air supplied is

Qτ

=

63.0 x 55 = 3465 cfm; Try the 12x24x8 unit of table 4-3

Q/unit=9oO[O'1 /O.35]1Ι2=481 cfm; n =4755 l481 = 7 .2

modules

Use 8 modules [Note: The M-24 media could also be used]

4-27

Q, = (25 - 15) / 0.S = 12.5 cfm / person

Φ.

4-28

= 15 +

12.5 =

27 .5 cfm / person

Filter location is B, figure 4-9

Use Eq. 4-12, solve for RQ.. since RQΓ = { -QoEv[C,

_(1_Et)Co] + N}/ (EvEfcS)

RQr={-2OOxO.85[180-(1-0.8)0.0]+(10x150x35'32)]l (0.85 x 0.8 x '180) where Co

RQr

=

=

0.0

185 ft3 / min or cfm

for-testing or instruοtional puφoses only tο Exοerpts from this work may be reproduced by instructors for distribution on a not-ibr-profit basis of this νork beyond ιhαι permitted ιrαnslαιiοn or reprodiction Αny other adopted. been has the textbook students enroΙΙed in οourses fοr which is unlιτνν.ful' oνner copyrighι of ιhe permissiοn the Αcι ιiithout by Sectiοns Ι 07 or 1 08 οf the Ι 976 tJnited Sιαtes Copyrιght

76

Qo

4-2g

= 2OO

cfm, Qs

:

(1S5 + 2OO) = 385 cfm

Solve Εq' 4-11 for RQ, RQr

=

(_Qo)(Eu)(cr)+ N / ΕrEiC,

RQr = [ (-20) (0.65) (220) + (125) (35.32 ft3/m3)] (0.65X0.7)(220)

-Ψ9-*_!1!5^, ' =(0.65)(0.7)(220)

RQ.

4-3o

= 15.53 cfm/person

For filter location A, use Εq. 4-1 1, solving for RQ,

RQr

=

(-QoEvCs

+

N) / (EvEfCs)

RQr = t (-2OO (0.85) 180) + (10 x 120 x 35.32 ft3/m3 (0.85 x 0.8 x '180

RQr

4-31

=

)l

t

)

183 cfm, Qo

= 2OO

cfm;

d,

=

383 cfm

(a) This type of space will require a high ventilation (supply air)

rate to handle the load, air cleanliness is not the main criterion.

Therefore, a low efficiency filter with low pressure drop is acceptable. From table 4-2, assume occupancy will be about 30 persons / l

OOO

ft2. So the total design occupancy is 90

persons. Τhe design will be based on this occupancy although the cooling requirements may dictate a larger supply air rate. 1br distribution οn a not-fοr-prοfit basls for testing οr instructiοnal purposes only to ιhαt permitted students enrolΙed in courses Γor which the textboοk has been adopted. Αny οther reprοducιion or trαnsΙαtion of ιhis νork beyond by Secιiοns 1 07 or ] 08 of the Ι 97 6 [Jniιed Stαtes Cοpyright Αcι νιthout the permissiοn οf ιhe copyright oνner is unlανful.

Exceφts from this work may be reproduced by instruοtors

77

A ''fresh air'' balance on the filter d," = (20

gives

Φ,

=

(Q" _ Qo) / Ef

-15) / 0.5 = 10 cfm / person recirculation rate

4-31 (continued)

Φ, :

1O +

1

5 =25 cfm / person supply rate

Qτ = 25 x 90 = 2250 cfm total supply rate Net face area, (b)

Αf

= 2250

/ 35o

= 6'43 ft2

A higher efficiency would reduce the total amount of air and

reduce the required face area. However this is not desirable in this case. First the filter system would have to be enlarged to handle the greater amount of air. A lower filter efficiency could be used and still maintain the required air quality. For example, suppose the load dictates 4000 cfm instead of 2250 cfm, then for 90 PeoPle

Φ,

=

4ooo / 90 = 44'4 cfm / person

Using a minimum of 15 cfm / person of outdoor air. Qr. = 44.4

8. Et

4-32

(a)

-

15 = 29 '4 cfm / Person

29.4: (20 - 1s) / Er = 5 Ι29'4 : 0'17 or 17oλ required =

Q=

(Q, / v) 60

(i|.

l5x

-ir)

Εxοerpts frοm this work may be reproduοed by instructors for distribution on a not-for-profit bι studeπs enrolled in courses for which the textbook has been adopted. Αny oιher reprοduction ο by Sections ] 07 or ] 08 of ιhe 1976 Uniιed SιαteS Copyrighι Αcι ν)ithouι ιhe permission οfιhe cop;

I

225 people

75F

RH=5ο%

125,ο0ο

78

Υ x13 ft3

/ Ιba

Φ. = (125,oOo x13)

Qs (b)

=

Ι [

60 x (28 - 1e.4

3,149 cfm

Φ,

= Φo

Φ,

=

= 15 x225

3,375 cfm

-D (c)

)]

Q. must be 3,375

cfm, find

50 52

ne\Λ/

Supply air condition

125,000 = (3,375 I 13) 60 (28 - i.)

i'

= 28

-

(125,000 x 13 ) / ( 3,375x 60)

=

20 Btu i lba

Locate new condition on chart aS Sho\Λ/n' Coil must cool oDA down to this new condition.

Exceφts frοm this work may be reproduοed by instruοtors Γor distribution on a not-tbr-profit basis for testing οr instructional purposes only to students enroΙΙed in courses for which the textbook has been adopted. Αny οιher reproduction or ιrαnsιation of ιhis work beyond thαι permitted by Sections 107 οr ]08 ofιhe )976 Uniιed SιαιeS Cοpyright Αct withοut the permissiοn οfιhe cοpyright oνι]ner Β unlα:wful.

Exοeφts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instruοtional purposes only to students enrοlled in courses for which the textbook has been adopted. Αny other reproduction or trαnslαtiοn of this work beyond thαt permitted by Sections Ι07 or ]0B of the Ι976 United Stαtes Copyright Αct without the permission of the copyright owner is unlαwful. Requests for permission or further informαtion should be αddressed to the Permission Depαrtment' John Wiley & Sοns, Ιnc, ] ] ] Riνer Street, Hoboken, NJ 07030.

Chapter 5

5-'1

Χ4 =O.8 (Btu

(a)

k = CΔx =0.2

(b)

k = 1 .14 x 0.1 = 0.114 W / (m-C)

_

in) / ( hr - ft2

_

l

5-2

(a)

C=k

5-3

(a)

R = 1lC= 1/0.055 = 18.3 (ft'-

hr- F)/

R' = R l A= 1 l CA= 18.3/ 1ο0

=

Δx=o.3o / 5.5 = o.o55 Btu / 1ft2-nr_ (b) C = O.O43 / ο. 14 = .307 W/ (m2 _ C)

F)

11

Btu

0.183 (

hr-F)/Btu

(b) R ='1 I .307 = 3.26(m'-C)/W R'=3.26l9.3=0.35C/W 5-4

ΣRi , Rgyp =1ΙC=1l3'1=o'32 Rbtd = !0.33 = 3.03; Rair = 0.68

R=

R = 0.68

+

0.32+ 3.03

+

R-0.68

R=0.68

0.32+ 0.68

R=0.32

R = 5.03 (hr- ft2- F) / Btu

5-5

tnb Γ2

ιn2

+ η R'= 2πk| 2πkoL AssumeL=1ft

kι:0'2

Btu - in' t(ft2

_hr_F);

kp =314 Btu-in

tσe _hr_F)

81

lnside Surface (7 m/s

0120

)

=

overalI Τhermal Resis.

0.652 m2clW

Between Frame

Αt Framinq

Outside Surface

0.17

0.17

4 in. Face Brick

0.65

0.65

Sheathing

1.32

1.32

lnsulation

1

5-10

1.0

4.27

2x4 stud

Gypsum board

0.32

0.32

lnside surface

ο.68

0.68

14.14

7.41

Τotal

UA:U1Α; +U1Α1, U = UiAi Αr

A

=

14'5

A'

16 16^nd A =lΞanα

/Α + U1Α1/A

υ=L R

0 o77Btu , =ΓΨ "14.14) ΓΨ " +1= : 7.41J " -)116 110 5-11

/ (r',..-tt'

-r) A

An ordinary walt with ε

=

O.9 has a unit resistance of 0.68.

highIy reflective wall, ε

=

O.05, has a unit resistance of 1'70.

Assume radiation heat transfer is zero for reflective wall. Τhen the resistance due to convection alone is approximately

Rc=1.7; hc=1/Rc=0.59; h.*r:1/0'68 = 1'47 Frac. Conv. = ha lh. *,

=

0.59 I 1'47 = 0'4

by instructors fοr distribution on a not-for-profit basis for testing or instructional purposes only tο has been adopted' Αny οther reprοclucliοn the texibook *nιοh

from this work may be reproduοed Εxοerpts "'.Φ;.;;;1l;l,,.o,rr.r'r-

o^:.:::Ei::::":!,::iχ:,i''Ψ'y''o"!:*!

-,

-

-

,

82

RzQ x 4\ Rr (2 x 6) wind. 0'17 0'17 1. Outside surface, 15 mPh 0'79 0'79 2. Siding 1'32 1'32 3. Sheathing 11'0 19'0 4. lnsulation, 4'27 2x4 6.7 2x6 0'32 0'32 5. GYPsum wall board 0'68 0.68 6. lnside surface 18'55 28.98 Total

5-12

Αssume 15 mph

Ut

=

0.035 Btu / (hr - ft2 - F)

υ2 __O.o54 Btu / 1ιlr - ft2 - F1 % DiffereΠCΘ = [o'O5-4r_0-035) ι'' )'

5-13

0.0sη

[

ool = 35'2

Air space will be near the indoor temperature with small Δt across the air Space.

Use t."rn R

5-14

1.oz(rrr

50 F and Δt

=

=

10 F and read

-f( -r) I Atu

Assume tr"rn R

5-15

=

=

=

50

F;

3.55 (hr - ft2 _F) /

[Tabte

S-3a]

or 0.18

m2clW

Δt = 10 F

Btu

or 0.62

(.2

_c/W)

[Τabte 5-3a]

qc/Α = U"Δt Find U for highly reflective surfaces because radiation will be minimal. This will give a good approximation for the convection component. From Table 5-2a,l1orz', heat flow down

83

Uc=1/R = 1l(2x4.55) =

0'11

q./Α or

" --l' (2x0

U^ _

= o.625; q.

/Α

0.625(63 - 43) = 12'5 W/m2

=

/ a '.4l 'l

Γ, _ '4 |l

_[ l' l _ι1οo]

l( =σ'n'Lι1oO]

Q/A..

for ε1= t2=O'9, (q/A),.

= O.1

-]'

E : 0.82,

713 x9.s2 t635

l'

σ' o

_s.τol

m

Radiation heat transfer is about 10 times greater'

5-16

U*

= O.O7 Btu / (hr - tt2

Ud = O.4O Btu / (hr - ft2

-

- F)

Uwin = O'81 Btu / (hr - ft2

Ad

Α*

=

11

g'

- F)

17 '78 ft2; Awin =25'0 ft2;

= 117 '2

20'

f(

Parallel heat flow Paths

UΑ

ι

Iι I_

=

5-17

=

U*A*

+

U6Α6

+

U*;nΑ*in

(o.O7 x117'2) + (O.4 x 17.78) + (0'81 x25'0) 117.2

O.3O Btu /(hr -

q/Α

f( _F)

or about

1

'72\Ν t(m' _

c)

= U(ti _to)

a not-for-profit basis foι testing or instructiona] purposes onιy to Exοerpts from this work may be reproduced by instruοtors for distribution on reprολucιion or ιrαnsιαtion of ιhis work beyond ιhαι peπnitted Αny οιher uJo|t.α. students enrolled in courses tbr wbich the texibook has been : : -'-ι.41-^

^^^'',.:/Ιa!

^''''aΔv;"

"'-Ι6!$ιΙ

84

From Table 5-4b, construction 2, R = 8.90 (hr

Assume insulation does not

fitl the

-

ft2

-

F) / Btu

airspace'

tt' - f)/Btu and Remove R for metal bath and plaster of 0.47 (f''ι. ) \"' add R for acoustical tile and insulation' Ceiling, R" = 1 / 0'8; insulation'

q/Α = o'o48 (72 - 5) =

1

9!+=oe71wr(m2-c) \ 0.1761

Ud = 2.27 Uwin =

-a)'

Wl(m2

4'62w1(m2 -

/

Table 5-8

")'

Table 5-5b

Αw

=

35 m2;Αwin =8m2;Αd = 2m2

UΑ

=

U*A*

+

U6

1x35) u _ Q.e7

+

+

Αα

+

-Fi

3'22 Bτυ / (hr _ ft2)

From Table 5-4a, Construction

Uw=

1'00;

0'048 Btu / (hr - ft2

R1e61=20.68; U = 1/R =

5-18

R='1

UwinΑwin

(2.27x2)

+

35

(4'62x8)

-_

2.16 w

I

(m2

- c)

5-19 U = O.14 Btu / (hr

_

ft2 _ F1τable5 _ 4a, Construction No. 2

R=1ΙO.14 = 7'14, Rn =7'14_(1 to'44) + (1/0.55) =6.69 Un = 0.15 Btu/(h r

5-20

-

ft2

- F) or about ο.85 W

l

1m2-c1

Αssume Ηardwood, k = 1.25 (Btu-in) / (hr - ft2 - F) Winter Summer Rι =

0.68

Ri = 0.68

not-for_profit basis for testing or ιnstructionaι purposes only to ExcΘφts from thts work may be reproduοed by instruοtors for distribution on a reproλuction or ιrαfisιαtion of ιhis ινork beyond thαι permiιιed students enτolled in courses for which the teΧtboοk has been adopted. Αny οther

85 Rα =

'1

'375

l

0.25 R, = 2.03

1.25

Rα = 1 '375

l

1'25

Ro = 0.17

Ro =

R* = '1.95

-F)

U, =0.49 Btu / (hr-ft2

U* =0.51 Btu /(hr-ft2-F)

Both values are greater than the value given in

Τable 5-8 of O.39 Btu / (hr _

5-21

Computed: Ri Ri

-

+RgaRo=0.96=R

:

Utub

-r);

\Ν

l(m2

\

From Table

_c); /'

5-5

Rn

l l,.Jfl

=

Table 5-5a

U=1.08 Btu / 1nr - ft2 - F)

-f( -r)

=U-R', ' ., = ++ 1.og

Uw

computed

Table 5_5b; Same result

(b) Αssume tr"rn : 50 F; Δt = Ras = 1.ol (nr

-c)

or 5.92 wl(m2

_r);

1.O4 Btu / (hr - tι2

or 5.91

5-23 (a)

F), but acceptable.

= 0.68, Rs = 0.03 (estimate); Ro = 0.25

U = 1.04 Btu / (hr - tt2

5-22 (a)

ft2

1

10 F

r atu

.O1

o.o89 Btu /(hr - tt2 _

_ ο.o29 Btu lJιur\ιll/ (hr --ιLft2 -ν'νLζ)

= 1'94, Un = O.52 Btu/1ιlr-ft2-F)

11

_F) _| l

or O.51 W/(m2 _ c)τaοle 5-9

(-'_c) -)

ιJlv'l\J vYl1ιιl or O'16 W/

τante 5-1o

(b) Q=UΑ(ti _tg); tr= t"ur-A

Εxοerpts from this work may be reproduοed by instruοtors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in οourses for which the textbook has been adopted. Αny οther reproducιion or trαnslαtion οf ιhis νork beyond thαt permitιed ..-_''r-] ^_':'_^'-

!

^'.'.ι ^'.4'ι

86 tavg

A

=35'8

F

:22

F

(2'44

(12

C)

5-23 (continued)

=

R"

=

R1,

Rrin =

5-25

*

)

0.ο89 (4 x 20 x7) (72 - 1 3'8) = 2,900 Btu / hr or 0.85 kW

qn = 0.029 (20

5-24

Figure 5-7

tι:72re2c)

tg__35'8-22 = 13.8; q*

C) Table 5-11}Chicaαo.lllinois

Un =

x20) (72 - 13.8) = 675 Btu /hr or 0.2 kW

0.029 Table

5-10

1l 0.48 Table 5-1a (Fibrous Pad) 2.08

36.6

Re

=

#"+

U"

=

0.027 Btu / (hr - ft2 - F) or 0.16 Wl(m'z - C)

(a) R*=: "0089+11+

=

(1 t \

g.1) =

22.6

U* = 0.044 Btu/(hr-ftu-F) or 0.25 w1m2-c)

Rfι

=

+ (1 tO.4s): -: 0029

36.6

\

Un = 0.027 Btu/(hr - tt2

-f)

or 0.155

Wl(m'z-c)

(b) Refer to problem solution 5-23 Q* = 0.044 (4 x20 x7) (72 - 13.8) = 1434 Qn =

5-26

0 027 (20 x20) (72 - 13.8) = 629 Btu / hr or 0.18 kW

Rins =

C

=

Btu / hr or 0'42k\^Ι

#

=

4'1z

O.24Btu/(hr

(rrr

-'f(

_f( _r)l

εtu

-F) or 1.36 Wl(m'z - c)

Excerpts fτom this work may be reproduced by instruοtors for distribution on a not-for-prοfit basis for testing οr instructional purposes only to

studentsenrolledincoursesforwhichthetextbοokhasbeenadopted. 'l-''':'''

bνSectiοη.s1[]7ny ιnρn{t]ιo 1076'ΙΙb;l.)ζl-l-"/-^4'-')-ι"}

ΑnyοιherreproductionοrιrαnsΙαtionofιhisνοrkbeyondιhαιpermiιιed :"'''.Ι^''Ω'Ι

l

λ / .*'

87

Then from Fig. 5-8, U' Q=

5-27

U'P

=

0.85 Btu/(hr

- to) = 0.85 x 300172 -101

(ti

tι =

72 F (22 C) Assumed

R"

= R5 +

Rfi,.,,

-f(

=R1

R1η, Ub

=

=

-F) or 1.47 W(m-C)

15,8'10 Btu/hr or 4.63 kW

o.ο52 Btu / 1nr -tt2 _

11 Table

5-9

+Ru6*Rqyp=(5.0) + 0.0 +(1 12.22) = 7.22

1 +7.22=26.5 R^- 0.052 5-28

o.o38 Btu / (hr -ft2 _F) or o'22ννl(m'z_c)

U"

=

Ub

: 1.14

Rn

=

from Table 5-9

++ 1.14

O.7 + (1

t

12'6)= 1.66 (m2-c)Λ//, Un = 0.60 W(m2-

c)

or

Un = O'1OO Btu / 1nr -ft2

-F)

This does not account for the walls above grade.

5-29

U = 0'16 Τable 5-'10 (no finish)

Rn:++(t69)+ ' 0.16 \ Un

5-30

(1 t4.6) =6.611m2-c1

= 0.15 W/(m2-C) or 0.027 Btu / (hr

q/A=

Un(ti

-ts)=

(ti

= ti _UnR1(ti

- r)

-tt)/Rt=(tt-t)lR2

Rl=Rgyp+R1nr+R1, R1 = tl

-ft2

lw

(112.6)+ 0.7 +0.12=

0.90

_tg) =20 - [1.05 x 0.9 (2ο - 10)]

EΧcerpts frοm this work may be reproduced by instructors for distribution on a not-for_pΙofit basis fοr testing or instruοtional purposes only to students enτolled in courses for which the textbook has been adopted . Αny οther reproduction or trαnsltιιion of ιhis νοrk beyοnd ιhαt permiιιed '.''-:-ι'ι ^'"-^-:' "'-ι5υ{ιi ' hιi 9λ"};.-" ιηa ^'' ]Λ9 ^.}L- ιoaA ι L-;.^) c'-''^'

. . .

88

t1= 14.6 C or 58F R2

=

Rrr, +Ri

=

(

I 12.6) +0'12='20

tz = 20_ [O.60 x O'2 (2o-1ο)] = 18.8

5-31

q/A=Un

(ti

c

or 65.8 F

_tg)=(tι _t'')/R1; Rl=Ri +Rc

= 0.12 + (1 I 4.6) -- 0.34 tι=20_ (O.15) (O.34) (20-10) = 19.5 c or 67 F

5-32

C

= 0.2 Btui(hr-ft2-F);

U'P

8=

U'ni

=

(t1

_to);

Figure 5-8

Ui, = 0.81 Btu/ (hr-ft-F) or 1.a

1.37 Btu / (hr - ft -F)

(a) q/P = 0.81 (70 - 5) = 52'7 (b) q/P = 1.37 (70 - 5) = 89'1

5-33

Q=

W(m-C)

Δt / R'

;

R'=

Eq. 5-25; L>>Ζ'

Btu / (hr - ft) or 50.7 W/m Btu / (hr - ft) or 85.6 W/m

L=

100ft

2πkL

. Γzoo x121Γ^ tn(12x1OO/2x3o)l

_'nL-o--@J _3"100Ι12) R'

= 8.

12

x

1O-3 thr - F) / Btu

Which neglects the resistance

of pipe wall.

purposes only to for distribution on a not_for-profit basis for testing or instruοtional Excerpts from this work may be reproduοed by instructors οf this νork beyond ιhαt permitιed οr ιrαnslαιioi reprολucιiοn oιher Αny uJo|i"a. υ"., hu, '' : '1-"4'ι Students enΙollod in courses fbr ινh'iοh the textbook "' ^' L- Q.nl;n-"

ιnη ^.' 'Λο

70-42 - =3,4488tu 6' 8.12 x 1O-'

/ hr;

q/L

= 34.488tu/(hr-ft)

1W/m orq=1.01 kW; 9=33 L 5-34

Q=Δt /R'

R'g

=

=4.98x10-3 C/W

2π (1.4) 100

film and the tube wall' Neglect resistance of the inside

.

'

O=

5-35

60-5

----------c =11.04kw 4.98 x 10--

Moisturewillmovetowardtheinside.Locatethevapor

insulation' retardent on the outer side of the retardent is placed on the The insulation will beοome wet if the plywood would probably \Λ/arp inside or left out entirely and the

and rot.

5-36 (a) Q/A = Uo(ti Ro = O'68

+

_to)=(tι _t1)/R1 =(ti

O'45 +1

1+

1'O +O'8 + O'17 = 14'1

Uo

=

_F1 O'o71 Btu /( ιrr - ft2

Rr

=

0'68

R2

=

O'68

+ +

O'45 = 1'13( hr O'45 +11

_t)lR2

f(

-F) / Btu

-12'13(hr

- ft2

-F)

/ Btu

tt=tι_R1Uo(tι_to)=7O_(1.13xO'O71)(7O-1O)=65.2F purposes only to basis for ιesting or instructiona' for distribution οn a noι-for-proΓlι ιyt.pern:ted insιructors by reproduced be lnay Excerpιs Γrom ιhis \νοrk j:*tbook has *9,i'"Ιl'Ji."o"'i,-:';;;;;;;;;;;"Ζii"111i117; γ,:y:γ:'::'|"*o students enrolled in courses foΙ ^γηι:ι lh.

90

tz=70-(12'13 x 0.071) (70-10) = 18'3 F of air to surfaces

(b) At 70 Fοο, 3O%o R.Η. and possible leakage

1

or 2-

tdp=37F<65F-Πocondensationexpected (c)Since|z=lS.3Fismuchlessthanthedewpoint,condensation would oοcur' 1' Place vapor retardent at the location of interface

5-37

Assume infiltration is negligible ufAf (ti - t") = UwA*(t. - to)+ U'P(t. - to)

=

rilcp(tc - to)

+(U*Α* +U'P+rhcp)to',ti=72F; (UtAt + U*Α* + U'P + rhc, )

*ιc __UrAtt,

Αssume 1.5 in. of wood floor, Pine;

=1; 'Rf

U,

Rwooο =

to =

1o F

'1'5l0'8 =1'88

Rt =O'92+1'88+O'92 =3'72;Ut=0'27

UΛr= 0'27 x 30 x 60 = 484 Btu/(hr-F)

rr -a'R* "*_Rw

νv

=0.68+(6/15)+ O'17 =1'25',U*

=

O'80

U*A* = O.8O x 2(30 + 60)2 -- 288 Btu/(hr-F)

U'P=1.8x(30+60)2=324 ,h.p

=

20 x 0.075 x 60 x o '24

x72 + ιc_-484

+ (288 + 324 +

= 21

'6 Btu/(hr-F)

21'6)10

484+288+324+21'6

5-38

=

36.85 F

(a) Q/A = U(ti -to) = (tr-to)/R1;

(construction 2) + Uz=0.112', Rr=O.17 + 0.33 + 4'17 2'22 = 6'89

only to basis foτ 1e^sting or lnstructional purposes ' by instructors for distribution on a not-for-pτofrt permitιed ιhαt beyond νork ExcΘrpts from this work may be rοproduced ιhis of t.d. ,ι-ny otlr", ,rproEu.ιiοn οr ιrαnslαιioλ ι-'", υ"* students enτolled in courses fοr whiοh the textbo"t "J"p

91 t1 =

(6.89 x 0.112) (72-0) = 55.6 F

(b) U : Rr

0.211 (construction 1)

= 0.17 + 0.33

ti =to+R1U(ti

(c)

+2'22

=

2'72

-to)=O + (2.72x0.211)(72-0) = 41.3F

lf room air leaks into the air space for the case of no roof deck

insulation (b) there could be some condensation since

t6p =

50

F at72 F and 45% RΗ. With the insulation, no condensation would be exPected.

5-39

(ti

-to) / Rr

= (ti

-ti)

/ R1

R, = 4'5 or O.79 1m2-cyw ; Τable 5-4a (Const. No.1) Rτ

=

0.68

+

O.45 + O.94 =2.O7 1nr-tt2_F1/Btu or 0.365 1m2-cμru

Between Furring and block

tl

= ti _

ft,'

tdp = 9.5

_ to) = 22_

ffiιr'+17)

= 3.98 C or 39

F

C, Assuming room air can diffuse into the air space,

condensation likely will form on the concrete block surface. Therefore, place vapor retardant on inside surface of gypsum

board. Use foil backed retardent. Retardent must not touch concrete blocks!

purposΘS onιy Exοerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instruοtional ,.,-r.':'!!i|',?! .o,,'",'to. *ni:| j:;boοΙ has been adopted.

;;#;;;Ι;ln

lhΘ

o'l

"!!1:

to

_

":::::i:::::":!,*r::!'Ρ'n! 'y,'o*Ψ*o

,

92

5-40

U1A1(ti-tn) + UaΑ+(ti-tn) = 2U3Α3(tn-to) + lJzAzftn-to)

_ (UlAl + UzΑz )ti + 2UgΑgto + UzΑzto '' 2υsΑg + UzAz * UlΑl * UqAι

,^

UrAr

=

UzAz

0.09 x 8 x 20 = 14.4',

=

U3Α3

=

0.09x3x8

= 1.08

UιAι=0.09x3x20=5'4

0.09 x 8.54 x20 = 15.4',

_(4.4+15.4)70+(2 x 0.8 x o) + (15.a x o) rL6= -S5.8F " (2 x 1.08) + 15.4 + 14.4 + 5.4 Place water pipes in this space with some caution. 5-41

Uf

Αf (ti-tb ) = (U*Α* + UυtΑt

tg = turg

Ut =

-Α - 37.6 _23

1

&

;

Rf

=

)(tυ -tg ) ; ti =

72 F

= 14'6 F or B C

(2 x 0.92) + (1 5/0.8)

+

2.1 = 5.82

Carpet and Fibrous pad assumed, Ur = 0.172 Btu/ (hr U*

=.164 Table

tb=

ltΑrti

U1Α1

+(

5_

U*Α*

9;

Uot =

0.029 Τable

5

ft2 - F)

_ 1o

+ Ubf Αf )tg

+ U*Α* *UυrΑr

x72) r, l'b-_(0.172x400

+ (.1G4 x 80 x 7 + 0.029 x 4OO)1a.6

to = 30.3 F or -0.95 C

Exοerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instruοtional purposes only to students enrolled in courses fοr whiοh the textbook has bοen adopted. Αny oιher reproλucιion or ιrαnslαιiοi qf ιhis νork beyond ιhαι nermiιιed by Seαiοns Ι 07 οr ! 08 n! n." t o7A t Ιr;lDs <}-t-^ .^^',-]-Ll λ οt 1η: ιΙ^^...'Ι- -

basis fοr by instructοr9 jor distribution on a not-for-prοΓrt Excerpts frοm this work may be reproduοed been has textbook enrοlled in οourses for whiοh the testing or instructiοnal purposes on1y to studeπs or ]07 Sections of this work beyond thαt permitted by adopted. Αny other reproduction or trαns,lαtion unlαwfuΙ' is Αct withλut the permission of the copyright owner Ι0B οf the ]976 Unitei Stαtes Copyright John should be'αddressed ti the Peimission Depαrtment' Requests fοr permissioln or furthir"infοrmαtion iiria sonλ, nr, ] 1 ] Riνer Street, Hoboken' NJ 07030'

CHAPTER

6

6-1

PSYC may be used to find the Refer to Table B-1 . The computer program RΗ' humidity ratio from t66 and assumed 100% Wind Direction, deg. CCW from N

OR

11

6

140

0.0

(b) Milwaukee, Wl

-2

13

290

0.0

ΑL

-9

4

10

0.0

24

12

340

0.003

(e) Αlbuquerque, NM

18

I

360

ο.0

Charleston, SC

28

7

20

0.003

(a) Pendleton,

(c) Anchorage,

(d) Norfolk,

(f)

VA

6-2 possible condensation on inside Design relative humidity is determined by (which is the maximum dewof glass. Find glass surface temperature poi;t temperature of the inside air allowed)' q/A = U(t' - to) = Cr(tr - b) tι= 72"F; t., = glass Surface temperature U = 0.65 Btu/(hr-ft2-F)' Table 5-5a

111

c1 uhi Cι = 1'172

, hi = 1.46 Btu/inr-ft2-f) Btυl (rrr-ft2-F1

94

tr=

Uti + to (Cr

cl

- U)

lndoor

City

Tου,

oF

tr=

Outdoor Του,

oF

top,

oF

Design or Max.

RH-%

72

-10

35.5

26.2

72

23

50.2

46.1

72

b

42.6

34.6

72

3

41.3

32.9

72

39

57.3

59.9*

(g) Boise, lD

72

-'16

32.8

23.6

Rapid City

72

I

44.0

36.5

(a) Caribou, ME (b) Birmingham, (c) Cleveland, (d) Denver,

ΑL

oΗ

CO

(e) San Francisco,

*

CΑ

RH = 60 o/o would probably be uncomfortable RΗ = 40 to 50% would be more realistic

6-3

Assume that the weather strip does not change the conveοtive heat loss. From Figure 6-2, Cp = 0.3. Using Eq. (6-7b) with the air density of 0 'F, the pressure difference due to wind is

ο: ΔP. =

[o

.086ψ!\'( ls*ot * |.467 fi

ft')\

mph) (o.rnr.o

z.(y.rrΙbm_ ft)

ι

ΔP* =

l'\

lbf_s')

ι

t"'tt ) lbf l ft')

0'037in'wg

Αssuming slight stack effect, ΔP

^y

0.04 in. water

Using Table 6-1 and Fig. 6-1, puφoses only to Exceφts from this work may be reprοduced by instruοtors for distτibution οn a not-for-profit basis Γor testing or instructional

ΑnyοtherreproducιionorιrαnslαιionofιhisνοrkbeyondthαtPermiιιed studentsοnrolledincοursestbrwhichthetextbookhasbeenadopted. by Sections Ι07 or Ι08 ofthe ]976 United Stαtes Copyright Αctνιthout the permιssiοn ofιhe copyrighι oνner is unlaννful-

95

Loose fit with non-\Λ/eather-stripped, K - 6; a lL = O.75 cfm/ft Loose fit with weather-stripped, K= 2; Q/L = O'24 cfmtft Total length of crack, ι = [(3 x 3) + (2 x 5)] x 9 = 171 ft Using Ll2for calculation, then Q,, = 0.75 x 17112= 64.1 cfm, Q, = 0'24x17112= 2O'5 cfm

I Now Q, =

Q,r-8,,

Q,,

rh cr(t1

-

to)

-At-4, Qt

=

v

cp(t;

-

tr)

64.t-20.5 64.r

= 0.68

or a reduction of 68% in sensible heat loss.

Also, (Kl

- K)lΚ1=

β-2)16 = 0.67 or 670/o Reduction.

6-4

From Fig. 6-2, Cp = 0'52tor windward wind' Assuming standard sea level air density, the pressure difference due to the wind speed of 13 m/s is

ΔP., =

z.( ι.okg

Βry] s) _*

_ 53.6Pα

t i/-s'

(a) From Table 6-2, K = 1 for tight-fitting. Then, from Fig. 6-1, Q/L = 0.60 L/m-s Q = 0.60 x (0.9 + 2'0) x2 = 3l8_LΔ

Αssuming that the wind speed and wind direction are the Same as the given conditions for the bank at Rapid City, SD, the heating load (at -20'6 basis for testing or instructional puφoses only to EΧcerpts from this work may be reproduοed by instructors foι distτibution on a not-fbr-prοfit or ιrαnslαιion οf ιhis ννοrk beyond ιhαt permitιed reprοiucιion Αny oιher been adopted. has the textbook fbr which in οourses students enτol1ed owner is unlαwful' copyrighι permission οfιhe ιhe ΑcιΙνilhouι rJnited Copyright States ]08 ofthe Ι976

by Sections

107

or

--σ6.C outdoor temp. and 72'C indoor temp.) due to the door infiltration can be calculating using Eq. (6-2b) as:

a,

=

(:

+

ε*

ο. o ο

1

Τ)

.#l

('

ι

r

r,

* }r)rr,

_

1_zo.o1)"

c=

393.tW

(b) From Table 6-2, Κ = 2 for average-fitting' Then, from Fig. 6-1, Q/L = 1'25 L/m-s

Q

=

'1

.25x (0.9

+ 2.0)

OO'+)

α, _(', .zs* 0

x 2=7-25--Lls

('.rrfil ι rrrn?)o,

_

e20.6))" C = 819 0W

(c) From Table 6-2, K = 6 for average-fitting' Then, from Fig. 6-1, Q/L = 3.40 L/m-s

e

= 3.40

α,

=(lν

x (0.9

τ2*

+ 2.0)

x2

=

19f2-Lls

('.#l ιrr, *}a)o'_e20.6))"C '001c)

_2227'6W

6-5

From Figure 6-2, Cρ = 0.52' Using Eq. (6-7b) with the standard air density, the pressure difference due to wind is

o'sz (o

Λ p _ ι\ΓΙυ-

oτ

οs#)

κl

(zz-en- ι

( b^_f) z.| ιz.17':::-_!lbf _s'z) ι

r2

#)'

(

ι

o.rnro :n:γ

r,

^\ lbflft')

|

ΔP'

=

0't35in'wg

Neglecting stack effect and pressurization, ΔP - 0.135 in' water From Table 6-1, K = 2 for average-fitting with non-weather-stripped. From Fig. 6-1, Q/L = 0.60 cfm/ft. L" = [(3 x2.5) + (2 x 4)]x3 = 46.5 ft

puφoses only tο on a not-for_profit basis for testing or instruοtronal Excerpts from this work may be reproduced by instruοtors for distribution

studentsenrοlledinοourses1brwhiοhthetextbookhasbeenadopted. Αnyotherreproducι'ιonorιrαnsιαιionοfthisτνοrkbeyondιhαtpermiιted permisiiοn ofιhe cοpyrighι oνner is unlαwful' by Sectiοns 107 or ]08 oftnn isri {Jniιed SιαιeS Copyrighι Αcιτνιιhοuι ιie

Q

= 0.60 x

46.5 = 27.9 cfm.

b-b (a) The wind effect is assumed to be independent of height and pressure differences due to wind are the same as those given in Ex. 6-1. 3'd

Floor: ΔP"/Cο

= 0.037;

orientation ΔP,

Windward 0.03

0.03 0.03

Sides

Leeward gth

Fιoor:

ΔP,

= 0.037 x 0.8 = 0.03 in.

ΔP* 0.066 -0.066 -0.033

ΔP./Co _ -0.100; ΔP,

-

ΔPτ ο.ο96 -0.036 -0.003

-0.100 x 0.8 = -0.08 in. water ΔPτ -0.ο14

ΔP, 0.066 Windward -0.08 -0.066 -0 08 Sides -0.08 -0.033 Leeward orientation ΔP.

water

-0.'146 -0.1 13

(b) For Bitlings, MT, design conditionS are to = -7oF, tι= 72"F, Φι= 28o/o. From Table 6-3, K = 0.66 for conventional οurtain wall.

Αir will infiltrate on windward side only on 3'd floor. Windward - 3E floor QiA = 0.15 cfm/ft2; Q = 0.15(120 x 10) = 180 cfm Then 9 = (1 80 x 60/1 2.4)(0.24)(72 - (-7)) = 16,514 Btu/hr " Q,. = (180 x 60/12.4)(0.005 - 0.000)1060 = 4,616 Btu/hr Qt = Q, + 8r. = Ζ*1-3oBtu1hΙ [3'd Floor] gth

Floor

- All exfiltration Qt

=

on this floor.

oοΞtuΔr

19th

Flooη

puφoses only tο EXοerptS from this wοrk may be reproduced by instructors for distributioIr on a not-for-pro1it basis for testing or instruοtional ιhαι permitted students enrolled in courses tbr which the textboοk has been adopted. Αny οιher reprotluctiοn οr trαnslαtion of this νork beyond by Secιions Ι07 or ] 08 of ιhe Ι 976 t'Jnited Sιαtes Copyrighι Αcι 1υiιhouι ιhe permission οf the cοpyrighι οlυner is unΙιrννful.

986-7

Doors: Double vestibule type ΔPτ = o.146 in' water, assume 1/8 in' οracks Q/L = 16 cfm/ft tFιg. 6-7] , L -- 32 ft [Ex' 6-2] Q = 16 x 32 x0'7 = 358 cfm (Assume30o/oreductionforvestibuledoors)

(a) Windward

6-7 (Cont.)

Side Doors: Double vestibule tYPe ΔPτ = -o'o52 in' water' 1/8 in' cracks Q = O.O (negative pres' dιff') (b) Windward: ΔPτ = O146 in' water' K Q/A = o.2o cfm/ft2 [Fig. 6-6]

=

0'66 lΤable 6-3]

A=120x10=1200ft2 Q

= 0.29(12ο0) =

in' water' K = O'66 [Table 6-2] = O.O (negative pres' dιff')

Sides: ΔPτ

8

240 cfm

= -O'O52

UULeeward: ΔPτ = O'047 in' water, K Q/Α = O.1Oo cfm/ft2, A = 1200 ft2

Q

= O'66

[Fig' 6-6]

= O.1οO(1200) = 120 cfm;

Total infiltration for the walls is 8* = 240 + O.O + 120 = 360 cfm (c) Totat infiltration is sum for doors and walls' From Εx.6-2'for leeward door' Q = 179 cfm' Then the total door filtration is

Qo=358+179=537cfm(neglectinfil.duetotraffic). Andthetotalwallinfiltrationis360cfm,thenQτ=897cfm only to basis for testing oΙ instructional puφoses instτuctors foτ dlstribution on a not-for-profit beyond ιhcιt permiιιed Εxcerpts tiom this wοτk may be reproduced by νork textbook has been uaoρt,i 'ιny ornr, ,rprort|urιιon'o''trαnsΙαtιon-ξthιs students enιolled in οourses tυ. *ι-'λ}, the per-ission of ιhe cοpyright oινner is unΙανfuΙ' Sιαιes Cοpyrιgh' 'ι'i"rii'i"'' to:ii'inιir-d ιn' o7 ] 08 or 1 07 Sectiοns by 'i"

9V

70oF For Charleston, WV: to = 1 1oF' ti = q, = (897 x οοll \ 'τz)(o'z4)(70-11) = 65'ο25 Btu/hr q. = (897 x 60/1 1'72)(1060)(O'OO5 - O OOO) = 24'338 Btu/hr q = q" + 9. = 89,363 Btu/hr

6-8

pressure difference due (a) Assuming standard sea level air density, the

to the wind sPeed of 20 m/s is (

o.o,urtψ\'(zo*pt *1.461ιL:\ /

,,_

|υ.Ι9/+ tbfl

UUWindward: ΔP*=O197xO25=ooo: Leeward: ΔP* = O"197 x (-0'5) = -0'099 in

fi,J

=o 197in,water

χ- } E:lι, n*l:

AssumPtions: _ 40oF 1) temperature differencΘ, tι to' = zi tιle neutral pressure level is at floor 9' 3) the floor height is 12 ft', and 4) Cο = 0'80' Then, from Fig.6-5,

Floor Floor Floor Floor

ΔP" = O'13 x O'8O = 0'104 in' water 1: h = 108 ft., ΔPr/Co = O.13, and in' water 5: h = 60 ft., ΔP, = 0'065 x O'8O = 0'052 -0'068 in' water 15: h = 72ft., ΔP, = -O'O85 x 0'8O = -0j28 in' water 20: h = 132ft., ΔPS = -Q'160 x O'80 =

Leeward

Windward

puφoses only to basis fοr testing οr instructional by instruοtoτs foτ distribution on a not-for-profit ιhαι permιtted Excerpts tiom this work may^bο reproduοed the textbook *' γγ119iJ]1'^"'Ιi";:i:j!i::;:;:;:':""":jfi,':i:,i{:x:λx#,.beyond unlαwful is students enrolled ιn couΙSes tor whiοh owner coρyrighι ιhe of

;}i::j;μιfii,Z'i'"fλ!"in:ii'ini,a s**,

copy,ign,'a;,\iιiio"i''Ι,

& r *--,-,

i,.ι'i-n

100

Flnοr 1

5

15 20

ΔP* 0.049 0.049 0.049 0.049

ΔP" 0.1 ο4 0.052 -0.068 -0.128

ΔPτ 0.1 53 0.101

-0.019 -0.ο79

ΔP" 0.104 0.052 -0.068 -0.128

ΔP* -0.099 -ο.099 -0.ο99 -0.099

ΔPτ 0.005 -0.047 -0.167 -0.227

(b)

6-8 (Cont.)

.*

& \λd tl

#

-o.2

ΔR. *E*-lΛfiΠ$ffard

lnfiltration

ο.1

0.0

-ο.1

Ε.2

ilt. ιlrateι ".-,t

-

LEθΛrard

from 1't to 13th Floor Leeward Sides, 1't Floor onlY

- Windward Sides,

Fιoor Exfiltration _ Wind\Λ/ard Sides, from 14th to 2oth Leeward Sides, from 2nd to 2Oth Floor 1rt floor, lnfiltration on all sides windows

(c)

-

through doors, walls and fixed

for tight fitting. Windward Walls: from Table 6-3, K = 0'22 f rorn f ig. 6-6, Q/A = O.OB cfm/ft2'

A=(1οο+60)1 2=1920f(

Q

= O.O8

x 1920 = 154 cfm

puφoses only to basis for testιng or instruοtional by instructοrs for distributιon on a not-for-profit beyond ιhαι Permiιιed τυork EΧceφts fιom this work may be reproduced ιr*rΙαιiοn-ξιhιs textbook ι,u, υ..n uαipi"J . Αny other reproa|ur'r,in-o, students enrolled in courses for whiοh the of ιhe cοpyright owner is unlανυful' Stαtes Copyrιght Αct iιthouι ιie permisiιon by Sections Ι 07 οr ] 08 o7 rnr'i 9ii'i'ieιΙ

1υ1

for tight fitting' Leeward walls: from Table 6-3, K = 0'22 From Fig. 6-6, Q/A = O'OO5 cfmlft2' A = (1OO + 60)12 = 1920 f(

Q =0.005x1920= 10cfm

for 1/8 in' crack' windward Doors: from Fig. 6-7, Q/L = 17 cfm/ft 6-8 (Cont.) For vestibule doors, assume a 35% reduction' Q/L = 17 x 0.65 = 1 1'05 cfm/ft L = (3 x 6.75) + (2 x 6) = 32'25 ft Q = 11.05 x 32.25 = 356 cfm cfm/ft for 1/8 in' crack' Leeward Doors: from Fig. 6-7, Q/L = 1.5 Γor vestibule doors, assum e a 35o/o reduction' Q/L = 1.5 x 0.65 = 0'975 cfm/ft Φ = O.975x32'25 = 31 cfm

Then,totalinfiltration(neglectingtrafficeffect)is + 551 cfm' Qtot = 154 + 1O + 356 31 = negative pressure differentials for (d) and (e) lnfittration rate is zero due to

the 1Sth and 2οth floors' 6-9

For Minneapolis, MN:

to = -1

1oF'

t1

= 70"F

'

[Note:Δt=70-(-11)=81oFisinconsistentwithProblem6-8where to be minor] error is assumed Δt = 40"F was used; however, an

floor' (a) From Prob. 6-8, Qτ = 551 cfm for 1't Btu/hr q, = (5s1 , dotl 2'15)(0'24)(70 - (-11)) = 52'896 o.ooo) = 11,537 Btu/hr o, = lbSt x 60/1 2.15)('1060)(0.O04 Qt= Qr* Qr. =64,433 Btuihr

(b) and (c) qt = O'O due to zero infiltration puφoses only to basis foτ'testing οr instruοtional by instruοtoΙs for distribution on a not-for-profit ιhαt permiιιed beyond wοrk ιhis Exceφts frοm this wοrk may be reprodυοed o' oon'ιotionλf the ,.oυ" oi i'"r'i""n uαopt.a. ,q"y "rn}')ir"ai.iion students enrol'ed in courses for whiοh of the copyrighι owner is unlα:wful' permιsiιon tie t]nited SιαιeS δ"iyriii by Sectiοns 107 οr 1 08 οf the Ι 97 6 'i,thout

ii

102 6-10

For Des Moines, lA:

to =

-4oF, ti = 70oF'

windows, doors, Transmission heat loss (negtecting infiltration) through walls, and roofs can be determined by Eq. 5-19 as: q = UA(tι -to)

2= 144ft2;

Windows: A

= (3 xa)1

Doors:

A

= (3 x 6.75)1

wails:

A

From Table 5-5a, U = 0.55 Btu/(hr - ft2 -'F1; q = 0.55(144)(70 - (-4)) = 5,861 Btuihr

2=243ft2', From Table 5-8, U = O'28 Btu/(hr - ft2 - 'F); (assume panel with metal storm door) q = 0 28 (243)(70 - (-4)) = 5,035 Btu/hr

64)21- 144- 60.75 = 1395.25ft2', From Table 5-4a, U = 0.14 Btu/(h r -ftz - 'F); = 8[(36 +

Q=

O.

14(1 395. 25)(70

-

(-4)) = 14,455 Btu/hr

Roof/Ceilinq: A = 36 x 64 = 2304 ft2', Ξrorn Example 5-3, U = O.83 Btu/(hr _ ft2 _ 'F1; q = 0.083(2304)(70 - (-4)) = 14,151 Btu/hr

can be determined Transmission heat loss through the slat-on-grade floor bY Eq 5-23 as: Q = U'P(ti -to)

Floor:

p = (36 +64)2=200ft; R-value u' = o.8o Btu/(hr - ft - F), from Fig' 5-8 (assume insulation of 5.4 (hr - ft2 -'F)/ Btu and d = 2ft)' q = 0.8(2OOX7O - (-4)) = 11'840 Btu/hr heat losses; Finally, total transmission heat loss is the sum of all for testing or instruοtional puφοses only to instruοtors for distribution on a not-for-profit basis Exοerpts tiom this work may be reproduced by ιhis νork beyοnd ιhαι permitιed trαnsιαtionω or reproaγι'ιo" Αny oιher . uαφi.J t'u, υ."n students enrolled in οοurses tω wιιiοn the tοxtboοk is unlανυful' oνner copyrιghι ιhe of per.isiion tie copyrιgnt by Secιions ] 07 or ] 08 o7,n, i o)i inιted Sιαtes

iriiiiiouι

103 qt 6-1

=

5l.ΞΖBtuΔt

1

standard air density, From Figure 6-2, Cp= 0'52. Using Eq' (6-7b) with the the pressure difference due to the wind of 15 mph is (0

LP,

-

5η[ 0.0765ψy\fr' )

ι

ΔP*

ι

z.(nlιlbry__s2 _

=

lbf

0.058 in. water

pressurization' thus For a low-rise building, neglect stack effect and ΔP1 = 0.058 in. water and doors' From TabIes 6-'1 and 6-2, Κ = 1 for tight-fitting windows From Fig. 6-1, Q/L = O.'13 cfm/ft' ft L" = [(3 x 3) + (2 x a)]x3 + (3+6.75)x2x3 = 109'5 e = 0.13 x 109.5 = 142 dm, 9s

=

Qr=

Qt=

(14.2 x 60/1 2.15)(0'24)(70

-

(-4)) = 1,245 Btu/hr

(ιι'zx60/12'15)(1ο6OXο.Oο5_0.00ο)=372Btulhr Q, + Qr. = 1,617 Btu/hr

6-12 For Ηalifax, Nova Scotia:

to = 2oF, ti =

70oF'

Refer to Problem 6-10 for other data'

Windows: q

= 0.55(1 44)(70

-

2) = 5'386 Btu/hr

Doors:q=0.28(243)(70-2)=4'627Btulhr

wails:

13,283 Btu/hr Q = 0.14(1 395. 25)(70 - 2) = κootrcuιιιno: q = O.083(23O4)(70 -2)= 13'004 Btu/hr q = O.8(2OοX70 _2) = 10'880 Btu/hr

Γ*r,

or instructional puφoses only to for dlstribution οn a not-fbr-profit basis for testing EΧceφts from this work may be reproduced by instruοtοrs ιhis νork beyond thαι permitted ιrαnslαtionω or reproλ3''ιo' other Αny . t u, υ.", uαφi.J students enrolled in οourses fbr which the textbοok unΙαwful' is οwner copyrighι ιhe of per*isiion /ii i"i.d Sror", copyrιsnt ii 'Ιiiiouι ιi" by Sectiοns ] 07 or 1 0B q 'n""i

Total:

104 qt =

4Ζl€oBtu/hr

6-13 oF; ti 70"F = Memphis, TN; to = 21 R*= 0.92+ 1.55 + 0.99 + 1.77 +0.17 = 5.4 (Tables 5-1 a'5-2a) U* = 115.4 = O.'185 Btu/(hr - ft2 - 'F) Us = 0.81 Btu/(hr - f( - "F) (Table 5-5a) Αs = 6xax3 = 54 ft2 n* = (40xl O)-54 = 346 ft2 q,= 0.185 x 346 x (70 - 21) = 3,136 Btu/hr qs = 0.81 x 54 x (70 - 211= 2,143 Btu/hr Qtotrl =

5zβ auk!

6-14

Concord, NH; to = -2F, ti = 70oF R*= 5.4-0.99+ 3.0=7.41 U* = ο.135 Btu/(hr _ ft2 - "F) Us = 0.81 Btu/(hr -f( - 'F) (From problem 6-13) q* = 0.135 x 346 x [70 - (-2)]= 3,363 Btu/hr qn = 0.81 x 54 x (72) = 3,149 Btu/hr Qtotrι =

οβ1ΖBuer

6-15

lnstructor suPPlies solution.

6-16

(a) q=9s vs

(i'

-

iu)

puφoses only to for distτibution on a not-for-profit basis for testing or instructional Exceφts from this work may be reproduοed by instΙuctοrs thαι permitιed beyond νork οf ιhis or trαnsΙαtion reprοiuctιοn udo|ted ' Any other students enτolled in οourses tbl which the textbook t-'as b"en is unlcnνful' oνner cοpyrighι Act iιthouι tie permisiion of ιhe by Secιiοns Ι 07 or ] 08 ο7 *e l ol i inιιed Stotn, Copyrιght

(280'00Ο)(14.6) qν":-=:-_ .Ο^= _ v! _21.8)60 =^251 i._iu 19(S

(b)

β2.7

q = rh cp(t, -

.

Q'

=

t..)

=

qν,

9ε vs

cp(ts

-

105 cfm

tr)

(250,00Ο)(14.6)

;r=δ=

5'633 cfm (O24X1 15_7O)ω =

6-17

SHF

=

-100'999 = -3.O3 = 33, ooo ===, - oo, ooo)

4st Q./(9. + α")

(1

1

Locate states, and οondition line and heating pΓocess on psychometric chart. _ t') Q. = rh cp(t, _ tr) or Γh . = q '/cr(t, Γh. = 100,000/(0'24 x20) = 20,833 lbm/hr Q. = Γh. X vr/6O = 20,833 x 14.05/60 Q" = 4,878 cfm or about 4,900 cfm

EΧοeφtS frοm this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instruοtional puφoses onιy to permiιιed students enτolled in οourses for whiοh the textbοok has been adopted. Αny other reproduction or trαnslαιion of this work beyοnd ιhαt by Secιions 107 or ] 08 of the Ι 97 6 Uniιed Stαtes Cοpyright Αcι ιι,ithout the permission οf ιhe copyrighι oινner is unlnιful.

106

t

ffi

p n

i

Excerpts frοm this work may be reproduοed by instruοtors for distribution on a not-for-profit basis for tΘSting or instructional puφoses only to

Αnyotherreproductionortrαnslαtionοfthisνorkbeyondthαιpermitted studentsenτοlledinοoursesforwhichthetextbookhasbeenadopted. ] 07 or 108 of ιhe Ι976 [Jnited SιαteS Cοpyright Αct νiιhout the permission of the copyright oνner i3 unlωνful.

by Sections

Exοeφts from this work may be reproduced by instruοtοrs for distribution οn a not-for-profit basis for testing or instructional purposes only to students enrοlΙed in courses for which the textbook has been adopted. Αny οther reproduction οr trαnslαtion οfthis wοrk beyond thαt permitted by Sections ]07 or Ι0B of the Ι976 United Stαtes Cοpyright Αct without the permissiοn οf the copyright owner is unlαwful. Requests for permission or further informαtion should be αddressed to the Permission Depαrtment, Jοhn Wiley & Sοns, Ιnc, ] Ι ] Riνer Street, Hoboken, NJ 07030'

Chapter 7 7-1

First, find longitude from Table B-1a

Then, convert Daylight Saving Time to Local Standard Time using Eq. 7-5 Next, determine the equation of time from Table 7-2 Finally, determine Local Solar Τime using Eq. 7-6 Τhe following table Summarizes the solutions of the problem.

Location Norfolk,

VA

Lincoln, NE Casper, WY Pendleton, OR London, UK

7-2

Longltuαe' 76.2 96.75 106.47 118.85 0.45

Standard Meridian, oνν

75 90 105 120 0

Daylight Savings Τime 9:00:00 ΑM 1:00:00 PM

10:00:00 ΑM 3:00:00 PM 7:00:00 PM

Local

^.-"":' Eouation δtanοarα oτ -' llme ilme

:

LocalSolar llme

ΑM -2'41 min 7:52:47 A|'Λ PM -2.41min 11:30:35ΑM 9:00:00 AM -2'41 min 8:51:43 ΑM 2:00:00 PM -2'41min 2:02:11PΜ 6:00:00 PM -2.41min 5:55:47 PM 8:00:00

12:00:00

Ηour angle (negative for morning and positive for afternoon) can be determined by

h:Ι5* (LST _l2) (a) h = 15*(8.19 -'12:00) = 15*(-3.683) = -55.25 deg. (b) h = 15*(10:03 - 12'.00) = '15*(-1 .950) = -29.25 deg.

(c) h = 15*(15:46- 12:00)= 15*(3.767)= 56.50deg.

107

(d) h = 15"(12.01 - 12:00) ='15*(0.017)= 0.25 deg'

7-3

Αt sunset and sunrise, β =0"; sin(B)

From Eq. 7-8; οos(/)

ο

=ρ

οos(h). οos(δ)

- _sin(/).

sin(δ)

οos(h,,) = οos(ft,,) = _ tan(/)' tan(δ)

The following table summarizes the solutions of the problem.

Location MT orlando, FL Anchorage, AL Honolulu, Ηl Billings,

Latitude,'N

'i"J;::',3" Cos(h)

45.8 28.43 61.17 21.35

20.6 20.6 20.6 20.6

^ffi::. -0.3865 112.7 -0.2035 101.7 -0.6829 133.1 -0.1469 98.4

',|1:ff ',tι''i AM AM 3:07 ΑM 5:26 AM

4:29 5:13

7:30 AM 6:46 AM 8:52 AM

6:33 AM

Note earlier sunrise at greater latitudes 7-4 /=

33.0 deg. N

h = 15.(9-12) = -45.0 deg.

on Sep 21,

δ = 0.0 deg.

From Eq. 7-8; sin(B)

: οos(/).

οos(ft). οos(δ)

+

sin(/). sin(δ)

= 0.593

β = 36'37 deg'

From Εq' 7-11 Φ

cos{ υυs ψ = ;

sin δοos/ _ cos δsin / οos h

β

= -0.478

= 118.57 deg. (clockwise from north)

-"o,

nn a nnt-fnr_nrnfit hqqis fοr testinρ or instructional puφoses only to Exοerpts from this work may be reDroduced bv instnlctοrq f'οr rliqtrih,lfinn

108

7-5

At sunris e, β

_ 0"; sin(B)

From Eq. 7-8; οos(/) οos(h,. )

(a)

-

=

'

g

οos(h)

'

οos(δ) _ _sin(/)

' sin(δ)

_tan(/)' tan(δ)

June 21'' δ-- 23'45 deg; / = 58 deg'

-8'93 hours cos h = -O'6942; h = -133'96 deg' or

Sunrise is at 3:04 AM (Solar Time) From Eq. 7-11; Φ

λ__ sin

^^^ ν"oo

δ cos /

οos δ sin / οos h

-

cos

β

= O.751

north) = 41.33 deg. (clockwise from

(b) cos h

Dec21'' δ= -23'45 deg; =

/

= 58 deg'

0.6942; h = -46'04 deg' or -3'07 hours

Sunrise is at 8:55 AM (Solar Time)

. sinδcos/ οosΦ=" From Εq'7-11;

_οosδsin lcosh

=-0'751

from north) Φ= 138.67 deg. (ctockwise

7-6

Maximumsolaraltitudeangle,βwilloccuratsolarnoon,h=0

]09From Eq.

_ βû*:90 _ Min\, u,)

7-1O,

From Table 7-2, |δ'u"l = 23.45

(a)

Denver,

CO:

I

= 39.75 deg. N.

For north latitude, / is positive and greater than |δrrr| so we need largest positive value of δ. From Table 7-2, δ'u" = 23'45 deg. and hence β'", = 73.70 deg. Therefore, maximum solar altitude angle occurs at solar noon on June 21.

(b)

Lansing, Ml: l= 42.77 deg. N.

For north latitude, / is positive and greater than |δrrr| so we need largest positive value of δ. From Table 7-2, δ'u"= 23.45 deg. and hence β'u"= 70.68 deg. Therefore, maximum solar altitude angle occurs at solar noon on June 21.

(c)

Sydney, Αustralia:

/=

33.95 deg. S.

For south latitude, / is negative and largest negative rralue of δ.

|/|

is greater than lδ'u"l So

\Λ/e

need

From TabΙe 7-2, δ'u, = -23'45 deg. and hence β'", = 79.50 deg. Therefore, maximum solar altitude angle occurs at solar noon on Dec 21. 7-7

Longitude'. Lt= 100 deg. W

Local Standard Time. LCT = 3:30 pm

on Nov 21, Eoτ

= '13.8 min

instruοtional puφoses οnly to Excerpts from this ινorktnay be reproduced by instruοtors for distτibution on a not-fοr-profit basis for testing or

l . _-__t!^lj.-^^''-^^^

''^'^-^).'^1:^'"^-|-âclntinnnfthj.sνοrkbeνondιhαιDermiιιed

110

Using Eq. 7-6, pm.

LSf

= 15.50

-

(100-90).4 /60+ '13.8/60 = 15.063 Hr or 3:04

Latitude: l = 37 '5 deg. Ν

Ηour angΙe: h = 15-('15.063-12)

on Nov 21,

=

45.95 deg.

δ = -19.8 deg.

Using Eq' 7-8 to calculate solar altitude, β= 21.36 deg. Then using Εq.7-11 to calculate solar azimuth; from north) Surface azimuth;

ψ = 12+180 = 192

Φ

=

226'56 deg. (clockwise

deg. (clockwise from north)

Finally, using Εq.7-12 to calculate wall-solar azimuth

y= 1226.56-1921= 34.56 deg. 7-8 Using Εq' 7-13b to calculate angle of incidence for a vertical surface θ = 39.92 deg.

Using Εq' 7-13a to caIculate angle of incidence for an inclined surface For surface tilt = 70", θ= 32.30 deg'

7-9 For Ottawa, Ontario on July 21,

Longitude. Lt= 75.67 deg. W Latitude: Ι = 45.32 deg. N

ΕXοeφtS from this work may be reproduced by lnstruοtors for distribution οn a not-for-profit basis fοr testing or instructional purposos only to studerrts enrolΙed in cοurses for whiοh the tcxtbook has been adopted. Αny οιher reρroductioη or ιrι1fisΙαιiοn οf ιhis work beyond ιhαt ρermitιed

Equation of Τime: ΕoT

= -6.2 min

Dec|ination: δ= 20.6 deg. (a) Eastern Daylight Savings Time:

EDSI

=

4:00 pm

Using Eq. 7-6, LS7- = 14.852 Hr or 2'.51 pm. Hour angΙe: h = 15*(14'852-12)

=

42.78 deg'

Using Eq. 7-8 to calculate solar altitude, β= 47"16 deg. Using Εq'7-13c to calculate angle of incidence for a horizontal surface, θ= cos-1(sin(47.16)) = 42'84 deg. (b) At

sunset, β cos(fr

):

=

0 and sinp

=Q

_tan(/). tan(δ)

Hour angle: h = 112.34 deg.

Solar time at sunset:

LSf

= 12 + hl15 = 19.49 hr or 7:29 pm.

Εastern Daylight Savings Τime can be calculated by

7-10

(L'

ΕDST

=

- ΕoT

+L

EDSΓ

= 19.49 +(75.67-75)-(4/60)-(-6.2/60)+1

=

LST

+

_ ΕSη(4min/deg-t4l)

20.638 hr or 8:38 pm.

For Philadelphia, PA on July 21, Longitude'' Lι= 75.25 deg. W Latitude: / = 39.88 deg. N Equation of Time: ΕoT = -6.2 min Declination: δ= 20.6 deg. Eastern Daylight Savings Time: EDSI = 10:30 am Εxοerpts from this work may be reproduοed by instruοtors for distribution on a not-for-profit basis fοr testing or instruοtional puφoses οnly to students enrol]ed in courses for which the textbook has been adopted. Αnyl other renro,l1ιft;Λh ^v lνΛf.l^ιi^,^ ^{lL]d 'ιl^yL λo1'^') tι1nt ^trmιtted

112

Using Eq. 7-6, LS7Γ = 15.852 Hr or 3:51 pm. Hour angle: h = 15-(15.852-12) = 57.78 deg.

Using Eq. 7-8 to calculate solar altitude, β = 49'42 deg.

Using Εq' 7-11 to find solar azimuth;

Φ


(a) Using Εq' 7-13c to calculate angle of incidence for a horizontal surface, θ = cos-1(sin(49.42)) = 40.58 deg.

(b) For vertical surface facing southeast, Surface Tilt; α = 90 deg., and Surface azimuth; ψ= 135 deg. (clockwise from north)'

Using Εq' 7-12 to calculate wall-solar azimuth , y= |114'3-135ι

=

20.7 deg'

Using Εq' 7-13b to calculate angle of incidence for a vertical surface, θ=

cos-l(cos(49.42)cos(20 7ο)) = 52'52 deg'

(c) For inclined surface faοing south, Surface Tilt; α = (90-40) = 50 deg., and Surface azimuth', ψ= 180 deg. (clockwise from north)'

Using Εq' 7-12 to calculate wall-solar azimuth, y= l114.3-180|

=

65.7 deg.

Using Εq.7-13a to calculate angle of incidence for an inclined surface, d = cos-1 (cos(49.42)cos(65.70)sin(50)+sιn(+g.42)cos(50))

=

46.'t 1 deg.

7-11 7-12

7-13

For Calibou, MΑ on July 21,

instructional puφoses only to . Exοerpts from this work may be reproduοed by instruοtors f,or distribution οn a not-Γor-profit basls for testing or ιhαl permιιted reοrοιlu"',--:':::':::!*::ι::]'-υg:,oηd Αny other has been adοpted. the texδook for which in courses enrolled students

113

Longitude' Lt= 46.87 deg. W Latitude: l= 68.02 deg. N Equation of Time: EOT = -6.2 min

Declination: δ= 20.6 deg. Solar Parameters; Α = 346'4 Btu/hr-ft2 or 1093 Wm2, B = 0.186, and C = 0.138 Eastern Daylight Savings Time: ΕDSr = 2:00 pm Surface Tilt; α = 60 deg. Surface azimuth, SW; ι/ = 225 deg' (clockwise from north)

Using Eq. 7-6, LSI = 14'72 Ηr Hour angle: h = 15.(14.72-12)

= 41

.58 deg.

Using Eq. 7-8 to calculate solar altitude, β = 36.04 deg.

Using Εq'7-11 to find solar azimuth;

Φ__

230.2 de9. (clockwise from north)

Using Εq.7-12 to calculate wall-solar azimuth, y= 5'2 deg' Using Εq' 7-13a to calculate angle of incidence, θ = 7.45 deg.

Using Εq' 7-15 and clearness number of 1, 796.75 Wm2

Gryρ =

252'51 Btu/hr_ft2 or

Using Εq' 7-16a, Gρ = 250.28 Btu/hr-ft2 or 790'03 Wm2 Using Eqs, 7-18 and 7-2O, Gα= 26'13 Btu/hr-ft2 or 82'46 Wm2 Therefore, total clear sky irradiation is276.51 Btu/hr-ft2 or 872.49 Wm2 7-14

Given lnformation: Date: June 21 Longitude' Lt= 96'0 deg' W Latitude: / = 36.0 deg. N Εquation of Τime: ΕoT = -1 .4 min Declination: δ= 23.45 deg.

puφoses only to EΧοerpts fiom this wοrk may be reprοduοed by instruοtors for distribution on a not-for-profit basis Γor testing or instructional οf ιhis wοrk beyond ιhαι permiιted students enrolled in οourses fbτ whiοh the textbook has been adopted. Αny οther reρrολucιion οr ιrαJ'lsιalion

114

Solar Parameters; Α = 346.1 Btu/hr-ft2 or 1092 Wm" ' B = 0' 1 85, and C = 0.137 Central Daylight Savings Time: CDSI = 8:00 pm Surfaοe Τilt; α = 90 deg. Surface azimuth, SW; ι/ = 225 deg' (clockwise from north) Reflectance from water; Ps = 0'25 Using Eq. 7-6,

LSf

= 18.58 Hr

Hour angle: h = 15-(18.58-12) = 98'65 deg'

Using Eq. 7-8 to calculate solar altitude, β

Using Εq' 7-15,

G,νρ =

= 7 '02

deg'

76'24 Btuihr-ft2 or 240'5 Wm2

lrradiation reflected from the ground can be determined by

Gn: PrF.r(sinβ +C)G'o where F'ncan be determined from Εq'7-24' ThereforΘ, Gκ = 2'47 Btu/hr-ft2 or 7'8

7-15

Wm'

Given lnformation: Date: lιΛar 21 Latitude: / = 56.0 deg. N Equation of Time: ΕoΤ = -7'5 min Declination: δ= 0.0 deg. Solar Parameters; Α = aoε.g Btu/hr-ft2 or 1164\Nlm', B = 0'149' and C = 0.109 Local Solar Time: LSf = 12:00 Pm Surface Tilt; α = 90 deg' Surface Azimuth, S; ψ= 180 deg' (clockwise from north) Clearness number; CN = 0'95 Diffuse Reflectance from Sno\Λ/; ρn = 0'7 Hour angle: h = 0.0 deg. puφoses onιy to distribution on a not-for_profit basis for testing or instruοtional Excerpts from this work may bο reproduοed by instructors for of ιhis ιοrk beνοfrr] .hot nermiΙtρi ιrοnsιαtion or ,eοroλucιion οthei adopted. has been 'qnν students enrolled in courses for which the textboοk

115

Using Eq. 7-8 to calcuΙate solar altitude, β= 34'0 deg' Using Εq' 7-11 to find solar azimuth;

Φ


Using Εq' 7-12 to calculate wall-solar azimuth,

7r=

Using Εq' 7-13b to calculate angle of incidence, Using Εq.7-15, Gtvρ Using Εq'

7

=

0.0 deg.

θ=

34.0 deg.

268.5 Btu/hr-ft2 or 847 '1 \Νlm2

-16a, Gρ = 222'6 Btu/hr-ft2 or 702'3

Using Εqs' 7-21 and 7-22, Gα

= 33.O Btu/hr-ft'

Wm'

or 104. 1 \Νlm2

lrradiation reflected from the ground can be determined by

G^: PrF'r(slnβ+C)G'o where F*n can be determined from Εq.7-24' Therefore, GR = 62'8 Btu/hr-ft2 or 198.1 \Νlm'

7-16

Given lnformation: Date: Aug 2'1 Latitude: I = 32.0 deg. N Equation of Time: ΕoT = -2'4 min Declination: δ= 12.3 deg. Solar Parameters; Α = 350.9 Btu/hr-ft2 or 1107 \Λllm2, B = 0'182, and C = 0.134 Local Solar Time: LSf = 10:00 am Surface Tilt; α = 45 deg. Surface azimuth, SW; ι/ = 225 deg. (clockwise from north) Diffuse Reflectance from ground; ρn = 0.3 Hour angle: h = -30.0 deg.

Using Eq. 7-8 to calculate solar aΙtitude, β = 56.1 deg. Exοerpts from this wοrk may be reproduοed by instruοtors for distribution on a not-for-pιofit basis for testing or instructional puφoses only to students enrolled in οourses for which the textbοok has been adopted. Αny οιher renrοdυcιiοn or trαnsΙαtion οf thi't τνοyk hcνοnι] thηl nPrmittρ/]

116

Using Εq.7-11 to find solar azimuth; Φ= 118.7 deg. (clockwise from north) Using Εq.7-12 to calculate wall-solar azimuth, r= 106.3 deg. Using Εq.7-13b to calculate angle of incidence, Using Εq' 7-15,

Gruo =

θ=

61.5 deg.

281'8 Btu/hr-ft2 or 889'1 Wm2

Using Εq'7-16a, Gρ = 134'4 Btu/hr-ft2 or 424'0 Wm2 Using Eqs. 7-18 and 7-2o, Gα= 32'2 Btu/hr-ft2 or

10'1

'7

Using Eqs. 7-23 and7-24, GR = 11'9 Btu/hr-ft2 or 37 '7

\ΝΙm"

\'ΙΥlm2

',11.9) = 178.6 Btu/hr-ft2 or Using Eqs. 7-25 , Gt= (34.4 + 32.2 + = (424.0 + 101 .7 + 37.7) = 889.1 Wm'z

7-17

The following results are determined from a computer program employing equations in the book from Eqs. 7-6 to 7-26' Following tables summarize input and output data calculated for southwestfacing vertical window at32 deg. N latitude, 90 deg. W longitude, for all daylιght hours of a clear day on July 21with ground reflectance of 0.2 and clearness number of 1. lnput Data

Longitude 90 Standard Meridian 90 -6.2 EOT 32 Latitude Declination 20.6 225 Surf Azimuth 90 Surf Tilt 346.4 A 0,186 B 0,138 c cN1 0'2 RΗoG

deg deg min

deg deg deg deg Btu/hr-ft2

to on a not_for-profit basis for testing or instructional puφoses only Exοerpts from this work may be reproduοed by instructors for distribution nr trn'ζΙntinn n{1|1jq aιnrΙ' holnι'] th'ι Δ29Φ;fio) fρn/^.h'"tinn othcr Αnν heen adonted. has textbοοk students enrοlled in courses tbr ιvhich the

117

Output Data

cDsr 7.00 8.00 9.00

LsI 5.90 6.90 7.90 10.00 8.90 1.00 9.90 1

12.00 13.00 14.00 15.00 16.00 17.00 18.00 19.00

*Unit

o

. Φ,. -91.55 9.50 71.57

o

"

Gruo* Go* Ga* Gρ* G,* 153.43 151.90 112.19 O.OO 6.97 3.40 10.37 -76.55 21.78 78.63 146.37 140.64 209.84 o.OO 13.03 10.68 23.71 -61.55 34.38 85.69 139.31 128.74 249.18 o.OO 15.47 17.51 32.98 -46.55 47.09 93.60 131.40 116.76 268.71 o.OO 16.69 23.39 40.07 -31.55 59.65 104.24 120.76 104.98 279.23 o.OO 17.34 27.g5 45.2s 10.90 -16.55 71.33 123.59 101.41 93.63 284.65 o.OO 20.57 30.90 51.46 1 .90 -1.55 78.52 172.69 52.31 83.01 286.52 34.88 24.03 32.03 90.94 12.90 13.45 73.44 229.79 4.79 73.4s 285.30 81.05 27.54 91.28 139.87 13.90 28.45 62.18 252.83 27.83 65.62 280.70 115.85 30.36 28.70 174.91 14.90 43.45 49.71 264.52 39.52 60.08 271.44 135.41 31.69 24.45 1g1.54 15.90 58.45 37.00 272.79 47.79 57.55 254.30 136.46 30.69 18.81 185.97 16.90 73.45 24.37 279.93 54.93 58.44 220.69 115.51 26.33 12.15 153.99 17.90 88.45 12.00 286.94 61.94 62.60 141.60 65.16 1s.97 4.90 86.03 h,

β,

Ψ,

θ,

1

of lrradiation is Btu/hr-ft,

7-18 Using the developed program, following tables summarize input and output data caΙculated for south-facing Surface tilted at 45 deg. on Apr 21 in Louisville, KY. lnput Data

Longitude Standard Meridian

EoΤ

Latitude

Declination Surf Αzimuth Surf TiΙt

85.73 90 1.1 38.18 1.6 180 45 1

A

358.2

B

0.1

deg deg min

deg

deg deg

deg

Btu/hr-ft2

64 0.12

c

CN RHOG

1

0.2 Output Data

tSI 1.0

2.0 3.0

h,

o

-165.0 -150.0 -'135.0

β'. -38.3 -32.9 -24.8

Φ," 18.8 35.7 49.8

θ," Gruo* 161.2 164.4 0.0 144.3 150.0 0.0 130.2 135.4 0.0 ψ,"

ιJD

Gα*

0.0 0.0 0.ο

0.0 0.0 0.0

ιJR ^*

0.0 0.0 0.0

ιra ^*

0.0 0.0 0.0

Exοerpts from this work may be reproduοed by instructors for distributiοn on a not-for_profit basis for testing or instruοtional purposes only to students enrolled in courses for whiοh the textbοok has been adοnted. Αnll nthρy "o^.^)"

118

4.0 -120.0 _15.1 61 .5 5.0 -105.0 _4.3 71.6 6.0 -90.0 7.1 80.8 .0 -75.0 18.9 s9.9 8.0 -60.0 30.6 99.7 9.0 -45.0 42.0 111.3 10.0 -30.0 52.3 126.8 11 .0 -15.0 60.2 149.3 12.0 0.0 63.4 180.0 13.0 15.0 60.2 210.7 14.0 30.0 52.3 239.2 15.0 45.0 42.0 248.7 '16.0 60.0 30.6 260.3 17 .0 75.0 18.9 270.1 18.0 90.0 .1 279.2 19.0 105.0 -4.3 288.4 20.0 120.0 -15.1 298.5 21 .0 135.0 _24.8 310.2 22.0 150.0 _32.9 324.3 23.0 165.0 -38.3 341.2 24.0 180.0 _40.2 360.0

o.o o.o o.o o.o 99.2 91.4 95.7 o.o

118.5 120.7 1ο8.4 106.0

90.1 80.3 68.7 53.2 30.7 0'ο 30.7 53.2 68.7 80.3 90.1 99.2

7

7

*Unit

of lrradiation is Btu/hr-ft,

108.4 1

18.5

130.2 144.3 161.2 180.0

o.o o.o 9.8

76.8 215.8 49.2 22.1 62.5 259.6 120.0 26.6 48.4 280.3 186..1 28.7 35.1 291.1 238.3 2g.8 23J 296.5 271.5 30.4 18'4 298'2 282.9 3o.5 23.7 296.5 271.5 30.4 35.1 291.1 238.3 2g.8 48.4 2S0.3 186.1 28.7 62.5 259.6 120.0 26.6 76.8 215.8 49.2 22.1 91.4 95.7 o.o 9.8 106.0 o.o o.o o.o 120.7 o.o o.o o.o 135.4 o.o o.o o.o 150.0 o.o o.o o.o 164.4 o.o o.o o.o 175.2 o.o o.o o.o

0.0 0.0 0.7 2.8 4.8 6.5 7.8 8.6 8.9 8.6 7.8 6.5 4.8 2.8 0.7 0.0 0.0 0.0 0.0 0.0 0.0

0.0 0.0 10.5 74.1 151.4

221.3 275.9 3,10.5

322.3 310.5

275.9 221.3 1s1.4 74.1

10.5 0.0 0.0 0.0 0.0 0.ο 0.0

7-19 Using the developed program,.following tables summarize input and output data calcuΙated for an east-facing windδw, 3 ft. wide by 5 ft. high, with no set baοk on a clea r Jul21 day in-Boise, lD. Longitude

Ιnput Data

Standard Meridian

ΕoT

Latitude

Declination Surf Azimuth Surf Titt

A B

116.22 120 -6.2 43.57 20.6 90 90 346.4

deg deg min

deg

deg deg deg

Btu/hr-ft2

ο.'186

c

CN RHOG

h, o

LSr 5.0 -,105.0 6.0 -90.0

0.1

38

1

0.2

Output Data

β,.

Φ,'

ψ,"

3.8

65.0 74.8

25.0

14.0

15.2

o Guo* Go" 25.3 21.5 19.5 20.6 160.9 .150.6 θ,

Gα*

3.6

27.4

GR* Gt* e 0.4 23.5 352.4 6.'1 184.1 2761 .3

Exοerpts from this work may be reproduced by instructors for on a not-for-profit basis for testing or instruαionalρμmosesonl}, rnich the texthn.ι. hqc h-.- ".1l'_'Ι:9*':'

ilffi;;;;;l#;':il;:'r;

to

119

7.0 8.0 9.0 10.0 1.0 12.0 13.0 14.0 15.0 16.0 17 .0 18.0 19.0 1

*Unit

oE

7

-20

-75.0 -60.0 -45.0 -30.0 -15.0 0.0 15.0 30.0 45.0 60.0 75.0 90.0 105.0

24.7 35.6 46.2 56.1 63.8 67.0 63.8 56.1 46.2 35.6 24.7 14.0 3.8

84.4 94.8 106.9 123.0 146.7 18ο.0

213.3 237.0 253.1 265.2 275.6 285.2 295.0

25.3 35.8 48.6 62.1 76.0 90.0

5.6 4.8 16,9 33.0

222.0 251.6 267.7 276.8

281.6 283.0 104.0 281.6 117 '9 276.8 131.4 267.7 144.2 251.6 154.7 222.0 159.4 160.9 154.7 21.5

56.7

90.0 123.3

147.0 163.1 175.2

.185.6

195.2

205.0

of lrradiation is Btu/hr-ft2

200.7 203.9 177.2 129.6

68.2 0.0 0.0 ο.0 0.0 0.0 0.0 0.0 0.0

12.3 249.9 18.1 260.6 23.0 236.3

36.8 38.5 36.1

31.4 26.2

21.5 17 .5

17 '2

16.6

.15.6

13.8 10.0 1.3

26.8 29.2 30.0 29.2 26.8 23.0 18.1 12.3 6.1 0.4

187.8 123.6

3747.9 3908.9 3544.7 2817.2

51.4 46.6 44.0 39.7 33.7 26.1 16.1 1.8

18s3.6 771.7 699.7

659.8 594.8 505.9 391.9 241.6 26.7

is the rate at which solar energy strike the window in Btu/hr

Given lnformation: Latitude:

l= 32.47 deg.

Surface azimuth, S;

ζz-=

N

180 deg. (clockwise from north)

Windowwidth;W=4ft. Windowheight; H=6ft. Setbackdistance; b=1 ft. (a) On

April2l

Declination: δ= 12.3 deg. Local Solar Τime: LSr= 9:00 am

Hour angle: h = 15.(9-12) = -45.0 deg.

Using Eq. 7-8 to calcu|ate solar altitude, β = 43'82 deg' Using Εq' 7-1'1 to find solar azimuth; Φ= 106'27 deg. (clockwise from north) Using Εq' 7-12 to calculate wall-soΙar azimuth, y= 73'73 deg. X

Using Εqs. 7-28 to 7-30 to calcuΙate shaded dimensions, ! t

r-

Χ = (1 ft.)-tan(73.73) = 3.43 ft. y = (1 ft ).tan(43.82)/cos(73.73) = 3.42 ft.

v

6'

Shaded area can be calculated by

4' Εxο.erpts from- this work may be reproduced by instructors 1br distribution on a not-Γor-profit basis for testing or instructional puφoses only tο students enrol1ed in courses fbr which the textbook has been adopted. Αny other reprοducιion or ιrαnslα.tiοi of ιhis wοrk beyδni thαι permiιιed

bySections]07οr]08ofιheΙ976UniιedlδιqaeγC-apyrighιΑcιw,iιhouιιheDern&1!!Mοf1hecιwιlrjsh1

oνue-r

j'9g!!-ιg+&!.

."!:'.''' a

'-7

120

Α,h =W

*

H

_ (W

-

x)

*

(H _ y) = 22'52

ft2

7-20 (Cont.) Therefore, the percentage of the window that is shaded is 93"8%. (b)

On July

21

Declination: δ= 20.6 deg. Local Solar Time. LSl. = 12:00 pm

Hour angle: h = 15*(12-12) = 0.0 deg.

Using Eq. 7-8 to calculate solar altitude, β = 78.13 deg. Using Εq.7-11 to find solar azimuth; Φ= 180.0 deg. (clockwise from north) Using Εq' 7-12 to calculate wall-solar azimuth, r= 0.0 deg. Using Eqs. 7-28 to 7-30 to calculate shaded dimensions, x = (1 ft.)*tan(0.0) = 0.0 ft. y = (1 ft.)*tan(78.13)/cos(0.0)

=

4.76 ft.

Shaded area can be calculated by

Α,h=W*H_(W _x)*(Ι-Ι_y)

= 19.03ft2

Τherefore, the percentage of the window that is shaded is 79.3%. (c) On Sep 21

Declination: δ= 0.0 deg. Local Solar Time: LSl. = 5:00 9m

Hour angle: h = 15*(17-12) = 75.0 deg.

Using Eq. 7-8 to οalculate solar altitude, β= 12.61 deg. Using Εq'7-11 tofind solarazimuth; Φ=261.81 deg. (clockwisefrom north) Excerpts from this work may be reprοduced by instruοtors for dlstribution on a not-for_profit basis for testing or instruοtiona] puφoses only

studentsenrolledincoursesforwhiοhthetextbοokhasbeenadopted. ΑnyotherreproducιionortrαnsΙαιionοfthisνοrkbeyondthαιpermιιted ;a ''"Ι^''t\'1 by Secιions Ι07 or ]08 ofιhe Ι976 [JnitedιSιαιesCοnνriqhl Α'| \υithn1ιt thο hovh'i'CiΛ' n{tho '^---:nιot '1ιlv'-

to

.

121

Using Εq' 7-12 to calculate wall-solar azimuth , y= 81.81 deg. 7

-20 (Cont.)

Using Εq.7-28 to calculate the horizontally shaded dimension,

X,

x = (1 ft.).tan(81.81) = 6.95 ft.

Since x is greater than W, the window is completely shaded. Therefore, the percentage of the window that is shaded is 100%'

7-21

Given: Problem 7-20 with a tong 2 ft overhang located 2ft above the top of the window. For this problem, bo for overhang is the sum of the overhang depth and the setback; henοe, bo = /+] = 3 ft. (a) Τhe vertically shaded dimension on the window due to the overhang can be calculated bY:

lo=botanβlcosy_!o-. where 1rr-, is the distance of the overhang above the window. Therefore, Υo = (3 ft.)*tan(43.82)/cos

Sinοe

η

(73'73) - 2 _ 8'27 ft'

is greater than H, the window is completely shaded.

Therefore, the percentage of the window that is shaded is 100%. (b) Similafly, yo= (3 ft.)- tan(78.13)/cos(0.0)

- 2=

12'27 ft'

Since y, is greater than H, the window is completely shaded' '100%' Therefore, the percentage of the window that is shaded is

basis 1br testing or instruοtional puφosΘs onιy to Exοeφts from thls work may be reproduced by instruοtοrs for distribution on a not-for-profit or ιrαnslαιiοn ο| this νork beyond ιhαt permiιιed reprολucJioι Αny οιher aJopted. has been texibook whiοh the fbr in courses students enrolled '1 l^'-!'1 ' ^ b,ySectiοns Ι07 οr Ι08of the 1q76τΙ-ito)-qf'''"?""";*13

122 the setback' there is no (c) Since the window is completely shaded due to need to calculate Yo'

7

-22

setback' Given. Problem 7-2owith 6 in. setback instead of 1 ft' (a) Using Eqs. 7-28 to 7-30 to calculate shaded dimensions,

x = (0.5 ft.)*tan(73.73) = 1'71 fL , = (o.s ft.)-tan(43.82)/οos(73'73) = 1'71 ft'

Shaded area can be calculated bY Α,n =W

*

H

_ (W

_x)* (H _ y) = 14.19 fi2

is 59'1%' Therefore, the percentage of the window that is shaded (b) Using Eqs. 7-28 to 7-30 to calculate shaded dimensions,

γ = (O.5 ft.)-tan(0'0) = ο'0 ft' , = (ο.s ft.)*tan(78.13)/cos(0'0)

_-

2'38 ft'

Shaded area can be calculated bY Α,l, =W

*

H

_ (W _'T )

*

(H _ y) = 9'52

ft2

shaded is 39'7%' Therefore, the percentage of the window that is

(c)UsingEqs.T-28to7-3Otocalculateshadeddimensions, x = (0.5 ft.).tan(81'81) = 3'48 ft' , = (O.S ft.)-tan(12.61)/cos(81 'S1) = 0'79 ft'

Shaded area can be calculated Α,h =W

*

bY

H _(W _.χ)* (H

_

y)

=

21'27

ft2

or instruοtional puφoses onΙy to for djstribution on a not-for-profit basis Γor testing Exceφts from this work may be reproduced by instructors

studentsenrolledinοoursestoιwhichtnetexδookhasbeen Ιo7AΙΙbi+^)c'-!"^^f''";..1'! hιι'Sρrtinnrl07nrl0Rn{tι1o

λiy'nrrrrproλur^ιιoλ-o,ιrαnslαιiοiofιhisνorkbeyondthαtρermitted

^l"i'r'i ''',.'"''^""i'-''""'''''-"'^-^{i"^^^'^'';-ι^1

'''-''"iο"-ι^'{"l

123 Therefore, the percentage of the window that is shaded is 88'6%'

7

-23

on December 21, Declination: δ= -23'45 deg. Using the same procedure

as described in Problem 7-20, the following table summarizes the calculated data.

Solar Surface- '"Jlff#:"' :"^::j Hour Solar Azimuth, Solar Dimension Angle, Αltitude, :,ol1' ""; o o Time, Azimuth,' f\ZιlΙluιl '' " 1x1, ft hr 1'37 8.OO -6O.O0 9.98 126.22 53 78 ',136.52 o'95 43 48 9.OO -45.00 1g.4g 0'60 1o,oo -3o.oo 27.17 148.96 31'04 o'29 1.OO -15.00 32.27 163.69 16'31 o oo O'OO 18o.OO 34.08 12.00 O.OO 29 0 16'31 196.3',1 32.27 13.00 15.00 o 60 14.00 3O.OO 27.17 211'04 31'04 0 95 43'48 48 22g 19.49 15.00 45.00 1'37 53'78 23g.78 16.00 60.0O 9.98 '1

7

-24

7

-25

7

-26

7

-27

ι"#::γ

Shaded

Dimension Area, (y)' ft

o'30 o'49 0'60 0'66 0'68 0'66 0'60 0 49 0 30

ft2

8'98 7 '18 5'65 4'20 2'71 4'20 5 65 7 18 8 98

%Shaded

Αrea 37

'4

29'9

23',5 17 1

',1

17

'5 ',3

'5

23'5 29',9 37

'4

This problem is similar to ProblemT-21 but the overhang depth is 3 ft in instead of 2ft. Since the window in ProblemT-21 is completely shaded all all cases. Τhe window in this problem is also completely shaded in cases since the overhang depth is greater in this problem'

Eqs' First, we need to know angle of incidence and solar irradiation. Using problem), 7-8 to 7-26 (or a computeiprogrφ developed for previous puφoses only to on a not-for-profit basis for testing or instructional Exοeφts 1iom this work may be reproduced by instructors for distτibution

studentsenroΙledincoursestbrwhichthetextbookhasbeen

adopted' Αnyotherreproλucιiοnοrtrαnslαιioλofthisworkbeyondιhαιpermiιted

124

incidence angle and solar irradiation on a southwest-facing window for Boise, lD on a clear July 21 day at 3:00 pm solar time are Angle of Incidence θ= 52.4 deg', Direct Solar lrradiation: Gp = 163'4 Btu/hr-ft2' Diffuse Solar lrradiation: Ga + Gκ = 34'5 + 23'0 = 57 -3 Btu/hr-ft', and Total Solar lrradiation: G1 = 163'4 + 57 '3 = 220'9 Btu/hr-ft2' Then, the area of the glazing and of the frame is calculated to be 12'44 and 2.56 ft2, resPectivelY.

ft2

From Table 7-3, solar heat gain coefficients for the glazing system lD 21c are SHGGgo(52.4") = 0.548 and SHGGgα= 0'52' be From Table 5-2, the outside surface conductance may be estimated to 4.0 Btu/hr-ft2-'F.

From Τable 5-6, the U-value for the fixed, double glazed window having aluminum frame with thermal break utilizing metal spacers is 1.13 Btu/hrft2-'F.

From Table 7-1, solar absorptance of the aluminum frame (assuming the window is not a nev/ one) is 0.8.

Αssuming the window with no setback (Ar,u'" = frame can be calculated using Eq' 7-31 as:

Aruπ),

the SHGC for the

SHGGr = 0.8*(1 '1314'0) = 0'226' Then, using Εq.7-32, the total solar heat gain is Qsnc = (0.548.1 2.44 + 0.226*2.56)*163'4 + (0.52*1 2.44 + 0.226*2.56)-57.5 = 1613'68 Btuihr' 7

-28

glazing From Table 7-3, the glazing transmittance and absorptances for the system lD 21c are

puφoses only to distribution on a not-for-prοfit basis for testing or instruοtional Excerpts from this work may be reproduοed by instructors for of ιhιs wοrk beyond ιhαι permitted ιrαnsιαtιοi or reproλuction Αny oιher uJopt"α. has υ.", tΘx;book the students enrοl]ed in οourses for whiοh

125 7

-28 (Cont.)

Tρ6(52'4") = O.+1 56, Αtwοβ2Α) = O'140, Αfzυοφ2'4) = 0'1524' Ta= O'40, -Af ια= o.'13, and Αfza= O.15.

Using Eq. 7-35, total transmitted solar heat gain is Qrru"', = (0'4156-163.4 + 0.4ο*57.5)-12'44 = 1130.9'1 Btu/hr.

Using Eq. 7-36, total solar heat gain absorbed by the glazing is 8 oroo,, = =

[1

63'4*(0"1 4+0'

1

524) + 57'5*(0"1 3+0' 1 5)]-1 2'44

794.64 Btu/hr.

From Table 5-5a, the U-value for the center of glass is 0.42 Btu/hr-ft2-"F. Similar to the previous problem, the outside surface conductance may be estimated to be 4.0 Btu/hr-ft2-'F. Then, the inward flowing fraction for glazing layer 1 can be calculated by:

Nt=0.42 14.0=0.105 From Table 5-2a, the inside surface conductance may be estimated to be 1.46 Btu/hr-ft'-"F.

The conductance from the inner pane to the outdoor air can be calculated by:

,11

flo'2=

1

1 1=-1

U hi 0.42

=o'59Btu/hr-ft2-'F

ι.46

Then, the inward flowing fraction for glazing layer 2 can be calculated by: Nz= 0.42 / 0.59 = 0.71

puφoses only to Exοerpts 1iom this work rnay be reproduοed by instruοtors for distribution on a not-for-profit basis for testing or instructional permitted students enrolled in οourses fbr which the textboοk has been adoρted' Αny oιher reproduction οr lrαnslαtiοn of ιhis wοrk beyond thαι

126 7-28 (Cont.)

Using Εq. 7-38, the inward flowing fraction of the gΙazing system is N=

63.4*(0.'1 0S*0. 14+0.7 1*0.1 524) + 57.5*(ο.'1 05*0. 13+0.71-0. 1 5)] l 220 '9

[1

= 0.122

Using Eq. 7-39 and the SHGGr calculated from the previous problem, the solar irradiation absorbed by the frame is Qoroo,f = (163.4 + 57.5)*2.56*0.226 = 127 .80 Btu/hr.

Using Εq' 7-40, the total absorbed solar heat gain of the fenestration system is Quruo,ur =

794.64*0.122 + 127 .80 = 224.75 Btu/hr.

The total solar heat gain is then Qsrc = 1130.91 + 224.75 = 1355.66 Btu/hr. 7

-29

From Table 7-4, lAC for a lighted-color Venetian blind installed on a residential double-pane window is 0.66.

Using Εq' 7-41, the total solar heat gain is Q

7

-30

suc = φ'226-2'56-220'9) + [0.548*12.44*163.4 + 0.52* 12.44*57 .5]*0.66 = 1108.48 Btu/hr.

From Table 7-6, for a Ιighted-color Venetian blind, shade transmittance, reflectance, and absorptance are 0.05, 0.55, and 0.40, respectiveΙy.

Using Εq' 7-42, the transmitted solar heat gain is

127

7-30 (Cont.) Qrroo = 0.05*1130.91 = 56.55 Btu/hr.

Using Εq' 7-43, the absorbed solar heat gaΙn is 4or"" = 224.75 + 0.40*1 130.91

+ 0.55*1 130.91 *0.122*(0.'13+0.I S)

=

7-31

698.36 Btu/hr.

From Table 7-3, solar heat gain coefficients for the glazing system lD 5b are

SHGGgo(52'4")

=

0.6256 and SHG Ggα = 0.60.

Similar to Problem 7-27 , SHGGr = 0.226.

Then, using Εq'7-32, the total solar heat gain is Qsμc = (0.6256-12.44 + 0'226*2.56)*1 63.4 + (0.60.12.44 + 0.226*2.56)*57.5 = 1828.64 Btu/hr.

7

-32

From Table 7-3, the glazing transmittance and absorptances for the glazing system lD 5b are TDθ(52'4") = 0.5332, 'Af ,οoβ2'4) = 0.1924, 1froθβ2'4) = O'12, Ta= 0.51, -Atro= 0.19, and -Arzd= 0.11.

Using Εq. 7-35, total transmitted solar heat gain is Qrsac,g = (0.5332*'163.4 + 0.51*57.5)*1

2.44 = 1448.64 Btu/hr.

Using Eq. 7-36, total solar heat gain absorbed by the glazing is λ

Q

πllc'g =

[1

63.4*(0.1924+0'12) + 57.5-(0.

1

9+0.

1

1)1-12.44

7

-===Ξ

-32 (Cont.) =

849.60 Btu/hr.

From Table 5-5a, the U-value for the center of gtass is 0.55 Btu/hr-ft2-"F.

similar to the previous problem, the outside surface conductance may be estimated to be 4.0 Btu/hr_ft2_.F. Then' the inward flowing fraction for glazing layer 1 can be caΙculated by:

Nz=0.SS /4.0=0.1375 SimiΙar to the prevΙous probtem, the inside surface conductance may be estΙmated to be 1.46 Btυ/hr-ft2-"F.

The conductance from the inner pane to the outdoor air can be calcuΙated by.

ho,z=t\ =T-] 1 (]_τ O55 - 1-46

= 0.88 Btu/hr-ft2-"F

Τhen' the Ιnward flowing fraction for gtazing layer

2 canbe

calcutated by:

/vz=0.55/0.gg=0.625 Using Εq. 7-38, the inward flowing fraction of the glazing system is N = []-63 4-(O.1375-ο. 1924+o.625*0'12) + 1375*0' 19+0'625*0' 1 1 l 22o'g )] =

'Jr';Δ''

Τhe solar irradiation absorbed by the frame is the Same as the previous probΙem, and is equaΙ to 127 .80 Btu/hr'

Using Εq' 7-4O, the totat absorbed solar heat gain of the fenestration system is

7 -32 (Cont.)

+ 127 '80 = Qo'"o'gt = 849'60*0'1OO

212j6 Btu/hr'

The total solar heat gain is then + Qwc = 1448'64

212J6

= 1661'4 Btu/hr'

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instruοtionaΙ purposes only to students enrolled in courses for which the textbook has been adοpted. Αny other reprοduction οr trαnslαtion of this wοrk beyond thαt permitted by Sections ] 07 or ] 0B of the 1976 United Stαtes Copyright Αct without the permission of the cοpyright owner is unlαwful. Requests for permission or further infοrmαtion should be αddressed to the Permission Depαrtment, Jοhn Wiley & Sons, Ιnc, ] Ι ] Riνer Street, ΙΙoboken, NJ 07030.

Chapter 8 B-1

a) The heat gain is generally gΓeater than the cooling load during the morning hours M/hen sunlight first strikes a building and the internal loads first begin. Heat is being stored in the building structure, furnishings, etc. b) Late at night when occupants are not present, lights and equipment are off and solar radiation is zero, the building gives up stored heat to the air, which the equipment removes as cooling load. The heat gain may be quite small, zero, or negative. c) At some time during the day, probably early evening, as heat gain is

decreasing, and equilibrium condition can be established when heat gain and cooling load are equal. Or, some interior zones, where the cooling load is driven only by internal heat gains may reach equilibrium if the heat gain remains constant for a number of hours.

8-2 MultipΙe design conditions should be checked, including peak dry bulb along with mean coincident \Μet buΙb, and peak wet bulb along with mean coincident dry bulb temperature.

8-3

ΑSHRΑE 90.1 specified the2'5% design conditions, which roughly

corresponds to the 1% design conditions in the current Handbook of Fundamentals and the textbook. Location Norfolk, VA

Outdoor

Outdoor

WB,'F

DB.'F

91

76

75

DB.'F

lndoor

lndoor RH. % 50

Elevation, ft

Latitude,

30

36.90

ON

131

8-4 into the tayer library' The select materials; some may need to be entered here' (Note that not resulting wall construction ,pp"rrt as shorrun using a density of 120 everything is specified exacity, .o that a student ιυit1yt"i υiicκ wiιι get a different set of CTF coefficients') -"..-"...-τns1] 0 3

5nΟ

5α]

130 "18 3.

for room mode, we After running the HvacloadExplorer program in execute obtain the following CTF coefficients for this wall. zn,

n

xn, Btu/h-ft2-"F

Yn, Btu/h-ft2-"F

Btuih-ft2-"F

0

4.276507

0 ο0ο445

0.642344

1

-5.36497

0.ο1 1581

-0.98287

0.638772

2

1.141149

0.01 1845

0.376555

-0.02179

3

-0.027 59

0.001 134

-0.01 101

4

-7

.7Ε-05

0.0ο0017

-5E-06

Φn

B-5

except that the RThis problem is solved in the Same \Λ/ay aS Problem 8-4, 13 insulation is changed to 5.5" thick R-19 insulation. or instructional puφoses only to for dtstribution on a not-for-profit basis for testing Εxοerpts fιom this work may be reproduοed by instΙuοtοrs ιhis work beyοnd thαι permιιιed oαnsΙαtionΖf up-iurrιλn-o, iny oιh* t,u, υ"., uai-pt.α' students enrolled in courses tor whiοh the teΧtbook unlανful' is owner cοpyrighι the of copyrιg|rt 'a|ι rr"ιiiλuι ιn, prr^ι''ιλn by Secιions 107 or ] 08 o7 ιne ii;i' initid Sror^

132

The folΙowing CTF coefficients are obtained: n

xn, Btu/h-ft2-"F

Yn,

Ζn,

Βtu/h-ft2-'F

Btu/h-ft2-"F

0

4.277384

0.000071

0.644513

1

-5.95084

0.004622

-1.08666

0.779066

2

1.847897

0.008936

0.510931

-0.10021

3

-0.16027

0.001835

-0.05401

0.001435

4

0.001331

0.00004

0.000734

Φn

8-6

Αgain, this problem follows the procedure of the last two problems. The thickness of the roll roofing must be estimated, and the conductivity chosen to match the overall conductance. (k=thickness*conductance)

The following CΤF coefficients are obtained: Εxceφts from this work may be reproduced by instruοtors for distτibution on a not-for-profit basis for testing or instruοtional puφoses only to students eι'ιrolled in courses for whiοh the textbook has been adopted. Αny oιher reproduction or ιrαnsιαιion of this ινork beyond ιhαι permiιted by Secιiοns ] 07 οr 1 08 of ιhe Ι97 6 Uniιed Sιαtes Cοpyrighι Αcι w ithοuι ιhe permissiοn οf ιhe copyrighι oνner is unlανιful'

133

n

xn, Btu/h-ft2-'F

Btu/h-ft2-'F

Btu/h-ft2-'F

0

1.014657

0.006092

0.644513

1

-1.09939

0.029838

-0.6816

2

0.126521

0.006044

0.079104

3

0.000256

0.000071

0.00003

Yn,

Ζn,

Φn

0.150594

8-7

ln this case, a reasonable value for the resistance of the air-space must be selected. For the air^-space, an R-value of 1 is chosen; thus conductivity is set to '12 Btu-in lhr- ft2- F, and the thickness Ιs set to 12 in Density anα bp are set to zero and 0.24, respectively.

Νntε: Ιayers listed fιnm tοp tο bοfiom :epr*s*ni {rα:n thg nutsiιJ* tο inside ot lhe sur{εce

The following CΤF coefficients are obtained: n

xn, Btu/h-ft2-'F

Btu/h-ft2-'F

Btu/h-ft2-'F

0

1.014651

0.00468

0.654471

1

-1.12785

0.027234

-0.71129

2

0.1 51 609

0.00674

0.095526

3

0.000351

ο.0001ο6

0.000053

Yn,

zn,

Φn

0.1 781 59

Exc-erpts from this work may be reproduced by instruοtors for.distribution on a not-for_profit basis for testing or instruοtional puφoses only to students enrolled in courses Γor which the textbook has been adopted' Αny οther reproλcιion or trαnsl(]ιiοn of ιhis νοrk beyontl thαι permiιιed by Sectiοns ]07 οr ]08 ofιhe 1976 Uniιed StαιeS Cοpyrighι Αcιw'ithouι ιhi permissιλn οfιhe cοpyrighι ονner ιi unlαινful'

134

8-8

Again, reasonable values must be assumed for the density of the acoustical tile and the specific heat of the limestone concrete.

The foΙlowing CTF coefficients are obtained: Yn,

Zn'

3.162792

Btu/h-ft2-'F 0.002232

Btu/h-ft2-'F 0.2851 '16

1

-3.76069

0.01895

-0.38995

0.71ο366

2

0.633425

0.007779

0.137 459

-0.01912

3

-0.00642

0.000149

-0.0ο352

n

xn, Btu/h-ft2-'F

CI

Φn

8-9

First, app|y the exterior οonvective heat transfer correlation, Equation 818a, to determin Θ h"' Assume the 15 mph wind is windward on the surface, which results in h" = 2.3 Btu/(h-ft2-F). Estimate the sky temperature as 10.8 R below the outdoor ambient temperature = 546.87 R. Then estimate the effective sky temperature for a vertical surface from Εquation 8-25 tsky,o=

cos (90/2)λry +(1-cos(90/2))f,

=

550.0 R

EXceψtS from this work may be reproduοed by instructors for distτibution on a not-for_profit basis for testing or instructional puφoses only to students enrolled in οοurses for which the textbook has been adopted. Αny oιher reproduction or trαnsιαιion of this wοrk beyοnd ιhαι permitιed by Sectiοns Ι 07 οr Ι 08 οf the 1976 tJnited Stαtes Copyrι1hι Αcι \νitlιout ιhe permission of the copyrighι oιιner is unlαινfuΙ.

136 Declination Surf Αzimuth Surf Tilt

20.6 270 90

Α

346.4

B

0.1

c

CN

RΗoG

deg

deg deg Btu/hr-ft2

86 0.1 38 1

0.2 Output Data

MDST LSI h, o 1 .00 23.79 176.83 2.00 0.79 -168.17 3.00 1.79 -153.17 4.00 2.79 -138.17 5.00 3.79 -123.17

β,

"

-34.27 -33.23

-28.80 -21.65 -12.54

6.00 7.00 8.00 9.00

4.79 -108.17 -2.11 5.79 -93.17 9.19 6.79 -78.17 21.05 7.79 -63.17 33.22 10.00 8.79 -48.17 45.49 11.00 9.79 -33.17 57.51

12.00 13.00 14.00 15.00 16.00 17.00 18.00 19.00 20.00 21.00 22.00 23.00 24.00

*Unit

10.79 11.79 12.79 13.79 14.79 15.79 16.79

-18.17

68.46

-3.17 75.28 11.83 72.19 26.83 62.36 41 .83 50.63 56.83 38.41 71 .83 26.17 17 .79 86.83 14.15 18.79 101 .83 2.58 19.79 1 16.83 -8.27

20.79 '131.83 21.79 146.83 22.79 161.83

-18.00 -26.06 -31.74

of lrradiation is Βtu/hr-ft2

Φ,'

Ψ,o

θ,o

356.41 86.41 87.03 13.26 256.74 101.06 28.82 241.18 114.99 42.20 227.80 128.63 53.39 216.61 141.58 62.87 207.13 152.80 71

.22

198.78

79.02 190.98 86.89 183.11 95.77 174.23

107.55 162.45 127.34 142.66 168.24 101.76

218.87 51.13 245.60 24.40 259.78 10.22

269.56 0.44 277.71 7.71 285.45 15.45 293.49 23.49 302.43 32.43

312.83 42.83 325.24 55.24 339.93 69.93

159.17 156.37 146.65 134.23 120.81

106.97 92.97 78.94 65.01 51.37 38.42

27.20 20.83 23.63 33.35 45.77 59.19 73.03

* ιr/vD

ιJD ^*

ιrd ^*

ιJR ^*

0.00 0.00 0.00 0.00 0.00 0.00

108.07 206.38 246.69 266.87 277.85 283.62 285.80

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00 6.71

0.00 0.00 0.00 0.00 0.00 0.00 3.22

5.53 0.00 0.00 0.00 0.00

5.07 0.00 0.00 0.00 0.00

.51 0.93 ο.00 0.00 0.00 0.00

6.19 0.10 0.00 0.00 0.00 0.00

284.93 54.68 280.80 1 18.63 272.32 170.01 256.78 201.20 227.20 202.07 161 .85 151.27

ιrt ^* 0.00 0.00 0.00 0.00 0.00 0.00 9.93 23.08 32.24 39.29 44.53

12.82 10.26 15.32 16.92 16.57 22.71 17.25 27.27 17.61 30.29 47.91 20.83 31.59 52.42 25.38 31.06 111.11 30.63 28.75 178.01 35.51 24.81 230.32 38.43 19.50 259.13 37.19 13.16 252.42 27

184.97 6.10 0.00 0.00 0.00 0.00

The hourly dry-bulb temperature is calculated using Equation 8-2. Here, the hour nearest to the local solar time has been used to determine the temperature. A spreadsheet is used to obtain the solution. lteration is accomplished by simpΙy pasting the calculated values of Io" back into the " Io", estimated" column. lnput Data U-Value Solar absorotivitv Ihermal emissivitν

0.1

Btui(h-ft2-F)

0.8

0.9

Exceφts from this work may be reproduced by instruοtors for distrrbution on a not-for-proΓit basis for testing or instructional puφoses only tο students enrolled in courses for whiοh the textbook has bοen adopted. Αny other reproducιion or trαnsιαtion οf ιhis wοrk beyond thαι permitted by Secιions 107 or ]08 ofthe Ι976 United Stαtes Cοpyriqht Αctιliιhοuι ιhe permissiοn οfthe cοpyright oνner is unlανful.

137 Τis

72.O

F

Peak temperature Dailv Ranqe Mean Wind Soeed

96.0 25.4 10.0

F F mph Output Data

Clock Time 1.00

2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.ο0 11.00 12.00 13.00 14.00 15.00 1 6.00 17 0ο 18.00 19.00 20.00 21.0ο 22.00 23.0ο 24.00

Local

Solar Time

23.79 0.79 1.79 2.79 3.79

4.79 5.79 6.79 7.79 8.79 9.79 10.79 11.79 12.79 13.79 14.79 15.79

16.79 17.79 18.79 19.79 20.79 21.79 22.79

Outdoor Dry-bulb Temp. (F)

75.2 73.9 72.6 71.6 70.9 70.6 71.1

skv

Τemp. (F)

Effective

skv

Temp.

64.4 63.1

61.8 60.8

(F)

67.5 66.3 65.0 64.0

60.1

63.2

59.8

63.0 63.5

60.3 61.6 63.9

72.4 74.7 78.0

67.2

8'1.8

71 .0

74.1

86.1

75.3 79.4 82.4 84.4 85.2 84.4 82.7 79.9 76.6 73.3 70.5

78.5 82.5 85.6 87.6 88.4 87.6 85.8 83.0 79.7 76.4 73.6

67,9 65.9

71 .1

90.2 93.2 95.2 96.Ο

95.2

o?ξ

90.7 87.4 84.1

81.3 78.7 76.7

64.7

67.0 70.3

69.1

To",

hc

estimated (Btu/(h-

hrsky

hrg..o

(Btu/(h

(Btu/(h-

(F)

ft'-F))

-fr-F))

ft'-F))

74.60 73.15 71.94 70.72

1.58 1.58

69.75 69.03

'1.58

0.46 0.46 0.45 0.45 ο.45 0.45 0.45 0.46 0.47 0.48 0.49 0.5ο

0.47 0.47 0.46 0.46 0.46 0.46 0.46 0.47 0.48 0.49 0.50

0.51

0.52 0.54 0.58

69.'18

74.55 78.99 83.53 88.42 93.24 98.33 109.05 131.23 148.77 159.92 162.49 153.00 118.72 84.81

81.65 78.97 76.55

'1.58

1.58 1.58 1.58 1.58

1.58 1.58 1.59 '1.59

1.59 1.59 1.61

1.62 1.62 1.62 1.62

1.60 1.58 1.58 1.58 1.58

0.53 0.57 0.60 0.62 0.62 0.60 0.54 0.49 0.48 0.47 0.47

ο.51

0.61

ο.63 ο.63

To",

calculated from 8-24 (F)

73.14 71.93 70.72

69.75 69.02 68.78

(Btu/(htt')) 0.1

1

-0.01

-0.13 -0.23

72.34 77.55 82.47 87.66 92.76 97.77 102.84 122.43

-0.30 -0.32 0.03 0.56 1.05 1.57 2.08 2.58 3.08 5.04

142.11

7.01

155.98 162.15

8.40 9.02 8.65 6.64

158.51

0.55 0.50 0.49 0.48

138.39 86.48 81.63 78.96 76.54

0.48

74.60

0.6'1

9conduction

1.45

0.96 0.70 0.45 0.26

8-11

This problem uses the same solution procedure as Problem 8-10. lnput Data

Longitude 116.22 Standard Meridian 105 EOT -6.2 Latitude 43.57 Declination 20.6

deg deg min deg deg

Exceφts from this woτk may be reproduced by instructors fbr distrrbution on a not-for_profit basis for testing or instructional puφoses only to students enrolled in οourses for whiοh the textbook has been adopted. Αny other reproduction οr trαnsιαtion of this ινοrk beyond ιhαι permitted by Secιions ] 07 or ]08 of ιhe Ι 976 United Sιates Cοpyright Αcι ν,ithout ιhe permission of the cοpyrighι oνner is unlαwfuΙ.

138 Surf Αzimuth Surf Tilt

A B

c

CN RHOG

180 90

deg deg

346.4

Btu/hr-ft2

86 38

0.1 0.1 1

0.2 Output Data

MDST LSr h, " β, " 1.00 23.15 167.23 -24.77 2.00 0.15 -177 .77 -25.80 3.00 1.15 -162.77 -23.91 4.00 2.15 -147.77 -19.34 5.00 3.15 -132.77 -12.59 6.00 4.15 -117.77 -4.21 7.00 5.'15 -102.77 5.31

8.00 6.15 -87 .77 9.00 7.15 -72.77 10.00 8.15 -57 .77 .0ο 9.15 -42'77 12.00 0.15 -27 .77 1 1

'1

13.00 14.00 15.00 16.00 17.00 18.00 19.00 20.00 21.00 22.00 23.00 24,00

*Unit

11.15 -12.77

12.15 2.23 13.15 17.23 14.15 32.23 15.15 47 .23 16.1 5 62.23 17 .15 77 .23 18.'15 92.23 19.15 107 .23 2015 122.23 21.15 137.23 22.15 152.23

15.60 26.32 37 .17

47 '76 57 .42

64.68 66.95 62.91 54.71

44.67

33.95 23.10 12.48

2.38 -6.85 -14.79 -20.95

of lrradiation is Btu/hr-ft'z

Φ,

"

Ψ,

o

" 152.14 154.11 150.59 143.20 133.87 123.75 θ,

Grρ*

Go*

0.00 0.00 17.66 162.34 0.00 31.94 148.06 0.00 44.76 135.24 0.00 56.15 123.85 0.00 66.47 113.53 0.00 76.20 103.80 103.29 173.45 0.00 85.92 94.08 93.66 227.72 0.00

346.83 166.83

2.32

177

'68

96.42 83.58 108.99 71.01 125.92 54.08 151 .07 28.93

0.00 ο.00 0.00 0.00 0.00 0.00 113.42 46.47

254.61 22.70 269.45 58.92 277.79 87.76 281.98 105.54 185.34 5.34 283.00 1 10.30 217 .50 37.50 281.09 101.56 239.78 59.78 275.81 80.22 255.07 75.07 265.88 48.70 266.84 86.84 248.28 11.34 277.01 97.01 215.64 0.00 286.67 106.67 106.26 146.49 0.00 296,51 116.51 116.49 3.97 0.00 307 .11 127.11 126.80 0.00 0.00 318.90 138.90 136.76 0.00 0.00 332.16 152.16 145.67 0.00 0.00

84.88 77.37 71.58 68.02 67.06 68.82 73.09 79.44 87.38 96.45

Gα*

0.00 0.00 0.00 0.00 0.00 0.00

2.89 10.77 16.45 20.78

24.56 27.57 29.47

29.99 29.04 26.78

ιJR ^

*

0.00 0.00 0.00 0.00 0.00 0.00 1.07 7.06

13.24 18.90 23.67 27.24 29.38 29.95 28.90

^*

Lz1

0.00 0.00 0.00 0.00 0.00 0.00 3.96 17,83 29.69

62.38 107.15 142.58 164.39 170.24 159.50

26.32'133.31

23.50

22.36 17

.29

94.57

19.55

48.18

5.19 0.07 0.00 0.00 0.00

26.46 14.28 0.32 0.00 0.00 0.00

15.02

9.10 0.25 0.00 0.00 0.00

11,44

lnput Data U-Value Solar absorptivitv Γhermal emissivih Tis Peak temoerature Dailv Ranqe Vlean Wind Soeec

0.1

Btu/(h-ft2-F)

0.9 ο.9 72.0 96.0 30.3

F

1

1.0

F F

mph Output Data

Exceφts from this work may be reproduοed by instructors fbr distribution on a not-for-profit basis for testing or instructional puφosΘs only to students enrolled in courses for which the textboοk has been adopted. Αny other reprοducιion or ιrαnslαtiοn of ιhis work beyond thαι permiιιed by Sectiοns ]07 οr Ι08 ofιhe 1976 Uniιed Stαtes Cοpyright Αctτιiιhout the permission ofιhe cοpyright oνner is unΙαwful.

139

Clock Time

Local

Solar Time

Outdoor Dry-bulb Temp. (F)

23.15 0.15

73.0 71.2

3.00

1.15

69.6

4.00

2.15 3.15

68.'1

1.00 2.O0

5,00 6.00 7.00 8.00 9.00

4.15

10.0ο 1 1.0ο

12.00 13.00 14.ο0 15.00 1 6.00 17.00 18.00 19.00 20.ο0 21.00 22.00 23.00 24.00

5.15 6.15 7.15 8.15 9.15 10.15 11.15 12.15 I 3.15 14.15 1 5.15 16.15 17.15 18.15 '19.15

20.15 21 15

22.15

66.9 66.0 65.7 66.3 67.8 70.5

74.5 79.0 84.2 89.ο 92.7 95.1

96.0 95.1

93.0 89.6 85.7 81.8 78.4 75.4

skv

Temp. (F)

Effective

Skv

Temp.

62.2 60.4 58.8 57.3 56.'1

55.2 54.9 55.5 57.0 59.7 63.7 68.2 73.4 78.2 81.9 84.3 85.2 84.3 82.2 78.8 74.9

(F)

65.3

63.5 62.0 60.5

h"

To"'

estimated (Btu/(h-

tt'-ε))

71.15 69.41

1.72 1.72

67.96

1.72 1.72

0.45 0.45 ο.45 0.44 0.44 0.44 0.44 0.44 0.45 0.46 0.49 0.52 0.55 0.57 0.58 0.58 0.57 0.54 0.52

0.46 0.46 0.46 0.45 0.45 0.45 0.45 0.45 0.46 0.47 0.50 0.53 0.56 0.58 0.59 0.59 0.58 0.55 0.53 0.52 0.50 0.49 0.48 0.47

64.1

58.7 60.2 62.9 66.9 71.4 76.5 81.4

68.60

I

74.15 81.31

99.67 116.02

85.0

129.01 137.21 '139.67

87.5 88.4

127.91

71 .0

78.1 74.1

67.6 64.6

70.8 67.8

(Btu/(h-

-ft'?-F))

58.1

82.0

(Btu/(h

ft'-F))

58.4

87.5 85.3

hrgrα

(F)

66.51 65.35 64.48

ξoa

hrsky

136.52 114.37 100.24 93.64 85.63 79.56 76.37 73.47

1.72 1.72 1.72 1.72 1.72 1.73 1,74 1.74 1.75 1.75 1.75 1.75 1,74 1.73 1.73 1.72 1.72 1.72 1.72 1.72

0.51

0.49 0.48 0.47 0.46

To.,

calculated

fromS-24 (F)

71.15 69.41

67.96 66.51

65.35 64.48 65.5'1

70.67 75.96

Qconduction

(Btu/(hft2))

-0.09 -0.26 -0.40 -0.55 -0.67 -0.75 -0.65 -0.13 0.40

89.08

1.71

106.51

3.45

121.05 131.75 137.47 137.47 132.07 121.67 107.09 98.57

4.90 5.97 6.55 6.55 6.01

4.97 3.51

91.61

2.66 1.96

83.43 79.56 76.37 73.47

0.76 0.44 0.15

114

B-12

This problem is solved in the same manner as Example 8-2. The results (conduction heat fluxes for each hour in Btu/(hr-ft2)) may be summarized in tabular form as: Hour 1

2 3

Day

1

0.312 0.463 0.508

5

0.494 0.454

6

0.410

7

0.387

B

0.402

9

0.473

4

2 1.126 0.954 0.804 0.673 0.56't 0.475 0.426 0.426 0.487

Day

Day 3

1126 0.954 0.804 0.673 0.561

0.475

0.426 0.426 0.487

Exceφts from this work may be reproduοed by instructors for distrrbution οn a nοt-for-profit basis for testing or instructional puφoses only tο students enrolled in courses for whiοh the textbook has been adοpted. Αny οther reproduction or trαnslαtion οf ιhis νork beyond ιhαι permiιιed by Secιiοns Ι07 or Ι08 οfιhe Ι976 Uniιed Stαιes Cοpyright Αct wiιhouι the permission ofthe copyrighι oνner is unlανful.

140 10

0.612

11

0.820

12 13 14 15 16 17 18 19

1.089 1.399

0.620 0.825 1.092 1.401

1.715

1

1.998

1.999

2.223

20

2.362 2.405 2.352 2.215

21

22 23 24

1.327

.716

2.223

2.362 2.405 2.353

0.620 0.825

1.092

-

1.401 1

.716

1.999

2.223

2.362 2.405 2.353

2.016

2.215 2.016

2.215 2.016

1.786

1.786

1.786

1.551

1.551

'1.551

1.327

1.327

Because the wall is Ιightweight, the results converge rapidly. 8-13

This problem is soΙved in the Same V/ay aS the previous problem. Note that the additional insulation substantially reduces the conduction heat flux, as expected. The resuΙts (conduοtion heat fΙuxes for each hour in Btu/(hr-ft2)) may be summarized in tabular form as: Hour

Day

1

2

0.203 0.329

3

0.379

4

11

0.380 0.356 0.323 0.298 0.296 0.328 0.407 ο.535

12

13 14 15 16 17 18 19

1

5 6 7 o

I 10

Day 2 0.915 0.778 0.659 0.554

Day 3 0.915 0.778 0.659 0.554

0.464

0.464

0.390 0.340 0.322

0.340

0.344

0.344

0.390 0.322

0.417

0.417

0.542

0.711

0.542 0.715

0.925

0.927

1.154 1.373

.156 1.374 1.559

0.927 1 .156 1.374 1.559

1.692 1.758 1.754

1.692 1.758 1.754

1.559 1.691

1.757

1.754

'1

0,715

20

1.685

'1.685

1.562 1.406

1.685

21

1.562

1.562 1.406

22

'1.406

Exc-erpts from this work may be reproduced by instructors for.distribution οn a not-for-profit basis for testing or instruοtional purposes only to students enrolled in οourses fοr which the textbook has been adopted. Αny other reprοdλCιion or ιrαnslαtion of ιhis lιork beyond thαι permiιιed (Jniιed by Sections ] 07 οr ] 08 οf ιhe 1 976 Sιαιes Cοpyright Αcι ιυithouι thi permissιλn ο7 ιhe copyrighι ονner ιi unlαwfut'

141 23 24

1.237

1.070

1.237 1.070

1.237

1.070

8-14

The solution to this problem is similar to that of Problem 8-9, except that to estimate the maximum possible surface temperature, the surface may be assumed to be adiabatic, and U is then zero. Also, the surface-to-ground radiation coefficient is zero, and no correction is necessary for the sky temperature, as the surface is assumed to be horizontal. Assume the wind is windward, h"= 1.3 Btu/(h-ft2-F). Then, the final converged answer for the surface temperature is: hr,sky=

1.361

Btu/(h-ft2-F)

fr"=

201.0 F

B-'15

From Table 8-2, heat gain for occupants that are "Seated, very light work" have 245 Btulhr (72 W) sensible heat gain, and '155 Btu/hr (45 W) latent heat gain. The sensible portion is assumed to be 70% radiative/ 30% convective.

The sensible heat gain from people is72 Wperson x 30 people The radiative portion is 0.7 x 2160 = 1512νν Τhe convective portion is 0'3 x2160

=

=

2160 W.

'

648 W.

The latent heat gain from people is 45 Wperson x 30 peoPΙe = 1350 W. The sensibΙe heat gain from lighting is '1 .5 ννft2 x 4OOO sq. ft. = 6000 W; 20o/o is assumed to enter the plenum space directly, leaving 4800 W which is assumed to be 59% radiative I 41% convective. The radiative portion is 0.59 x 4800 = 2832\'tΥ. The convective portion is 0.41 x 4800 = '1968 W. The sensible heat gain from equipment is 1 ννfi( x 4OOO sq. ft. = 4O0O W, which is assumed to be 20o/o radiative I 80% convective. (Note this Εxcerpts fiom this work may be reproduοed by instruοtοrs for distribution on a nοt-fbr_profit basis for testing or instruοtional puφoses only to students enrolΙed in courses for which the textbook has been adopted. Αny οιher reproductιοn οr trαnslαtiοn of this work beyοnd ιhαt permitted by Sectiοns l07 οr 108 ofιhe Ι976 Uniιed Sιαιes Copyrighι Αcιlνithoul the permission ofthe copyright οwner is unlαwfuΙ.

142

assumption is based on the assumption that most of the equipment is fancooled. Students are likely to make varying assumptions.) Τhe radiative portion is 0'2x 4000

=

800 W.

The convective portΙon is 0.8 x 4000 = 3200 W. The total sensible heat gain is 2160

+

4800

The radiative portion is 1512 + 2832

+

+

The convective portion is 648 + 1968

4000 = 10960 W.

800 = 5144 W.

+

3200

=

5816 W.

The total latent heat gain is 1350 W. 8-16

From Table 8-2, heat gain for occupants that are involved in "Sedentary \ι/ork" is275 Btu/hr (81 W) sensible heat gain, and 275 Btulhr (81 W) latent heat gain. The sensible portion is assumed to be 70o/o radiative/ 30% convective.

The sensible heat gain from people is 81 Wperson x 35 people

=

2835 W.

Τhe radiative portion is 0'7 x 2835 = 1984.5 W. The convective portion is 0.3 x 2835 = 850.5 W. The latent heat gain from peopΙe is 81 Wperson x 35 peoPle

=

2835 W.

Τhe sensible heat gain from lighting is '15 \,ΙΥlm2 x 75O m' = 11250 W; 50% is assumed to enter the plenum Space directly, Ιeaving 5625 W that is assumed to be 59% radiative I 41% convective. The radiative portion is 0.59 x 5625

=

The convective portion is 0.4'l x 5625

3319 W. =

2306 W.

The sensible heat gain from office equipment is 7000 W, which is assumed to be 20o/o radiative I 80% convective. (Note this assumption is based on EXοeφtS from this work may be reprοduced by instruοtοrs for distribution on a not_for-profit basis for testing or instruοtional purpοses only to students enrolled in οourses fοr which the textbοok has been adopted' Αny oιher reproduction or ιrαnsιcιιiοn ofιhis work beyond ιhαι permitted bySectiοns Ι07 οr ]08οf the Ι976UniιedSιαιesCopyrighιΑcιy,ithoutιhepermissionof ιhecopyrighιοwnerisunlατνful.

143 the assumption that most of the equipment is fan-cooled. Students are likely to make varying assumptions.)

The radiative portion is 0.2x 7000 = 1400 W. The convective portion is 0.8 x 7000 = 5600 W. The total sensible heat gain is 2835 + 5625 + 7000 = 15460 W.

The radiative portion is 1984.5 The convective portion is 850.5

+

3319 + 1400

+

2306 + 5600

=

6703.5 W.

=

8756.5 W.

The total latent heat gain is 2835 W. 8-17

Heat gain to the space = 0.8 x 6000 W

=

4800 W

Problem 8-18 At 4.00 p.m., 70 people are present. Assuming "seated, light offiοe \Mork", the sensible heat gain per person is245 Btu/hr (72νν) and the latent heat gain per person is 200 Btu/hr (59 W).

Sensible heat gain

=

245 Btu/hr/person x 70 people = 17150 Btu/hr.

Latent heat gain = 200 Btu/hr/person x 70 people = '14000 Btu/hr. Αt 6:00 p.m., no one is present; sensible and latent heat gains are O Btu/hr.

8-19

Excerpts fiom this work may be reproduοed by instructοrs for distribution on a not-fοr_prοfit basis for testing or instruοtional purposes onΙy to students enrolled in courses for which the textbook has been adοpted. Αny οιher reproduction or trαnsιαιion of this work beyond thαt permitted by Sectiοns 1 07 or ]08 of the 1 976 United Sιαtes Copyright Αcι ννithout ιhe permission of ιhe cοpyrighι οινner is unlανfuΙ.

144 First, compute the properties of the corresponding fictitious surfaces, using Eqns 8-35, 8-36, 8-37. ResuΙts are shown in the shaded table entries, beΙow.

1

2 3

4 5

Surface

Area (ft')

Νorth roof

639.7 639.7

South roof West wall East wall Αttic floor

84.0

84.0 1176.0

0.9 0.9 0.9 0.9 0.9

A-ε

T(F)

575.8 575.8 75.6 75.6 1058.4

122

λ

A-ε-T

70241.8 82332.6 7711.2

1983.7 1983.7 2539.4

92

6955.2

2539;4,

95

100548.0

1'4;47.',4'

143 102

τ;lF)

εl]f

0',9

,

1,10.6

1ο3.9 113,8

0.9 0.9 0.9 0:9

114.1

128,4

Τhen, compute the radiant interchange factor and radiation heat transfer coefficient using Eqns. 8-38 and 8-39. Using Eqn. 8-40, estimate the radiative heat flux from each Surface (Q,.uα), then determine the radiative heat transfer from each Surface (Q,"rα). Then, compute the total radiative heat transfer from all surfaοes = -69,769.5 Btu/hr. Divide by the total Surface area, 2623.4 ft2, to get the baΙancing factor, -26'6 Btu/(hr-ft'1, which must be subtracted from the previously caΙculated heat flux from each surface to determine the "balanced" radiation heat flux from each surface (q,rοlbal). Multiply by the area to determine the radiation heat transfer from each Surface (Q,"α/bal)' Cheοk to see that they noνv Sum to zero. Qraο

Trus (R)

Suι"face

Fit

I

North roof

0.872

576.0

11.4

2 3

South roof West wall East wall Αttic floor

0.872

583.1

11.9

567.6 562.7 571.4

1

4 5

ο.897 0.897 0.832

hri

(hr-ft2))

11.2 1.0

10.6

129.7 463.7 -132.6 -242.5 -355.3

qr"6/bal

Q,uo

(Btu/

(Btu/hQ

82950.9 296624.7 -11140.0 -20368.1 -417837.0

(hr-ft')) 156.3

(BtΨ

Q,u6lbal (Btu/hr)

99964.1

490.3

313637.9

-'106.0

-215.9

-8906.1 -18134.2

-328.7

-386561.8

8-20 First, compute the properties of the corresponding fictitious surfaces, using Eqns 8-35, 8-36, 8-37. Results are sho\Mn in the shaded table entries, below.

Surface 1

North roof

Area (m2)

120.7

ο.9

A-ε

T(C)

Α-ε-Τ

108.7

43

4672.9

'

A,

t1

Ti(e)

372:7,

0.:,9,

38.3

Exοerpts from this work may be reproduοed by instruοtors for distribution on a not_for-profit basis for testing or instructional puφoses only to students enro]led in courses for which the textbook has been adopted' Αny other reproducιiοn or ιrαnslαιion of ιhis ινork beyond ιhαt permiιted by Sectiοns l 07 οr Ι 08 of the Ι 976 Uniιed Sιαtes Copyrighι Αcι withοuι ιhe permission of ιhe copyrighι olυner is unlaνvfuΙ.

145

4

South roof West wall East wall

5

Attic floor

2 2

120.7 18.0 18.0 216.0

0.9 0.9 0.9 0.9

108.7 16.2 16.2

194.4

50 36 38 32

5433.6

372.7

583.2

475t.:5

0.9 ,0.9

615.6

4Ι5.5

0.;9

6220.8

277.5

ο.,9

36.0 39;6 39.5 45,3

Τhen, compute the radiant interchange factor and radiation heat transfer coefficient using Eqns. 8-38 and 8-39. Using Eqn. 8-40, estimate the radiative heat flux from each Surface (9,"rα), then determine the radiative heat transfer from each Surface (Q,"α). Then, compute the total radiative heat transfer from all surfaces = -3027.9 W. Divide by the total surface area, 493.5 m2, to get the balancing factor, -6.1 Wmz, which must be subtracted from the previously calculated heat flux from each surface to determine the "balanced" radiation heat flux from each surface (q,u6lbal). Multiply by the area to determine the radiation heat transfer from each surface (Q,"α/bal). Check to see that they no\M Sum to zero.

1

2 J

4 5

Surface

Fir

Tuus (K)

hrt

(Wm'\

North roof South roof West wall East wall Attic floor

0.872

313.8 316.2 310.9

6.'1

b_J 6.1

28.6 87.2 -22.0

31 1.9

6.2

-9.4

0.872

0.897 0.897 0.835

31 1.8

57

Qraα

^

-76.2

Q,"o

(il4

3459.0 10534.5 -395.4 -168.4 -16457.6

qru6/baΙ

Q,,ο/bal

(Wm')

(w) 4199.9

34.8 93.4 -15.8 -3.2

11275.4

-285.0 -58.0 -15132.3

-70.1

8-21

The solution procedure is identicaΙ to that of Problem 8-19, except the emissivities for surfaces 1 and 2 are 0.1. Fictitious surface properties are shown in the first table.

4

Surface Area (ft') North roof 639.7 South rool 639.7 West wall 84.0 East wall 84.0

5

Attiο floor

1

2

J

1176.0

A-ε

ε 0.1 0.1

64.0 64.0

0.9 0.9 0.9

75.6 75.6 1058.4

T(F)

A-ε-T

122

143 102

7804.6 9148.1 7711.2

92

6955.2

2539.4

100548.0

14:47.'.4

oξ

At 1983.7

ε:1

T^,(F)

1983:7,

0.6 ο.6

2539;4

0'5

97.6 96.6 98.6 99.2

ο.5

t 0,2

11'3.3

Τhe total radiative heat transfer from a|l surfaces = -3476.1 Btu/hr. The balancing factor is -1. 3 Btu/(hr-ft2).

Surface

Fir

Τ*s (R)

hri

9rao

(Btu/

Q,rα

(Btu/hr)

Qrrα/bal

(Btu/

Q,,a/bal (Btu/hr)

Exceφts from this work may be reproduced by instruοtors for distributiοn on a not-for-profit basis for tΘsting or instructional puφoses only tο students enrolled in courses Γor which the textbook has been adοpted. Αny other reprοducιion or ιrαnslαιion of ιhis νork beyond thcιι peιmitted by Secιions ]07 or ]08 οfιhe Ι976 Uniιed Sιαιes Copyright Αctνiιhοut the permission ofιhe cοpyrighι οwner is unlcrνυful.

146 (hr-ft')) 1

2 3

4 5

North roof

0.098

South rooi West wall Εast wall Attic floor

0.098

0.874 0.874 0.222

569.5 579.5

1.2

560'ο

10.5 10.3 2.7

555.3 563.8

1.3

(hr-ft'))

30.3 60.8 35.5 -74.1

-49.8

19378.4

31.6

38903.3

62.1

2984.3 -6220.9 -58521.2

36.9 -72.7 -48.4

20226.0 39751.0 3095.6 -6109.6 -56963.0

Note that the radiative heat fluxes from surfaces 1 and 2, and to surface 5 are significantly lower. (The heat flux incident on surface 5 has been reduced by s5%.) Τhe catch is that "in real life", everything else does not remain the same. ln particular, the temperatures would change significantly.

8-22 The solution procedure is identical to that of Problem 8-20, except the emissivities for surfaces 1 and 2 are 0.1. Fictitious surface properties are shown in the first table. Surface 1

2 J

4 5

North roof

South roof West walΙ East wall Attic floor

Area (m2)

A-ε

r(c)

A-ε-T

372,7 .475.5

120.7 120.7 18.0 18.0

0.1 0.1

12.1 12.1

43 50

0.9

Jb

0.9

16.2 16.2

519.2 603.7 583.2

0.9

194.4

38 32

615.6

216.0

6220.8

,Λ

ε',Ι

Tτ(c)

372'7'

ο.6

33;6

0'6

33ι'2

,,33.9

,475,,5'

0,5 0.5

277,,5

o'.2

41.1

33,8

The total radiative heat transfer from all surfaces = -341.5 W. The balancing factor is -0.7 Wm2. Surface 1

2

J 4 5

North roof

South roof West wall East wall Attic floor

Fy

0.098 0.098 0.870 0.870 0.241

9raα

Trrs (K)

hrl

/Wm')

311.4 314.8

0.7 0.7 5.8 5.8

6.3

308.1

309.0 309.7

1.6

^

Q,"ο (h4

765.2

qru6/bal

Q.,a/bal

7.0 12.3

848.7 1490.7

(Wm')

117

1407.1

12.1

217.6

12.8

24.6

443.7 -3175.0

25,3

-14.7

-14.0

(w)

230.0 456.1

-3025.6

8-23

Exοerpts frοm this wοrk may be reproduced by instruοtors 1br distribution on a not_for-prοfit basis for testing or instructional puφoses οnΙy to students enτolled in courses for which the textbook has been adopted. Αny οlher reprοducιiοn or ιrαnslαΙion οf ιhis νork beyond ιhαι permilted by Sections ] 07 or 1 08 οf the Ι 976 Uniιed Sιαtes Copyrighι Αct νιthout the permissiοn ofιhe copyright ονner is unΙαυful.

147

Convective heat transfer coefficients are determined from Table 8-8. Coefficients for the pitched roof surfaces are based on the "Sloping - 45 degrees" surface position. Α more sophisticated approach would involve interpolation. The resuΙts are summarized below. Surface

Area

(ft2)

τ(F)

1

North roof

639.7

122

2

South roof

639.7

143

3

West wall East wall Attic floor

84.0 84.0 1 176.ο

102

4 5

92 95

Surface Position Sloping 45 deqrees Sloping 45 deοrees Vertical Veftical Horizontal

Direction of

Heat Flow

h" Btu/(hr-ft2-F)

(Btu/(hr-ft"))

Downward

o.42

15.54

Downward

0.42

24.36

Horizontal Horizontal

0.56 0.56 0.18

9.52 3.92 1.80

Downward

9""onu""tion

8-24

Convective heat transfer coefficients are determined from Table 8-8. Coefficients for the pitched roof surfaces are based on the "Sloping - 45 degrees" surface position. A more sophisticated approach \Mould involve interpolation. The results are Summarιzed below. Surface

Area

(m2)

r(c)

1

Νorth roof

120.7

43

2

South roof

120.7

50

J

West wall Εast wall Attic floor

18.0 18.0 216.0

36 38 32

4 5

Surface Position Sloping 45 deqrees Sloping 45 deqrees Vertical Vertical Horizontal

Direction of Heat Flow

(wm2-K)

hc

Qt'"onu""1on

Downward

2.39

33.46

Downward

2.39

50.1 9

Horizontal Horizontal

3.18 3.18 1.02

22.26

Downward

rιΛ//m')

28.62 3.06

B-25 Exceφts from this wοrk may be reproduοed by instructors for distribution on a not-Γor-profit basis for testing or instructional puφoses only to students enrolled in οourses for which the textbook has been adopted. Αny οther reprοducιiοn or ιrαnsιαιion of ιhis wοrk beyond ιhαt permιtted by Sectiοns ]07 οr ]08 οfthe Ι976 United Stαιes Copyrighι Αcιll,iιhout ιhe peιmission ofιhe copyrighι οlνner is unlcrννful.

r_ 148 First, the solar irradiation on the window is obtained in the same manner as the solution for Problem 7-17. The following tables show results for the west-facing window. lnput Data

Longitude 101.7 deg deg Standard Meridian 90 -6.2 min Eoτ Latitude 35.23 deg Declination 20.6 deg 270 deg Surf Αzimuth deg 90 Surf Tilt A 346.4 Btu/hr-ft2 0.186 B 0.138 c cN1 RHOG 0.2 Output Data

o θ, " G,νρ" Go* Ga* Gr* " Ψ, β, " Φ, " 15.00 13.12 16.75 69.25 229.59 40.41 74.35 283.92 76.59 27.06 30.47

cDSr Lsr

h,

The layer absorptances of the double-pane (lD5a) can be found from Τable 7-3 as: GDirect,outer: 7f ,774 dοg) σDirect,inner

:'dr(7 4deg)

: O.13 : 0.06

\Mindo\Λ/

134.12

v/ith '1l8 in. sheet glass

(χdiffuse,outer:

.fiyaφrn_ 0.11

acΙiffuse'inne,

.fz'aixur,

:

G,*

:

0.07

Then, the solar radiation absorbed by each pane of the double-pane window may be determined by (neglecting incident solar radiation from the inside): Q"itob,o,b,d,outer,

Q"itob,o,b,d,n*',,"

0.13(76.5g) + O.11(57.53) : 16.29 εtu/(hr_ft^2) :0.06(76.5g) + 0.07(57.53) : 8.62Βtνl(hr-ft2) i, θ

j,ρ:

8-26 Exοeφts 1iom this work may be reproduced by instruοtors for distribution on a not-for-profit basis for testing or instructional puφoses only to students οnrolled in courses for which the textbook has been adopted. Αny oιher reproducιion or trαnsιαtion οf this ''νork beyond ιhαt permitιed by Secιions ]07 οr ]08 ofthe Ι976 (Jniιed SιαteS Cοpyright Αcιwithοuι ιhe permissiοn οfιhe copyrighι owner is unlαινful.

149 First, the solar irradiation on the window is obtained in the same manner as the solution for Problem 7-17 ' Τhe following tables show results for the west-facing window. lnput Data

Longitude 108.53 deg deg Standard Meridian 105 -6.2 min EOT 45.8 deg Latitude Declination 20.6 deg 270 deg Surf Αzimuth deg 90 Surf Τilt Btu/hr-ft2 346.4 A 0.186 B 0.138 c cN1 0.2 RHOG Output Data

MDST LSr h, " Φ,' Ψ, o θ, o Grvo* Gρ" Gd* Gπ* β,' 15.00 13.66 24.92 57.57 227.35 42.65 66.77 277.89 109.60 29.57 27.29 The layer absorptances of the double-pane (lD5a) can be found from Table 7-3 as: &Direct,outer:

.ir(67 deg) : 0.L27

aDirect,inner: 7t1167 deg)

: 0.073

\Λ/indo\M \Λ/ith

ddffise'outer: σc]iffuse,inner:

Gt*

166.46

1/8 in. sheet glass

-fι,aι1urr: 0.11 .lz,aιρr, : 0.07

Τhen, the Solar radiation absorbed by each pane of the double-pane window may be determined by (neglecting incident solar radiation from the inside): Q"

it ob,o,bnd,outer'

Q"ιt ob,o,bud,inner,

j, ο : 0.Ι27 (|0g.6) + 0. 1 1 (56.s6) j, θ : 0.073(109.6) + 0.07(56.86)

: :

Btu/(hr-ft2) 1 1.98 Btu/(hr-ft2)

20.1 7

8-27

RεsULTs BY ΤΗΕ Ηts ΜΕΤΗΟD UsιΝG T}*|Γ ΗVΑCεXpLORΕR pRΟBRΜΑ ARΕ l-{lcl-lΕR Τι-.|ΑΝ RΕ$ULΤ$ βY ΤΗf; RΤS ΜETl*iΟD Usl ΝG Τl-tΕ spRΕΑD$ι*| HΕΤ.

8-28 Εxcerpts ftom this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional puφoses onΙy to permiιιed students enrolled in οourses Γor which the textbook has been adopted. Αny οther reprοducιiοn or ιrαnslαtiοn οf ιhis νork beyond thαι by Secιions ] 07 or 1 08 of ιhe Ι976 United Stαιes Cοpyrilht Αct without the permissiοn ofthe cοpyright oνner is unΙαννful.

150

The RTS method is used to obtain the cooling load results for this problem. The following table shows total cooling loads and cooling loads due to window heaigains for both low-e and regular double-pane windo\Λ/S' Αιl cooling loads due to other heat gains are the same as those shown in Example 8-16. Low-E Windows

Hour

(from ExamPle 8-16) Window Window Total

SΗG

Conduοtion (Btu,hr)

(Btu,&r)

1

186

364

2

146

2

6

110 79 58 52

299 246

7

(Btu/hr)

Regular Windows Windοw Conduction

4418 3843

228 179

420

4516

345

3921

3352

135 97 72 63 75

284

3414 2989

112 178

537

2940 2623

61

tJo 254

2419 2465

8

92

465

2737

9

145

710

8'190

10

215

978

9562

toc

11

300

1247

1

JOO

12

389

1492

13

469

14

14250

477 574 654

15 16

533 577 593

1694 1 833 1897

0883 12143 13275

17

585

1787

1B

553 503

1624

19

20 21

22 23 24

444 386

188'1

139'1

089 839

1

327

674

274 228

546

445

Total

(BtuAr)

(Btu,4lr)

167

5

SΗG

(Btu,&ιr)

203

4

Window

5007 5486 15701 1 0635

1

1

9550 8460 7477 6588 5777 5057

707 726 717

677 616 544 472 401 336

279

234 193 159

2662

293

2518 2829

820 1129 1437 1720 951

1

2111

2185 2167 2060 1

873

'1604

1256 968 777

629 513

2452

8333 9761 11141

12458 I

3637

14648 15425 5905 16105 11008 9877 8727

1

7692 6765 5922 5176

As shown in the above table, usιng the regular \Μindo\Μ Would resuιt in slighily higher cooling loads than using the low-e window. The following 'hcrease in cooling loads due to changing the type of figure illustrates the w]ndow from the low-e window to the reguιar window.

puφoses onιy to on a not-for_profit basis fοr testing or instruοtional Exceφts from this work may be reproduοed by instruοtors for dlstributiοn thαι permiιted beyοnd νork ιhιs ιrαnslαtionλf or Αny olher reprodλc'ilon students enrolled in courses tbr which the textbook has been λ"p"ο copyrighι οwner is unlαwful' permission ofιhe thδ Αctιiιthout Copyright Sιαtes Ι976 Uniιed ]07 or 108 ofιhe by Sections

151

Cooling Load Comparison 18000 '15000

αl

12000

o

9000

dΦ cr)

:Ξ

o o ζ)

6000 3000 0

't0 '13

'16

Tirne, Hour

8-29 SolutΙon to be provided by an instructor. 8-30 First, the solar irradiation must be determined and is the same as that shown for Problem 8-10. Τhen, the hourly dry bulb temperature is calculated using Εquation 8-2. Here, the hour nearest to the local solar time has been used to determine the temperature. Finally, the sol-air temperature is calculated using Εquation 8-63 with the thermaΙ radiation correction term being zero for a vertical surface.

Local lnsolation Outdoor Solar (Btu/h- Drybutb Clock Time Time ft2) Τemp (F) 1.00 23.79 0.00 75.2 2.00 0.79 0.00 73.9 3.00 1.79 0.00 72.6 4.00 2.79 0.00 71.6 5.00 3.79 0.00 70.9 6.00 4.79 0.00 70.6

Sol-air Temp (F)

75.2 73.9 72.6 71.6 70.9 70.6

Exceφts from this work may be reproduοed by instruοtors for distribution on a not-for-profit basis fοr testing or instruοtional puφoses only to students enrolled in courses flοr whiοh the tΘΧtbook has been adopted. Αny oιher reprοduction or trα.nslαtion οf this νork beyond ihαι permiιιed by Sectiοns ] 07 or ] 08 of the 1 976 tJnited Stαtes Copyright Αct y'iιhouι the peιmission of ιhe copyright ονner ii untαwful.

152 7.00 8.00 9.00 10.00 1

1.00

12.00 13.00 14.00 15.00 16.00 '17.00 '18.00

19.00 20.00

21.00 22.00 23.00 24.00

5.79

9.93

71.1

73.1

6.79

23.08

77.0

7.79 8.79 9.79 10.79 11.79 12.79 13.79 14.79 15.79 16.79 17.79 18.79 19.79

32.24 39.29 44.53 47.91 52.42

72.4 74.7 78.0

20.79 21.79 22.79

81 .1

85.8 90.7 95.7 100.6 115.4 130.8

8'1.8 86.'1

6.'10

90.2 93.2 95.2 96.0 95.2 93.5 90.7 87.4

0.00 0.00 0.0ο 0.00

84.1

84.1

81.3 78.7 76.7

81.3 78.7 76.7

111 .11

178.01

230.32 259.13

252.42 184.97

142.1 147.1

143.9 127.7 88.6

8-31

This problem uses the same solution procedure as Problem 8-30. Note that the solar irradiation is the same as that shown for Problem 8-1 1. Clock Time

Local

Solar Time

1.00 2.00 3.00

23.15

4.00

2.15

5.00 6.00 7.00 8.00 9.00 10.00 11.00 12.00 13.00

3.15

14.00

15.00 '!6.00 17.00 18.00 19.0ο

0.1 5

1.15

4.15 5.15

6.15 7.15 8.1 5

9.15 10.15 11.15 12.15 1 3.15 14.15 15.'1 5

16.15 17.15

20.00 21.00

'18.15

22.00

20.15

19.15

Ιnsolation

Outdoor Drybulb

(Btu/h-ft2) Temp

0.00 0.00 0.00 0.00 0.00 0.00 3.96 17.83 29.69 62.38 107.15 142.58 164.39 170.24 159.50 133.31 94.57 48.18 26.46 14.28 0.32 0.00

(F)

Sol-air Temp (F)

73.0 71.2

73.0 71.2

69.6

69.6

68.1

68.1

66.9 66.0 65.7 66.3 67.8 70.5 74.5 79.0 84.2 89.0 92.7

66.9 66.0 66.3 69.0 72.3 79.9 90.6 100.4 108.8 114.6 1 16.6

95.1

115.1

96.0

110.2 102.3 96.9 91.8 85.7 81.8

95.1

93.0 89.6 85.7 81.8

Exοerpts from this work may be reproduοed by instruοtοrs for distributiοn οn a not-for-profit basis for testing or instruοtional puφoses only to students enrolled in οourses Γor which the textbook has been adopted. Αny other reproduction οr ιrαnsιαtion of ιhis νork beyond thαι permiιιed by Sectiοns ] 07 οr ] 08 οf ιhe Ι 97 6 Uniιed Sιαtes Copyright Αct τν ithouι the permission of the cοpyright oινner is unΙαννfuΙ.

153 21.15 22.15

23.00

24.00

0.00 0.00

78.4 75.4

78.4 75.4

8-32 This problem uses the simiΙar Solution procedure as Problem 8-3O. First, the solar irradiation is determined for the flat roof using the procedure described in Chapter 7. The resuΙts are shown below. Ιnput Data

Longitude Standard Meridian

EOT Latitude

Declination Surf Azimuth Surf Τilt

106.62 105 -6.2 35.05 20.6 0 0 346.4

Apar Bpar

deg deg min deg deg deg deg

Btu/hr-ft2

86 0.1 38 0.1

Cpar CN RHOG

1

0.2

Output Data

o MDST Lsr h, o β,. Φ, " Ψ, 1.00 23.79 176.83 -34.27 356.41 356.41 2.00 0.79 -168.17 -33.23 13.26 13.26 3.00 1 .79 -153.17 -28.80 28.82 28.82 4.00 2.79 -138.17 -21.65 42.20 42.20 5.00 3.79 -123.17 -12.54 53.39 53.39 6.00 7.00 8.00 9.00

4.79 -108.17 -2.11 5.79 -93.17 9.'19 6.79 -78.17 21.05 7.79 -63.17 33.22 10.00 8.79 -48.17 45.49 1.00 9.79 -33.17 57 51 1

12.00 10.79 -18.17

11.79 -3.17 14.00 12.79 11.83 '13.00

15.00 16.00 17.00 18.00 19.00 20.00 21

26.83 41.83 56.83 71.83 17 .79 86.83 18.79 101 .83 13.79 14.79 15.79 16.79

.00 19.79 16.B3 1

22.00 20.79 131.83 23.00 21.79 146.83

68.46 75.28 72.19 62.36 50.63 38.41

26.17 14.15

2.58 -8.27 -'18.00

-26.06

θ,

o

Gruo* Go* Gα* Gπ*

0.00 0.00 0.00 0.00 0.00 62'87 62'87 92'11 0.00 71 .22 71.22 80.81 108.07 124.27 123.23 1 18.80 111.65 102.54

0.00 0.00 0.00 0.00 0.00 0.00 17 .26

79.02 79.02 68.95 206.38 74.12 86.89 86.89 56.78 246.69 135.17 95.77 95.77 44.51 266.87 190.31 107.55 107 .55 32.49 277 .85 234.37

127.34 168.24 218.87 245.60 259.78 269.56 277

'71

285.45 293.49 302.43 312.83 325.24

21.54 283.62 263.81 14.72 285.80 276.42 17.81 284.93 271.28 27.64 280.80 248.75 39.37 272.32 210'52 51.59 256.78 159.55 277 '71 63.83 227 .20 10ο.20 285.45 75.85 '161.85 39.s6 293.49 87.42 5.53 0.25 302.43 98.27 0.00 0.00 312.83 108.00 0.00 0.00 325.24 116.06 0.00 0.00 127.34 168.24 218.87 245.60 259.78 269.56

0.00 0.00 0.00 0.00 0.00 0.00

14.91 28.48 34.04 36.83 38.34 39.14 39.44 39.32 38.75 37.58 35.44 31

.35

22.33

0.76 0.00 0.00 0.00

Gt*

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.ο0 0.00 0.00 32.17 0.00 102.60 0.00 169.21 0.00 227.14 0.00 272.71 0.00 302.95 0.00 315.86 0.00 310.60 0.00 287.50 0.ο0 248'10 0.00 194.98 0.00 131 .56 0.00 61.90 0.00 1.01 0.00 0.00 0.00 0.00 0.00 0.00

Excerpts from this work may be reproduced by instruοtors for distribution on a not-fbr-profit basis 1br testing or instruοtional puφoses only to students enτolled in courses Γor which the textbook has been adopted. Αny other reprοducιiοn or trαnslαιion οf ιhis work beyοnd ιhαt permiιted by Secιions ]07 οr Ι08 of ιhe Ι976 Uniιed Stαtes Cοpyright Αcι1ι.ithοut ιhe permission οf the copyrighι oνner is unlωΦl.

154 24.00 22.79 161.83 -31.74 339.93 339.93 121.74

*Unit of lrradiation is Βtu/hr-ft2

0.00

0.00

0.00

0.00

0.00

Then, the sol-air temperature is determined using Equation 8-63 with the thermal radiation correοtion term being 7 "F for a horizontal surface.

Clock Time 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.0ο 9.00 10.00 11.00 12.00 13.ο0 14.00 15.ο0 16.00 17.00 18.00 19.00 20.00 21.00 22.00 23.00 24.00

Local Solar Time 23.79 0.79 1.79 2.79 3.79 4.79 5.79 6.79 7.79 8.79 9.79 10.79 11.79 12.79 13.79 14.79 15.79 16.79 17.79 18.79 19,79 20.79 21.79 22.79

lnsolation (Btu/htt2) 0.00 0.ο0 0.00 0.00 0.00 0.00 32.17 102.60 169.21

227.14 272.71

302.95 315.86 310.6ο 287.50 248.10 194.98 131.56

61.90 1

.01

0.00 0.00 0.00 0.00

Outdoor Drybulb Τemp (F) 75.2 73.9 72.6 71.6 70.9 70.6 71 .1 72.4 74.7 78.0 81.8 86.1 90.2 93.2 95.2 96.0 95.2 93.5 90.7 87.4 84.1 81.3 78.7 76.7

Sol-air Temp (F)

68.2

66.9 65.6 64.6

63.9 63.6 70.5 85.9 101.5 116.4 129.3 139.7 146.3 148.3 145.7 138.6 127.2 112.8 96.0 80.6 77.1

74.3 71.7 69.7

8-33

For hour 15, Equation 8-64 is used to find the conduction heat flux. Q"conduction

= .0052 x (151 '2-74) + .001 44 x (138'1-74)

+

.00645 x (120.3-74)...

=

2.897 Btu/(hr-ft2)

ExcΘrpts from this work may be reproduοed by instruοtors for distributiοn on a not-for-profit basis for testing or instruοtional purposes only tο students enrοlled in courses for whiοh the textbοok has been adopted. Αny οther reprοducιion or ιrαnslαιiοn of ιhis νork beyond thαι permιιιed by Secιiοns 107 οr 108 ofthe Ι976 (Jniιed SιαιeS Copyrighι Αcι|ι)iιhouι the permission ofιhe cοpyright oνner is unlανful.

155 8-34

Equation 8-64 is used to find the conduction heat flux for each hour. q" (Btu/(hr-

Hour

Hour

q" (Btu/(hr-

ft2))

ft2))

1

1.835

13

0.828

2

1.824

14

0.798

J

1.772

15

0.791

4

1.693

16

0.810

5

1.595

17

0.861

6

1.486

'18

0.948

7

1.372

19

1

.071

8

1.259

20

1.225

ο

1.149

21

1.396

10

1.047

22

1.563

11

0.956

23

1.704

12

0.882

24

1.797

8-35

For hour 12, Εqυation 8-64 is used to find the conduction heat flux. Q"conduction

x (143.9-72) + 0.044510 x 0.047321 x (1 21 .4-72)...

= 0.0061 92

=

7

(1

34.3-72)

+

.028 Btu/(hr-ft2)

8-36 Equation 8-64 is used to find the conduction heat fΙux for each hour.

Ηour

q" (Btu/(hrft2))

Hour

ft2))

1

0.674

13

1.050

2

0.401

14

1.544

3

0.1

99

15

2.012

4

0.051

16

2.409

q" (Btu/(hr-

Exceφts from this work may be reproduced by instruοtοrs for distribution on a not-for-profit basis for testing or instructional purposes only to students enτolled in οourses fοr which the textbook has been adopted. Αny oιher reprοclucιιοn or trαnsιαιιοnλf ιhιs work beyond ihαt peιmiιιect by Sections Ι 07 or ] 08 οf ιhe 1 97 6 tJnited SιαιeS Cοpyrighι Αcι Ι4,ithout the permission οf the copyrighι oνner ιi unlαwful.

156 5

-0.058

17

2.694

6

-0.138

18

2.841

7

-0.197

19

2.834

8

-0.232

20

2.671

9

-0.209

21

2.361

10

-0.075

22

1.936

11

0.194

23

1.466

12

0.583

24

1.031

8-37

Equation 8-64 is used to find the conduction heat flux for each hour with soΙ-air temperatures calculated in Problem 8-31. q" (Btu/(hrft2))

Hour

q" (Btu/(hrft2))

1

3.492

13

-0.055

2

3.147

14

0.1

Hour

65

3

2.758

15

0.563

4

2.348

16

1.112

5

1.937

17

1.754

6

1.536

18

2.417

7

1.154

19

3.026

I

0.796

20

3.510

9

0.472

21

3.823

10

0.1

99

22

3.958

11

0.00ο

23

3.931

12

-0.098

24

3.765

8-38

Using the simplified approach, the solution procedure is the same as that of ProbΙem 7-27. First, we need to know angΙe of incidence and Solar irradiation. Αssuming a west-facing window, the incidence angle and solar Ιrradiation for Albuquerque, NM on a c|ear July 21 day at 3:OO pm solar time are (see solution in Problem 8-10 for reference)

Angle of lncidence θ = 65.0 deg., Direct Solar lrradiation. Gρ = 1 18.6 Btu/hr-ft2, Exοerpts from this work may be reproduced by instructors for distribution on a not-fοr-profit basis for testing or instructionaΙ purposes onΙy to

studentsenrolledincoursesforwhichthetextbookhasbeenadopted. Αnyotherreprοdicιionοrιrαnsιαtionλ1ιhιsνorkbeyoidihαιpermiιιed by Secιions ]07 or 108 of ιhe Ι976 United Sιαιes Cowrighι Αcι withοuι thb permissιλn of the copyrighι oνner ii unlανful.

157

Diffuse Solar lrradiation: Ga a Gκ = 30.6 + 28.9 = 59'4 Btu/hr-ft2 Then, the area of the glazing and of the frame is calculated to be 27 .2 and 4.8 ftz, respectively.

ft2

From Table 7-3, solar heat gain coefficients for the glazing system lD 5b are SHGG9ο(65") = 0.515 and SHGGsα = 0.60. From Τable 5-2, the outside surface conductance may be estimated to be 4.0 Btu/hr-ft2-'F. From Table 5-6, the U-value for the fixed, double glazed window having aluminum-clad wood/vinyl frame with insulated spacers is 0.48 Btu/hr-ft2-"F. From Τable 7-1, solar absorptance of the vinyl frame painted white is 0.26.

Αssuming the window with no setback (Ar,r'" = frame can be calculated using Eq. 7-31 as: SHGG1= 0.26*(0.4814.0)

Asuπ),

the SHGC for the

= 0.031.

For an unshaded window, the total solar heat gain is calculated using Eq. 7 -32 as Qsμc = (0.51 5*27 '2 + 0.031-4.8)-'1 '18.6 + (0.60-27 '2 + 0.031*4.8)*59.4 = 2657 '2

Btυlhr'

B-39

This problem uses the same solution procedure as the previous problem. Assuming a south-facing window, the incidence angle and solar irradiation for Boise, ]D on a clear Jυly 21 day at 3:00 pm solar time are (see solution in Problem 8-1 '1 for reference) Angle of lncidence' θ = 68.8 deg., Direct Solar lrradiation: Gρ = 101.6 Btu/hr-ft2, Diffuse Solar lrradiation: Ga + Gκ = 29'0 + 28.9 = 57 '9 Btu/hr-ft2, and Exοeφts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional puφoses only to students enro11ed in courses for ινhich the textbook has been adopted. Αny oιher reproducιion or ιrαnslαιion of ιhis work beyοnd ιhαι permitted by Secιions ]07 οr ]08 οfthe Ι976 Uniιed Sιαιes Copyright Αcιwiιhout ιhe permissiοn οfthe copyrighι o'τιner is unlανfuΙ.

t

'158

The area of the glazing and of the frame is the same as that calculated in Problem 8-38. From Table 7-3, solar heat gain coefficients for the glazing system lD 29a are SHGG,ο(68.8") = 0'408 and SHGGsα= 0.57. From Table 5-6, the U-value for the fixed, triple glazed window having aluminum-clad wood/vinyl frame with insulated spacers is 0.44 Btu/hr-ft2-"F. The outside surface conductance and solar absorptance of the frame are assumed to be the same as those in Problem 8-38.

Assuming the window with no setback (Ar,u'" = frame can be calculated using Eq. 7-3'1 as: SHGGr = 0.26*(0 .4414.0)

Asuπ),

the SHGC for the

= 0.029.

For an unshaded window, the total solar heat gain is calculated using Eq. 7-32 as. Qsuc = φ'408-27 '2 + 0.029-4.8)*101.6 + (0.57*27.2 + 0.029*4.8)-57.9 = 2047.4 Btu/hr.

8-40

First, determine conduction heat gain by multiplying fluxes from Problem 833 by the surface area, 8OO ft2. Then, from Table 8-20, select the radiative/convective split to be 63%137o/o Apply the split to determine the convective and radiative heat gains. Then, apply Equation 8-67 to the radiative heat gains to determine the radiative cooling load. Sum the radiative cooling load and the convective heat gain to get the cooling load.

Excerpts Γrom this work may be reproduοed by instructors fοr distribution on a not-for-profit basis for testing or instruοtional purposes only to students enrolled in courses for whiοh the textbook has been adopted. Αny other reproducιion or ιrαnsΙαtion οf this τυork beyond ιhαt permiιιed by Secιions ] 07 or ] 08 of ιhe Ι 976 United StαιeS Cοpyrighι λcι νithοut ιhe permissiοn of the copyright owner is unlατνful.

'159

Hour

Conduction Heat Gain

Convective

(Btu/h0

HG

Radiative

HG

1

5462.3

2021.0

3441.2

2

4899.6

3ο86.7

3 4

4334.5 3796.4 3300.5 2854.1 2460.6

1812.8 1603.8 1404.7 1221 2 1056.0 910.4

8

2123.8

785.8

9

1854.2 1673.6 1598.0 1629.0 1759.4 1983.0

686.1 619.2 591.3 602.7

5

b 7

10 11

12

13 14 15

2318.0 2803.7 3450.5 4215.9 5016.0 5741.3

16 17 18 19

20

6266.5 6473.8

21

22

6345.1 5971.2

ZC

24

651.0 733.7 857.6

2903.7 2813.1

2730.7 2391.7 2079.3

2702.9

1798.1

2339.0 2222.4

1550.2 1338.0 1168.2 1054.4 1006.7 1026.3 1 108.4 1249.3 1460.3 1766.3

1037.4 1276.7 1559.9 1855.9

2173.8 2656.0

2124.3

3617.0

23'18.6

3947.9 4078.5 3997.4

2395.3 2347.7 2209.4

Radiative Cooling Load

3160.1

376'1.9

2583.0 2460.2

2113.0 2014.0 '1930.6

1868.0 1829.3 '1815.8

1827.8 1868.9 1947.4 2068.9

Cooling Load 4924.7 4625.9 4306.7 3987.7

3681.4 3395.0 3132.9

2898.9 2700.1 2549.9

2459.2 2432.1 2466.8 2561.5

2726.6 2984.8 3345.5

2230.3

3790.2

2419.2 2614.7 2789.4 2913.3 2969.4 2961.2

4275.2 4739.0

5108.0 5308.6 5317.1

5170.6

Cooling Loads and Heat Gains 7ο00.0 60ο0.0 5000.0

Ιt Φ

4000.0

Θ

G

Ι

3000.0

ω

2000.0

6)

oδ

o

J

1

000.0 0.0

Exοeφts fiom this work may be reprοduοed by instructors for distributron on a not-fοr-profit basis fοr testing or instruοtional puφoses only to students enrolΙed in οourses for which the textbοok has been adopted. Αny οιher reprοductiοn or trαnsιαtion of this work beyond ιhαι permitted by Secιions ] 07 οr ] 08 of ιhe Ι 976 LΙniιed Sιates Copyright Αcι νithout ιhe permission of ιhe copyrighι οlνner is unlcrνful.

16ο

8-41

First, determine conduction heat gain by multiplying fluxes from Problem 835 by the surface area, 1OOO ft2. Then, from Τable 8-20, select the radiative/convective split to be 84oλl16o/o' Apply the split to determine the convective and radiative heat gains. Then, apply Εquation 8-67 to the radiative heat gains to determine the radiative cooling load. Sum the radiative cooling load and the convective heat gain to get the cooling load.

Hour

Conduction Ηeat Gain

Convective

Radiative

1

2227.2

356.4

1870.8

2

1338.2

627.7

4

57.3 -397.4 -746.0 -935.8 -610.8 507.2

214.1 100.4 9.2 -63.6 119.4 -149.7 -97.7

1124.1

J

81.2 370.2 730.7 1124.5 1512.8 1857.6

426.1

5 6 7

o 9

10

2313.5

11

4567.1

12

7028.4 9455.0

13 14

22

116ο9.7 13293.3 1 4350.1 14672.7 14222.5 13ο18.3 11142.9 8809.6 6593.1

z3

4782.8

24

3353.'1

15 16 17 '18

19

20 21

HG

(Btuihr)

2126.9 2296.0 2347.6 2275.6 2082.9

HG

527.3 48.2 -333.8 -626.7 -786.0 -513.1

4530.5 4222.8 3938.6 3676.2 3436.0 3228.9 3120.5 3179.7 3417.1

12054.1

651 1.5

12325.1

6849.5 7024.8

409.5

1ο54.9

s538.2

765.3

4017.6

536.5

2816.6

1

5220.7

4744.6

4316.0 4893.4

I

Cooling Load

4864.4

1943.3 3836.3 5903.8 7942.2 9752.2 I 1 166.3

1946.9 1Ο935.4 9360.0 7400.1

1782.9


3808.3

5486.8 6042.8

7018.0 6824.5

6467.5 6042.8 5622.2 5228.0

4323.2 3947.7 3612.7 3316.7 3079.2

3022.8 3260.8 3787.2

4539.0 5440.6 6406.2

7344.4 8169.7 8807.6 9197.1 9300.4 9101.0 8607.3 7877.0 7097.7

6387.5 5764.5

or instructional puφoses only to Exc€φts frοm this work may be reprοduοed by instructors for distribution οn a not-for-profit basis for testing permitted or trαnsιCιιion of thιS work beyond ιhαt students enτo]led in courses for which the textbook has been adopted. Αny oιher reprοducιion οfthe Ι976 Uniιed Stαtes Copyrighι Αctwiιhout ιhe permission οfιhe cοpyright oνner is unlαwful.

η s;"iι'n'

]07 οr ]08

161 cooling Loads and Ηeat Gains

L

1

6000.0

1

4000.0

1

2000.0

'10000.0

fl

αl

--r-- Conduction Heat Gain

8000.0

(!

(, (! Φ !

(Btu/h0

-*x* Cooling Load

6000.0

οδ

Φ

o

J

911131517192123

8-42 First, determine conduction heat gain by multiplying fΙuxes from Problem 836 by the surface area, 1200 ft2 ' Ther , from Τable 8-2o, seΙect the radiative/convective Split to be 84γoμe% Αpply the split to determine the convective and radiative heat gains. Then, apply Equation 8-67 to the radiative heat gains to determine the radiative cooling load. Sum the radiative cooling load and the convective heat gain to get the cooling load. Conduction Heat Gain (Btιl/hr)

Convective

HG

Radiative ΗG

809.2 481.0

129.5

679.7

2

77.0

404.1

3

238.5

4

38.2 9.8

200.4 51.4

5

61.2 -69.3

11 .1

-58.2

6

-166.0

7

-139.5 -199.0

10

-237.0 -278.7 -250.9 -90.4

11

233.3

12

700.2

-26.6 -37.9 -44.6 -40.2 -14.5 37.3 112.0

Hour 1

o

I

-234.1

-210.8 -75.9 196,0 588.2


Cooling Load

1203.0 105.3 1018.3 941.3 872.8 811.4 756.2 707.9 674.0 667.9

1332.4 1182.3 1056.5

699.7

737.1

771.3

883.3

1

951.1

861.7 784.9 718.3

663.3 633.8 653.4

Excerpts Γrom this work may be reproduοed by instructοrs for distribution on a not_for_profit basis for testing or instruοtional puφoses only to students enrolled in οourses for which the textbook has been adopted. Αny other reproducιion or trαnslαιion of this work beyοnd ιhαt permiιted by Secιions 107 οr 108 ofthe Ι976 United Stαtes Copyright Αcινηithοut the permissiοn ofthe copyrighι oνner is unlcrwful.

162 13 14

15 16 17

18 19

1260.3 1852.6 2414.9

201.6 386.4

1058.6 1556.2 2028.5

2890.5 3233.3

462.5 517.3

2428.0 2716.0

3409.7 3401.3

545.6

2864.2 2857.1 2692.2 2380.2

3205.0 2833.6 2323.3

20 21

22 23 24

1759.4 1237.6

296.4

544.2 512.8 453.4 371.7

281.5 198.0

1951 .6

1477.9 1039.6

877.1

1007.4 1149.7 1290.4 1416.6 1516.5 1580.7 1602.7 1579.7 1513.7 1417.5 1309.3

1078.8 1303.8 1

536.1

1752.9 1933.9

2062.0 2124.9 2115.5

2033.0 1885.5 1699.0 1507.3

Gooling Loads and Heat Gains 4000.0 3500.0 3000.0 αt Φ

Θ

2500.0

(τ

Φ

1500.0

cδ

'1000.0

Ι

_--ο_ Conduοtion Heat Gain

2000.0

(Btu/hr) *_α_*

Cooling Load

ιl1

o

J

500.0 0.0 -500.0

8-43

First, the hourly Soιar heat gains are determined using the same solution procedure Shoν1/n in Problem 8-38. The results are Sho\Λ/n below. Note that the SolaΓ irradiation on the window is the Same aS that shown in Problem 8-10. Also, note that the calculated Soιar gain at 3:00 p.m. is slightly different from that shown in Problem 8-38 due to rounding errors. lnput Data

Glass Area Frame Αrea

27.2 4.8

ft', ft2

Excerpts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instructional purposes οnly to students enrolled in courses for whiοh thο textboοk has been adopted. Αny οιher reproducιiοn or ιrαfiSlαιiοn of this work beyond thαι permiιιed by Secιions ]07 οr ]08 ofthe Ι97 i United Stαtes Copyrιghι Αct wiιhout the permission οfιhe copyright oνner is unΙαwfuΙ.

163 Diffuse SHGC Angular SHGC - 0' Angular SHGC - 40' Angular SHGC - 50' Angular SHGC - 60" Angular SHGC - 70" Αngular SHGc - 80" Frame sΗGc

Clock Time 1.00

2.00 3.00

4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00 12.00 13.00 14,00 15.00 16.00 17.00 18.0ο 19.00

20.00 21.00 22.00 23.00 24.00

lnc. Αngle, ' 87.03 101 .06 114.99 128.63 141.58 152.80 159.17 156.37 146.65 134.23 120.81 106.97 92.97 78.94 65.01 51.37 38.42 27.20 20.83 23.63 33.35 45.77 59.19 73.03

0.6 0.7 0.67 0.64 0.58 0.45 0.23 0.03'1

Dir

Output Data Diff

lrradiation,

Btu/hr-ft2 0.00 0.0ο 0.00 0.0ο 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 54.68 18.63 170.01 201.20 202.07 151.27 5.07 0.0ο 0.00 0.00 0.00 1

lrradiation,

Btu/hr-ft2 0.00 0.00 0.00 0.00 0.00 0.00 9.93 23.08 32.24 39.29 44.53 47 .91 52.42 56.44 59.38 60.32 57.93 50.35 33.70 1.03 0.00 0.00 0.00 0.00

Solar Heat Gain, Btu/hr 0.00 0.00 0.00

0'00 0.00 0.00 163.51

380.04 530.96 647.01

733.28 788.98 863.28 1314.44 2657 .05

3940.06 4657.12 4594.55

3393.40 111.74 0.00 0.00 0.00 0.00

ln the original RΤS methodology, two types of radiant time faοtors v/ere utilized to convert So|ar heat gains into cooling Ιoads. The Solar.-RTS was used to convert the beam transmitted solar gain whiΙe the NonsoΙar-RΤS V/aS used to convert all other Solar gains' However, to simpΙify the calculations, only one RTS (Nonsolar-RTs) is used in this edition. Since the calculated Solar heat gains include both transmitleΞ aιd absorbed Solar gains, the recommended radiative and convectivΞsplits shown in Table 820 would not be applicable. For this problem, it is assumed that the radiative fraction of the combined solar heat gain is about 0.9. Therefore, Exοeφts from this wοrk may be reproduced by instructors fοr distribution on a not-for-profit basis for tΘSting or instructional purposes only tο

studentsenrolledincoursesforwhichthe textbookhasbeenadopted. Αnyοtherreproductionοrtrαnslαtionοfιhisνοrkbeyondιhαιpermiιιed by Sectiοns 107 or ]0B ofthe Ι976 Uniιed Stαtes Copyrighι Αcιwιιhοuι ιhe permissiοn ofιhe cοpyright oνner is unlατνfuΙ.

164 the radiative/convective split is 90%110%. Then, apply the split to determine the convective and radiative heat gains and apply Equation 8-67 to the radiative heat gains to determine the radiative cooling load. Αnd, finally, sum the radiative cooling load and the convective heat gain to get the cooling load.

Solar

Heat Gain

Convective HG

Radiative

0.0 0.0 0.0 0.0 0.0 0.0 38.0

0.0 ο.0 0.0 0.0 0.0 0.0 147.2 342.0

53.1

16

0.0 0.0 0.0 0.0 0.0 0.0 163.5 380.0 531.0 647.0 733.3 789.0 863.3 1314.4 2657.1 3940.1

265.7 394.0

17

4657.1

465.7

777.0 1 183.0 2391.3 3546.1 4191.4

4594.6 3393.4

459.5

4135.1

339.3

111.7 0.0 0.0 0.0 0.0

11.2

3054.1 100.6 0.0 0.0 0.0 ο.0

Hour 1

2 3

4 5

6 7

I 9 10 11

12 13 14

15

18 19

20 21

22 23 24

(Btu/hr)

16.4

64.7 73.3 78.9 86.3 131.4

0.0 0.ο 0.0 0.0

HG

Radiative Cooling Load 141 0 91.6 59.8 39.2 25.9

Cooling Load 141.0

91.6 59.8 39.2 25.9

17.2

17,2

477.9

87.6 215.2 339.6

582.3

450.5

103.9 253.2 392.7 515.2

660.0

543.8

617.1

710.1

615.8 687.3 932.5 1661.8

694.7 773.6 1064.0 1927.5

2566.6

2960.6

3304.7 3630.8 3261.0 1624.6

3770.4

4090.3 3600.4 1635.8

912.1

912.1

549.4 343.2 218.7

549.4 343.2 218.7

Exοerpts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instructional puφoses onιy to permiιιed students enrolled in courses }or whiοh the textbook has been adopted. Αny other reproducιiοn or ιrαnsιαιion of ιhis νοrk beyond thαι by Secιions 1 07 or ] 08 of the 1 97 6 Uniιecl Sιαιes Copyrιghι Αcι |uithοuι the permissiοn οf ιhe copyright owner is unΙαινful.

-165

Cooling Loads and Heat Gains 5000.0

4500.0 4000.0 3500.0

I

Φ

3000.0

_ο*x-

(!

Θ 2500.0 G

Φ J.

Φ (!

Cooling Load

2000.0 1500.0

o

J

Soiar Heat Gain (Btu/hr)

1

000.0 500.0 0.0

B-44

This problem uses the same solution procedures as Problem 8-43. Note that the solar irradiation on the window is the same as that shown in Problem 8-11. Also, note that the calculated Soιar gain at 3:00 p.m. is slightly different from that shown in Problem 8-39 due to rounding errors. ThΞ radiative/convective split of 90%l10% is also used for this problem. lnPut Data

Area Frame Area Diffuse SHGC Angular SΗGc - 0" Angular SHGC - 40' Αngular SHGO - 50' Angular SHGC - 60' Angular SHGC - 70' Angular SHGC - 80" Frame SHGC Glass

27.2 4.8

ft' ft2

0.57 0.68 0.65 0.62 0.54 0.39 0.18 0.029

OutPut Data

Clock

lnc.

Time

Angle,'

1.00 2.00 3.00

152.14 154.11 '150.59

Dir lrrad,

Btu/hr-

ft2 0.00 0.00 0.00

Diff lrrad, Btu/hr-

Solar Heat Gain,

f1'

Btu/hr

0.00 0.00 0.00

0.00 0.00 0.00

or instructional puφoses only to EΧοeφtS from this work may be reprοduοed by instructors fbr distribution on a not-for-profit basis for testing ιhis νork beyond thαι Permiιιed οf trαnslαιiοn or reproiucιιon other Αny adopted. has been the textbook students enrolled in courses fbr which is unlωυful. by Sectiοns Ι 07 or Ι 08 οf the ] 97 6 (]niιed Stαtes Copyright Αct wiιhouι the permission οf the copyrighι oινner

166 4.00 5.00 6.00 7.00 8.00 9.00 10.00 1

1.00

12.00 '13.00

14.00 15.00 16.00 17.00 18.00 19.00 20.00

143.20 133.87 123.75 113.42 103.29 93.66

0.00 0.00 0.00 0.00 0.00 0.00

84.88

22.70

77.37 71.58

58.92 87.76 105.54 1 10.30 101.56 80.22 48.70 11.34 0.00 0.00 0.00 0.00 0.00 0.00

68.02 67.06 68.82

Ηour

Heat Gain (Btιl/hr\

Convective

Radiative

0.0 0.0 0.0 0.0 0.0 0.0

ο.ο 0.0 0.0 0.0 0.0 0.0 6.2

0.0 0.0 0.0 0.0 0.0 0.0 55.7 251.0 418.0 612.7 1025.8 1549.2 1926.2 2029.8 1842 1 1395.9 880.3 533.2 372.6

1

2 3

4 5

o 7

I I

HG

61.9 278.9

27.9

464.4

46.4

10

680.7

68.1

11

114.0

214.0 225.5

to

1139.7 1721.4 2140.2 2255.4 2046.8 1551.0

17

978.1

97.8

18 19

592.5

59.2

413.9

41.4 22.3 0.5 0.0

12 13

14 15

20 21

22 23 24

223.4 5.0 0.0 0.0 0.0

172.1

204.7 1

55.1

----ΑΔ 0.0

1139.73 1721.36 2140.21

58.85 59.94 57.95

22.00 23.00 24.00

Solar

6'1.91

278.89 464.39 680.73

54.81

73.09 79.44 87.38 96.45 106.26 116.49 126.80 136.76 145.67

21.00

0.00 ο.00 ο.00

0.00 0.00 0.00 3.96 17.83 29.69 39.68 48.23

ΗG

201.1

4.5 0.0 0.0

..--o.0

2255,35

53.1 0

2046.81 1551.04

45.86

978.1 5

36.84 26.46 14.28 0.32 0.0ο 0.00 0.0ο

592.46 413.94

223.45 4.97 0.00 0.00 0.00


Cooling Load

356.3 341.8 328.6 316.4 305.0

356.3 341.8 328.6 316.4 305.0

294.1

294.1

297.0 338.9 391.2 457.1 583.2 763.0 935.3 1046.5 1067.8

303.1

1149.3 1272.0 1272 4

366.8

437.6 525.2 697.2 935.1

991 0

1146.1

853.6 723.9

783.1

633.6

675.0

551.8 465.4

465.9

421.3 393.5 373.0

951.4

574.1

421.3 393.5 373.0

Exceφts from this wοrk may be reproduοed by instructors Γor distribution on a not-for-profit basis for testing or instruοtional puφoses only to students enrolled in courses for whiοh the textbοok has been adoρted' Αny oιher reprοducιion or ιrαnslαιιon of ιhis νork beyond thαι permιιιed by Secιions ]07 or ]08 ofιhe Ι976 Uniιed Stαtes Cοpyrighι Αcιlνithout lhe permissiοn ofιhe copyright oνner is unΙcrννful'

167


2000.0

L

Φ I

500.0

--l_Solar

Ηeat Gain (Btu/hr) ,--x- Cooling Load

ιE

Θ

(E

Ι

Φ

1

Φ

000.0

Φ

o

J

500.0

0.0

8-45 lnternal

Hour

Heat Gain (w)

I

200.0 200.0 200.0 200.0 20ο.0 200.0 200.0 2000.0 2000.0

2 3

4 5

6 7

δ

I

10

2000.0

11

2000.ο 2000.0 2ο00.0 2000.0

12 13 14 15

2000.0

16

2000.0

17 18

2000.0

19

200.0

2ο00.ο

Convective

HG

Radiative HG

100.0 100.0 100.0 100.0 100.0 100.0 100.0 1ο00.0 1000.0 1000.0

100.0 100.0 100.0 100.0 100.0 100.0 100.0 1000.0 1000.0 1000.0

10οο.ο

'10οο.ο

1000.0 1000.0 1ο00.0 1000.0 1000.0 1000.0 1000.0 100.0

1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 100.0


Cooling Load

125.3

225.3

116.6 111.O

216.6 211.0

107.3 105.0

207.3 205.0

103.4

102.4 566.7 753.8

203.4 202.4 1566.7 1753.8

85'1.1

1

906.9 940.8

1906.9 1940.8

(Vψ

851 .1

962.1

1962.1

975.6 984.3 990.0 993.6 996.0 532.6

1975.6 1984.3 1990.0 1993.6 1996.0

632.6

EXceφtS from this wοrk may be reprοduced by instructors fοr distribution on a not-fοr-profit basis for testing or instructional puφοses only to students enτolled in οourses for which thθ teΧtbook has been adopted. Αny οιher reproduction or trαnsΙαtion of this τνοrk beyond ιhαι permitted bySectiοns Ι07 or Ι08of ιhe ]976UniιedStαtesCοpyrightΑctwithοutthepermissiοnοf

thecοpyrighιownerisunlcrννful.

'168 200.0 200.0 200.0 200.0 200.0

20 21

22 23 24

100.0 100.ο 1ο0.ο 100.0 100.0

100.0 100.0 100.ο 100.0 100.0

346.2 249.2 193.6

446.2 349.2

159.9

293.6 259.9

138.7

238.7


2000.0

Ξ '6

1500.0

o

+*x-

(!

Ι (Σ)

lnternal

l-1eat

Cooling Load

Gain (W) (\Λ/)

oδ (υ

o

J

8-46

Hour 1

2 3

lnternal Heat Gain

rw)

200.0

200.ο 2ο0.ο

4

200.0

5

200.0

6

200.0

7

200.0 2000.0 2000.0 2000.0 2000.0 2000.0 2000.0 2000.0

I o

10 11

12 '13

14

Convective

Radiative

HG


Cooling Load (w)

100.0 1οο.0 100.0 100.0 100.0 100.0 100.0 1000.0 1000.0 1000.0 1000.0 1ο00.ο 1000.0

100.0

405.'1

505.1

392.9 381.7 371.2 361.3

492.9 481.7

HG

10οΟ.0

'100.0 100.0

100.0 100.0 100.0 100.0 1000.0 1000.0 1ο00.0 1000.0 10ο0.ο 1000.0 1000.0

352.1

343.3 518.9 562.8 59ο.6 612.3 630.8

471.2 461.3 452.1 443.3 15'18.9

1562.8 1590.6 1612.3 1

630.8

647.1

1647.1

661.9

1661.9

Εxοeφts from this work may be reproduced by instΙuctors for distribution on a not-for-profit basis for testing or instructional puφoses only to students enrolΙed in οourses for which the textbook has been adopted' Αny oιher reprοductιοn or ιrαnsl(tιion of ιhis work beyond thαt permiιιed by Sectiοns Ι 07 or ] 08 of the 1 976 Uniιed Stαtes Copyrιghι Αcι'ννiιhouι ιhe permissionbf the cοpyrighι οwner is unΙαlνful.

'169

Γ

2000'ο 2000.0

15 16

2000.0

17 18 19

2000.0 200.0 200.0 20ο.ο 200.0 200.0 200.0

20 21

22 23 24

100ο.0 1000.0 1000.0 1000.0 100.0 100.0 100.0 100.0 100.0 100.0

1000.0 1000.0 1000.0 1000.0 100.0 100.ο 1ο0.ο 100.0 100.0 100.ο

675.5

1675.5

688.2 700.0

1700.0

711 .1

1711 .1

537.6 495.7

637.6

469.9 449.9 433.2 418.4

'1688.2

595.7 569.9 549.9 533.2 518.4

Gooling Loads and Heat Gains 2500.0

2ο00.0

1500.0

1000.0

500.0

; 0.0

:.

.::l

---ο-

lnternal Fleat Gain (W1

*-s*

οooling Load (W)

&-x** *

L-.*

ΕΧcerpts tiom this ινork may be reproduοed by instructοrs for distributiοn on a not_for-profit basis fοr testing or instruοtional puφoses only to students enrolΙed in οouτses 1br whiοh the textbook has been adopted. Αny oιher reproduction or trαnsιαιιon of this work beyond ιhαι permiιιed by Secιiοns ]07 or ]08 οfιhe Ι976 LΙnited StαιeS Cοpyright Αcι νiιhouι ιhe permiSSiοn οfιhe copyrighι oνner is unlαwful'

170

8-47 Gomparison of LW and

Ξ Ξo(!

Mνι/ 1

Zone Responses

2000.0

o

σ, 1500.0

--r-- lnternal Heat Gain β) MW1 Zone Clg. Ld αv)

:=

o o

o

τ, (τ,

_--α.*

Ι

-_.*- ΗW Zone Clg. Ld. (W)

1000.0

aι'

'6

Θ

(τ,

α)

ι*. \Η \

50ο.ο

_

5**+-

ι ξ--'. l)* ι ':'* Ι^__x

-t-

11

'13

15

17

19

21

23

Hour

As shown in the figure, there is a signifrcant difference in the response of the two Zones, with the ΗW zone having substantiatly more damping and time delay.

Exceφts f?om this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instructional puφoses only to students enrolled in courses fοr whiοh thΘ textbook has been adopted. Αny oιher reproclucιiοn or ιrαnsιαιionbf ιhιs work beyoλd ihαι permiιιed by Sections ]07 οr ]08 ofιhe }976 United Stαιes Cοpyright Αctwithοut the permissiοn οfιhe cοpyrighι οwner ιi unlανful'

171

8-48

Assumptions applied to each heat gain are discussed in the solution to ProbΙem 8-15. The equipment heat gain is assumed to be continuous. The total convective and radiative heat gains are determined in the folΙowing table. Τhe Ιatent cooΙing Ιoads are equivalent to the latent heat gains shown in the last column.

Name:

Radiative Fraction:

ξ99p]9_ Liοhtino

Eοuipment

0.7

0.59

0.2

Ηeat Gain (\Λ/)

Ηeat Gain /w)

Heat Gain (w)

1

0

0

2

U

0

3

0

0

4

ο

5

0

6

0

0 0 0

7

ο

0

o

2160 2160

4800 4800 4800 4800 4800 4800 4800

lloυr

I 12 13

2160 2160 2160 2160

14

2160

15

2160 2160 2160

10 11

16

17 18

0

48CI0

4800 4800 4800

19

0

20

U

0 n

21

ο

0

22 23 24

0 0 0

0 0

4000 4000 400ο 4000 4000 4000 4000 4000 400ο 4000 4000 4000 4000 4000 4000 4000 4000 4000 4000 4000 4000 4οο0 4000 4000

Total

Total

Latent from

Τotal

Radiative

Convective

People

Heat

Heat

Heat Gain rιΛΛ

Heat Gain (w)

320C)

0

320ο 3200 3200 3200 3200 3200

0

Gain (W)

Gain (W)

4000 4000 4000 4ο00 4000 4000 4000

800 800 800 800 800

0960 0960 1 0960 1 0960 1 0960 1 0960 1 0960 1 ο960 1 0960 1 0960 8800 1 1

4000 40ο0 400ο 400ο 4000 4000

800 800 5144 5144 5144 5144 5144 5144 5144 5144 5144 5144 3632 800 80ο 800 800 800 800

0 0 0 0 n

581 6

350 350 1 350 1 350 1 350

581 6

1

581 6

1

5816

1

581 6 581 6 581 6

5816

581 6

58l 6

1

1

350

35ο 350 1 350 1 350

5168 3200

0

32ο0 3200 3200 3200 32ο0

0

0 0 0 0 0

The sensible loads are then determined from the radiative and convective heat gains using Equation 8-67 and the radiant time factors from Τable 821, as shown in the next table. ΕΧceφts liom this work may be reproduced by instructors 1br dlsιribιltlon οn a noιtbr-pτofit basis for testing or instructiona] purposes onΙy to it-ι οοιlrses fοr ιhich the teΧιbook has bοen adοptοd. lπ;l οιΙιer reprοdιιction or trαnslcιιioιι cf ιhis ιι'ork beyond ιhαι ρerfrιiιted by Sectiοlιs ] 07 οr Ι08 οf ιhe Ι 97 6 Uniιed StαιeS Copyrighι Αcι ινiιhouι ιlιe ρernιission οf the cοpyrighι ο'!'ner iS unlαιν.ful' sιuderrts en:-olled

172

lnternal

Hour 1

2 3

4 5

6 7 8 9 10 11

12 '13

14

15 16 17 18

Heat Gain (w)

Convective ΗG

Radiative

4000.0 4000.0 4000.0 40ο0.0 4000.0 4000.0 4000.0 10960.0 10960.0 10960.0 10960.0 10960.0 10960.0 1096ο.0 10960.0

32ο0.0 3200.0 3200.0 3200.0 3200.0 3200.0 3200.0 5816.0 5816.0 5816.0 5816.0 5816.0 5816.0 5816.0 5816.0

'10960.0

58 1 6.0

10960.0

5816.0 5168.0 3200.0 320ο.0 3200.0 3200.0 3200.0 3200.0

800.0 800.0 800.0 800.0 800.0 800.0 800.0 5144.0 5144.0 5144.0 5144.0 5144.0 5144.0 5144.0 5144.0 5144.0 5144.0

88ο0'0 4000.0 4000.0 4000.0 4000.0 4000.0 4000.0

'19

20 21

22 23 24

HG

3632.0 800.0 8ο0.0 800.0 800.0 800.0 800.0


Cooling Load (w)

2218.4

5418.4 5362.5

2162.5 2110.6 2062.1 2016.7 1973.8 1933.3 2782.4

5310.6 5262.1 5216.7 5173.8

5133.3 8598.4

2996.'1

8812.1

3131.2 3237.5

8947.2

3328.ο 3408.1 3480.7 3547.6 3609.6 JOO /. /

3413.3 2799.1

2625.5 2512.4 2422.9

9053.5 9144.0 9224.1 9296.7

9363.6 9425.6 9483.7 8581.3 5999.1 5825.5

5712.4 5622.9

2346.8

5546.8

2279.4

5479.4

Cooling Loads and Heat Gains 120ο0.0 10000.0

Ξ '6

8000.0

Θ

Φ Φ

Ξ

oδ π,

o

6000.0 4000.0

J

2000.0 0.0

ι

instructional purposΘs only to Εxοeφts fiom this woτk may be reproduced by instructofs for distrlbution on a not-for-profit basis for testing or of this νοrk beyond ιhαt permiιted trαnsιαιion or reprοducιion Αny other adopted' has been textbοok which thΘ foτ in οouτses enrolled students by Secιions Ι07 or Ι08 οfthe ]976 ιJnited Stαtel Cοpyrighι Αct\'ιithout the permissiοn ofιhe cοpyrighι owner is unΙαlυfuΙ.

173 8-49

Assumptions applied to each heat gain are discussed in the solution to Problem 8-'16. The equipment heat gain is assumed to be continuous; the lighting heat gain is assumed to occur from 8 a.m.-6 p.m. The total convective and radiative heat gains are determined in the following table. The latent cooling Ιoads are equivalent to the latent heat gains shown in the last column.

People

Liohtinο

Eουioment

Fraction:

0.7

ο.59

0.2

Hour

Heat Gain (w)

Heat Gain (w)

Name:

Radiative

1

0

0

2

0

3

0

ο 0

4

0

0

5

U

0

6

0

0

7

0

0

9 't0

2835 2835 2835

11

2835

12

2835 2835 2835 2835 2835

5625 5625 5625 5625 5625 5625 5625 5625 5625 5625 5625

I

13 14

15 '16

17 18 19

2835

20

0

0 0

21

0

0

22 23 24

0 0

ο ο

0

0

0 0

Heat Gain rw) 70ο0 7000 7000 7000 7000 7000 7000 7000 7000 7000 700ο 7000

7000 7000

7000 7000

7000 70ο0 7000 7000 7000 7000 7000 7000

Latent from

Τotal

Total

Radiative

Total Convective

Peοοte

Heat

Heat

Ηeat Gain

Heat Gain (w)

Gain (W)

Gain (W)

70οο 7000

1400 1400 1400 1400 1400 1400 140ο

7ο00

7000 70ο0 7000 7ο00 5460 1 5460 1 5460 1 5460 1 5460 1 5460 1 5460 1 5460 1 5460

'15460

6703.25 6703.25 6703.25 6703.25 6703.25 6703.25 6703.25 6703.25 6703,25 6703.25

12625

4718.75

1

400

7000

1

7000 7000

1400 1400

7000 700ο

7000

1

400

1400 1400

/\ΛΛ

5600 5600 5600 5600 5600 560ο 5600 8756.75 8756.75 8756.75 8756.75 8756.75 8756.75 8756.75 8756.75 8756.75 8756.75 7906.25 5600 5600 5600 5600 5600 5600

0 0

0 0 0 0

0

2835 2835

2835 2835 2835 2835 2835

2835 2835 2835 0

0 0 0

0 ο 0

The sensible loads are then determined from the radiative and convective heat gains using Εquation 8-67 and the radiant time factors from Τable 821 , as shown in the next table. EXceφtS fτοm this work may be reproduοed by instruοtors for distribution on a not_for_prοfit basis Γor testing or instructional puφoses only to students enro|led in courses for whiοh the textboοk has been adopted. Αny οιher reproduction or ιrαnslαιiοn οf this νork beyond thαt permitιed by Secιions ]07 or ]08

ofthe Ι976 LΙniιed

Stαtes Cοpyright Αctlυithοut the permissiοn

οfιhe cοpyright oνner is unlανful.

174

lnternal

Hour 1

2 3

4 5

6 7

8 9

10 11

12 '13

14 15

ιo 17 '18

19

20

Heat Gain (w)

Convective ΗG

Radiative

7000.0 7000.ο 7000.0 7000.0

5600.0 5600.0 5600.0 5600.0 5600.0 5600.0 5600.0 8756.8 8756.8 8756.8 8756.8 8756.8 8756.8 8756.8 8756.8 8756.8 8756.8 7906.3 5600.0 5600.0 5600.0 5600.0 5600.0 5600.0

1400.0 1400.0

7000.ο 7000.0 7000.ο 15460.0 15460.0 15460.0 15460.0 15460.0 1546ο.0 15460.0 15460.0 15460.0 15460.0 12625.0 7000.0 7000.0

7000.0 7000.0 7000.0

21

22 23 24

7000.0

HG

1

400.0

14ο0.0 14ο0.0 1

400.0

140ο.0

6703.3 6703.3 6703.3 6703.3 6703.3 6703.3 6703.3 6703.3 6703.3 6703.3 4718.8 1400.0 1400.0 1400.0 1400.0 1400.0 1400.0


Cooling Load (w)

2888.2

8488.2 8430.6 8377,5 8327.8 8280.7

2830.6 2777.5 2727.8 2680.7 2635.8 2592.7 3814.6 4292.6

4539.8 4694.4

4806.9 4897.4 4975.3 5045.3 5109.7 5170.2 4754.8 3825.2 3446.7 3246.7 31 19.9 3027.2 2952.4

8235.8 8192.7 12571.3 13049.3 13296.5 13451.2 13563.6 13654.2 13732.1

13802.0 13866.5

'13926.9

12661.0 9425.2 9046.7 8846.7 8719.9 8627.2 8552.4

Cooling Loads and Heat Gains 18000.0 16000.0

Ξ '6

Θ

(!

q)

Ξ oE IE

o

J

14000.0 '12000.0

10000.0 8000.0

---G- lnternal

*x*

Fteat Gain (W;

Cooling Load (νγ)

6ο00.0 4000.0 2000.0 0.0

basis for tΘsting or instruοtional puφoses only to Εxcerpts tioιτ this work may be reproduced by instΙuctors fοτ distribution οn a not-for-profit reprodλc'tiοn or trαnsιαιιon of thιs νork beyond ιhαt permitted students enrolled in οourses tbr which the textbook has been adopted. Αny other thecopyrightoνnerisunΙαινful' iysr,itιon, 107 οr ]0Bοf the 1976ΙJnitedStαtesCopyrighιΑctlνithoutthipermissiοnof

175 8-50

Heat gain to the space = 0.8 x 6000 W = 4800 W, assumed 59% radiative, 41o/o convective, from Τable 8-20' The sensible loads are then determined from the radiative and convective heat gains using Equation 8-67 and the radiant time factors from Table 8-21, as shown below. Τhere are no latent cooling loads. lnternal

Ηour

Heat Gain (w)

Convective ΗG

Radiative HG

0.0 0.0 0.0 0.0 0.0 1968.0 1968.0 1968.0 1968.0 1968,0 1968.0

0.0 0.0 0.0 0.0 0.0 2832.0 2832.4

3

0.0 0.0 0.0

4

0.0

5

0.0

6

18

4800.0 4800.0 4800.0 4800.0 4800.0 4800.0 4800.0 4800.0 4800.0 4800.0 4800.0 4800.0 48ο0.0

't9

0.0

20

0.0 0.0 0.0 0.0 0.0

1

2

7 8 9

10 11

12 13 14

15 16 17

21

22 23 24

'1968.0

1968.0 1968.0 1968.0 1968.0 1968.0 1968.0 0.0 0.0 0.0 0.0 0.0 0.0

2832.0


Cooling Load (w)

79.8 52.4 34.7

79.8 52.4 34.7

23.1

23.1

'15.6

15.6

1473.9

3441.9 4028.5

2060.5 2365.5

4333.5 4508.5

2832.0

2540.5 2646.7

2832.0

2713.2

4614.7 4681.2

2832.0

2755.6

4723.6

2832.0 2832.0 2832.0

2782.9 2800.5

4750.9 4768.5

2812.0

2832.0

2819.5

4780.0 4787.5 4792.3

2832.0

2832.0 2832.0 0.0 0.0 0.0 0.0 0.0 0.0

2824.3

2827.5 1366.5 777.9

471.8 296.0 189.3 122.5

4795.5 1366.5 777.9

471.8 296.0 189.3 122.5

A plot showing the lighting heat gain and resuιting cooling loads follows.

puφoses onΙy toExcerpts frοm this work may be reproduced by instructors for distribution οn a not-for-pro1it basis 1br testing οr instruοtional this ινork beyοnd ιhαt permitιed students enτolΙed in courses tbr ινhiοh the tsxtbook has been adopted. Αny οther reproclucιion or trαnsιαtιon οf by Secιionι 107 οr ]08 ofthe Ι976 United Sιaιes Copyri?hι Αcι1|,ithouι the permission ofιhe copyrighι olνner is unlανful.

t-

176

Cooling Loads and Heat Gains

6000.0

5000.0

Ξ ,6

4000.0

--{-

Θ (υ

Ι

Φ

3000.0

|nternal Fleat Gain (\Λl)

___x* Cooling Load

(\Λ/)

oδ ιl,

o

J

57

ι-

9

11 13 15 17 19 21

23

Hour

8-5'1

The schedule described in problem 8-18 is reduced to the number of peopιe present per hour in the table below. Assuming "Seated, light office *ork'', the sensible heat gain per person is 245 Btu/hr (72νν) and the latent heat gain per person is 2bO Btu/hr (59 W). lnternal heat gains from occupants are assumed to be 70o/o radiative. The latent cooling loads are equivalent to the latent heat gains shown in the table. The sensible loads are then determined from the radiative and convective 8heat gains using Equation 8-67 and the radiant time factors from Table 21 , as shown below.

As is readily evident from the plot, the heavyweight zone significantly damps the response to the heat gains'

on a not-fοr-profit basis for testing οr ιnstructional purposes only to Exοerpts from this work may be reproduced by instructors for distribution οther reprοiucιιon or ιrαnslαιion of ιhis work beyond ιhαι permiιιed . Αny uαopt.α υ..n t-,u, textbook which the fbr StudΘnts enΙolιed in courses permissiοn οfthe copyrighι owner is unlανful' ιhi ΑcιΙyithouι by Secιiοns ]07 or ]08 ofιhe Ι976 United SιαιeS Copyri?ht

177

People

Hour

οresent

1

0

2

0

3

0

4

0

5

0

6

0

7

0

9

40 40 60 60 60 70 70

I

0

10 11

12 13

14 '15

16 17 18 19

70 10

20

0

21

0

22 23 24

0

0 0

0 0

lnternal

Heat Gain (w) 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

2880.0 2880.0 4320.0 4320.0 4320.0

5040.0 5040.0 5040.0 720.0 0.0 0.0 0.0 0.0 0.0 0.0 ο.0

Convective

HG

0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 864.0 864.0 1296.0 1296.0 '1296.0

1512.0 1512.0 1512.0 216.0 ο.0 0.0 ο.0 0.0 0.0 0.0 0.0

Radiative

HG

0.0 0.0 0.0 ο.0 0.0 0.0 0.0 0.0

Latent heat gain

Radiative Cooling Load 20.6

Cooling Load (w) 20.6

0

12.5 7.8 5.0 3.4 2.5 2.0

12.5 7.8 5.0 3.4 2.5 2.0

ο

(\Λ/)

0 0 0

0 0

1.6

1.6

0

2016.0 2016.0

1021.8 1484.5

3024.0 3024.0 3024.0 3528.0 3528.0 3528.0 504.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

2233.7 2593.8 2784.5

1885.8 2348.5 3529.7 3889.8

2360 2360 3540 3540 3540 4't30 4130 4130

3143.9

4080.5 4655.9

33'17.9

4829.9

3410.8 1931 .3

4922.8 2147.3

1010.7 552.7 309.2 175.8

1010.7 552.7 309.2 175.8

101.1

101.1

0

58.8 34.6

58.8 34.6

0

590 0 0 0 0

0

Cooling Loads and Heat Gains 6000.0 5000.0

Ξ 'Ξ

4000.0

Θ π,

Φ

Ξ

οδ π,

o

3000.0

--*-

*-s*

lnternal Heat Gain (W1 Cooling Load (W)

2ο00.0

J

1000.0

puφoses only to Exceφts from this work may be reproduοed by instructors fοr distribution on a not-for-profit basis for testing or instruοtional ιhαt permitιed students enrolled in course s for which the textbook has been adopted. Αny oιher reproducιiοn or ιrαnsιαιion of ιhιs νork beyond by Secιiοns ] 07 or ] 08 οf the Ι 97 6 (Jnitecl Sιαιes Copyrιghι Αcι Ι|iιhοuι the permissiοn οf ιhe copyright οwner is unlανfuΙ.

178

Solution to be provided by an instructor.

basis for instru:lor:.fo, distribution on a not-for-profit has been by reprοduced be may wοrk Excerpts from this for which the textboοk on1y ω stuαeλs enrο1led i, "or.r"s ]07 or 10B testing or instruοtional purposes ινork beyo'nd thαt permitted by Sections o}thιs owner is unlαwful' adopted. Αny other repiodλction "y γ""ititi"λ the permission-o/'λu copyrιgit Αct of the 1976 (Jnitecl itotus Copyright 'ιiinλ, ιhe Piimiision Depαrtment, John ιnfοrmαtionilrouιa i" oaarrrruiΙo h'uouurt, for permission or further

'';i,;;;'';:"{,-i", il l Riνir

Street' Hoboken' NJ 07030'

CHAPTER 9 average degree day is 6283' From Table 9-1, the number of From Fig. 9-1, Co = 0'60 Using Εq' 9-2,

f-

_ tu l hr)(}'60) (24hr l dαy)(6283' F dαy)\2-^2-^1000Β = 438,727 std ft3 (0 η(7ο_ 12"tr)(10ο \Βtu l stdff)

gas Or F = 438.7 mcf of natural

f-

= 102,867 kw-hr

(24hrldαy)(6283'F_dαy)(Ζ-^?5,000Βtulhr)(0'60) l kW _ hr) (1 J) ql o _Γz' r)βaL2Βtu

0'287 $Elec = 102,867(0.1 O) = $1 $Gas = 438.7(4.5) = $1,974

SΕΙec_$Gαs $Gαs

|0287_ι974 ι914

=

4.2

5'2 times as much' or the electric cost is about Source energy using eιec'

_ (1 02,s 67 k-t4l

=

=

a

lΨ

!u_

@at)(000stdff = 1063'6 mcf

Source energy using gas

hDβ

438'7 mcf

l

kY-' hr)

lmΦ

>_Ξ-

ΕSΕ,_ΕSG

1063.6

_438'7

- more1.42Source energy' =

"o,"ff=-tiU That is eleο. heat uSeS 242% 9-3

DU' The following are information for Washington' degree day is 4224' From Table g-l,1he number of average From Fig. 9-1 , CD= O'62' is 20 "F' From Table B-1a, the outdoor temperature

Forenergyefficientfurnace,assume85o/oeffiοiencyfactor, Using Εq' 9-2,

-, F ,-,= 04hr l d αν)(4224"

"

_ dαy)(|20'000Βt u

l

hr)(0'62\ 'o'=177,468stdft3

20"trX1O00Βtu l stdft3)

Or F = 177 .5 mcf of natural gas

-10.85X7O_ 9-4

Load Profiles

+5 Qro" = ato

1

-

ξ

120,OO0=a(20)+b g=2(gQ)+b

80000

160000

tηoooo

αt_ l z000ο

€6

j

12ο,ooQ =(20 _60)a

a = -3OOO, b = 180,000 3'000 to Q uo" = 180,000 -

E'

Ξ3

lοοοoο soooo 6oooo 4oooo 2OoOo 0

30

4o

Outdoor TemP, "F

9-5 8o. = Qrn - Qιnt = 18O'oOO _ to ξo" = 160,000 3,000

-

3'Oo0

to

-

20'0ο0

or instructional puφosΘS only to on a not-'br-pro'it basis for testing instruοtοrs for distribution by reprοduced be may work *is wor-k beyond ιhαt permiιted this Exοeφts from Α-ny oιher *p-a""riλλ"]r'i,o^ιoιι*7 the tοxtbook r-,* υ*. oνner is unlανful' studοnts enrolled in courses to. wι,λι-' "α"pi.λ. iλ'! copyright ιhe of irrλ'*iλ" l'oii'iλιi'rλi,'λ,r, copyrιgn, 'a}iΙ:i';;u' by Secιions ]07 or 108 o7 m,

186

9-6

lιl

lll

Group

lVVVl

Sunday Monday

Tuesday Wednesday Thursday Friday Saturday

shift

shift 2

1

1481012162024

Hour

Αssumes Sunday and Saturday in shift 2 See Table 9-2 shift 1 hrs in Group ea. qp

shift 1 Days in ea. qp.

hrs ea. qp

Total hrs in ea. οp.

Frac. of shift 1 hrs ea. gp

Frac. of hrs in ea. qp.

Total

shift

1

I

0

0

0

28

0.0

'1.0

il

0

0

0

28

0.0

1.0

Πt

2

5

1ο

28

0.36

0.64

IV

4

5

20

28

0.71

0.29

V

4

5

20

28

0.71

0.29

VI

4

5

20

28

0.71

0.29

instructional puφoses only tο EΧceφts from ιhis wοrk may be reproduοed by instruοtors for distribution οn a not-for-profit basis for testing or wοrk beyond ιhαt permitιed οf this ιrcιnsιαιion or reprodiction Αny other adoptedbeen has students enrolled in courses for whrch the textbook ιhecopyrighιowneriSunlανfuΙ. bySecιiοns ]07 οr ]08of the Ι976(JnitedStαιeSCopyrightΑcινιthoutthepermissionof

lo/

9-6 (Cont.) Bin.

Temp

lII

Shift t hrs in each Group

III

shift

IVVVI

96 77 67 47 48 38 28 17

68 82 94 62 65 54 36 17

12

16

69 58 86 87 99 75 66 28 22

6

1

4

11

1

0

0

1

62 57 52 47 42 37 32 27 22

0

0

0

0

0

0

0

0

ο

0

0

0

0

0

34 38 49 36 35 32 27

0

0

10

0

0

I

17

0

0

12

0

0

I

hrs 267

255 296 232 247 199 157 72 58 22 2

TotaΙ:

shift 2 hrs

375 346 388 337 420 422 347 157 113 73 16

1807

2994

9-7

The procedure is the same as Problem 9-6. Use appropriate bin data from Αpp. B in last step as per Table 9-3. 9-8

Refer to Εxample 9-2, insert shift hours of Problem 9-6 in column 2 and 3 of Table 9-5 and recalculate. 9-9

Reconstruct Τable-9-3 for the appropriate city to obtain Shift A and Shift B hours. lnsert the hours in columns 2 and 3 of Table 9-5 and recalculate.

Exοeφts from this work may be reproduced by instructors for distribution on a not_for-profit basis fοr testing or instruοtional purposes only to students enτo11ed in courses for whiοh the textbook has been adopted. Αny other reproduction or ιrαnslαtion of this work beyond ιhαι permιtted by Secιiοns ] 07 or 1 08 of ιhe 1 976 Unιιed SιαιeS Cοpyrighι Αct ινithοut ιhe permission of the copyrighι οlνner is unlανful.

188

9-10

Reconstruct Table 9-3 using the shift hour fractions from Problem 9-6 and bin hours and temperatures for the appropriate city. lnsert the shift hours in column 2 and 3 of Table 9-5 and bin temperatures in column 1 and recalculate.

9-1

1

Solution furnished by an instructor.

'ξ/'''

Exοeφts from this work may be reproduced by instructors for distribution on a not-for_profit basis for testing or instructional puφoses only to students enrolled in courses for which the textbook has been adopted. Αny οther reproduction οr trαnsιαtion of this 'ιιοrk beyond ιhαt permitted by Secιions ] 07 or 108 ofιhe Ι976 [Jniιed Sιαιes Copyrighι Αcι ινithout the permission ofthe copyrighι oνner is unlανful.

Exοeφts from this work may be reproduced by instructors for distributiοn on a not-for-prοfit basis fοr testing or instruοtiοnal puφoses only to students enrolled in courses fοr which the textbook has been adοpted. Αny other reproduction or trαnslαtion of this wοrk beyond thαt permitted by Sections 107 or Ι 0B of the 1976 ΓJnited Stαtes Copyright Αct without the permission οf the copyright ov)ner is unlαwful. Requests for permission or further informαtion should be αddressed to the Permission Depαrtment, John Wiley & Sons, Ιnc, ] ] ] Riνer Street, Hoboken, NJ 07030'

1O-1. (a)

',*

CHAPTER

1O

o+ + ρΖι_Pz+'Ψ

+

ρZz+

ρWp +

=

-80 (ft

-

neglect

(!

iιz;

Wp =

-Η, g/g/"

ρ

tt, Vl

=

VziZι=Ζz

lbf)/lbm

Pz = Pl + ρWp = 20 + 62.4(80)1144 = 54.7 psig Ps = Pz - PΖs_ U:)rg= 54.7 _ (62'4 X 50)/1 44 _ (20 x 62'41144)

Pg= 54.7 _ 3ο.3

=

Pι=Pz- PZι-((izg_

24'4 psig (1't)g+

=54.7

ffiX25)

28.7 psig

(#)(20+

(b) Neglecting the pump, the pressure or head required for this pipe is:

ΔP

= 28'7 - 20

psi or ΔP

=

8.7 psi

Δ Η 20.1 ft.

87I I

Γ

lo t-

20

lο lσ lο) |Ξ Ξ

Ηl

This is the characteristic for only part

of the total system.

oa

0L_

10-2.

Note:

1

= Hz

* Ηο +

150 (.

trZz=

(8 x 2.31)

-

250 + 30 + 300

=

'15) =

190

Ηl

Pι

'10-3.

= 98.5 ft of water =

42'6 psig ε 294 kPa

Jt

Ηl_Ηz=Hp+ !,t+Ζz=0 H^ =

-

('-2,

=

970 kPa

=

-25- 300

= -325ft: H^ = 325ft of head

100

10-4.

(Ι

+ΙΙ

)+ 12 + 1b

Ξ(σ

ο Ξ5o 75

10-5. (a)Ηl=alΦι+Zι Hz=ΔzQz+Ζz

H2 ιL

U

z2

ο.)

Hq

I

(σ

-c.

Series Connection

Ql=Qz;SumΗ

z

a

a

Εxcerpts from this work may be reproduced by instructors 1br distribution on a not-for-proΓit basis for testing or instruοtiοnal puφoses only to students enτolled in οourses for ιvhiοh the textbook has been adoρted' Αny οιher reproducιion or ιrαnsΙαιion of ιhis νοrk beyond ιhαι permitted by Sections ]07 or Ι08 ofιhe ]976 United Sιαιes CopyrighιΑctwithouι the permissiοn οfιhe copyrighι oνner is unΙανfuΙ.

H=Ηlr

191

H=Hl *Hz Η=Qz(aι+Az)+Gι+zz) Parallel Connection

Hl = Ηz, Ζι=ZzorZ=0 Sum Q

Ql= Q

Ê,Φz= Yar

=Q

l

+Q

Ê Yaz

r= JH_Ζ(Jil q * ^[-νη) .

a

6ra*M)' 10-6.

t

+Ζ

t=f!Y1 ' D2g ; V = 6. 3ft/sec, L = 3OO ft, D =

4.026 12

Referring to Figures 10-2a and 1 0-2b 62'4(1 .04)6'30(4'026 l12) Re = ρνD _ = 75,696 (2'7 11490) μ ε = 0.00015 ft; Table 10-1

ε/d=o.ooo1

12 5" 4.026

ε/D = 0.00045; f = (,=

0'022 Fig.

/3OO) 0.022 \'

10-1

(6.30)2

(4.026 t 12) 2(32.17)

= 12.6 ft of water

:

= 12.1 ft of 30o/o E.G.sol.

38 kPa

-'.--ΞExcerpts frοm this work may be reproduοed by instructors for djstribution on a nοt-for-profit basis for testing or instruοtional puφoses only to students enrolled in courses Γor which the textbook has been adopΙed' Αny οther reproduction or ιrαnsιαtion of ιhis νork beyond ιhαι permitted by Secιions ] 07 or 1 08 of the ] 976 United Sιαtes Cοpyright Αct νithouι ιhe permission of ιhe copyright οwner is unlανful.

192

10-7.

(a)

so

Ξ(σ

925

J-

100

125

150

gpm

(b)

Qn

(c)

28ft

(d)

10-8.

=

48 gpm; Qs

60 gpm Qs = 32 gpm Qc = 41 gpm

Qn

=

24 gpm; Qc

= 32 gpm;

=

ο_ 125 gpm

}

Refer to Problem 10-7; Ζι

(a) Hn= uoQi, θn= Hs = auQ3, Θs = Hc = a.Q3, θc =

Ζz _ 0

9= 50' ai +

9= a6

30'

= +=+ aa 45'

=O.9O8O

= o'o278 =o.O'148

a2

H(^n/o!o8

(b) Q

=

+

JΤt

02

ozτa + J.roιua)2

= 100 gpm; H = O.OO155 x (1OO),

-

645.06

= 0.001

55Q2

1S.5 ft

Excerpts from this work may be reproduced by instruοtors for distribution on a nοt-for-prοfit basis for testing οr instructional purposes only to students enro]Ιed in courses for which the textbook has been adopted' Αny other reproduction or ιrαnslαtion of this work beyoncl ιhαι permiιιedbySectiοns Ι07 or l08 of ιhe ]976LlniιedSιalesCοpyrightΑctιι]iιhouιιhepermissionοf ιhe copyrighιονnerisunlανfuΙ.

193

Hn = Hs = Ηc = 'l5.5 ft

Qn=

JΠ/rΑ

Qg=

.'m

Qc =

Jss/oJl4g

(c) From

(a)

=.,/l55/O.OO8 =44gpm

lo'o2?8

above. H

Qn=

= 32.4 gom = 0.001

55(12q2 = 24.2 tt

=55qpm =

29.5 gpm

Qc = J%2toβ148

=

4O.4 qpm

e Q

23'6 οpm

Jπ2loβ278

Qg =

1o-9.

=

= 125 gpm

r|2g"(P, '--L ρ

= CαA

_Pr)l''''

D2

v'vv = O.55 'D1 =84.8 154.1

-]

aSSume Cα = 0.638 using Fig. 10-9;

Ar= -4 ζ (O.os4s)2 = O.00565

Pι_Pz Q

V C6

ρ

m2

= O.O98(13.55)9.s = 13.O13 J/kg

= 0.63S(0.0056 5)t2(13,013)11/2 =

z=

^:

3.26 m/s; Re

-

0.638 From Fig.

0.0184 m3/s x 292 gpm

999(3'26)(0:085) 1.4x10-3

= 1.98 x'105

10-9.

Τherefore the original assumption is satisfactory. Excerpts from this work may be reproduced by instruοtors for distribution on a not-for_profit basis for testing or instructional puφoses only to students enrolled in courses tbr which the textbook has been adopΙed' Αny οther reprοduction or ιrαnslaιιon of ιhis νori beyond-thαι permitted by Secιions ] 07 or Ι 0B of ιhe ] 976 United Sιαιes Cοpyrighι Αct wiιhοuι the permissiοn of ιhe copyrighι ονner is unΙαwful.

ι

194

1o-1o. (a)V.. = Γ2g" (Po,

_ρ,ll"' _|z x s?'ιτ

x o.os x 144 ι0 +91x 144)

L-" [ ρ )] L

Table A-1a;

ρ

1t2

=3.97 ft/sec

= 59'83 lbm/ft3

(b)rir = ρVA(O. 821= 59.83 x 3.97

r+][qΨ''l2 12 ) ι4rι

*

o.εz

rh = 39 lbm/sec or 140,674 lbm/hr

10-11

(a) 40

#

Ξ(σ

35

(b)

Read from Fig. 10-11a at 35 ft and '125 gpm, W. = '1.6 HP

(c)

Q

(d)

Τhis is actually out of the operating range of the pump and the

Φ T2s

= 180 gpm, H, = 20 ft; 1.8 ΗP

efficiency is very low. ln situations like this there is a danger of overloading the pump motor; however, that does not appear to be a problem in this case since the motor is probably a 2HP model. Εxcerpts from this work may be reprodιιced by instruοtors for distribution on a not_fοr-profit basts for testing or instruοtional purpοses only to students enrοlled in οourses for whiοh the textbοok has been adοpted. Αny other reprodιιctiοn οr ιrαnslαιion οf ιhis work beyond thαt permiιιed by Sections ]07 or Ι08 οfιhe 1976 Unιιed Stαιes CopyrightΑct1|iιhout ιhe permissiοn οfthe copyrighι oινner is unlcrlυful.

195

10-12. NPSHR

=

NPSHA

=

ft (Figure 10_1 1b) [e"e") -2,-r,- f&9"] \ ρg / ι ρg) NPSHΑ

=

20

to ( 20='uΧ 13.55x62.4 -Ζ"_2_ t_l'soτxι 44 12

\

62.2

Ζ"=32'85_20_2_1'17

=

'10-13. (a) 231 gpm, ξp = 73'4% W, = 12* tlP

(b) 225 gpm ηo = 73 '3o/o W, = 12- ΗP

10-14.

(a)

η,

= 73.3oλ;

W,

= 11. 5

az.z

9.68 ft; (2.)rr, = 9.68 ft

200

140 100

HP 185

(b)

225 gpm; '185 ft of head

149

ηr=73.5%;Ws=14ΗP

(c)

ηo= 73o/o,

W.

= 14 ΗP

225

231

gpm

10-15 From Problem 10-14b, the original system defined by 225 gpm and 149

ft of head and would operate at 242 gpm, 173 ft of head and require 14.4 HP with the 7in impellor. Τhen,

EΧcerpts 1iom this work may be reprοduced by Ιnstruοtors for distribution on a not-for-profit basis for testing or instruοtiοnal puφoses οnly to students enrolled in οοurses for which the textbook has been adopled' Αny οιher reproducιion or ιrαnsιαιion οf ιhis wοrk beyοnd thαι permiιted by Sections ] 07 or Ι 08 of ιhe Ι 97 6 United Sιαtes Copyrighι Αcι ινithouι ιhe permission of ιhe cοpyright οlυner is unlαwful. _.ι _Ι

196

ΓPln= 35οo[+.l =3o37 \242 )

" Π3

H^ =

w^

"

1

(go37

)'

= l3O ft

ι3500/

4.4(s037)' ι3500,

=

e.4 HP

The Εfficiency wouΙd not Chan ge,74.2oλ

1o-16

=

Dn

"r(#)=

Hn = *,

η

1.

(Ψ)'

.[Ψ)u

7(o e7)

=

173

=

6 8 in

(o'g44)= 163 ft

= 14'4(o'g17)

= 13 2

ΗP

The Εfficiency would not Chan ge,74.2oλ

10-17. Uses Fig.

1

0-20 or program PIPE

(a) 25 gρm; 1 % in., V < 4 ftlsec,2in dia. or less (b) 40 gpm; 2in. V < 4 fVsec,2in dia. or less (c) 15 gpm; 1Υιin., v < 4 fUsec, 2in dia. or less (d) 60 gρm;2% in', '(,'1< 4 pVsec; dia > 2 in. (e) 2OO gpm; g %in., .(,'1< 4 ftllOO pUsec; dia > 2in.

(η

2ooo gpm; 8 in., /1 slightly

>

4 ftl1oo

ft

Exοerpts 1iom this work may be reprοduced by instructors fbr distribution on a not-fοr-profit basis for testing or instructional puφosos only to students enrolΙed in courses tbr which the textbook has been adopted' Αny οιher reproduction or trαnsιαιion of this νork beyond ιhαι permiιιed by Secιions ] 07 or Ι 08 of ιhe ] 97 6 Unιted Sιαιes Copyrighι Αcι ν ithοuι the permission of ιhe cοpγ,ι'ighι ονner is unlαwful.

197

1ο-18. (a) K = 30 ft, ft = 0.019; K = 0.57 (Table

V

(b) K

10-2; Figure 10-22a) = 3.82 fVsec; ! r = 0'57(3'822lβ2.2x2)= O.13 ft

= 340 ft, ft= 0.017;

V

= 5.0

f/sec; !

r

K

= 5.78

= 5.78 x 5.02t132.2

x2)

(c) K= 60ft, ft= 0.018; K = 1.08 V = 6.5 ft/sec; ! r= 1.08 x 6.52t(2x32.2) '10-1e.

!r=

2.31

(#)'

=

2.24 ft

= 0.71 ft

= 10.8 ft of water or 4.7 psi.

10-20. Assume com. stl. pipe Q

= O.O3

mt/s = 108 m3/hr, size pipe for about 4 mllOO m

From Fig. 10-20, use 5 inch pipe, lD = 130 mm a, f -_ ..25 ml100 m; [1= (3'251100)200

nt λ'

ΔPg = 35 kPa

Γ o-99-1' ' | "

For strainer. ΔP"

1 0.00722 J

=

ι7

=

'27

6.5 m of water or 63.7 kPa

kPa

Then for the pump: ΔPp = 63.7 + 35 +

'17

.3 + 3(1000)(9 .807)11000 = 145'4

kPa

Ηp = 145.419'807 = 14.8 m

Q

= O.O3 mu/s

-

30 L/s

10-21. Size the pipe using Fig. 10-20 or program PIPE. Fitting equivalent lengths found using Fig. 10-22a; 10-22b and Τable 10-2' Program PιPE could be used to solve the complete problem including fitting losses. Data for hard calculations are summarized below: Εxcerpts from this work may be reproduced by instruοtors for distribution on a not-for-pro1it basis for testing or instruοtional puφoses οnly to students enrolΙed in οourses Γor whiοh the textbook has been adopted' Αny other reproducιion or ιrαnsΙαtion of ιhis νork beyond ιhαt permitted by Sections ]07 or Ι08 ofιhe Ι976 Uniιed SιζlιeS Cοpyright Αcιlνiιhοuι ιhe permission ofιhe cοpyrighι oνner is unΙανful.

L

198

lO(3)

1ot3] Θ

Θ

ro(3)

I

/o\

\:-/

o(3)

]

Sec. No. 1

5

6 7

4 10 2 3

I 9

ch

qpm

stze

tn.

120 3

70 2.5 40 2 40 2

90 2.5 120 3

5ο 2 50 2 30 1.5 30 1.5

120

5ι5)

it

Le

ιt

ft./'100 ft

ft.

ft.

3.38

45

1.5

1.5

3.64

'15

0.6

0.6

3.'1

24

0.7

3.'1

13

0.4

5.84

27

1.6

1.6

3.38

42

1.4

1.4

4.7

22

1

4.7

26

1.2

6.3

28

1.8

6.3

13

0.8

Coil

Con.

Valve

ft.

ft.

11.4

Τotal ft.

12.1

12.0

10.0

12.4

11.0

10.0 14.4

11.2 16.2

15.0

15.8

20

Exοerpts lrom this work may be reproduced by instruοtors for distribution on a not-1br-profit basis for testing or instructional purposes only to students enrol]ed in courses Γor whiοh the tΘXtbook has been adopted' Αny οιher reproducιiοn or trαislαtiοn of ιhis νori beyoncl ιhαι permitted by Secιions ] 07 or Ι 08 of ιhe ] 976 (Ιnιted SιαιeS Copyrighι Αct wiιhοut the peimission οf ιhe cοpyrighι owλer is unΙcnνful.

199

The head losses for the three parallel runs are approximately the

same.

For run (1-5-6-Z-4-10), Hp = 49.6 ft For run (1-5-S-g-10), Hp = 55.5 ft For run (1-2-3-4-10), Hp = 46J ft Therefore, a pump should be selected to provide about 56 ft of head at 120 gpm.

10-22. 500 gpm, Use 5 inch pipe; !'f

V

= 8.0

=

4.17 fil1OO

ft

ftlsec

Length of pipe = 160 + 3O + 12 = 202 6-5 in elbows = ,lS

ft

ft

(Figure 10_22)

3-5 in gate valve = 12 ft 1-5 in gΙobe valve = 130 ft; Total equivalent length = 419 ft

. 4.17(41e\ /, = -Ι_1}-J "/ = For strainer: !

|

!

Τhen Hp =

17

For cond

"o

17

'5 ft of water

2.31 = e.24ft of water "= = 20 ft of water

[#)'

'5 + 9.24 + 20 + (3o

_

12) = 64.7 ft at 5OO gpm

10-23'Use Εq. 10-33

Exοerpts frοm this work may be reproduοed by instructors for distribution on a not-Γor-profit basis fbr testing or instructional puφosοs only to students enrolled in οourses fοr which the textbook has been adopted Αny οιher Ιeprοdactιon or ιrq'nsιαιion of ιhis νork beyοnd thαι permiιιed by Sections ] 07 or Ι 08 of ιhe 1 976 United Sιαιes Cοpyright Αct νiιhλut ιhe peimission of the cοpyτ.ighι owner is unlανfuΙ.

L

200 ,,_

".

_ -

6oott*:;*g I

-,l) - 3x6 sxlo-u (1,0

gz.οgο

10-24' Use Eq. 10-34 v, Vr-_8.Ζ-gal

=

-oull

= 19.4 9a1. = 74 L

og.οgο ]

Ψξg L[ 0'ο1ο0zz

_,'l _ 3x6.5x1

οoo[[9

=

)

o_u (1,, o _

ou1l

'-___-_-J

ι '_ 69:ο%,

33 L

10-25. Use Eq.

10-33 tl = 60oF, P2= 50 psig, P1 = 20 psig, v1 = O.O16ο53 ft3/lbm

vz = O.O1 6772 ft3/lbm, tz = 220"F

Vτ=

= _11_5_gal.

-

435 L

10-26

(a) Use Eq.

'10-16

P,+ PτPzι= Pzgzz +ρνv+ρg

'g"9cι

''

=

ff

lL

Τ 24o

9c

(zz-zι)+ ρw

+

P9nLE

9cl

-

Exοerpts from this work may be reproduced by instruοtors for distribution on a nοt_for-profit basis for testing οr instructional purposes only to students enrolled in οourses for which the textbook has been adopted,' Αny other reproductιon or ffαλιαιion of ιhis wori beyond-ιhαι permiιιed by Secιiοns ]07 οr ]08 οfιhe Ι976 L]niιed Sιαιes CopyrightΑctνΙιhouι ιhe peimission οfthe copyrighι owλer is untιιυful'

Ι

tΙ,

=

#

e4o)

#(60)

+

-

#

es1=Be psis or

61

2 kpa201

(b)

P,'+ρg!= P2+ρg2 9c - --gc Pz=Pι-

t

(zι-zz)=89.o

Pz = -15 psig

=

-effi

-i03 kpa or about o absorute

(c) No, makeup water is not available to overcome a pressure of gg psig. However, the domestic water system probabry has a

booster pump.

10-27

(a)

Pι

=

=

Pz *

(zz-zι) + ρνν * ρ9ι, o* gc 9c

62.4(240) 5+ 144

62.4(60)

144

.

62.4(25) 144

'n0.8

Pr = 93.8 psig or 647 kpa

(b) η =Pz+ Pl

!9 9c

(zz-zι)

= 109 psig or 752

=

5 *62|(,2ra0) = 5 + 104 =.t09 psig

kPa

Excerpts from this work may be repτoduced by instructors for distributiοn οn a not-for-profit basis for testing or instructiοnal puφoses only to students enrolled in courses for which the iextbook t-,u. υ".n permiιιed by Sections ] 07 or Ι 08 of the ] 976 1dγνωj Ατ-ιl λrir, Ιrpr"a""rιon or ιrαnslαιion οf ιhis work beyond ιhαι United Stαιes copyrighi Αcι withλuι ιhe p,e1miιsιon of ιhe copyright ονner is unlαινful.

202 (c) This location is at Ιeast workable. Ηowever the pressure at the pump is still very high. The domestic service water

pressure would have to be boosted to a higher pressure at the 2oth

floor.

10-28

(oo , sz) Qo

*Q.

=

+

(a,

x +o)=

(ao x

ιτ)

Qα= 100

Solve Simultaneous 57Qb + (O x 1OO) - 4OQb = 100 x47

ab = (b) (c)

Ψ 17

Q.

= 41.2 say 41gpm

= Q.. = 1OO -

41 = 59 gpm

Size all pipe for 100 gpm D = 3 in. from Fig.

'1

0-20 or PIPE

10-29. (a) Each chiller requires 600 gpm. Since chiller 2 is partially loaded must have the full flow of 600 gpm. Therefore, Q"p = 1200 - 750 = 450 gpm

(b) (150 - 60)

(c) LR

+ (450

x 42)

=

600 ts, ts

=

46.5 F

= 150/600 = 0.25

(d) Main pipe to and from sec. Circuits: D = 8 in. com.

stl.

Excerpts from this work ιnay be reproduοed by instructors for distribution on a not-for_profit basis fbr testing or instructional puφoses only to students enτοlled in courses for whiοh the textbook has been adopted' Αny οther reproductiοn or trαnslαιion of this work beyond ιhαι permitιed by Sectiοns ]07 or Ι08 οfιhe ]976 United Snιes CοpyrighιΑcιιυiιhοuι the permissiοn ofιhe copyright oνner is unlαννfuΙ'

it

203

Dns = Dco = O in. com. sfl. (S in. a litile small) D"p =

6 in. com. sfl.

Dsc = DRo = 6 Ιn. com. stl. ( could be 5 in. but easier to make all 6

in.)

(e) Rpm,

(η ιW or

10-30. (a)

= ΓPIΠl

=

050l12oo)

*#=1-ff

= 35OO(75o/12o0) =

2188

=,l_[ffiJ' =, (##)

= O 756

75o/o

Q"n = 12OO

-750

=

450 gpm

(450x42)+ (750x60) = 53.3 F 1200 Both chillers receive the same temp. water

(b) Qrtr

+ Qztz = Qsts; ,. =

(c) Load

ratios are the same:

LR= 10-31.

##=0628or63%

Εxcerpts from this work may be reproduοed by instructors fοr distribution on a not-for_profit basis fοr testing or instruοtionaΙ puφoses only to students enrolled in courses for whiοh the textbook has been adopted. Αny οιher )eproclucιion or ιrαnslαιiοn of this ιυοrk beyoncl thαι permitted by Sectiοn's Ι 07 or 108 of ιhe ] 976 ΙJnitecl Stαtes Cοpyrιghι Αcι wιthλuι ιhe peλιssion of the copyright ονner is untωνful.

204 10

(13)

2

2Ο(6) 3

*{s) T'r^;n ι }iJJυα' ^

!

**ntΓ*if Φr 1*{S}

u?

ι$

Ω*{s} s 2Ο(s) 8

BaΙ*n*s ve,v* {typi*aii

**rnnr** pip*

Note: Piping is type L copper Αll |engths are total equivalent lengths

10-31.

FΙow rate Coil opm (L/s) A 40(2.5) B 40(2.5) c 50(3.2) (continued)

Section No.

gpm 130 90 50

1-2

2-3 3-4 ^,îl

Dia.

C

in.

3

2% 2

' he reproduced

Lost head ft (m) Coil Con. valve 12(3.7) 10(3) 15(4.6) 12(3.7) 18(5.5) 15(4.6)

ir ftl1ο0' 3.7 4.8 5.0

L"

!.r ft

ft

60 20 30

2.2

r;

)

355

by instruοtors for distribution on a not_Γor-profit basis fortiting or instructionai puφoses onΙy to -ιhαι hiοh the textbook has been adopted' Αny oιher λeprοduction οr ιrαislαιiοn of ιhis wori beyond ''/ United Sιαιes Cοpyrighι Αcι ιιithouι ιhe permission of ιhe copyv.ight oνλer is unlωυful.

205

Con. C 4-5 Com. oi 2-6 Coil A Con. A

130

3-7

Coil B Con. B 7-8

3.7

0

3 3

40

2

3.4

15 1.5 0.0 39.2 1.0 12

40 30

10

40

2

3.4

10

80

2%

3.9

20

23

0.5 15 12

ft

ft

0.8 28.3

ft

(tote

(total)

(totat)

Circuit 1-2-3-4-5-1 is the path of greatest lost head. From Fig. 40 ftof head and 130 gpm the 7 in., 1750 rpm model which

choose at froduces about 43

ft of head.

1O-1 1

Ξp^q 2ΟΟ 2ΟΟ lbu] 2(rΟ)3(rΟ} t p,ih-*:y +ο* PumP {12)l 4Q t1i

r

f

ι*

{}

Ιs*ianc* vxlγ* {typi*ai}

*hill*r

p ιl Π]$} $

*οntr*i . valv*

ιJ

{

typi*xi}

10-32. Notes: PΙpe is schedule 40, commercial steel. Αll lengths are total equivalent lengths exc|uding control valves.

Circuit

Α B

c

Flow rate gpm (L/s) 60(3.8) 70(4.4\ 70(4.4\

Control valve head loss ft (m) 40(12) 5ο(15) 50(15)

'' be reproduοed by instruοtors for distribution on a not-for_profit basis for testing or instructional puφoses on1y to "hiοh the textboοk has been adopted. Αny oιh-er )eproducιiοn or ιrαλhιiοn o7 tnπ beyond thαι '976 United Stαtes Cοpyrighι Αcι ννithouι the peλission of ιhe copyright ονner 'orκ is unlαwful.

206

Section

if

Dia.

gpm

No.

in.

200 140 70

4 3 2%

4-1

200

4

2-5 Con. A 5-4

60

2%

1-2 2-3

3-4 Con.

L"

(.t

ft

ft

ft/'l00' 2.4 4.2 3.5

200 200 240

2.4 2.5

400 240

4.8 8.4 8.4 50

9.6

81 .2

ft

(total)

54.4

tt

(total)

51.4

ft

(total)

o

40 140 70

3-6

3 2%

200 40

4.2 3.5

Con. B

8.4 1.4

50

Circuit 1-2-3-4-1 has the largest head loss of alΙ paths. Select pump for 2OO gpm at 81 ft of head. From Fig. 10-1 1, use: 5Υ' in., 3500 rpm model. Will operate at 96 ft at 200 gpm. 10-33

(a)

qst = 20

/1

_

x 12,000 x2= 480,000 Btu

gst

ρc, (t,.-t.)

480000 62.4(1) (60-45)

=

512.8 ft3

orQ=3,8369a1

(b) Vol

= 513 ft3 10.2 ft

ora Space

8

ftx 8 ftx 8 ft ora cyΙindrical tank

8 ftdia. x

10-33. (continued) Solution - Sl: (a)

q

=

Qg1

=

(352_280) (2)= 144k\^l-hr= mc, (trt.) _ Qρcp(t-t.)

Qst ρο, (t1ts )

' 1le reproduοed

144 x3600 = 14 m3 980(4. 184) (1 6-7)

by instruοtors for distribution on a not-for-profit basis for testing or instruοtional purposes only to '.'hiοh the textboοk has been adopted. Αny oιher reproduction οr trαnslαtion of ιhis wοrk beyond thαt ^76 (Ιnιιed Sιαιes Copyrighι Αcινιιhout ιhe permission οfthe cοpyrighι oιυner is unlανful'

207

(b) Vol.

=

2Αmx2.4mx2.4m

10-34. Solutions may/can Vary. Α typical solution

(a)

(b)

Use 2 chillers of '15 tons total capacity in a reverse return system similar to figure 10-32. The piping would be routed overhead around the complex with supply and return running parallel, starting and returning to the equipment room. Total flow rate is

Qτ =16 x2'25

(c)

is:

=

Estimated length 3600 ft

36 gpm Using PIPE or Fig 10-21; Dia. = 2in'

=225x4x2= 1s00ft.

Τotal Eq. Length=2x 18OO

=

Assuming an average Ιoss of about 2.5 ft1100 ft; The pump head required would be: Hp = 2.5 x 3600/100 = g0 ft with flow rate of 36 gpm

'10-35 Solutions may vary (a)Figure 10-34 is a schematic of what the system wouΙd be. However, there would be 3 chillers and the secondary piping would be routed in a square fashion around the outside of the parking garage in reverse return. (b)Τhe primary system would appear as in Figure 10-34 with the

common pipe as shown because of the expected variable and light load at night.

(c)The tertiary circuits would be as shown in Figure 10-34 and piped in a reverse return manner. (d)For each building: Excerpts Γrom this work may be reproduοed by instructors Γor distribution on a not-fοr-profit basis for testing or instruοtiοnal puφoses only to students enrolΙed in οourses for whiοh the textboοk has been adopted' Αny other reprοductiοn οr ιrαislαιion of ιhis νork' beyοnd thαt peιmitted by Sections Ι 07 or Ι 08 of the ] 976 tJnited Sιαιes Cοpyrighι Αcι 1ψiιhouι the permission of ιhe cοpyrighι owner is unΙα:wful.

208

Qi

1500 x 't2000

600 gpm

4x500(60-45) = QΤ=4x600 =2400 gpm =

(e) Dia. = 10

in., Figure 10-20 or plpE

10-36. ;ControΙ valve (Typical)

I

Y φ a_Air Vent (Typical) ,+9*Π4 Heating Device (Typical) r-€

T<-TypicatTrap

IE

ι o r--5

erξι--] !- - -<-'-' -ξ2 < ;-

10-37

'

Possible

Vacuum Breaker on each Heatinq Device

Αssume boiler pressure of 2'O psig with ΔP/L = 2'O oz or 0'125 psi/1OO ft' (TabΙe 10-4a). Τhen, ΔP = o'125x 175l1OO = 3'5 ozor 0'22 psi ΔP is about Υzthe alΙowabΙe from Table 10-4a'

Assume boiler pressure of 1.o psig with ΔP/L Ξ 0'125 psi/1Oo ft. (TabΙe 10-4a)' Then ΔP = 0'125 x 175l1OO = 0'22 psi which is near the maximum in Τable 10-4a' Either boiler pressure could be used, but select2.0 psig to be conservative.

Εxc-erpts fiom this work may be reproduced by instructors fοr distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in cοurses for whiοh the textboοk has been adοpted. Αny oιher reproducιiοn or ιrαnsιαιion of ιhis wοrk beyoncl thαt permiιted by Sectiοns Ι 07 or 108 of ιhe ] 97 6 (]nitect Stαιes Cοpyrigh} Αcι νith'οuι ιhe peimission of ιhe cοpyright ονλer is untωνfuΙ.

209

From Figure 10-48a at 850 lb/hr; ΔP/L = 0.'125 ρsil100 ft, and boiler pressure of 2.0 psig: Pipe diameter = 4 in., with steam velocity of 4,00ο ftlmin at zero psig. Correct velocity to 2'0 psig (Fig. 10-49a)

V

= 3,800 ftlmin

10-38. For each

unit at full load: = 283 lb/hr

ft.

Pipe size depends on slope of line, Τable 10-5a. For slope of 1/8 to Υι in./ft, D = 1 in. nominal specify slope of % in./ft (conservative).

10-39. Refer

to Table '10-5a.

The available head is = 2 x'100/110 = 1'82ftl1ο0 ft. Then at 850 lb/hr of condensate flow, D = 1 in. nominal is adequate.

10-40

(a)

q = rhcp(t,-t,) =

ga##(1)(6s

- 42)

^F^ gpm - 1200x 1200x7'48 -= λ1'250 = 60 X 624(654η

ι'ι ^

-..-

(b) Αssuming no changes in the temperatures, the total flow rate would be:

-P Ξ9t 1l25O) 1200 \

Φ., =

'/

=

937.5 or 938 gpm

Τhe chillers could share the flow:

-=469gpm (J, ^-938

'2=

and be above their minimum flow of

jι

-

Qmin

469 625

=

70o/o.

0.75 or or Ts%

EΧcerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instruοtional puφoses only to students enrolΙed in οourses for whiοh the textbook has been adopted. Αny οther reproducιiοn or ιrαnslαιion οf ιhis νork beyond thαι permitted by Secιions ]07 οr Ι08 οfthe E976 LΙniιed Stαtes Copyright Αct\νithout ιhe pe'ιmission οfιhe copyrighι οwner is unΙαlvful.

210

Τhis is probably the best way to operate. Τhere would be no bypass

flow and the parallel pumps couΙd operate at:

RPM,

=

*x 625

35OO =

2,626

(c) At 6ο% full load, again assuming no change in temperature, the totaΙ flow rate would be:

Qp = 0.6(1 250)

=

750 gpm

This is too much flow for one chilΙer and not enough for two chillers at minimum flow of 875 gpm or 438 gpm each. Τherefore, both chilΙers will have to operate at least at 438 gpm each and some ftow

bypassed equal to:

Φυ,

=

875 - 75O = 125 gpm

The pumps could both be slowed to:

RPMP = '

,9'? 1250

(d) Αt 25% of

fuΙΙ

ι35oο) = 2'450

capacity, again assuming the temperatures do not

change, the flow rate for the load is:

d,

= 0.25(1250) = 313 gpm

which is less than the minimum flow rate for even one chiller. Τherefore, shut down one chiller and operate the other at least its minimum flow of 438 gpm. The bypass flow would be: Qop

=

438 - 3]3 = 125 gpm

One pump would be shut down and the speed reduced for the other pump to:

RPM,'

=

=2,453 *(3sOO) 625

Excerpts 1iom thrs work may be reproduced by instructors for distributiοn on a not-for-profit basis for testing or instructional puφoses only to students enrolled in οοurses for whiοh the textbook has been adopted. Αny other reproclucιion οr ιrαislαιion of this work beyοnd'thαt permitted by Sections ] 07 or Ι 08 of the ]976 Uniιed Sιαtes Cοpyrighι Αcι withοuι ιhe permission of ιhe copyright owner is unlανfut.

Excerpts from this work may be reproduοed by instructors for distribution on a not-fοr-prοfit basis for testing or instructional pulposes only to students enrolled in courses for which the textbοok has been adopted. Αny other reproduction or trαηslαtion of this work beyond thαt permitted by Sections ]07 or Ι0B of the Ι976 United Stαtes Copyright Αct without the permission of the copyright iwner is unΙαιυful. Requests for permission οr further informαtion should be αddressed to thi permission Depαrtm'ent, John Wiley & Sons, Ιnc, ] Ι ] Riνer Street, Hobοken' NJ 07030.

GHAPTΕR 11-1

.

(a) Using Eq.

,

1

'1-1b

1=L959ll-,

=

11

ζJΑ"

O.353 '-Ao'ν= Ψ= s5ο

-=ug-ô)

ft2;

X5o

=

' 50.,/0.353

(b) Q,

=

CQov"/v,, C = 2; Εq. 11-2a

1:

=

68.5 ft;

x1ρρ

Αssumed K

= 34.2ft; xlso = 22.8ft

(Q,)so = 2(3OO)850/SO = 10,2OO ft3/min (Q,)loo = 6Oο(85O)/1OO = 5,1O0 ft3/min (Q,)rso = 6OO(850)/150 = 3,400 ft3/min

11_2'

Using Εq. 11-3 t, _ t, = 0.8(to _ tι.)

(V,/Vo), Δt,

= 0.8(100-75)

(Δt,)so = 0.8(28)50l1100 = 1'02 F (Δt*)loo =

2'04 F

(Δt )lso = 3.06 F

11-3.

50 ftlmin throw = 24

-

6 = 18 ft

=6

V,ll'tοO

212

From Eq. '1 1-'1 and Qo = Vo x Αo; Assume K

Qo_

Xζ

Q"

λ;=1l3Kro'ffi

*v* 18 x 5o =1.13K.13x6 '1

Αny combination shown would

be

acceptable. The size would depend

on the available total volume

flow

rate of air and the size of the space.

11-4.

Q,

=

Q oC V

Vo=

Q o/Ao

"lV,

,

= ιzstl L-

r9)'l=

L4 \ιz1

D^

δ

29 39 49 59 118

ln.

ft.

a

4

0.25 0.33

5

0.4t7

6

0.500 1.00

J

t2

r-

οsο fVmin

1

6, x =

12ft

636 x 6(1 .i q'γ4 t12 = 159 ft/min ^l+x(O.S)2

'

Qr= 125x2x6361159 11-5.

= 132.7

Eg. 11=2a

V,= V"(1.13)Kl lx ; K=

vr=

=6

= 1000 cfm

A ceiling type diffuser system has the ability to handle large quantities of air because the air is discharged radially and

diffuses the high velocity jet in a short distance. 11-6. (a) A perimeter type system would be necessary to achieve a

satisfactory heating performance. Αny other type of system would lead to a cold and drafty floor.

ExοeΙpts Γrom this wοrk may be reproduced by instructors for distribution on a not-fοr_profΙt basis for testing or instruοtional puφoses only to students enrolled in οοurses for which the textbοok has been adopted. Αny other reprοduction or trαnsι{ιιion οf this work beyοnd ιhαι permiιted by Sections ]07 οr l08 ofthe Ι976 tJnited Stαtes CopyrighιΑctνiιhoutιhe permission ofιhe copyrighιονneris unlαιllfuΙ.

(b)

An overhead type system would be preferred because of the greater need for cooling during the summer and less need

213

for heat durΙng the winter.

11-7

'

Α perimeter type system would be the best choice. This type

system is required to do a good job of heating. Α spreading jet should be used when heating and a nonspreading jet should be used when cooling.

11-8'

Some kind of overhead system wouΙd be preferable since cooΙing would be the dominant mode of operation. However, ceiling diffusers with radial discharge woutd not be required due to a low volume of circuΙated air' Α high side walt type of system or ceiling diffusers with discharge in only one or two directions with a Ιarge throw would be preferred. This would give the maximum air motion with a smaΙl amount of circulated air.

11-9.

10 in. round diffuser, TabΙe 11-4;650 cfm

lnterpolation between 600 & zoo cfm is required

NC=0.5x(21 -17)+17=19 x5o = 0.5(1 1-10) + 10 = 10.5 ft

p=oo62(ffi)'=oo73in

wq

11-10.

For 150 cfm/ft, ΔPo

= O.08

x (150116712 = O.ο65 in' wg.

Excerpts from this work may be reproduced by instructors for distributiol on a not-for-profit basis fοr testing or instructionaΙ puφoses only to students enrolled in courses for which the textbοοk has been adopted Αny oιh,er Ιλaurtιon or ιrαnslαιion οf this wοrk beyοnd thαι permitted by Secιions ] 07 or Ι 08 of ιhe ] 976 ('|nited Sιαιes Cοpyright Αct w iιhλuι ιhe p,e{m|ssιon of ιhe copyrighι oνner is untιnνfut.

214

Throw values are for a 4 ft active length then X5ο =

21 _ 0.6(4) = '18.6

ft;

Τhe uncorrected NC for a 10 ft length is NC

= 23

_ 0.6(5) = 20.

For a length of 6 ft the correction is -2.

Corrected: NC

-2

= 20

= 18

11-11. Model 28, 4-48 T-Bar; Table '1 1-6, 270 cfm lnterpolate:

NC

=

xso =

0.7 (36

-

32) + 32 = 34.8 or 35

l0

0.7(11 -10) *

p=O.r[

, ^-^'t2

+l )

ι245

= 10'7 or 11 ft

=o.13in.wq

-

11-12' From Table 11-1, L = 12 ft' Then Q=

from Τable 11-2 at

40 Btu/(hr-ft21, lxuo/L1rr*

=1.3 and the range is1.2 - 1.8, and X5s = 1.3

x12= 15.6ft

A good solution would be to use the 4 in. size with '150 cfm/ft. with uncorrected throw of 18 ft and

NC

= 19.

The corrected throw is: Xso ='18 x 0.85 = 15.3 ft and

P

=

0.057rΨ)' ι139

)

NC = 19 - 4= 15

= O.066 in. wq

-

11-13. (a) Room char. Length

= 14 ft,

Table

1 1-'1

(x5ρ/L)rr, = 0.8, Table 11-2

Range of xso/L

= 0.5

to 1.5; xso = 0.8 x 14= 11.2ft

EΧοerpts fτom this work may be reproduced by instΙuctoΙS fοτ distribution on a not-for-profιt basis for testing or instruοtional purposes only to students enrolled in courses for which the textbook has been adopted. Αny other reprοducιιon or ιrαnsιαιion οf this ιυork beyond thαt permiιted by Secιions ] 07 or Ι 08 οf the ] 976 ιJniιed Stαtes Copyrighι Αcι lυ iιhouι the permission οf the cοpyrighι οwner is unlαwful.

215

The best choice would be a 12 in. size with 600 cfm

(b)

13.

xso =

14'3 --'-'--νν':9 Q) = 14.3ft;x5e/L 80'ι 14

ΔPo = o.o81

11-14. (a)

(ffi)'

=

= 1.02(in the range)

0.0g6 in. wg., NC =

22+ffιol

=24.5

Room char. Length is 26 ft,Table 11-2 (x5ρ/L),,, = 1.6 (Table 11-2); range of (x5g/L) = 1'2

- 2'3

26 = 41.6 ft; Q/diff = 60012 = 3OO οfm From Τable 11-5, the 18 x 4, 14 x 5, or

Xso = 1.6 x

12 x 6 sizes may be acceptable aΙthough the throw is

less than desired. Xso

= 31 ft

Xso/L = 31.6126 = 1.2 (barely in the acceptable range)

(b)

Xso =

31- ft (zero defΙection)

NC =

11-15

22*,

^po:

o.o6e

(#)'=

0.065 in. ws.

lt is good practice to keep the core veΙocity below 5OO ftlmin.

A

solution is the 18 x '1 2; Table 11-7

ΔPo- -o.O45

[,Ψ)' ι535/

= -O.O57 in. wg

65 NC= 21 + fZl 135' ι =24 Note that static pressure and ΔPo are negative.

11-16. Guidelines: 1-Place diffusers under or between double windows. 2-Select throw using the ΑDPl procedure. Characteristic length Εxοerpts from this work may be reprοduοed by instΓuctoΙS for distribution on a not-for-proflt basis for testing or instruοtional puφoses only to students enrolled in courses for which the textbook has been adopted' Αny oιher reproduction or ιrαislαtiοn of ιhis wori beyond'thαt permiιιed by Secliοns ] 07 or Ι 08 οf ιhe Ι 976 Uniιed Stαtes Copyrighι Αct νithout ιhe permission οf ιhe cοpyrιghι owλer is unlαwful.

216 =

floor to ceiling.

3-Noise criteria (Nc) should usuaΙly be less than 30. 4-Be Sure that the totaΙ pressure required is compatible with the pressure characteristics of the system. For example, a smalΙ commercial system may have a fan that produces only

about 0.6 in. wg. total pressure while a large commerciat system may operated at2-5 in. wg. total pressure. The diffusertotal

pressure Ιosses should be no more than abo ut 10% of the fan total pressure. 5- Use data from Table 11-17

'

1

1-3

GuideΙines:

1-center diffusers in square or nearly square spaces. Divide large or irregular spaces into imaginary square spaces and pΙace a diffuser in each Space. Select throw using ΑDPl procedure. 2-Τry to obtain a balance between many small diffusers versus a few very rarge diffusers to be cost effective. ?_ vι

4_ t

See Problem 11-16

5- Use data from Table 11-4

11-18. Guidelines: 1-Locate diffusers about 12 in. below ceiling on inside waΙls. Setect throw using ADpl procedure. 2-Τhe jet may be spread with this type diffuser. However, more than one diffuser should be used where the room width is at least two times the room depth. 3-

Iζ. See problem 11-16

Exοerpts frorn this work may be reproduοed by instructors for distributiοn on a not-fοr-prοfit basis for testing or instruοtional purposes only tο students enroΙ]ed in courses for which the textbook has been adopted Αny oιl'r, Ιrpioauαιοn or ιrαnslαriοn of this νοrk beyond ιhαι peιmiιιed by SeCιionS Ι07 or Ι08 οfιhe ]976 (]niιed Sιqtes Copyrigh'ι Αcιlνiιhλuι ιhe peimιssiοn οfιhe copyrighι ονner is untανυful.

v&

217

45- Use data from Table

1

'1-S

Excerpts from this work may be reproduοed by instruοtors for distribution on a not-for-profit basis for testing or instruοtional puφoses onιy to students enrolled in courses for whiοh the textbοok has been adopted' Αny other reproduction or trαλlωion of this work beyond'thαι permitted by Secιions ]07 or Ι08 ofthe ]976 (]niιed Stαtes CopyrighιΑcιιυithouι the permission οfιhe cop7righι owλer is unlανful.

218 -ffi

11-19. Guidelines: '1-Locate grilles in ceiling near the inside wall.

2-Noise criteria (NC) should be less than 30. 3-The negative static pressure should be held to minimum, especially for light commercial systems with small fans.

core velocities of less than 5oo ftlmin will usually yield a quiet system with a reasonably low-pressure loss. Higher velocities and pressure ross may be tolerated with heavy commercial applications. 4- Use data from Tale 11-7.

11-20.

H-24, Mod 28 at 89 cfm each

(a)

+l+

<-l-+ +l-+

2Ψ'-windows

*l-+ +t-e +l-r

€l+ +l-,

(b)

X1oo=

3ft; L =7

ft; X19ρ/L

€Ι_+

=3l7

= O.43; o.3 < (x19ρ/L) < 1.o 0.06 in. V/g.; ΝC = 20

89 cfm/diffuser; ΔPo Νote: other acceptable soΙutions also exist.

11-21. SimiΙar to

11-2O - Diffusers shouΙd throw air towards the windows - Arrange to obtain uniform air motion - Might use diffusers with short throw around exposed walls with larger units in the interΙor.

11-22. L = 9 ft; Table'l i-1 Exοeιpts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instruοtional puφoses only to stιιdents enrolled in courses for whiοh the textbook has been adopled,' Αny othe, )eprocturtιon or trαnsιαtion of ιhis νork beyond ιhαι peιmiΙιed by Secιions ] 07 or Ι 08 οf the ] 97 6 tJniιezι Sιαιes Copyrιgh'ι Αct νιthλut ιhe peλission οf the cοpyrighι oνner is unlαwful.

219 X59/L = 0.9; Τable 11=2, straight Vanes ( Assume light load for a secondary system)

xso=0.9x9=8.'1

ft

Α solution: 9-4 ft length diffusers with 50 cfm/ft, 2 in. size, Τable 11-3, x = 8.5 ft (no correction required); NC = 15 - 4 = ΔPo =

β0l4q2 x O.036

11

= O'047 in. wg.

Place 3 diffusers on each exposed wall

11-23. Use

4-'12 in. size from Τable 11-4

650 cfm/diffuser; L Room Load = 18

=

20

ft

80

Btu

= (hr _ ftΖ )

78

x5sil

= 0.8,

Table 11-2

x = 16 ft (desired) Xactual

=

(650 - 630)

F5

_

63οi

(7 _ '15) + 15 = 15.5 ft

xact_15.5_., : -fl = 0.78 (in acceptable ^ NC = 27;

11-24' Use

ΔPo=O 105

rΨ]'= ι630/

O.112 in. wg.

14-H-48, Model 28 diffusers from Τable 11-6;

229 cfmldif. as shown. L =

range)

= 20 ft, xlse/L

0.3 and acceptable range is 0.3 to 1.0.

Εxcerpts 1iom this work may be reproduced by instruοtors for distribution on a not-for-prοfit basis fοr testing or instructional puφoses only to students enrolled in οourses for which the textbook has been adopted' Αny οther reproductiοn or trαnsιαιion of ιhis νork beyond-thαι permitιed by Secιions ] 07 or Ι 08 of ιhe ] 976 United Sιαtes Copyrighι Αcι νithouι ιhe permissiοn of ιhe copyrighι oνner is unlιrνιful'

220

Desired throw: xloo = 0.3 x 20

= 6 ft.

Actual throw:

'1

Xact =

6.5, TabΙe

Xr"t/L = 6'5Ι20 = 0.33,

NC

=

'1-6

o.K.

80

29, ΔPo = 0.095 in. wg.

ltLr 78

11-25. Refer to Problem 11-23, Q

=

2600 cfm; refer to Table 11-7.

Αssume a Ιay-in ceiling with 2 ftx 4 ft tiles. Τo assure a quiet return, limit NC to about

20. Use 2-24 in. x24 in. grilles

with

1300 cfm each. Nc < 25, ΔPo = -0.048 in. wg.

11-26. Refer to Problem Assume

a2ftx4

11

-24,3200 cfm.

ft lay-in ceiling.

Use 24 in. x 24 in. size from Τable 11-7

'

Using three units,

cfm/grille = 320013 = 1067; Nc < 2ο ΔPo = -0.033 + 0.006 = -0'027 in. wg.

Excerpts fiom this work may be reproduοed by instruοtors Γor distribution on a not-for-profit basis for testing or instruοtional puφoses only to students emοlled in οourses for which the textbook has been adoρted' Αny οιher reproduction or trαnsΙαtion of ιhis work beyond ιhαι permιtted by Secιions ] 07 or Ι 08 of the ]976 tJnited Sιαtes Cοpyright Αcι νithouι the ρermission of ιhe copyright oνner is unlανfuΙ.

Exοerpts from this wοrk may be reproduced by instructors fοr distribution on a not-for-profit basis for testing or instruοtional puφoses οnly to students enrοlled in courses fοr whiοh the textbook has been adopted. Αny other reproduction or trαnslαtion of this work beyond thαt permitted by Sections ]07 or Ι 0B of the t 976 United Stαtes Copyright Αct without the permission of the copyright owner is unlαwful. Requests for permissiοη or further informαtion should be αddressed to the Permission Depαrtment, John Wiley & Sons, Ιnc, ] ] ] Riνer Street, Hοboken, NJ 07030.

CHAPTER 12-1.

(a) W.

= rt (Pl

=

(b) r,= (c)

V

_ Pz)lρ Αssume standard air = o.60

HP - O.45 kW

=#ffi+

= O 54

or 54oλ

ξffi

P. = 1.9 =

-

'84=2,381ft/min,

12-2

Qz =

a,

ΔP.l

ΔPoz =

ΔΓ

W,

=

Q381l4ooq2

=

ο.35 in wg

2000x'1.55 = 0.49 6350

(c)

=

0'4911'1 = 0'44 or

abovet P,

#ffi

=

ΔP,z

Pu =

0.35 = 1.55 in wg

Ιs = Ws/Wrr,

(d) From

2

Q(Pl -P2) _ 2ooο(1.9) 6350 6350

= 2Ooo/O

W"

1

=

2ooo

[Hffi)'

(ffi#)

".[ffi)=

\iν, [RPM, ιRPM1

]'

'

= 1.55 in wg

= 1.55

(κpv")2

=l

44o/o

=24oocrm

(ffi#)'

= 1,133 L/s

= 223in wg ρ 555 Pa

ι o( -ι1οOo/ -*Ψ9\ =z'τιin '1

(12o0lu ι1000/

=

wg

=682Pa

l.9 HP = 1'42k\Ν

221

12-3.

(a,b) Qz =

Q (750i900)

=

0.833 Q

Poz= PO1(75ol9OO)2 = ο'694

HPz- ΗP1(75oi9oo)3

r

Pο1

= 0.579

HPl

80ο RPM

7OO

RPM

a

Po

HP

a

Po

HP

6,000

2.3

2.75

5,250

1.76

1.84

10,000

1.87

3.5

8,750

1.43

2.34

14,000

1.15

3.45

12,250

0.88

2.31

12-3 2.4

6

Ξ 2.0

5

.Ξ

4

σJ

I

'1.6

Φ

3 (L

Ξ

Φ 1.2

α E Ξo

Po = 1-30 in. wg Φ = 9625 cfm ΗP = 2.34

Ξ

U)

Φ

(d)

0.8

2

t- 0.4

1

0.0

02

468111214

0

cfm x 10-3

12-4. Since pressure in in. wg. is plotted on the ordinate instead of head the pressure must be adjusted to reflect the barometric pressure at 5280 ft elevation. Po = (Po)rtο(ρ/P.tα) = (Po).tα(Po/Po,'tα)

also, W = W.tα(ρ/Ps*o)= Po,,tα =

Wr16(P6/P5'r16)

14.696 psia; Po = O.491(29'42 _ 0.0009 x 5280); Eq. 3-4

or instruοtional puφosοs only to Excerpts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing of this work beyοnd ιhαt trαnslαιion or reproduction Αny other has been adopted. the tixtboοk for wtriοh in courses enrolled students is unΙανful' oνner copyrighι permissiοn οf ιhe Αcι the ν)ithouι Sιαιes Cοpyrigh} Ι976 LΙnitec] Ι of ιhe permiιιed by Sectiοns ] 07 οr 08

222

Po = 12.112 psia

Co the

Τhen in Denver,

ne\M characteristics

may be obtained by

computing Po and W at various volume flow rates from Fig. 12-8. Po = (Po),ro(2.112t14.696) = 0'824(P6)916 ?Πd

W - 0-824

Wstο.

Q cfm '10,000

6,000

14,000

Po

W

Po

W

Po

W

Sea Level

2.3

2.75

1.87

3.5

1.15

3.45

Denver

1.9ο

2.27

1.54

2.88

0.95

2.84

(w",0 - w)1oo 3.5 - 2.88 :. (b) ΔW=ff= _ 35 .

ΔW

= 18% (decrease)

12-5. Refer to Problem P6 = (99.436

-

12-4 for explanation.

0.10 x 1618) = 83.256 kPa

Po = (Po)stο = (83.256/101 '325) = 0.822(Po).16 \Λ/ = Wstο

x0'822

(a)

Q

m3/min

180

155

125

W

Po

W

Po

Po

W

Sea Level

400

1

350

320

1

600

260

2000

Αlbuquerque

329

1110

263

1315

214

1644

οnly to Excerpts from this work may be reproduοed by instruοtors fοr distribution on a not-for-profit basis for testing or instruοtional puφoses wοrk beyond ιhαι students enrolled in οourses fοr whiοh the textbook has been adopted. Αny oιher reproducιion or ιrαnsιαιiοn of ιhis permiιted by Sections ]07 or Ι08 ofιhe Ι976 Uniιed Stαιes Copyrighι Αctwiιhout ιhe permission ofιhe copyright oνner is unlαινful.

223

(1600--1315) ΔW = 10O = 17'ro/odecrease 1 600

(b)

12-6. (a)

This is at the limit of the good selection range. lt would be better to choose a different fan.

(b) Α near perfect match with the fan capable of producing about 1.85 in. V/g. totaΙ pressure at 10,000 cfm.

(c) A bad application and

out of the recommended range.

Would probably be unstable.

12-7

[From Fig. 12-9]

(a) No, fan (b) No,

is too small.

not a good application, fan is too large.

(c) YeS, near perfect appΙication; moderate fan speed, high efficiency.

12-8.

150 m3/min, 4OO

Pa

[From

Fig 1 2-10J

The fan would be acceptable and is reasonable. η1 =

12-9.

55%; RPM = 85O;

W,

= 185o

(a) At 1418 cfm = 1420 cfm, Ve P" =

P.

[+Ψl' ι4005,

= 0.88

-

=

o.zs

W = 2OOO ft/min

in. wg., Po =

0.25 = 0,63 in. wg.

-

P,

+

P,

= O.88 in. wg.

518 in. wg.

EΧcerpts from this work may be reproduced by instructors 1br distribution οn a not-for-profit basis for testing or instructional puφoses only to students enrolled in οourses for whiοh the textbook has been adopted. Αny οther reproducιion or ιrαnslαιion of this νοrk beyοnd thαι permitted by Sections ] 07 οr Ι 08 of the ] 976 United Stαtes Copyright Αcι νithοuι ιhe permission of ιhe copyright ονner is unlαιιful.

224

(b) From Τable

12-1 in col. For 5/8 in. wg.

The rpm is 1092 and power is 0.39 ΗP

0.94 O.BB

\o10e2

rPm

0.80

1240

1420 cfm

12-10' (a)

ΔPo = 3.0 + 0.3 + 0'20 = 3.50 in. wg. desired

(b)

fanr.ι ,'t {z' P

svstem with svstem eifect factor

3.

'osystem wlthout system

ι

3.CI

iffu"

I

.

't3,500

Q cfm

(c)

13,500 to 14,000 cfm

12-11. /2istem, actual

βε,,f-οr whiιih tl

οr ]08 οfthe ]9

effect faοtor

setection without system effect factor

15,00ο

225

System eff. Factor 12-12' D" = (4 x 12 x

=

610

- 430

=

.1gO

pa

16lπ)112 = 15.6 in

Αssume blast area ratio

= 0.7,

Table 12-3

\/ V = 4ο00(1 2 x 161144) = 5333 fVmin " One eff. Duct length = 5.3 diameters, table 12_2 or L" = 5.3 x '15.6 = 83 in. % Εff ' Duct length = 100 x 30/83 = 36

Elbow in position C, Fig. 12-13 EΙbow loss factor = 0'79, Τable 12-5 ΔPo = 0.79(5333l4o0q2 = 1.40 in. wg.

12-13' V

ι = 4OO

0l

|π

x 142l14 x 1aa)]

=

4,276 ftlmin

Duct length = 28 in.; R/D = 10.5114 = o.7s; L/D = 29114 = 2.0 Elbow and duct loss factor = 1.O, Τable 12-6 ΔPo

-

1'0(4276l4ooq2 = 1'14 in. wg.

12-14' Blast area ratio

=

0.7, TabΙe 12-3

D" = (4 x 20 x 2Olπ)1l2 = 22'6 in.

V

= 1O,OOO

x

1441(20 x20) = 3,600 fVmin

L" = 3.6 dia., Table 12-2 L/Le

-

(0122.6)13.6 = 0.12

Co = 0.4,

Table 12-4

ΔPo = 0. 4(3600l4ooq2 = O.32in. wg.

12-15' D" = (4 x 12 x 12lπ)1l2 = 13.5 in. Exοerpts from this work may be reproduced by instructors fοr distribution on a nοt-for-profit basis for testing οr instructiona1 puφoses onΙy to students enrolled in cοurses for whiοh the textbook has been adopted. Αny οιher reproductiοn or tαλhfiοn οf ιhis wοri beyond ιhαι permiιιed by Seclions ] 07 οr t 08 οf ιhe 1 97 6 Uni ιed Sιαtes Copyrιghι Αcι ν iιhout the permission of the copyright oνner is unlαwful.

226

V"

= 25001(12x 121144) = 2500 ft/min

One eff. Duct length

=

2.5 diameters,Table 12-2

L = 2.5 x 13.5 = 33.9 or 34 in. 12-16. From Problem 12-15, Vr= Ve = 2500 ft/min (assumed) ΔPo = C"(v /4005)2; Co = O.16/(250O t400q2 = O'41

From Τable 12-6, LlH = 4'3 Length = 4.3 x 12 = 51.5 in.

12-17. (a) Τhe design condition and the observed condition are on nearly the same system characteristic. Therefore, it is probable that the fan is not running at the desired speed of about 920 rpm but at a lower speed of about 6'10 rpm.

(b) The fan is operating near the 920 rpm characteristic but something related to the duct system has changed.

Possibly a damper is closed, a duct has collapsed or some other obstruction is present.

(c)

Both the system and the fan characteristic have

changed. The duct system has probably become fouled or slightly damaged is some way while the fan speed has decreased slightly due to \Μear and tear.

1.5x5000 = 1.62 12-18' Wsn,l = 16 HP; Wsh,2 Ξ 6350x0.73

%

Diff

= rco(16_,!62) 16

ι

)

= +οO%

[decrease from 1 to 2!

puφoses only to Εxοerpts from this work may be reprοduced by instruοtors for distτibution on a not-for-profit basis lor testing or instruοtional ιhis work beyond thαt students enrolled in courses for whiοh the textbook has been adopted. Αny other reprοductiοn or ιrαnslαιion of permiιιed by Sections ] 07 οr Ι 08 of ιhe 1 976 United Sιαιes Cοpyrighι Αcι Ι|ithοuι ιhe ρerfiιissiοn οf ιhe cop1'ν'ighι oνner is unΙανfuΙ.

227

12-19. (a) Assume 15,000 cfm is an equivalent value for the day. Forfull load point 1: Wr

= 16

x0.746x24=286.5 kwh

For part load cond.: Wp

= 6.7

x

ιW

=

(28q'9_ ]20) x 286.5

'1OO

=

0]46 x24

58o/o

= 120.Okwh

(decrease)

(b) No, the fan would be forced to operate to the left of the maximum

pressure and would probably be unstable.

12-20' W l = 28.5 ΗP; W z=

ΔW =

(28'?:17 '5)

12-21. (a) Wr

W,

ιW (b) W,

28.5

=

17

1OO

'5

HP

= 39o/o (decrease)

28.5 x 0.746 x24 = 510 kwh

'0 xO'746 x24 (510_:183)

= 27

= =

(static po\Μer used)

510

1OO

27 x0.746x24

ΔW =

(510_l_483) 51ο

=

1OO

=

483 kwh (vanes assumed "λ open)

= 5'3% (decrease) 483 kwh = 5'3% (decrease)

(a) and (b) essentiaΙly the same.

12-22' The actual inside dimensions are 10 x 8 in. or D"

=

9.8 ιn., Table 12-7

For duct, unlined, ΔPo/L = 1 .8 in. wg./100 ft (Fig. 12-21)

O

-

2ooox144 1 0x8

= 3600 fvmin

From Fig. 12-23, roughness corr. Factor = 1.51, then for the Exοerpts from this work may be reproduοed by instruοtors fοr distribution on a not_for-profit basis for testing or instructional puφoses only to students οnτolled in courses Γor which the textbook has been adopιed' Αny other reprοduction or ιrαnsιCιιion of this ινork beyond thαt permiιιed by Secιions ]07 or Ι08 ofιhe ]976 United Sιαιes CοpyrightΑcιwithouι the permissiοn ofιhe copyright olνner is unlαwful.

228

lined duct, ΔPo/L = 1 .8(1 .5'11 = 2'72 in' wg./100 ft. and ΔPo

-

50 x2'7211ο0 = '1.36 in. wg. or about 338 Pa

12-23. ΔPo = (ΔPo)rl X

P/P.l,

Pυ = 0.491(29'42

-

ρ _]!}Psι Pο,rz

0.0009 x 5000) = 12.236 psia

ΔPo = 1.36(12'236114.696) = 1'13 in. wg. or about 282 Pa 12-24.

Q

=

600 cfm

| tl

()Θ

-=t'

Dz = Ds = 10 in'', A2lA1= 0.6 = Α3/Aa

ΔPd/L = 0.185 in. wg./100 ft; Fig' 12-21 ΔPzs = 0.1 85 x20l100 = ο.037 in. wg.

For contraction, AzlAι = 0.6, Coz=

0'21

For expansion, A+/As = 1'67, Cρa = O.80

Υz=

600

]

Table 12-gA

'

Vs= - = 11OOfUmin; Υι=Vgx ff ι(ιo1'

=660ft/min

4112 )

ΔPιz= 0'21(11}ol4oO5)2

= O.O16 in. wg.

ΔPοo = 0.80(660/4o05)2 = a'O22 in. wg.

ΔPo = 0.037 + 0.016 + 0'022 = 0.075 in. wq.

Exοerpts Γrom this work may be reproduοed by instructors for distributiοn on a not_for_profit basis for testing or instructional puφoses only to students enrolled in courses for which the textbook has been adopted. Αny oιher reprοducιιon or ιrαnsιαιion of this νork beyοnd thαι permitted by Secιions ] 07 or Ι 08 οf the ] 976 United Stαιes Copyrighι Αct νithout ιhe permission of the copyright oιυner is unlανful.

12-25' (a)

229 Bellmouth, Co = 0'2i Αbrupt, Co = 0.5; Table 12-109 & 10Α ΔPos = O'2(1ooo/4oo5)2 = O.o125 in' W9.

- 3.1 Pa

ΔPon = O.5(1oOo/4oO5)2 = O.0313 in. wg.

x7'8Pa

o/o

Diff .

(b) ΔPoa

-

(0'0313 - 0'0125) (1oo) = lsoo/o 0 0125

= O'2(40O0l40oq2 = O.2O in. wg.

ΔPon = O.5(4OOol4ooq2 = O.50 in. wg' o/o Diff.

-

(0.5

- 0.2) (1oo) =

o.2\/

ε 50 Pa

x

124 Pa

15Oo/o

12-26. Table 12-8a, Co = 0.25 = 1200l[(πla)x(1 4ln)2! = 1122'5 ftlmin ΔPo O.25(1122.5l4oO5)' _ O.O2 in. wq. V

o

=

also

Vo = 0'6l|@lξ(0.35)2] = 6'24 mls ΔPo

-

12-27. (a)

0.25(6.2411.2q2 = 5.8 Pa Co = 0.15, Table 12-8b

Vo ΔPo

or

= 25OO

-

x

1441(16 x 16) = '1406 ftlmin

o.15(1 40614005)2 = O.O'185 in. wq.

Vo = 1.21(0.4 x 0.4)

=

7.5 m/s

ΔPo = 0'15(7 '5l1'2q2 = 5.1 Pa

(b) Co = 1 '2 Τable 12-8C ΔPo = 1.2(140614005)2 = O.148 in. wq.

or

ΔPo = 1'2(7.5l1'29)'= 40.6 Pa

Excerpts from this work may be reproduοed by instruοtors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in οοιπses for whiοh the textbook has been adopted. Αny other reproduction or trαnslαtion of ιhis νοrk beyond ιhαι permiιted by Sectiοns ] 07 οr l 08 οf ιhe ] 97 6 Uniιed Sιαιes Copyrighι Αcι wiιhout the ρermission οf ιhe cοpyrighι oνner is unlανful.

230

a/a

12-28.

c=

25O18OO = 0.3125

Aυ/Ac=(6112)2=O.25 Co = 0.345, Τable 12-11A

Vυ = 25Ol|(πl4)(6t2)2]= 1273 ft/min or

Vυ = 0'12[(ila)(ο.15)2]

=

6.8 m/s

ΔPoο = 0.345(1 273t4o05)2 = O.O35 in. wg.

or

ΔPoο = 0.345(6 '8t1'2q2 = 9.6 Pa

Qr/Qο=55O/80O=0.6875 ΑS/Ac =

(0112)'= 0.694

C" = 0.135, Table 12-11A

or

V.

= 550l|(πla)(0l12)'! = 1οo8 ftlmin

V,

= o.26ll(ila)(0.25)21 = 5.3 m/s

ΔPo" = O.135(1oO8/4οο5)2 = O.OO9 in. wq.

or

12-29.

ΔPo" = O.135(5.3/1 2972 = 2'3 Pa

From Problem 12-28

a/aο

= O.31 25;

A/A"= 0'25

Vo = 1273 fVmin or 6.8 m/s Cο = 0.93, Τable 12-11F ΔPoo = 0.93(1 273l4oo5)2 = O.O94 in. wq.

or

ΔPoο = O.93(6.811'2q2 = 25.8 Pa

Qr/Qc

C,

V,

= 0.6875; Αr/A" = 0.694

= 0.135;

Table 12-118

= lOOB fVmin or 5.3 m/s

ΔPo, = O.135(1oOs/4oo5)2 = O.OO9 in. wq. Excerpts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instruοtional puφoses only to students enrolled in οourses for whiοh the textbook has been adopted. Αny οther reproduction or trαnsιαtion of ιhis νork beyond ιhαι permiιιed by Secιions ] 07 or Ι 0B of ιhe ]976 United Stαιes Copyrighι Αcι wiιhouι ιhe permissiοn οf ιhe copyrighι oνner is unlα:lνful'

or

231

ΔPo, = o.135(5.3/1 29)2 = 2'3 Pa

12-30' (a) Ao/Al

vr

=

ΑoVo

= 6.0, θ = 180 deg., Co = 37'4,

Table 12-98

o?9-0"'^!o = 2,ooo ft/min (18x18) =

ΑlVl,

Vo =

1,

Αo

x 2,O0O = 2000/6 = 333 fVmin

ΔPo = 37.4(333l4o0q2 = 0.260 in. wq.

(b) Co = 14.35 (Table 12-98) ΔPo = 14.35(333t4ooq2 = O.O99 in. wq.

12-31.

ao/Q"

= 5OO/1OOO = O.5o

Αο/A" = (8Ι12)'= o'444 ΑS/Αc = (8/1 2)' = 0'444

Q,/Qc

(a)

= 500/1OOO = O'5

12-12A lnterpolation required 1 Table 12-12A l ASHRΑE Duct Fitting

Cυ = 0.755, Table

C" = O.2'15,

Vο = 50Ot|πt4)(8t2)2]= 1432 ft/min

or

or use Data

- v,

Vυ= O.24l|(πl4χo.2)'] =7'64mls ΔPoυ = 0.755(1 4g2l4o05)2 = O.O97 in. wο.

or

ΔPoo = O'755(7 .64112972 = 26'5 Pa

ΔPo, = O'215(1432I4OO5)2 = O.O28 in. wq.

or

ΔPo, = O'215(7.641129)2 = 7 '54 Pa

Excerpts f,rom this wοrk may be reproduced by instructors Γor distributiοn οn a not-for-profit basis for testing or instructional puφoses only to students enrolled in courses fοr which the textbook has been adopted. Αny oιher reproductiοn or trαnsιαtiοn of ιhis work beyond ιhαι permiιιed by Secιιons ] 07 or Ι 08 of ιhe ] 97 6 tJniιed Sιαιes Copyrighι Αcι νithοuι the permission of ιhe copyright ονner is unlιrwful.

233

1

55

0.'135

0.07 4

0.50

2(0.26)

1020

0.032

0.034

0.140

2

72

0.055

0'ο40

0.16

2(0.26)

630

0.004

0.013

0.057

14

3

20

0.050

0.010

2.0

0.17

550

0.038

0.003

0.051

13

12-33' L = D x Co/f; f

=

0.019, ΤabΙe 12-13

Bellmouth: L = 1 x0.210.019 = 10.5 Αbrupt Int.: L = 1 x 0.5/0.019

Q

35

=

ft

26.3 ft

= 1OOO x πl4 = 785 cfm; ΔPo/L =

O.12in'wg'l1oo ft, Fig' 12-21

ΔPog =

0.12x 10'5l100

= 0.0126 in. wg. or about 3.'1

ΔPon =

0.'12x26'3l100

= 0.0316 in. wg. or about

Pa

7'9 Pa

12-34' From ProbΙem 12-26' Co = 0.25, D = 14 in' Lu = DxC olfi f = 0.017 ,

γ'= -

Q

lx

12

0'25 0.017

= 17

Table 12-13

'2ft

= 12OO cfm; ΔPο/L = 0.'13 in. wg./1OO ft

ΔPo = 0.13 x 17 '2l100 = 0.022 in. wg. or about 5.6 Pa

Note: Most of following duct sizing problems can be solved with the computer program, DUCT.

12-35. From Figure 12-36.

Excerpts frοm this wοrk may be reproduced by instructors fοr distribution on a not_for-profit basis fοr testing or instructional puφoses only tο students enrolled in οourses for whiοh the textbook has been adopted' Αny οther reproduction or ιrαnsιαtion οf ιhis νork beyοnd ιhαι permιιιed by Sections ] 07 οr ] 08 οf ιhe Ι 976 United Sιαtes Cοpyright Αcι Ι|'ithοuι the permissiοn of ιhe copyright oνner is unlανful.

234

Estimate Tota] Εquivalent Length of Run 1-2-3 to be approximately 132 ft, Τable 12-14' Then ΔPo/Le = (0.13 x 1 00)1132 = 0.'10 in.

wg./100 ft size ducts using Figure 12-21 and record the

actual ΔPo/L from Figure 12-21'

Section

Le

a

D

ΔP/L

ΔPn

No.

ft

cfm

in.

in. wq.

in. wg.

100 ft

0.084

0.ο38

220

I I

ο.090

0.014

71

100

6

0.083

0.059

4

55

80

5

0.14

0.077

5

55

120

6

0.125

0.069

1

45

300

2

16

3

Run 1-2-5 actually has the greatest lost pressure. ΔPlη = 0.038 + 0'077

=

0.115 in. wg.;

ΔPιzs = 0.038 + 0.014 + 0.059 = 0.121 in. wg.

ΔPlzs = 0.038

+

0.014 + 0.059 = 0.1'1

12-36. The design pressure loss is (0.25

-

'1

in.

\Λ/g.

0.1 ) = 0.1 5 in. wg. (for supply ducts)

Assume the run with the largest equivalent length

is:

1-2-3-4-5, Le = 185 ft

Τhen for design: ΔPo/Le

-

(0'15 _ 0'03) x 1OO = 0.065 in. wg./'1oO ft 185

Exοerpts fτom thls wοrk may be reproduοed by instructors for distribution on a not-for-profit basis for testing or tnstruοtional puφoses only to students enrolled in courses fοr whiοh the textbook has been adοpted. Αny oιher reproduction or trαnsΙαιion of this νork beyond ιhαι permitted by SecιΙοns ]07 οr ]08 οfιhe Ι976 United Sιαιes CopyrightΑcιιι,iιhοuι the permissiοn οfιhe copyι'ighι ονner is unlnνful.

235 Section Θ has a total flow of 845 cfm. Τherefore, the maximum velocity in section Θ wjl! be about 800 fVmin if a 14 in. duct is used.

Exοerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional puφoses only to students enτolled in οourses for which the textbook has been adοpted. Αny other repsφluction or ιrαnsιαιion of ιhis νork beyond thαι permitted by Sections ] 07 or Ι 0B of the 1976 Uniιed Snrcs Copyright Αcι wiιhout ιhe permis\ o/*e copyright oνner is unlαwful.

\

236 12-36. (continued) (a)

Section

Le

a

D

ΔP/L

ΔPn

No.

ft

cfm

in.

in. wο.

in. wg.

100 ft 1

88

845

14

0.065

0.ο57

2

18

595

12

0.07ο

0.013

3

16

395

12

0.034

0.0054

4

17

275

9

0.065

0.0'1

5

46

125

7

0.065

0.030

6

51

250

I

0.060

0.031

7

43

200

8

0.072

0.031

I I

49

120

7

0.065

0.032

49

150

I

0.040

0.020

1

With the equal friction method, every branch should have a damper for

balancing purposes. Αctual total pressure loss: ΔPo

-

ΔP1 + ΔP2 + ΔP3 + ΔPa + ΔPu + ΔPοs

ΔPo = 0'146 in' wg'

Note that run 1-2-3-4-7 actuaΙly has the greatest loss in total pressure but the difference is not significant. Use ΔPo = 0.15 in. wg.

(b)

Sizing of the longest run, 1 -2-3-4-5, is the same as (a) above where ΔPo/L = 0.065 in. wg./100 ft. Construct a new table as follows:

EΧcerpts fiοm this wοrk may be reproduοed by instructors tbr distribution on a not-for-pro1lt basis for testing or instruοtional puφoses only to students enrolled in οourses fοr which the textbook has been adopted. Αny oιher reproducιiοn or ιrαnslαtiοn of ιhis work beyond ιhαt

petmiιιed by Secιions Ι07 or 108 ofιhe Ι976 Uniιed Sιαιes Copyrι4hιΑcι'!ιιhouι ιhe permission οfthe copyrighι oνner is unΙιrwfuΙ.

---------------MAIN DUCT RUN

237

BRΑNCH DUcτS

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(e)

(10)

(1 1)

(12)

(1

Sec

Le

cfm

DJwxh

ΔP

v

ΔPo

ΣΔPo

Br.

ΔPι

Le

ΔPi

οfm

No.

ft.

in.

L

fpm

(2)(5)

Σ(7)

Seο

ΔPoot

ft.

L

No.

-(8)*

('10)100

ΔPα

(11)

100

3)

(14)

(1

5)

D"/wxh

v

in.

fpm

I

BB

845

14

0.065

800

0.057

0.057

f)

0.ο39

51

0.076

250

9

550

2

'18

595

12

0.070

760

0.013

0.070

7

0.036

43

0.084

200

I

570

J

16

?oξ

12

0.034

500

0.005

ο.ο75

8

0.035

49

0.071

120

7

500

4

17

275

I

0.065

600

ο.01'1

ο.086

9

0.020

49

0.041

150

I

420

5

46

125

7

ο.065

500

0.ο3

0.116

ffuser G 0.030

0,146

The left 8 columns are the same as (a) above. The branches, 6-7-8-9, are sized to balance in the right hand 7 columns.

(c)

Equal Friction Method

-- Design Procedure -Sysιem type: Supp}y Duct Sizing Method: Equal Friction Rounding Method: Round Nearest

-- Ean Selection -Knoι\Ιn

Fan Parameter: F'an TotaΙ Ρressure :

Fan Αlrflow: Fan or Externa1 Total Pressure:

B45.0 cfm Ο.25Ο in. Ο . ΟΟ0 in.

External Total Pressure:

Ο.25Ο in.

Coi1 Lost PreSsure: Εi1ter Lost Pressure: 0 . 000 in. Misc. Lost PreSSure: 0.ΟΟΟ in.

ΑΗU AΗU

0.

250 in.

wg

wg wg wg wα wg

Pressure for Supply System: 0.150 in. wg - or 60.0 for Return System: 0.1Ο0 in. wg - or 40.0

AHU Pressure

%

z

-- Lost Pressure from Αir HandJ_ιng Unit to Diffuser -Diffuser ΙD Q Tota1 Delta P (cfm) (1n. wg) 71

125.0

A.L20

Excerpts from this work may be reproduced by instruοtοrs for distribution οn a not-for-prοfit basis for testing or instruοtional puφoses only to students etrrolled in οourses for which the textbook has been adopted. Αny other reproducιion or ιrαnsιαιion οf this wοrk beyond thαt permitted by Sections ] 07 οr 108 οf the ] 976 United Stαιes Copyrighι Αcι ''νiιhout ιhe permission οf ιhe cοpyrighι oνner is unlανful.

238

22

t6 30 34

Total

150.0

120.a 200.0 250.A B45.

0.154

0. 130 0. t25

0. Ι21

Ο

-- Calculated Fitting

VaΙues --

ΙD Fitt1ng Type

1 2 3 4 5 6

Air Ηandling Unit Straight Duct

7.0

15Ο.0 1s0.0 15Ο.0 15Ο.0

561.3 561.3 561.3 561.3

?n 7.Ο 7.0

L24.0 ι20.0 120.0

449.0 449.a 449.0

8.0 8.0

2Ο0.0 2AA. 20Ο.0

573.0 573.0 573.0

9.0 9.0 9.0

250.0 250.0 250.0

565.9 565.9 565.9

Tee / νΙΥe

14. Ο 72 .4

9 Straight Duct / wye

1Ο Tee

11 Straight Duct 12 Tee /

,νΙye

Straight

Duct

Straight

Duct

Εlbow Elbow

1Δ

main branch

Straight

EΙbow

main branch

10.0 8.0

72 .0

72 .0

10.0 9.0

main branch

1

il?}l.n coΙnmon

'7.0 1 .0

7.0 1.0

'1

.0 1 .0

Duct

ΕΙbow

Diffuser,/ Gri11e

Elbow

.0

10.0 9.0 1.0 ?n 9.0 7.0

coΙτιmon

Diffυser / GrΙΙ Straight Duct Straight

a

14.0

Elbow

Ε]-boν'τ

Ο

common

Diffuser / GrilΙe Straight Duct

Elbow

Ο

oΔ

coτnmon

Duct

Diffuser / GritΙ Straight Duct

Elbow

EΙbow

Dlffuser / GrilΙ

Velocity (ftlmin)

19a.4 151 .6 124.2 573.0 151 .6 124.2 622.5 449.0 '724.2 622.5

ConicaΙ Contraction Εlbow Elbow

Q (cfm) 845.0 845.0 845.0 845.0 845.0 595.Ο 250.0 B45.Ο 595.0 395.0 20Ο.0 595.0 395.0 215.0 L20.0 395.0 215.0 L25 .0 150.0 215.0 125 .0 L25 .0 L25 .0 L25 .0

0.0

14. 14.

7 Straight Duct B Tee / wye

13 14 15 16 17 18 19 20 21 22 23 24 25 26 21 28 29 30 31 32 33 34

Dia. ( in)

]

2η

n

15Ο.0

120 .0

2Ο0.0

250.0

Delta

(in.

P wg)

0.0

Ο.000

19A.4 '790.4 19A.4

0. Ο13 0. Ο06

'79Ο.4

'75'7.6

565.

9

461 .1

561.3 622.5 461 461 461 461

.1 .1 .'7

.1

0.013

AP

(in.

/L

wg)

0.06415

0.006 0. Ο05

0.018 0.

ΟΟ7

0.004

0.011_61

0.01_7

0.007 0.003

0.08259

0.007 0.002 0.009

0.07138

0.008 0.002 0.002 0.003 0.030 0.006 0.005 0.016 0.005 0.040 0.002

0.05817

0. Ο16

0. Ο11 0. Ο03

Ο.05B17 Ο. Ο8082 Ο. ΟB0B2

0.05405

Ο.036

Ο.0Ο3

0.010 0.005 0.040 0.003 0.013 0.004

0. ο7106

0.06004

0. Ο50

(c) Balanced Capacity Method -- Design Procedure

--

EΧceφts from this wοrk may be reproduced by instruοtors for distribution on a not_for-profit basis for testing οr instruοtional puφoses only to students enτolled in οourses for whiοh the textbook has been adopted' Αny other reproducιion or ιrαnsιαtion of ιhis νοrk beyond ιhαι permiιΙed by Sectiοns ] 07 or l 08 οf the 1976 United Stαιes Copyright Αcι withοuι ιhe permissiοn of ιhe cοpyright oνner is unlανful.

____

239

System type: Supply Duct Sizing Method: Balanced Capacity Rounding Method: Round Nearest

-- Fan Selection -Known Fan

Fan Αirflow: Fan or External Total Ρressure:

CoiΙ Lost Pressure: E'iΙter Lost Ρressure: Misc. Lost PreSSure :

ΑHU AΗU ΑHU

0.250 in.

Parameter: Fan Total Pressure :

Εxterna1 Total Pressure:

B45.0 cfm

in. in. in. in.

wg wg wg

Ο.25Ο in.

wg

Ο.250 0.Ο00 0.000 0 . 0Ο0

wg

ι^rq

Pressure for SuppΙy System: 0'150 in. wg - or 60.0 Pressure for Return System: 0.1Ο0 in. wg - or 40.0

% %

-- LoSt Pressure from Αir Ηand1ing Unit to Diffuser _Diffuser ΙD Q Tοtaι Delta P (cfm) (in. wq) L] 22 26 30 34 Total --

125.0 150.0 720.0 200.0 250.0

0.138 Ο.154

0.145 0.140 0.141

845.0

Cal cu-Lated

Fitting Values --

ΙD Fitting Type

Dia. (in)

0. Ο 14.0 14.0 14.0 14.0 1'2 .0 8.0 common 14.Ο 72.a 7 Straight Duct 10.0 8 Tee / \NΥe main branch 1 .0 common 72.0 10.0 9 Straight Duct 9.0 10 Tee / Wye main branch 6.0 common 10.0 9. Ο 11 Straight Duct 6.0 L2 Tee / Wye main branch 1 .0 coΙnmon 9.0

1 2 3 4 5 6

Air Handling Unit Straight Duct Conicaf Contraction E}bow trlbow main Tee / Wye branch

Q (cfm) 845.0 845.0 845.0 845.0 845.0 595 . 0 250.0 845.Ο 595.0 395.0 200.0 595.0 395.0 215.0 720.0 395.0 215 '0 \25.0 150.0 21 5.0

Velocity (ftlmin)

DeΙta P (in. wg)

ΔP/L

(in.

wg)

0.0 '790.4 190.4 190.4 '190.4 '7 57 .6 116.2

0.00

.6 124.2 1 48.4 15'1 .6 -124.2 622'5 6lL.2

Ο.0Ο7

0.07167

0.007

0.08259

622.5 636.6 561.3

0.0Ο7

0.07138

'790.4 '751

124.2

622.5

0.013 0.013

Ο.06415

Ο.006

0.006

0 . 005

0.016

0.004 0.016 0.0Ο3 Ο.013

0.003 0.009

Exceφts from this rνork may be reproduced by instructors for distribution οn a nοt-for-profit basis for testing or instruοtional puφοses only to students enrolled in courses for which the textbook has been adopted. Αny οther reproductiol1 or trαnsιatιon οf ιhis work beyond thαt permitted by Sections ] 07 οr Ι 08 οf the t 97 6 Uniιed Sιαtes Copyrighι Αct ||1ιthout ιhe peιmissiοn of the copyrighι oνner is unΙcτwful.

-

240

13 Straight

6.0

Duct

0.72304 L4 Elbow

15 Sιraight Duct 76 Εlbοw l1 Diffuser / Grille 8 19 2a 27 22 23 24 25 26

1

2"7

28 29 30 31 32 33 34

Sιraight

Duct

Straighι

Duct

Elbow

Elbow

Diffuser / Grille

Elbow

Straight

Elbow

Straighι

Elbow

L25 .0 L25 .0

7.0 /.u 7.0 1.0

150.0 150.0

1rη 1,25

6.0

Duct

6.Ο

6.0

Diffuser / GrilΙe El-bow

6.0 6.0 6.0

1.4 1.0 7.0

Duct

Diffuser / Grille

8.0 8.0 8.0

Εlbow

Strarght Duct

Elbow

Diffuser,/ Grille

n

.0

15Ο.0

150.0 1"20 .0 120.0 t2a .0 120.0 200. Ο 200.0 200.0 200. Ο 250.0 25Ο.0 250.0 250.0

0.016

636.6

125 .0

636.6 636 .6 636 .6

0.004 0.005 0.007

561.3 561.3 561.3 561.3

0.006 0.00s 0.016 0.005 0.040 0.004 0.023 0.006 0.036 Ο.005

6tt .2 6L7.2 6L1.2 148.4

Ο.030

n

nT q

'1

ΔQ

"7

48 .4

0.009

'716.2 1L6 .2 'Ι 16 .2

0.005

Δ

0.72344

0.08082 Ο.08082

0.ΙL427

0.13629

0. Ο40 ^ ^41

0.007 0.050

0.10661

12-37. ΔPos + ΔPon = ο.70 _ 0.35 = 0.35 in. wg.

ΔPos = 0.65(0.35) = 0.23 in. wg. ΔPoκ

^y

0.35 _0'23 = 0.12 in. wg.

Τhe method of Solution is similar to Problem 12-36' An acceptable solution follows: (a)

Longest run

-

1

-2-3-4-5-6-7-8-9-1

1

-13

The summation of equivalent lengths may vary v/ith designers. ΔPo/Le

=

(0.23

- 0.03) 100

217

=

0.092 in. wg./'100 ft

Size all Suppιy ducts for this pressure loss per unit length.

basis for testing or instructional puφoses only tο EΧcΘrpts from this work may be reproduced by instructοrs for distτibution on a not-for-profit or ιrαnslαtion οf ιhis work beyond ιhαι reproductiοn Αny other adopted' has been tΘXtbook the for whiοh in courses students enrolled copyright oνner is unlατιful' ιhe peimission of ιhe Αct νiιhout Copyrigh't Stαtes 1976 Uniιed pe,rmιιed by Sections ] 07 οr l δε o7 ,ιr,

241

1

1200

16

150

2

1

050

14

100

3

850

14

100

4

750

12

75

5

650

12

100

b

550

12

200

7

475

10

8

225

I

9

175

7

10

50

5

11

125

6

12

50

5

13

75

6

14

50

5

basis for testing or instruοtional puφoses only ω Εxοerpts frοm this work may be reproduced by instructors for distribution on a not-for-profit or ιrαnsl(tιion of ιhis work beyond thαι reproducιion Αny oιher adopted. has been textbook the for wt'iοh in courses students enrolled copyrighι oνner is unΙανful' peimissiοn ofιhe the Αct ιlithouι Cοpyrigh't Stαιes United Ι976 permiιιed by Secιiοns ] 07 or l δε i7 in,

----=-

c

12-42. (continued)

Ξ

9,rE

Γ..-

Ιt*

lo IO

Ξ o Θ c c{

O

ιο (o

(o

ιo

c{

(o

Ιιf, |Ξ

Φ

N'

ιο

ιΩ

ιo

ιο

rf

ιο

ιο

cηE ι-O

Θ ιf)

o o c.l

Θ Θ

o o Θ o

ιc)

Θ

ιο

ιο Ν

Γ--

Ν

ιo

lcs

la

I; |Ξ l(J

ol Θl ^

Ξ ο Ξ ()

ΘΙ Ξl !.t

z

Ξ

tα]

Φ

o

(o

Φ

Θ

ci

ιΓ)

ιΙ.J o Θ O σ)

ιΓ) C.l

c;

Cη Γ.-

lο)

Φ

Ι\-

c)

Θ

o

Γ.t

ιο

Γ.-

ιο

ιa)

(r)

οn

Θ

o (o

τΦ

Θ

Θ

ci

c!

Lο

c!

Φ

Φ

c.l

(η

C.l

Φ

l> tt.lg lΞ l= lΦ lΝ lω ]ο lcΛ lο lΦ IL

C.l

ιo

t-

o

Ξ*-*uΦ{

ο-

€Φ

Ξ οE o O

-l-

= Ξ o Φ

ιΓ)

Φ

Φ

co

c;

ci

ci

ci Θ

ci

oδ Ε2

co

Cη

σ)

(\

Θ

c\l

αl

\r

o

c{

δds

ti-

ιΓ)

vO

(o

σ) (o

Cη

Φ

σ)

ιo

(o

c)

o o o ci ci o Θ

ε*"3Ξ

Ι-

o

o

Φ cj

^l^

Θ

Γ-*

cο

Θ Θ

sσ) o

ο.l

c;

Θ ci (o

<)

c;

c.l

ο Φ

o ci

Θ ci

Θ

c)

Φ

Θ

Γ.-

σ)

o

Cη

Φ

I()

lc

l-c lσ) l-c I

σ,

ο.l ο.l

v ιΓ) o c! o Θ (η cf) o o $o Θ Θ o I ο Θ ci c;

lo lo lU)

= o o Φ

(o cη I

c{

.c

c

Ξ o ι-

U)

U' (σ

(Ι)

-o (σ

z !cr>a

Ξ t t--

O

Ξ ο

z Ξ

Σ

Θ σ)

O O cO

c{

Φ

Θ Γ-

Θ Θ Φ

o o ιο o Φ Γ.-

Θ (o Γ.t

o ο Γ..-

Θ ιο $

Θ (Ι)

o Ξ

c

Φ

oXl -

σ)

cf) Θ cO Θ -i ci Θ O ci

c;

Ξci.Ξ

$

οo

Φ

$

(o

c{

σ) o

(η

ci

c;

c{

c!

Θ

σ)

οo

οo

(o

o Θ Θ o Θ ιo ιο

ιΓ)

()

Θ

Θ

ιο

Φ

()

c.l

C.l

Φ

(o

Φ

οo

u.

ιo

ιΓ)

$

(η

\r

τ

Γ--

c.l

ιf)

_ο

Φ

Cη

c{

$

Cη

(η

Φ

O

Φ

ο q α (o

Θ -O (σ

(o

!-

_ο (σ

Ω

.c

Ξ(υ

Θ

ιf)

cf)

c{

.Ξ J

ζΣ

t

(υ

U)

o

Ξ (σ ο

Ι

(σ t#

o

.l

F

σ)

Ξ

lo l-o

-.l

Ξci.Ξ

(/)

Ε (σ

c) o o Θ Θ ιo o o ο o Ν Ν Γ.-

Ν

cr)

v

ιr,

Φ

Φ +

Γ-

σ)

cf)

o

F

(υ

ο-

E (σ

-ο

ο *

12-37. (continued) ΔPo for the longest run will be about 0'22 in. wg' for the above sizes"

Τherefore, size the return system for a pressure loss of (0.35 _0'22)

-

or 0.13 in. wg.

For the return system:

(L")r* x 230 ft,

('1

r

-

2r

-

3r)

then ΔPo lL"= r0'13-__0'05'] ''oo = O.O35 in. wg./'1oo ft

ι230)

Using the equal friction method: Section

a

D

L"

No.

cfm

in.

ft

1r

1200

18

115

0.038

0.044

2r

800

'16

70

0.033

0.023

3r

400

12

44

0.ο36

0.0'16

4r

400

12

14

0.036

0.005

5r

400

12

28

0.036

0.010

ΔP"/L

ΔPo in. wg.

Return system is the same for parts (a) and (b). ΔPo for return = 0.133 in.

\Mg.

(c) Equal Friction Method -- Desiqn Procedure -System type: Supply Duct Sizing Method: Equal- Γriction Rounding Method: Round Nearesι

-- Ean SeΙection -Known Εan Parameter: Εan Total Pressure :

0.700 in.

wg

244 Εan Airflow: Fan or ExternaΙ Total Pressure:

1000.0 cfm

Externa1 Total Pressure:

0.350 in.

0.700 in. Coil Lost Pressure: 0.25Ο in. F'iΙter Lost Pressure: 0 . 10Ο in. Μisc. Lost ΡresSure: 0.0Ο0 in.

ΑHU AΗU AHU

wg wg wg wg ι^rg

Pressure for SuppΙy System: 0.228 in. wg - or 65.0 Pressure for Return System: 0.123 in. wg - or 35.0

%

%

-- Lost Ρressure from Αir Ηand1ing Unlt to Diffuser -Diffuser ΙD Q Total DeΙta Ρ (cfm) (in. wg) 2'7 30 34 38 56 60 63 61 12 15 Total

75.Ο 75. Ο 75.0 s0.0 75.0 100.0 100.0 100.0 200.0 150.0 1000.

0.221 0.201

0.2Ι1

0.272 0.171 0.158 0.171 0.L42 0.2a2 0.131

Ο

-_ Calculated Fittinα Values -ΙD Εitting Type

1 2 3 4

Dia. (in)

Αir Ηandling Unlt Ο.0 Conical- Contraction 14.0 Straight Duct 14.0 Tee / wye main l2.a branch 1 .0 coΙτΙnon 14.0 5 Straight Duct 12.0 6 Tee / Wye main 12 .0 branch 1 '0 coΙnmon L2 '0 7 Straiqht Duct l2.0 B Tee / vfrze main 10.0 branch 6.O common L2.0 9 Straight Duct 10.0 10 Tee ,/ Wye main 10.0 branch 6.0 coΙπnon 10.0 11 Straight Duct 10.0 L2 Tee ,/ Wye main 9. 0 branch 6.0 conτnon 10.0 13 Straight Duct 9. 0

Q (cfm)

VeΙocity (ftlmin)

1000.0 0. Ο 1000.0 935.4 1Ο00.0 935.4 850. Ο 082.3 15Ο. Ο 561.3 1000.0 935.4 850.0 L082.3 650 . 0 82'7 .6 2a0.0 1 48.4 B50.0 1082.3 650.0 B2'7 .6 550.0 1008.4 1OO.O 5O9.3 650.0 821.6 550.0 1008.4 450.0 825.! 100.0 509.3 550. Ο 1008.4 450.0 825.7 350 . Ο 192 .2 100.Ο 509.3 450.0 825.L 350 . 0 1 92.2

Delta P (in. wg)

(in. ^P/Lwg)

0.000 0.011

0.007 O. O1O

0.08745

0.061

0.003

0 . 008*<10>

0.13819

Ο.058

0.005

Ο.0Ο9 Ο.Ο37 0.

ΟΟ6

O. O1O*<10>

0.08429

0.15164

0.064

0.004 O .

ΟΟ5

0.

014

0.10485

a.024

0. 11082

Excerpts from this work may be reproducοd by instruοtors for distribution on a not_for-profit basis for testing or instructional puφoses only to students enrolled in οοurses for whiοh the textbook has been adopted. Αny οther reproductiοn of trαnslαtion of ιhis νork beyοnd thαι permiιιed by Secιions Ι07 or l08 οfthe Ι976 Uniιed Stqιes Copyright Αcι'|ι)iιhout the permission ofthe cοpyrighι oνner is unlcrwful.

14 Tee 15 L7 18 19 20 27 22 2? 24 25 26 2'l 28 29 3Ο 31 32 33 34 35 36 31 38 53 54 55 56 51 58 59 6Ο 6L 62 63 64 65 66 61 6B 69 70 11 12 13 14 15 16

main 8. Ο branch 5.0 common 9.0 Straight Duct B.0 ΕΙbow B.0 Straight Duct 8.0 Tee ,i Wye main B.0 branch 4.0 common 8.0 Straight Duct B.0 7.0 Tee / Wye main branch 5.0 coΙnmon B.0 1 .0 Straight Duct Tρρ / Ιn1lzo main 5.0 branch 5.Ο coΙnmon 1.a Εlbow 5. Ο Straight Duct 5. Ο Rectangular Transition 5. Ο Diffuser / GrilΙe Straight Duct 5.0 Rectangular Transition 5. Ο Diffuser / Grille Elbow 5.0 5. Ο Straiqht Duct RectanguΙar Transition 5.0 Dif fuser / Gril-le Elbow 4.0 4.0 Strarght Duct Rectangular Transition 4.0 Dif fuser ,/ cril1e Elboιv 5.0 Straiqht Duct 5.0 RectanguΙar Transition 5.0 Diffuser / cril1e Ε1bow 6.0 Straight Duct 6.Ο Rectangular Transition 6.0 Diffuser / cri]ιe Straight Duct 6. Ο Rectangular Transition 6. Ο Diffuser / Gri1Ιe E1bow 6.0 Straight Duct 6.0 Rectangular Transition 6.0 Diffuser / Gril]_e '7 .A Straight Duct 1.0 Elbow 1 .0 Straight Duct Rectangu1ar Transition 1 .0 Diffuser / Gritle Straight Duct 1 .0 Rectangular Transition 7.0 Dif fuser ,/ Grille Elbow 8.0 f νν

,/

/

Wye

νγli

ν

215.a 75.Ο 35ο.Ο 215.0 215.0 215.0 225.a 5Ο. Ο 215.4 225.a 150.0 75.0 225.0 150.0 75.0 75.0 150.0 75.0 75.0 75.0 75.0 75.0 75.0 75.0 75. Ο 75.0 75.0 75.0 50.0 50.0 5Ο.0 50. Ο 75.0 75.0 75.0 75.0 100.0 100.0 1Ο0.0 100.0 100.0 100.0 1Ο0.0 100.0 1Ο0.0 100.0 1Ο0.0 200.0 200.0 20Ο. Ο 20Ο. ο 20Ο. Ο 150.0 150.0 150.0 215.0

* - De}ta P ιfas computed using

.8 550.0 "781

192.2

.8 1B1.B 181 .B 644.6 573.0 l1i .B 644.6 561.3 550. Ο '7B1

644.6

561.3 550.0 550.0 561.3

0.019

0.0Ο6

0.7268'7

0.013

0.L268'7

0.0Ο8

Ο.0Ο9*<10>

a

.021

Ο07

0.0880Ο

0.018

0.08082

0.

0.003 0.011 0.002 0.007

550.0 550.0 210.a

0.0Ο5

550.0 210.a

0.0Ο5

55Ο.0 550.0 21a.0 573.0 573.0 180.0 550.0 550. Ο 210.0 509.3 509.3 360.0 5Ο9.3 36Ο.0 509.3 509.3 360.0 148.4 148.4 48 ' 4 120.0 "7

561.3 54Ο.0 181

.B

the fitt1ng

245

Ο. Ο05

0.006

0.072

0.030

0.0Ο7

Ο.11869

0.11869

0.025 0.0Ο6

Ο.018

0.11869

0.017

0.16916

0.005 A.025 0.007 Ο.010

0.020 0.006

0.01B

0.11869

0.008

a.0122l

0.072

a.0B221'

0.005 0.030 0.004 0.003 0.045 Ο.003

0.032 0.0Ο4

0.008

0.003 0. Ο45

a.08221-

0.02'7

0.l.3629

.a21

0.1-3629

0.032

0.08082

0.013 0

0.001 0.045 0.001 Ο.020

0.005

equivalent lenqth

Balanced Capacity Method Exοeφts from this wοrk may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enτοlled in οourses Γor whiοh the textbook has beοn adopted. Αny οther reproduction or trαnslαιion of ιhis νοrk beyond thαt pernιiιιed by Secιions ]07 or l 08 οf the ]976 Uniιed Stαtes Copyright Αcι νιιhout ιhe permissiοn οf ιhe cοpyrighι owner is unlcrνιful.

246 -- Desiqn Procedure -Note that almost alΙ branch ducts need a damper to increase the diameter and reduce ve1οcity. System type: Supply Duct Sizing Method: Balanced Capacity Rounding Method: Round Nearest

-- Fan Selection -Known F'an Parameter: Ean Tοta1 Pressure

:

0.700 in.

Εan Αirflow: Fan or Εxternal Total Pressure:

100ο.Ο cfm

0.25Ο in. 0 . 1Ο0 in. 0.000 in.

wg wg wg wq

External Total Ρressure:

0.350 in.

wg

Ο.7ΟΟ in.

CoiΙ Lost Pressure: Filter LoSt Pressure: Misc. Lost Pressure:

ΑHU

AHU Pressure ΑHU Pressure

wg

for Supply System: 0.228 in. wg - or 65.0 for Return System: 0.L23 in. wg - or 35.0

?

?

-- Lost Pressure from Αir Handling Unit to Diffuser -Diffuser lD

TotaΙ Delta Q (cfm) (in. wq)

2'7 30 34 3B 56 60 63 61 12 15 TotaΙ

75.0 75.0 75.0 50.0 75.0 100.0 100. Ο 100.0 2Ο0.0 150.Ο

P

0.221 0.243

0.271 0.2L2

0.232 0.219 0.285 0.25'7

0.202 0.17Ο

10Ο0.0

-- CaΙcuΙaιed Γiιting Vafues -ΙD Εltting Type

Dia. (in)

1 2 3 /

Ο.0 .0 14. Ο 72.a 6. Ο 14.0 l2.0 !2.a 1.a

Αir Ηand1ing Unit Conical Contraction Stralght Duct .Γaa / τlτ.,o main branch coττunon 5 Straight Duct 6 Tee / wye main branch

14

Q (cfm) 10Ο0.0 1000.0 1000.0 850.0 15O . O 1000.0 850. Ο 650.0 20Ο.Ο

VeΙocrty (ftlmin) 0.0 935.4 935.4 L082.3 1 63 .9 935.4

1-082.3 821.6 '748.4

DeΙta P

(in. wg) (in. ^P/Lwg) 0.000 0.011

0.007 0.010

0.08745

O. O57

0.0Ο3

0.13819

Ο.OΟ8*<1O>

0.058

Excerpts from this work may be reproduοed by instruοtors for distribution on a nοt_for-profit basis fοr testing or instruοtional puφoses only to students enrοlΙed in οourses fοr whiοh the textbook has been adopted. Αny οther reProduction or ιrαnsΙαιiοn of ιhis work beyond ιhαι permiιιed by Secιiοns Ι07 or ] 08 οf the Ι 97 6 United SιαιeS Copyrighl Αct νiιhouι ιhe peιmission of ιhe cοpyrighι olνner is unΙcrννful.

7 Straight Duct B Tee / wye 9 Straight Duct 10 Tee / Wye 11 Strarght 'rΙ99^^

coΙnmon

main

branch coΙnmon

/ rîr7^ νlyν

maln

branch

branch

LB

L9 20

2L

Duct

Straight

Duct

Tee / Wye

Straiqht

Duct

/ l1lrΣ^ ιγ ] ν

22

Straight Duct Tee / Wye

24 25 26 21 28 29 30 31 32 33 34 35 36 31 38 53 54 55 56 51 58 59 60 67 62 63 64 65 66 61 68

Elbow

23

ι145.9

L2 .0

10.0

Duct

Straight

10Ο.0

10.0

coΙnmon

branch

/

Elbow

Δo

1_2

main

coΙτιΙnon

L1

ι082.3

10.0 10.0

13 Straight Duct main 74 Tee / Wye ι5

.0

85Ο.0 650. Ο

72 .0

coΙnmon

main branch

coτnmon ηa1n

branch common

4.Ο

1Ο.0

9.0 4.0 10.0 9.0 8.0 4.0 on 8.0 8.0 8.0 8.0 4.ο 8.0 8.0 7.0

5.Ο o.u

coτΙυnon

1.4 4.0 5.0 ?n

Rectangufar Transition

5.0 5.0

Straight

main branch

Duct

Diffuser / crille Straight Ducι Rectangular Transition Diffuser / Gritle

tr1bow

Straight

Ducc

Rectangular Transition

Diffuser / Grille

Elbow

Straight Ducι Rectangular Transition Diffuser / Grille

Elbow

Straighι Sιraighι

4.0

4.Ο

ηn

5.Ο trΛ

4.0 4.0 ΔΓ\

Duct

4.4 4.0 4.0

Duct

4.0 4.4 4.0

Recιangufar Trans1tion Diffuser / GriΙ1e ΕΙbow

5.Ο

Rectangular Transition Diffuser / GriΙ]e

Straight Ducι Rectanqular Transition Diffuser / Grille

4.0

Scraight Duct Rectangular Transition

4.Ο

ΕΙbow

Diffuser / crille Straight Ducι

4.Ο Δol

4.0

7.Ο

650.0 550.0 450.0 100.0 55Ο.

Ο

450.0 350.0 1Ο0.0 450.0 35ο. ο 275.0 75.0 350.0 t1\ a

ο' tr Λ

215.0 225.0 50.0

215 .0 aa tr Δ

150.0 75.0 225.0 15Ο.0 75.Ο 75.Ο 150.0 75.0 75.0 ?tr

Λ

75.0 75.0 75.0 75.ο

7η

n

821 .6 1008.4 821 .6 10ΟB.4 825 .7 1145. 9

1ο0B.4 825.L 192.2 1145. 9 825.1 -7q2 2 181 .8 859.4 1A) 181 181 181

)

.8

.8 .8 644 .6

573.0 81 .8 644 .6 561.3 550.0 644 .6 561.3 859.4 550.0 561.3 55Ο.0 550.0 210 .0

'7

859.4 214 .0 550.

Ο

75.0 75.0

550.0 214 .0

5Ο.

573.0 573.0 180.0

75.Ο 50.

Ο Ο

50.0 50.0 75.0 75.0 75.Ο 'tr

Λ

1Ο0.0 1Ο0.0 10Ο.

Ο

100.0 1ο0.0 10Ο.0 100.0 100.0 10Ο.0 100.0 100.0 2ο0. Ο

859.4 859.4 210 .0 1145.9 1145.9 360.0 L1_45

.9

360.0

1145.

9

360.

Ο

7L45 .9

148.4

247 0.005 0.009 0.039

0.08429

0.006

0.15164

0.004 0.005 0.031

Ο.104B5

0.014 0.005 0.017

0.11082

0.006 0.008 0.013

0.L2681

0.010*<10> a .062

0.!268'7

0.009*<10> 0 .021

0.0Ο7

Ο. Ο88Ο0

0.018

0.08082

0.003 0.011 0.007 0.007

Ο.006

0.at2

0.11869

9.02!

0.35266

0.00s 0.030 0

.024

0.025 Ο.006

0.018

0.11869

0.017

Ο.16916

Ο.053

0.35266

Ο.0Ο5 0. Ο25 0. ΟΟ7

0.010 0.020 0.015 .424 0.03Ο 0 .021 0

0.060

0.59144

0.090

0.59144

0.042 0.045 0

.042

0.032 Ο a2'1

0.06Ο

0.59'744

a.021

0.!3629

0.042 0.045

Exοerpts from this work may be reprοduced by instructors for distribution on a not-fοr-profit basis for testing or instruοtional puφoses only to students enrolled in courses for which the textboοk has been adopted. Αny other reprοducιiοn or ιrαnsιαtion οf this νork beyοnd thαι permitted by Sections ] 07 οr l 08 οf the 1 976 United StαιeS Copyrighι Αcι'withouι ιhe permission of the cοpyrighι owner is unlανυful.

248 70 lt 12 13 14 15 16

']48.4 200.a 0.013 1.0 69 EΙbow 148.4 0.02'1 0.1-3629 1 .0 200.0 Straight Duct 120.0 0.001 2Ο0.0 RectanguΙar Transition 1 .0 0.045 200.0 Diffuser / Gril-le '7 150.0 0.069 0.17139 6. Ο 63.9 Straight Duct 150.0 0.007 540.0 Rectangular Transition 6. Ο Ο.020 150.0 Diffuser / Griι1e '7B'7.B 0.0Ο5 2'75.a B.0 EΙbow * - Delta Ρ was computed using the fitting equivalent Ιength

Return Ducts, Equal Friction Method -- nηοj ^_ D:ocedure L ]

vνυΙYlΙ

__

System type: Return Duct Sizing Method: Εqual Friction Rounding Method: Round Nearest

-- Fan SeΙection -Known Εan Parameter: Εan Tota1 Pressure

Fan ΑirfΙow: Fan or External Tοtal Ρressure:

Coil Lost Pressure: ΕiΙter LoSt Pressure: Misc. Lost Pressure:

AΗU AΗU ΑΗU

Εxternal Tota1 Pressure:

:

1200,0 cfm Ο.7ΟΟ in' Ο.25Ο in.

0.700 in.

0.100 in. 0.000 in.

wg wq wg wg

0.350 in.

wq

wg

Pressure for SuppΙy System: 0.228 in. wg - or 65.0 Pressure for Return System: 0.123 in. wg - or 35.0

% %

-- Lost Ρressure from Αir Ηandling Unit to Diffuser -Diffuser ΙD Q TotaΙ Delta P (in. wg) (cfm) 11 14 l"7

400.0 400.0 400.Ο

0.125 0.113 0.096

Tota1 120Ο.0 _- CaΙcu1ated F'itting VaΙues -_ TD

Fitting Type

Αir Handllng Unit Rectangular Transition Straight Ducι main Tee / \ηye branch coπlmon 5 Straight Duct main 6 Tee / wye branch coΙnmon

1 2 3 4

Dia. (in) 0. Ο 18.0 1B.0 1'2.a 16.0 18.0 16. Ο 12.0 \2.0 16.0

Q (cfm) 1200.0 72A0.0 L200.a 400.0 800.0 1200.0 800.0 4ΟΟ.0 400.0 800.0

Velocity (ftlmin) Ο.0 679.1 619.1 509.3 573.0 679.1

0 509.3 509.3 573 '

573.0

LP/L Delta P (in. wg) (in. wg) Ο. ΟΟ0

0.001

0.002

0.03512

0.006

0.03a24

0.026 0.026 0.0]-5

0.013

Exceφts from this work may be reproduced by instruοtors for distribution on a not-for-profit basis for testing or instructiοnal puφoses only to students enrolled in οourses for whiοh the textbook has been adopted. Αny other reproducιion or ιrαnslαιion of ιhis νοrk beyond thαι permitιed by Sectiοns ] 07 or Ι 08 of ιhe 1976 United Sιαtes Cοpyright Αcι νithouι the permission οf ιhe copyright oνner is unΙανfuΙ.

7 Stralght Duct 0. Ο3476 B Elbow

12 .0

400.0

509.3 400.0

0.007 509.3

9 Straight Duct 10 Rectangular Transition 11 Diffuser / criΙle 72 Straight DucL 13 Rectangular Transition L4 Diffuser / GrilΙe 15 Straight Dυct 16 Rectangular Transition 11 Diffuser / Gri-ι1e

L2 .0 L2 .0

4Οο.0 4Ο0.0 4Ο0.0

s09.3 l-00.0

40Ο.0

100.0

0.001 0.014 0.050 0.001 0.014 0.050 0.003 0.014

Ο. Ο03

12 .0

L2 .0 L2 .0

L2.A L2 .0

400.0

5Ο9.3

400.0 400.0 400.0 400.0

509.3 100.

Ο

249 0

.034'7

6

0.03476 0.03476

0.05Ο

12-38. The three branches from the plenum must be designed as close as

possibιe for the Same preSSUre ]oss. Start \Ι/ith B since it appears to be more extensive than Α or c.

For B: ΔPo/Le =[ For

Α:

ΔPo/Le

=(

For

C: ιP"/L"

=(

0.18

- 0.025

145

0.18-0. 025 142 0.18

-

0. 025

104

'100 = 0.107 in. wg./100 ft '100 = 0.109 in. wg./10ο ft

100 = 0.149 in. wg./100 ft

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instruοtional puφoses only to students enrolled in courses for whiοh the textbook has been adopted. Αny other reproductiοn or trαnsιαtion of ιhis wοrk beyond ιhαι permiιted by SecιioιιS ] 07 or ] 08 οf the Ι 97 6 United SιαιeS Copyrighι Αct νithouι the peιmission of ιhe cop1',ι"ighι oνner is unlcrνful'

250

12-38. (continued) BRΑNCΗ DUCTS

MA|N DUcτ RUN

(11)

(12)

(13)

(14)

ΔPi

Le

ΔPi

cfm

De

v

Sec

ΔPo6+

ft.

L

in.

fpm

Νo.

-(8)+

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(e)

(1

Seο.

Le

cfm

D"

ΔP

v

ΔPo

ΣΔPo

Br.

No.

ft.

in.

L

fpm

(2)(5)

Σ(7)

100

actual *r"

0)

(1 0)1

(1

5)

00

(1 1)

-ΔPα

1

B

8

44

500

12

.057

650

.025

025

14

0.094

55

0.171

125

b

660

o

22

375

10

.085

700

.019

.044

10

0.ο75

52

0.144

200

7

760

11

25

175

7

0.1

630

o28

.072

12

0.047

38

0.124

75

5

550

IJ

54

100

o

.ο87

520

.047

0.119

Tot

145

.025

0 144

1

50

40ο

2

19

300

Α

c

1

ΔPο

25

200

48

100

Tot.

142

15

56

225

17

48

100

Tot.

104

10

o

.095

760

ο.048

0.048

7

0.084

38

0.221

100

ξ

750

.ο92

700

0.018

.ubb

b

0.ο66

48

0.1

38

100

b

510

4

0.046

q7

0.ο8'1

'100

o

500

to

0.041

38

0.1 ο8

125

ο

610

ο8

59ο

0.02

.095

530

0.046

ΔPο 7 ^

.086 .

t5z

0.025

157

17

850

0.095

.095

.085

510

0.04'1

'136

0.025

161

ΔPο

Note that the resulting total pressures losses turn out to be: (ΔPo)a = 0.144 in.wg., (ΔPo)n = 0.157 in. wg.; (ΔPo)c = 0.161 in.

\Μg.

Within the accuracy of the calculation these are appΓoximately equal' It

may be necessary to use a damper in branch B, sec. 8.

12-38. Branch A, Balanced Capacity Method __

Γ)Αq

ννJf

i

α- pηocedure -Y11

!

l

System type: Supply Duct Sizing Method: Balanced Capacity Rounding Method: Round Nearesι __ Εan Sefection -_ Known Εan Parameter: Plenum Tota1 Ρressure

Fan Αirflow:

:

0.1B0 in.

wg

4Ο0.0 cfm

Excerpts from this work may be reproduced by instructors for distribution on a nοt-for-profit basis fοr testing or instruοtional puφoses οnly to students enrolled in οourses 1br whiοh the textbook has been adopted' Αny other reproduction or trαnsιαιion of this νork beyοnd ιhαt permittecι by Secιions ]07 or Ι08 ofthe ]976 Uniιed Sιαtes Cοpyright Αcινiιhouι ιhe permission ofιhe cοpyrighι owner is unlαwful.

Fan or External Total Coil Lost E'iΙter Lost Misc. Lost

0

251

wg 1^rg

wg \^rg


0.269 in.

1^rg

Pressure for Supply System Pressure for Return System

0.180 in. 0.089 in.

wg wg

AΗU ΑHU ΑΗU

.269 1n. 0.000 in. 0.000 in. 0.Ο00 in.

Pressure ΡreSSure Pressure Pressure

or or

6'7

0%

33 0%

Lost Ρre$-sure from Αir Handling Unit to Diffuser -Diffuser ΙD

1

ν^

(cfm)

r^-^1 ΙOιd-L

Ο 100.0 100. Ο 100.0 100.

15 19 23 29

Total

(in.

ΙD Εitting Type

Values -Dia. (in

)

0.Ο

Stralght Duct Tee / wye main

qn

Elbow

branch

coΙnmon

7 Straight Duct 8 Tee / wye main

branch

coΙnmon

9 Straiqht Duct 10 Tee ,/ Wye main

branch coτnmon

Straighι

Duct

Straight

Duct

E1bοw

RecιanguΙar Transition

Diffuser / Grille

EΙbow

Sιraight

Duct

9.ο 9.0 9.Ο

8.0 5.0

on

8.0 1.4 trΛ

8.0

'7.0

ξn

5.0 1.0 5.0 5.0 5.0 5.0

ζn

RectanquΙar Transition 5.0

Diffuser / cr1lιe

Efbow

Straiqht

wg)

0.171 0.155 0.141 0.187

Αir Handllng Unit Conlcal Contraction Straight Duct

11 72 13 1'4 15 16 t'7 18 79 2A 2L 22 23 24 2-Ι 28

n r

400.0

-- Calculated Fittlng

1 2 3 4 5 6

n^l ιreΙLd-^

Duct

5.0 trΛ

RectanguΙar Transition 5.0 Diffuser / cri]_ιe

5.0 Sιraight Duct ButterfΙy Damper 5.Ο RectanguJ-ar Transition 5.Ο

a

(cfm) 4ΟΟ.

40Ο.

Ο Ο

400.0 40Ο.

Ο

400.0 300.0 100.0 400.0 3Ο0.0

200.0 100.0 300.0 200.0 100.0 100.0 200.

Ο

100.0 100.0

1ΟΟ.0

10Ο.0 10Ο.

Ο

100.0 100.0 100.0 100.0 1ο0.

Ο

100.0 100.0 100.0 1Ο0.

Ο

10Ο.Ο

100.0

Velocity ( ftlmin)

Delta P (in. wg)

Ο.0 905.4 905.4 905.4 905.4

0.000 0.011 0.003 0.011 0.014 0.006 0.039

Qξo /

733.4 onξ / 859.4

t48.4

7?? I a ξo 1 148.4 133.4 1))

Λ

148.4 133.4 1aa

Λ

1))

Λ

360.

Ο

7??

Δ

a 1a

Λ

36Ο.0

??? aa1

Δ Λ

360.0 111 1" 1

Λ Λ

360.0

(in. ^P/Lwg)

0.

14157

0.14157

0.018

0.14878

0.016 0.004

0.1_3629

0.005 0.020

0.

Ο]_3

0.020 0 .0!2 0.016 0.010 0.025

0.1_9911

0.016 0.010 0 .425 0.007 0.016

0.Ι99'71

0.0Ο7

0. Ο10

a.025 0.016 0.058

0

.1997'7

0.L9911 0.3-991"7

0. Ο10

Exοeφts from this work may be reproduοed by instruοtors for distributiοn on a not_for_profit basis for testing or instructional purposes only to students enrolled in οourses for which the textbook has been adopted. Αny οther reproducιion or trαnslaιiοn of ιhis work beyond ιhαι permiιted by Secιions ] 07 or Ι 08 οf the ]976 Uniιed SιαιeS Copyrιghι Αct'wiιhouι ιhe permissiοn of ιhe copyrighι oνner is unlανfuΙ.

252

29 Diffuser / Grille

100.0

0.025

Branch B, Balanced Capacity Method /t"

Desιgn Procedure _-- -\. System tyρe: Supply Duct Sizing Method: Balanced Capacity Rounding Method: Round Nearest

-- Εan Sefection Κnown Ean

__

Parameter: Plenum Total PressurΘ :

Fan Αirflow: Fan or Εxternal Total Pressure:

AHU AΗU AHU

Coil Lost Ρressure: Γi1ter Lost Pressure: Misc. Lost Pressure: External- Total Pressure:

500.0 cfm Ο.18Ο in.

000 in. 0.000 in. 0.Ο00 in. 0.

0.180 in.

0.180 in.

wg

wg wg wg Wg

wg

Pressure for SuppJ_y System: Ο.18Ο in. wg - or 1OO.O Pressure f or Return System: 0 . 000 in. \^Ιg - or .ο

% %

-- Lost Ρressure from Αir Ηandling Unit to Diffuser -Diffuser ΙD Q TotaΙ De]ta P (crm1 (1n. wg) 13 L6 19 23 Total

10Ο.0 75.0 200.0 125.a

0.185 0.179 0.\42 Ο.15B

5Ο0.0

-- Calculated Εittinα VaΙues -ΙD Εitting Type 1 2 3 Λ 5 6 7 8 9

A1r Ηandling Unrt ConicaΙ Contraction Straight Duct 'Γaa / τι;l',o main

Dia. (in)

0. Ο 10.0 10.0 9.0 branch 6.O coπιrrion 1Ο.0 Straight Duct 9.0 Tee / wye main 6.ο branch 6.0 coΙnmon 9.Ο Elbow 6.0 Straight Duct 6.0 Tee / wye main 4.0 branch 5.0 coΙnmon 6.0

Q (cfm)

Velocity (ftlmin)

500.0 5Ο0.0 500.0 375.0 Ι25.O 500.0 375.0 2ΟΟ.0 175.Ο 375.0 175.0 175.0 75.Ο 1ΟΟ.0 175.Ο

0.0 916.1 976.1 848.8 636.6 916.7

B4B.B 1O18.6 891.3 84B.B

891.3 891.3 859.4 '733.4 B91.3

Delta P (in. wg) Ο. OOO Ο. O11

(in. ^P/Lwg)

0. Ο18 O.Ο06

0.72'723

0.019

0.725"75

O.o42 Ο.OO9

0.019 0.008

O.014 0.OO6

0.22175

0.020

Εxceφts from this work may be reproduοed by instruοtors fοr distribution on a not-1br-profit basis for testing or instructional puφoses only to students enτolled in οourses for which the textbοοk has been adopted. Αny οther reproducιion or ιrαnslαιion of ιhis νork beyond thαι permiιιed by Sections ]07 οr Ι08 οfιhe 1976 Uniιed Stαtes Copyright Αcιννιιhouι ιhe permission οfthe copyrighι olνner is unΙιrwfuΙ"

10 11 \2 13 L4 15 16 71 18 19 20 27 22 23

5.0

El-bow

ξn Straight Duct RectanguJ-ar Transition 5.0

Diffuser,/ Gril1e Straight Duct Rectangular Transition Diffuser / GriΙ1e Straight Duct RectanguΙar Transition Diffuser,/ Griιle Straight Duct Butterfly Damper Rectangular Transition Diffuser / critle

'733.4

100.0 100.0 100.0

1?2'

75.0 75.0 75.0 200.0 200.0

6.0 6.0

859.4

210 .0

1018

636.6 636.6 450.0

1-25.0 1oξ

Λ

L2s.A

Branchffilanced Capacity

.6

120 .0

200 .0 125 .4

b.u 6.0 6.0

Δ

360.0

10Ο.0

4.0 4.0

0.007 0.028 0.010 0 .025 0 .028 0 .025 0 .025 0.041 0.014 0.025 0 .0L2 0.044 0.005 a

253 0.1_99'71

0.35266 0

.29022

0.72304

.02s

Method

-- Design Ρrocedure -System type: Supply Duct Sizing Method: Balanced Capacity Rounding Method: Round Nearest

-- Εan Se1ecιion -Knoιtn Ean Parameter: Plenum F.an

Εan ΑirfΙοw: Pressure: Pressure: Pressure: Ρressure:

Ο.

ΕXterna1 Tota1 Pressure:

0.

or External Total ColΙ Lost Filter Lost M1sc. Lost

AΗU ΑHU AHU

Total Pressure :

Ρressure for Supply System: Pressure fοr Return System:

0.180 in.

225.0 cfm

180 0Ο0 0 . 000 0.000

in. in. in. in.

wg

18Ο in.

wq

0.

0.180 0.

Ο0Ο

wg

wg wg \^Ιg

ln. in.

or or

\.^/g

wg

10Ο.0 z .C)

z

-- Lost Pressure from Αir Ηandlinq Unit to Diffuser Diffuser ΙD a Totat DeΙta (cfm) (in. wg) 10Ο.0 725.0

1-4

1B

Total

aatr

0.186

Λ

-- Ca1cu1ated Fittinα

ΙD Fitting

Ο.191

Type

Values -Dia. (in

)

1 Air Handling Unit 2 Conical Contraction

0.0 Ο

8.

a

(cfm)

225.A .0

225

Velocity (

ftlmin) O. 644

O .6

Delta P (in. wq)

AP

(1n.

/L

wg)

O. OOO

O . OO5 Excerpts from this work may be reproduced by instruοtors for distribution on a not-for-profit basis for testing or instructional puφoses only to students enrolled in courses Γor which the textboοk has been adopted. Αny οther reproducιion οr ιrαnslαιion of this νori beyond-ιhαι permiιιed by Secιions ] 07 or Ι 08 of ιhe 1 976 ιJnited Stαtes Copyrighι Αct ιιiιhout ιhe permissiοn of the copyrighι oινner is untαwfuΙ'

254

3 Stralght

0.08800

Duct

4 Butterfly Damper 5 Straight Duct 6 Elbow 7 Straight Duct 8 Elbow 9 Straight Duct 1Ο Tee / Wye main

branch

11 72 13 74 15 71 18

Straight

Efbow

Duct

coΙτ'ιmon

RectanguΙar Transitj-on

Diffuser / Grille Sιraight Duct Rectangular Transition Diffuser / GriΙle

8.0 8.0 8.0 8.0 8.0 8.0 8.0 6.0 6.0

8.Ο

6.0 6.Ο

6.0 6.0 6.0

225 .0

22\

Λ

na tr n

225 .0 225 .0 C1ζ

^

.0 1Ο0.0 1_25

225.0 1ΟΟ.

Ο

100.0 10Ο.0 100.0 14tr

^

L25 .0 L25 .0

644

644 .6 644 .6

644 .6 644 .6 644 .6

644 .6 636.6 509.3 644 .6 509.3 509.3 360.0

636.6 450.0

.6

0.003

0.!12

0.003 0.004 0.006 Ο.005 Ο.006

0.003 0.011

0.08800 0.08800 0.08800

oo7

o. 0822L

0.010

0.12304

o.

0.003 0.003 0.02s

0.00s 0 .025

12-39. Solution follows Example 12-14 closely. 12-40 Solution follows Exampte 12-14 closely

Exceφts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instruοtional puφoses only to students enrolled in courses for which the textbook has been adopted. Αny οιher )eproducιion or ιrαλhιion of ιhis wοrk beyond ιhαι permiιιed by Secτiοns ] 07 or Ι 08 of ιhe 1976 (]niιed Stαtes Copyrighι Αcι ιιiιhouι ιhe peλission of the cοpyrighι

oιιλer ii

unΙαw'ful.

-

255

12-41.

SF 2

F

1

Po=o

c

M

E

A

1

8,,

2

Supply fan: ΔPo

S

=

4 in. wg.

Return fan: ΔPo = 1.75 in. wg. 12-42.

SF

4

2 S

Space Pressure S

1

Pυ=0 -1

A

-2

le

Fan, ΔPo = 5.75 in. wg. Exοorpts from this work may be reproduοed by instruοtors for distribution on a not-for-profit basis for testing or instructional puφoses only to students enτolled in οourses for which the textbook has been adopted. Αny other reproduction or trαnsιαti()n οf this work beyond ιhαι permitιed b) Sections ] 07 or Ι08 of the 1 976 Uniιed Stαtes Copyright Αct lνiιhout ιhe permission ofthe copyrighι owner is unΙawfuΙ.

257 12-43. 4

2

RF

1

Pυ=0

c

M

Ε

S

-1

-2

Supply fan: ΔPo

=

4 in. wg.

Return fan: ΔPo = 1.75 in. wg. 12-44. 6

SI 4

1

Pυ=0 -1

-2

Excerpts from this work may be reproduced by instructors for distribution on a not-for-prοfit basis for testing or instructional purposes onΙy to students enrolled in οοurses for whiοh thΘ tΘxtbook has been adopted. Αny other reprοduction or ιrαnslαιion οf ιhis νοrk beyond thαι peιmiιted by Secιions ] 07 οr ] 08 of ιhe Ι 976 Uniιed Stαtes Copyright Αcι νithout ιhe permission of the cοpyrighι oινner is unΙαw/ul'

258 Fan, ΔPo = 5'75 in. wg. 12-45.

(a)

Αssume a reasonable duct velocity of about '1200 fpm. ΔPo/L in. wg./100 ft. and D" = 18 in. (may be converted to

20x14

= 0.095

in. for

example) For the duct: ΔPο

-

(0.095 x 40)1100 = 0.038 in. wg.

For elbows: Co = O.15; ΔP" = 2x o.15(118O/4OO5)2 = 0'026 in. wg. For damper: Co = O.52, ΔPα = 0'52(118o/4OO5)2 = 0.045 in. wg. For grille: ΔP, = 0'25 in. wg For expansion: Vo = V.'(A.'/Αo) = 118Ol2 =59O fpm ΔPu

overall: ΔPo ΔPo

(b)

-

1'2(59ol4oo5)2 = 0.026 in. wg.

= 0.038 + 0.026 + 0.045 +

-

0'25

+

0.026

=

ο.385 in. wg.

For 18 in. duct with 1,ο00 cfm, ΔP/L = 0'027 in. wg./10ο ft For duct: ΔP6 = 0'027 x 401100 = 0.01 1 in. wg. For elbows: ΔP" = 2x0'15(59ol4og5)2 = 0.006 in. wg. For griΙle: ΔP, = O'25(1Qoo/2oοο)2 = 0.063 For expansion: Vo = 59012 = 295 fpm

ΔP" = 1'2(295l4oO5)2 = O.OO7 in. wg. Εxοerpts frοm this work may be reproduced by instruοtors Γor distribution on a nοt-for-profit basis for testing or instruοtional puφoses only to students enrolled in courses for which the textbook has been adopted- Αny oιher reproduction or trαnsιαιion of this νork beyond ιhαι permitted by Sections Ι07 οr 108 οfthe ]976 Uniled Sιαtes Copyrighι Αcιwiιhοuι ιhe permissiοn οfthe copyrighι oνner is unlωνful.

For damper: ΔP6 = 0.385 _ (0.01

259 '1 +

0.007 + ο.063 + 0.007)

=

ΔP6 = 0'297 in' wg. = Co"(59O lAooqz

(c)

Co" = 0.29710.022 = 13.7

12-46. Equal Friction Method Note that a damper has been inserted in duct 6 (Nο. 34 betow) to cause an increase in duct diameter from 8 to 9 1n. with a consequent decrease in velocity to an acceptable 1evel.

-- Deslgn Procedure -System type: Supply Duοt Sizing Method: trqua1 Εr1ction Rounding Method: Round Nearest

-- Fan Selection -Known Fan Parameter: E'an Tota1 Pressure

Εan

Fan Αirflow:

or Ext.ernaΙ Total Coil Lost Εilter LoSt Misc. Lοst

AΗU AΗU AΗU

:

ο.900 in.

845.0 cfm 0.9ΟΟ in.

Pressure: Pressure: Pressure: Pressure:

Ο.500 in. 0.100 in. 0.050 in.

wg wq wg wα


0.250 in.

wg

wg

Pressure for Supply System: 0.150 in. wg - or 60.0 Pressure for Return System: Ο.10Ο in. wg - or 40.0

% %

-- Lost Ρressure from Αir Ηand1ing Unit to Diffuser -Diffuser ID Q TotaΙ Delta Ρ (cfm) (in. wq) 19 24 28 32 38 Total

150.0 L25.0 120.0 200.0 25Ο.0

Ο.141

0.128 a.123 0.11s 0.143

845.0

-- Ca1cu1ated Fittinα Values -TD Εitting Type

Dia. (in)

Q (cfm)

Velocity (ftlmin)

Delta P (in. \^/g) (in. ^P/Lwg)

EΧοerpts from this work may be reproduced by instruοtors for distributiοn on a not_fοr_profit basis for testing or instruοtional puφoses only to students enrolled in courses fοr which the textbοok has been adopted. Αny oιher reproducιiοn or ιrαnslαιion of this work beyond ιhαι permitted by Secliοns ] 07 or 1 08 of the ] 976 Uniιed Sιαιes Copyrighι Αct wiιhοuι the permissiοn οf ιhe cοpyrighι ονner is unlωυful.

260 1 2 3 4 5 6

Αir ΗandΙing Unit Conical Contraction Straight Duct Elbow

Straight

Elbow

0.0

14.0 14.0 14.0 14.0

845.0 845.0 845.0 845.0 845.0 845.0

190 .4

main

14.0 ι2 .0

845.0 595.0

15'7 . 6

0. Ο05

coΙτιmon

74 .0

main

10.

190.4 151 .6

0. Ο07

0

0.0Ο7

0.08259

0.0Ο7

0.07138

L4 .0

Duct

7 Straight Duct 8 Tee / vlye 9 Straight Duct 10 Tee / Wye

on

branch

12.A

branch

8.0

12 .0

co]πnon

11 Straight Duct: L2 Tee ,/ Wye main

branch coΙτιmon

13 Stralght Duct 1-4 Tee / Wye main

21-

22 23 24 25 26 21 28 29 JU 31 32 3 3 34 35 36 3'/ 38

Straight

ΕΙbow

Duct

10.

Ο

9.0 7.0 10.0

branch

15 16 11 18 L9 20

Ο

coΙττnon

Strai ght Duct Rectangular Transition

Diffuser,/ Grltle Sιraight Duct

EΙbow

Straight Ducι Rectanqular Transition Diffuser / Gri1Ιe EΙbow Straight Ducc Rectangular Transition Diffuser,/ crill_e Εlbow

SιraΙqhι ι]uct Rectaigular Transition Diffuser,/ Grille

E]_bοw

Stralght Ducι Butterfly Damper Straight Duct RectanguΙar Transition Diffuser / Gri1le

9.Ο

1.0 1.0 9.0 1.0 1.0 1.4 1.0 1.4 1.0 7.0 1.0 1.0 7.0 7.0 8.0 8.0 B.Ο

9.0 9.0 9.0 oΛ 9.Ο

250 .0

845.0

ξoζ

n

395.0 200.0 595.0 395.0

215 .0

12A.A 395.Ο

215.0 150.0

L25 .0

15Ο.0 15Ο.0

150.0 150.0

0.0

190 .4 190 .4

190.4 190 .4

190.4 56η

q

1'Δ

573.0' -151 .6 '1 'Δ '\ G22 449 .0

124 .2 622 .5

561.3 46'7 .'7

561.3 561.3 561.3 450.0

150. Ο 125 .0 L25 .0

46'7

12η

375.0

1atr Λ rZJ . υ T

rξ

n n

L20 .0

72A.A 720 .0

120.0 200.0 20Ο.

Ο

240.A 200.0 250.0 250.0 250.0 250 .0

250.

Ο

.1

461 .1 Δ61 1

449.A

449 .0

36Ο.0

573.0 573.0 600.0

565.9 565.9 ξ6η

q

565.9 750.0

0.000 0.008 0.003

0.06415

0.005

0. Ο6415

0.006

Ο. Ο6415

0.0Ο6

Ο.006

0.018 0.004 0.017

0.003 0.016 0.

.01 L61

ΟΟ3

0.011 0.006 0.005 0.016 0.004 0.040 0.013

Ο.0Ο2

0.002 0.002 Ο.030 0.002 0.011 0 .002 Ο.036 0.003 0.010 0.000 0.04Ο

0. ΟΟ3 Ο. Ο13 0

.024

0.001 0.001 0.

0.08082 0.0BΟ82

0.05817 0.05817

0.05405

0.07106

0.06Ο04

0.06004

Ο5Ο

Balanced Capacity Method Note that dampers have been inserted in ducts 6 and 7 (No. 31 and 36 below) to cause an increase in duct diameter and a consequent decrease in velocity.

-- Design Procedure -System type: Supply Duct Sizing Method: Balanced Capacity Rounding Method: Round Nearest

Exοerpts frοm this work may be reproduοed by instruοtors for distribution οn a not-for-profit basis for testing or instructional purposes only to students enτoΙΙed in courses for which the teΧtbook has been adopted. Αny οther reproducιion or trαnlslαιion of ιhis ινork beyond ιhαι permiιιed by Secιions ] 07 or Ι 08 of ιhe 1 976 United Stαtes Cοpyrighι Αcι νιιhout ιhe permission of the cοpyrighι owner is unlα:lνful.

261

_- Εan Selection -Known Γan

Parameter: Fan Tota1 Pressure :

Εan Airflow: Fan or Εxternal Total Pressure:

Coil Lost Pressure: Εi1ter Lost Pressure: Misc. Lost Pressure:

ΑΗU AΗU AΗU

Externa1 Total Ρressure:

0.9OO in.

845.0 cfm

0.900 Ο . 500 0.100 0.Ο50

in. in. in. in.

wg wg wg wα

0.250 in.

wg

wg

Pressure for Supply System: 0.150 in. wg _ or 60.O Pressure for Return System: Ο.100 in. wg - or 4O.O

? %

_- Lost Pressure from Αir Ηandling Unit to Diffuser -Diffuser ΙD Q Total De]ta P (cfm) (in. wg) 19 24 28 33 38 Total

150.0 725.0 720.0 200. Ο 2s0.0

0.141 0.728 0.140 0.140 0.131

845.0

-- Calculated Fittlng Values -ΙD Εitting Type 1 2 3 4 5 6 7 8

Α1r ΗandΙing Unit Conical Contraction

Dia. (in)

Ο Ο Straight Duct 14.0 Elbow 14 . 0 straight Duct 14.0 Ειbow 14.0 Straight Duct 14.ο Tee / wye main 72.A branch 10.0 coΙnmon 14.0 9 Stra1ght Duct 72.0 10 Tee / Wye main 10.0 branch 8.0 coΙτunon ι2.0 11 Straight Duct 10.0 12 'Tοο / τιι',o main 9.0 branch 6. O coπ]Ιnon 10.0 13 Stralqht Duct 9. Ο \4 τee / Wye main 1 .0 branch '7 .0 co]nmοn 9.Ο 15 Straight Duct 1 .a 0. 14.

Q VeΙocity (cfm) (ftlmin) O O O 0 845.0 B45.O 845.0 595. O 250.0 845.Ο 595.Ο 395. O 20Ο.0 595.0 395. Ο 21 5.O L2a.O 395.0 2']5.O 15O. Ο L25.0 2'75.0 150.0 B45. B45. 845. 845.

De1ta P

(in. wg)

O. O 19O.4 '79a.4 '7 90 .4 i90.4 190.4 19O.4 j5j .6 458.4

Ο.020

151 .6 j24.2 573.0

O. OO4 O. O17

190.4

151 .6

124.2 622.5 6Ιl.2

'724.2

622.5 561.3 46i .i 622.5

561.3

O. OOO Ο. OΟ8 O. OO3 O.

OO6

(in. ^P/L wg) O. Ο6415

o.oo5

0.06415

O.OO6

O.O6415

O.OO6

O. OO5

O.ΟΟ7

0.0'716'7

OO7

O. O8259

O.

O.OO3

O. O13 O. OO7 O. OO3

O. O713B

OO6

Ο. O8Οs2

0.011 O.

Εxceφts from this work may be reproduced by instructors for distribution on a nοt-for_profit basis for testing or instruοtional purposes only to students enrolled in courses for which the textbook has been adopted. Αny other reprοducιion or ιrαnsιαιion of this work beyοnd ιhαι permiιted by Sectiοns Ι 07 or Ι a8 of the ] 976 Uniιed Stαιes Copyright Αcι wιthout ιhe peιmission οf the copyrighι oνner is unΙαιvful.

262

j 1-6 Elbow 1-1 Stralght Duct i .0 1B Rectangular Transj-tion i.0

19 20 2ι 22 23 24 25 26 21 28 29 30 31 32 33 34 35 36 3-l 38

Diffuser / Grilte Straiqht Duct Ε1bow Straight Duct Rectangular Transition Diffuser / eriΙle El-bow Straight Duct Rectangular Transition Diffuser / Grill_e Elbow Butterfly Damper Straight Duct Rectangu1ar Transitiοn Diffuser / Gri11e E1bow Butterfly Damper Straight Duct

Reclangular Transit1on

Diffuser,/ critle

j.A

1.O

1.0

i.0 6. 6.

0

0

6.0 8.0 8.0 8.0 8.

Ο

1Ο.0 10. Ο

10.0 1Ο.0

.0

15ο.0 150.0 150.0 150.0 12η n 125.0 1t\ cl 725 .0

125.0 120 .0 120 .0

120.0 120 .0 200 .0 200. Ο 200 .0 200.0 200.0 250.0 250.0 250.A 250. Ο 250 .0

561.3 561.3 450.0

0.016

0.08082

461 .'7 461 .'7 461 ."7

0.013 0.002 0.002

0.05817

375.

Ο

677 .2

1 2 36Ο.0 G1

0.0ο5

0.004 0.040

0.002 Ο.030

0.004

0.023

0.1\42-7

0.010

0.07106

573.0 573.0

0.007 0.036 0.003 0 .024

600.0

0.00Ο

ξ??

n

458 .4

0.040 0

.002

458 .4

0.0L6

750.

0.001 0.050

458.4

Ο

Ο.05817

0.008

0.03599

Exοeφts frοm this work may be reproduced by instructors for distribution on a not-for_profit basis for testing or instructional puφoses only to students enrolled in courses for which the textbook has been MoptΘd. Αny οther reproduction or ιrαislαιion οf ιhN wλri beyond ιhαι peΙmiιιed by Secιiοns ] 07 or Ι 08 of the 1976 United Stαtes Cοpyrighι Αcι νiιhοut ιhe peimission of the cοpyrighι oνvλer is unlαννfut.

Exceφts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instructional purposes only tο students enrolled in οourses for whiοh the textbook has been adopted. Αny other reproduction οr trαnslαtion of this wοrk beyond thαt permitted by Sections ]07 or Ι0B of the Ι976 United Stαtes Copyright Αct without the permission of the cοpyright o'ι)ner is unlαwful. Requests for permission or further infοrmαtion should be αddressed to the Permission Depαrtment, John Wiley & Sons, Ιnc, ] ] ] Riνer Street, ΙΙοboken, NJ 07030.

CHAPTER

13

13-1. From Ξq' 13-2

h-= tr* _

c-

Α (C*

)

=

r.,,.

lb* _ t2 (lο* , n,

Now C and W are related by

C

|% Wρ" ψ* = lba fto

=

Eq

)

=

ft'

ft2

_hr

13-14

lb*/ft3

Τhe density of dry air must be used. Then from Εq. ,1 3-17 hο =

hrP"' =

-ft3 X9 =

ftz _ hr--

ft3

]b, _hr

ft2

Consider Eq. 13-13 which is dimensionless

h

Btu

P3Cp,hη -

η-zι-

ftz _ hr _

ft3 lb,F

F |b, Btu ^-Ξ-^---;-

ft2

- hr

ft3

-

|

Clearly dimensionless when C* is used.

13-2.

using Eq. 13-'18,

h Le2t3 1; hα = = = -η- = ,,'9. = 41'7lba/(hr cpahα Cp, 0.24

also h, h6

: 0.057 kgal(m'-

13-3. hd k

= hα/ρ" = 41 .7lO'O75 = 555.6 ft3/1hr

=

0.61s p"oa7

s)

_ ft')

-

ftr)

ΚΞo

0,075x100x60x(1 t14_

=

oβ44

261

= g52

k = 0.0147 Btul(ft-hr-F) (Τable A-4a) n=

Ot.O,::!,

x 0.61 5(852)047 \ ' = 2.S9Btu/(hr-ft2-F) (1t12)

h6 = h/cpa

h,

13-4. Nu

=2.5910.24= iO.B lba/(ft2-h11

= h6/ρ" = 1O.8/0 .O75 = 144

=

0.023 Re0.8 pro 3 0r h

0.023 (k/D) Reo.s pro

=

and h6 = h/cpa, assuming Le = Re = ρ V olμ; νL=

V

= 600/

0.044lbmlft-hr; ρ

Re =

(il$

=

ff lσt'-hr) 3

'1

= 471 ft/min or 28,260 ftthr

0.075 (TabΙe A-4a)

0'075Ι?8'260x1 = 48,170 0.044

Pr = 0.7; k = 0.01 47 btulhr-ft-F (TabΙe A-4a) h = O.O23

ΨΨ

(48,17o)os(O.7)o., = 1.7 Btu/(hr-ft2-F)

1

hα = 1'710'24 =

7'1 lba/(ft2-hr)

h* = h6/ρ^=7'1l0'075 = 95 ft3/(ft'-hr)

13-5. 43,560

ft2 = 1

acre;

Γi'l

* = hdΑ(Wr, - W-)

Use j factor anaΙogy, h/crh6 = Le2l3

Αssume: Le

Thenho-

= 0.85; Cρ + 0'24 E

'

o.24(o.8q2t3

Using chart 1: WW*

=

=23.221bal(ft2-hr)

0.013 Ιbu/lba

- 0.0223lb"ilba

(assume sat. air at 80 F)

Excerpts from this work may be reproduced by instruοtors fbr distribution on a nοt_for-profit basis for testing or instructiοnal purposes οnly to students enroΙled in οourses Γor rνhiοh the textbook has been adopted. Αny oιher reproductiοn o, ιroλιotion of ιhis νοrk beyond ιhαι pemitιed by Sections ] 07 or Ι 08 of ιhe ] 976 Uniιed Sιαιes Copyrighι Αct y]iιhout ιhe permission of the copyright owner is unlανful.

__.Ξ_ 262 Γh, = (23'22)1ooo(43,56OXo.o223 fr,,, = 9,415,ooO lbr/hr

13-6. Use analogy of Eq.

Ξ

-

o.o1 3)

19 gpm/acre

13_1g

Ql= hαΑ(W- - Wr)irgi Q. = hΑ(t- - t*)

q = qr + Q., W,

ho= h

coLe2l

Q/Α

=

13-7.

=

W-

= O.O'1 10;

Chart

1

144Ε\- 9(1.15)=,- = 3 \ l' ν'il 024(σ82τπ =49'23lba/(ft2-hη |

49.23(0.o 11-0.00765)1o65 = 176 Btu/(hr _ ft')

Q./A = 9(1

q/Α

= 0.00765;

.1

5)(75-50) = 25g Btu/(hr-ft2)

435 Btui(hr _

ft2)

= 1'37 kWlm2

Qr= rh,(i*-i,,)

Γh, = hdΑ(WV, - W-) h6 = h/(cr"Lezu); cp, = hα =

W,

'1

=

0.24 Btul(lba _ F);

'510'24 = 6'25lbal(hr

0'0223lb,/lba

=

Q1

ft')

1

6.25(300 x 15O)(O. 0223

Γh, = 3,572lbWhr

1

W* = o.0096 lb,/Ιba

i- = 28.4 Btu/lba; Chart Γh,

-

Le2t3 =

-

0.0096)

= 3,572(1,05o) = 3,75O,600 Btu/hr oΓ

x 1,o99 kW

Αny water on the deck and occupants neglected.

13-8'

lt is assumed that the blanket is folded in half over the clothes line with one side exposed to air. ho

=

h

cp

Le-2t3

=

18.g7tbai(hr -ft ) #(0.g3)-2l3 =

Excerpts frοm this work may be reproduced by instructors for distributiol on a nοt-for-profit basis for testing or instructionaΙ puφoses only to students enrolled in οourses for which the iextboοk t-'u, υ""r, adopted. Αι-ιsl oιher Ιrj)a""rnn or ιrαnslαιion of rhis νork beyοnd thαι permitted by Secιions ] 07 or ] 08 οf the Ι 976 |Jniιecl Sιαιes copyrιgh't ιcι νιιnλut ιn" po7'iriοn οf the copyrighι oνner is unlανυful.

Γh*

=

hdΑ(Wb

Γhr, =

}

W,); Wn = 0.0312;W^= 0'0152

-

'ΔΘ=m*/[hdΑ(Wv_W")]

- 4)

(16

Δθ=

263

I 8.

87(56)(0. 03 1 2

Δθ = 42'6

min

-

0.01 52)

= 0.71 hr

Say 45 min.

13-9. The procedure is the same as example 13-1 except that the energy balance line A-B will have a positive slope and

tl

ι=75"F, tlz

Ans: 68162"F;

=

90"F

.4 ft2:4.8 ft

17

13-'10. The solutions to this problem closely follows example 13-2.

Ans: 77169"F; 17.4

Αns.

31126

C;

ft2;

4.8 ft

'1.6 m2; 3 m

13-11. The procedure is the Same aS example 13-2 exοept that the energy

balance line A-B will have a negative slope and the inlet and outlet water temps. are reversed.

Αns: 71t69"F;

17

'4 ft2;5.5

ft

50

13-

hαam =5510.24=229.2

45 Ε

_ο

h1/h6 = -3.05

40

=#,

J

οο

|ι

o-

30

Ι_

25

τ Εxοerpts from students enrοl]

permiιted by Sι

35

ιlJ

20

12.

A. =

4000 x

rh

r/G,

0.071

6O/1OOO = 17 ft2

'l I

-JI

t t2t

ι

Y 60

70

80

90

100

or instruοtional puφoses only to |αιion οf ιhis wοrk beyond thαt ghι οwner is unlαwful.

X

264 Υ

:

2'7, Then

L = Gry/hdθm = 10ο0 x2'71229'2 =

11 .8 ft

13-13. The solution to this problem closely follows example 13-3 13-14.

Ans: 1.4 to

1.5

13-15. Solution of this problem follows example '13-4 closely. '13-'16.

Αns:

5OO ft2; 12'2 ft

13-17. Extrapolate the 72 F wb curve in Fig. 13-9. The largest cooling tower modeI ''M'', iS not Ιarge enough to handle 2ο00 gpm.

Therefore use two towers of 1000 gpm each. Select the model "L" which ls rated at about 1100 gpm.

13-'18 See example 13-3; the cooΙing tower must be larger. 13-19. See example '13-3; the cooling tower must be larger.

13-20. (a) Model B or C using Fig. (b) Cooling Range

=

t

1

'13-9

-t. 2= 100-85 = 15 F

Exοerpts Γrom this work may be reproduοed by instructors for distribution on a not-for-profit basis fοr testing or instruοtional purposes only to students enrolled in courses for which the textbook has been adopted' Αny οther reprοduction or trαnsιαtion of ιhis νοrk beyοnd thαι peιmitted by Secιiοns ]07 or l 08 of ιhe ] 976 United Sιαιes Cοpyright Αcι the permission ofιhe copyrighι owner is unlαινful'

'yiιhouι

265

Αpproach = t. z-twol = 85 _ 76 = 9 F Tower capacity =

q

13-21 q

-

Q

= (200 x 60 x 8.33)(1)(15) = 1,499,400 Btu/hr

5OO qpm x Δt; qpm = = 50 '=??o,oô 500x1 0

50 qpm/ton= (250,000 /1

Note:

!n this

case,

'1

-^^. =3.0 5,000)

ton = 15,000 Btu/hr

Cold water temperature; t" From Fig. 13-7;

t*o =

= 70

-

10 = 60 F

42 F

13-22. Αlbuquerque, NM; tψ6 = 64 F (Τable B-1a) (a) From Fig. 13-7; cold water temperature gpm/ton

=

2'5i

73 F;

water temperature Ξ (73 + 10) 83 F

\Λ/arm

(b) Charleston, SC;

=

tψ6 =

79 F (Table B-1a)

From Fig. 13-7; cold water temperature = 84 F, gpm =

ΨΨ 15,000

x2'5= 83 gpm (a &

13-23' (a) tons = ''?9?99o 15,000 (b) gpm/ton

=

Ψ 80

=

80, gpm/ton = 24OΙ8O= 3.O, maximum

= 4'o',max. t,5

13-24. Model G, nominal rating

-

b)

=

two =

65 F

600 gpm & 250 tons (Table 13-2).

Using Figure 13-9; assume gpm is constant. Exοerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructiοnal puφoses only to students enrolled in coursοs for which the textbook has been adopted. Αny other reprοduction or ιrαnslαtion of this work beyond thαι permitted by Sections ] 07 οr Ι0B οf the ]976 Uniιed Stαtes Copyrighι Αcι'|ι)ithout ιhe permission of the cοpyright owner is unlrrwful.

72F

266

With cooting range of (97 Max.

two =

two =

85) = 12

76 F (Figure 13-9)

With cooling range = 15, Max.

-

tu,

= 100 F

74 F (Figure 13-9)

Exοeφts from this work may be reproduced by instruοtors for distribution on a not-for_profit basis for testing or instructional puφoses only to students enτolled in courses for whiοh the textbook has been adopted. Αny οther reproducιion or trαnslαιion of ιhis νork beyond thαt permitted by Sections ]07 or ] 0B οf ιhe Ι976 United Sιαtes Cop1ι'ight Αcι without the peιmission of the copyrighι owner is unΙττννful.

-------_-=\

---

basis for Excerpts from this work may be reprοduced by instructors for distribution on a not-for_profit textbοok has been testing or instruοtional puφbses orty to students enrοlled in courses for whiοh the Sections 107 or by permitted thαt work beyond this of adoptΞd. Αny other reprοiuction oi trαnslαtion is unlαwful' owner copyright the permission of the withοut ΙOti οf the lbτo υnιeλ Smtes Copyright Αct Depαrtment, Permission the to be αddressed Requests for permission or furtλer ffirmαtion should Join Wiley & Sons, Ιnc, ] Ι ] Riνer Street' Hoboken' ΝJ 07030'

CHAPTER 14

120-60 14-1. (a) P - 200 60

= 0.43

-

*=

200-180 120

- 60

= 0.33

F = 0.985 {Fig. 14-191 (1so - 6_q)_:(299_ 120) LMTD _ (180 - 60) ln

(200 -120)

LMTD

(b) C"

= 98.7oF

=

(fr

=

5490 Btu/hr-F

"r)",,

cn = cc(1,2

-

*uo

= 50oo

tu1)/(t*z

-

= '16,500 Btu/hr-F

(c)

Cn

=

53.35(520)

t*r) = 5490(1 20

φ'24)

-60y(200

-

180)

(rh cp)* = (Q pcp)* = 16,500 Btu/hr-F

=

16'500

Q=

(d) q

(29'92x0'491x144)

(6OlΧ1)

=2η5ft3/hr; Q

-

275(7.48) = 34 gpm 60

UAF(LMTD)

UA =

Cι-.,

(t*z--t*l) _ 16,500(2ο0

0.985(e8.7)

F(LMTD)

(e) NTU

=

NTU

=

UA Cmin

UΑ

Cc

3390 = 0.62 5490

_ 180)

UA Cair

=

3390 Btu/hr-F

^-

267

(f) \/

120-60 "= 200-60

0.43

268

14-2' (a)

q

UΑF(LMTD) = (rh cp),i,(1 10-50) Γh, = 4000 x 14'7 x 144l(53.35 x 510) = 311'2 lbΙmin or 18,672|b/hr Q = 18,672(0.24)(110-50) = 268,874 Btu/hr q = (rh cr,,)('l80 - tr.,o) = Q5 x 8.33)(1)(180 - tho)60

tι.o =

=

1

-

80

=?9y+

25x8.33x60

= 158.5 F or 159 F

p= 110-50 =0.46: *='180-159 =0.35

-50

180 F = 0.98; Fig. 14-1

LMrD=S# 't[

1

10

- 50

=88

^J

A = 9/(UF x LMΤD)' ='!u!!!ô== = 312 10x0.98x88

(b)

Cair = 18,672(0'24) = 4481 = Cmin Cwat = 25 x (60.817 .48) x 60 = 12,193 =

Cr*

,

= o.3T

=

110 180 -

NTU

5o

50

= 0.7,

=

0.401'

,μ=

Γh air

4481 crr, 12,193 Cmin

Fig. 1 4-18

UA/C,1. = O'7i

14-3. (a)

ft2

= 32οo x 60

x

0'7\!481 ,10

::'::'!!-

53.35x555

Cair=Cmin=3294 1ox3oo NTU = 0.91 3294 ε = 0.615, Fig. 14-18 at C.inlCrr, = 0

= 314

ft2

= 13,726lb/hr 125

T

refriο.

l

l

125

atr.

+ Exοerpts from this work may be reproduced by instructors for distributiοn on a nοt-for-profit basis for testing or instruοtional puφoses only to students enrolled in courses for whiοh the textbook has been adopted. Αny oιher reprοduction or trαnslαtiοn of this νork beyond ιhαι permitted bν Secιιons ] 07 or Ι 08 οf ιhe Ι 97 6 LΙniιed Sιαιes Cοp1τ'ighι Αct Ιψiιhouι the permissiοn of ιhe copyrighι oνner is unlcτwfuΙ.

r_.14-3.

269 (continued)

0.61s = ,':"=-'^? 125

113'5'F

t,o =

(b)

- 95

, tco = t"o = 0.6'1

s(125- gs)

Q = Cui,(tao - tai) - rh,.irg = 3294(11 3.5 3294(1 13.5 - 95) lrl,. = ------------.6s.s = 928 tb/hr

.

14-4' (a) ,= Γηl κy

-]

L

lm =

(1.0

-

+ g5

95)

90(0.00 8 l 12) J

L

- 0.5) 18.26= 0.76

12

R/r = 1l0.5 = 2.0i η = ο.8, Fig' 14-4

(b)

,.,

=

mrΦ

,_ '

taηh(mrl)

-

[E -,][.'+ O.35rnBj

\mrφ) \r )\ ,ο =

ξs =

Λ^

1+ A

1

14-4 the answers are the same

(1 - η); η = O.78 from proplem 14-4

1

Uo ho?ro J-

1.243

tanh(0.9454) =0.7g 0.9454

ξs = 1-0.9(1-0.78)

'1

=

(18.26)(0 '5112)1.243 = 0.9454

(c) Within readability of Fig.

14-5.

r/

1

200x(1/9)

=

0.80

Δx

k(Α, /Αo = 0.17;

)

1

1

hi(Αi /Αo

Assumes Α;

=

(0.01 5112)

) 10x0.8 Αo and

k.opp",.

(100x1) = 100

Exοerpts 1iom this work may be reproduced by instruοtors for dιstribution on a not-for-profit basis for testing or instructional puφoses only to students enrolled in courses for which the textbook has been adopted. Αny οther reprοducιion or ιrαnslαιiοn οf ιhis wοrk beyond thαι permiιιed by Secιions ]07 or Ι 08 of ιhe 1976 United Sιαιes Cοpyright Αcι withouι ιhe permissiοn οf ιhe cοpyright oνner is unlαlνful.

270

The second term may be neglected Uo = 5.9 Btu/(hr-ft2-F)

14-7.

'1

Uo=!10x0.8

1

100(1/9)

=0.133

Uo = 7.5 Btu/(hr-ft2-F)

14-8. ,' -

,

=

tanh(m/) m!.

z"sτ _1''' =64.18 v =| κy ] l za1o.'t l]

Γeη-l''' L

6x1 o_3

[

mI = 64.18(6

x 1O-')

= 0.385;

η

14-g' ns=1 -η)_1-o.s5(1 *(1

=

m-1

0.953

_O.95)

η. = 0.96

Ar =

2ΗLWP,

Where P"

=

mm2;

Α,

= LW mm2;

A

=

Ar + (LW _ tLWPS)

fin pitch in fins/in. and L = W =

1

Δ =- 2HLWP' + LW _ tP, _2x6x0'47 +1_0'16x'47 - '1l' lo+ '1AΛ' Αo ,ΗL\Λ/P' 2x6x'47

+=\:γ--J^ u 1400 57(0.96)

= O.O19;

U = 52.3 W(m' _ c)

Excerpts from this rνork may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instructional puφoses only to students enrolled in οourses for whiοh the textbook has been adopted. Αny οther reprοducιion or trαnsιαιiοn οf ιhis work beyond thαι permitted by Secιions ] 07 or l 0B of ιhe ] 976 United Stαtes Cοpyrighι Αcι lνiιhout ιhe permission of the cοpyright owner is unlαwful.

271 14-11

.

η=

tan h(m r/

)

'Φ= [}_,][,+O35.(})] R" = 1.2s ψ(β - 0'2)'''; m = r

Γ

mrΦ

znl1l2

Lπ]

*=},β=;;L>M (a) Diml

=;'=Ψ

=

0.56 in

b = 1.35 in

' rΓrg)' 2Lι2l*or1'''

Dim, =

l

=

jιιo.uu)'* (1'35)'1'''=

O.73

Then L = Dimz = 0.73in.; M = Dimr = 0.56 in.

0'56 =1'75:3=0'73 ψ=(0.64 t2) 0.56 R" r

=1.3

t2=2.22 = 1.27(1.Ts)(L3-0.3)1

φ = (2.22-- 1)t1 + O.35ln(

2.22)!= 1'56;

mrΦ

-

n_ '

tanh(0.762) =0.869 0.762

.

=

16.33(0.32112)1.56 = 0.631

Γ

2x10

L e0(0.01/ 12)

f'''

= 16.33

(b) Dim.-a-12.5mm '2

' !2'lZZt + 12.5'ltt2 = 12.65 mm

Dim2 =

L = Dimz = 12.65 mm M = Dimr = 12.5 mm

t=12'5 =2.s:g= R" r

5

b=22mm

12'65 12.5

= 1.27(2.s)(1.01 2

-

=1.012

0.g)1t2 = 2.69

Excerpts fiom this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instruοtional puφoses on1y to students enτolled in οourses for whiοh the textbook has been adopted' Αny οther reproducιιon or ιrαnsιαtion ofιhis lνοrk beyond thαι permitted by Sections ] 07 οr ] 08 οf ιhe Ι 97 6 United Sιαιes Copyrighι Αcι y,ithοuι the permission of ιhe copyright owner is unlωνful.

ft-1

272

-

1)t1 + 0.35 ln(2.69)] = 2.26 E'\vv _^ -_

2.69

Γ zxoε ll=66.67m-1

Φ =

m=l

1170(0.00018)_j mrQ = 66.67(0.005)2.26 = 0.753 _ tanh(O.753)

'

11 =

0.753 η = 0'85

14-12. + tube wall resistance v-_ -1-, h; (Α, 7no ;ι neslecting Uo= hoΠo _----{

-

(a) 1o=1

1

%

(b) ro

=

%

14-13. (a)

1o10€6

=1

1=

*(1 1

:^

0,.068ro s3

x

x

= 0.120; Uo =

οoo(l o)

1

Rct = 2.222

'12

*

-0.9(1 -0.81)

Rct = 4.15

(b)

) -η)-1_o.9(1 _O.s4)=0.86

*

=

0.83

^4^

1a.+rl o1

0 64( ι

10-6

8.60 Btu/(hr-ft2-F)

= 17.8; Uo

1t12

0.010-

= 0.056 kW(m2

- c)

1)'

ο.010

)

]""' 10-a 1hr-ft2-r1/εtu

fins/in = 0.472 fin/mm

rc(1t0 Rct=3.913x10-7

72_

ι 0.18 0.18

uott=

1)' )

1o

1

.oe3 x 1o-41m2

- cyw

l 14-14' Re= ρνD;ρ=60.6 μ

lbm/ft3

Excerpts frοm this work may be reproduοed by instruοtors fbr distribution on a not-for-profit basis for testing or instructional purposes only to students enτolled in οourses for which the textbook has been adopted. Αny οther reproductiοn οr ιrαnsιαιion ofthisνorkbeyond thαt peιmitted by Secιions ] 07 οr Ι 0B of the ] 976 Uniιed Sιαιes Copyι'ighι Αcι νψiιhοuι the Permissiοn of ιhe copyrighι oνner is unlανfut.

273

Table C-2

Dr = 0.545112 = 0.0454 ft;

V

=

x+

ι 2'5 e 'lA=

=3.44fVsec

7 '48^ 60(π t4)(o'o454)2

μ = o.93 lbm/(ft

Re =

-

hr) = 2'58 x 1O-a lbm/(ft-sec) Τable A-1a

60.6(3.44)0'0454 2.58x10-a

=

36,6g3i Re = 36,700

(L/D),in = 410'0454 = 88 ft Pr = 2'43 (Pr = crμ/k)

ιlo '',u k

h

14-15'

= O.ο23

= O.O23

κ"Bu Pro';

k = 0.383 Btu/(hr-ft-F)

,Ψf''='J. (36,7oo)o'12.431o' = 1,136

(0.0454)

ρ = 1'o1(62.4) = 63.02 LBM/FT2 [Fig.

μ= 0.7l1490

=

4'7 x

'1o-4

1

Btu/(hr-ft2-F1

0-2a]

lbm/ft-sec [Fig.

1O-2b]

Cp = 0.93 Btu/lbm-F [Fig. 14-ε];

K = 0.93 Btu/lbm-F [Fig. 14-9]

fVsec [ProbΙem M-1aJ; D = 0.0454 ft [Problem 14-14]

Υ

= 3.44

'045Δ Re=ffi=20,940 63.02(3.4 4)0

(L/D),1n = 88

ft

[Problem 14-14]

o _Cpβ _4'7x1o41sοoo)o.93 _ΕΕ.)

lr---ν''νia-

h

14-16'

k

ο.285

= O.O23 ,!o:',u=u], (2ο,94O)ou15'521o'=

(0.0454)

ρ = 1.O45

x 62.4 = 65'21

μ = 1 '3l1490 = 8'725 x co = 0.81

;k= 0.22; V

lbm/ft2

1O-a

690 Btu/(hr-ft2-F)

;

lbm/(ft-sec)

= 3.44 ft/seC,

D = 0.ο454 ft

Re=#=11,670 Exοerpts from this work may be reproduοed by instructors for distribution on a not-for-proΓit basis for testing or instruοtional puφosos onιy to students enτolled in courses for which the textbook has been adοpted. Any οιher reproduction or trαnsιαtion of this νork beyond thαt permitted by Secιions ] 07 οr 1 08 οf ιhe 1 976 United SιαιeS Copyrighι Αcι withοuι the permissiοn of ιhe copyrighι oινner is unlαwful.

274 ,,"

h

=

_ 0.81x8.725x1 0-43600 0.22

.1

= ..16

1 ,670)0.(.1 ..t6)0. 0.023 !:?3=). (i 0.0454 '

14-17. Use hydraulic dia. Dη = 4rn =

a(AJP)

=

209 Btu/(hr-ft2-F1

for rectangular channel =

4(3/8) = 1.5 in. = 0.125 tt

ρ = 62.4lbm/ft3 [Table A-1a]; μ = 3.45 lbm/(ft-hr) [Table A-1a] u':o!!_),Ψ Re = -'-?? = 32,556

(3.45l3600) cp = 1.003 Btu/lbm-F [Table a-1a] k = 0.338 Btu/(ft-hr-F); Pr = 3.45 x 1 .003/0.338 =

(a) For cooΙin9, h

= O.O23

(b) For h

14-18.

Dr, =

h = O.O23

9Ψ 0.125

!D

'10

R"o'Pro'

(32,556)0r11o.21o. = 5O9 Btu/(hr-ft2-F)

heating;

= O.O23

0.125

9Ψ 0.125

(32,556)0s11O.21o. = 642Btu/(hr-ft2-F)

ft [From problem

14-171

ρ = 62.4(1.o45) = 65'2lbm/ft3 [Fig. μ = 3.5/149O = 2.35 x

1o_3

1

O-2a]

lbm/ft-sec [Fig

'

10-2b]

Re= c,

=

0.89 btu/(lbm-F) [Fig. 14-8]

k = 0.28 Btu/(ft-hr-F) [Fig. 1a-g];

Pr

=

2.35 x 1 o-3136oο)(o.s9) l0 '28 = 26.9

Excerpts Γrom this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instruοtional puφoses οnly to students enrolled in courses for whiοh the textbook has been adopted. Αny other reproducιion or ιrαnsιαιion of ιhis work beyond ιhαι permitιed by Sections ] 07 or Ι 08 of ιhe 1 976 United Sιαιes Cοpyrighι Αcι y,ithouι ιhe permission of the cοpyright owner is unlανυful.

275

(a) Cooling h

=

0.023

(b) Heating: h

=

14-1g' (a) Re = ρΥD _

μ

Assume L/D

Φ

Then

k

('13,900)0t126.910t = 28s Btu/(hr-ft2-F)

#

=

π = 9 9?: 0.o12

285 '""

(26'9):1 (26'910'e

=

396 Btu/(hr-ft2-F)

99ο'2(1 '5)(0'012) _ ^''' =29'905 >

o'uffi

60

,

o.o23 Reo.8Pro4, Pr =

[o'sgοxlo3x+'lεzl

ι

(O.637X31 ,157)o s(3.91)o

h = 8287 W(m2

-

C)

=

6373

)

=

e.gl

4

8.29 kw(m',- C)

Data from Figures 10-2a, 10-2b, 14-8 and 14-9.

(b) Re =

(1

'ο28)999(1 '5)(0'012) 15,4OO = 1.2x10-3

.2x10-3x3.7x10-3 ,r=ffi=8.9 1

:o.o23 h = Ξ'Ξ]Ξ (ο.5o)(15,4oO)08(8.9)oo = 5140 W(m2 _ c) 0.012

= 5.1

14-20. (a) Re =

4 k\Νl(m2

- c)

62'4(0'5)(0'3!112) = 922< 2soo (3.45l3600)

hD .1.86tReP. D .,lls [ra)o 'o : Αssume ( n)o'o l.vvιl.-,, = τ L, ιr',,l,,1oΦLΙl,,ιλJ

=,

Εxοerpts from this work may be reproduοed by instructors for distribution on a nοt-for-prοfit basis for testing or instruοtional purposes only to students enτolled in courses Γor whiοh the textbook has been adopted. Αny oιher reproductiοn or trαnslαιion of this work beyond ιhαι permitted by Secιiοns ] 07 or Ι 08 of the ] 976 Uniιed Sιαιes Copyrighι Αct |νiιhout ιhe permission of ιhe cοpyright οlιner is unlcrlνfuΙ.

276

,,=ξ1# π

p22(1o.ol(ffi),"'

sff#

=

=1O'4 =

(b) Same procedure as part a using data for

66 Btu/(hr-ft2-F)

30o/o

ethylene

glycol from Figures 10-2a, 10-2b,14-8 and 14-9.

992.2(0.1.0)x103 = 1519 653 There is a question about the flow regime. lt is probably

14_21. Re

-

in the transition region. Assume it is laminar and use

7

Eq. 1 4-24 and assu me | -U-

10.14

I

= 1.

[ρ'J

Pr

-

o.653X'1ο_3(4.182) = 4'34 0.63

1 89φΞ3) τn=-O.O1 1519(4'34)rΨ']]1/3 ι'--_\'ι 3 )'

14-22. Use average values for

G.' =:o;(rh."9'v )"us=

Α.''

(rh,),us

-

Gu and

G.

0.912;(Gr)",s

-

328

W(m'- c)

and Eq. 14-26.

= ln't,,u z(o.sε9)'

=237.81bm/(ft2-hr)

4\12)

(1 + 0.1)12 = 0.55 lbm/hr

(Gu),us = 0.55/4" =

290'6 lbm/(ft2-hr)

DG, _ r0.589)r?ΞΞ)

= 12 3

lιt ι 12 ,ι 0.95 / DGu _r0 589)rΞ9ψ) = 15 0.95 12 / βι ι ,ι

H=

=

138(pr.,,.

[#j"'L?

E)"102

Εxcerpts from this work may be reprοduced by instruοtors for distribution on a not-fοr-prοfit basis for testing or instruοtional purposes οnΙy to students enrotled in οourses for whiοh the textbook has been adopted. Αny other reprοduction or trαnsιαιιon οf ιhis work beyond thαι permiιιed by Sections ] 07 or Ι 08 οf the ] 97 6 Uniιed Sιαtes Copyrighι Αct νilhout ιhe permission οf the copy'ighι oνner is unΙαινfuΙ.

277

pρ'(

e")"'

o )"'

.,u.o1, 61

=

βι lρ") Pr' _ 0.95(1.001)

= 1oo8

ι0.0135/

=2.48 0.384 i,, = 1001 Btu/lbm Δt ρ 80 = 1'160 _ s0); twall Ξ

6

0.384 -

13.8 - - ο.58 g

=

h

t12

QΑ8)

80'F (Using water outside 2 '4n' Γ 1φ]-l"u '' Uυδl' L,1

888 Btu/(hr-ft2-F)

=

14-23. Use average values of G. and A"= "4 ζ (O.o15)'= 1'767 x (G.

,(uη]

)"us =

(ο.s8)o. 126x1o_3 Ιz 1.767x10-a 0.126x10-3

(Gu)"ug =

1O-a

(1 + O.

1

Gu and

,,,oo8]o

"

Eq. 14-26

mt =

2) t 2

1.767x10-a

the tubes)

-

s)

= 0.399 kg/(m2

-

0.314 kg/(m2

s)

DGr _ ο 015(0.311) = 12'1

βι

0.390x10_"

DGu

βι

( ρr)"' |ο"

Pr.'

-

Δt =

45C

)

O.o15(0.39_9

=

O'39Ox1

Ι2 l1

L0'219J 9jΙ1' 19x1 03 = 2'46; O.39Ox1o_3

0.665

= (73

[

gzο

= 1024

ig = 2326kJ/kg

_ 28); liquid water assumed outside tubes

h=13gφj95)(2'46)1t3lffi].,u..o24)o2=5022W(m2-c) h

= 5.02 kw(m2

- c)

14-24' Use Εq. 14-28 At inlet x : 0.20; at outlet 10oF superheat p,ειJ*FsnRιε 1"Q,psi?i k,':. Ω if,'$-"t.nω puφoses only to from this wοrk may ,l"rΓj],' ξ,ο,Jhm/fu wοrk beyοnd ιhαι

R_22

1

Excerpts students enrolled in οourses'for which tbe textbook has been adopted. Αny oιher reproduction or trαnsιαιιon of this permιtted by Sections ]07 or Ι08 ofιhe 1976 United Sιαιes CοpyrightΑcι\υithout ιhe permission ofιhe copyright ονner is unlαννful.

278

Since X" Σ 1.O; Cl = 8'2 x 1O-3; n = O.4 Αssume tube wall thickness of 0.016 in. Τhen D; = 0.375 - 2(0.0161 = 0.343 in.

π-g|

Ai=

'4

= 6'417

x

1o-4 ft2

80 rh= A j 6.417x10-a

rJ = -:-

= 124,700 lbm/(ft2

-

hr)

= 0'52lbm/(ft-hr) at 30'F (sat. temp. at 70 psia) GD 124,700(0.343 I 12) = 6855 =

μ.

-

βι

0.52

30'F (sat. temp.); Table Α-3a

k = 0.056 Btu/(hr-ft-F) at i1g = 88'5 Btu/lbm

(0.o56)

h

=

8.2x 1O-3

h

=

779 btu/(hr-ft2-F)

(O.343,

Γrοεssl, (ττaφ'qaa'sβz'ιτ))lo 5(32l 7) r)L'ooccΓ ι J.]

o

14-25. Use Equation '1 4-28 R-22;G

ξ

= 2OO kg/(m2

-

s); Dr = 8.5 mm; L = 2 m; P,= 210 kPa

= 30%, Xe = 100%

Tsat =

-24C a|210 kPa abs. Pres. [Table A-3b]

μ. = O.27O x 1O-3 N-s/m2 [Table Α-3b]; extrapolate k. = 0.107 W(m - c) [Table A-3b] iβ= 223 kJ/kg [Table A-3b] GD=200(0.0085_)

βι

Cι

0

= 8.2

'270x10_3

x

6 = 8'2x

'1O-3;

1O-3

=6296

n = O.4

(o'1O7)Γ 0.0085 L'

h = 4106 W(m2 _ C)

=

β296\2

7(223)1 000

Ie 2x9.80 7

4.11 kννlm',_

. r0.

J]

c

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profιt basis for testing or instruοtional puφoses only to students enrolled in οoursοs for which the textbook has been adopted. Αny other reprοducιion or ιrαnsιαιion οf ιhis work beyond thαι permitlecl by Sectiοns ] 07 or Ι08 οf ιhe ] 976 tJnited Stαtes Cορyright Αct \νiιhout ιhe permissiοn οf the copyrighι oνner is unΙπwfuΙ.

14-26.

" f !!1 D2g

lr, =

279 ,

Assume isothermat

Re = 36,700; ProbΙem 14-14; smooth tubes f =0.022, Fig. Dr = lr,

14-27'

0.0454 ft;

=o.o2z*

" f !ζ, D2g

l ι. =

V

10-1;L=(6x6)+(5xl)=41

V

= 3.44 fUsec,

*lt!!l,t

-!1- 2x32.17 0.0454

ft

Problem 14-14

=3.65ft

Αssume isothermal

=O.5ft/sec; f =64/Re =641922=O.OOg

L = (10 x 10) + (9 x 1.5) = 113.5 ft; Di = 0.34 in. lι. = 0.06 n*11?^'?x12, -

0.34

14-28.

Ψ',

2x32.17

=

1'07 ft

Refer to Fig. 14-10 Load/circuit = 10 x 1 2,000110 = 1 2,000 Btu/hr Length/οircuit = (6 X 5) + (5 x 0.75) = 33.75 ft

(a) ΔP. /L = 0.10

ΔP.

= 0.10 x 33.75

(b) ΔP. /L = 0.04 CF

Gt,^

CF

=

1.25 Fig. 14-10 4.22 psi

x 1'25

=

x 1'25

= 1.7 psi

psi/ft

= 1.25

ΔP.

14-29.

psi/ft;

= 0.04 x 33.75

= 18OO lbm/(hr-ft'); t, =

7O"F,tz= 120"F

(a) Figure 14-12 Excerpts frοm this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instructional puφoses only to students enrοlΙed in courses fοr which the textbook has been adopted' Αny other reprοduction or trα'nsιαιioi of ιhis νork beyλni thαι permitted by Sectιons ] 07 or Ι 08 of the 1 976 Uniιed Sιαtes Copyrighι Αcι νiιhout ιhe pemission οf the copyrighι owner is unlαwfuΙ'

280

Re, =

G'Xo

μ

, Gc =

G"

lbm/(ft2-hr) σ - ΨΨ 0.56= 3214 ''_'-''\-

4.6 x 10-2 lbm/(ft-hr) at t = 95oF (Table A-4a)

ψ=

3214(1 .083 t12)

Re,

=

4.6x10-2

6306

j = 0.0091;f = 0.021 Fig. 14-12

I

=

jG" c,

=

8.91 Btu/(hr-ft2-F)

(b) Re6

=

Pr-2l3 = O.OO91 (3214)0.24(O.T)-2t3

6306 x 0.52511.083 = 3057; assumes expanded tubes

plus fin collars.

A 4xoxrα _ 4 1.25x1.083x0.56

At πD6D

π 0'0152x0 '525x12

JP

= (3057)-04(1O.OS)-015 =

j=

9.

= 1O.O8

0.0285 (Eq.

14-39)

1x1ο-3 Fig. 14-14; h = O.OO91(3214)(0.24)(0.7γzl!

= 8.9 Btu/(hr-ft2-F)

D*=

0.525x10 08 = 0.904 (1.25 - 0.525)

1+

116.7

Using Eq. 14-44.

" FP= (3o57)-0 2'[αoo+.1 (

0.525,0

I 1.25 - 0.525

t

4(

L-

\67

o o06) ) ]"

ι |

'zs _,,l_o

10.e04

0.173 f

=

4.2 x 1O-2 or

f

=

0.042 (Note that f may be in error up to

!35o/o (Figure 14-15)

Exοerpts lrom this work may be reproduοed by instructors for distributlon on a not-for-proflt basis for testing οr instructiοnal puφoses only to students enrolled in courses for whiοh the tΘxtbook has been adopted. Αny οιher reprοducιιon or ιrαnsιαιιon of ιhis work beyond thαι permitted by Sectιons ] 07 οr Ι 08 οf the 1 976 Uniιed Stαtes Copyrighι Αcι wiιhout ιhe permission of the copyrighι owner is unlαwful.

I

u

_

-

14-30. (a) G"

=

4.5 kg/(m'-.);

tp =

2OC; Re =

Gc(0'0275)

281 '

μ

μ = 1 8'2 x 10-6 Ν-s/m2 1Table A-4b]

Re =

4'5(0'027Ξ) 18.2x10-o

=

6800;

Cp = 1.OO5

kJ/kg-C

j = 0.009; f = 0.020 [Fig. 14-121

jG" copr-2t3 = o.oog0(4.5)1.005(0.7)-2t3

h

=

6

= 0.051

-

0.051 kJ/(m2-s-c)

kw(m2-c)

(b) See problem 14-29b for prcedure

14-31. Use Eq. 14-42 or 14-45

l,n -ρ, ' 'i=

eι_,l*'+4l

el-Γr,r *

o2{ 2gρ-ρ1L\,*" )lo''_' )_,

#!S#L 53.35(530)

η λ-,

= 0.O75 lbm/ft3; ρ2 = O.068 lbm/ft3

P,, = (ρl + ρ2)l2= 0'072lbmlft3

A _ αΥ

_147(1.083112)5 = 118'45

Aο σAf, Where V

= Αt.

(1)0.56

xL;

A1,"

= 1 ft2;L = 5 x'1.083/1 2 = 0'451

β214)2

* ιo.u6), )r9Ψ _ l)* ''=2(32'17)(0'072)(0.075)(3600)Ζ _Γ(, /\0.068 ) L\

lh

o 042(1 1s 5)(0 075)l

o

072 -l

h = 12.2ft of air

ΔPo =

Ω,

(1i::?)Α;'z

=

0 18 in. wg

Εxοerpts from this work may be reproduced by instruοtors for distribution οn a not_for-profit basis for testing or instruοtional puφoses only to students enrolled in courses Γor whiοh the textboοk has been adopted. Αny other reproducιion or ιrαnsιαιion οf ιhis wοrk beyond thαι permιtted by Sections ] 07 or Ι 08 of the ]976 United Stαtes Copyright Αcι νiιhout ιhe permissiοn οf the cοpyrighι owner is unΙανful.

282

14-32. Use Equation 1 4-42 or 14_45 ti = 10C; t2 = 30C, ffΑ" = 1 18.5 from problem 1O1x103

oι =

ffi1

pm =

1.203 kg/m3

t'

=

= 1.244 kg/m3; Pz= 1.161 kg/m3;

ffi[(,'. \

o.O19(1 18.5)/ I

r,

(0 56)2

)(#-

1).

1μ!1

1.203 )

= 1.67 m of air

ΔPo = 1

6,

(#lι',rool

or ΔPo = 1'67(1'244)9'807

14-g3. Re = G.Dr, μ

κe=lffi

2700(0.0101)

From Fig. 14-16;j =

G" co j

= 2'O8mm of wg. =

20.4 Pa

, Dι' = O.O101 ft

Αt 65.F, μ = 4.39 x

h

14-31

Pr-2t3

=

1O-2

Ιbm/(ft-hr) [Table A-4a]

=621 0.013, f = 0.053

= 2700(0.24)O.013(0.72)-2t3 =

Where co= 0'24 Btu/(lbm-F); Pr

14-34' Use Eq.

O.

S Btu/(h r-ft2-F)

Τable A-4a

14-33; tυ = 65 F

14.6x144

'o Ξ -ff 53.35(535) Pm =

= O.72 From

1

(ρl + ρ")l2

= =

0'074lbmlft3; ρ2= 0.077 lbmlft3

0.076 lbmift3

Εxοerpts from this work may be reproduced by instructors for distribution on a not-for_profit basis for testing or instructional puφoses only to students enrolΙed in οοurses fοr which the texδook has been adοpted. ,lny otlrr', iφr:oΙλ'-"ιiλn o, ιronrtotιon of ιhis νork beyoncl ιhαt permiιιed by SeCιionS ] 07 or Ι 08 of ιhe Ι 97 6 Uniιed Sιαιes Copyrιghι Αcι νιιnout inte p,eλλsion of the cοpyrighι oτι,ner is untcιwful.

28s

A 4L -ψI2 π=π= oοlοl

= 132

Assume a contraction ratio of 0'5 Then Κi -- 0 '32', Κ" = 0 '27 Fig ' 1 4-17 ΔPo

π +

2

_ _

e7oq2

2β2.17

(#-

ΔPo/Pg1

)1

1)

4'6(1 44)(0.07

-o

053(1

_ t(o.32 * l- os2)

4)(3600)'

3r)ιffi#] _ r

_ (o 5)2

_'''1#'\

=4'126x10-a

0'17 in' wg' ΔPo = 4'126x 1o-4114.6)(1 44)(12)t62'4 =

14-35. (a) Coil DescriPtion' Type of coil = Refrigerant condenser coil Tube pattern = Staggered plate-fin-tube Material

=

Αluminum fins with copper tubes

Refrigerant type = Refrigerant 134

Finned side fluid = air of Hg Finned side air pressure = 29'92 inches

Face area

-- 4'44 square feet

Height of heat exchanger = 20'0 inches Width of heat exchanger = 32'0 inches

Numberofrowsoftubesintheairflowdirection=4 Numberoftubesperro\Λ/=16circuitsontubeside=4 inches Fin pitch = 8 fins/inch Fin thickness = '006 Vertical tube spacing = 1'250 inches

puφoses only to basis for testing or instructional by instructors for distribution on a not-fbr-profit ιhis νork beyond ιhαι of ΕxcerDts from this work may be reproduced λny- o,rnr, ,rρroλur'ioλir-ιron,Ιαιιoi ,ι-'. ιexibook ιr, υ..nrjopi.α*' oνner is unlωυful' sιudenιs enτolled in courses ι", *-L1.λ ih" prr^ιr,rio' oj'n" copyrighι i,o*, ] 07 ''ιhout permiιιed by Sections

* ,oi'iiiλ)jili'bnii

ciiilii, iri

284

Horizontal tube spacing = 1'083 inches Tube outside diameter = .500 inches Tube wall thickness = .016 inches lnside tube fouling factor = 'OOOO BTU-HR-SQFT-F

'/6C

14-95. (continued)

Btu/hr Total heat transfer rate = -48783'2 -48783'2 Btu/hr Sensible heat transfer rate =

Entering air conditions: Dry bulb temPerature

Face velocitY

=

=

95'0 F

650'00 FPM

CFM Air volume flow rate = 2888'9

Leaving air conditions: F Dry bulb temPerature = 111'4

Tube side conditions: 125'0 F Refrigerant saturation temperature = of water Air pressure loss = '393 inches PSI Tube side Pressure loss = 1'07 Fin efficiency =

'3gg

Surface effectiveness = '824

Tubesideheattransfercoefficient=388.88tu/hr-SQFT-F Finnedsideheattransfercoefficient=12'5Btu/hr-SQFT-F Mean temperature difference

=

-20'7 F

(b) Yes

14-36. Coil DescriPtion' Type of coil

=

water or brine solution

coil Tube pattern = staggered plate-fin-tube tubes Material = aluminum fins with copper

Tube side fluid

=

water ι:;i:i'b:;,ff:;;i!:'ii:i1r,*"Ιir;?j':ii,:!;;iiii:i!jΙ:'i#ii!*'' permι

:i:.J:fJ:f;':T:J$E::',i:;i1:i1i:'*iJ$ιι'{ij:{..Ti i ii6 t'Jniιed Sιαιes Copyrighι Ι 07 ρermiιιed by Secιiοns "''iοi')i 'i

Αcι

w

ilhouι ιhe

286

Finned side fluid

=

air

Finned side air pressure = 29.92 iches of Hg

Face area

=

5.56 square feet

Height of heat exchanger = 20.0 inches Width of heat exchanger = 40.0 inches Number or

ro\Λ/S

of tubes in the air flow direction = 2

Number of tubes per Fin pitch

=7

ro\Μ =

16 Circuits on tube side

=

4

fins/inch Fin thickness Ξ .008 inches

Vertical tube spacing = 1.250 inches Ηorizontal tube spacing = 1.083 inches Tube outside diameter = .500 inches Τube wall thickness = .016 inches lnside tube fouling factor = .0000 Btu-hr-SQFT-F Diameter of inlet pipe/header = 1.0 inch(s)

Total heat transfer rate

=

-95759.1 Btu/hr

Sensible heat transfer rate

=

-95759.1 Btu/hr

Entering air conditions: Dry bulb temperature Ξ 7ο.ο F

Face velocity

=

650.00 FPM

Air volume flow rate

=

3611.1 CFM

Leaving air conditions: Dry bulb temperature = 94.6 F

Tube side conditions: Entering fluid temperature = 150.0 F

Leaving fluid temperature = 128.2 F Excerpts from this work may be reproduced by instructors for distribution on a not-for_prοfit basis for testing or instructional purposes only to 'work beyond thαι students enrolled in courses for whiοh the textbook has been adopted . Αny oιher reprοduction οr trαnsΙαtion of ιhis permitted by Sections l 07 οr Ι 08 of ιhe ] 976 t]niιed Stαιes Copyrighι Αct wιthouι ιhe permissiοn οf ιhe copyrighι oιυner is unΙαwful'

Τube side fΙuid velocity

=

287

4.00 FPS

Cooling or heating liquid flow rate

=

9.0 GPM

Air pressure loss = .187 inches of water Tube side head loss Fin effiοiency

= 8.20

- .831

feet or water

Surface effectiveness = .846

Tube side heat transfer coefficient = '1368.8 Btu/hr,SQFT-F Finned side heat transfer coefficient = 14.4 Btu/hr-SQFT-F Mean temperature difference

14-37. From problem

=

-56.8 F

14-29, Re = 6306 (based on xp)

and h5 = 8.9'1 btu/(hr-ft2-F) :jn _.1 = r -1280 Nr(Re)-1 2 (Eq. 14-42) jι

i

then Ψ _ 1_ 1280 x 5 (Re)-1'2

iι

=

i 1_(8x12so)(6306)_1 1 and Jξ_ , ^

js

1

now hε/hs

_ (5x1 2s0)(63o 6l_ι 'z = jε/js

1_ 6400 Re_1 '2

_

=

o g ιJ.ιr, 872

= 0'872

hε = 0'872(8.91) ha =

7.77 Btu/(hr-ft2-F)

14-38. From problem 14-30, Re jn

js-

1- 128oNr(Re)l 1- 64OORe-1

2

2

jο _ 1_(6x128ο)(680)] 1- 64OO(6801-r z

js

=

6800 and h5 = 51W(m'- C)

[From sotution

2

to

14-371

= 0.96

Excerpts from this work may be reproduοed by instructors for distribution on a nοt_for_profit basis for testing or instructional puφoses only to students enrolled in cοurses for which the textbook has been adopted. Αny οther reprοducιion or ιrαnslαιion of ιhis τνork beyond thαt permιιιed by Sections ]07 οr Ι08 ofthe ]976 United Stαtes CοpyrighιΑcιwithοuι the permιSsion ofιhe copyright o'ννner is unlαwful.

288

ιl ur rr s = jo/js = 0.96

- C)

hο = 0'96 x 51 = 49W(m2

14-39' Re, = ρ%xo

μ

or o.o49 kW(m2

rΨ

(1.os3/12) x 60 = 8225 = =o':1:= 0.0445 0.54

j = 0.0095 (Fig. 14-12)

ιl - jcco

- c)

ο.o73,

Pr-2l3 = O.OO85 x

ffi

x60x

o-24(o'η-2t3

h = 10.5 Btu/(hr-ft2-F) 0.0123 - 0.0092 .1 .033 x 1O-a,) Using Chart c.. = v ι_ _ 55

85

0.0123-0.0063 .^^ νz-- 85_q5 Cavg

=

= 1.5 x

1O-a

)

1

or

PSYC

Use Eq. 1 4-7o &14-73',k = 1zε*s}η;

1'27 x1O-a;

From Table 5-1a. M2 =

Ψr

##l, -Ψ]

=

1'27

ψ(β- o.3)"' =

'ιr [E'_ l)[l /ι

*= Φ

= (1 '265

MrΦ = 22.7

'

14-40' For

-

o.3)1t2

= 1'265

+ 0.35 ln(1.265)] = 0'287

xO'287

=

0'285

g7.4o/o = 0.g74 or

1 *

(1 -

80/67oF;

tοp =

=

(1

ft-1

r)

xΨ

mrΦ

#

M = 22'7

O.35ιn&l

_ 1)[1

tanh(mrl)

Πm-_--'ξ,s

+

= u, 3;

η,)

=

f _O.94(1 _ o'g74)=

O.98 or 98%

60oF

surface temperature must be equal to or less than 60"F' puφoses only to for distribution on a not-for-profit basis for testing or instruοtionaι Exοerpts from this work may be reproduced by instructors wοrk beyond thαι of this trαnsιαιiοn οr reproλucιion Αny οιher uJopt.α. bee, students enrolled in οourses for whiοh the texδook has νιιhout ihe permission of the cοpyright owner is unlα'wful' permitιed by Secιions Ι 07 * l οε'"iinr''i ii'6-Unirrd Srorω cipyr,ιghι 'aci

289 Moisture would condense at the base of the fin on the tube outer surface if it condenses at all' Let t* be this temPerature.

9

=

UiAi(t,-

tr) = h6η9A(i,

-

ir)

1 =-1' *,1*,Αr=Ai UιAi hiAι kΑ, :_- '9o]:^ ,nα ' ""- 1U,=!*41= hi k 1000 12x190

where --"-'_

= O.OO1

Ui = l OOO Btu/(hr-ft2-F); where k"opp", = 190 Btu/(ft-hr-F) t* = tr

.

ffi(i,-

i*) = 50 +

Αssume a value for tr, read

ffi

i* from

(31.7

-

i*)

chart 1 and compute t* to

check assumption' Assume t* = 55'8'F then i' = 23'7 Btu/lba and the calculated t* checks O.K. Therefore moisture will condense at the base of the fin and on some portion of the fin. There will probably be no condensation near the outer edge of the fin.

14-41. For 27Ι19 g,

tοο =

15 C

Solution is similar to problem 14-40 UiAi(t, _ tr) = h6η6(i3 _ i*)

1

=

1

+0'0005

ui 53 t*=ti

-

ffi

58

= O.O1888;

(i"_i*)

Uι= 52.98

=14'3-,g4,,* β4.2_i,)

*H

For t* = 16, i* = 45. Checks O.K. puφoses only to on a not-for-profit basis for testing or instruοtiοnal Εxcerpts fτorτ this work may be reproduοed by instruοtors for distribution reprοduction or ιrαnsιαιion οf ιhis νοrk beyond thαl Αny oιher adopted. has been the textiook which 1br in οourses students enrοlΙed ιhe copyright ονner is unlαwful' ihe permissiοn prr^ιt'rca υy srctiοns ] 07 or l οε ij rni l ozο LΙniιed Sιαtes cip)rιgnt 'ιci lνιthοuι

ξ

290

is There will be no condensation because the tube outside wall

greaterthanthedewpointtemperatureoftheair. 14-42" This problem is intended for computer solution because considerable iteration is required'

Coil DescriPtion: Type of coil

=

water or brine solution

Tube pattern = staggered plate-fin-tube coil Material = aluminum fins with copper tubes Tube side fluid

=

Finned side fluid

water =

air

Finned side air pressure = 29'92 inches of Hg

Face area = 12.50 square feet Height of heat exchanger = 30'0 inches Width of heat exchanger = 60'0 inches; W = 2H Number of rows of tubes in the air flow direction = 5

Number of tubes per

ro\Λ/ =

24

Circuits on tube side = '12

Finpitch=12fins/inchFinthickness=.008inches Vertical tube spacing = 1'250 inches Horizontal tube spacing = 1'083 inches Tube outside diameter = '500 inches Tube wall thickness = '016 inches

lnsidetubefoulingfactor=,OOO0Btu-hr-SQFT-F Diameter of inlet pipeihead er = 2'5 inch(s) or instructional purposes only to for distribution on a not-for_profit basis for tΘsting Εxοerpts from this work may be reproduced by instructors ιhis work beyond thαι of trαnsιαιion or iy oιη" has υ*,' students enrolled in courses fbιwhich the texibook "p'oλucιιon "J"pi"α. 'q οf the copyrighι owner is unlανful' permission ihe coiir|ιgn, S,o'r, 'a"i permitιed by Sectiοns ] 07 * οε'"iiλr''i ii'6-inι,"d 'ιthout

l

291

14.42. (continued) Total heat transfer rate = 232885.0 Btu/hr

Sensible heat transfer rate = 164919.4 Btu/hr Entering air conditions: Dry bulb temperature = 80.0 F Wet bulb temperature Ξ 68.0 F

Enthalpy = 32'3 Btu/LBMΑ Humidity ratio = 83.3 grains/LBMA

Face velocity

=

550.00 FPM


=

6875.0 CFM

Comment: coil is 34.3 percent dry Leaving air conditions: Dry bulb temperature = 57.4 F

Wet bulb temPerature = 57.1 F EnthalPY = 24'4 Btu/LBMA Ηumidity ratio = 68.7 Grains/LBMA


Leaving fluid temperature = 62.4 F Τube side fluid velocity

=

4.00 FPS

Cooling or heating liquid flow rate = 26.9 GPM

Air pressure loss = .774 inches of water Τube side head loss = 14.09 feet of water Fin efficiency = .gg9


Tube side heat transfer coefficient = 822.3 Btu/hr-SQFT-F Finned side heat transfer coefficient = 10.0 Btuihr-SQFT-F ExοeΙpts from this work may be reproduced by instructors 1br drstribution on a not-for-pro1lt basis for testing or instructional purposes only to students enrolled in courses fοr whiοh the textbook has bοen adopted. Αny oιher reproducιion or τrαnsιαtion of ιhis wοrk beyond thαι permitled by Sections ] 07 οr Ι 08 of ιhe ] 976 tJniιed Sιαtes Copyright Αct without ιhe permissiοn οf the cοpyrighι ονner is unΙανfuΙ.

292

14-43. Coil DescriPtion. Τype of coil = Direct expansion Tube pattern Material

=

=

Staggered circular-fin-tube coil

Aluminum fins with copper tubes

Refrigerant tYPe = refrigerant22

Finned side fluid

=

air

Finned side air pressure = 29.92 inches of Hg

Face area

= 10.31 square feet

Height of heat exchanger = 24.8 inches Width of heat exchanger = 60.0 inches Number of rows of tubes in the air flow direction = 4 Number of tubes per ro\Μ =

Fin pitch = 12

16

Fins/inch

Circuits on tube side = 16

Fin thickness Ξ .014 inches

Vertical tube spacing = 1.500 inches Horizontal tube spacing = 1.300 inches

Tube outside diameter = .625 inches Tube wall thickness = .022 inches lnside tube fouling factor = .0000 Btu-HR-SQFT-F Total heat transfer rate

=

241221.0 Btu/hr

Sensible heat transfer rate = 162201'8 Btu/hr Entering air conditions: Dry bulb temperatuΓe = 82.0 F purposΘS only to EΧceφts fiοm this work may be rοprοduced by instructors for drstribution on a not-for-profit basis for testing or instιuοtional of this νork beyοnd thαt students enrolled in courses for which the texibook has been adopted. Αny oιher reproducιion or trαnsιαtiοn oνner is unlανful' permitted by Secιiοns ] 07 or Ι 08 οf ιhe ] 976 (Ιnited Sιαιes Cοpyrιght Αct \υιιhouι the permissiοn of ιhe copyrighι

293 Wet bulb temPerature

=

67.0 F

Enthalpy = 31.4 Btu/LBMA Ηumidity ratio = 74'9 Grains/LBMΑ

Face velocity

=

500'00 FPM


=

5156.3 CFM

Comment: Coil is .0 Percent drY Leaving air conditions: Dry bulb temPerature = 52'2 F Wet bulb temperature Ξ 50.7 F

Enthalpy = 20.6 Btu/LBMA Humidity ratio = 52.1 Grains/LBMA

Tube side conditions. Refrigerant saturation temperature = 35.0 F

Air pressure loss = .623 inches of water Tube side pressure loss = 1.36 PSI Refrigerant quality entering/leaving evaporator = '29 Enthalpy change in evaporator = 62'75 Btu/LBM Fin efficiency =

.679


Tube side heat transfer coefficient

=

375.5 Btu/hr-sQFT-F

Finned side heat transfer coefficient = 9.3 Btu/hr-SQFΤ-F

14-44. Coil DescriPtion: TYPe of coil = Steam

Tube pattern Material

=

=

Triangular plate-fin-tube coil

Aluminum fins with copper tubes

Τube side fluid = Steam Finned side fluid

=

Air

Finned side air pressure = 29'92 inches of Hg basis for testing or instructional purposes only to Exceφts frοm this work may be reproduced by instructors for distτibution on a not-for-profit or trαnslαιiοfi ofthiswοrkbeyοnd ιhαt reprοducιion Αny οther Students enroιled in courses for whiοh the texibook has been adopted. ihe permission of ιhe copyrighι owner is unlιrννful' permitted by Sectiοns Ι 07 οr l οε iiin, ] 9t76 (Jnitecl Stαtes Copjrιght ,ιci τιithout

294

Face area = 12.00 square feet Height of heat exchanger = 24'0 inches Width of heat exchanger = 72'0 inches Number of rows of tubes in the air flow direction = 2 Circuits on tube side = 16 Number of tubes per ro\M =

16

Fin thickness = .006 inches

Finpitch=8Fins/inch

Vertical tube spacing = 1.500 inches Ηorizontal tube spacing = 1'299 inches

Τube outside diameter = '625 inches Tube wall thickness = .0'18 inches lnside tube fouling factor = .0000 Btu-hr-SQFT-F Total heat transfer rate

=

-554503'6 Btu/hr


=

-5545ο3'6 Btu/hr

Entering air conditions: Dry bulb temperature Ξ 60.0 F

Face velocitY

=

750'00 FPM

Αir volume flow rate

=

9000.0 cFM

Leaving air conditions: Dry bulb temperature Ξ 116'1 F

Tube side conditions: Steam temPerature

= 227

'1 F

Steam saturation pressure

=

5'000 PSIG

Air pressure loss = '269 inches of water Fin efficiency = .738 surface effectiveness = '756 purposes only to on a not-for_profit basis for tοsting or instruοtional Exceφts from this work may be reproduοed by instruοtors 1br distribution τυork beyond thαι οf ιhis ιrαnslαtion or reprολucιion Αny οιher students enrolled in οourses tbr w*ch the texibook has b..n uJo|t"d. ονner is unΙανful' isizο LΙniteιl Sιαιes cοpyrιglιt ,ιci νιthout the permission of ιhe copyright permitted by Sections ] 07 οr l οε^"i

'n,

295

14-44. (continued) Τube side heat transfer coefficient = 2051.7 Btu/hr-SQFT-F Finned side heat transfer coefficient = 14.6 Btu/hr-SQFT-F

14-45. Coil DescriPtion: Type of coil

=

Water or brine solution

Tube pattern = Staggered plate-fin-tube coil Material = Aluminum fins with copper tubes

Τubesidefluid=3Oo/oethyleneglycolsolution Finned side fluid

=

Αir

Finned side air pressure = 29'92 inches of Hg

Face area

= 5.56

square feet

Height of heat exchanger = 20'0 inches Width of heat exchanger = 40'0 inches Number of rows of tubes in the air flow direction = 2 Circuits on tube side = 4 Number of tubes per ro\Λ/ =

16

Fin pitch

=

7

Fins/inch

Fin thickness = .008 inches

Vertical tube spacing = 1'250 inches Horizontal tube spacing = 1'083 inches Tube outside diameter = .500 inches Tube wall thickness = .016 inches

Insidetubefoulingfactor=.OOO0Btu-hr-SQFT-F '1'0 inch(s) Diameter of inlet pipe/header =

Total heat transfer rate

=

-9ο610'1 Btu/hr


=

-90610'1 Btu/hr

purposes only to distribution on a not-for-profit basis for testing or instructional EXceφtS fiom this work may be reproduοed by instructors tbr beyond ιhαt νork of ιhis or ιrcιnsιαιion reproλucιion Αny other aJopteα. students enrolled in οourses fbr which the texibook has been copyrighι oνner is unlωυfuΙ' i szο (Jniιed SιαιeS cip)rιgΙιt 'ιci νιιhοuι ihe permission of ιhe permiιted by Secιions ] 07 or l οε^"i

'n,

296 Entering air conditions: Dry bulb temperature = 70.0 F

Face velocity

=


650.00 FPM =

3611.1 cFM

Leaving air conditions: Dry bulb temperature = 93.3 F

Tube side conditions: Entering fluΙd temperature = 150.O F

Leaving fluid temperature = 128.4 F Tube side fluid velocity = 4.00 FPS Cooling or heating Ιiquid flow rate

= 9.O

GPM

Αir pressure loss = '186 inches of water Τube side head loss ='10.13 feet of water FΙn efficiency

-

.83'1


Tube side heat transfer coefficient

= 796.O

Btu/hr-SQFτ-F

Finned side heat transfer coefficient = 14.4 Btu/hr-seFT-F Mean temperature difference = -57.6 F Τhere is a 5 percent reduction in capacity and increased pressure loss on the tube side.

14-46. Coil Description: Type of coil

=

Τube pattern Material

=

Water or brine solution

=

Αluminum fins with copper tubes

Tube side fluid Exc€φts

Staggered plate-fin-tube coil = 30oλ

ethylene glycol solution

from this work may be reproduced by instruοtors for distribution on a nοt-for-profit basis Γor testing or instruοtional purposes only to

studentsenroΙIedincoursesfοrwhiοhthetextbookhasbeenadopted.

Αnyοtherreprολucιionοrιrαnsιαtioiofιhislυorkbeyλnithαι

permitιed by Sectiοns ] 07 or Ι 08 of ιhe 1 976 United Sιαιes Cοpyrighι Αct

ιυ

ithοuι ihe permission of ιhe copyright olνner is unlανful.

Finned side fluid

=

297

Air

Finned side air pressure = 29.92 inches of Hg

Face area = 12.50 square feet Height of heat exchanger = 30.0 inches Width of heat exchanger = 60.0 inches Number of rows of tubes in the air flow direction = 5 Number of tubes per

Fin pitch = 12

ro\Λ/ =

24

Fins/inch

Circuits on tube side = 12

Fin thickness Ξ .008 inches

Vertical tube spacing = 1.250 inches Horizontal tube spacing = 1.083 inches Τube outside diameter = .500 inches Τube wall thickness = .016 inches lnside tube fouling factor = .0000 Btu-hr-SQFT-F Diameter of inΙet pipe/header = 2'5 inch(s)

Total heat transfer rate = 211374.2 Btu/hr

Sensible heat transfer rate = 155955.9 Btu/hr Entering air conditions: Dry bulb temperature Ξ 80.ο F

Wet bulb temperature Ξ 68.0 F EnthalPY = 32'3 Btu/LBMA Ηumidity ratio = 83.3 Grains/LBMA

Face velocity

=

550.00 FPM


=

6875.0 cFM

Comment: Coil is 43.6 percent dry Excerpts from this rγοrk may be reproduced by instructors for distribution on a not-for-pro1'it basis for testing or instructional purposes only to students enτolled in courses for whiοh the textbook has been adopted. Αny oιher reproductiοn of trαnsι.]tiοn ofthis τιork beyond ιhαt peιmiιιed by Secιions ] 07 or Ι 08 οf ιhe l 976 Uniιed Sιαιes Copyrighι Αct |νiιhout ιhe permission of the copyrighι oνner is unlαwfuΙ.

298

14-46. (continued) Leaving air conditions: Dry bulb temperature Ξ 58.6 F Wet bulb temperature Ξ 58.2 F

Enthalpy = 25'1 Btu/LBMA Humidity ratio = 71.4 Grains/LBMΑ


Leaving fluid temPerature Τube side fluid velocity

=

=

62.0 F

4'00 FPS

Cooling or heating liquid flow rate = 26'9 GPM

Αir pressure loss

= .756

inches of water

Tube side head loss = 18.13 feet of water Fin efficiency =

.699


Tube side heat transfer coefficient = 476.4 Btu/hr-sQFT-F Finned side heat transfer coefficient = 10.0 Btu-hr-SQFT-F The capacity is reduced by about 9 percent, the pressure loss on the tube side is increased and the leaving air temperatures have increased by about 1 degree.

14-47. Check Examples 14-1 through 14-5

Coil Description: Type of Coil

=

Water or Brine Solution

Τube Pattern = Staggered Plate_Fin-Tube Coil Material

=

Αluminum Fins With Copper Tubes

basis for testing or instruοtional purpοses οnly to Excerpts 1iorτ this wοrk may be reproduced by instructors fbr distribution on a not-for-profit reproducιion or ιrαnsιαtion οf ιhis work beyond thαι students enroΙled in οourses for whiοh the texibook has been adopted. Αny oιher ]976 Uniιed Sιαιes Copyrighι ,ιci νithouι ihe permission of ιhe copyright oνner is unlιτwfuΙ' permitιed by Sectiοns ] 07 or l oε i7

'n,

Tube Side Fluid

=

Finned Side Fluid

Water =

Air

Finned Side Air Pressure Face Area

=

299

=

0. FT. of Εlevation

2.17 Square Feet

Ηeight of Heat Exchanger = 12.5 ]nches Width of Heat Exchanger = 25.0 linches Number of Rows of Tubes in the Αir Flow Direction = 5 Number of Tubes Per Row =

Fin PΙtch = 8

Fins/lnch

10

Circuits on Τube Side = 5

Fin Thickness = .006 lnches

Vertical Tube Spacing = 1.250lnches Horizontal Tube Spacing = 1.083 lnches

Tube Outside Diameter = .525 lnches Tube Wall Thickness = .015 lnches

lnside Tube Fouling Factor = .0000 ΗR-FTΛ2_F/Btu Diameter of Inlet Pipe/Ηeader = 1.3 lnch(s)

Total Heat Τransfer Rate = -133026.9 Btu/HR

Sensible Heat Transfer Rate = -133026.9 BtuiHR Entering Αir Conditions: Dry Bulb Τemperature = 50.0 F

Face Velocity = 950.00 FPM Αir Volume Flow Rate= 2061.6 cFM Leaving Air Conditions: Dry Bulb Temperature = 107.6 F

Tube Side Conditions: Entering Fluid Temperature = 150.0 F EXcerpts fiοm this work may be reproduced by instruοtοrs Γor drstribution on a not-for-profit basis for testing or instruοtional purpοses only to students enrolled in courses for whiοh the textboοk has been adopted. Αny οther reproducιιon or trαnsιαιion of this ιιork beyond thαι permitιed by Sections ] 07 οr Ι 08 of ιhe Ι 976 United Sιaιes Cοpyrighι Αct withοuι ιhe permission of ιhe copyright oνner is unlα:ινful.

300

Leaving Fluid Temperature = 128.3 F Τube Side FΙuid Velocity

=

4.00 FPS

Cooling or Heating Liquid Flow Rate = 12.5 GPM

Αir Pressure Loss = 1.13'1 lnches of Water Tube Side Head Loss Fin EfficieΠCy

=

= '750

6.77 Feet of Water

Surface Effectiveness = '771

Tube Side Ηeat Transfer Coefficient = '1 354.2 Btu/hr-SQFΤ-F Finned Side Ηeat Τransfer Coefficient = '19.5 Btu/hr-SQFΤ-F Mean Temperature Difference

=

-58.5 F

Τhe above results show that a 5 row coil would easily satisfy the specified requirements. Τhe manual calculation of the

examples are very conservative.

Exοerpts from this work may be reproduced by instruοtors for distribution on a not-for-profit basis for testing or instruοtional purposes only to students enrolled in courses for which the textbook has been adopted. Αny oιher reproducιion or trαnslαtion οf this wοrk beyond ιhαι permitted by Secιions ] 07 or Ι08 of the ] 976 Uniιed StαιeS Copyrighι Αct 1νiιhout ιhe permission of ιhe copyrighι owner is unlανful'

Exceφts from this wοrk may be reproduced by instruοtors for distribution on a not-for-proΓrt basis fοr testing or instruοtional purpδsο, on1y to students enrolled in courses for whiοh the textbook has been ]07 or ]08 adoptΞd. Αny οther repiodiction οr trαnslαtion of this work beyond thαt permitted by Sections is unlαwful. o7 ilιe lιzο (]nited Smtes Copyright Αct without the permission of the copyright owner iequests for permission or furtier-idormαtion should be αddressed to the Permission Depαrtment, Jοhn Wiley & Sons, Ιnc, 11Ι Riνer Street, Hoboken, ]\ΙJ 07030.

Chapter 15

:

COP

15-1

qe/\rv;

UseP-idiagram

_νV = i+_ig =119.5_105.5

= - 14.0 Btu

/

lbm

p

14

tlooF

Q"=il -iι=42'5-119'5 =

ooF

4

-77 Btu / lbm

ooF

g"=-Q"+\M=77-14 =

63 Btu / lbm

(a)

COP=63 114=4.5

(b)

CoP"".n.1 = 50ο

0r=

4.5 7.14

(c)

ψ= Q"

coP

=

then

(d)

=

l (570- 5οο)

0.63 or 63

=

7'14

%

10 (12,000) _26,667 Btu 4.5

HP

10'5

ton 10 = =

Qe = ΓhQe oΓ Γh =

1.O5 and

1ο.5

HP N

7.8 kW

Ψ =O.22 kW

10 (12,000)

63

lhr =

=

1905 lbm / hr = 0.24 kg /s

300

(e) W

(f)

15-2

=

PD

10.5 HP from (c) above = 7.8 kW

rhv.

=

W

=

_νv

-

=ig

_W / rir

:

398.4 + 2'5 / 0.ο5 = 448'4 kJ / kg

-i4:260'3 -

w

=

-i4

9e

:

i3

=

9c -\M

(b) COP

15-3

O.O1 m3 / s

i+_iο

Qc = i1

(

x

2.5 kW; rh = 0.05 kg / s

=

(a) 8c=il_iη, i+

1905_ι0'68) = 21'6 ft3 / min 60

398.4

448'4

=

-188'1 kJ / kg

- 448'4: -50

kJ / kg

- 188.1-50 : 138.1 kJ

/ kg

= 138.1 I 50 = 2.76

coP

)carnot =

(c)

iι

(d)

0r = 2.76 /

#*:3.86

= 448'4 kJ / kg from part (a)

3.86 = 0 '72 or 72

o/o

R- 134a; Energy Balanοe rh1i1 +

mηiι =m2i2+ rh5i5

is =

+iι _iz

Ps

i.l

= P+

= 44'94

+

,

rh1 =

fi2=

rh3 = rhη = rhs

101 '54 _ 37.98

!

:16.6 Psia

w=i6-is,So=Ss

-r0

: 108'5

Btu / Ibm

100 F sat. liquid

P1: 138.8 Pjsla 5

F

sat. vapor 80 F Excerptsfromthiswοrkmaybereproduοedbyinstructorsfordistributionona.not-for-profitbasisfortestingorinstructionalpurposesonlyto ,l -_...__^ll^r;^^^llfaAqfnru,lιiοhthetextbοοkhasbeenadonteιl. Αnνotherreυroducιionortrαnslαιiono{thiswοrkb?νοbdthηι^2v":1'^)

_

301 io

:'127 Btu / lbm lChart

\Ι/

:

3]

127 - 108.5 = 18.5 Btu / lbm

Qe = i+

_iz =

_i3 : iι

1o1 '54

_

37'98 = 63'6

15-3 (continued)

HP

^t ton rhge HP

ton

=

P

t9

(18.5) 778 (12,ooo) =1.37 (63.6) 60 (33,000)

Γ.

_ 0ν =| 1+ C

15-4

132

TT8

L

c^ rr..1l"'

\PoJl

I

n = k = 1.17; 1

v3 vb

ln= 0.855

r80

P./P6:180120:9 v3

v6= ?2.38 -1I:

(a)

nu

=|ι+

uuu] O.O3 - o.03 (9)o

,..

PD

rh PD

0.75 2.14

Πν

Γ,

=

2

o.9o [chart 4 tab le Α3a]

: ftua or 'v

(b)

P

]1ι1

=

PD

0u lb

v3

0.90

=

0.75

=

0.15

/ ft3

* o.15_ o.15 (91o'εss

0.90

Exοemtsfrοmthiqrνnrl-tra\,L--__-^j,'^^lL-.:__}_'.â^.^.--f,j_+-:!_,'+:^-_-___^}ε^--.nfi'+λ-"iafΛf

tροt;fσΛrinctrlotinnql

nllrnnsεsnnlvt^

rh /

PD =

0.15 t 2-14 = O'07 lb / ft3

(e\ fr, - tro = 1- o'07 = o.8o or 80 % \v" 0.35 rha (d) Power is directly proportional

to the mass flow rate

therefore, Power compares as in (c) above'

15-5

0ν =o'7o

4

-

cyl

Pι

Pι

=

3" bore, 4" stroke, 800 rPm

49]

psia (chart 3)

= 138.8

Πν =

2 qsoF

psia (chart 3)

frΥ2, / PD; Υ2a

=

1

'o4 ft3 / Ibm;

Table Α-2a @ 55 F / 52 psia _ Qιz = rh (i2 η)

a PD= GπlΦ() β00) = 52.36 1728

ft3 / min

iι __iι = 46 Btu / tbm iz = 112 Btu / lbm

(iz_,.,l 1ιz=ηPt" Υ2a

or 15-6

=

Ξ#f4

(.ε'46)

=

2g26 Btu / min

912=139,560 Btu/hr = 11'6tons

R-22, assume suPerheat = 20 F _.

Ξ--

__^fi+

L-.].

f^f tea1inσ nr instnlctiοnaι nurooses onlv to

303

Subcooling = 10 F

t.

=

130 F, te

=

(a)

45 F

3t

120 E

P 90

-3 \

P Φ

'15-6 (Continued)

ι|1

tl:130-10=120F

iι:iz:46 (b)

15-7

4\'d

35

Btu / lb

Q" = 144,000 Btu / hr (Fig. 15-7)

Wc

(c)

Ρ;.} b*

te

=

=

14.8

32.5

kW = 50,498

F

Btu / hr

(Fig. 1 5-τ); W

=

13.3 kW

Refer to Fig ' 15-7 cΑP = 133,000 Btu/hr te= 47 F

^

l1

.'jL''L:^-.^_-^^+f^r-nrnfithasisfοrtestinsorinstructional purposesonΙyto

304

Evapl a),( c) t-

Ξ φ

α)

o () o o (d

Ξ

ο-

-Ξ

(τ,

ζ)

ι_

Φ

Ξ

o

(

orator 15-8

b)

o(a)( c)

io

Tι

Design

Pointo

-o(

-c

Ξ Φ

o o o

13.7

Ξ Υ

;

=

co-

'δ

(ι)

ω

o α

Ξ

o(σ

o

Measured Operating Points Excerpts from this work may students enrolled in courses by Secιions ]07 οr ]08 οf ιhe 1976 United

ιhis work

Sι*eΝwff#{&ψeff&Etrβwib,gΦeruεgffi'"'^''n'''

-

only to ιhαι permiιιed

-305

Pe = 69

psia; te

=

30

F

(Τable A-4)

(a) Τhe condensing temperature is still about

'1

'15

F, but the

evaporating temperature is low, about 30 to 31 F.

1_*=0.36

(b)

(qα

(c)

lt appears that the evaporator is not loading the compressor.

_q) / aα

=

305

or 36 % low

Check for proper air fΙow over evaporator. Fan speed may be low or an obstruction exists.

15-9

Suction valve,

ΔP:2

Discharge valve, ΔP '10

psia

G

= 4 psia

F S.H. in intake man. and cyl.

Piston clearance=5oλ'

Γ1+C_.ι r"']"nl γ"

lzτ s

sοsia ,-Ι

sat R-22 νao'

450F

P=90.73psia

ι

ηu=|

1

rPoJ.] ]

uo

ExcΘrpts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to l students enrolΙed in cοtlrses fnr whinh tLΔ +__}L^^l'- L^^ L^^- ^j^_ι'

V3 = Vg at 45

vo

=

F;

vg = 0.604 ft3 / lbm (Τable Α-3a)

0.66 ft3 / lbm (Chart 4 at 55 F / 89 psia)

Pc=275+4= 279 psia;

Pb = 90.73

-2= 88.73 psia

n:1.16, C=0.05

(a) (b) ri'r

- ry, (PD) / v. :

(c)

16l /o οo+\

(Ξ+\1/1

L "_v'vι'ιs&rcJ

o" =| 1+ O.O5 _ o.O5

(ο.838) 20 / 0.604

ψ=!Ψ, W=*o",[[ *

=

1J9(ss.73) 0.16'

P.

]ι

:

.n-1 n

I Pυ)

.*

27.75|bmlmin N 0.2 kg /s I

-11 ]

l, 2'79 ,oi6

)l 'ο _.' '|(8εzs)

(144) (o .66) l (

1

0,466 ft - lbf

/lbm

0.80(33,ο00)

'ν

ou

I

x 31 kJ/kg

w _27.75(10,4666) _ '15-10 (a)

I 1

15-9 (cont inued) =

0'εeε

+

.,l

Γ

L

=|l+ O.04

_ o.o4

1.1.0

(Ψ

'u'

Note: Αn average value

HP x

)'''o] c>f

8.2 kW

ι*l

k assumed and 2 psi p res. loss

Excerpts from this work may be reproduοed by instruοtors for distribution on a not_for-profit basis for testing or instruοtional purposes only to ctrr.lan+.

^--îr-

i

307

assumed in suction header and valve. Tv = 0.90,

Γh =

(PD) ηu

lνz

β*p=zoo

19

irδ

o6o

= 9'4 (ο.90) l 0'74

t"

'55

\l

I I

60.oF

1.44 lbm / min

rh ='l

oor

tzι6. Γ

w

=

I

-

(53) 44 (o'77) \ /Lι53/ -_1!l [Ψ'] (1'4_1) \ /

vιl =

1'4_1

*

1

fr* 0m

=

9491 (11.44) / (0.9 x778)

QH=ψ+Qι=(155x60) +

3O,OOO

=

=94 91( ft _ lbη / lbm

1t55

Btu / min

I

= 39,3ooBtu/hr

15-10 (Continued)

or qH

=

655 Btu / min

W23 =i2

9Η = ia

(b)

-i3,

_i3i

iZ

=iZt

W = 133 +

iη = 9μ +iu =

j9491

778

_#+ 11.44

=125.2Btu / Ιbm

+125'2= 68 Btu / lbm

lteration is required P3 will decrease with the lighter load but Pz is (P3 / P2) t'n will be about the same

as part

also lower and

(a);

ν2 l ν6 will be about constant. Then Tv : Constant.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis -+,,f,^-+-.-.---rl

j

flοr testinρ οr l'nstnlctiοnaΙ nιlrnοses

onlv to

308 However , Υ2= O.85 and rh =

: (PD)ην l νe

=

(9.4) 0.87 / 0.85

9.62 lbm / min.

ΠoWW W=

(1

=

9491 (48)(1 03) : 53 (0.77) 778

14.78 Btu / tbm

4.78) (9.62) (601 = 8530 Btu/hr;

Qn = 8530 + 24,000 = 32,530 Btuihr

Which assumes Ps I Pz is constant and 2 psi pres. loss in the valve. ie = iz

t w :83.5

+

14'78 --94'3 Btu / lbm

iι =iz_ 9ιz = 111_24,000 / 9'62 (60) = 69'4 Btu / lbm

,,

=

#

(50) = 188

psia

18

Ps 41.9 .83.5 9 4.3 t

15-11

Reduced air flow reduces the load on the evaporator. Without suction pressure control the evaporator pressure will decrease until condensate will freeze and completely block the

evaporator air flow. Liquid refrigerant will return to the

purposes only to Εxοerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional

309

compressor and eventually cause the compressor suction valve to faiΙ. lnstall an evaporator pressure regulator set to maintain a

15-12

pressure such that the temperature of the evaporator surface will not decrease below the freezing point for water. lnstall a suction pressure regulator on the compressor inlet.

15-'13

Τhe regulator shouΙd be set to limit the suction pressure to a level compatible with the compressor capacity.

15-14 (a) Using chart 2 with the construction shown, the final temperature is 9ο F

(b)

L=!=0.18s 'v m - (ν 20

or i

18.5

15-15

o/o

vapor (Use chart 2)

tb ammonia/lb sol.

Using chart 2 X3

=

0.495; ts

=

125 F

28ooF

l8ooF

0

TzΞτ7 l6TT-sia 0.25

.495 χ+

Excerpts from this rπork may be reproduced by instructors for distribution on a not-for_profit basis for testing or instruοtionaΙ purposes only to students enrolled in courses fbr whiοh the textbook has been adopted. Αny other reproduction or trαnslαιiοn of this work beνond ιhαΙ nemilto) by Sectiοns ]07 or ]08 οfιhe !Q76 [Ιnifa'] Slnlo" r'^'.'-':'-ι'' ι ''

310

15-16

(coP)ma, = Ψ:fd: \ _ - /Ιlιaλ Tg(To _

Te=75

+

460

=

535

T^ = '9 -

ξ )'

R;

To

180

=]00

+

+

460

:

64OR;

460 = 560 R

(COP)ma x = 2.675

15-17

Refer to Chart 5 for saturated vapor at 10 mm hg.

Vapor must first be condensed to sat. liquid at Q" = irs, i'n =

'10 mm hg.

Table Α-1a;

1ο64.8 Btu/lbmu for 1 lbm of vapor oΓ 9. = 1064.8 Btu

at 50 F, P = 0.178 psia or 15-17 (continued)

P = 10 mm hg Locate point I at x Locate point s atx

= =

0; P

= 10 mm hg

0.6 ; P = 10 mm hg

ffru=1;m.=5

ms5mv mm6ms E

lv = ;5- tΠS = i ob

x 45'5

Ξ 37.g

(depends on scale used)

(a) x = 0.50 (b)

Q, =

i, -

io =

-50

-

(-70) = 20 Btu / lbm of solution

Exοerpts Γrom this work may be reproduced by instruοtors for distribution on a not-fοr-profit basis for testing or instruοtional purposes only to students enτolΙed in οourses for which the textbook has been adopted. Αny oιher reproducιion or ιrαnslαιion of ιhis νοrk beyond thαι pemιιιed by Sectiοns Ι07 or ]08 ofιhe ]976 tΙnited Sιοιρs Canιιriohι Δ?r1υ!a!"^''a n'^ ' '

311

Qtot

: 1064.8

+ (6

x 20) = 1 ,184.8 Btu

i

lm

ib

70δ

\m-1 h9

χ+

0.5

0.6

0.8

Excerpts from this work may be reproduοed by instruοtors for distribution on a not_for_profit basis foτ testing or instructional purposes only to students enτolled in οourses for whiοh the textbook has been adopted. Αny other reproduction or trαnslαtion of ιhis ιιork beyond thαι permitιed bν Sections ]07 or ]08 οfιhe ]976 [kιitad Stntρc Γnh1'';-L} ι^l''':'1''' ' ''

311

ADDENDUM to Solutions ManuaΙ fοr McQuiston, ΙΙVAC 6e ProbΙem 6-10 For the floor, it is unοlear what2 in. vertiοal edge insulation means (whether 2 in. is the thiοkness of insulation or the depth of the edge insulated). . The solution assumes that the insulation has R-value of 5.4 hr-ft2-oF/Btu and the depth of the edge is 2 ft. For the door, Table 5-8 in the 6th edition does not have U-value for the wood storm door and there are three types of the wood door with 1 % in. thickness. ο The solution assumes that the doors are panel doors with metal storm doοr; henοe, its U-value is 0.28 But/hr-ft'-'F.

ΡrobΙem 7-9 The standard time zone for ottawa, ontario is Εastern Standard Time instead of Central Standard Time. The solution uses Eastern Standard Time.

.

Prοblem 7-14 For the specified loοatiοn, the sunset oοοurs before 9:00 p.m. CDST on June 21. ο The solution uses 8:Ο0 p.m. CDST instοad of 9:00 p.m.

Prοblems 8-25 and 8-26 Both problems do not specifu the window orientation. . The solutions assume the west-facing window for both prοblems.

Tabte 8-20 Reοommended radiative and conveοtive fraοtions for solar heat gains should be revised sinοe the 6th edition uses the SHGC values in the calοulation of the (οombined) solar heat gain for the RΤS methοd.

Example 8-16 Τhe example actually uses 90%/10% of radiative/οonveοtive split of the cοmbined solar

heatgain. Ηowever,thetext (page270) says 100%/0Υoforthetransmittedsolarheatgain and 630Λ1370Λ for the absorbed sοlar heat gain.

L--**-

312

ProbΙems 8_25 and 8-26 The solutions for both problems use 90%11,0% for the combined solar heat gain.

ExampΙe

9_1

The οalοulation for this example should be

"

=

('1Ι11?2Ψoi19.']!o:9Φ _ Ι22'606 (0.ssx70- 0)(1000)

(Changing 13 to24 and 122790 to 122606).

McQuiston HVAC Analysis Design 6th Solutions

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