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REFRI DATION

AIR CON ~N IONING

3rd Edition

Hipolito B. Sta. Maria

{(dimu;i Book Store QUAD ALPHA CENTRUM BLDG. 125 Pioneer s... Mandaluyong City

.l...

Contents

!'//!I!ishcd b»

r(dimtd

Book Store

Preface

VB

Chapter

1

PHILIPPINE COPYRIGHT, /I.)X9, 2001 bv Hipolito B. si« Mario

2

Third Editioll, 2001

Refrigerating Machine and

Reversed Carnot Cycle,

.

Simple Vapor Compression

Refrigeration Systems

1

12

3

Actual Vapor Compression Cycle

4

Multipressure Systems

5

Cascade Systems

6

Air-Cycle Refrigeration

7

Steam-Jet Refrigeration

8

Air Conditioning

9

Cooling Tower................................................ 154

.

57

)

1

, ( I

All rights reserved. No part of this book may be

reproduced in anvjorm or by an\' meall.\', except brief quotations

review. without permission in writing [rom the Author.

{or

II

Printed II\'

Kaluvaun /'/(\\ .'Ilktg. Ell!.. 1.)6 Kalavaan St., l iilnnau

Q//I':OI/ ( '1/ v

""-.

10

Drying

11

Product Load

. .

90

.................................

98

.

107

.

113

,.

'" '"

78

'"

'"

166

II/C.

ISBN 971-08-6/72-7

181

-Preface

This book has a two-fold purpose: to serve primarily as a college text and as a reference material for the practicing engineer. The author attempts to present refrigeration and air conditioning in the simplest possible way without sacrificing quality. Principles and concepts are discussed thoroughly using mostly basic thermodynamics, followed by illustrative problems arranged in the order of increasing difficulty. Additional exercises are also provided at the end of each chapter to further enhance the learning process. Answers are given to all the problems so that the student himself can check the accuracy of his solutions. For the student to get the most from this book, he should solve all the problems at the end of each chapter conscientiously. This is the best way to study and learn any technical subject.

)

1,

The Author

....

~

--Chapter

1

Refrigerating Machine and Reversed Carnot Cycle SI units SI or the International System of Units is the purest form and an extension and refinement of the traditional metric system. )

1 l

Unit of Force The unit of force in the SI system is the newton, denoted by the symbol N. 1 kg f = 9.8066 N 1 lb f = 4.4484 N 1 lb, = 0.4536 kgf

Unit of Pressure The SI unit of pressure is the pascal and is denoted by the symbol Pa. 1 Pa

= 1 Nzm"

1 bar = 100 kPa 1 psi = 6895 Pa 1 atm =.101.325 kPa = 1.033 kg/em" = 14.696 lb zin"

........

2 Refrigeration and Air Conditioning

Refrigerating Machine and Reversed Carnot Cycle 3

Unit of Energy and Work

Heat Engine and Refrigerating Machine

The SI unit of work is the newton meter denoted by Nm or joule denoted by J.

A system operating in a cycle and producing a net quantity of work from a supply of heat is called a heat engine and is represented as shown in Fig. 1-1.

1 Nm = 1 J

1 cal

= 4.187 J

Hot reservoir (source of heat)

1 Btu = 252 cal = 1055 J ~

Unit of Power

@-+W

The SI uni t of power is the watt, denoted by the symbol W. It is defined as tho rate of doing 1 Nm of work per second. 1W

= 1 ,Us

1 hp

=

1~

0.746 kW

= 4~.4

~

Btu/min

Unit of Refrigerating Capacity

Cold reservoir (receiver of heat)

Fig. 1-1. Representation of a heat engine The standard unit of refrigeration is ton of refrigeration or simply ton denoted by the symbol TR.

W

=.

QA

One ton of refrigeration is the rate of cooling required to freeze one ton of water at 32°F into ice at 32°F in one day.

e

=

Q

In English units

where

1 TR = (2000 lb) (144 Btu/lb) 24 h 1 TR

= 12,000 Btu/h = 200 Btu/min

In Metric units

-

QR

W _ QA - QR QA A

QA

= the heat supplied

QR

= the heat rejected

W

= the net work

e

= the thermal efficiency

1 TR = (12,000 Btulh) (252 callBtu) 1 TR = 3024 kcallh

=.

50.4 kcallmin

In Sl units 1 Tl{

= (200

Btu/rnin.) (1055 J/Btu)

I TH = ~11 k.L'min = 3.52 kW

~

A refrigerating machine will either cool or maintain a body at a temperature below that of its surroundings. A refrigerating machine may be represented by the diagram shown in Fig. 1-2. The machine R absorbs heat QA from the cold body at temperature T 1 and rejects heat QR to the surroundings at temperature T 2 and, during the process, requires work W to be done on the system.

4

Refrigeration and Air Conditioning

Refrigerating Machine and Reversed Carnot Cycle ;)

T2

surroundings

The Carnot power cycle consists of the following reversible processes: process a-b: isothermal expansion, 1'" = T = 1'] b process b-e: isentropic expansion, SI> = S, process c-d: isothermal compression, T = T = T, c

~

ri

process d-a: isentropic compression, Sri = S, cold

body Fig. 1-2.

\{l"prl"~entation of

QA

cOP =

----w- =

where

QA

W = QA

QA

Q,,_.Q

,

= the heat absorbed from the cold body or

The performance of a heat engine is expressed by its thermal efficiency. The performance of a refrigerating m achin e is expressed by the ratio of useful result to work, called energy ratio or coefficient of performance (COPl.

T

e

e

_ 1'] ~ T.

T

]

where

Q A = the heat added by the hot body

1']

= the heat rejected to the cold body

c y

s

('arnot Cyril'

= the thermal efficiency

= the temperature of the Source of heat

T 2 = the temperature of the sink or receiver of heat

d

I·:l

,"I

(1'1 - 1',,) ISh - S) r 1 (SSa )

'b

= W

Q

e

'CJ~ l'li~

:!'!,

1

W = the work done

a

....

= (1' _. T ) (S - S )

QR

The Carnot Cycle p'

QR

-

= 1'\ (S" - S) - 1'2 (SI> - S)

!\

refrigeration produced QR = the heat rojected to the surroundings W = the work done or mechanical energy consumed COP = the coefficient of performance

1

.s, - Sa)

= 1'"

1\

)

.s, - S)

1'1

QR = 1'2 (S,--S)

a refrigerating machine

W == Qi( -- Q\

Q\

=:

6

Refngeration and Air Conditioning


The Reversed Carnot Cycle

where

T

p c

T=:~r a

F'i~.

s /{"v('r'sed Carnot Cycle

I ·1.

= the heat absorbed from

QR

:::

W

= work done

the cold body

the heat rejected to the hot body

COP= the coefficient of performance

d

v

QA

The reversed Carnot cvcl« follows the same processes as in the power producing Carnot cycle, but the cycle operates in the counterclockwise or reverse direction. The reversed Carnot cycle consists of the followi ng pn)('('ss('s:

T[

= the refrigeration temperature

T2

= the temperature of heat rejection to the surroundings

(1-1) A refrigerating system operates on the reversed Carnot cycle. The higher temperature of the refrigerant in the system is 120°F and the lower is lOoF. The capacity is 20 tons. Neglect losses. Determine: (a) Coefficient of performance. (b) Heat rejected from the system in Btu/min. (c) Net work in Btu/min. (d) Horsepower,

SOLUTION

process a-b: isentropic compression, Sa == Sh T

process b-e: isothermal heat rejection, T h ::: T, ::: T 2 process c-d: isentropic expansion, Se = Sa process d-a: isothermal heat absorption, T d == T, == T, Q A =Tj(Sa- S) Q R == T 2

e

dJ

rs, - S)

~R

..

470 0 R

b

T'. = 10 + 460 = 470

,.

T,

QA::: 20 TR

==T2 (Sa - S a)

W = QR - QA

B

= T 2(Sa - Sri) - T[ ($a - So)

= (T

Til (Sa - Sdl

-

2

COP = Q A

:::

W

--..

COP

=

(T 2

T) T~ - T[

TI(Sa-Sa) - T I) (Sa- Sa)

= 120 + 460 :::

0R

580

0R


1;,1'11"'1;1111111 .u u] .\11' (1III1iJll.lllIllg

T I T! - T

( :I I

CO!' "

(b)

COP=~

=

470 ::i80 - 470

= 4.273

.1)

COP

W

=

COP

=

(:20) (200) 4,273

= 936

Btu/min.

QR = W + Q\ = 9:l6 + 4000 == 4936 Btu/min, (c)

W W

=

936 42.4

= :2:208 hp

(1-2) A refrigNation system operates on the reversed Carnot cycle. The minimum and maximum temperatures are minus 25 cC and plus 72 C. respectively. If the heat rejected at the condenser is 6000 k.l/rn in , draw the TS diagram and find, (a) power input required. n nd (b) tons of refrigeration developed. SOLUTION T

:Cjb

T] = -25 + 273 T,~ =

s

....

= 248

K

72 + 273 = 345 K

QR = 6000 kJ/min

a

~

-

248

345 - 248 = 2.557

j

T.

A = QW --~Q Q A

(~~

_ (COP) (QR) 1 + COP

W

=

{\ ~A

=- = 20 .44 211

= 936 Btu/rn in. , hi

(d)

T2

R

Q,\

W

T

COP

(2.557) (6000) = 4313kJ/min. 1 + 2.557

QR - QA = 6000 - 4313 = 1684 kJ/min. 4313

TR



PROBLEMS

(1-3) The power requirement of a Carnot refrigerator in maintaining a low temperature region at 238.9 K is 1.1 kW per ton. Find (a) COP, (b) Te;' and (c) the heat rejected.

(1) A reversed Carnot cycle is used for refrigeration and rejects 1,000 kW of heat at 340 K while receiving heat at 250 K. Determine (a) COP, (b) the power required, and (c) the refrigerating effect.

SOLUTION

Ans. (a) 2.778, (b) 264.7 kW (c) 735.3 kW

T

CQb

'. '1\

II!

T[ = 238.9 K

·238.9K

d

a

,/ (2) A reversed Carnot cycle has a refrigerating COP of 4. (a) What is the ratio Tmax/Tmin? (b) If the work input is 6 kW, what will be the maximum refrigerating effect, kJ/min and tons.

W = 1.1 kW/TR Ans. (a) 1.25,

QA

~ 1 T R or 3.52 kW

W

=1.1kW

Ans. 9230.7 kW

QA 3.52 kW =3.2 COP = W = 1.1 kW (b)

T[ COP = T - T 2

3.2

j

238.9 - T l-238.9

T 2 = 313.6K (c) Q n

~

= W + Q A = 1.1 + 3.52

6.82 TR

'(3) A reversed Carnot engine removes 40,000 kW from a heat sink. The temperature of the heat sink is 260 K and the temperature of the heat reservoir is 320 K. Determine the power required of the engine.

s

(a)

(b)

:=

4.62 kJ/s

I

/ Simple Vapor Compression Refrigeration System 13

Chapter

2

The pressure-enthalpy (ph) diagram and the temperature entropy (T's) diagram are the diagrams frequently used in refrigeration cycle analysis. Of the two, the ph diagram is the one most commonly used. The process which comprise the standard vapor-compression cycle are:

Simple Vapor Compression Refrigeration Systems

1-2, reversible and adiabatic compression from saturated vapor to the condenser pressure 2-3, reversible rejection of heat at constant pressure de superheating and condensation 3-4, irreversible expansion at constant enthalpy from saturated liquid to the evaporator pressure 4-1, reversible addition of heat at constant pressure in evaporation to saturated vapor

III

Refrigeration is that branch of science which deals with the process of reducing and maintaining the temperature of a space or material below the temperature of the surroundings. Refrigerant is the working substance in a refrigeration system.

, }

: I

l

Condenser

~

2 •

:=L

Analysis of the Vapor Compression Cycle 1.

Compressor

·.~w -1Jw-~

Expansion valve

J

~

w

2

ell

1:1

.::s

.S

~I

r.:l

oj

'"'"

.~~

~.1

;'"

Q,

l<

Evaporation

6'J$'

:.::

e::s

e8 8

~I

/

4

:2.

CondeI1i\er

Entropy, kJtq K

Fig 2-2. The p-h and T-s diagrams of standard vapor compression cycle

q, q,

=h =.

2 -

/Q

CIc

L / ' ~n,

".-{: Enthalpy, kJlkg

=

W m(h2 - hI) VI = mVI w ::: work of compression, kJ/kg h, ::: enthalpy entering the compressor, kJlkg h, ::: enthalpy leaving the compressor, kJlkg m ::: mass flow rate, kg/s W ::: theoretical compressor power, kJ/s or kW v l ::: specific volume of vapor, m 3/kg or Llkg VI::: volume flow rate, m 3/s or Lis

Fig. 2-1. Vapor compression refrigeration system

~

=h2-h i

h,

he at r'ejr-ct.ed , k.Jlkg

rl

--=

Qc ::: mrh 2-h) 3

c

~m


Simple Vapor Compression Refrigeration System 15

h,

=

h,

= enthalpy leaving the condenser, kJlkg

Qc

= total heat

3.

enthalpy entering the condenser, kJlkg

:). Coefficient of Performance (COP) The coefficient of performance is the ratio of the refrigerating «ffect to the work of compression.

rejected, kJ/s

Expansion valve

COp= h. - h, h, - h.

hI

(.;.

11

Efficiency ,

~

Y = ~ = D2LN 11 4

hg::14

, J

4.

h,

= enthalpy enu-ring the

h,

=

Y D = displacement rate of compressor D = diameter of cylinder (bore) L = length of stroke N = number of cycles completed per unit time N = (n ) (1) (number of cylinders) for single-acting compressor . N = (n ) (2) (number of cylinders) for double-acting compressor n = compressor speed

expansion valve, kJlkg

enthalpy leaving the expansion valve, kJlkg

Evaporator

/~

1kI~

~hl

7' qe

=h

I

/,Q.

m~

- h4

qe = h I - h3

r--.

7 Qe

= mth, -

A single-acting compressor makes one complete cycle in one rr-voluti on.

h)

Q e = m (h I - h) 3

ll!l('

A double-acting compressor makes two complete cycles in revolution. The actual volumetric efficiency n,"" is defined as n va

qe = refrigerating effect, kJlkg h, = enthalpy entering the evaporator, kJlkg hi = enthalpy leaving the evaporator, kJlkg Qe

=

volume flow rate entering compressor displacement rate of compressor

YI

refrigerating capacity, k,J/s

n VA. =-YD

III

.....

h, - h.

The power per kW of refrigeration is the inverse of the coefficient of performance, and an efficient refrigeration system has a low value of power per kW of refrigeration but a high coefficient of performance.

II

I

= -o-h"":;'l_-...."h"':';':l

Ihsplacement rate is the volume swept through by the pistons suction strokes per unit time .

t li.-i r

IIi


j{efrir.;l'ration and Air Conditioning

The mechanical efficiency n m is given by

The clearance volumetric efficiency n ve is given by

Pz ] n v e = 1 + c-c [ ~

Uk

[V2 ] = 1 + c-c [V2] ~ = 1-c ~-1

nm

_ WI

_ nIP

- Wn

-

BHP

where JHP = indicated horsepower of the compressor

c = per cent clearance Vc

indicated work brake work

= specific volume of vapor after isentropic

BHP == brake horsepower of the compressor

compression to P, k = 1304 for ammonia ~.

~

"t'

(2-1) A simple saturated refrigeration cycle for R-12 system operates at an evaporating temperature of -5°C and a condensing temperature of 40°C. Determine i a) the refrigerating effect per kg, (b) the work per kg, (c) the heat rejected at the condenser per kg, (d) the COP. For a refrigerating capacity of 1 kW, determine (e) the total heat rejected at the condenser, (f) the work, and (g) the volume flow rate.

SOLUTION

l:I .~
t:: ....

QI

50

'1:

....QI

S

.tQ

~

~

as

='

~

:J)

~

...

::l tn

Z)

-~--17,' 961kPo

III '" ...

~lkPa

Po. 10

0

3

2

.. 5 6 Diacharge-to-suction pressure ratio

/

40°C

!j

1 Enthalpy, kJlkg

Fig. 2-3 Clearance and actual volumetric efficiency of a Refrigerant 22 high-speed compressor.

h, h.

v ] = v g at --5°C == 0.06496 m'/kg

The adiabatic compression efficiency or simply compression efficiency nc is defined as: nc

...............

=

isentropic work of compression actual work of compression

= h, at 40°C = 238.5 kJlkg == h , at -5°C = 349.3 kJ/kg

h , = h at 961 kPa and (;11

j{t'f'rigerating effect

.h(

Wurk =

1('(

l l.-n t rr-jr-rt c-d

h,~

hi = =

:~72

==

S2

h - h,

.- 349.:3

equal to s] == 372 kJlkg

= 349.3 -- 238.5 = 110.8 k.Jlkg = 22.7 k.J/kg

he' - h, == :372 - 238.5 =

1:~;~.5

k,Jlkg

Simple Vapor Compression Refrigerat 11111


(d)

hi - hi COP = h -h 2

VI

=~ = 4.88

h f4

For a refr igcr at ing capacity of 1 kW (e)

III

=

;1)

II

(f)

W

i/

= (m i

t h.

- hi)

= (0.00903

Sl

= 368 k./lk:'

x 4 =

h 4 - h., h

h 4 - h f4 h g 4 - h.,

2~33.5 - 200 351.48 - 200 = 0.2212

kg/s) (372 - 238.5) kJ/kg Mass of flash gas = 0.2212 kg/kg

= (0.00903

kg/s) (372 - 349.3) kJ/kg

Q.,

. il) m=

=

h. - h,

= 0.20!i k.J/s (g)

equal to

h f at O°C = 200 kJ/kg

lg4

Q, = (rn) (h .. - h) = 1.206 k.I/»

S2

h g 4 = h g at O°C = 351.48 kJ/kg

1 kJ/s 110.8 kJ/kg = 0.00903 kg/s

Qe h I -h ,

= v g at O°C = 0.05539 mVkg

h, = h at 847.7 kPa and

'2'2.7

I

VI = (m ) (VI) = (0.00903 kg/s) (0.06496 mVkg) = 0.000!i866 mVs

(2-2) An air conditioning system of a high rise building has a capacity of :3!iO kW of refrigeration, uses R-12. The evaporating and condensing temperatures are O°C and 35°C, respectively. Determine the following: (a) Mass of flash gas per kg of refrigerant circulated, (b) Mass of R-12 circulated per second, (c) Volumetric rate of flow under suction condition, (d) Work of compression in kW, and (e) COP. SOLUTION

350 kJ/s (351.48 - 233.5) kJ/kg

=

2.97 k /s g

,(') VI = (m ) ('11) = (2.97 kg/s) (0.05539 mVkg ) = 0.1645 m"/:-\ 'r/l

W= (rn ) (h., - h.) = (2.97 kg/s) (368 - 351.48) kJ/kg = 49.06 kW

'I'!

_ h-h 1 'j COP- h,• - h 1

351.48 - 233,5 = 7.14 368 - 351.48

(2-3) A refrigeration system using refrigerant 22 is to hav« refrigerating capacity of 60 kW. The evaporating temperature I"; ~-10°C and the condensing temperature is 42°C. Determine. ';11 the volume flow rate of refrigerant at inlet to the compressor. . 1>1 the power required by the compressor, and (c) the fraction of \ .ipor in the mixture at entrance to the evaporator expressed 11I11h on a mass basis and a volume basis. ,I

<-;()LUTION 'f

III

k

_ ....._

~

7

2

t

..~

l6l0kPn

~

~

P<

1 Enthalpy, kJlkg

It I

= hi at

= :1!i 1 .48

kJ/kg

,, 42"C 41

354 kPa

'

L

, \ --lOGC

35""C = '233.5 kJ/kg

h, = h, at () C ~~.

:

Enthalpy, k.J1kg

h.

= h, at 4'2"(' = '2!i'2.4 k.J/kg

-zo I{efrigeration and Air Conditioning h.

= h g at -10 oe = 401.6 kJlkg

VI

= v g at -lOoe = 65.34 Llkg

h,

=h

Qe

(a) m== h _ h I

Simple Vapor Compression Refrigeration System :21

3

at 1610 Pa and

82

;()LUTION ',J I

equal to

SI

= 440 kJlkg

_ 60kJ/s - (401.6 _ 252.4) kJlkg = 0.4021 kg/s

VI = (rn ) (VI) == (0.4021 kg/s) (65.34 L/Kg)

= 26.27

~'I ~.

Lis

(c) x 4 V

4

~oc

208 kPa

1

h, -- h., h l - h f4 ._ 252.4 -' 188.4 == 0.3002 == - h - - == h - h - 401.6 - 188.4 f~4

==

V

f4

+

X

4V fg4

~-l

==

V

l4

f4

+

Enthalpy, kJ/kg

Xl (V g4 - V f4)

h, = h.at 24°e == 312.87 kJlkg

== 0.7588 + (0.3002) (6!"d4 - 0.7588) == 20.146 Llkg

h. == h g at -18°e

For 1 kg of refrigerant entering the evaporator, point 4, the volume of vapor is, (0.3002 kg/kg) (6534 Lzkg) == 19.615 Llkg 19.615 . 0 f vanor.volume vapor, vo ume basi aSIS == - - 0.9736 or 97.36% F raction 20.146

vI

= vg

h,

=

at -18°e

=

1439.94 kJlkg

= 0.5729 •

hat 974 kPa and

S2

m 3lkg

equal to

8

1

= 1657 kJlkg

I'i Rcfrigerat.ing effect == h j - h, == 1439.94 - 312.87

(2-4) A simple vapor-compression cycle develops 13 tons of refrigeration. Using ammonia as refrigerant and operating at a condensing temperature of 24°e and evaporating temperature of -18 c e and assuming that the compression are isentropic and that the gas leaving the condenser is saturated, find the following: (a) Draw the ph diagram, (b) refrigerating effect in kJ/kg, (c) circulation rate of refrigerant in kg/min, (d) power requirement, (e) volume flow in cubic meter/(min) (ton), in coefficient of performance, and (g) power per ton.

== 1127.1 k,Jlkg

"I

(~,

== 03 tons) (211 kJ/min. ton)

Qe

III

h , - hl

=

2743 kJ/min

2743 kJ/min 0439.94-312.87) kJlkg == 2.43 kg/min

,II \\' = (rn ) (h, - h.) == (2.43 kg/min) (1657 - 1439.94) kJlkg =

,. I

\

527.5 kJ/min or 8.79 kW

. 1 == (m ) (VI) == ..:.......----::"------::(2.43 kg/min) (0.5729 m 3lkg) -"' 13 13 tons

=

0.1071 m:3/ (min) (ton)

II

('01' __

"I

\\'

f

~.

. 4

.2

L

II

/1

== 15.44 kW

:/

-

~

(b) W== (rn) (h , - h.) == (0.4021 kg/s) (440 - 401.6) k.Jzkg

II

974kPa

-~

all

h j - h4 h,h1 , H 79 kW 1:1 t on s

1439.94 - 312.87 == 5.19

1657 -- 14:39.94

= O.G7()

k W/ton

......... 22



(2-5) A refrigeration system is to be used to cool 45,000 kg of water from 29°C to 18°C in 5 hours. The refrigerant is ammonia and the operation conditions are 616 kPa evaporating pressure and 1737 kPa liquefaction pressure. Determine (a) the coefficient of performance, (b) the quantity of cooling water in the condenser for an increase in temperature of 7°C, (c) the compressor power, anel (d) the volume of flow rate entering the compressor. The specific heat of water is 4.187 k-I/kg e Co.

Let m V' =

mass flow rate of cooling water circulated in the condenser Heat to water = Heat from refrigerant

mV'=

(c) W = (rn ) (h 2

.>

-

hi) = (0.1085 kg/s) (1620 - 1471.6) kJ/kg

= 16.10 kW lei)

V = (m ) (VI) = (0.1085 kg/s) (205.2 L/kg) = 22.3 L/s j

1737 kPa

as

(2-6) An ammonia simple saturation cycle operates between evaporator and condenser temperatures of -20°C and :j5°C, respectively. The system is to be used in producing 5000 kg of ice at -12°C from water at 29°C in 20 hours. Assuming losses to be 20 per cent of the heat to be absorbed from the water, determine (a) the mass flow rate, (b) the heat rejected at the condenser, and (c) the power required by the compressor. The specific heat of ice is 2.094 k.Izkg s C'' and the heat of fusion is :335 kJ/kg.

~

~". ~

'"'"

~

.l_616kP'

1

1

...............

- h 3) (0.1085 kg/s) (1620 - 410.4) kJ/kg (4.187 kJ/kg. CO) (7 CO)

= 4.48 kg/s

SOLUTION

I

= (m ) (h ,

(m) (c) (~t)

Enthalpy, kJ/kg

~OLUTION

h, = h, at 1737 kPa = 410.4 kJ/kg hi = h g at 616 kPa = 1471.6 k-Izkg VI

= ", at 616 kPa = 205.2 L/kg

h, = h at 1737 kPa and

S2

equal to

Sl

= 1620 kJ/kg

- h 2 = 1471.6 - 410.4 = 7 It:: ( a)\ COP = hi h - hi 1620 - 1471.6 . o 2

,

1352kPa

III

~

,, 35°C

~

4'

(b) Q = (45,000 kg) (4.187 kJ/kgeCO) (29 - 18) Co ::: 115.1 kJ/s c (5 h) (3600 s/h)

Q.. h-h , .,

115.1 kJ/s = 0.1085 kg/s (1471.6 - 410.4) kJ/kg

191 kPa

l -20°C

Enthalpy, kJlkg

.'

-! I

l{ef'rjgf~ratjoil

and Ai r Conditioning

h, = h. at 3f)'C


= 366.1 kJlkg

hI = h g at -20'C

,'-)()LUTION

= 1437.2 kJlkg

h , = hat 1352 kPa and

S2

equal to

51 =

1735 kJlkg

(a) Theoretical heat to be absorbed per kg of water

= (l kg) -I-

(4.187 kJ/kg -CO) 129- 0) Co

-I- (

·Cll

1 kg) (335 kJlkg)

~

J:

Actual heat to be absorbed = (1.20) (481.6) = 577.9 kJ/kg

=

72

~.

(1 kg) (2.094 kJ/kg-CO) [0 - (-12)] Co = 481.6 kJ

Qe

I

~

(500 kg) (5779 k.Jlkg) = 40.13 kJ/s

(20h)(3600s!h)

Enthalphy, kJlkg

Qe

40.13 kJ/s = 0.03747 kg/s h. - h, = (14:37.2 - :i66.1l kJlkg

m

(b) Qc = (rn)

(h

2

-

h 3 ) = (0.0:i747 kg/s) (1735 - 366.1) kJlkg

hi

= h, at 1200 kPa =- 346.1 kJ/kg

h.

= h, at 247 kPa = 1445.2 kJlkg


= 51.29 k.J/5

S2

equal to 5[ = 1675 kJlkg

1

(c) W =

(rn )

(h, - h) = (0.03747 kg/s) (1735 - 1437.2) kJlkg

la)

= 11.16 kW

[P]k = 1

n =1-1-c-c - 2 vc P

I

-I-

J

0.05 - (0.05) [1200]-;-;;;; 247

= 0.882 or 88.2o/r (2-7) An ammonia refrigerating cycle operates at 247 kPa suction pressure and 1200 kPa condenser pressure. Other data are the following: 28 kW refrigerating capacity 5% compressor clearance so« compression efficiency 75% mechanical efficiency 74'lr actu al vol u metri c efficiency

I

b l Actual work = isentropic work nc

= 287.3

Determine (a) the clearance volumetric efficiency, (b) the ideal and actual COP, (c) the mass flow rate of ammonia, and (d ) the brake work.

----......

h[

1675 - 1445.2 0.80

kJlkg

h 2 - hi = 1445.2 - 346.1 = 4.78 h, - h] 1675 - 1445.2

Actual COP

h[ - hI actual work

m= h _ h. [

lei)

-

nc

Ideal COP

Qe

1[")

h2

,J

1445.2 - 346.1 287.3

= 3.83

28 kJ/s (1445.2 - 346.1) kJlkg = 0.02547 kg/5

Indicated work = (0.02547 kg/s) (287.3 kJlkg) = 7.:i18 kW

26

Refrigeration and

All'

indicated work

Brake work


Co ndit.iuning

nm

7.318 0.75

Vw

= 9.76 kW

III W

(2-8) An ammonia simple saturation cycle operates at 247.1 kPa suction pressure and 1169 kPa condenser pressure. For a load of 161111 k.l/min, determine (a) the mVh of cooling water in the condenser if the water temperature is increased by 6 Co, (b) the compressor work, and (c) the piston displacement if clearance is 5'!r .

m

(81.32 kg/min) (60 min/h) 1000 k g/m"

w -~

= (m) (h, - h.) = (1.52 kg/min) (1677 = 352.3 kJ/min or 5.87 kW

vc

n vc

1169kPa

•• I

l

,2

:?;l

=1 + c -

=1 -

c[

247.1

Pj

[2 -1] = 1 - 0.05 [0.4878 v 0.15

c

-1]

Assume n vc = n v a D

... ;:l

n va

(m) (V 1 ) n va

'" Ul

£

247.1 kPa

~--_._------------:-::----_.

Enthalpy, kJlkg

= h , at 30

GC

h , = h g at -14 = v g at

= 341.77

GC

Qe h , - h,

kJlkg

= 1445.2 kJlkg

(1.53 kg/min) (0.4878 m1lkg) 0.8854

= 0.843 mvmi n =

(2-9) A six-cylinder, 6.70 x 5.70-cm, refrigerant 22 «ompreasor operating at 30 rls indicate a refrigerating capacity "f96.4 kW and a power requirement of19.4 kW at an evaporating t «mperature of 5°C and a condensing temperature of 35°C. ( 'ornpute (a) the clearance volumetric efficiency if the clearance volurne is 5 per cent, (b) the actual volumetric efficiency, and (c) t lu- compression efficiency.

:-;( )LUTION

-14°C = 0.4878 mJlkg

h, = h at 1169 kPa and V = V at 1169 kPa and z

S2

equal to

Sl

= 1677 kJlkg

S2

equal to

S1

= 0.15 m'lkg

1688 kJ/min = 1.53 kg/min (1445.2 - 341.77) kJlkg

(m)(hz-h,) (C) (L1tw )

=

~

... #

-----

J

(1.53 kg/min) (1677 - 341.77) kJlkg (4.187 kJlkg CO) (6 CO)

'" K 1.:r2 kg/min

Enthalpy, kJlkg

~

= 0.8874

0.8854 is the more accurate value, since 0.15 m:Jlkg is just approximation from the Mol lier diagram. V V = __ 1

rJ

m

[

1169] = 1 + 0.05 - (0.05) [ -.- 1..304

-P2 ] k

2

al'

(al m=

1445.2) kJlkg

= 0.8854

:111

VI

4.879 m:J/h

[

,(.) n

SOLUTION

h(

=

1.:

>2

28


hi = h at 35"C

= 243.1


kJlkg

SOLUTION

h , = h g at seC = 407.1 k.Izkg

= 40.36

VI = v g at 5'C

Llkg

h, = h at 13!)5 kPa and

8

2

equal to SI = 428 kJlkg

v 2 = V at 1355 kPa and S2 equal to SI = 20 Llkg (a)

n., = 1-

C[2 --1] v 2

Q, hi - hi

=

,

nva =

V

(c)

l

or 94.91 (Jc

(IT)

Enthalpy, kJlkg

\6.70 cm)2 (5.70 cm) (30 x 6 cycles/s) (4) (1000 cm3/Li

= :36.17

h 3 = h, 1204 kPa = 346.6 kJlkg

Lis

h. = h , at 291.6 kPa = 1450.2 kJlkg VI = v g at 291.6 kPa = 417.5 Lis

'13"'2 _~_._I- = 0.6558 or 65.5800

36.17

Actual work of compression =

h 2 = hat 1204 kPa and S2 equal to SI = 1653 kJlkg 1_9.4 kJ/s = 33.00 kJlkg

0.rJ878 kg/s

Isentropic work of compression = h 2- hI = 428 - 407.1

(ll' = (1.05) (rn.) (c.) (~tl) = (1.05) (102) (3.14) (8) = 2690 kJ/min or 44.83 kW

~

111

= 20.9 kJlkg . ffi 20.9 C ompression e iciency, n = - - = 0.63.339'c or 63.330;' c 33.00

(2-10) A 100 x 200-mm ammonia compressor with a compression efficiency of 80 per cent operates with a suction pressure of 291.6 kPa and a condenser pressure of 1204 kPa at 23 r/s. The refrigerant cools 102 kg/min of brine by 8 degrees in the brine cooler. The specific heat of the brine is 3.14 kJlkgeC". Electric input to the motor driving the compressor is 14.33 kW. Motor efficiency at this load is 92 per cent. Assuming 5 per cent of the useful refrigerating effect is lost by brine cooler from the room, determine the mechanical and volumetric efficiencies of thl' roru pr e ssor.

~.

>2

96.4kJ/s = 0.5878 kg/s 1407.1 - 243.1) kJlkg

LN =

+= [,

1

I

D2

I

='

VI = (m) (VI) = \0.5878 kg/s) (40.36 Llkg) = 23.72 Lis V =.2:.. J) 4

- --

~

e

=1- (0.05)[4?36]_1 20

= 0.9491 (b ) m =

~

=

44.83 k.]/s (1450.2 - 346.6) kJlkg

h, - h,

\\. = (rn) .

l nd irn te

(h 2

d

h) = (0.0406) (1653 - 1450.2) = 8.23 kW

-

war

= 0.0406 k /s g

k

8.23 8.23 = - - = - - = 10.29 kW n, 0.80

HI';"'I' work = output of motor = (14.33) (0.92) = 13.18 kW

\1, ' 1,.lllIcal efficiency = 11111

= 0.781 or 78.1%

(VI) = (0.0406) (417.5) = 16.96 Lis

~ D LN = I

i~:~~

!

TC

(10cm)2 (20 cm) (23 cycles/s) = 21 68 L/s (4) (1000 crnVl.) .


n va

VI

Simple Vapor Compression Refrigeration System 31 I 'iston speed = 2L (rpm)

16.95 = 0.7818 or 78.18% 21.68

= V D

(2-11) A four-cylinder refrigerant 12 compressor operates between evaporator and condenser temperatures of 4°C and 43°C. It is to carry a load of 20 tons ofrefrigeration at 1200 rpm. If the average piston speed is 213 m/min and the actual volumetric efficiency is 80 per cent, what should be the bore and stroke of the compressor.

2.314

X

(213 m/min) (100 em/m) (2) (1200) rev/min

L

=

VD

= ~

10 6 = ~ CD ern)" (8.875 em) (1200 x 4 cycles/min) 4

D = 8:315 em

~

~

,

----

-

k

7

2

~

I

£

I

~

Enthalpy, kJlkg

h, = h, at 43°C = 241.6 kJ/kg h.

= h at 4°C = 353.18 kJ/kg = v g at 4°C = 0.04895 m Vkg g

VI h, = h at 1034 kPa and S2 equal to SI = 372 kJ/kg m

Qe h, - h 4

(20 tons) (211 kJ/min • ton) (353.18 - 241.6) kJ/kg

= 37.82 kg/min

VI = (rn) (VI) = (37.82 kg/min) (0.04895 m 3/kg) = 1.851 m 3/min V D = ~lva =

10~8501

6

= 2.314 m 3/min or 2.314 x 10 em

3/min

em

D2LN

SOLUTION

'/

= 8.875

Simple Vapor Compression Refrigeration Syst"111 Ti


Effects of Operating Conditions

:1)

Refrigerating effect For the --10

Effects of Increasing the Vaporizing Temperature

oe

= hI -

(2 -12) A R-12 simple saturated refrigerating cycle operates at an evaporating temperature of -10 oe and a condensing temperature of 40'e. Show the effects of increasing the vaporizing temperature to 5':e.

evaporating cycle, the refrigerating effect h 4 = 347.1 - 238.5

=

108.6 kJlkg

For the 5°C evaporating cycle, the refrigerating effect

= h.. -

h 4,

= 353.6 -

238.5

= 115.1 kJlkg

I,) Effect on mass flow rate per ton

SOLUTION

For the -10 m

I~

III

k

"'-'''~A

ce

evaporating cycle, the mass flow rate,

211 h I -h 4

e

211

347.1 _ 238.5 = 1.943 kg/(min) (ton)

For the 5°C evaporating cycle, the mass flow rate,

772

~

t

l

m'=

I

J

211 353.6 _ 238.5 1.833 kg/(min) (ton)

Effect on compressor volume capacity For the -10 oe evaporating cycle, the volume flow rate,

Enthalpy, kJfkg

h,

= h. = h 4 = h.

VI

VI

= v g at -lOGe = 0.07665 m

h. = h g at -lOGe

= 347.1


S2

equal to

Sl

= 373 kJ/kg

Effect on

cor

equal to

eop

SI

= 371 kJlkg

hI - h 4 _ h 2 - h,

347.1 - 238.5 -'----373 - 347.1

=:

4.19

Ftlr the 5°e evaporating cycle,

eop

= hg at 5°e = 353.6 kJ/kg S2

,I,

F"r the -10 oe evaporating cycle,

kJ/kg

"r = vg at 5°e = 0.04749 ms/kg

h , = h at 961 kPa and

VI' =(rn') (vIJ = 0.833) (0.04749) = 0.087 m 3/(min) (ton)

3/kg

For the 5°e evaporating cycle (cycle I.' -2' -3'-4'-1')

h.

= (rn) (v 1) =(1.943) (0.07665) = 0.1489 m 3/(min) (ton)

For the 5°e evaporating cycle, the volume flow rate,

at 40 ce = 238.5 kJlkg

For the -10 ce evaporating cycle (cycle 1-2-3-4-1)

~

211 h h., - 4'

hI' - h 4 · = 353.6 - 238.5 = 6.61

h , - h.. 371 - 353.6


Simple Vapor Compression Refrigeration System :l5

(e) Effect on work per ton of refrigeration For the -10°C evaporating cycle, the work,

I

W = (m) (h 2-h l ) = (1.943) (373-347.1)

Effects of Increasing the Condensing Temperature (2-13) A simple saturated refrigerating cycle using R-12 as the refrigerant, operates at a condensing temperature of 40 nC and an evaporating temperature of -lOOC. Show the effects of increasing the condensing temperature to 50°C.

= 50.32 kJ/(min) (ton) or 0.839 kW/ton

r

SOLUTION For the 5°C evaporating cycle, the work, W'

,

= (rn) (h

2

,-

h.)

= (1.833) (371

- 353.6)


•

\.

2

~

e

(f) Effect on condenser performance

::l

:l

Qc

= (rn) (h, -

h)

= (1.943) (373

= 261.3 kJ/(min) (ton) or

219 kPa

"

- 238.5)

Enthalpy, kJlkg

4.355 kW/ton

For the 5°C evaporating cycle, the heat rejected at the condenser,

h,

= h g at -lOoC = 347.1 kJlkg

VI

==

Vg

at -10°C == 0.07665 mvkg

For the 40°C condensing cycle (cycle 1-2-3-4-1)

Q'c == (m') (h 2, - h) = (1.833) (371- 238.5)

= 242.9 kJ/(min) (ton) or

. ,U

tt.

For the -10°C evaporating cycle, the heat rejected at the condenser,

I

1219 kPa 961kPa

h, = h 4 == he at 40°C = 238.5 kJlkg

4.05 kW/ton

h, == h at 961 kPa and

Effects of increasing the vaporizing temperature (1) The refrigerating effect per unit mass increases.

S2

equal to

SI

= 373 kJlkg

For the 50°C condensing cycle (cycle 1-2'-3'-4'-1) h; == h 4 · = he at 50°C = 248.9 kJlkg

(2) The mass flow rate per ton decreases.


S2'

equal to

SI

== 378 kJlkg

(3) The volume flow rate per ton decreases. (4) The COP increases. (5) The work per ton decreases.

,)

I{( -frigerating

effect

Fill' the 40°C condensing cycle, the refrigerating effect

(6) The heat rejected at the condenser per ton decreases. = h , - h 4 == 347.1 - 238.5 = 108.6 kJlkg 1,'111"

t he

= hi

~

50°C condensing cycle, the refrigerating effect - hi = 347.1 - 248.9 = 98.2 kJlkg

36

Simple Vapor Compression Refrigl'l'atlOil


For the 4We condensing cycle, the heat rejcct.cd :11 condenser,

For the 40 e condensing cycle, the mass flow rate, C

Q, = (m) Ih 2

211 347.1 _ 238.5 == 1.943 kg/Gnin) (ton)

2~~ )

=

,,'

;347.12~1248.9

h)

-

= 11.94.3)

fill'

(37:3 - 238.5)

== 261.:3 k-L'(rn in I (ton) or 4.3:")5 kW/tul1

For the 50 e condensing cycle. the heat r ejcctr-d con de nser,

For the 50 0 e condensing cycle, the mass flow rate, m' = h

."

II Effect on condenser performance

(b) Effect on mass flow rate per ton

211 m == hi - h,

:-l\'!"il'

Q'c = (m') (h 2

== 2.149 kg/(min) (ton)

-

.r t

tIl'

h) = 12.149) (378 -- 248.9)


(c) Effect on compressor volume capacity l-Tf.. cts of increasing the condenser temperature

For the 40,ce condensing cycle, the volume flow rate,

(1) (2)

V!, = (m) (v I) = (1.943) (0.07665) == 0.1489 m:l/(mim) (ton)

,

1.3)

14) (.5 J (6)

For the 50°C condensing cycle, the volume flow rate,

VI' == (rn') (VI) = 2,149) (0.07665)

= 0.1647 m:/(min) (ton)

I

The The The The The The

refrigerating effect per unit m ass rlecn';l~'''. mass flow rate per ton increases.

volume flow rate per ton increases.

eop decreases.

work per ton increases.

heat rejected at the condenser per ton Iner,':I",'>

12-14) A 150 x 150-mm, twin-cylinder. -i n u!« ;1('[1111-;

12 compressor running at 25 r l « cu rri o« "

" Irigerating load of 39.4 kW while operating at 340 k Pu - uc t i on

rlld 1277 kPa discharge pressures. If thp discharge [)t"'';-'; 11 1'('

, "fTC' raised to 1397 kPa, at what speed should the co m prr.s sor

I., run to carry the same load assuming the volumetr-ic l,fficIPn('\

r.-m a in the same?

tr i g o r a n t

(d) Effect on eop

For the 40 De condensing cycle,

eop = h 1 - h 4 = 347.1 - 238.5 h 2-h 1 373-347.1

= 419 .

For the 50 0 e condensing cycle,

:IIL( TTl ON

eop = h 1 - h 4 , = 347.1 - 248.9 = 3.18 h 2,-h 1 378-347.1 (e) Effect on work per ton of refrigeration For the 40 0 e condensing cycle, the work,

w .. (m) (h, -

h)= 0.943) (393 - 347.1)

= 50.32 kJ/(min) (ton) or 0.839 kW/ton For the 50 0 e condensing cycle, the work,

.

~

1397kPa

lZr.1kPa

r; _I

t:

"'I

h , = 1\ at 340 kPa =

I

W' = (m') (h, - h.) = (2.149) (378 - 347.1) = 66.40 kJ /Cruin) (ton) or ~.107 kW/ton

~

Enthalpy. kJlq

3~)2,8

k,Jlkg

h,

=

h

= hi at 1:397 kPa = 2fi:"i.3 kJ/kg

l

h, at 1277 k Pa = 2.'11 k-I/kz

Simple Vapor Compression Refrigeration System :;~l

:18 Rcfriacrat.ion and Air ('()Ilditioning

Let ru

=

old

Ill;\';';

flow rate

For the saturated cycle (cycle 1-2-3-4-1)

m' = nvw m ass flow rate

= old comprt-ssor speed

= now c.nnpresaor speed

n n'

Qp

:30.4 k,J/s

m= h _ h

=

1m')

:39.4 k,J/s :i[)28 - 255.3) k-J/kg

h\- h ,

Iv

l

= h g at --5°C = 349.3 kJ/kg

at --5

DC

= 0.06496 m 3fkg

~J61 k Pa and

S2

equal to

SI

= 3';2 k-I/kg

"or the superheated cycle (cycle 1'-2'-3-4-1')

Q,

= 0.4041

kg/s

4

l

TIl

= 0.071 mVkg

hi' = h at-261 kPa and 15°C = 362 k,Jlkg h 2· = h at 961 kPa and 8 2, equal to 8 1 = 387 k.Jfkg

= ~rYL!2l (n ) Im'lln) n =---

= v at 261 kPa and 15°C

VI'

1m) Iv,)

)

~D2L(2) (n') 4

h,

g

h, = h at

= 0.3870 kg/s

volume flow rate entering compressor . ffici vo I umetnc e Icwncv. = . displacement rate of compressor

J

=V

(:3:")28 - 251) k.I/kg

'I]

111'

v[

=

'II

10.4041) (25)

0.:3870

For the saturated cycle, the refrigerating effect is = h, - h 4 = 349.:3 - 238.5 = 110.8 kJ/kg

Effects of Superheating the Suction Vapor (2-15) A simple saturated r efri gerat ing cycle for R-12 system operates at an evaporating temperature of -boC and a condensing temperature of 40"C. Show the effects of superheating the suction vapor from --5°C to IS-C.

Refrigerating effect

For the superheated cycle, (1) When superheating of suction vapor occurs inside the " frigerated space, useful cooling results and the refr'igeratirg II, -ct is =

SOLUTION

hi' - h 4 = 362 - 238.5 = 123.5 k,Jfkg

(2) When superheating of suction vapor occurs in the ,-Ilon line, no useful cooling results and the refrigerating I',·," is

'I

~.

-

-k----T--72'

-; h J

--

h4

= 349.3 -

238.5

= 110.8 kJfII:g

1II'd on mass flow rate per ton 'I'

'11

the saturated cycle, the mass flow rate, =

211 h[ _ h

211 = 349.3 _ 238.5 4

Enthalpy, kJ/kg

h

~

= h, = h, at 40"C = 238.5 kJfkg

= 1.904 kg/train) (ton)

·10 Refrigeration .uid Ail' Conditioning

Simple Vapor Compression Refrigeration

For the su pr-rhc-a t cd cycle,

(11 With useful cooling, the mass flow rate,

211

_

m' = h .- h I

..j

1

<_ h -. h

(2) Without useful cooling, COP:=

h~

211 362 _ 238.5 := 1.709 kg/(min) (ton) :349.3 - 2:38.5 := .1.+:{ 387 - :362

j

(2) Without useful cooling, the mass flow rate,

,211 111 = ~h4

:-)",1 \'In

211 = 349.3 - 238.5

1.904 kg/(min) (ton)

,,) Effect on work per ton of refrigeration For the saturated cycle, the work, W:= (rn ) (h, - h.) := (1.904) (372 - :349.31

(c! Effect on com pressor volume capacity

= 43.22 kJI (min) (ton) or 0.720 kW/ton

For the suturntvd cycle, the volume flow rate, V. I

:=

tm

:v

1= ,

For the superheated cycie, (1)

!I.904) (006496):= 0.1237 m1/(min) (ton)

W':=

For the superheated cycle,

t

j)

=

With useful cooling, the volume flow rate,

(2)

Vi = :=

(2)

With useful cooling, the work,

(m') (v,) =

(1.709) (0.071)

(rn i (v'r) =

(h 2 - hl'):= !I.709)

(387-:36~)

42.73 kJ/(min) t ton ) or

0.71·~

k\,'I'fj'l

Without useful cooling. the work, W":=

0.1213 m'/( min) (ton)

(rri')

(h 2

-

h,)

= (1.904)

(387 -

:{f)~~)

:= 47.6 kJ/ (min) (ton : or 079:-; k W/ton

Without useful cooling, the volume flow rate, V'1 =

(rn')

t,

Effect on condenser performance For the saturated cycle, the heat rejected at the condenser,

(1.904) (0.071)

Qc

= 0.1352 m ' /(m i n ) (ton)

= (m) (h 2 -

h.:

= (1.904) (372 - 238.5)

= 254.2 kJ/ (min) (ton) .01' 4.24 kW/ton (r1)

E fleet on COP

I'-I)r

hi - h 4 _ 349.3 - 238.5 Frl\' the saturated cycle, COP = i , _ h - 372 _ 349.3 :2

= 4.88

1

!I) With useful cooling, the heat rejected at the condenser, Q'c

For the saturated cycle, ( 1)

the superheated cycle,

'With useful cooling, COP =

h - h ~.' _ h ~.

1..

:= 4.94

;362 - 238.5 387 - 362

= (m') (h ,

- h) = !I.709) (387 - 238.51

= 253.8 kJI (min) (ton) or 4.23 kW/ton

(2) Without useful cooling, the heat rejected at the condenser, Q'c'= ( m l (h, - h,,):= (1.904) (387- 238.5) =- '2H'2.7 k,JI (min) t ton l or 4.71 kW/ton

~

........ Simple Vapor Compression Refrigerauon :--,

'i2 Refrigeration ann Air ('onditioning


Effects of supcrhc-atinj; the suction vapor When supe rhr-at i nu /)!"oduces useful cooling: (1) Tho ref'rig<'rating effect per unit mass increases. (2) 'I'hc m;t:·:s flow rate per ton decreases. (3) The vol u r.ic flow rate per ton decreases. ( ) The COl' incre-ases. (5) Th e work per tun decreases.

h J =: h 4 =: he at 40°C =: 238.5 k.L'kg

For the subcooled cycle (cycle 1-2-3'-4'-1)

h, =: h 4 = h[ at 30°C =: 228.5 k.Izkg

• :1)

When superhcat ing occurs without useful cooling: (6) The n,rri~Tr
Refrigerating effect For the saturated cycle, the refrigerating effect =: 349.3 - 238.5 =: 110.8 kJlkg

=h

I,

I

sarn o.

(7) (8) (9) (10) (11)

Tho TI1<' Thl' The The

Illa,;,..; now rate per ton remains the same. vol u.r.o now rate per ton increases. COP decreases. work per ton increases. heat 1'8.: ect ed at the condenser per ton increases.

II

For the subcooled cycle, the refrigerating effect =: h. =: 349.3 - 228.5 =: 120.8 kJlkg h I Effect on mass flow rate per ton

For the saturated cycle, the mass flow rate,

~"nec'"s

of

SL~)COOli._,~

the Liquid

(2--16) A simple saturated refrigerating cycle using R-12 as the refrigerant operates at an evaporating temperature of -5°C and a condensing temperature of 40 'C. Show the effects of st bcoolirig the liri.id from 40C to :30'C before reaching the ex )Lnsion valve.

m =:

211 hi - h 4

211

349.3 _ 238.5 =: 1.918 kg/t m i n ) It OI}1

For the subcooled cycle, the mass flow rate,

211

m' =: h - h , 1

211 349.3 - 228,5

=: 1.747 kg/Cm in

i I t on

:

SOLUTION I·:rfect on compressor volume capacity

r -,r

III

~

n_ --

k

72

j·'(lr the saturated cycle, the volume flow 'rate, VI =: (rn ) (VI) =: (1.918) (0.06496) =: 0.1246 mil (mini (Ion) I" ,I'

E ~

I'~ L 4

VI

261kPa

~

~

." t he saturated cycle, COP =:

ih - _hh

349.3 - 2:38.5 372 - 349.3

4 1

=: 4.88

Enthalpy, kJ/kg

=: v g at -5°C =: 0.06496 rrrvkg ~

111,('t on COP

2

hi=: 1\ at -5"C =: 349.3 kJlkg h.

the subcooled cycle, the volume flow rate, V, =: (rn ') (v 1 ) =: (1.747) (0.06496) =: 0.1135 m1/(minl (ton)

h at 9f)1 k Pa and .".' r-qu al to s, =: 372 kJlkg

I

'·1

f

hi - h h(' s u hcoo lr-d cycl«, COP =: h _ h 4 }

349.3 - 228.5 372 - 349.3

1

=

:l.:l2

...


42 Refrigeration ann Air Conditioning

Effects of super hua t i ng the suction vapor When superheat ing produces useful cooling: (l) The refrigP1'ating effect per unit mass increases. (2) The muss now rate per ton decreases. (3) The volu r.re fl ow rate per ton decreases. (c) The COP increases. (5) The work pc'r ton decreases. When superheating occurs without useful cooling: (6) The n{ri,~'crating effect per unit mass remains the same. n) The muss now ratc per ton remains the same. (8) The vol u n.o flow rate per ton increases. (9) The COP decreases. (10) The work per ton increases. (11) The heat re..'ected at the condenser per ton increases.


h,

= h , = h, at 40°C =

238.5 kJ/kg

For the subcooled cycle (cycle 1-2-3'-4'-1)

h., = h.

= h, at 30°C

= 228.5 kJ/kg

(a) Refrigerating effect For the saturated cycle, the refrigerating effect = 349.3 - 238.5 = 110.8 kJ/kg

= hi

For the subcooled cycle, the refrigerating effect = 349.3 - 228.5 = 120.8 kJ/kg

= h. -

- h,

h 4,

(b) Effect on mass flow rate per ton For the saturated cycle, the mass flow rate,

~":'ec~s

of

Sl..~)cooli._~;

the Liquid m

(2-16) A simple saturated refrigerating cycle using R-12 as the refrigerant operates at an evaporating temperature of -5°C and a condensing temperature of 40:JC. Show the effects of st. bcool in g the lirt.i d from 40 C to 30°C before reaching the ex.ian sion valve.

211 h. - h ,

211 349.3 _ 238.5

= 1.918 kg/(min) (ton)

For the subcooled cycle, the mass flow rate,

C

211

rn' = h - h , 1

211 349.3 - 228.5

= 1.747 kg/t m in) (ton)

SOLUTION (c) Effect on compressor volume capacity

r

as

-.r

1

7

For the saturated cycle, the volume flow 'rate, V l = (In) (VI) = (1.918) (0.06496) == 0.1246 m:3/ (min) (ton)

2

:?il

For the sub cooled cycle, the volume flow rate, VI = (rri') (VI) = (1.747) (0.06496) = 0.1135 m 3/(min) (ton)

~

oj

(d)

Effect on COP For the saturated cycle, COP

h - h4 h

=i

_

2

349.3 - 238.5 372 - 349.3

1

= 4.88

Enthalpy, kJ/kg VI

= vi! at -5 s e = 0.06496 mvkg

hi

= 1\ at -SoC:::: 349.:3 k,J/kg

11,

==

li n t

~)()1

k l'n .uid s , ('quill tq

h -h For the subcooled cycle, COP = h 1 _ h 4 ' 2

SI

= :172 kJlkg

1

349.3 - 228.5 372 - 349.3 = 5.32

44


(e) Effect on work per ton of refrigeration


SOLUTION

For the saturated cycle, the work,

W = (rn ) (h , - h.) = 0.918) (372 - 349.3)

ssoc

aLL

= 4:3.f);3 k.l/r min ) (ton) or 0.726 kW/ton

k

1460 kPa

,

For the subcoolrd cycle, the work,

WI = (m') (he - h.) = (1.747) (372 - 349.3)

\

'\

38°C 466kPa

= 39.66 kJ/( min) (ton) or 0.661 k Wzt.on

; (f)

Effect on condenser performance For the saturated cycle, the heat rejected at the condenser, Q = (rn ) (h, - h) = n.918) (372 - 238.5) =

Enthalpy, kJ/kg

256.1 k-L'(min) (ton) or 4.268 kW/ton

For the subcooled cycle, the heat rejected at the condenser, Q'c = (rn') (h , - h:3 ) = 0.747) (372 - 228.5)

= 250.7

h,

= h. at 33=C = 240.5 kJ/kg

hi

= h at 5°C and 466 kPa = 410 k.Izkg

VI

=

v at 5°C and 466 kPa


S2

= 53

L/kg

equal to

SI

= 440 k.I/kg

k.Iz(min) (ton) or 4.178 kW/ton

Effects of sub cooling the liquid (1) The refrigerating effect per unit mass increases. (2) The mass flow rate per ton decreases. (3) The volume flow rate per ton decreases. (4) The COP increases. (5) The work per ton decreases. (6) The heat rejected at the condenser per ton decreases.

112 kJ/s = 0.6608 kg/s (410 - 240.5) k-Izkg

(a) m=

V 1 = mv 1 = (0.6608) (53)

SUBCOOLING AND SUPERHEATING (2-17) A refrigerant 22, four cylinder, 85 x 70-mm, compressor operating at 28 rls, a condensing temperature of 38 nC, and an evaporating temperature of -2°C carries a refrigerating load of 112 kW. The motor driving the compressor has an efficiency of 92 per cent and draws 32.5 kW. There is a 5°C subcooling and 7°C superheating of the suction gas entering the compressor. Compute (a) the actual volumetric efficiency and (b) the compression efficiency.

= 35.02 Lis

= ~ D2 LN = (It) (8.5 cm)'2 (7.0 em) (28 x 4 x cycles/s)

V !l

4

(4) (1000 cmvl.) =

nva

44.49 Lis

VI VI)

(b) Isentropic work

= (rn ) (h 2 -

h.)

= (0.6608)

(440 - 410)

= 19.82 kW Actual work = (32.5) (0.92)

= 29.90 kW

= 19.82 = 0.6629 or 66.29(1r

Compression efficiency, n c

.L

•2

29.90

46



(2·-18) An ammonia compressor operates at an evaporator pressure of :316 kPa and a condenser pressure of 1514.2 kPa. The refrigerant is subconled ;) degrees and is superheated 8 degrees. A twin-cylinder compressor with bore to stroke ratio of 0.85 is to be used at 1:200 rpm.. The mechanical efficiency is 77% . For a load of 87.5 k \V, determine: (a) the quant.it.y of cooling water in the condenser if the increase in temperature is 7 degrees, (b) the bore and stroke, (c) the size of th.- driving motor, and (d) the quality of the r.-fr iger ant entering the evaporator.

= (In)

( o ) VI

Assume

(VI)

llva

= (0.07877 kg/s) (410 Llkg) == 32.3 Ll.s

= 100 ck, then, V D == V l= 32.3 L/s

= ~D2LN == !:TC) (0.85L cm)2 (L cm) (20x2 cycles/s) 4 (4) (1000 cm3/L)

V D

= 0.0227 L'L/s 0.0227L3 == 32.3

I

SOLUTION:

L

= 11.25 em

D = (0.85) (11.25) = 9.56 em

a:l ~

-

\

1514.2 kPa -~

(d W

== (rn)

(h, - h.) == (0.07877 kg/s) 0715 -1472) kJ/kg

:12

== 19.14 kW

,..ilcl

II

I I

/1

...

. P ower Input to compressor

39°C

= power output 0 f motor =

31G kPa

19.14 _ 0.77

= 24.86kW (d) x 4 --

Enthalpy, kJ/kg

h, = h, at 34°C = 361.2 kJlkg h. == h at 316 kPa and O°C == 1472 kJlkg VI

= v at 316 kPa and O°C = 0.41 m:'lkg

h , == h at 1514.2 kPa and

(a)

m=

Q,.

11="'Fl 11

=

S2

equal to

87.5 kJ/s (1472 _ 361.2) k.J/kg

SI

= 1715 kJlkg

= 0.07877 kg/s

361.2 - 163.2 = 0.1536 or 15.36% 1289.44

(2-19.) A 10-ton vapor compression refrigeration system is to evaporate R-12 at-20°C and to condense it at 40°C. The twin cylinder vertical compressor is to run at 900 rpm; it is single acting, has the bore equal to the stroke, and has 2% clearance. Vapor enters the compressor at -10 o e and liquid enters the expansion valve at 30 aC. (a) What refrigeration flow rate is required? (b) What would be the bore and stroke of the compressor? (c) What power is required by the compressor?

(m ) (h - h.) 2 n1 = _ _.....:0.-..._;';""

w (c ) (.6 t) w

== :L()4

~

kg/s

(0.07877 kg/s) (1715 - 361.2) kJlkg 4.187 kJ/kg. CO) (7 CO)

(d) What is the quantity of the cooling water required in the condenser for a 8-degree increase in temperature?

....l

,,-"ll

48

Simple Vapor Compt-ession Refrigeration System 49


= (rn ) (h , -

(c) W

SOLUTION

h.)

= (0.2947 kg/s) (383

- 348) kJ/kg

= 10.31 kW 960.7 kPa

-

CIS'

~

t

~

~

£

k

,,2

//

:~oC

(m) (h 2 - h) _ (0.2947 kg/s) (383 - 228.54) kJlkg (c ) L~ t \\ ) (4. 187 kJ/kg. CO) (8 CO)

(d) m ,

=

1.36 kg/s

Liquid-Suction Heat Exchanger

150.9 kPa

The functions of the heat exchanger are: 1. To ensure that no liquid enters the compressor, and 2. To subcool the liquid from the condenser to prevent bubbles of vapor from impeding the flow of refrigerant through the expansion valve.

Enthalpy, kJlkg

h, = h, at 30°C = :228.G4: kJlkg h,

.=

v]

= v at

liz

= hat 960.7

h at 150.9 kPa and -10"C

= 348

k.I/kg

150.9 kPa and -1 O"C = 0.12 mJ/kg

= 383 kJ/kg

kPa and

5;,

equal to

v z = v at 960.7 kPa and

S2

equal to s] = 0.021 mVkg

(a) m =

Qe

= (10 tons) (3.52 kW/ton) = 0.2947 k

h. - h, (b) n vc

V]

=1-

S1

(348 - 228.54) kJ!kg

c [ -v] -1] v2 •

/s

SOLUTION

g

For the saturated cycle: COP = 4.88 (See problem 2-1) For the heat exchanger cycle:

= 1-0.02 [0.12] - - -1 = 0.9057 0.021

= (rn) (v]) = (0.2947 kg/s) (0.12

ms/kg)

= 0.03536

(2-20) A simple saturated refrigerating cycle using R-12 as refrigerant operates at an evaporating temperature of -5°C and a condensing temperature of 40;C. Show the effects of installing a heat exchanger that would superheat the suction vapor from _ t)c,C to 15 c,C in the heat exchanger.

3

mvs

~

Assume n va = n vc VD

V

=~ = va

VD

~

0.03563 0.09057

= 0.03904

m

3/s

= ~ D2 LN = ~ (D m)" (D m) =

1413.7

D:3 =

or 2.3424 mVmin

(900x2 cycles/min)

1413.7 D'' mVmin 2.3424

D

= 0.1183 m or 11.83 em

L

=.:

II)n cm

Heat

exchanger

4

I

6

1

50



(2-21) A refrigerant 22 vapor compression system includes a liquid-to-suction heat exchanger in the system. The heat exchanger cools saturated liquid coming from the condenser from 32 to 21°C with vapor which comes from the evapor ator at 10°C. GC

•

-

~

_n --

I

-

>2

~

j

(a) Calculate the COP of the system w i th ou t the heat exchanger but with the condensing temperature a 32::>C and an evaporator temperature at -10°C. (b) Calculate the COP of the system with the heat exchanger. (c) If the compressor is capable of pumping 20 Lis measured at the compressor suction, what is the refrigerating capacity of the system without the heat exchanger? (d) With the same compressor capacity as in (c), what is the refrigerating capacity of the system with the heat exchanger?

Enthalpy, kJlkg

)

h) = h at 261 kPa and 1f)°C =3G2 kJ/kg h, = h at 961 kPa and

So!

SOLUTION

equal to 5! = 387 kJlkg ( a)

h, = hi at 40 c C = 238,f) k.I/kg Heat balance of t hl~ heat exchanger hI + h fi = h + h< J

1255kPa

Refrigerating effect = he - h, = h. - h, COP = hI - h~l _ 362 -- 238.5 h - h 1 - 387 - 360.,

2

~

1

=4.94

2

'\.

\" ~c

4'

3fi4.3 kPa

I

....

-,

-10°C

The coefficient of performance of the heat exchanger cycle differs than that of the saturated cycle by only

Enthalpy, kJlkg

4.94 - 4.88 = 1.23~ 4.88 Depending upon the particular case, the coefficient of performance of a heat exchanger cycle may be either greater than, less than, or the same as that ofa saturated cycle operating between the same pressure limits.

h, = he at 32 = 239.2 kJlkg h. = h g at --10 = 401.6 kJ/kg GC

GC

VI

h2

=V =h

COP

g

at -10

GC

= 65.34

at 1255 kPa and

= ~l2 =hh 3, =

L/kg S2

equal to

S1

401.6 - 239.2 __ - ,').17

= 433 kJlkg



PROBLEMS

(b)

/

~

I

(1) An ammonia vapor refrigeration cycle operates at an evaporator temperature of -16°C and a condensing temperature of 32°C. Determine the coefficient of performance (a) for an ideal saturation cycle, (b) for wet compression with saturated vapor leaving the compressor, and (c) if the vapor at suction to the compressor is superheated 6 degrees.

Ans. (a) 4.50, (b) 4.70, (c) 4.28 .15.5

/' (2) A standard vapor compression system produces 20 tons of refrigeration using R-12 as a refrigerant while operating between a condenser temperature of 41.6 C and an evaporator temperature of -25°C. Determine (a) the refrigerating effect in kJ/kg, (b) the circulating rate in kg/s, (c) the power supplied, (d) the COP, (e) the heat rejected in kW, and (f) the volume flow rate in Lis.

Enthalpy, kJJq

h, :::: hi at 32°C = 239.2 kJ/kg

= h. = hi at 21°C = 225.3 kJ/kg h, = h at -1 oce = 401.6 kJ/k CT b

h,

)

f'

h. + h , :::: 11] + h,

h.

=:

Ans. (a) 100.26 kJ/kg, (b) 0.7022 kg/s, (c) 25.69 kW,

239.2 + 401.6 - 225.3 == 415.5 kJ/kg

(d) 2.74, (e) 96.09 kW, (f) 92.18 L/s

v 1 = 72 L/kg


S2

h 6 -h_ COP = h - h. 0)

2

equal to

Sl

=

450 kJ/kg

401.6 - 225.3 ___----= 5.11 450- 41fi.5

-,

(c) Without the heat exchanger. VI

20 Lis

= ~ = 65.34Llkg = 0.3061 kg/s

ill

Q

= (rn ) (h. -

h

e

) 4

= (0.3061 kg/s) (401.6 -

239.2) kJ/kg

= 49.71 kW (d) With the heat exchanger. V1 m =-v1

=

20 Lis 72L/kg

= 0.2777 kg/s

Q,. = (rn ) (h, - h) = (0.2777 kg/s) (401.6 - 225.3) kJlkg

.....l

::: 4R.96 kW

(3) A manufacturing company is intending to use its water cooled condenser for its proposed cold storage room. The name plate of the condenser gives the following specifications: refrigerant ammonia condenser water inlet . 30°C condenser water outlet 40°C condensing temperature 35°C refrigerant flow 3.84 kg/min circulating water flow 120 kg/min (a) If the company decided to purchase a new compressor and evaporator, find the tonnage of the system and the temperature in the evaporator, (b) what is the COP? (c) find the theoretical hp required. Ans. (a) 19.73 TR; -8°C, (b) 4.83,

(c) 19.26 hp .

(4) A standard ammonia vapor-compression cycle developing 20 tons of refrigeration operates with a condensing temperature of 32°C and an evaporating temperature of -14°C. Calculate the


(a) refrigerating effect, (b) circulating rate of refrigerant, (c) theoretical power, (d) COP, (e) gallons per minute of cooling water in the condenser, if ~t is 8Co, (f) quality of the refrigerant entering the evaporator, and (g) temperature of the refrigerant leaving the compressor. (l gallon contains 8.33 lb. of water) Ans. (a) 1093.7 kJ/kg, (b) 3.86 kg/min., (c) 15.44 kW, (d) 4.56, (e) 40.6 gpm (5) Asingle-acting, twin-cylinder 15.25 x 15.25-cm. compressor receives saturated ammonia vapor at -18°C and discharges it at 1514.3 kPa. The system is used to produce ice at a final temperature of -10°C from water at 27°C. The compressor runs at 450 rpm with an actual volumetric efficiency of 78%. Assume compression and mechanical efficiencies to be 82% and 88%, respectively. Calculate the following: (a) capacity of the system in tons of refrigeration, (b) brake hp required by the compressor, (c) tons of ice produced per day if compressor runs only 20 hours a day. Assume losses to be 10% of the refrigeration required to produce the ice, and (d) ideal and actual COP. Ans. (a) 17.04 TR, (b) 32 hp, (c) 8.37 tons, (d) 3.4,2.79 (6) An R-12 standard refrigeration cycle operates at an evaporating pressure of 386 kPa and a condensing pressure of 1009 kPa. Show the effects of decreasing the vaporizing pressure to 270 kPa on the following: For a unit mass, (a) refrigerating effect, (b) COP, and (c) work. For a refrigerating capacity of 1 kW, (d) power, (e) mass flow rate, (f) heat rejected, and (g) volume flow rate. (7) An industrial plant has available a 4-cylinder, 76-mm bore by 102-mm stroke, 800 rpm, single-acting compressor for use with Refrigerant 12. Proposed operating conditions for the compressor are 38°C condensing temperature and 5°C evaporating temperature. It is estimated that the refrigerant will enter the expansion valve as a saturated liquid, that the vapor will leave the evaporator ata temperature of 7°C, and will enter the compressor at a temperature of 13°C. Assume a compressor volumetric efficiency of 70 per cent. Assume frictionless flow. Calculate the refrigerating capacity in kW for a system equipped with this compressor. Ans. 40.96 kW

-----..L


(8) A refrigerant 12 refrigeration system requires a load of 54 kW at an evaporator pressure of 270 kPa and a condenser pressure of 1009 kPa. The refrigerant is subcooled 10 degrees before entering the expansion valve and vapor is superheated 14 degrees before entering the compressor. A twin-cylinder compressor with stroke equal to 1.25 times the bore is to be used at a speed of 27 r/s. The volumetric efficiency is 84 per cent. Determine (a) the quantity of cooling water in the condenser for an l l-degree increase in temperature, (b) the bore and stroke, and (c) the compressor power. Ans. (a) 1.40 kg/s, (b) 8.58 x 10.73 ern, (c) 10.5 kW (9) A refrigerant 22 refrigeration system carries a load of 82 kW at an evaporator pressure of 354 kPa and a condenser pressure of 1460 kPa. The liquid refrigerant is subcooled by 4 degrees before entering the expansion valve and the vapor is superheated by 5 degrees before entering the compressor. The compressor operates at 28 r/s. The stroke-to-bore ratio of the twin-cylinder compressor is 1.20 and the actual volumetric efficiency is 82 per cent. Determine (a) mass flow rate of refrigerant, (b) mass flow rate of cooling water in the condenser for a 7-degree change in temperature, and (c) the bore and stroke. Ans. (a) 0.5025 kg/s, (b) 3.43 kg/s, (c) 9.2 x 11 em (10) A refrigerant 22 refrigerating system is operating with a condenser temperature of 42°C and an evaporating temperature of O°C. (a) If the liquid line from the condenser is soldered to the suction line from the evaporator to form a simple heat exchanger and if as a result of this the. saturated vapor leaving the evaporator is superheated 10 degrees, how many degrees will the saturates liquid leaving the condenser be subcooled? Determine (b) the volume flow rate, and (c) the compressor work. Ans. (a) 5.3°C, (b) 1.102 L/sITR, (c) 0.662 kWITR (11) In a conventional refrigeration cycle which uses Freon 12 as the refrigerant, the temperature of the evaporating fluid is -20°C. It leaves the evaporator as saturated vapor at -20°C and enters the compressor. The pressure in the condenser is 1.30 MPa. The liquid leaves the condenser and enters the expansion valve at 40"C. It is proposed to modify this cycle by adding a


heat exchanger that would superheat the suction vapor to 25°C. Compare the coefficient of performance of these two cycles. What is the temperature of the refrigerant entering the expansion valve with the heat exchanger? For a load of 50kW, determine the vol ume flow rate for both cycles. Ans. 2.79, 2.80, 12°C, 0.0522m 3/s, 0.0523 m

3/s

Chapter

3

Actual Vapor Compression Cycle As the refrigerant flows through the system there will be pressure drops in the condenser, evaporator, and piping. Heat losses or gains will occure depending on the temperature difference b e twee n the refrigerant and the surroundings. Compression will be polytropic with friction and heat transfer instead' of isentropic. The actual vapor compression cycle may have some or all of the items of departure from the simple saturation cycle as shown on the p-l.diagram in Fig. 3-1.

as

~

1 Enthal~y,

kJJkg

Fig. 3-1. Actual vapor compression cycle

.......l


1.) 2.) 3.) 4,) 5.) 6,) 7.) 8.) 9.) 10.) 11.) 12.)

Actual Vapor Compression Cycle 59

Superheating of the vapor in the evaporator, 10-11.

Heat gain in the suction line, 11-12.

Pressure drop in the sunction line, 12-13.

Pressure drop due to wire drawing at the compressor

suction valve, 13-1. Polytropic compression with friction and heat transfer, 1-2. Pressure drop at the compressor discharge valve, 2-3. Pressure drop in the delivery line, 3-4. Heat loss in the delivery line, 4-5. Pressure drop in the condenser, 4-6. Subcooling of the liquid in the condenser or subcooler, 6-7.

Heat gain in the liquid line. 7-8.

Pressure drop in the evaporator, 9-10.

Show the thermodynamic states at various points on p-h diagram and calculate: (a) Refrigerating capacity in TR assuming 2 per cent loss of useful refrigeration by heat gain from room in brine cooler (b) Rate of flow of ammonia (c) Compressor IHP and mechanical efficiency (d) Compressor volumetric efficiency (e) COP of the cycle (D Make a system heat balance. SOLUTION

~

The pressure drop in the evaporator is large and that in the condenser is small. Due to various pressure drops, the capacity of the plant is decreased and the power consumption per unit of refrigeration is increased. The COP of the actual cycle is reduced.

Q.

110°C

(3-1) The following data were taken from a test on a twin cylinder, single-acting 15 em x 20 ern, 320 rpm compressor ammonia refrigeration plant. Temperatures of refrigerant: After expansion valve Entering brine cooler Leaving brine cooler Entering compressor Leaving condenser Entering condenser Leaving condenser Entering expansion valve Pressures of refrigerant: Compressor discharge and condenser Compressor suction Brine circulation Temperature drop of brine in cooler Specific heat of brine Input to motor Motor efficiency at this load Compressor jacket cooling water Temperature rise of jacket water

.&.

-25°C -25°C -18°C -8°C 120°C 110°C 30°C 32°C 1352 kPa 132 kPa 102kg/min 7Co 3.14kJlkg (K) 18.8 kW 92 % 5 kg/min 8.9 C

3

Qc 5

<'-'lId alorap room

Brine pump

4

Expansion32"C valve

3O"-C

I

L

condenser

Q4-6

60


Actual Vapor Compression Cycle 61 (c)

lit as

I

~

,,

t

~

m

Qj

lit _.

120°C

£ 152.2 kPa

132kPa

I~

Energy balance about compressor

Enthalpy, kJ/kg

h, == h, at 30°C == 341.77 kJfkg

h, == h 6 == h, at 32°C == 351.5 kJfkg

h, == h at 132 kPa and-18"C = 1445 kJfkg h. == h at 132 kPa and -8°C == 1470 kJlkg

(a)

(rn) (hI)

= (m) (h.) + Qj W = (m) (h 2-h) + Qj W = ~ H +' Q.J

+ W

Change of enthalpy, ~H = (rn) (h 2-h l ) = (2.091) (1725 - 1470)

= 533.2 kJ/min

VI

==

V

at 32kPa and -8°C == 0.95 m 3fkg h, == h at 1352 kPa and 120°C = 1725 kJfkg

Heat of jacket water, Qj = (row) (c) (~ t w) = 186.3 kJ/min

h,

;=

h at 1352 kPa and 110°C == 1700 kJfkg

Compressor indicated work, W

e

= (1.02) (rn.)

(c.) (~tb)

= (1.02) (102) (3.14) (7)

= ~ H + Q.J = 533.2 + 186.3 = 719.5 kJ/min or 11.99 kW

Refrigerating capacity (from brine) Q

= (5) (4.187) (8.9)

Compressor IHP

= 11.99 = 16.07 hp 0.746

= 2287 kJ/min

Compressor brake work _ 2287 - 211

= 10.84 TR

= 17.3kW Mechanical efficiency

(b)

Refrigerating capacity (from refrigeratiqn)

Qe

= (m) (h, - h 6 )

2287 = (rn) (1445 - 351.5)

m = 2.091 kg/min

~

= output of motor = (18.8) (0.92)

= indicated work brake work

_ 11.39 - ~

=

or 69.31 %

0.6931



(d)

V 1 = (rn) (v 1) = (2.091) (0.95) = 1.986 m 3/min V0 = ~D2 LN =

4

.2:... 4

(0.15 m)" (0.20 m) (320x2 cycles/min)

= 2.262 m 3/min va

(e)

V Vo

1.986 2.262

=~=--=0.878or87.8%

n

COP = ~ = W

2287 kJ/min

= 3.18

!

I,

(D

Heat balance Heat input to the system (a) Compressor, W = 719.5 kJ/min (b) Evaporator, Qe = 2287 kJ/min (c) Pipe line from evaporator to compressor, Q7.1 = (m) (h 1-h 7) = (2.091) (1470-1445) = 53.3 kJ/min

(3-2) An ammonia compression plant is to be designed for a capacity of 50 TR. The system operates with a condensing temperature of 40°C and an evaporating temperature of -20°C. The other data are the following: Temperatures leaving evaporator entering compressor leaving compressor entering condenser leaving condenser entering expansion valve Wiredrawing suction valve discharge valve Compressor speed Mechanical efficiency Volumetric efficiency Stroke-to- bore-ratio

-10°C -5°C 95°C 85°C 33°C 36°C 20 kPa 40 kPa 400 rpm 80% 77% 1.3

(d) Pipe line from condenser to expallsion valve, Q4-5 = (m) (h 5-h 4) = (2.091) (351.5 - 341. 77) = 20.3 kJ/min Total heat input = 719.5 + 2287 + 52.3 + 20.3 = 3079.1 kJ/min Heat rejected by the system (a) Jacket cooling water, Q.J = 186.3 kJ/min

A twin-cylinder, double-acting compressor is to be used. The heat absorbed by jacket water is 30 per cent of the indicated work of the compressor. Determine (a) the bore and stroke (b) the brake work, (c) the heat lost or gained between compressor and condenser, and (d) the heat rejected from the condenser.

SOLUTION

(b) Pipe line from compressor to condenser, Q2-3 = (m) (h 2- h) = (2.091) (1725-1700) = 52.3 kJ/min (c) Condensing cooling water, Qc = (m) (h 3-h4) = (2.091) (1700-341. 77) = 2840.1 kJ/min

•

.~

j

180.74 kP. ~ _11.O·14_~.

=lO~ _

Total heat rejected = 186.3 + 52.3 + 2840.1 = 3078.7 kJ/min Enthalpy, kJlkg

h.

~

= hi

at 33"C

= 3fi6.3 kJ/kg

Actual Vapor Compression Cycle (-):-)


(b)

"

GH =

(rn )

(h, - h 9 ) = (9.69) (1650 - 1470)

=

1744.2 kJ/min

QJ = 0.30 W

190.74 kPa .9

_-'It-----,I-'--I

W = GH + QJ W

= 1744.2

+ 0.30 W

W = 2491. 7 kJ/min or 41.53 KW

1

170.74kPa

Brake work

=

indicated work mechanical efficiency

~

41.53 = 51.91kW 0.8 or 69.56 hp

I·

COMPRESSOR

h , = h.at 33°C = 356.3 kJlkg h 6 = h, = h, at 36°C = 370.96 kJlkg

h,

= hat 190.74 kPa and -10°C

= 1460 kJlkg

h, = h at 190.74 kPa and -5°C = 1470 kJlkg v

1

h,

= v at 170.74 kPa and h. equal to h; = 0.75 = h at 1557 kPa and 95°C = 1650 kJlkg

m

3lkg

h, = h at 1557 kPa and 85°C = 1628 kJlkg

Q (a)

m =h _ s

h

=

(50 tons) (211 kJ/min.ton) (1460 - 370.96) kJlkg

7

V = (rn ) (v 1

1

)

= (9.69 kg/min) (0.75 m 3lkg) = 7.268 m

V 1 _ 7.268 = 9.439 rrrVmin VD=n- 0.77 va

v _ It D2 LN D-

=:

D 2 (1.3 D) (400) (2) (2)

D = 0.1794 m or 17.94 em L = (1.3) (17.94) = 23.32 em

...

3/min

Ql-4

(d)

Qc

= (rn ) (h1- h 4 ) = (6.69) (1650-1628) = 213.2 kJ/min = (rn )

(h.l-h) = (9.69) (1628-356.3) = 12,323 kJ/min

(3-3) An ammonia compression plant is to be designed for a capacity of 30 TR. The cooling water temperature requires a condenser pressure of 1400 kPa and the brine temperature a pressure of 291.6 kPa in the brine cooler. The following temperatures will exist at the points designated: compressor suction O°C en tering condenser 90°C leaving condenser 32°C leaving evaporator -5°C Wiredrawing through compressor valves suction 35 kPa discharge 70 kPa A two-cylinder, vertical cern pressor is to be used at 120 m/min piston speed. mechanical efficiency 80% adiabatic compression efficiency 82°k volumetric efficiency 75 clr stroke-to-bore ratio 1.20 \

4

9.439

= 9.69 kg/min

(c)

Determine (a) the rate of circulation of ammonia, (b) the BHP of the compressor, (c) the heat rejected at the condenser, (d) the bore, stroke, and rpm of the compressor, and (e) the COP.

66



(c)

SOLUTION

Qc = (m ) (h,

- h.)

= (5.685 kg/min) (1645 -

351.5) kJ/kg

= 7353.5 kJ/min

(d)

r/

as ~.

un

t

_.

,

.,.

VI

n

va

£

6'

___

291.6kPa _~.6JtP.!

__

"~~l ,

"

O.C V

-lO·C Enthalpy, kJlkg

h~ = h 6 = h, at 32°C = 351.5 kJ/kg h, = h at 291.6 kPa and _5°C = 1465 kJ/kg

= h at 1470 kPa and sz equal to SI = 1745 kJ/kg = h at 1400 kPa and 90°C = 1645 kJ/kg

(30 tons) (211 kJ/min.ton) = 5.685 kg/min (1465 - 351.5) kJlkg

Q"

= h_-h o

(a)

m

(b)

Isentropic work

Actual work Indicated work

=

= h, -

h. = 1745 - 1475

= 270 kJ/kg

isentropic work _ 270 = 329.3 kJlkg n 0.82 c

= (5.685 kg/min) ~329.3 kJ/kg) = 1872.1 kJ/min or 31.2 kW =

31.2 0.746 = 41.82 hp

indicated work 41.82

. . = BHP of compressor = mechanical efficiency 0.80

= 52.28 hp

~

o

=~ 4

(rpm)

D

= 0.1965 m or 19.65 em

L

= (1.20) (19.65) = 23.58 em

rpm

(e)

60 120 m/min =-- m rpm (2) (rpm)

D2 LN

7t 60 3.639= (-4DZ) ( - ) (2) rpm

h, = h at 291.6 kPa and O°C = 1475 kJlkg

3/kg v I = v at 256.6 kPa and h. equal to h, = 0.48 m hz,

_ 2.729 = 3.639 m/min 0.75

-

piston speed (rpm)

L

" ,,-r ' C

h,

= (m ) (VI) = (5.685 kg/min) (0.48 mvkg) = 2.729 m/min

Vo

t;

.

VI

"1

_ piston speed 2L

COP

= 254

(30 tons) (211 kJ/min.ton) = 3.38 (31.2 kJ/s) (60 s/min)

(3-4) A simple ammonia compression system has a compressor with a piston displacement of 1. 76 m Vm in , a condenser pressure of 1557 kPa and an evaporator pressure of 227 kPa. The liquid is subcooled to 35°C and the vapor leaving the evaporator and entering the compressor is -6°C. The temperature of the vapor leaving the compressor is 90°C. The heat rejected from the ammonia to the compressor jacket is 5070 kJ/h. The volumetric efficie vcy of the compressor is 80%. Determine" (a) machine capacity in tons, (b) IHP of the compressor, and (c) COP of machine based on indicated work.

68



SOLUTION

""

(

J

.LV-V.'

...... _

\

,

1<

,

"

CII

90°C

~

A _:

2

'\.

i

i

II

£

(3-5) Saturated vapor Freon 12 refrigerant at 219.12 kPa leaves the evaporator and enters the compressor at minus 5°C. The refrigerant leaves the condenser as saturated liquid at 25°C and enter the expansion valve at 22°C. Heat rejected from the condenser amount to 74 KW. The work to the compressor is 55.5 kJlkg, while the heat lost from the compressor is 4.2 kJlkg. If 1.15 kJlkg of heat are lost in the piping between the compressor and condenser, calculate the refrigeration capacity in tons. SOLUTION

I'~ l "

'I

,

Enthalpy, kJ/kg ~

h , = h , = h. at 35°C = 366.1 kJlkg h,

=h

v1

==

at 227 kPa and -6°C

~

= 1465 kJ/kg

v at 227 kPa and -6°C = 0.55 m.3lkg

£

h , == h at 1557 kPa and 90°C = 1640 kJ/kg (a)

(n

V1

(V ) o

) v

2

CII

6 '

= (0.80) (1.76) = 1.408 m.3/min

2 56 k g/ mIn . 1 = 1.408 mvrnin m =V_ =. v1 0.55 m 3/kg

Q e

= (rn) (h 1-h4 ) 2813.2 211

(b)

= (2.56) (1465 - 366.1)

219.12 kPa

.c.'

t

Enthalpy, kJlkg

= 2813.2 kJ/min l.15kJ11rg

= 13.33 TR

4.2kJ/kg

~H

= (m)

Q].

= - - = 84.5 k.I/min

W

= Q + ~ H = 84.5 + 448 = 532.5 kJ/min or

(h, - h.)

= (2.56) (1640 -

5070 60

1465)

= 448 kJ/min 55.5kJ/kg

.

' 8.875 kW

]

IHP (c) COP

= 8.875

0.746

11:9 h

P

= refrigerating effect indicated work

h, :;:: h at 219.12 kPa and -5°C = 350 kJlkg _ 2813.2 - 532.5

5.28

= 223.65 kJlkg == h, at 22°C = 220.75 kJlkg

h 4 = h, at 25°C

h, :;:: h,

h. :;:: h g at 219.12 kPa = 347.13 kJ/kg I

..l



Energy balance about compressor, h, + 4.2 = h. + 55.5 h,

= 350

+ 55.5 - 4.2

= 401.3

SOLUTION Qc

kJ/kg

Energy balance about piping from compressor to condenser, h, + 1.15 = h,

I'

Expansion Q valve e

h, = 401.3 - 1.15 = 400.15 kJ/kg

w

1'1

r

....l

Energy balance about condenser, qc + h, = h, qc

= 400.15-223.65 = 176.5 kJ/kg

(a)

Qe

Q 74 kJ/s mass flow rate, m = -c= = 0.419 kg/s qc 176.5 kJ/kg

Refrigerating capacity

= (rn) (h 7-h6 ) = (0.419) (347.13-220.75)

= (100 tons) (3.52 kW/ton) = 352 kW

Actual work =

(b) Refrigerating load

= (m w) (c) (~tw) = (m w) (4.187) (22-2) = 83.74m w kW

= 52.95 kW

83.74m w = 352

= 15.04 TR 52.9 kW = 3.52 kW/ton

(3-6) A vapor compression refrigeration system is designed to have a capacity of 100 tons of refrigeration. It produces chilled water from 22°C to 2°C. Its actual coefficient of performance is 5.86 and 35 per cent of the power supplied to the compressor is lost in the form of friction and cylinder cooling losses. Determine:

Qe 352

= - - = 60 kW

actual COP 5.86

m w = 4.2kg/s of chilled water . the evaporator = 22 + 2 = 120C mean temperature of water In 2 From the steam table, the specific volume of water at 12°C, v f = 0.0010005m 3/kg Volume flow rate of chilled water

(a) -size of the electric more required to drive the compressor in kW, (b) volume flow rate of the chilled water in liters per second, and (c) the condenser cooling water required in kg/s for a temperature rise of lOCo.

= (4.2 kg/s) 90.0010005 m 3/kg)

(c) Qj

= 0.0042 m 3/s or 4.2 Lis = 0.35W = (0.35) (60) = 21kW

Energy balance about the system

Q c + Qj

= W + Q e

Qc + 21

= 60 + 352

o, = 391 kW

72



Condenser cool i ng water, me :::

Qr (c)(L\t) w \\

V0

391 (4.187)(10)

:::

41t

D2 LN :::

1t . "4 (0.25 m)" CO.30rn) (300x4 cycles/min)

::: 17.671 m Vmin

t'l

'h

iI,r

= 9.34 kg/s

VI

(3-7) A 4-cylinder, single acting ammonia compressor 25 ern x 30 ern operates at 300 rpm. Condenser and evaporator pressure are 1204 kPa and 227 kPa, respectively. The vapor from the evaporator to the suction of the compressor is dry saturated and there is no subcooling in the condenser. Clearance factor of the compressor is 2 per cent. Compression process is polytropic with n = 1.20. Determine (a) the refrigerating effect, (b) the horsepower required. (c) the heat rejected to compressor jacket water, and (d) the heat rejected at the condenser.

::: (nv e ) (Vo ) ::: (0.9397) (17.671) ::: 16.605 rn 3/min

VI rn :::

w

w

J'

VI

:::

16.605 m 3/min 0.52826 m Vkg

npiv i [(P2)~ n -1] 1-n PI

::: - -

:::

(1.20)(227)(052826) [( 1204)\~~~1 -1] 1-1.20 227

::: 230.66 kJ/kg

SOLUTION

::: 230.66 kJ/kg (31.43 kg/min) ::: 7247.3 kJ/min or 120.76 kW

CIS

::: 120.79 ::: 1619 h 0.746 . P

l204kPa

- ..

I

P

~

(c)

1.

V

2

I

.

227 kPa

41

, ... '" -16°C

h, = h, at 1204 kPa = 346.6 kJ/kg

::: h g at 227 kPa

v] :::

Vg

at 227 kPa

= 1442.6 ktl/kg

= 528.26 L/kg

(a) Refrigerating effect ::: h t-h 4 (b) n ::: 1 + c-c vc

:::

P ..2. 1204 ..l... [2] = 1 + 0.02-0.02 [ _ _]1.20 P. 227

n

::: 0.9397

1

1442.6 - 346.6 ::: 1096 kJ/kg

I

I

[_2];: : (0.52826) [ 12272]l.2O ::: 0 1315 m 3/kg PI 04'

h 2

::: h at 1204 kPa and 0.1315 rn 3/kg ::: 1610 kJ/kg

,1H

= (rn)

Enthalpy. kJlkg

h.

::: V

(h2 - h)

= (31.43 kg/min) (1610 -

1442.6) kJ/kg

::: 5261.4 kJ/min Qj (d) Q e

= W-dH = 7247.3-5261.4 ::: 1985.9 kJ/min or 33.1 kW = (m) (h2-h3 ) ::: (31.43) (1610-346.6) ::: 39,709 kJ/min or 661.8 kW

74-

, t

,~



(3-8) An ammonia compression system operates at a capacity of 150 tons. Condensation temperature is 35°C. The evaporation temperature in brine cooler is -24°C. Ammonia leaves the evaporator and enters the compressor at -8°C. Ammonia enters the expansion valve at 30°C. Were drawing through the compressor valves: suction 35 kPa, discharge 70 kPa, compression is polytropic with n = 1.22, volumetric efficiency is 75%. Calculate (a) IHP,(b) heat transferred to cylinder water jacket, (c) piston displacement, (d) heat transferred to condenser circulating water, and (e) COP.

h, = h at 1422 kPa and 0.1356 m 3Jkg = 1760 kJ/kg h,

= h, = 1760 kJ/kg

W

= nPlvl

n-l

[(..ELy~

I-n

W

SOLUTION

=

PI

(1.22)( 124.2)(1.0) 1-1.22

-1]

r( L

1422) \2~;1 -1] = 380.3 kJ/kg 124.2

= (28.18 kg/min) (380.3 kJ/kg) = 10,717 kJ/min or 178.6 kW

w•

IHP

J

~ i

~

pV U2

~

~;~ _

_0

·c

= 239.4 hp

= (m) (h 2-h 1 ) = (28.18 kg/min) (1760-1465) kJ/kg = 8313.1 kJ/min

-t,-), , 1 "

,

(b) ~H

178.6

~c

'"

-2~OC

Q.J =

W-~H = 10,717-8313.1 = 2403.9 kJ/min or 40.07 kW

(c) VI = (m) (v 1) = (28.18 kg/min) (1.0 m 3Jkg ) = 28.18 m 3/min

Enthalpy, kJlkg

h, = h, at 30°C = 341.8 kJ/kg h,

v I = v at 124.2 kPa and h. equal to h, = 1.0 m 3Jkg

Qe

050 tons) (211 kJ/min.ton) = 28.18 kg/min (1465-341.8) kJ/kg

h, -h 5 I

V 2

.....1

V8

= 37.57 m 3/min

•

= h at 159.2 kPa and -8°C = 1465 kJ/kg

h. = h, = 1465 kJ/kg

(a) m

28.18 VD =nVI - - = 075

I

= V [~] -;- = 0.0)[124.2]1.2'2 = 0.1356 m 3Jkg I

PI

1422

(d) Qc == (m) (h 3-h4 ) = (28.18) (1760-341.8) = 39,965 kJ/min or 666.1 kW (e) COP = h 6-hs =

1465-341.8 380.3

= 2.95

l



PROBLEMS (1) A Freon 12 refrigeration system, with a 10-TR capacity and a coefficient of performance of 3.23, operates with a condenser pressure of 984.5 kPa and an evaporator pressure of 150.8 kPa. The temperatures entering and leaving the compressor are -10°C and 80°C, respectively. The temperature entering the expansion valve is 34°C. The compressor is water jacketed and operateds with unknown pressure drops through the valves. Compute (a) the mass flow rate per minute, (b) the indicated work, and (c) the heat removed by the jacket water. Ans. (a) 18.43 kg/min, (b) 19.87 kW, (c) 270.5 kJ/min (2) A test of a 10-TR ammonia vapor compression refrigeration system gave the following results: condenser pressure 1600 kPa evaporator pressure 191 kPa temperature leaving evaporator coils -10°C temperature entering compressors O°C temperature leaving compressors 100°C temperature entering condenser 80°C temperature leaving condenser 38°C coefficient of performance 3.1 (A) Find heat lost or gained, (a) between evaporator coils and" compressor, (b) between compressor and condenser, and (c) to condenser water. ~

(B) 'Find, (a) temperature in the evaporator coils in saturated state, and (b) quantity of the vapor in the evaporator coils fol~wing expansion through valve. (C) water.

Find, (a) the work, and (b) the heat absorbed by jacket Ans. A: (a) 44.37 kJ/min, (b) 106.1 kJ/min, (c) 2399.5 kJ/min B: (a) -20°C (b) 14.62% C: (a) 11.34 kW, (b) 329.5 kJ/min

(3) A refrigerant 12 refrigerating system operates with a condensing temperature of 38°C and an evaporating temperature of -12°C. The refrigerant leaves the evaporator and enters the compressor at -2°C. The compressor is a 4-cylinder, V arrangement single acting type, direct driven by an electric motor at 875 rpm. The clearance is 5 per cent, the capacity is 40 tons, bore-stroke ratio is 1.25 and the compression is polytropic with n = 1.275. Determine (a) the bore and stroke, (b) the piston speed, (c) the indicated hp of the compressor. Ans. (a) 14.73 x 11.78 em, (b) 206.2m/min, (c) 50.3 hp (4) A food freezing requires 20 tons of refrigeration at an evaporator temperature of -30°C and a condenser temperature of 35°C. The refrigerant Freon 22 is subcooled 4°C before entering the expansion valve and the vapor is superheated 5°C before leaving the evaporator. The clearance is 4%, the suction and discharge valve pressure drops are 35 kPa and 50 kPa, respectively. Compression is polytropic with n = 1.08. The mechanical efficiency is 75%. An 11.5 x 11.5-cm, six cylinder single acting compressor running at 1000 rpm is used. Determine (a) clearance volumetric efficiency, (b) piston displacement, (c) total volumetric efficiency, (d) theoretical hp, (e) actual hp, (f) heat rejected during compression, (g) heat rejected to condenser cooling water, and (h) quantity of cooling water in the condenser if the temp~ature rise is 6°C. Ans. (a) 67.38%, (b) 7.167 m Vm, (c) 66.61%, (d) 35.86 hp (e) 47.85 hp, (f) 332 kJ/min, (g) 5492.3 kJ/min, (h) 218.6 kg/min

Refrigeration and Air Conditioning 79

Chapter

4

I

•

To compressor

Flash tank

Multipressure Systems )

;]

i'

.1

A multipressure system is a refrigeration system that has two or more low-side pressures. The low-side pressure is the pressure of the refrigerant between the expansion valve and the intake of the compressor. Amultipressure system is distinguished from the single-pressure system, which has but one low-side pressure.

Removal of Flash Gas A saving in the power requirement of a refrigeration system results if the flash gas that develops in the throttling process between the condenser and evaporator is removed and recompressed before complete expansion. The vapor is separated from the liquid by an equipment called the flash tank. The separation occurs when the upward speed of the vapor is low enough for the liquid particles to drop back into the tank. Normally, a vapor speed of less than 1 mls will provide adequate separation.

• To evaporator

Expansion valve

Fig. 4-1. Flash tank for removing flash gas during expansion process

Intercooling Intercooling between two stages of compression reduces the work of compression per kilogram of vapor. Intercooling in a refrigeration system can be accomplished with a watercooled heat exchanger or by using refrigerant. The watercooled intercooler may be satisfactory for two-stage air compression, but for refrigerant compression the water is usually not cold enough. The alternate method uses liquid refrigerant from the condenser to do the intercooling. Discharge gas from the low-stage compressor bubbles through the liquid in the intercooler. Refrigerant leaves the intercooler as saturated vapor. Intercooling with liquid refrigerant will usually decrease the total power requirements when ammonia is the refrigerant but not when refrigerant 12 or 22 is used.

Two evaporators and one compressor (4-1) Calculate the power required by a system of one compressor serving two evaporators. On evaporator carries a load of 35 kW at 10°C and the other a load of 70 kW at -5°C. A back-pressure valve reduces the pressure in the 10°C evaporator to that of the -5°C evaporator. The condensing temperature is 37°C. The refrigerant is ammonia. What is the COP?

80

Multipressure Systems Refrigeration and Air Conditioning 81

SOLUTION:

10

, ~

0e

evaporator

2

Condenser

mass flow rate, m 4 ==

3

35

35

h 5-h4

== 1471.6-375.9 == 0.0319 kg/s

-5°e evaporator

f

,..- Evaporator 35KW ]ODC

I

I I

.... 1A

6

,

Pressurereducing valve

I

8

or

Evaporator 70KW

-5OC

1

mass flow rate, m 7 == -..7..Q

h g-h 7

==

70

1456.2-375.9 == 0.0648 kg/s

Total mass flow rate, m 1 == m 6 + m s == m 4 + m

7

== 0.0319 + 0.0648 == 0.0967 kg/s

h& M6

~

hI ~

,

MI

4

f

.I

~

hS

3/

I /4

1

1

1432kPa

616 kPa

>2

//~C

MS

m.h. == m 6h6 et- m.h,

356 kPa

(0.0967) h,

== (0.0319) (1471.6) + (0.0648) (1456.2)

hI = 1461.3 kJlkg

Enthalpy, 1rJ1kg

= h, = h, 37°C = 375.9 kJlkg bs = h, = h g at 10°C = 1471.6 kJlkg


S2

equal to

SI

= 1665 kJlkg

h, = h,

h g = h g at -5°C = 1456.2 kJlkg

J

Power == (rn.) (h 2-h 1 ) == (0.0967) (1665-1461.3) == 19.7 kW

COP = ~ 19.7

= 5.33

~

fJ '

82



SOLUTION

Two compressors and one evaporator Two-stage compression with intercooling and removal of flash gas is often the ideal way to serve one low-temperature evaporator. This system requires less power than with a single compressor, and often the saving in power will justify the cost of the extra equipment. (4-2) Calculate the power required in an ammonia system which serves a 210-kW evaporator at -20°C. The system uses two-stage compression with intercooling and removal of flash gas. The condensing temperature is 32°C.

r.

,.----------------11 I

...

...1

4

Condenser 1--------olIlI----------. 320C 1 • i

PI = saturation pressure at -20°C = 190.74 kPa

P4

Intermediate pressure for minimum work, P2

2 1

-Y090. 74) 0239.6) = 486.3 kPa

=h

at 1239.6 kPa and S4 equal to S3

= 1615 kJ/kg

= h, = h, at 32°C = 351.5 kJ/kg

= h = h, at 486.3 kPa = 215.3 kJ/kg

g

mass flow rate through evaporator, m r ;:::

tercooler and flash tank

=

h, = h at 486.3 kPa and S2 equal to SI = 1563 kJ/kg

h 3 = h g at 486.3 kPa = 1465.5 kJ/kg

h, 6

= -YPIP4

h. . = h g at -20°C = 1437.2 kJ/kg

h4 h,

s

= saturation pressure at 32°C = 1239.6 kPa

210 h, - h,

mI

210 = 0.172 kg/s 1437.2 - 215.3

= m = m,

2

ma

Evaporator 210kW -3)OC

......-- m2 h2

ha

1239.6kPa I

~

it

£

I

k. 7 4

1'1

nJ!

7f---! I

, 32'»C

ms

.1 ~

I

190.74 kPa

h7

m7

Enthalpy, kJlkg

Mass balance about intercooler m 2 = m 7 and m, = m,

.....

Refrigeration and Air Conditioning 85 84

Mul tipressure Sy:-:tems

Encruy balance about intercooler mh, + mh, = m.h, +mlh:l (0.172) n563) + (m) (351.S) = (0.172) (215.3) + (m 3 ) (1465.5)

# (4-3) In an ammonia system, one evaporator is to provide 200 kW of refrigeration at -30°C and another evaporator is to provide 220 kW at SoC. The system uses two-stage compression with intercooling and is arranged as shown below. The condensing temperature is 40°C. Calculate the power required by the compressors. What is the COP of the system?

m:] = 0.208 kg/s

f

Low-stage power = (rn.) (h 2-h 1) = (0.172) (1563-1437.2)

I-~

4 r--

...

= 21.6kW

3

High-stage power = (rn.) (h 4 - h ) = (0.208) (1615-1465.5) =31.1kW

Evaporator 5°C 220KW

Total power = 21.6 + 31.1 = 52.7 kW

l

Two compressors and two evaporators

I'

The system which has two evaporators operating at different temperature is common in industrial refrigeration.

2

- - ,-

Intercooler and flash tank

-

7

Evaporator -30°C 200KW

1

1557kPa

L:

~

>

4

1 1+--51772 ~c

120 kPa

5~C

1

,

\

\

-30°C

-----Enthalpy, kJlkg

----l

~II~

86



mass flow rate through high-stage compressor, m, == m 6 + m = 0.204 + 0.218 = 0.422 kg/s

SOLUTION h, = h , at -30°C = 1422.9 kJ/kg h, = h at 517 kPa and

equal to

S2

SI

= 1630 kJ/kg

h,.) = h g at 5°C = 1466.8 kJ/kg

h,

=h

at 1557 kPa and

equal to

S4

SI

= 1625kJ/kg

=h, =hi at 5°C = 223.2 kJ/kg

COP

-30°C Evaporator ___ " 200 -- h -h 1422.9-223.2 ')()()

mass flow rate, m! ==

1

of

~ \

,

m7

R

= m , == m 2 == ill l

5°C Evaporator 200 200 mass flow rate, m 6 = - - - == 1466.8-390.8 == 0.204 kg/s h 3-h 6 Heat balance about intercooler

•

1. Dl_

..

las

---t

he

bor

.., mh, + m.h,

j

= rnh, + m.h,

390.6m + (0.167) (1630) == 1466.8m + (0.167) (223.2) m = 0.218 kg/s

\) =

(0.167) 0630-1422.9)

High-stage power = (m 3 ) (h 4 - h ) == (0.422) 0625-1466.8) = 66.76 k\V Total power == 34.59 + 66.76

hi) = h, = hi at 40°C = 390.6 kJ.kg h,

Low-stage power = (rn.) (h 2- h = 34.59 kW

==

= 4.142

= 101.4 kW

'.

iii:'

88

Multipressure

Sy~t('ms


PROBLEMS (1) In an ammonia refrigeration system, the capacity is 210 kW at a temperature of -20°C. The vapor from the evaporator is pumped by one compressor to the condensing pressure of 1431 kPa. Later, the system was revised to a two-stage compression operating on the cycle shown below with intercooling but no removal of flash at 555 kPa. (a) Calculate the power required by the single compressor in the original system. (b) Calculate the total power required by the two compressors in the revised system.

(2) An ammonia refrigeration system has a capacity of210 kW at an evaporating temperature of -20°C when the condensing pressure is 1431 kPa. (a) Compute the power requirement for a system with a single compressor. (b) Compute the total power required by the two compressors in the system shown below where there is no intercooling but there is flash-gas removal.

3

"

~

I

I

.1

.2 555kPa

5

a

5

Intercooler

1 •

-

-

Evaporator I 2

Evaporator 210kW I

-

210kW -20°C

f

-20OC

Ans. (a) 60.36 kW (b) 56.23 kW

Ans. (a) 60.35 kW, (b) 56.87 kW

1 •

I

Refrigeration and Air Conditio,ning 91 .J

SOLUTION

Chapter

5

(a)

Refrigerant 22 cycle

Cascade Systems

497.6kPa

as

!l

~

~~

::l gj

f'

/,• ,~

1

~

A cascade system combines two vapor-compression units, with the condenser of the low-temperature system discharging its heat to the evaporator of the high-temperature system. Cascade systems can normally furnish refrigeration down to about --100°C. There are two types of cascade systems, the closed cascade condenser and the direct-contact heat exchanger. In the closed cascade condenser fluids in the high-pressure and low-pressure system may be different but in the direct-contact heat exchanger the same fluid is used throughout the system. The direct-contact heat exchanger type is most common used. (5-1) Acascade refrigerating system uses refrigerant 22 in the low-temperature unit and refrigerant 12 in the high-temperature unit. The system develops 28 kW of refrigeration at -40"C. The refrigerant 12 system operates at -lODe evaporating and 38°C condensing temperature. There is a 10-degree overlap of temperatures in the cascade condenser. Calculate (a) the mass flow rate of refrigerant 22, (b) the mass flow rate of refrigerant 12, (e) the power required by the refrigerant 22 compressor, and (d) the power required by the refrigerant 12 compressor.

.c condenser

1/

105kPa

"' -, ODe

Enthalpy, kJlkg

h, = h, = h, at ODe = 200 kJlkg h)

= h g at -40

= 388.6 kJlkg

DC

h, = h at 497.6 kPa and Jill

Qe h 1-h4

S2

equal to s) = 427 kJlkg

28 kJ/s = (388.6-200) kJlkg

0.1485 kg/s (b)


914.2kPa WI

G

Expansion .I¥e

G)

~2

4

e

as

Hlp-pmsltte

'1

k

~

E~

::l 00 00

~

8

.Ie

219.1 kPa

1/ '~De " "

-lODe

"'2

to

Low-pressure tvaponlor

Enthalpy, kJlkg

h.

= h~ = hi at 38°C = 236.R

k,J/kg

92 Cascade Systems


h.:J = h g at -IOoe = 347.1 kJlkg

h, = h at 914.2 kPa and

S6

SOLUTION

equal to

85

= 372 kJlkg

(a) Ammonia cycle

(b) Refrigerant 12 cycle

"

~~ .,,,,~ .., MS

m5

ba

.

~

Energy balance about cascade condenser,

m.h, + rn.h, = m.h, + rn.h,

~ ).

~

" ooe

m;j(hs-h s ) = m 1(h2-h 3 )

m s(347.1-236.5) = (0.14.85) (427-200) m s = 0.3048kg/s (c) Refrigerant 22 cycle W 2 = m 1 (h 2 - h.) = (0.1485) (427 - 388.6) = 5.70kW (d) Refrigerant 12 cycle W 1 = m, (h 6 - h) = (0.3048) (372 - 347.1) = 7.59 kW

Enthalpy, kJlkg

h, = h, = h. at ooe = 200 kJ/kg h) = h g at -35°e = 1415.2 kJ/kg v) = v g at --35°e = 1212.5 L/kg h, = h at 430.4 kPa and S2 equal to s) = 1620 kJ/kg Qe 10kJ/s m) = h)-h 4 = (1415.2 _ 200) kJ/kg = 0.00823 kg/s V - -V)- D nva = 11.74

(m)) (v) n va

(0.00823 kg/s) (1212.5 L/kg) ~ ~ 0.85

-

Lis


1

(5-2) The events of a cascade, split-stage, low-temperature refrigerating cycle are as follows: A refrigerant 12 vapor compression system operates between -10oe and 33°e with saturated liquid on the high side of the expansion valve and saturated vapor entering the compressor whose volumetric efficiency is 88 per cent. An ammonia compression system operates between -35°e and ooe with saturated liquid entering the expansion valve and saturated vapor entering the compressor whose volumetric efficiency is 85 per cent. The refrigerant 1Z evaporator serves as the condenser for the ammonia. For the ideal cycle and 10 kW of refrigeration, find (a) the displacement rate of each compressor, and (b) the total actual power required for compression efficiencies of 85 per cent.

lI:l

~

~~

;j

rn rn

&:.

I

11

~

219.1kPa

,.

....

-lOOC

Enthalpy kJlkg

h, = h, = h, at 33°e = 231.5 kJ/kg h. = h g at -10 oe = 347.1 kJ/kg v . = v g at -10 oe = 76.65 L/kg h" :;: h at 80S.4 kPa and Sfi equal to :J

.j

s', =

370 k.Jlkg

94

Cascade Systems

Refrigeration and Air Cond it.ionmg 9.5

m.I h, - h s) = m l (h , - h.) m, (347.1-231.5) ,)

SOLUTION

= (0.00823) (1620-200) 7

m.:J = 0.101 kg/s

v= D

V_o

(rn.) (v .) :) ,)

n va

n

\";1

(0.101 kg/s) (76.65 Lis) = 8.80 Lis = 0.86

(a)

40°C High-pressure condenser

c:'

Expansion valve

r-C)

(b) Ammonia cycle Theoretic work -= mCh, - h.) = (0.00823) (1620 - 1415.2) ==

Actual work

3

1.69 kW

= 1.69/n c = 1.69/0.85 = 1.99 kW

Expansion


~ ~ ml -

-1r-C)

-so-c

Evaporator

Theoretical work == m,_, (h fi- h) == 2.:31

Actual work

I

vwve~

Compressor

__---l

= (0.101) (370-347.1)

kW

= 2.31/n c = 2.31/0.85 = 2.72 kW

Total actual work = 1.99 + 2.72 = 4.71 kW

1557kPa as

~

!1l

!1l

(5-3) In a certain refrigeration system for low temperature application, a two stage operation is desirable which employs ammonia system that serves a 30-ton evaporator at -30°C. The system uses a direct-contact cascade condenser, and the condenser temperature is 40°C. Find the following: (a) Sketch the schematic diagram of the system and draw the complete process on the ph diagram. (b) The cascade condenser pressure in kPa for minimum work. (c) The mass flow rate in the low and high pressure loops in kg/so (d) The total work.

~

.

6

0

~

~a

\

(432kPa)

P ll9.9kPa

--r~, ~C Enthalpy, kJlkg

(b) Intermediate pressure or minimum work ==-V P 1Pfi

= -V (119.9)(1557) = 432 kPa h. = hI' at -30°C = 1422.86 kJ/kg h , = h at 432 kPa and s~ equal to 8 1 = 1600 kJ/kg h, = h 4 = h. at 432 kPa = 200.47 kJ/kg h) = h g at 432 kPa = 1461.81 kJ/kg h, = h at 1557 kPa and Sfi equal to s, = 1638 ktJ/kg h == h, == h. at 1;');')7 k Pa == :~~)O,f)9 k,J/kg'

/11'


96 Cascade Systems

Q (30) (3.52) kJ/s 6 k / (c) m l = hl-l\ = (1422.86-200.47) kJ/kg = 0.08 4 g s Low pressure loop

j

m l = m 2 = m:, = m 4 = 0.0864 kg/s

High pressure loop

m s = m, = m? = m s

PROBLEMS (1) A two-stage cascade refrigeration system uses ammonia as the working substance. The evaporator is at -35°C and the high-pressure condenser is at 1514.2 kPa. The cascade condenser is a direct-contract type. The refrigeration load is 90 tons. Determine (a) the mass flow rate in the low-pressure loop, (b) the mass flow rate in the high-pressure loop, (c) the power required, (d) the COP, and (e) the quantity of the fluid entering the evaporator.

Energy balance about heat exchanger

Ans. (a) 0.2572 kg/s, (b) 0.341 kg/s,

m.h, + m.h, = m.h, + m.h,

(c) 116.08 kW, (d) 2.73, (e) 10.37lJt'

m.h, + m.h, = m.h, + m.h,

m s =

m l (h 2 - h,)

h)-hF\

J

I

m.o

I

a

J1 l

(d) Work of low pressure compressor, W LP = m l (h , - h.)

"

I

..1

(0.0864) (1600-200.47) = 0.1129 kg/s 1461.81-390.59

= (0.0864) (1600 -

1422.86)

(2) A two-stage cascade refrigeration system uses ammonia as the refrigerant. The mass flow rate in the high-pressure loop is 0.10 kg/so The condenser saturated temperature is 38°C and the evaporator temperature is -40°C. The cascade condenser is a direct-contact type. Determine (a) the refrigerating effect in tons, (b) the power required, and (c) the COP. Ans.la) 26.19 tons, (b) 37.22 kW, (c) 2.48

= 15.31 kW

Work of high pressure compressor, W HP

= m s (hfi- h) = (0.1129) (1638 - 1461.81) = 19.89 kW

Total work

= 15.31

+ 19.89

= 35.20 kW

6 6


Chapter

The air-cycle system is ideally suited for use in air-craft because it is light in weight and requires less space than the vapor-compression cycle. One disadvantage of the air cycle is that it is not as efficient as the vapor-compression cycle.

6

Air-cycle refrigeration may be designed and operated either as an open or a closed system. In the closed or dense-air system, the air refrigerant is contained within the piping or component parts of the system at all times and with the refrigerator usually maintained at pressures above atmospheric level. In the open system, the refrigerator is replaced by the actual space to be cooled with the air expanded to atmospheric pressure, circulated through the cold room and then compressed to the cooler pressure. Such a system is inherently limited to operation at atmospheric pressure in the refrigerator.

Air-Cycle Refrigeration

COOLER

p

r 3•

2

..

1

3

'r

Fig. ~l. Closed or dense

Fig,

~2.

~

1

..

Open-air system

air system

The air-cycle refrigeration system is the only air cooling process developed for wide commercial application in which a gaseous refrigerated is used throughout the cycle. Compression is accomplished by a reciprocating or centrifugal compressor as in the vapor-compression cycle, but condensation and evaporation are, of necessity, replaced by a sensible cooling and heating of the gas. An air cooler is used in place of a condenser and a refrigerator in a place of an evaporator. The expansion valve is replaced by an expansion engine or turbine.

v Fig. 6-3. Ideal air-refrigeration cycle Process 1-2: isentropic compression Process 2-3: constant pressure rejection of heat Process 3-4: isentropic expansion Process 4-1: constant pressure absorption of heat

s

100 Air-Cycle Refri gerat ion

Refri ge r a tion a nd Air Co ndi t ion ing 10 1

Analysis of the Air-Refrigeration Cycle (1 )

P olytropi c ex pa ns ion

( b)

n-1 nP-aV:1 W = - [ (- P4)n- - 1]

Refrigerat or

E

Refrigeratin g effect = (rn) (cp ) (T1- T), kW whe re m = ma s flow r ate of a ir, kg/s c p = 1.0 062 kJ/kg.K , specific heat of ai r at con stant pressure

= nP aV1 [(~)11_1] 1-n

T 2 ) , kW

Pa

(5) Net wo r k N et work

= ( m ) (c) (T:l -

Pa

n- 1

(2) Cooler Heat r ej ect ed

1-n

= comp ressor wor k -

ex pa nde r work

(6) Co effici ent of Perform anc e COP

(3) Comp r essor

= r efrig eratin g effect work

( a)

Is en tropi c compressi on

W = kP 1V 1 c 1-k =

l

(b )

(6- 1) An ai r r efrigerat ion system oper at ing on a closed cycl e is requir ed t o produce 50 kW of r efrigerati on with a cool er pressure of 1550 kPa and a r efr igerator pressure of 448 kPa . Le aving-air te m pe r a t u res a re 25°C for cool er a n d 5°C f or r efrigerator. As suming a t heo retic a l cycl e wi th ise n t r op ic co m p r es s io n a n d ex p a ns io n , n o cl earan ce a n d n o lo s se s , d et ermin e (a) th e mas s flow rat e , ( b) th e compress or di splacem ent , (c) t h e ex pa n de r displ a cement, a n d (d ) t h e COP.

[(p ~ ) k-1 - 1] PI

k

kmRT 1 [(p ~ . k - l ] 1-k ~) k - 1

P olytropic compression

[(~ n~l

np lV l

We =

1-n

PI

nmTR J =

[t ~

PI

l _ n

SOLUTION p

]

3

r=

-- ..

-1

)n:1 - 1 ]

(4) Ex pa nder

•

--

~

-..

>

1

(a) Isentropic expa n sion

v.. F. T

kp-'IY-1 [(P4 ] = - )~ k - 1 1- k P1 _ kmRT1 - 1-k

[(~) k: P1

1-1]

v PI = P2 = T1 = T, =

448kPa 1150 kPa 5 + 2 73 = 27 8 K 25 + 27 3 = 298 K

.,.-1'1'( 102 Air-Cycle Refrigeeration

P2

= T 1 [ -PH -]

T2

1550 1.4-1

[- ] 1.4 448

= 278

P4 k-1 448 1.4-1 P::3] k = 278 [1550] 1.4

= T3 [

T4

k-1 k


(6-2) A 35-kW refrigeration load is to be carried by an air refrigerating system with a discharge pressure of 1370 kPa and a refrigerator pressure of 345 kPa. Compression and expansion processes are polytropic with n = 1.37. The pressure at the start of expansion is 1344 kPa and the temperature is 32°C. Air temperature leaving the refrigerator is O°C and entering the compressor is 5°C. Clearance for both compressor and expander is 5%. Determine (a) the piston displacement of the compressor and expander, (b) the net work and, (c) the COP.

= 396 .3K

= 209.0 K

(m) (cp ) (T i - T 4 ) = 50

(a)

(m) (1.006~) (278

- 209) = 50

m = 0.720 kg/s

(b)

SOLUTION

Compressor pirston displacement without clearance, V =

mRTl Pi

1

(0.72) (0.28708) (278) 448

=

= 0.128

3

p

3/ m s

,

1344kPa -t---

(c) Expander pistvon displacement without clearance, V 4

=

(0.72) (0.28708) (209) mRT 4i4 = 448

P4

kmRT

P

«n w c =_lrC(_2) 1 - k L PI

k-1 k

\

= 0.0964 m 3/s

\

\

"

v

448

P2 = 1370 kPa

k-1

= kmRT [C'(~fk -1]

1 - k . P3

=

(0.~72)

= 345 kPA

PI

= 1344 kPa T 1 = 5 + 273 = 278 K

T 3 = 32 + 273 = 305 K

3

(1.4)

(0.28708) (298)

1-1.4

6

[(1550)\~~1 -1]

= 85.61 k\W

=

-

345kPa

-1

1-1.4

E

.

\

]

= (1.4) (0 .. 72) (0.28708) (278)

W

l370kPa -1----2

P3

448 1.4-1 [ (1550)--rr- -1]

T a-

=0 +

273

= 273 K

64.37 kW

Net work

= 8:,5.61 -

COP = _ 5~

64.37

= 21.24 kW

P2

T2

= T I [ -PI]

T

= T

= 2.35 4

3

.£=!.. n

1370

= 278 [ 345 -]

!.d1..::! 1.37

- 403 .5 K

p4 .!!=.! 345 !.d1..::! [ _ ] n = 305 [ _ ] 1.37 = 211.3K P3 1344

;~

(~,

I~.

~


104 Air -Cycle Refrigeration (m ) (c

( a)

ll

)

(T, - T)

= 35

n-1

W

(rn ) (1.0062) (273 - 211 .3)

m

E

= 35

1

1

[ P2]n PI

vc

= 1 + 0 .05 _ 0 .05[1370J 1.:37

345

= 0 .9132 (0 .564) (0 .28708 ) (278)

345

= 0.1 30 m 3/s

~ :~i~2

pi ston displacement =

= 0 .142 m

3/s

Expander .

1

1

[ P3]n

n vc: =l+c-c P 4

= 1 + 0.05 -0.05 [1344J1.37

345

.

= 0.9151 mRT 4 volume flow rate = - -

P4

= 0 .099 . diISP 1acement piston

(b) W c

=

mRT n

(0. 564 ) (0 .28 708) (2 11.3)

345

m 3/s

0 .099 = 0 .108 m 3/s = 0.9151 P

I

n-1

[(-i-)-n- ~lJ

= (1.37) (0. 56 4) (0 .28 708 ) (278) 1 - 1.37

= 75.21 kW

= 56 .20 kW

Net work

= 75.21 -

( c) COP =

mRT 1 volume flow rate = - -

Pl

P3

1-n

1.37-1 _ (1.3 7) (0 .56 4) (0 .28708) (305) [( 345 fT37-- 1] 1-1.37 1344

= 0 .564 kg/s

Compressor n =l+c-c -

= nmRT 3 [(~)n- -1]

[(1370) \1~;1_1J

34 5

:~

1 . 1

56.20

= 1.84

= 19 .01

kW

"

1'1,,(1'

106 Air-Cycle Refrigeration

PROBLEMS (1) An open air refrigeration system carries a load of 35 kW with a suction pressure of 103 kPa and a discharge pressure of 690 kPa. The temperature leaving the refrigerator is 5°C and that leaving the cooler is 30°C. The compression is polytropic with n = 1.33 and the expansion is also polytropic but with n = 1.35. Determine the power required and the COP.

Chapter

7

Steam-Jet Refrigeration

Ans. 23.77 kW, 1.47 (2) An air refrigeration system is required to produce 52.5 kW of refrigerati-on with a cooler pressure of 1448 kPa and a refrigerator pressure of 207 kPa. Leaving air temperatures are 29°C for cooler and 5°C for refrigerator. Expansion is. isentropic and compassion is polytropic with n = 1.34. Determine the COP.

/

IftZIdfLINB WA'I'D BITVBN

/

tnBAII MOZZL&

TOCONDmfBD

~

Ans. 1.47 1~;~~:::!::':~';-:=:t~"~::~E:"

I

/

BOO8I'DZJI:C'I'OJl SPRATS

IlADUP-- WATER

"_'_".'_16.".

WATEll LAVEL

COImloL

C3CtJLATD«JPUIIP / ' "

COOLKDARBA

Fig 7-1. Steam-Jet Refrigeration System

Steam jet refrigeration is a type of water evaporative refrigeration wherein water is used as the refrigerant. The principle of operation is based on the fact that water will boil or vaporize rapidly, at a relatively low temperature, if the pressure on the surface is reduced sufficiently.

~


Cascade System 109

A steam jet refrigeration is shown in Fig. 7-1. An ejector sucks or draws water from the surface of the water in the evaporator or flash chamber, causing the pressure in the evaporator to drop. The ejector reduces the pressure in the evaporator to a point at which the water will vaporize at the desired temperature. In vaporizing, it absorbs heat and cools the remainder of the water in the evaporator. The steam pressure at the ejector nozzle should be about 1030 kPa.

SOLUTION DaZZle ateam

he

lIlab-up

The evaporation of the water in the evaporator, reduces the temperature of the water in the evaporator. This cold water, 4°C to 21°C, is circulated by means of pumps, to the area to be cooled. Steam jet systems are used extensively in air conditioning, and for cooling of water in certain chemical plants for gas absorption. The cooling temperatures provided by the steam jet mechanism are usually between 4°C and 21°C. Temperatures below 4°e are impractical due to the danger of freezing.

·.,.t I "

(7-1) A steam-jet pump maintains a temperature of 5°C in the evaporator. The cooled water leaves at the same temperature and warms to 10 as it does its refrigeration. Make-up water is available at 2re. There are required 1.27 kg of nozzle steam per kg of vapor removed from the evaporator. For a 175-:[{W plant, compute the volume of vapor handled, the quantity of steam needed, and the quantity of chilled water.

13

vapor

water

-

moe

J,.

4

z

W8l1D.water

woe

z

-.

r---.-

- - - - - ----: .-. -=-:=-" - -- - .. --- -- --=--= --- '",":2f'~§ ~ ~__~::===0

-..::-

,

I I

gi!z.S~~ eYaporat.ar

~ water

PUDP

h

I

I I

:

: 17Gb

I I I I

, I I I

2

hI

I

I

cooled area

L_~

---.J

From the steam table

0e

h, = h, at 5°C = 20.98 kJ/kg h, = h, at 10°C = 42.01 kJ/kg

h 3 = h g at 5°C = 2510.6 kJ/kg v3

= v g at 5°C = 147.12 m 3/kg

h4

= h, at 2roC = 88.14 kJ/kg

Let z

= vapor removed from the evaporator, carried

by nozzle steam m

= mass flow rate of chilled water

Energy balance about the whole system, neglecting pump work, zh, + 175 = zh, 88.14z + 175

/2

,

= 2510.6z

z = 0.07224 kg vapor/s volume of vapor removed = zV:l = (0.07224) (147.12) = 10.63 mvs


kg steam kg vapor quantity of nozzle steam = (1.27 kg vapor) (0.07224 second ) = 0.0917 kg/s

Energy balance about the cooled area, mh, + 175 = mh, 20.98m + 175 = 42.01m m = 8.32 kg/s (7-2) A vacuum refrigeration system consists of a large insulated flash chamber kept at low pressure by a steam ejector which pumps vapor to a condenser. Condensate is removed by a condensate to an air vent. Warm return water enters the flash chamber at 13°C; chilled water comes out of the flash chamber at 5°C. Vapor leaving the flash chamber has a quality of 0.97 and the temperature in the condenser is 32°C. For 350 kW of refrigeration. (a) how much chilled water at 5°C does this system provide? (b) how much make-up water is needed? (c) how much vapor must the steam ejector remove from the flash chamber?

SOLUTION

~'i J

~

~

t

The diagrammatic layout of Problem 7-1 is used in this problem. From the steam table h, = h, at 5°C = 20.98 kJ/kg h, = h, at 13°C = 54.60 kJ/kg h, = h, at 32°C = 134.15 kJlkg h, = h rn + x 3h fg3 ='20.98 + (0.97) (2489.6) = 2435.9 kJ/kg v1

= X,1Vj(l = (0.97) (147.1> = 142.7 mVkg

Cascade System 111

(a) Energy balance about the cooled area

m h , + 350 = mh,

20.98m + 350 = 54.60m

m = 10.41 kg/s

(b) Energy balance about the whole system, neglecting pump work, zh, + 350 = zh 3.

= 2435.9z

134.15z + 350 z

= 0.1521 kg/s

(c) vapor removed

= zV 3 = (0.1521) (142.7) = 21. 7 kg/s

If" 112

Air Conditioning 113


PROBLEMS (1) There are removed 3.06 mvs of vapor from the water evaporator of a vacuum refrigeration system; the warm water enters the evaporator at 18°C and chilled water leaves at 10°C; make-up water enters at 29°C. Determine the refrigerating capacity.

Chapter

8

Air Conditioning

Ans. 68.99 kW (2) In a steam jet refrigeration system, the evaporator is maintained at 6°C. The chilled water leaves at the same temperature and warms to 11°C as it absorbs the load. Make-up water in available at 25°C. For a capacity of 50 tons, determine (a) the quantity of chilled water required, and (b) the volume of vapor removed. Ans. (a) 8.381 kg/s, (b) 10.07 m 3/s

Air conditioning is the simultaneous, control of temperature, humidity, air movement, and the quantiiity of air in a space. Moist air is a binary mixture of dry air and water vapor. Dry air is the non-condensing comtponents of the mixture, mainly the nitrogen and the oxygen. Vapor is the condensable component cofthe mixture, the water vapor or steam which may exist in a sarturated or superheated state. P, = PH + P,

where P,

= total

mixture pressure

P, = partial pressure exerted I by the dry air

P,

= partial pressure exerted l by

the vapor

Saturated air means that the vapor in the air is saturated. Unsaturated air is containing super-heated vapor. Humidity ratio (moisture content, r.mixing ratio or specific humidity) W is the ratio of the mass of vwater vapor to the mass of dry air.



Mixt ur e:

air: mass volume

mass

/

volume

~

temperature

temperature

•

/

t

~

=V

It

pressure = Pt

pressure

¥

~

~~

=m =T

Water vapor: =mv mass volume = Vv =V

Entropy, kJlkg K

Fig.8-2. Isothermal process from super heated vapor to saturated vapor.

Fig. 8-1. A mixture of dry air and water vapor

w = mass of water vapor mass of dry air

= mv rna

Ps V/RsT Pa V/RaT

o =[:~]t (8-1) Compute the humidity ratio of air at 65 per cent relative humidity and 34°C when the barometric pressure is 101.3 kPa. Solution

_ PsRa _ p§Ra 287ps _ - PaRs - (pt-Ps)R s - 461.5 (Pt-ps)

Pd = saturation pressure at 34°C = 5.318 kPa Ps = (0) (P d) = (0.65) (5.318) = 3.457 kPa

W=0.622~ Pt -Ps

R a = 287 Jlkg.K, gas constant of dry air

R, = 461.5 Jlkg/J, gas constant of water vapor Relative humidity 0 is the ratio of the partial pressure of water vapor in the air to the saturation pressure corresponding to the temperature of the air.

W=

Ps P, - P,

0.6~: ~z. -

(0.622) (3.457) = 0.022 kg vaporlkg 101 .3 - 3.457 drv air ryal

Enthalpy h of a mixture of dry air and water vapor is the sum of the enthalpy of the dry air and the enthalpy of the water vapor. Enthalpy values are always based on some reference condition, and the zero value of the dry air is chosen as air at O°C and the zero value of the water vapor is saturated liquid water at O°C. h = cp t + Whg kJlkg dry air

~···rAir Conditioning 117


where c = specific heat of dry air at constant pressure p

= 1.0062 kJfkg.K (approximately)

h = enthalpy of saturated steam at temperature t, kJfkg g

(8-2) What is the enthalpy of the air-vapor mixture in Problem (8-1)?

SOLUTION

Wet bulb temperature t w is the temperature of air as registered by a thermometer whose bulb is covered by a wetted wick and exposed to a current of rapidly moving air. Wet bulb depresssion is the difference between the readings of the wet and dry bulb thermometers. Dewpoint temperature t (Ip is the saturation t.emperature corresponding to the actual partial pressure of the steam in air or the temperature at which condensation of moisture begins when the air is cooled at constant pressure.

I

= cpt + Wh g = (1.0062) (34) + (0.022) (2563.6) = 90.61 kJfkg

Specific volume is the number of cubic meters of mixture per kilogram of dry air.

t&f

t

f

RT RT v = _p3 = _3_ m 3fkg of dry air 3

Pt-Ps

~

(8-3) What is the specific volume of an air-vapor mixture at 30 and a relative humidity of 40 per cent at 101.3 kPa pressure? 0e

Entropy,kJlkg . K

SOLUTION P, = saturation pressure at 30 Ps = (0) (Pd)

0e

= 4.241 kPa

= (0.40) (4.241) = 1.696 kPa

T = 30 + 273 = 303 K

v=

RT 3

P, - P,

I II

h g = h g at 34°e = 2563.6 kJfkg h

~I

Fig. 8-3

Degree of saturation u is the ratio of the air humidity ratio W to the humidity ratio W s of saturated air at the same temperature.

(287)(303) 3fkg = 101,300-1696 = 0.873 m dry air

Dry bulb temperature t d is the temperature of air as registered by an ordinary thermometer.

=

u

[;]t s

(0.622) (p) u=

Pt - Ps _ [ Ps ] [PI-Pel] (0.622) (p) - -p;Pt~Ps Pt-Pd

u

= 0 [:~Pel] I

Ps



(8-4) A mixture of dry air and water vapor is at a temperature of 21°C under a pressure of 101.3 kPa. The dew point temperature is 15°C. Find (a) partial pressure of water vapor, (b) relative humidity, (c) humidity ratio, (d) enthalpy, (e) specific volume, and (f) degree of saturation.

(f)

P...L Ws = 0.622 Pt-Pd (0.622) (2.4861) = 0.01565 kglkg

Ws = 101.3 _ 2.4861

SOLUTION u=

W 0.01064 = 0.6799 or 67.999t W s = 0.01565

Psychrometry is the study of the properties of mixtures of air and water vapor.

~

~

~... 8

Psychrometric properties are the properties of moist air.

8 ~

A psychrometric chart is a graphical representation of the thermodynamic properties of moist air.

Entropy, kJIkg . K

(a) P, = saturation pressure at 15°C = 1. 7044 kPa (b) Pr\

= saturation pressure at 21°C = 2.4861 kPa 0=

(c) W

~ _ 1.7044 P, - 2.4861 = 0.6856 or 68.5691:

= 0.622

Ps p _p t

(0.622) (1. 7044)

=-'.,

f"\.,

()

.,

rt

r« A

A

= 0.01064 kg/kg

5

(d) h g = h g at 2PC = 2540 kJ/kg Dry-bulb temperature

h = c p t + Whg = (1.0062) (21) + (0.01064) (2540) = 48.16 kJ/kg (e) T = 21 + 273 = 294 K

Fig. 8-4

R a = 287 Jlkg.K

P,

v

= Pt-Ps = 101.3 -

1. 7044

= 99.6 kPa

R;,T _ (0.287)(294) = 0.847 mvkg dry air 99.6

= -P-,,- -

Constant property lines on a psychrometric chart

't:l

en

'-<:

:::r " ""i

o

1"'5

8

(1)

e-e

:!.

1.lel

<:>

o

. ~

:r to '1 ..... Barometric pressure = 101.325 kPa

~..,

~ A'

.~~

~~<\

o

~

//S

7S

/10

70

lOS

I

100

95 90

85

60 II

r

Ss ~"\

----

-.........- ............

- ....

............

-

~

SO - .~

............ - ........

'It )

<'Is



Basic Processes in Conditioning of Air The common processes involved in air conditioning are at practically constant pressures and can be illustrated by lines on a psychrometric chart.

Sensible Heating This process involves heating at constant humidity ratio (moisture content). It is represented by a horizontal line from left to right on the psychrometric chart.

Air-flow

Ia I

DP - _

Heating coils

3

7~J

/5

2~O--"1

6/ ~ ~8

4

(8-6) Determine the quantity of heat required to raise 14 m vrnin. if air at 20°C and 80 per cent relative humidity to 350C. What is the final relative humidity? SOLUTION

VI

Fig. 8-7.

Basic psychrometric processes

= 20°C and 0 1 = 80%,

From the psychrometric chart at td, h, = 50kJ/kg

mass flow rate, m

= 0.847 m 3/kg

= 14 mVrn in . = 16.53 kg/min 0.84 7

m~3/kg

Process 0-1: Sensible heating Process 0-2: Sensible cooling Process 0-3: Humidifying Process 0-4: Dehumidifying Process 0-5: Heating and humidifying Process 0-6: Cooling and dehumidifying ~

Process 0- 7: Cooling and humidifying Process 0-8: Heating and dehumidifying The first two processes, sensible heating and cooling, involve only a change in the dry bulb temperature. Humidifying and dehumidifying involve a change in humidity ratio (moisture content). The last four processes involve both changes in temperature as well as in humidity ratio.

moe

ss-e

O 2 = 349c h, = 65.5 kJ/kg

Q = m(h 2-h I ) =

= 16.53 (65.5-50)

256.20 kJ/min. or 4.27 kJ/s

124



,,

Sensible Cooling This process involves cooling the air at constant moisture content. It is represented by a horizontal line from right to left on the psychrometric chart.

---DB

' ...... L--WB Airflow

DP-- ...... --I---DP

---Airout

Cooling coils jClIc3l

.(8-7) Determine the quantity of heat removed from 14 m' per minute of air when cooled from 37° dry bulb and 2PC wet bulb temperatures to 15°C. What are the initial and final relatives humidities?

..

Make-u p

water

Eliminator plates

External heater or cooler

Fig. 8-8. Air Washer

SOLUTION From the psychrometric chart at t d 1 = 37°C and t W1 = 21°C,

0 1 = 23% h. = 61 kJ/kg VI

= 0.889 m 3/kg

In an air washer, air flows through a spray of water as shown in Figure 8-8. As the air flows, it may be cooled or heated, humidified, or dehumidified , or simple adiabatically saturated, depending on the mean surface temperature of the water. The water is externally cooled or heated or simple recirculated by a pump. Make-up water is added to compensate for any loss in the case of humidification of air. Eliminator plates are provided to minimize loss of water droplets.

3/min.

. mass flow rate, m = 14 m 3/kg = 15 .75 k g/ min. 0.889 m

Heating and Humidifying

O2 = 85~ h, = 38 kJ/kg Q = m(hz.-h 1 )

= 15.75 (38-61)

= - 362.3 kJ/min. or -6.04 kJ/s

15°C

3'?C

Air can be simultaneously heated and humidified by either drawing it through a spray, or over trays of warm water. The water must be heated during the process in order to supply the latent heat needed for evaporation. Trays of warm water can be kept at constant temperature by steam. However, when warm water is sprayed into the air stream, as in an air washer, the water must be heated as it is circulated.

DB---j---r-- DB WB--

_---

---WB

Airflow _ _ ----"0

DP- __ .J.

-t-----DP

Air Washer

126



(8-8) How much heat and moisture must be added to 28 m' per minute of air at 24°C and 40 per cent relative humidity to raise it to 38°C dry bulb and 2T'C wet bulb temperatures?

SOLUTION From the psychrometric chart at td. = 24 DC and 0} = 40(!r hI = 43 kJlkg WI = 0.0075 kglkg V'

l

Air can be simultaneously cooled and dehumidified by being drawn over cold surfaces oj- through sprays of cold water. When air is cooled and dehumidified by being drawn through a spray chamber, the spray water is cooled outside of the spray chamber by means of cooler installed in the circulating pipe line. The water is constantly chilled as it is circulated. The temper.ature of the water rises while being sprayed because it absorbs sensible and latent heat from the air. However, the heat that the water absorbs is being continually removed by the cooler installed in the circulating pipe line.

= 0.852 m'(kg

mass now rate, m

=

28 mvrnin

. ' = 32.86 kg/min.

0.852 m"lkg

From the psychrometric chart at t d 2 h,

Cooling and Dehumidifying

= 38

DC

and t W2

= 27

DC,

When cooling coils are used, the temperature of the film of air in contact with the surface of the metal pipe wall determines the condition of the air flowing through the coil. The temperature of this surface film is always at some point between the temperature of the air and the temperature of the water, or other refrigeration, inside the coil.

= 84.5 kJlkg

W 2 = 0.0181 kglkg Q = m(h 2 - h.)

= 32.86

(84.5 - 43)

Moisture added

= 1364 kJ/min. or

22.73 kJ/s

DB--- ...............

oI

• _ ... DB

WB--- ........... _I __ -'WB

= mt W 2 - WI) = 32.86 (0.0181 - 0.0075) = 0.3483 kg/min. or 0.00581 kg/s

AirFlO\1l ---1IIr IF

DP - - - ................... ---DP

Cooling coils

2

40* 24°C

38°C

(8-9) How much heat and moisture must be removed to cool 28 m' per minute of air from 35 DC dry bulb and 26 DC wet bulb temperatures to 21 DC and 50 per cent relative humidity?

-r"

Air Conditionii.ig 129


Cooling and Humidifying

SOLUTION From the psychrometric chart at t d 1 = 35°C and t WI = 26°C, h. = 80kJ/kg

Air can be simultaneously cooled and humidified. This fiction takes place in an ordinary air washer in which the same water is continuously recirculated, but is neither heated not co~,led.

WI = 0.0175 kg/kg

INITIAL STATE OF AIR

VI = 0.896 m '/kg'

/ . 28 m 3/min. = 31 .25 k gmin. mass fl owrae,m=· t 0.896 rrr'/kg From the psychrometric chart at t d 2 = 21°C and O2 = 50%,

FINAL STATE OF AIR

DB I i ' I I ---- ...............T I I I

"ll

"t" ... _

~-.!!j-

~

- - - -

----____

I

'"

,. --

-

I

I

I

I

-:0 --- ---_--i::t~j=-~~'!B_-------

---~~~

--r-

DP -

10

,0

b

AIR WASHER

'III

I

I

I

I

I I

II I i I I I .L.l.L.L

DP

PERFECT

WASHER

h, = 40.5 kJ/kg W 2 = 0.0077 kg/kg

Fig. 8-9. Constant water circulation without heating or cooling

Q = m (h 2- h.)

= (31.25) (40.5-80) = -1234.4kJ/min. or-20.57 kJ/s Moisture removed = m(W 2-W 1) = (31.25) (0.0077-0.0175) = 0.306 kg/min. or 0.0051 kg/s

Approximate path

21°C

35°C

Figure 8-9 shows the changes occuring in the dry bulib and dew point temperatures of the air flowing through an air wrasher in which the same water is constantly recirculated without; being either heated or cooled. In a perfect washer, the final dr~ bulb and dew point temperatures of the air would be equal to each other, and equal to initial wet bulb temperature of the ai:r. The efficiency of an air washer is defined as the ratio of the actual drop in dry bulb temperature to the maximum theoreticaJ drop in dry bulb temperature that could take place if the washer were 100 per cent efficient, and the air emerged saturated. In such a case, the final dry bulb temperature of the air would be eq ual to its initial wet bulb temperature. . (8-10) Air at 33°C dry bulb and 19°C wet bulb temperatures is cooled and humidified by passing it through an air washer in which the water is continuously recirculated. The air leaves the air washer at 23°C dry bulb temperature. Determine the moisture added per kg of dry air. What is the efficiency of the air washer?

130



From the psychrometric chart at t d2 = 41°C and t

SOLUTION

W2

= 19°C,

W 2 = 0.0048 kg/kg Moisture removed

W l = 0.0081 kg/kg

=W

j

-

\V 2 = 0.0070-0.0048

= 0.0023 kglkg

W 2 = 0.0122 kg/kg Moisture added = W 2 - W 1 = 0.0122 - 0.0081

= 0.0041 kg/kg

1

Air washer efficiency 23°C

33°C

33-23 33--19 = 71.43%

2

Chemical Dehumidifying Air can be dehumidified by passing it over chemicals that have an affinity for moisture. Usually in so doing the moisture is condensed and gives up its latent heat, raising the dry bulb temperature of the air. The air leaves drier and warmer. The wet bulb temperature may increase or decrease. Units employing such chemicals are used in some comfort air conditioning installations but mainly for industrial air conditioning. Since the leaving-air temperature is usually higher than wanted, it is necessary to add a sensible cooling process to get the desired final air condition.

24°C

4l o e

Mixing Air When air at one condition is mixed with air at another condition, the condition of the final mixture can be shown on the psychrometric chart by a point on the line joining the points indicating the original condition. The exact location of the final point will depend upon the masses of the original air quantities.

(8-11) Air at 24°C dry bulb and 15°C wet bulb temperatures enters a dehumidifier and leaves at 4PC dry bulb and 19°C wet bulb tem·peratures. How much moisture has been removed per kilogram of dry air? 11\1

SOLUTION From the psychrometric chart at t d l

= 24°C

W = 0.0070 kg/kg j

and t Wl

= 15°C,

-----

rr



The following equations apply to the mixture of two quantities of air. Subscripts 1 and 2 refer to the two streams of

air and subscript 3 refers to the mixture.

For the second stream t d2 = 25°C

m3 = m l + m2

W 2 = 0.011 kg/kg

m.h, = m.h. + m.h,

3

3/s V2 0.20 m == 0.233 kg/s m 2 == - V 2 == 0 • 858 m 3Jkg

m 3 W3 = m.W. + + m 2W2

mlW I +m 2W2 W3 = - -m- -

m 3 == m I + m 2 = 0.606 + 0.233 == 0.839 kg/s

3

m 3td3 = mlt dl + m 2td2

mIt dl + m 2td2 t d3 = - - m ---

_ mlt dl + m 2tdp2

t d3 -

m3

3

W _ mIW I + m 2W 3 -

m 3tdp3 = mlt dpl + m 2tdp2

= m.t., I m+ m2t d 2 P

V2 = 0.858 rrr'/kg'

VI _ 0.50 m 3/s m 1 == -VI- 0.825m3/kg = 0.606 kg/s

m.h, + m.h, h 3 = - -m- -

t dp3

t W2 = 18°C

m 3

2

==

0.606(15) + 0.233(25) == 17.8 0.839

== 0.606(0.0085) + 0.233(0.011)

0.839 ::: 0.0092 kg/kg

P

3

From the psychrometric chart at t d3 = 17.8°e and W 3 = 0.0092 kg/kg

(8-12) One half cubic meter per second of air at 15°C dry bulb and 13°C wet bulb temperatures are mixed with 0.20 m' per second of air at 25°C dry bulb and l8°e wet bulb temperatures. Determine (a) the dry bulb, wet bulb, and dew

point temperatures of the mixture, and (b) the enthalpy of the

mixture.

t W3 == 14.6°e t dp3 == 12.5°e

h, == 41kJ/kg

SOLUTION

From the psychrometric chart: For the first stream t dl = 15°C

t"l

= 13°C I

WI = 0.0085 kg/kg

VI = 0.825 m 3/kg

J

,.'r

J

0e

134



Alternate Solution t d:J --

m]t rl ] + m 2td2 m

(113 (rn.) (35) + (2/3) (rn.) (26)

m3

:3

2 = 29°C

1

17.8°C 25°C

15°C

tw3 = 14.6°C ~

=12.B'C

50~

h3 =41kJkg 26°C

(8-13) Outdoor air at 35°C dry bulb and 24°C wet bulb temperatures is to be mixed with room air at 26°C dry bulb temperature and 50 per cent relative humidity. The final mixture

is to consist of one-third outdoor air and two-thirds return air from the room. Find the resulting dry bulb and wet bulb temperatures of the mixture.

29°C

36°C

t WJ = 20.7°C

Al tern ate sol u tion: Divide line 1-2 into three equal parts. Point 3 is one-third of the line from point 2. Then read td, and t"x

SOLUTION t d ] = 35°C

Cooling-Load Air Quantities

t Wl == 24°C

The quantity of air circulated must be adequate to handle the cooling load as the air warms up to room temperatures frorn its suppy temperature. The lower the supply temperature, the less quantity which must be circulated, but the minimum temperature is determined by the system arrangements, the necessity of avoiding drafts and cold regions, the ceiling height, and the throw required.

me

t d2 = 26°C

O2 = 50%

1

ml

=T m 3

m2

=T m

2

3

..

"

136 Refr igeration a nd Air Condi tio ning

Air C o n d it io n ing 1:37

_m

~

-.m &pplyair

•

supply air

Fig . 8- 11. La te nt hea t load .

Moisture pic k ed up = (W 2- W 1 ) k g/kg dry ai r Fig. 8-10. Se nsible heat load

Q 5 = me P (t 2- t 1)

Qs = 1.0062 wh ere Q s

( rn ) (t 2- t 1)

where W 2 = humidi ty rati o of inside air, kg/k g W I = humidity rati o of su pp ly air, kg/kg The laten t h eat of stea m (wa t e r va por) in the a ir condit ion ing is appr oxim atel y 2500 k.J/k g .

= se ns ible h eat load , kJ/s or kW QL

m = ma ss flow rat e of su pply ai r, k g/s cp

Q L = 25 00 (rn ) (W 2- W 1 ) kJ/s or kW Q L = lat ent h eat load , kW

=s pec ific h eat of th e supply ai r = 1.0062 k.I/kg e C", a pprox im ately

t2

= m(W 2-W 1 ) kg/s

Qr

= in s ide sp a ce temperature t o be maintain ed ,

°C dry bulb

t 1 = supply a ir t emperature en te ring space, °C dry bulb'

BUpplyalr

_

) F ig. 8- 12. Tot al hea t load

Q T = Q s + QJ. Q T = m t h .-.h 1 )

---..m

"f' 138 Refr iger ation a nd Air Conditioning

where QT

Air Cond itio ning 139

= total heat load , kW

(c)

= (25 00)

(rn ) (W 2 - WI )

= (25 00 ) 0 2 .49) (0 .0 114-0.0 10 1)

h , = enthalpy of inside air, kJ/kg

= 40 .59 kW

hi = enthalpy of s u pply air, kJ/kg

(d ) S HR = -

Sen sible h eat ratio or se n sible heat factor is the ratio of th e se n sible heat load to the total h eat load.

-~

Qs

Q l.

Q " Q s + QL

~ l. ~

88 88 + 40. 59

Various Methods of Handling the Air Supplied to a Conditioned Space

SHR = Q + Q - Q s

L

T

(A)

(8- 14) An auditorium is to be maintained at a temperature of 25 °C dry bulb and 19°C wet bulb t emperatures . The sensible heat load is 88 kW and 58 kg per hour or moisture mu st be r emoved . Air is supplied to the auditorium at 18°C. Determine (a) th e mass flow r ate of s u pply air, (b) the r elative humidity, d ew point, and wet bulb t emperatures of the s u pp ly air, (d) th e se ns ib le heat ratio.

The u s e of all outside air with no r ecirculati on i s un economical unl ess the outside conditions ar e clos e in t emperatures and humidity to th e in sid e conditi on s maintain ed. Recirculation is impracticabl e in spac es wh ere obj ection abl e odors ari s e . (8- 15) In a s p a ce, th e sen sible h eat load is 13 .5 kW and the latent h eat load is 3.4 kW. Outsid e air is at 32 nC and 50 per cent r elative humidity. Th e spac e is t o b e m aintain ed at 25 °C dry bulb and 18°C w et bulb temp eratures. All outside air is suppli ed with r eheater to satisfy th e s p a ce conditions. The condition ed air leav es the su pp ly fan at 17°C . Determin e (a) the refri geration load , (b) the capacity of the s u pp ly fan , and (c) th e h eat supplied in the r eheater.

SOLUTION ( a)

Qs

= (1.006 2) ( m ) (t 2 - t l)

88 = (1.0062 ) ( rn ) (25

~

18)

m = 12.49 kg/s (b) Moi sture picked up =

58 = 0.0013 kg/kg (3600X12.49)

From the psychrometric chart at t 2 W 2 = 0 .0114 kg/kg.

= 25°C and tW

2

= 19°C,

All Outside Air Supplied, No Recirculation

SOLUTION

Humidity ratio of supply air, WI = W 2 - 0.0013 = 0.0114 - 0.0013 = 0.0101 kg/kg From the psy chrometric chart at t l = 18°C and WI = 0 .0101 kg/kg,

0 1 =77% t ,lpl

t WI

= 14.2°C = 15 J:;oC

...

m - -~ Qnside Air

32-<:

~

4

50"4 RH

) = (1. 006 2)( m) (t~- t ) 1:3 .5 = (1.0062)( m)(25 Qs

m = I .GEl k g/ s

17 )

J

// Air Cond it ioning 141

140 Refri gera ti on a nd Air Conditi oning

Point 4: At t. = 25"C a n d t w • W. Point 3: Ql. = (2 500)

( rn )

(B) Recirculated and O utsid e Air Supp lied

= 18"C

= 0.0101

kglk g

Recirculatin g part of th e air is more eco no m ica l t h a n u sin g a ll outs ide air, a s th e t emp e rature of th e r ecircul at ed a ir is u sually mor e favorab le than th at of th e outsid e a ir . In t h e or d in ar y con d it ion er, th e air m u st be cooled t o a r el ativ e ly low te m pe ra t u re to condense th e s u r p lus humidity, a n d wh en t.his te m pe rat u re is too low for com for t a ble deli very to th e cond it ioned s pace , so m e r eh eating of th e dehumidified air is n ece ss ary.

(W. - W)

3.4 = (2500) (1.6 8 ) (0.0101 - W ;)

W 1 = 0.009 3 kglk g At t~

= 17°C a n d W ~ = 0.0093 kglkg v~

= 0.835

m'lkg (8-16 ) A s pa ce to be air condit.ioned ha s a sens ible h e at load of 36 kW a n d a latent h e at load of 10 .2 kW. Th e s pace is to be maintain ed at 26 "C and 50 per ce n t rel ativ e humidity. Outsid e a ir is a t 32"C dry bu lb and 24"C wet bulb tempera tures . The con d it ione d air will enter th e space at 19"C. If 45 per cent of th e supply air is fres h air and th e r est is r ecircul at ed a ir, find ( :1) th e volum e flow rate of fresh ai r at su pp ly con d itions , (b ) th e a p pa rat us d ew point, and (c ) th e refri gerati on load .

h, = 40.5 kJlkg Point 2: At W .l

= W ~ = 0.00 93 kglkg a n d sat u r a ted, h, = 37 kJlkg

Po int 1: At t 1 = :32°C a n d 0 1 = 50 Ck , h i = 71 kJlkg (a ) Refriger ati on loa d

= mf h . -

h2)

= (1. 68 ) (71

- 37)

= 57 .12 kW

SOLUT IO N

= m V:1 = ( 1.68) (0.8 35) = 1.40 m 3/s he at s u pp lie d = m (h:1 - h) = ( 1.68) (4 0.5-37) = 5.88 kW

(b) Fan ca pacity (c)

4

Reci'cutated Nt

-

Concf~ioned

] SpaaI 4

.....

"'o-~

32"C

26"C

50% lUI ~".

f)

24"C 1

2

Q" = ( 1.006 2 ) ( m : (t , - t Il 36 = (1 .0062 ) t m ) (26 - 19 )

)

m = 1) . 11 k g/s



Point 4: At t 4

= 26°C and 0 4 = 50%

= (2.81) (53

h, = 53 kJ/kg

= 116.4 kW

= (2500) (m) (W 4

-

W 3)

W,) = 0.0098 kg/kg

h. = rnOh i + rn rh 4 o rn

Refrigeration load

= 19°C and W 3 = 0.0098 kg/kg

- 39) + (2.30) (72.5 - 39)

(2.30) (72.5) + (2.81) (53) 5.11

= (m) (hi) - h 2 )

;;:: (5.11) (61.7 - 39) = 116 kW

v 3 = 0.841 rn 3/kg

mass flow rate of outside air, rn o = (0.45) (m) = (0.45) (5.11) = 2.30 kg/s

(a) Volume flow rate of outside air, Vo = (rn.) (v 3)

= (2.30) (0.841) = 1.93 Point 2: At W 2

(b)

1

rn 3/s

= W 3 = 0.0098 kg/kg and saturated h,

= 39

kJ/kg t 2

h)

Another solution:

10.2 = ((2500) (5.11) (0.0106 - W 3 )

At t 3

= (m) (h, -h 2 ) + (rn o) (h, -

W 4 = 0.0106 kg/kg

QL

Point 3:

(c) Refrigeration load

= 14°C

Apparatus dew point, t d p 2 = 14°C

Point 1: At t l = 32°C and t WI = 24°C h, = 72.5 kJ/kg mass flow rate of recirculated air, rn r = rn - rn o = 5.11 - 2.30 = 2.81 kg/s

••~~----=__~7

~

3

, ..... 4

(to conditioned space

r

144 Refri gerati on a nd Air Conditioning

Air Co nd it ion ing 14 5

(8 -17 ) An a ir con d itio ned th eat er is t o be ma in t a in ed a t 26.7°C db t emperature a nd 50 Ck relati ve humidity. Th e ca lcu lated t otal se n sible heat load in th e th eater is 126 ,240 kcal /h and latent heat loa d is 82,92 0 k ca l/h. Th e a ir mixture at 28 .9"C db and 22 .2 uC wb t emperature is cool ed to 17 .22°C db a n d 15°C wb te m pe ratu re by chi lled water coolin g coils and del ive re d as supply air to t he th eater. Calculate the t on s of r efrigeration r equ ired .

Q T = m t h , - h) 875 ,753 = m (54 .5 - 4 2 ) m = 70 ,060 k glh Refri geration loa d

= m th , _ h. ) = (70 ,06 0 ) (65

- 42 ) (2 11) (6 0)

2

SOLUTION

= 127 .3 TR

ReciltUlared M

-

Theater 3

....

____::::

26 .7°0 50%RH

Outside M

n

(8- 18) An au ditorium is to be mai ntained a t a t emperature of 26"C dry bu lb and 50 lk r el a t ive humidity. Air is t o be s u pplie d at a t emperature n ot lower t h an 15°C dry bulb . Th e sens ible h eat ga in is 110 kW a nd th e latent h eat ga in is 3 7 .5 kW. Tak e ventillatin g ai r a s 25 'k by wei ght of th e air fr om th e room a nd is at 3 5"C d ry bu lb a n d 60 £,1, r e lativ e hu midity. Deter min e th e refri geration ca pacit y in ton s . S OLU T IO N

Point 1: At t l

= 28.9"C and

t WI

4/

= 22 .2"C

Recirculated Air

r

Wl

Auditorium 4

ITlr

hi Point 2: At t 2

kJlkg

mo 1

= 17.22"C a nd t W2 = 15°C h,

Poi nt 3: At t 1

= 65

ss-e

.... ::;:

Outside ~r

""''' m

2G°C

~

5O% RH

~iJ:r-----

60%RH

= 4 2 kJ/k g

= 26 .7"C and

01

= 50
RI-I

Q, = ( 1. 006 2 ) (rn ) (t . - t j )

h , = 54 .5 kJlkg

110

Q T = 126,240 + 82 ,920 = 209,160 kca llh

m

= 9 .93 8

(rn) (26 - 15 )

k g/s

O~t sid e ai r supp ly, m o = (0 .25 ) (9.938 ) = 2 .485 k g/s

= (209 ,16 0 kc a l/h ) (4.18 7 k.Izkcal )

= 875 ,753 kJIh

= (1. 0 062)

Point 4:

A

t . = 26 "C and 0 .

h . = 53 k,Jlkg

= 50 'ir

RH W 4 = 0 .0 106 kglkg

,

146 Re frigeratio n a nd Air Conditioning

Air Condit ionin g 14 7

QI, = (25 00 ) (m) (W 4 - W3)

Point 3:

37 .5 = (2500) (9.93 8) (0 .0106 - W~ )

W I = 0 .00909 kglk g

At t = 15°C and j

W~

= 0 .009 09 kglk g

h, = 38 kJlk g

= 35°C a nd 0 1 = 60% RH hi = 90 kJlkg

a ir, m = m - m o = 9.938 - 2.4 85 = 7.453 kg/s

P oint 1: At t l Return

Refrigeration load = mil (hi - h ) + m , (h , - h ) = (2 .485) (90 - 38 ) + (7.453) (53 - 38)

= 241 kJ/s =

24 1 3.52

= 68 .5 TR

1 ~

(C)

Recircul at ed Air with Extern al -Byp a ss System

....

In the bypass sy ste m , part of the r ecir culated a ir is con t r oll ed by damp er ac ti on in or de r t o make it bypa ss th e cond iti on er. This warm e r byp a ss a ir is u s ed t o reheat th e a ir lea vin g t h e refrigerat ed coils t o a mor e s u itable t emperature for distributi on at th e g r i ll e ou t le t s . Thi s m eth od produ ce s e cono m i ca l perform an ce as com pa re d with sy st e ms wh ere rehea t mi ght be req u ir ed. (8- 19 ) In a n au di to r iu m , t h e se ns ible h ea t en ter ing is 96 kW and th e lat ent h eat is 38.4 0 kW. Ou t side air is a t 33°C dry bul b and 25°C wet bulb t emperatures . Th e s u pply ai r en te rs a t 18°C and the a u di to r ium is t o be maintained a t 25°C dry bu lb a n d 18°C wet bulb t emperat~l !:~ .S_. If 20 per cent of th e fan ca pacity is fresh air and the by pass system is used, d etermine (a ) th e app aratus d ew poin t , and (b) t he refrigerati on ·loa d.

An other solu t ion : h ~

S OLUTION

mOh! + m rh ~

m (2 .485 ) (90 ) + ( 7.453) (53) 9 .938

= 62.25 kJ/kg

-

Refrigerati on load

= m (h l

Conditioned. Space

h. )

-

= (9.938)

4

(62.25 - 38 )

= 241 k.I/ s or 68 .5

1

mo ~

TR

......

......

25°C lS"C Outodde Air

33"C

1

25°C

Q,

= ( 1.0062) (m ) ( t

4

-

t.1 )

96 = ( 1.0062) (rn ) (25 - 18 )

a~--To a uditori u m

J

m = 13 .63 kg/s

14H Refrigeration and Air Conditioning


Point 4: At t 4 = 25°C and t \\4 = 18"C h 4 = 50.5 kJlkg

Solving equation (2) and (3) simultaneously,

t 111'4 = 14.2°C

y

= 0.509

,;;

W4 = 0.0101 kglkg Point 3:

QI, = (2500)

(m )

c = 0.491

(W 4

-

Vi,)

Substituting in either equation (2) or (3)

38.4 = (2500) (13.631 (0.0101 - W,J WI

t2

= 0.0090 kglkg

Apparatus dew point = 10.7°C

At t, = 18°C and Vi, = 0.0090 kglkg trip, =

(a) Let y c

= 10.T'C

(b)

= t W2 = t d p 2 = 10. 7°C

Pain t 2: t 2

12.5°C

h, = 31 kJ/kg

= per cent air bypassed

= 33°C and tw I = 25°C

Point 1: At t l

= per cent air conditioned

h,

m o = (0.20) (m )

me = (0.491) 4

ill r

y

2

c

= rn e -

rna

(rn )

( 1)

.~\ \

6.69 kg/s

2.73 = 3.96 kg/s

nl 0 h ]

+ rn r h 4

rn

::J

(2)

('

(2.73) (76) + (3.96) (50.5) /

(3)

kW

Another solution:

h.=

yt dp4 + ct dp 2 = t dp3

= 12.5

=

-

h 2)

= (2.73) (76 - 31) + (3.96) (50.5 - 31)

yt 4 + ct 2 = t 3

14.2y + ct 2

= (0.491) (13.63)

= 6.69 -

= 200.1

25 y +ct 2 = 18

= (0.20) (U~.63) = 2.73 kg/s

Refrigeration load = (rna) (hI - h 2 ) + (m) (h 4 3

y+c=l

= 76kJ/kg

~ 60.9 kJ/kg

6.69

150

'f t


Refrigeration load = (m)

~

~

(h, - h;2)


PROBLEMS

u

= (6.69) (60.9 -31)

(1) Calculate the specific volume of an air-vapor mixture in cubic meters per kilogram of dry air at the following conditions t = 32°C, W = 0.016 kg/kg, and Pt == 100 kPa.

= 200 k\V

Ans. 0.899 cu mlkg 11'1

1

(2) Moist air at a dry bulb temperature of 25°C has a relative humidity of 50 per cent when the barometric pressure is 101.4 kPa. Determine (a) the partial pressures of water vapor and dry air, (b) the dew point temperature, (c) the specific humidity, (d) the specific volume, and (e) the enthalpy.

Ans. (a) 1.5855 kPa, 99.81 kPa; (b) 13.9°C, (c) 0.00988 kg/kg, (d) 0.857 cu m/kg, (e) 50.32 kJ/kg (3) Air at a temperature of 33°C has a relative humidity of 50 per cent. Determine (a) the wet bulb temperature, (b) the dew point temperature, (c) the humidity ratio, (d) the enthalpy, and (e) the specific volume. Ans. (a) 24.5°C, (b) 21.4°C, (c) 0.016 kglkg, (d) 74 kJ/kg, (e) 0.887 cu mlkg (4) How much heat is required to raise the temperature of 0.50 cu m per second of air from 19°C dry bulb and 15°C wet bulb to 36°C? What is the final dew point temperature? Ans. 10.42 kJ/s, 12.3°C How much heat must be removed to cool 30 cu m per (5) minute of air from 34°C dry bulb and 18°C dew point to a wet bulb temperature of 19°C? What is the final relative humidity? Ans. 7.62 kJ/s, 83% (6) ow much heat and moisture must be added to 15 cu m per inute of air at 21°C dry bulb and 30 per cent relative humi Ity to raise it to 37°C and 40 per cent relative humidity? Ans. 13.39 kJ/s, 0.00333 kg/s

I

0'.

152



(7) How much heat must be removed to cool 50 cu m per minute of air at 29°C dry bulb and 21°C wet bulb temperatures to 16°C dry bulb and 14°C wet bulb temperatures? How much moisture was removed?

. I

,AIL

'OJ

Ans.

20.5~)

k.J/s,

O~OO~315

kg/s

(8) Air at 32°C and 20 per cent relative humidity is cooled and dehumidified by means of an air washer until the relative humidity becomes 90 per cent. How much moisture was added per kg of dry air. What was the air washer efficiency and the dew point temperature of the leaving air?

An s. (a) 8.18 m Vs, (b) 13.3°C, (c) 4B7.6 k W

(13) A store to be maintained at 2f,oC and 50 per cent relative humidity has a sensible heat load of 18.90 k W and a latent heat load of 6.30 kW. Outside air is at :32°C dry bulb and 23°C wet bulb temperatures. The conditioned air enters at 17°C If30 per cent of the supply air is fresh air and the bypass system is used, detennine (a) the refrigeration required, and (b) the volume of the bypass air at supply condition.

Ans. 0.0057 kg/kg, 92%, 16.5°C

(9) A stream of outdoor air is mixed with a stream of return air in an air conditioning system that operates at 101 kPa pressure. The flow rate of outdoor air is 2 kg/s, and its condition is 35°C dry bulb temperature and 25°C wet bulb temperature. The f1 ow rate of return air is 3 kg/s, and its condition is 24°C and 50 per cent relative humidity. Determine (a) the enthalpy of the mixture, (b) the humidity ratio of the mixture, and (c) the dry bulb temperature of the mixture.

An s. (a) 41.1 kW, (b) 0.819 mVs

Ans. (a) 59 kJ/kg, (b) 0.012 kg/kg, (c) 28.5°C (10) .An auditorium is to be maintained at 25°C dry bulb temperature and 50 per cent relative humidity. The supply air enters the auditor-ium at 17°C. The sensible and latent heat loads are 150 k\V and 61 kW, respectively, Determine the wet bulb temperature, relative humidity, and volume flow rate of the supply air.

',Cl

Ans. 14"C. 72%. 15.54 m 3/s (11) In a certain space to be air conditioned the sensible

and latent heat loads are 20.60 kW and 6.78 kW, respectively.

Outside air is at :-l3°C thy bulb and 24°C wet bulb temperatures. The space is to be maintained at 25 with a relative humidity not exceeding 50 per cent. All outside air is supplied with reheater. The conditioned air enters at 18°C. Determine (a) the refrigeration load required, (b) the capacity of the supply fan, and (c) the heat supplied in the reheater.

(12) An air conditioned auditorium with a capacity of 1000 persons is to be maintained at 24°C dry bulb temperature and 55 per cent relative humidity. The sensible and latent heat loads are 115 kW and 42 kW, respectively, The conditioned air enters the auditorium at 1 T'(': For proper ventilation, 40 per cent of the supply air is fresh air and the rest is recirculated air. Outside air is at 34°C and 50 pPf cent relative humidity. Determine (a) the volume flow rate of recirculated air, (b) the apparatus dew point, and (c) the refrigeration load,

1·1

QC

'. ~

-I

Ans. (a) 106.6 kW, (b) 2.44 m 3/s, (c) 16.06 kW

~

..

Air Conditioning 155 I

Chapter

wh.-r.-

9

==

Ill,

III

I

I

mass of water entering

= mass of water leaving

Illi I

111

III ,

=

mass of dry air entering

~.li" II,

= mass of dry air entering

hi

Cooling Tower

~

!II'I

h.' = enthalpy of air leaving, kJlkg

~

1')11 1

IIIII

11,

hi

A cooling tower is a device commonly used to cool condenser water in power and refrigerating plants.

:=

enthalpy of water entering, kJ/kg

1

h, = enthalpy of water leaving, kJlkg

11,1[1

Mass bal an co

warm m3 waterin - - -

III

ha. I te

1

I

+ rna + mW, = m.) + m" + m"W 2

water evaporated r -

3

2 rna

--\

/

\

~ W2

\

/

I

\

/

whe-re

Arrout

= m 1 - m.) = m;1(W 2 - WI)

\V I = humidity ratio of air entering, kglkg W 2 = humidity ratio of air leaving, kg/kg

\

/

, ~' WI' 1

11 1,1

j

':1 '

I

11

I1 I

\

The cooling effectiveness of a cooling tower is

1

I

1

'1

IIII11

Airin - - ) Ola

cooling tower efficiency

ht

=

II ~

1':1

W.

where

t,

=

temperature of water entering

rn4

t 4 = temperature of water leaving

t.

t WI = wet bulb temperature of air entering

Fig. 9-1. Diagrammatic cooling water

~ 1

1

mh, + mh , - m.h, + m" h , =

!~~

I,II'

Energy balance

mh, - mh,

1

math, - h.) I

I

:1 " ~

(9-1) In a cooling tower 28.34 mv/rnin of air at 32°C db and 24°C wb enter the tower and leave saturated at 29°C. (a) To what temperature can the air stream cool a spray of water which enters at 38°C, with a flow of 34 kg/min of water? (,bb)))Iow many kg per hour of make-up water is needed to co~ate for the water that is evaporated?

11'

~I I

II Ili l l

I

III

l

jill I

''-'

156

\



Energy balance

SOLUTION

rn.h; - mh, = mil (h, - h.)

ID:i = 34 kglmin.

,

(34) (159.21) -

38°C

33.68h~ =

(32.06) (95 - 72.5)

3

h, =

~

~\

/

2

(b) Make-up water = rna (W 2 \

II /

:rJOC 24°C

t 4 = 33.2°C

\\

/

-

kJ/kg

I••t;

100% RH

\

/ 1

139.~1

\

I

rna

2,9OC

= D.3206 \

-

W,) = (32.060) (0.0256 - 0.0156)

kg/min or 19.24 kg/h

(9~2)

Water enters a cooling tower at 52°C and leaves at 38°C. Air enters at 29°C and 50~ relative humidity and leaves in a saturated condition at 40°C. Determine (a) the mass and volume of air needed per kg of water entering, and (b) the quantity of water that can be cooled by 1 cu m per second of air. Atmospheric pressure is 102 kPa.

r-~=====~ --- -=~ __-_-t:---=------= =-"'1 - ,-- -,:-_---:..-...ij

" ---t.& ~

SOLUTION Point 1: At tdl

= 32°C and t WI = 24°C

103

VI = 0.884 m 3/kg

h. = 72.5 kJ/kg WI = 0.0156 kg/kg

ma

=

VI VI

I.;~, 3/min

28.34 m 0.884 m Vkg

= 32.06

'Point 2: At t d 2 = 29°C and 0 1 = 100% RH

h, = 95 kJ/kg W 2 = 0.0256 kg/kg Point :5: h,

2

kg/min

)

\

/

'11

/ /

~.~

<>. \

40°C

100%RH

\

\

/

;

rna

= h, at 38°C = 159.21 kJ/kg

m, = 34 kg/min

1

-------~--=---

60

(a) Mass balance m , - m 4 = mil (W 2

-

WI)

m:! - m .. = (32.06) (0.0256 - 0.0156)

:34 -- m .. = 0.3206

m 1 = :3:3.68 kg/min

mol

t I.", l.

P01~

At t dl = 29°C and 0 1 h , = 61 kJ/kg W 1 = 0.0127 kg/kg

I

3ErC

= 50~

RH

VI = 0.874 ml/kg

..

'y )

~

158 Refr igeration 'a nd Air Conditioning

Point 2:

Air Conditio ning 159

t rl 2 := 40°C, 0 2 = 100 % RH

Volum e of ai r

P.1 2 = saturation pressure -at 40 °C

= 7.378 kPa

(b) VI = 1 m:l/s of air

= (1.00) (7.378) = 7.37 8 kPa

Ps2 = (0 2) (Pd2)

(O.622 )(PS2) W 2 = p t - p :;2 '

CO.622X7.378) = 0.04 85 kg/kg = 102-7 .378

"

Point 3:

h;

52°C

= 21 3.51 kJlkg

Point 4:

h, = h , a t 38°C

= 159 .21 kJlkg

(1 )

'.H ~ -'/

J

m:J-m 4 = rna (W 2 - W I)

(1 )

= (1.144)(16 5.11- 6 1) = 119 .1

(3)

= 01 " (W 2 - W I)

m3

-

01 4

rn 3

-

01 4 = (1. 144) (0.048 5 - 0 .01 27 ) ( 4)

Solving equ a tions (3) and (4) simultaneou sl y,

Mass balance

(0. 0485-0.0 127)

0.03 58 rna + m 4 =1

159. 21014

m.1 -m 4 = 0.04096

21 351-1 59 .21m 4 = rna (165 .11-61 )

= 21 3 .51

-

= rna (h , - h I)

M as s balance

m.h, - mh . = rna (h , -hI)

S olvin g equ at ions

1 m:J/s _ 0 .87 4 m 3lkg = 1.144 kg/ s

-;:--

2 13 .5 101 3- 159 .21m 4

Energy b alance

= rna

-

2 1:3.51m 3

1 kg

1 - m4

VI

m.h, m .h ,

= (1.0062 ) (40) + (0.0485) (2574.4) = 165 .11 kJlkg

104 .11rn " + 159 .21m 4

V

_1_ -

En ergy balance

h, = (cp ) (t) + (W) (h g )

( a) m 3 =

=

n1

h g2 = h g at 4 0°C = 2574.4 kJlkg

=h, at

= rn"v I = (0 .5518 ) (0,874) = 0.4 823 rn'' a ir/kg water

01.1

(9- 3) Fifty gall on s per minute of 'water en t e r s a coolin g to wer at 46 °C. Atm ospheric air a t 16 °C db a n d 55 % RH en t er s th e tower at 2. 85 cu rn per secon d and leav es a t 32 °C sa t u ra t ed . Determine (a) th e volume of water that lea ves the towe r, and (b) th e ex it t emperature of the water.

(2)

and (2) simu lt a n eous ly, rna = 0 .5518 kg air/kg water

1,

. ~

= 2.073 kg water/s

)

ITI,

160


SOLUTION


Energy balance

150 gpm

m3

we

8

~

I

2

\

\

I I

..

32°C

100., RH

~

(3.122)

(J

92.62) - 3.036h 4 = (3.442) (111 - 32)

I

h

1~~

\

I

\,

Volume of water leaving

l~~~{t -~

= 55% RH

01

1

h.:

=

32 kJ/kg

v j = 0.828 m

= (3.036 kg/s) (0.0010031 m'Vkg) 3.045 LIs

(b) t 4 = 2Fl.9"C

t...

~

= 16°C_and

25.9°C

= 0.003045 m3/s or

4

At t d 1

108.51 kJ/kg

Specific volume of water at 25.9 c,C = 0.0010031 mvkg

-"""-'

Point 1:

=

t4 =

\

rna 6

(h, - h.)

\

I

1/

= m"

~

1

I

rnh, - mh,

/

3/kg

I-I

(9-4) Water at 55°C is cooled in a cooling tower which has an efficiency of 65o/c. The temperature of the surrounding air is 32°C dry bulb and 70% relative humidity. The heat dissipated from the condenser is 2,300,000 kJ/h. Find the capacity in liters per second of the pump used in the cooling tower.

\V j = 0.0056 kg/kg

m =

2

=

V1

a

SOLUTION

2.85 m:l/s = 3.442 k /s 0.828 m 3/kg g

Q =2,300,000 kJ/h

'"/,,,

Point 2:

At t d2 = 32°C and

02

= 100% RH

W2

h 2 = III kJ/kg Point 3:

= 0.0307 kg/kg

h, = h, at 46°C = 192.62 kJ/kg (50 gal/min) (3.7854 L/gaD (60 s/min) (0.0010103 m Vkg) (1000 L/m:3)

;~

•

ll,fi

m,., I

CbndeosEr

)

3

fi50C

l i/i

2

,li

;,: ~

fj

-

m 4 = rna (W 2

-

Cooling tower

32°C

~

(a) Mass balance

m3

1

'J

= 3.122 kg/s

70%RH

Wl)

4

3.122 - m 4 = (3.442) (0.0307 - 0.0056)

t.

mol = 3.036 kg/s

Pump

...i

..

'P Air Conditioning 16;3


Point 1:

At t d l = 32°C and 0 1

= 70% RH

SOLUTI< )0:

m.

t w l=27.4°C cooling tower efficiency = tt 3 ~ tt 4 3

3

wI

2

55 - t 1 55 - 27.4

0.65

t 4 = 37.l

BOOKW Cooling

oC

t~

Engine

Heat balance about condenser, (m w ) (c w ) (t, - t 4 ) =

1

Git:oerator

.320C

Q

70%RH

(m w ) (4.187) (55 - 37.1) = 2,300,000

"

m, = 30,688 kg/h or 8.524 kgls Specific volume of water at 37.PC

= 1.0067

Pump L/kg Point 1:

Capacity of the pump = (8.524 kg/s) (1.0067 L'kg)

At

t ril

= 32°C and 0 1 = 70(k RH

t W 1 = 27.4°C

= 8.58 Lis

Temperature of approach

An atmospheric cooling tower is to provide cooling for the jacket water of a four stroke, 800 kW Diesel generator. The cooling tower efficiency is 60% at a temperature of-approach of 10cC. If ambient air has a relative humidity of 70% and dry bulb temperature of 32°C, determine the cooling water supplied to the diesel engine in liters per hour. Generator efficiency is 97%. (9-5)

=t

4

t W 1

-

10=t 4 - 2 7 . 4 t4

.

. .

= 37.4°C t.,

cooling tower efficiency = - '-

t,-

0.60

tl

t4

t'\l

t 1 - 37.4 t, .- 27.4

= 52.4°C

Brake power of engine = power input to generator

J f

1

AI

800

= 0.97

= 824.7 kW

1"

Air Conditioning 165 164 Refrigeration and Air Conditioning

PROBLEMS

Assuming the following diesel engine heat balance:

(1) 1\ nu-cb anical-draft cooling tower receives 115 m" per second of atmospheric air at 103 kPa, 32°C dry bulb temperatnre, 55% RII and discharges the air saturated at 36°C. If the tower receive» 200 kg/s of water at 40°C, what will be the exit tempernturv of the cooled water?

Useful work == 30% Cooling loss = 25%

· d . Q 824.7 kW H eat supp1ie to engme, A = 0.30 = 2749 Heat absorbed by cooling water = (0.25) (2749) = 687.25 kW

Ans. :31.2°C '~

J,, ,

(ro) (cJ (t, - t 4 )

= 687.25

"

I

(m w ) (4.187) (52.4·- 37.4) = 687.25

(2) I n a cooling tower water enters at 52°C and leaves at 27°C. Air at 29"C and 47% RH also enters the cooling tower and leaves at 46"C fully saturated with moisture. It is desired to determine (a) the volume and mass of air necessary to cool one kg of water, and (b) the quantity of water that can be cooled with 142 cu ill per minute at atmospheric air.

m; = 39,384 kg/h Ans. (a) 0.5742 cu m, 0.66 kg; (b) 247.3 kg specific volume of water at 37.4°e = 1.0069 L/kg Volume of cooling water = (39,384 kg/h) (1.0069 Lfkg) = 39,656 Lih

, 1

,

~,~I

~ I

.

'~

~

:1 ,,1

"

.

,\

AJ

(3) A cooling tower receives 6 kg/s of water of 60°C. Air enters the tower at 32°C dry bulb and 27°C wet bulb temperatures and leaves at 50°C and 90 per cent relative humidity. The cooling efficiency is 60.6 per cent. Determine (a) the mass flow rate of air entering, and (b) the quantity of make-up water required. Ans. (a) 3.253 kg/s, (b) 0.1818 kg/s

".

Drying 167

Chapter

SOLUTION

10

Gross weight = 10kg Bdw = 8 .5 kg

Drying

After the material have been processed:

iI

weight of moisture = (r ega in) (Bdw) = (0.20) (8.5)

Hygroscopic materials are those substances which are particularly variable in the moisture content which they can possess at different times . Most of the organic materials, such as wood , leather, foodstuffs , paper, tobacco, cloth, and hair are quite hygroscopic. Bone-dry weight Bdw or dry weight is the final constant weight reached by a hygroscopic substance after being dried out. Regain is the hygroscopic moisture content of a substance expressed as a percentage ofthe bone-dry weight ofthe material. . Regaln

=

= 1.70 gross weight

+ weight of moisture

= 8.5

+ 1.70

= 10 .2 kg 10.2 Weight of product per kg of original material = - - 10

weight of moisture bone- d ry weig . ht

= 1.02 kglkg

Moisture content is usually expressed as a percentage of the gross weight of the body, and may refer to both hygroscopic and purely surface moisture. Moisture content = weight of moisture gross-weight

Gross weight

= Bdw

kg

= Bone-dry weight + weight of moisture

(l0-1) A 10-kg sample from a batch of material under test is found to have a bone-dry weight of 8.5 kg. This material is processed and is then found to have a regain of 20 per cent. How much weight of product appears for each kilogram of original material? I,V

•

~j

.~

t

)

;I~


Drying 169 Table 10-1 Moisture in Hygroscopic Materials Moisture is expressed as regain, per cent of the dry weight of the material

----------------------Per Cent Relative Humidity 40

50

60

70

80

4.50 6.50

6.20 8.00

8.80 9.90

11.10 12.40

14.50 15.40

Oak-tanned Newsprint, wood pulp White bond, rag Solid tire Cigarette Timber (average)

13.60 4.70 5.50 0:44 13.30 7.60

16.00 5.30 6.50 0.54 16.00 9.30

18.30 6.10 7.50 0.66 19.50 11.30

20.60 7.20 8.80 0.76 25.00 14.00

24.00 8.70 10.80 0.88 33.50 17.50

Steam activated

26.20 0.81 15.20 15.70 5.20 5.50 7.20 4.30 1.90 8.00 10.80

28.30 29.20 30.00 1.24 1.46 1.03 17.20 18.80 20.20 18.50 20.80 22.80 8.10 5.90 6.80 9.50 7.90 6.60 9.90 11.60 8.50 7.00 6.10 5.10 3.60 2.40 3.00 8.90 10.20 11.90 12.80 14.90 17.20

31.10 1.67 21.50 24.30 10.00 11.50 13.60 8.40 4.30 14.30 19.90

Material

Description

Bread Flour

White

Leather Paper Paper Rubber Tobacco Wood Activated charcoal Domestic coke Silica gel Cotton Cotton Cotton Hemp Linen Rayon Silk Wool

;I f,1

~

(10-2) Raw cotton has been stored in a warehouse at 29°C and 50 per cent relative humidity. (a) The cotton goes through a mill and passes through the weaving room kept at 31°C and 70 per cent relative humidity. What is the moisture in 200 kg of cotton? (b) For 200 kg of cotton from the warehouse, how many kilograms should appear in the woven cloth, neglecting lintage and thread losses?

J

SOLUTION

;!

'i,

(a) From Table 10-1, the regain of raw cotton at 50% relative humidity is 6.6%. weight of moisture = (regain) (Bdw) = (0.066) (Bdw)

gross weight = Bdw + weight of moisture

Absorbent Cloth Raw fibers Manila and sisal-rope Table cloth Fibre Skein Skein

200 = Bdw + (0.066) (Bdw) Bdw

= 187.6 kg

moisture in 200 kg = 200- 187.6 = 12.4 kg or moisture = (0.066) (187.6) = 12.4 kg

(b) From Table 10-1, the regain of finished cotton (cloth) at 70% relative humidity is 8.1%. Gross weight = Bdw + weight of moisture

:l';'

:~.

~~

= Bdw + (regain) (Bdw) = 187.6 + (0.081) (187.6)

.t\ \

= 202.8 kg

,

'41

Drying 171


= h, = 291.4 kJ/kg

h,

Point 3:

(10-3) Eight hundred kg of dry air per hour is saturated at 46°C, then heated to 106°C, passed through an adiabatic dryer and discharged at 61°C. Determine each of the following: (a) total kg of water in air before heating, per hour (b) final kg of water in air when discharged, per hour (c) water evaporated from material in the dryer, per hour (d) relative humidity of air leaving the dryer (e) dew point of air leaving the dryer (f) percentage of saturation of air leaving the dryer (g) number of cu m of air per second before dryer

h

'l

jO(.J

=

h g at 61°C

= 2611.3 kJlkg

h, = c pt 3 + W 3hg3 291.4 = (1.0062) (61) + (W 3 ) (2611.3) I

,I

j~]t

s

W 3 = 0.08809 kg/kg

(a)

Mass of water in air before heating,

I

SOLUTION

= m WI = (800 kg/h) (0.0688 kglkg) = 55.04 kg/h

mw (b)

Mass of water in air at discharge = mW 3

= (800 kg/h) (0.08809 kg/kg) = 70.47 kg/h m =800 kWh

x-{

beater

1 i I

2 W2 ·lO(f'C ,

Dryer

3 Wa 6P

(c)

-r--

,

= (800 kg/h) (0.08809 -

WC

100~ RH

= 46°C, 0 1 = 100% RH

Pd1 =

PS 1

-

2

)

= (01) (P d1) = 10.098 kPa (0.622) (10.098) 101.325 _ 10.098 - 0.0688 kglkg

h g 2 = h g at 106°C = 2685.4 kJlkg

= cpt 2 + W 2h g2

Pt - Ps3

(W3 ) (Pt)

,

Ps3

I

0

\

Point 2: W 2 = W 1 = 0.0688 kglkg

= 20.881 kPa

(0.622) (Ps3)

saturation pressure at 46°C = 10.098 kPa

W1

saturation pressure at 61°C

W3 =

,,\1

h,

3

(d) Point 3: Pd3 =

Point 1: t d1

= m(W W 0.0688) kglkg = 15.43 kg/h

Water evaporated from the material

3

0.622 + W 3 = Ps3 Pd3

=

(0.08809) (101.325) 0.622 + 0.08809

= 12.57 kPa

12.57 20.881 = 0.602 or 60.2%

.~ ~'f

., ~ \'

= (1.0062) (106) + (0.0688) (2685.4)

(e)

Dew point of air leaving the dryer = saturation temperature at 0.012~7 MPa = 50.5°C.

(f)

Percentage of saturation of air leaving the dryer u

3

r

~ ~t

~t

-

-

Pd3] Ps3

= 0.602

[ 101.325 - 20.881] 101.325 - 12.57

= 291.4 kJlkg

I

\

= 0.5456 or 54.56%

Drying 173


(g)

SOLUTION

Volume of air before drying, V2 = mv 2

w = (0.622) (PS2)

2

Pt - Ps2

1

_ (W2) (Pt) _ (0.0688) (101.325) _ kP

Ps2 - 0.622 + W2 0.622 + 0.0688 - 10.091 a

_ v2 -

(R) (T2) n - n _

_ (0.287) (106 + 273) _ 3 - 101.325 _ 10.091 - 1.192 m /kg

V2 = mV2 = (800 kg/h) (1.192 m 3/kg) or 0.265 m 3/s

• di

~

:1

33"C 60%

aa

y

4WC.60UH

_

~~ ~

38°C' 65%

na

...

Heating

W, ~~ I 93"co--I •

f6

Chamber

= 953.6 m 3/h

(10-4) A tower-type moisture dryer is to deliver 1000 kg/h of cassava flour with 2% residual moisture of 20% in the feed. The air to be heated in heating chamber is a mixture of fresh air at 33°C db and 60% RH, and hot humid air from dryer at 49°C db and 60% RH. The air mixture at 38°C db and 65% RH is heated by a steam coil to 93°C. The dryer is properly insulated so that moisture absorption can be considered adiabatic. Compute: (a) Required flow of heated air mixture to dryer. (b) Capacity of forced draft fan on dryer. (c) Heat in kcallh for heating the air mixture in heating chamber. (d) Percentage by weight of fresh air in mixture.

"

3

I

~

l>r)w 6

m'lI moiBture

moisture

Ws .1000 kwh

Point 2:

w~

Ws

At t d2 = 38°C and O2 = 65% RH h 2 = 109 kJ/kg W2

Point 3:

W3

= 0.0275 kglkg

= W 2 = 0.0275 kg/kg h g.3= h g at 93°C = 2664.9 kJ/kg h.'3 = c p t 3 + W 3hg3

= (1.0062) (93)

,1)lr

1

+ (0.0275) (2664.9)

\(

= 166.86 kJ/kg

\

'~

1 .. 1,~

.~'

.;f .~

"

Point 4:

Pd4 = saturation pressure at 49°C = 11. 75 k Pa Ps4

= (0 4) (Pd4) = (0.60) (11.75) = 7.05 k

W _ (0.622)(P S4) _ (0.622)(7.05) 4P t - Ps4 - 101.325 _ 7.05

Pa

= 0.0465 kg/kg

' ''1''

174 Refri geration a nd Air Conditioning

Dryin g 175

W 6 = Bdw + (0 .02) (W 6 )

Point 6 :

( l O-f» ) 1\ dri er is to b e d esi gned t o r edu ce the moi sture content of cop r a from 1) 2 per ce n t t o 6 per cent. Atm ospheri c air nC a t 3 1 d ry bu lb and 24 °C wet bulb temperatures , is h eated by ste a m coi Is t o 8 8 °C before e n ter in g t h e dri er. Steam e n t e rs th e r eh catur a t a pressure of 133 .9 kPa and 95 pe r cent quality. Th e a i r l eav es th e drier a t 40 °C w ith a relativ e h umidity of 80 p e r cent. Co p r a e n t e rs the dri er at th e rate of 1.30 k g/so D e t erm in e (a ) till' mass flow rate of air and (b ) the mas s flow rate of s t e a m .

Bdw = (0 .98) (W 6 ) = (100 ) (0 .98) = 980 kglh W. = Bdw + (0 .20) (WJ

Point 5 :

~

~

W _ Bdw .'i - 0 .80 (a)

980 = 0 .80 =1225 k glh

Mass balance about dryer

SOLUT ION

maW a + W.'i = m aW4 + W 6

ma

Ws - W6 W4

W~

-

Saturated liquid (1l8lIU.IIIIld)

1225 - 1000

0.0465 - 0 .0275

= 11,842 k glh or 3 .29 kg/ s

m"

1

si-c (b )

24° e

(0 .62 2) ( p sa) = 0 .02fJ75 kg/kg W a = 101.325 - Ps:J

IDa

~I

2

Wz

I Dr ier

88"C II

Rchcatcr

6% moisture

W5

Psa = 4 .29 kPa

V

a

RaTa RaTa (0 .287 ) (9 3+2 73 ) = 1. 083 mVkg

= --= - - Pa Pt - Ps3 = 101.325 - 4.29

m a = m a ss fl ow r ate of a ir m s = m a ss fl ow rate of st ea m

Ca pa cit y of fan = m ava = (3 .29 kg/s ) ( 1.083 rn'Vkg )

= 3 .fJ63

Point 1:

At t ,I I = 3 1°C and t " , = 24 °C

m ~/s

WI = 0 .0 16 kg/kg (c)

Q = rna (h a - h 2 ) = (3 .29 ) ( 166 .86 - 109) = 190.36 kcal/h -

(d )

(190.3 6 kJ/s ) (36 00 s/h) = 163,672 kcallh

-.

=1

(1)

33 x+4 9y= 38

(2)

X+Y

~

hi

""

i .',,,

,

.f

~

Point 2:

= 72 kJ/k g

t d2 = 88 °C, W ~ = W I = 0 .016 k g/kg

h ~~ = b, a t 88"C = 26 56 .9 kJ .k g

h 2 ':'\~t 2 + W2h g2 = ( 1\ 006 2 ) (88 ) + (0 .0 16) (2656. 9 )

Solving equ a t io ns (1 ) and (2) s im u ltaneous ly,

= 131.06 kJ/k g

x = O,6R7f) or 68 .75 c;( fr esh a ir I

to

3

~

W3

4O"C 80%RH

4 6~

moisture

W. = 1.30 kglB

Drying 177 176


Point 3:

t d3 P

= saturation pressure at 40°C = 7.3'Z5 kPa

P

d3

S3

W3

( 10-6 \ The temperature of the air in a dryer is maintained constant by the use of steam coils within the dryer. The product enters the- dryer at the rate of one metric ton per hour. The initial moist.urr- content is 3 kg moisture per kg of dry solid and will be dried to a moisture content of 0.10 kg moisture per kg of dry solid. Air .-n ter s the dryer with a humidity ratio of 0.016 kg moistu rt' per kg of dry air and leaves with a relative humidity of 100 f!r while the temperature remains constant at 60 aC. If the total prrssu r« of the air is 101.3 k Pa, determine:

0 3 = 80%

= 40°C,

= (03) (Pd3) = (0.80) (7.375) = 5.900 kPa _ (0.622) (P s3 ) Pt - P

(0.622) (5.900) Ik 101.325 _ 5.900 - 0.0385 kg g

_ -

s3

Point 4:

Bdw

= (0.48) (W 4 ) = (0.48) (1.30) = 0.624 kg/s

Point 5:

W5

= Bdw + (0.06) (W 5)

W5

Bdw _ 0.624:= 0.664 kg/s = 0.94 - 0.94

(a) Total required amount of air in kg/h under entrance conditions. (b ) Capacity of the forced draft fan to handle this air in cu m/min.

SOLUTION

(a) Mass balance about drier

m WI

= fia W

fia W z + W 4

fi a(W 3 - W Z) = W 4 fi a(0.0385 0.0160)

-

3

EIre

+ W 5

3 kgmoisture kg dry solid

x 6

~:,

!~

fia (h z - h.)

h, :::: h f6 +

h fg6

h, :::: h.,

W. Point 1:

Ws

WI

:=

6

_ (0.622) (p,) WI P t - PsI P sl =

(h g6

-

h f6 )

(0.95) (2688.3 - 452.85)

:::: 2123.7 kJlkg (rn ) (2123.7)

:=

(28.27) 031.06 - 72)

11

ill"

= 0.7862 kg/s

= 1000 kwh

= 0.016 kg/kg

h, - h, = hf6 + x 6h rg6 - hf6 :::: X

3

0.10 kg moisture

kg dry solid

:&

= mh, + mh,

:=

100% RH

0.664

,

h.)

1

Dryer

(b) Energy balance about reheater

-

0000

2·

4

fia = 28.27 kg/s

fi/h 6

W2

W 5

= 1.30 -

mh, + ruh.

steam

VI

=

W IP t

(0.016) 001.3) 0.622 + 0.016

0.622 + WI

= 2.540 kPa

Ra T l

RaT]

.(0.287) (60 + 273)

Pa

P. - PsI

101.3 - 2.540

= 0.968m:Jlkg

"

Drying 179


Point 2:

PROBLEMS

t d2

= 60°C,

Pd2

= saturation pressure at 60°C = 19.94 kPa

Ps2 = (0 2)

O2 = 100% RH

( 1I Tuhacco in warehouse, held at 30°C and 40 per cent relut iv.: h u m i di ty, is placed in a room at 32°C and 70 per cent rclut iv. liu m id ity preparatory to being worked on. For each 50 kg (If I (IiI; II'CO moved from the warehouse, what is its bone-dry weight" What is the actual weight of this quantity of tobacco after ~I;,vlng in the working room?

(Pd2) = 19.94 kPa

W = (0.622) (PS2) = (0.622) (19.94) = 0.1524 kg/kg 2 101.3 - 19.94 P, - Ps2

An s. 44.13 kg, 55.16 kg Point 3:

Bdw = 250 kg

Point 4:

. MOIsture content =

(~) Air enters an adiabatic drier at 6 mls through a 2-m diarm-t rr d uct at 29°C dry bulb and 22°C wet bulb temperatures. It is lit-at I,d to 80°C before reaching the material to be dried and leaves t I}(' drier at 44°C and 80 per cent relative humidity. The matr-r i a l c-n t.crs the drier with a moisture content of24 per cent and It'av('s with a moisture content of 8 per cent. Determine (a) the mass of water removed per kg of dry air, (b) the volume flow rate of air o n tering the reheater, (c) the kg of water evaporated per sr-ro nd , (d) the mass flow rate of material leaving the drier, and (t') t.h« heat requirement of drier per kg of water evaporated.

0.10 = 0.0909 or .9.09% 1.0 + 0.10

W4 = (0.0909) (W) + Bdw

W

= 4

Bdw 1 _ 0.0909

=

250 A nAn

= 275 kg/h

(a) Mass balance about dryer mW 1 + W 3 = mW 2 + W 4

W.3 - W4

m = W

2-W1

Ans. (a) 0.0343 kg/kg (b) 18.85 cu mis, (c) 0.74 kg/s, (d) 3.507 kg/s, (e) 1496 kJ/kg water =

1000 - 275 0.1524 - 0.016

(b) Capacity of fan = mv = (5315 kg/h) (0.968 m 1 3600 s/h

= 5315 kg/h

3/kg)

= 1.43

m

3/s

"

(~'"

(3) A drier is to be designed to reduce the water content of a certain material from 55 per cent to 10 per cent. Air at 29°C dry bulb temperature and with a humidity ratio of 0.005 kg/kg is heated to 50 0 e in a reheater before entering the drier. The air leaves the drier at 38°C with 70 per cent relative humidity. On the basis of 1000 kg of product per hour, calcuiate (a) the volume flow rate of air entering the reheater, and (b) the heat supplied in the reheater.

Ans. (a) 9.59 cu mls (b) 234.8 kW (4) A dryer is to deliver 1000 kg/h of palay with final moisture content of 10%. The initial moisture content in the feed is 15% at atmospheric condition with 32°C dry bulb and 21°C wet bulb. The dryer is maintained at 45° while the relative humidity of the hot humid air from the dryer is 80%. If the steam pressure supplied to the heater is 2 MPa, determine the following:

'i

.".


(a) (b) (c) (d) (e)

Palay supplied to the dryer in kg/h. Temperature of the hot humid air from the dryer in DC. Air supplied to dryer in cu m/h. Heat supplied by the heater in kW. Steam supplied to heater in kg/h.

Chapter

Ans. (a) 1058.8 kg/h, (b) 45°C, (c) 1348.4 cu m/h, (d) 5.40 kW, (e) 10.28 kg/h

11

Prodnet Load Wlu-n a product enters a storage at a temperature above the ternpr-r.u lin' of the space, the product will give off heat to the space until it ('ools to the space temperature. The heat gain from the

product is computed by the following equations: Q

=Q 1 + Q2 + Q3

where Q = product load, kJ Q 1 = heat to cool from entering temperature to freezing temperature, kJ Q 2 = heat to freeze, kJ Q 3 = heat to cool from freezing temperature to final storage temperature, kJ Q 1 = (rn) (c b ) (t e - t r) Q 2 = (rn) (h.) Q 3 = (rn) (c) (t r - t s ) I'

~:l

where m = mass of the product, kg

cb = specific heat above freezing, kJ/kg. Co ca = specific heat below freezing, kJ/kg. Co t e = entering temperature, °C

t, = freezing temperature, °C t s = storage temperature, °C h. = product latent heat, kJlkg


Product Load 183

11-1) Compute the heat to be removed from 110 kg of lean beef if it were to be cooled from 20°C to 4°C, after which it is frozen and cooled to -18°C. Specific heat of beef above freezing is given as 3.23 k.I/kg.C", and below freezing is 1.68 k-J/kg.C''. Freezing point is -2.2°C, and latent heat of fusion is 233 kJ/kg.

specific heat of water heat of fusion of ice SOLUTION Q 1 = (c l )

SOLUTION

4.2292 kJ/kg.Co 335 kJ/kg

(~

t) = (1.9387 k.I/kg.C") [0 - (-4) ] Co ;::

Q 2 = 335 kJ/kg Q I = (110 kg) (3.23 k.I/kg.C") [ 20- (-2.2) ]CO = Q 2 = (110 kg) (233 kJ/kg)

7,888 kJ

(~

= 2,920 kJ Q= 36,438 kJ

(11-2) A refrigeration storage is supplied with 10,000 kg of fish at a temperature of 20°C. The fish has to be cooled to -10°C for preserving it for a long period without deterioration. The cooling takes place in 10 hours. The specific heat offish is 0.7 above freezing point and 0.3 below freezing point which is -3°C. The latent heat of freezing is 55.5 kcal/kg. Find the refrigerating capacity of the plant in tons. Q I = (10,000 kg) (0.7 kcal/kg.C") [20 - (-3)]Co= 161,000 kcal Q 2 = (10,000 kg) (55.5 kca/kg)

= 555,000 kcal

Q3 = (10,000 kg) (0.3 kcal/kg.C") [-3 - (-10)] Co Q

= 21,000 kcal = 737,000 kcal

= 24.35 tons

= 335.0 kJ/kg

t w ) = (4.2292 k-J/kg.C") (7 - 0) Co

= 29.6 kJ/kg

Heat from ice

= 25,630 kJ

Q3 = (110 kg) (1.68 kJ.CO) [-2.2- (-18)] Co

Q = (737,000 kcal) (4.187 kJ/kcal)

Q:J = (cw )

Heat load

>

7.8 kJ/kg

;:: 372.4 kJ/kg

(1.30) (114 kg) (3.35 k.I/kg.C") (29 - 7) Co

;:: 10,992 kJ 10,922 kJ ;:: 29.33 kg/24 h mass of ice;:: 372.4 kJ/kg

(11-4) Four hundred kilograms of poultry enter a chiller at 6°C and are frozen and chilled to a final temperature of -16°C for storage in 12 hours. Compute the product load. specific heat above freezing specific heat below freezing latent heat freezing temperature

3.18 k.I/kg.C" 1.55 k-I/kg.C'' 246 kJ/kg -3°C

SOLUTION

(10 h) (3,600 sth) (3.52 kW/ton) (11-3) A mass of ice at -4°C is needed to cool 114 kg of vegetables in a bunker for 24 hours. The initial temperature of the vegetables is assumed to be 29°C. It is also assumed that the average temperature in the bunker is 7°C, within the 24-hour period. If the heat gained per hour in the bunker is 30% of the heat removed to cool the vegetables from 29°C to 7°C, what would be the required mass of ice? specific heat of ice specific heat of vegetables

1.9387 k-I/kg.C" 3.35 k.I/kg.C"

Q I = (400 kg) (3.18 k-I/kg.C") [6 - (-3)] Q 2 = (400 kg) (246 kJ/kg) Q3 ::: (400 kg) (1.55 k-I/kg.C'') [(-3) - (-16) ] Co

;:: 11,448 kJ ::: 98,400 kJ ::: 8,060 kJ Q::: 117,908 kJ

117,908 kJ ::: 2.73 kW Q::: (12 h) (3,600 s/h)

'fJ'"

"

(

187


Table 1 (continued)

Table 2 Refrigeration 11 1"'''1)( -rt ics of liquid and saturated vapor

Enthalpy, kJ/k1

t, DC

P,kPa

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

886.57 915.03 944.18 974.03 1004.6 1035.9 1068.0 1100.7 1134.3 1168.6 1203.7 1239.6 1276.3 1313.9 1352.2 1391.5 1431.5 ·1472.4 1514.3 1557.0 1600.6 1645.1 1690.6 1737.0 1784.3 1832.6 1881.9 1932.2 1983.5 2035.9 W89.2 2143.6 2199.1 2255.6 2313.2

46 47 48 49

SO 51 52 53 54 55

Entropy, kJ/k,· K

Itf

hI

298.527 303.300 308.081 312.870 317.667 322.471 327.284 332.104 336.933 341.169 346.614 351.466 356.326 361.195 366.072 . 370.957 375.851 380.754 385.666 390.587 395.519 400.462 405.416 41fr.382 n5.362 420.358 425.369 430.399 435.450 440.523 445.623 .50.751 455.913 461.112 466.353

1410.41 1411.11 1481.17 1482.53 1483.18 1483.81 1484.42 1485.01 1485.59 1486.14 1486.67 1487.18 1487.66 1488.13 1488.57 1488.99 1489.39 1489.76 1490.10 1490.42 1490.11 1490.98 1491.21 1491.41 1491.58 1491.72 1491.83 1491.88 1491.91 1491.89 1491.83 1491. 73 1491.58 1491.38 1491.12

If

1.34452 1.360S5 1.37654 1.39250 1.40843 1.42433 1.44020 1.45604 1.47185 1.48762 1.50337 1.51908 1.53477 1.55042 1.56605 1.58165 1;59722 1.61276 1.62828 1.64317 1.65924 1.67470 1.69013 1.70554 1.72095 1.73635 1.75174 1.76714 1.78255 1.79798 1.81343 1.82891 1.84445 1.86004 1.87571

"

vf

5.3626 5.3512 5.33'9 5.3286 5.3175 5.3063 5.2953 5.2843 5.2733 5.2624 5.2516 5.2408 5.2300 5.2193 5.2086 5.1980 5.1874 5.1768 5.1663 5.1558 5.1453 5.1349 5.1244 5.1140 5.1036 5.0932 5.0827 5.0723 5.0618 5.0514 5.0409 5.0303 5.0198 5.0092 4.9985

1.6426 1.6466 1.6507 1.6541 1.6588 1.6630 1.66i2 1.6714 1.6757 1.6800 1.6844 1.6888 1.6932 1.6977 1.7023 1.7069 1.7115 1.7162 1.7209 1.7257 1.7305 1.7354 1.7404 1.7454 1.7504 1.7555 1.7607 1.7659 1.7712 1.7766 1.7820 1.7875 1.7931 1.7987 1.8044

v, 144.578 140.214 136.006 131.950 128.037 124.261 120.619 117.103 113.708 110.430 107.263 104.205 101.248 98.3913 95.6290 92.9579 90.3743 87.8748 85.4561 83.1150 80.8484 78.6536 76.5276 74.4678 72.4716 70.5365 68.6602 66.8403 65.0746 63.3608 61.6971 60.0813 58.5114 56.9855 55.5019

t,OC

P, kPa

hf

-30 -25 -20 -15 -10 -8

9.24 12.15 15.78 20.25 25.71 28.20 30.88 33.76 36..86 40.18 41.92 43.73 45.60 47.54 49.53 51.60 53.73 55.93 58.21 60.55 62.97 65.4 7 68.04 70.70 73.43 76.25 79.15 82.14 85.21 H8.38 91.64 94.99

174.25 178.53 182.81 187.09 191.39 193.11 194.83 196.55 198.27 200.00 200.86 201. 73 202.59 203.46 204.32 205:19 206.05 206.92 207.79 208.65 209.52 210.39 211.26 212.13 213.00 213.87 214.74 215.61 216.48 217.35 218.22 219.10 219.97 220.84 221.72 222.59 223.47 224.34 225.22 226.10 227.85 229.61 231.37 233.13 234.90

-6 -4

!

f

~

/

\

-2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28' 29 30 32 34 36 38 40

n.44

101.98 105.62 109.37 113.21 117.16 121.22 125.38 134.05 143.18 152.78 162.87 173.46

Specific volume, L/kg

Entropy, kJ/kg· K

Enthalpy. kJ/kg


h

x

373\1 37611 37866 381.12 38371 384.80 385.82 386.84 387.86 388.89 389.40 389.91 390.42 390.93 391.44 391.95 392.46 392.97 393.47 393.98 394.49 395.00 395.51 396.02 396.52 397.03 397.54 398.04 398.55 399.05 399.56 400.06 400.57 401.07 401.57 402.07 402.57 403.08 403.58 404.08 405.07 406.07 407.06 408.05 409.04

Sf

0.90099 0.91824 0.93517 0.95179 0.96813 0.97459 0.98100 0.98738 0.9937.1 1.00000 1.00313 1.00625 1.00936 1.01246 1.01555 1.01863 1.02170 1.02476 1.02782 1.03086 1.03389 1.03692 1.03994 1.04294 1.04594 1.04893 1.05191 1.05488 1.05785 1.06080 1.06375 1.06669 1.06961 1.07254 1.07545 1.07838 1.08125 1.08414 1.08702 1.08989 1.09561 1.10130 1.10696 1.11259 1.11819

Sg

1.72074 1.71447 1.70885 1.70377 1.69922 1.69753 1.69592 1.69438 1.69291 1.69150 1.69082 1.69018 1.68951 1.68888 1.68826 1.68766 1.68707 1.68650 1.68594 1.68539 1.68486 1.68434 1.68383 1.68333 1.68285 1.68238 1.68193 1.68148 1.68105 1.68062 1.68021 1.67982 1.67942 1.67905 1.67868 1.67832 1.67798 1.67764 1.67731 1.67699 1.67638 1.67581 1.67527 1.67476 1.67429

vf

vg

0.62466 0.62894 0.63331 0.63777 0.64234 0.64419 0.64606 0.64795 0.64985 0.65178 0.65275 0.65372 0.65470 0.65568 0.65667

1581.77 1225.53 960.954 761.949 610.466 560.196 514.840 473.883 436.764 403.130 387.493 372.593 358.366 344.792 331.859 319.500 307.698 296.427 285.648 275.347 265.483 256.063 247.037 238.396 230.130 222.205 214.614 207.332 200.361 193.665 187.245 181.089 175.166 169.485 164.034 158.786 153.754 148.903 144.246 139.768 131.305 123.462 116.135 109.430 103.151

0.65.7t>6

0.65866 0.65966 0.66067 0.66168 0.66270 0.66327 0.66475 0.66578 0.66682 0.66786 0:66891 0.66997 0.67102 0.67209 0.67316 0.67424 0.67532 0.67641 0.67750 0.67860 0.67971 0.68082 0:68194 0.68307 0.68533 0.68763 0.68995 0.69230 Q.69468

'f'

J 188 Refrigeration and Air Conditioning

189

l

,

Table 2 (continued)

Table 3 Refrigeration 12: properties of liquid and saturated vapor

I

,I

Enthalpy, kJ/tg

1,1

-I

t,OC

P,tPa

hf

hg

If

Ig

vf

45 50 55 60 70 80

202.28 234.64 270.83 311.10 405.15 519.21

239.32 243.75 248.21 252.68 261.68 270.79

411.49 413.93 416.34 418.73 423.42 427.98

1.13206 1.14576 1.15929 1.17267 1.19898

1.67324 1.67237 1.67165 1.67109 1.67031 1.66992

0.70074 0.70700 0.71346 0.72014 0.73421 0.74937

1.2~479

Enthalpy, kJ/kg


Entropy. kJ/kg· K

vg

t,OC

~J

~O

89.2884 77.6428 67.8040 59.4543 46.2114 36.3872

,

-55 -50 -45 -40 -35 -30 -28 -26 ..24 -22 -20 -18 -16 -14 -12 -10 -9 -8 -7 ~

-5 -4 -3 -2 -1 0 1 2 3 4 5 6

1 8 9 10 .11 12 13 14 15 16 17 18 19 20

~1 .Ja

"\

P, kPa

22.62 29.98 39.15 50.44 64.17 80.71 100.41 109.27 1.18.72 128.80 139.53 150.93 163.04 175.89 189.50 203.90 219.12 227.04 235.19 243.55 252.14 260.96 270.01 -279.30 288.82 298.59 308.61 318.88 329.40 340.19 351.24 363.55 374.14 386.01 398.15 410.58 423.30 436.31 449.62 463.23 477.14 491.37 505.91 520.76 535.94 551.45 567.29

Entropy, kJ/kg' K


f

hg

sf

'«

vf

vg

146.463 150.808 155.169 159.549 163.948 168.369 172.810 174.593 ! 76.380 178.171 179.965 181. 764 183.567 185.374 187.185 189.001 190.822 19"1 :734 192.647 193.562 194.477 195.395 196.313 197.233 198.-154 199.076 200.000 200.925 201.852 202.780 203.710 204 ..6 42 205.575 206.509 207.445 208.383 209.323 210.264 211.207 212.152 213.099 214.048 214.998 215.951 216.906 217.863 218.821

324.236 326.567 328.897 331.223 333.541 335.849 338.143 339.057 339.968 340.876 341.780 342.682 343.580 344.474 345.365 346.252 347.134 347.574 348.012 348.450 348.886 349.321 349.755 350.187 350.619 351.049 351.477 351.905 352.331 352.755 353.179 353.600 354.020· 354.439 354.856 355.272 355.686 356.098 356.509 356.918 357.325 357.730 358.134 358.535 358.935 359.333 359.729

0.77977 0.79990 0.81964 0.83901 0.85805 0.86776 0.89516 0.90244 0.90967 0.91686 0.92400 0.93110 0.93816 0.94518 0.95216 0.95910 0.96601 0.96945 0.97287 0.97629 0.97971 0.98311 0.98650 0.98989 0.99327 0.99664 1.00000 1.00335 1.00670 1.01004 1.01337 1.01670 1.02001 1.02333 1.02663 1.02993 1.03322 1.03650 1.03978 1.04305 1.04632 1.04958 1.05284 1.05609 1.05933 1.06258 1.06581

1.61373 1.60552 1.59810 1.59.142 1.58539 1.57996 1.57507 1.57326 1.57152 1.56985 1.56825 1.56672 1.56526 1.56385 1.56250 1.56121 1.55997 1.55938 1.55897 1.55822 1.55765 1.55710 1.55657 1.55604 1.55552 1.55502 1.55452 1.55404 1.55356 1.55310 1.55264 1.55220 . 1.55176 1.55133 1.55091 1.55050 1.55010 1.54970 1.54931 1.54893 1.54856 1.54819 1.54783 1.54748 1.54713 1.54679 1.54645

0.63689 0.64226 0.64782 0.65355 0.65949 0.66563 0.67200 0.67461 0.67726 0.67996 0.68269 0.68541 0.68829 0.69115 0.69407 0;69703 0.70004 0.70157 0.70310 0.70465 0.70622 0.70780 0.70939 0.71099 0'.71261 0.71425 0-.71590 0.71756 0.71924 0.72094 0.72265 0.72438 0.72612 0.72788 0.72966 0.73146 0.73326 0.73510 (,.73695 0.73882 0.74071 0.74262 0.74455 0.74649 0.74846 0.75045 0.75246

637.911 491.000 383.105 302.683 241.910 195.398 159.375 147.275 136.284 126.282 117.167 108.847 101.242 94.2788 87.8951 82.0344 76.6464 74.1155 71.6864 69.3543 67.1146 64.9629 62.8952 60.9075 58.9963 57.1579 55.3892 53.6869 52.0481 50.4700 48.9499 47.4853 . 46.0737 44.7129 43.4006 42.1349 40.9137 39.7352 38:5975 37.4991 36.4382 35.4133 34.4230 33.4658 32.5405 31.6457 30.7802

h

~ 190 Refrigeration and Air Conditioning

191

Table 3 (continued)

Table 4 Refrigeration 22: properties of liquid and saturated vapor

Enthalpy, kJjkg

Entropy, kJjkg' K

-----t,OC

j,

I'

11\

,J

j

~I:1 Ii ~II

I,

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 52 54 56 58 60 62 64 66 68 70 75 80 85 90 95 100

P,kPa

583.47 599.98 616.84 634.05 651.62 669.54 687.82 706.47 725.50 744.90 764.68 784.85 805.41 826.36 847.72 869.48 891.64 914.23 937.23 960.65 984.51 1008.8 1033.5 1058.7 1084.3 1110.4 1136.9 1163.9 1191.4 1219.3 1276.6 1335.9 1397.2 1460.5 1525.9 1593.5 1663.2 1735.1 1809.3 1885.8 2087.5 2304.6 2538.0 2788.5 3056.9 3344.1 ------

hf

-

219.783 220.746 221.712 222.680 223.650 224.623 225.598 226.576 221.557 228.540 229.526 230.515 23.1.506 232.501 233.498 234.499 235.503 236.510 237.521 238.535 239.552 " 240.574 241.598 242.627 243.659 244.696 245.736 246.781 241.830 248.884 251.004 253.144 255.304 257.486 259;690 261.918 264.172 i66.452 268.762 271.102 277.100 283.341 289.879 296.788 304.181 312.261 ---~.

hg

sf

Sg

360.122 360.514 360.904 361.291 361.676 362.059 362.439 362.817 363.193 363.566 363.937 36'4.305 364.670 365.033 365.392 365.749 366.103 366.454 366.802 367.146 367.487 367.825 368.160 368.491 368.818 369.141 369.461 369.777 370.088 370.396 370.997 371.581 372.145 372.688 373.210 373.707 374.180 374.625 375.042 375.427 376.234 376.777 376.985 376.748 375.887 374.070

1.06904 1.07227 1.07549 1.07871 1.08193 1.08514 1.08835 1.09155 1.09475 1.09195 1.10115 1.10434 1.10753 1.11072 1.11391 1.11710 1.12028 1.12347 1.12665 1.12984 1.13302 1.13620 1.13938 1.14251 1.14575 1.14894 1.15213 1.15532 1.15851 1.16170 1.16810 1.17451 1.18093 1.18738 1.19384 1.20034 1.20686 1.21342 1.22001 1.22665 1.24347 1.26069 1.27845 1.29691 1.31637 1.33732

1.54612 1.54579 1.54547 1.54515 1-.54484 1.54453 1.54423 1.54393 1.54363 1.54334 1.54305 1.54276 1.54247 1.54219 1.54191 1.54163 1.54135 1.54107 1.54079 1.54051 1.54024 1.53996 1.53968 1.53941 1.53913 1.53885 1.5'3856 1.53828 1.53199 1.53770 1.53712 1.536'51 1.53589 1.53524 1.53457 1.53387 1.53313 1.53235 1.53153 1.53066 1.52821 1.52526 1.52164 1.51708 1.51113 1.50296

Specific volume, L/kg vf

Enthalpy, kJ/kg

vg

i.rc

P, kPa

hf

hg

~O

37.48 49.47 64.39 82.71 104.95 131.68 163.48 177.76 192.99 209.22 226.48 244.83 264.29 284.93 306.78 329.89 354.30 367.01 380.06 393.47 407.23 421.35 435.84 450.70 465.94 481.57 497.59 514.01 530.83 548.06 565.71

134.763 139.830 144.,959 150.153 155.414 160.742 166.140 168.318 170.507 172.708 174.919 177.142 179.376 181.622 183.878 186.147 188.426 189.571 190.718 191.868 193.021 194.176 195.335 196.497 197.662 198.828 200.000 201.174 202.351 203.530 204.713 205.899 207.089 208.281 209.477 210.675 211.877 213.083 214.291 215.503 216.719 217.937 219.160 220.386 221.615 222.848 224.084

379.114 381.529 383.921 386.282 388.609 390.896 393.r38 394.021 394.896 395.762 396.619 397.467 398.305 399.133 399.951 400.759 401.555 401.949 402.341 402.729 403.114 403.496 403.876 404.252 404.626 404.994 405.361 405.724 406.084 406.440 406.793 407.143 407.489 407.831 408.169 408.504 408.835 409.162 409.485 409.804 410.119 410.430 410.736 411.038 411.336 411.629 411.9J8

'

-----------_.- .. ----.

-----_.

0.75449 0.75655 0.75863 0.76073 0.76286 0.76501 0.76716 0.76938 0.77161 0.77386 0.77614 0.77845 0.78079 0.18316 0.78556 0.78799 0.79045 0.79294 0.79546 0.79802 0.80062 0.80325 0.80592 0.80863 0.81137 0.81416 0.81698 0.81985 0.82277 0.82573 0.83119 , 0.83804 0.84451 0.85121 0.85814 0.86534 0.87282 0.88059 0.88870 0.89716 0.92009 0.94612 0.97621 1.01190 1.05581 1.11311

29.9429 29.1327 28.3485 27.5894 26.8542 26.1422 25.4524 24.7840 24.1362 23.5082 22.8993 22.3088 21.735'9 21.1802 20.6408 20.1173 19.6091 19.1156 18.6362 18.1706 17.7182 17.2785 16.8511 16.4356 16.0316 15.6386 15.2563 14.8844 14.5224 14.1 ?01 13.4931 lZ.8509 12.2412 11.6620 11.1113 10.5872 10.0881 9.61234 9.15844 8.72502 7.12258 6.82143 6.00494 5.25759

-55 -50 ~5

f

i "

-40 -35 -30 -28 -26 -24 -22 -20 -18 -16 -14 -12 -10 -9 -8 -7 ~

-5 -4 -3 -2 -1 0 1 2 3 4 5

6

3

4.563~~ 3.90280

\,

~

7 8 9 10 11 12 13 14 15 16 17 18 19 20

5~n.78

602.28 621.22 640.59 660.42 680.70 701.44 722.65 744.33 766.50 789.15 812.29 835.93 860.08 884.75909.93

Entropy, kJ/kg • K sf

0.73254 0.75599 0.77919 0.80216 0.82490 0.84743 0.86976 0.87864 0.88748 0.89630 0.90509 0.91386 0.92459 0.93129 0.93997 0.94862 0.95725 0.96155 0.06585 0.97014 0.97442 0.97870 0.98297 0.98724 0.99150 0.99575 1.00000 1.00424 1.00848 1.01271 1.01694 1.02116 1.02537 1.02958 1.03379 1.03799 1.04218 1.04637 1.05056 1.05474 1.05892 1.06309 1.06726 1.07142 1.07559 1.07974 1.08390


's 1.87886 1.86389 1.85000 \.83708 1.82504 1.81380 1.80329 1.79927 1.79535 1.79152 1.78779 1.78415 1.78059 1.77711 1.77371 1.77039 1.76713 1.76553 1.76394 1.76237 1.76082 1.75928 1.75775 1.75624 1.75475 1.75326 1.75279 1.75034 1.74889 1.74746 1.74604 1.74463 1.74324 1.74185 1.74047 1.73911 1.73775 1.73640 1.73506 1.73373 1.73241 1.73109 1.72978 1.72848 1.72719 1.72590 1.72462

vf

vg

0.68208 0.68856 0.69526 0.70219 0.70936 0-.71680 0.72452 0.72769· 0.73092 0.73420 0.73753 0.74091 0.74436 0.74786 0.75143. 0.15506 0.75876 0.76063 0.76253 0.76444 0.76636 0.76831 0.77028 0.77226 0.77427 0.77629 0.77834 0.78041 0.7824Q 0.78460 0.78673 0.78889 0.79107 0.79327 0.79549 0.79775 0.80002 0.80232 0.80465 0.80701 0.80939 0.81180 0.81424 0.81671 0.81922 0.82175 0.82431

537.152 414.827 324.557 256.990 205.745 166.400 135.844 125.563 116.214 107.701 99.9362 92.8432 86.3546 80.4103 74.9572 69.9478 65.3399 63.1746 61.0958 59.0996 57.1820 55.3394 53.5682 51.8653 50.2274 48.6517 47 .13~4. 45.6757 44.2702 42.9166 41.6124 40.3556 39.1441 37.9759 36.8493 35.7624 34.7136 33.7013 32.7239 31.7801 30.8683 29.9874 29.1361 28.3131 27.5173 26.7477 26.0032

", 192


I ~ I:I

Table 4 (continued) Enthalpy, kJjkg t,OC

P, kPa

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47

935.64 961.89 988.67 1016.0 1043.9 1072.3 1101.4 1130.9 1161.1 1191.9 1223.2 1255.2 1287.8 1321.0 1354.8 1389.2 1424.3 1460.1 1496.5 1533.5 1571.2 1609.6 1648.7 1688.5 1729.0 1770.2 1812.1 1854.8 1898.2 1942.3 2032.8 2126.5 2223.2 2323.2 2426.6 2533.3 2643.5 2757.3 2874.7 2995.9 3316.1 3662.3 4036.8 4442.5 4883.5

~·8

49 50 52 54 56 58 60 62 64 66 68 70 75 80 85 90 95

Entropy. kJjkg • K

"f

hg

Sf

Sg

Vf

225.324 226.568 227.816 229.068

412.202 412.481 412.755 413.025 413.289 413.548 413.802 414.050 414.293 414.530 414.762 414.987 415.207 415.420 415.627 415.828 416.021 416.208 416.388 416.561 416.726 416.883 417.033 417.174 417.308 417.432 417.548 417.655 417.752 417.838 417.983 418.083 418.137 418.141 418.089 41 '.978 417.802 417.553 417.226 416.809 415.299 412.898 409.101 402.653 38'6.708

1.08805 1.09220 1.09634 1.10048 1.10462 1.10876 1.11290 1.11703 1.12116 1.12530 1.12943 1.13355 1.13768 1.14181 1.14594 1.15007 1.15420 1.15833 1.16246 1.16659 1.17073 1.17486 1.17900 1.18315 1.18730 1.19145 1.19560 1.19977 1.20393 1.20811 1.21648 1.22489 1.23'333 1.24183 1.25038 1.25899 1.26768 1.27647 1.2i535 1.29436 1.31758 1.34223 1.36936 1.40155 1.45222

1.72334 1.72206 1.72080 1.71953 1.71827 1.71701 1.71576 1.71450 i.71325 1.71200 1.71075 1.70950 1.70826 1.70701 1.70576 1.70450 1.70325 1.70199 1.70073 1.69946 1.69819 1.69692 1.69564 1.69435 1.69305 1.69174 1.69043 1.68911 1.68777 1.68643 1.68370 1.68091 1.67805 1.67511 1.67208 1.66895 1.66570 1.66231 1.65876 1.65504 1.64472 1.63239 1.61673 1.59440 1.54712

0.82691 0.82954 0.83221 0.83491 0.83765 0.84043 0.84324 0.84610 0.84899 0.85193 0.85491 0.85793 0.86101 0.86412 0.86729 0.87051 0.87378 0.87710 0.88048 0.88392 0.88741 0.89097 0.89459 0.89828 0.90203 0.90586 0.90976 0.91374 0.91779 0.92193 0.93047 0.93939 0.94872 0.95850 Q.96878 0.97960 0.99104 1.00317 1.01608 1.02987 1.06916 1.11810 1.18328 1.28230 1.52064

~30.324

231.583 232.~47

234.115 235.387 236.664 237.944 239.230 240.520 241.814 243.114 244.418 245.727 247.041 248.361 249.686 251.016 252.352 253.694 255.042 256.396 257.756 259.123 260.497 261.877 263.264 266.062 268.891 271.754 274.654 277.594 280.577 283.607 286.690 289.832 293.038 301.399 310.424 320.505 332.616 351.767

'S){j [w '::lWnrOA :l!J!:l::ldS

Specific volume, Ljkg

o

o o

or.

o o

o o

C"!

~

0 0 ~

0 0

v:

o o o

o o

0 ooog 00000 ~r-:~~....:

"'!

N

o

N-

V

g

25.2829 24.5857 23.9107 23.2572 22.6242 22.0111 21.4169 20.8411 20.2829 19.7417 19.2168 18.7076 18.2135 17.7341 17.2686 16.8168 16.3779 15.9517 15.5375 15.1351 14.7439 14.3636 13.9938 13.6341 13.2841 12.9436 12.6122 12.2895 11.9753 11.6693 11.0806 10.5214 9.98952 9.48319 9.00062 8.54016 8.10023 7.67934 7.27605 6.88899 5.98334 5.14862 4.35815 3.56440

2.55133\

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7

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Temperature, 0('

~:""'--:;Ii"=

lL- ~ -r-r/

7

,-~-r:,'\,'

Z,-~r..- ""- -+-'bcb

370

380

-20

V- 17 V

390

I

0

20

7

40 41"0

I

~/

Enthalpy, kJ/kg

400

l.L--~~

!'

420

I

1/

1300

I

320

-r

330

T

340

350

360

o

320

330

340

.,

If

\f

III

\

...-

....

350

.. ~

-l

A_~'-·-I' - - - - -f I

kJ/k.

370 EnUlalpy.

360

•

380

An

430

I

/

~;

v

/

/

390

400

410

~,f-v=FV 60 80

v

V

1/

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./

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[7/

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1

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/

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V

V

/

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L

V

/'

" /'

L

~(J(J~~~

450

r

100

420

420

100

/

I 1./ 5

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{

.....1......a

'J.

iz,

~

I

430

0.300

O.ISO

_--_

..:::

..........~

~_.

...........

4.000

3.000

2.000

1.500

0.(,00 > 0,700 ;~ ·G 0.800 :.; 0.900 if] 1.000

0.500

:.;

~'l

'-C

......

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crq

S'

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c-t

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0.300

O.~OO

0.150

0.100

1 ~ 0

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....._----...-.........

/

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l---t-" '\, C\

440

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JIt.::.~~=-t7

-

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--_.. t:iZ:'Et1< ,V

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460

V/'/'

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80

~

410

1/

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400

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390

Q: ......~

rr A

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I

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380

,,+-~~A:±="-+-'~:#:;~l[

-yr

1

u// V!..-I1I .... 1L&oo-'"7\ , ~.-.,..,~

10!-

I

II

370

_......

Figure 3 Pressure-enthalpy diagram of superheated refrigerant 12 vapor.

200

300

.---

310

_... -'

I I I I I I I I I I i·' 1'-1 1

300

..

/

./

60

t -r 7 / ~ f--7' 17 7 / L".,L- ~ V / L,-..."L.'--"7~/'\,~(J V / . / -'- ~r---7 / II _JA--, 7

-30

-40

40

l) f-----f---+---+----+---+---

10

~

I

"/, \

..A

,/

1.78

I

450

V/'// ~/ ./,/ / / /' /' ,/

Y V\

/\

/'

1.76

/ / / 'bt>." / 20 I-----+_-+~--+_--+-_ _+_-+_-+__------,~r--~~_""'.---:=__A_~ '7" / . J - - ~,1'\,' ./ ./ V 1-----1_-1-_--+_-+-_-+-_+--_-"10 J --7' 7 ~-/ ~ / / "" -, -r="[7' / 7 / ~-'bD _~/' / // ~ 1---7'~/ ,-~-.,.~'-"/ / q~

L

10

'/

/

"-"

! (J/

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L ~;.--.:\ '\'b f----I--+--+--+--+---+--+-----jf---+--+If..--7-/---4~=..v-'"#;i!"':":::::+-+-~"L-J--+7

~400

j

\

7"~-<'\t>. -~

#_++~.f_--+-----,L7____11_c==-'-/:..H-._:;;JfL=J,..~-b-7L+---+' 30 I----j---+---+--+---+--+---+--l----I~ 7 ",I

::;

~

\

\L-4-~

V

I

I

1.74

440

430 1.72

LLU-~L--,,~ X f----+-_+--+_-+---II---+-_+----+_-+----+_-+--_+-.30~J,..V-+!L~-==J..l\-_\,.o<'.''\,~m___+_,1- 1

/\...

/'

1.70

420

,\(J:-r

1.68 7 -v

410

Entropy, kJ/kg: . K

I

400

Figure 2 Pressure-enthalpy diagram of superheated refrigerant 11 vapor.

'J

~ 40

:.;

~

I

I

f----~-_+_-__+_-+_-I___~-____+_-__+_-+_-I____+--+_-___+_ 40,.. ~I-

360

350

100 1 1/ ...L-~...1-·~·11 90 I___---+--/----+---+---+--+--+----lf..----+---+-_l_ 0 f-----+_--+-_+-_f---_+_--+-_+------,f---_I_--+--+----c-~A~~____.-~/~A+_-~A+_T /' 80 , _--+-_-+----+_--+-_+----+_--+-_+----I-_---+-__I+/-c=l.l..l/:..-""'.-£-=7=----++-~_+_+__+4' \D ~ / / 'v: 70 r 10 / :-,/ hO I-----+--/---+---+---+--+--+-----,f----+-_I_

:00

240

340

_ - ....

~

.~

350

200

300

400

500

600

360

360

370

T

T

380 390

400

Entropy, k J /kg . K

380

Temperature, "('

370

400

390

I

410

I

4\0 440

430 Enthalpy kJjkg

420

440

450

w

~ ...... _.:

-

~_ ... ~~----.,...- ~ ... - _ . , . . -

_

po

b

-. _.-

~

-

_.p'';''''''

I

•

.-

=-.

-

~'".;;:,

_._

460

I ~ X'(!z::t:SJl4=="'"K \ A

470

I

,'. 470

1.76 1.78 1.80 1.82

460

. -._-_.

450

"1 - - - - - , ) .

430

1.70 1.72 '\ Ii-I

420

::::;:;:;;;;;;:;.:;:;;=====:=..;:...:.....,._.~.---_

~. 4 Pr~ss~re-enthalPY diagram of superheated refrigerant 22 vapor.

"-'

~ c:

'lJ

'"

c, ..><:

350

1000 900 800 700

--"J~~....

-,....-----.

,

480

480

490

490

::l

E

v

E

cc ~

0.300

0.200

0.IS0

O.OllO 0.100

Vl

B c,

c:

OJ

o.oso "0 >

0.070

0.060

O.OSO

0.040

(TQ

5'

§

e-: c.

::l

o

~

~

?

P

::l

OJ

o' ::l

~

~

CD

iQ'

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0;

;.c

~

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Ref Aircon1A Reviewer

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