ch07.pdf

Chapter 7

1. Evaluate the perfor performance mance of an Alternate Altern ate E nergy Tech Technologies nologies AE-21E AE-21 E flat plate solar collector (Figure (Figu re 7.10) 7.10) for 21 February Febru ary at 40-degrees 40-de grees north north latitude. latitud e. The wat wat er inlet temperature temperature is 60 F, and the ambient temperat temperature ure varies varies from f rom 25 25 F at 7 am (solar time) to 45 F at at solar noon to 35 deg degrees rees at 5 pm (solar (solar time). The collector is inclined inclined at 40 degrees from the horizontal. Metrics of interest include the hourly efficiency, the hourly solar energy collect col lected, ed, and the percent of solar energy captured. Hour-by-hour flat plate solar collector analysis. 40o N. Latitude 21 February Determine the performance of a Alternate Energy Technologies AE-21.

Enter information about the flat plate solar collector (obtained for SRCC web site). 2

Area := 20.73 ⋅ ft

AreaTot := 1 ⋅ Area

Inte Interc rcep eptt := 0.66 0.66

Slope :=

−1.123 ⋅

2

Area AreaTo Tott = 20.7 20.73 3 ⋅ ft

F := R

BTU 2

hr ⋅ ft

⋅F

Setup vectors for I (solar irradiation), Tin, Tamb . Obtained from weather data base. ORIGIN

≡

i := 1 .. 11

1

⎛ 21 ⎞

⎛ 60 ⎞

122

⎜ ⎟ ⎜ 205 ⎟ ⎜ 267 ⎟ ⎜ 306 ⎟ ⎜ ⎟ I := ⎜ 319 ⎟ ⋅ ⎜ 306 ⎟ ⎜ ⎟ ⎜ 267 ⎟ ⎜ 205 ⎟ ⎜ 122 ⎟ ⎜ ⎝ 21 ⎠

60

BTU 2

ft

⋅ hr

⎜ ⎟ ⎜ 60 ⎟ ⎜ 60 ⎟ ⎜ 60 ⎟ ⎜ ⎟ Tin := ⎜ 60 ⎟ ⋅ F ⎜ 60 ⎟ ⎜ ⎟ ⎜ 60 ⎟ ⎜ 60 ⎟ ⎜ 60 ⎟ ⎜ ⎝ 60 ⎠

⎛ 25 ⎞ 30

⎜ ⎟ ⎜ 35 ⎟ ⎜ 40 ⎟ ⎜ 43 ⎟ ⎜ ⎟ Tamb := ⎜ 45 ⎟ ⋅ F ⎜ 45 ⎟ ⎜ ⎟ ⎜ 43 ⎟ ⎜ 41 ⎟ ⎜ 38 ⎟ ⎜ ⎝ 35 ⎠

Calculate the solar collector efficiency.

1

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯ → Tin − Tamb ⎞ ⎛ 0.66 1.12 BTU η := 0.66 − 1.1233 ⋅ ⋅ 2 I hr ⋅ ft ⋅ F ⎝ ⎠ η i :=

ηi

if

ηi > 0

0 otherw otherwise ise

η=

1

0

2

0.384

3

0.523

4

0.576

5

0.598

6

0.607

7

0.605

8

0.588

9

0.556

10

0.457

11

0

1

Calculate the useful energy harvested.

⎯⎯⎯⎯ → Quseful := ( AreaTot ⋅ I ⋅ η )

Quseful

=

1

0

2

970.786

3

2.223·10 3

4

3.187·10 3

5

3.791·10 3

6

4.015·10 3

7

3.837·10 3

8

3.257·10 3

9

2.362·10 3

10

1.157·10 3

11

0

⋅

BTU hr

The daily harvested energy is the sum of the hourly harvested energy. 9

QTotal :=

∑

(Q

useful

i= 1

i

⋅ 1 ⋅ hr )

kWh := 1000 ⋅ watt ⋅ hr

QTotal

Find the percent of incident solar energy captured. 9

∑ percent :=

i= 1

(Q

useful

i

⋅ 100) percent = 56.521

9

∑

i= 1

(AreaTot ⋅ Ii)

=

6.929 ⋅ kWh

2. Evaluate the performance of a Heliodyne Gobi 408 flat plate solar collector (performance characteristics available from the SRCC web site) for 21 January at 32degrees north latitude. The water inlet temperature is 58 F, and the ambient temperature varies from 25 F at 7 am (solar time) to 45 F at solar noon to 35 degrees at 5 pm (solar time). The collector is inclined at 32 degrees from the horizontal. Metrics of interest include the hourly efficiency, the hourly solar energy collected, and the percent of solar energy captured. Hour-by-hour flat plate solar collector analysis. 32o N. Latitude 21 January Determine the performance of a Heliodyne Gobi 408.


Area := 32.23 ⋅ ft


Intercept := 0.70

Slope :=

−0.70 ⋅

2

AreaTot = 32.23 ⋅ ft

BTU hr ⋅ ft

2

⋅F


≡

i := 1 .. 9

1

⎛ 106 ⎞ ⎜ 193 ⎟ ⎜ 256 ⎟ ⎜ 295 ⎟ ⎜ ⎟ I := ⎜ 308 ⎟ ⋅ ⎜ 295 ⎟ ⎜ ⎟ ⎜ 256 ⎟ ⎜ 193 ⎟ ⎝ 106 ⎠

BTU ft

2

⋅ hr

⎛ 58 ⎞ ⎜ 58 ⎟ ⎜ 58 ⎟ ⎜ 58 ⎟ ⎜ ⎟ Tin := ⎜ 58 ⎟ ⋅ F ⎜ 58 ⎟ ⎜ ⎟ ⎜ 58 ⎟ ⎜ 58 ⎟ ⎝ 58 ⎠

⎛ 25 ⎞ ⎜ 30 ⎟ ⎜ 35 ⎟ ⎜ 40 ⎟ ⎜ ⎟ Tamb := ⎜ 45 ⎟ ⋅ F ⎜ 43 ⎟ ⎜ ⎟ ⎜ 41 ⎟ ⎜ 38 ⎟ ⎝ 35 ⎠


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ → Tin − Tamb ⎞ ⎛ BTU η := 0.70 − 0.70 ⋅ ⋅ I 2 hr ⋅ ft ⋅ F ⎝ ⎠

⎛ 0.482 ⎞ ⎜ 0.598 ⎟ ⎜ 0.637 ⎟ ⎜ 0.657 ⎟ ⎜ ⎟ η = ⎜ 0.67 ⎟ ⎜ 0.664 ⎟ ⎜ ⎟ ⎜ 0.654 ⎟ ⎜ 0.627 ⎟ ⎝ 0.548 ⎠

F := R

⎛ 1.647 × 103 ⎜ 3.723 × 103 ⎟ ⎜ ⎟ ⎜ 5.257 × 103 ⎟ ⎜ ⎟ 3 ⎜ 6.249 × 10 ⎟ ⎜ ⎟ 3 Quseful = ⎜ 6.655 × 10 ⎟ ⋅ ⎜ ⎟ ⎜ 6.317 × 103 ⎟ ⎜ ⎟ 3 5.392 10 × ⎜ ⎟ ⎜ 3⎟ 3.903 10 × ⎜ ⎟


⎯⎯⎯⎯ →

Quseful := ( AreaTot ⋅ I ⋅ η )

BTU hr

3

⎝ 1.873 × 10 ⎠ The daily harvested energy is the sum of the hourly harvested energy. 9

QTotal :=

∑

(Q

useful

i= 1

i

⋅ 1 ⋅ hr )

kWh := 1000 ⋅ watt ⋅ hr

QTotal


∑ percent :=

i= 1

(Q

useful

i

⋅ 100) percent = 63.376

9

∑

i= 1

(AreaTot ⋅ Ii)

=

12.021 ⋅ kWh

3. An office building in Meridian, MS is to employ Alternate Energy Technologies AE21E flat plate solar collectors (Figure 7.10) for space and hot water heating. The collectors will be mounted facing south tilted at the same angle as the latitude. The estimated monthly load is provided in the following table: Month

Load (1000 MJ) 108 100 80 60 20 15 15 15 15 25 60 95

January February March April May June July August September October November December

The solar system is to provide all the hot water heating requirements for the summer months and a yearly solar heating fraction of 50 percent. (a) How many AE-21E solar collectors are needed to meet the specifications (b) If natural gas cost $13 per million Btu, how much is saved by the system? ORIGIN

≡

1

Set up all input parameters on a monthly basis using the range variable i: i := 1 .. 12

Define month in terms of days.

6

MJ := 10

⋅J

⎛ 31 ⎞ ⎜ 28 ⎟ ⎜ 31 ⎟ ⎜ 30 ⎟ ⎜ ⎟ ⎜ 31 ⎟ ⎜ 30 ⎟ month := ⎜ ⎟ ⋅ day ⎜ 31 ⎟ ⎜ 31 ⎟ ⎜ 30 ⎟ ⎜ ⎟ ⎜ 31 ⎟ ⎜ 30 ⎟ ⎝ 31 ⎠

⎛ jan ⎞ ⎜ feb ⎟ ⎜ mar ⎟ ⎜ apr ⎟ ⎜ ⎟ ⎜ may ⎟ ⎜ jun ⎟ ⎜ jul ⎟ ⎜ ⎟ ⎜ aug ⎟ ⎜ sep ⎟ ⎜ ⎟ ⎜ oct ⎟ ⎜ nov ⎟ ⎝ dec ⎠

The monthly-average conditions from NREL are used to obtain solar and weather data for the location, Meridian, MS, in this case. The load must come from building information. I := 1 ⋅

kW ⋅ hr 2

m

Irradiation (daily)

⎛ 3.6 ⎞ ⎜ 4.4 ⎟ ⎜ 5.0 ⎟ ⎜ 5.6 ⎟ ⎜ ⎟ ⎜ 5.6 ⎟ ⎜ 5.6 ⎟ I := ⎜ ⎟⋅ 5.4 ⎜ ⎟ ⎜ 5.5 ⎟ ⎜ 5.2 ⎟ ⎜ ⎟ ⎜ 5.2 ⎟ ⎜ 4.1 ⎟ ⎝ 3.5 ⎠

I = 3.6 ⋅

⋅ day

MJ 2

m

⋅ day

Load (monthly)

⎛ 108 ⎞ ⎜ 100 ⎟ ⎜ 80 ⎟ ⎜ 60 ⎟ ⎜ ⎟ ⎜ 20 ⎟ ⎜ 15 ⎟ Load := ⎜ ⎟ 1000 ⋅ MJ 15 ⎜ ⎟ ⎜ 15 ⎟ ⎜ 15 ⎟ ⎜ ⎟ ⎜ 25 ⎟ ⎜ 60 ⎟ ⎝ 95 ⎠

3.6 ⋅ MJ 2

m

Average Temperature C

⋅ day

⎛ 7.2 ⎞ ⎜ 9.4 ⎟ ⎜ 13.7 ⎟ ⎜ 17.8 ⎟ ⎜ ⎟ ⎜ 21.8 ⎟ ⎜ 25.6 ⎟ Tamb := ⎜ ⎟C 27.2 ⎜ ⎟ ⎜ 27.0 ⎟ ⎜ 24.1 ⎟ ⎜ ⎟ ⎜ 17.8 ⎟ ⎜ 13.1 ⎟ ⎝ 9.1 ⎠

Insert solar collector characteristics: Number := 107 FR τα := 0.66

W

FRUL := 6.37

2

m

2

A := 1.926Number ⋅ m

⋅C

Implement the f-chart analysis usi ng the definiti ons of X and Y.

Compute X and Y:

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ →

A ⋅ month⎤ ⎡ X := 0.97FRUL ⋅ ( 100 ⋅ C − Tamb) ⋅ Load ⎦ ⎣

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ →

⎛ Y := A ⋅ 0.96 ⋅ 0.97FR τα ⋅ I ⋅ ⎝

month ⎞ Load ⎠

1

X

=

1

1

2.931

1

0.471

2

2.791

2

0.562

3

3.679

3

0.883

4

4.522

4

1.277

5

13.335

5

3.958

6

16.371

6

5.107

7

16.553

7

5.089

8

16.598

8

5.183

9

16.701

9

4.742

10

11.214

10

2.94

11

4.78

11

0.935

12

3.263

12

0.521

Y=

Calculate f (the solar fraction per month) based on the f-chart correlation:

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ → 2 2 3 f := ( 1.029 ⋅ Y − 0.065 ⋅ X − 0.245 ⋅ Y + 0.0018 ⋅ X + 0.0215 ⋅ Y ) 1

f := i

<1 i

f if f i

1 otherwise

If solar will more than satisfy the load, set f = 1. f =

Determine the monthly load supplied by solar system.

⎯⎯ →

Load solar := ( f ⋅ Load )

1

0.258

2

0.337

3

0.518

4

0.702

5

1

6

1

7

1

8

1

9

1

10

0.951

11

0.496

12

0.28

Estimate the yearly solar fraction.

1

Load solar =

1

2.783·104

2

3.371·104

3

4.143·104

4

4.212·104

5

2·104

6

1.5·104

7

1.5·104

8

1.5·104

9

1.5·104

10

2.378·104

11

2.974·104

12

2.655·104

12

∑ percent :=

i= 1

(Load

solar i

⋅ 100)

12

∑

Load

i

i= 1

⋅ MJ

percent = 50.193

Part (a): The number of collectors required is 107. The resulting solar fraction is 50.2 percent. 12

TotalSolar :=

∑

Load solar

i

i= 1

Savings :=

TotalSolar 6

10

⋅ 13

TotalSolar

=

4

8.477 × 10 3

Savings = 3.76 × 10

⋅ BTU

Part (b): The total saving for the system is $3760 per year.

⋅ kWh

4. Work problem 2 for a New England, a Midwest, and a Pacific-northwest locations. Contrast and discuss the results for the different locations. Problem 2 addresses a single day performance of a specificied solar collector. The weather data provided in the textbook is for increments of 8 degrees in latatude with no other geographic location. For this problem different locations will be assessed by evaluating the performance at 32, 40, and 48 degrees. Weather data will be shown for the different latitudes, and the results placed in a table. In order to provide a basis for comparison, the inlet water temperature and the air temperature distribution was not changed.

Hour-by-hour flat plate solar collector analysis.

21 January

Determine the performance of a Heliodyne Gobi 408.


Area := 32.23 ⋅ ft


Intercept := 0.70

Slope :=

2


F := R

BTU

−0.70 ⋅

2

hr ⋅ ft

⋅F


≡

i := 1 .. 9

1

⎛ 106 ⎞ ⎜ 193 ⎟ ⎜ 256 ⎟ ⎜ 295 ⎟ ⎜ ⎟ I := ⎜ 308 ⎟ ⋅ ⎜ 295 ⎟ ⎜ ⎟ ⎜ 256 ⎟ ⎜ 193 ⎟ ⎝ 106 ⎠

BTU 2

ft

32 degrees

⋅ hr

⎛ 74 ⎞ ⎜ 171 ⎟ ⎜ 237 ⎟ ⎜ 277 ⎟ ⎜ ⎟ I := ⎜ 391 ⎟ ⋅ ⎜ 277 ⎟ ⎜ ⎟ ⎜ 237 ⎟ ⎜ 171 ⎟ ⎝ 74 ⎠

BTU 2

ft

⋅ hr

40 degrees

⎛ 19 ⎞ ⎜ 132 ⎟ ⎜ 206 ⎟ ⎜ 249 ⎟ ⎜ ⎟ I := ⎜ 364 ⎟ ⋅ ⎜ 249 ⎟ ⎜ ⎟ ⎜ 206 ⎟ ⎜ 132 ⎟ ⎝ 19 ⎠

⎛ 58 ⎞ ⎜ 58 ⎟ ⎜ 58 ⎟ ⎜ 58 ⎟ ⎜ ⎟ Tin := ⎜ 58 ⎟ ⋅ F ⎜ 58 ⎟ ⎜ ⎟ ⎜ 58 ⎟ ⎜ 58 ⎟ ⎝ 58 ⎠

BTU 2

ft

⋅ hr

48 degrees


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ → Tin − Tamb ⎞ ⎛ BTU η := 0.70 − 0.70 ⋅ ⋅ I 2 hr ⋅ ft ⋅ F ⎝ ⎠

η i := η i

if

ηi > 0

0 otherwise

⎛ 25 ⎞ ⎜ 30 ⎟ ⎜ 35 ⎟ ⎜ 40 ⎟ ⎜ ⎟ Tamb := ⎜ 45 ⎟ ⋅ F ⎜ 43 ⎟ ⎜ ⎟ ⎜ 41 ⎟ ⎜ 38 ⎟ ⎝ 35 ⎠

⎛ 0 ⎞ ⎜ 0.552 ⎟ ⎜ 0.622 ⎟ ⎜ 0.649 ⎟ ⎜ ⎟ η = ⎜ 0.675 ⎟ ⎜ 0.658 ⎟ ⎜ ⎟ ⎜ 0.642 ⎟ ⎜ 0.594 ⎟ ⎝ 0 ⎠

⎛ 0 ⎞ ⎜ 2.346 × 103 ⎟ ⎜ ⎟ 3 4.129 10 × ⎜ ⎟ ⎜ 3⎟ 5.212 10 × ⎜ ⎟ ⎜ ⎟ Quseful = 7.919 × 103 ⋅ ⎜ ⎟ ⎜ 5.279 × 103 ⎟ ⎜ ⎟ 3 ⎜ 4.264 × 10 ⎟ ⎜ ⎟ 3 ⎜ 2.527 × 10 ⎟


⎯⎯⎯⎯ →


⎝

0

BTU hr

⎠


QTotal :=

∑

(Q

useful

i= 1

i

⋅ 1 ⋅ hr )

kWh := 1000 ⋅ watt ⋅ hr

QTotal

=

9.283 ⋅ kWh


∑ percent :=

i= 1

(Q

useful

i

⋅ 100) percent = 62.36

9

∑

(AreaTot ⋅ Ii)

i= 1

Latititude 32 40 48

QTotal 10.75 kWh 11.37 kWh 9.38 kWh

percent 56.68 63.03 62.36

Under the stated assumptions, the energy collected varies with latitude while the percent of solar energy captured is 57-62 percent.

5. Work problem 2 using several different flat-plate solar collectors from the SCCR results reported on the SRCC website. Discuss the effect slope and intercept for the conditions of problem 2. Hour-by-hour flat plate solar collector analysis. 32o N. Latitude 21 January The only change in the solution for Problem 2 is that performance information on several different solar collectors will be used. The SRCC data base is used. Enter information about the flat plate solar collector (obtained for SRCC web site). NC-32. 2

Area := 34.20 ⋅ ft


Intercept := 0.50

Slope :=

2


BTU

−0.90 ⋅

hr ⋅ ft

2

⋅F

Heliodyne Gobi 408. 2

Area := 32.23 ⋅ ft


Intercept := 0.70

Slope :=

2


BTU

−0.70 ⋅

hr ⋅ ft

2

⋅F

AES A E-21. 2

Area := 20.73 ⋅ ft


Intercept := 0.66

Slope :=

−1.123 ⋅

2

AreaTot = 20.73 ⋅ ft BTU

hr ⋅ ft

2

⋅F


≡

1

⎛ 106 ⎞ ⎜ 193 ⎟ ⎜ 256 ⎟ ⎜ 295 ⎟ ⎜ ⎟ I := ⎜ 308 ⎟ ⋅ ⎜ 295 ⎟ ⎜ ⎟ ⎜ 256 ⎟ ⎜ 193 ⎟ ⎝ 106 ⎠

i := 1 .. 9

BTU 2

ft

⋅ hr

⎛ 58 ⎞ ⎜ 58 ⎟ ⎜ 58 ⎟ ⎜ 58 ⎟ ⎜ ⎟ Tin := ⎜ 58 ⎟ ⋅ F ⎜ 58 ⎟ ⎜ ⎟ ⎜ 58 ⎟ ⎜ 58 ⎟ ⎝ 58 ⎠

⎛ 25 ⎞ ⎜ 30 ⎟ ⎜ 35 ⎟ ⎜ 40 ⎟ ⎜ ⎟ Tamb := ⎜ 45 ⎟ ⋅ F ⎜ 43 ⎟ ⎜ ⎟ ⎜ 41 ⎟ ⎜ 38 ⎟ ⎝ 35 ⎠

F := R

⎛ 0.31 ⎞ ⎜ 0.497 ⎟ ⎜ 0.559 ⎟ ⎜ 0.591 ⎟ ⎜ ⎟ η = ⎜ 0.613 ⎟ ⎜ 0.603 ⎟ ⎜ ⎟ ⎜ 0.585 ⎟ ⎜ 0.544 ⎟ ⎝ 0.416 ⎠


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ → Tin − Tamb ⎞ ⎛ η := Intercept + Slope ⋅

⎝

⎠

I


⎛ 682.038 ⎞ ⎜ 1.989 × 103 ⎟ ⎜ ⎟ 3 2.967 10 × ⎜ ⎟ ⎜ 3⎟ 3.617 10 × ⎜ ⎟ ⎜ ⎟ Quseful = 3.911 × 103 ⋅ ⎜ ⎟ ⎜ 3.687 × 103 ⎟ ⎜ ⎟ ⎜ 3.107 × 103 ⎟ ⎜ ⎟ 3 ⎜ 2.175 × 10 ⎟

⎯⎯⎯⎯ →


η i :=

ηi

if

ηi > 0

0 otherwise

⎝

914.836

BTU hr

⎠


QTotal :=

∑

(Q

useful

i= 1

i

⋅ 1 ⋅ hr )

kWh := 1000 ⋅ watt ⋅ hr

QTotal = 6.755 ⋅ kWh


∑ percent :=

i= 1

(Q

useful

i

⋅ 100) percent = 55.374

9

∑

(AreaTot ⋅ Ii)

i= 1

Collector NC-32 Gobi 408 AET AE-21

QTotal

percent

8.35 kWh 12.02 kWh 6.76 kWh

41.484 63.376 55.374

The NC-32 has the smallest intercept of the three collectors.

6. An office building in a specified location is to employ Heliodyne Gobi 408 flat plate solar collectors for space and hot water heating. The estimated monthly load is provided in the following table: Month January February March April May June July August September October November December

Load (1000 MJ) 130 120 100 70 25 20 20 20 20 25 70 105

The solar system is to provide all the hot water heating requirements for the summer months and a yearly solar heating fraction of 50 percent. (a) How many Gobi 408 solar collectors are needed to meet the specifications (b) If natural gas cost $9 per million Btu, how much is saved by the system? This problem was set up to permit an instructor to select the location. The problem will be worked using Meridian, MS weather data. To select another city, use the NREL data base for weather information. ORIGIN

≡

1

Set up all input parameters on a monthly basis using the range variable i: i := 1 .. 12

6

MJ := 10

⋅J

⎛ 31 ⎜ 28 ⎟ ⎜ 31 ⎟ ⎜ 30 ⎟ ⎜ ⎟ ⎜ 31 ⎟ ⎜ 30 ⎟ month := ⎜ ⎟ ⋅ day ⎜ 31 ⎟ ⎜ 31 ⎟ ⎜ 30 ⎟ ⎜ ⎟ ⎜ 31 ⎟ ⎜ 30 ⎟ ⎝ 31 ⎠

Define month in terms of days.

⎛ jan ⎜ feb ⎟ ⎜ mar ⎟ ⎜ apr ⎟ ⎜ ⎟ ⎜ may ⎟ ⎜ jun ⎟ ⎜ ⎟ ⎜ jul ⎟ ⎜ aug ⎟ ⎜ sep ⎟ ⎜ ⎟ ⎜ oct ⎟ ⎜ nov ⎟ ⎝ dec ⎠

The monthly-average conditions from NREL are used to obtain solar and weather data for the location, Meridian, MS, in this case. The load must come from building information. I := 1 ⋅

kW ⋅ hr 2

m

⋅ day

Irradiation (daily)

⎛ 3.6 ⎞ ⎜ 4.4 ⎟ ⎜ 5.0 ⎟ ⎜ 5.6 ⎟ ⎜ ⎟ ⎜ 5.6 ⎟ ⎜ 5.6 ⎟ I := ⎜ ⎟⋅ 5.4 ⎜ ⎟ ⎜ 5.5 ⎟ ⎜ 5.2 ⎟ ⎜ ⎟ ⎜ 5.2 ⎟ ⎜ 4.1 ⎟ ⎝ 3.5 ⎠

I = 3.6 ⋅

MJ 2

m

⋅ day

Load (monthly)

⎛ 130 ⎞ ⎜ 120 ⎟ ⎜ 100 ⎟ ⎜ 70 ⎟ ⎜ ⎟ ⎜ 25 ⎟ ⎜ 20 ⎟ Load := ⎜ ⎟ 1000 ⋅ MJ 20 ⎜ ⎟ ⎜ 20 ⎟ ⎜ 20 ⎟ ⎜ ⎟ ⎜ 25 ⎟ ⎜ 70 ⎟ ⎝ 105 ⎠

3.6 ⋅ MJ 2

m

Average Temperature C

⋅ day

⎛ 7.2 ⎞ ⎜ 9.4 ⎟ ⎜ 13.7 ⎟ ⎜ 17.8 ⎟ ⎜ ⎟ ⎜ 21.8 ⎟ ⎜ 25.6 ⎟ Tamb := ⎜ ⎟C 27.2 ⎜ ⎟ ⎜ 27.0 ⎟ ⎜ 24.1 ⎟ ⎜ ⎟ ⎜ 17.8 ⎟ ⎜ 13.1 ⎟ ⎝ 9.1 ⎠

Insert solar collector characteristics: Number := 62 FR τα := 0.70

FRUL := 4.2

W 2

m

⋅C

Implement the f-chart analysis usi ng the definiti ons of X and Y.

Compute X and Y:

2

A := 2.994Number ⋅ m

⎯

⎯ ⎯ ⎯ ⎯ → A month ⋅ ⎡ ⎤ X := 0.97FRUL ⋅ ( 100 ⋅ C − Tamb) ⋅

⎣

Load

⎦

⎯ ⎯ ⎯ ⎛ Y := A ⋅ 0.96 ⋅ 0.97FR τα ⋅ I ⋅

⎝

⎯ →

month ⎞ Load ⎠

1

X=

1

1

1.446

1

0.374

2

1.381

2

0.447

3

1.748

3

0.675

4

2.302

4

1.045

5

6.336

5

3.025

6

7.292

6

3.659

7

7.373

7

3.646

8

7.393

8

3.713

9

7.439

9

3.398

Y

=

10

6.66

10

2.809

11

2.433

11

0.765

12

1.754

12

0.45

Calculate f (the solar fraction per month) based on the f-chart correlation:

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ → 2 2 3 f := ( 1.029 ⋅ Y − 0.065 ⋅ X − 0.245 ⋅ Y + 0.0018 ⋅ X + 0.0215 ⋅ Y ) 1

f := i

<1 i

f if f i

1 otherwise

If solar will more than satisfy the load, set f = 1. f =

Determine the monthly load supplied by solar system.

⎯⎯ →

Load solar := ( f ⋅ Load )

1

0.261

2

0.327

3

0.482

4

0.692

5

1

6

1

7

1

8

1

9

1

10

1

11

0.506

12

0.307

Estimate the yearly solar fraction.

1

Load solar =

1

3.399·104

2

3.921·104

3

4.816·104

4

4.847·104

5

2.5·104

6

2·104

7

2·104

8

2·104

9

2·104

10

2.5·104

11

3.543·104

12

3.224·104

12

∑ percent :=

i= 1

(Load

solar i

⋅ 100)

12

∑

Load

i

i= 1

⋅ MJ percent = 50.69

Part (a): The number of collectors required is 62. The resulting solar fraction is 50.7 percent. 12

TotalSolar :=

∑

Load solar

i

i= 1

TotalSolar

Savings :=

6

10

⋅9

TotalSolar

Savings

=

=

5

1.021 × 10 3

3.135 × 10

⋅ BTU

Part (b): The total saving for the system is $3135 per year.

⋅ kWh

ch07.pdf

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