A

c where M = actual mass of storage capacity (kg).

Although, in Eq. (11.48), / is included on both sides of equation, it is relatively easy to solve for / by trial and error. Since the $, /-charts are given for various storage capacities and the user has to interpolate, the use of Eq. (11.48) is preferred, so the actual charts are not included in this book. The /-charts are used in the same way as the /-charts. The values of $max Y, and X need to be calculated from the long-term radiation data for the particular location and load patterns. As before, /L is the average monthly contribution of the solar energy system, and the monthly values can be summed and divided by the total annual load to obtain the annual fraction, F.

Example 11.12

An industrial process heat system has a 50 m2 collector. The system is located at Nicosia, Cyprus (35°N latitude), and the collector characteristics are FrUl = 5.92 W/m2-°C, Fr(to)„ = 0.82, tilted at 40°, and double glazed. The process requires heat at a rate of 15 kW at a temperature of 70°C for 10 h each day. Estimate the monthly and annual solar fractions. Additional information is (to)„ = 0.96, storage volume = 5000 L. The weather conditions, as obtained from Appendix 7, are given in Table 11.11. The values of the last column are estimated from Eq. (2.82).

Table 11.11 Weather Conditions for Example 11.12

Month

H (MJ/m2)

Ta (°C)

Kt

H0 (MJ/m2)

January

8.96

12.1

0.49

18.29

February

12.38

11.9

0.53

23.36

March

17.39

13.8

0.58

29.98

April

21.53

17.5

0.59

36.49

May

26.06

21.5

0.65

40.09

June

29.20

25.8

0.70

41.71

July

28.55

29.2

0.70

40.79

August

25.49

29.4

0.68

37.49

September

21.17

26.8

0.66

32.08

October

15.34

22.7

0.60

25.57

November

10.33

17.7

0.53

19.49

December

7.92

13.7

0.47

16.85

As can be seen, the values of Ho are slightly different from those shown in Table 2.5 for 35°N latitude. This is because the actual latitude of Nicosia, Cyprus, is 35.15°N, as shown in Appendix 7.

Solution

To simplify the solution, most of the results are given directly in Table 11.12. These concern RB, given by Eq. (2.108); HD /H, given by Eqs. (2.105c) and (2.10d); R, given by Eq. (2.107); rn and rd,n, given by Eqs. (2.84) and (2.83), respectively, at noon (h = 0°); RBn, given by Eq. (2.90a) at noon; HD/H, given by Eqs. (11.32); and Rn, given by Eq. (11.31).

Subsequently, the data for January are presented. First,we need to estimate (Ta)/(Ta)n. For this estimation, we need to know S and then apply Eq. (11.10) to find the required parameter. From Eqs. (3.4a) and (3.4b),

594 Designing and Modeling Solar Energy Systems Table 11.12 Results of Radiation Coefficients for Example 11.12

Month

RB

HD /H

R

rn

rd,n

RB,n

Hd/H

Rn

Jan.

1.989

0.40

1.570

0.168

0.156

1.716

0.590

1.283

Feb.

1.624

0.36

1.381

0.156

0.144

1.429

0.505

1.225

March

1.282

0.36

1.162

0.144

0.133

1.258

0.469

1.119

April

1.000

0.35

0.982

0.133

0.123

1.074

0.450

1.018

May

0.827

0.29

0.867

0.126

0.116

0.953

0.336

0.955

June

0.757

0.25

0.812

0.122

0.112

0.929

0.235

0.921

July

0.787

0.25

0.834

0.124

0.114

0.924

0.235

0.939

Aug.

0.921

0.27

0.934

0.130

0.120

1.020

0.276

1.008

Sept.

1.160

0.29

1.103

0.140

0.129

1.180

0.316

1.117

Oct.

1.503

0.34

1.316

0.152

0.141

1.400

0.432

1.216

Nov.

1.885

0.36

1.548

0.164

0.153

1.648

0.505

1.311

Dec.

2.113

0.42

1.620

0.171

0.159

1.797

0.630

1.285

From Figure 3.24, for a double-glazed collector,

Therefore,

(Ta)D = (Ta)n X 0.87 = 0.96 X 0.87 = 0.835 (Ta)G = (ra}„ X 0.57 = 0.96 X 0.57 = 0.547

These values are constant for all months. For the beam radiation, we use Figures A3.8(a) and A.3.8(b) to find the equivalent angle for each month and Figure 3.24 to get (Ta)/(Ta)„. The 12 angles are 40, 42, 44, 47, 50, 51, 51, 49, 46, 43, 40, and 40, from which 12 values are read from Figure 3.24 and the corresponding values are given in Table 11.13. The calculations for January are as follows:

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Solar Panel Basics

Solar Panel Basics

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