## Load Heat Exchanger Size Correction

The size of the load heat exchanger strongly affects the performance of the solar energy system. This is because the rate of heat transfer across the load heat exchanger directly influences the temperature of the storage tank, which consequently affects the collector inlet temperature. As the heat exchanger used to heat the building air is reduced in size, the storage tank temperature must increase to supply the same amount of heat energy, resulting in higher collector inlet temperatures and therefore reduced collector performance. To account for the load heat exchanger size, a new dimensionless parameter is specified, Z, given by (Beckman et al., 1977):

(UA)l where eL = effectiveness of the load heat exchanger.

(m cp )min = minimum mass flow rate-specific heat product of heat exchanger (W/K).

(UA)l = building loss coefficient and area product used in degree-day space heating load model (W/K).

In Eq. (11.15), the minimum capacitance rate is that of the air side of the heat exchanger. System performance is asymptotically dependent on the value of Z; and for Z > 10, the performance is essentially the same as for an infinitely large value of Z. Actually, the reduction in performance due to a small-size load heat exchanger is significant for values of Z lower than 1. Practical values of Z are between 1 and 3, whereas a value of Z = 2 was used for the development of the f-chart of Figure 11.2. The performance of systems having other values of Z can be estimated by multiplying the dimensionless parameter Y by the following correction factor:

0.139

Equation (11.16) is applied in the range of 0.5 < Z < 50. The load heat exchanger size correction factor can also be determined from Figure 11.4 directly, obtained by plotting Eq. (11.16) for this range. Dimensioness parameter Z FIGURE 11.4 Load heat exchanger size correction factor.

Example 11.4

If the liquid flow rate in Example 11.2 is 0.525 L/s, the air flow rate is 470 L/s, the load heat exchanger effectiveness is 0.65, and the building overall loss coefficient-area product, (UA)l, is 422 W/K, find the effect on the solar fraction for the month of November.

### Solution

The minimum value of capacitance needs to be found first. Therefore, if we assume that the operating temperature is 350 K (77°C) and the properties of air and water at that temperature are as from Tables A5.1 and A5.2 in Appendix 5, respectively,

Cair = 470 X 0.998 X 1009/1000 = 473.3W/K C_t„„ = 0.525 X 974 X 4190/1000 = 2142.6W/K

Therefore, the minimum capacitance is for the air side of the load heat exchanger. From Eq. (11.15), eL (m Cp)

0.729

The correction factor is given by Eq. (11.16):

0.139

0.139

0.729

0.93

From Example 11.2, the value of the dimensionless parameter Y is 0.50. Therefore, Yc = 0.50 X 0.93 = 0.47. The value of the dimensionless parameter X for this month is 1.76, which from Eq. (11.13) gives a solar fraction f = 0.323, a drop of about 2% from the previous value.

Although in the examples in this section only one parameter was different from the standard one, if both the storage size and the size of the heat exchanger are different than the standard ones, both Xc and Yc need to be calculated for the determination of the solar fraction. Additionally, most of the required parameters in this section are given as input data. In the following example, most of the required parameters are estimated from information given in earlier chapters.

Example 11.5

A liquid-based solar space and domestic water heating system is located in

Nicosia, Cyprus (35°N latitude). Estimate the monthly and annual solar fraction of the system, which has a total collector area of 20 m2 and the following information is given:

1. The collectors face south, installed with 45° inclination. The performance parameters of the collectors are FR(ja)n = 0.82 and FRUL = 5.65 W/m2-°C, both determined from the standard collector tests.

2. The flow rate of both the water and antifreeze solution through the collector heat exchanger is 0.02 L/s-m2 and the factor FR /FR = 0.98.

3. The storage tank capacity is equal to 120 L/m2.

4. The (ia)/(Ta)n = 0.96 for October through March and 0.93 for April through September.

5. The building UA value is equal to 450 W/K. The water to air load heat exchanger has an effectiveness of 0.75 and air flow rate is 520 L/s.

6. The ground reflectivity is 0.2.

7. The climatic data and the heating degree days for Nicosia, Cyprus, are taken from Appendix 7 and reproduced in Table 11.4 with the hot water load.

Table 11.4 Climate Data and Heating Degree Days for Example 11.5

Month

H (MJ/m2)

Ta (°C)

Clearness

Heating °C

Hot water

index kT

degree days

January

8.96

12.1

0.49

175

3.5

February

12.38

11.9

0.53

171

3.1

March

17.39

13.8

0.58

131

2.8

April

21.53

17.5

0.59

42

2.5

May

26.G6

21.5

0.65

3

2.1

June

29.2G

29.8

0.70

0

1.9

July

28.55

29.2

0.70

0

1.8

August

25.49

29.4

0.68

0

1.9

September

21.17

26.8

0.66

0

2.G

October

15.34

22.7

0.60

1

2.7

November

1G.33

17.7

0.53

36

3.G

December

7.92

13.7

0.47

128

The loads need to be estimated first. For the month of January, from Eq. (6.24):

= 450(W/K) X 24(h/d) X 3600(J/Wh) X 175(°C-days) = 6.80 GJ

The monthly heating load (including hot water load) = 6.80 + 3.5 = 10.30GJ. The results for all the months are shown in Table 11.5.

 degree days Dh (GJ) Dw (GJ) L (GJ) January 175 6.80 3.5 10.30 February 171 6.65 3.1 9.75 March 131 5.09 2.8 7.89 April 42 1.63 2.5 4.13 May 3 0.12 2.1 2.22 June 0 0 1.9 1.90 July 0 0 1.8 1.80 August 0 0 1.9 1.90 September 0 0 2.0 2.00 October 1 0.04 2.7 2.74 November 36 1.40 3.0 4.40 December 128 4.98 3.3 8.28 Total = 57.31

Next, we need to estimate the monthly average daily total radiation on the tilted collector surface from the daily total horizontal solar radiation, H. For this estimation, the average day of each month is used, shown in Table 2.1, together with the declination for each day. For each of those days, the sunset hour angle, hss, is required, given by Eq. (2.15), and the sunset hour angle on the tilted surface, h'ss, given by Eq. (2.109). The calculations for the month of January are as follows.

From Eq. (2.15), hss = cos_1[-tan(L)tan(6)] = cos_1[-tan(35) tan(-20.92)] = 74.5°

From Eq. (2.109), h'ss = min{hss, cos-1[-tan(L - |3)tan(6)]}

= min{74.5°,cos-1[-tan(35 - 45)tan(-20.92)]} = min{74.5°,93.9°} = 74.5°

- [0.505 + 0.00455(hss - 90)]cos(115£r - 103) 0.775 + 0.00653(74.5 - 90)

- [0.505 + 0.00455(74.5 - 90)]cos(115 X 0.49 - 103) 0.38

- _ cos(L - P)cos(8)sin(ftSs) + (n/180)hs'ssin(L - (3)sin(8)

_ cos(35 - 45)cos(— 20.92)sin(74.5) + (n/180)74.5sin(35 - 45)sin(-20.92) cos(35)cos(—20.92) sin(74.5) + (n/180)74.5sin(35)sin(-20.92)

 1 + cos(45) + 0 2 1 - - cos(45) 2 2

And finally, Ht = RH = 1.62 X 8.96 = 14.52 MJ/m2. The calculations for all months are shown in Table 11.6.

 Month N 6 (o) hss (o) Ks (o) Hd /H Rb R Ht (MJ/m2) Jan. 17 -20.92 74.5 74.5 0.38 2.05 1.62 14.52 Feb. 47 -12.95 80.7 80.7 0.37 1.65 1.38 17.08 March 75 -2.42 88.3 88.3 0.36 1.27 1.15 20.00 April 105 9.41 96.7 88.3 0.38 0.97 0.96 20.67 May 135 18.79 103.8 86.6 0.36 0.78 0.84 21.89 June 162 23.09 107.4 85.7 0.35 0.70 0.78 22.78 July 198 21.18 105.7 86.1 0.34 0.74 0.81 23.13 Aug. 228 13.45 99.6 87.6 0.34 0.88 0.90 22.94 Sept. 258 2.22 91.6 89.6 0.33 1.14 1.07 22.65 Oct. 288 -9.6 83.2 83.2 0.34 1.52 1.32 20.25 Nov. 318 -18.91 76.1 76.1 0.36 1.94 1.58 16.32 Dec. 344 -23.05 72.7 72.7 0.38 2.19 1.71 13.54

We can now move along in the f-chart estimation. The dimensionless parameters X and Y are estimated from Eqs. (11.8) and (11.9):

10.30 X 109

0.67

10.30 X 109

The storage tank correction is obtained from Eq. (11.14):

0.25

0.89

Then, the minimum capacitance value needs to be found (at an assumed temperature of 77°C):

Cair = 520 X 0.998 X 1009/1000 = 523.6W/K Cwater = (0.02 X 20) X 974 X 4190/1000 = 1632W/K

Therefore, the minimum capacitance is for the air side of the load heat exchanger.

_ Zl (mCp) min _ 0.75 X (523.6) _ . 0_ Z — - — - — 0.8 7

The correction factor is given by Eq. (11.16): -0.139

0.139

0.87

0.94

Therefore, and

When these values are used in Eq. (11.13), they give f = 0.419. The complete calculations for all months of the year are shown in Table 11.7.

 Month F Fc f fL January 2.53 0.67 2.25 0.63 0.419 4.32 February 2.42 0.76 2.15 0.71 0.483 4.71 March 3.24 1.21 2.88 1.14 0.714 5.63 April 5.73 2.24 5.10 2.11 1 4.13 May 10.49 4.57 9.34 4.30 1 2.22 June 11.21 5.38 9.98 5.06 1 1.90 July 11.67 5.95 10.39 5.59 1 1.80 August 11.02 5.59 9.81 5.25 1 1.90 September 10.51 5.08 9.35 4.78 1 2.00 October 8.37 3.53 7.45 3.32 1 2.74 November 5.37 1.72 4.78 1.62 0.846 3.72 December 3.09 0.78 2.75 0.73 0.464 3.84 Total = 38.91

From Eq. (11.12), the annual fraction of load covered by the solar energy system is:

11.1.2 Performance and Design of Air-Based Solar Heating Systems

Klein et al. (1977) developed for air-based systems a design procedure similar to that for liquid-based systems. The f-chart for air-based systems is developed for the standard solar air-based solar energy system, shown in Figure 11.5. This is the same as the system shown in Figure 6.12, drawn without the controls, for clarity. As can be seen, the standard configuration of air-based solar heating system uses a pebble bed storage unit. The energy required for the DHW is provided through the air-to-water heat exchanger, as shown. During summertime, when heating is not required, it is preferable not to store heat in the pebble bed, so a bypass is usually used, as shown in Figure 11.5 (not shown in Figure 6.12), which allows the use of the collectors for water heating only.

The fraction f of the monthly total load supplied by a standard solar air-based solar energy system, shown in Figure 11.5, is also given as a function of the two parameters, X and Y, which can be obtained from the f-chart given in Figure 11.6 or from the following equation (Klein et al., 1977):

f = 1.040F - 0.065X - 0.159F2 + 0.00187X2 - 0.0095F3 (11.17)

Air-to-water heat exchanger

Collector array

Collector array rAAAn

Pre-heat tank

Hot water supply

DHW Auxiliary

Cold water supply

Building

Hot water supply

DHW Auxiliary Warm ► air supply

Building

FIGURE 11.5 Schematic diagram of the standard air-based solar heating system.

Air systems

Range: 0 < Y< 3; 0 < X < 18 Mh s = 0.25 m3/m2

Air systems

Range: 0 < Y< 3; 0 < X < 18 Mh s = 0.25 m3/m2 6 8 10 12 X = Reference collector loss/Heating load

FIGURE 11.6 The /-chart for air-based solar heating systems.

6 8 10 12 X = Reference collector loss/Heating load

FIGURE 11.6 The /-chart for air-based solar heating systems.

Example 11.6

A solar air heating system of the standard configuration is installed in the same area as the one of Example 11.2 and the building has the same load. The air collectors have the same area as in Example 11.2 and are double glazed, with FrUl = 2.92 W/m2-°C, FR(Ta)n = 0.52, and (ra)/(Ta)n = 0.93. Estimate the annual solar fraction.

### Solution

A general condition of air systems is that no correction factor is required for the collector heat exchanger and ducts are well insulated; therefore, heat losses are assumed to be small, so FR /FR = 1. For the month of January and from Eqs. (11.8) and (11.9).

0.19

### 35.2 X 109

From Eq. (11.17) or Figure 11.6, f = 0.147. The solar contribution is fL = 0.147 X 35.2 = 5.17 GJ. The same calculations are repeated for the other months and tabulated in Table 11.8.

 Month H (MJ/m2) Ta (°C) L (GJ) X Y f fL January 12.5 10.1 35.2 0.70 0.19 0.147 5.17 February 15.6 13.5 31.1 0.69 0.24 0.197 6.13 March 17.8 15.8 20.7 1.11 0.45 0.367 7.60 April 20.2 19.0 13.2 1.63 0.78 0.618 8.16 May 21.5 21.5 5.6 3.84 2.01 l 5.60 June 22.5 29.8 4.1 4.54 2.79 l 4.10 July 23.1 32.1 2.9 6.41 4.18 l 2.90 August 22.4 30.5 3.5 5.44 3.36 l 3.50 September 21.1 22.5 5.1 4.03 2.10 l 5.10 October 18.2 19.2 12.7 1.74 0.75 0.587 7.45 November 15.2 16.2 23.6 0.94 0.33 0.267 6.30 December 13.1 11.1 33.1 0.74 0.21 0.164 5.43 Total load = 190.8 Total contribution = 67.44

It should be noted here that, again, the values of f marked in bold are outside the range of the f-chart correlation and a fraction of 100% is used because during these months, the solar energy system covers the load fully. From Eq. (11.12), the annual fraction of load covered by the solar energy system is

Therefore, compared to the results from Example 11.2, it can be concluded that, due to the lower collector optical characteristics, F is lower.

Air systems require two correction factors, one for the pebble bed storage size and one for the air flow rate, which affects stratification in the pebble bed. There are no load heat exchangers in air systems, and care must be taken to use the collector performance parameters FRUL and Fr(toi)„, determined at the same air flow rate as used in the installation; otherwise, the correction outlined in Chapter 4, Section 4.1.1, needs to be used.