## Example

A residential building located in 35°N latitude is well insulated and has a direct gain passive solar energy system. Estimate the fraction of heating load supplied by the solar energy system for December and the auxiliary energy required, given the following information:

Window area = 10 m2.

Effective heat capacity of the building, Cb = 60 MJ/°C.

Allowable temperature swing = 7°C.

Low set point temperature = 18.3°C.

U value of window without night insulation = 5.23 W/m2-°C.

Building UA, excluding direct gain window = 145 W/°C.

Degree days for December, estimated at a base temperature of 18.3°C = 928 °C-days

Mean ambient temperature, Ta = 11.1°C. Monthly average daily total radiation H = 9.1 MJ/m2. Monthly average (Ta) = 0.76.

Solution

The thermal load needs to be calculated first. The UA for the building, including the direct gain window, is

Because the latitude of the building under consideration is the same as the one used in Example_11.12, rn = 0.171, rdn = 0.159. _

From Table 2.5, Ho = 16.8 MJ/m2, which from Eq. (2.82) gives KT = 0.54. For December^ hss = 72.7°; from Eq. (11.32a),_ HD/H = 0.483; from Eq. (2.105c), Hd /H = 0.35; and from Eq. (2.108), RB = 2.095. Using (3 = 90° (vertical surface) and assuming a ground reflectance of 0.2, RBn = 1.603, Rn = 1.208, R = 1.637. From Eq. (11.66),

_ NSAr _ NHR(Ta)Ar _ 31 X 9.1 X 106 X 1.637 X 0.76 X 10

L L 15.82 X109

! = (UA)h_(Tb - Ta) = 197.3(18.3 - 11.1) = 186 9 W/m2 tc (Ta)Ar 0.76 X10

The next parameter that we need to calculate is !>. From Eq. (11.33), _ Itc _ 186.9 X 3600

A = 2.943 - 9.271Kt + 4.031K2 = 2.943 - 9.271 X 0.54 + 4.031(0.54)2 = -0.888

B = -4.345 + 8.853Kt - 3.602K2 = -4.345 + 8.853 X 0.54 - 3.602(0.54)2 0.615

1.208

1.637

0.566

CbATb

CbATb

Ht (ta)Ar\$ HR(t«)Ar\$ 9.1 X 1.637 X 0.76 X 10 X 0.566 From Eq. (11.71b),

P = [1 - exp(-0.294F)]0 652 = [1 - exp(-0.294 X 6.55)]0 652 = 0.902

From Eq. (11.71a), f = min{PX + (1 - P)(3.082 - 3.142\$)[1 - exp(-0.329X)], 1} = min{0.902 X 0.222 + (1 - 0.902)(3.082 - 3.142 X 0.566)

6.55

11.4.2 Collector Storage Walls

The thermal analysis of collector storage walls is presented in Section 6.2.1, Chapter 6, where a diagram of the wall and the thermal gains and losses are given. The unutilizability concept, developed by Monsen et al. (1982), can also be applied in this case to determine the auxiliary energy required to cover the energy supplied by the solar energy system. Again here, two limiting cases are investigated: zero and infinite capacitance buildings. For the infinite thermal capacitance case, all net monthly heat gain from the storage wall, Qg, given by

Eq. (6.52), can be used. The monthly energy balance of the infinite capacitance building is given by

where Lm = monthly energy loss from the building (kJ), given by Eq. (6.45).

For the zero thermal capacitance case, which applies to both the storage wall and the building structure, maximum auxiliary energy is required. The collector storage wall in this case acts as a radiation shield that alters the amplitude but not the time of the solar gains to the building. The monthly energy balance of the zero capacitance building is given by

The dumped energy, QD, can be determined by integrating QD, the rate at which excess energy must be removed to prevent the room temperature from reaching a value above the high thermostat set temperature. The rate of dumped energy, QdD , is the difference between the rate of heat transfer through the collector storage wall into the building and the rate of heat loss from the building structure, given by

where Uk = overall heat transfer coefficient of the thermal storage wall including glazing, given by Eq. (6.50) (W/m2-°C).

For the case of the zero thermal capacitance collector storage wall, an energy balance gives

where U0 = overall heat transfer coefficient from the outer wall surface through the glazing to the ambient, without night insulation (W/m2-°C). Solving Eq. (11.76) for Tw gives

Substituting Tw from Eq. (11.77) to Eq. (11.75) gives

UkAw

This equation can be integrated over a month to give QD by assuming that (Ta) and Ta are constant and equal to their mean monthly values (Ta) and Ta:

 Uo (UA) + 1 lUk

where Itc is the critical radiation level, which makes QD equal to 0, given by

It should be noted that the summation in Eq. (11.79) is the same as the summation in daily utilizability !>, given by Eq. (11.30); therefore, Eq. (11.79) becomes

The solar fractions corresponding to the limits of performance of the collector storage wall systems, given by Eqs. (11.73) and (11.74), are

z Lm + Lw J' Uo + Uk where X is the solar-load ratio given by

Two parameters are then required: the storage capacity of the building, Sb, and of the storage wall, Sw. The building storage capacity for a month is given by (Monsen et al., 1982):

where

Cb = effective building storage capacitance (J/°C).

ATb = allowable temperature swing, the difference between the high and low thermostat settings (°C).

The storage capacity of the wall for the month is given by (Monsen et al., 1982): Sw = cpA p(ATw )N (11.86)

where cp = heat capacity of the wall (J/kg-°C). p = density of the wall (kg/m3). 6 = thickness of the wall (m).

ATw = one half of the difference of the monthly average temperatures of the outside and the inside surfaces of the wall (°C).

The heat transfer through the wall into the heated space, Qg, given in terms of at„„ is

Solving Eq. (11.87) in terms of ATW and substituting into Eq. (11.86), pc„62Qg

w 2kAt

A dimensionless parameter called the storage-dump ratio needs to be specified. It is defined as the ratio of a weighted storage capacity of the building and wall to the energy that would be dumped by a building having zero capacitance, given by (Monsen et al., 1982):

The solar fraction here is given by f = 1 - rQ^usr (11.90)

A correlation of the solar fraction, f, was developed from simulations as a function of fi and Y (Monsen et al., 1982):

f = min{Pf. + 0.88(1 - P)[1 - exp(-1.26f,)],1} (11.91a)

where

It should be noted that here the X dimensionless parameter is not used in the correlation for the solar fraction. The auxiliary energy required for a month is given by

The monthly auxiliary energy requirements are then added to obtain the annual auxiliary energy needs of the building.

The steps to follow to estimate the annual performance of the collector storage wall are:

1. Estimate the absorbed solar radiation for each month.

2. Estimate the loads Lm and Lw from Eqs. (6.45) and (6.46), respectively, by taking into account internal heat generation, if it exists.

3. Estimate the heat gain across the collector storage wall, Qg, using Eq. (6.52).

4. Estimate daily utilizability and the energy dump that would occur in a zero capacitance system, QD, from Eq. (11.81).

5. Estimatef, Sb, and Sw from Eqs. (11.82), (11.85), and (11.88), respectively.

7. Finally, estimate the monthly fraction, f, and auxiliary energy, Qaux.