Exam pl e ll l

The building of Example 11.15 is fitted with a collector storage wall. All the problem data of Example 11.15 apply here with the following additional information about the collector storage wall: Density = 2200 kg/m3. Heat capacity = 910 J/kg-°C. Wall thickness, w = 0.40 m.

Loss coefficient from the wall to the ambient, Uo = 4.5 W/m2-°C. Overall heat transfer coefficient of the wall including glazing, Uw =

2.6 W/m2-°C. Thermal conductivity of the wall, k = 1.85 W/m-°C.

Estimate the solar fraction for December and the auxiliary energy required for the month.

Solution

Initially, we need to calculate the loads Lm and Lw. From Eq. (6.45),

As the room temperature is the same as the base temperature, (DD)h = (DD)R. From Eq. (6.46),

(note from Chapter 6, Section 6.2.1, h = 8.33 W/m2-°C). From Example 11.15, R = 1.637. From Eq. (2.107),

Ht = RH = 1.637 X 9.1 = 14.9 MJ/m2 From Eq. (6.51),

_ 14.9 X 106 X 0.76 + (2.97 X 18.3 + 4.5 X 11.1) X 24 X 3600

31.50C

Therefore, the total heat transferred into the room through the storage wall is given by Eq. (6.52):

Uo

(UA)

+ 1

2.97

UkAwNS $

1.637

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