Question 7.7.1: Thermal Resistance of Building Wall Engineers must be able t...

Thermal Resistance of Building Wall

Engineers must be able to predict the rate of heat loss through a building wall to determine the heating system’s requirements. The wall cross section shown in Figure 7.7.5 consists of four layers: an inner layer of plaster/lathe 10 mm thick, a layer of fiberglass insulation 125 mm thick, a layer of wood 60 mm thick, and an outer layer of brick 50 mm thick. For the given materials, the resistances for a wall area of 1 m² are R_{1} = 0.036, R_{2} = 4.01, R_{3} = 0.408, and R_{4} = 0.038°  C/W.
Suppose that T_{i} = 20°C and T_{o} = −10°C. (a) Compute the total wall resistance for 1 m² of wall area, and compute the heat loss rate if the wall’s area is 3 m by 5 m. (b) Find the temperatures T_{1}, T_{2}, and T_{3}, assuming steady-state conditions.

7.7.5
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a. The series resistance law gives
R = R_{1} + R_{2} + R_{3} + R_{4} = 0.036 + 4.01 + 0.408 + 0.038 = 4.492 °C/W
which is the total resistance for 1 m² of wall area. The wall area is 3(5) = 15 m², and thus the total heat loss is
q_{h} = 15 \frac{1}{R} (T_{i}  −  T_{o}) = 15 \frac{1}{4.492} (20 + 10) = 100.2  W
This is the heat rate that must be supplied by the building’s heating system to maintain the inside temperature at 20°C, if the outside temperature is −10°C.
b. If we assume that the inner and outer temperatures T_{i} and T_{o} have remained constant for some time, then the heat flow rate through each layer is the same, q_{h} . Applying conservation of energy gives the following equations.
q_{h} = \frac{1}{R_{1}} (T_{i}  −  T_{1}) = \frac{1}{R_{2}} (T_{1}  −  T_{2}) = \frac{1}{R_{3}} (T_{2}  −  T_{3}) = \frac{1}{R_{4}} (T_{3}  −  T_{o})
The last three equations can be rearranged as follows:
(R_{1} + R_{2})T_{1}  −  R_{1} T_{2} = R_{2} T_{i}
R_{3} T_{1}  −  (R_{2} + R_{3})T_{2} + R_{2} T_{3} = 0
−R_{4} T_{2} + (R_{3} + R_{4})T_{3} = R_{3} T_{o}
For the given values of T_{i} and T_{o}, the solution to these equations is T_{1} = 19.7596, T_{2} = −7.0214, and T_{3} = −9.7462°C

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