Linear Algebra : Calculating heat loss through a wall
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Engineers use the concept of thermal resistance R to predict the rate of heat loss through a building wall in order to determine the heating system's requirements. This concept relates the heat flow rate q through a material to the temperature difference ∆T across the material: q = . This relation is like the voltage-current relation for an electrical resistor: i = . So the heat flow rate plays the role of electrical current, and the temperature difference plays the role of the voltage difference v. The SI unit for q is watt/meter2. A watt is 1 joule/second.
The wall shown in Figure 4(a) 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. Figure 4(b) shows the electrical equivalent circuit.
Figure 4 (a)
Figure 4 (b)
If we assume that the inner and outer temperatures Ti and T0 have remained constant for some time, then the heat energy stored in the layers is constant; thus the heat flow rate through each layer is the same. Applying conservation of energy gives the following equations:
The thermal resistance of a solid material is given by R = where D is the material's thickness and k is the material's thermal conductivity. For the given materials, the resistances for a wall area of 1 meter2 are R1 = 0.036, R2 = 4.01,R3 = 0.408, and R4 = 0.038 K/watt.
Suppose that Ti = 20 and To= Find the other three temperatures and the heat loss rate q in watts/meter2. Compute the total heat loss rate in watts if the wall's area is 10 meters2.
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