The inner and outer surfaces of a 0.3 m thick 400?cm X 700?cm north wall of an e

Question

The inner and outer surfaces of a 0.3 m thick 400?cm X 700?cm north wall of an electrically heated home are maintained at temperatures of 40oC and 22oC, respectively. The rate of heat transfer through that wall is 1159 W. (a) Determine the thermal conductivity of that wall, which is made of brick, in W/m.K. If the temperatures of the surfaces are maintained constant for a period of 8 h, (b) determine the total heat loss through the wall for that period, and (c) the cost of that heat loss if the cost of electricity is 0.07/kWh.

Explanation / Answer

RATE OF HEAt transfer H = dQ/dt = KA(T2-T1)/L is the formula. where K is coffeicient of thermal conductivity

A = area

L = thickness

T2-T1 = change of temp

re arranging the terms K = Power * L/A (T2-T1)

K = 1159 * 0.3/4*7 * (40-22)

K = 0.689 W/mK

b. power = heat loss/ time

then heat loss dQ = power * time = 1159* 8*60*60= 3.337 *10^7 J

c. 0.07 1/KWhr = 0.07*1/1000 = 0.00007 1/Whr

cost = 0.00007 * 1159 = 0.08113/hr

cost = 0.08113* 8h = $0.649

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