Explanation Given Information: Steel wall thickness L s = 0.635 cm . The thermal conductivity of steel wall, k s = 15.1 W-m − 1 -K − 1 . The thermal conductivity of brick wall, k b = 0.72 W-m − 1 -K − 1 . The inner temperature T i = 300 ° C . The outer temperature is T o ≤ 40 ° C . The heat transfer rate is the rate by which an amount of heat flows from one place to another place. The rate of heat transfer is calculated with the following formula. Q = − k A T o − T i L Here, Q is the rate of heat transfer, k is thermal conductivity A is the area, L is the length of the wall. T o is the outer surface temperature, and T i is the outer surface temperature. Calculation: At steady-state, a temperature reduction of 0.7°C takes place over the layer of steel. T i − T m = 0.7 o C Here, T m is the temperature at the intermediate layer. The inner temperature of the system T i is 300 ° C . 300 o C − T m = 0.7 o C T m = 299.3 o C Let the thickness of the wall which is made of brick is L b . The heat transfer from both the layer (the layer of steel Q s and the layer of brick Q b ) is equal. Q s = Q b With the help of a formula, write the equation as follows for the same cross-sectional area. − k s T j − T m L s = − k b T o − T m L b Put the values in the above expression. ( − 15

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Chapter 2, Problem 2.36P

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Calculate the thickness of the brick wall and the rate of conduction of the wall surface.

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