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Heat Loss by Conduction

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**A plane brick wall, 25 cm thick, is faced with 5 cm thick concrete layer. If the temperature of the exposed brick face is 70**^{o}C and that of the concrete is 25^{o}C, find out the heat lost per hour through a wall of 15 m x10 m. Also, determine the interface temperature. Thermal conductivity of the brick and concrete are 0.7 W/m.K and 0.95 W/m.K respectively.

Calculations:

Heat loss Q = (T_{a} - T_{b}) / R

Where Ta = 70^{o}C; Tb = 25^{o}C; and

R = R_{brick} + R_{concrete}

R_{brick} = L_{brick}/(Ak_{brick})

R_{concrete} = L_{concrete}/(Ak_{conctrete})

A = 15 x 10 = 150 m^{2}

R_{brick} = 0.25/(150 x 0.7) = 2.381 x 10^{-3} ^{o}C/W

R_{concrete} = 0.05/(150 x 0.95) = 3.509 x 10^{-4} ^{o}C/W

R = 2.381 x 10^{-3} + 3.509 x 10^{-4} = 2.732 x 10^{-3} ^{o}C/W

Q = (70 - 25) / (2.732 x 10^{-3}) = 16471.4 W = 16471.4 J/sec = 59.3 MJ/hr

Heat loss per hour = 59.3 MJ

At steady state, this is the amount of heat transferred through the brick wall or concrete per hour.

i.e.,

Q = A k_{brick}(T_{a} - T_{i})/L_{brick} = 16471.4 W

150 x 0.7 x (70 - T_{i}) / 0.25 = 16471.4

70 - T_{i} = 39.2

T_{i} = 70 - 39.2 = 30.8^{o}C

Interface temperature T_{i} = 30.8^{o}C

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Last Modified on: 01-May-2024

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