Heat Loss by Conduction

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^oC and that of the concrete is 25^oC, 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.

Heat loss Q = (T_a - T_b) / R

Where Ta = 70^oC; Tb = 25^oC; and

R = R_brick + R_concrete

R_brick = L_brick/(Ak_brick)

R_concrete = L_concrete/(Ak_conctrete)

A = 15 x 10 = 150 m²

R_brick = 0.25/(150 x 0.7) = 2.381 x 10^-3 oC/W

R_concrete = 0.05/(150 x 0.95) = 3.509 x 10^-4 oC/W

 R = 2.381 x 10^-3 + 3.509 x 10^-4 = 2.732 x 10^-3 oC/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^oC

Interface temperature T_i = 30.8^oC