Question

A reactor’s wall 320 mm thick, is made up of an inner layer of fire brick (k = 0.84 W/m°C) covered with a layer of insulation (k = 0.16 W/m°C). The reactor operates at a temperature of 1325°C and the ambient temperature is 25°C. 

(i) Determine the thickness of fire brick and insulation which gives minimum heat loss. 

(ii) Calculate the heat loss presuming that the insulating material has a maximum temperature of 1200°C. 

If the calculated heat loss is not acceptable, then state whether addition of another layer of insulation would provide a satisfactory solution.

Answer 1

Given : t₁ = 1325°C ; t₂ = 1200°C, t₃ = 25°C ; 

LA + LB = L = 320 mm or 0.32 m 

∴ LB = (0.32 – LA) ; ...(i) 

kA = 0.84 W/m°C ; 

kB = 0.16 W/m°C. 

(i) LA : ; LB : 

The heat flux, under steady state conditions, is constant throughout the wall and is same for each layer. Then for unit area of wall,

Considering first two quantities, we have

∴ Thickness of insulation L_B = 320 – 114.6 = 205.4 mm.

(ii) Heat loss per unit area, q :

Heat loss per unit area, q = \(\cfrac{t_1-t_2}{L_A/k_A}\) = \(\cfrac{1325-1200}{0.1146/0.84}\) = 916.23 W/m².