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# Answer to Question #134593 in Mechanical Engineering for kumanan premsing

Question #134593
A reactor wall, 320mm thick is made up of an inner layer of fire brick (k = 0.84 W /m 0C) covered with a layer of insulation (k = 0.16 W /m 0C ). The reactor operates at a temperature of 1325 0C and the ambient temperature is 25 0C, .
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 12000C.
1
2020-09-23T10:54:06-0400

Given information

Reactor's wall = 320 mm thick

Thermal conductivity", k_1 = 0.84 W\/m.c"

Ti=1325 o C

Ta=25 o C

T inter=1200 o C

The mode of heat transfer from the wall is conduction. The amount of heat loss takes place from reactor wall is equal to the amount of heat that passes through fire wall and insulation.

Therefore "q=(t_1-t_3)\/(R_1+R_2) =(t_1-t_2)\/(R_1)=(t_2-t_3)\/(R_2)"

"R_1=L_A\/k_1\n; R_2=L_B\/k_2"

"(1325-25)\/{(L_A\/0.84)+(L_B\/0.16)}=(1325-1200)\/(L_A\/0.84)"

"L_A +L_B=0.32"

"L_B=0.32-L_A"

"(1300)\/{(L_A\/0.84)+(0.32-L_A\/0.16)}=(125)\/(L_A\/0.84)"

"(1300)\/2-5.06L_A=(105)\/(L_A)"

"L_A=0.1147m=114.7mm"

"L_B=0.32-L_A =0.32-0.1147=0.2053 m=205.3mm"

"Q=q=(1325-25)\/(0.1147\/0.84)+(0.2053\/0.16)"

"Q=q=915.7 W\/m^2"

Hence , the thickness of fire brick and insulation which gives minimum heat loss are 0.1147 m and 0.2053 m and the heat loss is "915.7 W\/m^2"

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