Heat Transfer Solved Sample Problems

Sample Problem – Heat Transfer by Conduction across a composite wall

Problem Statement

Determine the overall heat transfer by conduction per unit area occurring across a furnace wall made of fire clay. Furnace wall has a thickness of 12" or a foot. The wall is insulated from outside. Thermal conductivity values for the wall and insulation materials are 0.1 W/m·K and 0.01 W/m·K, respectively. The furnace operates at  6500C. Average ambient temperature outside the furnace wall is 300C and allowable temperature on the outer side of insulation is 800C. If the air side heat transfer coefficient is 0.4 W/m2·K, calculate the minimum insulation thickness requirement.


The sample problem can be solved by following the steps given here. First the maximum possible heat transfer rate from furnace wall to the atmosphere is calculated. Then based on this maximum possible rate, minimum requirement of insulation thickness can be estimated.


Refer to EnggCyclopedia's article about heat transfer coefficients, for relation between heat transfer rate and the individual heat transfer coefficients between wall and air.

Q/A = hA×(T2-TA) – Air side heat transfer rate in W/m2

Hence, Q/A  = 0.4×(80-30) = 20.0 W/m2

This is the maximum limit of heat transfer rate through the furnace wall and insulation.


The conductive heat transfer through a flat wall is described in EnggCyclopedia's article on conduction.

For a flat wall,

Q/A = k1×(T1-Ti)/L1 = k2×(Ti-T2)/L2


(T1-Ti) = (Q/A)×(L1/k1)... (1)

and (Ti-T2) = (Q/A)×(L2/k2)... (2)

(1) + (2) gives,

(T1-T2) = (Q/A)×(L1/k1 + L2/k2)... (3)

From this equation we can say that for composite walls with layers of different materials, the overall heat transfer rate can be represented as,

(T1-T2) ÷ (Q/A) = (L1/k1 + L2/k2) = heat transfer resistance

The inverse of heat transfer resistance represents conductive heat transfer coefficient, given by,

Conductive heat transfer coefficient =1 / (L1/k1 + L2/k2) = k1k2/(L1k2+L2k1)


Maximum allowable heat transfer rate represents minimum insulation thickness requirement. Hence, Q/A = 20.0W/m2

So, in equation (3), all the variables are known except for L2. Hence this equation can be solved to determine L2.

The solution to equation (3) is L2 = 0.25 m or 10 inches.

This is the minimum insulation thickness requirement for furnace wall.

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