Question #115217
A furnace wall consists of 250 mm firebrick,125 mm insulating brick, and 250 mm building block.The inside wall is at a temperature of 600°C and the atmospheric temperature is 20°C.The heat transfer coefficient of the outside wall is 10 W/m2k, and the thermal conductivities of the fire brick, insulating brick, building brick are 1.4, 0.2, 0.7 W/m k respectively. Neglecting radiation, calculate the rate of the Heat Loss per unit wall surface area, and the temperature of the outside wall suface of the furnace.
[ANS:460 W/m2;60°C]
1
Expert's answer
2020-05-13T09:45:56-0400

According to Fourier's law we can calculate the heat loss:


Q=kAΔTδ, q=QA=ΔTδ/k=TiTfRth= =TiTfδ1/k1+δ2/k2+δ3/k3+1/α=460 W/m2.Q=-kA\frac{\Delta T}{\delta},\\ \space\\ q=\frac{Q}{A}=-\frac{\Delta T}{\delta/k}=\frac{T_i-T_f}{R_{th}}=\\ \space\\=\frac{T_i-T_f}{\delta_1/k_1+\delta_2/k_2+\delta_3/k_3+1/\alpha}=460\text{ W/m}^2.

The temperature of the outside wall surface of the furnace can be found if we use the same equation without term 1/α1/\alpha:


Tout=Tiq(δ1/k1+δ2/k2+δ3/k3)=66.1C.T_{out}=T_i-q(\delta_1/k_1+\delta_2/k_2+\delta_3/k_3)=66.1^\circ\text{C}.

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