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Question

Air at 20C20^\circ C with a convection heat transfer coefficient of 25W/m2K25 W/m^2 \cdot K blows over a horizontal steel hot plate (k=43W/mK).(k = 43 W/m \cdot K). The surface area of the plate is 0.38m20.38 m^2 with a thickness of 2 cm. The plate surface is maintained at a constant temperature of Ts=250CT_s = 250^\circ C and the plate loses 300 W from its surface by radiation. Calculate the inside plate temperature (Ti).(T_i).

Solution

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Applying the energy balance for the plate

Q˙cond=Q˙conv+Q˙radkATiTsL=hA(TsT)+Q˙rad430.38Ti2500.02=250.38(25020)+300\begin{align*} \dot{Q}_\text{cond}&=\dot{Q}_\text{conv}+\dot{Q}_\text{rad}\\ kA\dfrac{T_i-T_s}{L}&=hA\left(T_s-T_\infty \right)+\dot{Q}_\text{rad}\\ 43*0.38*\dfrac{T_i-250}{0.02}&=25*0.38*\left(250-20 \right)+300 \end{align*}

Hence, the inside plate temperature is

Ti=253.04 oC\color{#c34632}{T_i=253.04\ ^o\text{C}}

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