Question

The outer surface of the grill hood shown in Fig. P2.52 is at 47℃ and the emissivity is 0.93. The heat transfer coefficient convection between the hood and the surroundings at 27℃ is 10 w/m^2·K. Determine the net rate of heat transfer between the grill hood and the surroundings by convection and radiation, in kW per m² of surface area.

Answer 1

Calculate the convection heat transfer.

$$ \begin{aligned} &\frac{Q_{c}}{A}=h\left(T_{t}-T_{0}\right) \\ &\frac{Q_{c}}{A}=10(47-27) \\ &\frac{Q_{c}}{A}=200 \mathrm{~W} / \mathrm{m}^{2} \end{aligned} $$

Calculate the radiation heat transfer.

$$ \begin{aligned} &\frac{Q_{r}}{A}=\in \sigma\left[T_{2}^{4}-T_{0}^{4}\right] \\ &\frac{Q_{r}}{A}=0.93\left[5.67 \times 10^{-8}\right]\left[(47+273)^{4}-(27+273)^{4}\right] \\ &\frac{Q_{r}}{A}=0.93\left[5.67 \times 10^{-8}\right]\left[(320)^{4}-(300)^{4}\right] \\ &\frac{Q_{r}}{A}=125.8 \mathrm{~W} / \mathrm{m}^{2} \end{aligned} $$

Calculate the net heat transfer due to convection and radiation.

$$ \begin{aligned} &\frac{Q}{A}=\left(\frac{Q_{c}}{A}\right)+\left(\frac{Q_{r}}{A}\right) \\ &\frac{Q}{A}=200+125.8 \\ &\frac{Q}{A}=325.8 \mathrm{~W} / \mathrm{m}^{2} \end{aligned} $$