electrically heated plate

logistic_guy

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The diameter and surface emissivity of an electrically heated plate are \(\displaystyle D = 300 \ \text{mm}\) and \(\displaystyle \varepsilon = 0.80\), respectively.

\(\displaystyle \bold{(a)}\) Estimate the power needed to maintain a surface temperature of \(\displaystyle 200^{\circ}\text{C}\) in a room for which the air and the walls are at \(\displaystyle 25^{\circ}\text{C}\). The coefficient characterizing heat transfer by natural convection depends on the surface temperature and, in units of \(\displaystyle \text{W}\text{/m}^{2} \cdot \text{K}\), may be approximated by an expression of the form \(\displaystyle h = 0.80(T_s - T_{\infty})^{1/3}\).

\(\displaystyle \bold{(b)}\) Assess the effect of surface temperature on the power requirement, as well as on the relative contributions of convection and radiation to heat transfer from the surface.
 
The diameter and surface emissivity of an electrically heated plate are \(\displaystyle D = 300 \ \text{mm}\) and \(\displaystyle \varepsilon = 0.80\), respectively.

\(\displaystyle \bold{(a)}\) Estimate the power needed to maintain a surface temperature of \(\displaystyle 200^{\circ}\text{C}\) in a room for which the air and the walls are at \(\displaystyle 25^{\circ}\text{C}\). The coefficient characterizing heat transfer by natural convection depends on the surface temperature and, in units of \(\displaystyle \text{W}\text{/m}^{2} \cdot \text{K}\), may be approximated by an expression of the form \(\displaystyle h = 0.80(T_s - T_{\infty})^{1/3}\).

\(\displaystyle \bold{(b)}\) Assess the effect of surface temperature on the power requirement, as well as on the relative contributions of convection and radiation to heat transfer from the surface.

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\(\displaystyle \bold{(a)}\)

Power \(\displaystyle =\) convection \(\displaystyle +\) radiation

\(\displaystyle P = q_{\text{conv}} + q_{\text{rad}} = Ah(T_s - T_{\infty}) + A\varepsilon\sigma(T^4_s - T^4_{\text{sur}})\)

The formula looks complicated but once you know what each variable represents, it becomes a piece of cake. Don't forget to convert the temperatures to \(\displaystyle \text{Kelvin}\) and use \(\displaystyle \text{SI}\) units. Also no need to mention that \(\displaystyle \text{Stefan-Boltzmann Constant} \ (\sigma)\) will be used.

\(\displaystyle P = \pi\left(\frac{0.3}{2}\right)^2\bigg[0.8\left(473.15 - 298.15\right)^{4/3} + 0.8\left(5.67 \times 10^{-8}\right)\left(473.15^4 - 298.15^4\right)\bigg] = 190.71 \ \text{W}\)
 
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