electrically heated plate

[logistic_guy]

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.

 

[khansaheb]

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.

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[logistic_guy]

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

Be careful khan. This thread contains radiation!

 

[khansaheb]

Be careful khan. This thread contains radiation!

Nothing new ... Radiation is everywhere....

 

[logistic_guy]

Nothing new ... Radiation is everywhere....

Wrong. There is no radiation in one's butthole when washed with mentholated shampoo like Head & Shoulders and water!

 

[logistic_guy]

$\displaystyle \bold{(a)}$

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.

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