Question

Consider a 30-cm-diameter hemispherical enclosure. The dome is maintained at 600 K, and heat is supplied from the dome at a rate of 65 W while the base surface with an emissivity of 0.55 is maintained at 400 K. Determine the emissivity of the dome.

Answer 1

Answer:

$\epsilon_2=0.098$

Explanation:

Diameter $d=30cm=0.3m$

Temperature $T=600k$

Rate of supply $r=65W$

Emissivity of base surface $\in_b =0.55$

Temperature at base $T_b=400k$

Generally the equation for Area of base surface is mathematically given by

 $A_b=\frac{\pi}{4}d^2$

 $A_b=\frac{\pi}{4}0.3^2$

 $A_b=0.0707m^2$

Generally the equation for Area of Hemispherical dome is mathematically given by

 $A_h=\frac{\pi}{2}d^2$

 $A_h=\frac{\pi}{2}0.3^2$

 $A_h=0.1414m^2$

Since base is a flat surface

 $F_{11}+F_{12}=1$

 $F_{11}=0$

Therefore

 $F_{12}=1$

 $A_b=0.0707m^2$

Generally the equation for Net rate of radiation heat transfer between two surfaces is mathematically given by

 $Q_{21}=-Q_{12}$

 $Q_{21}=\frac{\sigma(T_1^4-T_2^4)}{\frac{1-\epsilon}{A_b\epsilon_1} +\frac{1}{A_bF_{12}} +\frac{1-\epsilon_2}{A_h*\epsilon_2} }$

Where

 $\sigma=5.67*10^{-8}$

Therefore

  $65=\frac{(5.67*10^{-8}(400^4-600^4))}{\frac{1-0.55}{0.0707*0.55}+\frac{1}{0.0707}+\frac{1-\epsilon_2}{0.1414*\epsilon_2}}$

 $\epsilon_2=0.098$

 $\epsilon_2 \approx 0.1$

Therefore  the emissivity of the dome is

 $\epsilon_2=0.098$