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adell [148]
3 years ago
5

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.
Physics
1 answer:
SIZIF [17.4K]3 years ago
3 0

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

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