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

`\epsilon_2=0.098`

Step-by-step 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=(\pi)/(4)d^2`

`A_b=(\pi)/(4)0.3^2`

`A_b=0.0707m^2`

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

`A_h=(\pi)/(2)d^2`

`A_h=(\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)=(\sigma(T_1^4-T_2^4))/((1-\epsilon)/(A_b\epsilon_1) +(1)/(A_bF_(12)) +(1-\epsilon_2)/(A_h*\epsilon_2) )`

Where

`\sigma=5.67*10^(-8)`

Therefore

`65=((5.67*10^(-8)(400^4-600^4)))/((1-0.55)/(0.0707*0.55)+(1)/(0.0707)+(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`