Two very long concentric cylinders of diameters D1 = 0.42 m and D2 = 0.5 m are maintained at uniform temperatures of T1 = 950 K and T2 = 500 K and have emissivities ε1 = 1 and ε2 = 0.55, respectively. Determine the net rate of radiation heat transfer between the two cylinders per unit length of the cylinders. (Use σ = 5.67 × 10-8)

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Netta00 Netta00 · Answer 1 · 2019-11-13T09:24:13+01:00

Answer:

Q=33.34 KW/m

Explanation:

Given that

D₁=0.42 m

A₁= π D₁ L

For unit length

A₁= π D₁ = 0.42 π m²

D₂=0.5 m

A₂= 0.5 π m²

ε₁= 1 ,ε₂= 0.55

T₁=950 K ,T₂ = 500 K

[tex]Q=\dfrac{\sigma (T_1^4-T_2^4)}{\dfrac{1-\varepsilon _1}{A_1\epsilon _1}+\dfrac{1-\varepsilon _2}{A_2\epsilon _2}+\dfrac{1}{A_1F_{12}}}[/tex]

F₁₁+F₁₂= 1

F₁₁= 0

So, F₁₂= 1

[tex]Q=A_1\dfrac{\sigma (T_1^4-T_2^4)}{\dfrac{1-\varepsilon _1}{\epsilon _1}+A_1\dfrac{1-\varepsilon _2}{A_2\epsilon _2}+1}[/tex]

[tex]Q=0.42\pi \dfrac{5.67\times 10^{-8}(950^4-500^4)}{\dfrac{1-1}{1}+0.42\pi\times \dfrac{1-0.55}{0.5\pi\times 0.55}+1}[/tex]

Q=33.34 KW/m