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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)

1 Answer

1 vote

Answer:

Q=33.34 KW/m

Step-by-step 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


Q=(\sigma (T_1^4-T_2^4))/((1-\varepsilon _1)/(A_1\epsilon _1)+(1-\varepsilon _2)/(A_2\epsilon _2)+(1)/(A_1F_(12)))

F₁₁+F₁₂= 1

F₁₁= 0

So, F₁₂= 1


Q=A_1(\sigma (T_1^4-T_2^4))/((1-\varepsilon _1)/(\epsilon _1)+A_1(1-\varepsilon _2)/(A_2\epsilon _2)+1)


Q=0.42\pi (5.67* 10^(-8)(950^4-500^4))/((1-1)/(1)+0.42\pi* (1-0.55)/(0.5\pi* 0.55)+1)

Q=33.34 KW/m

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