Question

Question 2 3 points Save Ans Two very long concentric cylinders shown in the figure are maintained at uniform temperatures. Determine the net rate of radiation heat transfer between the two cylinders per unit length (in Watt). Assume D4 0.3 m, T = 453 K and E1 = 05 D -0.5 m T: - 500K $=0.7

Answer 1

For radiation heat transfer between two very long concentric cylinders

(from heat transfer data book)

oA(T- T 1 82 A1 A2 F12 1 12 E2

o(TD,L) (T- T3 12 1 -e2D 1 E2

therefore

$q_{12}=\frac{5.67\times 10^{-8}(\pi\times 0.3\times L)(453^{4}-500^{4})}{\frac{1}{0.5}+\frac{1-0.7}{0.7}(\frac{0.3}{0.5})}$

$\frac{q_{12}}{L}=\frac{5.67\times 10^{-8}(\pi\times 0.3)(453^{4}-500^{4})}{\frac{1}{0.5}+\frac{1-0.7}{0.7}(\frac{0.3}{0.5})} = -482.72 W/m$

note: the negative sign shows the heat transfer from outer cylinder (2) to inner cylinder (1)

therefore

$\frac{q_{21}}{L}= 482.72 W/m$