# hwk13_sol - F F11 = 0 therefore F12 =1 98953 1 010472 21 22...

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ME 410 Spring 2008 Homework 13 Due: April 22, 2008 Determine the temperature of a 200 W cylindrical heating element of diameter 2 cm by length 1 m suspended inside a parallelepiped oven of dimensions 1m by 1m by 1m. The oven walls are at 500 K and emissivity of 0.85. The heating element has emissivity of 0.92. Consider the heating element to be surface 1,and the surrounding parallelepiped to be surface 2 Set-up Radiosity equations 4 2 2 2 22 1 21 2 2 2 12 1 11 1 1 1 ] [ ] [ T J F J F J J F J F J G J q σ ε ρ + + = + = = Note: ρ 2 =1- ε 2 =.15 View factors need to be found, 2 surfaces, 4 view factors, 1 is unique. 22 21 12 11 F F F

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Unformatted text preview: F F11 = 0 therefore F12 =1. 98953 . 1 010472 . 21 22 12 2 1 21 = − = = = F F F A A F Setting up radiosity equations ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ′ ′ = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − − 4 2 2 4 2 2 1 2 1 22 2 21 2 12 11 / 200 1 1 T DL T q J J F F F F π Inverting and solving for J’s D L ME 410 Spring 2008 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 84 . 3524 94 . 6707 2 1 J J Solve for T 1 K J F J T 45 . 592 4 1 1 2 12 1 1 1 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = σ ε ρ...
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## This note was uploaded on 07/25/2008 for the course ME 410 taught by Professor Benard during the Spring '08 term at Michigan State University.

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hwk13_sol - F F11 = 0 therefore F12 =1 98953 1 010472 21 22...

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