Thermodynamics HW Solutions 928

# Thermodynamics HW Solutions 928 - 685 K 273 30 K 400 K W/m...

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Chapter 12 Radiation Heat Transfer 12-37 Two concentric spheres are maintained at uniform temperatures. The net rate of radiation heat transfer between the two spheres and the convection heat transfer coefficient at the outer surface are to be determined. Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. Properties The emissivities of surfaces are given to be ε 1 = 0.1 and ε 2 = 0.8. Analysis The net rate of radiation heat transfer between the two spheres is () () W 1669 = + × = + = 2 4 4 4 2 8 2 2 2 2 1 2 2 1 4 2 4 1 1 12 m 4 . 0 m 15 . 0 7 . 0 7 . 0 1 5 . 0 1 ] K 400 K 700 )[ K W/m 10 67 . 5 ]( m) 3 . 0 ( [ 1 1 ) ( π ε σ r r T T A Q D 2 = 0.8 m T 2 = 400 K ε 2 = 0.7 D 1 = 0.3 m T 1 = 700 K ε 1 = 0.5 T surr = 30 ° C T = 30 ° C ε = 0.35 Radiation heat transfer rate from the outer sphere to the surrounding surfaces are W
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Unformatted text preview: 685 ] ) K 273 30 ( ) K 400 )[( K W/m 10 67 . 5 ]( m) 8 . ( )[ 1 )( 35 . ( ) ( 4 4 4 2 8 2 4 4 2 2 = + − ⋅ × π = − σ ε = − surr rad T T FA Q & The convection heat transfer rate at the outer surface of the cylinder is determined from requirement that heat transferred from the inner sphere to the outer sphere must be equal to the heat transfer from the outer surface of the outer sphere to the environment by convection and radiation. That is, W 9845 685 1669 12 = − = − = rad conv Q Q Q & & & Then the convection heat transfer coefficient becomes ( ) [ ] C W/m 5.04 2 ° ⋅ = ⎯→ ⎯ π = − = ∞ h h T T hA Q conv K) 303-K (400 m) 8 . ( W 984 2 2 2 . & 12-25...
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## This note was uploaded on 01/24/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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