Thermodynamics HW Solutions 511

# Thermodynamics HW Solutions 511 - Chapter 5 Numerical...

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Chapter 5 Numerical Methods in Heat Conduction t T T r r r C r T T r k T T T T h r r r i i i i i i Δ Δ + = Δ + + + + Δ + + 6 1 6 6 56 plate 6 5 56 plate 4 6 4 plate 6 6 56 ) 2 / ]( 2 / ) ( 2 [ ) ( ) 2 ( ]} ) 273 ( ) 273 [( ) ( ){ 2 / ]( 2 / ) [( 2 δ πρ δπ σε π where k C, kW/m 1504 ) ( C, kW/m 3439 ) ( 3 steak 3 plate ° = ° = p p C C ρ steak = 1.40 W/m. ° C, ε steak = 0.95, , h α steak 2 m/ s 093 10 6 . if = 187 kJ/kg, k plate = 401 W/m. ° C, , and ε plate 2 s 117 10 6 plate = 0.90, T = 20°C, h =12 W/m 2 . ° C, δ = 0.01 m, Δ x = 0.005 m, Δ r = 0.0375 m, and Δ t = 5 s. Also, the mesh Fourier number for the steaks is τ steak 2 m /s)(5 s) m == × = Δ Δ t x 2 6 2 0 005 0186 (. ) . The various radii are r 4 =0.075 m, r 5 =0.1125 m, r 6 =0.15 m, r 45 = (0.075+0.1125)/2 m, and r 56 = (0.1125+0.15)/2 m. The total amount of heat transfer needed to defrost the steaks is mV steak 32 kg /m )[ (0.075 m) m kg = ρπ (( . ) 970 0 015 0 257 ] . QQ Q m C T m h total, steak sensible latent steak if steak kg)(1.55 kJ /kg. C)[0 -(-18 C)]+ kg)(187 kJ /kg) = 55.2 kJ =+ = + ° ()( ) Δ 0 257 0 257 The amount of heat transfer to the steak during a time step i is the sum of the heat transferred to the steak directly from its top surface, and indirectly through the plate, and is expressed as
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