Problem 2 ots 1-D Heat Transfer with Convection. The solid material is subjected to a temperature differential that can be described using the mixed boundary conditions. Use four identical elements to determine the steady-state temperature distribution through the solid. Both ends of the surface are subjected to heat loss by convection (Nodes 1 and 5). Assume the cross-sectional area is the same for all elements, A=0.01 m?. Insulated, 1-d heat transfer in x-dir Gas 1: h1=2000 W/(m²K) Gas 2: hs=5000w/(m²K) T1°=110°C=383.15 K Solid surface: T5°=81°C=354.15 K +x k=150 W/(m K) 0.1 m

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Chapter2: Steady Heat Conduction
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Problem 2 pts 1-D Heat Transfer with Convection. The solid material is
subjected to a temperature differential that can be described using the mixed
boundary conditions. Use four identical elements to determine the steady-state
temperature distribution through the solid. Both ends of the surface are
subjected to heat loss by convection (Nodes 1 and 5). Assume the cross-sectional
area is the same for all elements, A=0.01 m?.
Insulated, 1-d heat transfer in x-dir
Gas 1: h1=2000 W/(m?K)
Gas 2: h5=5000 W/(m²K)
T1°=110°C=383.15 K
Solid surface:
T50=81°C=354.15 K
+x
k=150 W/(m K)
0.1 m
Transcribed Image Text:Problem 2 pts 1-D Heat Transfer with Convection. The solid material is subjected to a temperature differential that can be described using the mixed boundary conditions. Use four identical elements to determine the steady-state temperature distribution through the solid. Both ends of the surface are subjected to heat loss by convection (Nodes 1 and 5). Assume the cross-sectional area is the same for all elements, A=0.01 m?. Insulated, 1-d heat transfer in x-dir Gas 1: h1=2000 W/(m?K) Gas 2: h5=5000 W/(m²K) T1°=110°C=383.15 K Solid surface: T50=81°C=354.15 K +x k=150 W/(m K) 0.1 m
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