Consider a slab of face area A and thickness L. Suppose that L = 20 cm,...
Homework Answers
Pcond=kA.(Th-Tc)/L
where k is the thermal conductivity of copper (401 W/m•K), A is the cross-sectional area (in a plane perpendicular to the flow), L is the distance along the direction of flow between the points where the temperature is TH and TC.
k=401 W/m•K
L=79 cm2=79x10-4 m2
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Consider a slab of face area A and thickness L. Suppose that L = 20
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