The steady-state temperature distribution in a one-dimensional slab of thermal conductivity 50 * w/(mK) and thickness 50 mm is found to be T = a + b * x ^ 2 , where a = 2000C , b=-20000 c/ m^ 2 , T is n^ C and x in meters. (a) What is the heat generation rate in the slab? (b) Determine the heat fluxes at the two wall faces
The steady-state temperature distribution in a one-dimensional slab of thermal conductivity 50 * w/(mK) and thickness 50 mm is found to be T = a + b * x ^ 2 , where a = 2000C , b=-20000 c/ m^ 2 , T is n^ C and x in meters. (a) What is the heat generation rate in the slab? (b) Determine the heat fluxes at the two wall faces
Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Kreith, Frank; Manglik, Raj M.
Chapter1: Basic Modes Of Heat Transfer
Section: Chapter Questions
Problem 1.35P: A section of a composite wall with the dimensions shown below has uniform temperatures of 200C and...
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The steady-state temperature distribution in a one-dimensional slab of thermal conductivity 50 * w/(mK) and thickness 50 mm is found to be T = a + b * x ^ 2 , where a = 2000C , b=-20000 c/ m^ 2 , T is n^ C and x in meters. (a) What is the heat generation rate in the slab? (b) Determine the heat fluxes at the two wall faces
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