(a) Explanation Given: Mechanical power: P m e c h Electric power: P e l e c Length: L Thermal conductivity of copper shaft: k s Thermal conductivity of square pad: k p Coefficient of convection: h h Width of pad: W Thickness of pad: t Assumptions: System at the steady-state condition. Heat conductions through pad The shaft is one dimensional. Properties are constant. Radiation is negligible. Calculations: According to the conservations of energy yields, P e l e c − P m e c h − q h − q p − q s = 0 ... (1) We know that, Heat transfer through housing is q h = h h A h ( T h − T ∞ ) . Heat transfer through pad is q p = k p W 2 ( T h − T ∞ ) t . And Heat transfer through shaft is q s = M cosh m L − θ L θ B sinh m L Here, θ L = 0 , M = ( π 4 D 3 h s k s ) 1 2 ( T h − T ∞ ) and m L = ( 4 h s L 2 k s D ) 1 2 By putting the values in equation, we get: q s = ( π 4 D 3 h s k s ) 1 2 ( T h − T ∞ ) tanh m L ( 4 h s L 2 k s D ) 1 2 From equation (1): P e l e c − P m e c h − q h − q p − q s = 0 P e l e c − P m e c h − ( h h A h ( T h − T ∞ ) ) − ( k p W 2 ( T h − T ∞ ) t ) − ( ( (b) To determine The value of housing temperature, T h

6th Edition

ISBN: 9780470501962

Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine

Publisher: Wiley, John & Sons, Incorporated

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Chapter 3, Problem 3.117P

(a)

To determine

The expression for (Th−T∞) .

(b)

To determine

The value of housing temperature, Th

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Chapter 3, Problem 3.116P

Chapter 3, Problem 3.118P

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Introduction to Heat Transfer

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The expression for ( T h − T ∞ ) .

Solution Summary: The author explains that the expression for (T_h) is derived from the equation. Heat conductions through pad The shaft is one dimensional.