Thermodynamics HW Solutions 80

Thermodynamics HW Solutions 80 - Chapter 2 Heat Conduction...

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Chapter 2 Heat Conduction Equation 3 7 ft Btu/h 10 7.820 × = π = π = = W 1 Btu/h 412 . 3 ft) 12 / 15 ]( 4 / ft) 12 / 08 . 0 ( [ W 1000 ) 4 / ( 2 2 wire L D G V G g Similarly, heat flux on the outer surface of the wire as a result of this heat generation is determined by dividing the total rate of heat generation by the surface area of the wire to be 2 5 ft Btu/h 10 1.303 ×
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This note was uploaded on 01/14/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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