Thermodynamics HW Solutions 692

# Thermodynamics HW Solutions 692 - Chapter 8 Internal Forced...

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Unformatted text preview: Chapter 8 Internal Forced Convection 96 . 1 Pr C J/kg. 4206 /s; m 10 326 . / C W/m. 675 . ; kg/m 3 . 965 2 6- 3 = ° = × = ρ μ = υ ° = = ρ p C k Analysis (a) The mass flow rate of water is kg/s 0.9704 m/s) 8 . ( 4 m) (0.04 ) kg/m 3 . 965 ( 2 3 = π = ρ = V A m c & The Reynolds number is 062 , 98 /s m 10 326 . m) m/s)(0.04 (0.8 Re 2 6 = × = υ = − h m D V which is greater than 4000. Therefore, the flow is turbulent and the entry lengths in this case are roughly m 4 . m) 04 . ( 10 10 = = ≈ ≈ D L L t h which are much shorter than the total length of the pipe. Therefore, we can assume fully developed turbulent flow in the entire pipe. Assuming the copper pipe to be smooth, the Nusselt number is determined to be 1 . 277 96 . 1 062 , 98 023 . Pr Re 023 . 3 . 8 . 3 . 8 . = × × = = = k hD Nu h and C . W/m 4676 ) 1 . 277 ( m 04 . C W/m. 675 . 2 ° = ° = = = Nu D k h h h i which is much greater than the convection heat transfer coefficient of 15 W/m 2 . ° C. Therefore, the convection thermal resistance inside the pipe is negligible, and thus the inner surface temperature of the...
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## This note was uploaded on 01/24/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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