Thermodynamics HW Solutions 650

Thermodynamics HW Solutions 650 - m 20 . m) 02 . ( 10 10 =...

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Chapter 8 Internal Forced Convection 8-40 Water is to be heated in a tube equipped with an electric resistance heater on its surface. The power rating of the heater and the inner surface temperature are to be determined. Assumptions 1 Steady flow conditions exist. 2 The surface heat flux is uniform. 3 The inner surfaces of the tube are smooth. Properties The properties of water at the average temperature of (80+10) / 2 = 45 ° C are (Table A-9) Water 10 ° C 3 m/s 80 ° C (Resistance heater) L D = 2 cm 91 . 3 Pr C J/kg. 4180 /s m 10 602 . 0 / C W/m. 637 . 0 kg/m 1 . 990 2 6 - 3 = ° = × = ρ μ = υ ° = = ρ p C k Analysis The power rating of the resistance heater is kg/s 132 . 0 kg/min 921 . 7 ) /min m 008 . 0 )( kg/m 1 . 990 ( 3 3 = = = ρ = V m W 38,627 = ° ° = = C ) 10 80 )( C J/kg. 4180 )( kg/s 132 . 0 ( ) ( i e p T T C m Q The velocity of water and the Reynolds number are V m c V A == × = (8 / ) (. / . 10 60 002 4 0 4244 3 m/ s m) m/s 3 2 π 101 , 14 /s m 10 602 . 0 m) m/s)(0.02 (0.4244 Re 2 6 = × = υ = h m D V which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly
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Unformatted text preview: m 20 . m) 02 . ( 10 10 = = D L L t h which is much shorter than the total length of the duct. Therefore, we can assume fully developed turbulent flow in the entire duct, and determine the Nusselt number from 79 . 82 ) 91 . 3 ( ) 101 , 14 ( 023 . Pr Re 023 . 4 . 8 . 4 . 8 . = = = = k hD Nu h Heat transfer coefficient is C . W/m 2637 ) 79 . 82 ( m 02 . C W/m. 637 . 2 = = = Nu D k h h Then the inner surface temperature of the pipe at the exit becomes C 113.3 = = = e s s e e s s T T T T hA Q , 2 , C ) 80 )]( m 7 )( m 02 . ( )[ C . W/m 2637 ( W 627 , 38 ) ( & 8-12...
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This note was uploaded on 01/22/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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