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Unformatted text preview: Chapter 8 Internal Forced Convection 853 Hot air enters a sheet metal duct located in a basement. The exit temperature of hot air and the rate of heat loss are to be determined. Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the duct are smooth. 3 The thermal resistance of the duct is negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm. Properties We expect the air temperature to drop somewhat, and evaluate the air properties at 1 atm and the estimated bulk mean temperature of 50 C (Table A15), 7228 . Pr C J/kg. 1007 /s; m 10 797 . 1 C W/m. 02735 . ; kg/m 092 . 1 2 5 3 = = = = = p C k Analysis The surface area and the Reynolds number are m 9.6 m) m)(12 2 . ( 4 4 = = = aL A s D A p a a a h c = = = = 4 4 4 0 2 2 . m 509 , 44 /s m 10 797 . 1 m) m/s)(0.20 (4 Re 2 5 = = = h m D V which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly m . 2 m) 2 . ( 10 10 = = h t h D L L which is much shorter than the total length of the duct. Therefore, we can assume fully developed turbulent flow for the entire duct, and determine the Nusselt number from 2 . 109 ) 7228 . ( ) 509 , 44 ( 023 . Pr Re 023 . 3 . 8 . 3 . 8 . = = = = k hD Nu h and C . W/m 93 . 14 ) 2 . 109 ( m 2 . C W/m. 02735 . 2 = = = Nu D k h h The mass flow rate of air is kg/s 0.1748 m/s) (4 0.2)m )(0.2 kg/m 092 . 1 ( 2 3 = = = V A m c In steady operation, heat transfer from hot air to the duct must be equal to the heat transfer from the duct to the surrounding (by convection and radiation), which must be equal to the energy loss of the hot air in the duct. That is, Q Q Q E = = = conv,in conv+rad,out hot air Assuming the duct to be at an average temperature of T s , the quantities above can be expressed as : Q conv,in  =  = = 60 ln 60 ) m 6 . 9 )( C . W/m 93 . 14 ( ln 2 2 ln s e s e i s e s i e s i s i T T T T Q T T T T T T A h T A h Q : Q conv+rad,out ( 29 [ ] 4 4 4 4 2 8 2 2 2 4 4 ) 273 10 ( ) 273 ( ) . W/m 10 67 . 5 )( m 0.3(9.6 + C ) 10 )( m C)(9.6 . W/m 10 ( ) ( K T K T Q T T A T T A h Q s s o s s o s s o + +  =  + = : E hot air C ) C)(60 J/kg. 7 kg/s)(100 1748 . ( ) (  =  = e i e p T Q T T C m Q This is a system of three equations with three unknowns whose solution is C 3 . 33 and , , = = = s e T T Q C 45.1 W 2622 Therefore, the hot air will lose heat at a rate of 2622 W and exit the duct at 45.1 C. 854 "!PROBLEM 854" "GIVEN" 832 Air 60 C 4 m/s L = 12 m = 0.3 Air duct 20 cm 20 cm T = 10 C Chapter 8 Internal Forced Convection T_i=60 "[C]" L=12 "[m]" side=0.20 "[m]" Vel=4 "[m/s], parameter to be varied" "epsilon=0.3 parameter to be varied" T_o=10 "[C]" h_o=10 "[W/m^2C]" T_surr=10 "[C]" "PROPERTIES" Fluid$='air' C_p=CP(Fluid$, T=T_ave)*Convert(kJ/kgC, J/kgC)...
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This homework help was uploaded on 03/19/2008 for the course ME 410 taught by Professor Benard during the Spring '08 term at Michigan State University.
 Spring '08
 BENARD
 Heat Transfer

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