Thermodynamics HW Solutions 695

Thermodynamics HW Solutions 695 - C . W/m 17 . 10 ) 31 . 32...

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Chapter 8 Internal Forced Convection 8-73 Hot exhaust gases flow through a pipe. For a specified exit temperature, the pipe length is to be determined. Assumptions 1 Steady operating conditions exist. 2 The inner surface of the pipe is smooth. 3 Air is an ideal gas with constant properties. 4 The pressure of air is 1 atm. Properties The properties of air at 1 atm and the bulk mean temperature of (450+250)/2 = 350 ° C are (Table A-15) L D = 15 cm Exhaust gases 450 ° C 3.6 m/s 250 ° C T s = 180 ° C 6937 . 0 Pr C J/kg. 1056 /s m 10 475 . 5 C W/m. 04721 . 0 kg/m 5664 . 0 2 5 - 3 = ° = × = ° = = p C k υ ρ Analysis The Reynolds number is 9864 /s m 10 475 . 5 m) m/s)(0.15 (3.6 Re 2 5 = × = = D m V which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly m 1.5 = m) 15 . 0 ( 10 10 = D L L t h which is probably 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 31 . 32 ) 6937 . 0 ( ) 9864 ( 023 . 0 Pr Re 023 . 0 3 . 0 8 . 0 3 . 0 8 . 0 = = = = k hD Nu Heat transfer coefficient is
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Unformatted text preview: C . W/m 17 . 10 ) 31 . 32 ( m 15 . C W/m. 04721 . 2 = = = Nu D k h The logarithmic mean temperature difference is C 2 . 148 450 180 250 180 ln 450 250 ln ln = = = i s e s i e T T T T T T T The rate of heat loss from the exhaust gases can be expressed as [ ] L L T hA Q s 25 . 710 ) C 2 . 148 ( m) 15 . ( ) C . W/m 17 . 10 ( 2 ln = = = & where L is the length of the pipe. The rate of heat loss can also be determined from [ ] kg/s 0.03603 = /4 m) (0.15 m/s) )(3.6 kg/m 5664 . ( 2 3 = = c VA m & W 7612 C ) 250 450 )( C J/kg. kg/s)(1056 03603 . ( = = = T C m Q p & & Setting this equal to rate of heat transfer expression above, the pipe length is determined to be m 10.72 = = = L L Q W 7612 25 . 710 & 8-57...
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