FTFS Chap19 P094 - Chapter 19 Forced Convection Review...

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Unformatted text preview: Chapter 19 Forced Convection Review Problems 19-94 Wind is blowing parallel to the walls of a house. The rate of heat loss from the wall is to be determined. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Re cr = 5 × 10 5 . 3 Radiation effects are negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm. Properties Assuming a film temperature of T f = 10 ° C for the outdoors, the properties of air are evaluated to be (Table A-22) 7336 . Pr /s m 10 426 . 1 C W/m. 02439 . 2 5- = × = ° = υ k Analysis Air flows along 19-m side. The Reynolds number in this case is [ ] 6 2 5 10 792 . 7 /s m 10 426 . 1 m) (8 m/s ) 3600 / 1000 50 ( Re × = × × = =- ∞ υ L V L which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow. Using the proper relation for Nusselt number, heat transfer coefficient is determined to be [ ] C . W/m 78 . 30 ) 096 , 10 ( m 8 C W/m. 02439 . 096 , 10 ) 7336 . ( 871 ) 10 792 . 7 ( 037 . Pr ) 871 Re 037 . ( 2 3 / 1 8 . 6 3 / 1 8 . ° = ° = = =- × =- = = Nu L k h k L h Nu o L The thermal resistances are 2 m 24 = m) m)(8 3 ( = = wL A s C/W 0014 . ) m C)(24 . W/m 78 . 30 ( 1 1 C/W 1408 . m 24 C/W . m 38 . 3 ) 38 . 3 ( C/W 0052 . ) m C)(24 . W/m 8 ( 1 1 2 2 2 2 2 2 ° = ° = = ° = ° =- = ° = ° = = s o o s value insulation s i i A h R A R R A h R Then the total thermal resistance and the heat transfer rate through the wall are determined from W 122.1 = ° °- =- = ° = + + = + + = ∞ ∞ C/W 1474 . C ) 4 22 ( C/W 1474 . 0014 . 1408 . 0052 . 2 1 total o insulation i total R T T Q R R R R 19-90 R R R T T Air V ∞ = 50 km/h T = 4 ° C L = 8 m WALL T ∞ 1 = 22 ° C Chapter 19 Forced Convection 19-95 A car travels at a velocity of 60 km/h. The rate of heat transfer from the bottom surface of the hot automotive engine block is to be determined for two cases. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Re cr = 5 × 10 5 . 3 Air is an ideal gas with constant properties. 4 The pressure of air is 1 atm. 5 The flow is turbulent over the entire surface because of the constant agitation of the engine block. 6 The bottom surface of the engine is a flat surface. Properties The properties of air at 1 atm and the film temperature of (T s + T ∞ )/2 = (75+5)/2 = 40 ° C are (Table A-22) 7255 . Pr /s m 10 702 . 1 C W/m. 02662 . 2 5- = × = ° = υ k Analysis The Reynolds number is [ ] 5 2 5 10 855 . 6 /s m 10 702 . 1 m) (0.7 m/s ) 3600 / 1000 60 ( Re × = × × = υ =- ∞ L V L which is less than the critical Reynolds number. But we will assume turbulent flow because of the constant agitation of the engine block....
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This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.

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FTFS Chap19 P094 - Chapter 19 Forced Convection Review...

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