Thermodynamics HW Solutions 724

Thermodynamics HW Solutions 724 - + + = Ra Nu 2 2 m 25 . 94...

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Chapter 9 Natural Convection 9-31 A thick fluid flows through a pipe in calm ambient air. The pipe is heated electrically. The power rating of the electric resistance heater and the cost of electricity during a 10-h period are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local atmospheric pressure is 1 atm. Properties The properties of air at 1 atm and the film temperature of ( T s + T )/2 = (25+0)/2 = 12.5 ° C are (Table A-15) Asphalt L = 100 m D =30 cm T s = 25 ° C ε = 0.8 T sky = -30 ° C T = 0 ° C 1 - 2 5 K 003503 . 0 K ) 273 5 . 12 ( 1 1 7330 . 0 Pr /s m 10 448 . 1 C W/m. 02458 . 0 = + = = = × = ° = f T k β υ Analysis The characteristic length in this case is the outer diameter of the pipe, m. 3 . 0 = = D L c Then, 7 2 2 5 3 -1 2 2 3 10 106 . 8 ) 7330 . 0 ( ) /s m 10 448 . 1 ( ) m 3 . 0 )( K 0 25 )( K 003503 . 0 )( m/s 81 . 9 ( Pr ) ( × = × = = c s L T T g Ra () [] () [] 29 . 53 7330 . 0 / 559 . 0 1 ) 10 106 . 8 ( 387 . 0 6 . 0 Pr / 559 . 0 1 387 . 0 6 . 0 2 27 / 8 16 / 9 6 / 1 7 2 27 / 8 16 / 9 6 / 1 = + × + =
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Unformatted text preview: + + = Ra Nu 2 2 m 25 . 94 ) m 100 )( m 3 . ( C . W/m 366 . 4 ) 29 . 53 ( m 3 . C W/m. 02458 . = = = = = = DL A Nu L k h s c and W 287 , 10 C ) 25 )( m 25 . 94 )( C . W/m 366 . 4 ( ) ( 2 2 = = = T T hA Q s s & The radiation heat loss from the cylinder is W 808 , 18 ] ) K 273 30 ( ) K 273 25 )[( .K W/m 10 67 . 5 )( m 25 . 94 )( 8 . ( ) ( 4 4 4 2 8 2 4 4 = + + = = surr s s rad T T A Q & Then, kW 29.1 = = + = + = W 094 , 29 808 , 18 287 , 10 radiation convection natural total Q Q Q & & & The total amount and cost of heat loss during a 10 hour period is kWh 9 . 290 h) kW)(10 1 . 29 ( = = = t Q Q & $26.18 = = /kWh) kWh)($0.09 9 . 290 ( Cost 9-25...
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