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Unformatted text preview: Chapter 9 Natural Convection 932 A fluid flows through a pipe in calm ambient air. The pipe is heated electrically. The thickness of the insulation needed to reduce the losses by 85% and the money saved during 10h 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 Insulation will drop the outer surface temperature to a value close to the ambient temperature, and possible below it because of the very low sky temperature for radiation heat loss. For convenience, we use the properties of air at 1 atm and 5 ° C (the anticipated film temperature) (Table A15), 1 2 5 K 003597 . K ) 273 5 ( 1 1 7350 . Pr /s m 10 382 . 1 C W/m. 02401 . = + = = = × = ° = f T k β υ Analysis The rate of heat loss in the previous problem was obtained to be 29,094 W. Noting that insulation will cut down the heat losses by 85%, the rate of heat loss will be W 4364 W 094 , 29 15 . ) 85 . 1 ( insulation no = × = = Q Q The amount of energy and money insulation will save during a 10h period is simply determined from kWh 3 . 247 h) kW)(10 094 . 29 85 . ( , = × = ∆ = t Q Q saved total saved $22.26 = = ) kWh / 09 . )($ kWh 3 . 247 ( = energy) of cost t saved)(Uni Energy ( saved Money The characteristic length in this case is the outer diameter of the insulated pipe, insul insul c t t D L 2 3 . 2 + = + = where t insul is the thickness of insulation in m. Then the problem can be formulated for T s and t insul as follows: ) 7350 . ( ) /s m 10 382 . 1 ( ) 2 3 . ( K] ) 273 )[( K 003597 . )( m/s 81 . 9 ( Pr ) ( 2 2 5 31 2 2 3 ∞ × + = = insul s c s t T L T T g Ra υ β ( 29 [ ] ( 29 [ ] 2 27 / 8 16 / 9 6 / 1 2 27 / 8 16 / 9 6 / 1 7350 . / 559 . 1 387 . 6 . Pr / 559 . 1 387 . 6 . + + = + + = Ra Ra Nu m) )(100 2 3 . ( C W/m. 02401 . insul s c c t L D A Nu L Nu L k h + = = ° = = π π The total rate of heat loss from the outer surface of the insulated pipe by convection and radiation becomes ] ) K 273 30 ( )[ .K W/m 10 67 . 5 ( ) 1 . ( + ) 273 ( 4364 ) ( ) ( 4 4 4 2 8 4 4 + × = + = + = ∞ s s s s surr s s s s rad conv T A T hA T T A T T hA Q Q Q σ ε In steady operation, the heat lost by the side surfaces of the pipe must be equal to the heat lost from the exposed surface of the insulation by convection and radiation, which must be equal to the heat conducted through the insulation. Therefore, ] 3 . / ) 2 3 . ln[( K ) m)(298 C)(100 W/m. 035 . ( 2 W 4364 ) / ln( ) ( 2 tank insul s o s insulation t T D D T T kL Q Q + ° = → = = π π The solution of all of the equations above simultaneously using an equation solver gives T s = 281.5 K = 8.5 ° C and t insul = 0.013 m = 1.3 cm ....
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 Spring '08
 BENARD
 Heat Transfer, Insulation

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