This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chap 13 Heat Exchangers Review Problems 13111 Hot oil is cooled by water in a multipass shellandtube heat exchanger. The overall heat transfer coefficient based on the inner surface is to be determined. Assumptions 1 Water flow is fully developed. 2 Properties of the water are constant. Properties The properties of water at 300 K ≈ 25 ° C are (Table A9) 14 . 6 Pr /s m 10 894 . / C W/m. 607 . 2 6 = × = = ° = ρ μ υ k Analysis The Reynolds number is 771 , 43 /s m 10 894 . ) m 013 . )( m/s 3 ( Re 2 6 = × = = υ D V m which is greater than 10,000. Therefore, we assume fully developed turbulent flow, and determine Nusselt number from 245 ) 14 . 6 ( ) 771 , 43 ( 023 . Pr Re 023 . 4 . 8 . 4 . 8 . = = Nu and C . W/m 440 , 11 ) 245 ( m 013 . C W/m. 607 . 2 ° = ° = = Nu D k h i The inner and the outer surface areas of the tube are A D L A D L i i o o = = = = = = π π π π ( . )( ) . ( . )( ) . 0 013 1 0 04084 0015 1 004712 m m m m m m 2 2 The total thermal resistance of this heat exchanger per unit length is C/W 609 . ) m 04712 . )( C . W/m 35 ( 1 ) m 1 )( C W/m. 110 ( 2 ) 3 . 1 / 5 . 1 ln( ) m 04084 . )( C . W/m 440 , 11 ( 1 1 2 ) / ln( 1 2 2 2 2 ° = ° + ° + ° = + + = π π o o i o i i A h kL D D A h R Then the overall heat transfer coefficient of this heat exchanger based on the inner surface becomes C . W/m 40.2 2 ° = ° = = → = ) m 04084 . )( C/W 609 . ( 1 1 1 2 i i i i RA U A U R 1387 Outer surface D , A , h , U Inner surface D i , A i , h i , U i Chap 13 Heat Exchangers 13112 Hot oil is cooled by water in a multipass shellandtube heat exchanger. The overall heat transfer coefficient based on the inner surface is to be determined. Assumptions 1 Water flow is fully developed. 2 Properties of the water are constant. Properties The properties of water at 300 K ≈ 25 ° C are (Table A9) 14 . 6 Pr /s m 10 894 . / C W/m. 607 . 2 6 = × = = ° = ρ μ υ k Analysis The Reynolds number is 771 , 43 /s m 10 894 . ) m 013 . )( m/s 3 ( Re 2 6 = × = = υ D V m which is greater than 10,000. Therefore, we assume fully developed turbulent flow, and determine Nusselt number from 245 ) 14 . 6 ( ) 771 , 43 ( 023 . Pr Re 023 . 4 . 8 . 4 . 8 . = = Nu and C . W/m 440 , 11 ) 245 ( m 013 . C W/m. 607 . 2 ° = ° = = Nu D k h i The inner and the outer surface areas of the tube are A D L A D L i i o o = = = = = = π π π π ( . )( ) . ( . )( ) . 0 013 1 0 04084 0015 1 004712 m m m m m m 2 2 The total thermal resistance of this heat exchanger per unit length of it with a fouling factor is C/W 617 . ) m 04712 . )( C . W/m 35 ( 1 m 04712 . C/W . m 0004 . ) m 1 )( C W/m. 110 ( 2 ) 13 / 15 ln( ) m 04084 . )( C . W/m 440 , 11 ( 1 1 2 ) / ln( 1 2 2 2 2 2 2 , ° = ° + ° + ° + ° = + + + = π π o o o o f i o i i A h A R kL D D A h R Then the overall heat transfer coefficient of this heat exchanger based on the inner surface becomes C . W/m 39.7 2 ° = ° = = → = ) m 04084 . )( C/W 617 . ( 1 1 1 2 i i i i RA U A U R 1388 Outer surface...
View
Full
Document
 Spring '06
 Rajadas

Click to edit the document details