Nusselt number for a prescribed geometry hL f(x,Rex, Pr) Lo oca khk f f(re, Pr) Average dP (For a prescribed geometry is known) =o Many convection problems are solved using Nusselt number correlations incorporating Reynolds and prandtl numbers The Nusselt number is to the thermal boundary layer what the friction coefficient is to the velocity boundary layer ou dP x Re Re L dx Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 6 Nusselt number for a prescribed geometry (For a prescribed geometry, is known) Many convection problems are solved using Nusselt number correlations incorporating Reynolds and Prandtl numbers • The Nusselt number is to the thermal boundary layer what the friction coefficientis to the velocity boundary layer. ( ) (Re ,Pr) Average k hL ,Re ,Pr Local k hL f * f L L Nu f Nu f x = = = = * * dx dP = = = = * * * 0 * * 2 ,Re , Re 2 2 dx dP f x y u V C L L y s f
Heat transfer coefficient, simple example Given 100%C. Experimental measurements ofar Air at 20C flowing over heated flat plate temperatures at various distances from the surface are as shown Experimental measurements EXperinental incasurcnicnts 4r,8o=6.2mm 2 mm ayly-o Ay L,2 mm Air I rnrn Su plate 2U 50 100°C Find: convective heat transfer coefficient. h Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 7 Heat transfer coefficient, simple example • Given: Air at 20ºC flowing over heated flat plate at 100ºC. Experimental measurements of temperatures at various distances from the surface are as shown • Find: convective heat transfer coefficient, h Experimental measurements
Heat transfer coefficient, simple example Solution aT Recall that h is computed by From Table a-4 in appendix, at a mean fluid temperature T +t (average of free-stream and surface temperatures) 20+100)/2=60°C the air conductivity k is =0.028 W/m-K Temperature gradient at the plate surface from experimental data is -66.7K/mm =-66, 700 K So. convective heat transfer coefficient is -0.028×(-66700) 80 W =23.345 m K Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 8 Heat transfer coefficient, simple example • Solution: Recall that h is computed by • From Table A-4 in Appendix, at a mean fluid temperature (average of free-stream and surface temperatures) the air conductivity, k is 0.028 W/m-K • Temperature gradient at the plate surface from experimental data is -66.7 K/mm = -66,700 K/m • So, convective heat transfer coefficient is: 0 = − − = T T y T k h s y f x m K W 23.345 80 - 0.028 ( 66700) 2 = − h = 2 m T Ts T + = T C m = (20 +100) 2 = 60