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 EngineeringHeat 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