Heat transfer coefficient,simple example ·Solution: OT Rccall that h is computed by Ts-To From Table A-4 in Appendix,at a mean fluid temperature 7n7。+7 2 (average of free-stream and surface temperatures) Tm=(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-66.7 K/mm =-66,700 K/m So,convective heat transfer coefficient is: h、 -0.028×(-66700) 80 W =23.345 m2KHeat 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