Air at 60 degree C leaves a furnace and passes through a 12m

Air at 60 degree C leaves a furnace and passes through a 12-m long, 20-cm times 20-cm square duct at an average velocity of 4 m/s. If the duct has a constant surface temperature of 33 degree C, determine the exit temperature of the air and the rate of heat, transfer.

Solution

convective heat transfer coefficient of air, h_c= 10.45-v+10v^1/2 , Where v is the velocity of air.

in this case h_c= 10.45-4+(10*4^1/2) = 26.45W/m²⁰c.

Rate of heat transfer, q_c= h_cA (t_{surface -} t_air ) , Where t_air is the temperature of air, and t_surface is the temperature of surface of duct. and A =area of duct= 12*(0.2*0.2)=0.48m²

q_c=26.45*0.48*(33-60)=-342.792Watts , negative sign indicates heat transfers from air to duct surface or air is losing heat.

So, The Rate of heat transfer from air to duct=342.792W.