Answer to Question #207487 in Mechanical Engineering for amam

Question

Air is heated in a 4cmu 4cm square duct at uniform surface flux of 590 The mean air velocity is 0.32 m/s. At a section far away from the inlet the mean temperature is 40 . The mean outlet temperature is 120 . Determine the required length and maximum surface temperature.

Answer 1

Mean temperature $\bar{T_m}=\frac{T_{mi+T_{mo}}}{2}=\frac{40+120}{2}=80^oC$

$L=\frac{\rho s \bar{u}c_p(T_{mo}-T_{mi})}{4q''_s} = \frac{0.9996*0.04*0.32*1009.2(120-40)}{4*590}=0.4378 m$

Reynold's number, $Re= \frac{\bar{u}De}{v}= \frac{0.32*0.04}{20.92*10^{-6}}=611.9$

The flow is laminar, hence the Nusslet number , $Nu=\frac{h}{k}=3.608* \frac{0.02991}{0.04}=2.7 W/m^{2 \space \space\space 0}C$

$T_s (L)=T_{mi}+q''_s[\frac{4L}{\rho s \bar{u}c_p}+\frac{1}{h(L)}]= 40+590[\frac{4*0.4378}{0.9996*0.04*0.32*1009.5}+\frac{1}{2.7}]=338.5^0 C$