Answer to Question #227803 in Chemical Engineering for kaa

Question

a horizontal tube carrying hot water has a surface temperature of 355.4 K and the outside diameter of 25.4 mm. The tube is exposed to room air at 294.3 K. Calculate the nature convection heat loss for a 1 m length of pipe. if the tube is replaced with two horizontal plate and the bottom plate is hotter than the upper plate. compared the resulted value of natural convection heat loss using tube and plate

Answer 1

Calculating the Rayleigh number

$= (\frac{g \beta D^3 (T_s-T_{\infin})}{v^2})(Pr)\\ = (\frac{9.81*( \frac{1}{324.85}) 0.0254^3 (355.4-294.3)}{(18.39*10^{-6})^2})(0.7035)\\ =62896.88 <10^{12}$

Calculating the Nusselt number

$N_{uD}=\{0.60+ \frac{0.387Ra_D^{1/6}}{[1+(\frac{0.559}{Pr})^{Pr}]^{8/27}} \}^2=6.8885$

Calculating the heat transfer coefficient

$Nu_D= \frac{hD}{k}\\ 6.8885= \frac{h*0.0254}{28.13*10^{-3}}\\ h= 7.6288W/m^2k$

Calculating the heat loss

$Q= hA(T_s-T_{\infin})\\ Q= h(\pi DL)(T_s-T_{\infin})\\ Q= 7.6288(\pi *0.0254*1)(355.4-294.3)\\ Q= 37.19 W$