Mech Engr - Heat Transfer across Cylindrical Tube
Start with Fourier's Law of Heat Conduction
ref1
[tex]
\renewcommand{\vec}[1]{\mbox{\boldmath $ #1 $}} <br />
\vec{Q} =-k \bar{\nabla} T[/tex]
For this geometry (cylindrical tubing) by Fourier's Law,
ref2
[tex]Q=k A \left (\frac {\Delta T}{\Delta r} \right )[/tex]
Heat Transfer Across Length of Cylindrical Tubing
[tex]\mbox {\Huge Q= $\frac {2 \pi k L (T_i-T_o)}{ln (\frac{r_o}{r_i}) }$ }[/tex]
[itex]k[/itex] - thermal conductivity of material [BTU/(hr-ft-deg F)]
[itex]L[/itex] - length of tube (ft)
[itex]T_i[/itex] - temperature along inside surface of tube (deg F)
[itex]T_o[/itex] - temperature along outside surface of tube (deg F)
[itex]r_o[/itex] - outside tube radius (ft)
[itex]r_i[/itex] - inside tube radius (ft)
[itex]Q[/itex] - heat transfer (BTU/hr)
Heat Flux - Heat Transfer Rate per Unit Area
ref3
[tex]Q^{''} = \frac {Q}{A} \ \ \ \ \ \ \ \left ( \frac {BTU}{hr \cdot ft^2} \right )[/tex]
For this geometry
[tex]A = 2 \pi r_o L \ \ \ \ \ \ (ft^2)[/tex]