Question 1.5: Consider heat transfer through a tube wall (R1, T1 and R2, T......
Consider heat transfer through a tube wall (R_1, T_1 and R_2, T_2) with length L and of conductivity k, where T_1 > T_2. Derive the wall temperature profile T(r) based on a heat flux balance and find an expression for the heat transfer rate \dot{Q}.
| Concepts | Assumptions | Sketch |
| • Steady 1-D heat balance (as shown) | Steady 1-D (radial) heat conduction |
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| • OR | • Very long tube, | |
| • Reduced heat transfer equation (Sect.2.5) | • i.e. T=T(r) only | |
| • Fourier’s Law q=-k\frac{dT}{dr} | • Constant properties | |
| • Heat flow rate: \dot{Q}=qA | • No internal heat generation |
• In cylindrical coordinates: q_r=-k\frac{dT}{dr}
• Heat flow balance for a thin shell (i.e., the wall) R_1 ≤ r ≤ R_2 is:
qA\big|_r -(qA)\big|_{r+dr}=0\\• Expanded (see truncated Taylor series in App. A):
-\frac{1}{\mathrm{\bf{r}}}\frac{\mathrm{d}}{\mathrm{d}\mathrm{{r}}}(\mathrm{{r}}{\mathrm{{q_r}}})=0~\mathrm{\bf{or}}\,\frac{1}{r} \mathrm{\bf{\frac{d}{\ d r}}}(kr\frac{dT}{dr} )=0
Thus, solve
\frac{d}{dr}(r\frac{dT}{dr} ) =0 subject to T = T_i at r = R_i ; i=1,2
The general solution is T(r) = C_1 \ln r +C_2 , so that with the B.C.’s invoked,
T(r)=\frac{T_1-T_2}{\ln (\frac{R_1}{R_2} )}\ln (\frac{r}{R_2} ) +T_2
Recalling that {\dot{Q}}=qA=-kA{\frac{\mathrm{dT}}{\mathrm{dr}}}={\cancel {\displaystyle{C}}}, with A_i =2R_i πL we have
\dot{Q}=\frac{2\pi k L}{\ln(\frac{R_{2}}{R_{1}})}(T_{1}-T_{2})
Graph:
Comments:
The solution represents heat loss through a pipe wall of thickness R_2 − R_1 without internal/external convection.
