Question 7.7.4: Water at 120°F flows in a copper pipe 6 ft long, whose inner ......
Water at 120°F flows in a copper pipe 6 ft long, whose inner and outer radii are 1/4 in. and 3/8 in. The temperature of the surrounding air is 70°F. Compute the heat loss rate from the water to the air in the radial direction. Use the following values. For copper, k = 50 lb/sec-°F. The convection coefficient at the inner surface between the water and the copper is h_{i} = 16 lb/sec-ft-°F. The convection coefficient at the outer surface between the air and the copper is h_{o} = 1.1lb/sec-ft-°F.
Assuming that the temperature inside the pipe wall does not change with time, then the same heat flow rate occurs in the inner and outer convection layers and in the pipe wall. Thus the three resistances are in series and we can add them to obtain the total resistance. The inner and outer surface areas are
A_{i}=2\pi \ r_{i}L=2\pi\left({\frac{1}{4}}\right)\left({\frac{1}{12}}\right)6=0.785\,\mathrm{ft}^{2}A_{o}=2\pi \ r_{o}L=2\pi\left(\frac{3}{8}\right)\left(\frac{1}{12}\right)6=1.178\;\mathrm{ft}^{2}
The inner convective resistance is
R_{i}={\frac{1}{h_{i}A_{i}}}={\frac{1}{16(0.785)}}=0.08~{\frac{\mathrm{sec-°F }}{\mathrm{ft-lb}}}R_{o}={\frac{1}{h_{o}A_{o}}}={\frac{1}{1.1(1.178)}}=0.77\,{\frac{\mathrm{sec-°F }}{\mathrm{ft-lb}}}
The conductive resistance of the pipe wall is
R_{c}=\frac{\ln\left(\frac{r_{a}}{r_{i}}\right)}{2\pi \ L k}=\frac{\ln\left(\frac{3/8}{1/4}\right)}{2\pi(6)(50)}=2.15\times10^{-4}\ \frac{\mathrm{sec-°F }}{\mathrm{ft-lb}}Thus the total resistance is
R=R_{i}+R_{c}+R_{o}=0.08+2.15\times10^{-4}+0.77=0.85~{\frac{\mathrm{sec-°F}}{\mathrm{ft-lb}}}The heat loss from the pipe, assuming that the water temperature is a constant 120° along the length of the pipe, is
q_{h}={\frac{1}{R}}\,\Delta T={\frac{1}{0.85}}(120-70)=59\,{\frac{\mathrm{ft-lb}}{\mathrm{sec}}}To investigate the assumption that the water temperature is constant, compute the thermal energy E of the water in the pipe, using the mass density ρ = 1.94 slug/ft³ and {{c}}_{p} = 25,000 ft-lb/slug-°F:
E=m \ c_{p} \ T_{i}=\left(\pi \ r_{i}^{2}L\rho\right)c_{p} \ T_{i}=47,624\,\mathrm{ft}-\mathrm{lb}Assuming that the water flows at 1 ft/sec, a slug of water will be in the pipe for 6 sec. During that time it will lose 59(6) = 354 ft-lb of heat. Because this amount is very small compared to E, our assumption that the water temperature is constant is confirmed.