SOLUTION In this example, the concepts of PtD are applied in conjunction with the concepts of internal forced convection and steady onedimensional heat conduction. The inner pipe surface temperature T_{s, i} is determined using the concept of internal forced convection. Having determined the inner surface temperature, the outer surface temperature T_{s, o} is determined using one-dimensional heat conduction through the pipe wall and insulation.
Assumptions Steady operating conditions exist. 2 Radiation effects are negligible. 3 Convection effects on the outer pipe surface are negligible. 4 One dimensional heat conduction through pipe wall and insulation. 5 The thermal conductivities of pipe wall and insulation are constant. 6 Thermal resistance at the interface is negligible. 7 The surface temperatures are uniform. 8 The inner surfaces of the tube are smooth.
Properties The properties of saturated water vapor at T_{b}=\left(T_{i}+T_{e}\right) / 2=320^{\circ} C are c_{p}=7900 J / kg \cdot K , k=0.0836 W / m \cdot K , \mu=2.084 \times 10^{-5} kg / m \cdot s, and \operatorname{Pr}=1.97 (Table A-9). The thermal conductivities of the pipe and the insulation are given to be k_{\text {pipe }}=15 W / m \cdot K and k_{ ins }=0.95 W / m \cdot K, respectively.
Analysis The Reynolds number of the saturated water vapor flow in the pipe is
Re =\frac{4 \dot{m}}{\pi D_{i} \mu}=\frac{4(0.05 kg / s )}{\pi(0.05 m )\left(2.084 \times 10^{-5} kg / m \cdot s \right)}=61,096>10,000
Therefore, the flow is turbulent and the entry lengths in this case are roughly
L_{h} \approx L_{t} \approx 10 D=10(0.05 m )=0.5 m (assume fully developed turbulent flow)
The Nusselt number can be determined from the Gnielinski correlation:
Nu =\frac{(f / 8)( Re -1000) Pr }{1+12.7(f / 8)^{0.5}\left( Pr ^{2 / 3}-1\right)}=\frac{(0.02003 / 8)(61,096-1000)(1.97)}{1+12.7(0.02003 / 8)^{0.5}\left(1.97^{2 / 3}-1\right)}=217.45
where
f=(0.790 \ln Re -1.64)^{-2}=0.02003
Thus, the convection heat transfer coefficient for the saturated water vapor flow inside the pipe is
h=\frac{k}{D_{i}} Nu =\frac{0.0836 W / m \cdot K }{0.05 m }(217.45)=363.58 W / m ^{2} \cdot K
The inner pipe surface temperature is
T_{e}=T_{s, i}-\left(T_{s, i}-T_{i}\right) \exp \left(-\frac{h A_{s}}{\dot{m} c_{p}}\right) \quad \longrightarrow \quad T_{s, i}=271.52^{\circ} C
\text { where } \quad A_{s}=\pi(0.05 m )(10 m )=1.571 m ^{2}
the thermal resistances for the pipe wall and the insulation are
R_{\text {pipe }}=\frac{\ln \left(D_{\text {interface }} / D_{i}\right)}{2 \pi k_{\text {pipe }} L}=\frac{\ln (0.06 / 0.05)}{2 \pi(15 W / m \cdot K )(10 m )}=1.9345 \times 10^{-4} K / W
R_{\text {ins }}=\frac{\ln \left(D_{o} / D_{\text {interface }}\right)}{2 \pi k_{\text {ins }} L}=\frac{\ln (0.105 / 0.06)}{2 \pi(0.95 W / m \cdot K )(10 m )}=9.3753 \times 10^{-3} K / W
where D_{o}=0.06 m +2(0.0225 m )=0.105 m
The total thermal resistance and the rate of heat transfer are
R_{\text {total }}=R_{\text {pipe }}+R_{\text {ins }}=9.5688 \times 10^{-3} K / W \text { and } \dot{Q}=\frac{T_{s, i}-T_{s, 0}}{R_{\text {total }}}=\dot{m} c_{p}\left(T_{i}-T_{e}\right)
Thus, the outer surface temperature is
T_{s, o} =T_{s, i}-R_{\text {total }} \dot{m} c_{p}\left(T_{i}-T_{e}\right)
=271.52^{\circ} C -\left(9.5688 \times 10^{-3} K / W \right)(0.05 kg / s )(7900 J / kg \cdot K )(350-290)^{\circ} C
=44.7^{\circ} C
Discussion The insulation thickness of 2.25 cm is just barely sufficient to keep the outer surface temperature below 45^{\circ} C. To ensure the outer surface to be a few degrees below 45^{\circ} C, the insulation thickness should be increased slightly to 2.3 cm, which would make T_{s, 0}=41^{\circ} C.
