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Question : A 10 -m-long metal pipe kpipe =15 W / m.K has an inner diame...

A 10 -m-long metal pipe (kpipe =15W/m⋅K) \left(k_{\text {pipe }}=15 W / m \cdot K \right) has an inner diameter of5 cm an outer diameter of 6 cm is used for transporting hot saturated water vapor at a flow rate of 0.05 kg / s (Fig.8-33). The water vapor enters and exits the pipe at 350∘C 350^{\circ} C  and 290∘C 290^{\circ} C, respectively. In order to prevent thermal burn on individuals working in the vicinity of the pipe, the pipe is covered with a 2.25- cm thick layer of insulation (kins =0.95W/m⋅K) \left(k_{\text {ins }}=0.95 W / m \cdot K \right)  to ensure that the outer surface temperature Ts,o T_{s, o}  is below 45∘C 45^{\circ} C. Determine whether or not the thickness of the insulation is sufficient to alleviate the risk of thermal burn hazards.

 

 

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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 Ts,i T_{s, i}  is determined using the concept of internal forced convection. Having determined the inner surface temperature, the outer surface temperature Ts,o 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 Tb=(Ti+Te)/2=320∘C T_{b}=\left(T_{i}+T_{e}\right) / 2=320^{\circ} C  are cp=7900J/kg⋅K,k=0.0836W/m⋅K,μ=2.084×10−5kg/m⋅s 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 Pr⁡=1.97 \operatorname{Pr}=1.97  (Table A-9). The thermal conductivities of the pipe and the insulation are given to be kpipe =15W/m⋅K k_{\text {pipe }}=15 W / m \cdot K  and kins=0.95W/m⋅K k_{ ins }=0.95 W / m \cdot K, respectively.

Analysis The Reynolds number of the saturated water vapor flow in the pipe is

Re=4m˙πDiμ=4(0.05kg/s)π(0.05m)(2.084×10−5kg/m⋅s)=61,096>10,000 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

Lh≈Lt≈10D=10(0.05m)=0.5m 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=(f/8)(Re−1000)Pr1+12.7(f/8)0.5(Pr2/3−1)=(0.02003/8)(61,096−1000)(1.97)1+12.7(0.02003/8)0.5(1.972/3−1)=217.45 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=(0.790ln⁡Re−1.64)−2=0.02003 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=kDiNu=0.0836W/m⋅K0.05m(217.45)=363.58W/m2⋅Kh=\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

Te=Ts,i−(Ts,i−Ti)exp⁡(−hAsm˙cp)⟶Ts,i=271.52∘C 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

 

 where As=π(0.05m)(10m)=1.571m2 \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

Rpipe =ln⁡(Dinterface /Di)2πkpipe L=ln⁡(0.06/0.05)2π(15W/m⋅K)(10m)=1.9345×10−4K/W 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

 

Rins =ln⁡(Do/Dinterface )2πkins L=ln⁡(0.105/0.06)2π(0.95W/m⋅K)(10m)=9.3753×10−3K/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 Do=0.06m+2(0.0225m)=0.105m D_{o}=0.06 m +2(0.0225 m )=0.105 m

The total thermal resistance and the rate of heat transfer are

Rtotal =Rpipe +Rins =9.5688×10−3K/W and Q˙=Ts,i−Ts,0Rtotal =m˙cp(Ti−Te) 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

Ts,o=Ts,i−Rtotal m˙cp(Ti−Te) T_{s, o} =T_{s, i}-R_{\text {total }} \dot{m} c_{p}\left(T_{i}-T_{e}\right)

 

=271.52∘C−(9.5688×10−3K/W)(0.05kg/s)(7900J/kg⋅K)(350−290)∘C =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∘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∘C 45^{\circ} C. To ensure the outer surface to be a few degrees below 45∘C 45^{\circ} C, the insulation thickness should be increased slightly to 2.3 cm, which would make Ts,0=41∘C T_{s, 0}=41^{\circ} C.

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