Consider a horizontal thin-walled metal pipe 5 cm OD surrounded by a 5 cm layer of insulation of thermal conductivity 0.1W/m°C. The pipe outer surface is known to be at a temperature of 350 K. The ambient medium is still air at 300 K. Estimate the heat loss from a meter length of pipe (Fig. 16.14).

Question

Consider a horizontal thin-walled metal pipe 5 cm OD surrounded by a 5 cm layer of insulation of thermal conductivity [latex]0.1 W / m ^{\circ} C[/latex]. The pipe outer surface is known to be at a temperature of 350 K. The ambient medium is still air at 300 K. Estimate the heat loss from a meter length of pipe (Fig. 16.14).

Accepted Answer

As in Example 16.5, both the outer surface temperature of the insulation and the heat transferred across the insulation layer are unknown. An iterative method is required to solve the problem. To start the iteration, we assume a value of [latex]T_{2}=315 K[/latex].We use the correlation recommended by McAdams for obtaining the convective heat transfer coefficient.

Most of the data is shown in the figure itself. Air properties are taken at the ambient temperature of 300 K.

Kinematic viscosity: [latex] u=15.89 imes 10^{-6} m ^{2} / s[/latex]

Thermal conductivity: [latex]{k}=0.0263 W / m ^{\circ} C[/latex]

Prandtl number: [latex]P r=0.7[/latex]

Isobaric expansion coefficient: [latex]\beta=\frac{1}{300}=0.0033 K ^{-1}[/latex]

Both the Rayleigh number and the Nusselt number are based on [latex]D_{2}=2 r_{2}=2 × 0.075 = 0.15 m[/latex]. The Rayleigh number is then calculated as

[latex]R a_{D_{2}}=\frac{9.81 imes 0.0033(315-300) 0.15^{3}}{\left(15.89 imes 10^{-6} ight)^{2}} 0.7=4.59 imes 10^{6}[/latex]

Flow is laminar and we use Eq. 16.59 to get the average Nusselt number as

[latex]\overline{N u}_{D}=0.53 R a_{D}^{\frac{1}{4}}, 10^{4}

[latex]\overline{N u}_{D_{2}}=0.53 imes\left(4.59 imes 10^{6} ight)^{\frac{1}{4}}=24.53[/latex]

The convective heat transfer coefficient is then given by

[latex]\bar{h}=\frac{\overline{N u}_{D_{2}} k}{D_{2}}=\frac{24.53 imes 0.0263}{0.15}=4.3 W / m ^{2 \circ} C[/latex]

Using resistance concept, we may calculate a better value of [latex]T_{2}[/latex] as

[latex]T_{2}=\frac{T_{1}+T_{\infty} \frac{\bar{h} r_{2}}{k_{i}} \ln \left(\frac{r_{2}}{r_{1}} ight)}{1+\frac{\bar{h} r_{2}}{k_{i}} \ln \left(\frac{r_{2}}{r_{1}} ight)}=\frac{350+300 \frac{4.3 imes 0.075}{0.1} \ln \left(\frac{0.075}{0.025} ight)}{1+\frac{4.3 imes 0.075}{0.1} \ln \left(\frac{0.075}{0.025} ight)}=311 K[/latex]

We may run through the calculation with this value to get the following:

[latex]R a_{D_{2}}=3.366 imes 10^{6}[/latex] , [latex]\overline{N u}_{D_{2}}=22.7[/latex] , [latex]\bar{h}=3.98 ext { W.m }^{2}{ }^{\circ} C[/latex]

A better value for [latex]T_{2}[/latex] is then obtained as [latex]T_{2}[/latex] = 311.7K, at which point we stop the iteration. The corresponding heat loss per meter of pipe is given by

[latex]Q=\pi D_{2} \bar{h}\left(T_{2}-T_{\infty} ight)=\pi imes 0.15 imes 3.98(311.7-300)=21.9 W[/latex]

Question 16.7: Consider a horizontal thin-walled metal pipe 5 cm OD surroun...

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