A thin-walled pipe of 38 mm OD carrying high-pressure steam at a temperature of 120^◦C is insulated by a high temperature insulating material of thermal conductivity equal to 0.106 W/m^◦C. The insulation is eccentric with an eccentricity of 50 mm and the diameter of the outer surface of the insulat

Question

A thin-walled pipe of 38 mm OD carrying high-pressure steam at a temperature of 120°C is insulated by a high temperature insulating material of thermal conductivity equal to 0.106 W/m°C. The insulation is eccentric with an eccentricity of 50 mm and the diameter of the outer surface of the insulation is 204 mm. Determine the heat loss per meter from steam. The outer surface of the insulation has a mean temperature of 44 °C. If the same heat loss takes place with insulation having no eccentricity, what will be the outer diameter of the insulation? Assume that there is no change in the temperature values.

Accepted Answer

Case (a) Eccentric insulation:
Using the notation of Fig. 5.29, we have

[latex]D = 0.204 m, d = 0.038 m, z = 0.05 m[/latex]

The conduction shape factor is calculated, using Eq. 5.131 as

[latex]S=\frac{2\pi }{\cos h^{-1} \left[\frac{D^{2} +d^{2}-4z^{2} }{2Dd} ight] }[/latex] (5.131)

[latex]S=\frac{2\pi }{\cos h^{-1} \left[\frac{0.204^{2} +0.038^{2}-4 imes 0.05^{2} }{2 imes 0.204 imes 0.038} ight] } =4.52[/latex]

The other pertinent quantities are specified as

Inner surface temperature: [latex]T_{1} = 120°C[/latex]
Outer surface temperature: [latex]T_{2} = 44°C[/latex]
Insulation thermal conductivity: [latex]k = 0.106 W/m°C[/latex]

Heat loss from steam per meter length of pipe is then given by

[latex]q = kS(T_{1} − T_{2}) = 0.106 × 4.52 × (120 − 44) = 36.4 W/m[/latex]

Case (b) Insulation layer without eccentricity:
If the concentric insulation loses the same amount of heat per meter of pipe length, the temperatures are the same and the thermal conductivity of the insulation material is the same, then the conduction shape factor has to be the same as in the previous case, i.e., S = 4.52. However, in this case the shape factor is given by Eq. 5.133. Hence the outer diameter of insulation is given by

[latex]S=\frac{2\pi }{\ln \left[\frac{D}{2d} ight] } =\frac{2\pi }{\ln \left[\frac{R}{2r} ight] }[/latex] (5.133)

[latex]D=de^{\frac{2\pi }{S} }=0.038 imes e^{\frac{2\pi }{4.52} }=0.153m[/latex]

Thus there is much saving of insulation material if eccentricity is avoided!

Question 5.9: A thin-walled pipe of 38 mm OD carrying high-pressure steam ...

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