A cylindrical heating element of diameter 0.1 m and length 2 m is driven tightly into a hole drilled in the ground. Heat is produced in the heater electrically at a rate of 100 W. The heater surface temperature is measured at ten locations along the length of the heater. It is found that the averag

Question

A cylindrical heating element of diameter 0.1 m and length 2 m is driven tightly into a hole drilled in the ground. Heat is produced in the heater electrically at a rate of 100 W. The heater surface temperature is measured at ten locations along the length of the heater. It is found that the average temperature is 84.9°C. The ground surface temperature is measured some distance away from the hole and is found to be 25.5 °C. Estimate the thermal conductivity of the soil.

Accepted Answer

Nomenclature used here is according to that in Fig. 5.30. The given data is written down as

Diameter of heating element: [latex]D = 0.1 m[/latex]
Length of the heater: [latex]L = 2 m[/latex]
Heat generated by the heater: [latex]Q = 100 W[/latex]
The surface temperature of heater: [latex]T_{1} = 84.9 °C[/latex]
The surface temperature of soil: [latex]T_{2} = 25.5 °C[/latex]

We calculate the conduction shape factor using Eq. 5.134 as

[latex]S=\frac{2\pi }{\ln \left(\frac{4L}{D} ight) }[/latex] (5.134)

[latex]S=\frac{2\pi }{\ln \left(\frac{4 imes 2}{0.1} ight) }=1.434[/latex]

Hence the thermal conductivity of the soil is given by Eq. 5.135 as

[latex]k=\frac{Q/L}{S(T_{1} -T_{2} )}[/latex] (5.135)

[latex]k=\frac{100/2}{1.434 imes (84.9 -25.5)} =0.587W/m°C[/latex]

Question 5.10: A cylindrical heating element of diameter 0.1 m and length 2...

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