The temperature below the ground is affected by the daily variations of temperature above it. The period of the daily variation may be taken as 24 h. The amplitude of the variation of the temperature at the surface is taken as the unit. The material of the top layers is known to be gravelly sand wi

Question

The temperature below the ground is affected by the daily variations of temperature above it. The period of the daily variation may be taken as 24 h. The amplitude of the variation of the temperature at the surface is taken as the unit. The material of the top layers is known to be gravelly sand with a thermal diffusivity of [latex] 1.403 × 10^{-7} m^{2}/s[/latex] . Determine the depth at which the temperature amplitude is just equal to 5% of the amplitude at the surface. What is the time lag at this depth? Make a plot of the response at a depth of 0.1 m below the surface. Assume that the problem may be treated using the semi-infinite solid model and that there is no internal heat source below the ground.

Accepted Answer

Since the problem involves the time scale of the order of hours the problem is solved using time in hours. The thermal diffusivity needs to be converted to [latex]m^{2}/h[/latex]. This may be done as follows.

[latex]α = 1.403 × 10^{-7} × 3600 = 5.05 × 10^{-4} m^{2}/h[/latex]

Let [latex]x_{5}[/latex]% be the depth at which the amplitude is 5% of that at the ground level. The period of oscillation is given as T = 24 h. The circular frequency of the oscillation ω is then given by

[latex]\omega =\frac{2\pi}{T} =\frac{2\pi }{24}=0.262rad/h[/latex]

We then have

[latex]\frac{T}{T_{s} } =0.05=e^{-\sqrt{\frac{\omega }{2\alpha } }x_{5/100} }[/latex]

Taking natural logarithms, we get

[latex]x_{5/100} =-\frac{\ln 0.05}{\sqrt{\frac{\omega }{2\alpha }}} =-\frac{\ln0.05 }{\sqrt{\frac{0.262}{2 imes 5.05 imes 10^{-4} } } } =0.186m[/latex]

The time lag at this depth is calculated using Eq. 5.46 as

[latex]t^{*} =\frac{1}{ω} \sqrt{\frac{ω}{2\alpha } }x=\frac{x}{\sqrt{2\alpha ω} }[/latex] (5.46)

[latex]t^{*}_{5/100} =\frac{x_{5/100} }{\sqrt{2\alpha ω} } =\frac{0.186}{\sqrt{2 imes 5.05 imes 10^{-4} imes 0.262 } } =11.434h[/latex]

Now consider the state of affairs at a depth of x = 0.1 m. The amplitude at this depth is given by

[latex]\frac{T_{0.1m} }{T_{s} } =e^{-0.1 imes \sqrt{\frac{0.262}{2 imes 5.05 imes 10^{-4}} } } =0.2[/latex]

The time lag at this depth is calculated as

[latex]t^{*}_{0.1m}=\frac{0.1}{\sqrt{2 imes 5.05 imes 10^{-4} imes 0.262} } =6.147h[/latex]

[latex]\frac{T_{0.1m} }{T_{s} }=0.2sin[0.262(t-6.147)][/latex]

where t is in h. This is plotted in Fig. 5.10 along with the input, i.e., the temperature variation at the surface.

Question 5.4: The temperature below the ground is affected by the daily va...

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