Study Question Set 05

More on Heat Fluxes

1. At 11:30 in the morning, we measure a soil heat flux density \(H_{G(\rm{5cm})}\) of \(25\,\rm{W}\,\rm{m}^{-2}\) using a heat flux plate installed at 5 cm depth. Calculate the soil heat flux density at the surface \(H_{G(\rm{0})}\), if the soil’s heat capacity in the layer from 0 to 5 cm depth is \(2 \,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1}\) and the temperature in the same layer changed from 24.8\(^{\circ}\rm{C}\) at 11:00 to 25.3\(^{\circ}\rm{C}\) at 12:00.

Tip

Assume a uniform and linear warming rate of the soil: \[ \frac{\Delta T}{\Delta t} = \frac{25.3^{\circ}\rm{C} - 24.8^{\circ}\rm{C}}{1\,\rm{h}} \\ = 0.5\rm{K}\,\rm{h}^{-1} = 1.38 \times 10^{-4}\rm{K}\,\rm{s}^{-1} \]

Rearranging from the Lecture 8, equation 6, we can estimate heat flux at the surface \(H_{G(\rm{0})}\):

\[ \begin{align} H_{G(\rm{0})} &= H_{G(\rm{5cm})} + C\, \frac{\Delta T}{\Delta t} \Delta z \\ &= 25,\rm{W}\,\rm{m}^{-2} + 2 \,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1}\, \times 1.38 \times 10^{-4}\rm{K}\,\rm{s}^{-1} \times 0.05\,\rm{m} \\ &= 25\,\rm{W}\,\rm{m}^{-2} + 13.8\,\rm{W}\,\rm{m}^{-2} \\ &= \underline{38.8\,\rm{W}\,\rm{m}^{-2}} \\ \end{align} \]

2. For the same soil, at 20:30 in the evening, we measure a soil heat flux density \(H_{G(\rm{5cm})}\) of \(-12\,\rm{W}\,\rm{m}^{-2}\). Calculate the soil heat flux density at the surface \(H_{G(\rm{0})}\), if the temperature in the layer from 0 to 5 cm depth changed from 7.5 at 20:00 to 7.0\(^{\circ}\rm{C}\) at 21:00.

Tip

Now we have a cooling rate (i.e. negative change of temperature over time): \[ \frac{\Delta T}{\Delta t} = \frac{7.0^{\circ}\rm{C} - 7.5^{\circ}\rm{C}}{1\,\rm{h}} \\ = -0.5\rm{K}\,\rm{h}^{-1} = -1.38 \times 10^{-4}\rm{K}\,\rm{s}^{-1} \]

Similar to the first example, the heat flux at the surface is: \[ \begin{align} H_{G(\rm{0})} &= H_{G(\rm{5cm})} + C\, \frac{\Delta T}{\Delta t} \Delta z \\ &= -12\,\rm{W}\,\rm{m}^{-2} + 2 \,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1}\, \times \left( -1.38 \times 10^{-4}\rm{K}\,\rm{s}^{-1} \right) \times 0.05\,\rm{m} \\ &= -12\,\rm{W}\,\rm{m}^{-2} + (-13.88\,\rm{W}\,\rm{m}^{-2}) \\ &= \underline{-25.9\,\rm{W}\,\rm{m}^{-2}} \\ \end{align} \]

3. What is meant by heat sharing?

Tip

Heat sharing refers to how the soil and the atmosphere (or two other materials) share in accepting sensible heat (i.e. \(R_n\) - \(LE\)) during the daytime and share in releasing it at night. Heat sharing is determined by the ratio of the thermal admittances of the two materials (Lecture 11, slides 13): \[ \frac{H}{H_g} = \frac{\mu_a}{\mu_s} \]

where \(\mu_s\) is the thermal admittance of the soil and \(\mu_a\) is the thermal admittance of the atmosphere.

4. Calculate the sensible heat flux \(H\) at 11:30 for the example in Question 1, if the soil’s thermal conductivity is \(k = 0.27\,\rm{W}\,\rm{m}^{-1}\,\rm{K}^{-1}\) and the atmospheric thermal admittance \(\mu_a\) is \(\approx 5000 \,\rm{J} \,\rm{m}^{-2} \,\rm{K}^{-1} \,\rm{s}^{-1/2}\).

Tip

Rearrange the heat sharing equation from above:

\[ H = H_G \times \frac{\mu_a}{\mu_s} \]

We can calculate \(\mu_s\) by:

\[ \mu_s = \sqrt{k\,C} \\ = \sqrt{0.27\,\rm{W}\,\rm{m}^{-1}\,\rm{K}^{-1} \times 2 \times 10^{6} \,\rm{J}\,\rm{m}^{-3}\,\rm{K}^{-1}} \\ = 735 \,\rm{J} \,\rm{m}^{-2} \,\rm{K}^{-1} \,\rm{s}^{-1/2} \]

Hence:

\[ H = H_G \times \frac{\mu_a}{\mu_s} \\ = 38.8\,\rm{W}\,\rm{m}^{-2} \times \frac{5000 \,\rm{J} \,\rm{m}^{-2} \,\rm{K}^{-1} \,\rm{s}^{-1/2}}{735 \,\rm{J} \,\rm{m}^{-2} \,\rm{K}^{-1} \,\rm{s}^{-1/2}} \\ = \underline{283\,\rm{W}\,\rm{m}^{-2}} \]