Study Question Set 04

Ground Heat Flux

The specific heat of water is $c_p = 4.18 kJ kg^{-1} K^{-1}$. Calculate the heat capacity $C$ of water.

Tip

$C = \rho c_p$, where $\rho$ is the density of the material. For water we know that $\rho = 1 \times 10^3\, \rm{kg}\, \rm{m}^{-3}$ (i.e. $1\,\rm{kg} = 1 \ell = 10 \times 10 \times 10\, \rm{cm}$), therefore $4.180 \times 10^{3}\, \rm{J}\, \rm{kg}^{-1}$ K$^{-1}$ $\times$ $10^3\, \rm{kg} \, \rm{m}^{-3}$ = .

Calculate the heat capacity $C$ of a dry mineral soil with a porosity of 55%. Use values from the table in Lecture.

Tip

$P$ = 55% for a dry soil means $\theta_a = 0.55$ and $\theta_{m} = 1 - \theta_a = 0.45$: \[ \begin{eqnarray*} C &=& \theta_m\,C_m + \theta_a\,C_a \\ &\approx& \theta_m\,C_m \\ &=& 0.45 \times 2.1 \,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1} \\ &=& \underline{0.945\,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1}} \end{eqnarray*} \]

Subscripts $a$ and $m$ refer to air and mineral matter, respectively. Note, the second term is small compared to the first one ($\theta_a\,C_a = 0.45 \times 0.0012 \,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1} = 0.00066 \,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1}$) and can be neglected.

Calculate the heat capacity $C$ for the same soil if it is completely saturated.

Tip

$P$ = 55% for the saturated case means $\theta_w = 0.55$ and $\theta_{m} = 1 - \theta_w = 0.45$:

\[ \begin{eqnarray*} C &=& \theta_m\,C_m + \theta_w\,C_w \\ &=& 0.45 \times 2.1 \,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1} + 0.55 \times 4.18 \,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1} \\ &=& 0.945\,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1} + 2.299\,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1} \\ &=& \underline{3.24\,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1}} \end{eqnarray*} \]

Calculate the heat capacity $C$ of a partly saturated soil with $P = 50\%$, $\theta_a = 0.30$ and an organic to mineral ratio of 1.5. Again, use the table in Lecture.

Tip

$P = 50\%$ and $\theta_a = 0.30$ means $\theta_{w} = P - \theta_a = 0.20$. An organic to mineral ratio of 1.5 (3/2) means $(1-P) = 0.5$ is made up of 0.3 $\theta_{o}$ and 0.2 $\theta_{m}$:

\[ \begin{eqnarray*} C &=& \theta_m\,C_m + \theta_o\,C_o + \theta_w\,C_w \\ &=& 0.3 \times 2.{5}\,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1} + 0.2 \times 2.{1} \,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1} \\ && + 0.2 \times 4.18 \,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1} \\ &=& \underline{{2.0}\,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1}} \end{eqnarray*} \]

If you increase the soil volumetric water content $\theta_w$ of any soil by 0.1, how does the heat capacity $C$ of the soil change?

Tip

$\Delta \theta_w = 0.1$:

\[ \begin{eqnarray*} \Delta C &=& \Delta \theta_w\,C_w \\ &=& 0.1 \times 4.18 \,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1} \\ &=& \underline{0.418\,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1}} \end{eqnarray*} \]

$C$ of the soil will increase by $0.418\,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1}$.

Assume we have a soil with $C = 2\,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1}$ and we measure a soil heat flux density $Q_G$ of $+100\,\rm{W}\,\rm{m}^{-2}$ all going to the first 10 cm of the soil, how fast would the layer 0-10 cm heat up?

Tip

The warming rate of a material is defined by:

\[ \begin{eqnarray*} \frac{\Delta T}{\Delta t} = \frac{1}{C}\, \frac{\Delta Q_G}{\Delta z} \end{eqnarray*} \]

$Q_G$ at 0 cm depth (surface) is $+100\,\rm{W}\,\rm{m}^{-3}$, $Q_G$ at 10 cm must be zero because there is no energy distributed to lower layers, i.e. only topmost 10 cm experience heating, hence $\Delta Q_G = 100\,\rm{W}\,\rm{m}^{-2} - 0\,\rm{W}\,\rm{m}^{-2} = 100\,\rm{W}\,\rm{m}^{-2}$ (same as $\,\rm{J}\,\rm{s}^{-1}\,\rm{m}^{-2}$):

\[ \begin{eqnarray*} \frac{\Delta T}{\Delta t} = \frac{100\,\rm{J}\,\rm{s}^{-1}\,\rm{m}^{-2}}{2 \,\rm{MJ}\,\rm{m}^{-3}\,\rm{K}^{-1} \times 0.1 \rm{m}} = 0.0005 \,\rm{K}\,\rm{s}^{-1} = \underline{1.8 \,\rm{K}\,\rm{h}^{-1}}. \end{eqnarray*} \]

Write Fourier’s law and explain it briefly.

Tip

Fourier’s Law: $Q_{G} = -k\,\Delta T/\Delta z = -k(T_{2}-T_{1})/(z_{2}-z_{1})$. It states that the flow rate of heat conducted through a solid material (or still fluid) is proportional to the temperature gradient.

In a dry and uniform mineral soil with a porosity of 55%, we measure soil temperatures $T_1$ at 2 cm and $T_2$ at 6 cm. $T_1 = 20 \rm{^{\circ}C}$, $T_2 = 18.5 \rm{^{\circ}C}$. Calculate the soil heat flux density $Q_G$ assuming a thermal conductivity of $k = 0.27\,\rm{W}\,\rm{m}^{-1}\,\rm{K}^{-1}$.

Tip

Use Fourier’s Law:

\[ \begin{eqnarray*} Q_G = - k\, \frac{\Delta T}{\Delta z} \end{eqnarray*} \]

You insert the thermal conductivity of $k = 0.27\,\rm{W}\,\rm{m}^{-1}\,\rm{K}^{-1}$:

\[ \begin{eqnarray*} Q_G &=& - 0.27\,\rm{W}\,\rm{m}^{-1}\,\rm{K}^{-1}\, \frac{20\rm{^{\circ}C}-18.5\rm{^{\circ}C}}{0.02\,\rm{m} - 0.06\,\rm{m}} \\ &=& - 0.27\,\rm{W}\,\rm{m}^{-1}\,\rm{K}^{-1}\, \frac{1.5\rm{K}}{-0.04\,\rm{m}} \\ &=& \underline{10.1\,\rm{W}\,\rm{m}^{-2}} \end{eqnarray*} \]

At 5 cm depth we measure a soil heat flux density $Q_G = 20 \,\rm{W}\,\rm{m}^{-2}$ and simultaneously a temperature gradient of -0.5 K$\,\rm{cm}^{-1}$. Calculate the thermal conductivity $k$.

Tip

Again we use Fourier’s Law, but rearranged:

\[ \begin{eqnarray*} - Q_G \frac{\Delta z}{\Delta T} = k \end{eqnarray*} \]

We can directly plug-in the inverse of the gradient $\Delta T/\Delta z = -0.5\, \rm{K}\,\rm{cm}^{-1}$ into $\Delta z/\Delta T$: \[ \begin{eqnarray*} k &=& -20\,\rm{W}\,\rm{m}^{-2} \times -0.02\, \rm{K}\,\rm{m}^{-1} \\ &=& \underline{\textcolor{red}{+} 0.4\,\rm{W}\,\rm{m}^{-1}\,\rm{K}^{-1}} \end{eqnarray*} \]

For a soil with a specific heat $c_p =$ 1.8 kJ kg$^{-1}$ K$^{-1}$, a density $\rho =$ 1.4 Mg m$^{-3}$, and a thermal conductivity $k = 0.4\,\rm{W}\,\rm{m}^{-1}\,\rm{K}^{-1}$, calculate the thermal diffusivity $K$.

Tip

The thermal diffusivity $K$ tells us how quickly temperature waves propagate down into the soil, and $K$ is defined by:

\[ \begin{eqnarray*} K &=& \frac{k}{C} = \frac{k}{\rho\,c_p} \\ &=& \frac{0.4\,\rm{J}\,\rm{s}^{-1}\,\rm{m}^{-1}\,\rm{K}^{-1}}{1.4\times 10^{3}\, \rm{kg}\, \rm{m}^{-3}\,1.8 \times 10^{3}\,\rm{J}\,\rm{kg}^{-1}\,\rm{K}^{-1}} \\ &=& \underline{0.16 \times 10^{-6}\,\rm{m}^{2}\,\rm{s}^{-1}} \end{eqnarray*} \]

Note the fine - but important - difference between the symbol $k$ for thermal conductivity (Latin k’) and the symbol $K$ for thermal diffusivity.

Calculate the thermal admittance $\mu$ for the same soil.

Tip

The thermal admittance $\mu$ is strictly speaking a surface property. It defines how well a surface can accept or release heat.

\[ \begin{eqnarray*} \mu &=& \sqrt{k\,C} = \sqrt{k\,\rho\, c_p} \\ &=& \sqrt{0.4\,\rm{J}\,\rm{s}^{-1}\,\rm{m}^{-1}\,\rm{K}^{-1} \times 1.4 \times 10^{3}\, \rm{kg}\, \rm{m}^{-3} \times 1.8 \times 10^{3}\,\rm{J}\,\rm{kg}^{-1}\,\rm{K}^{-1}} \\ &=& \sqrt{1.008 \times 10^{6} \,\rm{J}^{2} \,\rm{m}^{-4} \,\rm{K}^{-2} \,\rm{s}^{-1}} \\ &=& \underline{1004 \,\rm{J} \,\rm{m}^{-2} \,\rm{K}^{-1} \,\rm{s}^{-1/2}} \end{eqnarray*} \]

Assume we know a soil’s dry fraction of organic material ($f_o = $25%), and its bulk density (_s = 1.4 $\,\rm{Mg}\,\rm{m}^{-3}$). What is the mass of organic ($M_o$ ) and mineral material ($M_m$) contained in one cubic metre of this soil?

Tip

$M_m$ and $M_o$ is the mass of mineral and organic material, respectively, in one $\rm{m}^{3}$ of soil.

The total mass of the dry soil $M$ in one cubic-metre is given by the bulk density (_s = 1.4 $\rm{Mg}\,\rm{m}^{-3}$):

\[ M = M_m + M_o = \rho_s \times 1\, \rm{m}^3 = 1.4\,\rm{Mg}\,\rm{m}^{-3} \times 1\, \rm{m}^3 = 1.4\,\rm{Mg} \]

$f_o$ is the organic mass fraction (given: $f_o = 0.25$) which is the mass of organic material to the total mass

\[ f_o = \frac{M_o}{M} = \frac{M_o}{M_o + M_m} = 0.25 \]

solving for $M_m$ and $M_o$:

\[ M_m = M \times (1 - f_o) = 1.4\,\rm{Mg} \times (1 - 0.25) = 1.4\,\rm{Mg} \times) 0.75 = 1.05 \,\rm{Mg} \]

\[ M_o = M \times f_o = 1.4\,\rm{Mg} \times 0.25 = 0.35\,\rm{Mg} \]

Using the values of specific heat for organic ($c_o$) and mineral ($c_m$) given in the table of Lecture 10, slide 5, calculate the composite heat capacity ($C_s$) for the dry soil in Question 12.

Tip

Using the mass of organic and mineral material contained in one cubic metre (determined in Question 12), we can formulate the densities of organic ($\rho_o$) and mineral material ($\rho_m$) in the same soil:

\[ \rho_o = \frac{M_o}{1\rm{m}^{3} \times \theta_o} \]

\[ \rho_m = \frac{M_m}{1\rm{m}^{3} \times \theta_m} \]

Where $(1\rm{m}^{3} \times \theta_o)$ is the volume of organic material in one cubic metre, and $(1\rm{m}^{3} \theta_m)$ is the volume of mineral material in one cubic metre.

Lecture 10, slide 5 provides $c_m = 0.8 \rm{J}\,\rm{kg}^{-1}\,\rm{K}^{-1}$ and $c_o = 1.9 \rm{J}\,\rm{kg}^{-1}\,\rm{K}^{-1}$. Using $C = \rho c$ allows us to then determine the heat capacity of organic ($C_o$) and mineral material ($C_m$) in this soil:

\[ C_o = \rho_o \,c_o \]

\[ C_m = \rho_m \,c_m \]

The composite heat capacity of the dry soil ($C_s$) is the sum of the compound heat capacities weighted by the respective volume fractions:

\[ C_s = \theta_o C_o + \theta_m C_m \]

replacing $C_o$ by $\rho_o \,c_o$ (and same for $C_m$) then gives:

\[ C_s = \theta_o \rho_o \,c_o + \theta_m \rho_m \,c_m \]

replacing $\rho_o$ by $\frac{M_o}{1\rm{m}^{3} \theta_o}$ (and same for $\rho_m$) then gives:

\[ C_s = \theta_o \frac{M_o}{1\rm{m}^{3} \theta_o} \,c_o + \theta_m \frac{M_m}{1\rm{m}^{3} \theta_m} \,c_m \]

Note that then $\theta_o$ and $\theta_m$ cancel out:

\[ C_s = \frac{M_o}{1\rm{m}^{3}} \,c_o + \frac{M_m}{1\rm{m}^{3}} \,c_m \]

Inserting the values:

\[ C_s = \frac{ 0.35\,\rm{Mg}}{1\rm{m}^{3}} \, 1.9\, \rm{J}\,\rm{kg}^{-1}\,\rm{K}^{-1} + \frac{1.05 \,\rm{Mg}}{1\rm{m}^{3}} \,0.8 \, \rm{J}\,\rm{kg}^{-1}\,\rm{K}^{-1} = \]

\[ 0.665 \, \rm{J}\,\rm{m}^{-3}\,\rm{K}^{-1} + 0.84 \, \rm{J}\,\rm{m}^{-3}\,\rm{K}^{-1} = 1.505 \, \rm{J}\,\rm{m}^{-3}\,\rm{K}^{-1} \]

We learn from this exercise that generally we can rewrite the composite heat capacity of a soil ($C_s$) from using heat capacities of compound substances and volume fractions:

\[ C_s = \theta_o C_o + \theta_m C_m + ... \]

to using specific heat and known mass ($M_o$, $M_m$, etc.) of the compound substances in a soil volume $V_s$:

\[ C_s = \frac{M_o}{V_s} \,c_o + \frac{M_o}{V_s} \,c_m + ... \]

where $V_s$ is the volume of the soil, and $M_o$ and $M_m$ is the mass of organic and mineral material in the same volume $V_s$. This has the advantage of avoiding assuming any specific denisty of organic and mineral material, which is difficult to determine practically.