Study Question Set 07

Covariances

(@)You measure a covariance \(\overline{w^{\prime}T^{\prime}} = -0.031\,\rm{m}\,\rm{s}^{-1}\,\rm{K}\). Average air temperature is 10 \(^{\circ}\)C and the density is 1.230,

\[ \begin{eqnarray*} H &=& C_a \, \overline{w^{\prime}T^{\prime}} = \rho_a \, c_p \, \overline{w^{\prime}T^{\prime}} \\ &=& 1.230\,\rm{kg}\,\rm{m}^{-3} \times 1010 \,\rm{J}\,\rm{kg}^{-1}\,\rm{K}^{-1} \times -0.031 \,\rm{m}\,\rm{s}^{-1}\,\rm{K} \\ &=& -38.5 \,\rm{J}\,\rm{m}^{-2}\,\rm{s}^{-1} = \underline{ - 38.5\,\rm{W}\,\rm{m}^{-2}} \end{eqnarray*} \]

This is likely a night-time situation, when the surface is cold and that atmosphere is warmer. The profile of potential temperature is likely increasing with height. Excess sensible heat (i.e. \(\rho_a c_p T\)) is transported from higher levels of the atmosphere down to the layers close to the surface, as indicated by the negative sign of \(H\).

(@)You measure a covariance \(\overline{w^{\prime}\rho_v^{\prime}} = 1.73 \times 10^{-4}\, \rm{kg}\,\rm{m}^{-2}\,\rm{s}^{-1}\). \(\rho_v\) is the water vapour density in \(\rm{kg}\,\rm{m}^{-3}\). Average air temperature is 30 \(^{\circ}\) C. Calculate \(LE\).

Use the latent heat of vaporization for water, \(L_v\), form Lecture. Rounding to the nearest whole number, we’ll get:

\[ \begin{eqnarray*} LE &=& L_v \, \overline{w^{\prime}\rho_v^{\prime}} \\ &=& 2427 \times 10^{6} \,\rm{J}\,\rm{kg}^{-1} \times 1.73 \times 10^{-4}\, \rm{kg}\,\rm{m}^{-2}\,\rm{s}^{-1} \\ &=& 420 \,\rm{J}\,\rm{m}^{-2}\,\rm{s}^{-1} = \underline{420\,\rm{W}\,\rm{m}^{-2}} \end{eqnarray*} \]

(@)Determine the Bowen ratio \(\beta\) if \(\overline{w^{\prime}T^{\prime}} = -0.121\,\rm{m}\,\rm{s}^{-1}\,\rm{K}\) and \(\overline{w^{\prime}\rho_v^{\prime}} = 1.21 \times 10^{-4}\, \rm{kg}\,\rm{m}^{-2}\,\rm{s}^{-1}\). Average air temperature is 20 \(^{\circ}\) C and density is 1.188 \(\rm{kg}\,\rm{m}^{-3}\) .

The bowen ratio is \(\beta = H / LE\). Analogous to Question 1, calculate \(H\):

\[ \begin{eqnarray*} H &=& \rho_a \, c_p \, \overline{w^{\prime}T^{\prime}} \\ &=& 1.188\,\rm{kg}\,\rm{m}^{-3} \times 1010 \,\rm{J}\,\rm{kg}^{-1}\,\rm{K}^{-1} \times 0.121 \,\rm{m}\,\rm{s}^{-1}\,\rm{K} \\ &=& 145.1 \,\rm{W}\,\rm{m}^{-2} \end{eqnarray*} \]

Analogous to Question 2, calculate \(LE\):

\[ \begin{eqnarray*} LE &=& L_v \, \overline{w^{\prime}\rho_v^{\prime}} \\ &=& 2.453 \times 10^{6} \,\rm{J}\,\rm{kg}^{-1} \times 1.21 \times 10^{-4}\, \rm{kg}\,\rm{m}^{-2}\,\rm{s}^{-1} \\ &=& 296.8 \,\rm{W}\,\rm{m}^{-2} \end{eqnarray*} \]

Hence

\[ \begin{eqnarray*} \beta &=& \frac{H}{LE} = \frac{145.1\,\rm{W}\,\rm{m}^{-2}}{296.8\,\rm{W}\,\rm{m}^{-2}} = \underline{0.49} \end{eqnarray*} \]

(@)Given is \(LE\) = 240 W m \(^{-2}\) at 20 \(^{\circ}\) C air temperature. Determine the covariance \(\overline{w^{\prime}q^{\prime}}\), where \(q\) is the specific humidity (in g water vapour per kg air ,i.e. g kg \(^{-1}\)).

2452 ::: {.callout-tip collapse=“true”} The relation between specific humidity \(q\) and water vapour density \(\rho_v\) is: \(\rho_v = \rho_a q\), where \(\rho_a\) is the density of the (moist) air. Because \(\rho_a\) is not changing significantly it can be taken out of the averaging operator:

\[ \begin{eqnarray*} LE &=& L_v \, \overline{w^{\prime}\rho_v^{\prime}} \\ &=& \, L_v\, \overline{w^{\prime}(\rho_a q)^{\prime}} \\ &=& \rho_a \, L_v\, \overline{w^{\prime}q^{\prime}} \end{eqnarray*} \]

rearrange:

\[ \begin{eqnarray*} \overline{w^{\prime}q^{\prime}} &=& \frac{LE}{\rho_a \, L_v}\\ &=& \frac{240\,\rm{J}\,\rm{s}^{-1}\,\rm{m}^{-2}}{1.188\,\rm{kg}\,\rm{m}^{-3} \times 2.453 \times 10^{6} \,\rm{J}\,\rm{kg}^{-1} } \\ &=& 8.24 \times 10^{-5}\,\rm{m}\,\rm{s}^{-1}\,\rm{kg}\,\rm{kg}^{-1} \\ &=& \underline{0.0824\,\rm{m}\,\rm{s}^{-1}\,\rm{g}\,\rm{kg}^{-1}} \end{eqnarray*} \]

:::

(@)Over a rice paddy you measure a covariance between vertical wind and methane concentration \(\rho_{\rm{CH}_4}\) in \(\rm{\mu g}\,\rm{m}^{-3}\) of \(\overline{w^{\prime}\rho_{\rm{CH}_4}^{\prime}} = 10 \,\rm{m}\,\rm{s}^{-1}\,\rm{\mu g}\,\rm{m}^{-3}\,\). Determine the mass flux density between surface and atmosphere.

This covariance is already a mass flux density, because its units are mass per square meter and per second if sorted properly:

\[ \begin{eqnarray*} F_{\rm{CH}_4} &=& \overline{w^{\prime}\rho_{\rm{CH}_4}^{\prime}} \\ &=& 10 \,\rm{m}\,\rm{s}^{-1}\,\rm{\mu g}\,\rm{m}^{-3}\\ &=& \underline{10 \,\rm{\mu g}\,\rm{m}^{-2}\,\rm{s}^{-1}}\\ \end{eqnarray*} \]