Study Question Set 06

Turbulence Statistics

  1. During an hour, you measure air temperature \(T\) every 10 minutes according the table below. Calculate the following terms:
Minutes \(T\)
10 \(12.6^{\circ}\textrm{C}\)
20 \(11.2^{\circ}\textrm{C}\)
30 \(11.9^{\circ}\textrm{C}\)
40 \(13.1^{\circ}\textrm{C}\)
50 \(12.0^{\circ}\textrm{C}\)
60 \(11.8^{\circ}\textrm{C}\)
  1. \(\overline{T}\)
  2. \(T^{\prime}\) at 40 min
  3. \(T^{\prime 2}\) at 20 min
  4. \(\overline{T^{\prime}}\)
  5. \(\overline{T^{\prime 2}}\)
  6. \(\overline{T^{\prime}}^2\)

  1. \(\overline{T}\) is the temporal average of \(T\): \[ \begin{eqnarray*} \overline{T} &=& \frac{1}{6} \sum_{t=0}^{5} T(t) \\ &=& \frac{1}{6} \left(12.6 + 11.2 + 11.9 + 13.1 + 12.0 + 11.8 \right) \\ &=& \underline{12.1^{\circ}\textrm{C}} \end{eqnarray*} \]

  2. \(T^{\prime}\) is the deviation of a single measured value from the temporal average of the time series (\(\overline{T}\)). For the time step at 40 min we can write:

\[ \begin{eqnarray*} T(40\textrm{min}) &=& \overline{T^{\prime}} + \overline{T}(40\textrm{min}) \\ T^{\prime}(40\textrm{min}) &=& T(40\textrm{min}) - \overline{T} \\ &=& 13.1^{\circ}\textrm{C} - 12.1^{\circ}\textrm{C} \\ &=& \underline{1\, \textrm{K}}\\ \end{eqnarray*} \]

  1. \(T^{\prime 2}\) is the squared deviation for a a single measured value from the temporal average of the time series. For the time step at 20 min we can write: (\(\overline{T}\)):

\[ \begin{eqnarray*} T^{\prime 2}(20\textrm{min}) &=& (T(20\textrm{min}) - \overline{T})^2 \\ &=& (11.2^{\circ}\textrm{C} - 12.1^{\circ}\textrm{C})^2 \\ &=& (-0.9^{\circ}\textrm{C})^2 \\ &=& \underline{0.81\, \textrm{K}^2}\\ \end{eqnarray*} \]

  1. \(\overline{T^{\prime}}\) is the average deviation in the time series. By definition, this term must be always zero, i.e \(\overline{T^{\prime}}=0\) (or more general \(\overline{a^{\prime}} = 0\), where \(a\) is any parameter or term). Let’s check this:

\[ \begin{eqnarray*} \overline{T^{\prime}} &=& \frac{1}{6} \sum_{t=0}^{5} (T(t)-\overline{T}) \\ &=& \frac{1}{6} \left( (12.6-12.1) + (11.2-12.1) + (11.9-12.1) \right. \\ && \left. + (13.1-12.1) + (12.0-12.1) + (11.8-12.1) \right) \\ &=& \frac{1}{6} \left( 0.5 - 0.9 -0.2 + 1.0 - 0.1 - 0.3 \right) \\ &=& \underline{0\,\textrm{K}} \end{eqnarray*} \]

  1. \(\overline{T^{\prime 2}}\) is the the variance of the time series of \(T\). It is defined as the average of the squared deviations of all time steps. This term is necessarily zero\footnote{Note for those who have taken statistics: You might be used to the , where you divide by \(N-1\), not \(N\):

\[ \begin{eqnarray*} \overline{T^{\prime 2}} &=& \frac{1}{N-1} \sum_{t=0}^{N-1} (T(t)-\overline{T})^2 = \frac{1}{5} \sum_{t=0}^{5} (T(t)-\overline{T})^2 \\ \end{eqnarray*} \]

However, in atmospheric turbulence studies, we prefer the biased variance (divided by \(N\), as shown in the question above and the lecture slides where we divide by \(N\)). The biased variance is a good estimation of the dispersion of a sample of observations, but not necessarily the best measure of the whole population of possible observations.

\[ \begin{eqnarray*} \overline{T^{\prime 2}} &=& \frac{1}{6} \sum_{t=0}^{5} (T(t)-\overline{T})^2 \\ &=& \frac{1}{6} \left( (12.6-12.1)^2 + (11.2-12.1)^2 + (11.9-12.1) ^2\right. \\ && \left. + (13.1-12.1)^2 + (12.0-12.1)^2 + (11.8-12.1)^2 \right) \\ &=& \frac{1}{6} \left( 0.25 + 0.81 + 0.04 + 1 + 0.01 + 0.09 \right) \\ &=& \underline{0.37\,\textrm{K}^2} \end{eqnarray*} \]

  1. \(\overline{T^{\prime}}^2\) is the temporal average of the deviations squared. Note the fine (but essential) difference of the overbar that does not include the square, and because \(\overline{T^{\prime}} = 0\) (see (d)) also its square is zero:

\[ \begin{eqnarray*} \overline{T^{\prime}}^2 &=& \overline{T^{\prime}} \times \overline{T^{\prime}} \\ &=& 0\,\textrm{K} \times 0\,\textrm{K} \\ &=& \underline{0\,\textrm{K}^2} \end{eqnarray*} \]

  1. Simplify the following terms. \(T\) is temperature, \(p\) is pressure, \(q\) is absolute humidity, \(u\), \(v\), \(w\) are the longitudinal, lateral and vertical wind components.

  1. \(\overline{5}\)

  2. \(\overline{8v}\)

  3. \(\overline{\overline{T}\overline{p}}\)

  4. \(\overline{\overline{u}}\)

  5. \(\overline{q^{\prime}}\)

  6. \(\overline{T^{\prime}} \times \overline{w}\)

  7. \(\overline{3q^{\prime}}\)

  8. \(\overline{w^{\prime} \times \overline{u}}\)

  9. \(\overline{\overline{T}p}\)

  10. \(\overline{wT}\)

  1. The average of a constant is equal the constant itself \[ \begin{eqnarray*} \overline{5} = \frac{1}{1}\sum_{i=0}^{0} 5 = 5 \end{eqnarray*} \]

  2. The average of a constant times a value is equal the constant times the averaged value:

\[ \begin{eqnarray*} \overline{8v} = 8 \times \overline{v} \end{eqnarray*} \]

  1. Similarly we can regard temporal averages as constants (they are not changing over time):

\[ \begin{eqnarray*} \overline{\overline{T}\overline{p}} = \overline{T}\overline{p} \end{eqnarray*} \]

  1. The average of an averaged value is equal the averaged value

\[ \begin{eqnarray*} \overline{\overline{u}} = \overline{u} \end{eqnarray*} \]

  1. Similar to 1(d):

\[ \begin{eqnarray*} \overline{q^{\prime}} = 0 \end{eqnarray*} \]

  1. Again, the first term is zero - so is it’s product:

\[ \begin{eqnarray*} \overline{T^{\prime}} \times \overline{w} = 0 \times \overline{w} = 0 \end{eqnarray*} \]

\[ \begin{eqnarray*} \overline{3q^{\prime}} = 3 \times \overline{q^{\prime}} = 3 \times 0 = 0 \end{eqnarray*} \]

\[ \begin{eqnarray*} \overline{w^{\prime} \times \overline{u}} = \overline{w^{\prime}} \times \overline{u} = 0 \times \overline{u} = 0 \end{eqnarray*} \]

  1. Combining the results from (c) and (e):

\[ \begin{eqnarray*} \overline{\overline{T}p} &=& \overline{T} \times \overline{(\overline{p}+p^{\prime})} \\ &=& \overline{T}\overline{p} + \overline{T} \times \underbrace{\overline{p^{\prime}}}_{=0} \\ &=& \overline{T}\overline{p} \end{eqnarray*} \]

  1. In a first step apply Reynold’s decomposition to both, \(w\) and \(T\). One of the resulting terms is a covariance (\(\overline{w^{\prime}T^{\prime}}\)) which is not zero (see lecture):

\[ \begin{eqnarray*} \overline{wT} &=& \overline{(\overline{w}+w^{\prime}) \times (\overline{T}+T^{\prime})} \\ &=& \overline{\overline{w}\overline{T}} + \underbrace{\overline{\overline{w} T^{\prime}}}_{=0} + \underbrace{\overline{w^{\prime}\,\overline{T}}}_{=0} +\overline{w^{\prime}T^{\prime}}\\ &=& \overline{w}\overline{T} + \underbrace{\overline{w^{\prime}T^{\prime}}}_{\textrm{Covariance}} \end{eqnarray*} \]

  1. Calculate the following parameters if \(\overline{u} = 4\, \textrm{m}\, \textrm{s}^{-1}\), \(\overline{v} = 0\, \textrm{m}\, \textrm{s}^{-1}\), \(\overline{w} = 0\, \textrm{m}\, \textrm{s}^{-1}\), \(\sigma_u = 0.4\, \textrm{m}\, \textrm{s}^{-1}\), \(\sigma_v = 0.2\, \textrm{m}\, \textrm{s}^{-1}\), and \(\sigma_w = 0.1\, \textrm{m}\, \textrm{s}^{-1}\).

  1. \(I_u\)

  2. \(I_w\)

  3. \(\overline{w^{\prime 2}}\)

  4. \(\overline{u^{\prime 2}} + \overline{v^{\prime 2}} + \overline{w^{\prime 2}}\)

  5. MKE/m

  6. \(\overline{e}\)

  1. \(I_u\) is the turbulence intensity of the longitudinal wind component \(u\). \(I_u\) has no unit.

\[ \begin{eqnarray*} I_u &=& \sigma_u / \overline{u} \\ &=& 0.4\,\textrm{m}\,\, \textrm{s}^{-1} / 4 \,\textrm{m}\,\, \textrm{s}^{-1} \\ &=& \underline{0.1} \end{eqnarray*} \]

  1. \(I_w\) is the turbulence intensity of the vertical wind component \(w\). \(I_w\) has no unit.

\[ \begin{eqnarray*} I_w &=& \sigma_w / \overline{u} \\ &=& 0.1\,\textrm{m}\,\, \textrm{s}^{-1} / 4 \,\textrm{m}\,\, \textrm{s}^{-1} \\ &=& \underline{0.025} \end{eqnarray*} \]

  1. \(\overline{w^{\prime 2}}\) is the variance of the vertical wind component \(w\). The variance is the square of the standard deviation \(\sigma_w\):

\[ \begin{eqnarray*} \overline{w^{\prime 2}} &=& \sigma_w^2 \\ &=& (0.1 \,\textrm{m}\,\, \textrm{s}^{-1})^2 \\ &=& \underline{0.01 \,\textrm{m}^2\,\, \textrm{s}^{-2}} \end{eqnarray*} \]

\[ \begin{eqnarray*} \overline{u^{\prime 2}} + \overline{v^{\prime 2}} + \overline{w^{\prime 2}} &=& \sigma_u^2 + \sigma_v^2 + \sigma_w^2 \\ &=& (0.4 \,\textrm{m}\,\, \textrm{s}^{-1})^2 + (0.2 \,\textrm{m}\,\, \textrm{s}^{-1})^2 + (0.1 \,\textrm{m}\,\, \textrm{s}^{-1})^2 \\ &=& 0.16 \,\textrm{m}^2\,\, \textrm{s}^{-2} + 0.04 \,\textrm{m}^2\,\, \textrm{s}^{-2} + 0.01 \,\textrm{m}^2\,\, \textrm{s}^{-2} \\ &=& \underline{0.21\, \textrm{m}^2\,\, \textrm{s}^{-2}} \end{eqnarray*} \]

  1. \(MKE / m\) is the mean kinetic energy per unit mass (lecture 12, slide 18), i.e.

\[ \begin{eqnarray*} MKE / m &=& \frac{1}{2}(\overline{u}^2+\overline{v}^2+\overline{w}^2)\\ &=& \frac{1}{2}(4^2+0^2+0^2)\\ &=& \frac{1}{2}(16)\\ &=& \underline{8\, \textrm{m}^2\,\, \textrm{s}^{-2}}\\ \end{eqnarray*} \]

  1. \(\overline{e}\) is the turbulent kinetic energy per unit mass (lecture 12, slide 18) i.e.

\[ \begin{eqnarray*} \overline{e} &=& \frac{1}{2}(\overline{u^{\prime 2}}+\overline{v^{\prime 2}}+\overline{w^{\prime 2}})\\ &=& \frac{1}{2}(\sigma_u^2+\sigma_v^2+\sigma_w^2)\\ &=& \frac{1}{2}\left( (0.4 \,\textrm{m}\,\, \textrm{s}^{-1})^2 + (0.2 \,\textrm{m}\,\, \textrm{s}^{-1})^2 + (0.1 \,\textrm{m}\,\, \textrm{s}^{-1})^2 \right) \\ &=& \frac{1}{2}(0.16 \,\textrm{m}^2\,\, \textrm{s}^{-2} + 0.04 \,\textrm{m}^2\,\, \textrm{s}^{-2} + 0.01 \,\textrm{m}^2\,\, \textrm{s}^{-2}) \\ &=& \underline{0.115\, \textrm{m}^2\,\, \textrm{s}^{-2}} \end{eqnarray*} \]