19.3: Turbulence Statistics
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“If I had only one day left to live, I would live it in my statistics class — it would seem so much longer.”
– Anonymous. [from C.C Gaither (Ed.), 1996: Statistically Speaking: A Dictionary of Quotations, Inst. of Physics Pub., 420 pp].
For air pollutants emitted from a point source such as the top of a smoke stack, mean wind speed and turbulence both affect the pollutant concentration measured downwind at ground level. The mean wind causes pollutant transport. Namely it blows or advects the pollutants from the source to locations downwind. However, while the plume is advecting, turbulent gusts act to spread, or disperse, the pollutants as they mix with the surrounding air. Hence, we need to study both mean and turbulent characteristics of wind in order to predict downwind pollution concentrations.
19.3.1. Review of Basic Definitions
Recall from the Atmospheric Boundary-Layer (ABL) chapter that variables such as velocity components, temperature, and humidity can be split into mean and turbulent parts. For example:
\(\ \begin{align} M=\bar{M}+M^{\prime}\tag{19.1}\end{align}\)
where M is instantaneous speed in this example, \(\ \bar{M}\) is the mean wind speed [usually averaged over time (≈ 30 min) or horizontal distance (≈ 15 km)], and M’ is the instantaneous deviation from the mean value.
The mean wind speed at any height z is
\(\ \begin{align}\bar{M}(z)=\frac{1}{N} \sum_{i=1}^{N} M_{i}(z)\tag{19.2}\end{align}\)
where M_{i} is the wind speed measured at some time or horizontal location index i, and N is the total number of observation times or locations. Use similar definitions for mean wind components \(\ \bar{U}, \bar{V}\), and \(\ \bar{W}\).
Smoke plumes can spread in the vertical direction. Recall from the ABL chapter that the ABL wind speed often varies with height. Hence, the wind speed that affects the pollutant plume must be defined as an average speed over the vertical thickness of the plume.
If the wind speeds at different, equally spaced layers, between the bottom and the top of a smoke plume are known, and if k is the index of any layer, then the average over height is:
\(\ \begin{align} \overline{\bar{M}}=\frac{1}{K} \sum_{k=1}^{K} \bar{M}\left(z_{k}\right)tag{19.3}\end{align}\)
where the sum is over only those layers spanned by the plume. K is the total number of layers in the plume.
This works for nearly horizontal plumes that have known vertical thickness. For the remainder of this chapter, we will use just one overbar (or sometimes no overbar) to represent an average over both time (index i), and vertical plume depth (index k).
The coordinate system is often chosen so that the x-axis is aligned to point in the same direction as the wind vector points, averaged over the whole smoke plume. Thus,
\(\ \begin{align}\bar{M}=\bar{U}\tag{19.4}\end{align}\)
There is no lateral (crosswind) mean wind \(\ (\bar{V} \approx 0)\) in this coordinate system. The mean vertical velocity is quite small, and can usually be neglected \)\ (\bar{W} \approx 0 \), except near hills) compared to plume dispersion rates. However, \(\ u^{\prime}=U-\bar{U}, \quad v^{\prime}=V-\bar{V}\), and \(\ w^{\prime}=W-\bar{W}\) can be non-zero, and are all important.
Recall from the ABL chapter that variance σ_{A}^{2} of any quantity A is defined as
\(\ \begin{align} \sigma_{A}^{2}=\frac{1}{N} \sum_{k=1}^{N}\left(A_{k}-\bar{A}\right)^{2}=\frac{1}{N} \sum_{k=1}^{N}\left(a^{\prime 2}\right)=\overline{a^{\prime 2}}\tag{19.5}\end{align}\)
The square root of the variance is the standard deviation:
\(\ \begin{align}\sigma_{A}=\left(\sigma_{A}^{2}\right)^{1 / 2}\tag{19.6}\end{align}\)
The ABL chapter gives estimates of velocity standard deviations.
Sample Application (§)
Given an x-axis aligned with the mean wind U = 10 m s^{–1}, and the y-axis aligned in the crosswind direction, V. Listed at right are measurements of the V-component of wind.
- Find the V mean wind speed and standard deviation.
- If the vertical standard deviation is 1 m s^{–1}, is the flow isotropic?
t (h) | V (m s^{–1}) |
0.1 | 2 |
0.2 | –1 |
0.3 | 1 |
0.4 | 1 |
0.5 | –3 |
0.6 | –2 |
0.7 | 0 |
0.8 | 2 |
0.9 | –1 |
1.0 | 1 |
Find the Answer
Given: V speeds above, σ_{w} = 1 m s^{–1}, \(\ \bar{U}\) = 10 m s^{–1}
Find: \(\ \bar{V}\) = ? m s^{–1}, σ_{v} = ? m s^{–1}, isotropic (yes/no) ?
Assume V wind is constant with height.
Use eq. (19.2), except for V instead of M:
\(\ \bar{V}(z)=\frac{1}{N} \sum_{i=1}^{N} V_{i}(z)=(1 / 10) \cdot(0)=\underline{\mathbf{0ms^{-1}}}\)
Use eq. (19.5), but for V: \(\ \sigma_{V}^{2}=\frac{1}{N} \sum_{k=1}^{N}\left(V_{k}-\bar{V}\right)^{2}\)
= (1/10) · [4 + 1 + 1 + 1 + 9 + 4 + 0 + 4 + 1 + 1] = 2.6 m^{2} s^{–2}
Use eq. (19.6)
\(\ \sigma_{v}=\left[2.6 \mathrm{m}^{2} \mathrm{s}^{-2}\right]^{1 / 2}=\underline{\mathbf{1 .61ms^{-1}}} \)
Use eq. (19.7): ( σ_{v} = 1.61 m s^{–1}) > ( σ_{w} = 1.0 m s^{–1}), therefore Anisotropic in the y-z plane (but no info on σ_{u} here).
Check: Units OK. Physics OK.
Exposition: The sigma values indicate the rate of plume spread. In this example is greater spread in the crosswind direction than in the vertical direction, hence dispersion looks like the “statically stable” case plotted in Fig. 19.2. By looking at the spread of a smoke plume, you can estimate the static stability
19.3.2. Isotropy (again)
Recall from the ABL chapter that turbulence is said to be isotropic when:
\(\ \begin{align} \sigma_{u}^{2} \approx \sigma_{v}^{2} \approx \sigma_{w}^{2}\tag{19.7}\end{align}\)
As will be shown later, the rate of smoke dispersion depends on the velocity variance. Thus, if turbulence is isotropic, then a smoke puff would tend to expand isotropically, as a sphere; namely, it would expand equally in all directions.
There are many situations where turbulence is anisotropic (not isotropic). During the daytime over bare land, rising thermals create stronger vertical motions than horizontal. Hence, a smoke puff would loop up and down and disperse more in the vertical. At night, vertical motions are very weak, while horizontal motions can be larger. This causes smoke puffs to fan out horizontally at night, and for other stable cases.
Similar effects operate on smoke plumes formed from continuous emissions. For this situation, only the vertical and lateral velocity variances are relevant. Fig. 19.2 illustrates how isotropy and anisotropy affect average smoke plume cross sections.
19.3.3. Pasquill-Gifford (PG) Turbulence Types
Table 19-2a. Pasquill-Gifford turbulence types for Daytime. M is wind speed at z = 10 m. | |||
M (m s^{–1}) | Insolation (incoming solar radiation) | ||
---|---|---|---|
Strong | Moderate | Weak | |
< 2 2 to 3 3 to 4 4 to 6 > 6 |
A A to B B C C |
A to B B B to C C to D D |
B C C D D |
Table 19-2b. Pasquill-Gifford turbulence types for Nighttime. M is wind speed at z = 10 m. | ||
M (m s^{–1}) | Cloud Coverage | |
---|---|---|
≥ 4/8 low cloud or thin overcast | ≤ 3/8 | |
< 2 2 to 3 3 to 4 4 to 6 > 6 |
G E D D D |
G F E D D |
During weak advection, the nature of convection and turbulence are controlled by the wind speed, incoming solar radiation (insolation), cloud shading, and time of day or night. Pasquill and Gifford (PG) suggested a practical way to estimate the nature of turbulence, based on these forcings.
They used the letters “A” through “F” to denote different turbulence types, as sketched in Fig. 19.3 (reproduced from the ABL chapter). “A” denotes free convection in statically unstable conditions. “D” is forced convection in statically neutral conditions. Type “F” is for statically stable turbulence. Type “G” was added later to indicate the strongly statically stable conditions associated with meandering, wavy plumes in otherwise non-turbulent flow. PG turbulence types can be estimated using Tables 19-2.
Early methods for determining pollutant dispersion utilized a different plume spread equation for each Pasquill-Gifford type. One drawback is that there are only 7 discrete categories (A – G); hence, calculated plume spread would suddenly jump when the PG category changed in response to changing atmospheric conditions.
Newer air pollution models do not use the PG categories, but use the fundamental meteorological conditions (such as shear and buoyant TKE generation, or values of velocity variances that are continuous functions of wind shear and surface heating), which vary smoothly as atmospheric conditions change.
Sample Application
Determine the PG turbulence type during night with 25% cloud cover, and winds of 5 m s^{–1}.
Find the Answer
Given: M = 5 m s^{–1}, clouds = 2/8 .
Find: PG = ?
Use Table 18–2b. PG = “D”
Check: Units OK. Physics OK.
Exposition: As wind speeds increase, the PG category approaches “D” (statically neutral), for both day and night conditions. “D” implies “forced convection”.