The shear stress tau is a downward flux of turbulent momentum tau = rho.u*2. The difference in horizontal momentum per unit volume between heights z1 and z2 is rho.u(z2) - rho.u(z1). Now we define the resistance of a layer to momentum transfer rD(z1, z2) to be: the difference in momentum density across the layer divided by the momentum flux across the layer, as follows
rD(z1, z2) = rho.[u(z2) - u(z1)]/tau = [u(z2) - u(z1)]/u*2.
This resistance has units s/m.
In a similar way we obtain the following expression for the thermal resistance rH(z1, z2) of the layer to heat flux to be: the difference in enthalpy density across the layer divided by the heat flux across the layer, as follows
rH(z1, z2) = rho.CP[T(z1) - T(z2)]/H,
where H is the upward heat flux (W/m2).
The evaporative resistance rv(z1, z2) of the layer to the water vapour flux is given by the difference in water vapor density across the layer divided by the evaporative flux across the layer as follows:
rv(z1, z2) = 0.622rho[e(z1) - e(z2)]/PE,
where E is the evaporative flux (kg/m2s).
Since the same turbulent motions produce all these resistances, we assume
rD(z1, z2) = rH(z1, z2) = rv(z1, z2) = [u(z2) - u(z1)]/u*2.
This example shows how to calculate the resistance to turbulent transport using the wind speeds at two heights above the surface. This resistance is then used to calculate the heat flux and the water vapour flux through the boundary layer using the temperatures and the vapour pressures at the two heights. The data for the calculations are given in Table 11.
| Height z | u(z) | T(z) | e(z) |
|---|---|---|---|
| 1 m | 4.0 m/s | 20.5°C | 1.93 kPa |
| 2 m | 4.8 m/s | 19.8°C | 1.87 kPa |
Using the wind speeds we get u* = k[u(z2) - u(z1)]/ln(z2/z1) = 0.462m/s and rD(z1, z2) = [u(z2) - u(z1)]/u*2 = 3.75s/m.
Using this resistance to calculate the heat flux, we get
H = rho.CP[T(z1) - T(z2)]/rH(z1, z2) = [1.2kgm-3×103Jkg-1K-1(20.5°C - 19.8°C)]/3.75sm-1 = 224 W/m2.
Using the same resistance to calculate the evaporation rate, we get
E = [0.622rho[e(z1) - e(z2)]/P.rv(z1, z2) = 0.622×1.2kgm-3(1.93kPa - 1.87kPa)]/100kPa×3.75sm-1 = 1.19×10-4kgm-2s-1.
This is equivalent to 0.43mm liquid water evaporated per hour. Since the latent heat of vaporization of water Lv at 20°C is 2453kJ/kg, this is an energy flux
LvE = 2453kJkg-1×1.19×10-4kgm-2s-1 = 292 W/m2.
Values of the temperature T(z0) and vapor pressure e(z0) at the surface are difficult to measure directly, but they can be estimated from the heat and water vapor fluxes respectively at a given height z.
From rH(z0, z) = rho.CP(T(z0) - T(z))/H we have
T(z0) = T(z) + rH(z0, z).H/rho.CP,
where rH(z0, z) = (u(z) - u(z0))/u*2 = u(z)/u*2, because u(z0) = 0.
Similarly, from rv(z0, z) = 0.622rho[e(z0) - e(z)]/PE we have
e(z0) = e(z) + rv(z0, z)P.E/0.622rho,
where rv(z0, z) = u(z)/u*2.
Using the data in Table 11 we have z = 1m, u(z) = 4.0m/s, u* = 0.462m/s. Therefore
rH(z0, 1m) = rv(z0, 1m) = 4.0ms-1/(0.462ms-1)2 = 18.7sm-1,
which gives
T(z0) = 20.5°C + (18.7sm-1×224Wm-2)/(1.2kgm-3×1005Jkg-1K-1) = 24.0°C.
and
e(z0) = 1.93kPa + (18.7sm-1×100kPa×1.19×10-4kgm-2s-1)/(0.622×1.2kgm-3) = 2.23 kPa.
The total energy leaving the surface by turbulent transport is
H + LvE = 224 W/m2 + 292 W/m2 = 516 W/m2.
This should be the same as the net radiation flux absorbed by the surface, assuming the conduction of heat into the ground below the surface can be neglected.
Bowen's ratio B is defined to be the sensible heat flux divided by the latent heat flux. In this example we have
B = H/LvE = 224Wm-2/292Wm-2 = 0.77,
which is typical for a vegetated surface. Over a water surface B is usually about 0.1, and B = -1 over a water surface when the net radiation is zero. Over dry areas and deserts B is large and positive (in the range 2 to 10).
The drag coefficient CD(z) at height z is defined by:
CD(z) = tau/[rho.u2(z)] = (u*/u(z))2.
It follows from the wind profile equation that
CD(z) = k2/[ln(z/z0)]2.
Therefore, the value of CD(z) at a standard height (e.g. 10m) depends only on the roughness length z0. The drag coefficient is a useful parameter for studying the transfer of heat and water vapour through the boundary layer by turbulence.
At standard height 10m the drag coefficients for different surfaces are:
Rough grass CD(10m) = 0.00335 Field crops CD(10m) = 0.00754 Forest, town CD(10m) = 0.0302
Since the resistance rD(z0, z) between the surface and a standard height z is u(z)/u*2, it follows that the conductance for turbulent transport from the surface to height z is
u*2/u(z) = u(z)CD(z).
Therefore, we can write the heat flux from the surface as
H = rhoCp[T(z0) - T(z)]u(z)CD(z),
and the water vapor flux from the surface as
E = (0.662rho/P)[e(z0) - e(z)]u(z)CD(z).
These expressions are useful for calculating heat and water vapor fluxes for various wind speeds u(z) at height z over a surface with drag coefficient CD(z) when the temperature and the vapor pressure at height z0 are known.
The logarithmic wind profile, and the relations derived from it, apply only in a neutrally stable surface layer.
The kinetic energy per unit volume of air due to vertical turbulent movement is
(KE)turb = (½)rho(w')2 = (½)rho.L2(du/dz)2,
where L is the mixing length. If the air at height z with density rho is carried to height z+L with surrounding air density rho+L[d(rho)/dz], then it is subject to a buoyancy force per unit volume
Fbuoy = L[d(rho)/dz]g,
where g is the acceleration of gravity. Integrating from height z to height z+L we find the kinetic energy per unit volume due to buoyancy at height z+L is
(KE)buoy = (½)L2[d(rho)/dz]g.
But differentiation of the equation of state gives d(rho)/dz = -(rho/T)(dT/dz). Therefore
(KE)buoy = -(½)rho.L2(g/T)(dT/dz) per unit volume.
Richardson's Number Ri is the ratio
Ri = -(KE)buoy/(KE)turb = (g/T)(dT/dz)/(du/dz)2.
The sign of Ri depends on whether the temperature rises or falls with height. Turbulent transport is strongly inhibited by a temperature inversion.
If Ri > 1, turbulent motions are rapidly damped out. Even Ri > 0.2 causes laminar flow in the surface layer. If Ri < -1, free convection is the dominant mechanism of transport. Even Ri < -0.1 is sufficient to amplify turbulence in the boundary layer.
The range within which the theory of turbulence, friction velocity, and the logarithmic wind profile is strictly true is very small, namely
-0.01 < Ri < +0.01,
but the theory of turbulence can be used when Ri is not too far outside this range.
For the data given in Table 11 we have
Ri = (g/T)(dT/dz)/(du/dz)2 = [9.8ms-2(19.8 - 20.5)°Cm-1]/[293K(4.8 - 4.0)ms-1m-1] = -0.04,
so the theory of turbulence is a reasonably good approximation.
By R. H. B. Exell, revised 2004. King Mongkut's University of Technology Thonburi.
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