The rate of change of pressure P with height z is:
dP/dz = -rho.g,
where g = 9.8 m/s2, rho = P/RTv, and Tv is the virtual temperature. This gives the hydrostatic equation:
dP/P = (-g/RTv)dz.
Integrating from the surface (z = 0, P = P0) we get the pressure at height z:
P(z) = P0exp(-gz/R<Tv>),
where <Tv> is the mean virtual temperature in the layer.
For dry air in an adiabatic process
CPdT = RTdP/P.
Since dP/P is given by the hydrostatic equation we get for the dry adiabatic lapse rate GAMMA:
dT/dz = -g/CP = -GAMMA = 0.975×10-2 K/m,
or about 1°C per 100m height.
For unsaturated moist air the adiabatic lapse rate is almost the same as the dry adiabatic lapse rate.
A volume V of dry air with density rho has weight rho.Vg. The upward force due to the displacement of surrounding air with density rho' is rho'Vg. Therefore the net buoyant force F per unit mass is:
F = (rho' - rho)Vg/rho.V = (T - T')g/T',
where T and T' are the temperatures of the parcel of air and the surrounding air respectively. For moist air we should use virtual temperatures:
F = (Tv - Tv')g/Tv'.
Let the actual lapse rate of the surrounding air be
gamma = -dT'/dz.
If gamma < GAMMA, then in a vertical displacement of the air we have
-dT'/dz < dT/dz, or dT < dT'.
Suppose initially T = T'. After an upward displacement, T < T' and the buoyant force is downwards. After a downward displacement, T > T' and the buoyant force is upwards. These results show that the air is stable.
If gamma = GAMMA, then the stability of the air is neutral.
If gamma > GAMMA, then the air is unstable.
Let theta' be the potential temperature of the surrounding air. Then the stability criteria can be written:
If d(theta')/dz > 0, then the air is stable.
If d(theta')/dz = 0, then the air is neutral.
If d(theta')/dz < 0, then the air is unstable.
Let GAMMAs be the pseudoadiabatic lapse rate of vertically displaced saturated moist air. It is always true that GAMMAs < GAMMA. The different classes of stability with actual lapse rates gamma are as follows (See Fig. 9):
gamma < GAMMAs: Always stable.
gamma = GAMMAs: Stable when unsaturated, neutral when saturated.
GAMMAs < gamma < GAMMA: Stable when unsaturated, unstable when saturated.
gamma = GAMMA: Neutral when unsaturated, unstable when saturated.
gamma > GAMMA: Always unstable.
Fig. 9. Diagram of lapse rates and classes of stability.
It can be shown that when a thick layer of air is lifted, for example by flow over a mountain, the following processes occur:
(a) The lapse rate in dry air approaches the dry adiabatic lapse rate - stable air becomes less stable, and unstable air becomes less unstable.
(b) The lapse rate in moist air can be made stable, neutral, or unstable, depending on the lapse rate of the wet bulb potential temperature thetaw:
If d(thetaw)/dz > 0, then the air is convectively stable (moisture condenses at the top of the layer first).
If d(thetaw)/dz = 0, then the air is convectively neutral.
If d(thetaw)/dz < 0, then the air is convectively unstable (moisture condenses at the bottom of the layer first).
If the wind is strong, turbulent mixing of the air near the ground makes the surface layer neutrally stable. Then the wind speed u has a vertical profile given by:
du/dz = A/z,
where A is independent of the height z, but depends on u at a standard height and on the nature of the surface (See Fig. 10). Integration gives u(z) = A.ln z + B, where B is a constant chosen to make u = 0 at height z0. The height z0 is called the roughness length of the surface. Now B can be writtten B = A.ln(z0), and the wind profile is written
u(z) = A.ln(z/z0).
Fig. 10. Wind speed profile at the surface.
Table 8 gives examples of roughness lengths for different surfaces. The value of z0 is approximately one tenth of the size of the objects causing the roughness.
|Rough grass||0.01 m|
|Field crops||0.1 m|
|Forest, towns||1.0 m|
Since A is the same at different heights z1 and z2, we have
A = u(z1)/ln(z1/z0) = u(z2)/ln(z2/z0).
Solving for z0 we get
z0 = exp[(u(z2).ln z1 - u(z1).ln z2)/(u(z2) - u(z1))] .
Given: z1 = 1m, z2 = 2m, u(z1) = 4.0m/s, u(z2) = 4.8m/s. Then z0 = 0.031m. This is a typical value for rough grass.
Examples of wind profiles for roughness length 0.1 m and different wind speeds at height 10m are shown in Table 9 and Fig. 11.
|Roughness length z0=0.1m|
Fig. 11. Wind speed profiles for the same roughness length z0=0.1m and different speeds at height 10m.
Examples of wind profiles for different roughness lengths and the same wind speed at height 10m are shown in Table 10 and Fig. 12.
|Wind speed 5m/s at height 10m|
Fig. 12. Wind speed profiles for different roughness lengths and the same wind speed 5m/s at height 10m.
The wind profile equation can be derived using turbulent boundary layer theory. The mixing length L in turbulent flow is the distance a lump of air moves until it mixes with the surrounding air.
Suppose a lump of air at height z+L with horizontal velocity u(z+L) falls to height z and mixes with the surrounding air with horizontal velocity u(z). The momentum lost per unit volume is
rho(u(z+L) - u(z)) = rho.u' = rho.L.du/dz,
where u' is the turbulent part of the velocity.
Observations show that L is proportional to z:
L = kz,
where k is a universal constant (von Karman's constant)
k = 0.40.
Now the wind profile equation becomes
u' = kz.du/dz.
Suppose w' is the downward speed due to turbulence. Then the momentum lost per unit horizontal area per unit time creates a horizontal shearing stress tau (a horizontal force per unit horizontal area) given by:
tau = rho.u'w'.
Since rho.u' is the momentum density of the turbulence, and w' is the downward speed of the momentum density, the shearing stress tau is the downward flux of turbulent momentum.
The friction velocity u* is defined by
tau = rho.u*2.
When the turbulence is the same in both the horizontal direction and the vertical direction we assume
u* = u' = w'.
Now the wind profile equation can be written
du = (u*/k)dz/z.
Comparison of this with the wind profile equation given earlier shows that u*/k = A, and we can write the wind profile equation as:
u(z) = (u*/k)ln(z/z0).
To calculate u*2 from u(z1) and u(z2) we eliminate z0 from the equations:
u(z1) = (u*/k)ln(z1/z0) and u(z2) = (u*/k)ln(z2/z0).
u* = k[u(z2) - u(z1)]/ln(z2/z1),
where k = 0.40.
Given: z1 = 1m, z2 = 2m, u(z1) = 4.0m/s, u(z2) = 4.8m/s. Then u* = 0.462 m/s.
By R. H. B. Exell, revised 2004. King Mongkut's University of Technology Thonburi.
Back to Home Page