250
WILLIAM B. LARGE
evaporate or are suspended and there is effectively a latent cooling of
the lower atmosphere. The significance of this effect remains outstanding
because of the lack of a direct measurement standard.
Three practical ways computing the fluxes are: A) shift the wind,
temperature and humidity to 10m and neutral stability so that neutral
10m coefficients can be used directly, B) shift the coefficients to the
height, and stability of the atmospheric state variables. and C) shift the
temperature and humidity to the height of the wind, z u , then shift the
coefficients to this height and to the atmospheric stability. The details of
(C) follow, but if the atmospheric state variables are given at the same
height, it just becomes equivalent to a particularly efficient version of
(B). This efficiency can also be achieved by shifting the temperature and
humidity to the wind height off-line. The calculations are most efficient
when the wind height, z u , equals the 10m reference height of the transfer
coefficients.
The iterative procedure for (C) is as follows:
1) Assume θ(z u ) = θ(z θ ) and q(z u ) = q(z q ) and compute the virtual
potential temperature, θ v = θ(z u ) (1. + .608q(z u )). Then make a first
guess of neutral stability and U N = |∆
U | to give the transfer coefficients
(Fig. 5). The initial turbulent scales are then computed as:
u
∗ =
ρ
−1
a
| τ | =
C D |∆
U |
θ
∗ =
Q H
ρ a c p u ∗ =
C H
√
C D
[θ(z θ ) − SST ]
q
∗ =
E
ρ a u ∗ =
C E
√
C D
[q(z u ) − q sat (SST )]
(47)
2) Begin the iteration loop with estimates of the stability parameters
ζ u = z u /L, ζ q = z q /L and ζ θ = z θ /L, where L is the Monin-Obukhov
length (22):
ζ(z) =
κ g z
u ∗2
t ∗
θ v
+
q ∗
(q(z q ) + 0.608 −1 )
(48)
For each of the these, use (20) to find the integrals of the dimensionless
flux profiles of momentum, ψ m (ζ), and of heat and moisture, ψ s (ζ).
3) Shift the wind speed to 10m and neutral stability, and the temperature
and humidity to the wind height :
U N (10m) = |∆
U | (1 +
√
C D
κ
[ln(z u /10m) − ψ m (ζ u )] )
−1
(49)
θ(z u ) = θ(z θ ) −
θ ∗
κ
[ln(
z θ
z u
) + ψ h (ζ u ) − ψ h (ζ θ )]
(50)
WILLIAM B. LARGE
evaporate or are suspended and there is effectively a latent cooling of
the lower atmosphere. The significance of this effect remains outstanding
because of the lack of a direct measurement standard.
Three practical ways computing the fluxes are: A) shift the wind,
temperature and humidity to 10m and neutral stability so that neutral
10m coefficients can be used directly, B) shift the coefficients to the
height, and stability of the atmospheric state variables. and C) shift the
temperature and humidity to the height of the wind, z u , then shift the
coefficients to this height and to the atmospheric stability. The details of
(C) follow, but if the atmospheric state variables are given at the same
height, it just becomes equivalent to a particularly efficient version of
(B). This efficiency can also be achieved by shifting the temperature and
humidity to the wind height off-line. The calculations are most efficient
when the wind height, z u , equals the 10m reference height of the transfer
coefficients.
The iterative procedure for (C) is as follows:
1) Assume θ(z u ) = θ(z θ ) and q(z u ) = q(z q ) and compute the virtual
potential temperature, θ v = θ(z u ) (1. + .608q(z u )). Then make a first
guess of neutral stability and U N = |∆
U | to give the transfer coefficients
(Fig. 5). The initial turbulent scales are then computed as:
u
∗ =
ρ
−1
a
| τ | =
C D |∆
U |
θ
∗ =
Q H
ρ a c p u ∗ =
C H
√
C D
[θ(z θ ) − SST ]
q
∗ =
E
ρ a u ∗ =
C E
√
C D
[q(z u ) − q sat (SST )]
(47)
2) Begin the iteration loop with estimates of the stability parameters
ζ u = z u /L, ζ q = z q /L and ζ θ = z θ /L, where L is the Monin-Obukhov
length (22):
ζ(z) =
κ g z
u ∗2
t ∗
θ v
+
q ∗
(q(z q ) + 0.608 −1 )
(48)
For each of the these, use (20) to find the integrals of the dimensionless
flux profiles of momentum, ψ m (ζ), and of heat and moisture, ψ s (ζ).
3) Shift the wind speed to 10m and neutral stability, and the temperature
and humidity to the wind height :
U N (10m) = |∆
U | (1 +
√
C D
κ
[ln(z u /10m) − ψ m (ζ u )] )
−1
(49)
θ(z u ) = θ(z θ ) −
θ ∗
κ
[ln(
z θ
z u
) + ψ h (ζ u ) − ψ h (ζ θ )]
(50)