Geography Reference
In-Depth Information
Thus the net rate at which the pressure force does work due to the x component
of motion is
∂
∂x
(pu)
(pu)
A
−
(pu)
B
δyδz
=−
δV
A
where δV
δxδyδz.
Similarly, we can show that the net rates at which the pressure force does work
due to the y and z components of motion are
=
∂
∂y
(pv)
δV and
∂
∂z
(pw)
δV
−
−
respectively. Hence, the total rate at which work is done by the pressure force is
simply
−
∇·
(p
U
)δV
The only body forces of meteorological significance that act on an element of mass
in the atmosphere are the Coriolis force and gravity. However, because the Coriolis
force,
U
, is perpendicular to the velocity vector, it can do no work. Thus
the rate at which body forces do work on the mass element is just ρ
g
−
2
×
U
δV .
Applying the principle of energy conservation to our Lagrangian control volume
(neglecting effects of molecular viscosity), we thus obtain
·
ρ
e
U
δV
D
Dt
1
2
U
+
·
=−
∇·
(p
U
)δV
+
ρ
g
·
U
δV
+
ρJδV
(2.35)
Here J is the rate of heating per unit mass due to radiation, conduction, and latent
heat release. With the aid of the chain rule of differentiation we can rewrite (2.35) as
e
U
e
U
D (ρδV )
Dt
D
Dt
1
2
U
1
2
U
ρδV
+
·
+
+
·
(2.36)
=−
·∇
−
∇·
−
+
U
pδV
p
U
δV
ρgw δV
ρJ δV
where we have used
g
=−
g
k
. Now from (2.32) the second term on the left in
(2.36) vanishes so that
1
2
U
U
ρ
De
D
Dt
Dt
+
·
=−
·∇
−
∇·
−
+
ρ
U
p
p
U
ρgw
ρJ
(2.37)
This equation can be simplified by noting that if we take the dot product of
U
with
the momentum equation (2.8) we obtain (neglecting friction)
1
2
U
U
D
Dt
ρ
·
=−
U
·∇
p
−
ρgw
(2.38)
