CSU ATS601
2.6
Fall 2015
Total energy budget
So far, we have talked about conservation of mass and conservation of momentum. However, what about
conservation of total energy (not just internal)?! This has come into our thermodynamic equation, but what
about total energy? Specifically, we will discuss the total specific energy (i.e. energy per unit mass).
There are in principle three different contributions to the total energy of a fluid:
• kinetic energy: v2 /2
• potential energy:
= gz
• internal energy: I = I(↵, T )
The goal is to derive a general conservation law for the total energy of the fluid (Hint: it is conserved!).
To begin, we obtain an equation for the specific kinetic energy by taking v· of the momentum equation
(ignoring viscosity):
✓
◆
Dv
rp
D(v2 /2)
1
v·
=-r
)
= - (v · r)p - (v · r)
Dt
⇢
Dt
⇢
where the last step uses the trick that
1
D
= - (v · r)p ⇢
Dt
(x, y, z) is not a function of time, and thus,
this term to the left hand side:
D 2
v /2 +
Dt
1
= - (v · r)p
⇢
D
Dt
(2.109)
= (v · r) . Moving
(2.110)
From your high school physics course, you may recall learning that kinetic plus potential energy is conserved, however, this is only true in classic Newtonian mechanics (point masses). In this case, you would
not have the pressure term on the right-hand-side, and you would obtain this conservation law. However, for
a fluid, there is internal energy to be considered - namely, energy could get converted due to pressure work
- that is, the fluid expanding and contracting against a pressure force.
We want to express the pressure term in terms of the internal energy of the system I. Thus, we make use
(once again) of the 1st law of thermodynamics. As a reminder, 2.91 tells us that:
cv
DT
D↵
= Q̇ - p
Dt
Dt
(2.111)
but for an ideal gas with constant heat capacity, I = cv T and so,
D(cv T )
DI
D↵
=
= Q̇ - p
Dt
Dt
Dt
(2.112)
Applying chain rule, the continuity equation and product rule, we obtain:
DI
Dt
E. A. Barnes
D↵
p D⇢
= Q̇ + 2
Dt
⇢ Dt
p
1
= Q̇ - r · v = Q̇ - (r · (pv) - (v · r)p)
⇢
⇢
= Q̇ - p
26
(2.113)
(2.114)
updated 18:50 on Monday 14th September, 2015
CSU ATS601
Fall 2015
From 2.110, we see that we want an equation for the term on the right-hand-side, and that this term appears
in the equation directly above for
DI
Dt .
Thus:
1
1
DI
- (v · r)p = Q̇ - r · (pv) ⇢
⇢
Dt
(2.115)
Plugging this into 2.110 and defining the total specific energy as e = (v2 /2 +
De
1
= Q̇ - r · (pv)
Dt
⇢
or
+ I) leads to
@(⇢e)
+ r · [(⇢e + p)v] = ⇢Q̇
@t
(2.116)
The second equation comes from Vallis 1.186 and requires some extra steps. Integrating over a bounded
domain, say the whole atmosphere, with suitable boundary conditions (such that the divergence term integrates to zero), we get:
DE
@E
=
= Q̇,
Dt
@t
with
Z
E ⌘ (⇢e)dV
and
Z
(2.117)
Q̇ ⌘ (⇢Q̇)dV
V
V
In other words, the only sources or sinks for total energy are diabatic heating or cooling!
The fact that the divergence terms vanish when integrated over the atmosphere is a result of the
divergence theorem. This theorem stats that the volume total of all sinks and sources (the volume
integral of the divergence), is equal to the net flow across the volume’s boundary. Since we are not
allowing things to flow ”out” of the atmosphere, the integral of the divergence must vanish.
2.7
Hydrostatic balance
In-class discussion:
What is the mass of the atmosphere?
Radius of earth: a = 6.37 ⇥ 106 m
Surface area of earth: 4⇡a2 = 5.09 ⇥ 1014 m2
Surface pressure: 1.013 ⇥ 105 Pa or N/m2
Pressure = Mass ⇥ g/Area
Mass = 5.3 ⇥ 1018 kg
What can you infer from the above relation about the time rate of change of the globally integrated
surface pressure? E.g.
Z
Z
d
psfc d d =?
dt
(2.118)
longitude latitude
E. A. Barnes
27
updated 18:50 on Monday 14th September, 2015
CSU ATS601
Fall 2015
Discuss the hydrostatic paradox
Before making our set of equations more complicated by introducing rotation, it is worthwhile to discuss
a basic static balanced result in the non-rotating, Cartesian coordinate system. Static refers to a state where
accelerations are zero (equivalent to no-motion). Balance refers to a balance of forces.
The only force that currently appears in our horizontal momentum equations is the pressure gradient
force:
Du
Dt
Dv
Dt
=
=
@u
@u
1 @p
+ v · ru =
+ u@x u + v@y u + w@z u = @t
@t
⇢ @x
@v
@v
1 @p
+ v · rv =
+ u@x v + v@y v + w@z v = @t
@t
⇢ @y
(2.119)
(2.120)
Thus, under static conditions (acceleration equals zero), we get a rather trivial balance:
Du
=0
Dt
)
@p
=0
@x
and
@p
=0
@y
(2.121)
However, the vertical momentum equation is far more interesting:
Dw
@w
@w
1 @p
=
+ v · rw =
+ u@x w + v@y w + w@z w = -g
Dt
@t
@t
⇢ @z
(2.122)
In this case, we have both the pressure gradient force but also, the force of gravity. Setting the acceleration
equal to zero results in:
@p
= -⇢g
@z
(2.123)
This is the equation for hydrostatic balance (the “hydro” comes from hydraulic pressure), and it tells us that
in the case of no vertical acceleration, the pressure gradient balances gravity.
Constant density fluid: e.g. ocean
In the case of a fluid with constant density (e.g. to good approximation, the ocean), we can integrate
both sides w.r.t. z and obtain:
Z
Z
@p
dz = -⇢gdz
@z
)
p = p0 - ⇢gz
(2.124)
That is, pressure in a liquid decreases linearly with height.
Variable density fluid: e.g. ideal gas, the atmosphere
E. A. Barnes
28
updated 18:50 on Monday 14th September, 2015
CSU ATS601
Fall 2015
For an ideal gas in an isothermal atmosphere (T = T̂ = constant) and applying the ideal gas law we get:
1 @p
p @z
=
g
RT
@(ln p)
g
=@z
RT
)
p
=
p0
Zz
ln p - ln p0 = ln
)
p
g
z - z0
ln
= - (z - z0 ) = ,
p0
Hp
RT̂
z0
)
g
g
dz 0 = RT
R
)
-
Zz
@(ln p) 0
dz =
@z 0
z0
Zz
Zz
-
z0
g
dz 0 (2.125)
RT
1 0
dz
T
z0
with
Hp ⌘ RT̂ /g
(2.126)
(2.127)
Thus,
p = p0 e-(z-z0 )/Hp
(2.128)
This is called the barometric formula, where Hp is known as the pressure scale height. Thus, for an ideal
gas, the pressure decreases exponentially with height (not linearly, as in the case of a fluid).
Of course, the real atmosphere is certainly not isothermal! However, we may approximate T̂ as some
mean temperature (e.g. averaged over the depth of the troposphere). For a typical value of T̂ ⇠ 240K, a
typical scale height is Hp ⇠ 7 km. Similar types (but more complicated) formulas can be obtained for
temperatures that are a linear function of z, etc.
It is hydrostatic balance that relates pressure at a given level to the weight of the fluid above that level:
weight above per unit area ⌘
1
Z
0
(g⇢)dz = -
z
p(1)
Z
0
dp = -
p(z)
Z0
0
dp = -(0 - p(z)) = p(z)
(2.129)
p(z)
Note - a fluid that is not in hydrostatic balance need not obey this relationship! So, you may now wonder,
“how good of an approximation is hydrostatic balance?” And this may lead to yet another question of “how
do we even attempt to answer this question?!” The answer is that we will use scale analysis to try and
determine the importance of hydrostatic balance in the vertical momentum equation. (Note, we will later
see that this scaling is too naive in general).
2.8
2.8.1
Scale analysis (see Vallis 1.11 and 2.7)
Scaling of the vertical momentum equation
Dw
@w
@w
1
=
+ (u · rH )w + w
= - @z p - g
Dt
@t
@z
⇢
(2.130)
Looking first at the left-hand-side,
Dw
@w
@w W UW W 2
=
+ (u · rH )w + w
⇠
+
+
Dt
@t
@z
T
L
H
E. A. Barnes
29
(2.131)
updated 18:50 on Monday 14th September, 2015
CSU ATS601
Fall 2015
where
• rH : the horizontal (2-d) part of the del-operator
• W: scale of vertical wind (w ⇠ W)
• T : scale of time (t ⇠ T )
• L: scale of horizontal distance (x, y ⇠ L ⇠ 1000 km)
• U: scale of horizontal winds (u, v ⇠ U ⇠ 10 m/s)
• H: scale of vertical distance (z ⇠ H ⇠ 10 km)
We further assume that W scales with H and U scales with L, and that T represents an advective time-scale
(meaning it is a distance divided by a velocity; these assumptions are consistent with large-scale extratropical
dynamics). Then,
W H
⇠
U
L
)
W=
HU
,
L
T⇠
L
H
⇠
U W
)
Dw U2 H
⇠ 2 ⇠ 10-6 m/s2
Dt
L
(2.132)
This is then an approximate scale for the left-hand-side of the vertical momentum equation (2.122). On the
other hand, the right-hand-side scales as:
1 @p RTtemp 3 ⇥ 102 ⇥ 3 ⇥ 102
⇠
⇠
⇠ 10m/s2
⇢ @z
H
104
g ⇠ 10m/s2 ,
(2.133)
where we have used the ideal gas law in the form of p/⇢ = RTtemp . These terms are seven orders of
magnitude larger than the vertical acceleration term! Thus, from this simple scaling analysis, we can conclude that at least the basic state can be assumed to exist in hydrostatic balance and that it is an excellent
approximation. However, as indicated above, this scaling is too naive when considering large-scale flows.
We will return to this scaling for hydrostatic balance later on.
2.8.2
Scaling of viscosity (Reynolds number)
We now take a moment to go back to our momentum equations and justify the removal of the viscosity
term for large-scale motions. We will work from the horizontal momentum equations, and assume constant
density:
@u
1
+ (u · r)u = - rp + ⌫r2 u
@t
⇢
In terms of scales,
U
T
E. A. Barnes
U2 RTtemp
⇠
L
L
30
⌫
U
L2
(2.134)
(2.135)
updated 18:50 on Monday 14th September, 2015
CSU ATS601
Fall 2015
where we have used the ideal gas law in the form of p/⇢ = RTtemp , and unlike for the hydrostatic scaling,
the pressure gradients are in the horizontal direction only.
The ratio of the terms on the right-hand-side compared to the viscous term is:
U2 /L
UL
=
2
⌫U/L
⌫
(2.136)
(note that the two terms on the left-hand-side have the same scaling when T is advective, that is, T ⇠ L/U).
Re ⌘
UL
⌫
(2.137)
Re, the Reynolds number tells us how important inertial forces are compared to viscosity. Large values of
the Reynolds number suggest that the inertial forces dominate.
E. A. Barnes
31
updated 18:50 on Monday 14th September, 2015