Conservation of mass
If we imagine a volume of fluid in a basin, we can make a statement about the
change in mass that might occur if we add or remove fluid – it would be…
M w
 inflow rate – outflow rate
t
Now consider a small cube, fixed in space, through which air is flowing

The x direction mass flux is given by u kg m2 s 1

Using a Taylor series expansion about
the center point, the rate of inflow
through side A would be:

  x 

u

 u yz

x 
2 

and through side B would be:

  x 

u

 u yz

x 
2 

As with the basin, the rate of accumulation of mass in the cube must be:

M x 
  x 
  x 
  u   u yz   u   u yz
t
x 
2 
x 
2 


M x

 u xyz
t
x
M x

 u xyz
t
x
Similarly:
M y

 v xyz
t
y
M z

 wxyz
t
z
The net mass accumulation in the cube
would be:


M


  u   v   wxyz
t
y
z
 x

Dividing by the volume xyz





  u   v   w
t
y
z
 x






  u   v   w
t
y
z
 x

This equation can be written as:
 


   V
t
The flux form of the mass
continuity equation
Using the vector identity:
 

 
  V   V  V  
this equation can be written as:

 
 V      V  0
t
or:

1 d
  V  0
 dt
The velocity divergence
form of the mass
continuity equation
 


   V
t
The flux form of the mass
continuity equation

1 d
  V  0
 dt
The velocity divergence
form of the mass
continuity equation
For the special case of an incompressible fluid, density doesn’t change
following parcel motion so the continuity equation reduces to:

 V  0
The incompressible form
of the mass continuity
equation
Finally, let’s derive the mass continuity equation in pressure coordinates…
Consider a volume of air:
Using the hydrostatic equation
We can write our volume as:
and the mass of the parcel as:
V  xyz
p
  g
z
p
 z
 g
or
xyp
V  
g
M  V  
xyp
g
If we follow parcel motion, the mass of parcel should be conserved:
d
d  xyp 

M   0   
dt
dt 
g 
d
d  xyp 

M   0   
dt
dt 
g 
Divide both sides by
M
1 d
M   0   g d   xyp 
M dt
xyp dt 
g 
Apply chain rule to RHS
1  d x 
d y 
d p 


y

p


x

p


x

y0


xyp  dt
dt
dt

Recalling that
d x 
 u,
dt
d y 
 v,
dt
d p 
 
dt
The equation can be reduced, in the limit that the volume approaches 0, to be
u v 
 
0
x y p
The pressure coordinate form
of the mass continuity equation
The various forms of the mass continuity equation so far
 


   V
t
The flux form of the mass
continuity equation

1 d
  V  0
 dt
The velocity divergence
form of the mass
continuity equation

 V  0
The incompressible form
of the mass continuity
equation
u v 
 
0
x y p
The pressure coordinate form
of the mass continuity equation
Finally, substituting the geostrophic and ageostrophic wind:
u v 
 
0
x y p
u g  uag  vg  vag  


0
x
y
p
ug  
1 
f y
1 
vg 
f x
1  2 1  2 uag vag 





0
f xy f xy x
y
p
uag
vag



0
x
y
p


  Vag 
0
p
The ageostrophic wind form of
the mass continuity equation
The various forms of the mass continuity equation
 


   V
t
The flux form of the mass
continuity equation

1 d
  V  0
 dt
The velocity divergence
form of the mass
continuity equation

 V  0
The incompressible form
of the mass continuity
equation
u v 
 
0
x y p


  Vag 
0
p
The pressure coordinate form
of the mass continuity equation
The ageostrophic wind form of
the mass continuity equation
The pressure coordinate form is particularly enlightening
u v 
 
0
x y p
We can integrate this equation between the top of a column and the surface
 Vh t
t

  Vh p    


s
 Vh s


  Vh
 
Pt

   Vh

Ps
 t  s 


  Vh
 
Pt

   Vh

Ps
 t  s 
If there is no vertical air motion at the surface and there is convergence,
air will rise, a direct outcome of mass continuity. Vertical motion and
column divergence patterns are directly related.
Conservation of energy
We will consider mechanical energy, thermal energy, and total energy
The complete momentum equations
du uv tan uw
1 p
 2u



K 2
 2v sin   2w cos
dt
a
a
 x
z
dv u 2 tan vw
1 p
 2v



K 2
 2u sin 
dt
a
a
 y
z
dw u 2  v 2
1 p


 g  2u cos
dt
a
 z
Multiply (1) by u, (2) by v, and (3) by w to get energy equations
1 du2 u 2v tan u 2 w
u p
 2u



 uK 2
2 dt
a
a
 x
z
1 dv2 u 2v tan v 2 w
v p
 2v



 vK 2
2 dt
a
a
 y
z
1 dw2 wu 2  wv 2

2 dt
a

w p
 z
 2uv sin   2uwcos
 v2u sin 
 wg  2uwcos
1 du2 u 2v tan u 2 w
u p
 2u



 uK 2
2 dt
a
a
 x
z
1 dv2 u 2v tan v 2 w
v p
 2v



 vK 2
2 dt
a
a
 y
z
1 dw2 wu 2  wv 2

2 dt
a

w p
 z
 2uv sin   2uwcos
 v2u sin 
 wg  2uwcos
Add equations together: note that earth curvature and Coriolis
force terms all cancel!

d  u 2  v 2  w2 
1 
2 

   V  p  gw  Vh  K 2 Vh
dt 
2

z

Note that:
 gw   g
dz
d

dt
dt
Move this to left side


d  u 2  v 2  w2
1 
2 

     V  p  Vh  K 2 Vh
dt 
2

z



d  u 2  v 2  w2
1 
2 

     V  p  Vh  K 2 Vh
dt 
2

z

Kinetic Energy
Work done by PFG
Work done by Friction
Potential Energy
A change in total mechanical energy of a parcel of air must come
about by work done by the pressure gradient and frictional forces

Note that V is parallelto p in geostrophic flow so the first term on the LHS
is zero in geostrophic flow. Only the Ageostrophic wind component does
work.
The thermodynamic energy equation
First law of thermodynamics can be expressed as:
dT
d
Q  cv
p
dt
dt
or
dT
dp
Q  c p

dt
dt
where Q is the diabatic heating rate and  is the specific volume
Mechanical energy equation
 
 1 
d  u 2  v 2  w2
0  
    V  p  Vh  F
dt 
2
 
simplifying
friction notation
Total energy equation (add TEE and MEE)
2
2
2
 

 1 
dT
d

d
u

v

w

Q  cv
p
 
    V  p  Vh  F
dt
dt dt 
2
 
2
2
2
 

 1 
dT
d

d
u

v

w

Q  cv
p
 
    V  p  Vh  F
dt
dt dt 
2
 
Note that:
So:

1 
V  p  V  p

and
dp p 

 V  p
dt t
1 
 dp p 
V  p     

 dt t 
2
2
2
 


dT
d

d
u

v

w
dp
p
Substituting: Q
 c
p
 
    

 Vh  F
v
dt
dt dt 
2
dt
t

2
2
2
 


d
u

v

w

p
Q  cvT  p 
 
 Vh  F
dt 
2
t

The Energy Equation
2
2
2
 


d
u

v

w
p

Q  cvT  p 
 
 Vh  F
dt 
2
t







If flow is adiabatic Q  0 , frictionless F  0 and
p
 0 , then:
steady state
t
u 2  v 2  w2
cvT  p 
   const.
2
Special case of Bernoulli’s equation for incompressible flow
-For an atmosphere at rest, an increase in elevation results in a decrease in
hydrostatic pressure
Other implications
u 2  v 2  w2
cvT  p 
   const.
2
In accelerating flow over a hill the pressure difference between p2
and p1 must be > hydrostatic
Potential Temperature
Temperature a parcel of air would have if it were brought dry
adiabatically to a pressure of 1000 mb. “Dry adiabatically” implies
No exchange of mass or energy with the environment, and no
Condensation or evaporation occurring within the air parcel.
Potential temperature equation derived
from 1st law of thermodynamics:
dS  c p d ln T  Rd d ln p
For an adiabatic process, dS = 0
c p d ln T  Rd d ln p  0
S = Entropy
cp = Specific Heat at constant pressure
Rd = dry air gas constant
c p d ln T  Rd d ln p  0
Integrate equation from an arbitrary temperature and pressure to
the potential temperature q and a pressure of 1000 mb.
q
Rd
T d ln T  c p
1000
 d ln p
P
 q  Rd  1000

ln  
ln
 T  cp  p 
 1000

q  T 
 p 
Rd
cp