Atmospheric Dynamics examples sheet (GV) - suggested answers.
The following formulae are used in the answers:
A.Geostrophic wind: U g 
B.Gradient wind:
1
g
k   p  k   p z
f
f
U2

 fU  
 fU g ;
R
n
C.Thermal wind: (a)
D. Absolute vorticity
U g
g
 k   pT
z
fT
a  f 
R  0 for cyclonic curvature
(b)
U g
r
  k   pT
1np
f
U U

R n
E. Potential vorticity Q   (2    v).θ   g a
1(i)
θ
for synoptic scale
p
In the straight part of the flow, U  Ug
Thus, from A:   fU  1.46 10 4 sin(50  ). 100  0.0112ms 2
Find gradient wind in trough, from B.
310 
6 1
U 2  1.12 10  4 U  0.0112
 U 2  335U  33500  0
(0.0112 = -/n)
Solve quadratic equation, choosing smaller root (why?) gives U = 80.6 ms-1
Decrease in kinetic energy = gain in potential energy (per unit mass)
 2  1752 m s
    v
2
2 2
 horizontal displacement perpendicular to geopotential contours:


n
 1752
 0.0112
 156 km in the - ve n direction (i.e. equatorwar d)
Substitute in C(b) with  U g /  ln p  100 / ln( 1000 / 300) , since jet stream is at ~300 mb.
Then T ~
 3.2 105 Km1  3.2 K per 100 km
2. As above, with T  10
, where D is the depth of the atmospheric column, here 9 km.
50D
 U g  68 ms 1 . Air flowing into trough stretches
Substituting in C(b) to find U gives U ~
vertically giving upward vertical motion (near the ground), thus cooling of air and condensation
(cloud and rain).
1
3. Amplitude of wave is 9 so if centre of wave is at 50N, ridge is at 54.5 and trough at 45.5.
f at 45.5=1.04  10-4 s-1; at 50=1.12 .10-4s-1; at 54.5=1.19  10-4s-1
a) From A, using z  120
, U g  17.9ms -1
600000
b) Gradient wind in trough, from B with    2 10 3 ms 2 :
n
2
R


U  RfU 
0
n
Solving the quadratic equation gives U=16.1 ms-1
Gradient wind in ridge, from B:
U 2  RfU  R  0
n
Solving the quadratic equation gives U=21.8ms-1
c) There is no shear vorticity in this case. So, from D:
 a in trough  1.24 10 -4 s 1 ;  a in ridge  0.918 10 -4 s 1
4. Again use D, with no shear vorticity term.
R=912 km, from the formulae given: +ve in trough and –ve in ridge.
f in trough (47.75) = 1.0810-4 s-1 so  a =1.40910-4 s-1
f in ridge (52.25) = 1.1510-4 s-1 so  a =0.82110-4 s-1
Fractional change in depth of air column is calculated using E. In the straight part of the flow,
 a =f=1.1210-4 s-1.

 1.409
 1.26
Since a
is conserved, p( trough )
p(straight )
1.12
p
p(ridge )
 0.821
 0.73 .
p(straight)
1.12
5. For geostrophic wind (A), U=15 ms-1. In the trough, gradient wind is the solution of B:
U2 + fRU - fRUg= 0

U2 + 100U – 1500 = 0

U = 13.2 ms-1
In the ridge, similarly, U=18.4 ms-1
In the straight part of the flow Ua, the ageostrophic wind, is zero. In the trough it is –1.8 ms-1. So
the divergence in Ua in the flow entering the trough is  1.8
, approximately.
( 4)
The flow is convergent (HU < 0) so downward motion (stretching of air columns) is forced at
the tropopause. This persists throughout the troposphere.
Downstream of the trough, HU is similar in magnitude but opposite in sign, so the flow is
divergent near the tropopause and upward motion in the troposphere results.
The same result can be derived directly from E:  a increases from 10-4 s-1 (f) in the straight part
of the flow to ~ 1.365  10-4 s-1 in the trough (from D and using gradient wind for U with same
value of f). Conservation of PV demands that air columns stretch as  a increases, so forcing
downward motion. The opposite situation applies (obviously) downstream of the trough.
2