Atmospheric Dynamics MAQ32806
Chapter 4
Circulation, Vorticity
and Potential Vorticity
Exercise 4.1
3
N
2
2
1
x
= −16 × 10 20 m/s
b. =
̅ =
= −20 + 0 − 0 − 0 = −20 × 800 × 10
y
a. = ∙ = + − − 4
−16 × 10
=
= −2.5 × 10# 800 × 10
% %
0 − −20
−
= 0−
= −2.5 × 10# % %
800 × 10
Exercise 4.2
3
V
+
a. = & ∙ ' = ()
*
,
dλ
r
+
= 20 × 200 × 10 × *
,
= 20 × 200 × 10 × 2-
= 25.1 × 10 b.
= −25.1 × 10 Exercise 4.3
4
Use Bjerknes’ circulation theorem (Holton eq. 4.6):
− = −2Ω sin 453 ×
−
− 0 = −2Ω sin 453 × -) − -) = −2 × 7.292 × 10# sin 453 × - × 100 − 200 × 10
= 9.72 × 10 From new circulation C2 calculate tangential velocity:
= ( × 2-)
yields
9.72 × 10
( =
= 15.5 2 × - × 100 × 10
Exercise 4.4
5
Use Kelvin’s circulation theorem
(Holton 4.3):
:
<
= −
= − >?
'@<
:;
=
'@< = '@< = 0
for p = const.
,,#
,,,
:
= −>? '@< − > ? + 5 '@<
:;
,,,
,,#
= −>?'@
1005
1005
1005
+ > ? + 5 '@
= 287 × 5 × '@
= 7.16 1000
1000
1000
Exercise 4.5
6
a.
−25 × 2 − −30 × % %
=
−
=0−
600 × 10
% %
3.535
=−
= −5.89 × 10 600 × 10
b.
B =
B =
(D =
−35 − 25 × 0 + 25 × − 15 − 30 × = −22.2/
4
+ 0 − 30 ×
= −0.88/
4
D + D = 22.2/
%(
35 − 15
# =
=
3.33
×
10
% 600 × 10
%(
25 − 30
=
=
−8.33
×
10
% 600 × 10
Exercise 4.5 cont’d
7
EF =
EF
B %( B %(
−
(B % (B %
−0.88
−22.2
#
=
× 3.33 × 10 −
× −8.33 × 10
22.2
22.2
= −9.65 × 10 c.
= EF + GHIJ → GHIJ = − EF = −5.89 × 10 + 9.65 × 10
= +3.76 × 10 d.
GHIJ =
(
(
22.2
→ >E =
=
= 5.9 × 10 = 5900L
>E
GHIJ 3.76 × 10
Exercise 4.6
8
a.
5
10 = 0.60; 20 = 0.80/
= 2- × 0.20 × 0.80 − 2- × 0.10 × 0.60
= 2- × 0.1 = 0.628 3
b.
=
=
0.628
- × 0.20 − - × 0.10
= 6.67 c.
10 = 0.05; 20 = 0.03/
Net out:
Divergence:
2- × 0.20 × 0.03 − 2- × 0.10 × 0.05 = 2- × 0.001 NO;P;
2- × 0.001
=
=
0.0667
)OQ
- × 0.20 − 0.10
Exercise 4.6 cont’d
9
d. Use continuity equation:
%R
= −T ∙ U → R = −T ∙ U
S
%S
5
3
VWXY
,
,.
R = −T ∙ U S
RZ3[ = −0.0667 × 0.1 = −6.67 × 10 ,
= −6.67 Exercise 4.7
a.
10
20
v
0
250
500
b. Vorticity in cylindrical coordinates:
=
1 %
2-)
)& − )& cos
) %)
L
1
2-&
2-)
= ) +
sin
)
L
L
=
=
1%
1 %
) −
\
) %)
) %*
],
1
2-)
)&22-)
& − & cos
+
sin
)
L
L
L
Exercise 4.7 cont’d
11
b. Alternative: vorticity in natural coordinates:
%( (
%( (
=−
+
=
+
%@ >E %) >E
%
2-)
=
& − & cos
%)
L
)
1
2-&
2-)
+
== ) +
sin
)
)
L
L
c. At r = 250 km:
=
d.
e.
2- × 10
20
+
sin - = 8 × 10# 500 × 10
250 × 10
],
= 2-) = 2- × 250 × 10 × 20 = 31.4 × 10 ̅ =
=
2-) 2
40
=
=
= 1.6 × 10 -)
)
250 × 10
Exercise 4.8
12
:
a. It is ‘hidden’ in the
+ ` term, because:
:;
:
%
+ ` = −U ∙ a + ` − b
+`
:;
%<
b.
cd
cZ
= −
c def
cg
Last term:
−
c def
ch
j
%
%<
%b
%
−b
k
%
%<
%b
%
Only the k-component:
c d
c[
− +`
cH
cg
+
cJ
ch
−
ci cJ
cg c[
−
ci cH
ch c[
l
%b
%<
%b
%S
% %b % %b
−
%< % %< %
%b % %b %
=−
−
% %< % %<
Exercise 4.9
13
a. 43oN→ ` = 9.95 × 10# :
Use Holton (4.19): :; + ` = − + `
For L: = − + ` n
;
dW
pZ
1
= −n ;
+
`
dq
,
Z + `
'@
= −nΔ;
, + `
yields
`
≡0
;
yields
% %
+
% %
m
dW
'@ + ` = −nΔ;
dq
Z = −` + , + ` O mpZ
Z = −9.95 × 10# + 1 × 10# + 9.95 × 10# × O ×
= 3.06 × 10# ×,st ××,,
st
For H: Z = −9.95 × 10# + −1 × 10# + 9.95 × 10# × O × e×,
= −2.42 × 10# ××,,
Exercise 4.10
14
See previous exercise with f = 0
L:
Z = , O mpZ
= 1 × 10# × O ×
×,st ××,,
H: Z = −1 × 10# × O × ×,
st
××,,
= 1.19 × 10# = −0.84 × 10# For H (ζ0+f) and (ζ0) have opposite signs, hence the vorticity decreases
for f=0 where it increases for f>0.
Exercise 4.11
15
Conservation of potential vorticity:
u +`
nv
=u +`
n<
3.2 × 10# + 1 × 10 ×
yields
nv
n<
1
1
=
+
1
×
10
×
10 × 10
9 × 10
= 1.88 × 10# Exercise 4.12
a.
16
w = u + `
nv
−
n<
= 9.81 × 0 + 2 × 7.292 ×
10#
×
= 1.77 × 10 x Lu
b.
yields
c.
−35
×−
20 × 10
= 1.77PVU
sin 453
w = w
35
9.81 × + ` ×
=
1.77
×
10
23.5 × 10
+`
= | = 1.21 × 10 + ` = + 2 × 7.292 × 10# × sin 483 = 1.21 × 10
= 1.28 ×
10#
( 10
= =
>
>
yields
yields
= 1.28 × 10#
> = 782 × 10 = 782L
Exercise 4.12 cont’d
d.
cos } = → = > cos }
>
n = > − = > − > cos } = > 1 − cos }
x
e.
n = > 1 − cos }
3 × 111 = 782 × 1 − cos } → } = 553
f.
= >} = 782 × 10 ×
55
× - = 750 × 10 180
750 × 10
∆; = =
= 750 × 10 ≈ 0.86
Q
(
10
17
Exercise 4.13
18
10 km
5 km
1 km
+ ` + `
=
ℎ
ℎ
+ 2 × 7.292 × 10# × sin −403
0 + 2 × 7.292 × 10# × sin −403
=
5000
9000
yields
= −7.5 × 10# Exercise 4.14
1
-<
-<
1 %Φ
−`, &, cos
= &, cos
=−
 = −
`,
<,
<,
`, %
1 %Φ
1
 =
=
`, ƒL L cos L = ƒ cos L
`, %
`,
a.
b.
19
U0
p=0
c
0
p=p0
c.
% %
 =
−
= −ƒL sin L
%
%
π/k
2π/k
x