Atmospheric Dynamics MAQ32806
Chapter 6
Quasigeostrophic Analysis
Exercise 6.1
2
u = 42 m/s
u = 0 m/s
Exercise 6.1 cont’d
a. Use Holton (3.32):
= − = −
3
−2 × 7.292 × 10 × sin 30
1000
= 42 − 0 ×
ln
300
287.05
= −8.86 × 10! K/m
b.
15 − 45 N ≅ 30 × 111 × 10 m
∆ = 30 × 111 × 10 × −8.86 × 10! = 29.5K
ln
1000
300
Exercise 6.1
c. <∆T> ≈ 25 K
4
T = -5 K
T = -30 K
Exercise 6.2
a. ∆ZJan = 5060 m and
Use Holton (1.31):
5
∆ZJul = 5480 m
+ yields
) *
+
) =
ln
=
ln
*
+,
+,
JAN:
5060 × 9.81
1000
=
ln
= 249.5K
287
500
JUL:
=
5480 × 9.81
1000
ln
= 270.2K
287
500
∆<T>=270.2 - 249.5 = 20.7 K
b. TJAN = -20°C
TJUL = 0°C hence ∆<T> = 20 K
Exercise 6.3
6
a. Use the definition of potential temperature
+1
0=
+
ln 0 = ln +
yields ln 0 1 1
ln +1 − ln +
=
−
78
78
+
+ 78 +
... (1)
Gas law:
+9 = → =
+9 yields 9 + 9
= +
+ +
... (2)
Substitute (2) in (1):
ln 0 1 9
+ 9 1
=
+
−
+
+ 78 +
=
Hence from (3):
But:
9=−
Φ
+
25
34
; = −9
in (4):
1 1 9 1 1 9
1
+
−
=
+ 1−
+ 9 + 78 + 9 +
78 +
ln 0
9
9
=−
− 1−
+
+
78 +
... (4)
... (3)
Exercise 6.3 cont’d
;=−
7
9
9
− 1−
+
78 +
9=−
Hence from (4):
,Φ
+
−1
;=
78
+,
b. Use (1): ; = −9
ln 0
1 1
−
= −9
+
+
= 78 +
>
1
, 1
=−
−
=
+
78 +
78 +,
−
Φ 1
+ +
Φ
+
QED
Exercise 6.4
8
QG vorticity equation:
?@
= −@
?@ + − C@
? + − D ∙ F
A
B
@
In this case:
−1.7 × 10
= −@
?@ + − CG@
?@ + − 2 × 7.292 × 10 × sin 38 D ∙ F
12 × 3600
B
>
>
Hence: D ∙ H = 4.38 × 10! I Exercise 6.5
9
First, let’s prepare some ingredients:
Φ
M+
= H cos
cos N B − 7A
B
+
Φ
= − J
@ = −
1 Φ
=J
,Φ
M+
=
−
NH
cos
sin N B − 7A
B ,
+
,Φ
=0
,
1 Φ
M+
C@ =
= H cos
cos N B − 7A
B
+
1 ,Φ ,Φ
M+
?@ =
+
=
−NH
cos
sin N B − 7A
B , ,
+
Now use the QG vorticity equation:
?@
?@
?@
P
+ @
+ C@
+ C@ O = A
B
+
and fill in our ingredients:
Exercise 6.5 cont’d
(−N7) −NH cos
M+
M+
cos N B − 7A + J −N , H cos
cos N B − 7A +
+
+
M+
P
+0 + OH cos
cos N B − 7A = +
+
Hence:
Integrate:
P 1
M+
= H cos
cos N B − 7A N , 7 − J + O
+ +
+ H
P=
M
M+
N 7 − J + O sin
cos N B − 7A +
,
10
Exercise 6.6
11
QG thermodynamic energy equation:
Φ
Φ
−
+ FS ∙ T −
= σP
+
+
A
In this case:
−
1 P=
+ @
; A
B
Φ
−
+
VΦ
M
Φ
M+
=−
+ N H sin
sin N B − 7A
+
V+
+
+
Φ
M
M+
−
= − H7 sin
cos N B − 7A
A
+
+
+
Φ
M
M+
−
= H sin
cos N B − 7A
+
+
+
B
From problem 6.5:
@ = −
1 Φ
=J
Exercise 6.6 cont’d
12
Substitute in QG expression for ω:
Yields:
HM
P=
;+
P=
1 + @
; A
B
M+
J − 7 sin
cos N B − 7A +
b. From Problem 6.5:
P=
Hence:
+ H
M
N , 7 − J + O sin
HM
;+
Results in: 7 − J = −
J−7 =
O
, ,
M
N, +
;+ ,
+ H
M
M+
cos N B − 7A +
N, 7 − J + O
−
Φ
+