Atmospheric Dynamics MAQ�32806

Chapter 3

Elementary Applications

of the Basic Equations

Exercise 3.12

a. Use: �� = ���

dφ

a

dy

� = ���� =1����� =

1����2� �� =

2� ����

b. � ∙ �� = ����� +�����

�� = −������ ����� =

������

Hence: � ∙ �� = ��� −��

���� + �

��������

= −��������� +

��������� + �

����

���

1�

= � ����−1������ = −

�������� = −��

��

Expressions on a pressure surface

Exercise 3.23

a. Continuity equation (3.5):���� +

���� �

+ � �! = 0 → � �! = −$ !

ω=0

ω(p)

ω(p0)

p=0

p0

% % � = −% $ ! �!�

&

' �

&

! − 0 = −% ( − 2(!& ! �!�

&

! = − (! − (!& !

� !0

= −(! + (!& !

� = −(! 1 − !!&

b. Maximum if � ! �!⁄ = 0Hence:

� !�! = −( − 2(!!& → ! = 12!&

And *��+ = −( 12!& 1 −

12!&!& = −14(!&

Exercise 3.2 cont’d4

ω=0

ω(p)

p=0

½p0

p0

c.

Exercise 3.35

a. Holton (3.7): -� = −./�/�! �/ = ����0 → -� = 0

b. Isothermal: T = constant = T0

-� = 1.&��! −�.�! =

1.&��!

Exercise 3.46

V

n

r

�!�� < 0���

�!�3 > 0

V

r

�!�� < 0���

�!�3 < 0

n

Exercise 3.57

a. From (3.9):5650 = 0 → −

��� = 0

Φ does not change in the streamwise direction

Flow is parallel to Φ-isolines

b. From (3.10):5650 = 0 → 6 = ����0��0 → −�Φ�� = ����0.

The lateral Φ gradient does not change

565� = 0 → No horizontal wind shear

Exercise 3.68

1 = −6� = −10

2 × 7.292 × 10<= × sin 45& = −97 × 10BC = −97DC

1& ≜ 111.1DC → 194DC ≜ 1.74&Diameter of inertial circle = 2R = 194 km

45N

45+1.74=46.74N

a.

45N

45-1.74=43.26N

b.

Exercise 3.6 cont’d9

c.

45N

45+1.74/2=45.87N

45-1.74/2=44.13N

d.

45N

45+1.49=46.49N

45-0.26=44.74N

b

c

a45o

a=97cos45o=68 km

b=a+97=165 km » 1.49o

c=97-a=29 km » 0.26o

Exercise 3.710

n r

Fp

r0

p

p0

V Cyclostrophic balance:

FG = FH

−CI�!�� = C �3

Use gas law: ! = I1. → 1I =

1.!

Yields:−1.!

�!�� = �3 and �� = −�3

% 1! �! =

�1.% 3�3

J+&

�+�

→ ln!&! = �1. ×

12 3& �

! = !&L<'MJ+M/�OP

Exercise 3.811

If pressure gradient →0 then Vg→0

Use Holton (3.15b):

6�JQR = −�12 ± ��1�4 + �16� = −�12 ± �12 1 + 46��1

If Vg→0 then 1 + 46��1 ≈ 1 +26��1

Hence: 6�JQR ≈ −�12 ± �12 1 + 26��1 = �12 −1 ± 1 + 26��1

+ sign: 6�JQR ≈ 6� geostrophic wind

- sign: 6�JQR ≈ −�1 − 6� ≈ −�1 inertial flow

Exercise 3.912

u=25m/s=90km/hDu/Dt

va

a. Va

points to the left of DV/Dt

b. Use Holton (2.24):5�50 = ��Q → �Q = 1

9.3744 × 10<= ×80 − 100 × 10B

3600��Q = −16.46C�<*

Vg

α

c. �� = 25� + 16.46� = 30C�<*���X = �0�� 16.4625 = 33°

Direction = 270°-33°=237°

d. �� = −������

yields ���� =� ��� = 2.867 × 10<]CC<*

Exercise 3.9 cont’d13

u=25m/s=90km/hDu/Dt

Vg

α

vav⊥

e.�^�Q =

66� → �^ = 16.46 × 2530 = 13.7C�<*

_ = �^ × �`�!L = 13.7 × 2.867 × 10<] = 3.9 × 10<BC�<* = 0.39�C�<*

Motion is “uphill” → loss of Ekin → slowing down

Exercise 3.1014

N

y

x

850

500

a. 6P = 15� + 25� = 29.15C�<*

uT

vT

VT

α

β

b. X = �0�� 2515 = 59° → � = 59° − 45° = 14°

�P = 6P cos � = 29.15 × cos 14° = 28.3C�<*

�P = 6P sin � = 29.15 × sin 14° = 7.1C�<*

Exercise 3.10 cont’d15

� = 1.117 × 10<]�<*α

β

uT

vT

N

y

x

850500

VT

c. Holton (3.32):� .�� = −�P�1 `� !&

!* = −28.3 × 1.117 × 10<]

287 `� 850500cc

= −2.076 × 10<=d/C� .�� = +�P�1 `� !&

!* = 7.1 × 1.117 × 10<]

287 `� 850500cc

= +5.208 × 10<ed/C

� .�� = � .

���+ � .

���= 2.14 × 10<=d/C

Exercise 3.10 cont’d16

d. Temperature advection: −� ∙ �f = −� �.�� − ��.��

850 hPa: ��� = − −15 cos 45° × 5.208 × 10<e − 15 cos 45° × −2.076 × 10<== 5.524 × 10<= + 2.202 × 10<] = 2.75 × 10<]d/�

500 hPa: ��� = − 25 cos 45° × 5.208 × 10<e − 25 cos 45° × −2.076 × 10<=

= −9.206 × 10<= + 3.670 × 10<] = 2.75 × 10<]d/�

Exercise 3.1117

900-700 hPa

VT10 m/s

900 hPa

10 m/s

700 hPa

N

y

x

6P = 10 2 = 14.14C�<*

�f = 6P�1 `� !&!* = 14.14 × 1 × 10

<]287 `� 900

700cc= 1.96 × 10<=d/C

���L�0 �� = −� �.�� − ��.�� = −(10) × −1.96 × 10<= × cos 45° = 1.39 × 10<]

d� = 0.5d/ℎ

∇∇∇∇T

Exercise 3.11 cont’d18

VT

20 m/s

500 hPa

10 m/s

700 hPa

N

y

x

6P = 20 − 10 = 10C�<*

�f = 6P�1 `� !&!* = 10 × 1 × 10

<]287 `� 700

500cc= 1.04 × 10<=d/C

∇∇∇∇T

���L�0 �� = −� �.�� − ��.�� = −(10) × −1.96 × 10<= × cos 90° = 0

d� = 0d/ℎ

Exercise 3.11 cont’d19

δz=2250 m

500

600

700

800

900t=0 t=δt

Tu

Tl

$. = .j − .k = ΓR$� = 0.01 × 2250 = 22.5dUpper layer: no temp advection > no T change

Lower layer: temp advection T change = 0.5 K/h

$0 = 22.50.5 = 45ℎ

Exercise 3.1220

p δδδδ δδδδ ave ∆∆∆∆p δδδδ ave x ∆∆∆∆p ωωωω ωωωω

(hPa) (x10�5 s�1) (x10�5 s�1) (hPa) (x10�3 Pa/s) (x10�3 Pa/s) (hPa/h)

300 0.85 �140 �5.04

0.45 200 90

500 0.05 �230 �8.28

�0.10 200 �20

700 �0.25 �210 �7.56

�0.50 150 �75

850 �0.75 �135 �4.86

�0.90 150 �135

1000 �1.05 0 0

Use: ! = ! − ∆! + ∆!$̅

With: $̅ = $ ! + $ ! − ∆!2

Exercise 3.1321

Γo=& = 4 dDC = 4 × 10<B dC�.�0 = −2

dℎ = −5.55 × 10<]

d�

� = −10C� (�3�C0ℎLL��0)�.�� = 5

d100DC = 5 × 10<= dC

Use Holton (3.42): = -�<* �.�0 + �

�.��

With: -� = ΓR − Γ I�⁄ and ≈ −I�_ΓR = �

�� =9.811005 = 9.76 × 10<Bd/C

Combine: _ = −1ΓR − Γ

�.�0 + �

�.�� = −1

9.76 − 4 × 10<B × −5.55 − 5 × 10<]

= 0.183C� = 18.3�C�

Exercise 3.1422

p δδδδ δδδδ ave ∆∆∆∆p δδδδ ave x ∆∆∆∆p

(hPa) (x10�5 s�1) (x10�5 s�1) (hPa) (x10�2 Pa/s)

1000 0.9

0.75 150 11.25

850 0.6

0.45 150 6.75

700 0.3

0.15 200 3.00

500 0

�0.30 200 �6.00

300 �0.6

�0.80 200 �16.00

100 �1Sum = �1.00

�!p�0 = +1 × 10<� q� �⁄ = 36 q� ℎ = 0.36 ℎq� ℎ⁄⁄