James I.N. Introduction to Circulating Atmospheres

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A.2

Solutions

Chapter

2 389

Take the time average, noting that Q = Q while Q! = 0, and the required

result follows:

QR = QR + Q'R' or Q'R' =

QR-QR.

This expression is useful when handling a long time series of data, since it

only requires one pass through the data to compute the covariances.

2.2 If Q is conserved, then the value of Q for a parcel of fluid displaced a

distance r\ in the y-direction represents its value at its initial position y. It

follows that:

But, using Taylor's expansion:

Then assuming that the mean displacement is zero, it follows that:

0! = —*lQy

d SO Y\

= —= .

2.3 Using Taylor series expansions, write:

., Ax

.„

=QT

AxQ' + ^-Q" - ^-Q!" + 0(Ax%

where Q! = dQ/dx, etc. Then adding and re-arranging:

Using this, a finite difference substitution for the diffusion equation may be

written:

Since this is a linear equation, the error obeys the same finite difference

equation as the analytical solution; seek solutions to the error of the form

5 oc

mAt

™

/iV

€

™/

The condition for stability is that | e |< 1.

Substituting in, we have:

^! _

os(27r/AT)) e - 1 = 0.

390 Solutions

Problems

By inspection, this quadratic equation has distinct real roots

for e

whose

product is 1.

follows that for one root,

| e |>

1 and so there is instability,

no matter what At, Ax are chosen.

Using the same method,

is easy to verify that the substitution:

n —Sin

-^2\\ln+\

*ln

Sln-\h

while only first-order accurate, is stable for sufficiently small At. The slightly

more elaborate 'Dufort-FrankeF substitution:

is second-order accurate and stable for all At.

A.3 Solutions

Chapter

3.1 This

is a

straightforward application

Eqs. (3.3), (3.8) and (3.11). The

results are: Venus-1234 Earth days, Earth-38 Earth days, Mars-0.23 Earth

days.

Using data in Table

10.1,

these work out

5.1, 38, and 0.23 planetary

rotation periods

5.4, 0.105

and

3.3

10~

planetary years respectively.

We conclude that diurnal and seasonal effects will be large

Mars, only

seasonal effects will

large

Earth,

and

neither diurnal

nor

seasonal

effects will

significant

Venus. The Venus result

open

question,

since most

the dynamical activity

is in

the layers above the cloud tops,

at 30kPa.

this pressure

used

place

, then

value

4.3 Earth

days is obtained for

suggesting strong diurnal effects. See Chapter 10 for

more details.

3.2 The linearized, perturbed thermodynamic equations for the atmosphere

and the surface may be written:

respectively. Assuming that

ATt, vary

exp(<j£), we obtain two simul-

taneous homogeneous equations

for q,

whose determinant must

zero,

i.e.,

(**T **Tl\ i6(2^rjr

—

—=—

+ — U.

Solutions

Chapter

3 391

convenient

write this

terms

of q

SaT^/C

and q

4aT

;

the roots

this quadratic equation

are:

1/2

)

2(1

e)q

)

Assuming that q

, the two

roots

can be

simplified further,

and we

find:

= -q

or q = -(2 - e)q

The first root corresponds

to a

'fast' relaxation time, more

less equal

the

radiative equilibrium timescale introduced with

Eq.

(3.11), while

the

second represents

'slow' timescale characteristic

of the

surface.

3.3

The

thermodynamic equation

can be

written:

T _

+AT

i0Jt

Here, co

2TZ/T

. This first-order ordinary differential equation

straight-

forward

solve using standard methods, giving

t/z

)

is a

constant

integration which depends upon

the

initial conditions;

the

term proportional

to A

tends

zero

long times,

which case,

the

solution

sinusiodal oscillation

of T

about

with amplitude AT

/(l +ATL

)

and phase

lag

tan~

(27TT

If x

small compared

to t

, it

follows that

the time between

the

maximum equilibrium temperature

and the

maximum

actual temperature

approximately

Temperatures calculated

this

way,

with

time-dependent forcing

but no

dynamical effects,

are

sometimes called

'radiatively determined'

distinct from 'radiative equilibrium' temperatures.

3.4

The

circuit

consider consists

of two

adiabatic processes

and two

isobaric processes.

The

heat added during

isobaric process

is c

AT.

the

pole-equator temperature difference

at the

surface

is AT

then

the

upper level temperature difference must

(p/p

)

. Thus,

the

total heat

added during

the

circuit,

/ TdS, is

{l — (p/p

)

and the

mean rate

at which

it is

added

{l — {p/p

)

}hc This must balance

the

rate

of dissipation

kinetic energy,

Finally,

have

the

result that

/{c

— {p/p

)

}

= 18K.

This

rather less than observed,

although, given

the

crudity

the calculation, better agreement

perhaps

not

expected.

The

main shortcoming

is the

assumption that

all air

parcels

circulate between pole

and

equator:

the

average

air

parcel will undoubtedly

undergo

more limited circuit.

392 Solutions

Problems

A.4

Solutions

Chapter

4.1 Take Fig.

4.1

(a),

showing DJF. The typical poleward wind V in the upper

branch of the Hadley cell is

2.5

ms"

and the horizontal extent L of the cell

4100

km.

Hence the time taken for an air parcel to traverse the cell, L/V, is

around

days.

Rather than trying to measure the vertical velocities directly,

use continuity to show that the typical vertical velocity W ~

(Ap/L)V;

the

time taken to sink from the upper to the lower level is Ap/W, i.e. L/V.

The time taken for a complete circuit in the Hadley cell is therefore around

4L/V, i.e.

days.

4.2 This is a similar calculation, except that it is easier to estimate the

vertical velocities, rather than the horizontal velocities, from Fig.

4.1

(a).

typical vertical velocity in the Ferrel cell is 10~

Pas~

and so the time to

go from 30kPa to

100

kPa is

days. Hence the time for a complete circuit

would be

4Ap/W,

i.e.

324

days.

4.3 The Brunt-Vaisala frequency is written in finite differences:

030-0100

065

Since the diagrams are plotted with pressure as the vertical coordinate, use

the hydrostatic relationship to write:

Az ~ —Ap,

so that

h{p)9

Ap'

Take Fig. 4.2(a) for DJF, and read off some values of 0, as follows:

pkPa 0at3O°N 6 at 60°N

100

290

260

310

285

330

310

Noting h(p) = 3.90 x

10~

JK"

Pa"

kg"

, then N

= 1.5 x 10"

°N and 2.2 x 10-

°N.

4.4 Take Fig. 4.2(a) for DJF at

°N by way of example:

[u]

at 30kPa is

23ms-

and

[u]

at 100kPa is about 3ms"

. Thus d[u]/dp is 2.86 x lO^s"

The right hand side of the thermal wind equation, Eq. (1.53) is (h/f)8[6]/dy.

AA Solutions to Chapter 4 393

At 45 °N / is 1.03 x 10"

; other quantities required at the midlevel (65 kPa)

were calculated in problem 4.3. We find that (hA[6])/(fAy) = 2.85 x 1CT

which is in far better agreement with A[u]/Ap than we have any right to

expect!

4.5 From T

= ((l-a)S/<r)

1/4

, it is straightforward to calculate T

as 256 K

(globe), 173 K (pole) or 274 K (equator),so that

ATJT

= A6

= 0.39.

Hence, choosing H =

km, we find from Eq. (4.12) that Y = 3460 km. It is

not quite clear whether it would be preferable to use the radiative equilibrium

AO/6 or the observed A6/6 to estimate Y; the use of the equilibrium value

compensates for the heat extracted from the poleward edge of the Hadley

cell by the midlatitude eddies, ignored in the Held-Hou model.

4.6 The precipation rate P is 3000 mm per year, i.e. 3000 kg m~

year"

Thus the rate of heating is LP/3.2 x 10

Wm"

, ie 243 Wm"

. According to

the Held-Hou model, the heating rate for the symmetric Hadley cell will be

pPs(9

o ~~

9MO)/(S

E)' Using the values in the text, this works out at around

6Wm~

. To estimate the vertical velocity in the moist Hadley cell, convert

the heating rate into units of Ks"

by multiplying by

g/{c

This gives

2.3 x lO^Ks"

. Then w ~ £g/(T

) ~ 6.8 x lO^ms"

. From continuity,

a typical v ^ (L/H)w, ie 1.5 ms"

4.7 Assuming angular momentum conservation for a flow which emerges

with

[u]

= 0 at y = yo, the zonal wind at any other latitude will be:

within the Hadley cell, falling discontinuously to smaller values beyond

y = Y according to Held-Hou theory. Thus reading off values of

(f)

from

Fig. 4.7(a) and converting latitude into distance from the equator, y, we find

for </>o = 2°, y

= 2.73 x 10

m and [u]

is therefore 84.8 ms"

while for

(f)

= 4°, y

= 4.00 x 10

m and [u]

is 173.7 ms-

4.8 First deal with the units of s

. The function h(p) has units J K""

kg"

Pa"

, and dO/dp has units of KPa"

. So the units of s

= hd9

/dp

may be written as m s~

Pa"

, although there are a number of alternative

ways of expressing this combination! Now consider the section shown in

Fig. 4.2(a) for DJF, and take 45 °N as a representative midlatitude point.

Between

100

kPa and 30kPa, A0 is around 43 K. At 65kPa, h(p) is 3.9 x

10~

J K"

kg"

Pa"

. Hence s

- hAd/Ap works out as 2.4 x 10"

m s"

Pa~

4.9 From Eq. (4.31), we wish to estimate typical midlatitude values of the

terms (f/s

)[u

]

and (h/s

)[v

]

. A suitable value of s

was estimated

394 Solutions to Problems

in the last problem. Now, for the first term, simple finite differences enable

[u'v']

to be written as

[u'v']

mSLX

/(ApAy).

The maximum momentum flux is

at 25kPa, with a value of 40m

; take Ap = 75kPa and Ay = 2 x 10

We conclude that (f/s

)[u'v

]

is around 1.4 x 10~

Pam"

. In estimating

the thermal term, note that the section shows [v'T']; convert to

[v'O']

multiplying selected values by

(p/p

if you wish, but it really does not

make a significant difference. The maximum of [v'T'] is at 85kPa, with

smaller values above in the troposphere. Hence a mean

[v'0

]

is estimated

to be ll.4K.ms~

(see problem 5.2 for a quick way of estimating a vertical

mean) and [v'0']

~ b'^'lmean/^

' ^1^

simply Ap/A6 (see solution

4.8) and so (h/s

)[t7W]

is estimated to be 4.6 x lO^Pam"

. Given

the crudity of these estimates, we can only conclude that both forcings are

roughly comparable. Note that both forcings have the same sign. Write the

total typical forcing term as 2 x 10~

Pam~

. To estimate the typical

magnitude of the induced meridional wind

~ Aip/Ap, suppose the two

terms on the left hand side of Eq. (4.31) have comparable magnitude; then

2fAxp .

VPP

- ^T "

forcin

Hence [v] may estimated as roughly s

Ap/(2f

) x (forcing) which, using

values of/ and 5 for midlatitudes, works out at around

0.2

ms"

This is

reasonably consistent with the values shown in Fig. 4.1 for the winter Ferrel

cell.

4.10 Your picture should show a dipole of meridional circulation, with

poleward flow at the level of wave breaking and equatorward flow above

and below. The poleward flow creates acceleration f[v] which offsets the

deceleration due to wave drag. In the steady climatological state, f[v] must

exactly balance d[u]/dt in the absence of other processes, and so we can

estimate that the midlatitude d[u]/dt required to balance alms"

meridional

flow would be about 10ms"

day"

A.5 Solutions to Chapter 5

5.1 K = (u

)/2 = 1131kg"

. The rate of destruction of K is K/T

which is

2.6 x

10~

Wkg"

. Multiply by p

/g to convert this to Wm~

, which comes

out to be

2.65

Wm~

. This is only about 1% of the global mean insolation

243 W

which is why the heating due to friction at the base of the

atmosphere is not usually included in global energy budgets.

5.2 For DJF in the NH, [v'T'] has a maximum at 85kPa,

°N of

A.5

Solutions

Chapter

5 395

The average through the troposphere is less than this; one way

of estimating this is to read off the temperature fluxes at three equispaced

levels and then to apply Simpson's rule: the average of any quantity Q

is then given by (go + 4<2i +

Qi)/^-

Applying this formula, the depth

averaged [v'T

] at

°N is probably closer to lOKms"

. It has dropped to

2 K m s"

°N.

The distance between these latitude circles is 1.11 x

x25° = 2.8 x 10

m. Hence, d[7T]/dy = 8/2.8 x 10

= 2.9 x lO^Ks"

Multiply by c

to convert this to 2.9 x 10~

Wkg~

, i.e.

Wm"

. Similar

calculations can be performed for other latitudes.

5.3 From Fig. 5.5(a), the maximum [tiV] is 40m

°N; estimating

the average as in problem 5.2 yields 20m

. This falls to close to zero by

°N,

and so a typical [wV] in the subtropics is 20/(2 x 10

) = 10"

ms~

often conveniently expressed as 1.2 ms"

day"

. The depth averaged

[u]

9 ms"

(see problem 5.6), and so the spin up time is \u]/[u'v

]

, which is

9 x 10

s, i.e. about 10 days.

5.4 From Eqs. (5.6), the scales in the x- and y-directions (taken to be half

a wavelength) are {ng/f)(Z

)

and (ng/f)(zG/tfi)V

respectively. For

the given eddy at 45 °N, these work out as

1850

km and

2620

km repectively,

which have a ratio of

0.71.

From Eq. (5.10), the phase tilt is 22.5°.

5.5 Assuming that the amplitude and temperature flux do not vary with

height, this is a straightforward application of Eq. (5.17), from which we

deduce that

is 42

°,

that is, the phase difference between the

kPa wave

and 50kPa wave is 84°. The zonal wavelength is (27ig//)(Z

/i/

)

1/2

, that is,

3590

km.

Therefore the horizontal separation between the upper and lower

troughs is 3590 x (84/360) =

837

km.

5.6 AZ =

]/(2s

))

(h[Q

]/(-26

)).

From

the

definition

of iV

it follows that -8

= RT6N

/(pg

) = 4.7 x lO^KPa"

, while for typi-

cal midtropospheric pressures (say

62.5

kPa), h(p) = (R/p)(p/p

)

= 4.7 x

10"

JK^Pa^kg"

. From Fig. 5.15, AZ = 3.7 x 10

Jm"

, i.e., 3701kg-

Hence,

a global mean value of

]

1//2

is 8.6 K. If we assume that the tem-

perature varies with latitude according to Eq. (4.4), then

]

1//2

= A0/3 and

so A0 = 26 K. A typical

is related to AE in much the same way. Since AE

is observed to be 1.2 x

Jm"

, i.e., 1181kg-

, we have (0*

) - 24K

, i.e.,

a typical temperature fluctuation is 4.9 K.

The kinetic energies are more straightforward. From KZ = 4.7 x

J m"

that is, 461kg"

, the typical zonal wind must be 9.6 ms-

, and from KE =

7.5 x

Jm"

= 74Jkg"

, the typical eddy wind is 12.2ms"

5.7 This problem requires estimates of the terms p^fi/"!;*] and

396 Solutions to Problems

(h/s)

[9]

[lf¥l Take values for DJF at 45°N: [u]

is 15/2 x 10

7.5 x lO^s"

and the average

[u*v*]

is 20m

. Hence the magnitude

of the integrand of the first term is 1.5 x lO^Wkg"

, i.e. 1.53 Wm~

. The

factor

(h/s)

can be estimated from values given in problem 5.6 and comes

to 10JK^kg-

. A typical midlatitude

[6]

is -30/(3 x 10

) = -KT

Km"

while for

[v*6*]

take a value of lOKms"

. Hence the thermal term is

-lO^Wkg"

or -10.2 Wm"

. We conclude that the thermal (or 'baro-

clinic') conversion dominates. Both estimates are likely to be overestimates

since they are based on midlatitude values rather than global means, but the

ratio of the two terms is unlikely to change much with a better global mean.

A.6 Solutions to Chapter 6

6.1 Figure 6.4(a) shows two maxima and two minima around

°N,

sug-

gesting wavenumber 2 is the dominant wavenumber at

°N.

Now the

dimensionless wavenumber n is related to the dimensional wavenumber k

by:

k = n/(acos(/>),

(f>

being latitude. So we have k = 6.28 x 10~

. From the same diagram,

we note that the phase tilt between

100

kPa and 30kPa is around 40° of

longitude, i.e. 5 = 10° and Ap = 35kPa in Eq. (5.17). In order to invert

Eq. (5.17) and estimate [v*T ], we must estimate the amplitude A of the

steady waves; inspection of the diagram suggests that

150 m

might be a good

value. In that case,

[v*T*]

is around 6Kms"

. This is entirely comparable

with the tropospheric average for

°N which may be read off Fig. 6.5(a).

6.2 From Eq. (6.8),

Z =

To estimate the wavenumber, assume the ridge represents half a wavelength

in each direction, so that k = n/L

, I = n/L

, i.e. k = 3.14 x lO^m"

/ = 6.28 x lO^m"

and K = \/k

= 3.20 x lO^m"

. At

°N,

P = 2Qcos0/a = 1.62 x 10"

. Hence, Z = -1.15 x

l0-

h/H

Assume H

km;

then Z = 2.47 x 10~

. The negative sign indicates

that there is anticyclonic vorticity above the summit of the mountain. To es-

timate the typical meridional wind, write

where the streamfunction

amplitude V = -Z/K

= 2.41 x 10

. Then

7.6

ms"

A.6

Solutions

Chapter

6 397

6.3 When drag is included, the vorticity amplitude is

fU(h/H

)

Substituting values, we note that (i^Zc)"

= 0.72ms"

and (U - p/K

) =

13.42 ms"

; we anticipate that the effect of friction is not very large in this

case.

In fact, comparing | Z | and | Z

| and using the binomial theorem,

we find that the difference between the vorticity amplitude with and without

friction is only around 0.3%.

6.4 On the £-plane, K

= fi/U = (2Qcos0/a)/((7

cos(/)) = 2Q/(al/

This is independent of latitude

or y. Since / = \JK

—

, and k is

conserved following the ray, it follows that the direction of the ray does

not vary with y, i.e. it is a straight line. If this argument is recast in full

spherical geometry, it turns out that the rays follow great circle paths. For

the given flow, K

= 1.07 x 10~

. It follows that the longest propagating

wave has zonal wavenumber k = K

= 1.07 x lO^m"

, i.e. n = 4.82.

Zonal wavenumber 4 is therefore the largest propagating wavenumber at

45 °N, and meridional wavenumber / = 5.9 x 10~

. A packet with these

wavenumbers will propagate at an angle tan^Z/fc) = 34° to the zonal

direction.

6.5 The meridional component of the group velocity of a steady Rossby

wave is c

= 2(7 cos a sin a = 2Ukl/K

. For this case, k = 6.66 x lO^m"

and K

= p/U =

1.08xl0"

so that / =

-k

)

= 7.97xl0~

Hence, c

= 14.7ms"

, and the time taken to cover 20° of latitude, i.e.

2.2 x 10

m, is 1.5 x 10

s or 1.7 days.

6.6 The speed of a wave packet was given by Eq. (6.19): this is simply

dgs/dt (where d

/dt is the rate of change following the packet):

-f^ = 2[/cosa.

Since tana = ///c, it follows that:

l _ k

~dt " (l + l

)k~dt ~ ~Kl~dt'

Noting that / = (K

—

)

and that k is conserved by the packet, we find

that:

k d(K

)

dt 2K

l dy dt 2K

l dy

)

398 Solutions to Problems

Finally, noting that cos a =

k/K

sin a = l/K

and that c

= 2U cos a sin a,

we obtain the required second equation:

a _

[/cos

This formulation has certain practical advantages, principally that it avoids

the difficulty of integrating Eq. (6.24b) near the poleward turning point of

the ray, where /

—•

0 and k

—•

6.7 Take a value of

]

K m s"

as representative of the lower tropo-

sphere. Take / as lO^s""

and s

/h = 5.7 x lO^KPa"

(see problem 5.6).

Then the vertical component of the Eliassen-Palm flux is F2 = —hf[v'6']/s

i.e. 2.3 m

s^Pa"

The zonal flow acceleration is dF2/dp; assuming

is small

in the upper troposphere, with Ap ~ 70 kPa, this means that the acceleration

is 3.3 x 10~

ms~~

or 2.8

ms""

day"

. From Fig. 5.5, a typical [v'9

] is around

20m

s~~

which varies on a typical horizontal scale of 2 x 10

m. Hence the

acceleration due to eddy momentum flux convergence is around 10~

ms~

or 0.9 ms"

day"

. This is not negligible, but we must conclude that the eddy

acceleration is dominated by the vertical component of the Eliassen-Palm

flux.

6.8 From the definition of the Eliassen-Palm flux in pressure coordinates,

Eq. (6.73), it follows that:

The zonal mean thermal wind relationship, Eq. (1.53), can be written in the

same notation as [u]

h[6]

/f,

and so:

-^ h

F-V[H] = -[u]

[u*v'

Hence, taking the global average, we find the required result, that

^ = <F-V[ul).

(see Eq. (5.37)).

A.7 Solutions to Chapter 7

7.1 Note that a pure Kelvin wave has zero [u*v*] since v* = 0, and the

planetary and Rossby gravity waves all have zero

[u*v*]

since they have no

meridional phase tilt. However, the combination of a Kelvin wave and a