GEF3450Exercises for group session: Balanced Flow

Ada GjermundsenE-mail: ada.gjermundsen@geo.uio.no

September 22, 2016

Figure 1: Figure from http://www.surf-forecast.com

[...] This extreme generality whereby the equations of motion apply tothe entire spectrum of motions - to sound waves as well as cyclone waves -constitutes a serious defect of the equations from the meteorological point ofview. It means that the investigator must take into account modifications tothe large-scale motions of the atmosphere which are of little meteorologicalimportance and which only serve to make the integration of the equations avirtual impossibility.J. G. Charney

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Exercises from Joe’s GFD compendium, “Fluid Mechanics” by Kundu and Cohen, “Atmo-sphere, Ocean and Climate Dynamics” by Marshall and Plumb, “Atmospheric Science” byWallace and Hobbs, “Dynamic Meteorology" by Holton and “Geophysical Fluid Mechan-ics” by Weber.

1 Scaling

Had our theory been perfect and were we able to resolve all scales of motion,there would be no need for scaling. However, since our atmosphere and oceanare so complicated and we easily get lost in the Navier-Stokes equations, theusual procedure is to simplify them a bit. We use scaling to evaluate therelative magnitude of the different terms involved, to try our best to get ridof teeny-tiny terms (which hopefully simplifies the equations a bit).

Exercises:

1. Problem set 2: Scale the x-momentum equation for parameters typi-cal of the ocean.Assume:

• U = 10 cms−1

• W = 0.01 cms−1

• L = 100 km

• D = 5 km

Also use the advective time scale, T ∝ L/U and that sin(θ) ≈ 1. Showthat the geostrophic balance also applies with these scales. Note thatI haven’t given you the pressure scale ∆p/ρ. Can you estimate whatit is, given the above scaling? What if it were actually much less thanthis - what could you say about the motion?

2. Problem set 2: Scale the y-momentum equation.Assume:

• U = 1 ms−1

• W = 1 cms−1

• L = 100 m

• D = 0.5 km

Also use the advective time scale, T ∝ L/U and that sin(θ) ≈ 1.Which are the dominant terms? How big is ∆p/ρ? Finally, write theapproximate equation.

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2 Streamlines

The streamfunction ψ is defined for incompressible (divergence-free) flows intwo dimensions. The flow velocity components can then be expressed as

u = −∂ψ∂y

v =∂ψ

∂x(1)

If ψ = p/ρcf0, u and v represent the geostrophic velocities.

The streamfunction can be used to plot streamlines, which represent thetrajectories of particles in a steady flow.

Exercises:

1. A two-dimensional steady flow has velocity components

u = y v = x (2)

show that the streamlines are rectangular hyperbolas

x2 − y2 = const. (3)

Sketch the flow pattern, and convince yourself that it represents anirrotational flow in a 90◦ corner.

2. For streamfunctions ψ with the following functional forms, sketch thevelocity field:

i) ψ = my

ii) ψ = my + n cos(2πx/L)

iii) ψ = m(x2 + y2)

iv) ψ = myx

where m and n are constants.

3. For each of the flows in the previous exercise describe the distributionof vorticity.

4. Consider a velocity field that can be represented as

~V = −k̂×∇ψ (4)

where ψ is called the streamfunction. Prove that ∇H · ~V is everywhereequal to zero and that the vorticity field is given by

ζ = ∇2ψ (5)

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3 Geostrophic balance

Figure 2: Figure from UK Met Office

Let’s look at one of the most important force balances that drive the windsand the ocean currents -it is a force balance that is a foundation of e.g. syn-optic meteorology– the geostrophic balance.

Here’s how it works. We start with a pressure gradient–in figure 3, highpressure is at the bottom and low pressure is at the top. Naturally the airwants to flow from high pressure towards the low pressure, so it acceleratesin that direction. This force, that makes air move from high to low pressure,is called the Pressure Gradient Force, or the “PGF” as it is often abbreviated.The red arrow above represents the direction of this force–from high to lowpressure. The PGF only depends on the density of the air and the strengthof the pressure gradient–nothing else.

However, there’s another “force” at work on this rotating planet of ours–theCoriolis “force”. This isn’t really a true “force” but an apparent change inthe direction of the wind due to the fact that the earth is rotating out fromunder it. This effect depends on the Coriolis parameter “f” and the velocityof the wind (which is why I’ve abbreviated it as “fv” in the figure). Thefaster the wind goes, the stronger this Coriolis turning effect becomes. Inthe northern hemisphere, the Coriolis effect always acts to make the windsappear to turn to the right of their direction of movement–always. Simply afact of the direction the earth rotates.

So as the air accelerates due to the pressure gradient force, the velocity in-

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Figure 3: Figure from http://lukemweather.blogspot.no/

creases and consequently the Coriolis effect (which depends on velocity) alsoincreases. This turns the winds more and more to the right the faster theygo. Finally, it gets to a point where the tendency of the Coriolis effect toturn the winds to the right exactly balances the pressure gradient force, likewe see in figure above where the red PGF arrow is exactly balanced by theblue Coriolis effect (fv) arrow. We see that the wind is no longer movingfrom high to low pressure, but perpendicular to the pressure gradient.From Luke’s weather blog: http://lukemweather.blogspot.no/

Note that the exact same applies for the ocean, just replace wind with cur-rent.

The Geostrophic Equations:

ug = − 1

fρ

∂p

∂y(6)

−vg = − 1

fρ

∂p

∂x(7)

Exercises:

1. Problem set 2: What is the pressure gradient required to maintain ageostrophic wind at a speed of v = 10 ms−1 at 45◦N? In the absenceof a pressure gradient show that air parcels flow around circles in ananti-cyclonic sense of radius v

f . What is the name of the motion?

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2. Two moving ships passed close to a fixed weather ship within a fewminutes of one another. The first ship was steaming eastward at arate of 5 ms−1 and the second northward at 10 ms−1. During the 3 hrperiod that the ships were in the same vicinity, the first recorded apressure rise of 3 hPa while the second recorded no pressure changeat all. During the same 3 hr period, the pressure rose 3 hPa at thelocation of the weather ship (50◦N, 140◦W). On the basis of these data,calculate the wind speed and direction at the location of the weathership. (Hint: assume geostrophic conditions)

3. Problem set 2: Consider a low pressure system centered on 45◦S,whose sea level pressure is described by

p = 1000hPa−∆p · e−r2/R2, (8)

where r is the radial distance from the center. Determine the struc-ture of the geostrophic wind around this system; find the maximumgeostrophic wind, and the radius of the maximum wind, if ∆p =20 hPa, R = 500 km, and the density at sea level of interest is1.3 kgm−3. (Assume constant Coriolis parameter, appropriate to 45◦S,across the system.)

4. i) A typical hurricane at, say 30◦ latitude, may have low-level winds of50 ms−1 at a radius of 50 km from its center: do you expect this flowto be geostrophic?

ii) Two weather stations near 45◦N are 400 km apart, one exactly tothe northeast corner of the other. At both locations, the 500 hPa windis exactly southerly at 30 ms−1. At the northeastern station, the heightof the 500 hPa surface is 5510 m; what is the height of this surface atthe other weather station?(Hint: assume the wind to be geostrophic and parallel to lines ofconstant height (Z) on an isobaric surface, then ug = − g

f∂Z∂y and

vg = gf∂Z∂x )

5. Problem set 2: The Gulf stream flows northward along the coast ofthe United States with a surface current of an average magnitude of2 ms−1. If the flow is assumed to be in geostrophic balance, find theaverage slope of the sea surface across the current at a latitude of 45◦N.

6. Write down an equation for the balance of radial forces on a parcel offluid moving along in a horizontal circular path of radius r at constant

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speed v (taken positive if the flow is in the same sense of rotation asthe Earth).

Solve for v as a function of r and the radial pressure gradient, andhence show that:

i) if v > 0, the wind speed is less than its geostrophic value

ii) if |v| << fr, then the flow approaches its geostrophic value

iii) there is a limiting pressure gradient for the balanced motion whenv > −1/2fr.

(Hint: use the gradient wind balance equation)

7. Consider a circular tornado which rotates with a constant angular ve-locity ω around the vertical axes. In a distance r from the center ofthe tornado , the tangential velocity is ωr.

i) Make a sketch of the forces acting on a fluid parcel at a distancer from the center of the tornado. Assume the density is given by theequation of state for dry air.

ii) At a distance r0 from the center of the tornado, the pressure is p0.Show that the pressure at the center of the tornado is given by:

p = p0 exp(−ω2r20

2RT) (9)

iii) If the temperature is 288 K, and pressure and wind speed at 100 mfrom the center are 1000 hPa and 100 ms−1, respectively, what is thecentral pressure?

8. Problem set 2: An aircraft flying a heading of 60◦ (i.e. 60◦ to the eastof north) at air speed 200 ms−1 moves relative to the ground due east(90◦) at 225 ms−1. If the plane is flying at constant pressure, whatis the rate of change in altitude (in meters per kilometer horizontaldistance) assuming a steady pressure field, geostrophic winds and f =10−4 s−1?(Hint: geostrophic wind is parallel to lines of constant height (Z) onan isobaric surface (p=const), then ug = − g

f∂Z∂y and vg = g

f∂Z∂x )

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9. An airplane pilot crossing the ocean at 45◦N latitude has both a pres-sure altimeter and a radar altimeter, the latter measuring his absoluteheight above sea. Flying an air speed of 100 ms−1, he maintains alti-tude by referring to his pressure altimeter set for sea-level pressure of1013 hPa. He holds an indicated 6000 m altitude. At the beginning ofone-hour period, he notes that his radar altimeter reads 5700 m, andat the end of the hour he notes that it reads 5950 m. In what directionand approximately how far has he drifted from his heading?(Hint: assume the wind to be geostrophic and parallel to lines ofconstant height (Z) on an isobaric surface, then ug = − g

f∂Z∂y and

vg = gf∂Z∂x )

4 Thermal Wind

Figure 4: Differential thermal heating of the atmosphere can create verystrong winds like the jet streams. figure from NASA/Goddard Space Flight CenterScientific Visualization Studio

The thermal wind is oriented parallel to the temperature contours withcold air to the left and warm air to the right.

Meteorologists like to define something called the thermal wind. The thermalwind really isn’t a wind at all–it’s the difference in winds between two differ-ent levels (wind at top minus wind at bottom). The thermal wind is parallelto temperature contours (note! in the ocean the thermal flow is parallel todensity contours). This means that as you follow the thermal wind arrow,temperature (i.e. density in the ocean) is not going to change in that direc-tion. However, part two of the adage says that the thermal wind is alwaysoriented with colder air to its left and warmer air to its right. Because ofthis, we can label the region to the left of the thermal wind arrow as "cold"and to the right of the thermal wind arrow as "warm".

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Figure 5: Ada’s handwavy sketch of thermal wind

This fancy feature of the thermal wind makes it useful for diagnosing tem-perature structures based on winds, or the other way around: calculatewind profiles based on pressure or temperature measurements from differentheights in the atmosphere. It’s amazing how everything is connected!Most of the text is taken from Luke’s weather blog: http://lukemweather.blogspot.no/

The Thermal Wind Balance:

There are several different versions of the thermal wind equation. For the at-mosphere it is common to use the temperature gradient at constant pressuresurfaces to calculate the thermal wind

~ug∂p

= − g

fpk̂×∇pT (10)

~uT = ~u(p1)− ~u(p0) =R

fln[p0p1

]k̂×∇pT (11)

while in the ocean it is more common to consider changes in density (sincethe thermal wind is parallel to density contours in the ocean)

~ug∂z

= − g

fρck̂×∇ρ (12)

... but don’t panic if you see other versions - they all describe the samephenomenon:-)

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Exercises:

Exercises from “Fluid Mechanics” by Kundu and Cohen, “Atmosphere, Ocean and Cli-mate Dynamics” by Marshall and Plumb, “Atmospheric Science” by Wallace and Hobbs,“Dynamic Meteorology" by Holton and “Geophysical Fluid Mechanics” by Weber.

1. The mean temperature in the layer between 750 hPa and 500 hPa de-creases eastward by 3◦C per 100 km. If the 750 hPa geostrophic windis from the southeast at 20 ms−1, what is the geostrophic wind speedand direction at 500 hPa? Let f = 10−4s−1.

2. What is the mean temperature advection in the 750 − 500 hPa layerin problem 1?

3. Problem set 2: Assuming θ = 45◦N , so that f = 10−4 s−1, and thatg ∼ 10 ms−2:a) Assume the pressure decreases by 0.5 Pa over 1 km to the east butdoes not change to the north. Which way is the geostrophic wind blow-ing? If the density of air is 1 kgm−3, what is the wind speed?Note 1 Pa= 1kgm−1s−2).

b) If the temperature of air was 10◦ C throughout the atmosphere andthe surface pressure was 1000 hPa, what would the pressure at a heightof 1 km be?Note 1 hPa=100 Pa.

c) The temperature decreases by 1/5◦C over 1 km to the north butdoesn’t change to the east. What is the wind shear (in magnitudeand direction)? Assume the mean surface temperature is 20◦C. Is thisveering or backing?Hint: Use the pressure coordinate version of thermal wind and thenconvert the pressure derivative to a z-derivative.

4. Problem set 2: Thermal wind on Venus

We know (from various space missions) that the surface density onVenus is 67 kgm−3 and the height of the troposphere is 65 km. Also,the Venetian day is 116.5 times longer than our day(!) We want toestimate the wind velocity at the top of the tropopause.

a) Write down the thermal wind equation for the zonal wind, u, inpressure coordinates.

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b) Convert the pressure derivative to a z-derivative by assuming thedensity decays exponentially with height, with an e-folding scale (thescale height) equal to half the height of the tropopause.

c) If the temperature gradient in the northern hemisphere is −1.087 ·10−4 degrees/km, what is the zonal velocity at the tropopause? As-sume the temperature gradient doesn’t change with height and thatthe velocity at the surface is zero. Note R = 287 J/(kg K).

5. Suppose that a vertical column of the atmosphere at 43◦N is initiallyisothermal from 900 hPa to 500 hPa. The geostrophic wind is 10 ms−1

from the south at 900 hPa, 10 ms−1 from the west at 700 hPa, and20 ms−1 from the west at 500 hPa.

i) Calculate the mean horizontal temperature gradients in the two lay-ers 900 hPa to 700 hPa and 700 hPa to 500 hPa.(Hint: use the thermal wind eqn.)

ii) Compute the rate of advective temperature change in each layer.(Hint: DT

Dt = 0 following the wind.)

iii) How long would this advection pattern have to persist in order toestablish a dry adiabatic lapse rate between 600 hPa and 800 hPa?Assume that the lapse rate is constant between 900 hPa and 500 hPa,and that the 600 hPa to 800 hPa thickness is 2.25 km?(the dry adiabatic lapse rate is 9.8 ◦Ckm−1)

6. Problem set 2: Fig 6 shows the trajectory of a “champion" surfacedrifter, which made one and a half loops around Antarctica betweenMarch, 1995, and March, 2000 (courtesy of Nikolai Maximenko). Reddots mark the position of the float at 30 day interval.

i) Compute the speed of the drifter over the 5 years. (Hint: the dis-tance for one loop around Antarctica at the latitude where the drifteris located is approximately 22× 103 km)

ii) Assuming that the mean zonal current at the bottom of the oceanis zero, use the thermal wind relation (neglecting salinity effects) tocompute the depth-averaged temperature gradient across the Antarc-tic Circumpolar Current (ACC). Hence estimate the mean temperaturedrop across the 600 km-wide Drake Passage.(Hint: use the thermal wind equation: f ∂u∂z = g

ρref

∂ρ∂y , the density equa-

tion for fresh water ρ = ρref (1−αT [T−Tref ]) where αT = 2×10−4 K−1

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Figure 6: From “Atmosphere, Ocean and Climate Dynamics” by Marshalland Plumb

is the expansion coefficient and an average depth of 4 km)

iii) If the zonal current of the ACC increases linearly from zero at thebottom of the ocean to a maximum at the surface (as measured bythe drifter), estimate the zonal transport of the ACC through DrakePassage, assuming a meridional velocity profile u(y, 0) = U0 cos(πy2L)and that the depth of the ocean is 4 km. Let U0 be the speed youcalculated in i) and L be the width of Drake Passage. The observedtransport through Drake Passage is 130 SV. Is your estimate roughlyin accord? If not, why not?

7. i) From the pressure coordinate thermal wind relationship, (eqns. 3.45-46 in the gfd notes), and approximating

∂u

∂p' ∂u/∂z

∂p/∂z(13)

show that in geometric height coordinates

∂u

∂z' − g

T

∂T

∂y(14)

ii) The winter polar stratosphere is dominated by the “polar vortex”, astrong westerly circulation about 60◦ latitude around the cold pole, as

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depicted schematically in the Fig. 7. (This circulation is the subject ofconsiderable interest, because it is within the polar vortices - especiallythat over Antarctica in the southern winter and spring - that most ofthe ozone depletion is taking place.)

Assuming that the temperature at the pole is (at all heights) 50 Kcolder at 80◦ latitude than at 40◦ latitude (and that it varies uniformlyin between), and that the westerly wind speed at 100 hPa pressure at60◦ latitude is 10 ms−1, use the thermal wind relation to estimatethe wind speed at 1 hPa pressure at 60◦ latitude. (you need the gasconstant for air = 287 Jkg−1K).

Figure 7: From “Atmosphere, Ocean and Climate Dynamics” by Marshalland Plumb

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