# PHYS/OCEA 4511/5511: Atmospheric Dynamics I

Ian Folkins,

Three quizzes (30 percent), Assignments (20 percent), and final exam (50 percent). The number to letter conversion is A+(90.0-100), A(85.0-89.9), A-(80.0-84.9), B+(75.0-79.9), B(70.0-74.9), B-(65.0-69.9), C+(62.0-64.9), C(58.0-61.9), C-(55.0-57.9), D(50.0-54.9), F(below 50.0).

Three quizzes (25 percent), Assignments (20 percent), Project (10 percent), and final exam (45 percent). The number to letter conversion is as for undergraduate except anything below a B- is an F. (see FGS calendar for implications on programme continuation).

Find a paper to discuss and talk about it to the class. The grade will be half from the written submission (about 5 pages) and half from the talk. Your project should be a critical review, and discuss, for example, why you agreed or disagreed with the main arguments of the paper, how useful or important the argument was, and how it compared with other papers in the same area. Your written project should not include figures from the paper you are discussing.

# Textbook:

Mid-Latitude Atmospheric Dynamics, Jonathan Martin. It is not in the bookstore. You can buy it from me for 50 bucks, order it online, or get from a student that took the course last year.

# Synopsis:

The course material will be quite close to the first 5 chapters of thye text. There will be some additional material on boundary layer dynamics.

# Tips for Graduate Student Talks

(1) Is every word of every slide large enough to be seen by a person in the back of the room? A very common mistake to use label axes that are very hard to read, or to take plots from other papers where the labels are very small. Of course you can read it while sitting at your computer. Can you walk 5 m away from your computer screen and still read it? This is a good test. If not, add extra text to fix it, ar use a different image.

(2) Every talk should be built around a single core argument. What is the take-home message you want your audience to understand? It is good to start a talk with a puzzle/mystery/paradox or some other unexplained phenomenon to provoke the audience interest. What is the purpose of your talk? Why is this topic important? Why should the audience care? A very common beginner mistake is to try to say too much. Only include something if it supports your main argument.

(3) A good rule of thumb is 1 slide/minute, but leave extra time for questions during the talk, after, and for switchover. So for a 20 minute talk, at MOST 15 slides, usually less.

(4) Appeal to the audience on multiple levels - through simple pictures, mathematical reasoning, intuition. Don't just write arguments down and expect people to follow them, or to retain their interest.

(5) Who is your audience? What is their background? What vocabulary do they understand? The purpose of a talk is not to talk about your research, but to build a bridge to where THEY are, so you communicate something.

(6) Giving a scientific talk is mainly a public relations exercise, an advertisement for a paper, where the real content should be. Do not bore the viewer with details of the calculation, but address most important sources of uncertainty.

# Quiz Dates

Quiz 1: October 3, 2011 (Class 11)

Quiz 2: October 31, 2011 (Class 22)

Quiz 3: November 30, 2011 (Class 34)

# Final Exam: 10:00 am - 1:00 December 14, 221C

For each quiz and the exam, I will supply you a formula sheet, explanations, constants, etc. You are not permitted to bring additional materials to the quiz, other than a calculator.

# Material for Quiz 1

The emphasis will be on the lecture notes. However, you should be familiar with Chapters 1 and 2 from the text. You may find the discussion in Chapter 3 on scale analysis helpful.

# Material for Quiz 2

The emphasis will be on the notes and problem sets since the last quiz. We have covered up to Section 4.2 in the text. You should study the parts of the text that we covered in class. You are not required to be familiar with the detailed mathematics of Section 4.2; just what we covered in class.

# Assignment 1. Due: Monday September 19, 5:00.

1. A person is swimming with a snorkel that is 1 m long. Suppose that he expands his lung volume and reduces the pressure inside his lungs by 10%. Assume that the original air pressure in his lungs, and the pressure of the atmosphere is 1000 hPa and the density of the atmosphere is 1 kg/m3. Estimate the acceleration of the air in the snorkel. Note that 1 Pa = 1 N/m2. (Explanatory Note: you get a huge acceleration which is unrealistic. I don't think it is unrealistic to think you can expand your lung volume by 10 %. But is probably unrealistic to think you can do this and not let any additional air into your lungs, as would be required to reduce your pressure by 10%. It's hard to expand your lungs and keep your mouth closed.)

2. An elderly gentlemen is drinking a milkshake through a straw having a length L = 30 cm. The density of the milkshake is 1.2 times the density of water. Atmospheric pressure is 1000 hPa. Calculate the pressure in his lungs at which the milkshake will exit the straw and he will be able to drink. Assume the straw is vertical, and that 30 cm refers to the length of the straw that is above the level of the milkshake in the glass. (Note: I suppose you could do an experiment and estimate the amount by which people could reduce their lung pressure from atmospheric pressure by measuring the height of the tallest straw they could drink from.) Hint: you can first drink from a straw when you are first able to apply a tiny upward acceleration (i.e. assume zero).

3. Viscosity converts parcel kinetic energy to thermal energy (randomized kinetic energy or temperature), via momentum exchange. This is an idealized example to help explain how this occurs. Suppose that a 1 kg air parcel with a velocity of 20 m/s is adjacent to another 1 kg air parcel with a velocity of 10 m/s. After a diffusional exchange of momentum (viscosity), the 1 kg air parcels have velocities of 19 m/s and 11 m/s respectively. Note that this diffusional momentum exchange conserves the momentum of the sum of the two parcels.

(a) How much parcel kinetic energy was lost during this diffusional momentum exchange (in J)?

(b) By conservation of energy, and in the absence of any external energy sources or sinks, this reduction in parcel kinetic energy must be accompanied by an increase in thermal energy (kinetic energy of molecules). Assume that the increase in thermal energy is distributed equally between the two air parcels. Determine the increase in temperature of the two air parcels. (You can assume that this diffusional momentum exchange occurs at constant pressure, and use the specific heat at constant pressure for dry air, usually called cpd.)

# Assignment 2. Due: Monday September 26, 5:00.

1. 1.10 from the text. (This question is trivial; don't use in the future.)

2. 1.12 from the text. It is best to assume that the horizontal wind at the service station is zero. The rate of pressure change as measured in the car as it passes the service station is then equal to the local pressure change plus the contribution due to the motion of the car in the pressure field. This contribution is negative since the car is directed (south) to lower pressure. The pressure should therefore be decreasing more rapidly in the car than at the service station. The answer in the text is therefore incorrect: it gives a faster rate of decrease at the service station than in the car. (In the future: give a question where the advective component of the tendency comes from the motion of an air parcel, not a contrived situation like a car.)

3. 2.1 from the text. The answers are in the text. Your grade depends on how clearly you derive the answers. You ignore the effect of the atmosphere: the pressure in the fluid is equal to the hydrostatic pressure. (Hints: for (a) it may be easier to think of solving for the pressure at the bottom of the rotating pan. Assume the pressure in the fluid is given by the hydrostatic approximation, and ignore the pressure of the atmosphere. Under these assumptions, the pressure gradient acceleration is independent of depth. Also note that you are solving for an acceleration not a force, as it suggests in the text. In the (b), ignore the Coriolis Force. The only acceleration is the centripetal acceleration of a rotating fluid parcel.)

Answer for 2.1(c). The best way to calculate the volume of the container as it rotates is to imagine it is made up of rings of fluid, with each ring 2*pi*r in circumference, dr in thickness, and h(r) in height. The volume of the fluid is then V = Int(from 0 to r0) 2*pi*r*h(r)*dr. Use the solution for h(r). Then integrate over dr from 0 to r0. You should get V = pi*r0*r0*h0 + (pi*omega*omega*r0*r0*r0*r0)/4*g. You set this equal to m*r0*r0*z0 to get a relationship for h0 in terms of z0.

4. 2.4 from the text. Note that the two weights are for the passenger going in opposite directions in the train but at the same speed. You need the mass of the passenger. You can calculate this from the average (stationary) weight, and a constant value of g. (Note that to calculate the "real" mass of the person, you would have to be given the stationary weight of the person plus the actual "effective" gravity at the location. Remember that "effective" includes the rotational acceleration. You aren't really given this so just use a standard constant value for g.)

# Assignment 3. Due: Wednesday October 12, 5:00.

1. 2.2.

2. 2.10. In this case, balanced means a 3 way balance between the friction, pressure gradient, and Coriolis acceleration vectors.

3. 3.5(c) only. You have to assume that, along the axis of the jet, the flow is mostly zonal, and that the acceleration vector is mainly coming from changes in the zonal wind speed.

4. Estimate the Rossby number of the rotational circulation in a cup of tea.

# Assignment 4. Due: Wednesday October 19, 5:00.

1. 3.8.

2. 3.9. Don't answer the last part.

3. 3.10. The easiest way to solve this is to enforce conservation of momentum. Changing latitude changes the distance R from the rotation axis. Remember to use the total angular velocity.

4. From Holton: "Isolines of 1000 to 500 hPa thickness are drawn on a weather map using a contour interval of 60 m. What is the corresponding layer mean temperature interval?"

# Assignment 5. Due: Wednesday October 26, 5:00.

1. A 1 kg parcel of air is rising at a constant vertical velocity. The parcel is being heated by radiation at 0.1 W/kg. What is the vertical speed w of the parcel if the temperature of the parcel is constant? (Hint: use a relationship in your notes between the heating rate, dp/dt, and dT/dt. Also need to assume hydrostatic balance.)

2. 3.1 (a) and (b) from the text. You also have to assume frictionless adiabatic flow. (Notes: the solutions for (a) and (c) given on page 76 appear to be incorrect. Next year: ask to do (c) also.)

3. An air parcel with a temperature of 20 C at 1000 hPa is lifted dry adiabatically. What is its density when it reaches 500 hPa?

4. A high altitude balloon remains at a constant potential temperature as it circles the earth. The balloon is in the lower stratosphere where the temperature is independent of height, and equal to 200 K. If the balloon is displaced a small distance vertically from its equilibrium (netrally buoyant) altitude, what would be the period of the oscillation?

# Assignment 6. Due: Wednesday November 9, 5:00.

4.5. (i) Calculate mean horizontal temperature gradients in the two layers 900-700 hPa and 700-500 hPa.

In this question, many of you used vector formulas for relating the temperature gradient vector and the thermal wind. However, you are making life difficult for yourself by introducing extra steps; 99 percent of the time, you are better off using the component forms of the equations for calculations.

(dT/dy) between 900 hPa and 700 hPa: There are a variety of ways of doing this question. One way is to use equations (4.28). At 900 hPa: ug = 0 and vg = 10 m/s. At 700 hPa: ug = 10 m/s and vg = 0. In using (4.28), it is extremely important to note that the derivative is evaluated with p "up" going down. So in going from 700 hPa to 900 hPa, ug decreases by 10 m/s, so dug/dp = (-10 m/s)/(200*100 Pa). For p, you would use the average value of p between 700 hPa and 900 hPa, or 800 hPa. Therefore, dT/dy = [f*(800*100)/R]*[(-10 m/s)/(200*100 Pa)]. R = 287 and f = 9.9E-05. I get dT/dy = -1.38E(-05) K/m.

(dT/dx) between 900 hPa and 700 hPa: Again using (4.28), it is easy to rearrange to get (dT/dx) = -[(f*p)/R]*(dvg/dp). You can use the previous value for f*p/R. The change in vg in going from 700 hPa (0 m/s) to 900 hPa (10 m/s) is positive so dvg/dp is positive. Should again get dT/dx = -1.38E(-05) K/km.

(dT/dy) between 700 hPa and 500 hPa: Same approach as above should get dT/dx = 1.04E-05 K/km. dug/dp is just the negative of that used above for the 900 - 700 hPa layer, and use 600 hPa for the average pressure.

(dT/dx) between 700 hPa and 500 hPa: Same approach as above should get dT/dy = 1.04E-05 K/km.

(ii) Calculate the rate of geostrophic temperature advection in each layer. Do not do the third part of the question.

The mean rate of geostrophic temperature advection in each layer is -(ug,vg)*(dT/dx,dT/dy), where Vg = (ug,vg) refer to the mean layer geostrophic winds. For both layers, (ug,vg) = (5,5), and use you answers for the T gradient vector from above. Note the negative sign.

4.9 Assume the two systems are axisymmetric. By intensity is meant whether the tangential wind increases or decreases with height between the surface and 850 hPa. For each system, draw a cross section of how isobars between the surface and 850 hPa would vary with height. Your horizontal axis should be distance from the center of the low, or high.

Answer: I will draw the diagrams in class. However, one good way to think is that the thermal wind is parallel to isopleths of thickness with the warm air on the right. Therefore in both cases the thermal wind is counter clockwise (CCW). The surface wind is CCW for the low and CW for the high. Therefore the thermal wind adds to the surface wind for the low (more instense with height), and cancels the surface wind for the high (less intense with height).

4.12. Do all parts. This question seems to require that you make many assumptions.

(a) You have to work back from knowledge of the geostrophic wind at 850. What is the CA? What is the PGA? What is the density at 850 hPa? What is the PG at 850 hPa? What is the PG at 1005 hPa? What is the PGA at 1005 Pa? What is the Vg at 1005 hPa? Assume that the 1005 - 850 hPa layer temperature is equal to a straight average of the temperatures at 1005 hPa and 850 hPa. itself can be calculated exactly.

Most of you got 30 m/s, which is correct. The book has 26.2 m/s. This seems to be obtained by assuming that the mean layer that you get from the thickness expression can be set equal to the temperature at 850 hPa. However, it is more accurate (though also an approximation) to assume that = 0.5*(T_1005 + T_850), and solve fr T_850. The best solution is to set the PGA = CA using the expression for pressure gradient on a height surface (i.e. as in 3.36b), and use the Ideal Gas Law to express the density in terms of P and T. Use this constraint to solve for the pressure gradient. Then repeat the procedure at 1005 hPa, using the same pressure gradient, to solve for the geostrophic wind speed at 1005 hPa.

(b) Eq (3.51) is useful. By kinetic energy generation is meant the rate of kinetic energy increase due to acceleration by the PGA. It also looks like you have to assume the actual wind speed V is approximately equal to the geostrophic wind speed Vg (but the directions of the velocity vectors must be different). In principle, the actual wind speed should be used in the expression for KE generation by the PGA, but I don't see how to solve this. If using natural coordinates would need R, which is not given. Note that the cross-isobar angle means the angle between the pressure isobars and the wind vector, not the angle between the velocity vector and the PGA which you initially solve for.

Most of you got an angle close to 13 degrees, whereas the book has 14.81 degrees, which is obtained by using their slightly incorrect value for the geostrophic wind speed at the surface from (a). Basically use dKE/dt from pressure gradient work equals (-1/rho)*V*dot*PG = (-1/rho)*Vg*PG*cos(theta), where theta is the angle between the velocity vector and the pressure gradient vector, and assuming the actual speed is close to the geostrophic wind speed. Use your answer for the pressure gradient from above, and substract theta from 90 degrees to get 13 degrees. The book is not asking for the angle between the pressure gradient vector and the velocity vector (theta = 77 degrees), but for the angle the wind vector makes with the isobars.

(c) You can assume the rate of change in geopotential height is zero, and that there exists a steady state balance between KE generation by the PGA and KE loss due to friction.

(i) Note that the T advection must be negative to cancel the positive heating.

# Assignment 7. Due: Wednesday November 16, 5:00.

1. (3.2 Holton) The actual wind is directed 30 degrees to the right of the geostrophic wind. If the geostrophic wind speed is 20 m/s, what is the rate of change of wind speed (i.e.solve for dV/dt). Let f = 1.0E-04. Assume the geostrophic wind is pointing north.

Answer: The best approach is to use the natural coordinate expression (4.38a). Draw a diagram in which you have geopotential contours aligned in a north-south direction with high geopotential (warm air) to the east. The wind vector is pointing northeast, so cutting diagonally uphill across geopotential contours. You would therefore expect the parcel to be experiencing a deceleration. From the geostrophic wind, dphi/dy = 0, and dphi/dx = f*vg. You want to find dphi/ds, where ds is a segment alnog the parcel trajectory. from your diagram, you should show dx = ds*sin(30). So dV/dt = -sin(30)*dphi/dx = -0.001 m/s2.

Some of you tried to solve for the ageostrophic wind vector and then use Eq (3.41). However, you are not given the actual wind speed (only the direction), so it is impossible to solve for the ageostrophic wind. It is therefore also impossible to solve for the acceleration vector. So if you did solve for the acceleration vector, you must have made some additional assumption.

2. (3.4 Holton) (a) What is the geostrophic wind speed (m/s) on an isobaric surface for a geopotential height gradient of 100 m per 1000 km?

Answer: Straightforward application of \$4.39b). Almost all got 9.8 m/s. Note that speed is always positive. Also note you must convert 100 m of geopotential height to geopotential by multiplying by g.

In all cases below, assume the absolute value of R is 1000 km and f = 1.0E-04.

(b) What is the gradient wind speed for a regular low?

9 m/s.

(c) What is the gradient wind speed for an anomalous low?

108 m/s.

(d) What is the gradient wind speed for a regular high?

11 m/s.

(e) What is the gradient wind speed for an anomalous high?

89 m/s.

3. (3.10 Holton) The mean temperature in the 750 - 500 hPa layer decreases eastward by 3 C per 100 km. The 750 hPa geostrophic wind is from the southeast at 20 m/s.

(a) What is the geostrophic wind speed and direction at 500 hPa? Let f = 1.0E-04.

Answer: It helps to draw a diagram. Use (4.28) with dT/dy = 0, so only vg changes between 750 hPa and 500 hPa. Since dT/dx < 0, dvg/dp > 0, so vg should decrease in going from 750 hPa to 500 hPa. The decrease is -34.9 m/s, so vg goes from 14.14 m/s at 750 hPa to -20.76 m/s at 500 hPa. Since ug is unchanged, at 500 hPa (ug,vg) = (-14.14,-20.76) m/s. The speed is 25.1 m/s, and points 55.7 degrees south of westward.

(b) What is the mean rate of geostrophic temperature advection in the 750 - 500 hPa layer? Express in K/day.

Note: To find the average geostrophic temperature advection of the layer, you must use the average geostrophic wind 0.5*(Vg(750) + Vg(500)). Should get -1.5 K/hour, or -37 K/day.

In Future replace by: (3.10 Holton) The mean temperature in a 750 - 500 hPa layer decreases eastward by 3 C per 100 km. The 750 hPa geostrophic wind is from the southeast at 20 m/s.

(i) What are the (u,v) components of the geostrophic wind at 500 hPa? Assume that formulas like (4.28) in the text can be applied to finite pressure layers.

(ii) What are the (u,v) of the thermal wind of the 750 - 500 hPa layer?

(iii) What is the sign of the geostrophic temperature advection in the 750-500 hPa layer? (Don't have to do the calculation; can show graphically) Would expect veering or backing?

4. 4.17(a) from the text. The angle convention is CCW from north. For example, 0 degrees is pointing south, 90 degrees is pointing west, 180 degrees is pointing north, and 270 degrees is pointing east. Please convert your divergence value to /day (easier to understand than /sec).

# Assignment 8. Due: Wednesday November 23, 5:00.

1. (3.22 of Holton) The divergence of the horizontal wind at various pressure layers above a given station is as follows (units of divergence in E-05 per inverse s): (i) 0.6 in the 1000-850 hPa layer, (ii) 0.45 in the 850-700 hPa layer, (iii) 0.15 in the 700-500 hPa layer, (iv) -0.3 in the 500-300 hPa layer, (v) and -0.8 in the 300-100 hPa layer. Assume that the vertical velocity w = 0 at 1000 hPa, and that the atmosphere is isothermal with a constant temperature of 260 K. Find the vertical velocity w and the pressure velocity omega at the six levels: (i) 1000 hPa, (ii) 850 hPa, (iii) 700 hPa, (iv) 500 hPa, (v) 300 hPa, (vi) 100 hPa. The easiest way to solve this question is to "bootstrap" (or iterate) yourself up from the surface using the surface boundary condition.

2. (4.1 Holton) (i) What is the circulation about a square of 1000 km on a side for an easterly (i.e. westward flowing) wind that decreases in magnitude toward the north at a rate of 10 m/s per 500 km?

(ii) What is the mean relative vorticity in the square?

3. The surface pressure over both the land and ocean is 1000 hPa. The average 1000-900 hPa temperature over the ocean is 280 K. The average 1000-900 hPa temperature over the land is 290 K. You can assume f = 0.

(i) Estimate the rate of change of absolute circulation dC/dt of a loop where the two ends are 40 km apart.

(ii) Estimate an average geopotential height of the loop.

(iii) Let < v > refer to the mean tangential velocity along a rectangle where one end goes from 1000-900 hPa over the ocean, while the other end goes from 1000-900 hPa over the land. Suppose < v > = 0 at t = 0. Estimate < v > after 10 minutes.

4. 5.6 from the text (a) and (b).