Three quizzes (30 percent), Assignments (20 percent), and final exam (50 percent).
Three quizzes (30 percent), Assignments (15 percent), Project (10 percent), and final exam (45 percent).
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 perhaps how it compared with other papers in the same area (but you are not obliged to read other papers). Your written project should not include figures from the paper you are discussing. The paper you choose should have some relationship to Atmospheric Dynamics.
(Look at: 3, 5, 9 - 15, 18 - 23, 29, 32, 34, 61 - 63)
(Look at: 1 - 14, 18 - 29, 53)
(look at: 21, 23 - 29)
(Look at: 2 - 11, 16 - 17, 19 - 32, 40 - 48)
Mid-Latitude Atmospheric Dynamics, Jonathan Martin. It is not in the bookstore. You can buy it from me for 60 dollars, order online, or obtain from a student that took the course last year. It is the same text I use for Dynamics II. If it is in reasonable condition, I will buy back from you at the end of the term for 40 dollars.
The course material will be quite close to the first 5 chapters of the text. There will be some additional material on boundary layer dynamics.
(1) 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.
(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 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. It is an advertisement for the paper, where the real content should be. Do not bore the viewer with details of the calculations. But do address the most important sources of uncertainty.
Quiz 2: Wed Oct 28 (Class 23)
Quiz 3: Nov 26
All Lecture Notes and Assignments. Relevant parts of the Old Tests and Slides. Chapters 1 - 5 from the text, with the exception of Sections 4.2, 4.5, and 5.4.
For each quiz and the exam, I will supply you a formula sheet. You are not permitted to bring additional materials to the quiz, other than a calculator.
The emphasis will be on the lecture notes up to Thursday October 1 and the three assignments. You should be familiar with the lecture notes up to page 69, and with the relevant parts of the Old Tests below. What questions from the Old Tests are "relevant" and based on material we have covered, I hope you can determine from the context. However, if anyone has specific questions about which questions from the Old Tests they should be responsible for, I will put clarifications here.
The emphasis will be on the lecture notes, starting with discussion of the Thickness Relationship on page 70 (Oct 7, 2019), up to the thermal wind lecture on Thursday October 22, and on on Assignments 3 - 5. In the text, we have covered from Chapter 3 to Section 4.3. You are not responsible for the material on pages 85-88 related to the Montgomery stream function.
You should also be familiar with the relevant parts of the Old Tests. These are: "General" (some), "Scale Analysis", "Heating and Layer Thickness", "Energy Equation", "Conservation of Angular Momentum", "Kinetic Energy Generation", "Pressure Gradient Acceleration" (some), "Large Scale Shape ...", "Boundary Layer .." (but just on Frictionally Induced Convergence, so questions 1, 4, 7), and "Geostrophic Wind".
The emphasis will be on the lecture notes starting with thermal wind up to the divergence equation (pages 119 - 175 from the lecture notes). The main topics will be the thermal wind, natural coordinates, circulation, sea breeze circulations, vorticity, potential vorticity, and the divergence equation. The relevant assignments are 6 - 8. From the text, we have covered Sections 4.3, 4.4, (not 4.5), 5.1, and 5.2 up to page 128.
A person is breathing through a 1 m long horizontal tube. Suppose that he expands his lung volume and reduces the pressure inside his lungs by 0.01%. Assume that the original air pressure in his lungs and the pressure of the atmosphere is 1000 hPa, that the density of the air in the tube is 1 kg/m3. Estimate the acceleration of the air in the tube. Note that 1 Pa = 1 N/m2.
Question 2.
A bathtub is 2 m long. The water at one end is 1 cm higher than the other. What is the magnitude and direction of the pressure gradient acceleration in the tub. Assume the density of water is 1000 kg/m3.
Question 3.
Molecular viscosity converts parcel kinetic energy to thermal energy (randomized kinetic energy or temperature), via molecular diffusion. 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 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.)
(a) Suppose that the temperature at a weather station is 20 C. Consider this point the origin, and assume that the temperature decreases by 3 C for every 50 km north. There is no temperature variation in the zonal direction. Write down T(x,y).
(b) What is the temperature gradient vector at the station?
(c) If the wind is blowing from the northeast at 20 m/s and the air is being heated by radiation at a rate of 1 C/hour, what is the local rate of temperature change at the station (in C/hour)?
Question 2.
2.1 from the text. Show how you derive your equations. You need to use radial coordinates (cylindrically symmetric) coordinates. I.e. the radial coordinate should be r. Do not use x and y. Suggestions: Ignore the weight of the atmosphere. The water pressure at the bottom of the tank therefore equal to the weight per unit area of the overhead water column. Under these assumptions, the pressure gradient acceleration is independent of depth. The given answer to (a) has the dimensions of acceleration (same as g). You are actually solving for an acceleration, not a force as indicated. This is the radial acceleration. In (b), ignore the Coriolis acceleration. The only acceleration is therefore the centripetal acceleration of a rotating water parcel. For (c), you need to do a simple integral to find an expression for the volume of water in the container. You have to define the volume of a ring of fluid, with each ring 2πr in circumference, dr in thickness, and h(r) in height. Use your solution for h(r), and integrate over dr from r = 0 to r = r0. To get the final answer, you need to get a relationship for h0 in terms of z0.
Question 3.
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" or "total" gravity at the location. Remember that "effective" includes the acceleration due to the rotational acceleration of the earth. You aren't given the local value for the total gravitational acceleration, so just use a standard constant value for g. Also remember that effective gravity refers to the gravity at that location for a stationary air parcel, but does not include the corrections due to non-zero u or v.)
2.1(a). Compare two points separated by a radial distance dr. To find the pressure gradient acceleration along the horizontal radial direction, you need the pressure difference dp. This is given by dp = ρ*g*dh, where dh is the difference in the depth of the fluid above the two points (caused by the slope at the top). The inward pressure gradient acceleration required to provide the inward centripetal acceleration is PGA = -(1/ρ)dp/dr = -(1/ρ)*ρ*g*dh/dr = -g*dh/dr. Note that for h increasing with r (as you would expect for a rotating fluid), the PGA is negative (i.e. toward the axis of rotation).
2.1(b). The inward centripetal acceleration is -r*ω*ω. Set this equal to the PGA. -g*dh/dr = -r*ω*ω. This gives: dh = (1/g)*ω*ω*r*dr. Solve to get h(r) = h0 + ω*ω*r*r/(2*g), where h0 is the height at the center.
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 2πr in circumference, dr in thickness, and h(r) in height. The volume of the fluid is then V = ∫0 r0 2πrh(r)dr. Use the solution for h(r). Then integrate over dr from 0 to r0. You should get V = πr0*r0*h0 + (π*ω*ω*r04)/4*g. You set this equal to π*r0*r0*z0 to get a relationship for h0 in terms of z0.
2.2 from the text.
Question 2.
2.10 from the text. In this case, "balanced" means a 3 way balance between the friction, pressure gradient, and Coriolis acceleration vectors. Assume that the given angle refers to an angle between the pressure isobars and the velocity vector. For simplicity (and easier for me to mark), assume that the pressure isobars are zonal with higher pressure to the south. What you should find are the PGA and friction vectors with dimensions of acceleration. (Very odd in the text to refer to PGF as an acceleration). I think the easiest solution is to assume balance and write: PGA + CA + F = 0. (Note: this is a VECTOR equation with each acceleration having x and y components, though PGA is purely meridional with my assumed geometry). Note that the CA vector is calculated from the V vector and the latitude, and the F vector is then inferred from this balance equation. I am not sure that my convention for the angle is the same as the text, so might get different answers.
2.10 from the text. Assume high pressure is to the south so the PGA is directed north. Start from the VECTOR equation
PGA + CA + F = 0.
Then write an equation for the x and y components separately. The PGA is directed north, the CA is directed at right angles to the velocity, and the friction F is directed in the opposite direction to V. So,
PGA = (0,PGAy).
Let V refer to the parcel speed 15 m/s. Then
CA = (fv,-fu) = f(Vsin(25),-Vcos(25)).
Let Ff refer to the amplitude of the frictional acceleration. It must point in the opposite direction to the velocity, so we can write:
F = -Ff(cos(25),sin(25)).
The x components and y components of PGA + CA + F must sum to zero individually. The component of the PGA in the x direction is zero. Therefore:
f*V*sin(25) - Ff*cos(25) = 0.
Using V = 15 m/s and f = 9.4E(-5) gives Ff = 6.6E(-4) m/s2 for the amplitude of the frictional acceleration.
In the y direction:
PGAy - f*V*cos(25) - Ff*sin(25) = 0.
This can be solved using using V = 15 m/s, and the above value for Ff, to get PGAy = 1.56E(-3) m/s2, which is also the amplitude of the PGA.
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.
Question 2.
3.6. I think "temperature decreases at a constant rate", must refer to with respect pressure, since in this case the mean temperature of a layer, as used in the thickness equation, is just equal to the average at 500 hPa and 1000 hPa. Otherwise, this would not be true, and the question would be hard to solve.
Question 3.
3.8. You assume the geopotential height of the 1000 hPa surface is fixed.
Question 4.
3.9. Don't answer the last two parts, i.e., just explain why this value is reasonable.
Question 5.
3.10. The easiest way to solve this is to enforce conservation of momentum, where total momentum M is the sum of the part due to the earth's rotation and another part due to the zonal wind speed u. There is a formula in the text for M. You could also solve from a more basic expression for the angular momentum. Changing latitude changes the distance R from the rotation axis. Remember to use the total angular velocity (i.e. including the angular velocity due to the earth's rotation).
Question 6.
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?"
3.5(c). You must use the equation given in 3.5(b), show the direction of the acceleration vector, and explain the direction the cross product should point.
A 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 the formula sheet between the heating rate, dp/dt, and dT/dt. This is a version of the first law of thermodynamics that we did not derive. You also need to assume hydrostatic balance.)
Question 2.
3.1 (a), (b), and (c) from the text. You also have to assume frictionless adiabatic flow, that the temperature is constant, and that the kinetic energy comes entirely from the horizontal velocity. (Note: the solution for (a) given on page 76 is incorrect. I think the answer to (b) is correct. In (c) they ask for a "force vector", but give the answer as an acceleration, so there is a dimensional inconsistency. Their answer is also incorrect. This question should be treated as if they are asking for the nonhydrostatic (vertical) acceleration. You do not need the absolute value of surface pressure to solve the problem.)
Question 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? (Note: do not make any assumption about the scale height of the atmosphere, or the background temperature profile.)
Question 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 of the atmosphere is independent of height, and equal to 200 K. If the balloon is displaced a small distance vertically from its equilibrium (neutrally buoyant) altitude and the temperature change in the balloon is adiabatic, what would be the period of the oscillation?
3.1(c). Take the nonhydrostatic pressure difference (36.4 Pa), and divide by the vertical distance (10 m) and the density (1.3 kg/m3). Should get 2.8 m/s2.
4.5. This question becomes long and tedious if answered fully. Ignore the last sentence of this question, and just answer the following. Remember it is easier to use the component forms of the thermal wind equations, rather than the vector or cross product forms. Also, remember that something increasing to larger pressure (downward) has a POSITIVE derivative with respect to pressure, and vice versa.
(a) What is the thermal wind of the 900-700 hPa layer?
(b) What is the thermal wind of the 700-500 hPa layer?
(c) What are d⟨T⟩/dx and d⟨T⟩/dy of the 900-700 hPa layer (where ⟨T⟩ refers to the mean layer temperature)?
(d) What are d⟨T⟩/dx and d⟨T⟩/dy of the 700-500 hPa layer?
(e) What is the mean temperature tendency of the 900 - 700 hPa layer due to geostrophic temperature advection?
(f) What is the mean temperature tendency of the 700 - 500 hPa layer due to geostrophic temperature advection?
Question 2.
4.9 Assume the two systems are axisymmetric (circular sysmmetry). By intensity, the text means whether the tangential wind increases or decreases with height between the surface and 850 hPa. This is basically a question about using the thermal wind. For each system, also 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. (Think of what the surface wind should be in each case, and how it would be modified by the thermal wind as you go up.)
NOT NEEDED TO ANSWER PROBLEM BUT MAY HELP: In my view, there are two odd aspects to this problem.
(1) You have to assume that the central surface pressure of the low does not go below 1000 hPa. Otherwise the mean temperature between 1000 and 850 hPa can't be defined. But typically of course, most lows would have a central surface pressure below 1000 hPa. Strange!
(2) The solid lines of the diagram show geopotential. I assume this is geopotential at 850 hPa. It does not make sense to draw the geopotential at the surface, since you would typically assume the surface is Mean Sea Level where the geopotential is zero. (You could also define a geopotential on the 1000 hPa surface, again only if the central surface pressure of the low did not go below 1000 hPa. But for the sake of argument, assume they are showing the geopotential on the 850 hPa surface.)
On a pressure surface, the geopotential varies for two reasons. If the pressure at the surface is higher, the geopotential on a pressure surface will go up, because it implies the pressure thickness is bigger (and therefore in general the height of the pressure surface). If the mean temperature between the pressure surface and the surface is higher, the geopotential of the pressure surface should also go up. In the case of the low that is shown, the smaller surface pressure and colder temperatures near the surface, would both imply that you would expect the geopotential on the 850 hPa surface to decrease toward the center (as shown). In the case of the high that is shown, the temperature effect would be the same, but the higher surface pressure near the center would be associated with larger geopotential on the 850 hPa surface at the center of the low. Evidently, this effect ends up being stronger than the temperature effect. Of course, the higher you go up, provided the temperature gradient is maintained, you would expect the temperature effect to win out.
Question 3.
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 ⟨T⟩ is equal to a straight average of the temperatures at 1005 hPa and 850 hPa. I think 30 m/s is the correct answer, rather than 26.2 m/s as in the text. Note that you can not use the gradient wind expressions since you are given no information about radius of curvature R.
(b) Eq (3.51) is useful here. By kinetic energy generation is meant the rate of kinetic energy increase due to acceleration by the component of the PGA in the direction of motion, so in (3.51) could assume that the geopotential is constant and there is no friction. 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 for this. If using natural coordinates would need radius of curvature R, which is not given (so can't use). Note that the cross-isobar angle refers to 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. I believe the correct answer is close to 13 degrees, rather than 14.81 degrees as in the text. BY "kinetic energy generation", they mean KE generation by the PG acceleration, NOT the net KE generation by both the PG acceleration and friction.
(c) You can assume there exists a steady state balance between KE generation by the PGA and KE loss due to friction. I think you should get about -6.5E-04 N.
(d) I think the answer is about 276 km.
Question 4.
4.13 (a) Note that the T advection tendency must be negative to cancel the positive heating tendency. Do not do part (b). By "temperature advection" is meant "temperature advection tendency".
4.5, Temperature Gradients : Most students used the equations directly relating VT = (uT,vT) to the temperature gradients from the formula sheet. Here it is important that p0 be the larger pressure and p1 the smaller pressure, and the thermal wind be the geostrophic wind at the top pressure surface minus the geostrophic wind at the lower pressure surface. You can also use (4.28) from the text, in which case you will get slightly different answers.
(a)
Vg = (0,10) at 900 hPa, and Vg = (10,0) at 700 hPa.
VT = (10,0) - (0,10) = (10,-10).
VT = (10,-10) m/s for the 900 - 700 hPa layer
(b)
Vg = (0,10) at 500 hPa, and Vg = (10,0) at 700 hPa.
VT = (0,10) - (10,0) = (-10,10).
VT = (-10,10) m/s for the 700 - 500 hPa layer.
(c) For the 900 - 700 hPa layer:
d⟨T⟩/dx = -1.38E-5 K/km
d⟨T⟩/dy = -1.38E-5 K/km
(d) Note that ln(900/700) does not equal ln (700/500).
For the 700 - 500 hPa layer:
d⟨T⟩/dx = 1.04E-5 K/km
d⟨T⟩/dy = 1.04E-5 K/km
Answer for Temperature Advection : 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.
(e) For the 900 - 700 layer, I get:
d⟨T⟩/dt = 1.39E-4 K/s
(f) For the 700 - 500 layer, I get:
d⟨T⟩/dt = -1.04E-4 K/s
Answer to Question 2 (4.9).
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).
Answer to Question 3 (4.12).
(a) :
The book has 26.2 m/s rather than 30 m/s. This
seems to be obtained by assuming that the mean layer
(b) :
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). dKE/dt
from pressure gradient work equals (-1/ρ)*V*dot*PG =
(-1/ρ)*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 subtract 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) :
The sum of kinetic energy production (0.02) plus friction drag equals zero.
Friction drag is just -V*F, where V is the speed and F the frictional drag
force (Note no angle here). So the answer is F = 0.02/(30 m/s).
(d) : Use the pressure gradient from (a) and determine distance required
to get 10 hPa.
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. For simplicity, assume the geostrophic wind is pointing north.
(Note that it is not possible to calculate the ageostrophic wind or
real wind vector from the information given.)
Question 2.
(a) What is the geostrophic wind speed (m/s) on an isobaric
surface for a geopotential height gradient of 100 m per 1000 km?
Assume f = 1.0E-04.
In all
cases below, assume the absolute value of R is 1000 km and f = 1.0E-04.
Note that you must determine the sign of R as appropriate for each given
geometry.
(b) What is the gradient wind speed for a regular low?
(c) What is the gradient wind speed for an anomalous low?
(d) What is the gradient wind speed for a regular high?
(e) What is the gradient wind speed for an anomalous high?
Question 3.
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.
Let f = 1.0E-04.
(i) What are the (ug,vg) 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 is the geostrophic temperature advection in the 750-500
hPa layer? Express your answer in K/day.
Question 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 (from east),
180 degrees is pointing north, and 270
degrees is pointing east (from west). Please convert your divergence
value to /day (easier to understand than /sec).
The best approach is to use the natural coordinate expression
(4.38a). Draw a diagram in which the geopotential contours are aligned
in a north-south direction with high geopotential (warm air) to the
east (since the geostrophic wind points north).
The actual 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,
dΦ/dy = 0, and dφ/dx = f vg. You want to find dΦ/ds, where ds is
a segment along the parcel trajectory. From your diagram, you should
show dx = ds*sin(30). So dV/dt = -sin(30)*dΦ/dx = -0.001 m/s2.
Note: You could try 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 actual acceleration vector
(only the acceleration magnitude).
So if you did solve for the acceleration vector, you must have made
some additional assumption.
Answer to Question 2
(a):
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.
(b):
9 m/s
(c):
108 m/s
(d):
11 m/s
(e):
89 m/s.
Answer to Question 3
(i):
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 increase in the positive p direction from 500 hPa
to 750 hPa, or i.e.
should decrease in going from 750 hPa to 500 hPa. I get
dvg = -34.4 m/s (where for p I used the average of 500 hPa and 750 hPa).
So vg goes from 14.14 m/s at 750 hPa to -20.3 m/s
at 500 hPa. Since ug is unchanged, at 500 hPa
(ug,vg) = (-14.14,-20.3) m/s.
The speed is 25.1 m/s, and points 55.7 degrees south of westward.
This question could also be done by solving the integral
(integrating over dp), or equivalently, by
using the expressions for the thermal wind
involving ln(p0/p1) on page 5 of the formula sheet.
(ii):
First calculate the average geostrophic wind
of the pressure layer. Then take the dot product with the T gradient vector.
The answer is -4.2E-04 K/s (-1.5 K/hour
or -36.3 K/day).
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
Assume that the vertical velocity w = 0 at 1000 hPa, and that the
atmosphere is isothermal with a constant temperature of 260 K.
Assume that the heights of the p levels are fixed, so the
local derivate of p with respect to time for fixed z is zero.
Find the pressure velocity ω at the following
six levels. Keep ω in Pa/s.
(i) 1000 hPa
The easiest way to solve this question is to "bootstrap" (iterate)
yourself up from the surface using the surface boundary condition.
Question 2.
(i) What is the circulation about a square of 1000 km on a side for
an easterly (i.e. westward flowing)
(ii) What is the mean relative vorticity ξ in the square?
Question 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 circulation dC/dt of a loop
if the two ends are 40 km apart.
(ii) Estimate the 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.
The other end goes from 1000-900 hPa over the land.
Suppose < v > = 0 at t = 0.
Estimate < v > after 10 minutes.
Question 4.
(i) 5.6(a) from the text.
This is actually a strange question.
If you use the natural coordinate expression for vorticity ξ
given in (5.25),
you should get that the vorticity ξ is zero for all r for the given v(r).
This is because of a cancellation between the shear and curvature terms.
The circulation about any circular loop divided by the area
is 2A/r2,
implying that the average ξ within the circle is non-zero.
These results appear to be inconsistent.
However, if you use the definition of ξ as the circulation divided by the area
in the limit that the area goes to zero, the ξ at the origin
goes to infinity as r goes to zero.
So the average average ξ about the
circle is always coming from the singularity at the origin
(where the tangential
speed equals infinity).
(ii) 5.6(b) from the text. You have to do this question as a single loop and single
area.
(i)
If the local tendency of p with respect to time is zero, i.e. p(z)
is fixed, then ω = -ρgw.
So if w = 0 at p = 1000 hPa,
then ω = 0 at p = 1000 hPa also.
(ii)
(iii)
Use the same procedure for all other levels. Just
work your way up. 0.1575 Pa/s
(iv)
0.1875 Pa/s
(v)
0.1275 Pa/s
(vi)
-0.0325 Pa/s.
Answer to Question 2
(i)
The contributions to the circulation from the sides will be zero.
Let ub be the u component of the horizontal
wind at the bottom of the square.
Since the wind is westward,
ub < 0 (but the value of ub has no effect on the answer).
u(z) gets less negative in the positive y direction.
u(y) = ub + (10/(500*1000))*y = ub + (1/50000)*y
Let ut be the u component of the horizontal wind at the top of the square.
Then ut = ub + (1/50000)*1000*1000 = (ub + 20) m/s
Let L be the length of the side of a square. Then
C = ubL - utL =
ub*L - (ub + 20)L = -20L = -2*107
m2/s.
It is good practice to draw the wind configuration on a piece of
paper
and confirm to yourself using a pinwheeel that you would
expect negative circulation.
Note that a negative sign is used for the contribution to the
circulation from the top of the square,
since the loop integral
is going in the negative x direction.
(ii)
Just take answer from (i) and divide by the area.
Answer to Question 3
(i)
This is a straightforward application of the example in the notes.
Use the formula in the notes:
dC/dt = R*ln(p/(p-dp))*(Tl- To) = 302 m2/s2.
Here, p = 1000 hPa, dp = 100 hPa, To = 280 K, and Tl = 290 K.
(ii)
Use the thickness relationship to calculate the geopotential height
difference at the two ends and average.
You can assume the geopotential height at 1000 hPa is zero (i.e.
at sea level).
You should get 894.8 m on one end and 864 m on the other end, with an
average of 879 m.
(iii)
You use the formula in the notes.
d< v >/dt = [R*ln(p/p-dp)(Tl - To)]/2*(H+L) = 0.0037 m/s2.
After 10 minutes, < v > = 2.2 m/s
Answer to Question 4
(i)
"Annulus" means doughnut, but that is lighter gray here.
Assume this question is referring to the inner circle.
Just fill in the appropriate radius for 2A/r2 above.
(ii)
Assume that they are referring to the outer annulus here.
The circulation about any circle is C = (A/r)2πr = 2πA.
However, in doing the loop you have to do the inner circle in the
clockwise direction, so get a negative contribution.
The two contributions cancel out, so the circulation (and vorticity)
of any annulus is zero.
Assignment 7. Due: 5:00 November 13
Question 1.
Answers to Assignment 7.
Answer to Question 1
Assignment 8. Due: November 20 2020
Question 1.
(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) -0.8 in the 300-100 hPa layer
(ii) 850 hPa
(iii) 700 hPa
(iv) 500 hPa
(v) 300 hPa
(vi) 100 hPa.
wind that decreases in
magnitude toward the north at a rate of 10 m/s per 500 km?
Answers to Assignment 8.
Answer to Question 1
Use DIV = -[ω1000 - ω850]/[1000 hPa - 850 hPa].
Note that the derivative should be calculated with the p coordinate
increasing.
Using ω1000 = 0 from (i), you get
ω850 = 0.09 Pa/s.
Then calculate the
density ρ from T = 260 K and p = 850 hPa, and use ω = -ρgw
to get w.
Note that a positive divergence (outward flowing air) near the
surface is generating downward motion (positive ω) at the 850 hPa level,
as you would expect.
Note also that a non-zero ω is not inconsistent
with the assumption that p(z) is fixed,
since ω refers to the rate
of change of pressure FOLLOWING the parcel.