Three quizzes (30 percent), Assignments (20 percent), and final exam (50 percent). The number to letter conversion is A+(89.5-100), A(84.5-89.5), A-(79.5-84.5), B+(76.5-79.5), B(72.5-76.5), B-(69.5-72.5), C+(64.5-69.5), C(59.5-64.5), C-(54.5-59.5), D(49.5-54.5), F(below 49.5).
For graduate students, the weighting is quizzes (30 percent), assignments (15 percent), project (10 percent), and final exam (45 percent). The number to letter conversion is the same as for undergraduates, except that grades below B- are considered F. (For grades of F, see FGS calendar for implications on programme continuation).
We will cover most of Chapters 6 - 9 of Mid-Latitude Atmospheric Dynamics, by Jonathan Martin. We will also spend several lectures near the end of the course discussing Rossby and Kelvin waves. The parts of Chapters 6 - 9 that we DO NOT cover are: page 164 to the start of Section 6.4 (page 166), section 6.4.3 only slightly, Section 7.3 on the SG equations, the sections on PVeg on pages 226 - 229, Section 8.7, pages 278 - 179 on PV inversion, the derivation of (9.23) (i.e. the previous three pages), and Section 9.5.1.
Material to be Covered: Class notes until Friday January 27. First two assignments. "Previous Test Questions" up to and including "Chapter 6: Destruction of Thermal Wind Balance by Geostrophic Flow". Text: Starting with Section 5.4, Chapter 6 up to page 171, but not pages 164 - 166.
Material to be Covered: all material on Q vectors. This includes the class notes starting with "Geostrophic Paradox" (January 25, 2017). All class lecture notes up to March 3 (i.e no Chapter 8 material on March 5). From the text: Section 6.4 (except 6.4.3). Chapter 7: 7.1, 7.2, 7.4 (excluding parts on Qg pp 214 - 217), and 7.5 (excluding parts on PVeg pp 226 - 229). You are not responsible for 7.3. Assignments 3 and 4.
5.11 from the text. I think the easiest way to solve this question is to use potential vorticity. The PV of an incompressible fluid parcel is PV = η/h, where h is the thickness of the parcel. You can assume rigid body rotation. Answer the following parts:
(i) What is the relative vorticity ξ at time t, using the given tangential velocity and radius? Assume that the water is undergoing rigid body rotation at all times.
(ii) Assume that a water parcel has a mass equal to 1 kg What is the PV of a cylindrical parcel with radius 1 cm, the given latitude, and the ξ from (i). Note that the choice of the parcel mass will affect you your answer here for PV, but should not affect your later answer to part (iii). The important issue is that whatever mass you assume, you keep it constant.
(iii) Give the answer to (a). Use PV conservation, and determine the initial radius of the disk with ξ = 0. I get an answer slightly different from the answer in the text. Note that water parcels would tend to become elongated in the vertical as they "spin up" and approach the drain (due to the convergent flow), so that h will increase.
(iv) Give the answer to (b). Note that the water parcel is now starting out with some small negative ξ. The question is asking what this initial small negative ξ would be for the contracted and elongated water parcel at the later time with radius 1 cm to have ξ = 0. Do not answer (c). You then need to calculate the tangential velocity from ξ.
6.7 from the text. I think the most straightforward approach is to use (6.2b). Just go through the same arguments as you would with the Northern Hemisphere, but use your left hand rather than right hand in evaluating the cross product, due to the change in sign of f. The warm equatorward side of the jet would now be toward the top of the page. Show a cross-section of the transverse vertical ageostrophic circulations you would expect at the jet entry and jet exit regions (as in the notes), and indicate whether direct or indirect. Note that the prevailing upper tropospheric mid-latitude flow in the SH is eastward as in the NH. Do not change the direction of the jet.
6.9 from the text. Note that this is an unusual arrangement where the thermal wind of a layer (1000 - 850 hPa) is almost opposite to the geostrophic wind at its upper surface (850 hPa), so the geostrophic wind decreases with height from the surface. This also occurs in warm core cyclones (e.g. hurricanes). Presumably, a low pressure system at the surface is generating the local minimum in geopotential height, and therefore a cyclonic vortex. In 6.9a, technically you are looking for the maximum in relative geostrophic vorticity. You don't have enough information to determine the maximum in total vorticity. For 6.9c, the Sutcliffe Development Theorem can be viewed as stating (page 160), "Synoptic scale upward motions, the result of greater divergence aloft than near the surface in any air column, are forced by cyclonic vorticity advection by the thermal wind." Opposite for downward motion.
The question below is an application of the quasi-geostrophic vorticity equation, in which the local geostrophic vorticity tendency is expressed as a sum of the horizontal geostrophic vorticity tendency and the divergence forcing. The latitude is 45.
(i) Point A is 1000 km to the east of point B. Suppose that the geostrophic relative vorticity ξg = 0 at point A, and ξg = 2.0E-05 1/s at point B. The average geostrophic wind between the two points is ug = 30 m/s. Suppose the geostrophic relative vorticity pattern is fixed in space. What is the divergence at point A? Give your answer in /day.
(ii) Suppose that the divergence you have calculated in (i) is constant between 300 hPa and 200 hPa, but zero elsewhere. The pressure velocity has ω = 0 at 200 hPa (the tropopause). Calculate and plot the ω profile between 1000 hPa and 200 hPa. Use p as the vertical axis. Show your answer in hPa/day.
(iii) Calculate the total heating rate dT/dt rate at 500 hPa due to the vertical motion calculated in (ii). Assume that the diabatic heating is zero, and that ρ = 0.5 kg/m3. Convert your answer to K/day. Hint: You have to use a formula from first term.
6.9. (a) There is some flexibility here in what the right answer is. The low level geostrophic wind, following the geopotential height contours would generate positive geostrophic vorticity everywhere where there is cyclonic curvature. The cyclonic curvature looks to be larger at the north and south ends of the low, so there is likely the maximum. If you add the planetary vorticity (f), then the absolute vorticity max is on the north side of the low.
6.9 (b) The thermal wind would be parallel to thickness contours with the warm air on the right.
6.9 (c) I would argue for upward motion on the northeast side of the low, where you would appear to have strongest positive vorticity advection by the thermal wind, and downward motion on the northwest side where you would have negative vorticity advection by the thermal wind.
Answer to Question 2.
(i) The local tendency in geostrophic vorticity would be zero, so there would be a balance between geostrophic vorticity advection and the divergence term. Tou should get DIV = 0.518/day. Note that this is positive.
(i) Starting from ω = 0 at 200 hPa, ω would become more negative and reach a value of -52 hPa/day at 300 hPa. It would then be constant below 300 hPa.
(iii) Should get about -10 K/day. (Cooling since vertical motion is upward).
On a pressure surface, the geopotential increases quadratically with x, but is independent of y: Φ(x,y) = x2. Assume that the temperature decreases linearly in the positive y direction: T(x,y) = -y. (Note: There is no need to use natural coordinates in this problem.)
(i) Obtain expressions for the Vg, T gradient, and Q vectors.
(ii) Draw each vector at the following locations in the (x,y) plane: (0,0), (2,0), (0,2), (-2,0), and (0,-2).
(iii) Would you expect vertical motion with this configuration?
6.8 from the text. This figure uses Figure 6.3A in the text. Do NOT refer to Figure 6.8. However, I found it very difficult to draw Q vectors at all of the indicated locations. It is even harder to guess where the locations of Q vector convergence and divergence are. Therefore, do this only in 2 nearby locations where you think the direction of the Q vector is quite clear, and you can therefore diagnose the vertical motion in one place. At the two location where you diagnose Q, also draw the geostrophic wind, and the derivative of Vg along s.
7.2 from the text. Assume that the warm air is below and the cold air is on top. This enables you to draw the T gradient vector. Assume that the geostrophic vector wind field is equal to V(x,y) = (-y,x). For part (a), calculate the actual value of the Q vector at the point, assuming the absolute value of dT/dn = 1. It is actually probably easiest to go back to the original definition of the Q vector, rather than the natural coordinate form. Do not plug in values for R or p. Note also that there is a misprint in the text for the natural coordinate expression for the Q vector. There is an f instead of a p. For parts (b) and (c), you can treat θ as equivalent to T. For part (c), you should calculate F as given just after Eq. (7.11), with F1 and F2 defined on page 198. The angle β is defined on page 199. I guess they are asking whether the expressions for the geostrophic frontogenesis functions obtained in (b) and (c) are consistent. Note that the expression in part (c) is only valid for wind patterns where the divergence is zero.
7.8 (a). These would be the four quantities listed after Equation (7.8) in the text.
7.8 (c). You would use Equation (7.9b) in the text. But note that (7.9b) is written in an ambiguous manner (not clear which quantity is being squared). The formula sheet has a better version. In the three given equations Assume that the units of x and y are m; U and V are m/s, and T is K. Also assume that T and PT can be treated as equivalent. You need to know how to take the magnitude of the temperature gradient vector. Hint: you do not just add the x and y components.
Assume the geopotential varies with x and y, Φ (x,y) = -y2. Assume the temperature varies with x and y, T(x,y) = -x - 2*y. Note: the natural coordinate expression for the Q vector is mainly useful for graphical application. In mathematical type calculations where you are given variables as a function of x and y, there is usually no advantage to using the natural coordinate form, and it usually causes students to make errors. For example, in situations where the s vector is a function of position, it would be impossible to solve for the Q vector. You are not given f here: do not assume a value for f.
(a) Make a rough sketch (as in a contour plot) of the Φ(x,y) and T(x,y) patterns in the (x,y) plane.
(b) Calculate the geostrophic wind vector Vg (i.e. the general expression, not at special points).
(c) Calculate the temperature gradient vector, and its magnitude.
(d) Obtain an expression for the Q vector Q = (Q1,Q2).
(e) Would you expect induced vertical motion with this Q vector? Explain.
(f) Calculate D, F1, F2. You can assume that the geostrophic wind is a good approximation of the actual wind.
(g) Calculate the rate of change in the temperature gradient from the two dimensional geostrophic flow in this temperature configuration. I don't believe there is any benefit here in using the natural coordinate expression for frontogenesis (i.e. the one using beta) here. The problem with this expression, in calculations, is that in general beta is a function of position.
(b) Vg = (2y/f,0)
(c) (-1,-2) and sqrt(5)
(d) Q = (0,2R/fp)
(e) No, because the Q vector divergence is zero. Note that whether the Q vector is travelling up across T contours has to do with frontogenesis rather than vertical motion.
(f) D = F1 = 0 and F2 = 2/f
(g) You have to use the 2D Frontogenesis function. Looking for the rate of change in the T gradient, not T. Another problem here was using the natural coordinate express using beta, rather than the more general expression in x and y. For reasons I don't fully understand, people using the natural coordinate expression almost always got the wrong answer, perhaps because beta was incorrect. The correct answer is - 4/(sqrt(5)*f).
7.7(a). It is probably be helpful to photocopy the figure from the text.
Read the following case study.
(i) Print out Figure 3, and maybe expand the region around NS. On Figure 3, draw the rough outline of the maximum snowfall region from Figure 5.
(ii) On the map, indicate the location you would have expected ridge building associated with the release of latent heat from the snowfall.
(iii) On the map, indicate the region of expected trough deepening associated with low level cold advection.
(iv) From the geopotential height changes associated with (ii) and (iii), would you have expected increased PVA over the surface low? Explain.
(v) From a snowfall rate of 100 cm/day, calculate the geopotential tendency (in m/day) you would expect at 500 hPa under the following assumptions: (a) you divide the snow amount by 10 to get an effective liquid water amount, (b) all the heat is released between 1000 hPa and 500hPa, and (c) the surface pressure is constant. It might also help if you assume the mean 1000 - 500 hPa temperature prior to the snowfall is 255 K. (in future: maybe break into parts; see answer).
Do the problem on the handout in class (determining the approximate horizontal velocity vectors from isentropic contours on a pressure surface.). Problem 1, Chapter 3 of Lackman. Assume the pressure contours are fixed in space. This then automatically enables you to determine the sign of omega.
(v) Assume a 1m2 column of air. Within this column, within a day, 10 cm of water have converted from vapor to ice. 10 cm of water over 1 m2 is 100 kg. Multiply this by the sum of lv + lf (since get ice from vapor) to get 2.8E+08 J. Between 1000 and 500 hPa, you have 5000 kg/m2. Using the heat released in a day, and the specific heat of dry air, this would give a warming of 55 K. Suppose the initial average temperature between 1000 and 500 hPa is 255 K. Then the standard formula gives a geopotential height at 500 hPa of 5072 m. If this column is heated by 55 K to 310 K (obviously unrealistic), then the new geopotential height at 500 hPa would be 6167 m. This would correspond to a geopotential tendency of (6167 - 5072)m/day = 1095 m/day. This is much larger than you would ever observed, partly since it would be very tough to sustain a snow rate of 100 cm/day for a full day, and partly since much of the heat released would be advected downstream. However, just trying to establish the point that precipitation can increase the local geopotential height.
Answer to Question 3.
I would say A: weak sinking, B: sinking, C: weak, D: rising, E: weak rising. When air is directed toward lower pressure, would expect ω < 0, and rising, and vice versa. This is exactly true in the limit that the pressure isobars are stationary. This is a very simple question which doesn't exactly relate to the immediate course material.
(i) The rain rate at the surface is 50 mm/day. Assume that all of the latent heat from this rain formation is evenly distributed between 700 hPa and 300 hPa. Calculate the heating rate Q. Express your answer in J/kg*s.
(ii) If the temperature T = 260 K and p = 500 hPa, what is the potential temperature?
(iii) Given the heating rate Q from (i), use a formula from first term to find the potential temperature tendency at 500 hPa. In this formula, the heating rate is represented as Q with a dot over it. Express your answer in K/day.
(iv) Assume that the potential temperature tendency at 200 hPa is zero. Estimate the mean PV tendency between 500 hPa and 200 hPa. Assume that the relative vorticity is zero. Express your answer in PVU/day. Assume a latitude of 45 degrees. Assume that the relative vorticity at both pressure levels is zero.
(v) Assume that the potential temperature tendency at 900 hPa is zero. Estimate the mean PV tendency between 900 hPa and 500 hPa. Express your answer in PVU/day. Assume that the relative vorticity at both pressure levels is zero.
(vi) Would you expect the latent heating to increase or decrease the height of the tropopause? Explain. (Where the tropopause in mid-latitudes is defined as a PVU contour).
(i) 0.35 J/kg*s. The amount of rain that falls over 1 m2 in a day is 50 kg. Multiply by Lv to get J/day*m2. Divide by the kg/m2 between the two pressure levels (4082 kg/m2), and by the number of seconds in a day.
(ii) 317 K
(iii) 36.7 K/day
(iv) If the θ at 200 hPa is fixed, and the θ at 500 hPa is increasing, the stability between 500 hPa and 200 hPa will go down, and the PV of the layer will go down as well. Should get -1.23 PVU/day.
(v) In this case the stability between 900 hPa and 500 hPa will go up and the PV will go up. Should get a PV tendency of 0.9 PVU/day.
(vi) Since the PV of the upper troposphere goes down, you would expect the height of the 2 PVU contour to lower, so the tropopause should increase in height.
A shallow water wave is propagating in the x direction. The amplitude of the height displacement is given by a(x) = cos(2πx/λ). Make a plot of the amplitude variation in the x direction between x = 0 and x = λ. On this plot, also show the variation in (i) the hydrostatic pressure p at the bottom of the fluid layer, (ii) upward velocity w of the surface, and (iii) the depth averaged divergence. I am only interested here in the relative phases of these quantities, and these can be determined using your intuition only: no math is required. You must show plots of all four quantities, ideally using different colors. Note that the horizontal pressure gradient acceleration under the assumptions of the shallow water equations is independent of depth, so that the horizontal velocities are also depth independent.
In the shallow water equations, the pressure at any point in the fluid equals the hydrostatic pressure. However, in some situations, this assumption is invalid. A pressure sensor at the bottom of the ocean should not be able to detect instantaneously the pressure changes associated with waves at the surface (i.e. increased pressure under the crests and lower pressure under the troughs.) What is the physical process that partially "shields" hydrostatic pressure changes occurring near the surface and prevents them from being instantly fully detected near the bottom? Just looking for some reasonable speculation rather than mathematical explanation. At what locations would you expect the actual pressure at the bottom of the ocean to be larger than the hydrostatic pressure. Explain. Hint: the hydrostatic approximation is valid only if dw/dt = 0.
(i) The hydrostatic pressure p at the bottom of the fluid layer will be in phase with the vertical displacement a(x).
(ii) The largest upward velocities at the surface will occur when the vertical displacement of the wave is zero, in the direction of propagation, therefore at x = λ/4.
(iii) The maximum depth averaged convergence will occur when the upward velocity of the surface is largest (required by mass conservation: you need most rapid inflow of mass at the point where the rate of increase in the height of the surface is largest). Therefore the divergence will reach its most largest negative value at x = λ/4, and its largest positive value where the rate of decrease in the height of the fluid is fastest, at x = 3λ/4. Note also that the divergence must be zero at the locations where the height is extremal.
Answer to Question 2.
This question is a bit open ended. However, suppose you are at the bottom of the ocean under a wave crest. Then, near the surface, the water is accelerating downward. Note that w = 0 at the crest at the surface, but dw/dt reaches its largest negative value. You could say that the fluid is in partial "free fall". In this case, the fluid is not being held up by the pressure of the fluid below, to the same extent, and the real pressure will be less than the hydrostatic pressure. Conversely, under the troughs, the fluid is being accelerated upward most rapidly by the fluid pressure beneath, so that the real pressure must be larger than the hydrostatic pressure. Therefore, since the real pressure at the bottom of the ocean is reduced where the hydrostatic pressure is largest (under the crests), and the real pressure is increased at the troughs where the hydrostatic pressure is strongest, the amplitude of the oscillation in real pressure at the bottom of the ocean will be less than the oscillation in hydrostatic pressure.