Name:___________________________________ 24 August 2004



Name:___________________________________ 17 February 2022 ATMS 310 INDIVIDUAL EXAM#01Choose the single best answer in Questions (1) – (15), each question is worth four points. Questions (16) - (18) together are worth 40 points. EXAM#01 is worth 100 total points.(1) The SI name for pressure (Pascal) has the symbol ‘Pa’ and units [N m-2]. The ‘N’ in the Pascal units stands for ________ and has the basic SI (or MKS) units [________].(a) Nansen, kg m s-2(b) Nansen, m s-2(c) Newton, kg m s-2, LP#1, slides#7-8(d) Newton, m s-2(2) The governing equations of ATMS 310 are primarily concerned with the conservation of mass, ________, and ________.(a) entropy, enthalpy(b) entropy, energy(c) momentum, enthalpy(d) momentum, energy, LP#1, slide#13(3) In order to apply Newton’s 2nd Law on our rotating planet, we partly account for our ________ frame of reference by introducing the ________ Coriolis force.(a) inertial, apparent(b) inertial, fundamental(c) spinning, apparent, LP#1, slide#14(d) spinning, fundamental(4) The ________ force accounts for both fundamental and apparent forces.(a) Coriolis(b) frictional(c) gravitational, LP#1, slides#24-27(d) pressure gradient(5) Which of the following properties does NOT partly determine the strength of the Coriolis force?(a) divergence of the horizontal wind, LP#1, slide#32(b) latitude of object (e.g., air parcel)(c) rotation rate of earth(d) speed of the horizontal wind(6) A balance between the ________ pressure gradient and ________ forces is known as ________ balance.(a) horizontal, Coriolis, cyclostrophic(b) horizontal, gravitational, hydrostatic(c) vertical, Coriolis, cyclostrophic(d) vertical, gravitational, hydrostatic, LP#1, slide#34Name:___________________________________ 17 February 2022 ATMS 310 INDIVIDUAL EXAM#01(7) The local (________) rate of change of an atmospheric property (e.g., u, v, w, T) is the total (________) rate of change plus ________.(a) Eulerian, Lagrangian, advection, LP#2, slide#100, (b) Eulerian, Lagrangian, rotation(c) Lagrangian, Eulerian, advection(d) Lagrangian, Eulerian, rotation(8) When conducting a scale analysis of the equations of motion for weather disturbances on the synoptic scale, the horizontal velocity scale is three orders of magnitude ________ than the vertical velocity scale and the ________ Number is much smaller than one.(a) greater, Ekman(b) greater, Rossby, LP#2, slides# 32 and 38-41(c) smaller, Ekman(d) smaller, Rossby(9) In words, the ________ states, “…the change in internal energy of the system is equal to the difference between the heat added to the system and the ________ the system.”(a) First Law of Thermodynamics, angular momentum of(b) First Law of Thermodynamics, work done by, LP#2, slide#55(c) Vertical Momentum Equation, angular momentum of(d) Vertical Momentum Equation, work done by(10) The atmosphere shown in the vertical cross section above is statically stable. Hence, ?A is ________ than ?B and the environmental lapse rate is ________ than the dry adiabatic lapse rate.(a) greater, greater(b) greater, less, LP#2, slides#62-63(c) less, greater (d) less, less (11) An air parcel in adiabatic flow displaced upward from ?B in the atmosphere of Problem (10) will become ________ buoyant, and begin to oscillate at its ________.(a) negatively, buoyancy frequency, LP#2, slides#63, 65(b) negatively, inertial frequency(c) positively, buoyancy frequency(d) positively, inertial frequencyName:___________________________________ 17 February 2022 ATMS 310 INDIVIDUAL EXAM#01(12) The above illustration is a force balance schematic in the Northern Hemisphere of the horizontal pressure gradient force (P), Coriolis force (Co), and centrifugal force (Ce) for flow about a ‘regular’ high. This is an illustration of ________ flow wherein the actual (or observed wind, labelled ‘V’) is ________.(a) cyclostrophic, sub-geostrophic(b) cyclostrophic, super-geostrophic(c) gradient, sub-geostrophic(d) gradient, super-geostrophic, LP#3, slides#24-29(13) A ________ atmosphere is one in which the density depends only on the ________ so that ________ surfaces are also surfaces of constant density.(a) barotropic, potential temperature, isentropic(b) barotropic, pressure, isobaric, LP#3, slide#46(c) baroclinic, potential temperature, isentropic(d) baroclinic, pressure, isobaric(14) The CFL condition of the linear one dimensional advection equation requires the Courant number to be less than or equal to ________, which is only a ________ and not a ________ condition for computational stability in a numerical atmospheric model.(a) 1.0, necessary, sufficient, LP#3, slides#61 & 65(b) 1.0, sufficient, necessary(c) 1.5, necessary, sufficient(d) 1.5, sufficient, necessary(15) Given the horizontal flow pattern depicted above, the greatest contribution to relative vorticity comes from the ________ term, causing the paddle-wheel to spin ________ (a.k.a., ________ vorticity).(a) curvature, clockwise, anticyclonic(b) curvature, counterclockwise, cyclonic(c) shear, clockwise, anticyclonic(d) shear, counterclockwise, cyclonic, LP#4, slides#21 & 23Name:___________________________________ 17 February 2022 ATMS 310 INDIVIDUAL EXAM#01(16) 10 pointsThe horizontal map above depicts a region in the Northern Hemisphere (N. H.) in which layer-mean isotherms (thin line contours) are drawn, along with three vectors (thick lines without the vector ‘head’), and a ‘Warm’ label indicating the position of a warm air mass. Draw the vector heads and label the vectors on the map with either the Vg,low, Vg,upper, or VT labels to represent the lower-level geostrophic wind (Vg,low), the upper-level geostrophic wind (Vg,upper), and the thermal wind (VT) vectors. Finally, write in the space below whether the scenario depicted on the map above is one of warm air advection (WAA), cold air advection (CAA), or neither and give evidence supporting your conclusion.Horizontal temperature advection type:VT vector must be parallel to layer mean isotherms and point north (cold air to left of VT direction). There are two options from this point, [1] Vg,low is smallest magnitude vector and points toward northeast, and Vg,upper is longest magnitude vector and points north/ northeast. Winds turn CCW with height (from low to upper), hence, backing => CAA, or[2] Vg,upper is smallest magnitude vector and points toward southwest, and Vg,low is longest magnitude vector and points south/ southwest. Winds turn CW with height (from low to upper), hence, veering => WAAName:___________________________________ 17 February 2022 ATMS 310 INDIVIDUAL EXAM#01 (17) 15 pointsFigure 17. Geopotential height contours for the 1000 hPa isobaric surface oriented in the north-south direction and located 160 km apart. Location of station X is noted on the map and is positioned at 45oN.The observed 1000 hPa wind speed and direction at station X is 18 m s-1 and 330o, respectively. Balanced flow is observed at this moment in which the horizontal pressure gradient force (PGF), Coriolis force (Co), and frictional force (Fr) are in balance. Compute the zonal and meridional components of the three forces and plot their directions (along with the observed wind) assuming that Fr acts approximately in the opposite direction from the observed wind. Determine in this situation if the observed wind is sub-geostrophic, geostrophic, or super-geostrophic.Coriolis parameter = 10.28445x10^-5 s^-1 Ug = zero (east-west PGF only)Vg = (9.81 m s^-2)/(f) * (10 – 50)m/(160x10^3 m) = -23.85 m s-1U=-18 sin(330o) = 9.0 m s-1V=-18 cos(330o) = -15.6 m s-1Coriolis force = f V i + (-f U) j = [(10.28445x10^-5 s^-1)*(-15.6 m s-1)] i +[(-10.28445x10^-5 s^-1)*(9.0 m s-1)] = -1.603x10^-3 m s-2 i + -9.256x10^-4 m s-2 jPGF = -g*(dz/dx) i= - (9.81 m s^-2)* (10 – 50)m/(160x10^3 m)i = +2.4525x10^-3 m s-2iFr = (+2.4525x10^-3-1.603x10^-3) m s-2 i + 9.256x10^-4 m s-2 j= -8.4931x10^-4 m s-2 i + 9.256x10^-4 m s-2 jWinds are sub-geostrophic; winds point toward low pressure => |18| < |-23.85|Name:___________________________________ 17 February 2022 ATMS 310 INDIVIDUAL EXAM#01 (18) 15 pointsFigure 18. Zonal wind speed (pink contours, interval of 2 m s-1 ) and meridional wind speed (red contours, interval of 2 m s-1) at the 600 hPa level. Four “+” markers indicate positions 1o latitude or longitude due north, south, east, and west of the location of interest. The grid on the map background is plotted every 2o latitude or longitude and the weather station (not shown) is located at 47oN, 123oW.The 600 hPa level zonal wind components directly east and west of a weather station (not shown) are 18 and 21.5 m s-1, respectively, and the meridional wind components directly north and south of the weather station are ?5.0 and ?15.0 m s-1, respectively. Compute the 700-500 hPa layer mean divergence using the horizontal wind information at the 600 hPa level. Use the continuity equation in isobaric coordinates to determine the vertical motion [Pa s-1] at the 700 hPa level if vertical motion at the 500 hPa level is zero (no vertical motion at the 500 hPa level). Assume 1 degree longitude equals 111 km [cos(Φ)], where Φ is latitude. As usual, 1 degree latitude equals 111 km. Please use centered differencing in your horizontal divergence calculations.Hdiv => du/dx + dv/dy = [(18-21.5)m/s]/[2x111x10^3xcos(47)] + [(-5.0 – {-15})m/s]/[2x111x10^3] = +2.1928x10^-5 s-1 Hdiv = -(omeg700 – omeg500)/(70000 – 50000)Paomeg700 = (70000-50000)Pa*(-Hdiv) = ?0.4386 Pa s-1 {rising motion at 700 hPa}Name:_____________________________________Work to be included for Problem _____ of Exam _____ in class ATMS _____ ................
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