vis = 74 visibility = vis - 50, for 56 vis 80
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1 ATSC 201 Fall 2018 Assignment 11 Answer Key Total marks out of 69 Chapter 9: A1h, A7a, A8h, A10all Chapter 12: A1h, A7h, A8h, A11g, E21, E23 Chapter 9 A1h) Find the pressure "reduced to sea level" using the following station observations of pressure, height, and virtual temperature. Assume no temperature change over the past 12 hours: h) P = 88kPa, z = 1200m, Tv = 12C. Given: Pstn (kpa) = 88 zstn (m) = 1200 Tv (degc) = 12 Tv (K) = Find: PMSL =? kpa First, use eqn 9.2: where ϒsa = K/m and Tv(to - 12h) = Tv(to) = K av Tv* = K Use eqn. 9.1: where a = PMSL = 29.3 m/k kpa Check: Units ok. Physics ok. Discussion: At some stations, the PMSL may be located below the ground!
2 A7a) A8h) Photocopy the USA weather map in Fig. 9.19, and analyze it by drawing isopleths for: b) temperature (isotherms) every 5 F. Discussion: Any kind of isopleth (isotherms included) will never cross over itself. Using the Canadian weather map of Fig. 9.20, decode the weather data for the station assigned by your instructor: h). T = 17 degc Td = 13 degc vis = 74 visibility = vis - 50, for 56 vis 80 vis = 24 km P = kpa clouds = cumulus and stratocumulus sky covered by low clouds 8/8 cloud base height = m above ground level wind = calm Pressure change = rising Net change of 0.1kPa in the past 3 hr sky cover = overcast or 100% Discussion: Using a map with multiple station plots, we can easily draw the isobars, isotherms, cloud coverage, and precipitation regions. We can then use the analysis to infer where the low pressure centre and its associated cold and warm fronts are.
3 A10all) Discussion: Weather forecasters in training at environment canada have to draw diagrams like this every morning with the surface obs of that morning! Chapter 12 A1h) Identify typical characteristics of the following airmass: h) mt. Given: Airmass = mt = maritime Tropical. Characteristics: Humid and warm. Formed in the subtropical high belt over ocean. Discussion: From Fig. 12.4, mt airmasses would most likely form between deg N and deg S (rough estimation). A7h) Find the external Rossby radius of deformation at 60 latitude for a cold airmass of thickness 500m and Δθ ( C) of: h) 16. Assume a background temperature of 300K. Given: Δθ ( C) = 16 or 16K because it's a difference φ ( ) = 60 H (m) = 500
4 T (K) = 300 Find: λr =? m Use eqn. 12.5: where g = 9.8 m/s^2 To find fc use eqn 10.16: fc = 2*Ω*sinφ where Ω = 7.29E-05 /s fc = 1.26E-04 /s Assume a dry air mass so that Δθ ~= Δθv, and T ~= Tv. λr (m) = 1.28E+05 m km Check: Units ok. Physics ok. Discussion: above the external Rossby radius of deformation is where the jet stream would be. The fact that cold air masses cannot redistribute cold air to the equator means some other forces will try to redistribute this heat. A8h) Find and plot the airmass depth and geostrophic wind as a function of distance from the front for the cases of the previous exercise. Assume a background potential temperature of 300K. Given: Δθ ( C) = 16 φ ( ) = 60 H (m) = 500 T (K) = 300 From A7h the final spillage distance of the front a = λr = km. a (m) = 1.28E+05 y is the distance behind a. Use eqn 12.6:
5 g = 9.8 m/s^2 Assume a dry air mass so that Δθ ~= Δθv, and T ~= Tv. Use eqn 12.7: y (0-a)(km) Ug (m/s) h (km)
6
7 Check: Units ok. Physics ok.
8 Discussion: this h vs y plot confirms Fig b, where the depth of the cold air mass increases with distance into the cold air mass. Ug vs y also confirms Fig b; Ug is strongest at the air mass boundary (y=0) and then decreases in strength further back into the cold air mass. Negative values indicate Ug is blowing to the East. How geostrophic wind and depth of cold air changes behind a cold front h (frontal depth, km) h Ug Distance from frontal boundary (km) Ug behind the front (m/s) A11h) Plot dryline movement with time, given the following conditions. Surface heat flux is constant with time at kinematic rate 0.2 K*m/s. The vertical gradient of potential temperature in the initial sounding is ϒ. Terrain slope is s = Δz/Δx. g) ϒ (K/km) = 12, s = 1/200.
9 Given: ϒ = 12 K/km K/m s = FH = 0.2 K*m/s Plot Δx vs Δt (the dryline movement with time). From Sample Application, QAK = FH*Δt. Use eqn 12.15: Δt (hrs) Δx (km) Δx (km) Dryline position vs time Δt = time after Sunrise (hrs)
10 Check: Units ok. Physics ok Discussion: This plot does not take into account nighttime when convective turbulence ceases, and prevailing low altitude easterlies advect moist air back towards the west E21) Background: Recall that a frontal zone separates warmer and cooler airmasses. The warm airmass side of this zone is where the front is drawn on a weather map. This is true for both cold and warm fronts. Issue: AFTER passage of the cold front is when significant temperature decreases are observed. BEFORE passage of a warm front is when significant warming is observed. Question: Why does this difference exist (ie. AFTER vs BEFORE) for the passage of these two fronts? Solution: Fronts on a weather map are always drawn on the warm side of the surface frontal zone. For a warm front, the baroclinic zone (the region with the strongest horizontal temperature gradient indicated by the tightest isotherms) is ahead of the warm front. For this reason, you will feel the rapid warming before the warm front approaches.
11 For a cold front the baroclinic zone is behind the cold front, so you will feel the rapid cooling after the cold front has passed. E23) Suppose you saw from your barometer at home that the pressure was falling. So you expect a front approaching. What other clues can you use to determine if the approaching front is warm or cold. Discuss along with possible pitfalls in this method. 1) Winds: South/Southwesterly winds indicate a cold front. Southeasterly or near easterly winds indicate a warm front. Pitfall: Wind directions are also affected by surface features, such as buildings and trees if in an urban area, and mountains (valley winds) if in a region with complex terrain. 2) Clouds: Gradually increasing cloud density (from cirrus through altostratus, stratus, and then nimbostratus) means a warm front. Towering cumulonimbus clouds indicate a cold front. Pitfall: Hard to measure if not enough moisture. (Other reasonable answers are also accepted.)
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