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Dynamic Meteorology (Lecture 5)

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Dynamic Meteorology (Lecture 5) 10/7/11 Dynamic Meteorology (lecture 5) Topics Fronts and midlatitude cyclones Life-cycle of a mid-latitude cyclone Frontogenesis and frontolysis Q-vector Pressure as vertical coordinate Properties of thermal wind balance Ageostrophic wind Jetstreak ([email protected]) (http://www.phys.uu.nl/~nvdelden/dynmeteorology.htm) Conceptual model Green:clouds Figure 1.58. 1 10/7/11 Weathermap 2 10/7/11 Classical conceptual models of fronts Left panel: a cold front. Right panel: a warm front. (Source: Wikimedia Commons) The impression is given that less dense warm air is lifted by the advancing cold air. This is wrong! Life cycle of mid-latitude cyclone Starts with a “wave” in the “Polar Front”. This wave grows due to the instability of the thermal wind. a few days Figure 1.59. 3 10/7/11 polar front time Why this life-cycle? Let us first define what a front is in mathematical terms; Then study mechanisms that lead to the formation of fronts; Then study mechanisms that lead to the characteristic frontal morphology, seen in previous slides Questions about fronts • What is a front? • How are fronts formed? • Why are fronts associated with clouds • What is the relation between fronts and jetstreams? 4 10/7/11 What exactly is a front? A front separates two air masses Air masses are characterized by: potential temperature, humidity or potential vorticity. Why? Sometimes gradients of these quantities can be very sharp Potential vorticity on 320 K level (approx. 10 km) in northern hemisphere on 6 Oct. 2011, 12 UTC. The standard definition of front-intensity is in terms of temperature Definition of front-intensity A front separates two air masses Suppose: air masses are characterized by potential temperature, 2 Intensity of front is measure by (∇hθ) ⎛ ∂ ∂ ⎞ where ∇h ≡ ⎜ , ,0⎟ ⎝ ∂x ∂y ⎠ € € 5 10/7/11 Frontogenesis function, Q-vector 2 d(∇ hθ) = 2Q ⋅∇ hθ dt d∇ hθ Q-vector Q ≡ (Q1,Q2 ) ≡ dt € If Q-vector is perpendicular to isentrope, potential temperature gradient increases/decreases.€ If Q-vector is parallel to isentrope, potential temperature gradient changes direction (isentrope rotates), but front does not intensify Derivation of Q-vector equation y-component: 2 d ⎛ ∂θ ⎞ ∂θ d ⎛ ∂θ ⎞ ⎜ ⎟ = 2 ⎜ ⎟ dt ⎝ ∂y ⎠ ∂y dt ⎝ ∂y ⎠ ∂ ⎛ dθ ⎞ ∂ ⎛ ∂θ ∂θ ∂θ ∂θ ⎞ ⎜ ⎟ = ⎜ + u + v + w ⎟ = 0 (no heating!) ∂y⎝ dt ⎠ ∂y ∂t ∂x ∂y ∂z € ⎝ ⎠ ⎛ ∂ ∂θ ∂ ∂θ ∂ ∂θ ∂ ∂θ ∂u ∂θ ∂v ∂θ ∂w ∂θ ⎞ ⎜ + u + v + w + + + ⎟ = 0 ⎝ ∂t ∂y ∂x ∂y ∂y ∂y ∂z ∂y ∂y ∂x ∂y ∂y ∂y ∂z ⎠ € d ∂θ ∂u ∂θ ∂v ∂θ ∂w ∂θ ≡ Q = − − − dt ∂y 2 ∂y ∂x ∂y ∂y ∂y ∂z € € 6 10/7/11 Frontogenetic terms Arrows: streamlines/x- or y-axis C:Cold W:Warm d ∂θ ∂u ∂θ ∂v ∂θ ∂w ∂θ ≡ Q = − − − dt ∂y 2 ∂y ∂x ∂y ∂y ∂y ∂z Which of the two frontogenetic processes € shown in the figures is represented mathematically in the equation above? confluence tilting Illustration of frontogenesis Air parcel in a b a rotating fluid* A Suppose: A horizontal solid B B lines are isentropes on a pressure surface frontogenesis frontolysis c B A *flow field is not exactly circularly symmetric 7 10/7/11 Frontogenesis in baroclinic wave Model simulation: blue contours: absolute value temperature gradient on isobaric surface (864 hPa). Red arrows: wind-vector. (figure 1.63) 30 hours apart Labels in units of 10-5 K m-1 Frontogenesis in baroclinic wave Model simulation: Blue contours: geopotential height labeled in m (864 hPa) cyan contours: isotherms on isobaric surface labeled in °C Red arrows: Q-vector. (figure 1.62) 8 10/7/11 Q-VECTOR, POTENTIAL TEMPERATURE (cyan) and HEIGHT (blue) THICK CONTOURS: HEIGHT:1500.0 m; TEMPERATURE: 0.0 °C; CONTOUR-INTERVAL: HEIGHT: 50.0 m; TEMPERATURE: 5.0 °C L rotating isentropes H H frontogenesis Q-vector in midlatitude depression pe model run 3520 837hPa : |Q1|=5*10^-11 K m^-1 s^-1 (min. value:10^-11 K m^-1 s^-1) 60.00 hrs Pressure coordinate Pressure: p(x, y,z,t) ∂p ∂p ∂p ∂p Chain rule: dp = dt + dx + dy + dz ∂t ∂x ∂y ∂z € ∂p On isobaric surface: dp = 0 and = 0 ∂t € Without loss of generality: dx = 0 ∂p ∂p ∂p Therefore: € dy + dz = 0 ? dy − ρgdz = 0 ∂y ∂z ∂y € ⎛ ∂p⎞ ⎛ ∂z⎞ ⎛ ∂Φ⎞ ⎜ ⎟ = ρg⎜ ⎟ ≡ ρ⎜ ⎟ Φ ≡ geopotential ⎝ ∂y⎠ z ⎝ ∂y⎠ p ⎝ ∂y ⎠ p € € € € 9 10/7/11 Geostrophic & Hydrostatic Balance g ⎛ ∂z⎞ g ⎛ ∂z⎞ Geostrophic wind: vg = ⎜ ⎟ ;ug = − ⎜ ⎟ f ⎝ ∂x⎠ p f ⎝ ∂y⎠ p ∂p ∂z RT Hydrostatic balance: = −ρg ? = − ∂z ∂ln p g € ∂vg R ∂T ∂ug R ∂T Thermal wind equation: = − ; = € ∂ln p f €∂x ∂ln p f ∂y € Properties of the thermal wind *Horizontal derivatives with 1 ∂Φ 1 ∂Φ * pressure held constant Geostrophic wind: v = ;u = − g f ∂x g f ∂y ∂vg ∂vg R ∂T ∂ug ∂ug R ∂T Thermal wind equation: p = = − ; p = = ∂p ∂ln p f ∂x ∂p ∂ln p f ∂y € ∂v g R In vector notation: = − kˆ × ∇ T ∂ ln p f € R p The thermal wind: v ≡ v p − v p = − ∫ 1 (kˆ × ∇ T )d ln p T g ( 1) g ( 0 ) p0 € f The thermal wind is parallel to the isotherms! € 10 10/7/11 R p v ≡ v p − v p = − ∫ 1 (kˆ × ∇ T )d ln p T g ( 1) g ( 0 ) p0 f The thermal wind is parallel to the isotherms! Turning€ of the wind with height See problem 1.18 Cold advection: backing with height Warm advection: veering with height The thermal wind in a mid-latitude cyclone FIGURE 1.60. NOAA image in channel 4 (IR) made on March 3, 1995, at 0157 UTC. 11 10/7/11 The thermal wind in a mid-latitude cyclone 850 hPa The thermal wind in a mid-latitude cyclone 500 hPa 12 10/7/11 Ageostrophic wind dv dv ˆ v = v g + v a = − fkˆ × v − ∇ Φ = − fk × v a dt dt 1 ⎛ dv ⎞ Ageostrophic wind v = kˆ ×⎜ ⎟ a f ⎝ dt ⎠ € € € ⎛ ⎛ ⎞ ⎞ 1 ˆ ∂ 1 ˆ ∂v a ∂v v a = k ×⎜ ⎜ k × ∇Φ⎟ + + (v ⋅ ∇)v +ω ⎟ f ∂t f ∂t ∂p ⎝ ⎝ €⎠ ⎠ geostrophic wind: fkˆ v × g = −∇ Φ € Isallobaric wind Inertial-advective wind 1 ˆ v g = k × ∇ Φ f Isallobaric wind can sometimes be a very important contribution € € gray: isobars[hPa] solid: isallobars[hpa/3hr] Example 28 May 2000, 0600 UTC Thick: isallobaric wind Thin: geostrophic wind Isallobaric+geostrophic wind Observed wind c 13 10/7/11 Jetstreak dv 0 v < 0 1 ⎛ dv ⎞ < a v = kˆ ×⎜ ⎟ dt a ⎝ ⎠ dv f dt > 0 dt € € € € v a > 0 € PROBLEMS Problems 1.14, 1.15, 1.16, 1.17. Instead of the case given in the lecture notes, we will do these exercises for the case of yesterday (6 October 2011): a cold front passing over the Netherlands. Some of you will analyse the warm side of the front and others the cold side of the front. You will calculate quantities such as the geostrophic wind, the vorticity and divergence. 14 10/7/11 Weathermap Weathermap 15 10/7/11 Weathermap Weathermap 16 10/7/11 Weathermap 17 10/7/11 18 10/7/11 19 10/7/11 20 10/7/11 21 10/7/11 FORECAST FROM LAST WEEK 22 10/7/11 Data needed for these exercises We will use the radiosonde data from six stations: Ekofisk (01400, 56.5, 3.2), Scheswig (10035, 54.5, 9.5), Hannover (10238/ETGB, 52.5, 9.7), De Bilt (06260/EHDB, 52.1, 5.2), Essen (10410/EDZE, 51.4, 7.0) and Meiningen (10548, 50.5, 10.4) The following slides show the radiosonde-measurements plotted in a tephigram. Data needed for these exercises Hannover/Bergen Ekofisk Schleswig De Bilt Essen Meiningen 23 10/7/11 24 10/7/11 25 10/7/11 26 10/7/11 PROBLEM 1.14. Analysis of observations Compute the 500 hPa wind at the points (x0, y0)=(5°E, 54°N) and (x0, y0)=9°E, 51.5°N) on on 6 October 2011, 12 UTC. First decide which measurements you will use (see the upper panel of figure 1.54). Compute the 500 hPa isobaric height at the gridpoint (x0, y0) =(7°E, 50°N) and (x0, y0)=(8°E, 56°N). Also compute the isobaric height- and wind- gradients. We can use this information to compute the geostrophic wind at 500 hPa (problem 1.15) and the gradient wind (problem 1.16). The observations and positions of the measuring stations can be obtained from the following website: http://weather.uwyo.edu/upperair/sounding.html. PROBLEM 1.15. Geostrophic wind Estimate the geostrophic wind (ug, vg) on 6 October 2011, 12 UTC at the points (x0, y0)= (5°E, 54°N) and (x0, y0)=9°E, 51.5°N) from the gradient of the isobaric height (problem 1.14). Compare the geostrophic wind with the “actual” wind at the same point. PROBLEM 1.16. Non-linear balance and gradient wind Calculate the gradient wind from eqs. (1.136) and (1.137) at the points (x0, y0)=(5°E, 54°N) and (x0, y0)=9°E, 51.5°N) on the 500 hPa level on 6 October 2011, 12 UTC with the velocity- and height gradients calculated in problem 1.14. Compare the gradient wind with the geostrophic wind (problem 1.15). PROBLEM 1.17. Relative vorticity and horizontal divergence Estimate the relative vorticity and the horizontal divergence on the 500 hPa isobaric level on 6 October 2011, 12 UTC at the points (x0, y0)=(5°E, 54°N) and (x0, y0)=9°E, 51.5°N) with the velocity gradients calculated in problem 1.16.
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