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.

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Weathermap

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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.

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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?

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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 ⎠ €

€

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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 €

€

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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

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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)

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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 € €

€ €

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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!

€

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   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.

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The thermal wind in a mid-latitude cyclone

850 hPa

The thermal wind in a mid-latitude cyclone

500 hPa

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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

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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.

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Weathermap

Weathermap

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Weathermap

Weathermap

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Weathermap

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FORECAST FROM LAST WEEK

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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

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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. Compare the relative vorticity with the planetary vorticity. Also give an estimate of the Rossby number (section 1.23). Compare the horizontal divergence at the two points (eq. 1.138b).

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5 groups of 2: 3 doing the cold side, of which one is using the measurement of Nordeney and the other two of Schleswig (respectively, 700 and 500 hPa).

Other two do the warm side (respectively, 700 and 500 hPa).

Each group shortly presents the results next week (Friday afternoon, 14 October 2011)

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