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Unformatted text preview: M.Sc. in Computational Science Fundamentals of Atmospheric Modelling
´ Peter Lynch, Met Eireann
Mathematical Computation Laboratory (Opp. Room 30) Dept. of Maths. Physics, UCD, Belﬁeld. January–April, 2004. Lecture 5 Steady Vortical Flows 2 The TaylorProudman Theorem
In deriving the Shallow Water Equations, we made the assumption that the horizontal velocity is independent of depth. Although dynamically consistent, this may seem an artiﬁcial limitation. However, we will show that rotation acts as a constraint on the ﬂow, so that under certain circumstances, variations in the direction of the spin axis are resisted. Taking the curl of the geostrophic equation (f constant): × (k × V ) = 0 Let us compute the components in Cartesian coordinates: i j k ∂/∂x ∂/∂y ∂/∂z = ×(−v, u, 0) = −v u 0 − ∂u ∂v ∂u ∂v ,− , + ∂z ∂z ∂x ∂y = 0. Theorem.
For incompressible, inviscid, hydrostatic, geostrophic ﬂow on an f plane, the velocity is independent of height. Proof. We assume the density is constant. We also ignore variations of the Coriolis parameter f . We assume geostrophic and hydrostatic balance: 1 ∂p fV = k × p , = −gρ . ρ ∂z
3 Combining this with the continuity equation · V = 0 we get ∂u ∂v ∂w = = = 0. ∂z ∂z ∂z This result indicates that, under the assumptions of the theorem, the motion is quasitwodimensional, with velocity independent of depth. If the bottom is ﬂat, w ≡ 0. The result surprised G. I. Taylor, who wrote [Taylor, 1923]: “The idea appears fantastic, but the experiments . . . show that the motion does . . . approximate to this curious type”.
4 The CSU Spin Tank
The TaylorProudman Theorem can be demonstrated beautifully by means of spintank experiments. A Spintank or rotating dishpan has been constructed at Colorado State University to demonstrate various geophysical ﬂuid phenomena. It was built with a small budget ($3000) and is portable. A description of the spintank at CSU is given at http://einstein.atmos.colostate.edu/∼mcnoldy/spintank/ At this site, a number of experiments are described. There are several MPEG loops showing the results of these experiments. See also the article in Bull. Amer. Met. Soc., December, 2003.
This Journal is freely available online.
5 TaylorProudman Column Schematic diagram of tank TaylorProudman Column 6 7 8 Simple Steadystate Solutions 9 10 Polar Coordinates
The ﬂow will be assumed to be axially or circularly symmetric, and it is convenient to introduce (cylindrical) polar coordinates. Let R and θ be the radial distance and azimuthal angle. Let U and V be radial and azimuthal components of velocity. Axially Symmetric Steady Flow
We will study some steadystate or timeindependent solutions. We assume that the radial velocity vanishes: U ≡ 0. We assume a steady state: ∂V = 0, ∂t We assume axial symmetry: ∂V = 0, ∂θ ∂p = 0. ∂t ∂p = 0. ∂θ Deﬁnition:
Flow with f V > 0 is called cyclonic. Flow with f V < 0 is called anticyclonic. Exercise: Show that cyclonic motion spins in the same sense as the earth, and anticyclonic motion spins in the opposite sense.
11 12 Some Vector Calculus
For polar (cylindrical) coordinates, we have ∂Φ 1 ∂Φ Φ= i+ j ∂R R ∂θ 1 ∂ (RU ) 1 ∂V ·V = + R ∂R R ∂θ 1 ∂ (RV ) 1 ∂U k· ×V = − R ∂R R ∂θ Since we have assumed U ≡ 0, the divergence and vorticity become 1 ∂V δ ≡ ·V = =0 R ∂θ 1 ∂ (RV ) ∂V V ζ ≡k· ×V = = + R ∂R ∂R R For solid body rotation, V = ωR where the angular velocity ω is constant. Then the vorticity becomes ∂ (ωR) ωR ζ= + = 2ω . ∂R R 13 The Coriolis term is f k × V = f k×(U i + V j) = −f V i + f U j . The pressure gradient force is 1 1 ∂p 1 ∂p p= i+ j or ρ ρ ∂R R ∂θ The directions of the unit vectors change with θ: ∂i ∂j = j, = −i . ∂θ ∂θ The advection may be calculated: V = U i + V j, so ∂ V∂ (U i + V j) + V· V = U ∂R R ∂θ ∂U V ∂U ∂V V ∂V =U + i+ U + j ∂R R ∂θ ∂R R ∂θ ∂ j V 2∂ j ∂ i UV ∂ i +U 2 + + UV + . ∂R R ∂θ ∂R R ∂θ When we assume azimuthal symmetry and vanishing radial velocity, all terms except the last one vanish: V2 i. R This is the usual expression for the centripetal acceleration. V· V=−
14 Aside: Natural Coordinates
Let s be a unit vector parallel to the velocity V, and let n be a unit vector perpendicular to s, pointing to its left. g h=g ∂h 1 ∂h i+ j ∂R R ∂θ The balance of forces in the radial direction now becomes V2 ∂h + fV − g = 0. R ∂R
In the azimuthal direction, all terms vanish. ( ∗) We will look at several solutions of (*). But ﬁrst, we digress to consider Natural Coordinates. Let R be the radius of curvature of the streamline. If the centre of curvature is in the direction of n, we take R positive, and call the ﬂow cyclonic. For R < 0, we call it anticyclonic ﬂow. (NHS Convention)
15 16 We write the velocity as V = V s where V = ds/dt is the speed. Note that both V and s vary along the trajectory. Positive curvature (R > 0) means ﬂow turning to the left. Negative curvature (R < 0) means ﬂow turning to the right. ds n Exercise: Show that =. ds R
In a distance ∆s along the curve, the ﬂow turns through and angle θ = ∆s/R. The unit vector s turns through the same angle, while its length remains unchanged. Therefore θ = ∆s. Moreover, this gives ∆s 1 = ∆s R The direction of ∆s is the same as n for positive R and opposite for negative R. Therefore ds n =. ds R which is the desired result. The rate of change of s along the trajectory is ds n ds ds ds n =, so that = =V . ds R dt ds dt R Then the acceleration is dV d(V s) = = dt dt ds dV s+V dt dt = V2 dV s+ n dt R . The balance of forces perpendicular to V is: V2 ∂h + fV + g = 0. R ∂n (∗∗) Equation (**) is equivalent to the result obtained above using polar coordinates. 17 18 The results in polar coordinates and natural coordinates must agree. However, they appear diﬀerent, because the sign conventions are diﬀerent. To avoid confusion, we state the conventions here: In Natural Coordinates: • The quantity V (the speed) is always positive • The radius of curvature R may be positive or negative In Polar Coordinates: • The quantity V may be positive or negative • The radius R is always positive We will not use natural coordinates. They are described above because they appear in Holton’s text. Geostrophic Balance (Ro
The balance of forces perpendicular to V is: 1) V2 ∂h + fV − g = 0. R ∂R
For synoptic ﬂow, the centrifugal force is small: V2 R Thus, V = VGeos ≡ g ∂h f ∂R fV ⇐⇒ Ro = V fR 1; This is a state of geostrophic balance, as discussed already. The ﬂow is cyclonic around low pressure and anticyclonic around high pressure. 19 20 Digression: Tidal Range in Irish Sea Geostrophically balanced ﬂow around low and high pressure. Within the Irish Sea, the maximum tidal ranges occur on the Lancashire and Cumbria coasts, where the mean spring tides have a range of 8m. At Carnsore Point, the range is less than 2m.
22 21 Morecambe Bay
Morecambe Bay is broad and shallow and has a large tidal range of up to 10.5 metres. The Bay is always changing with waves, tides and currents moving sediment from the Irish Sea into the embayment. Muds and silts settle, forming banks and marshes, whose edges are shaped by wandering river channels.
Shoreline Managament Plan Cyclostrophic Balance (Ro
Recall the balance of forces perpendicular to V: ∂h V2 + fV − g = 0. R ∂R Let us assume the Coriolis term is negligible: 1) ∂h V2 ∂h . = 0; V = ± gR −g ∂R R ∂R This is cyclostrophic balance. For V to be real, we must have ∂h/∂R > 0, i.e., low pressure at the centre! The centrifugal force is much bigger than the Coriolis force: V2 R fV , which implies Ro ≡ V V = 2ΩL f R 1. Cyclostrophic balance is found in small vortices. It may be cyclonic or anticyclonic.
23 24 Example: A Toy Tornado
The centrifugal term is important in tornado dynamics. [1] Calculate the Rossby Number for typical scale values. [2] Calculate the variation of depth with radius. [2] Let us assume solidbody rotation, ω = V /r constant. This is not very realistic, but is done to simplify the analysis. The balance of forces (Centrifugal = Pressure Gradient) gives ∂h ∂h V2 −g =0 or ω 2R − g = 0. R ∂R ∂R This gives the pressure gradient: ∂h ω 2 =R ∂R g and we can integrate this from 0 to R to get ω2 2 R 2g which is a parabolic proﬁle. This completes the solution. h = h0 + [1] We take the velocity and length scales to be V = 30 m s−1 and L = 300 m. Assuming f = 10−4 s−1, we have V 30 Ro ≡ = −4 = 103 1. f L 10 · 300 Almost every tornado is found to rotate cyclonically. However, there are documented cases of rotation in an anticyclonic direction. Smaller vortices such as dust devils are found to rotate indiﬀerently in both directions.
25 Stir your coﬀee rapidly. Study the shape of the surface.
26 Exercise: Down the Plughole
A popular legend holds that water draining from a bath spins cyclonically, that is, counterclockwise in the northern hemisphere. Show that this is a myth.
We take the velocity and length scales to be V = 1 m s−1 and L = 1 m. Assuming f = 10−4 s−1, we have V 1 Ro ≡ = −4 = 104 1. f L 10 · 1 Thus, the Coriolis eﬀect is completely negligible at this scale: Coriolis force Centrifugal force Inertial Balance
Once more, the balance of forces is: ∂h V2 + fV − g = 0. R ∂R If the pressure force is neglected, we get V2 + fV = 0 , which implies V = −f R . R This is inertial ﬂow. Since, by convention, R > 0 we must have f V < 0. That is, inertial ﬂow is anticyclonic. The angular velocity is independent of radius: ω = V /R = −f . This implies solid body rotation. The period for revolution is 2π T= ≈ 17.5 hours at 45◦N . f This is the time taken by a Foucault Pendulum to rotate through 180◦ and is called a half pendulum day.
28 Experiments conﬁrm that the two spin directions are equally likely. The balance in the bath is cyclostrophic, not geostrophic. Extremely delicate laboratory conditions are required to achieve a preference for cyclonic rotation on the bathroom scale.
27 Inertial oscillations are relatively unimportant in the atmosphere, where pressure gradients are rarely negligible, but are more important in the ocean. They are normally superimposed on a background ﬂow, leading to cycloidal patterns such as shown below. Gradient Balance
When all terms are retained, we have Gradient balance: V2 ∂h + fV − g = 0. R ∂R Using the deﬁnition of geostrophic wind VGeos = (g/f )∂h/∂R, V2 + f (V − VGeos) = 0 . ( ∗) R We may solve this quadratic equation for the wind speed: 1 −f R ± f 2R2 + 4f RVGeos . V= 2 For a given pressure gradient, there are two possible solutions, one regular and one anomalous. Not all solutions are physically possible. N.B. A strong anticyclonic gradient ∂h/∂R 0 implies f VGeos 0 so that V is complex. Strong gradients are not found near high pressure centres in the atmosphere. this becomes 29 30 For cyclonic ﬂow the gradient wind speed is less than the geostrophic speed; for anticyclonic ﬂow it is greater. This follows from (*): V2 . V = VGeos − fR So, • Cyclonic ﬂow (f V > 0): V  < VGeos (Subgeostrophic) • Anticylonic ﬂow (f V < 0): V  > VGeos (Supergeostrophic) Exercise:
Show that, for a given wind speed, the pressure gradient is greater when the ﬂow is cyclonic, and less when the ﬂow is anticyclonic. Discuss the implications of this for synoptic analysis. Exercises
(1) Download a weather chart with isobars and winds. • Study how the ﬂow direction and speed depend on the pressure gradient. • Locate a similar map for the southern hemisphere. • Find a similar map for the tropics. How are the pressure and wind ﬁelds related near the equator? (2) Search for “Coriolis Force”. Look for animations that show the eﬀect of this deﬂecting force. (3) Search for an example of an anomalous tornado, that is, a tornado with anticyclonic circulation. Does it diﬀer in any way from the usual cyclonic tornadoes? (4) Visit the site describing the CSU Spintank http://einstein.atmos.colostate.edu/∼mcnoldy/spintank/ and study the movie loops there. For a more comprehensive discussion of gradient balance, read Holton, Chapter 3. 31 32 ...
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This note was uploaded on 01/31/2011 for the course MATH 21A taught by Professor Osserman during the Spring '07 term at UC Davis.
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