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Atmospheric Motion Dynamics Newton’s second law F m x Mass × Acceleration = Force m d2x =F dt 2 Thermodynamics Concerned with changes in the internal energy and state of moist air. TMD Lecture 2 Newton’s second law The Coriolis force Newton’s second law m F d2x m 2 =F dt x A line that rotates with the roundabout Ω A line at rest in an inertial system Mass × Acceleration = Force applies in an inertial frame of reference. But we like to make measurements relative to the Earth, which is rotating! To do this we must add correction terms in the equation, the centrifugal and Coriolis accelerations. Apparent trajectory of the ball in a rotating coordinate system Effective Gravity The centripetal acceleration/centrifugal force g is everywhere normal to the earth’s surface outward force Ω v2 = Ω2 r r Ω 2R R g* v g* g r Inward acceleration − Ω 2R R g v2 = −Ω2r r g = g* + Ω2R effective gravity g on a spherical earth Effective Gravity If the earth were a perfect sphere and not rotating, the only gravitational component g* would be radial. Because the earth has a bulge and is rotating, the effective gravitational force g is the vector sum of the normal gravity to the mass distribution g*, together with a centrifugal force Ω2R, and this has no tangential component at the earth’s surface. effective gravity on an earth with a slight equatorial bulge When frictional forces can be neglected, F is the pressure gradient force total pressure F = −∇ p T force per unit volume ⎛ du ⎞ ρ⎜ − 2Ω ∧ u ⎟ = F + ρg ⎝ dt ⎠ du 1 = − ∇pT + g − 2Ω ∧ u dt ρ per unit mass g = g * +Ω 2R This is Euler’s equation of motion in a rotating reference frame. The Coriolis force does no work Perturbation pressure, buoyancy force Define Ω the Coriolis force acts normal to the rotation vector and normal to the velocity. u is directly proportional to the magnitude of u and Ω. pT = p0 ( z) + p where dp 0 = − gρ0 dz p0(z) and ρ0(z) are reference pressure and density fields p is the perturbation pressure Euler’s equation becomes − 2Ω ∧ u g = (0, 0, −g) ⎡ ρ − ρ0 ⎤ Du 1 + 2Ω ∧ u = − ∇p + g ⎢ ⎥ ρ Dt ⎣ ρ ⎦ Note: the Coriolis force does no work because u ⋅ ( 2Ω ∧ u) ≡ 0 the buoyancy force Important: the perturbation pressure gradient − 1 ∇p ρ Mathematical formulation of the continuity equation for an incompressible fluid ⎛ ρ − ρ0 ⎞ and buoyancy force g ⎜ ⎟ are not uniquely defined. ⎝ ρ ⎠ ⎛ ρ − ρ0 ⎞ 1 But the total force − ∇ p + g ⎜ ⎟ ρ ⎝ ρ ⎠ Indeed v + δv w + δw δy δx is uniquely defined. ⎛ ρ − ρ0 ⎞ 1 1 − ∇p + g ⎜ = − ∇p T + g ⎟ ρ ρ ⎝ ρ ⎠ u δz v w u + δu The mass continuity equation Rigid body dynamics ∇⋅u = 0 Incompressible fluid Mass m Compressible fluid Force F ∂ρ + ∇ ⋅ (ρu ) = 0 ∂t Anelastic approximation ∇ ⋅ (ρ (z)u ) = 0 o x Newton’s equation of motion is: m d2 x dt 2 =F Problem is to calculate x(t) given the force F Fluid dynamics problems Ø The force field is determined by the overall constraints provided by – the requirement of continuity – the boundary conditions Ø In particular, the pressure field at any instant is determined by the flow configuration – I will now illustrate this with an example! – Let us forget about density differences and rotation for this example Fluid dynamics problems Ø The aim of any fluid dynamics calculation is to calculate the flow field U(x,y,z,t) in a given region subject to appropriate boundary conditions and the constraint of continuity. Ø The calculation of the force field (i.e. the pressure field) may not be necessary, depending on the solution method. LO A mathematical demonstration HI U isobars Du 1 = − ∇p ' Dt ρ HI Momentum equation ∇⋅u = 0 Continuity equation The divergence of the momentum equation gives: U HI LO ∇2p' = −∇⋅ (ρu ⋅∇u) pump streamlines This is a diagnostic equation! But what about the effects of rotation? Assumptions: inviscid, irrotational, incompressible flow buoyancy form Newton’s 2nd law vertical component mass × acceleration = force ρ Dw ∂p = − T − gρ Dt ∂z Put p T = p (z) + p′ o ρ = ρ (z) + ρ′ o Then Dw 1 ∂p′ =− +b ρ ∂z Dt where where dpo = −gρ o dz ⎛ ρ−ρ o ⎞ b = −g ⎜ ⎟ ⎝ ρ ⎠ Buoyancy force in a hurricane buoyancy force is NOT unique ρ (z) ⎛ ρ − ρo ⎞ b = −g ⎜ ⎟ ⎝ ρ ⎠ o it depends on choice of reference density ρo(z) but − 1 ∂pT 1 ∂p ' −g = − +b ρ ∂z ρ ∂z is unique ρ (z) o Initiation of a thunderstorm z tropopause negative buoyancy outflow θ = constant original heated air θ = constant positive buoyancy LFC LCL Τ + ΔΤ U(z) T inflow negative buoyancy positive buoyancy Some questions HI outflow p' Ø How does the flow evolve after the original thermal has reached the upper troposphere? Ø What drives the updraught at low levels? original heated air – Observation in severe thunderstorms: the updraught at cloud base is negatively buoyant! LO LFC – Answer: - the perturbation pressure gradient LCL HI inflow HI HI negative buoyancy The geostrophic approximation For frictionless motion (D = 0) the momentum equation is Choose rectangular coordinates: k = (0,0,1) z Ω = Ωk Du 1 + 2 Ω ∧ u = − ∇p Dt ρ Let Ro → 0 k perturbation pressure uh 1 2Ω ∧ u = − ∇ p ρ y x This is called the geostrophic equation We expect this equation to hold approximately in synoptic scale motions in the atmosphere and oceans, except possibly near the equator. velocity components u = (u,v,w), u = uh + wk uh = (u,v,0) is the horizontal flow velocity Take k ∧ 1 2Ω ∧ u = − ∇ p ρ The geostrophic wind (k ⋅ u)k = (0, 0 w) 1 2 Ω k ∧ ( k ∧ u ) = 2Ω [( k ⋅ u ) k − u ] = − k ∧ ∇ p ρ − uh uh = and s h p = (∂p/∂x, ∂p/∂y, 0) 1 k ∧ ∇h p 2Ωρ 0= ∂p ∂z uh = 1 k ∧ ∇h p 2Ωρ Ø The geostrophic wind blows parallel to the lines (or more strictly surfaces) of constant pressure - the isobars, with low pressure to the left. Ø Well known to the layman who tries to interpret the newspaper "weather map", which is a chart showing isobaric lines at mean sea level. Ø In the southern hemisphere, low pressure is to the right. This is the solution for geostrophic flow. Choice of coordinates Ø For simplicity, let us orientate the coordinates so that x points in the direction of the geostrophic wind. Ø Then v = 0, implying that ∂p/∂x = 0 . u=− Geostrophic flow pressure gradient force isobar low p 1 ∂p 2Ωρ ∂y Ø Note that for fixed Ω , the winds are stronger when the isobars are closer together and, for a given isobar separation, they are stronger for smaller |Ω|. u high p isobar Coriolis force (Northern hemisphere case: > 0) A mean sea level isobaric chart over Australia Note also that the solution and uh = 1 k ∧ ∇hp 2Ωρ 0= ∂p ∂z tells us nothing about the vertical velocity w. L Ø For an incompressible fluid, ∇ ⋅ u = 0 . Ø Also, for geostrophic flow, ∇h ⋅ uh = 0 . Ø then ∂w/∂z = 0 implying that w is independent of z. H H H If w = 0 at some particular z, say z = 0, which might be the ground, then w ≡ 0. The geostrophic equation is degenerate! Ø The geostrophic equation is degenerate, i.e. time derivatives have been eliminated in the approximation. Ø We cannot use the equation to predict how the flow will evolve. Ø Such equations are called diagnostic equations. Ø In the case of the geostrophic equation, for example, a knowledge of the isobar spacing at a given time allows us to calculate, or 'diagnose', the geostrophic wind. Ø We cannot use the equation to forecast how the wind velocity will change with time. Vortex flows: the gradient wind equation Ø Strict geostrophic motion requires that the isobars be straight, or, equivalently, that the flow be uni-directional. Ø To investigate balanced flows with curved isobars, including vortical flows, it is convenient to express Euler's equation in cylindrical coordinates. Ø To do this we need an expression for the total horizontal acceleration Duh/Dt in cylindrical coordinates. The case of pure circular motion with u = 0 and ∂/∂θ ≡ 0. The radial and tangential components of Euler's equation may be written v2 1 ∂p + fv = r ρ ∂r ∂u ∂u v ∂u ∂u v 1 ∂p +u + +w − − fv = − ∂t ∂r r ∂θ ∂z r ρ ∂r 2 ∂v uv 1 ∂p ∂v ∂v v ∂v +u + +w + + fu = − ∂r r ∂θ ∂z r ρr ∂θ ∂t The axial component is ∂w ∂w ∂w v ∂ w 1 ∂p +u + +w =− ∂t ∂r ∂z ρ ∂z r ∂θ Ø This is called the gradient wind equation. Ø It is a generalization of the geostrophic equation which takes into account centrifugal as well as Coriolis forces. Ø This is necessary when the curvature of the isobars is large, as in an extra-tropical depression or in a tropical cyclone. The gradient wind equation Write Force balances in low and high pressure systems 1 ∂p v 2 0=− + + fv ρ ∂r r terms interpreted as forces Cyclone V V Ø The equation expresses a balance of the centrifugal force (v2/r) and Coriolis force (fv) with the radial pressure gradient. Ø This interpretation is appropriate in the coordinate system defined by r and θ , which rotates with angular velocity v/r. Anticyclone LO HI PG HI CO CO CE LO CE PG The equation 0=− 1 1 r ∂p v = − fr + f 2 r 2 + ρ ∂r 2 4 1 ∂p v 2 + + fv ρ ∂r r is a diagnostic equation for the tangential velocity v in terms of the pressure gradient: r ∂p 1 1 v = − fr + f 2 r 2 + ρ ∂r 2 4 1 2 Ø In a low pressure system, ∂p/∂r > 0 and there is no theoretical limit to the tangential velocity v. Ø In a high pressure system, ∂p/∂r < 0 and the local value of the pressure gradient cannot be less than −ρrf2/4 in a balanced state. Choose the positive sign so that geostrophic balance is recovered as r → ∞ (for finite v, the centrifugal force tends to zero as r → ∞ ). Ø Therefore the tangential wind speed cannot locally exceed rf/2 in magnitude. Ø This accords with observations in that wind speeds in anticyclones are generally light, whereas wind speeds in cyclones may be quite high. Limited wind speed in anticyclones In the anticyclone, the Coriolis force increases only in proportion to v: => this explains the upper limit on v predicted by the gradient wind equation. V CO = fv HI CE = v2 r CO CE PG 1 2 End of L2