Lecture Notes for GEOF110

Transkript

1 ecture Notes for GEOF0 Chapter 8 to Geopotential = hours Geostrophy = 4 hours Thermal Wind, barotropic/baroclinic = hours Ilker Fer Guiding for blackboard presentation, following Pond & Pickard, Introductory Dynamical Oceanography ydrostatic Equilibrium Stationary : u = v = w = 0 r Remains stationary dv / dt = 0 Diverse forces (friction, tides etc.) F = 0 r v r r r r r r + v v = α p fk v gk + F t becomes x, y α = 0 ; α = 0 y p z α = g z Isobaric (p=constant) surfaces are horizontal. No pressure force to cause horizontal motions. dp = gρdz ydrostatic equation in differential form.

2 Inertial Motion Assume ) orizontal isobaric surfaces = = 0 y (i.e., there are no pressure gradients in the horizontal direction no slope of sea surface and no slope on pressure surfaces in the interior ocean) ) Vertical velocity w = 0 everywhere 3) No friction (F = 0) 4) orizontal velocity do not vary in space u=f(t); v=f(t) du = fv dt dv = fu dt where f = Ωsinφ du = fv dt dv dt = fu dt + = Solution : u = Asin ft+ Bcos ft du f u 0 Choose e.g., u = 0 at t = 0, B = 0 u = Asin ft du = facos ft = fv dt v= Acos ft 3 u = Asin ft v= Acos ft r r r v = ui + vj r ( sin cos ) v = v = u + v = A ft + ft r v = v = A u = v sin ft v= v cos ft Note:. orizontal velocity vector has constant magnitude. Integration constant is the horizontal speed.. In the Northern hemisphere (N), f>0, velocity vector rotates clockwise (CW) in time and turns 360 o after one inertial period: π T i = f In other words: Eq. of motion for a body in N travelling CW in a horizontal circle at constant linear speed v and angular speed f = Ωsinφ. For example, wind blows steady in one directions, then stops after water has a speed V. If there is NO friction, due to its inertia, the motion continues (hence inertial motion). In real world, there s friction decay due to friction. 4

3 Particle path et s (x,y) be coordinates of an individual fluid particle: dx u = = v sin ft dt dy v= = v cos ft dt Integrate : v x x0 = cos ft f v ( x x0) + ( y y0) = v f y y0 = sin ft f v t=0 =π/f Northern emisphere v B=v /f t=π/f Coriolis Force (x 0, y 0 ) t=3π/f v v v t=π/f Each particle moves in a circle with centre at (x 0,y 0 ), with radius B = v /f B = inertial radius. Since the speed is constant Centripetal acc. = Coriolis acc. V fv B = In the ocean, at mid-latitudes, f 0-4 /s For v = 0. m/s B =V /f= km For v = m/s B = 0 km 5 π π π Period of revolution is inertial period = T i = = = angular speed Ωsinφ { Ω sinφ one sidereal (or remember from: day 4443 Time = circumference/speed =πb/v =πv /fv =π/f) one pendulum day (T ) f inertial period = half a pendulum day 90N T i h 45N T i 7 h Equator T i = T i = ½ T p Sidereal day stjernedøgn Pendulum day pendeldøgn Inertial oscillations treghets-svingninger inertial motion is anti-cyclonic in both hemispheres Northern emisphere Cyclonic = CCW Anti-cyclonic = CW Cyclonic = CW Southern emisphere Anti-cyclonic = CCW 6

4 Particle path If the ocean, in addition to inertial oscillations, also has a uniform velocity V, say in y-direction: v v x = cos ft, y = sin ft+ Vt f f V < v V > v Particle path Equator Real ocean: Translation speed + decay due to friction. (Fig. from Wells) 7 Geopotential We raise a mass M a vertical distance dz against gravity. Ignore friction. Amount of work done (= potential energy gained) = dw=mgdz Define: Change in geopotential (dφ) is gdz MdΦ=dW = Mgdz This is potential energy change per unit mass joules/kg = m /s Using hydrostatic eq. (dp = -ρgdz) dφ = gdz = α dp Note, α(s,t,p)=α(35,0,p) + δ (specific volume anomaly δ<< α) Integrate between levels, z to z : dφ= gdz = αdp Geopotential distance between z and z ( ) α( 35,0, ) Φ Φ = g z z = P dp δdp O Φstd Φ 443 { Φstd STANDARD GEOPOT. DISTANCE f(p) GEOPOTENTIA ANOMAY f(s,t,p) Units: Energy/mass Joules/kg or m /s [not metre!] Φ << Φstd Φ (0 ) 8

5 For g = 9.8 m/s dz = m Φ = 9.8 J/kg For convenience: a common unit for the geopotential is the dynamic metre. In atmosphere: dynamic meter = Z = Φ/0 In the ocean; dynamic meter = D = -Φ/0 So that D D is similar to z z. Geopotential surface: evel surface. The surface to which the force of gravity is perpendicular. Isobaric surface: Equal pressure surface Imagine a stationary (u=v=w=0) lake, with no currents, waves and no wind. Say atmospheric pressure = 0. The lake surface is the isobaric surface for p = 0. it is also level surface. Deeper isobaric surfaces will also be level surfaces. 9 Pressure force p α n i z Isobarflate, p = konst. i Nivåflate, Φ = konst. g, loddlinje Sats: Ingen bevegelse; isobarflaten må være en nivåflate. Bevis: p Balanse i vertikalen α cosi = g n p Balanse i horizontalt α sin i = 0 i = 0 n sin i or. komp kan skrives: α sin i = α cosi = gtan i n n cosi For å få balanse må vi ha en tilsvarende kraft (F/m) mot høyre: gtani F/m 0

6 Vi fant at Coriolis-kraften var viktig. En måte å få balanse på er å ha en hastighet V inn i papirplanet (in the northern hemisphere) slik at F = fv = g tan i Geostrofisk likning m Balance between the Coriolis force and the pressure force This would permit us to get speed from slope of isobaric surface, i. We cannot measure p accurately, men måler ρ(s,t,p) og beregne p fra dp = - ρgdz. Forsatt vanskelig å måle i, fordi havflaten ikke er en nivåflate. Måler derfor bare relative vinkler. Slopes are small e.g., f = 0-4 /s, v = m/s, tani = 0-5 m/00 km. We determine the difference between i at level z and i at level z. We get the velocity at z REATIVE to that at z velocity shear dv/dz. Geostrophic wind speed in the atmosphere can be calculated accurately by direct measurements of pressure at known levels. z A z, p x i V 0 V B z, p a Surface p=p A =konst Φ 0 p=p Φ fv = g tan i fv = g tan i g f V V = g i i = a b Note depth, z < 0 and use z -z, the distance between Φ & Φ ( ) ( tan tan ) ( ) 4 = ( ) ( ) a b ( z z3) ( z z4) ( ) = ( ) ( ) z z4 z a z3 b = f V V g z z g z z 3 4 z ( )? ( ) g z z = g z z dz 3 3 z3 Remember, gdz = αdp z 4, p V i z 3, p b p=p Φ g ( ) ( ) p p z z αdp = αdp 3 p p p p g z z = α dp+ δ dp 3 35,0, p p p B ( ) p p g z z = α dp+ δ dp 4 35,0, p p p ( ) = ( ) ( ) f V V g z z g z z 3 4 A V V = dp dp = Φ Φ p p B A B A f δ δ f p p [ ]

7 Utledening av praktisk geostrofisk likning fra bevegelseslikningene: r r dv v r r r r r r = + v v = α p fk v gk + F dt t r dv r Assume = 0; w= 0; F = 0 dt 0 = α + fv Coriolis = - Pressure Force Force 0 = α fu y 0 = α g z orizontal components can be combined: p fv = α n V = u + v : horizontal speed : horz. pressure term direction of V n y We do not neglect the Coriolis force, although it is small. It is balanced by the other small term, pressure force. We can only neglect small terms if there are larger terms.. Pressure gradient is initiated. Fluid starts to move downgradient 3. Coriolis force acts to the right (N) Alternatively:. Wind stress piles water to one side with Coriolis deflecting the movement. Pressure gradient balances x α/ y fu u α/ v fv α/ n fv V 3 z : Pressure gradients on constant z (= constant Φ) cannot be measured. A δ p pb pa pb pa δz α = α = α δx on Φ δx δz δx on P pb pc δz δ p δz = α = α δz δx on P δz δx on P δz δgz δφ Φ = g = = δx δx δx on P on P on P on P i δx C B δz Forandringen i nivået på en isobarflate Isobarflate, p =p = konst. Nivåflate, Φ = Φ konst. p 0 α fv = Φ = + + fv on P Φ fv = Bruk denne likningen i Φ nivåer: fv = P P 4

8 Geopot.distance: ( z ) Φ Φ = g z = Φ Φ std Φ =Φ + Φ { std + Φ { fv fv ( v ) f ( p) f ( S, T, p) Φ Φ Φ Φ std = = + + on P on P 3 = 0 = on P f v Φ = Φ x ikewise, ( u ) f u ( V ) f V Φ = y Φ = n Where V and V are the horizontal velocity components perpendicular to direction of n at levels and. Midler horisontalt : ( ) = ( ) B Φ f V V f V V dn n A 0 ( ) ( ) = = ( Φ Φ ) dx B A 5 Thermalvind-likningene Derived to show how T variations in horizontal lead to vertical variations in geostrophic wind. Geostrofisk: Se på x-komp: 0 = α + fv = ρ fv z 0 = α fu = ρ fv y z z z p 0 = α g Av hydrost.: = ρg z z ρg = ρ fv z ρ f = ( ρv) g z ρ f = ( ρu) y g z x-komp: y-komp: orisontalt: g ρ ( ρv) = z f g ρ ( ρu) = z f y g ρ ( ρv ) = z f n Notes: We can take f out of / z, as it does not change with z. In (ρu)/ z: Vertical variation in density is much less than shear (ρu)/ z ρ u/ z (consistent with Bossinesq) Note: Again we determined the vertical variation of velocity (i.e., shear) For ρ/; ρ/ y : Use e.g. virtual potential temperature in atmosphere, sigma_t in the ocean (upper 000 m). ight water on the right rule (N) 6

9 Absolutte hastigheter a) evel of no motion (NM, klassisk) b) Kontiunitets-prinsipper c) Strømmålinger d) Overflate-topografi fra satellitt Typical deep NM is assumed due to blief of weak currents in deep ocean. This can be wrong as observations show strong currents- but they are localised and probably transient. On the average and long time scale deep NM is a good approximation. Dynamisk topografi: Plotter overflatens geopotensial-topografi relativt til en annen flate Vi fant: f( V V) = ( ΦB ΦA) Velg overflaten som det ene nivå, mange stasjoner A, B, C,. Gir da muligheten til å plotte et kart av Φ er i forhold til et annet nivå. Φ=konst. fast Steep slope large speed high slow 7 North Atlantic Current Kuroshio Current Gulf Stream ACC 8

10 Dynamic topography is usually based on deep NM. Surface currents are generally strong and will not be affecte d much. BUT total volume transport integrated over full depth can be wrong! Say in 4 km water, NM assumed at km, and upper km mean geostrophic velocity is 0 cm/s. Say the actual mean current below km is cm/s. Error in velocity estimate is than 0%. Geostrophic volume transport is 0.x000 = 00 m3/s per m. Actual volume transport is 0.x000+0.x3000 = 80 m3/s per m 80% error. We know speed goes to zero at the sea bed. Why not use bottom as NM reference layer? The velocity profile approaches zero at the bottom due to FRICTION. But friction terms are neglected in geostrophic equations. We cannot use geostrophy when friction is important. 9 Relasjon mellom isobar-flater og nivåflater Typical NM is about 000 m in Pacific, between m in Atlantic. d) SOPE CURRENT BAROTROPIC. This would give zero geostrophic velovcity (correct) but absoluite velocity is not zero. UNIKEY because T,S variations are likely to cause variations in isobaric slopes. usannsynlig 0

11 Relasjon mellom isobar-flater og isopyknaler Isopyknal: konstant tetthets-flater vis tettheten bare er en funksjon av trykket (f.eks. S og T konstante), ρ = ρ(p) BAROTROPT massefelt Isobarflater og isopyknaler er parallel. vis tettheten også er en funksjon av andre parametere kan tetthets og isbarflatene ha forskjellige helning BAROKINT massefelt

12 For the -layer flow in panel c of the previous figure, interface slope is about 00 times steeper than the sea-surface slope. Derive: z 0 ρ η ρ No flow d x p b = d 0 η 0 ρgdz = ρ gdz d ( 0) ( ) At bottom isobar slope: η b d ρ η = 0 = ( ρ ρ) ρ ρ 0 ρ 3 ρ gdz = ρg d ρg η d = ρ gd ρ gη+ ρ gd b d η = ( ρ ρ) g ρg Noflow 3 4

13 5 Some disadvantages of the geostrophic method: -Only relative currents -Selection of NM -NM reaches bottom as stations get shallower -Only mean values over stations with large distance apart (sometimes this can be a benefit, e.g. smooths short-term variations, such as internal waves) -Friction is ignored -Break-down near Equator where Coriolis force is so small that friction can be important. -Will include any effect of long-period transient currents. 6