Introduction to Atmospheric Physics and Dynamics
Introduction to Atmospheric Physics and Dynamics ATOC 4720
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Chapter 3 Properties of Fluids Solids Liquids and Gases Stress a measure of the internal forces in a body between its constituent particles Normal stress a stress resulting from tension or compression Shear stress a stress resulting from sliding Rigid material a material that shatters rather than deforms when a large enough stress is applied Plastic material a material that deforms continuously and irreversibly when a stress is applied Elastic substance a material that deforms continuously when a stress is applied but returns to its original shape when the stress is removed Fluid a material that deforms continuously when acted upon by a shear stress of any magnitude Le a fluid has no rigidity What are some examples of rigid plastic and elastic materials What are some examples of a fluid Viscous fluid a fluid which transmits shear stresses internally Thermodynamic Properties of Air lftwo samples of air can exist in contact with each other without a change in properties the two samples are said to have the same state What properties of the air define its state State variables for an ideal gas any two variables that do not depend on the mass of the system and that completely define the state ofthe system Equation of state an equation that relates properties of state to one another Ideal gas law 1 R a constant pT p pressure T absolute temperature p density R gas constant RRM where R is the universal gas constant 8314 J mol391 K4 and M is the molecular mass of the gas Composition of the Atmosphere How does the composition ofthe atmosphere vary in space The Dry Atmosphere What are the constituents ofthe dry atmosphere Each gas that makes up the atmosphere obeys the ideal gas law pn annT Dalton s Law the total pressure exerted by a mixture of gases that do not interact chemically is equal to the sum ofthe partial pressures of each constituent Partial pressure the pressure that would be exerted by a single constituent in a mixture of gases ifthat constituent were present alone and occupied the same volume as the mixture of gases p T2ann Molecular mass for dry air Md 28966 g mol391 210157 21011 11 Gas constant for dry air Rd quot quot p p M 28704 Jkg39lK39l Ideal gas law for dry air 1 deT Example The density of air on the roof of Duane Physics The Moist Atmosphere Why do we care about water in the atmosphere Water vapor variables Vapor pressure e The partial pressure exerted by watervapor molecules in a given volume of the atmosphere Specific humidity q Mass of water vapor contained in a unit mass of air Mixing ratio r The mass of water vapor contained in a unit mass of dry air Application of the ideal gas law to a moist atmosphere Virtual temperature p ded pvRvT depdpvRvT p p Rd T R R 1 V pd q qu The ideal gas law for moist air 1 deTV Virtual temperature TV The temperature dry air would need to have if it were to have the same density as a sample of moist air at the same pressure Tv1 qq11T How does the density of dry and moist air compare What does this imply about the relationship between temperature and virtual temperature Example Temperature vs virtual temperature on the roof of Duane Physics Additional details about moisture variables can be found in Chapter 11 Static Stability Air parcel a small mass of air which we can follow through the atmosphere First law of thermodynamics Q c dl adip dt dt dt Adiabatic process a process in which an air parcel exchanges no heat Q with its surroundings environment In an adiabatic process dQ 0 Potential temperature 6 the temperature an air parcel would have if brought to a reference pressure p0 usually 1000 hPa through an adiabatic process Potential temperature is conserved ie it does not change if an air parcel does not exchange heat with its environment Derivation of the equation for potential temperature conservation of heat Combine the ideal gas law and first law of thermodynamics to give IQ cPdT Ed P Using dQO in an adiabatic process we can integrate this equation CP R C 0 dT ldp P Cil T l Rd 11 up Use T6for pp0 to find C C CP19 1 7n n Rd po 3 c ilnB lnpo ilnT 1np Rd Rd cpRd Td p This gives us an equation for potential temperature Rd cp 6TV P Example The potential temperature on the roof of Duane Physics How does the pressure temperature and potential temperature of an air parcel change if the air parcel is lifted adiabatically in the atmosphere Adiabatic cooling and adiabatic warming Example Adiabatic lifting of air from Boulderto Leadville Forces acting on an air parcel that is displaced vertically Start with an air parcel with volume V at height 2 Assume that this air parcel has the same temperature as its environment Tz For both the air parcel and the environment at height 2 Rel5p T 6 pl How does the temperature and potential temperature of the air parcel change as it is lifted adiabatically from height 2 to height zdz R c pZdZ d p P0 Tparcelzdz 6z The temperature of the environment at height zdz is Rajcp Tzdz 6zdz P0 What force does this air parcel experience once it is displaced air displaced air parcel weight of weight of J F mass of parcel gpzdzv gpparcelzulzv V p parcel zdz We can rewrite the expression for this force using the ideal gas law and noting that V of the air parcel and displaced air are equal p p RdT LRdTparcelz dz J zdz LRdTparcelzdz J 1 1 T7 T z dz parcel zdz g T parcel z dz T z dz T parcel z dz T z dz 6 z dz 6 parcel z dz 6 z dz Noting that 6P3cezdz62 eel F gk z ZdZJ 6zdz af dz 39 9 dz N2dz BruntVaisala frequency N the frequency at which an air parcel will oscillate if displaced vertically and acted upon by the restoring force arising from the buoyancy of the parcel also known as the buoyancy frequency What property of the environment determines the magnitude ofthe buoyancy force Dry adiabatic process an adiabatic process in which no phase change of water occurs A dry adiabatic process does not refer to process in which no water vapor is present only one in which the water vapor that is present does not undergo any phase changesllll What happens to an air parcel that is displaced vertically by a dry adiabatic process if dBdz 0 dBdz gt 0 dBdz lt 0 Static stability classes Stable unstable and neutral Example Static stability and a radiosonde sounding from Denver What happens when water vapor changes phase in an air parcel Moist adiabatic process an adiabatic process in which water vapor changes phase in an air parcel Moist adiabatic processes are discussed in greater detail in Chapter 11 The Continuity Equation Continuity atmospheric mass is not created or destroyed Divergence air expands flows outwards Convergence air compresses For an air parcel with volume dXdydz and a spatially varying density p the mass of air is given by Malr M pdxdydz How can we express the time rate of change of mass in this volume of air Z V X ass in at x Mass out at xdx pxuxdydz pxdxuxdxdydz fffpdxdydz ffltpugtx ltpugtmdydz 5t ff my pvy2y 1de ff pwz pwzdzdxdy This can be rewritten to give the flux form of the continuity equation 5 fff Wde ff dxdydz mm mm mm 0 at 9x 9 dz The continuity equation is a conservation equation for atmospheric mass We can also express the continuity equation as 13 nai rvaip rwaip pal 0 6 r9y z 6 5y 6z 1 2 3 What do terms 1 2 and 3 represent physically If density does not vary in time then mm 5pv 5pw 0 5x 3y z If in addition density does not vary in space u 3v 3w p i i 7 0 5x 3y z i 5x 3y z This relationship is known as Dines compensation and requires that divergence at one location in a column of the atmosphere is exactly balanced by convergence at another location in the column Chapter 9 Extratropical Weather Systems Example Upper level and surface features of extratropical weather systems Rossby waves Short waves Surface features Fronts Margules Model frontal surface Copyright 2005 by John Wlley amp Sons Inc or related companies All rights reserved Features ofthis model The front is a tilted plane that separates two homogeneous geostrophic ows with different but uniform temperatures an densities Front is stationary no across 39ont llow Exclude 39iction Orientation of axes Boussinesq approximation across 39ont changes in T and 9 small Margules formula fv2 ampV1 p2 p1 gtans P P 2 2 Relates slope of front 5 to changes in density p and geostrophic wind v across the front How does the wind change across this front Example Margules formula and a real front Frontal Cyclones What causes a cyclone to develop cyclogenesis and to decay cyclolysis Cyclogenesis What physical process will cause surface pressure to decrease a Cold b C JT Kquot Cool V 1 v Cold Warm 39h quot nslhcm What is the vertical distribution of convergence and divergence in a mature cyclone What is the cause ofthis convergence and divergence At the surface Aloft In the QG framework divergence and convergence occurs only through ageostrophic flow How does the gradient wind compare to the geostrophic wind for flow around areas of low trough and high ridge pressure supergeostrophic gt divergence convergence y HI X 10 convergence divergence convergence divergence convergence divergence 0 Copyright 2006 by John WIley amp Sons Inc or related companies All rights reserved Convergence su bgeostrophic gt elevation km 01 X We expect to find divergence ahead of east of troughs What other atmospheric features can cause upper level divergence Jetstreaks Left Entrance Ageostrophic wind due to Left Exit Dlvergenee advective acceleration Convergence 1 9143 Va 7 Mg 7 f 9x Divergence Convergence HIGH R39ght En39mnce R39gh EX Where do we find areas of Copyright 2006 by John Wiley amp Sons lnc or related companies All rights reserved convergence and divergence relative to this jetstreak Rising and sinking motion associated with divergence and convergence in the jetstreak can be diagnosed with 077W u av dz ax 3y In what quadrants of a jetstreak will we have rising sinking motion Example Calculation of the ageostrophic wind and rising motion associated with a jetstreak Sutcliffe Development Theorem The net convergence in a column ofthe atmosphere is often a small difference between relatively large convergence near the ground and relatively large divergence aloft Therefore it is often difficult to evaluate whether there is net convergence or divergence in the column and thus cyclolysis or cyclogenesis Sutcliffe derived an equation to predict changes in vorticity at the surface that avoided the need to evaluate net divergence in a column ago 0 ago aCT aCT u i v 7 v v T 07x T 07y g0 T 1 2 3 639 WT 4ng 2VT Cgo Geostrophic vorticity at the surface CT Vertical change in vorticity vorticity shear ugo vg0 geostrophic Wind components at the surface LIT VT vertical change in geostrophic Wind geostrophic Wind shear This equation is usually evaluated between the surface and 500mb so the vertical change in vorticity CT and geostrophic wind uT and VT represent changes in vorticity and geostrophic wind between the surface and 500mb These terms can be evaluated using a map of 1000500mb thickness Term 1 of the Sutcliffe development equation At the center of a cyclone llng and vggeo so ann 6 63 2 rZuT Thermal steering principal The low pressure center will propagate in the direction of the wind shear ur and v7 with a speed proportional to the wind ear For u 00 and v900 u7u500mb and wax500mb and the low pressure center will tend to propagate in the direction of the 500mb ow Term 2 of the Sutcliffe development equation This term is referred to as the thermal vorticity advection term uT V7 and Q can be evaluated using a map of 1000500mb thickness ldealized map of 1000500mb thickness CTmax Cm fmm fmax convergence Copyrlght 2005 by John Wiley amp Sons Inc or related companies All rights reserved Where is the thermal vorticity advection term a maximum minimum Where would cyclone intensi cation decay be expected Term 3 Planetary vorticity contribution This term will be positive for northerly negative v90 and VT but tends to be smaller than the other two terms in this equation Limitations of the Sutcliffe Development Equation Neglects adiabatic warming and cooling due to rising and sinking ot39 n m IO Neglects diabatic heating Diagnostic equation only Does not work well with weak shear situations 59 Cyclolysis Cyclolysis occurs when convergence exceeds divergence in the column of the atmosphere above a surface low pressure center a Cold b C T b Cool V 1 V Cold Warm 39 h 39 ns inc or How has upper level divergence changed by panel c How has near surface convergence changed by panel c Chapter 7 Circulation and Vorticity Circulation Cg d Integration is erformed in a counterclockwise direction 9 Copyright 2006 by John Wiley amp Sane inc m relaled companies All ngms reserved C is positive for counterclockwise flow Kelvin s Circulation Theorem The rate of change of circulation can be expressed as DC D12 A 395 d5 9S ljidx95dx iction p x 5 What is the magnitude of each term in this equation Fora barotropic fluid density is a function only of pressure Lip S L 95pp195d 9591dp 1 9012 9a 0 around a closed circuit Geopotential term 35st nglt1gt 9 1gt2 31 0 around a closed circuit In a frictionless inviscid flow the friction term is also zero In a barotropic inviscid fluid the circulation is constant E o Dt Bjerknes Circulation theorem Changes in circulation can arise due to friction or baroclinicity The sea breeze circulation 925W U0 0 LQ lt2 lt3 lt2 lt H Ln L 0l m 09090040 John Wiley amp Sons Inc or related How does the temperature overthe land and overthe ocean vary during the course of a day What impact does this have on the thickness of an atmospheric column How does the atmosphere respond to this horizontal variation in thickness Example Calculating the circulation associated with a sea breeze Relative circulation Circulation in the atmosphere arises due to our rotating frame of reference Cabsolute Ceath Crelative where Cea h earth R9 x 271R sin I th z 2191 sin If Cabsome is conserved what does this imply about changes in Orea ve for meridional flow Vorticity For solid body rotation 5C 2m3rV 2V 2 7 20 5A 7151 5r The earth undergoes solid body rotation with an angularvelocity of w Qsin 1 so Ceath Sin f What is the sign of gem in the Northern and Southern hemispheres 39 h Wiley amp Sons Inc vi The circulation around ABCD can be calculated as C 9S d 98udxvdy 5CCAB CBC CCD CDA mix V g xj y u g yj x v y 31 BM idx 7 5x amp 5y By 5y QJLSA Bx By Then the relative vorticity is given by What is the sign of gfor clockwise and counterclockwise flovx What does this imply about the sign of gfor flow around low and high pressure centers Example Calculation of relative vorticity from a weather map Conservation of Potential Vorticity What conditions were required for constant circulation according to Kelvin s circulation theorem On a constant potential temperature 6 surface RdCp RdCp pLLamp pcv6 Po RT R6 p R6 This is analogous to a barotropic fluid Therefore on a constant potential temperature surface the pressure gradient term is zero and Kelvin s circulation theorem is satisfied DC absolute Dt 0 6 This implies that Cabsolm e C6 39 Cabsolute C6 f 3ng 5A 5A 6 f constant Consider an air parcel that is confined between two potential temperature surfaces 6 and 6 56 separated by pressure interval 5p The motion ofthis air parcel will be adiabatic The mass ofthe parcel is given by 5M p5z5A 5pg5A and must be conserved following the motion This gives 6Mg 6Mg X 6A 7 6p 6p 66 6A6Mg 66 66 6p 6A constant x g 317 Combining this result with 6A 9 f constant gives an expression for RossbyErtel potential vorticity P constant x g 9 f constant I7 P g 9 f constant What does the 6661 term represent Potential vorticity depends on the depth ofthe fluid and the absolute vorticity Example Conservation of potential vorticity and flow over the Rocky Mountains lt 47 Copyright 2006 by John Wiley amp Sons lnc or related companies All rights reserved Copyrighl 2006 by John Wiley 5 Sons Inc or related companies All ngms reserved Air Increase Decrease Increase Decrease column I6le l pll I6le return to de th o g ral value Change in 56 Decrease Increase Decrease Increase a return to original value Change in IC fl Increase Decrease Increase Decrease Sign of Q Positive Negative Positive Negative Iquot quot39 g quot 39 quot 39 lee side trough southward motion Change in Ifl Increase Decrease Increase Decrease L Copyright3932006 by John Wiley amp Sons Inc or related companies All rights resewed For westerly flow across a mountain range a lee wave will form downstream of the mountain Stretching a column of the atmosphere results in generation of cyclonic vorticity Shrinking of a column of the atmosphere results in generation of anticylonic vorticity The Vorticity Equation Using NavierStokes equations scaled for midlatitude weather systems we can derive an equation for the time rate of change of vorticity DIM l i a j gxgh Dt p ax 3y lap 1 l l 2 Dr p ax Dt p 3y 2 312gtDha 1apd vi 1 3y Dt 3y p axay 3y 3y 2 ilt2gtgtamp la pd fui 2 Dr ax p axay 3x 3x 239139gtamp al f uivi Dt ax 3y ax 3y ax 3y uivi f Dt ax 3y ax 3y u viuivi c at 9x 65 9x 65 936 y What causes the relative vorticity at a fixed location to change in time How does vorticity change for a nondivergent flow For quasigeostrophic flow Dg and avg Egg f f 936 3y Dg 5Cg ff az What does this equation tell us about changes in relative vorticity in a quasigeostrophic framework Chapter 2 Mathematical Methods in Fluid Dynamics Scalars and Vectors Scalar any quantity which can be fully speci ed by a single number Vector a quantity which requires both a magnitude and direction to be Jlly speci ed What are some examples of scalar and vector quantities Vector Notation a MKlT My Magnitude ofa vector i u u Direction of a vector Direction armadaMK Coordinate systems on the Earth z 79 localy quotupquot 39 y quotNORTH meridional x 739 EAST zcna For a coordinate system with xyz we use unit vectors 7 f and 1 Algebra of Vectors Addition and subtraction of two vectors graphic method D Addition Subtraction Addition of two vectors it and V uxzuyjuzk and vvxzvy1 v1k 12 7ux vxuy vyuZ v1l Subtraction of two vectors a and V Vux vxi uy vyjuz vz12 Multiplication of a vector by a scalar graphic method 1 1B 1 AB 2AB 25AB 05AB Multiplication of vector 12 by scalar 0 Eli cuxi cuyj cuzk How does the direction and magnitude ofa vector change due to multiplication by a scalar Multiplication of two vectors Scalar product or dot product of 12 and V 39V 14va uyvy uzvz lEHVlcos When will the dot product of two vectors be equal to zero What does this tell us about the direction of the two vectors relative to each other Z Vector product or cross product of a and V L xi uyvz uzvyi uzvx uxvzj uxvy uyvxk or i j39 k u x v uX uy uz 12X vy VI uyvz uzvy uzvx uxvz 39 uxvy uyvx g What is the direction of the vector that results from the cross product The right hand rue What is the magnitude of this vector Magnitude i H isinH When will the cross product be equal to zero Z Scalar and Vector Fields Field a quantity defined over a given coordinate space The field is a function of the three coordinates of position and also of time T I m11 An example of scalar and vector fields from a weather map Coordinate Systems on the Earth How do scalar and vector fields change when the coordinate system is changed How would vector i uj My change under a rotation of the coordinate system 39 ux coso My sin or My cosa ux Sinai Meteorologists traditionally define a coordinate system relative to the Earth What are the implications of this coordinate system accelerating through space Noninertial frame of reference Gradients of Vectors The vectors we consider in meteorology olten vary in space and time ie they are functions ofboth space and time We can show this for a wind velocity vector as Exyzt This vector in component form can be written as 121 yyzyt my I goo2 of hxyz 0 Written in this way we see that this vector consists ofzonal meridional and vertical components of the wind that vary in all three spatial directions Xyz and vary in time The variation ofthe wind vector with respect to any one of the independent variables can be written as a partial derivative 39i zia39igp k t 9t t 9t What does each term in this equation represent physically What if we considered the partial derivative of 2xyyzt fXyyzt7 0f 0k with respect to X or y Y y gt gt gt gt gt gt gt gt gt gt gt gt gt gt gt X X E LB 31 3U ax ay 0 3739 a 0 Eulerian and Lagrangian Frames of Reference Parcel A Location 0 TTA1 TTxoyozotgtb Eulerian frame of reference properties of the atmosphere are de ned as functions of both space xyz and time In this frame of reference we can consider the properties at some xed point 0 located at position xoyozo The temperature T at this point would then be given by TC Txgygzgz In the Eulerian frame of reference the wind vector 2 is o en written as II uxyzzf vxyzz wcyz 1 or as IIuvw Lagrangian frame of reference de ne properties of the atmosphere as functions of time and of a speci c parcel of air In the Lagrangian frame of reference we are now following a speci c mass of air through the atmosphere rather than considering different masses of air passing a xed point For this case we would de ne the temperature T of an air parcel A as Advection What processes can cause the air temperature at a xed location to change Advection the change in properties at a fixed location due to the replacement of the original air parcel at that location with a new air parcel with different properties 39 warmer 39 39 warmer strong warm advection weak cold advection no advection Warm advection warmer air is replacing cooler at a given location Cold advection cooler air is replacing warmer at a given location What determines the rate at which the temperature changes due to advection Mathematical description of advection In a Lagrangian frame of reference the temperature of an air parcel is only a function oftime and can be written as DTDt This is known as a substantial material or Lagrangian derivative DDt In an Eulerian frame of reference the temperature is a function of x y z and t Txyzt and as such we need to consider the partial derivative with respect to time oiat if we want to consider changes in temperature with time at a fixed location Eulerian derivative the rate of change of a quantity at a fixed point 332 The relationship between the Eulerian and Lagrangian derivatives can be found from DT 5T 7 11111 7 D2 670 5 6T 51 5x g 5y 5z dt dx dy dz DT dT dTDx dTDy dTDz 777777 gt D2 dt ax Dt dy Dt dz D2 dT dT dT dT 7 u 7 v7 W7 dt dx dy dz thus dtDt 7 dT DT dT dT dT 7 7 u v7 W7 dx dy dz The term in the parenthesis on the right hand side ofthis equation is the advection term Is the sign of the advection term consistent with the physical interpretation of advection shown in the figure on the previous page Chapter 8 Simple Wave Motions Weather systems can be understood physically as waves with particular 39 various 39 g quot e atmosphere Properties of Waves Example Pendulum Oscillation of pendulum satis es 6 6n cosvr r a 6 angle of pendulum SD amplitude of motion 21v v frequency i T period time to complete one sine M9 oscillation a phase constant in radians Cupynghi 2006 by John Wliey amp Sons Inc Vt phase or related companies An rights reserved What is the physical interpretation of each term in this equation Propagating Wave 6 6n coskx1ymz ewe a The phase ofthis wave is given by 1 lowr 1y rm 71 701 The phase depends not only on time t but also position xyz k I and mare wavenumbers in the x y and 2 directions Wavenumber IE Vector given by components k I and m 124nm Wavelength A Distance between wave crests We can identify wavelengths in each direction AX M and AZ Wavelength and wavenumber are related by ki 7 mi A A A x y Z rail A Phase speed c The rate at which the phase of the wave propagates in each direction eg the rate at which the crest ofthe wave propagates This relationship implies that waves with different wavenumbers k l and m and thus different wavelengths AX M and AZ will propagate with different phase speeds ox Cy and 02 and will spread out disperse with time Dispersion relationship The relationship between the frequency v and wavenumber llEl for a wave If v 0 lIEl the phase speed c is not a function of wavenumber k and the wave is nondispersive Group velocity Q The rate at which the energy of a wave travels The components of the group velocity are given by 0310316 g ak gy at g am Example Atm ospheric gravity wave A gravity wave is a wave in the atmosphere in which buoyancy is the restoring force where Nis the BruntVaisala frequency What is the phase speed ofthis wave in each direction cX and c What is the total phase speed c ofthis wave Note that the zonal cX and vertical c phase speeds are not components of a vectorwhose magnitude is the total phase speed c What is the group velocity ofthis wave in the zonal and vertical directions Perturbation Analysis How can we avoid the nonlinearities in the atmospheric equations of motion Assume that all variables are the sum of a basic state and a small departure from that basic state Require that the basic state of all variables satisfy the governing equations Assume that the perturbations are sufficiently small that products of perturbations can be neglected Example Xcomponent ofthe Quasigeostrophic QG momentum equation Du u 0 Dr fa u u u igugigvgigfua0 t x y Define the basic state and perturbation of each term in the equation as ugxyt ugxt ugxyt Macaw Wm Macaw Substitute these definitions into the QG momentum equation ug ufua u0 u u39 7 u 7 u39 au u39 7 u39 u39 7g Juiguigu Juigv Jv Jfuafua0 t t g x g x g x g x g y g y We can drop all terms that include products of perturbations to give ug uig 7 ug 7 uig g 7 uig f7 f 0 77u iu iu iv 7 u u t t g x g x g x g y a a By definition the basic state satisfies the governing equation so M u J gig ua 0 at 3x Using this gives a governing equation for the perturbation variables au39 au o39l g7 gur g M f 0 iu viu at gax gax gay a This equation is a linear equation there are no products of dependent variables Application of perturbation analysis to the QG vorticity equation D aw 8 Egg f f For this analysis we will assume that the vertical velocity W is zero which also implies that the flow is nondivergent so D AWL Dt 3y Betaplane approximation Assume that the Coriolis parameter 2 varies linearly with latitude ffo y where is given by 2900s 0 ay a pa reference latitude f0 2S2sin 0 a radius of Earth 637x1o6 m Define the basic state and perturbations as Cg5g g r ugugug VgVgVgVg where we have assumed that the basic state has no meridional flow Vg 0 ie the basic state is purely zonal flow Using this definitions and the betaplane approximation gives DC V0 Dr g 9 a 9 a 9 a 9 a 2 g g u u v v v 0 at at 9x 9x 9x 9x 8y 9y i a f a f f g g g E g g u g v g t 0 omit perturbation products at at 3x 3x 3x 65 3C 3C i 7 v 0 zonal bas1c flow 0 at 8 x g g A wavelike solution ofthis equation is u lAcoskx ly Vt viz kAcoskx ly W This solution is nondivergent L 0 as required by the assumption of W 0 y u39g and V can be used to find 59 31 au 2 2 g7xg 7yg Ak l s1nkxly vt The dispersion relation the relationship between v and k for this wave can be found by substituting Cg and v g into the perturbation form ofthe governing equation ag 7 ag 7 Mg 7 vg 0 v Ak2 12Coskxly w kug Ak2 12Coskx ly w Ak coskx ly v2 C vAk2 12 k gAk2 12 Ak 0 gtv 7 k212kug From this dispersion relation we note that the frequency is a function of wavenumber The phase speed is given by This equation is known as Rossby s formula The zonal phase speed is a function of wavenumber and wavelength and waves with different wavenumber will propagate at different speeds Therefore this wave is a dispersive wave What is the direction of propagation of these waves for Hg 0 How does the zonal phase speed of longer wavelength waves compare to shorter wavelength waves Planetary Waves The assumption of nondivergent flow used when deriving Rossby s formula is not very realistic for midlatitude weather systems but is a more reasonable assumption for larger planetaryscale flows known as planetary waves or Rossby waves Example of planetary waves on upper air weather map The phase speed of Rossby waves CX j Mg W is determined in part by the adveotion ofthe wave by the mean flow g and by the propagation ofthe wave We can consider three cases for Rossby waves 1 Short wavelength waves large k H lt g 2 Medium wavelength waves L W 3 Long wavelength waves small k gt W What is the direction of motion of the Rossby waves for each of these cases The critical wavenumber ks at which the Rossby wave will be stationary is given by ks 1l Hg The wavelength of a Rossby wave can be calculated as A 2nacos n where n is the number of waves present at latitude 1 Once we knowthe wavelength we can calculate the wavenumber and phase speed of the wave At 1 45 this gives A 27m n gtk2l 2quot A a Cx g lkl We can also determine the period T of the wave at 1 45 noting that 2 EQa T CX rm 2n2 x7 n 02 wErm X ZnZa n 902 21 9 By definition 9 21day so T n days Example Calculate the wavelength wavenumber phase speed and period of a Rossby wave from an upper airweather map What speed zonal flow is required for this wave to be stationary Forcing of Planetary Waves Three primary mechanisms for generating planetary waves 1 Topographic forcing 2 Thermal forcing due to distribution of oceans and continents 3 Nonlinear interactions with smaller scale disturbances Longterm means of 500 mb geopotential heights indicate that planetary waves are located preferentially downstream of meridionally oriented mountain ranges Troughs are found downstream ofthe mountains and ridges are located upstream of the mountains What does this imply about the force extered on the Earth by these waves Equation for topographically forced Rossby wave Start with QG vorticity equation D aw i 7 Dr Cg f f az In a barotropic fluid the wind and hence vorticity does not vary with height so we can integrate the QG voriticity equation from the surface of the Earth hXy to the height ofthe tropopause H which is assumed to be a constant height Dg D D H hxyDt gffWH Whf1 h gt iDg DJL 3 1 HDt gfth quotgax For simplicity we will assume that the surface ofthe Earth is made up of a sinusoidal mountain hxy ho sinkx ly We will also assume that the height of the mountain is much less than the depth of the tropopause hltltH A stationary wave solution ie it does not depend on time is given by u lAcoskx ly viz kAcoskx ly The amplitude of this solution A can be found by substituting the solution into the governing equation A E 2 E u H g Using ks Hg gives lo 1 147 H lit kfl What is the amplitude of this solution for W ks Chapter 5 Scale Analysis Scales Horizontal Type of motion length scale Time scale Cold Front 50 km 12 days Tornado 100 m minutes Midlatitude weather system 1000 km several days Cumulus cloud 1 km tens of minutes Surf 10 m seconds Planetary wave 10000 km weeks months European Alps foehn or Rocky Mountain Chinook 10 km hours Dimensional Homogeneity all terms in an equation have the same dimensions Example NavierStokes Equation iuiviw 77v 2 272 ax 9y dz W W W alt lapd 92 92 92 at 9x 9y dz pax Zstin I 29w cos What units will the ratio of any two terms in this equation have Physically what would such a ratio tell us Rossby Number ratio of acceleration of wind inertial term and Coriolis term in NavierStokes equation LT2 artat N ZQLT 29min 9 29wcos 1 What is the difference between L and L in the previous equations We can use a wind speed scale U and length scale L in place of a time scale T ailat LT2 Zstin 2 2wcos p N ZQLT 1 Nng R0 L 29L What is the magnitude ofthe Rossby number for a midlatitude weather system What does this imply about the role of the Coriolis force in shaping this type of weather system What would cause the Rossby number to be large What would this imply about the dynamics of that weather system Other nondimensional numbers t 2 Reynolds number Re mental au M N U L NE viscous 32 321 0721 vUL2 V 72 72 72 3x 3y dz Froude number39 Fr inertial Llat LZL E I gravity g g gL Scale Analysis Scale analysis a process of simplifying the equations of motion using typical scales for the phenomenon of interest Scale analysis of the Navier Stokes equation Why would we want to use scale analysis on the Navier Stokes equation What is an example of a nonlinearterm in the NavierStokes equation Typical scales for a midlatitude weather system Scale Symbol Magnitude Horizontal wind scale U 10 ms391 Vertical wind scale W lO39st391 Horizontal length scale L 106 m Vertical length scale 4 H 10 m depth oftroposphere Time scale LU T 105 s Kinematic viscosity n 10395 m2s391 Dynamic pressure scale pp 103 m2s392 Total pressure scale Pp 105 m2s392 Gravity g 10 ms392 Density variation scale 5pp lO392 We will define f 2 2sin and f0 29sin45quot 103 gtlt10 1 for our analysis We can then rewrite the Rossby number as R0 N L foL Scale analysis of the horizontal equations of motion 2 2 2 x eqn u z w i v a73a72 va7 2 2vsin 2 2wcos at 6x dy dz p ampx 036 y dz 9v 9v 9v 9v 1 a a2 a2 2 y eqn 7 u7v7 w7 7amp v 77 v7 2 2usin at 0 5 dy dz p y 0 dy dz U2 U2 UW 6 w w scale 7 7 7 7 7 7 U z W L L H L L2 H2 f0 f0 magnitude 10 4 10 4 10 5 10 3 10 16 10 12 10 3 10 6 Based on this scale analysis which terms can we neglect u v at 9x 65 p ax M v 107p7d 2 2usin at 3x 0quot 37 a7u lapd29vsin We can define a horizontal material derivative as Dh a a a 7 7 u7 v7 Dt at 3x y This allows us to write our simplified horizontal equations of motion as l29vsin Dt p ax th l 29usin Dt p Dhuh 1 319d 31 29XM 3x 65 h Scale analysis of the vertical equation of motion 9w 9w 9w 9w 1 9p 92w 92w 92w z eqn 7 u7v7 w7 77 v 2 72 V72 g 2 2ucos 9t 9 9 dz p dz 9 9y dz UW UW W2 P VW VW scale 7 7 7 7 72 72 g fOU L L H pH L H magnitude 10 7 10 7 104g 10 10 1015 10 10 3 Based on this scale analysis which terms can we neglect The Geostrophic Approximation Any flow in which R0 7 0 is termed geostrophic How small does Ro need to be for the flow to be considered geostrophic For midlatitude cyclones R0 UfoL 10 1 lfwe neglect all terms in the horizontal momentum equation that are at least one order of magnitude smaller than the largest terms we get ia Zstin 0 p ax l 29usin 0 p y This is referred to as the geostrophic approximation These equations are purely diagnostic we cannot use them to forecast the future evolution of the atmosphere Geostrophic wind the component of the total horizontal wind that exactly satisfies the force balance between the horizontal pressure gradient force and the Coriolis force i39 g pf 19y V L 3957 g x a Northern Hemisphere 52 fgt0 b Southern Hemisphere S f lt0 isobar low pressure isobar low pressure pressure gradient force gt23 3 quot gt0 pressure gradient force gt0 U9 U3 Coriolis force 2 2ugsinu fug lt0 Coriolis force 2 2ugsinrh fug lt0 isobar mgr pressure y isobar mgn pressure x Copyright 2005 by John Wiley amp Sons Inc or relaled companies All rights reserved Example Calculation of the geostrophic wind from a weather map Since upper air weather maps are often presented as constant pressure maps it is useful to rewrite the equations for the geostrophic wind expressed on a constant pressure surface 13 ug 7i f y 13 v7i g fer p p Cyclonic flow around an area of low pressure cyclone Anticyclonic fow around an area of high pressure anticyclone Why does the wind blow in a counterclockwise direction around areas of low pressure in the Northern hemisphere Why do midlatitude weather systems tend to move from west to east 3 b L PGF 700 hPa 850 hPa 180 ar V v 1000 hPa 9 Copynght 2006 by John Wiiey amp sons Inc or retaled companies Ali rights reserved Chapter 1 Anatomy of a Cyclone The Beast in the East 1517 February 2003 Extra tropical cyclone an area of low pressure outside ofthe tropics Other names for extratropical cyclones Cyclone Midlatitude cyclone Frontal cyclone Storm system Severe weather associated with this storm system Rain Floods Mudslides Tornadoes Heavy snow What were the atmospheric processes responsible for the formation and evolution of this storm as it passed over the United States Why were such diverse severe weather events as heavy snow and tornadoes reported with this one system See weather maps of this storm on the Applied Atmospheric Dynamics CD ROM Describing the Atmosphere Troposphere lowest layer of the atmosphere where temperature normally decreases with increasing altitude SI Systeme Internationale Units Length meter m Mass kilogram kg Time second 3 Temperature Kelvin K see Appendix 2 for additional derived SI units Atmospheric Temperature 2005 KendallHunt Publishing I Fahrenheit CeISIus Kelvm C F F C K 40 40 35 31 water boils sea level 0 212 Q 100 Q 373 30 22 25 13 highest temperature recorded 136 58 331 20 4 average body temperature 9amp6 37 39 310 3915 5 average room temperature 68 20 293 10 14 395 23 water freezes sea level 32 o 273 32 41 10 50 15 59 20 68 lowest temperature recorded 129 39 134 25 77 U U U 30 86 35 95 Temperature Conversions 0C 039 39 32 K C 27315 Atmospheric Pressure Horizontal and vertical variations in pressure give rise to atmospheric motions which are the focus of ATOC 4720 What is atmospheric pressure A molecular perspective 0 at equilibrium higher pressure exerted WI In SI Units for pressure Pressure is a force per unit area From Newton s second law Fma Force has units of kg m s392 or Netwons N Force per unit area has units of kg m391 s392 N m392 or Pascals Pa Other commonly used units of pressure 1hPa 100 Pa 1millibar mb 1 hPa In the atmosphere the observed pressure is also equal to the weight of air above the observation point Therefore pressure decreases with increasing altitude in the atmosphere Z Leadville 2990 m 700 hpa T305 1926 m 300 hpa Rapid City 965 m 900 hpa Seattle 87 m X 1000 hPa US 0in standard altitude Surface pressure the actual air pressure measured at the surface of the eanh Mean sealevel pressure MSLP surface pressure interpolated to zero elevation sea level Why do meteorologists use MSLP instead of surface pressure 1020 lsobar contour line of A constant pressure 1016 QB 1012 Station Reports What are the typical weather observations made at surface weather stations around the world Station Model Wind Speed Temperature 0051951 1 Sea Level Cam Wind 28 229 Pressure Significant gt4 CI d C 1 2 kts gt 1 4 cu over Weather x226 5 kts Wind Speed Dew POlnt and Direction 10 kts Temperature 50 kts Cloud Cover Significant Weather v C O Clear C l Light Rain gt x LIghtSnow quot Drizzle 0 Scattered o MOderate Rain Moderate Snow it Freezing Drizzle 7i 0 Broken 0 0 Heavy Rain aziz Heavy Snow 1 Freezing Rain 0 Overcast o 7 t Rain Shower Snow Shower Sleet Obscured 4 7 Thunderstorm F09 i Drifting Snow Dew point temperature the temperature to which a small volume of air must be cooled at constant pressure in order for that air parcel to become saturated Dew point temperature indicates the amount of moisture contained in the air What units are used for temperature and dew point temperature reported on station models Significant weather Cloud cover Wind direction the direction the wind is coming from Wind speed Units for wind speed SI m s39 Meteorological convention Knots kts 1 kt 051 m s1 Station model wind speed plotting convention Wind speed symbols shown in Figure 14 are additive Actual wind speed is within 2 kts of plotted wind speed Sea level pressure Decoding sea level pressure reported on station models If coded SLP is greater than 500 Put a 9 in front ofthe 3 digit coded SLP Insert a decimal point between the last two digits Add units of mb Example coded SLP 956 Decoded SLP 9956 mb lf coded SLP is less than 500 Put a 10 in front of the 3 digit coded SLP Insert a decimal point between the last two digits Add units of mb Example coded SLP 052 Decoded SLP 10052 mb The normal range of sea level pressure is 950 to 1050 mb Record high sea level pressure 1086 mb Tosontsengel Mongolia Record low sea level pressure 870 mb Typhoon Tip western Pacific Time and weather obsenations UTC Universal time coordinate GMT Greenwich Mean Time Z Zulu time Quick facts about Universal Time Coordinate UTC UTC is based on a 24 hour clock so add 12 to any times after 1259PM 6AM UTC would be written as 06 UTC 12 noon UTC would be written as 12UTC 6PM UTC would be written as 18UTC If UTC time is given as both hours and minutes it looks like this 215AM UTC would be written as 0215 UTC 1200 noon UTC would be written as 1200 UTC 1020PM UTC would be written as 2220 UTC UTC never switches from standard time to daylight savings time This means we need to change how we convert between UTC and Mountain time depending on whether we are on standard time or daylight savings time How do I convert from UTC to MST or MDT MST UTC 7 hours MDT UTC 6 hours How do I convert from MST or MDT to UTC UTC MST 7 hours UTC MDT 6 hours Air Masses and Fronts Air mass a large volume of the atmosphere with relatively uniform temperature and humidity AA AA A Arctic very cold and or run maritime tropical warm amt moist ontinental polar cold and dry continental tropical warm and dry mr maritime polar cold and moist n A Antarctlc very cold and dry r equatorial very warm and moist What conditions favor the formation of air masses What happens to an air mass when it moves away from its source region Front boundary between differing air masses AAAA Cold Front Fronts are defined based on the thermodynamic differences A Warm Front across the front and the direction of movement of the air A v A v Stationary Front masses on either side ofthe front A A Occluded Front Surface Weather Maps l5 Feb 2003 DO LTC Upper Level Weather Maps Since pressure always decreases with height and above any given spot on the earth each height has a unique pressure we can use pressure as a vertical coordinate Weather data on upper level weather maps are plotted on constant pressure surfaces rather than constant height surfaces Pressure surface an imaginary surface above the ground where the pressure has a constant value One of the key properties meteorologists are interested in when looking at a constant pressure map is the height of the constant pressure surface Flat surface 3000 700 mb A E D uniform temperature quotg at every level 3 0 constant sea level pressure 3200 m 3000 a ltitu do 2800 tropical latitudes 2002 KendallHunt Publishing Commonly Used Constant Pressure Maps surface The height of a constant pressure surface varies with the temperature ofthe column of air below the pressure surface Lower heights correspond to a colder column temperature Therefore we expect and do find lower constant pressure heights near the poles and higher heights in the tropics Pressure Approximate Approximate Level Altitude ft Altitude km 850 mb About 5000 ft About 15 km 700 mb About 10000 ft About 30 km 500 mb About 18000 ft About 55 km 300 mb About 30000 ft About 90 km 250 mb About 35000 ft About 105 km 200 mb About 39000 ft About 120 km Upper Air Station Model Code for heights Level Condition Code Example 850 last 3 digits 410 1410 m 700 last 3 digits 970 2970 m 030 3030 m 500 first 3 digits 558 5580 m 300 first 3 digits 900 9000 m 250 lt 10000 m first 3 digits 997 9970 m 250 gt 10000 m center 3 digits 054 10 40 rn 200 center 3 digits 176 11 760 m Height of pressure surface tem erature above sea level in gelsius gt391 O K meters coded for different levels Height change during past dewpolm 39gt 5 12 hours depresswn meters below 500 mb in Celsius decameters 500 mb and above wind direction wind speed Idewpoint depression temperature dewpoint temp 2002 KendallHunt Publishing What are the differences between surface and upper air station models A sample 500 mb upper level weather map E17 554 as 54 quotera 7 2 9 5 6 29L JVQ K m 49552 f 5 gig I A 1 1 39 7 39 9 l Trough region of low heights on a constant pressure map Ridge region of high heights on a constant pressure map Shortwave a small ripple in the height field Longwave a large ripple in the height field How does the height ofthe 500 mb surface vary from south to north on this map What is the relationship between winds and height contours on an upper level constant pressure map Does this relationship vary between the Northern and Southern hemispheres What wind direction should we expect to find in the midlatitudes if lower constant pressure surface heights are found nearthe poles and higher heights are found in the tropics lsotherm contour line of constant temperature Baroclinic temperature varies on a constant pressure surface Barotropic temperature is constant on a constant pressure surface The Structure of a Typical ExtraTropical Cyclone Extratropical cyclone characteristics Size Pressure Winds Polar front boundary between warm tropical and cold polar air masses Stages in the life cycle of an extratropical cyclone a Cold Kquot C 1T 00 V 1 V Warm What physical process can cause the surface pressure to decrease Convergence accumulation of atmospheric mass Divergence removal of atmospheric mass Divergence associated with the shortwave shown in Figure 19a leads to the formation of a low pressure center at the surface ofthe earth What is the direction of circulation around this area of low pressure How does this circulation alter the distribution of temperature What types of fronts form in response to this circulation Warm sector wedge of warm air ahead of an advancing cold front and behind a warm front How does the height ofthe 500 mb constant pressure surface change in response to the changes in the distribution of temperature near the surface Mature stage of a midlatitude cyclone Venical crosssection from A to B X J l m quot y I Cord air lt MyMayM lt cwd air PreCIpvtahon Overcast 7 Warm gt Cold air Cold air Where are the upper level ridge and trough located relative to the low pressure center at the surface during the mature stage of the cyclone s life cycle What determines how much the surface pressure decreases or increases See animation of mature midlatitude cyclone on class web page Where are the clouds and precipitation located relative to the fronts in this cyclone How does the speed of movement of the cold and warm fronts differ What happens when the cold front catches up to the warm front Occluded front a boundary that separates two cold air masses at the surface with warm air aloft D Colder Z Cross section 1 on panel C y Z Cross section 1 on panel C y across the occluded front across the occluded front quot Coldquot X Cross section 2 on panel C W through the point where the cold through the point where the cold and warm fronts meet and warm fronts meet Warm icofd V Cross section 3 on panel C u V Cross section 3 on panel C u 2005 KendallHunt Publishing How does the position ofthe upper level trough relative to the surface low pressure center change once the cyclone has occluded What causes the cyclone to weaken and dissipate The February 2003 Storm Revisited Mature stage 00 UTC 15 February 2003 see weather maps shown above Occluded stage and the creation ofthe Beast in the Eastquot 00 UTC 17 February 2003 Surface Weather Map 4 25 o 5 I 21055 Ztl x r 30 34 30 l 2H 17 Feb 2003 00 UTC 390 500 mb Weather Map I gt2254s 7 V T2545 A LL j 2 555 V i 5 WE 5 22 555 is 1 4M s 5 I 70 563 2 5 2 39 56770357 392 h 1 i 59 90 500 mb i7 Feb 2003 00 UTC Which low pressure center at the surface is in the most favorable position to intensify at this time Chapter 4 Fundamental Forces Newton s Second Law Fma In atmospheric science is typical to consider the force per unit mass acting on the atmosphere Force a a mass In order to understand atmospheric motion accelerations we need to know what forces act on the atmosphere Body and Surface Forces Body or volume force a force that acts on a volume ofthe atmosphere Surface force a force that acts locally upon a part of a fluid Forces in an Inertial Reference Frame Inertial reference frame a nonaccelerating frame of reference Gravity a force that arises from the mutual attraction between two objects I4I G gravitational constant 6673x103911 N m2 kg39z M product of the mass of the two objects being considered r distance between objects For a unit mass of air at the surface of the Earth M 59742x1024 kg mass of Earth r 637X IO6 m radius of Earth g 983 m s392 Geopotential 11gt work required to raise a unit mass to height 2 ltIgtz fgzdz 0 Note that g varies with 2 since it depends on the distance from the center of the Earth Geopotential height Z I 1z 9 Z Tfgdz go goo Pressure Gradient Force Example Realtime weather map n5p Z V X 4 gt The force exerted on the left face of this air parcel due to pressure is 17A 176sz The force exerted on the right face of this air parcel due to pressure is p 6p6y6z p lip g y z 19x The net force exerted by pressure on this air parcel is 9 l X z 3X By dividing by the mass of the air parcel pdx y z we get the force per unit mass due to changes in pressure ie the pressure gradient force We can write all three components ofthe pressure gradient force as 77717j F 191 ape it p x a 92 In what direction does this force act relative to locations with high and low pressure Example Calculation ofthe pressure gradient force from a weather map Viscous Force AreaA V V gt 3 fluid 3 quotI n a Copyright 2006 by John Wiley amp Sons lnc or related companies All rights reserved Each layer in this fluid is experiencing a shear stress due to differences in the speed of motion of adjacent layers of fluid fof 7 Mdz y coefficient of shear viscosity depends on fluid being considered How does the shear stress change between adjacent layers of the fluid What does this imply about the net force acting on a layer ofthis fluid For a net force to arise there must be a gradient in the shear stress The force that arises due to a gradient in the shear stress is called a viscous force and can be represented as p A of M dzu dzu dzu l 2 2 2 dx dy dz Kinematic viscosity coefficient v 3 l5x lO395 m2 s391 in the atmosphere p In three dimensions this force is written as dx2 dy2 dz2 A dzv d2v dva F v 2 7272 J dx dy dz d2w d2w dZWL dx dy dz r Hydrostatic Balance For an atmosphere that is at rest what forces act in the vertical direction 1d quot7175 de dp gt7 dz Pg This equation is known as the hydrostatic equation Hydrostatic balance a balance between the vertical pressure gradient force and the gravitational force Combining the hydrostatic equation with the ideal gas law gives p Z 5 1n 7 idz P0 RdT Using a layer average temperature we can write P z poe RAT where H This equation is known as the hypsometric equation and relates pressure and height Using the definition of geopotential we can also write gidz p RdT dltD Rdel P p AltD Rddelnp Po p gtAZ Rfde1np g0 P0 Thickness AZ the difference in geopotential height Z between two pressure levels What causes the thickness to increase or decrease Example Thickness and atmospheric soundings Forces in a Rotating Reference Frame Noninertial frame of reference a frame of reference which is undergoing an acceleration Why is a frame of reference fixed to the Earth a noninertial frame of reference Angular velocity of rotation ofthe Earth 9 2 2 7292gtlt10395s391 1 s1derea1 day 86164 sec Sidereal day the time required for the earth to make one complete revolution in an absolute fixed coordinate system that is with respect to the stars Centrifugal Force Rotating table experiment centrifugal force Copyright 2005 by John Wiley amp Sons lnc or related companies All rights reserved What happens to the object when it is placed on the rotating table What force is required to keep the object stationary on the rotating table quot1er Centrifugal force outward radial force Centripetal force inward radial force Is this object experiencing an acceleration when viewed in a frame of reference rotating with the table Is this object experiencing an acceleration when viewed from a xed frame of reference Which forces are acting in each frame of reference The centn Jgal force arises only in observations taken in a rotating frame of reference and is due to the acceleration ofthe frame of reference The centrifugal force and the Earth mRszlcosw 39 equator Copyright o 2005 by John Wiley amp Sons Inc or related companies All rights reserved What is the direction ofthe centrifugal force for an object on the Earth The centrifugal force is a body force and can be combined with the gravity force to give an effective gravity Effective gravity g 981 m s392 at sea level For atmospheric science applications we use effective gravity g rather than gravity g Therefore effective gravity should be used in place of gravity in all ofthe previous equations in this chapter The addition ofthe centrifugal force to the gravity force results in the effective gravity force not being directed towards the center ofthe Earth In reality the Earth is not a perfect sphere and the effective gravity is exactly normal to the surface ofthe Earth at all locations neglecting topography Therefore the effective gravity force acts only in the z direction This discussion has only considered an object at rest in a rotating frame of reference What happens when an object is in motion relative to a rotating frame of reference Coriolis Force For an object in motion relative to a rotating frame of reference we need to consider an additional centrifugal force changes in the relative angular momentum ofthe object What is the centrifugal force acting on an object moving towards the east u 2 A Cme mLQ 7 R R 2 m92R 2LmR mL2R R R The first term on the RHS ofthis equation is the same as the centrifugal force for an object at rest and is incorporated into the effective gravity The last term on the RHS of this equation is assumed to be small since R gtgtu The remaining term on the RHS of this equation can be written as CF motion 2mQu sin If ZmQucos 1 and is the centrifugal force due to the motion of the object that we need to consider in our study ofthe atmosphere This component of the centrifugal force arises only for zonal eastwest motion Conservation of angular momentum For a rotating object the angular momentum of the object is given by Ia ngjg The angular momentum lw ofthis object will remain constant What happens to the terms in this equation if the object is moving towards the North Pole vgt0 2 6M mR9mR 6R 9 R MR 42 2 2 Rj u R Q R Q 2R 6RQ R 6R As the object moves towards the pole its zonal velocity u increases Noting that 6R vsin p62 we get 6 Zstin p62 As at a 0 we then have du i Zstm dt Conservation of angular momentum results in a zonal acceleration for objects moving in a meridional northsouth direction Similarly conservation of angular momentum also results in objects moving in the vertical direction experiencing a zonal acceleration equal to Q 2 2wcos dt Conservation of angular momentum does not lead to an acceleration for objects moving in a purely zonal direction The accelerations due to the additional component of the centrifugal force and due to conservation of angular momentum for an object in motion in a rotating frame of reference can be combined to form a single force known as the Coriolis force M Zstin Zchos p 29M sin pf ZQucos 51 m Using vector notation and noting that Q Qcos pf Qsin pl the Coriolis force can also be expressed as FCoriolis 2Q X a m In what direction does the Coriolis force act relative to the motion of the air Does this differ in the Northern and Southern hemispheres Example The Coriolis force on the roof of Duane Physics The NavierStokes Equation The sum ofthe forces acting in each direction gives the acceleration experienced by an air parcel 2 2 2 amp lipv afgaizaig 29vsin 29wcos Dt pax ax 3y dz 2 2 2 E la7pv a7 L L 29usin Dt p 3y ax 3y dz Dw lap 32w 32w a2w 7 77v 7 7 7 29 cos Dt g paz LaxZ ay2 az2 u 1 What does each term in these equations represent These equations are known as the NavierStokes equations We can rewrite the NavierStokes equations in an Eulerian framework as 2 2 2 alu viuwiu laipvaizailaig 29vsin 29wcos at ax ay az pax 9x 9y 91 2 2 2 u v w laipvaiaiai at 3x 3y dz pay ax 3y dz 2 2 2 37Wuaiwvaiwwaiw g laipv aiya72vaigv 29MCOS at 9x 9y dz paz ax 9y dz Perturbation Pressure We can rewrite the vertical momentum equation by defining a reference pressure p0 that is in hydrostatic balance with a reference density p0 as L dz p08 The total pressure is then given by p 190z 192 Substituting this definition into the vertical component ofthe NavierStokes equation gives DW 3 072w 92w 072w i 7 777 29 cos th pg azp0z pd pvkaxZ ayz az2 p u p apd 32w 32w 92w 7 v777 ZQucos P8 P08 az p ax2 ay2 az2 p 1 2 2 2 DW p pog19pd V8399 9 9x 9y dz Dt p p dz Qme Chapter 6 Simple Steady Motion Natural Coordinate System Natural coordinate system a coordinate system in which one axis is U normal to and to the lelt of the wind 17 Copyright 2005 by John Wiley amp Sons Inc or related companies All ngms reserved vector is 1 Natural coordinate system notation V nd 7 V Coordinate locations mm Axes unit vectors The NavierStokes Equations in Natural Coordinates Material Derivative Based on this geometry amp l l A 67 d 641M w R an w M l l lbfl6i R ii d R LDJE Dz Dsm v R DV ADV Av iariwqi What is the physical interpretation of the two terms that makeup the parcel acceleration in this coordinate system R radius of curvature R is positive when center of curvature is to the le of the wind vector Coriolis force fVT7 What is the direction ofthis term in the Northern and Southern hemispheres Pressure gradient force la if 07417 p as an NavierStokes equations in natural coordinates 5 component lap DI p 075 2 n component L fV 107 R p 9n What terms have been neglected in this set of equations What do we know about vertical motion and forces for this system Balanced Flow Balanced flow a purely horizontal frictionless flow that is also steady state For balanced flow the Navier Stokes equations in natural coordinates reduces to 5 component i0 DI p 075 2 n component L l R pan What does this imply about the orientation of the flow relative to isobars 1 2 This system can be further simplified to LfV R p 311 What does each term in this equation represent Inertial Oscillations For flow with no pressure gradient the governing equation reduces to V2 FfV0 and V fR What is the time period required for the flow to complete one revolution about its center of circulation 2 HR V 2rt 2r 1 day f ZQsimJ 28in Cyclostrophic flow The Rossby number in natural coordinates can be expressed as V M 39 W What conditions result in a large value of R0 2 R0L R What does this imply about the importance ofthe Coriolis term in the governing equations For large R0 the governing equation reduces to VLHEA R pdn What is the physical interpretation of this equation Geostrophic approximation What value of R0 is required to for the geostrophic approximation to be valid In natural coordinates the geostrophic approximation can be written as i W p an The Gradient Wind Approximation What force balance needs to be considered when Ro 1 Gradient wind the component ofthe flow that satisfies an exact balance between the centrifugal force Coriolis force and pressure gradient force 2 L l0 R pan 1 2 2 2 Vh i f R job 2 4 p an The gradient wind can also be expressed in terms of the geostrophic wind 1 2 2 2 V fj Lf7R fRVg 2 4 Multiple solutions for V are possible with this equation see table 61 What requirements exist for the sign of V in these equations What other requirements are there that allow for a physical solution ofthis equation Table 61 Sign and magnitude of terms in the gradient wind equation for all possible flow regimes in the Northern Hemisphere Northern Hemisphere Cyclonic Anticyclonic Anticyclonic Cyclonic CCW flow CW flow CW flow CCW flow around L around H around L around H f R 971 9n 1 lt E or lt E or 2 2 E 2 I E 2 f R Rapjz awaysgt awaysgt 27 Imaginary for Imaginary 4 p 311 f2R2 R 3p for 7 lt if 4 p 311 f2R2 lt 4 p 311 fR 2 either root V positive for root only but f2R2 R 9p root only never 7 gt if 4 p 311 Physical solutions for the Southern hemisphere fR f2R2 R 3p 2 Cyclonlc CW flow around L R lt 0 V 7 7 774 2 4 p an 1 2 2 E Anticyclonic CCW flow around H R gt 0 V f R 5 2 4 p an Physical solutions for the Northern Hemisphere z z 7 CyolonicCCW ow aroundL R gt0 v if if R 1 4 pin 1 Z Z T AnticyclonicCW owaroundH Rlt0 V7 775L4 p Force balance for Northern Hemisphere gradient wind VNH a b Copyright 2005 by John Wiley 8 Sons Inc or related companies All rights reserved In order for the solution ofthis equation to be real for the anticyclonic case H gt 5 4 p 77 This requires a light pressure gradient near the center ofa high and thus also light winds No such limit exists for ow around low pressure What is the physical explanation for this limit The geostrophic wind can be written as a function of the gradient wind 2 L 1 R p 6n V2 fV fvg 0 gt Vg v1 What does this imply about the magnitude of the geostrophic wind relative to the gradient wind for ow around low and high pressure centers We can also express the geostrophic wind as a function of the gradient wind and the Rossby number v8 v1 RU What does this imply about the magnitude of the geostrophic wind relative to the gradient wind as R0 increases Example Comparison of observed gradient and geostrophic winds from a surface weather map What role does friction play in resultant altering the gradient wind balance win Copyright zoos by John Wiley amp sons in or related companies All rights reserved The Boussinesq Approximation Boussinesq approximation allows variations in density to give rise to buoyancy forces in the vertical momentum equation but have no impact on the horizontal force balance To apply the Boussinesq approximation we must define p00 constant reference density p0z vertically varying density consistent with hydrostatic pressure profile pXy horizontally varying density consistent with horizontally varying dynamic pressure pd Using these definitions the geostrophic equation becomes ng i in natural coordinates and P00 3quot fug i07pid fv i07 in an Earthbased coordinate system P00 3 g P00 x The vertical momentum equation is given by lo loo 1 917d g 77 poo poo 91 The buoyancy force can be rewritten as P Poz mg a g 00 T00 and can be combined with the vertical momentum equation to give La P00 91 The Thermal Wind Taking the horizontal derivatives of the vertical momentum equation and substituting these into the vertical derivative of the geostrophic equation gives 10 37 az f ay az f ax or m g 77 avg g 77 z az Tm ay az Tm ax What is the physical interpretation ofthis equation ow can we use this equation to explain the increase of westerly winds with height in the midlatitude troposphere Z stratosphere cold u y a a T lt 0 z X 3y troposphere hot 2 Copyright 2006 by thn Wiley amp Sons Inc or related companies All rights reserved Thermal advection Warm air advection WAA the wind blows from a region of warmer temperatures to a region of cooler temperatures Cold air advection CAA the wind blows from a region of cooler temperatures to a region of warmer temperatures We can use the thermal wind relationship to evaluate the change in geostrophic wind over a layer of depth Az For eastwest oriented isotherms this gives 9 A Ai dz A g 6T f 2 u z Azz g ee 19y gZAz gz AZOAZZ Warm and cold advection cases in the Northern hemisphere PAP TAT 39 TAT g H t t a warm advection b cold advection Copyright 2006 by John Wiley amp Sons lncv or related companies All nng reserved In what direction does the geostrophic wind turn for the warm advection cold advection case Veering wind turns clockwise with height Backing wind turns counterclockwise with height What would these cases look like in the Southern hemisphere For both hemispheres goldhair advection leads to cyclonic turning of the geostrophic wind with eIg Warm air advection leads to anticyclonic turning of the geostrophic wind with height Departures from Balance Quasigeostrophic flow a flow in which small departures from geostrophic flow occur For this type of flow Rois small but finite Derivation of the quasigeostrophic equations Start with horizontal momentum equation scaled for midlatitude systems Dh h 1 dpd 7 apd A D p iz7 Ax ax 9y J fk h and use the hydrostatic approximation for the vertical momentum equation Ageostrophic wind component of the wind that is not in geostrophic balance We will define R0 1 ug Assume that variations in density are negligible so du dv a aiw0 and JJ0 dx dy dz dx dy Note that the geostrophic wind is nondivergent du dv dw gt 1 7 7 0 dx dy dz dw dud dva 7 7 dz dx dy Only the ageostrophic component of the wind can cause divergence The terms in this equation scale as ug U lualROU lwlNW xyL ZNH How do the vertical and horizontal velocity scales compare Use the definition ofthe ageostrophic wind to rewrite the horizontal momentum equation 61 9 a 613 uivi fv t x y z p x uVw fvgfv a u ua u ua u ua u ua gax vgva gay w gaz fva 2 Neglecting terms that scale to less the UT gives u u u Ju iv 7g t g x g y fvd The full horizontal quasigeostrophic momentum equation is u u u igu igv Jfva t g x g y v v v igu igv 7g fua t g x g y or in vector notation Ageostrophic Flow The quasigeostrophic momentum equation can be rewritten in terms of the ageostrophic wind components Va iigl u 6amp4 a f at f g ax g 3y What is the direction ofthe ageostrophic wind relative to the acceleration vector The equations for the ageostrophic wind components can be expressed in terms of pressure gradients using the definition of the geostrophic wind 1 072p 1 u a pv a p 2 2 g 2 g Poof axat Poof ax axay 1 072p 1 u 072p V a2pl 2 2 g g 2 Poof ayat Poof L axay 3y J For a flow in which the time rate of change term is largest 1 LP Poof2 axat ip Poof2 ayat and U3 and v8 are referred to as the isallobaric wind lsallobar line of constant apdat What is the isallobaric wind for the example below 3p a o hPah 2 hPah 4 hPah I Pressure steady falling falling rapidly 2 2 Based on this map Q i lt 0 and Q axar 9x a 9ny Then for a Northern hemisphere location fgt0so u gt0 and v 0 How will the change in pressure with time in this example alterthe geostrophic wind What is the direction ofthe Coriolis force associated with this ageostrophic component of the Wind acceleration of the northerly wind allowing the ow toraccelerate towards a balanced geostrophic state by me advectwe acce erauon term W J Mivi f ax ay 1 1314 1314 V7Mivi quot f x s y z z 375 v x my nnf z l v pnnf My 9y What 5 me ageostropmc Wmd for thejetstreak examp e be ow Low 69 Va u g gt 9 Hi h 9 7070 On the eft swde On the tht swde WII XgtQSOVHgtO Jul27xlt0 SOVHltO ow on each swde of ms jet On ewmer swde of thejet me Comohs force assooaxed Wm me ageostropmc ow acce erates the ow towards a geoatropmc ba ance In general an ageostrophic wind directed towards low pressure will accelerate the flow in the direction of the geostrophic wind while an ageostrophic wind directed towards high pressure will decelerate the flow in the direction of the geostrophic wind Geostrophic Adjustment a process of restoring the flow to geostrophic balance What role does the ageostrophic flow play at the surface and 500mb in the example below WARM Z g 1 TAN Copyright e 2005 by John Wiley amp Sons Inc or related companies All rlghls reserved accelerate the geostrophic wind to bring the flow back towards geostrophic balance alter the vertical shear of the geostrophic wind to bring the flow back to thermal wind balance alter the horizontal temperature gradient through rising and sinking motion to bring the flow back to thermal wind balance Chapter 11 Clouds and Severe Weather Moist Processes in the Atmosphere Importance of water in the atmosphere Humidity Variables Mixing ratio r The mass of water vapor contained in a unit mass of dry air mV units kgkg d r mv mass of water vapor units kg md mass of dry air units kg Vapor pressure 6 The partial pressure exerted by watervapor molecules in a given volume of the atmosphere The vapor pressure can be calculated from a known mixing ratio as I units the same as units of in this e uation e H MVMd p P q My Molecular weight of water vapor 18016 g mol391 Md Molecularweight of dry air 28966 g mol391 a MvMd 0622 Alternately the mixing ratio can be calculated if the vapor pressure is known Specific humidity q Mass of water vapor contained in a unit mass of air mv units kgkg q mdmv The following equations allow for conversion between mixing ratio and specific humidity r7 7 q q 1r q I 1 Saturation A dynamic equilibrium between air and a water surface in which there are as many water molecules returning to the water surface condensing as there are escaping evaporating When air is saturated water vapor will condense to form liquid water Saturation vapor pressure es The partial pressure that would be exerted by water vapor molecules in a given volume of the atmosphere ifthe air were saturated Saturation vapor pressure can be calculated using a number of semi empirical equations One ofthe more widely used equations for es is Teten s formula b T T es e0 exp T T1 units hPa 2 e06112 hPa b 1727 K1 T127316 K T2 3586 K As the air temperature increases the amount of water vapor required for the air to become saturated increases at an exponential rate Similarly as air is cooled the amount of water vapor required for the air to become saturated decreases at an exponential rate How can air become saturated in the atmosphere An air parcel can become saturated by Adding more water vapor to the parcel Decreasing the temperature of the parcel Some common examples of air becoming saturated For an air parcel that contains a nonzero amount of water vapor the air parcel can be cooled sufficiently such that it eventually becomes saturated The temperature to which this air parcel must be cooled at constant pressure is known as the dew point temperature Dew point temperature Td The temperature to which an air parcel must be cooled at constant pressure in order for the air parcel to become saturated Dew point temperature is the most commonly reported humidity variable on weather maps The dew point temperature can be used to calculate the vapor pressure with b T T e e0 exp M units hPa Td T2 The constants e0 b T1 and T2 are the same as those used when calculating es from T Saturation mixing ratio rs The mass of watervapor contained in a unit mass of dry air ifthe air were saturated Relative humidity RH The ratio of the vapor pressure to the saturation vapor pressure an indication of how close the air is to being saturated RH 3 x100 z 1x100 units 1quot es S Saturation with respect to an ice surface A dynamic equilibrium between air and an ice surface in which there are as many water molecules returning to the ice surface depositing as there are escaping sublimating How will the saturation vapor pressure over an ice surface compare to that over a liquid water surface es 6 s K266 T l es is the saturation vapor pressure with respect to an ice surface Relative humidity can be calculated with respect to an ice surface as RHZ ix100 Si Unless otherwise indicated saturation and all related variables es rs RH is always with respect to a liquid water surface Thermodynamic diagrams The Skew T log p diagram This diagram provides a graphic representation of atmospheric thermodynamic properties and can be used to diagnose cloud formation and determine how air parcels will respond to vertical motion Information plotted on a skew T log p diagram Pressure black horizontal lines Units hPa or mb logarithm of pressure is used as the vertical axis on this diagram Temperature Black lines that slope up and to the right units C Also used to indicate dew point temperature Saturation mixing ratio Blue dashed lines Units 9 kg391 Also used to indicate mixing ratio Dry adiabat Lines of constant potential temperature cuned yellow lines that slope up and to the left Units C Moist adiabat Lines of constant wetbulb potential temperature curved black lines that slope up and to the left Units C Skew T log p diagram Height m 12280 9590 hPa 5840 temperature C saturallon Mlxlng ratio glng Copyright 2006 by John Wiley amp Sons lnc or related companies All rights reserved Example Skew T diagrams and moisture variables using Duane Physics weather station observations Determine the mixing ratio and saturation mixing ratio on the roof of Duane Physics using the skew T diagram and the reported temperature dew point temperature and pressure How does the relative humidity calculated from these values compare to the reported relative humidity Plotting radiosonde data on a skew T diagram Example KFWD 5 June 2005 00 UTC Fort Worth TX US KFWD 1 3283 9730 196 72249 Date0000Z 5 JUN 05 Station KFWD WMO ident 72249 Latitude 83 Longitude 9730 Elevation 19600 LEV PRES HGHT TEMP DEWP RH DD WETB DIR SPD THETA THE V THE W THE E mb m C deg knt K K K K gkg SEC 982 196 314 224 59 90 247 165 10 3061 3094 2985 3591 1763 2 925 728 260 200 70 60 217 175 21 3059 3089 2975 3542 1611 3 898 989 236 198 79 38 209 186 21 3060 3091 2977 3552 1640 4 850 1468 204 155 73 49 171 195 20 3075 3100 2960 3472 1314 5 837 1601 230 50 31 180 123 200 20 3116 3129 2922 3319 654 6 830 1675 232 52 31 180 124 203 21 3126 3138 2926 3334 669 7 771 2310 176 26 37 150 90 207 23 3132 3143 2923 3320 599 8 749 2557 160 60 21 220 52 201 23 3141 3147 2902 3246 326 9 700 3127 118 122 17 240 15 200 20 3156 3160 2895 3226 214 10 617 4170 46 134 26 180 28 247 28 3189 3193 2906 3262 220 11 544 5180 45 175 35 130 90 260 34 3197 3201 2905 3258 178 12 523 5490 63 313 12 250 123 253 29 3212 3213 2897 3231 053 13 500 5840 91 321 14 230 143 245 22 3219 3220 2899 3238 052 14 475 6232 123 333 16 210 167 235 25 3227 3228 2901 3245 049 15 433 6930 155 435 7 280 198 235 33 3273 3274 2912 3281 019 16 400 7520 201 441 10 240 234 235 39 3289 3289 2916 3296 019 17 300 9590 355 575 9 220 368 235 41 3353 3353 2932 3356 005 18 250 10830 455 635 11 180 461 245 43 3384 3384 2940 3385 003 19 235 11238 489 649 14 160 493 244 47 3393 3393 2942 3394 003 20 200 12280 541 691 14 150 544 240 54 3471 3471 2960 3472 002 21 188 12677 569 709 15 140 571 238 51 3488 3488 2964 3488 001 22 172 13239 597 727 17 130 599 235 47 3531 3531 2973 3532 001 23 150 14100 593 753 11 160 595 245 38 3679 3679 3002 3680 001 24 130 14990 605 775 9 170 607 245 50 3811 3811 3024 3812 001 25 100 16590 681 841 8 160 682 250 42 3962 3962 3047 3962 000 26 90 17220 691 841 10 150 692 254 30 4063 4063 3060 4063 000 27 70 18740 637 817 7 180 640 160 15 4481 4481 3107 4481 001 28 67 19008 641 841 5 200 644 163 15 4529 4529 3111 4529 001 29 50 20820 589 789 6 200 595 5 5 5047 5047 3153 5048 002 30 46 21345 583 803 4 220 590 45 5 5183 5183 3162 5184 001 31 41 22065 619 809 6 190 624 63 12 5267 5267 3167 5268 001 32 33 23435 539 759 5 220 555 92 13 5816 5816 3198 5819 004 33 30 24050 527 747 5 220 546 105 14 6010 6010 3207 6013 005 34 26 24977 491 721 5 230 521 110 12 6363 6363 3222 6370 009 35 20 26700 473 713 5 240 514 45 12 6914 6914 3242 6925 013 36 14 29080 433 693 4 260 501 70 18 7792 7793 3268 7815 024 37 13 29579 439 699 4 260 508 67 16 7938 7939 3271 7962 024 Archived radiosonde data like this can be found at http vortexplymoutheduuacalpIt uhtm Example KFWD 5 June 2005 00 UTC Height m 12280 9590 7520 5840 temperature C saturation Mixing ratio glkg Lifting condensation level The level at which saturation occurs as an air parcel is lifted Example Determine the lifting condensation level LCL for an air parcel lifted from the surface at Fort Worth TX on 5 June 2005 at 00 UTC As an air parcel is lifted dry adiabatically the temperature of the air parcel changes at the dry adiabatic rate the mixing ratio of the air parcel remains constant Example Determine the lifting condensation level for air lifted from the roof of Duane Physics Height m 12280 9590 pressure hPa 5840 temperature 90 saturation Mixing ratio gkg Copyright 2006 by John Wiley amp Sons lnc or related companies All rights reserved Does the temperature of an air parcel continue to decrease at the dry adiabatic rate when lifted beyond the lifting condensation level Once an air parcel is lifted beyond the lifting condensation level latent heat is released as water vapor condenses and the potential temperature of the air parcel increases Stability in the Moist Atmosphere What condition defines an unstable atmosphere for unsaturated conditions Dry adiabatic lapse rate Pd Rate of temperature decrease with height required to maintain a constant potential temperature 2 Fdi sll977Kkm110Kkm4 c 1004Jkg39 K p The dry adiabats on a skew T diagram indicate the dry adiabatic lapse rate and the combination of temperature and pressure required for an air parcel to maintain a constant potential temperature as it moves vertically in the atmosphere An unsaturated air parcel that is undergoing an adiabatic displacement will move parallel to the dry adiabats on a skew T diagram lfthe lapse rate ofthe environment F is greater than D then the environment is statically unstable and an unsaturated air parcel that is initially displaced upward will continue to accelerate upward Moist adiabatic lapse rate F5 The rate of temperature decrease with height of a rising saturated air parcel The moist adiabats on a skew T diagram indicate the moist adiabatic lapse rate and the combination of temperature dew point temperature and pressure required for a saturated air parcel to maintain a constant wetbulb potential temperature 6W as it moves vertically in the atmosphere A saturated air parcel that is undergoing a moist adiabatic displacement will move parallel to the moist adiabats on a skew T diagram How does the magnitude of Pd compare to the magnitude of F3 What are the static stability classes for Fgt Fd F Fd Pd gt Fgt F5 F F3 Flt F3 What does each ofthese cases look like on a skew T diagram Example Determine the temperature and dew point temperature of an air parcel lifted from the surface to 200mb at KFWD on 5 June 2005 00 UTC Note Above the LCL the air parcel temperature and dew point temperature will be equal the air parcel is saturated and will maintain a constant wet bulb potential temperature Level of free convection LFC or zLFc The height at which the temperature of a rising air parcel Tparceo first exceeds the temperature of the environment Tenv Equilibrium level EL or ZEL The height above the LFC at which the temperature of a rising air parcel is once again equal to the temperature of the environment Convective available potential energy CAPE The vertically integrated air parcel buoyancy between the LFC and the EL Graphically the CAPE is the area between the air parcel and environmental temperature cunes on a skew T diagram between the LFC and EL Mathematically the CAPE can be calculated as in T T parcel envjdz ZLFC 9 Convective lnhibition CIN The energy barrierto be surmounted before an air parcel reaches the LFC CAPE is a measure ofthe energy available to accelerate an air parcel vertically in a thunderstorm and the maximum vertical velocity that can be achieved in a thunderstorm is given by w 2CAPEO395 T values of CAPE O to 500 J 39 weak What is the maximum vertical velocity for each of these values of CAPE Example Indicate the location of the LFC and EL on a skew T diagram for KFWD on 5 June 2005 00 UTC Also indicate the areas on this sounding that represent the CAPE and CW Example Determine the temperature and dew point temperature of an air parcel lifted from the roof of Duane Physics to 200mb Skew T diagrams and flow over mountains Skew T diagrams can be used to determine changes in the temperature and humidity of air as it is forced to cross a mountain range Example Determine the temperature and dew point temperature of an air parcel as it crosses the Rocky Mountains from Grand Junction CO to Boulder CO Height m 12280 9590 7520 pressure hPa 5840 temperature DC saturation ermg ratio glkg Copyright 2006 by John Wiley amp Sonsr ine or related companies All rights reserved Air Mass Thunderstorms overshoot anvil M m a pressure hPa 4 5 km gt 9 km gt 4 i2 km cumulus mature dissipating Copyright lt2 2006 by John Wiley amp Sons Inc or related companies All rights reserved Cumulus stage Rising air parcel becomes saturated and reaches the LFC Freezing level Location at which the temperature drops below 0 C Entrainment Environmental air is mixed into the cloud through turbulent motions What impact does entrainment have on the temperature and dew point temperature in a cumulus cloud compared to the values indicated on a skew T diagram During the cumulus stage of an air mass thunderstorm no precipitation falls and the cloud consists of a single updraft Mature stage Cloud depth has increased often to the tropopause and precipitation begins to fall from the thunderstorm At this point the cloud is referred to as a cumulonimbus cloud Downdraft Sinking air in the thunderstorm What two processes cause the downdra to form in an air mass thunderstorm As precipitation falls from unsaturated air below cloud base the air begins to cool This cooled air will sink since it is colder than its environme downdraft u dra This rain cooled air is the p source ofthe cool gust of air one o en experiences gust front when standing near a Copyright 2005 by John Wiley 3 Sons Inc thunderstorm The leading or related companies All rights reserved edge of this rain cooled air is known as the gust front Eventually the downdra comes to dominate the thunderstorm and the storm enters the dissipating stage During what time of day do air mass thunderstorms typically develop What is the typical lifetime of an air mass thunderstorm Multicell thunderstorms Multicell thunderstorms form when the gust front from one thunderstorm li s moist conditionally unstable air to its LFC causing a new thunderstorm to develop Supercell Thunderstorms and Tornadoes Supercell A thunderstorm in which the updraft is rotating Mesocyclone The rotating updraft in a supercell thunderstorm How does the environment in which a supercell thunderstorm forms differ from the environment in which an air mass thunderstorm forms 100 200 9 300 l Pressure mb 0 h o o o o 600 700 standard height above sea level kilometers 800 900 o 1000 80 70 60 50 40 30 20 10 Tropopause l Temperature Vertical wind shear within conditionally Parcel Ascent unstable layer l Environmental lapse rate Dewpoint l nearly equals dry adiabatic lapse rate within dry layer aloft Temperatu re X I I Dry cold airmass l l i Capping I Inversion 1 4 Warm moist airmass gt 2005 KendallHunt Publishing 010 20 30 40 fffiff CAPE gt 1500 J kg391 Vertical wind shear gt 20 m s391 over the lowest 6 km of the atmosphere What role does the vertical wind shear play in supercell formation and lifecycle How does the lifetime of a supercell thunderstorm compare to the lifetime of an air mass thunderstorm Crosssection of a typical supercell thunderstorm anvil rear backsheared i Pfg w ll anking line 40 km 4 tro popa use bounded weak echo region vault dashedg m 1 fmk l i 3 6R 5 2 K 39 overshooting top H l I I l t small hail rain in hook wall cloudlarge small hail heavy tornado hail hail rain rain mix rain moderate light rain mammatus virga NE updraft downdraft jetstream 2005 KendallHunt Publishing What causes a supercell thunderstorm to rotate g TlLTlNG c 367 Mesocyclone sheared winds from 39 the south rotation tilted by updraft Height km 2002 KendallHunt Publishing Lowlevel jet A distinct wind maximum in orjust above the boundary layer The mesocyclone is associated with an area of lower pressure One way that a tornado may form in a supercell is known as the bottomup approach 3910km The generation of rotation in a tornado in the bottom up approach is similar to Circulation Buoyant air i l mg boundary L entering storm updraft dnven by buoyancy 39 difference g ahead of gust front 39 0 5 km how the rotation of a F0 d H k 39 mesocyclone is created due rwar an gust front A k to tilting of horizontal 00 m vorticity mesoc C one CerU atlon y 39 us This horizontal vortICIty can is tilted upward as it moves under the upperlevelmesocyclone I to create lowlevel mesocyclone whwhlsstretchedto o5km direction by the supercell Forward ank gush quot the circulation can be 00 km enhanced by vertical form tornado updraft and the intenSIty of 2005 KendallHunt Publishing stretching
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