Introduction to Atmospheric Dynamics
Introduction to Atmospheric Dynamics ATOC 5050
Popular in Course
Popular in Marine Science
This 154 page Class Notes was uploaded by Jon Johns on Thursday October 29, 2015. The Class Notes belongs to ATOC 5050 at University of Colorado at Boulder taught by John Cassano in Fall. Since its upload, it has received 15 views. For similar materials see /class/232056/atoc-5050-university-of-colorado-at-boulder in Marine Science at University of Colorado at Boulder.
Reviews for Introduction to Atmospheric Dynamics
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 10/29/15
Basic Conservation Laws Atmospheric motions are governed by three principals consenation of momentum consenation of mass consenation of energy These consenation laws can be applied to a control volume ofthe atmosphere at a fixed location Eulerian orto a control volume of the atmosphere that is moving with the flow Lagrangian What are the independent variables for the atmospheric field variables in Eulerian and Lagrangian frames of reference Total Differentiation D Total substantial or material derivative 3 t This derivative is the time derivative following the motion Lagrangian It will often be easiest to derive our consenation laws in a Lagrangian frame Local or partial derivative This derivative is the time derivative at a fixed location Eulerian How can we relate the time derivative in a Lagrangian frame to the time derivative in an Eulerian frame Consider the field variable temperature TXyzt The position of an air parcel is given by Xyz and is a function of time so X Xt y yt and z zt Following the air parcel the rate of change of temperature is given by We can relate the total change in temperature 6T to changes at a fixed location and as a function of position 5T 5 5x g 5y 5z higher order terms at 3x 65 dz Dividing by at and taking the limit as 61 goes to zero gives EE amp Dr 32 ax Dt ay Dt az Dt DT 5T where 11m 7 a 7 areo at Dr D2 072 936 z or UVT where Dr at U u7vjwl and VT j I 07x 07y az The local rate of temperature change is thus given by E 07f Dt UVT What does each term in this equation represent 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 Temperature advection L7 VT When will L7 VT be positive and negative 39 39 warmer strong warm advection weak cold advection no advection Warm air advection warmer air is replacing cooler air at a given location Cold air advection cooler air is replacing warmer air at a given location What determines the magnitude of the temperature advection Example Calculate temperature advection from a surface weather map We can write similar equations for other variables y DiQI7 Vg 6t Dt D6 For a dry adiabatic process what Is the value of E In this case what can cause Bto vary at a xed location Total Differentiation of a Vector in a Rotating Coordinate System Newton s second law can be used to derive an equation that describes conservation of momentum one ofthe basic principles of atmospheric dynamics but this law applies to motion in an inertial reference frame In order to apply this law in a noninertial reference frame we either need to consider apparent forces that arise due to the motion ofthe noninertial reference frame or we need to relate the acceleration vector in an inertial reference frame to the acceleration vector in a noninertial reference frame Considervector A It A7 A Ag in an inertial reference frame It AX ij39 All in a reference frame rotating with angular velocity 9 A I Let 1 be the total derivative in the inertial reference frame then A DA A DA A DaADAX7 yj W Dr Dt Dt Dt This can also be written in terms ofthe components of A in the noninertial reference frame as A A A D A A A DA A A D ADAXiAXD l yjA Daj7zkAZDak Dt Dt Dt Dt y Dt Dt Dt What do theterms AX Dal A Q Z Dak represent D2 y D2 D2 The total derivative in the noninertial reference frame is given by DA DAX DAy A DAZ A 7 E l j 7k Dt Dt Dt Dt In the rotating reference frame the change of f a is given by 6fa 16Aa 16 a 6 0A 01 I z Z This expression relates 6 to changes in longitude A latitude 1 and height 2 For solid body rotation ox got 51 0 and 52 0 so a aim 6A Dividing by 62 taking the limit as or 0 and noting that Q gives 613 57 3757 Daampampi 3 075 Da 0 Dt 0 Fromthis figure we note that the direction 11 is towards the center of rotation as t 6A0 of D As shown in this figure the vector D i F has components parallel to unit vectors j and IE Noting that 11 has a magnitude equal to 52 gives t D A A Q39sin chos Dt J I I Since Q 0f Qjcosq ngsimp D A 11sz Dt D A Similarly J ij and Sf 2xk A DA A DA A DA DAx Ax Dal y Ay Da Zk AZ Dak Dt Dt DI DI Di 1 DaA D A Q x A Dt Dt Vectorial Form of the Momentum Equation in Rotating Coordinates From Newton s second law 6 25 m Applied in an inertial reference frame this gives Duo 5 a Dt E m As seen in Chapter 1 we needed to include apparent forces in this expression when applying Newton s second law in a noninertial reference frame We can arrive at the same result by using the relationship between 1314 t and D714 derived above Dt What does each term above represent We can also apply the relationship between 1314 and A t t to Ua to get Replacing Ua with Ua U x on the RHS of this equation gives D U DUQgtltF aDt Dt moxooqom Dt Dt ox ox0oqom Dt Dt QgtltUQgtltF ox0ox0oqom Dt DaUa 2oxm92R Dt Dt where we have used the vector identity Q X 9 x17 9 gtltQ xi 4221 What do each of the terms in this equation represent Using this with Newton s second law gives DU DU A 2A Fade M729 U Q R 7P F Dt Dr X 2m g g ZQxU 92RPggi I What do each of the terms in this equation represent Rearranging terms gives pzoxmp gm Dt ZQXUle Fr Dt p This equation is the momentum equation and represents the conservation of momentum in the atmosphere in a rotating reference frame Component Equations in Spherical Coordinates It is often preferable to work with the equation of motion in component form rather than in vectorial form The coordinate system that is typically used is a spherical coordinate system with axes given by longitude A latitude and vertical distance above sea level 2 The unit vectors in this coordinate system point towards the east i north j and upk The three dimensional wind vector is given by A A A D D D Uui vjwkWIth usiyaixwaiz Dt Dt Dr The distances DX and Dy can be expressed as Dx 0 cos JDA Dy aD Where a radius ofthe Earth The direction of the unit vectors j and IS in this coordinate system are not constant and vary with position The acceleration of U is then given by DU Du Dv A Dw A D D DIS 7 71 7 w D1 D1 D1 D1 D1 D1 D1 The terms uampVampWD7k represent the change in direction of the unit D2 D2 D2 vectors as the air parcel moves and are given by Di Dt acosrJ jsin Ecos Ej MamJ is k Dr 0 0 DE uf vs iilij Dr 0 a The acceleration can be written in component form as DU Du uv uw s Dv u2 vw A Dw u2v2 A 7 7 7tan il 77tan ij 7 7 k Dt Dr 0 a Dr 0 a Dr 0 The forces acting on the air can be written in component form as Coriolis force 2 2 x U 29wcos Zstin p 29M sin ZQucos Ml Pressure gradient force Pg1Vp1apiipiip p pax pay paz Gravity glE Friction viscous force Fr Frxi ij Frzk The component form of the momentum equation is given by amp tan l29vsin 29wcos Frx Dr 0 0 pax 2 ampLmn l ip2ousin y Dr 0 a pay 2 2 V la g29ucos Frz Dt a p dz Cunature terms terms that arise due to the curvature ofthe Earth These terms are proportional to 1a These terms take into account the changing direction of the unit vectors as the air moves through the spherical coordinate system These terms are nonlinear ie they include products of the dependent variables and make it difficult to handle these equations fortheoretical analysis What other terms in these equations are nonlinear Scale Analysis of the Equations of Motion Scale analysis allows us to estimate the magnitude of the terms in our equations and determine which terms may be neglected Not only can this simplify our equations by removing unimportant terms but it can also allow us to filter certain types of motion from the equations The typical magnitude ofthe dependent variables can be specified based on observations For midlatitude weather systems we find 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 of troposphere Time scale LU T 105 s Kinematic viscosity v 10395 m2s391 Dynamic pressure scale pp 103 m2s392 Total pressure scale Pp 105 m2s392 Gravity g 10 ms392 Density variation scale 5pp lO392 The variation in pressure 6p is normalized by density p such that 6pp is approximately constant with height in the atmosphere despite the large change in p and p in the vertical direction The time scale LU is an advective time scale and represents the time required for a weather system to move distance L assuming that the system is moving at the same speed as the wind U The vertical wind scale W is difficult to measure for midlatitude weather systems but can be estimated based on the horizontal winds For midlatitudes 45 and f0 2Qsin 2Qcos 104 s391 Applying this scaling to the horizontal components of the equations of motion gives A B c D E F G x Eq 7291 sin 4 2911 cos 45 w 7 3 5 F x y Eq 2914 sin W Fry Scales U2 L fOU fOW U72 1 3 m H 104 10 3 106 108 10395 10 3 1012 The friction terms are many orders of magnitude smaller than all of the other terms in the equations and can be neglected with minimal error The two largest terms are the pressure gradient force and the Coriolis force These terms are in approximate balance within an error of 10 Geostrophic relationship blip 91 p p This relationship is a diagnostic relationship it cannot be used to predict changes over time ie it is not a prognostic relationship Geostrophic wind I78 the wind that exactly satisfies the geostrophic relationship Vg ugz vg A1 EkX V Mp 1wsiws b n MW M3 Note that V can be evaluated for any pressure field but will most closely approximate the actual wind for midlatitude synoptic weather systems In this case 73 will be within 1015 ofthe actual wind What is the direction ofthe geostrophic wind relative to the pressure field a Northern Hemisphere S2 fgt0 in Southern Hemisphere S2 f lt0 isobar low pressure isobar low pressure l 103 V 1 dpd ressure radient force 7 gt0 ressure radlent force gt0 P 9 p dy P 9 p dy Lig Us Coriolis force 2 2ugsinri fug lt0 Coriolis force gt2 2ugsinm fug lt0 isobar riign pressure y isobar high pressure X Copyright a 2005 by John Wlley amp Sons Inc or related companies All rights reserved 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 151 ug 7i f y 151 v7i g f x o Example Calculation of the geostrophic wind from a weather map Why do midlatitude weather systems tend to move from west to east Why does the wind blow in a counterclockwise direction around areas of low pressure in the Northern hemisphere a b L PGF u 700 hPa 850 hPa 1300 3 V v 1000 hPa Copyrlgm 2005 by John Wiley amp Sons Inc or related companies All rights reserved Approximate Prognostic Equations the Rossby number In order for the momentum equations to be used as prognostic equations we must retain the acceleration term DDt This gives D 119 fv vg D 119 g fltu uggt The acceleration term is one order of magnitude smaller than the pressure gradient force or the Coriolis force The ratio of the magnitude of the acceleration term DuDt U2L to the magnitude of the Coriolis force foU is known as the Rossby number U R057 foL The geostrophic approximation is most valid for small Ro O1 Hydrostatic approximation Scale analysis of the vertical momentum equation gives 2 Eq DwDt 2m cos u2 v2a 10 13pBz g Fz 2 Scales UWL foU U2 a PopH g vWH Ins 2 107 103 10 5 10 10 10 5 The two largest terms are the vertical pressure gradient force and gravity All other terms are at least 3 orders of magnitude smaller and thus the balance between the vertical pressure gradient force and gravity the hydrostatic approximation is accurate to O10393 This gives 1 p dz It is also necessary to show that the horizontally varying pressure is also in hydrostatic balance To do this we will de ne a standard pressure p0z which is the horizontally averaged pressure at each height and a standard density p0Z such that p0z and p0Z are in hydrostatic balance LE p0 dz With these de nitions the total pressure and density are given by pxyzt 0 z p xyzt pWMJ 90 z plwm Substituting this into the hydrostatic approximation 11 p z g gives wg0 p0p dz 1 511 pjfhisgives pop p0 p0 For Prpo ltlt 1 1 p 3pop 7 7 g po 1 az lapo 1311 31170 Mg 0 2 2 g 1 91 P0 91 po 91 P0 91 0 The sum ofthe first and last terms on the LHS of this equation is zero from the hydrostatic approximation for the standard state leaving LQL2L23LO p0 dz p0 dz p0 dz Scale analysis ofthis equation indicates that the last term on the LHS of this equation is two orders of magnitude smaller than the other terms so L3P p 9amp4 p0 dz p02 dz Using i g to substitute into the last term on the LHS gives P0 Z which indicates that the horizontally varying pressure is also in the approximate hydrostatic balance What does the approximate hydrostatic balance that exists in midlatitude weather system imply about our ability to use the vertical momentum equation to predict changes in the vertical velocity The Continuity Equation The second principal governing atmospheric motions is conservation of mass which is expressed by the continuity equation Eulerian derivation 2 Consider a volume axoyoz that is fixed in space Atmospheric mass can m2 flow into and out of this a a Pu ipu gt 82 gt Pu EPu 7 volume due to the Wind 3y 8x From this figure the mass flux into the left face is given by 8pu 5xl 8x Ej yaz l l and the mass flux out of the right face is given by 8pu 8x 2 pu Joy z The rate of mass change in the volume due to these fluxes is pu goo M y z at 3x 3x 2 3 pu Bx y z at 3x Considering all three faces ofthis volume gives 9x 9y dz 9pu 3pv 90W 32 5x5y5z Dividing by the volume 6X6y z gives the rate of change of density ip 900 900V 90W V pU at 9x 9y dz What conditions will cause VpU to be positive or negative This equation is referred to as the mass divergence form of the continuity equation and states that the local rate of change of density is equal to minus the mass divergence This equation can also be expressed as aip pV39U UVp at Dp A 7 vU Dt p lvj th This is the velocity divergence form ofthe continuity equation and states that the fractional rate of density change following the motion is equal to minus the velocity divergence When will V U the velocity divergence be positive or negative How does the volume of our control box change for each of these cases Lagrangian Derivation of the Continuity Equation This derivation is an alternate method of deriving the velocity divergence form of the continuity equation Consider a fixed mass of air 6M p6V poxoyoz Since SM is constant poxoyoz is also constant and changes in p are balanced by changes in 5x5y52 1 D6M 1 Dp5V 1 D6Viamp0 5M Dr p5V Dr 3V Dr th LD3V le W Dr p Dr Noting that 1 D3V 1 Ddx6y5z 1 D3x 1 D5y 1 D5z W Dr 6x6y6z Dr 5x Dr 5y Dr oz Dr The faces of the box move at the speed of the wind Dx Dx 5x v uABt uBSt Dx6x Dx D x Dr D 5 D 5 Similarly 6v y and 5w z Dt Dr D 5 D 5 5 5 This gives LL V 1 x y z 5iw W Dr Bx y z Dt 5x 5y oz In the limit 6V 9 0 D5V A g lt Lag m 5V Dt ax ay az lvj th which is the velocity divergence form of the continuity equation Scale Analysis of the Continuity Equation We will use the same method as was applied for the scale analysis ofthe hydrostatic equation Use p 1001 pixmz Notethat 10392ltlt1suchthatl zland 1511p51 P0 P0 P0P P0 P0 P0 Substituting this into the velocity divergence form of the continuity equation gives VU0 i Dpo Dp p0 D2 D2 Since p0p0ZI Dpo 3 4 Dr at 3x V wa W 3y az dz The expanded velocity form ofthe divergence equation is then 1 apua va wa K aiu aim0 pO at 9x 9y dz pO dz ax 9y dz Scale analysis ofthis equation indicates 1ap 501 p39U 210ms 1 if N 7 7 N 7 iN 6 10397 s1 pO at p T p0 L 10 r r r 71 lap Nga N LENlo Z 10 I S N 10 7 5 1 poaxpo 39y pL poL 10m 7 r 2 71 lap Nl jNLENlo zlo 4m 40 8571 pO az p H pO H 10 m 2 E3 eel 11041 1 10 6 Sal pO dz p H pO H 104 In 91 but al tends to cancel so 31 10396 s39l ax 3y ax 97w E 10392ms 1 6 1 4 10 5 dz H 10 m Keeping the largest terms from this scale analysis gives l aiw0 p0 dz ax 3y dz l v 0 P0 dz This can also be written as V pOU 0 This indicates that for synoptic scale motions the mass flux calculated using the basic state density p0 is nondivergent 9000 3000quot 9P0W 9x dz iechangesin 9y exactly balance This is similar to the idea of an incompressible fluid but for an incompressible fluid 0 p does not change following the motion 1 and the continuity equation 1130VU 0 reduces to VU 0 p t This result indicates that in an incompressible fluid the flow is nondivergent u av aw and changes in 777 exactly balance ax 9y dz In the atmosphere 0 p can change following the motion 1 but for purely horizontal flow W 0 the synoptically scaled continuity equation reduces to 1 V U 0 P0 dz V U 0 and the horizontal flow is nondivergent ie exactly balance X 7 The compressibility of the atmosphere only needs to be considered when there is vertical motion and then must be accounted for z Thermodynamic Energy Equation The third fundamental principal of atmospheric dynamics is the conservation of energy First Law of Thermodynamics the heat added to a system is equal to the change in internal energy plus the work done by the system This law applies to a system that is in thermodynamic equilibrium ie a system at rest Can this law be applied to the atmosphere that is in motion We will consider a fixed mass of air that we will follow through the atmosphere a Lagrangian control volume as our thermodynamic system but this system will not be in thermodynamic equilibrium because it is in motion For this system the total energy is equal to the sum ofthe internal energy the energy associated with molecular properties and the kinetic energy due to the macroscopic motion movement ofthe system For this system the heat added diabatic heating is equal to the change in the total energy plus the work done by the system This can also be expressed as the diabatic heating rate J is equal to the rate of change of the total energy plus the rate at which work is done on the system Using e internal energy per unit mass 1717 kinetic energy per unit mass The total energy in the Lagrangian control volume with density p and volume 6V is given by p e lUU5V 2 The rate at which a force does work is equal to the dot product of the force and velocity vectors What forces act on the atmosphere Rate of work done by the pressure gradient force Rate of work FU Since p FA then F pA The rate at which work is done by L the pressure in the X direction is Wu i s c WUquot given by pu y z 8x For the control volume the rate at which pressure does work on the volume is given by puA 6y6z puB 6y6z Noting that puB can be expressed as 31 5x ME MA The rate of work done by the pressure in the X direction is 9pu 9x 5V Similarly the work done by the pressure in the y and z directions is 9pv 5V 9pw 5V 9y az and the total rate of work done by the pressure is VpU5V Rate of work done by the friction force We will neglect the work done by the friction force based on synoptic scaling arguments Rate of work done by the Coriolis force The Coriolis force is given by 2 2gtltU This force is perpendicular to Uand thus ZQxUU 0 so the Coriolis force does no work on the atmosphere Rate of work done by gravity pgUav Conservation of energy for the Lagrangian control volume gives D peU39U5V Dt VpUpg05VpJav Use ofthe chain rule on the term expressing the rate of total energy change gives 1A A 1A A D U39U 5V 7 39 pe2 De U U 1 A Dpav p5V 67U39U Dt Dr 2 Dr De ctj p5V Dt Dividing by 6V and noting that gU gw gives Damp101a V39pU pgwpJ Dt 1A A D DEU U piep U39Vp pV39U pgwpJ D1 D1 Take the dot product of U with the momentum equation and neglecting the friction term gives lt12U2Qx0 UvpUg 0100 1 DI U Vp wg DGUU A p D U Vppwg This equation represents the balance of mechanical energy due to the motion of the fluid rate of change of kinetic energy following the motion Subtracting this from the conservation of energy equation gives De A 7 V U J P Dr P P This equation represents the thermal energy balance the rate of change of internal energy following the motion The mechanical energy balance can be rewritten using the definition of geopotential dd gdz This equation is known as the mechanical energy equation The sum ofthe kinetic energy and the gravitational potential energy geopotential is the mechanical energy This equation states that the rate of change of mechanical energy following the air parcel is equal to the rate at which the pressure gradient force does work on the air parcel The thermal energy equation can be rewritten by noting that 1V 1 L2amp p p Dt Dt where we have used the continuity equation to replace VU Then De A p Dr pV UpJ De Dt E pampJ Dt Dt De Doc Jri47 Dt p0 VUJ p Note that e ch so De Doc 7 p7 Dt Dt Dch amp Dt Dt DT Doc CV p 7 D1 D1 J Thermodynamics ofthe Dry Atmosphere Comparison of thermodynamic equations from Holton and Wallace and Hobbs Holton Wallace and Hobbs JCv p dqchTpdoc Dt Dt d dT d JCp aamp 61 6p W Dr Dt Eic Edam mac Luge Dr T Dt Dr T T p Ds J D1n9 dq d9 iicp dsicpi Dr T Dr T 6 Scale Analysis of the Thermodynamic Energy Equation D1n6 J 6 p Dr T Define 6 60z 6 xyzt where 90 is the basic state potential temperature and 6 is the deviation from the basic state We will assume that 3 ltlt 1 so 6 ltlt 90 and z i 90 90 9 90 This gives 1 D60 D6 L 60 Br Br cPT 1 360 360 360 360 36 36 36 36 J 7 u7v7w7 u v w 7 90 3t 3x 3y 3z 3t 3x 3y 3z CPT Since 60 602 0 and 3x 3x 3y 1 390 36 36 3639 3939 J 7 w77u7v7w7 7 cpT 6O 3z 3t 3x 3y az We will assume that amp ltlt ago so we can neglect wag 3 z 6z 1 36 36 36 w d60 J 7 u v 77 7 60 31 3x 3y 90 dz cpT 1 36 36 36 d In 60 J 7 u v7 w 7 90 32 3x az cpT In the absence of cloudsprecipitation and away from the surface ofthe Earth is 1 deg C day391 in the troposphere C p Why would the heating rate given above differ in the presence of cloudsprecipitation or near the surface of the Earth 6 4 deg C away from the surface of the Earth T 99 99 99 T 9 9 9 T 9 77 v 7 7U7 7 7 90 at 07x y 9 LU L L 9 L T 6 14degC 1 6U110ms 67m4dengay Noting that Edi rd r de w d760wFd F 60 az What are typical values for Fand I in the troposphere wFd r WFd r 10392 m s 14 deg C km391 4 deg C dayquot Since 7 is less than the otherterms in the thermodynamic energy p equation 1 96 96 36 7 u v 7 0 60 a 3x ay 60 az 36 96 96 Z60 u 7 v W7 a 3x 3y az K Z90 0 According to this equation what physical processes can cause local time variations of 9 What is the typical sign of 2760 in the atmosphere z What does this imply about the impact of vertical motion on local changes in 9 Atmospheric Thermodynamics First Law of Thermodynamics A system possesses macroscopic kinetic and potential energy as well as internal energy u due to the kinetic and potential energy of its molecules or atoms q W U2 U1 where q thermal energy heat received by a system J W external work done by the system J U1 u2 internal energy of system before and after change J Changes in internal kinetic energy are manifested as changes in temperature Changes in internal potential energy are caused by changes in the relative positions of molecules due to forces that act between the molecules This relationship can be expressed in differential form as dq dW du where dq differential increment of heat added to the system dW differential increment of work done by the system du differential increase in internal energy of the system Both of these equations are statements ofthe first law ofthermodynamics Work done by a gas Consider a gas contained in a cylinder that is fitted with a piston Cylinder Working substance 9 Pressure gt 3 0 Volume lt F Piston The volume V and pressure p exerted by the gas can be shown on a pV diagram If the gas expands increase in V it will push the piston a distance dx and will do an amount of work dll given by dW Fdx The force exerted by the gas is equal to pA where A is the crosssectional area of the piston dW pAdx pdV The work done by the gas dll is equal to the pressure exerted by the gas multiplied by the change in volume d of the gas and is indicated by the shaded area under the curve on the pV diagram When the gas passes from state A to state B the work done is given by V def2pdV V1 When V1 gt V2 the gas is compressed work is done on the gas and Wlt0 When V2 gt V1 the gas expands work is done by the gas and Wgt0 The work done in going from volume V1 to volume V2 depends on the path of integration and as such is not an exact differential Dividing by the mass of the gas gives the specific work 0 2 fdwfpdoc ordWpda a1 and the first law ofthermodynamics is then expressed as dq du pda Specific Heats Specific heat The ratio of the heat added to a system to the change in temperature of the system dqdT The units for specific heat are J kg391 K391 The value of the specific heat depends on howthe material gas changes as heat is added Specific heat at constant volume cv a in dT v const At constant volume a gas does no work and the first law of thermodynamics reduces to dq du Then cv can be expressed as cv and du cvdT dT vconst Using this expression for cv the first law of thermodynamics is dq cvdT pda lnternal energy u is a function of state and changes in internal energy depend only on the initial and final states and not on the path taken between these states quot2 T2 Therefore fdu M2 u1fcvdT ul T1 Specific heat at constant pressure op Defined as 017 dq dT pconst In this case work is done by the gas since as heat is added to the gas the gas expands dW pdV Therefore some ofthe heat added to the gas goes into doing work and for a given change in temperature more heat must be added to a gas for a constant pressure process than for a constant volume process and cpgtcv The relationship between 0 and CP can be found by taking the first law of thermodynamics dq cvdT pda and rewriting it as dq CvdT 00906 adp Combining this with the derivative of the ideal gas law pa RT dpa dRT RdT gives dq chT RdT adp 0 RdT adp For a constant pressure process adp O and EC cVR dT pconst p This can be used to express the first law ofthermodynamics as dq cpdT adp The value of 0p and cv for dry air are cp 1004 J kg391 K1 cv 717 J kg391K391 Enthalpy Enthalpy is a function of state defined by haupa Since u p and a are functions of state enthalpy is also a function of state Differentiating the definition of h gives db du dpa Using du cvdT and dq cvdT dpa adp gives dq dh adp which is another form ofthe first law of thermodynamics From dq cpdT adp we see that db cpdT and in integrated form h OPT From this equation we note that h corresponds to the heat required to raise the temperature of a material from O to T K at constant pressure When heat is added to air at constant pressure dq db cpdT and the enthalpy increases Enthalpy is also commonly referred to as sensible heat Special Processes Referring to the first law ofthermodynamics expressed as dq cvdT pda or dq cpdT adp we can consider a number of special processes lsobaric process dp O dq cpdT 6quotch7 6quotdu CV CV Isothermal process dT O dq ocdp pdoc dw lsochoric process doc O dq chT dbl Adiabatic process do 0 cpdT adp chT pdo An adiabatic process is one in which a material undergoes a change in its physical state eg pressur e volume or temperature without any heat being added to it or withdrawn from it Consider isothermal and adiabatic compression illustrated on a pV diagram 4 Pressure C Adiabat B lsotherm A Volume 4 For both processes V and a decrease For the isothermal process shown by curve AB this implies that p must increase For the adiabatic process shown by curve AC the internal energy and thus temperature increases For the same mass of gas at the same volume points B and C the sample also have a adiabat AC on the pV diagram is steeper than the isotherm AB Air parcel a small mass of air that is thermally insulated from its environment such that its temperature changes adiabatically and that has exactly the same pressure as its environment Dry Adiabatic Lapse Rate Consider an air parcel undergoing an adiabatic change in pressure with no phase change of any water substance in the air parcel For this air parcel the first law of thermodynamics can be written as cpdT ocdp cpdT dp P Since this change in pressure implies a change in elevation dT RT dp c 7 if dz p dz From the hydrostatic equation dp dz pg Combining these equations give dT pgRT E Cpp g Cp 981 m 5392 1004 J kg39l K1 98 x103 K m 1 98 K km39l This decrease in temperature with altitude is referred to as the dry adiabatic lapse rate 11 rd E LT i98Kkm1 dry parcel Cp In general a lapse rate 1 always refers to the rate of decrease of temperature with height z The lapse rate in a column of air referred to as the environmental lapse rate need not be equal to the dry adiabatic lapse rate and typically averages 67 K km 1 in the troposphere What is the sign ofthe lapse rate if T decreases with height T increases with height T is constant with height Potential Temperature 6 the temperature an air parcel would have if it were expanded or compressed adiabatically from its existing pressure and temperature to a standard pressure p0 1000 hPa An expression for potential temperature can be derived by considering the first law of thermodynamics for an adiabatic dq 0 process cpdT ocdp Using the ideal gas law to replace a gives cpdT RTQ P LTJLP R T p This expression can be integrated from p0 where T0 6 by definition to p This equation for potential temperature is known as Poisson s equation For an adiabatic process the potential temperature of an air parcel is conserved ie it remains constant Typically this equation is applied to dry air so R R and cp cpd This gives 1 1 Lpw0m 0 cm 1004Jkg39 K p Example What is the potential temperature on the roof of Duane Physics How does the pressure temperature and potential temperature of an air parcel change ifthe air parcel is lifted adiabatically in the atmosphere Adiabatic cooling and adiabatic warming Example What would the temperature of an air parcel be if it were lifted from the roof of Duane Physics to Nederland by an adiabatic process Thermodynamic Diagrams Thermodynamic diagram a chart whose coordinates are variables of state These diagrams provide 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 Pseudoadiabatic Chart Stiive Pressure p hPa w s i 00 l 400 T gt 600 800 I FIG 14 St ve dlagram 39 1000 100 200 I 300 4 0 Temperature T K Skew T log P Height m 12280 9590 7520 pressure 5840 temperature C Information plotted on a dry 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 Dry adiabat Lines of constant potential temperature curved yellow lines that slope up and to the left Units C Example Plot weather observations from ATOC weather station on a skew T log P diagram Height m 12280 9590 pressure hPa 5840 temperature C Realtime example Stuve and Skew T soundings Atmospheric Thermodynamics Atmospheric Composition What is the composition of the Earth s atmosphere Gaseous Constituents ofthe Earth s atmosphere dry air Fractional Concentration by Constituent Molecular Weight Volume of Dry Air Nitrogen N2 28013 7808 Oxygen 02 32000 2095 Argon Ar 3995 093 Carbon Dioxide COZ 4401 380 ppm Neon Ne 2018 18 ppm Helium He 400 5 ppm Methane CH4 1604 175 ppm Krypton Kr 8380 1 ppm Hydrogen H2 202 05 ppm Nitrous oxide N20 5603 03 ppm Ozone 03 4800 001 ppm Watervapor is present in the atmosphere in varying concentrations from 0 to 5 Aerosols solid and liquid material suspended in the air What are some examples of aerosols The particles that make up clouds ice crystals rain drops ect are also considered aerosols but are more typically referred to as hydrometeors We will consider the atmosphere to be a mixture of two ideal gases dry air and watervapor called moist air Gas Laws Equation of state an equation that relates properties of state pressure volume and temperature to one another Ideal gas equation the equation of state for gases pVmRT p pressure Pa V volume m393 m mass kg R gas constant value depends on gas J kg391 K4 T absolute temperature K This can be rewritten as m 7RT p v ppRT p density kg ms or as pZRT m paRT a specific volume volume occupied by 1 kg of gas m3 kg391 Boyle s Law for a gas at constant temperature V oc 1p Charles Laws For a fixed mass of gas at constant pressure V oc T For a fixed mass of gas at constant volume 1 oc T Mole mol grammolecular weight of a substance The mass of 1 mol of a substance is equal to the molecular weight ofthe substance in grams m quot7 M n number of moles m mass of substance g M molecular weight g mol391 Avogadro s number NA number of molecules in 1 mol of any substance NA 6022x1o23 mor1 Avogadro s hypothesis gases containing the same number of molecules or moles occupy the same volume at the same temperature and pressure Using the ideal gas law and the definition of a mole gives pV nMRT Using this form ofthe ideal gas law with Avogadro s hypothesis indicates that MR is constant for all gases This constant is known as the universal gas constant R Rquot MR 83145 J K391 mor1 With the universal gas constant the ideal gas law becomes pV nRT Boltzmann s constant k gas constant for 1 molecule of any gas 67 A Using Boltzmann s constant the ideal gas law can be written as p n0kT nNA n0 number of molecules of gas per unit volume 7 Application ofthe ideal gas law to dry air Pd dedT or pdad RdT pd pressure exerted by dry air pd density of dry air Rd gas constant for dry air ad specific volume for dry air Rgtllt Rd 7 where Md Md apparent molecular weight of dry air 2897 g mol391 Emz Eml md r i Md 1 1 mi nl 2M 2 l m mass of ith constituent of dry air n number of moles of ith constituent of dry air Example Calculate the gas constant for dry air Example Calculate the density of air on the roof of Duane Physics Why is the calculated density not exactly correct Application ofthe ideal gas law to individual components of air Each gas that makes up the atmosphere obeys the ideal gas law Pi piRiT For water vapor the ideal gas law is e pVRVT or eocv RVT e pressure exerted by water vapor vapor pressure pv density of water vapor av specific volume of water vapor RV gas constant for water vapor Example Calculate the gas constant for water vapor Dalton s law of partial pressure the total pressure exerted by a mixture of gases that do not interact chemically is equal to the sum of the partial pressure of the gases P TEpiRi Partial pressure pressure exerted by a gas at the same temperature as a mixture of gases if it alone occupied all ofthe volume that the mixture occupies Example The pressure in a hurricane is observed to be 950 mb At this time the temperature is 88 deg F and the vapor pressure is 25 mb Determine the density of dry air alone and the density of water vapor alone Virtual Temperature How does the gas constant vary as the molecular weight ofthe gas being considered changes The molecular weight ofdry air is greater than the molecularweight of moist air ie one mole of dry air has a larger mass than one mole of moist air What does this imply about the gas constant for moist air compared to dry air For dry and moist air at the same temperature and pressure which will have the smaller density As the amount of moisture in the air changes the molecular weight ofthe moist air will also change causing the gas constant for the moist air to vary The density of moist air is given by m m pdpv where p density of moist air md mass of dry air mv mass of moist air V volume p density that mass m of dry air would have if it occupied volume V p39v density that mass m of water vapor would have if it occupied volume V p and p39v can be considered partial densities analogous to partial pressures Using the ideal gas lawthe partial pressure of dry air pd and water vapor e can be calculated as pd pLZRdT and e pLRVT From Dalton s law the total pressure exerted by the moist air is P Pd e pLszT pLRvT Rewriting in terms of the density of moist air p gives 7 0 RdT RVT P 6L RdT RVT L1 1 RdT p v L1 31 e RdT p where s b0622 RV Md This equation can be rewritten in terms of the virtual temperature TV as 7 or R T p RdTv p p d v where TV 5 T 1 71 P The virtual temperature is 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 How does the magnitude of TV compare to the magnitude of T Example The pressure in a hurricane is observed to be 950 mb At this time the temperature is 88 deg F and the vapor pressure is 25 mb Calculate the virtual temperature in the hurricane How does the observed temperature compare to the virtual temperature Why does it differ from the temperature in this way Calculate the density of dry air with a pressure of 950 mb and the actual density ofthe moist air in the hurricane Which density is greater The Hydrostatic Equation At any point in the atmosphere the atmospheric pressure is equal to the weight of all of the air lying above that point Therefore atmospheric pressure decreases with increasing height in the atmosphere This results in an upward directed pressure gradient force For any mass of atmosphere there is downward directed gravitational force cgggiismo mlt Hydrostatic balance the upward area a directed pressure gradient force is exactly balanced by the downward directed gravitational force Pressure p 6p Lr quot 39 Pressurezp Derivation of the hydrostatic equation Consider a thin slab of the atmosphere with depth 62 and unit cross sectional area 1 m2 If the density of this air is p then the mass of the slab is given by m p621 m2 The gravitational force acting on this slab of air is m9 9p521 m2 where g is the acceleration due to gravity 981 m s392 The change in pressure between height 2 and z z is 6p What is the sign of 6p The upward directed pressure gradient force due to the decrease of pressure with height is given by 6p For hydrostatic balance 6p 9152 and in the limit as 62 a 0 Zip gp which is the hydrostatic equation z This can also be written as 6 pg 9z and integrated from height 2 to the top of the atmosphere to give P oo fap fgpaz pz z W fgpaz What is the pressure at a height of z w This is the mathematical expression that indicates that the pressure at any height in the atmosphere is equal to the weight of all ofthe overlying air Geopotential Geopotential d5 the work that must be done against the Earth s gravitational field to raise a mass of 1 kg from sea level to a height 2 ie the gravitational potential per unit mass The force acting on a unit mass ofthe atmosphere at height 2 is g The work required to raise this mass from z to zdz is gdz and 1 dd 5 gdz 7dp ocdp p What are the units of geopotential Integrating this equation gives the geopotential 452 at height 2 Iz f gdz 0 By convention 450 O m2 s392 Since the equation for 452 is an integral of an exact differential the value of 452 does not depend on the path taken to get to height 2 Geopotential height Z I 1z Za ifgdz g0 goo where go is the globally averaged acceleration due to gravity at the Earth s surface 981 m s2 Since 9 go in the lower atmosphere Z 2 as shown below 2 Z ms 100 9847 950 500 4636 843 Thickness Using the ideal gas law we can eliminate p in the hydrostatic equation LP az pg RdTV Using the definition of ddgt gives dd gdz RdTV dip P Integrating this between geopotentials 451 and 452 gives 12 P2 f dd f RdTV d1 11 P1 p2 2 1 RdfTvd p1 Dividing by go gives P1 22 21 fTvdl 30 p2 P The difference 22 Z1 is referred to as the thickness between pressure levels p1 and p2 Hypsometric Equation In an isothermal atmosphere the thickness equation reduces to R T 22 21 M1119 H 111p1 80 P2 P2 where H is the scale height defined by R T H A 293Tv go The scale height is the efolding depth for pressure the depth over which the pressure decreases by a factor of 1e Example Calculate the scale height for the Earth s atmosphere Rearranging the thickness equation gives 22 Zl H P2 P1 exp Example Calculate the pressure in Boulder Elevation 1660 m for an isothermal atmosphere with a temperature of 250 K if the sea level pressure is 1013 mb In a nonisothermal atmosphere the thickness equation can be integrated by using a mean virtual temperature TV TTvd n p TTV Q p P2 P2 V P1 p1 E rom rggtigsonde f III p P2 2 I7 RdTv 293T T 80 V Virtual temperature TVK This gives the hypsometric equation 22 Z1 I71n P2 Example Calculate the thickness between the 1000 and 500 mb levels for an atmosphere with TV 250 K How will the thickness of this layer change if TV increases Example What is the thickness of the 1000 to 500 mb layer in the tropics with TV 280 K and in the polar regions with TV 230 K Upper Level Weather Maps Since pressure always decreases with height in the atmosphere 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 such as a sea level pressure map Pressure surface an imaginary surface where the pressure has a constant value One ofthe key properties meteorologists are interested in when looking at a constant pressure map is the geopotential height of the constant pressure surface Consider a layer ofthe atmosphere between sea level and the 700 mb constant pressure surface F39at surface The thickness ofthis 3000 700 mb I layer andthus A A geopotential height of g V the 700 mb surface 13 n39fgtrtvfmfvffure varies with the a temperature of the i column of air below 0 constant sea level pressure mb surface surface 3200 Lower heights E correspond to a 1 3000 colder column 3 temperature g 2800 Therefore we expect and do nd lower 0 constant pressure heights near the poles 2002 KendallHunt Publishing and higher heights In the tropics Commonly Used Constant Pressure Maps 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 10540 m 200 center 3 digits 176 11760 m Height of pressure surface above sea level t rgzgi e gt 1 0 meters coded for different levels Height change durin ast dewpomt gt g p 12 hours depreSSIon meters below 500 mb in Celsius decameters 500 mb and above wind direction wind speed ldewpoint depression temperature dewpoint templ 2002 KendallHunt Publishing What are the differences between surface and upper air station models Some upper air weather map terms 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 Example Realtime constant pressure weather maps How does the height ofthe 500 mb surface vary from south to north 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 near the poles and higher heights are found in the tropics Reduction of Pressure to Sea Level Since pressure decreases with altitude the pressure measured by a weather station in the mountains will be less than the pressure measured by a weather station at sea level Because meteorologists are interested in horizontal changes in pressure the pressures measured at weather stations across the Earth need to be interpolated to a common height which is typically sea level This interpolation can be done by solving the hypsometric equation for p1 the pressure at sea level and noting that Z O m e In go P2 22 iRdTv 111plj go P2 57022 ex 7 P1 P2 P RdTv J 22 2 In this equation 22 is the elevation ofthe weather station and p2 is the pressure measured by this weather station Example Calculate the sea level pressure based on the current weather observation from the ATOC weather station Atmospheric Thermodynamics Water Vapor in Air Why do we care about water in the atmosphere We have already discussed vapor pressure e as one way to describe the amount of watervapor present in air but meteorologists have several other measures of water vapor that they use Mixing ratio W The mass of water vapor contained in a unit mass of dry air mV units kgkg d WE mv mass of watervapor units kg md mass of dry air units kg Specific humidity q Mass of watervapor contained in a unit mass of air mV q 5 units kgkg mdmv The following equations allow for conversion between mixing ratio and specific humidity W W q1w q 1 q The vapor pressure can be calculated from the mixing ratio by noting that the partial pressure exerted by any gas in a mixture of gases is proportional to the number of moles ofthat gas in the mixture mV quotV MW p ndnv quot14 my Multiplying this expression by Aquot gives md mV m w d e p Mw w Md md The mixing ratio or specific humidity can also be calculated from the vapor pressure using 86 86 e 7 q 116 p p 1 e The virtual temperature can also be calculated in terms of mixing ratio rather than vapor pressure as TVE T T T w 1 1 1 Le 81W p w Example The pressure in a hurricane is observed to be 950 mb At this time the temperature is 88 deg F and the vapor pressure is 25 mb Calculate the mixing ratio and specific humidity of the air in this hurricane The mixing ratio specific humidity and vapor pressure are all absolute measures of the amount of water vapor in the air Each ofthese variables varies directly with the amount of water vapor present Saturation Vapor Pressure Consider a closed box with pure water at the bottom of the box Assuming that the air in the box initially contains no water vapor how will the amount of water vapor in the air change over time What impact will this change in the amount of water vapor have on the vapor pressure 7 e 7 93 Water Water a Unsaturated b Saturated The air in the box is said to be unsaturated if the rate of condensation of water vapor is less than the rate of evaporation from the water surface 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 if the air were saturated The saturation vapor pressure depends only on temperature and increases as temperature increases What is the physical explanation for this temperature dependence of the saturation vapor pressure The mathematical basis for this relationship between Tand es will be explored later in the chapter We can consider an analogous situation with ice instead of liquid water in a closed box m o or o How will the saturation vapor pressure over an ice surface 4 o 028 differ from that over a liquid 0 24 surface 020 a esi saturation vapor pressure 0 5 5 with respect to an ice surface DE mm M 0 957 gt esiT o Saturation vapor pressure es over pure water hPa w o n n 7 0 40 730 720 710 10 20 30 40 Temperature C Using the idea of saturation we can define additional saturation variables Saturation mixing ratio ws The mass of water vapor contained in a unit mass of dry air if the air were saturated md P ex P mvs mass of water vapor in air that is saturated with respect to a liquid water surface units kg md mass of dry air units kg Saturation specific humidity qs Mass of water vapor contained in a unit mass of air if the air were saturated 5 ex 7 pUda The saturation humidity variables saturation vapor pressure saturation mixing ratio and saturation specific humidity all depend on temperature Height m 12280 9590 7520 5840 3127 1992 969 temperature C saturation ermg ratio gkg Information plotted on this skew T log p diagram is the same as for a dry skew T log p plus Saturation mixing ratio Blue dashed lines Units g kg391 Also used to indicate mixing ratio How does ws vary with temperature at constant pressure How does ws vary with pressure at constant temperature What two changes can occur to cause air to become saturated in the atmosphere 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 with respect to a plane surface of pure water Dew point temperature is the most commonly reported humidity variable on weather maps Remember Temperature always determines the amount of moisture required for air to become saturated while dew point temperature always indicates the ACTUAL amount of water vapor in the air Relative humidity RH The ratio ofthe mixing ratio to the saturation mixing ratio an indication of how close the air is to being saturated RH a 1 x100 z 3x100 units W 6 Relative humidity will be high when W W5 6 es q qs 0r Td T Relative humidity will be low when W ltlt W5 e ltlt es q ltlt qs or Td ltlt T Relative versus absolute measures of humidity Height m 12280 9590 7520 5840 3127 1992 969 temperature C saturation Mixing ratio gkg Example Skew T diagrams and moisture variables using ATOC 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 Lifting condensation level LCL the level to which an unsaturated but moist parcel of air must be lifted adiabatically in order to just become saturated with respect to a plane surface of pure water How does the potential temperature and mixing ratio of an air parcel change as it is lifted adiabatically Example Determine the LCL for an air parcel lifted adiabatically from the roof of Duane physics Height m I 12280 9590 7520 5840 3127 1992 969 temperature C saturation ermg ratio gkg Latent heats During a phase change of a system eg ice to liquid water the molecular configuration of the molecules in the system is altered resulting in a change in the system s internal energy This change in internal energy can result without a change in the molecular kinetic energy and thus temperature of the system The heat associated with this change in internal energy is referred to as the latent heat Latent heat of melting Lm heat given to a unit mass of material to convert it from the solid to the liquid phase without a change in temperature Latent heat of freezing heat released from a unit mass of material when the material is converted from the liquid to the solid phase without a change in temperature Lm 334X 105 J kg391 forwater at 1013 hPa and a temperature of 0 deg C The latent heat of freezing has the same value as the latent heat of melting Latent heat of vaporization Lv heat given to a unit mass of material to convert it from the liquid to the vapor phase without a change in temperature Latent heat of condensation heat released from a unit mass of material when the material is converted from the vapor to the liquid phase without a change in temperature Lv 225x1o6 J kg391 for water at 1013 hPa and 100 deg c The latent heat of condensation has the same value as the latent heat of vaporization Wet bulb temperature TW The temperature to which air may be cooled by evaporating water into it at constant pressure until saturation is reached How does the wet bulb temperature differ from the dew point temperature For a sample of moist air undergoing an isobaric process the first law of thermodynamics gives dq cpdT where we have neglected the mass of water in the sample Evaporation of mass of water dW requires an addition of heat dq given by dq Lvdw These equations give cpdT LvCW which describes the change in temperature when mass of water dW evaporates in the wet bulb process This equation can be integrated to give T va dT fdw pwx TTw LV WWs Cp T T LV w wS w cp This equation can then be solved for the wetbulb temperature TW Tw T hws w Cp Saturated Adiabatic and Pseudoadiabatic Processes Does the temperature of an air parcel continue to decrease at the dry adiabatic rate when lii39ted 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 ofthe air parcel increases lfthe latent heat released as water vapor condenses remains in the rising air parcel and all ofthe condensate remains in the air parcel this process may still be considered adiabatic and reversible and is referred to as a saturated adiabatic process lfthe latent heat released as water vapor condenses remains in the air parcel but the condensate immediately falls out ofthe air parcel the process is no longer strictly adiabatic since the condensate will carry some heat out ofthe parcel and the process is referred to as a pseudoadiabatic process The amount of heat carried by the condensate is small compared to the heat carried by the air in the parcel so saturated adiabatic and pseudoadiabatic processes are virtually identical The changes in temperature pressure and saturation mixing ratio for a pseudoadiabatic process can be found by considering the first law of thermodynamics c LKRQ T p T p Noting that for a saturated process as Tdecreases Ws will decrease and water vapor will condense releasing latent heat The latent heat dq released by this condensation is given by dq Lvdws Substituting this into the first law of thermodynamics gives Lvdw5c dlRdl T P T p This equation indicates the relationship between changes in temperature pressure and saturation mixing ratio that occur in a pseudoadiabatic process Moist adiabats curved black lines that slope up and to the left on a skew T diagram graphically represent this relationship Helght m 12280 9590 hPa temperature C saturation Mixmg ratio gkg Copyright c 2006 by John Wiley at Sons inc or related companies All rights reserve A saturated air parcel that is undergoing a moist adiabatic or pseudoadiabatic displacement will move parallel to the moist adiabats on a skew T diagram Example Determine the temperature and dew point temperature of an air parcel lifted from the roof of Duane Physics to 200mb Saturated adiabatic lapse rate F5 rate of decrease of temperature with height for an air parcel that is rising or sinking under saturated adiabatic conditions This is also referred to as the moist adiabatic lapse rate How does the magnitude of Pd compare to the magnitude of 1 Equivalent potential temperature 1 the potential temperature an air parcel would have if all ofthe water vapor in the parcel were condensed To derive an expression for 1 start with the first law of thermodynamics expressed as Lvdws dT dp 0 7 R7 T P T p Taking In of Poisson s equation gives 1116lnT lnp 1npO Cp Cp Differentiation of this equation gives 56611151 9 T cpp 6 366 6112 Lvdws d9 0 7 T p 6 L d6 Noting that LV dws zd LVWS gives cpT cpT l d 5 CPT 6 Integrating this equation from an initial state of T W5 and 6 the air parcel state at the LCL to a final state in which all ofthe water vapor has been condensed and wST a 0 and 6 62 gives j lfd WxT LCPTJ 66 L Viwsln62 ln6lni cpT 9 This can be solved for the equivalent potential temperature 1 to give LVWS 6 6 e CXPL CpT J Note If the air parcel is initially unsaturated then the values of Ws and T used to calculate 1 are the saturation mixing ratio and temperature of the air parcel at the lifting condensation level As will be shown below He can be easily determined using a skew T log p diagram The wet bulb temperature TW can be estimated using a skew T log p diagram by noting the temperature of an air parcel that is brought moist adiabatically from the lifting condensation level to the initial pressure Wet bulb potential temperature 6W The temperature an air parcel would have if brought down moist adiabatically from the lifting condensation level to a pressure of 1000 mb Both SW and 69 are conserved for both dry and saturated pseudoadiabatic processes Using a skew T log p diagram one can relate T Tu TW 6 69 and 8N as shown below 400 400 8 D 5 9 600 600 3 9 D 800 800 1000 1000 Adiabatic liquid water content 5 The amount of water condensed during a pseudoadiabatic process This is given by the change in ws between the lifting condensation level and the level of interest Example Using a skew T log p diagram determine the following quantities w ws RH 6 LCL TW 6W 69 based on the current weather observation from the ATOC weather station Height m 12280 9590 7520 pressure hPa 5840 temperature C saturation Mixing ratio glkg Copyright 2006 by John Wiley amp Sons inc or related companies All rights reserved How do the values of T Ta and TW compare for an unsaturated air parcel For a saturated air parcel How do the thermodynamic properties of an air parcel change when the air parcel is lifted above its LCL and then descends dry adiabatically Example Determine T Ta w ws RH and 6for an air parcel that crosses the Rocky Mountains from Grand Junction CO to Boulder CO You may assume that all condensate immediately precipitates from the air parcel above the LCL Height m 12280 9590 pressure hPa 5840 temperature 9C saturation Mixing ratio gkg Copyright 2006 by John Wiley amp Sane inc or related companies All ngms reserved Static Stability Environmental and Parcel Temperatures As discussed above the temperature of an air parcel that is displaced vertically can be determined using a skew T diagram if it is assumed that the air parcel experiences adiabatic changes during the displacement The rate of decrease of the air parcel temperature with height is given by either the dry 11 or saturated 1 s adiabatic lapse rate In addition to the air parcel temperature we need to consider the temperature of the environment surrounding the air parcel The rate of decrease of the environmental temperature with height is given by the environmental lapse rate 1 One way to measure the environmental temperature is with a radiosonde Example Plotting radiosonde data on a skew T diagram KFWD 5 June 2005 00 UTC Fort Worth TX US KFWD l 3283 9730 196 72249 Date0000Z 5 JUN 05 Station KFWD LEV PRES HGHT TEMP DEWP RH DD WETB DIR SPD THETA THE V THE W THE E W K K K mb m C C C C deg knt K gkg SEC 982 195 314 224 59 90 247 155 10 3051 3094 2985 3591 1753 2 925 728 250 200 70 50 217 175 21 3059 3089 2975 3542 1511 4 850 1458 204 155 73 49 171 195 20 3075 3100 2950 3472 1314 5 830 1575 232 52 31 180 124 203 21 3125 3138 2925 3334 559 7 771 2310 175 25 37 150 90 207 23 3132 3143 2923 3320 599 8 749 2557 150 50 21 220 52 201 23 3141 3147 2902 3245 325 9 700 3127 118 122 17 240 15 200 20 3155 3150 2895 3225 214 10 517 4170 45 134 25 180 28 247 28 3189 3193 2905 3252 220 11 544 5180 45 175 35 130 90 250 34 3197 3201 2905 3258 178 12 523 5490 53 313 12 25 0 123 253 29 3212 3213 2897 3231 053 13 500 5840 91 321 14 230 143 245 22 3219 3220 2899 3238 052 14 475 5232 123 333 15 210 157 235 25 3227 3228 2901 3245 049 15 400 7520 201 441 10 240 234 235 39 3289 3289 2915 3295 019 17 300 9590 355 575 9 220 358 235 41 3353 3353 2932 3355 005 18 250 10830 455 535 11 180 451 245 43 3384 3384 2940 3385 003 20 200 12280 541 591 14 150 544 240 54 3471 3471 2950 3472 002 23 150 14100 593 753 11 150 595 245 38 3579 3579 3002 3580 001 24 130 14990 505 775 9 170 507 245 50 3811 3811 3024 3812 001 25 100 15590 581 841 8 150 582 250 42 3952 3952 3047 3952 000 Archived radiosonde data like this can be found at httpvortexplymoutheduuacalplt uhtml Example KFWD 5 June 2005 00 UTC Height m 12280 9590 7520 5840 temperature C saturation Mlxi g ratio glkg Example Consider an air parcel lifted adiabatically from the surface to 200 mb with the same initial temperature and dew point temperature as the environment Compare the temperature of this air parcel to the temperature of the environment at a 900 mb b 800 mb and c 500 mb Stattc Stabthty m Unsaturated Atr Constder the Sttuatton shown betow where rlt Fa How tht the temperature of the atrparcet cnange wt 5 htted from pomt 07 How does the temperature of the atr parcet at tntS hew Height What does tntS dtfference m temperature trnpty about dtfferences m denstty of the atr parcet and enW onrnentat arr TA 8 Temperature gt Howwoutd tntS anatyStS cnange tftne atr parcet were forced to stnk from pomt 07 An atrparcet tnat t5 teSS dense tnan ttS envtronrnentWtH nSe due to a buoyant force wntte an aH parcet tnat 5 more dense tnan ttS envtronrnent W t Stnk due to a buoyant force What 5 the dtrectton of the buoyant force actmg On the atrparcet m the exampte above wt 5 dtsptaced up down from ttS phgmat posttton at 07 Consider an air parcel with volume Vat height 2 Assume that this air parcel has the same temperature as its environment Tz and same volume as the environmental air it displaces At height z z the air parcel has density pparceimz and the displaced environmental air has density pmz The buoyant force per unit mass acting on the air parcel is given by air displaced air parcel weight of weight of J F mass of parcel gpzdzv gpparcelzwlzv V p parcel z dz szlz pparcelzdz gl l pparcelzdz We can rewrite the expression for this force using the ideal gas law p p Rdedz LRdTparcelzdZ J B r P LRdTparcelzdZ J 1 1 T T zdz parcel z dz g T parcel zdz T zdz T parcel zdz T zdz The air parcel will experience an upward directed buoyant force when it is warmer than its environment and a downward directed buoyant force when it is cooler than its environment How does the temperature of an air parcel that is displaced upward compare to the temperature of its environment when FltF FgtF Height Height TA TB TB TA Temperature gt Temperature gt b F I d What does this imply about the direction of the buoyant force acting on this air parcel Static stability classes Stable unstable and neutral Stable Unstable Neutral A f 3 B l A B rquot Q quot3 If the lapse rate of the environment F is greater than I then the environment is statically unstable and an unsaturated air parcel that is initially displaced upward will continue to accelerate upward We can also considerthe buoyant force as a function of potential temperature For both the air parcel and the environment at height 2 How does the temperature and potential temperature of the air parcel change as it is lifted adiabatically from height 2 to height zdz Rdc 6Zpzdz p P0 T parcel zdz The temperature ofthe environment at height zdz is RdCp P Tzdz 6zdz zdz J P0 Using this the buoyant force can be rewritten in terms of potential temperature as FB gw 6zdz Noting that Hparceizdz622 FB g 61 6zdzj 6zdz i6 dz 6 dz 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 05 we 6 dz What property ofthe environment determines the magnitude of the buoyancy force What happens to an air parcel that is displaced vertically by a dry adiabatic process if dedz envimnmem 0 dedz envimnmem lt 0 dedz envimnmem gt 0 It can be shown see text that lldli de T dz 3P Fd U Which allows Nto be expressed in terms ofthe dry and environmental lapse rates as g 05 N rd 4 A typical value of N in the atmosphere is l2x lO392 s391 This gives a period of oscillation of 17 2 z 8 min Oscillations of this type can give rise to gravity waves Gravity waves can be excited by ow over topography as shown below For a ow with wind speed U over a series of ridges separated by distance L the topography will force an atmospheric oscillation with a period given by Example Assume that the topography of a region is characterized by ridges that are 10 km apart in the direction ofthe ow If the environmental lapse rate is 5 deg C km391 and the temperature is 20 deg C nd the wind speed U that would create an orographically driven atmospheric oscillation that matches the period of the buoyancy oscillation Static Stability in Saturated Air For a saturated air parcel the rate of temperature decrease with height is given by the saturated adiabatic lapse rate 1 In the situation where an air parcel may be either unsaturated or saturated the stability classes discussed above need to be modified to account for the difference between the dry and saturated adiabatic lapse rate What are the static stability classes for an air parcel in an environment with Fgt Pd F II gt F gt I F I Flt 1 While it is useful to consider static stability criteria based on comparisons of the environmental lapse rate to the dry and saturated adiabatic lapse rates it is often easier to visualize these stability classes on a skew T log p diagram 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 skewT diagram between the LFC and EL Mathematically the CAPE can be calculated as in T T parcel envjdz ZLFC 9 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 2 CAPE Typical values of CAPE 0 to 500 J kg391 Very weak instability 500 to 1500 J kg391 Weak instability 1500 to 2500 J kg391 Moderate instability 2500 to 4000 J kg391 Strong instability gt4000 J kg391 Extreme instability What is the maximum vertical velocity for each ofthese values of CAPE Convective lnhibition CIN The energy barrier to be surmounted before an air parcel reaches the LFC Example KFWD 5 June 2005 00 UTC Height m I 12280 9590 7520 5840 3127 1992 969 temperature C saturation ermg ratio gkg Example Indicate the location ofthe LPG 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 CIN Circulation and Vorticity Example Rotation in the atmosphere water vapor satellite animation Circulation a macroscopic measure of rotation for a finite area of a fluid Vorticity a microscopic measure of rotation at any point in a fluid Circulation is a scalar quantity while vorticity is a vector quantity Circulation Theorem Circulation is defined as the line integral around a closed contour in a fluid ofthe velocity component tangent to the contour CE 930 6 l7 cosadl By convention C is evaluated for counterclockwise integration around the contour When will C be positive negative Consider a fluid in solid body rotation with angularvelocity 9 For a circular ring of fluid with radius R V QR d RCUL Then the circulation is C 95ng1 2mm What is the relationship between C and angular momentum in this case What is the relationship between C and angular velocity in this case C can be used to describe the rotation in cases where it is difficult to define an axis of rotation and thus angular velocity How will C change over time Take total derivative of C in an inertial reference frame DaCaamp A A DaUa A A Dad7 Dz Dt lt U dl 36 Dz dl U Dz But 45wlf isjust the line integral of the acceleration of wind z DaU le VIgt Dt p where we have neglected the friction force do not need to consider the Coriolis force for an inertial reference frame and have expressed the gravitational force as a gradient ofthe geopotential D8121 DaUa and Noting that 1 41x then DZ A Ddi A A 1 A A Ua UDU7 DUU 0 951D gr 112w Using this the change in absolute circulation following the motion is D C 1 A A 7 a EV dl VltIgtdl D 95 p P 95 The line integral of the gravitational force is given by gSVqgtdigS gIEdig gdzgS dlt1gt0 Then the change in circulation is given by DC 1 A 1 a V dl d D 95 p p 515 p p What is the physical interpretation of each term in this equation The sea breeze circulation 925W U0 0 QIA HMO lt3 lt3 lt H John Wiley amp Sons Inc or related n Nunatm uu r What are the physical mechanisms that lead to sea breeze formation Example Calculate the change in circulation associated with a sea breae For a barotropic fluid p pp and gB ldp 0 so p Dd Cd DI and absolute circulation is conserved following the motion This is known as Kelvin s circulation theorem What is the relationship between absolute and relative circulation Ca absolute circulation Ce circulation due to the rotation ofthe Earth C relative circulation C6 Ce C The circulation due to the rotation of the Earth is given by Ce g Ue df where U2 9 x7 is the velocity due to the rotation of the Earth Stokes theorem gSFd ffv xF dA A where r is the unit vector normal to the area A and the direction of r is defined by the righthand rule for counterclockwise integration of the line integral Using Stokes theorem we can rewrite our equation for Ce as Ce g UedfffVgtltUe dA Using the vector identity VerVxQxfVxQxRQVR2 2 For a horizontal plane is directed in the vertical direction and Ver ZQ 2 2sin f where we have used 2 Qcos 15 Qsin pl Then Ce ffVer dA2 2sin A2 2Ae A where A6 is the area projected onto the equatorial plane and A6 Asin Then C Ca C6 Ca 29Asin Ca 29Ae The relative circulation C is the difference between the absolute circulation Ca and the circulation due to the rotation of the Earth Ce Taking DDt of this equation gives DC DCa QDAE Br Br Br DC ggdP29 DAe Dt p Dt For a barotropic fluid this reduces to Dr Dt LC 29 DAB Integrating from an initial state 1 to a final state 2 gives C2 C1 29A22 Ael 2 2A2 sin 2 A1 sin bl What is the physical interpretation ofthis result Vorticity Vorticity a microscopic measure of rotation at any point in a fluid Vorticity is defined as the curl of the velocity V x 7 Absolute vorticity a V an Relative vorticity a V x U z j E a3VxUi i 3 dx dy dz L V W dw av all aw dv du A 7 il 7 7j 7 ik dy dz dz dx dx dy It is typical to consider only the vertical component ofthe vorticity vector Vertical component of absolute vorticity n 5 IE V x U Vertical component of relative vorticity C 5 IE V x U g 1 x y Under what conditions will g be positive What is the relationship between absolute and relative voriticity Vertical component of planetary vorticity k V x02 ZQsin q f Then n f The absolute vorticitiy is equal to the sum of the relative and planetary vorticity What is the relationship between vorticity and circulation Consider the circulation for flow around a rectangular area 6X6y u3 C 8y l 8v 3v 3 v 3 x x u gt 8x 6C u6xv 6x6y u y 6x v6y 8x y 6C amp ciu 616 ax y 6CC6A 6C C 39a We can also relate circulation and vorticity using Stokes theorem and the definition of vorticity CgSUdf CffVgtltU dA Cff dA This indicates that the average vorticity is equal to the circulation divided by the area enclosed by the contour used in calculating the circulation For solid body rotation C 297113 and g 29 In this case the vorticity C is twice the angular velocity 9 Vorticity in Natural Coordinates From this figure the circulation is 5C V6s d6s V 6n6s 5C Vd6s g n s d85 From the geometry shown in this gure d6s 636n where 65 is the angular change in wind direction This then gives 6C V636n n s 9n 5C Va3 W6n6s 6s an 66 nds 6s an In the limit as 6n6s gt 0 6116s as 911 5c Vaf5 av Noting that amp i then as RS V av CR5z9n What is the physical interpretation ofthis equation Can purely straigthline flow have nonzero vorticity ll 9 Where would this occur in the real 39 world 9K Can curved flow have zero vorticity b Potential Vorticity Kelvin s circulation theorem states that circulation is constant in a barotropic fluid Can we expand this result to apply for less restrictive conditions Density can be expressed as a function ofp and 6 using the ideal gas law and the definition of 6 Rcp Rcp piandeTamp gtT6 RT 19 p5 Then RC l RC RC cvc RC ppSp pl ppSpp Upsp Rngfp R6 R6 p On an isentropic surface p is only a function of p and the solenoidal term is given by E dpjp 0C gide39CvCp 0 Then for adiabatic motion a closed chain of fluid parcels on an isentropic surface satisfies Kelvin s circulation theorem DC DC 295Asin Dt Dt For an approximately horizontal isentropic surface C z 6614 where 9 is the relative vorticity evaluated on an isentropic surface Kelvin s circulation theorem is then given by DC 295mm D5A 6 29 sin D5A 6 f DI DI DI and 6A 6 f Const following the air parcel motion Consider an air parcel of mass SM con ned between two Hsurfaces From the hydrostatic equation gig pg and SM pSV pSZSA z 6M 6l6A 8 For SM conserved following the air parcel motion SM Jip A Const 8 Then SA 6 Constg 6p 6p 66 6 where Const 6x 66 Using this expression to replace SA in SA 9 f Const gives Constg 2Ce f Const Ertel s Potential Vorticity P is defined as 56 PE 7 C6 5 5p Ertel s potential vorticity has units of K kg391 m2 s391 and is typically given as a potential vorticity unit PVU 10396 K kg1 m2 s391 What is the sign of P in the Northern hemisphere For adiabatic frictionless flow P is conserved following the motion The term potential vorticity is used for several similar quantities but in all cases it refers to a quantity which is the ratio of the absolute vorticity to the vortex depth For Ertel s potential vorticity the vortex depth is given by How will C6 f change ifthe vortex depth increases decreases Potential vorticity for an incompressible fluid For an incompressible fluid p is constant and the mass M is given by M ph A where h is the depth of the fluid being considered Rearranging gives 6A M ph And the potential vorticity is given by 4710 As above the potential voriticity is constant for frictionless flow following the air parcel motion For a constant depth uid f n Const absolute vorticity is conserved following the motion This puts a strong constraint on the types of motion that are possible Consider a zonal ow that has 0 0 initially so no f0 Since f n Const then at some later time 5 f no f0 For a ow that turns to the north fgt f0 and C f0 flt 0 For a ow that turns to south flt f0 and C f0 fgt 0 Can these conditions be satisfied for both easterly and westerly ow C gt 0 lt o WESTERLY FLOW EASTERLY FLOW 1 NOT CONSERVED x 1 CONSERVED lt0 ggto For a fluid in which the depth can vary potential vorticity rather than absolute voriticity is conserved Consider westerly flow over a mountain a I3 I51quot I W I W I return to Decrease Increase Decrease Increase return to original value Increase Decrease Increase Decrease For easterly flow over a mountain 90 59 Y X b Aircolumn Decrease Increase Decrease Increase depth Return to 6le 6le 6le original value Changein 59 Increase Decrease Increase Decrease g Return to original value Changein I fI Decrease Increase Decrease Increase Sign of 2 Zero Positive Negative Positive Resulting westward northward southward southward motion Changein Return to Increase Decrease Decrease IfI original value As was the case for flows in which absolute vorticity was conserved the flow responds differently for easterly and westerly flow Vorticity Equation We are interested in being able to predict the time rate of change of vorticity without the constraints of adiabatic motion To do this we will derive the vorticity equation Since vorticity is defined as C 0 x y We will subtract 3 of the u momentum equation from g of the v y x momentum equation a avuavvavwjvfu 1ap Z x t x y p y u u u u l p 7 iuiviwi 77 y t x y z p x This gives u v w al vi t x y z x y x z y z y 1 M 103 p2 x y y x Noting that Dik wiwi w vi Dt t x y z y RAE EWEWE Dt t x y z the vorticity equation becomes D f fau v w v 3W3 1 p p p p 77 77 Dt x y x z y z 2 p x y y x This equation indicates that changes in absolute vorticity following the motion is due to a Convergence term Tilting term Solenoidal term What is the physical interpretation of each of these terms How will the vorticity change under the influence of convergence Physical interpretation of the tilting term v What is the sign of Q and Q 936 k in this example Does this lead to an increase C or decrease in absolute 1 vorticity The solenoidal term is the microscopic equivalent of the solenoidal term from the circulation theorem From the circulation theorem the solenoidal term is given by 43 45W di p Applying Stokes theorem S F ffV xFfsz to this gives A 36andf ffanVp12dA A Using the vector identity V x an Va gtlt Vp gives 45an ffVa xvpI dA A Expansion of the cross product gives VaprE a a a a w Since a p39li3 M M a p39li3 9y 9y pzay thisgives aaapa a 1apap 1 apap ax 3y 3y ax p2 ax 3y p2 3y ax Then ffVapr12dA iza a i23 3 A A p 3366 p ayax Dividing this by the area gives the soenoida term in the vorticity equation Vorticity equation in isobaric coordinates ag VV f wx fV7l Wwa 9t ap 912 What do each of the terms in this equation represent n isobaric coordinates there is no soenoida term Scale Analysis of the Vorticity Equation We can simplify the vorticity equation through scale analysis Consider the scales for midlatitude synoptic weather systems scale Horizontal scale Horizontal scale Mean Fractional fluctuation Z i 329 sin 32951 a 9y 9y 9y 3 cosZ COS a a a What is the magnitude of 3 in the midlatitudes g scales as where lt is used since and may partially cancel N X What is the relative magnitude of g compared to f0 ltL 5R0 10391 f0 N foL so g is often an order of magnitude smaller than f0 the planetary vorticity Using this we can approximate the convergence term as u av u av fwcayfwcay ie we can neglect the relative vorticity contribution to the convergence term This approximation is not valid near the center of intense areas of low pressure where C f 1 and then both relative and planetary vorticity must be considered in the convergence term What is an example of a type of flow where we cannot neglect g Scale analysis ofthe terms in the vorticity equation gives 2 N L N1010 5 2 a 3x 3y L WE WU 10 11 5 2 az L vi U 10 10 5 2 9y ff WJJOLU 10 9 5 2 3x 3y N Lw iu 24041 ax az 9y dz N L H 12 lt5 57PL2N1041 5 2 p axay ayax Np pL Note that lt is used forthe last three terms in this list since portions of each ofthese terms may partially cancel What is the largest term in the vorticity equation In order for the vorticity equation to be satisfied f ZVJ 103910 s 2 lt10399 s 2 x y which implies that the flow must be quasi nondivergent 10 6 s 1ltg10395 5 1 3x 3y L How does the magnitude of the divergence compare to the magnitude of the relative and planetary vorticity 10396 s 1 u 3v 2 2 77 lt710 R0 ax ayJfON10 4 s 1 6 1 lt710581104R0 ax 3y N10quot s39 The horizontal divergence is two orders of magnitude smaller than the fand one order of magnitude smaller than g Retaining the largest terms 103910 s392 in the vorticity equation gives at 3x 3y 3y DiCf fauav ua vagvaf fuv x y Dt ax 3y where Dhaaa 7 12 Dr at 3x 3y What is the physical interpretation ofthe scaled vorticity equation Near the center of intense low pressure systems where g and fhave similar magnitude the scaled vorticity equation becomes Di f all Dt C fax y Why can cyclonic systems become more intense than anticyclonic systems This is consistent with our analysis ofthe gradient wind equation where we found that the intensity of high pressure systems was limited while the intensity of low pressure systems was not Is the scaled vorticity equation appropriate in the vicinity of fronts Near fronts L lO5 m and W lO391 m s391 Using these scales indicates that the vertical advection tilting and solenoidal terms in the vorticity equation may all be as large as the divergence term and thus must all be retained Vorticity in Barotropic Fluids Barotropic Rossby Potential Vorticity Equation Consider an atmosphere that is Barotropic p pp lncompressible 0 Dr With a depth hXyt 22 Z1 where 22 and Z1 are the heights of the upper and lower boundaries of the atmosphere For an incompressible fluid the continuity equation is V39U0 in u 3x 65 az Then the scaled vorticity equation becomes Dh f i Q iw Dt 39 Cfaxoiy faz Approximating the vorticity C by the geostrophic vorticity Cg where g ax 3y gives Dh f aw quth In a barotropic fluid Vg does not vary with height and the equation above can be integrated from height Z1 to 22 to give z2 f z2 f Dgt 3zf gfaw Z1 Z1 Dh gf DI Z2 Z1 g fW12 Wz1 Dh gf h DI Cg fW12 Wz1 Noting that w a gives Dt thz D Z wz1 and wz2 Dr SO thz th1 Dh Z2 z1 th lezl wlzll39 WTE Then Dh gf Dh hTlggfl 1 Dl gfl gf Dz h Dz Dhln gfDhlnh Dz Dz Dz Rossby potential vorticity Cg fh The Rossby potential vorticity is conserved following the motion Barotropic Vorticity Equation For a barotropic incompressible fluid with purely horizontal flow W O the vorticity equation becomes Dh f Dl Cfo7z 0 and the absolute vorticity C f is conserved following the motion A more general result is that for a flow that has nondivergent horizontal winds in Q 0 3x 65 Dh f0 Dt For a nondivergent flow the wind can be represented by a streamfunction 1pxy such that 071 VwEIEXV1p ialp 9y ifu7vf 3x J Example Show that with this definition ofthe streamfunction the horizontal wind is nondivergent The relative vorticity is then given by The vorticity equation can then be written as Dt DhV21Jf 3V2 a 2 T7VW39VV 1pfgt0 aVZIJ 3 42 WWIHf What is the physical interpretation ofthis equation This equation can be solved numerically to predict the time evolution of 1p V and Q can then be calculated from 10 The flow in the midtroposphere 500 mb is often nearly nondivergent and this equation can be used to produce a reasonable prediction of the time evolution of the flow at this level Baroclinic Ertel Potential Vorticity Equation Ertel potential vorticity is given by M P 5 C6 f g 9p and is conserved following the motion for adiabatic frictionless flow We want to derive an equation for the time evolution of P in the presence of diabatic effects and friction This equation can be derived by rewriting the horizontal momentum equation in isentropic coordinates We can use 6 as a vertical coordinate in a stably stratified atmosphere since aeaz gt0 and 6 is a monotonic function of height The vertical velocity in isentropic coordinates is defined as 9529 Dr When will 9 0 For an air parcel with mass 6M 5M p5A5Z 5A ip 5A al66 051456 g g 36 where a l is the density in Xy6 space 8 Then the governing equations in isentropic coordinates are given by Horizontal momentum equation in isentropic coordinates av V v 1P 7V at 6 2 A A aV A k V 67 F J C6 X r where 1P cpT I is the Montgomery streamfunction Continuity equation in isentropic coordinates 309 96 La a V6 a7 Hydrostatic equation in isentropic coordinates RCp p all 7 p E Hltpgtcp Cbi where His the Exner function sentropic vorticity equation D C f A A A JV gt 6fV639Vk39V6XFr 9wi D a A where 3 3 V V6 is the total derivative followrng the horizontal motion 1 t on an isentropic surface The continuity equation can be rewritten as 50 1 O1V6V O2 D2 076 Ertel potential vorticity equation N A 309 E VV6P Dt at a 076 a39llEV6 gtlt Fr 9 39 where Ertel s potential vorticity is given by P 6 fa What is the physical interpretation ofthis equation For adiabatic frictionless flow 0 and P is conserved following the t motion 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 R MK My Magnitude ofa vector i Direction of a vector Direction armadaMK Coordinate systems on the Earth 2 R localy quotupquot 39 y quotNORTH meridional x 739 EAST zcna For a coordinate system with xyz we use unit vectors 7 7 and 1 Algebra of Vectors Addition and subtraction of two vectors graphic method D Addition Subtraction Addition of two vectors 2 and V u luyjuzk and v XlVyjVzk VuX vxzfuy vyjuzvzlE Subtraction of two vectors 2 and V 7ux vxf uy vyquotuz VZE Multiplication of a vector by a scalar graphic method 1B AB 2AB 25AB 05AB Multiplication of vector u by scalar 0 c1 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 u and 72 39V 14va uyvy uzvz l HVlcosG 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 u and 17 X1 uyvz uzvyz uxvz uzvx uxvy uyvx k or i j39 k u x v uX uy uz 12X vy 121 uyvz uzvy uxvz uzvx 39 uxvy um 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 isine 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 rotated How would vector i uj My change under a rotation of the coordinate system 39 ux coso My sin 0 My cosa ux Stud Meteorologists traditionally define a coordinate system relative to the Earth 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 2xyzt This vector in component form can be written as 121 yyzyt my I gyyz 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 912 eff 19g 4 ah A i i 7 7k t 9t 9t 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 Elementary Application of the Basic Equations For many applications it is preferable to work with the equations that govern atmospheric motions using an isobaric coordinate system Basic Equations in lsobaric Coordinates Horizontal Momentum Equation We will neglect the curvature terms friction force and Coriolis force due to vertical motion In height coordinates the horizontal momentum equation is then given by XV 1Vp Dt p where 7 uf v is the horizontal wind vector Replace the pressure gradient force term with 1Vpvp p ax 3y In an isobaric coordinate system D a a a a i7u a Dr at 9x 65 9p D where w a l Dt How does this differ from the expression for a total derivative in a height based coordinate system What is the physical interpretation of a The horizontal momentum equation in an isobaric coordinate system is DV A A 7 gtltV V I D fk The geostrophic relationship in isobaric coordinates is chg Vpl Vgil xvplt1gt f A e A 1071 lad Vgugl ng 7l iij f 9y f 9x One advantage of the isobaric coordinate system is that the geostrophic wind does not depend on density and thus a given geopotential gradient results in the same geostrophic wind at all heights in the atmosphere For constant fthe divergence ofthe geostrophic wind is zero Example Show that VP 7g 0 Continuity Equation We will derive the continuity equation conservation of mass using a Lagrangian control volume 6V 6X6y z of constant mass 6M p x y z Use the hydrostatic equation to replace 62 5p az Pg oz Jig Pg Then 5M 5x5yfip g 1 D5M g D5x5y5pg 0 5M Dr 3365pr Dr 1 D5x 1 D5y 1 D5p 5 Dr a Dr Dt 5M 5v 5w 7 0 5x 5y 5p In the limit as 5x5y5p a 0 an 07v aw 777 0 9x 65 9p How does this form of the continuity equation compare to the continuity equation used in a height coordinate system Thermodynamic Energy Equation Start with cp a J First Law of Thermodynamics t t DT 9T aT 9T Noting that amp 0 and 7 ui v7 mi gives Dt Dt at 3x 3y 3p W W W W c iuiviwi ocwJ p at 3x 3y 3p Using g g Zia and defining a static stability parameter Sp as 3p cp 6 3p SP 7 7 77 opp 317 9317 gives 3T 3T 3T J iui 7 S 07 at 3x 3y p cp In the static stability parameter the term 6661 in isobaric coordinates is equivalent to aeaz in height coordinates The static stability parameter can be rewritten as S Zi6rdr p 93p pg Under what conditions is Sp positive What does this imply about the sign of 6661 and how Hvaries in the vertical From this expression for Sp we note that Sp depends on 1p and thus will increase rapidly with height in the atmosphere Balanced Flows Many atmospheric flows can be understood by assuming a simple force balance In the next section we will consider steady state flow with no vertical component Natural Coordinates Natural coordinate system a coordinate system in which one axis is always tangent to the horizontal wind 7 and a second axis is always normal to and to the left of the wind 5 In this coordinate system the vertical unit vector is IE Coordinate locations in natural coordinates nz ln natural coordinates the horizontal wind vector is given by V Vi V is the horizontal wind speed and is de ned to be nonnegative D3 D2 where s is the distance along the curve followed by the air parcel sxyt The acceleration of the wind in natural coordinates is given by E DW v D2 D2 D2 D2 We will use this expression when writingAthe full equations of motion but Dt first we must derive an expression for 3 t Considerthe geometry for an air parcel moving along a curve a distance of as From this geometry we can derive an expression for D Dr is 6f a 61p T l tl IRI ltl pa E IRI R is the radius of curvature following the air parcel motion R is taken to be positive when the center of curvature is to the left of the flow in the positive 6 direction Dividing this expression by as and taking the limit as as a 0 gives 51 67W D D s39E where we have noted that the direction of g is parallel to 16 S This can now be used to express as Dr Ds Dr R 0101 71 Then the acceleration of the wind in natural coordinates is given by DV ADV vi itiin Dt Dr R What is the physical interpretation of the terms in this equation Pressure Gradient Force in Natural Coordinates Coriolis Force in Natural Coordinates gtlt 7 fV Component form ofthe horizontal momentum equation in natural coordinates DV 31 5 component 7 7 Dr 33 v2 31 n com onent 7 7 p R W art For flow parallel to height contours Q 0 so 0 and the wind speed is constant as Dr Geostrophic Flow in Natural Coordinates For geostrophic flow the Coriolis force and pressure gradient force balance 3CD ng 8 n P if W g f 8n ciao3a Co 4 0 V2 For this balance 0 2 Under what conditions will V gt 0 Example Calculate the geostrophic wind from a constant pressure map Inertial Flow What force balance will occur if there is no horizontal pressure gradient V2 O R W What is the path followed by an air parcel for this force balance Solving for the radius of curvature for this flow gives RK f What is the sign of R in the Northern and Southern hemispheres What does this imply about the direction of the flow The time required for an air parcel to complete one revolution around a path of radius R is given by 2 HR 2 71V J V 2i f lZQsin Jl lsin Jl lday Period 2 Where is this type of motion most likely to be observed Cyclostrophic Flow The Rossby number is the ratio of the acceleration ofthe wind to the Coriolis force In natural coordinates the Rossby number is given by VZRL Ro fV fR When will Ro be large What does this imply about the importance ofthe Coriolis force in the horizontal momentum equation For this case the horizontal equation of motion becomes 22 R an What force balance does this represent The wind that satisfies this equation is known as the cyclostrophic wind 05 vie an A real solution ofthis equation requires Raa220 n Can cyclostrophic flow occur around v both low and high pressure centers P Ce L ltigt Rgt0IltO C e gt V 04gt Rlt0 gtO an Example A tornado is observed to have a radius of 600 m a wind speed of 130 m s39 What is the Rossby number for this flow Is a cyclostrophic balance appropriate for these conditions What is the pressure at the center of the tornado assuming that the pressure at a distance of 600 m from the center is 1000 mb Gradient Wind Approximation What type of flow will occurwhen Rois small What form ofthe horizontal momentum equation needs to be considered when Ro1 Gradient wind a horizontal frictionless flow that is parallel to the height contours on a constant pressure map For the gradient wind the horizontal momentum equation is V 2 R an This quadratic equation can be solved for the gradient wind V fR f2R2 alt1gt 5 V i 7 R7 4 an 1 31 The gradient Wind equation can also be written using V9 to replace ai n to give 05 fR f2R2 V 7 7 R 2 4 ng Multiple solutions exist for these equations but not all ofthe solutions are physically reasonable What are the requirements for a physically reasonable solution to the gradient wind equation Northern Hemisphere CCW flow CW flow CW flow CCW flow around L around H around L around H f R Q an 2 2 0 ltB or ltB or f R Rad alwaysgti 2 alwaysgti 2 2 Imaginary 2 Imaginary 4 an for for 2 2 2 2 if R Re if R we 4 an 4 aquot 2 either root V ositive for root onl 2 2 root onl never p y butif f gt R V n Type of flow Regular Anomalous Anomalous No physical Positive Root low High Low solution Type of flow No physical Regular No physical No physical Negative Root solution High solution solution Baric flow pressure gradient and Coriolis force act in opposite directions Antibaric flow pressure gradient and Coriolis force act in the same direction Which gradient wind solutions are baric antibaric Cyclonic flow flow in which Rfgt O Anticyclonic flow flow in which Rflt 0 Which gradient wind solutions are cyclonic anticyclonic Regular Regular Low High a b P gt P L C e gt gt Ce Co V V c a Anomalous Anomalous Low High For the anomalous low and anomalous high cases the gradient wind solution does not reduce to the geostrophic solution for R a 00 The gradient wind equations forthe regular low and regular high cases in the Northern Hemisphere are 2 2 0 5 Cyclonic CCW ow around L R gt 0 V E f R R 2 4 9n 2 2 05 Anticyclonic CW ow around H R lt 0 V R n Example Calculate the gradient wind from an upper air weather map For both physically reasonable solutions for flow around a high pressure center the gradient wind solution requires that 2 2 fi gt R 4 an This requires that as R decreases 3amp2 the pressure gradient must also It decrease What is the physical reason for this limit What does this imply about the weather that would be experienced near the center of a high pressure center Of the balanced flow cases considered above the gradient wind solution will provide the best estimate of the actual wind since it makes the fewest assumptions Using the gradient wind equation with the pressure gradient force expressed in terms of V9 allows us to derive an expression for the ratio of the geostrophic wind to the gradient wind V i11 v fR Under what conditions will Vg gt VVg lt V In this case the geostrophic wind overestimates underestimates the actual wind speed What is the physical explanation for this The ageostrophic wind is defined as gV g and is the part ofthe wind that is not in geostrophic balance Recall that the geostrophic wind is nondivergent so divergence can only occur through the ageostrophic portion of the wind What does the difference between Vg and V imply about the ageostrophic wind in troughs and ridges What does this imply about the location of areas where converegence or divergence are occurring Example Convergence and divergence on a constant pressure map Trajectories and Streamlines Trajectory the path followed by an air parcel over time The radius of curvature R used in the previous section is the radius of curvature of the air parcel trajectory Streamline a line that is everywhere parallel to the flow at a given time For the balanced flow cases considered above the height contours are streamlines since the flow is parallel to the height contours Often we will estimate the radius of curvature from height contours on a constant pressure map How does the radius of curvature estimated in this way differ from the radius of curvature of the air parcel trajectory Using this geometry 5s R53 i lt 5s R R where 3 angular wind direction 3 distance along path and R radius of curvature Following the motion of an air parcel this gives is i where RT is the radius of curvature of a trajectory Ds RT At a fixed time this geometry implies 0 1 where RS is the radius of curvature of a streamline S S Noting that EL D1 D5 D1 RT and using the definition of the total derivative DDt gives Dill w Dt at as LEL RT 32 RS L1 at RT R 35 where 7 is the turning of the wind over time at a fixed location It 33 For steady state conditions E 0 and RT R5 For nonsteady state cases RT 96 Rs Consider the case of an eastward moving trough Figure 422 Streamlines and trajectories in an eastwardmoving upper trough Thin solid lines rep resent streamlines of the ow at some initial time while the dashed lines represent streamlines at some i later time The bold arrows AD AC and AB represent the trajectories of air parcels moving slower than the wave at the same speed as the wave and faster than the wave respectively Thermal Wind How will horizontal temperature gradients alter the height and thus slope of upper level constant pressure surfaces snumh sssssss eyes 39 gilt H F a 800mb U 3 Zerowind 109031227zu Using the hypsometric equation we can relate the thickness 62 between two pressure surfaces eg 1000 and 500 mb to the layer average temperature oz lnp 5 Assuming a flat lower pressure surface we will find that the upper pressure surface slopes down towards the cold air This slope results in a pressure gradient force quD This pressure gradient force can be used to calculate the geostrophic wind What does the change in slope with height imply about changes in the geostrophic wind with height For the case illustrated above u9 0 at all levels and v9 0 at 1000 mb and is positive at upper levels We want to derive a relationship between the change in the geostrophic wind with height and the horizontal temperature gradient This relationship is known as the thermal wind relationship Take 3 ofthe equation for the geostrophic wind to get an equation that P expresses the change in geostrophic wind with pressure ie the change in geostrophic wind in the vertical direction 9p f3y3p fay apf 9xapfax whim M Wgi62 1gtii 9p 9p From the hydrostatic equation 9p az pg ip g9z alt1gt p l 9p p p Then aug 1aRT 1aRT 9p fay pp 9p fax p aug R3T avg R3T pii 7 ire 7 07p f 07y p 07p f 07x p aug R 3T avg amp f 3y p 91111 f 335 p In vector form this gives the thermal wind equation an R alnp f ExVPT Thermal wind VT is the vector difference between the geostrophic wind at two pressure levels VT VgP1 VgP0 Integrating the thermal wind equation from pressure level pg to pressure level 7 gIves In A In R A f6Vgf 7kaPT6lnp 9 W n A A A R A Vgpl mpn e VT 7k xVPT1n j f pm or in component form R KT pa 5 7 7 1 i e mm mm f y I RaT pp v Ev p v p 771n7 r A I A a A ax l w where T is the layer average temperature What is the physical interpretation of these equations ow can we use these equations to explain the increase of westerly winds with height in the midlatitude troposphere Z stratosphere cold u y E a T lt 0 l 3 Z X r y troposphere hot UZl Copyrlght zone by John Wiley 8 Sons Inc or related companies All rights reserved What is the relationship between the thermal wind and thickness lines on weather maps Integrate the hydrostatic equation RT from p0 to p1 P P 31an T ap P P0 P0 13p1 lt1gtp0 11 10 RltT1n11 RltTln1 11 10 gZ1 ZO ng RTlnpO P1 where ZT is the thickness between pressure levels p1 and p0 From this we see that thickness 27 is proportional to the layer mean temperature Substituting this into the thermal wind equation gives MT alt1gt1 ltIgtogt VT ialtlt1gt1 ltIgtogt f 9y f 9x T 5324 VT 324 f 9y f 9x or in vector form A A l 1A A VT kXVltTgt1nLIZiTJikXVlt 1 DokXVZT This equation indicates that the thermal wind is parallel to thickness lines or isotherms of layer mean temperature on a weather map What is the direction of VT relative to warm and cold air in the Northern Southern hemisphere Cold To ST Cold air advection CAA Val the wind blows from a region v To of cooler temperatures to a 9g v region of warmer T temperatures To 81 Worm 0 Cold To ST Warm air advection WAA VT the wind blows from a region lt Vq1 T O of warmer temperatures to a region of cooler temperatures To8T Worm Backing wind turns counterclockwise with height Veering wind turns clockwise with height In what direction does the geostrophic wind turn for the warm advection cold advection case Using this information we can estimate the layer averaged temperature advection based on a single wind pro le estimate the geostrophic wind at upper levels based on the geostrophic wind and temperature eld at a lower level Example Determine whether CAA or WAA is occurring using a radiosonde observed wind profile Barotropic and Baroclinic Atmospheres Barotropic density depends only on pressure p pp From the ideal gas law p L RT We find for a barotropic atmosphere that T Tp VPT 0 ie temperature is constant on constant pressure surfaces 0 and 3x 3y 39 VT 0 The geostrophic wind does not vary with height in a barotropic atmosphere Baroclinic density depends on both temperature and pressure For a baroclinic atmoshphere the geostrophic wind can vary with height Vertical Motion What is the relationship between a and W waampaipV39Vpwaip D 32 az where l7 Vg Va is the horizontal wind and i7 ltlt Vg Va 10 Vg for midlatitude weather systems Noting that V39Vp7g 39Vp7a 39Vp but Vg LIE pr so pf Vg Vp0 and V39Vp7a Then wa Va 39Vp waip at az 31 A 39 wiV 39V w at a P Pg where we have used the hydrostatic equation to replace i z Scale analysis ofthis equation for midlatitude weather systems gives 91 10 mb day391 at V 39Vp 1 ms3911 Pa km3911 mb day 1 a ng 100 mb day391 So w ng CU wz i As noted in Chapter 2 W is of the order of l cm s391 10392 m s4 for mid latitude weather systems and is difficult measure accurately W can be estimated using the continuity equation kinematic method or the thermodynamic energy equation adiabatic method Kinematic Method For the kinematic method of estimating wwe will integrate the continuity equation in isobaric coordinates between pressure levels ps and p 3w pap wltpgtwltpsgt i i gtpltp psgt39 where lt indicates a pressure weighted vertical average Using a z ng and z is the height of pressure level p and 2S is the height of pressure level ps gives PZ8WZ pzsgwzs p ax 3y 10Z pzg ax 3y p In order to estimate Wz we will need to estimate the value ofthe horizontal divergence of the wind 30 31 I kix 9y p Vp V Recall that Vp 7g 0 so Vp 7 Vp 47 but m N 01li7gl This means that even a small error in the observed wind will result in a large relative error in Va and thus VP W7 is difficult to calculate accurately Adiabatic Method lfwe assume that the diabatic heating rate J is small then the thermodynamic energy equation reduces to W aT o39lT iuivi S w0 at 3x 3y p and 1 aT aT 6T w i iui 7 Sp at 3x 3y 1 aT 6T 6T pgwi iuivi Sp at x a 1 6T 6T aT 7 iuivi pgSp at 3x 3y We can use this equation to estimate W based on the local rate of temperature change horizontal temperature advection u v at 3x y and the stability Sp This equation can be difficult to use over a large area as g can be t difficult to estimate This equation will not give reasonable results in areas of large diabatic heating such as in clouds Another method of estimating a and vertical velocity will be presented in Chapter 6 Surface Pressure Tendency Changes in surface pressure ps can be used to indicate the approach of low pressure systems Using the continuity equation in isobaric coordinates we can derive an expression for the surface pressure tendency 955 t Integrating the continuity equation from the surface to the top of the atmosphere p 0 and assuming that w0 0 gives 0 u av pt wpsfaxay ap fV39V 9p Px p 0 Using Dp 3p A 3p w EisJV39V wi p5 D at p az and noting that W 0 over flat ground V39Vp7g 39Vp7a 39Vp Vg Vp0 and 7 39szO a gives px 07L r at V vap This equation indicates that the rate of surface pressure change is equal to the total convergence V i7 into the column above the surface Consider a flow with only a zonal component What will cause V7 to be positive negative What impact will this have on the surface pressure In practice it is difficult to use this equation to predict changes in the surface pressure since it is difficult to accurately calculate V7 from observations and it is often observed that regions of convergence at one level in the atmosphere are often nearly compensated for by areas of divergence at other levels This equation can still be used to provide a conceptual understanding of the mechanisms that can cause surface pressure to vary Example Development of a thermal low lt High gt Sea level
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'