Download presentation

Presentation is loading. Please wait.

Published byEvangeline Sutton Modified about 1 year ago

1
THE FEDERAL UNIVERSITY OF TECHNOLOGY, AKURE DEPARTMENT OF METEOROLOGY MET 302 ATMOSPHERIC DYNAMICS (Lecture notes prepared by Mr. N. O. Nnoli using some lecture notes, handouts, course outlines of Prof. J. A. Omotosho and other meteorological books) LECTURER: MR. N. O. NNOLI COURSE COORDINATOR: MR. S. OGUNGBENRO

2
COURSE OUTLINE 1.0 Introduction - Fundamental Forces 1.1 Definitions, Physical Dimensions and units 1.2 Atmospheric Scales and Conservation Laws 1.3 Real Forces - Pressure Gradient, Gravitational and Friction 1.4 Apparent Forces - Centrifugal, Gravity and Corilois 2.0 The Basic Conservation Laws 2.1 Fundamental Physical laws of conservation 2.2 Eulerian and Lagrangian Reference Frames 2.2 Total and Partial (local) Derivatives of Quantities 2.3 Total Derivatives of a Vector in Rotating System 2.4 The Vectorial Form of the Momentum Equation in Rotating Coordinates 2.5 The Component Equations in height Spherical Coordinates 2.6 Scale Analysis of the Equations of Motion, Geostrophic, Prognostic and Hydrostatic Equations ;

3
2.7 Continuity Equations - Eulerian and Lagrangian Derivation; Divergence, Convergence & Vertical Motion 2.8 Thermodynamic Energy Equation 2.9 Primitive Equations 2.10 Transformation Equations (height to pressure coordinates) 3.0 Elementary Applications of Basic Equations 3.1 Geostrophic, Gradient and Cyclostrophic Winds; Streamlines and Trajectories 3.2 Thermal Wind - Derivation and Uses 3.3 Thermal Wind and Jet Streams - Mid-Latitude and Tropical Situations 3.4 Barotrpic and Baroclinic Atmospheres 4.0 Circulation and Vorticity 4.1 Circulation Theorem 4.2 Application to Land and Sea Breezes 4.3 Vorticity (in Cartesian and Natural Coordinates) 4.4 Vorticity Equation - Derivation and Discussion for Mid-Latitude and Tropical Cases

4
5.0 Instability Processes in the Atmosphere 5.1 Static Instability 5.2 Conditional/Convective Instability 5.3 Conditional Instability of the Second Kind 5.4 Baroclinic Instability 5.5 Barotropic Instability 5.6 Atmospheric wave motions Grading System Continuous Assessment {Attendance (65% Attendance minimum)– Assignments – 40%} Examination – 60% Examination Malpractice is not spared but the victim should face disciplinary panel of the University and be rusticated. Be warned!

5
1.INTRODUCTION 1.1 Definitions Dynamic Meteorology – is the field of meteorology that is concerned with atmospheric motions that are associated with weather and climate. In contrast to synoptic meteorology, it employs analytical approaches based on the principles of fluid dynamics. With the increasing sophistication of methods of weather analysis and forecasting, the distinction between synoptic and dynamic meteorology is rapidly diminishing. The only really pure dynamic meteorologist is one who does not know what a weather map looks like, and the only pure synoptic meteorologist is one who does not make use of any of the equations that govern atmospheric motions. Both species are approaching extinction.

6
In the study of atmospheric motions, in dynamic meteorology, the discrete molecular nature of the atmosphere can be ignored, and the atmosphere can be regarded as a continuous fluid medium, or continuum. The various physical quantities which characterise the state of the atmosphere – pressure, density, temperature and velocity – are assumed to have unique values at each point in the atmospheric continuum. Moreover, these field variables and their derivatives are assumed to be continuous functions of space and time. 1.2 Physical Dimensions and Units The fundamental laws that govern the motions of the atmosphere are expressed in terms of physical quantities (field variables and coordinates) which depend on four dimensionally independent properties : length, time, mass and thermodynamic temperature.

7
The dimensions of all atmospheric field variables may be expressed in terms of multiples and ratios of these four fundamental properties. To measure and compare the scales of atmospheric motions a set of units of measure must be defined for the four fundamental properties. The International System of unit (S.I.) will be used exclusively. The four fundamental properties are measured in terms of the S.I. base units shown below (Table 1.1): Table 1.1 – S. I. Base Units Property Name Symbol Length meter (metre) m Mass kilogram (kilogramme) kg Time second s Temperature Kelvin K

8
All other properties are measured in terms of S.I. derived units which are units formed from products and/or ratios of the base units. For example, velocity has the derived units of meter per second (m/s). A number of important derived units have special names and symbols. Those commonly used in dynamic meteorology are shown below (Table 1.2): Table 1.2: Derived Units with Special Names Property Name Symbol Frequency Hertz Hz (s -1 ) Force Newton N (kgm/s) Pressure Pascal Pa (Nm -2 ) Energy Joule J (Nm) Power Watt W (Js -1 )

9
In addition, the supplementary unit designating a plane angle – radian (rad) – is required for expressing angular velocity (rads -1 ) in the S.I. system. 1.3 Atmospheric Scales The atmosphere is characterized by phenomena whose space and time scales cover a very wide range. The space scales of these features are determined by their typical size or wavelength, and the time scales by their typical lifetime or period. These features range from small-scale turbulence (tiny swirling eddies with very short life spans) to jet streams (giant waves of wind encircling the whole Earth). In reality, none of these phenomena is discrete but part of a continuum, therefore it is not surprising that attempts to divide atmospheric phenomena into distinct classes have resulted in disagreement with regard to the scale limits.

10
Most classification scheme use the characteristic horizontal distance scale as the sole criterion. A reasonable consensus of these schemes gives the following scales and their limits: Table 1.3: Atmospheric Scales Planetary Scale Large Scale (Synoptic Scale) Meso-ScaleMicro Scale > 7000km1000 – 7000km100 – 1000km< 100km Long Waves in the upper troposphere and stratosphere, Rossby Waves, Jet Streams Cyclones and Anticyclones, Fronts, Air Masses, etc. Cloud Clusters, Squall lines, Land and Sea- Breezes, hurricanes, etc. Cumulus Clouds, Tornadoes, Planetary Boundary Layer Turbulence, etc

11
Wave Number, N The wave number, k of a wave is k = 2π/L where L is the wave length. In meteorology, we multiply 2π/L by R, where R is the radius of the Earth = 6.37 X 10 6 m, therefore, Wave number, N = kR = 2πR/L ……………………… 1.1 N = 1, 2, 3, 4 are Planetary waves (ie. L = 10,000 – 40,000 km) N = 6, 7, 8, 9, 10, etc are Synoptic waves. N > 50 are meso-scale waves Assignment 1: Find wave numbers of (a)African Easterly waves (L = 3,000km) (b)Squall lines (L ~ 400km) (c)Cyclone Waves (L ~ 7,000km)

12
1.4 Fundamental Forces The motions of the atmosphere are governed by the fundamental physical laws of conservation of mass, momentum and energy. Newton’s second law of motion states that the rate of change of momentum of an object referred to coordinates fixed in space, equals the sum of all the forces acting. For atmospheric motion of meteorological interest, the forces, which are of primary concern, are pressure gradient force, gravitational force and friction. These are Real Forces. If, as is the usual case, the motion is referred to a coordinate system rotating with the Earth, Newton’s second law may still be applied provided that certain Apparent Forces – Centifugal and Coriolis Forces – are included among the forces acting.

13
The gravity force or effective gravity (which is also an apparent force) is the vectorial sum of the gravitational force and the centrifugal force Real Forces Pressure Gradient Force Air tends to flow from High to Low pressure. The variation of pressure in the horizontal although several orders of magnitude less than that in the vertical, is more important to the practicing meteorologist. For, unlike the large pressure gradient (change of pressure with distance) built into the vertical by gravity and hence (conversely) balanced by gravity, any horizontal gradient of pressure represents a state of imbalance which must give rise to horizontal air motion, i.e. wind. Wind speed increases as pressure gradient increases.

14
The spacing of isobars on surface weather chart and contours on constant-pressure charts, therefore, give a rough indication of wind speeds. When they are close together, the pressure gradient is strong and wind speeds are high and vice-versa. Air motion or wind contains water vapour. Wind with sufficient water vapour and flowing towards low pressure rises up, cools and the water vapour humidifies, saturates and condenses and clouds then form. These clouds grow in the presence of condensation nuclei and later fall as rain, Meteorologists are noted for rain forecasting and this explains why horizontal gradient of pressure is more important than vertical gradient of pressure.

15
Near the sea-level the magnitude of vertical pressure gradient force (∂p/∂z) is roughly 10hPa per 100 meters, whereas gradients of 10hPa per 100km in the horizontal produce winds in excess of hurricane force. Derivation of Pressure Gradient Force Horizontal pressure gradient force is the force per unit mass that tends to drive a parcel of air across (perpendicular to) isobars from high to low pressure. To derive the horizontal pressure gradient force using natural coordinates, we consider a small block of fluid with dimensions δn, δs, and δz as shown below, Fig The coordinate system is chosen so that the s axis is oriented parallel to the local isobar, the n axis points in the direction of higher pressure, and the z axis points upward, parallel to the line in which local gravity acts.

16
δsδs z δnδn n p + δp Higher pressure Lower pressure δzδz p Fig. 1.1: Horizontal pressure acting upon an infinitesimal block of fluid

17
The pressure force exerted by the surrounding air upon the left hand n-face of the block is given by +pδs δz, where p is the average pressure on this face of the block. The positive sign indicates that the pressure force is directed to the right. There is an almost identical, but opposite directed force on the other side of the block given by – (p+ δp) δs δz, where the pressure increases by the small increment δp as n increases from the left to the right-hand side of the block. The negative sign indicates that the pressure force is directed to the left. If the dimensions of the block are very small in comparison with the scale of the pressure fluctuations, we can write δp ≈ (∂p/∂n)δn where the partial derivative notation indicates that the derivative is taken at constant ‘s’ and ‘z’.

18
The net ‘n’ component of the pressure force on the block is simply the vectorial sum of the forces on the two ‘n’ faces, which is +pδs δz +[ - (p+ δp) δs δz] = pδs δz - (p+ δp) δs δz = pδs δz - pδs δz - δp δs δz = - δp δs δz = - (∂p/∂n) δn δs δz [since δp ≈ (∂p/∂n)δn] The negative sign indicates that the force is directed towards lower values of ‘n’ that is from higher pressure toward lower pressure. Dividing by the mass of the block (ρδn δs δz), where ρ is the density of the air, we obtain the ‘n’ component of the pressure gradient force per unit mass, P n = - (1/ρ)(∂p/∂n) ……………………………………… 1.2 Since the pressure forces on the ‘s’ face of the block exactly cancel, eqn. 1.2 gives the magnitude of the total horizontal pressure gradient force.

19
The force is directed perpendicular to the isobars on a horizontal surface and down the pressure gradient, that is, from higher pressure toward lower pressure. In similar manner, it can be demonstrated that the vertical component of the pressure gradient force is given by, P z = - (1/ρ)(∂p/∂z) ………………………….1.3 If the coordinate system (x, y, z) is chosen so that x and y each is either perpendicular or parallel to the isobars, then the x, y, and z components of pressure gradient force per unit mass could be derived as, P x = - (1/ρ)(∂p/∂x) ……………………………..1.4a P y = - (1/ρ)(∂p/∂y) ……………………………….1.4b P z = - (1/ρ)(∂p/∂z) ……………………………….1.4c

20
So that the total pressure gradient force per unit mass is, in vectorial form: P = - (1/ρ)( ∇ p) ……………………………… It is important to note that this force is proportional to the gradient of the pressure field, not to the pressure itself. In pressure coordinates(x, y, p), the horizontal pressure gradient force can be expressed in terms of the gradient of geopotential or geopotential height on pressure surfaces. Consider two points Q and R, separated by the vertical distance δz and the horizontal distance δn in the plane normal to the isobars, as indicated in Fig We will assume that the separation between the two points is small enough that the spatial gradients of p are uniform (which is equivalent to assuming that the pressure surfaces are parallel planes with equal spacing for equal increments of δp).

21
The difference in pressure between the two points is given by, δp = (∂p/∂n) δn + (∂p/∂z) δz Fig. 1.2 : Relation between horizontal and vertical pressure gradients and the slope of the surfaces of constant pressure. Now, if the two points lie on the same pressure surface δp=0, and in the limit as the separation of the points approaches zero, δn dn and δz dz, i.e. δp = (∂p/∂n) δn + (∂p/∂z) δz z n p-δp p p+δp p+2δpQ R δzδz δnδn

22
i.e. 0 = (∂p/∂n) δn + (∂p/∂z) δz (at constant pressure) i.e. - (∂p/∂n) δn = (∂p/∂z) δz i.e. - (∂p/∂n) dn = (∂p/∂z) dz (in the limit) i.e. - (∂p/∂n) = dz (∂p/∂z) dn (at constant pressure) Thus, at any point, the ratio of the horizontal pressure gradient force to the vertical pressure gradient force is equal to the local slope of the pressure surface passing through that point. The slope of pressure surface in large-scale atmospheric disturbances rarely exceed 1 part in Therefore, it is evident that in these systems the horizontal pressure gradient force is typically at least three orders of magnitude smaller than the vertical pressure gradient force.

23
Introducing the hydrostatic equation, ∂p/∂z = - ρg into the last expression, we obtain. - (∂p/∂n) = dz - ( ρg) dn (at constant pressure) i.e. – (1/ρ)(∂p/∂n) = – g(dz/dn) p In (x, y, p) coordinates (dz/dn) p and ∂z/∂n are identical. Incorporating this identity intdzo the above expression and making use of equation 1.2, we have, P n = - (1/ρ)(∂p/∂n) = – g (∂z/∂n) …………………… 1.6 Furthermore, noting from the mathematical definition of geopotential and geopotential height that gdz = dφ = g o dZ ………………………………………………1.7 where z = geometric height, Z = geopotential height and g o = globally averaged acceleration due to gravity = 9.8ms -2

24
the above equation 1.6 can be expanded in the form, P n = - 1/ρ (∂p/∂n) = – g (∂z/∂n) = – ∂φ/∂n = – g o (∂Z/∂n) …….1.8 If the coordinate system (x, y, p) is chosen so that x and y each is either perpendicular or parallel to the isobars then the x and y components of pressure gradient force per unit mass could be derived to be, P x = - (1/ρ)(∂p/∂x) = – g (∂z/∂x) = – ∂φ/∂x = – g o (∂Z/∂x) ………1.9a P y = - (1/ρ)(∂p/∂y) = – g (∂z/∂y) = – ∂φ/∂y = – g o (∂Z/∂y) ………1.9b In vector notation, the horizontal pressure gradient force assumes any of the forms, P n = - (1/ρ)( ∇ h p) = – g ( ∇ h z) = – ∇ h φ = – g o ( ∇ h Z) …………………1.10 where the so called horizontal gradient operator ∇ h ( ) denotes a vector oriented normal to the isobars, height contours, or geopotential contours, pointing toward higher values.

25
The minus signs indicate that the horizontal pressure gradient force points in the direction opposite to ∇ h ( ); that is toward lower values of p, z, φ or Z, respectively. It is implicitly understood that ∇ h p is taken at constant height, z, while ∇ h z, ∇ h φ and ∇ h Z are taken at constant pressure, p. Density no longer appears explicitly in the pressure gradient force – distinct advantage of the isobaric system. The advantage of equations 1.8 – 1.10 is that g unlike ρ varies only a little with height. At upper levels, the pressure coordinate forms of equations 1.8 – 1.10 are used extensively in place of equations 1.3 – 1.5. Example: 1 At a certain point, the horizontal gradient of sea-level pressure is 5hPa per 100km. Estimate the slope of the 1000hPa surface at this point (Assume that the air density, ρ = 1.25kgm -3 ).

26
Solution The pressure gradient at sea-level is – ∂p/∂n= 5hPa/100km = 500Pa/100000m The pressure gradient force P n = – 1/ρ(∂p/∂n) = 500Pa/(1.25X100000)m The pressure gradient at 1000hPa is – ∂Z/∂n (to be determined) The pressure gradient force P n = – g o ∂Z/∂n, where g o = 9.8ms -2 Since 1000hPa is close to sea-level, it is assumed both P n are equal Sea-level Isobars 1000hPa height contours p1p1 p2p2 Z2Z2 Z1Z1

27
i.e. P n = – (1/ρ)(∂p/∂n) = – g o (∂Z/∂n) i.e. P n = – 500/(1.25X100000) = – 9.8 (∂Z/∂n) i.e. ∂Z/∂n = – 500/(1.25X X 9.8) meter/meter i.e. ∂Z/∂n ≈ 4.08 X10 -4 meter/meter = 40.8m per 100km Example hPa H 1000hPa o A 28 o N L

28
In the figure above derived from the surface weather map below, determine the total horizontal pressure gradient force at the position A, (i) using height natural coordinates and (ii) using height Cartesian coordinates. Comment on the two answers. (Take 1° latitude = 111.2km and 1° longitude = 111.2cosΦkm, where Φ = latitude, density, ρ = 1.25kgm -3 ) Solution Using natural coordinates, we have, Pressure gradient force, P n = - (1/ρ)(∂p/∂n) ≈ - (1/ρ)(Δp/Δn) = - {(1002 – )x100}/(1.25x4x 111.2x 10 3 )N/kg ≈ 9.71x10 -4 Nk/g Using Cartesian coordinates, we have, P x = - {(1003 – 1006)}x100/(1.25x4x111.2x10 3 cos28) N/kg ≈ 6.11x10 -4 N/kg P y = - {(1007 – )}x100/(1.25x4x111.2x10 3 )N/kg ≈ x10 -4 N/kg ∴ P = √ (P 2 x + P 2 y ) = √{( 6.11x10 -4 ) 2 +(-8.09x10 -4 ) 2 }N/kg ≈ 1.0x10 -3 N/kg The two answers, P n ≈ 9.71x10 -4 Nk/g and P ≈ 1.0x10 -3 N/kg are same.

29
AoAo

30
Gravitational Force Newton’s law of universal gravitation states that any two elements of mass in the universe attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance separating them. Thus, if two mass elements M and m are separated by a distance r ≡ |r| (with the vector r directed toward m as shown in Fig. 1.4) then the force exerted by mass ‘M’ on mass ‘m’ due to gravitation is F g = – GMm/r 2 (r/r) ………………………………………… M m r Fig. 1.4: Two spherical masses whose centres are seperated by a distance r.

31
where G is a universal constant called the gravitational constant. The law of gravitation as expressed in eqn actually applies to hypothetical ‘point’ mass since for objects of finite extent r will vary from one part of the object to another. However, for finite bodies eqn may still be applied if |r| is interpreted as the distance between the centre of mass of the bodies. Thus, if the earth is designated as mass M and m is a mass element of the atmosphere, then the force per unit mass exerted on the atmosphere by the gravitational attraction of the earth is, F g = g* = – GM/r 2 (r/r) …………………………………………… Friction Friction is the mechanical force of resistance which acts when there is a relative motion of two bodies in contact.

32
In meteorology, the effects of friction are important in the flow of air over the earth’s surface and also when there is wind shear (the local change of wind velocity in a specified direction normal to the wind direction). Throughout most of the atmosphere frictional forces are sufficiently small that, to a first approximation, they can be neglected. A notable exception is the so called planetary boundary layer or friction layer (corresponding to the lowest 1km of the atmosphere) where the flow over a stationary underlying surface gives rise to a frictional drag force which is comparable in magnitude to the other terms in the horizontal equations of motion. The frictional force in the equations of motion represents the collective effect iof all scales of motion in exchanging momentum between air parcels and the environment to achieve a balance.

33
The vertical exchange of momentum are the most important; they always act to smooth out the vertical profile of the wind. The effects of surface friction visible on the synoptic scale are a decrease of surface wind speed relative to that appropriate to the pressure gradient and a frictional outflow of surface air from higher to lower pressure. The magnitude of these effects increase with the surface roughness and decrease with increasing height within the friction layer.

34
1.4.2 Non-Inertial reference frames and Apparent Forces In formulating the laws of atmospheric dynamics it is natural to use a GEOCENTRIC reference frame which is fixed with respect to the rotating earth. Newton’s first law of motion states that a mass in uniform motion relative to a coordinate system fixed in space will remain in uniform motion in the absence of any forces. Such motion is referred to as INERTIAL motion, and the fixed reference frame is an INERTIAL, or ABSOLUTE frame of reference. It is clear, however, that an object at rest with respect to the rotating earth is not at rest or in uniform motion relative to a coordinate system fixed in space. Therefore, motion which appear to be inertial motion to an observer in the rotating earth reference frame is really accelerated motion. Hence, the rotating frame is a NON-INERTIAL reference frame.

35
Newton’s laws of motion can only be applied in such a frame if the acceleration of the coordinates is taken into account. The most satisfactory way of including the effects of coordinate acceleration is to introduce ‘apparent’ forces in the statement of Newton’s second law. These apparent forces are the inertial reaction terms which arise because of the coordinate acceleration. For a coordinate system in uniform rotation, two such ‘apparent’ forces are required, the centrifugal and Coriolis forces. The Centrifugal Force The Centrifugal force arises in meteorology in two main ways. Firstly, the centrifugal force acts on the earth and atmosphere due to the rotation round the earth’s axis. Secondly, air moving on a curved path with respect to the earth’s surface is subject to a centrifugal force acting outward from the center.

36
To derive the expression for centrifugal force per unit mass, we consider a ball of mass ‘m’ which is attached to a string and whirled through a circle of radius r at a constant angular velocity ω. From the point of view of an observer in fixed space the speed of the ball is constant, but its direction of travel is continuosly changing so that its velocity is not constant. To compute the acceleration, we consider the change in velocity δV which occurs for a time increment δt during which the ball rotates through an angle δθ as shown below, Fig. 1.5 V V δVδV δθδθ ω δθδθ Fig. 1.5: Centripetal Acceleration r

37
Since δθ is also the angle between the vectors V and V + δ V, the magnitude of δV is |δV| = |V| δθ i.e. |δV|/ δt = |V| δθ/ δt In the limit as δt 0 δV is directed towards the axis of rotation and δt dt and δV dV i.e. |dV|/dt = |V|dθ/dt i.e. dV/dt = |V|dθ/dt (– r/r) But |V| = ωr and dθ/dt = ω, so that dV/dt = – ω 2 r Also, |V| = V = ω r ∴ ω = V/r ∴ ω 2 = V 2 /r 2 ∴ dV/dt = – ω 2 r = – V 2 /r 2 r …………………………………… where V = the linear velocity of the ball ω = the angular velocity of the ball r = radius of curvature of the path, and t = time

38
Therefore, viewed from fixed coordinates the motion is one of uniform acceleration directed towards the axis of rotation, and equal to the square of the angular velocity times the distance from the axis of rotation (or the square of linear velocity divided by the distance from the axis of rotation). This acceleration is called the centripetal acceleration. It is caused by the force of the string pulling the ball. If the motion is observed in a coordinate system rotating with the ball, the ball is stationary. However, there is still a force acting on the ball, namely the pull of the string. Therefore, in order to apply Newton’s second law to describe the motion relative to this rotating coordinate system we must include an additional apparent force, the centrifugal force, which just balances the force of the string on the ball. Thus, the centrifugal force per unit mass is equivalent to the inertial reaction of the ball on the string and just equal and opposite to the centripetal acceleration.

39
Gravity Force, Gravity or Effective Gravity The force per unit mass, called gravity or effective gravity or gravity force and denoted by the symbol g, is actually the vectorial sum of the true gravitational attraction g* that draws all elements of mass toward the centre of mass of the earth and the much smaller apparent force, the centrifugal force, that pulls all objects outward from the axis of planetary rotation, as indicated below, Fig Ω g* Ω2RΩ2R Ω2RΩ2R R g α g sinα g* = g Ω 2 R = 0 g = g* - Ω 2 R Fig. 1.6: The relationship between gravitational and gravity forces.

40
Thus, the weight of a particle of mass, m, at rest on the earth’s surface which is just the reaction force of the earth on the particle will generally be less than the gravitational force mg* because the centrifugal force partly balances the gravitational force. It is, therefore, convenient to combine the effects of the gravitational force and centrifugal force by defining a gravity force g such that g ≡ g* + Ω 2 R ……………………………………… 1.14 Therefore, except at the poles and the equator, gravity is not directed toward the centre of the earth. As indicated in Fig. 1.6, if the earth were a perfect sphere, gravity would have an equatorial component, gsin α, parallel to the surface of the earth.

41
The earth has adjusted to compensate for this equatorial force component by assuming the approximate shape of a spheriod with an equatorward “bulge” so that g is everywhere directed normal to the level surface. As a result, the equatorial radius of the earth is about 21km larger than the polar radius. In addition, the local vertical which is taken to be parallel to g does not pass through the centre of the earth except at the equator and the poles.

42
Coriolis Force The Coriolis Force is the force per unit mass that results from the rotation of the earth and acts on a moving particle with respect to the earth to deviate it. It is a three dimensional vector which is given in vector notation by the expression 2Ω x V where Ω is the earth’s angular velocity and V is the wind velocity. If the planetary rotation is clockwise (or counterclockwise) as viewed from space, the Coriolis force is directed to the left (or right) of the velocity vector. Thus, when viewed from above, the Coriolis force is directed toward the right of the velocity vector in the Northern Hemisphere and toward the left in the Southern Hemisphere. The Coriolis force is directed radially outward from the axis of rotation if the zonal motion is in the same sense as the planetary rotation (u > 0, westerly flow) and radially in ward toward the axis of rotation if zonal motion is in opposite sense to the planetary rotation (u < 0, easterly flow).

43
Coriolis force also arise in the case of radial motion toward or away from the axis of rotation. The existence of such a force can be deduced from the conservation of angular momentum. Coriolis force, which acts perpendicular to the velocity vector, can only change the direction of travel and not the magnitude. The mathematical form for the Coriolis force due to motion relative to the rotating earth can be obtained by considering the motion of a hypothetical particle of unit mass which is free to move on a frictionless horizontal surface on the rotating earth. If the particle is initially at rest with respect to the earth, then the only forces acting on it are the gravitational force and the apparent centrifugal force due to the rotation of the earth. The sum of these two forces defines the gravity, which is directed perpendicular to the local horizontal. If the particle is now set in motion in the eastward direction by an impulsive force, the centrifugal force on the particle will be increased since the particle is now rotating faster than the earth.

44
If Ω is the magnitude of the angular velocity of the earth, R the position vector from the axis of rotation to the particle, and u the eastward speed of the particle relative to the groind, the total centrifugal force could written as, (Ω+u/R) 2 R = Ω 2 R + 2ΩuR/R + u 2 R/R 2 …………………………… 1.15 The first term on the right is just the centrifugal force due to the rotation of the earth. This is, of course, included in gravity. The other two terms represent deflecting forces which act outward along the vector R (that is, perpendicular to the axis of rotation). For synoptic scale motion u<< ΩR, and the last term may be neglected, in a first approximation. The remaining term in 1.15, 2ΩuR/R, is the Coriolis force due to relative motion parallel to the latitude circle. The Coriolis force can be divided into components in the vertical and meridional directions, respectively, as indicated in Fig. 1.7 below.

45
Φ Φ R 2ΩuR/R 2 Ωu cos Φ 2 Ωu sin Φ Ω Fig. 1.7: Components of Coriolis force due to relative motion along a latitude circle Therefore, relative motion along the east-west coordinate will produce an acceleration in the north-south direction given by, (dv/dt) Co = - 2 Ωu sin Φ ………………………… 1.16 and acceleration in the vertical given by, (dw/dt) Co = 2 Ωu cos Φ …………………………. 1.17

46
where (u, v, w) designate the eastward, northward and upward velocity components, respectively, Φ is the latitude and the subscript Co indicates that this is the acceleration due only to the Coriolis force. A particle moving eastwards in the horizontal plane in the Northern Hemisphere will be deflected southward by the Coriolis force, whereas a westward moving particle will be deflected northward. In either case, the deflection is to the right of the direction of motion. The vertical component of the Coriolis force (1.17) is ordinarily much smaller than the gravitational force so that its only effect is to cause a very minor change in the apparent weight of an object depending on whether the object is moving eastward or westward. The above treatment is the Coriolis force due to relative motion parallel to latitude circle. We now consider Coriolis force in the case of radial motion toward or away from the axis of rotation.

47
Suppose now the particle initially at rest on the earth is set in motion equator-ward by an impulsive force. As the particle moves equator-ward it will conserve its angular momentum in the absence of torques in the east-west direction. Since the distance of the axis of rotation R increases for a particle moving equator-ward, a relative westward velocity must develop if the particle is to conserve its absolute angular momentum as shown in Fig. 1.8 below. Ω Fig, 1.8: Horizontal trajecxtories of air parcels that develop a zonal motion component as a consequence of the conservation of angular momentum.

48
Thus, letting δR designate the change in the distance to the axis of rotation for a southward displacement from a latitude Φ o to latitude Φ o +δΦ (note that δΦ < 0 for an equator-ward displacement), we obtain by conservation of angular momentum. ΩR 2 = (Ω + δu/(R+δR))(R+δR) 2 …………………………………… 1.18 where δu is the eastward relative velocity when the particle has reached latitude Φ o +δΦ. Expanding the right hand side, we have, ΩR 2 = Ω(R+ δR) 2 + δu(R + δR) 2 /R+ δR ΩR 2 = ΩR 2 + 2ΩRδR + ΩδR 2 + δu(R+δR) Neglecting second order differential, we have, 0 = 2ΩRδR + δu(R+δR) i.e. – δuR(1+δR/R) = 2ΩRδR i.e. – δu(1+δR/R) = 2ΩδR Now, δR/R --> 0 since R >> δR δu = – 2ΩδR

49
Now, considering Fig. 1.9 below, we have, R ΦoΦo Φ o + δΦ δΦ ΦoΦo a a ΦoΦo δR= - δycosα= - δysin Φ o = - aδ Φ sin Φ o - δy=- aδ Φ Ω δy = aδ Φ Fig. 1.9: Relationship of δR and δy=aδ Φ for an equatorial displacement. δy=aδΦ ∴ - δy=- aδΦ where a is the radius of the earth. ∴ δR= - δycosα= - δysinΦ o but α+Φ o = 90 ∴ α = 90 – Φ o δR = - aδΦcos α = - aδΦ sinΦ o ∴ δu = - 2R δR = 2R ΩaδΦsin Φ o α

50
Now, dividing through by the time δt and taking the limit as δt 0, we have, (du/dt) Co = 2Ωa(dΦ/dt)sinΦ o (du/dt) Co = 2ΩvsinΦ o, where v = adΦ/dt is the northward velocity component. Similarly, it is easy to show that if the particle is launched vertically at latitude Φ o, conservation of absolute angular momentum will require an acceleration in the zonal direction equal to –2ΩwcosΦ, where w is the vertical velocity. Thus, in general case where both horizontal and vertical relative motions are included, we have, (du/dt) Co = 2ΩvsinΦ – 2ΩwcosΦ ……………………………… Again, the effect of the horizontal relative velocity is to deflect the particle to the right in the Northern Hemisphere. This deflection force is nsgligible for motions with time scales that are very short compared to the period of the earth’s rotation.

51
Thus, the Coriolis force is not important for the dynamics of individual cumulus clouds, but is essential to the understanding of longer time scale phenomena such as synoptic scale systems. The Coriolis force must also be taken into account when computing long-range missile or artillery trajectories. As an Example 1.1: suppose that a ballistic missile is fired due eastward at 43°N latitude (2ΩsinΦ = s -1 at 43°N). If the missile travels 1000km at a horizontal speed u o = 1000m/s, by how much is the missile deflected from its eastward path by the Coriolis force? Solution We use the equation 1.16 for the solution of this problem i.e. (dv/dt) Co = - 2 Ωusin Φ This gives the deflection due to Coriolis force in the north-south direction. Since the southward displacement is required, then eqn will be integrated twice with respect to t thus,

52
∫dv = ∫– 2Ωusin Φ dt ∴ v = –2Ωu o sin Φ where it is assumed that the deflection is sufficiently small so that we may let u = u o be constant. Integrating again (from zero to t o ) to get the southward displacement, we have, ∫vdt = ∫ yo y1+ δy dy = –2Ωu o ∫ o to tdtsin Φ Thus, the total displacement is, δy = – (2 Ωu o t o 2 sin Φ) /2 = – Ωu o t o 2 sin Φ δy = –7.29x10 -5 x1000x(1000) 2 xsin 43 meters where t=displacement/velocity=1000km/1000m/s= 1000x1000meters/1000ms -1 = 1000second ∴ δy ≈ –4.97x 10 4 m = –49.7x10 3 meters ≈ – 50km Therefore, the missilr is deflected southward by 50km due to the Coriolis effect.

53
2. 0 THE BASIC CONSERVATION LAWS Atmospheric motion are governed by three fundamental physical principles: conservation of mass, conservation of momentum, and conservation of energy. The mathematical relations which express these laws may be derived by considering the budgets of mass, momentum, and energy for an infinitesimal control volume in the fluid. Two types of control volumes are commonly used in fluid dynamics. In the eulerian frame of reference the control volume consists of parallelopiped of sides δx, δy, δz whose position is fixed relative to the coordinate axes. Mass, momentum, and energy budgets will depend on fluxes due to the flow of fluid through the boundaries of control volume. (This type of control volume was used in the derivation of the pressure gradient force).

54
In the lagrangian frame, however, the control volume consists of an infinitesimal mass of “tagged” fluid particles; thus, the control volume moves about following the motion of the fluid, always containing the same fluid particles. The largrangian frame is particularly useful for deriving conservation laws since such laws may be stated most simply in terms of a particular mass element of the fluid. The eulerian system is, however, more convenient for solving most problems because in the system the field variables are related by a set of partial differential equations in which the independent variables are the coordinates x, y, z, t. In lagrangian system, on the other hand, it is necessary to follow the time evolution of the fields for various individual fluid parcels. Thus, the independent variables are x o, y o, z o, and t, where x o, y o, z o, designate the position which a particular parcel passed through at a reference time t o.

55
2.1 Total and Local derivative of a Field Variable The total derivative or substantive derivative or largrangian derivative of a field variable is the rate of change of that field variable following the motion of individual air parcels along their three-dimensional trajectories through space. The local derivative or eulerian derivative is the rate of change of the field variable at a fixed point (it is merely the partial derivative with respect to time). Equations having total derivatives with respect to time are convenient for dealing with problems in which it is necessary to follow particular air parcels that have some special property (for example, emissions from a smokestack). However, for most problems in atmospheric dynamics, it is not necessary to keep track of individual air parcels; it is sufficient to know the distribution of certain dependent variables (velocity V, pressure p, temperature T, etc.) as continuous functions of three spatial coordinates and time.

56
For such problems it is convenient to transform the equations so that the time derivative (∂/∂t) refers to changes taking place at fixed points in space. To derive a relationship between the total derivative and the local derivative it is convenient to refer to a particular field variable, temperature, for example. Suppose that the temperature measured on a ballon which moves with the wind is T o at the point x o, y o, z o and time to. If the balloon moves to the point x o +δx, y o +δy, z o +δz in a time increment δt, then the temperature change recorded on the balloon δT may be expressed in a Taylor series expansion as, δT = (∂T/∂t)δt+ (∂T/∂x)δx+ (∂T/∂y)δy+(∂T/∂z)δz+(higher order terms) Dividing through by δt and taking the limit δt 0, we obtain, dT/dt = ∂T/∂t+(∂T/∂x)(dx/dt)+(∂T/∂y)(dy/dt)+(∂T/∂z)+(dz/dt) where dT/dt = Lim δt -->0 δT/δt

57
is the rate of change of T following the motion. If we now let dx/dt = u, dy/dt = v, and dz/dt = w then u, v, w are the velocity components in the x, y, z directions, respectively, and we have, dT/dt = ∂T/∂t + (u∂T/∂x + v∂T/∂y + w∂T/∂z) ……………… i.e. ∂T/∂t = dT/dt – (u∂T/∂x + v∂T/∂y + w∂T/∂z) ……………… 2.2 In pressure Cartesian coordinates, we have, ∂T/∂t = dT/dt - (u∂T/∂x + v∂T/∂y + ω ∂T/∂z) ………………..2.3 where ω = dp/dt In natural pressure coordinates, we have, ∂T/∂t = dT/dt – V ∂T/∂s – ω ∂T/∂p ……………………………… 2.4 Using vector notation these expressions may be written as, ∂T/∂t = dT/dt – V∙ ∇ T (in height coordinates) ∂T/∂t = dT/dt – V H ∙ ∇ H T - ω ∂T/∂p (in pressure coordinates) …2.6 It should be noted that

58
– V ∂T/∂s = -V| ∇ T | cosβ = – V H ∙ ∇ H T where β is the a βngle between V and ∇ T, VH = ui + vj (the horizontal velocity), ∇ H = i ∂/∂x + j ∂/∂y (horizontal del operator), ∇= i ∂/∂x + j ∂/∂y + k ∂/∂z (three dimensional del operator), V = ui + vj + wk (the velocity vector). The term – V∙ ∇ T ts ca lled the temperature advection. For example, if the wind is blowing from a cold region toward a warm region – V∙ ∇ T will be negative (cold advection) and the advection term will contribute negatively to the local temperature change. Thus, the local rate of change of temperature equals the rate of change of temperature following the motion (that is, the heating or cooling of individual air parcels) plus the advective rate of change of temperature. The relationship given for temperature in equation 2.1 holds for any of the field variable (pressure, density, temperature, etc.). Furthermore, the total derivative can be defined following a motion field other than the actual wind field.

59
For example, we may wish to relate the pressure change measured by a barometer on a moving ship to the local pressure change. Example: 2.1 :The surface pressure decreases by 0.3kPa/180km in the east-ward direction. A shop steaming eastward at 10km/hr measures a pressure fall of 0.1kPa/3hr. What is the pressure change on an island which the ship is passing? Solution: If we take x-axis oriented eastwards, then the local rate of change of pressure on the island is ∂p/∂t = dp/dt - u ∂p/∂x where dp/dt is the pressure change observed by the ship and u the velocity of the ship. Thus, ∂p/∂t = – 0.1kPa/3hr – (10km/hr)( – 0.3kPa/180km) = – 0.1kPa/6hr Thus the rate of pressure fall on the island is only half the rate measured on the moving ship. If the total derivative of a field variable is zero, then that variable is a conservative quantity following the motion.

60
The local change is then entirely due to advection. Field variables that are approximately conserved following the motion play an important role in dynamic meteorology. Example 2.2: A cold front has just passed a station and the temperature is 10°C and falling at a uniform rate of 3°/hr. The wind is blowing straight out of the north at 40km/hr, and the vertical velocity is zero. At a station located 100km to the north, the temperature is – 2°C. Estimate the time rate of change of the temperature of air parcel as they move southward behind the front. Solution: We begin by expanding the total derivative of T w.r.t. time. dT/dt = ∂T/∂t = u ∂T/∂x + v ∂T/∂y + w ∂T/∂z but the wind is blowing straight out of the north and the vertical velocity is zero, ∴ u = 0 and w = 0. i.e. dT/dt = ∂T/∂t + v ∂T/∂y, We need dT/dt

61
On the basis of the information given in the problem we have no choice but to assume that the meridional wind velocity and the temperature gradient are uniform in the vicinity of the two stations. Substituting these into the equation, we obtain, dT/dt = – 3°C/1hr + (- 40km/hr){( - 2) – (+10)}°C/100km Since v = - 40km/hr (from the north) i.e. dT/dt = ( )°C/hr = 1.8°C/hr (warming) We conclude that the air is warming as it moves southward because the temperature at the station is not falling as fast as it would be if horizontal advection were the only process acting. Assignment 2.1: What is the rate of change of temperature in a thermometer shelter if the wind blows from the east at a speed of 5.2m/s and the temperature decreases from west to east by 5°C/100km? Assume that the air mass is moving isothermally. Is the change an increase or a decrease?

62
Assignment 2.2: The rate of change of pressure following the motion of a parcel of air is given by dp/dt = ∂p/∂t + u∂p/∂x + v∂p/∂y + w∂p/∂z (a) Given that in the tropics: ∂p/∂t = 3hPa/24hr, u ~ v ~ 10m/s, ∂p/∂x ~ ∂p/∂y ~ 5hPa/1000km, w ~ 1cm/s, ∂p/∂z ~ 1hPa/10m Show that dp/dt = ω ≈ w∂p/∂z ≈ – ρgw. Given that in the middle latitudes, ∂p/∂t = 3hPa/3hr, u ~ v ~ 10m/s, ∂p/∂x ~ ∂p/∂y ~ 10hPa/1000km, w ~ 1cm/s, ∂p/∂z ~ 1hPa/10m Show also that dp/dt = ω ≈ w∂p/∂z ≈ – ρgw.

63
2.2 Total Derivative of a Vector in Rotating System The mathematical statement of Newton’s second law of motion is called the momentum equation. Since momentum is a vector quantity, the momentum equation is a vectorial equation. Now, we are to transform the momentum equation from an inertial reference frame to a frame rotating with the earth. We then expand the resulting equation into its components in spherical coordinates and show how scale analysis can be used to simplify the component equations for meteorological problems. Total Differentiation of a Vector in a Rotating System The transformation of the momentum equation to a rotating coordinate system requires a relationship between the total derivative of a vector in an inertial reference frame and the corresponding total derivative in a rotating system. To derive this relationship, we let A be an arbitrary vector whose Cartesian components in an inertial frame are given by,

64
A = I A x + j A y + k A z and whose components in a frame rotating with an angular velocity Ω are A = I’ A’ x + j’ A’ y + k’ A’ z Letting d a A/dt be the total derivative of A in the inertial frame, we write, d a A/dt = i dA x /dt + j dA y /dt + k dA z /dt = i’ dA’ x /dt + j’ dA’ y /dt + k’ dA’ z /dt +di’/dt (A’ x ) + dj’/dt (A y ) + dk’/dt (A z ) Now, I’ dA’ x /dt + j’ dA’ y /dt + k’ dA’ z /dt ≡ dA/dt is just the total derivative of A as viewed in the rotating coordinates (that is the rate of change of A following the relative motion). Furthermore, since i’ may be regarded as a position vector of unit length, di’/dt is the velocity of i’ due to its rotation. Thus, di’/dt = Ω x i’ and similarly dj’/dt = Ω x j’ and dk’/dt Ω x k’ and can be written as

65
d a A/dt = dA/dt + Ω x A …………………………………… 2.7 which is the relationship. 2.3 Vectorial form of the Momentum Equation in Rotating Coordinates In an inertial reference frame Newton’s second law of motion may be written symbolically as d a V a /dt = Σ F ………………………………………………. 2.8 The left-hand side represents the rate of change of the absolute velocity V a following the motion as viewed in an inertial system. The right-hand side represents the sum of the real force acting per unit mass. It was earlier found through simple physical reasoning that when the motion is viewed in a rotating coordinate system certain additional apparent forces must be included if Newton’s second law is to be valid. The same result may be obtained by a formal transformation of coordinates in 2.8.

66
In order to transform this expression to rotating coordinates we must first find a relationship between V a and the velocity relative to the rotating system, which we will designate by V. This relationship is obtained by applying eqn. 2.7 to the position vector r for an air parcel on the rotating earth: d a r/dt = dr/dt + Ω x r …………………………………………… But d a r/dt ≡ V a and dr/dt ≡ V; therefore 2.9 may be written as V a = V + Ω x r …………………………………………………………2.10 which states simply that the absolute velocity of an object on the rotating earth is equal to its velocity relative to the earth plus the velocity due to the rotation of the earth. Next we apply eqn. 2.7 to the velocity vector V a and obtain, d a V a /dt = dV a /dt + Ω x V a …………………………………………2.11 substituting from 2.10 into the right-hand side of eqn gives d a V a /dt = d/dt(V + Ω x r) + Ω x (V + Ω x r)

67
d a V a /dt = dV/dt + d/dt(Ω x r) + Ω x V + Ω x (Ω x r) = dV/dt + d/dt(Ω) x r + Ω x dr/dt + Ω x V+ Ω x (Ω x r) Since Ω is assumed a constant vector ∴ d/dt(Ω) = 0, dr/dt = V. ∴ d a V a /dt = dV/dt + Ω x V + Ω x V+ Ω x (Ω x r) Now, from the vector identity, A x (B x C) = (A ∙ C)B – (A ∙ B)C ∴ Ω x (Ω x r) = (Ω ∙ r)Ω – (Ω ∙ Ω)r Ω ∙ r = 0 Since Ω is perpendicular to r, Ω ∙ Ω = Ω 2 ∴ Ω x (Ω x r) = - Ω 2 R Since r is taken as R, the vector perpendicular to the axis of rotation, with magnitude equal to the distance to the axis of rotation, so that we now have, d a V a /dt = dV/dt + 2Ω x V - Ω 2 R ……………………………… Equation 2.12 states that the acceleration following the motion in an inertial system equals the acceleration following the relative motion in a rotating system plus the Coriolis acceleration plus the centripetal acceleration.

68
If we assume that the only real forces acting on the atmosphere are the pressure gradient force, gravitational force and friction, we rewrite Newton’s second law in eqn. 2.8 with the aid of 2.12 i.e. d a V a /dt = Σ F = – (1/ρ)( ∇ p) + g* + F also d a V a /dt = dV/dt + 2Ω x V - Ω 2 R ∴ dV/dt + 2Ω x V - Ω 2 R = – (1/ρ)( ∇ p) + g* + F ∴ dV/dt = – (1/ρ)( ∇ p) – 2Ω x V + g* + Ω 2 R + F But g* + Ω 2 R = g = the gravity force ∴ dV/dt = – (1/ρ)( ∇ p) – 2Ω x V + g + F ……………………………… where F designates the frictional force. Equation 2.13 is the statement of Newton’s second law for motion relative to a rotating coordinate frame. It states that the acceleration following the relative motion in the rotating frame equals the sum of the pressure gradient force, the Coriolis force, the effective gravity, and friction.

69
This is vectorial form of the momentum equation in rotating coordinates. It is this form of the momentum equation which is basic to most work in dynamic meteorology. 2.4 Component Equations in Spherical Coordinates For purposes of theoretical analysis and numerical prediction, it is necessary to expand the vectorial momentum equation 2.13 into its scalar components. Since the departure of the shape of the earth from sphericity is entirely negligible for meteorological purposes, it is convenient to expand 2.13 in spherical coordinates so that the (level) surface of the earth corresponds to a coordinate surface. The coordinate axes are then (λ, Φ, z) where λ is longitude, Φ is latitude, and z is the vertical distance above the surface of the earth. If the unit vectors i, j, k are now taken to be directed eastward, northward and upwards, respectively, the relative velocity becomes,

70
V = iu + jv + kw where the components u, v, and w are defined as follows, u ≡ r cosΦ dλ/dt, v ≡ r dΦ/dt, w ≡ dz/dt …………………… This could be illustrated with Fig. 2.1 below, a z Φ Φ dλdλ rcos Φ rcos Φdλ=dx r r Φ a dΦdΦ z dy = rd Φ ∴ dy/dt=rd Φ/dt dz/dt = w, dz=dz 2.1a2.1b Fig. 2.1a and b: Velocity components u, v, w in spherical coordinates. r = a + z

71
Here, r is the distance to the centre of the earth, which is related to z by r ≡ a+z, where a is the radius of the earth. Traditionally, the variable, r, in 2.14 is replaced by the constant,a. This is a very good approximation since z << a for the regions of the atmosphere with which the meteorologist is concerned. For notational simplicity, it is conventional to define x and y as eastward and northward distance, such that dx = acos Φ d λ and dy = ad Φ. Thus, the horizontal velocity components are u ≡ dx/dt, and v ≡ dy/dt in the eastward and northward directions, respectively. The (x, y, z) coordinate system defined in this way ia not, however, a Cartesian coordinate system because the directions of the i, j, k unit vectors are not constant, but are functions of position on the spherical earth. This position dependence of the unit vectors must be taken into account when the acceleration vector is expanded into its components on the sphere. Thus, we write,

72
dV/dt = d/dt (iu + jv + kw) i.e. dV/dt = idu/dt + jdv/dy + kdw/dt + udi/dt + vdj/dt +w dk/dt 2.15 In order to obtain the component equations, it is necessary first to evaluate the rates of change of the unit vectors following the motion. We first consider di/dt. Expanding the total derivative as in eqn. 2.1 and noting that i is a function only of x (that is, an eastward directed vector does not change its orientation if the motion is in the north- south or vertical direction), we get, di/dt = ∂i/∂t + u∂i/∂x + v∂i/∂y + w∂i/∂z ∴ di/dt = u∂i/∂x δiδi i i i+δii+δi δλ δxδx Fig. 2.2: Longitudinal dependence of the unit vector i.

73
From Fig. 2.2, we see that, ∂i/∂x ≈ δi/δx, δx = acosΦδλ, δi = iδλ ∴ δi/δx = iδλ/acosΦδλ = i/acosΦ ∴ | δi|/δx = 1/acos Φ ∴ Lim | δi|/δx = |∂i/∂x| = 1/acos Φ δx 0 and that the vector ∂i/∂x is directed towards the axis of rotation. Thus, as illustrated in Fig. 2.3, we have, Ω Φ Φ a - kcos Φ δiδi jcos Φ Fig. 2.3: Resolution of δi northward and vertical components ∴ di/dt = u ∂i/∂x = (u/a cos Φ )(j sinΦ – k cosΦ) ……………………

74
Considering now dj/dt, we note that j is a function only of x and y. i.e. dj/dt = ∂j/∂t + u∂j/∂x + v ∂i/∂y + w ∂j/∂z dj/dt = u ∂j/∂x + v ∂i/∂y Φ δλ l l j j+δjj+δj δjδj l = a/tan Φ δλ Φ δxδx Ω δΦ a j J+δjJ+δj δjδj δy=aδΦ Ω Fig. 2.4: The dependence of unit vector j on longitude Fig. 2.5: The dependence of unit vector j on latitude

75
Now, considering u ∂ j/ ∂ x, we first take ∂ j/ ∂ x, Now, ∂ j/ ∂ x ≈ δj/δx With the aid of Fig. 2.4, we see that, δj = jδλ ∴ δj/δx = jδλ/δx Also, from Fig. 2.4, we see that δλ = δx/l But, l = acosΦ/sinΦ = a/tanΦ (from a particular rt. ∠ ed. ∆ ∴ δλ = δx/(a/tan Φ ) ∴ δλ/δx = 1/(a/tan Φ ) = tanΦ/a Since the vector ∂ j/ ∂ x is directed in the negative x direction, we have, ∂ j/ ∂ x ≈ δj/δx = – i δλ/δx =– (tan Φ i)/a ∴ ∂ j/ ∂ x = – (tan Φ i)/a ∴ u ∂ j/ ∂ x = – (u tanΦ)i/a Considering v ∂ j/ ∂ y, we first take ∂ j/ ∂ y, Now, from Fig. 2.5, it is clear that for northward motion |δj|=δ Φ. But δy =a δ Φ and δj is directed downward, so that, δj/δy = – kδ Φ /aδ Φ = – k/a ∴ ∂ j/ ∂ y = – k/a ∴ v∂ j/ ∂ y = – vk/a Hence, dj/dt = u ∂ j/ ∂ x + v∂ j/ ∂ y = – (u tanΦ)i/a – vk/a …………. 2.17

76
Considering now, dk/dt, we note that k is a function of only x and y. i.e. dk/dt = ∂k/∂t + u ∂k/∂x + v ∂k/∂y + w∂k/∂z i.e. dk/dt = u ∂k/∂x + v ∂k/∂y δλ a a δxδx k k k+δkk+δk δkδk δΦ a a δyδy k k δkδk k+δkk+δk Fig. 2.6: The dependence of the unit vector k on longitude Fig. 2.7: The dependence of the unit vector k on latitude Now, δx = aδy ≠acosΦδλ. Unit vector k is from centre of the earth to the atmosphere and not from the axis of rotation.

77
Now, from Fig. 2.6, it is clear that for eastward motion | δk|= δλ. But δx =aδλ and δk is directed eastward, so that, | δk|/ δx = δλ/aδλ = 1/a ∴ ∂ k/ ∂ x = i/a ∴ u∂ k/ ∂ x = ui/a From Fig. 2.7, it is clear that for northward motion | δk|=δΦ. But δy = a δΦ and δ k is directed northward, so that, | δk|/ δy = δΦ/ a δΦ = 1/a ∴ ∂ k/ ∂ y = j/a ∴ v∂ k/ ∂ y = vj/a ∴ dk/dt = u∂ k/ ∂ x + v∂ k/ ∂ y = ui/a + vj/a ……………………… 2.18 Now, substituting eqns to 2.18 into the left hand side of eqn. 2.13, we have, dV/dt = idu/dt + jdv/dt + kdw/dt + u{(u/acos Φ)( jsinΦ – kcos Φ)} + v(– uitanΦ/a – vk/a) + w(ui/a + vj/a) dV/dt = idu/dt– uitanΦ/a + uwi/a + jdv/dt + u 2 ( jsinΦ)/acos Φ +vwj/a + kdw/dt – u 2 kcos Φ /acos Φ – v 2 k/a dV/dt = (du/dt – uitanΦ/a + uwi/a)i + (dv/dt + u 2 tan/a +vw/a)j + {dw/dt – (u 2 + v 2 )/a}k ……………………………………………. 2.19

78
Equation 2.19 is the spherical-polar coordinate expansion of the acceleration following the relative motion. We must turn to the component expansion of the force terms in eqn The Coriolis force is expanded by noting that Ω has no component parallel to i, and that its components parallel to j and k are 2ΩcosΦ and 2ΩsinΦ, respectively. This could be shown in Fig. 2.8 below. Ω Φ Φ Φ Ω j k Ωsin Φ Ωcos Φ Fig. 2.8: Diagram showing the resolution of Ω in the j and k directions. Ω has no component in the i direction since it is perpendicular to i direction.

79
Thus, using the definition of the vector cross-product, we have, i j k –2Ω x V = – 2Ω 0 cos Φ sin Φ u v w cos Φ sin Φ 0 sin Φ 0 cos Φ = – 2Ω{ I v w – j u w + k u v } = – 2Ω{(w cos Φ-v sin Φ) I – (0 – usin Φ)j+ (0 - u cos Φ) k ∴ –2Ω x V = (– 2Ωw cos Φ+ 2Ω v sin Φ) I – (2Ωusin Φ)j+ (2Ωucos Φ) k The pressure gradient force may be expressed as, – 1/ρ ( ∇ p)= -1/ρ(i ∂p/∂x + j ∂p/∂y + k ∂p/∂z) ……………………… and gravity is conveniently represented as g = gk ………………………………………………………………… 2.22

80
where g is a positive scalar (g ≈ 9.8ms -1 ) at the earth’s surface. Finally, friction is expanded in components as, F = i Fx + j Fy + k Fz …………………………………………………………… 2.23 Therefore, substituting equations 2.19 – 2.23 into 2.13 and equating all terms in the i, j, and k directions, respectively, we have, du/dt –uvtanΦ/a+uw/a = -(1/ρ)∂p/∂x+2ΩvsinΦ -2ΩwcosΦ + F x …2.24 dv/dt + u 2 tanΦ + vu/a = -(1/ρ)∂p/∂y –2 ΩusinΦ + F y dw/dt – (u2+v 2 )/a = -(1/ρ)∂p/∂z - g - 2 ΩucosΦ + F z Accn. Curvature term p.g.f. grav. Coriolis term Friction which are the eastward, northward, and vertical component momentum equations, respectively.

81
The terms proportional to 1/a on the left hand sides in eqns – 2.26 are called the curvature terms because they arise due to the curvature of the earth. It can be shown that when “r” is replaced by “a” as done here – the traditional approximation - the Coriolis terms proportional to cosΦ in 2.24 and 2.26 must be neglected if the equations are to satisfy angular momentum conservation; because the curvature terms are non-linear terms (i.e. they are quadratic in the dependent variable), they are difficult to handle in theoretical analyses. Fortunately, as will be shown later, the curvature terms are un- imporant for mid-latitude synoptic scale motions. However, even when the curvature terms are neglected eqns – 2.26 are still non-linear partial differentialbequations as can be seen by expanding the total derivatives into their local and advective parts:

82
du/dt = ∂ u/ ∂ t + u ∂ u/ ∂ x + v ∂ u/ ∂ y +w ∂ u/ ∂ z with similar expression for dv/dt and dw/dt. In general the advective acceleration terms are comparable in magnitude to the local acceleration. It is primarily the presence of non-linear advection processes that makes dynamic meteorology an interesting and challenging subject. 2.5 Scale Analysis of the Equations of Motion Scale analysis or scaling is a convenient technique for estimating the magnitudes of various terms in the governing equations for a particular type of motion. In scaling, typical expected values of the following quantities are specified: The magnitude of the field variable The amplitude of fluctuations in the field variable and The characteristic length, depth and time scales on which these fluctuations occur.

83
These typical values are then used to compare the magnitudes of various terms in the governing equation to see if the terms are negligible for motions of meteorological concern. Elimination of terms – which are small – on scaling considerations not only has the advantage of simplifying the mathematics but in some cases has the very important property of completely eliminating or filtering, an unwanted type of motion. The complete equations of motion 2.24 – 2.26 describe all types and scales of atmospheric motions. Sound waves, for example, are a perfectly valid solution to these equations. However, sound waves are of negligible importance in meteorological problems. Therefore, it will be a distinct advantage if, as turns out to be true, we can neglect the terms which lead to sound wave type solutions and filter out this unwanted class of motions.

84
In order to simplify eqns – 2.26 for synoptic scale motions we define the following characteristic scales of the field variables based on observed values for middle latitude synoptic systems: u = v = U ∼ 10m/s horizontal velocity scale w = W ∼ 1cm/s = 0.01m/s vertical velocity scale x = y = L ∼ 10 6 m length scale ( ∼ 1/2π wavelength) z = D ∼ 10 4 m depth scale ∂ p/ρ ≈ ∆p/ρ ∼ 10 3 m 2 s -2 horizontal pressure fluctuation scale t = x/u = y/v = L/U ∼ 10 5 s time scale The horizontal pressure fluctuation ∆p is normalized by the density ρ in order to produce a scale estimate which is valid at all height in the troposphere despite the approximate exponential decrease with height of both ∆p and ρ. Note that ∆p/ρ has units of geopotential. Referring eqn. 1.8 we see that indeed the magnitude of the fluctuation of ∆p/ρ on a surface of constant height must equal the magnitude of the fluctuation of geopotential on an isobaric surface.

85
The time scale here is an advective time scale which is appropriate for pressure systems which move at approximately the speed of the horizontal wind, as is observed for synoptic scale motions. Thus, L/U is the time required to travel a distance L at a speed U. It should be pointed out here that the synoptic scale vertical velocity is not a directly measurable quantity. However, we show later that the magnitude of W can be deduced from knowledge of the horizontal velocity field. We can estimate the magnitude of each term in eqns and 2.25 for synoptic scale motions at a given latitude. It is convenient to consider disturbances centered at latitude Φ o = 45 ° N, and introduce the notation: f o = 2 Ω sinΦ o = 2 Ω cosΦ o ≈ s -1 The radius of the earth, a = 6.37x10 6 m = 0.637x10 7 m ≈ 1x10 7 m

86
Table 2.1: Scale Analysis of the Horizontal Momentum Equation ABCDEF X-component momentun equation du dt -2 Ω vsinΦ+2 Ω wcosΦ +uw a - uvtanΦ a = - 1∂p ρ∂x Y-component momentum equation dv dt +2 Ω usinΦ –+vw a + u 2 tanΦ a = - 1∂p ρ∂y Scales of individual terms U2LU2L foUfoUfoWfoWUW/aU 2 /a ∆pρL∆pρL Magnitude of the terms (N/kg or m/s /10 6 = = = = /10 7 = /10 6 = 10 -3

87
Table 2.1 shows the characteristic magnitude of each term in eqns and 2.25 based on the above scaling considerations. Frictional terms are not included because for the synoptic time scale frictional dissipation is thought to be of secondary importance above the lowest kilometer. 2.6 The Geostrophic Approximation and the Geostrophic Wind It is apparent from Table 2.1 that for mid-latitude synoptic scale disturbances the Coriolis (term B) and the pressure gradient force (term F) are in approximate balance. Therefore, retaining only these two terms in eqns and 2.25, we obtain as a first approximation the geostrophic relationship – 2 Ω vsinΦ = – fv ≈ – 1/ρ(∂p/∂x) …………………………………… a 2 Ω usinΦ = fu ≈ – 1/ρ(∂p/∂y) …………………………………… b where f ≡ 2 Ω sinΦ is called the Coriolis parameter. i.e. v G = (1/ρf) (∂p/∂x)……………………………………………………..2.27c u G = – (1/ρf) (∂p/∂y) ……………………………………………………..2.27d

88
The geostrophic balance is a diagnostic expression which gives the approximate relationship between the pressure field and horizontal velocity in synoptic scale system. The approximation, eqn. 2.27, contains no reference to time and therefore cannot be used to predict the evolution of the velocity field. It is for this reason that the geostrophic relationship is called a diagnostic relationship. By analogy to the geostrophic approximation (2.27) it is possible to define a horizontal velocity field, V G ≡ iu G + jv G, called the geostrophic wind which satisfies eqn identically. This geostrophic wind is that horizontal equilibrium wind associated with a two-way balance of horizontal pressure gradient force and the horizontal component of Coriolis force. Thus, in vectorial form, V G ≡ k x (1/ρf) ∇ p ……………………………………………… In height natural coordinates, V = – (1/ρf) ∂p/∂n (‘n’ is to the left, towards low pressure)..2.28b V = (1/ρf) ∂p/∂n (‘n’ is to the right, towards high pressure).2.28c

89
Thus, knowledge of the pressure distribution at any time determines the geostrophic wind. It should be kept clearly in mind that eqn always defines the geostrophic wind; but for large scale motions should the geostrophic wind be used as an approximation to the actual horizontal wind field. For the scales used in Table 2.1 the geostrophic wind approximates the true horizontal velocity to within 10-15% in mid-latitudes. 2.7 Approximate Prognostic Equations: The Rossby Number To obtain prediction equations, it is necessary to retain the acceleration (term A) in eqns and The resulting approximate horizontal momentum equations are: du/dt – fv = – (1/ρ)∂p/∂x …………………………………………… 2.29 dv/dt + fu = – (1/ρ)∂p/∂y …………………………………………… 2.30

90
Our scale analysis showed that the acceleration terms eqns and 2.30 are about an order of magnitude smaller than the Coriolis force and pressure gradient force. The fact that the horizontal flow is in approxiimate geostrophic balance is helpful for diagnostic analysis. However, it makes actual application of these equations in weather prognosis difficult because acceleration (which must be measures accurately) is given by the small difference between two large terms. Thus, a small error in measurement of either velocity or pressure gradient will lead to very large errors in estimating the acceleration. A convenient measure of the magnitude of the acceleration compared to the Coriolis force may be obtained by forming the ratio of the characteristic scales for the acceleration and the Coriolis force terms, (U 2 /L)/f o U.

91
This ratio is a non-dimensional number called Rossby number after the Swedish meteorologist Carl Gustav Rossby ( ) and is designated by Ro ≡ U/f o L Thus, the smallness of the Rossby number is a measure of the validity of the geostrophic approximation. 2.8 The Hydrostatic Approximation A similar scale analysis can be applied to the vertical component of the momentum equation Since pressure decreases by about an order of magnitude from the ground to the tropopause, the vertical pressure gradient may be scaled by P o /H where P o is the surface pressure and H is the depth of the troposphere. The term in 2.26 may then be estimated for synoptic scale motions and are shown in Table 2.2 below. As with the horizontal component equation we consider motions centered at 45 ° latitude and neglect friction.

92
The scaling indicates that to a high degree of accuracy the pressure field is in hydrostatic equilibrium, that is, the pressure at any point is simply equal to the weight of a unit cross-section column of air above that point. Table 2.2: Scale Analysis of the Vertical Momentum Equation ABCDE z component momentum equation dw dt – 2 ΩucosΦ –(u 2 +v 2 ) a = – 1 ∂p ρ∂z – g Scales of individual terms UW L f o UU2LU2L PoρHPoρH g Magnitude of the terms (m/s) 10 1, = = = = = 10

93
However, in the above analysis, it is not sufficient to show merely that the vertical acceleration is small compared to g. Since only that part of the pressure field which varies horizontally is directly coupled to the horizontal velocity field, it is actually necessary to show that the horizontally varying pressure component is itself in hydrostatic equilibrium with the horizontally varying density field. To do this it is convenient to first define a standard pressure p o (z), which is the horizontal averaged pressure at each height, and a corresponding standard density ρ o (z), defined so that p o (z) and ρ o (z) are in exact hydrostatic balance: (1/ρ)dp o /dz ≡ – g ………………………………………………… We may then write the total pressure and density fields as p(x,y,z,t) = p o (z) + p’(x,y,z,t) …………………………………………2.32a ρ(x,y,z,t) = ρ o (z) + ρ’(x,y,z,t) …………………………………………2.32b where p’ and ρ’ are perturbations from the standard values of pressure and density.

94
For an atmosphere at rest, p’ and ρ’ would thus be zero. Using the definition (eqn. 2.31) and (eqn. 2.32) and assuming that ρ’/ρ o is much less than unity in magnitude so that (ρ o + ρ’) -1 ≈ ρ o (1– ρ’/ρ o ) we find that – (1/ρ)∂p/∂z – g = – 1/(ρ o + ρ’)∂/∂z(p o + p’) – g = – (ρ o + ρ’) -1 ∂/∂z(p o + p’) – g = – [ρ o (1+ ρ’/ρ o )]-1(∂p o /∂z+ ∂p’/∂z) – g = – [ρ o -1 (1+ρ’/ρ o ) -1 ] (∂p o /∂z+ ∂p’/∂z) – g ≈ – ρ o -1 (1- ρ’/ρ o )(∂p o /∂z+ ∂p’/∂z) – g = – ρ o -1 (1- ρ’/ρ o )∂p o /∂z - ρ o -1 (1- ρ’/ρ o )∂p’/∂z – g = – ρ o -1 ∂p o /∂z +(ρ’/ρ 2 o )∂p o /∂z – (ρ o -1 ) ∂p’/∂z +(ρ’/ρ 2 o ) ∂p’/∂z – g But ∂p o /∂z = – ρ o g and ρ’∂p’/∂z ≈ 0 ∴ – (1/ρ)∂p/∂z – g = – ρ o -1 (– ρ o g) + (ρ’/ρ 2 o )(– ρ o g) – ρ o -1 ∂p’/∂z – g- = g - ρ’/ρ o g - ρ o -1 ∂p’/∂z – g ∴ – (1/ρ)∂p/∂z – g = – ρ o -1 (ρ’g + ∂p’/∂z) ………………………………….2.33

95
For synoptic scale motions, the terms in eqn have the magnitudes, ρ o -1 ∂p’/∂z ∼ [∆p/ρH ] ∼ 10 3 /10 4 ms -2 ∼ ms -2 ρ’g/ρ o ∼ ms -2 Comparing these with the magnitudes of other terms in the vertical momentum equation (Table 2.2), we see that to a very good approximation the perturbation pressure field is in hydrostatic equilibrium with the perturbation density field so that ∂p’/∂z + ρ’g = 0 ………………………………………………… 2.34 Therefore, for synoptic scale motions, vertical acceleration are negligible and the vertical velocity cannot be determined from the vertical momentum equation. Nevertheless, it is possible to deduce the vertical motion field indirectly from the continuity equation (kinematic method) and from thermodynamic energy equation (adiabatic method) to be treated later.

96
3.0 THE BASIC EQUATION AND GEOSTROPHIC WIND The approximate horizontal momentum equation 2.29 and 2.30, in height Cartesian coordinates, i.e. du/dt – fv = – (1/ρ) ∂p/∂x dv/dt + fu = – (1/ρ) ∂p/∂y may be written in vectorial form as, dV H /dt + fk x V H = – (1/ρ) ∇ p ……………………………………. 3.1 where V H = iu + jv is the horizontal wind velocity vector. In order to express 2.35 in isobaric coordinate form, we transform the pressure gradient force using eqs. 1.9a and 1.9b, i.e. P x = - (1/ρ)(∂p/∂x) = – g (∂z/∂x) = – ∂φ/∂x = – g o (∂Z/∂x) P y = - (1/ρ)(∂p/∂y) = – g (∂z/∂y) = – ∂φ/∂y = – g o (∂Z/∂y) to obtain, dV H /dt + fk x V H = – ∇ p Φ ……………………………………………. 3.2 where ∇ p is the horizontal gradient operator applied with pressure held constant.

97
Since p is the independent vertical coordinate, we must expand the total derivative as follows: d/dt = ∂/∂t + dx/dt (∂/∂x) + dy/dt (∂/∂y) + dp/dt (∂/∂p) = ∂/∂t + u (∂/∂x) + v (∂/∂y) + ω (∂/∂p) …………………… 3.3 Here ω = dp/dt (usually called ‘omega’ vertic ωal motion) is pressure change following the motion, which plays the same role in the isobaric coordinate system as w ≡ dz/dt plays in height coordinates. From eqn. 3.2, we see that the isobaric coordinate form of the geostrophic relationship is fV G = k x ∇ p Φ = k x ∇ p Z ………………………………………………… 3.4 One advantage of isobaric coordinate system is easily seen by comparing eqns and 3.4. In the latter equation density does not appear. Thus, a given geopotential gradient implies the same geostrophic wind at any height, whereas a given horizontal pressure gradient implies different values of the geostrophic wind depending on the density.

98
Furthermore, if ‘f’ the Coriolis parameter, is regarded as a constant the horizontal divergence of the geostrophic wind at constant pressure is zero i.e. ∇ p ∙ V G = 0. The geostrophic wind speed in Cartesian pressure coordinates (scalar form) could be wrtten as, u G = – (g o /f) ∂Z/∂y = – (1/f) ∂ Φ /∂y …………………………….. 3.5a v G = (g o /f) ∂Z/∂x = (1/f) ∂ Φ /∂x ………………………………...3.5b In natural coordinates it could be written as, V G = – (g o /f) ∂Z/∂n = – (1/f) ∂ Φ /∂n = – (1/ ρ f) ∂p/∂n..…..3.6 The geostrophic wind equations in height Cartesian, height natural, pressure Cartesian and pressure natural coordinates are given in Table 3.1. The balance of horizontal forces that results in the geostrophic wind fields is illustrated in Fig. 3.1 which applies to the Northern Hemisphere. Assignment 3.1: Show that ∇ p ∙ V G = 0 using equations 3.5a and b.

99
Height Cartesian Coordinates (for use at sea-level) Pressure Cartesian Coordinates (for use at 850hPa and above) x – component velocity u G = – (1/ρf) (∂p/∂y) u G = – (g o /f) ∂Z/∂y = – (1/f) ∂ Φ /∂y y-component velocity v G = (1/ρf) (∂p/∂x) v G = (g o /f) ∂Z/∂x = (1/f) ∂ Φ /∂x Height Natural Coordinates (for use at sea-level) Pressure natural Coordinates (for use at 850hPa and above) Total horizontal velocity V G = – (1/ρf) ∂p/∂n ( ‘n’ increases to wards the left, towards low pressure) V G = – (g o /f) ∂Z/∂n = – (1/f) ∂ Φ /∂n (‘n’ increases to wards the left, towards low pressure) Total horizontal velocity V G = (1/ρf) ∂p/∂n (‘n’ increases to wards the right, towards high pressure) V G = (g o /f) ∂Z/∂n = (1/f) ∂ Φ /∂n (‘n’ increases to wards the right, towards high pressure) Table 3.1: Summary – Geostrophic Wind Equations

100
p+ δp p p - δp p – 2 δp L ∙ PnPn C VGVG Fig. 3.1: The geostrophic wind and its relation to the horizontal pressure gradient force, P n and the Coriolois force, C. In order for Coriolis and pressure gradient forces to balance, the geostrophic wind must blow parallel to the isobars leaving low pressure to the left and high pressure to the right in the Northern Hemisphere.

101
It may be noted that, in either hemisphere, the geostrophic wind around a centre of low pressure is always in the same sense as the earth’s rotation. Such flow is said to be cyclonic. Geostrophic flow around high pressure centres is said to be anti-cyclonic. Geostrophic balance is a state which nearly holds (almost) everywhere in the atmosphere but does not exactly hold anywhere. This remark serves to emphasize both the usefulness of the geostrophic wind concept and its restrictiveness. It allows a good estimate to be made of wind speed when no direct observation of winds are available (i.e. north of 8 ° N and south of 8 ° S, above the planetary boundary layer, and away from curved flow of low’s and high’s). This practical use in the estimation of wind velocity will be demonstrated. The geostrophic approximation is also a useful simplifying assumption in theoretical work.

102
However, it implies no acceleration or vertical motion or pressure change and its usefulness is therefore limited in the study of large-scale pressure systems and rain-producing mechanisms. The method of using eqn. 3.6 to deduce wind speed from isobar spacing may be illustrated as follows: Fig. 3.2 shows a supposed distribution of mean-sea-level isobars drawn at 4hPa intervals and at different distances apart. The geostrphic wind equation 3.6 is given by V G = – (1/ρf)∂p/∂n XYZ ∙∙∙ ABC 100 km200 km 1004 hPa1008 hPa1012 hPa Fig. 3.2: Pressure gradient and calculated geostrophic winds

103
The values of ∂p/∂n at A, B, and C are (as always in calculation in meteorology) approximated by finite differences, ∆p/∆n. Now, f = 2ΩsinΦ, Ω = Earth’s angular velocity = 7.29x10 -5 s -1, Φ = latitude = 60 ° N, ρ = 1.2kgm -3, (∆p) A = ( )hPa = - 4hPa = – 400Pa, (∆n) A = 100km = 100x10 3 m ∴ V G (A) = – [(1004 – 1008)x100]/(1.2x2x7.29x10 -5 sin60x2x10 5 ) m/s =26.4m/s ≈ 52.8knots. Similarly, V G (B )= 13.2m/s and V G (C) = 17.6m/s. 3.1 Wind and Pressure near the Equator As the equator is approached, (within 8 ° N and 8 ° S) the geostrophic balance between pressure and wind, which in general holds well in higher latitudes, breaks down because the deflecting force caused by the Earth’s rotation about the vertical tends towards zero there. Tropical cyclones move away from the equator; the horizontal pressure variations close to the equator are small and such pressure fields as exist do not control the air motion in a simple way.

104
The prevailing conditions are of light, variable winds (‘doldrums’). These facts show how vital the Coriolis force is in the development of the large scale pressure and wind fields which are observed over most of the earth. Geostrophic balance also breaks down within sharp troughs and ridges where the balance of three forces instead of two are the case. The balance of three forces in this case is known as the gradient wind balance which is addressed below. 3.2 Gradient Flow and the Gradient Wind Approximation Within sharp troughs in the middle latitude westerlies the observed velocities are often sub-geostrophic by as much as 50% even though the streamlines still tend to be oriented parallel to the isobars. These large departures from geostrophic balance are a consequence of the large centripetal acceleration associated with the sharply curved flow within such regions.

105
In order to illustrate how these accelerations alter the balance between the Coriolis force and the pressure gradient force, it is convenient to represent them in terms of an apparent centrifugal force. The three-way balance between pressure gradient and Coriolis and centrifugal forces is shown in Fig. 3.3 for cyclonic and anti- cyclonically curved trajectories. In both cases, the centrifugal force is directed outward from the center of curvature of the air trajectories (denoted by dashed lines) and has a magnitude given by V 2 /R T, where R T is the local radius of curvature. The strength of the centrifugal force increases with wind speed and decreases as the radius of curvature increases. In the cyclonic case, the centrifugal force reinforces the Coriolis force so that, in effect, a balance of forces can be achieved with a wind velocity smaller than would be required if the Coriolis force were acting alone.

106
L H C V 2 /R T PnPn RTRT C PnPn p- δ p p p+ δ p V gr Fig. 3.3: The three-way balance between the horiziontal pressure gradient force, the Coriolis force and the centrifugal force, in flow along curved trajectories (------) in the Northern Hemisphere.

107
Thus in this case, it is possible to maintain a sub-geostrophic flow parallel to the isobars. For the anti-cyclonic curved trajectory, the situation is just opposite: the centrifugal force opposes the Coriolis force and, in effect, necessitates a super-geostrophic wind velocity in order to bring about a three-way balance of forces. Gradient flow is one associated with a three-way balance between Coriolis force, the centrifugal force, and the horizontal pressure gradient force. Like geostrophic flow, pure gradient flow can exist only under very special circumstances. The wind associated with a three-way balance between pressure gradient, Coriolis and centrifugal forces is called the gradient wind. Gradient wind at any point is just the wind component parallel to the isobars which satisfy the so called gradient wind equation for cyclonic and anti-cyclonic cases:

108
V 2 /R T + fV = – (1/ρ)∂p/∂n ………………………………………………. 3.7 i.e. V 2 + fR T V + (R T /ρ)∂p/∂n = 0 Using the general formula for quadratic equation to solve this equation, we have, V = – fR T /2 ± (f 2 R 2 T – (R T / ρ ) ∂p/∂n) ½ …………………………… Since the above equation 3.7 takes account of the centrifugal force due to the curvature of the parcel trajectories, the gradient wind is often a better approximation to the actual wind than the geostrophic wind. For the combination of sharp anti-cyclonic curvature and strong pressure gradient it is not possible to achieve a three-way balance of forces, that is, P n > fV – V 2 /R T ……………………………..……………………. 3.9 It is no coincidence then, that the combination of sharp troughs and tight pressure gradients is not uncommon, whereas sharp ridges are rarely if ever observed in conjunction with tight pressure gradients.

109
3.4 Cyclostrophic Flow In small scale system, the radius of rotation may be of the order of tens of meters as compared with the usual systems portrayed on weather maps in which the radius of rotation may be measured in thousands of kilometers. For this reason, the centrifugal term becomes much larger in such systems. Since the centrifugal term varies as the square of the speed and the Coriolis term as the first power of the speed, the Coriolis force is negligible even for high speeds. In such situations a balance of forces is to be sought between pressure gradient and centrifugal effects alone: i.e. V 2 /R T = – (1/ρ)∂p/∂n ……………………………………………… 3.10 The flow will always be around a low pressure centre and since the Coriolis effect is negligible the sense of rotation may be either cyclonic or anti-cyclonic as shown below Fig. 3.4.

110
L V 2 /R T V V L PnPn PnPn Fig. 3.4: The force balance in cyclostrophic flow P n designates the pressure gradient, V 2 /R T the centrifugal forces. This type of motion is called cyclostrophic and is to be expected in low latitudes where ‘f’ is small. Cyclostrophic flow is therefore defined as the markedly curved flow resulting from the two-way balance between the centifugal force and the pressure gradient force, Coriolis force being negligible. If the equation 3.10 is solved for V, we obtain the speed of the cyclostrophic wind: R > 0, ∂p/∂n < 0 R 0

111
V = [ – (R T /ρ) ∂p/∂n] ½ ……………………………………………… 3.11 The cyclostrophic balance approximation is valid provided that the ratio of the centrifugal force to the Coriolis force is large i.e. V/fR T which is the Rossby number is large. As an example of cyclostrophic scale motion we consider a typical tornado. Suppose that the tangential velocity is 30ms -1 at a distance of 300m from the center of the vortex. Assuming that f ≈ s -1 we obtain, Rossby number, R o = V/fR T ≈ 10 3 Therefore, the Coriolis force can be neglected in computing the balance of forces for a mature tornado. However, the majority of tornadoes in the Northern Hemisphere are observed to rotate in a cyclonic sense (anti-clockwise). Thus, in the initial stages of development of a tornado the Coriolios force serves to deflect air parcels accelerating towards the center of low pressure to the right so that a cyclonic circulation is established about the pressure minimum.

112
Smaller scale vortices, on the other hand, such as dust devils and water spouts, do not have preferred direction of rotation. The flow near the center of hurricane is usually close to cyclostrophic balance, whereas in typical mid-latitude disturbances it is usually closer to geostrophic balance. Both geostrophic and cyclostrophic balance may be viewed as special cases of gradient wind balance that obtain in the limit when one of the forces in the three-way balance becomes negligible in comparison to the other forces. 3.5 The Effect of Friction In deriving the expression for the speed of the geostrophic wind and the gradient wind, we have ignored friction. Sufficiently far from the Earth’s surface it is small enough to be ignored, but close to the ground the frictional force cannot be neglected. A rough value for the thickness of the frictional influence is 1km.

113
Within this lowest kilometer the frictional retardation is continually tending to make the wind speed sub-geostrophic, so that the Coriolis force is not quite strong enough to balance the horizontal pressure force, which is acting to push the air across the isobar from high towards low pressure (or from high towards low geopotential height). The excess of pressure gradient force over the Coriolis force pushes the parcel of air towards lower pressure. Equilibrium is established with the frictional force equal and opposite to the resultant of the other two forces – pressure gradient and Coriolis forces. The resultant wind blows at an acute angle across the isobar toward the low pressure as against being parallel in the geostrophic case. The three-way balance of forces required for a uniform flow (dv/dt = 0) in the Northern Hemisphere in the presence of friction is illustrated in Fig. 3.5 below.

114
PnPn V F C α α p-2 δ p p- δp p p+ δp L H Fig. 3.5: The three way balance of forces (pressure gradient, P n, Coriolis, C, and Friction, F) required for uniform flow in the presence of friction in the Northern Hemisphere. The light lines represent isobars in a horizontal plane.

115
The pressure gradient force P n is directed normal to the isobars. The Coriolis force C is directed to the right of the velocity vector V and the frictional force is directed opposite to the velocity vector. The angle between V and V G (denoted by α ) is determined by the requirement that C be just large enough to balance the component of P n in the direction normal to the velocity vector. The large the frictional force the larger the angle α and the more subgeostrophic the wind speed is. The observed tendency for the winds to blow parallel to the isobars above the lowest kilometer is evidenced of the smallest of the frictional force at these levels. Now, from Fig. 3.5, and considering the triangle of forces involving P n, C and F, we have, F = P n sin α ………………………………………………………………… C = fV= Pn cos α ……………………………………………………………3.13 ∴ F/fV = P n sin α /(P n cos α) = tan α ………………………………… 3.14

116
Example 3.1 At a certain station located 12N, the surface wind has the speed of 4.5m/s and is directed across the isobar from high to low pressure. If the magnitude of the frictional drag force is 9.5x10 -5 N/kg, calculate the magnitude of the pressure gradient force and the cross-isobaric angle. (Take Ω= 7.29x10 -5 s -1 ). Solution: The required equation for the solution is, ∴ F/fV = P n sin α /(P n cos α) = tan α ∴ F = fV tan α i.e. 9.5x10 -5 = 2x7.29x10 -5 x4.5sin12 tan α ∴ tan α = 9.5x10 -5 /(2x7.29x10 -5 x4.5sin12) = ∴ α = tan-1{9.5x10 -5 /(2x7.29x10 -5 x4.5sin12)} = 34.9 ° The pressure gradient force is got by substitution of this value into the equation, F = P n sin α, P n = F/sin α = 9.5x10 -5 /sin34.9 N/kg i.e. P n = 1.6x10 -4 N/kg

117
3.6 Ageostrophic Wind, V ag This is also the vector difference between the geostrophic and the actual winds, i.e. the actual wind vector V, minus the geostrophic wind vector V G. It is the geostrophic departure. See Fig. 3.6 below. Ageostrophic wind is of fundamental importance in that it is necessarily associated with convergence or divergence and vertical motion in the atmosphere. This will be dealt with later. Surface friction, pressure gradient change with time (isallobaric wind) and air movement on a curved path (gradient wind) are examples of conditions in which there is an ageostrophic wind. V VGVG V ag Fig. 3.6: Relationship between actual wind, V, geostrophic wind V G, and ageostrophic wind V ag.

118
Isallobaric wind: is that part of the ageostrophic wind which is associated with local time change of the pressure gradient. The isallobaric wind rarely exceeds 5m/s but is at times important in its effects on horizontal convergence and rainfall. 3.7 Variation of Pressure Gradient with Height Application of hydrostatic equation in the form – ∆ p = (pg/RT) ∆ z shows that where the temperature is high the pressure decreases slowly with height, and where it is low the pressure decreases rapidly. At higher levels the pressure gradient is the result of the pressure gradient at sea level plus a contribution due to the horizontal gradient of the average temperature in the layer up to the particular level. The change in pressure gradient as one goes upward that is due to the horizontal variation of temperature is demonstrated in Fig. 3.7

119
WARM COLD WARM COLD 1000hPa 500hPa 300hPa 200hPa 100hPa Tropopause Fig. 3.7: Schematic changes of slope of isobaric surfaces in middle latitude resulting from the average distribution of temperature in the northern hemisphere. The mean isotherms are assumed to be normal to the slide parallel to latitude lines.

120
This figure is a vertical cross-section in which the temperature increases from left to right (from North to the Equator) at all levels in the troposphere. On the left, where the air is cold, the pressure decreases more rapidly than on the right, where the air is warmer. Thus, the distance to the position where the pressure is 500hPa is smaller on the left than on the right and, similarly, for the distance from 500hPa to 300hPa. The slope of the isobars in the cross-section increases with height and so does the pressure difference at constant elevations. Thus, horizontal temperature gradient leads to a pressure gradient at upper levels. Since, on the average, cold air lies on the poleward side at all levels in the troposphere (Fig. 3.7) there is horizontal pressure gradient force directed from low to high latitudes and increasing with height, upwards to the tropopause. Thus, the westerly component of wind increases with height (low pressure being to the left of the wind in the Northern Hemisphere).

121
Above the tropopause the temperature gradient is generally in the opposite sense (not near the winter pole) so that the westerly component initially decreases with height and may change to easterly. Since the mean isotherms are assumed in Fig. 3.7 to be parallel to the latitude lines, i.e. normal to the slide, then the general rule applies that the shear vector in geostrophic flow (thermal wind) is parallel to the mean isotherms of the layer concerned and has low temperature on the left in the northern hemisphere. It is also clear that the larger the horizontal gradient of temperature the bigger is the difference of slope of adjacent isobaric surfaces and the stronger is the shear through a given layer. The geostrophic shear vector has low temperature on the right there because of the reversed sense of the relationship between pressure and wind direction in the southern hemisphere. However, this is easily seen to imply the same direction of shear (westerly in the troposphere, easterly above) as in the northern hemisphere.

122
3.8 Vertical Shear of the Geostrophic Wind: Thermal Wind The equations for the rate of change of the geostrophic wind component with height are derived using the isobaric coordinate system. Recall that in isobaric coordinates the geostrophic wind components are, u G = – (1/f) ∂Φ/∂y = – (go/f)∂Z/∂y and …………………………… v G = (1/f) ∂Φ/∂x = (go/f)∂Z/∂x ……………………………………… 3.16 where the derivative are evaluated with pressure held constant. Also with the aid of ideal gas law we can write the hydrostatic equation in terms of the geopotential. Now, ∂p/∂z = -ρg ∴ ∂z/∂p = – 1/(ρg) i.e. g∂z/∂p = – 1/ρ i.e. ∂ Φ /∂p = – 1/ρ = – α = – RT/p ……………………………………… 3.17 Differentiating, eqns and 3.16 with respect to pressure, we have,

123
∂u G /∂p = ∂/∂p{ – (1/f)∂Φ/∂y} and ∂v G /∂p = ∂/∂p{ (1/f)∂Φ/∂x} ∂u G /∂p = ∂/∂y{- (1/f)∂Φ/∂p} and ∂v G /∂p = ∂/∂x{(1/f)∂Φ/∂p} Applying eqn. 3.17, we have, ∂u G /∂p = ∂/∂y{(1/f) RT/p} and ∂v G /∂p = ∂/∂x {- (1/f) RT/p} i.e. ∂u G /∂p = R/(fp)(∂T/∂y) p – RT/(fp 2 )(∂p/∂y) T ∂v G /∂p = – R/(fp)(∂T/∂x) p + RT/(fp 2 )(∂p/∂x) T ∴ ∂u G /∂p = R/(fp)(∂T/∂y) p and ∂v G /∂p = – R/(fp)(∂T/∂x) p i.e. p∂u G /∂p = ∂u G /∂(lnp) = R/f (∂T/∂y) p ……………………………… p∂v G /∂p = ∂v G /∂(lnp) = – R/f (∂T/∂x) p …………………………… Or in vectorial form, ∂V G /∂(lnp) = – (R/f)k x ( ∇ T) p …………………………………………… Equations 3,20 is the relationship for the vertical wind shear or vertical shear of the geostrophic wind (i.e. the rate of change of geostrophic wind w.r.t. lnp.

124
Designating the thermal wind vector by V T, we may integrate eqn from pressure level p o to pressure level p 1 (p 1 < p o ) to get, p 1 V T = V G (p 1 ) – V G (p o ) = – R/f (k x ∇ T) p dlnp …………………… p o Letting T m denote the mean temperature in the layer between pressure p o and p 1 and x and y components of the thermal wind are thus given by, p 1 u T i + v T j = – R/f{(k x ∇ T m ) lnp } = – R/f{k x (i∂T m /∂x + j∂T m /∂y)ln(p 1 /p 2 )} p o i.e. u T i + v T j = – R/f{(j∂T m /∂x– i ∂T m /∂y) ln(p 1 /p o )} i.e. u T i + v T j = – (R/f) j (∂T m /∂x) ln(p 1 /p o ) + (R/f) i (∂T m /∂y) ln(p 1 /p o ) i.e. u T i + v T j = (R/f) j (∂T m /∂x) ln(p o /p 1 ) – (R/f) i (∂T m /∂y) ln(p o /p 1 ) Equating components, we have,

125
u T = – (R/f)(∂T m /∂y) p ln(p o /p 1 ) …………………………………… 3.22 v T = (R/f)(∂T m /∂x) p ln(p o /p 1 ) …………………………………… 3.23 In natural coordinates, we have, V T = – (R/f)(∂T m /∂n) p ln(p o /p 1 ) …………………………………… n is directed to the left – towards low pressure. u T and v T in eqns and 3.23 are the components of the thermal wind. Alternately, we may express the thermal wind for a given layer in terms of the horizontal gradient of the geopotential difference between the top and bottom of the layer: u T = – (R/f)ln(p o /p 1 )(∂T m /∂y) p = – (1/f)∂/∂y(RT m ) ln (p o /p 1 ) = – (1/f)∂/∂y {g o (Z 1 – Z o )} = – (1/f) ∂/∂y( Φ 1 – Φ o ) v T = (R/f) ln(p o /p 1 )(∂T m /∂x) p = (1/f)∂/∂x(RT m ) ln (p o /p 1 ) = (1/f)∂/∂x {g o (Z 1 – Z o )} = (1/f) ∂/∂x( Φ 1 – Φ o )

126
i.e. u T = – (1/f)∂/∂y {g o (Z 1 – Z o )} = – (1/f) ∂/∂y(Φ 1 – Φ o ) ……… v T = (1/f)∂/∂x {g o (Z 1 – Z o )} = (1/f) ∂/∂x(Φ 1 – Φ o ) …...……3.26 In natural coordinates, we have, V T = – (1/f)∂/∂n {g o (Z 1 – Z o )} = – (1/f) ∂/∂n(Φ 1 – Φ o ) ….……3.26 n is directed to the left, towards low geopotential. where Φ 1 – Φ o ≡ δ Φ = g o δ Z = RT m ln (p o /p 1 ) = the thickness of the layer between p o and p 1 measured in units of geopotential. So, we see that the thickness is proportional to the mean temperature in the layer. Hence, lines of equal δ Φ(isolines of thickness) are equivalent to the isotherms of mean temperature in the layer. Note also that δZ ≡ δ Φ/g o gives the approximate thickness in geopotential height as shown in eqns, 3.25,3.26 and 3.27 where Z 1 – Z o = δ Z = RT m /g o lnp o /p 1. Comparisons of eqns. 3.25,3.26 and 3.27 shows that the thermal wind for a layer bears the same relationship to the thickness δ Z of the layer as the geostrophic wind does to the height of a pressure surface

127
Thus, the thermal wind is parallel to lines of constant thickness, with low values of thickness to the left in the Northern Hemisphere and the reverse in the Southern Hemisphere. From eqn and 3.26 and considering that ∆Z= RT m /g o ln(p o /p 1 ) it follows that for two specific pressure levels (say 700hPa and 500hPa), each value of mean temperature corresponds to a specific value of thickness. Thus, the thermal wind “blows” parallel to the mean isotherms with low temperature to the left (N.H.). This is essentially the reason for calling the upper level wind vector minus the lower level wind vector the thermal wind. The hydrostatic equation may also be written in terms of mean density: ∆Z = (p o – p 1 )/ρ m g which shows that thickness lines are also lines of mean density of the isobaric layer. The thermal wind equation is an extremely useful diagnostic tool which is often used to check analyses of the observed wind and temperature fields for consistency.

128
It can also be used to estimate the mean horizontal temperature advection in a layer as shown in Fig. 3.8 below. It is clear from vector form of eqns and 3.26 that V T = (g o /f) k x ∇ (Z1 – Zo) = (1/f) k x ∇ ( Φ – Φ ) = (g o /f) k x ∇ {RT m /g o ln (p o /p 1 )} = R/f ln (p o /p 1 ) k x ∇ T m ……………………………………………… 3.27 which also shows that the thermal wind blows parallel to the isotherms (and lines of constant thickness) with warm air to the right facing downstream in the Northern Hemisphere. Thus, as illustrated in Fig 3.8a a geostrophic wind which turns counterclockwise with height (backs) is associated with cold air advection (flow across the isotherms from colder region to warmer region). Conversely, as shown in Fig. 3.8b clockwise turning (veering) of the geostrophic wind with height implies warm advection by the geostrophic wind in the layer.

129
It is therefore possible to obtain a reasonable estimate of the horizontal temperature advection and its vertical dependence at a given location solely from data on the vertical profile of the wind given by a single sounding. Cold Warm V G (o) V G (1) VTVT V G (o) V G (1) VTVT T o - δ T ToTo T o + δ T Fig. 3.8: The relationship between turning of the geostrophic wind and temperature advection: (a) backing of the wind with height; (b) veering of the wind with height. (a) (b)

130
Alternately, the geostrophic wind at any level can be estimated from the mean temperature field provided that the geostrophic velocity is known at a single level. Thus, for example, if the geostrophic wind at 850hPa is known and the mean horizontal temperature gradient in the layer hPa is known, the thermal wind equation can be applied to obtain the geostrophic wind at 500hPa. Fig. 3.9 shows the thermal wind in the vicinity of the developing cyclone. The light solid arrows represent geostrophic wind vectors at the 1000hPa level and the heavy solid arrows represent geostrophic wind vectors at the 500hPa level.

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google