1

Lecture 2 The Basic Conservation Laws

Control volumes + total differentiation

Control volume scale of molecules << δV << the scale of motion.

Eulerian framework: Control volume consists of a parallelepiped of sides δx,δy,δz ,whose position is fixed related to the coordinates.

Lagrangian framework: 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 Lagrangian frame is particularly useful for deriving conservation laws, and the

conservation of a quantity Q can be expressed as DQDt

= 0 . The Eulerian frame is,

however, more convenient for solving most problems because in that system the fieldvariables are related by a set of partial differential equations in which the independentvariables are the coordinates x, y, z , and t.

The conservations contain expression for the rates of change of density, momentum, andthermodynamical energy following the motion of particular fluid particle (i.e., inLagrangian frame). In order to apply these laws in the Eulerian frame it is necessary toderive a relationship between the rate of change of a field variable following the motionand its rate of change at a fixed point. The former is called the substantial, the total, orthe material derivative (denoted by D / Dt ). The latter is called the local derivative,merely the partial derivative w.r.t time.

These two frameworks are related by the concept of total differentiation.

Consider the temperature measured on a balloon that moves with the wind and supposethat this temperature is T0 at the point x0 , y0 , z0 and time t0 . If the balloon moves to thepoint x0 + δx, y0 + δy, z0 + δz in a time increment δt , then the temperature changerecorded on the balloon, δT , can be expressed in a Taylor series expansion as

δT =δTδt

⎛⎝⎜

⎞⎠⎟δt + δT

δx⎛⎝⎜

⎞⎠⎟δx + δT

δy⎛⎝⎜

⎞⎠⎟δy + δT

δz⎛⎝⎜

⎞⎠⎟δz + (higher order terms)

The rate of change of T following the motion becomes in the limit δt→ 0dTdt

=∂T∂t

+∂T∂x

⎛⎝⎜

⎞⎠⎟DxDt

+∂T∂y

⎛⎝⎜

⎞⎠⎟DyDt

+∂T∂z

⎛⎝⎜

⎞⎠⎟Dzdt

If letDxDt

≡ u, DyDt

≡ v, DzDt

≡ w,

then, using vector notation,

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∂T∂t

=DTDt

− U ⋅∇T

where U = iu + jv + kw is the velocity vector. The 2nd term on the rhs is called thetemperature advection.

Rate of change of a vector in a rotating frame(Note the notation is slightly different from the other text.)

Fig. 2.1

Consider first a vector of constant length relative to an inertial frame at a constant angularvelocity Ω . Then, in a frame rotating with that same angular velocity it appearsstationary and constant. If in a small interval of time δt the vector C rotates through asmall angle δλ then the change in C, as perceived in the inertial frame, is given by (seeFig. 2.1)

δC = C cosϑδλm (2.1)where the vector m is the unit vector in the direction of change of C, which isperpendicular to both C and Ω . But the rate of change of the angleλ is just, bydefinition, the angular velocity so that δλ = Ω δt and

δC = C Ω sinϑm δt =Ω ×C δt (2.2)using the definition of the vector cross product, where ϑ = (π / 2 −ϑ ) is the anglebetween Ω and C. Thus

DCDt

⎛⎝⎜

⎞⎠⎟ a

=Ω ×C (2.3)

where the left-hand side is the rate of change of C as perceived in the inertial frame.

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Now consider a vector A that changes in the inertial frame. In a small time δt the changein A as seen in the rotating frame is related to the change seen in the inertial frame by

(δA)a = (δA)R + (δA)rot (2.4)where the terms are, respectively, the change seen in the inertial frame, the change due tothe vector itself changing as measured in the rotating frame, and the change due to therotation. Using (2.2) and (δA)rot =Ω × A δt and so the rate of change of the vector A inthe inertial and rotating frames are related by

DADt

⎛⎝⎜

⎞⎠⎟ a

=DADt

⎛⎝⎜

⎞⎠⎟ R

+Ω × A (2.5)

This expression implies that viewing from the absolute (inertial) frame, the rotation of avector should give rise to an extra rate of change than viewing from the rotating frame.

Note that for a scalar B , DaBDt

=DBDt

.

Conservation of momentum (Velocity and acceleration in a rotating frame)In an inertial frame, Newton’s law may be written as

DaUa

Dt= Fi

i∑

In Section 1.5 we found through simple physical reasoning that when the motion isviewed in a rotating coordinate system certain additional apparent forces must beincluded if Newton’s second low is to be valid. Here we show that the same results maybe obtained by a formal transformation.

In order to transform the expression above to rotating coordinate, we must first find therelationship between Ua and the velocity relative to the rotating system, which isdenoted as U . This relation is obtained by applying (2.5) to the position vector r:

DarDt

=DrDt

+Ω × r

thus,Ua = U +Ω × r

Now apply (2.5) to the velocity vector and obtainDaUa

Dt=DUa

Dt+Ω × Ua

=DDt(U +Ω × r) +Ω × (U +Ω × r)

=DUDt

+ 2Ω × U − Ω2R

where Ω is assumed to be constant and R is a vector perpendicular to the axis ofrotation, with a magnitude equal to the distance to the axis of rotation.

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If we assume that the real forces acting on the atmosphere are the pressure gradient force,gravitation, and friction, the Newton’s 2nd law can be written as

DUDt

= −2Ω × U −1ρ∇p + g − 1

ρ∇τ

where the centrifugal force has been absorbed into the gravitation, the friction force isexpressed as the gradient of the shearing stress.

Spherical Coordinates: (λ, φ, r)Suppose coordinate direction unit vectors i, j, k are pointing eastward, northward andupward, respectively, rotating uniformly with the Earth. Then, the components ofvelocity vector are

u ≡ r cosφ DλDt, v ≡ r Dφ

Dt, u ≡ Dz

Dt(r = a + z)

For notational simplicity, it is conventional to define x and y as eastward and northwarddistance, such that Dx = acosφDλ and Dy = rDφ . Thus the horizontal velocity

components are u ≡ DxDt, v ≡ Dy

Dt.

(Read Holton pages 31-37 to figure out yourself how the following set of equations on aspherical coordinate are derived, especially those curvature terms. Note the typos in page32 and equation 2.13)

DuDt

−uv tanφa

+uwa

= −1ρ∂p∂x

+ 2Ωvsinφ − 2Ωwcosφ + Fx

DvDt

+u2 tanφ

a+vwa

= −1ρ∂p∂y

− 2Ωu sinφ + Fy

DwDt

−u2 + v2

a= −

1ρ∂p∂z

− g + 2Ωu cosφ + Fz

The terms proportional to 1/a on the left hand side are called curvature (metric) terms;they arise due to the curvature of the Earth surface. In addition to these nonlinearcurvature terms, the terms hidden in the total derivatives also contain nonlinear terms:

DuDt

=∂u∂t

+ u ∂u∂x

+ v ∂u∂y

+ w ∂u∂z

This is why atmospheric dynamics is a very challenging subject, which makes it aninteresting one too.

Some basic operators in spherical coordinate:

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∇Φ =i

r cosφ∂Φ∂λ

+ j1r∂Φ∂φ

+ k ∂Φ∂r

∇ ⋅V =1

r cosφ∂u∂λ

+∂(vcosφ)

∂φ⎛⎝⎜

⎞⎠⎟

k ⋅∇ × V =1

r cosφ∂v∂λ

−∂(u cosφ)

∂φ⎛⎝⎜

⎞⎠⎟

∇h2Φ =

1r2 cos2φ

∂2Φ∂λ2

+ cosφ ∂∂φ

cosφ ∂Φ∂φ

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

Conservation of Mass—Continuity EquationEulerian derivative (business as usual)

mass flux rate per area in x-direction: ρu

inflow of mass through the left-hand face: ρu − ∂∂x(ρu)δx

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outflow of mass through the right-hand face: ρu + ∂∂x(ρu)δx

2Net flow into box:

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ρu − ∂∂x(ρu)δx

2⎡⎣⎢

⎤⎦⎥δyδz − ρu + ∂

∂x(ρu)δx

2⎡⎣⎢

⎤⎦⎥δyδz

= −∂∂x(ρu)δxδyδz

Generalizing to 3-dimensional, the net rate of mass inflow is −∇ ⋅ ρU per unit volume,

which is must be balanced by net accumulation rate ∂ρ∂t

, thus

∂ρ∂t

+∇ ⋅ ρU = 0

This is the mass divergence form of the continuity equation.

Alternatively,∂ρ∂t

+∇ ⋅ ρU =∂ρ∂t

+ ρ∇ ⋅U + U ⋅∇ρ =DρDt

+ ρ∇ ⋅U = 0

So, 1ρDρDt

+∇ ⋅U = 0

It states that the fractional rate of increase of the density following the motion of an airparcel is equal to minus the velocity divergence.

Conservation of Mass following the fluid (Lagrangian Derivation)

Consider a “material volume” fluid enclosed by a “material surface” which is defined as asurface through which no mass flows. The mass δM , of the control volume is

δM = ρδxδyδzwhere we assume the shape of the material surface to be parallelepiped. Because mass isconserved following the motion, we have

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1δM

DDt(δM ) = 1

ρδVDDt(ρδV ) = 1

ρDρDt

+1δV

DDt(δV ) = 0

while 1δV

DDt(δV ) = 1

δxDDt(δx) + 1

δyDDt(δy) + 1

δzDDt(δz)

uA =DDt

x0 −δx2

⎛⎝⎜

⎞⎠⎟

uB =DDt

x0 +δx2

⎛⎝⎜

⎞⎠⎟

uB − uA =DDt

δx = δu

Likewise for v and w.

limδV→0

1δV

DδVDt

= limδ x→0

1δx

DδxDt

+ limδ y→0

1δy

DδyDt

+ limδ z→0

1δz

DδzDt

=∂u∂x

+∂v∂y

+∂w∂z

So1ρDρDt

+∇ ⋅U = 0 (2.31)

as before.

Approximations related to continuity equation:

Incompressible: DρDt

= 0; or ∇ ⋅U = 0

Anelastic (weak Boussinesq) approximation: ∇ ⋅ ρ0U = 0

Boussinesq approximation: ∇ ⋅U = 0 The Boussinesq approximation ignores allvariations of density of a fluid in the continuity and momentum equations, except whenassociated with the gravitational (or buoyancy) term. But this should absolutely not allowone to go back and use (2.31) to say that Dρ / Dt = 0 ; the evolution of density is givenby the thermodynamic equation in conjunction with an equation of state.

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CONSERVATION OF ENERGY—FIRST LAW OF THERMODYNAMICS

CHANGE OF INTERNAL ENERGY=HEAT ADDED-WORK DONE by the system

cvDTDt

= J − p DαDt

where cv is the specific heat at constant volume.This is a static statement, does it also hold for fluid in motion?

For fluid in motion, we must account for kinetic energy of the system:e = cvT is internal energy per unit mass (due to temperature, i.e., molecular

motion)K= (1 / 2)U ⋅U is kinetic energy

Therefore, total thermodynamic energy of (Lagrangian) control volume of fluid, δV , is

E = e + 12U ⋅U⎡

⎣⎢⎤⎦⎥ρδV

To compute work done onto the control volume, take the product of resultant of externalforces and velocity vector:

W = Fi ⋅Ui∑

Recall that the forces are pressure force, gravity, Coriolis force, friction.For the Coriolis force,

WCoriolis = (−2Ω × U) ⋅U = 0Coriolis force is always perpendicular to the direction of motion and can do no work.For the gravity,

Wgravity = ρg ⋅UδV = −ρgw δVTo derive the word done by PF (pressure force), consider parallelepiped control volume

W A,BPF = (pu)A − (pu)Bδyδz

but (pu)B = (pu)A +∂∂x(pu)⎡

⎣⎢⎤⎦⎥Aδx +

so, W A,BPF = −

∂∂x(pu)⎡

⎣⎢⎤⎦⎥Aδxδyδz

In general, the total rate at which work in done to the system by PF isWPF = −∇ ⋅ (pU)δV

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Conservation of EnergyIn summary,

DDt

ρ e + 12U ⋅U⎛

⎝⎜⎞⎠⎟δV⎡

⎣⎢⎤⎦⎥= −∇ ⋅ (pU)δV + ρg ⋅UδV + ρJδV (2.35)

Here J is the rate of heating per unit mass due to radiation, conduction and latent heatrelease.

Expanding

ρδV DDt

e + 12U ⋅U⎛

⎝⎜⎞⎠⎟+ e + 1

2U ⋅U⎛

⎝⎜⎞⎠⎟D(ρδV )Dt

= −U ⋅∇pδV − p∇ ⋅UδV − ρgw δV + ρJδV (2.36)

Recall the continuity equation gives D(ρδV )Dt

= 0 , thus

ρ DDt

e + 12U ⋅U⎛

⎝⎜⎞⎠⎟= −U ⋅∇p − p∇ ⋅U − ρgw + ρJ (2.37)

This can be further simplified by taking ρU ⋅ (momentum equation),

ρ DDt

12U ⋅U⎛

⎝⎜⎞⎠⎟= −U ⋅∇p − ρgw ----mechanical energy equation (2.38)

Subtracting (2.38) from (2.37), we obtain,

ρ DeDt

= − p∇ ⋅U + ρJ ----thermal energy equation (2.39)

Using the definition of geopotential, gw = g DzDt

=DΦDt

, we have

ρ DDt

Φ +12U ⋅U⎛

⎝⎜⎞⎠⎟= −U ⋅∇p (2.40)

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This is referred to as the mechanical energy equation. The sum of kinetic energy plus thegravitational potential energy is called the mechanical energy. Thus, following themotion, the rate of change of mechanical energy per unit volume equals the rate at whichwork is done by the pressure gradient force.

Noting that

−1ρ∇ ⋅U = −

1ρ2

DρDt

=DαDt

,

and that for dry air the internal energy per unit mass is e = cvT , wherecv = 717 J kg

−1 K −1 is the specific heat at constant volume, we obtain the familiar formof thermodynamic energy equation from (2.39):

cvDTDt

+ p DαDt

= J (2.41)

Thus the first law of thermodynamics indeed is applicable to a fluid in motion. Thesecond term on the lfs, representing the rate of working by the fluid system, represents aconversion between thermal and mechanical energy. It is this conversion process enablesthe solar heat energy to drive the motion of the atmosphere.

Thermodynamics of the dry atmosphereRecall the equation of state: Pα = RT , and take total derivative,

p DαDt

+α DpDt

= R DTDt

and substitute it into the thermodynamic energy equation, we obtain

cpDTDt

−α DpDt

= J

Dividing through by T,

cpD lnTDt

− R D ln pDt

=JT

≡DsDt

This equation gives the rate of change of entropy per unit mass following the motion for athermodynamically reversible process (the system has to be heated very slowly). Areversible process is one in which the system is in an equilibrium state throughout theprocess. Thus the system passes at an infinitesimal rate through a continuous successionof balanced states that are infinitesimally different from each other. In such a scenario,the process can be reversed, and the system and its environment will return to the initialstate. For such a process, the entropy defined above is a field variable that depends onlyon the state of the fluid.

Note: Reversible does not necessarily mean adiabatic. A diabatic process could bereversible.

Potential TemperatureFor an ideal gas undergoing an adiabatic process (reversible process in which no heat isexchanged with the surroundings), the first law gives

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cpD lnT − RD ln p = 0So, we have a perfect differentials: cpD lnT − RD ln p .

Integrate from ( p, T ) to ( ps ,θ ): cp D lnT 'T

θ

∫ = R D ln p 'p

ps∫ , then cp lnθT

⎛⎝⎜

⎞⎠⎟= R ln ps

p⎛⎝⎜

⎞⎠⎟

.

θ = T psp

⎛⎝⎜

⎞⎠⎟

R /cp

--Poisson’s equation

Potential temperature is the temperature that a control volume of dry air initially atpressure p and temperature T would have if it were expanded or compressed adiabaticallyto a standard pressure ps . We usually take ps to be 1000 hPa.

The way potential temperature is defined should lead immediately to

cpD lnθDt

= cpD lnTDt

− R D ln pDt

=JT

≡DsDt

Thus, for reversible processes, fractional potential temperature changes are indeedproportional to entropy changes. A parcel that conserves entropy must flow along anisentropic (constant θ ) surface.

Annual mean zonal mean temperature (T) distribution

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Annual mean zonal mean equivalent potential temperature (θe ) distribution

13

Adiabatic Lapse RateThe lapse rate is the rate of temperature change with height. We have

lnθ = lnT +Rcpln ps

p= lnT +

Rcp(ln ps − ln p)

d lnθdz

=d lnTdz

−Rcp

d ln pdz

butdpdz

= −ρg ⇒d ln pdz

=−ρgp

sod lnθdz

=d lnTdz

+Rcp

ρgp

but pρ= RT , so

1θdθdz

=1TdTdz

+Rcp

gRT

=1T

dTdz

+gcp

⎛

⎝⎜⎞

⎠⎟

orTθdθdz

=dTdz

+gcp

If dθ / dz = 0 , i.e., potential temperature is constant w.r.t height,

0 = dTdz

+gcp

⇒ −dTdz

=gcp

≡ Γd

Γd is the dry adabatic lapse rate and approximately constant throughout the lower

atmosphere, Γd =9.811004

= 9.8K km−1

Static StabilityTθdθdz

=dTdz

+gcp

= Γd − Γ

If Γ < Γd , θ increases with height. An air parcel undergoes an adiabatic displacementfrom its equilibrium level will be positively buoyant when displaced downward andnegatively buoyant displaced upward so that it will tend to return to its equilibrium leveland the atmosphere is said to be statically stable or stably stratified.

A measure of static stability is

N 2 =gθdθdz

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N is also called buoyancy frequency or Brunt-Vaisala frequency. Its link to the frequencyof buoyancy oscillation can be derived as the following.

Suppose a parcel of fluid in a hydrostatic environment is pushed away from itsequilibrium position. We have

dp0dz

= −ρ0g

for the environment. The vertical acceleration of the parcel, dwdt

, may be written as

DwDt

=D2δzDt 2

= −g − 1ρ∂p∂z

where p and ρ are the pressure and density of the parcel. In the parcel method it isassumed that the pressure of the parcel adjusts instantaneously to the environmentalpressure during the displacement, so p = p0 . This condition must to be true if the parcel isto leave no disturbance to its environment. Thus with the aid of hydrostatic equation, wehave

D2δzDt 2

= −g − 1ρ∂p0∂z

= −g + 1ρρ0g = g

ρ0 − ρρ

⎛⎝⎜

⎞⎠⎟

but p = ρRT and p0 = ρ0RT0 = p so ρ0T0 = ρT , so T0 (ρ0 − ρ) = ρT − ρT0

then ρ0 − ρρ

=ρT − ρT0

ρT0=T − T0T0

Also

θ = T psp

⎛⎝⎜

⎞⎠⎟

R /cp

and θ0 = T0psp0

⎛⎝⎜

⎞⎠⎟

R /cp

= T0psp

⎛⎝⎜

⎞⎠⎟

R /cp

so,T − T0T0

=θ0 −θθ0

Thus

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D2

Dt 2(δz) = g T − T0

T0= gθ −θ0

θ0Now using Taylor expansion to approximate θ0 at δz :

θ0 (δz) = θ0 (0) +dθ0dz

δz

If the parcel is initially at level z=0 where the potential temperature is θ(0) , then isdisplaced adiabatically to δz . Because the potential temperature of the parcel θ isconserved, thus the potential temperature of the parcel is θ(δz) = θ0 (0) and hence

θ(δz) −θ0 (δz) = θ0 (0) −θ0 (δz) = −dθ0dz

δz

Therefore,

D2

Dt 2(δz) = gθ −θ0

θ0= −

gθ0

dθ0dz

δz = −N 2δz (2.52)

where N 2 =gθ0

dθ0dz

. Equation (2.52) has a general solution of the form δz = A exp(±i N t) .

Therefore, if N 2 > 0 , the parcel oscillates about its initial level with a frequency of N.For averaged tropospheric stratification, N ≈ 1.2 ×10−1s−1 so that the period of abuoyancy oscillation is about 8 min.

In the case of N = 0 , examination of (2.52) indicates that no acceleration force will existand the parcel will be in neutral equilibrium with at its new level. However, if N 2 < 0(potential temperature decreases with height), the displacement will increaseexponentially with time, thus statically unstable. We thus arrive at the familiargravitational or static stability criteria for dry air:

N 2 > 0 statically stableN 2 = 0 statically neutralN 2 < 0 statically unstable

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Typical Vertical profiles of T ,θ,θe in the Tropics

θe is the temperature (T) the parcel would reach if all the water vapor in the parcel wereto condense, releasing its latent heat, and then the parcel was brought adiabatically tosurface pressure ps .

θe ≈ θ exp Lcqs / cpT( ) (9.40)

where Lc is latent heat of condensation, qs is the saturation mixing ratio.

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(from Marshall and Plumb, 2008)

Homework:

We know the dry adiabatic lapse rate Γd = −9.8K / km and the typical lapse rate of the

troposphere Γ = −dTdz

= 6.5K / km . Please estimate the frequency (or period) of the

buoyancy oscillation in the typical troposphere. You may assume the temperature to beT=280° Kelvin.