Download Main equations/relations from Lecture 1:

Lecture 2:
Clarifications of lecture 1:
Hydrostatic balance: Under static conditions, only gravity will work on the fluid.
Why doesn't all the fluid contract to the ground? Pressure builds up and resists the
compression. The balance between this pressure force and gravity yields the
hydrostatic balance relation.
What actually causes the resistance is the vertical gradient in pressure – the pressure
acts equally in all directions but it decreases with height, so at height z it is a certain
value and at height z+δz it is slightly lower, which will cause the fluid to want to
∂p
move upwards. But gravity will pull it down and balance will ensue:
= − ρg .
∂z
Hypsometric relation: relates a layer thickness to the vertically integrated (over logpressure) temperature. Can be in terms of:
p1
T ( p) dp
Height: z ( p 2 ) − z ( p1 ) = R ∫
g p
p2
p
R 1
dp
Geopotential height: zφ ( p 2 ) − zφ ( p1 ) =
T ( p)
∫
g o p2
p
p1
Geopotential: φ ( z 2 ) − φ ( z1 ) = R ∫ T ( p )
p2
dp
p
Lecture 2: Buoyancy, stability, convection and gravity waves
T1-radiative equilibrium and observed temperature profile: The jump at the
surface, and as we will see later also the tropospheric lapse rate are unstable and will
lead to convection.
T2-schematic radiative equilibrium and observed temperature profile.
The troposphere is in Radiative-convective equilibrium.
T3-schematic of Radiative-convective equilibrium.
Understanding this convection is the topic of this class.
In atmosphere we have convection in moist compressible fluid. Much more
complicated than a pan of water. We will start from simple to more complex.
Why does convection occur?
Horizontally uniform heating of water from below will force a spatially uniform
temperature decreasing with height. Even though light fluid is on bottom, the
gravitational force can be balanced by pressure gradient, according to hydrostatic
balance law. Zero net force. So why does convection develop? How come it is
spatially varying even if balanced state is horizontally homogeneous?
Æ Instability:
Lets look at the following example of a ball on a hill-valley:
Component of gravitational acceleration along the slope: -gdh/dx
Ball at point A, where dh/dx=0 is in equilibrium stable. What if displaced an
increment δx?
Slope at xA+δx: dh/dx(xA+δx)≈dh/dx(xA)+(d2h/dx2)A δx=(d2h/dx2)A δx
The equation of motion at δx: d2(δx)/dt2=-gdh/dx=-g(d2h/dx2)A δx
Solution: δx(t ) = C + e σ +t + C − e σ −t
⎛ d 2h ⎞
σ ± = ± − g ⎜⎜ 2 ⎟⎟
⎝ dx ⎠
At the peak (d2h/dx2) A<0 so the solution is exponentially growing- a small
perturbation will grow. Instability, convection.
At the trough (d2h/dx2) B>0 so the solution is oscillatory.
Energetics: At the trough, the perturbed ball moves uphill- we need to do work since
its potential energy increases, and it won't happen. In the absence of an external
energy source. At the ridge, the ball moves downhill, decreasing potential energy, and
increasing kinetic energy. Motion will continue.
Convection in water:
Buoyancy: A light object in water will bounce back up. What about a fluid parcel?
Fluid parcel: small but finite size, which is thermally isolated from environment and
always at same pressure as environment.
Assume ρ=ρ(T).
Above point A the fluid is the same in 1,2 and p (parcel). Hence pressure at A1, A, A2
is the same.
Above point B fluid column is lighter, and pressure is lower.
The forces acting on the parcel: (subscript E for environment and p for parcel)
Gravity – depends on the parcel mass: Fg=-gρpδAδz
Pressure gradient force- depends on the environmental pressure:
Ft+ Fb =-δpEδA= gρEδA δz
Net force on parcel: Fg +Ft+ Fb = g(ρE−ρp)δA δz
Divide by the parcel mass ρpδA δz to get the acceleration, which is a reduced form of
gravity: Buoyancy: b≡Ftot/mp = -g(ρp−ρE)/ ρp
If ρp<ρE, the parcel is positively buoyant: b>0, and it will accelerate upwards.
If ρp>ρE, the parcel is negatively buoyant: b<0, and it will accelerate downwards.
Lets look at a specific density profile, which we assume depends only on temperature
(incompressible water): ρ ref=ρ(1- α[T-Tref])
Assume that a parcel which is similar to the environment at z1 starts rising, to height
z2=z1+ Δz, and when it rises it does not exchange energy with the surroundings, thus
its temperature remains constant.
At z1: ρp= ρE(z1), Tp=TE(z1)
At z2: ρp= ρE(z1), Tp=TE(z1) but ρE= ρE(z2)= ρE(z1)+ Δz dρE/dz
The parcel's buoyancy is: b(z2)=-g(ρp−ρE)/ρp=g/ρp dρE/dz Δz =g/ρE(z1) dρE/dz Δz
The buoyancy is postivie/negative/0 if dρE/dz is postivie/negative/0, and the parcel
will rise more/sink back/remain still.
For the equation of state above, b>0,=0,<0 if dT/dz <0, =0, >0.
In the absence of heating and damping, a parcel will be unstable if the density
increases with height.
Mathematically: the equation of motion for the parcel's displacement Δz:
d2(Δz)/dt2=b= g/ρE(z1) dρE/dz Δz≡-N2 Δz N2≡-b/Δz =-g/ρE(z1) dρE/dz is the buoyancy
frequency which we will return to later.
if dρE/dz>0, N2<0 and the solution is exponential (instability):
Δz = Δ 1e
−N 2t
+ Δ 2e−
−N 2t
If dρE/dz<0, N2>0 and the solution is oscillatory: Δz = Δ 1 cos( Nt ) + Δ 2 sin( Nt )
We will discuss this case in a bit.
Energetics view: The energetics argument of the ball on the hill can't be applied to
the parcel, because we are dealing with a continuous fluid, and when the parcel rises,
another parcel has to go back down. Thus, we need to examine the initial and final
states of the entire fluid, and see what happened to the potential energy:
Lets assume the perturbations occurs with a parcel at z1 rising to z2, while a parcel at
z2 descends to z1 and besides that all the fluid remains unchanged.
The initial potential energy: PEinitial=g(ρ1z1+ ρ2z2).
The final potential energy: PEfinal=g(ρ1z2+ ρ2z1)
Their difference: PEfinal- PEinitial=-g(ρ2-ρ1)(z2-z1)≈ -g dρE/dz (z2-z1)2
This yields the same condition – if dρE/dz>0 the parcel's potential energy will be
reduced, and its kinetic energy will increase, meaning it will continue moving.
If dρE/dz<0 the parcel's potential energy will increase, on expense of its kinetic
energy, and the parcel will stop moving.
Note: energetics considerations only show if instability is plausible, but we need an
analysis like the one above to determine if instability will arise or not.
Water heated from below:
We assume a constant known heating rate per meter squared of H (W/m2).
This heating will be carried upwards by convection, at a rate H.
The heat content of a unit volume of fluid is: ρCT where C is the specific heat of
water. The heat flux (amount of heat transported across a unit area) is then ρCTw,
where w is vertical velocity.
W is highly variable in convection, hence we need to integrate over time (a few
convective cycles) and space (over a few convective cells).
Assume that over the domain and some time, 1/2 the fluid is moving upward with
temperature T+ΔT, and half the fluid is moving downward with temperature T. The
net integrated flux is: <H> = 1/2 ρC(T+ΔT)w- 1/2 ρCTw=1/2 ρCΔT.
How do we get w? – from energy considerations: Assume all kinetic energy comes
from release of potential energy: ΔKE =−ΔPE= g Δρ Δz
ΔKE= ρ /2 (u2+ v2+ w2)≈ 3/2 ρw2 Æ w2≈2/3 g Δρ/ ρ Δz
For the equation of state above: ρref=ρ(1- α[T-Tref]) , Δρ=ρ ref αΔT
Plugging in w, and then in <H> we get:
<H>=1/2 ρref C(2/3 g αΔz)1/2 ΔT3/2
Dry convection in a compressible atmosphere:
We need to account for compression due to pressure changes as the parcels
rise/descend.
Assume that a parcel at z1 has the environmental pressure and temperature:
p1=pE(z1), T1=TE(z1), ρ1=p1/(RT1)
The parcel rises to z2 adiabatically. Its pressure decreases like the environment. What
happens to T?
Consider a parcel of unit mass: ρV=1. Its exchange of energy with the surroundings
follows the law:
δQ= δU+ δW where δQ is the heating from outside, δU is the added internal energy
and δW is the work done by the parcel on its surroundings.
δQ= CvdT+pdV where Cv is the specific heat per unit volume (choose this since if
dV=0, all δQ goes to a change in T.
Want to find dT/dp, so change from dV to dp using the ideal gas law:
dV=d(1/ρ)=-1/ρ2dρ, dρ=dp/(RT)-p/(RT2)dT
Æ pdV= -p/(ρ2RT) dp + p2/(ρ2RT2) dT =RdT-dp/ρ
Æ δQ= (Cv+R)dT-dp/ρ = CpdT - dp/ρ
Note: Cp>Cv, meaning temperature will change less for a given heating when
pressure, rather than volume is kept constant. This is because some of the energy goes
towards work due to volume changes.
For adiabatic displacements, δQ=0, so CpdT = dp/ρ
Where the variables are taken to be the parcel's.
From the hydrostatic relation, dp=-ρEgdz, so dT/dz= - ρE/ρ g/Cp
For small displacements, ρ≈ρE, so that their ratio can be taken to be 1, and we get the
dry adiabatic lapse rate: Γd=-dT/dz=g/Cp
For Cp=1005J/Kg/K, Γd≈10K/km.