atmospheric science 4320 / 7320

Atmospheric Science 4320 / 7320

Anthony R. Lupo

Day one

The orders of magnitude of PGF, CO, and Ro for a synoptic scale disturbances:

Synoptic scale disturbance: 2000 - 6000 km (space) 1 - 7 days (time)

Day one

Typical values:

Day one

Mesoscale disturbances: 10 - 2000 km, 1h - 1d

Day one

Hurricane

Day one

Tornadoes: ][

Day one

The horizontal Equation of motion for inviscid flow in terms of the geostrophic departure vector.

geohageo

Vkfpdt

Day one

The horizontal acceleration vector is always normal and to the right of the geostrophic departure vector

It’s magnitude is directly proportional to the magnitude of the departure vector.

Day one

And here:

depghh

VkfVkfpdt

VVkfpdt

Day one

Then consider a case in which there are negligible spatial variations in Vgeo (spatially uniform PF) such that along the trajectory, Vg = constant

then, dt

dtVd agaggh

Day one becomes:

The time rate of change of the departure is always 90 degrees to the right of the Vag, and it’s magnitude is dependent on the magnitude of V departure!

agag Vkf

Day one

The departure vector is of constant magnitude but rotates with time to the right - clockwise - with an inertial period of

Day one The Inertial Oscillation Is a

consequence of Unbalanced flow! Thus, it is an oscillation whose frequencey is tied to the Coriolis parameter.

This is a “gravity-type” wave that is emitted from an unbalanced flow situation. More appropriately, this is a “Kelvin” type wave or oscillation.

Day one

Interial oscillation influential in:

Intense convection (MCSs)

Unbalanced jet streaks associated with highly curved flow and rapidly developing cyclones

Day one/two

Balanced flow in the friction layer.

Model I assume:

FRPGFCF

Day two

Thus the horizontal equations of motion become

The equation of motion in natural coords.

0ˆ1 FVfkp frh

Day two

Thus for balance we must have for the s-component,

the frictional force balances the downstream component of the PGF.

sF ˆ1

Day two

Then for the n-component;

the CF balances the normal component of the PGF in Cartesian coordinates

nVfnnp

fr ˆˆ1

Day two

The equation (BL)

ˆ1ˆ1

Day two

The diagram

W/out Friction With Friction

Day two

Does our assumption for FR above hold up?

Balanced friction layer flow omitting the assumption of Fr acting directly opposite the horizontal velocity.

Day two

The balance equation (Eq 1)

FVfkp hfrh

Day two

Thus we must determine by observation and calculation the direction of the friction vector:

PGF can be observed the coriolis force can be calculated

frVkfCF

Day two and Fr can be determined from

equation 1 as a residual:

Diagrammatically, you’ll get a result similar to our other one, at least it will be “CEFGW”

FVfkp frh

Day two

Friction in terms of the ageostrophic velocity vector

but recall that by geostrophy,

0ˆ1 FVfkp frh

geoh Vfkp

Day two

0ˆˆ FVfkVfk frgeo

geofr VVkfF

Day two

so in this context,

ageogeofr

Day two Note of caution: some folks adhere to this

position, that Vageo is equal to the difference between the Vfric and Vgeo.

We have shown this mathematically to be the case. We also know that Vgeo is non-divergent in it’s most basic form.

Thus what we have stated above is that Vfric is COMPLETELY the divergent part of the wind in a balanced flow.

Day two

However, you can also show mathematically, by Helmholtz partitioning that Vgeo contains a rotational and divergent part, and Vageo contains a divergent and rotational part (Keyser et al., 1989, MWR; Loughe et al., 1995 MWR, May, or Lupo, 2002, MWR )

Day two Example:

Where “chi” is the velocity potential.

Day two

Diagramatic Example of Vfr development

Start with a system in geostrophic balance.

Day two

Wind adjusts such that PGF is balanced by CO, and is steady state.

Introduce a Frictional decelerating force (Fric.) (opposite the wind)

Day two

Friction reduces Vh to a value less that Vgeo which then reduces the CF, to a value less than that of the PGF resulting in a net force in the normal direction across the isobars towards the lower pressure causing air to turn to it’s left.

Day two

Thus final balance can be achieved if Vfr adjusts so that:

Day two Properties of balanced flow in the friction

A) Horizontal non accelerating flow

B) Vfr < Vgeo or subgeostrophic wind speeds

C) Vfr has a component across the isobars from high to low

D) cross isobaric component is proportional to the Frictional force

Day two Friction layer convergence and

divergence results in vertical motions.

This will have important implications for when we examine the Vorticity, Omega Equations, etc....

Day two

Low pressure (surface convergence) gives upward motion. However, with regard to a developing system where vertical motion is forced aloft, the friction acts against cyclone development.

Day two

High pressure (surface divergence) gives downward motion. However, with regard to a developing system where vertical motion is forced aloft, the friction acts against anticyclone development.

Day three Gradient Flow

(Holton 65 - 68 and Hess p 180 ff)

Gradient flow is unbalanced frictionless flow along a curved path at a constant speed and parallel to the isobars (isoheights)

Day three

Recall we observe that Vh tends to be parallel to the isobars in straight and curved flow regions.

Assumptions:1)

hh ˆ2

Day three Normal component of centripetal acceleration

only. Rc (Radius of curvature) is positive for cyclonic flow and negative for anticyclonic flow.

4) We’ll also assume that Rc = Ri, or the curvature of the flow is the same as the curvature of the isobars.

5) acceleration = PGF - CF, normal acceleration due to unbalanced flow between PGF and CO.

Day three

The gradient wind equation

The horizontal equation of motion reduces to only the normal (n) component equation. (No (s) component, since there is no coriolis component in that direction)

Day three The expression:

so, In scalar form

Ri = radius of curvature of the isobars

nVfnnp

gr ˆˆ1

gr Vfnp

Day three

Equation above in form of (see it?):

where Vgr:

npRRffR

V iiigr

cbxax 2

Day three

Recall: We must solve for the gradient wind by either “completing the square”, i.e., add 0.5 the “b” term and squaring, then adding to both sides. Alternatively, we can use the quadratic formula:

Day three

Examine the possible solutions of the Gradient Wind (Vgr) equation.

In agreement with polar coordinate convention, counterclockwise (clockwise) flow is a positive (negative) value of Vgr.

Day three

Consider positive root:

As the pressure gradient goes to zero, Vgr goes to zero

npRRffR

V iiigr

Day three

Oh yeah? Show me!

Day three

But, if the pressure gradient force > 0, then Vgr is positive. (Observed Low, counterclockwise flow)

cyclonicV

PosRffR

Day three/four

Diagram!

Day three/four

If the pressure gradient force < 0, then Vgr is negative (Observed High, clockwise flow)

icanticyclonV

Day four

Diagram!

Day four

Consider negative root of Vgr:

As the pressure gradients go to zero, Vgr approaches -f Ri (anticyclonic motion with no pressure gradient limit). This is the motion of the Inertia Circle.

Day four

The Math!

npRRffR

Day four

If the pressure gradient force > 0 (Low pressure), Vgr is negative (clockwise flow around a Low)

PosRffR

npRRffR

Day four Thus, this is mathematically possible,

but not observed on synoptic and larger-scales. Clockwise tornado?

Now, if the pressure gradient force < 0 (High pressure), get two solutions:

Day four

1) anomalously strong anticyclonic (clockwise) flow

(but “radical term” stays +)

npRRffR

V iiigr

Day four

2) impossible cyclonic (counterclockwise flow) around high pressure.

npRRffR

Day four

Consider the Common Case (positive roots):

If pressure gradient is (+) or Low Pressure:

The radical: 04

npRRf ii

Day four never becomes negative, so there’s no

theoretical pressure gradient limit to the strength of low pressure.

Hurricanes and tornadoes can have incredible wind speeds and pressure gradients!

Also, their radius of curvature can be quite small. E.g., hurricane eye can be as small as 2 km!

Day four

There is a limit on how strong a “high” can get. The radical:

cannot be less than zero thus,

npRRf ii

Day four

Or, you can look at this equation and determine that there are no limits on the Radius of curvature for low pressures, while, for a high pressure there is a limit which the Radius of curvature cannot fall below.

Day four

Compare Geostrophic and Gradient Wind Values

First, Vg as a special case of Vgr.

Vgr:gr

gr Vfnp

Cent. PGF CO

Day four

In the case of geostrophic flow, the radius of curvature goes to infinity, and:

Day four

Gradient Flow vs. Geostrophic Flow

Gradient wind equation:

gr Vfnp

Day four

Substitute the geostrophic component into the gradient wind equation.

We get:

grgeoi

gr VVfR

Day four

Now Cyclonic Gradient Flow:

Note PGF > Co

and Ri = Rc > 0

so,geo

grgr V

Day four So,

(1) is now a “correction term”

Thus Vgr < Vgeo (subgeostrophic)

grgr V

Day four Anticyclonic Gradient Flow

CO > PGF

Ri = Rc < 0

Thus, Vgr > Vgeo

grgr V

Day four/five

Gradient Flow imbalance and Adjustments in a ridge trough system, containing a region in the ridge where |Ri| < |Ri min|

(see diagram and handout)

Day five Divergence / Convergence patterns associated with

gradient flow through a ridge and trough system with uniform isoheight contour spacing.

Uniform contour spacing : Geostrophic flow everywhere is the same, thus in “the straightaways” Vgeo = Vgr.

This depends on the assumption that f = fo though, because when f varies, some of the divergence/convergence is the result of the variance of f. (Recall, I said we could show this via Helmholtz partitions also). Though we can assume based on our drawings that this divergence is small.

Day five

Convergence: Since winds in ridge are super-geostrophic, we get speed convergence between ridge and trough line. Convergence aloft implies divergence at the surface and high pressure. Can maintain or develop high pressure.

Day five/six

A map:

Day five/six

Divergence: Since winds in trough are sub-geostrophic, we get speed divergence between ridge and trough line. Divergence aloft implies convergence at the surface and low pressure. Can maintain or develop low pressure.

Day six

Eulerian flow

1. assumptions

onlyPGFd

VVa hs

Day six

Equation of Motion becomes

Eulerian flow types (there are 3!):

Type 1 (straight isobars):

dtVd sh

Day six

Type II (tangential and centripetal acceleration) (curved isobars):

sh ˆ2

Day six

Type III (Cyclostrophic – closed isobars):

Day six Cyclostrophic flow (Eulerian flow

type 3)

Assumptions:

1)Flow is horizontal (2- dimensional) 2) Friction = 0 3) PGF >> CO of Ro >> 10

Day six

4) flow is parallel to the isobars 5) acceleration is due to

centripetal acceleration only (PGF = Centripetal force)

Day six

Thus the equation of motion in natural coordinates reduces to:

cyc ˆ1

Day six

in scalar form,

V icyc

Day six Thus, we must now examine the

behavior of the radical term, which is negative, but only possesses two real solutions:

Cyclonic

Anticyclonic 0,0

Day six

The diagrams:Cyclostrophic - cyclonic Cyclostrophic anticyclonic

Day six We can only interpret such flows

as having a low pressure center, no high pressure centers, else we get imaginary roots.

This type of flow reasonably approximates the eyewall of intense hurricanes, tornadoes, and dustdevils.

Day six

Hurricanes are all cyclonic disturbaces

95 - 99% of tornadoes are cyclonic

but dustdevils tend to be distributed 50 - 50.

Day six Inertial Flow

A horizontal-frictionless flow in which CO is the dominant force:

Assume: V3 = Vh (2 dimensional flow)

Coriolis force dominant.

Day six

Horizontal equation of motion in natural coordinates reduces to:

Day six Since the Co is always normal and to the right

of Vh, the acceleration is normal and to the right of Vh.

Diagrammatically:

We have a constant Vh being turned anticyclonically (toward the right), thus motion similar to the inertial oscillation, even though each was brought about in a slightly different manner.

Day six

The Inertial period with latitude ():

anticyclonic!

0ˆˆ2

cyc RfV

RnVfnR

Day six (These are

observed in oceanic circulations – Inertial Gyres)

Latitude (degrees)

Time (hours)

15 46.2

30 23.9

45 16.9

60 13.8

75 12.4

90 12.0

Day seven Acceleration of the horizontal

winds and the development of geostrophic departures.

Recall from earlier:

The Navier - Stokes equations for frictionless flow

Day seven

The equation:

From the definition of geostrophic flow

hhh Vkfp

geoh Vkfp

Day seven

and we get:

where Vageo = V - Vgeo

hgeoh VkfVkf

ageoh Vkf

Day seven

so, if we solve for the ageostrophic wind,

this is the most common formulation so we’ll use it.

V hageo

Day seven

Here’s an alternative (Trenberth and Chen, 1988–JAS, Lupo, 2001).

then recall that the acceleration is perpendicular and to the right of Vageo.

V dageo

Day seven

Let’s expand the total derivative on the RHS into its natural components (s,n,z,t)

Vd hhh

Day seven the approximate Vh as Vgeo (nothing

magic here, just an approximation.

then substitute 2 into 1 and get,

(A) (B) (C)

Vd geogeoh

V geogeoh

geoageo

ˆˆˆ

Day seven Term (A) The ageostrophic component due

to local changes in the wind with time.

Term (B) The Ageostrophic component due to horizontal advection

Term (C) The ageostrophic component due to vertical advection or vertical motion

Day seven

Term A The isallobaric wind.

and by definition (recall our substitution earlier)

V geoageo

V hgeo ˆ

Day seven

then taking the derivative with time,

Day seven Where ( ) is the local pressure

tendency

lines of constant pressure tendency are called isallobars

is the isallobaric gradient

Day seven

is the isallobaric component of the wind.

fV hageo 2

Day seven The ageostrophic component of the

wind has a component due to the pressure field change, which is proportional to the magnitude of the isallobaric pressure gradient.

This ageostophic component is normal to the isallobars and directed toward pressure fall regions.

Day seven Using typical values for pressure

changes in the atmsphere, and gradients of pressure changes over say, 1000 km.

The Isallobaric component of the ageostrophic wind is on order of 5 - 10 m/s. Very significant!

Day seven/ eight

Term B: the ageostrophic component due to horizontal advection

Vageo;

VV geoh

VV geo

Day eight

Local geostrophic imbalance, “geostrophic adjustment”, as we go from one geostrophic state to another.

Consider first straight isobars (diffluent) :

Day Eight

Last page courtesy of Uccellini and Kocin (1987)

Vgeo slows down (since it is proportional to PGF), thus locally CO exceeds PGF and Vgeo turns toward the right (anticyclonic curvature). (Jet Exit!)

Day Eight

3-D Jet streak (J.T. Moore)

Day Eight

Rising or sinking air (Univ. Illinois)

Day Eight

Day eight

Consider then confluent isobars

Vgeo speeds up (since it is proportional to PGF), thus locally PGF exceeds CO and Vgeo turns toward the left (cyclonic curvature).

Day eight

Consider this component for typical synoptic scales: (it’s on order of 1 m/s)

If the Isobars were straight, then Vageo would be 0, as would the partial V / partial s term.

Day eight

Consider Curved isobars

curved isobars, but uniformly spaced, we covered with the gradient wind. (where the Vageo is forced by centripetal acceleration)

Day eight

You could consider the two together, curved isobars and diffluent or confluent flow.

ageogeogr VVV

Day eight

Term C: The Ageostrophic component due to vertical motions.

With vertical motions, we must have shear as dictated by:

kV geoCageo

VV geo

Day eight

The vertical shear of the geostrophic wind is of course, the thermal wind!!

Consider Vh or Vgeo of constant direction, but increasing with height:

Day eight

Like This

Day eight

Or consider Vgeo of constant magnitude but veering with height (thus w > 0)

Day eightOr consider Vgeo of constant magnitude but veering with height (thus w > 0)

Day eight

Vh = super geostrophic for veering (+) with upward motion (+) or backing (-) with downward (-) motion

Vh = subgeostrophic for backing (-) with upward (+) motion or for veering (+) with downward (-) motion.

Day eight We can consider the two combined

for our typical warm air advection situation (veering w/height)

Vh has a component across isobars toward lower pressure

Day eight

backing w/ height is cold air advection

Cross isobaric flow toward higher pressure.

Term C using typical synoptic scale values is on order of 2 m/s

Day eight/nine

The Isobaric Coordinate system:

Until now, we have worked our basic equations in the x,y,z,t coordinate system, let’s work some of our basic concepts in the x,y,p coordinate system.

Day nine

Day nine Consider a small portion of an

isobaric sfc in the x -z plane

Consider the change in variable Q along two paths at the same instant in time:

322131 QQQ

Day nine

The picture:

Day nine

x varies z varies

(1) (2)

tyxtzytpy QQQ ,,,,,,

Day nine

Q tzy ,,

Q tyx ,,

Day nine

tzytpy

Day nine

then since z/x is the slope of the isobaric surface in x,

tpytyxtzytpy xz

,,,,,,,,

Day nine

In the y - z plane

tpytyxtzytpy yz

,,,,,,,,

Day nine

Then let’s put in vector form, thus,

tpytyx

tzytpy

ˆˆˆˆ

Day nine or,

where,

= The gradient of Q on a p-surface

= The gradient of Q on a z-surface

= The gradient of z on a p-surface

QQ pzp

QpQzzp

Day nine

The horizontal pressure gradient force:

In (x,y,z,t) coords:

pPGF h1

Day nine

substitute Q = Pressure

Day nine

and get:

Since there is no pressure gradient on pressure surface!

Day nine/ten

but from hydrostatic balance:

zgpPGF ph 1

Day ten

Why do we talk about PGF on a constant pressure surface? We can represent mass field in terms of pressure surfaces or height surfaces. Since these fields are “sloped” there is a component acting on them due to gravity. Recall definition of geopotential surface!

Day ten

The Inviscid 2 -D Navier-Stokes equation in x,y,p

We hammered on the geostrophic wind in lab!

hph Vkf

Day ten

The 3-D Del operator in the x,y,p,t coordinate system

The total derivative in x,y,z,t and x,y,p,t

ˆˆˆ3

Day ten

The vertical coordinate of Newton’s 2nd Law: Hydrostatic Balance

Day ten

now Q equals z;

In other words, NO Change!

Day ten Forms of the Hydrostatic equation

We can rewrite in other forms, to make it more consistent with coordinate system!

Day ten

The equation of continuity in (x,y,p,t)(Our cube)

Day ten

Total mass in the left face:

substitute: to get:

dzdyu11

dpdygu1

Day ten

Total mass out of the right face:

dzdyu22 dpdygu2

Day ten

Net Flux of mass in/out of cube

dxdydpxu

dpdygu

Day ten

Thus in y direction

and the “p” dimension.

dxdydpyv

dxdydpp p

Day ten Add ‘em up!

Now since hydrostatic balance holds, there is NO mass change, since the total mass between two isobaric surfaces remains the same! Thus, mass is conserved in x,y,p system. (Sutcliffe and Godart (1942)).

gdxdydp

Day ten

In vector form:

But commonly expressed as:

Day ten

Compare to x,y,z form:

Day ten

As we saw earlier in the semester, we can use continuity to calculate vertical motions (velocity).

What is the relationship between w and ?

hho dpVp

Day ten

where,

pVpVpV ageogeoh

Day ten

Recall:

and substitute to get:

pVBABA g

Day ten

Typical orders of magnitude

Term A 10 hPa day-1

Term B 1 hPa day-1

Term C 100 hPa day-1

Day ten/11

Where,

wdtdp ,

Day 11

Note the sign convention via hydrostatic balance!

Upward motion +w -

Downward motion -w +

Day 11 Let’s look at typical values for synoptic and mesoscales:

Synoptic Scale: w

Typical 1 cm s-1 10-3 hPa s-1

Strong Low 10 cm s-1 10 mb s-1

Mesoscale:

Cu 1 m s-1 100 mb s-1

Cb 10 m s-1 1 mb s-1

Day 11 Advantages of the Isobaric Coordinate system (A

summary)

On an isobaric surface, isolines of temp (T) are also isolines of density, since P is const. unlike on a z-surface. Isotherms are also isentropes!

The geostrophic wind does not depend on gradients of pressure and density and “f” variations, it now depends on height gradients and ‘f’ variations.

The thickness is directly proportional to temperatures (Thermal wind and hysometric equations).

Day 11 The vertical shear of the Vg is also

related to temperature gradients (thermal wind)

The equation of continuity is much simpler.

density no longer appears explicitly in N-S or Continuity equations.

Day 11 The isentropic and sigma coordinate

systems

We define a variable Q(x,y,,t)

Recall to go from x,y,z,t to x,y,p,t, we examined some geometry. Let’s cut to the chase, since the derivations the same:

Day 11

Here’s the equation:

Note we see a pattern developing, for a general conversion to x,y,z,t to x,y,c,t

where c = any vertical coordinate:

Day 11

The expression:

QQ cch

Day 11

So, now we must transform the N-S and Continuity equations to isentropic and sigma coordinates:

isentropic: (let c = and Q = Pressure)

pPGF 1

Day 11

we get;

Day 11

zgppz 11

Day 11

Now hold on,

and thus (after product rule),

R pp c

Day 11

and using some math/algebra (first break up exponent):

and, what is the gradient of theta on a theta surface?

R pp c

Day 11

Let’s cancel some things and use the Eqn of state:

Day 11

One more workout:

Day 11

now substitute into Eqn (1) and we get,

gzTczgTcp ppz 1

Day 11

Wait! Isn’t M = CpT + gz, dry static energy (a potential surface in a dry atmosphere), or the Montgomery streamfunction.

Day 11/12

Now, in sigma coordinates, where we assume sigma is a terrain following coordinate, and we specify a top (modeler’s coordinate!).

Q = Q(x,y,,t) topsfc

Day 12

Thus, the horizontal PGF is:

this will not simplify any more!

Day 12

The Navier Stokes equations in Theta and sigma coordinates:

Vkfpdt

VkfMdt

Day 12

The geostrophic wind! (PGF = CO)

Theta coords

Day 12

Sigma Coords

Day 12

The 3-d del operator in the general (c) theta () and sigma () coordinates:

ˆˆˆ

Day 12

The total derivative in the general (c), theta () and sigma () coordinates:

Day 12

Hydrostatic Balance: Recall general form or chain rule:

Then: general coordinate:

Day 12

Now hydrostatic balance in theta coords:

Eq (1)

Day 12

becomes (How?);

apply chain rule;

Day 12

substitute Eq. (1) into RHS.

Now use the Ideal Gas Law on RHS:

Day 12

And then;

gzTcTcTc

Day 12

In sigma coordinates:

Equation 1

Day 12

take derivative w/r/t p:

topsfc

topsfc ppp

Day 12/13

Substitute this into Equation (1):

topsfc

Day 13 Continuity equation:

In (x,y,,t), Let’s take a different approach this time, instead of looking at our “cube”, let’s look as some math instead:

Continuity in (x,y,z,t)dtd

Day 13

We can examine a “generalized” form of the divergence relationship.

VV ccchh

Day 13

Now we know that:

w = dz / dt

Day 13

take the partial of equation (3) w/r/t “c” (See prod. Rule on RHS?)

Day 13

Then, by chain rule:

Day 13

And substituting (4) into (2) and using (5), (3) and eventually into (1):

cccchh

Day 13

Now A has the form of ??? (term A is the RHS of the final form above)

AdtAd 1ln

Day 13

A generalized continuity equation:

Day 13

In theta coordinates:

Day 13

In sigma coordinates:

use the hydrostatic relationship:

Day 13

and get:

Vppdtd

Day 13

Which becomes:

ppVppppt

Vppdtd

tshtsts

Day 13

Let’s talk about fundamental Kinematic Concepts

In lab, we talked about divergence, which is a scalar quantity:

Day 13

We can prove that divergence is the fractional change with time of some horizontal area A.

Day 13

and, then

yxdtAd

Day 13

atmospheric science 4320 / 7320

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