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:
3
4
1
110
10
mkg
sf
sm
Vh
1.0
101
10
10
23
23
23
fL
VRo
sm
np
smkg
np
sm
Vf
h
h
Day one
Mesoscale disturbances: 10 - 2000 km, 1h - 1d
3
4
1
110
10
mkg
sf
sm
Vh
0.2
101
10
10
22
22
23
fL
VRo
sm
np
smkg
np
sm
Vf
h
h
Day one
Hurricane
3
5
1
1105
50
mkg
sxf
sm
Vh
10
1051
105.2
22
23
fL
VRo
sm
xnp
sm
xVf
h
h
Day one
Tornadoes: ][
3
4
1
110
50
mkg
sf
sm
Vh
5000
51
105
2
23
fL
VRo
sm
np
sm
xVf
h
h
Day one
The horizontal Equation of motion for inviscid flow in terms of the geostrophic departure vector.
geohageo
hhh
VVV
Vkfpdt
dV
ˆ1
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:
agh
agghh
depghh
Vkfdt
Vd
VkfVkfpdt
dV
VVkfpdt
dV
ˆ
ˆˆ1
ˆ1
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
Vd
dt
Vd
dt
Vd
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
dt
Vd
ˆ
Day one
The departure vector is of constant magnitude but rotates with time to the right - clockwise - with an inertial period of
f 2
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:
0
0
dt
Vd
FRPGFCF
VF
h
h
Day two
Thus the horizontal equations of motion become
The equation of motion in natural coords.
0ˆ1 FVfkp frh
nnp
ssp
PGF
nVfCO
sFF
fr
ˆ1
ˆ1
ˆ
ˆ
Day two
Thus for balance we must have for the s-component,
the frictional force balances the downstream component of the PGF.
ssp
sF ˆ1
ˆ0
Day two
Then for the n-component;
the CF balances the normal component of the PGF in Cartesian coordinates
nVfnnp
fr ˆˆ1
0
Day two
The equation (BL)
Fkf
pkf
V he
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)
dtVd
FVfkp hfrh
0ˆ1
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
ph1
frVkfCF
ˆ
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
ˆ1
Day two
Friction in terms of the ageostrophic velocity vector
but recall that by geostrophy,
0ˆ1 FVfkp frh
geoh Vfkp
ˆ1
Day two
thus,
or,
0ˆˆ FVfkVfk frgeo
geofr VVkfF
ˆ
Day two
so in this context,
Ffk
V
VkfF
VVV
ageo
ageo
ageogeofr
ˆ
ˆ
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 )
dt
dV ag
ag
,
Day two Example:
Where “chi” is the velocity potential.
h
d
r
dr
Vp
and
fgz
where
V
kV
VVV
N
22
ˆ
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
layer
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)
2) 3)
hVV
0F
nR
V
dt
Vd
c
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
nR
Vgr
i
gr ˆˆ1
ˆ2
gri
gr Vfnp
R
V
1
2
Day three
Equation above in form of (see it?):
where Vgr:
npRRffR
V iiigr
42
22
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:
aacbb
r2
42
2
2,1
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
42
22
Day three
Oh yeah? Show me!
0np
42
22ii
gr
RffRV
Day three
But, if the pressure gradient force > 0, then Vgr is positive. (Observed Low, counterclockwise flow)
0,0
iRnp
!,42
22
cyclonicV
PosRffR
V
gr
iigr
Day three/four
Diagram!
Day three/four
If the pressure gradient force < 0, then Vgr is negative (Observed High, clockwise flow)
0np
!,
...42
22
icanticyclonV
RffRV
gr
iigr
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!
igr
iigr
iiigr
fRV
fRfRV
npRRffR
V
22
42
22
Day four
If the pressure gradient force > 0 (Low pressure), Vgr is negative (clockwise flow around a Low)
gr
iigr
iiigr
V
PosRffR
V
npRRffR
V
.42
4222
22
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
42
22
Day four
2) impossible cyclonic (counterclockwise flow) around high pressure.
npRRf
npRRffR
V
ii
iiigr
4
4222
22
Day four
Consider the Common Case (positive roots):
If pressure gradient is (+) or Low Pressure:
The radical: 04
22
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,
04
22
npRRf ii
npRf
npRRf ii
1
4
42
22
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
i
gr Vfnp
R
V
1
2
Cent. PGF CO
Day four
In the case of geostrophic flow, the radius of curvature goes to infinity, and:
geogr
gr
Vnp
fV
COPGF
orR
VR
1
0,2
Day four
Gradient Flow vs. Geostrophic Flow
Gradient wind equation:
gri
gr Vfnp
R
V
1
2
Day four
Substitute the geostrophic component into the gradient wind equation.
We get:
np
fVgeo
1
grgeoi
gr VVfR
V
2
Day four
Now Cyclonic Gradient Flow:
Note PGF > Co
and Ri = Rc > 0
so,geo
i
grgr V
fR
VV
2
Day four So,
(1)
(1) is now a “correction term”
Thus Vgr < Vgeo (subgeostrophic)
geoi
grgr V
fR
VV
2
Day four Anticyclonic Gradient Flow
CO > PGF
Ri = Rc < 0
so,
Thus, Vgr > Vgeo
geoi
grgr V
fR
VV
2
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
!.
0.
0.
.
onlyPGFd
COc
Frb
VVa hs
Day six
Equation of Motion becomes
Eulerian flow types (there are 3!):
Type 1 (straight isobars):
dtVd
dtVd sh
Day six
Type II (tangential and centripetal acceleration) (curved isobars):
nRV
dt
Vd
dt
Vd
c
sh ˆ2
Day six
Type III (Cyclostrophic – closed isobars):
nRV
dt
Vd
c
h ˆ2
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)
nRV
dtVd
ac
h ˆ2
Day six
Thus the equation of motion in natural coordinates reduces to:
nnp
nR
V
i
cyc ˆ1
ˆ2
Day six
in scalar form,
or,np
R
V
i
cyc
1
2
npR
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
iRnp
0,0
iRnp
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:
nVfnR
V
or
nVfnR
V
hc
cyc
hc
cyc
ˆˆ
ˆˆ
2
2
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
cchc
cyc RfV
RnVfnR
V
sin
12
22
ffR
RVR
c
c
h
ci
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
dtVd
ˆ1
geoh Vkfp
ˆ1
Day seven
and we get:
where Vageo = V - Vgeo
hgeoh VkfVkf
dtVd
ˆˆ
ageoh Vkf
dt
Vd ˆ
Day seven
so, if we solve for the ageostrophic wind,
(1)
this is the most common formulation so we’ll use it.
dt
Vd
fk
V hageo
ˆ
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.
2
ˆ
ffk
V dageo
Day seven
Let’s expand the total derivative on the RHS into its natural components (s,n,z,t)
z
Vw
s
VV
t
V
dt
Vd hhh
hh
Day seven the approximate Vh as Vgeo (nothing
magic here, just an approximation.
(2)
then substitute 2 into 1 and get,
(A) (B) (C)
z
Vw
s
VV
t
V
dt
Vd geogeoh
geoh
z
Vw
fk
s
VV
fk
t
V
fk
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.
If,
and by definition (recall our substitution earlier)
t
V
fk
V geoageo
ˆ
pfk
V hgeo ˆ
Day seven
then taking the derivative with time,
and,
pfk
tt
Vh
geo
ˆ
tp
fk
t
Vh
geo
ˆ
Day seven Where ( ) is the local pressure
tendency
lines of constant pressure tendency are called isallobars
so,
is the isallobaric gradient
tp
tp
h
Day seven
and,
is the isallobaric component of the wind.
tp
fV hageo 2
1
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;
s
Vk
f
VV geoh
Bageo
ˆ)(
s
VV geo
ageo
Day eight
Local geostrophic imbalance, “geostrophic adjustment”, as we go from one geostrophic state to another.
Consider first straight isobars (diffluent) :
0s
Vgeo
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).
0s
Vgeo
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
s
VV h
dep
Day eight
Term C: The Ageostrophic component due to vertical motions.
With vertical motions, we must have shear as dictated by:
z
V
fw
kV geoCageo
ˆ
)(z
VV geo
ageo
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
Or
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
Now:
dtdp
vdtdy
udtdx
,,
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
322131 QQQ
Day nine
The picture:
Day nine
Or
x varies z varies
(1) (2)
tyxtzytpy QQQ ,,,,,,
Day nine
but,
(1)
(2)
xxQ
Q tzy ,,
zzQ
Q tyx ,,
Day nine
so,
xz
zQ
xQ
xQ
or
zzQ
xxQ
Q
tzytpy
,,,,
Day nine
then since z/x is the slope of the isobaric surface in x,
tpytyxtzytpy xz
zQ
xQ
xQ
,,,,,,,,
Day nine
In the y - z plane
tpytyxtzytpy yz
zQ
yQ
yQ
,,,,,,,,
Day nine
Then let’s put in vector form, thus,
tpytyx
tzytpy
jyz
ixz
zQ
jyQ
ixQ
jyQ
ixQ
,,,,
,,,,
ˆˆ
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
zzQ
QQ pzp
QpQzzp
Day nine
The horizontal pressure gradient force:
In (x,y,z,t) coords:
pPGF h1
Day nine
substitute Q = Pressure
zzp
pp
then
zzQ
pph
pph
Day nine
and get:
Since there is no pressure gradient on pressure surface!
zzp
p ph
Day nine/ten
but from hydrostatic balance:
then,
gzp
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
dtVd
ˆ
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
kp
jy
ix
ˆˆˆ3
dtdp
pQ
yQ
vxQ
utQ
dtdQ
dtdz
wzQ
wyQ
vxQ
utQ
dtdQ
p
z
,
,
Day ten
The vertical coordinate of Newton’s 2nd Law: Hydrostatic Balance
zp
pQ
zQ
Day ten
now Q equals z;
In other words, NO Change!
gzp
gpz
zz
1
Day ten Forms of the Hydrostatic equation
We can rewrite in other forms, to make it more consistent with coordinate system!
pTR
p
pTR
pzg
pzg
vd
vd
)3
)2
1)1
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
gdp
dz
dpdygu1
Day ten
Total mass out of the right face:
But,
dzdyu22 dpdygu2
dxxu
uup
12
Day ten
Net Flux of mass in/out of cube
dxdydpxu
thus
dxdydpxu
dpdygu
dpdygu
p
p
11
Day ten
Thus in y direction
and the “p” dimension.
dxdydpyv
p
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)).
0
gdxdydp
pyv
xu
Day ten
In vector form:
But commonly expressed as:
033 V
pVhh
Day ten
Compare to x,y,z form:
dtd
V
1
33
zw
Vhh
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 ?
p
po
hho dpVp
Day ten
Well,
where,
zp
wpVtp
dtdp
hh
pVpVpV ageogeoh
Day ten
Recall:
and substitute to get:
A B C
pVBABA g
,cos
zp
wpVtp
dtdp
hageo
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
Thus,
Where,
gw
dtdz
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:
zzQ
QQh
Day 11
The expression:
zzQ
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;
But,
zzp
ppz
gzp
Day 11
so,
(1)
then,
zgppz 11
pc
R
o
pp
T
Day 11
Now hold on,
and thus (after product rule),
pc
R
o
p
pT
Tp
pT
p
p
p
p
c
R pp c
R
ooc
R
o
p
1
Day 11
and using some math/algebra (first break up exponent):
and, what is the gradient of theta on a theta surface?
TT
T
p
pTp
p
p
p
p
p
p
c
R pp c
R
oo
o
c
R
o
p
)(
2
Day 11
See?
Let’s cancel some things and use the Eqn of state:
TT
ppc
R
p
1
TT
pTcp
111
Day 11
One more workout:
Tcp
and
Tcp
p
p
1
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.
Mpz 1
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
top
pp
pp
Day 12
Thus, the horizontal PGF is:
this will not simplify any more!
ppz
11
Day 12
The Navier Stokes equations in Theta and sigma coordinates:
hh
hh
Vkfpdt
Vd
VkfMdt
Vd
ˆ1
ˆ
Day 12
The geostrophic wind! (PGF = CO)
Theta coords
M
fk
V
VkfM
g
h
ˆ
ˆ
Day 12
Sigma Coords
fk
pfk
V
Vkfp
g
h
ˆˆ
ˆ1
xfxp
fv
yfyp
fu
g
g
11
11
Day 12
The 3-d del operator in the general (c) theta () and sigma () coordinates:
kjy
ix
kjy
ix
kc
jy
ix
ˆˆˆ
ˆˆˆ
ˆˆˆ
3
3
3
Day 12
The total derivative in the general (c), theta () and sigma () coordinates:
QQdtd
QVtQ
dtdQ
QQdtd
QVtQ
dtdQ
cQ
ccQ
dtdc
QVtQ
dtdQ
hh
hh
hh
Day 12
Hydrostatic Balance: Recall general form or chain rule:
Then: general coordinate:
gzc
cp
pc
gzc
Day 12
Now hydrostatic balance in theta coords:
Eq (1)
and;
p
pg
z
pc
R
o
pp
T
Day 12
becomes (How?);
apply chain rule;
tp
pcR
tT
Tt p
11
p
pcRT
T p
11
Day 12
substitute Eq. (1) into RHS.
Now use the Ideal Gas Law on RHS:
pcRT
T p
11
T
cTTcT
T pp
11111
Day 12
And then;
MTc
and
gzTcTcTc
p
ppp
Day 12
In sigma coordinates:
Equation 1
zg
p
and
gzc
cp
Day 12
Now:
take derivative w/r/t p:
topsfc
top
pp
pp
topsfc ppp
1
Day 12/13
Substitute this into Equation (1):
ts
topsfc
pp
then
pp
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:
(1)
Continuity in (x,y,z,t)dtd
V
1
33
Day 13
We can examine a “generalized” form of the divergence relationship.
)2(
33
cz
dtd
zc
zzc
c
VV
zw
VV ccchh
Day 13
Now we know that:
w = dz / dt
(3)cz
czVtz
dtdz
w hhc
Day 13
take the partial of equation (3) w/r/t “c” (See prod. Rule on RHS?)
)4(2
2
cz
ccz
cc
zcV
cz
Vcz
tdtdz
ccw
chhc
Day 13
Then, by chain rule:
(5)zc
cw
zw
Day 13
And substituting (4) into (2) and using (5), (3) and eventually into (1):
cc
cz
dtd
cz
Vdtd
cc
cz
dtd
cz
Vzw
V
cc
cz
ccV
tzc
Vzw
V
cc
cchh
cccchh
11
1
Day 13
Now A has the form of ??? (term A is the RHS of the final form above)
dtdA
AdtAd 1ln
Day 13
Then:
cc
Vcz
dtd
and
cc
Vcz
dtd
dtd
cc
cc
ln
ln1
Day 13
A generalized continuity equation:
cc
Vcz
dtd
cc
ln
Day 13
In theta coordinates:
or
V
zdtd
ln
V
pdtd
ln
Day 13
In sigma coordinates:
use the hydrostatic relationship:
Vz
dtd
ln
ts pp
Day 13
and get:
Vppdtd
tsln
Day 13
Which becomes:
V
ppVppppt
or
Vppdtd
tshtsts
tsln
Day 13
Let’s talk about fundamental Kinematic Concepts
In lab, we talked about divergence, which is a scalar quantity:
hh V
Day 13
We can prove that divergence is the fractional change with time of some horizontal area A.
dtdA
AVhh
1
Day 13
and, then
dtyd
xdtxd
yyxdt
AdA
and
dtyxd
yxdtAd
A
11
11
Day 13