ANSWERS to Problem Set 5
METR 3123, Atmospheric Dynamics II
1. A vertical cylindrical column of air at 30° N with radius 100 km expands to
twice its original radius. If the air is initially at rest, what is the mean
tangential velocity at its perimeter after expansion?
Consider a horizontal slice through the column (so the slice is tangent to the
earth's surface):
Initial radius ri = 100 km
Final radius rf = 200 km
= 2!105 m
= 105 m
Based on the information given, this problem is calling out for use of Kelvin's
circulation theorem to get the circulation. Then use the circulation to get the mean
tangential velocity. The Kelvin circulation theorem (relative form of it) is:
C(t)+ 2!Ae(t) = const
where A is the "shadow" area, that is, the projection of the actual area A on the
e
equator. In class we saw that for a horizontal (tangent to earth surface) area A,
we can relate A to A by A = A sin! where ! is latitude and the brackets
e
e
denote a spatial average. So the circulation theorem becomes
(1)
C(t)+ 2!A(t) sin!(t) = const .
To find the constant, use the initial conditions. No motion initially so the initial
circulation is 0. And the initial area A is the area of the circle when its radius was
100 km. And its initial latitude is 30° N.
! const = 2!!(105 m)2 sin 30! .
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Apply this in (1), get:
C(t)+ 2!A(t) sin!(t) = 2!"(105 m)2 sin 30! .
Rearrange it to get C(t) as
C(t) = !2!A(t) sin !(t) + 2!!(105 m)2 sin 30!
Evaluate it at the "final time" t f , i.e., right after the expansion. Since there's
nothing in the problem to suggest there's been a change of latitude, you may keep
the latitude the same.
C(t f ) = !2!!(2!105 m)2 sin 30! + 2!!(105 m)2 sin 30!
$
'
= 2!! sin 30! &(105 m)2 " (2#105 m)2 )
%
(
Use ! =
2!
= 7.2!10"5s!1
24!3600s
and sin 30! = 0.5
! C(t f ) = 2!7.2!10"5s!1 " !!0.5!1010 m2 (1" 4) = "6.85!106 m2s"1
Since the curve is circular, the wind along the curve is the tangential velocity (in a
!
cylindrical coordinate system), that is, u = v!!ˆ. We want the average of v! over
the perimeter of the circle. Start with "ave = integral/interval" with the integration
interval being the 2! radians
of the circle. Then relate the resulting expression to
! !"!
the circulation C = "# u !dl (because we know it). Recall that for a tiny chunk of
!"
!
! !"!
2!
circle, dl = rd! !ˆ . So C = "# u !dl = "! v!!ˆ !r d! !ˆ = "0 v!r d! .
2!
2!
0
0
1
1
So v! = ! v" d" =
2!
2!r
C
! v" r d" = 2!r
So at the final time we have:
v!(t f ) =
C
6.85"106 m 2s!1
=!
= !5.5ms!1
5
2"rf
2"(2"10 m)
(negative, so anticyclonic flow).
2
2. At t = 0, an air column at 60° N extends from sea level up to a fixed
tropopause at 10 km above sea level (ASL). At this time the air column has
zero relative vertical vorticity. The air column moves until it is over a
mountain barrier 2.5 km ASL at 45° N. What is the relative vorticity of the
column when it is over this mountain barrier? There's nothing in the problem
to suggest that baroclinic effects are important, so you can assume the flow is
barotropic (and use the barotropic potential vorticity theorem).
The barotropic potential vorticity theorem says that the ratio of the absolute
vertical vorticity (sum of relative vertical vorticity ! and earth vertical vorticity f)
to the thickness of the air column H is conserved. So it doesn't change as the
column moves around.
!(t)+ f (t)
= const
H(t)
(1)
(for the column)
So, using the initial data given, we calculate the const as:
const =
!(0)+ f (0) 0 + 2! sin 60! 2! sin 60!
.
=
=
H(0)
10km
10km
Apply it in (1) and rearrange the resulting equation to get relative vertical vorticity:
2! sin 60!
!(t) = const !H(t)! f (t) =
!H(t)! f (t) .
10km
Now evaluate it when the column is over the mountain top, noting that since the
height of the lower surface has increased to 2.5 km, the thickness of the column
has been reduced to 10 km - 2.5 km. So
!column =
over mtn
2! sin 60!
!(10km ! 2.5km ) ! 2! sin 45!
10km
= 2! #%sin 60!(1" 0.25)" sin 45! &(
$
'
use ! = 7.2"10#5s!1 (from last prob)
= 2!7.2!10"5s"1 #$%0.866!0.75 " 0.707 &'(
= ! 8.3"10!6 s!1
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3. A ring of air surrounding the earth is initially 100 km ASL at a constant
latitude of 60° N (i.e., the ring is the 60° N latitude circle at 100 km above sea
level). This ring of air is initially at rest with respect to the earth's surface. To
what latitude must this ring of air be displaced to (i.e., with no change in its
height) in order to acquire an easterly component of 10 m s!1 with respect to
the earth's surface?
This problem is begging for the Circulation Theorem (relative version).
Consider a "capping" area A (piece of spherical surface) bounded by the ring we're
interested in (which is the 60° N latitude circle at 100 km above sea level):
!
The projection of A onto equatorial plane (plane ! to ! ) is:
Ae = !R2 , where R = r cos! = (a + z ) cos!
2
! Ae = ! (a + z ) cos2 "
Circulation theorem (relative version) says:
C + 2!Ae = const , for a material curve,
or:
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(*) C f + 2!Aef =Ci + 2!Aei
Subscript "f" means "final". Subscript "i" means initial (t = 0).
Initial relative circulation: Ci = 0 since ring is at rest with respect to earth.
Initial Ae : Aei = !Ri2 , where Ri is initial radius (perpendicular to rotation axis),
Ri = (a +100km)cos60! . So Aei = !Ri2 = !(a +100km)2 cos2 60! .
! !"! 2"
Final relative circulation: C f = ! u f !dl = ! v f Rf d! = "10ms"1Rf 2" , where
0
v f is final zonal wind (note that it's negative since flow is easterly), and
Rf = (a +100km)cos!f is final R. Compared to initial state, height above the
earth doesn't change, but latitude does change, so Rf = (a +100km)cos!f .
! C f = "10ms!1 2! (a +100km)cos!f
Final Ae is: Aef = !(a +100km)2 cos2 "f
Put it all together in (*) to get:
!10ms!1 2! (a +100km)cos"f + 2" !(a +100km)2 cos2 "f
= 2"!(a +100km)2 cos2 60!
This is a quadratic equation for the unknown cos!f . Put it in a neater form by
dividing this equation by 2!!(a +100km)2 . Get:
cos2 !f !
10ms!1
cos!f ! cos2 60! = 0
"(a +100km)
Now plug in numbers. Use a = 6.37!106 m, " =
quadratic eqn for cos!f
2!
= 7.27!10#5 s#1 ,
24!3600s
and cos60! = 0.5 (so cos2 60! = 0.25 ). So the quadratic equation becomes:
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cos2 !f ! 0.02126 cos!f ! 0.25 = 0 .
Solve it using quadratic formula [ ax 2 +bx +c = 0 ! x =
"b ± b2 " 4ac
]. Get:
2a
0.02126 ± (0.02126)2 + 4!0.25
cos!f =
= 0.01063 ± 0.5001
2
So there are two possibilities:
cos!f = 0.01063 + 0.5001 = 0.51074
or
cos!f = 0.01063 ! 0.50011 = !0.48947
Must choose the positive solution. It's the only solution that makes sense. Why?
Final latitude !f must range somewhere between south pole ( !f = !90! ) and
north pole ( !f = 90! ). So !90! " !f " 90! . For this range of !f , cos!f is always
positive. So we must reject the negative solution in favor of the positive solution.
! cos!f = 0.51074
Take inverse cosine (making sure your calculator is set on degrees, not radians)
! !f = cos!1(0.51074) = 59.290
So, the ring must be displaced southward to latitude 59.290 (shift of 60! !59.29!
= 0.71! ) to get it to spin with an easterly speed of 10 m/s.
4. Consider the same initial scenario as in problem (3) but this time determine
the height the ring must be displaced to (i.e., with no change in latitude) in
order to acquire the same easterly component as in (3).
Consider same diagram as in problem 3, but this time we keep latitude fixed
and move along the radial (vertical) direction. So z changes. z f is final height of
ring above the surface. !f = !i = 60!N .
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Again, Circulation theorem says:
(*) C f + 2!Aef =Ci + 2!Aei
(exactly as before, in prob 3)
Initial relative circulation: Ci = 0
Initial Ae : Aei = !Ri2 = !(a +100km)2 cos2 60!
Final relative circulation: C f = !10ms!1 2! (a + z f )cos 60!
Final Ae : Aef = !Rf 2 = !(a + z f )2 cos2 60!
So (*) says:
!10ms!12! (a + z f )cos 60! + 2!!(a + z f )2 cos2 60!
= 2!!(a +100km)2 cos2 60!
The unknown ( z f ) always appears in the combination a + z f . So lets focus on
obtaining a + z f
and then subtract a from it. Divide above equation by
2!!cos2 60! . Get the quadratic equation,
(a + z f )2 !
10ms!1
(a + z f )! (a +100km)2 = 0
!
" cos 60
Evaluate the coefficients using a = 6.37!106 m, " = 7.27!10#5 s#1 .
(a + z f )2 ! 2.751"105 m(a + z f ) ! 4.186"1013 m 2 = 0
Use quadratic formula to solve for a + z f :
a + zf =
2.751!105 m ±
(2.751!105 m)2 + 4!4.186!1013 m 2
2
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! a + z f = 137550m ± 6471392m
subtract a from both sides (and use its
numerical value)
z f = 137550 m ! 6.37"106 m ± 6471392m
So we get two solutions:
+ root:
z f = 238942m ! 239km
- root:
z f = !1.27"107 m = !1.27"104 km = !12,700km
Note that if we were to add the radius of the earth a = 6.37!106 m = 6,370km to
the second root we'd still get a negative value for the radius (sum of z and a),
which is physically impossible. So we have to reject the second root. So the final
height of the ring is:
z f ! 239km .
This represents an upward displacement of the ring by about 139 km.
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