Download CONTENTS: I. Sea/Land Breeze II. Pressure Gradient Force III

Science A-30
Section Handout 4
March 4-5, 2008
CONTENTS:
I. Sea/Land Breeze
II. Pressure Gradient Force
III. Angular Momentum and Coriolis Force
IV. Geostrophic Flow
Relevant formulas and conversions:
Scale Height: H = kT/(mairg ) = R'T/g
Barometric Law: P(z) = P(zo)e-z/H
P = ρ(k/mair)T = ρR'T
Pressure gradient force: Fpg = m × ΔP
ρ
Δx
1 day = 86400 seconds
Circle: 360 degrees = 2π radians
Angular velocity: ω = 2π/t
Angular momentum: L = m r2 ω
Coriolis force: Fc = 2 m Ω v sin(λ)
Coriolis deflection: Δy = [ Ω (Δx)2 / v ] sin(λ)
I. Sea/Land Breeze
Onshore sea breeze
Warm land, cool water
Offshore land breeze
Cool land, warm water
ƒ
During the day, land heats up faster than the sea. This makes the scale height, H larger over
the land than over sea. Pressure at some altitude over land is greater than pressure over sea at
same altitude, which induces flow from high pressure (land) to low pressure (sea).
ƒ
Now there is less mass over land than over the sea. This makes the pressure at the surface
higher over the sea. At the ground, air flows from high pressure (sea) to low pressure (land).
ƒ
Conservation of mass completes the circulation.
ƒ
During nighttime, land cools off faster than the sea. The flow reverses. At the ground, air
flows from inland towards the sea.
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II. Pressure Gradient Force
Pressure gradient force (Fpg) is the force due to a pressure difference (ΔP) across a horizontal distance
(Δx) (directed from high pressure to low pressure).
ΔP
Pressure gradient =
(units: N m-2 / m = N m-3 Æ force on 1 m3 of air)
Δx
Pressure gradient force = Fpg =
m
ρ
×
ΔP
Δx
(units: kg (m3 kg-1) (N m-3) Æ N Æ force)
III. Coriolis Force
Coriolis force is an apparent force caused by the rotation of the earth. It causes moving objects to
appear to follow curved trajectories when viewed from a rotating reference frame such as Earth.
Consequently, freely moving objects on the surface of the Earth appear to be deflected to the right in
the Northern Hemisphere and to the left in the Southern Hemisphere.
Angular velocity (ω) is the amount of angle an object rotates per unit time, or simply one rotation (2π
radians) divided by the time it takes to make one rotation:
2π
ω=
, where T is the time required for one rotation (units: radians sec-1 or sec-1)
T
Stationary objects on the Earth’s surface will make one rotation per day, so their angular velocity is the
same as that of the Earth (represented by Ω):
2π
Ω=
= 7.3 × 10 −5 radians sec-1
(units: radians sec-1 or sec-1)
86,400 sec
Angular momentum (L) describes the relationship between the distance to the axis of rotation, r and the
angular velocity, ω. Ignoring external forces such as friction, angular momentum is conserved.
L = m r2 ω
(units: kg m2 sec-1)
A way to visualize the Coriolis Effect is to think about it in terms of angular momentum. Conservation
of angular momentum dictates that freely moving objects will preserve its value of L. Consider an
object traveling from the equator to the North Pole. Its r decreases. To keep L constant, the object’s
angular velocity must increase, causing the object to rotate faster than the ground underneath it
therefore appearing to veer eastward. Conversely for an object moving from the North Pole to the
equator, r will increase, so its angular velocity must decrease to keep L constant. Therefore, the object
will rotate slower than the land underneath it, appearing to veer westward.
Magnitude of the force (Fc) and acceleration (ac) due to Coriolis:
Fc = 2 m Ω v sin (λ )
(units: kg (s-1) (m s-1) Æ kg m s-2 Æ N Æ force)
F
a c = c = 2 Ω v sin (λ )
(units: (s-1) (m s-1) Æ m s-2 Æ acceleration)
m
• λ = latitude (at the equator λ = 0 and at the North Pole λ = 90º);
• m = mass of the object on which Coriolis force is acting;
• v = velocity of the moving object.
Deflection due to Coriolis Force:
Δx 2 Ω sin(λ )
Δy =
v
(units: (m2) (s-1) (m s-1)-1 Æ m Æ distance)
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IV. Geostrophic Flow
Geostrophic Balance: The Coriolis force and the pressure gradient force on an air parcel balance each
other, meaning they have the same magnitude but opposite signs.
Geostrophic Balance: The Coriolis force and the
pressure gradient force on an air parcel balance each other (Note: below we
cancel mass on both sides and equate acceleration).
2 Ω v sin(λ) =
1 ΔP
ρ Δx
When in geostrophic balance, the air will always flow parallel to the isobars
(contours of constant pressure).
• The closer together the isobars are (change in pressure per unit distance),
the stronger the pressure gradient force and the faster the wind will blow.
• In Northern Hemisphere, wind will blow with higher pressure to its right,
lower pressure to its left (opposite in Southern Hemisphere).
Northern Hemisphere
F Coriolis
PGF
H
F Coriolis
L
PGF
H
L
Pressure Gradient + Coriolis + Friction: In reality, winds below an altitude
of 1 km will be subjected to a frictional force with the earth’s surface.
• This is because the presence of objects on the ground, whether mountains, trees, or buildings, will
slow the wind down.
• Friction decreases the magnitude of the velocity and therefore the Coriolis force.
• Friction does not change the magnitude of the pressure gradient force.
• Therefore, wind blowing close to the ground surface is deflected towards low pressure, as the
pressure gradient force is larger.
Winds around highs and lows
Convergence: air is moving toward low pressure at the
surface:
• Once it reaches the center, the air will rise
• As the air rises, it will cool resulting in condensation,
clouds and rain.
Divergence: air is moving away from high pressure at the
surface:
• Mass will move out of the center of the high towards
lower pressure.
• Air will descend from above to replace the loss of
mass.
• As the air sinks, it will warm up and its relative
humidity will decrease -- sunny and fair weather.
Note: In order to close the circulation lows and highs at the surface are associated with the reversed
flows aloft, highs and lows, respectively.
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Example 1: Sea/land breeze
The temperature on a beach is 25ºC and 21ºC over the ocean approximately 4 km away. The surface
pressure over the beach and over the ocean is 1010 mb.
a) Calculate the mass densities (kg m-3) of the surface air over land and over the ocean.
P
R' T
Over the beach (T = 25o C) :
ρ=
ρ=
1010 mb
100 Pa N m -2
J
×
×
×
= 1.179 kg m -3
-1
-1
(287.5 J kg K ) × (273 + 25) K 1 mb
Pa
Nm
Over the ocean (T = 21o C) :
ρ=
1010 mb
100 Pa N m -2
J
×
×
×
= 1.195 kg m -3
-1
-1
(287.5 J kg K ) × (273 + 21) K 1 mb
Pa
Nm
b) Calculate the local scale heights (km) over the beach and over the ocean. Assume the given
temperatures are representative of the air column overlying these locations.
R' T
g
Over the beach (T = 298 K)
Scale height = H =
(287.5 J kg -1 K -1 ) × (298 K) kg m 2 s -2
×
= 8733 m = 8.733 km
9.81 m s -2
J
Over the ocean (T = 294 K)
H=
H=
(287.5 J kg -1 K -1 ) × (294 K) kg m 2 s -2
×
= 8616 m = 8.616 km
9.81 m s -2
J
c) Calculate the pressure at 3 km altitude over both locations in mb.
Use the barometric law and the scale height from above to calculate the pressure at 3 km height :
Over the beach (H = 8.733 km) :
3 km ⎞
⎛ z⎞
⎛
P(3 km) = P(0)exp⎜ - ⎟ = 1010 mb × exp⎜ ⎟ = 716 mb
⎝ H⎠
⎝ 8.733 km ⎠
Over the ocean (H = 8.616 km) :
3 km ⎞
⎛ z⎞
⎛
P(3 km) = P(0)exp⎜ - ⎟ = 1010 mb × exp⎜ ⎟ = 713 mb
⎝ H⎠
⎝ 8.616 km ⎠
d) Calculate the pressure gradient (mb km-1) at 3 km elevation from the coast to 4 km offshore. What
is the wind direction at 3 km altitude? How about at the surface?
Pressure gradient =
ΔP (Pressure over beach) - (Pressure over ocean) (716 − 713) mb
=
=
= 0.75 mb km -1
Δx
Distance between locations
4 km
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Example 2: Coriolis Force
a) Where is the Coriolis effect zero?
b) Where is the Coriolis effect the greatest?
Example 3: Angular Momentum
A circular space station in outer space (no friction) with a radius of 40 m is rotating three times per
minute to induce artificial gravity for its occupants.
a) In order to decrease the gravity felt by its passengers, should the space station increase or decrease
its angular velocity?
The angular velocity would need to be decreased for the gravity to decrease.
The radius of the space station would need to be increased for the gravity to decrease.
b) Calculate the final angular velocity if the space station magically increases its radius to 55 m.
Express your answer in units of rotations per minute and radians per minute.
Conservation of angular momentum :
L initial = L final
2
2
ωinitial = m rfinal
ωfinal
m rinitial
ωfinal = ωinitial ×
2
rinitial
(40 m) 2
=
(
3
rotations
per
minute)
×
= 1.6 rotations per minute
2
rfinal
(55 m) 2
1 rotation = 2π radians
1.6 rotations
2π
×
= 10 radians per minute
minute
1 rotation
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Example 4: Pressure Gradient Force, Coriolis Force and Geostrophic Flow
The geostrophic velocity of a 1.0 kg mass of imaginary air parcel with a mass density of 0.801 kg m-3
at 45°N latitude is 12 m s-1.
a) What is the pressure gradient force acting on this air parcel?
Since we are assuming geostrophic balance, pressure gradient force equals the coriolis force (calculate
Coriolis Force) :
Fpg = Fc = 2 m Ω v sin (λ ) = 2 × (1.0 kg) × (7.3 × 10 −5 radians per second) × (12 m s -1 ) × sin( 45o ) = 0.00124 N
b) If the air parcel was instead over Boston (42°N), what would be its velocity assuming the same
pressure gradient force from above was acting on it.
Fc = 2 m Ω v sin (λ )
Use the value for Fc above, to solve for v :
v=
Fc
0.00124 N
=
= 12.7 m s -1
-5
-1
o
2 m Ω sin(λ ) 2 × 1.0 kg × 7.3 × 10 radians sec sin(42 )
c) Near the ground, friction force plays a significant role. Does friction increase, decrease or have no
impact on the pressure gradient force?
Friction does not affect the pressure gradient or pressure gradient force.
d) Does friction increase, decrease or have no impact on the Coriolis Force?
Friction decreases the velocity of a moving object, therefore decreases the Coriolis Force.
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