Assignment 12 Answer Key Chapter 17: A6g, A8g, A12g, … · Assignment 12 Answer Key Chapter 13:...
Transcript of Assignment 12 Answer Key Chapter 17: A6g, A8g, A12g, … · Assignment 12 Answer Key Chapter 13:...
ATSC 201 Fall 2017
Assignment 12 Answer Key
Chapter 13: A3g, A7g, A14g, A16g
Chapter 17: A6g, A8g, A12g, A13g, A17g, A20g, A35g
Chapter 13
A3g)
Given: Δz_mtn = 1.6 km
Δz_T = 12 km
φ = 45 deg
Find: A = ? km
Use eqn 13.3:
where fc/β = Rearth*tan(φ) from page 444 below eqn 13.3
Rearth = 6371 km
fc/β = 6371 km
A = 849.466667 km
Check: Units ok. Physics ok.
Discussion: The higher the mountains, the shorter the
column of air will be, and with this greater change, relative vorticity will
need to decrease more to conserve potential vorticity. This decreasing
relative vorticity is what initiates the Rossby wave.
A7g)
Given: φ = 60 °
Δz_mtn = 2 km 2000 m
Δz_T = 12 km 12000 m
M = 50 m/s 0.05 km/s
Find: Rc = ? km
Firstly, need to find fc at location c:
Can assume that the ridge after location c is at the same latitude as location a.
Therefore can use equation 13.4 to find the amplitude of the Rossby wave.
where R_earth (km) = 6371
(m) = 6371000
A = 1061.83 km
1061833.33 m
Then, use eqn 13.2:
β = 1.14E-08 1/m*s
Ω = 7.29E-05 /s
Can then rearrange eqn 13.3 to solve for fc at location c:
fc@c = A*β*Δz_T/Δz_mtn = (1/s) = 7.29E-02 1/s
Next, we can find potential vorticity, knowing that ΔzTa = ΔzTc, with eqn. 13.6:
where fca = 2*Ω*sinφ = 1.26E-04 1/s
ζp = 1.05E-08 1/(m*s)
Knowing that potential vorticity is constant, we can rearrange 13.5 to solve for
Rc:
So Rc = M/(ζp*ΔzTc - fc@c)
Rc = -6.87E+02 m
-0.69 km
Check: Units ok. Physics ok.
Discussion: The radius of curvature at location c must be positive because it
is turning cyclonically to keep potential vorticity constant. The radius of
curvature equation at location b (13.7) does not apply at location c because
the Coriolis force is smaller (flow is closer to equator).
A14g)
Given: P = 85 kPa
W = 50 m/s
Find: ω = ? Pa/s
Use eqn. 13.14:
From Table 1- 5 estimate ρ
ρ = 1.006 kg/m^3 g = 9.8 m/s^2
ω = -492.94 Pa/s
Check: Units ok. Physics ok.
Discussion: The term omega, by definition, is the change of pressure with
time. So, when omega is negative, this means that you are going higher
up as pressure decreases with height.
A16g)
Given: D = 5.00E-05 /s
Δz = 2 km 2000 m
Find: Wmid = ? m/s
Use eqn. 13.15:
Wmid = 0.10 m/s
Check: Units ok. Physics ok.
Discussion: Due to the conservation of mass (continuity equation) if there is
horizontal divergence, then there must be vertical motion to fill in the air that
is leaving horizontally.
Chapter 17
A6g)
Given: α = 40 °
CD = 0.002
h = 5 m
Tve = 20 °C 293 K
Tvp = 15 °C
Find: Ueq = ? m/s
Use eqn. 17.9:
where:
Δθv = Tvp - Tve = (K) = -5
g = 9.8 m/s^2
Ueq = 16.39 m/s
Check: Units ok. Physics ok.
Discussion: The equilibrium downslope wind speed is the wind speed
at which the drag balances buoyancy
A8g)
Given: Tv = 20 °C 293 K
d = 700 m
Δθ = 4 K
Find: MSBF = ? m/s
Use eqn. 17.11:
where k = 0.62
g = 9.8 m/s^2
MSBF = 6.00 m/s
Check: Units ok. Physics ok.
Discussion: The sea breeze front is the boundary between two different
airmasses caused by the different specific heat capacities of water and land.
During the day, it will take longer for the water to warm up comparing
to land. This difference causes cooler marine air to form, and the
temperature difference induces a pressure difference which drives
the sea breeze.
A12g)
Given: g/Tv = 0.0333 m/s^2/K
Δθv = 4 K
h = 50 m
Find: co = ? m/s
Use eqn: 17.15:
co = 2.58 m/s
Check: Units ok. Physics ok.
Discussion: Wave do exist in the atmosphere, and behave similarly to water
waves.
A13g)
Given: M = 15 m/s
Find: Fr = ?
From eqn. 17.16, for interfacial waves:
Also for interfacial waves, eqn 17.18:
From the previous question:
cg = co = 2.58 m/s
Fr = 5.81
This flow is supercritical.
Check: Units ok. Physics ok.
Discussion: Air that is supercritical is moving so fast that noinformation can
travel upwind, so the upwind air does not 'feel' the effects of downwind flow
constrictions until it arrives at the constriction.
A17g)
Given: g/Tv = 0.0333 m/s^2/K
Δθv = 4 K
h = 300 m
Find: M_gap_max = ? m/s
Use eqn. 17.25:
Mgap_max = 6.32 m/s
Check: Units ok. Physics ok.
Discussion: Any wind speeds faster than this will likely cause
turbulence and a drag force that will subsequently slow down the winds.
Also, note that in this scenario, we are assuming that the gap is short
enough that the Coriolis force is small enough to be neglected.
A20g)
Given: g/Tv = 0.0333 m/s^2/K
M = 5 m/s
ΔT/Δz = 5 °C/km 0.005 K/m
Find: λ = ? m
Use eqn. 17.30 to find the natural wavelength:
Use eqn. 5.4a to find the Brunt-Vasala frequency:
where Γd = 9.8 °C/km
0.0098 K/m
NBV = 0.0222 /s
λ = 1415.13 m
Check: Units ok. Physics ok.
Discussion: NBV only exists in a statically stable environment.
A35g)
Tnew = 17 °C +/- a couple deg
Td_new = -4 °C +/- a couple deg
Use eqn 4.14b to find RH:
r @95kPa = 3 g/kg
rs @ 95kPa= 13 g/kg
RH@95kPa= 23.08 %
Discussion: This type of precipitation process over mountains often
occurs in our region (the west coast of British Columbia). With low-
pressure systems that bring in a lot of moist marine air, the east side
of the Vancouver Island (such as Tofino) often receives much more
precipitation than the Lower Mainland. The air that makes it to the
Lower Mainland is often much drier