Assignment 12 Answer Key Chapter 17: A6g, A8g, A12g, … · Assignment 12 Answer Key Chapter 13:...

9
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.

Transcript of Assignment 12 Answer Key Chapter 17: A6g, A8g, A12g, … · Assignment 12 Answer Key Chapter 13:...

Page 1: Assignment 12 Answer Key Chapter 17: A6g, A8g, A12g, … · Assignment 12 Answer Key Chapter 13: A3g, A7g, A14g, ... Use eqn. 13.14: ... airmasses caused by the different specific

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.

Page 2: Assignment 12 Answer Key Chapter 17: A6g, A8g, A12g, … · Assignment 12 Answer Key Chapter 13: A3g, A7g, A14g, ... Use eqn. 13.14: ... airmasses caused by the different specific

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:

Page 3: Assignment 12 Answer Key Chapter 17: A6g, A8g, A12g, … · Assignment 12 Answer Key Chapter 13: A3g, A7g, A14g, ... Use eqn. 13.14: ... airmasses caused by the different specific

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

Page 4: Assignment 12 Answer Key Chapter 17: A6g, A8g, A12g, … · Assignment 12 Answer Key Chapter 13: A3g, A7g, A14g, ... Use eqn. 13.14: ... airmasses caused by the different specific

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)

Page 5: Assignment 12 Answer Key Chapter 17: A6g, A8g, A12g, … · Assignment 12 Answer Key Chapter 13: A3g, A7g, A14g, ... Use eqn. 13.14: ... airmasses caused by the different specific

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

Page 6: Assignment 12 Answer Key Chapter 17: A6g, A8g, A12g, … · Assignment 12 Answer Key Chapter 13: A3g, A7g, A14g, ... Use eqn. 13.14: ... airmasses caused by the different specific

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)

Page 7: Assignment 12 Answer Key Chapter 17: A6g, A8g, A12g, … · Assignment 12 Answer Key Chapter 13: A3g, A7g, A14g, ... Use eqn. 13.14: ... airmasses caused by the different specific

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:

Page 8: Assignment 12 Answer Key Chapter 17: A6g, A8g, A12g, … · Assignment 12 Answer Key Chapter 13: A3g, A7g, A14g, ... Use eqn. 13.14: ... airmasses caused by the different specific

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.

Page 9: Assignment 12 Answer Key Chapter 17: A6g, A8g, A12g, … · Assignment 12 Answer Key Chapter 13: A3g, A7g, A14g, ... Use eqn. 13.14: ... airmasses caused by the different specific

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