ATSC 201 Fall 2021

Assignment 8 Answer KeyChapter 5: A5, A6, A10c, A11c, A14c, A16c, A17c Chapter 14: A3c,

A7c(1), A8c, A11c, A16c, A39c

**All thermo diagrams indicated can be found at the end of this document.

CHAPTER 5

A5)

See tephigram

A6)

r = 0.95 g/kg

θ = 62 °C

Tw = -26 °C

θw = 22.5 °C

Using thermodynamic state of the air parcel given in the previous exercise, plot it on the end-of-chapter large thermo diagram specified by your instructor (if none specified, use the skew-T diagram). Hint: Use a copy of a blank thermo diagram so you can keep the master copy clean.Instructor specified to use tephigram only.Use data from A4c) P(kPa) = 35, T(degC) = -25, Td(degC)= -30.

For one assigned air-parcel state from question A4, find each of the following using only a Skew-T or Tephigram to answer these questions. Do not solve equations (except for relative humidity)i) mixing ratioii) potential temperatureiii) wet-bulb temperatureiv) wet-bulb potential temperaturev) saturation mixing ratiovi) LCLvii) relative humidityviii) equivalent potential temperature

Check: Td less than T. Makes sense.Discussion: The dew-point temperature point on a tephigram also shows us the water-vapour mixing ratio of the air parcel at that height. The temperature point, on the other hand, shows us what the maximum mixing ratio of water vapour the air parcel can hold for that given temperature and pressure.

rs = 1.4 g/kg

LCL = 32.5 kPa

RH = 67.86 % RH = r/rs*100

θe = 22.5 °C

A10c)

Td = -61 °C

rs = 0 g/kg

T = -26 °C

*below -38C ri = 8 g/kg rL = rT-rs

rT = 8 g/kg

A11c)

LCL = 82 kPa

Use the data below as an air-parcel initial state, and plot it on the specifiedlarge end-of-chapter thermo diagram. Assume this air parcel is then lifted to a final height where P = 25kPa. Find the final values of the following variables Td, rs, T, rL and rT:c) P = 95 kPa, T = 20 degC, Td = 10 degC, (Skew-T)

Check: Wet-bulb temperature is between the dew-point and temperature. Ok.Discussion: The convenience of a tephigram is evident in this problem. We don't need to apply several equations like last HW to get some of the important variables. For example, the LCL would require us to do several steps ofcomputation.

Check: Physics ok. Discussion: As the air parcel is lifted from the surface, its temperature decreases dry adiabatically. Notice that as the parcel is being lifted past saturation, its saturated mixing ratio decreases. The saturated mixing ratio decreases because colder air can hold less water vapour. At temperatures below about -38°C, all hydrometeors are in the form of ice crystals.

Using the same thermo diagram as specified in exercise A10, find the pressure at the LCL for an air parcel that started with the conditions as specified in that exercise.

A14c)

See Skew-T

A16c)

Given: P0 = 90 kPa

p1 = 80 kPa

Te @90 kPa 21 deg C 294.15 K

Tp @90 kPa 21 deg C 294.15 K

Find: F/m ? m/s^2

Plot the above sounding on a large thermodynamic diagram from the end of this chapter. Use solid dots (of red color if possible) for the temperature and mark an open circle (blue color if available) for the dew points. Draw a solid (red, if possible) line to connect the temperature dots, starting from the bottom up, and draw a dashed (blue, if possible) line for the dewpoints.c) Tephigram

Suppose that you create an air parcel at the pressure-level (kPa) indicated below, where that parcel has the same initial thermodynamic state as the sounding at that pressure. Then move that parcel up through the environmentto the next higher significant level (ie. next lower pressure). What is the value of the buoyant force/mass acting on the parcel at its new level? c) 90.

Check: Physics ok. Discussion: Colder air can hold less water vapour, and thus it eventually reachessaturation and cannot hold any more water vapour. As the parcel continues to rise, itgets even colder moist adiabatically. The "excess" water vapour then condenses intoliquid droplets and forms cloud. This begins at the lifting condensation level. Notice that as we go higher up, the saturation mixing ratio decreases. The difference between the mixing ration from the surface (rT) and the saturated mixing ratio is the amount of liquid mixing ratio the parcel has due to condensation.

Check: looks reasonableDiscussion: The temperature line shows us the environmental temperature sounding of the atmospheric column. The dew-point temperature line shows us how much moisture is in the atmospheric column. At each level with data, the closer the temperature and dew-point temperature are, the more saturated it is.

Use eqn. 5.3b:

where |g| = 9.8 m/s^2

F/m= 0 m/s^2

A17c)

Given: P_bot 70 kPa

P_top 60 kPa

r_bot 1.1 g/kg 0.0011 g/g

r_top 1.1 g/kg 0.0011 g/g

T_bot 12 °C 285.15 K

T_top 2 °C 275.15 K

Find: Nbv = ?

Pbv = ?

Use eqn. 1.21:

where

a= 0.61 g/g

Tv_bot = 285.341336 K 12.1913357 °C

Tv_top = 275.334626 K 2.18462565 °C

ΔTv = -10.00671 K -10.00671 °C

Given the sounding data: Find NBV and PBV for an air parcel that starts in the middle of the layer indicated below, and for which its initial displacement and subsequent oscillation is contained within that one layer. Use the layer with a bottom pressure-level (kPa) of: c) 70

Check: Units ok. Physics okDiscussion: As the air parcel rises dry adiabatically from 90kPa to the next significant level (P=80kPa) we see that the temperatures of the parcel and the environment are equal, hence the parcels buoyancy acceleration is 0.

Use eqn. 1.26a:

where

a= 29.3 m/K

Tv_av = 7.18798065 °C 280.337981 K

Δz = 1266.17871 m

Use eqn 5.4a:

where |g| = 9.8 m/s^2

Γd = 9.8 K/km 0.0098 K/m

NBV = 0.05085507 rad/s

Use eqn 5.5:

PBV = 123.550819 s

2.05918031 min

CHAPTER 14

A3c)

Check: Units ok. Physics okDiscussion: This parcel will oscillate between the layers approximately every 2 minutes.

Given the following prestorm sounding. Plot it on a thermo diagram (use a skew-T). Find the mixedlayer height, tropopause height, LCL, LFC, and EL, for an air parcel rising from the surface. Use a surface (P = 100 kPa) dew-point temperature (°C) of:c) 20

Mixed layer height = 90 kPa, 0.9 km

Tropopause height = 25 kPa, 9.3 km

LCL = 88 kPa, 1 km

LFC = 83 kPa, 1.5 km

EL = 17 kPa, 10.1 km

A7c(1)

# of blocks= 1 of 0.5km* 1°C 0.5

1 of 1km* 4°C 4

1 of 2km* 12°C 24

1 of 2km *17°C 34

1 of 2km *14°C 28

1 of 1km *5°C 5

total area = 1*0.5 +1*4 +1*24 +1*37 + 1*28 +1*5

Total area= 95.5 °C/km 95500 °C*m

Use eqn. 14.3: CAPE = |g|*Total area/ avgTe

where

where |g| = 9.8 m/s^2

av_Te = -16 °C 257.15 K

CAPE = 3639.51001 J/kg

For the sounding from exercise A3, find the value of surface-based CAPE, using:(1) the height (z) tiling method. Assume Tv ≈ T. Also, on the plotted sounding, shade or color the CAPE area.

Check: Units ok. Physics okDiscussion: As expected, the top of the mixed layer is somewhere within the inversion layer of the vertical temperature profile (The inversion layer is between 92 kPa and 88 kPa).

Check: Units ok. Physics okDiscussion: Surface based CAPE indicates the amount of available potential energy for the thunderstorm to grow. If the air parcel from the surface reaches and goes beyond the level of free convection.

A8)

0.7* 2°C

1* 5°C

1* 11°C

1* 15°C

2* 20°C

1* 18°C

1* 16°C

1* 9°C

0.5* 2°C

Total area = sum of above

= 116.4 °C*km

= 116400 °C*m

Use eqn. 14.3: CAPE = |g|*(total area)/avgTe

where

where |g| = 9.8 m/s^2

av_Te = -15 °C 258.15 K

CAPE = 4418.82626 J/kg

Start with the sounding from exercise A3. Assume that the forecast high temperature for the day is 2°C warmer than the surface temperature from thesounding. Find the value of surface-based CAPE. Assume Tv ≈ T.

Check: Units ok. Physics okDiscussion: We see that with a greater surface temperature, the CAPE for the sounding increases. This explains why many convective thunderstorms occur in the late afternoon when surface heating in the daytime allows the surface temperature to reach its maximum.

A11c)

MLCAPE = 1500 J/kg

Find: Thunderstorm/tornado intensity and uncertainty

A16c)

Given: SB CAPE = 1500 J/kg

Find ω_max= ? m/s

use eqn. 14.7:

ω_max= 54.77 m/s

Use eqn. 14.8:

ω_max_likely = 27.3861279 m/s

Forecast the thunderstorm and/or tornado intensity, and indicate the uncertainty (i.e., the range of possible storm intensities) in this forecast, given aML CAPE (J·kg–1) value of:1) 1500

Discussion: There is a large amount of uncertainty in this method, as can be seen with the dark shadings in Fig 14.42. This speaks to the difficulty of trying to forecast thunderstorms as well as the uncertainty that all these averaging methods bring.

Estimate the max likely updraft speed in a thunderstorm, given a SB CAPE (J·kg–1) value of: c) 1500

Check: Units ok. Physics ok. Discussion: If we plug in the SB CAPE we found in A8, the most likely maximum updraft speed reaches almost 50m/s or 180km/hr!

From Table 14-1, a MLCAPE of 1500 indicates a moderately unstable atmosphere, therefore a moderate thunderstorm is likely.

From Figure 14.42, a MLCAPE of 1500 J/kg most likely indicates a supercell thunderstorm with no tornado. It is likely that a supercell will form, and unlikely that a tornado will form.

A39f)

From Table 14-7:

BRN = 70

BRN Shear = 30 m^2/s^2

CAPE ML = 1205 J/kg

CAPE MU = 1850 J/kg

CIN = no value

EHI = 0.5

LCL = 1.47 km

SCP = 1.1

0-6km Shear 15 m/s

SRH = 70 m^2/s^2

effective SRH = 60 m^2/s^2

STP = 0

Discussion: The different stability indices help forecasters to predict the

likelihood of thunderstorm and tornadic activity.

What value of the Bulk Richardson Number, BRN-Shear, ML CAPE, MU CAPE, CIN, Energy Helicity Index, Lifting Condensation Level, Supercell Composite Parameter, 0-6 km Shear, Storm-Relative Helicity, effective Storm-Relative Helicity, and Significant Tornado Parameter would you anticipate for the following intensity of thunderstorm (CB)? f) supercell with EF2- EF5 tornado

–50 –40 –30 –20 –10 0 10 20 30 40 50

Temperature (°C)

Pre

ssur

e (k

Pa)

100

90

80

70

60

50

40

30

20

15

T =

–60°

C iso

ther

m

–70°

C

–80°

C

–90°

C0.1 0.15 0.5 0.80.2 0.3 0.4 1.0 1.5 2.0 3 4 5 6 8 10 15

20g/kg

30

40

50g/kg

Mixing Ratio (g/kg)

Dry Adiabat θ = 50°C

60°C

100°C80°C

120°C 140°C

Moist A

diabat θW

0

1

2

3

4

5

6

7

8

9

10

11

12

Hei

ght (

km)

TephigramCopyright © 2015 by Roland Stull.Free copies of this page permitted, from “Practical Meteorology: An Algebra-based Survey of Atmos. Sci.”, via Creative Commons CC-BY-NC-SA/4.0International License.• State lines are thin.• Process lines arethick, and are labeledwith temperature wherethey cross P = 100 kPa.(θW = wet-bulb potential temp.)

Tθ θW

°C

–50 –40 –30 –20 –10 0 10 20 30 40

Temperature (°C)

Pre

ssur

e (k

Pa)

100

90

80

70

60

50

40

30

20

–60°

C isot

herm

T

–70°

C–8

0°C

–90°

C

810

1520

2530

4050

(g/k

g)

0.1 0.15 0.2 0.3 0.4 0.6 0.8 1.0 1.5 2 3 4 6

Mixing Ratio (g/kg)

50°C

100°C Dry Adiabat θ

60°C

70°C

80°C

90°C

Moi

st A

diab

at θ

W

Hei

ght (

km)

00.5

11.5

2

2.5

3

4

5

6

7

8

9

10

11

12Tθ θW

°C

Skew-T Log-P DiagramCopyright © 2015 by Roland Stull.Free copies of this page permitted, from “Practical Meteorology: An Algebra-based Survey of Atmos. Sci.”, via Creative Commons CC-BY-NC-SA/4.0International License.• State lines are thin.• Process lines arethick, and are labeledwith temperature wherethey cross P = 100 kPa.(θW = wet-bulb potential temp.)

–50 –40 –30 –20 –10 0 10 20 30 40

Temperature (°C)

Pre

ssur

e (k

Pa)

100

90

80

70

60

50

40

30

20

–60°

C isot

herm

T

–70°

C–8

0°C

–90°

C

810

1520

2530

4050

(g/k

g)

0.1 0.15 0.2 0.3 0.4 0.6 0.8 1.0 1.5 2 3 4 6

Mixing Ratio (g/kg)

50°C

100°C Dry Adiabat θ

60°C

70°C

80°C

90°C

Moi

st A

diab

at θ

W

Hei

ght (

km)

00.5

11.5

2

2.5

3

4

5

6

7

8

9

10

11

12Tθ θW

°C

Skew-T Log-P DiagramCopyright © 2015 by Roland Stull.Free copies of this page permitted, from “Practical Meteorology: An Algebra-based Survey of Atmos. Sci.”, via Creative Commons CC-BY-NC-SA/4.0International License.• State lines are thin.• Process lines arethick, and are labeledwith temperature wherethey cross P = 100 kPa.(θW = wet-bulb potential temp.)

–50 –40 –30 –20 –10 0 10 20 30 40

Temperature (°C)

Pre

ssur

e (k

Pa)

100

90

80

70

60

50

40

30

20

–60°

C isot

herm

T

–70°

C–8

0°C

–90°

C

810

1520

2530

4050

(g/k

g)

0.1 0.15 0.2 0.3 0.4 0.6 0.8 1.0 1.5 2 3 4 6

Mixing Ratio (g/kg)

50°C

100°C Dry Adiabat θ

60°C

70°C

80°C

90°C

Moi

st A

diab

at θ

W

Hei

ght (

km)

00.5

11.5

2

2.5

3

4

5

6

7

8

9

10

11

12Tθ θW

°C

Skew-T Log-P DiagramCopyright © 2015 by Roland Stull.Free copies of this page permitted, from “Practical Meteorology: An Algebra-based Survey of Atmos. Sci.”, via Creative Commons CC-BY-NC-SA/4.0International License.• State lines are thin.• Process lines arethick, and are labeledwith temperature wherethey cross P = 100 kPa.(θW = wet-bulb potential temp.)

–50 –40 –30 –20 –10 0 10 20 30 40

Temperature (°C)

Pre

ssur

e (k

Pa)

100

90

80

70

60

50

40

30

20

–60°

C isot

herm

T

–70°

C–8

0°C

–90°

C

810

1520

2530

4050

(g/k

g)

0.1 0.15 0.2 0.3 0.4 0.6 0.8 1.0 1.5 2 3 4 6

Mixing Ratio (g/kg)

50°C

100°C Dry Adiabat θ

60°C

70°C

80°C

90°C

Moi

st A

diab

at θ

W

Hei

ght (

km)

00.5

11.5

2

2.5

3

4

5

6

7

8

9

10

11

12Tθ θW

°C

Skew-T Log-P DiagramCopyright © 2015 by Roland Stull.Free copies of this page permitted, from “Practical Meteorology: An Algebra-based Survey of Atmos. Sci.”, via Creative Commons CC-BY-NC-SA/4.0International License.• State lines are thin.• Process lines arethick, and are labeledwith temperature wherethey cross P = 100 kPa.(θW = wet-bulb potential temp.)

–50 –40 –30 –20 –10 0 10 20 30 40

Temperature (°C)

Pre

ssur

e (k

Pa)

100

90

80

70

60

50

40

30

20

–60°

C isot

herm

T

–70°

C–8

0°C

–90°

C

810

1520

2530

4050

(g/k

g)

0.1 0.15 0.2 0.3 0.4 0.6 0.8 1.0 1.5 2 3 4 6

Mixing Ratio (g/kg)

50°C

100°C Dry Adiabat θ

60°C

70°C

80°C

90°C

Moi

st A

diab

at θ

W

Hei

ght (

km)

00.5

11.5

2

2.5

3

4

5

6

7

8

9

10

11

12Tθ θW

°C

Skew-T Log-P DiagramCopyright © 2015 by Roland Stull.Free copies of this page permitted, from “Practical Meteorology: An Algebra-based Survey of Atmos. Sci.”, via Creative Commons CC-BY-NC-SA/4.0International License.• State lines are thin.• Process lines arethick, and are labeledwith temperature wherethey cross P = 100 kPa.(θW = wet-bulb potential temp.)