Download 501 09 Quiz #3 solutions

1. 10 points each
1. List which of these quantities is conserved under the processes described below.
(a) mixing ratio, (b) saturation mixing ratio, (c) potential temperature, (d) equivalent
potential temperature, (e) relative humidity, (f) wet bulb potential temperature.
(d,e,f ) lifting of saturated air: Theta e, Theta w conserved, RH = 100%
(a, b, e) isothermal expansion: ws (T) conserved so w conserved, regardless of whether
the parcel is saturated. If w and ws are conserved, then RH is conserved too.
( a,c, d, f) adiabatic expansion: w and theta conserved; Theta e, Theta w too.
( d, e, f ) saturated ascent in a cloud: same as lifting of saturated air
( d,f ) cooling air to its wet bulb temperature by evaporating water into it: we can get
there by lifting the parcel to its LCL and descending moist adiabatically, conserving
theta-e and theta-w. Note that w and RH are not conserved.
Consider an air parcel at a pressure of 1000 hPa.
For what relative humidity is the potential temperature the same as the equivalent
potential temperature? 0%
For what relative humidity is the potential temperature the same as the wet bulb
potential temperature? 100%
Can the equivalent potential temperature ever be the same as the wet bulb potential
temperature? (yes, no; if yes, under what conditions) Yes but only of the air parcel is
so cold that the saturation mixing ratio is virtually zero.
If the atmosphere were to warm, the (observed lapse rate, the dry adiabatic lapse rate,
the moist adiabatic lapse rate) would certainly decrease. It isnʼt clear how the lapse rate
wound change. The dry adiabatic lapse rate certainly wouldnʼt change. The moist
adiabatic lapse rate would certainly decrease because saturation mixing ratios would be
higher and hence rising air would warm more as a result of the release of latent heat;
i.e., the difference between gamma sub d and gamma sub w would increase.
If the moist adiabatic lapse rate were defined for phase change from water vapor to ice
rather than water vapor to liquid water, the moist adiabatic lapse rate would be (steeper,
the same, less steep). The release of latent heat would be larger, so and hence rising
air would warm more as a result of the release of latent heat; i.e., the difference
between gamma sub d and gamma sub w would increase, and the moist adiabatic
lapse rate would be less steep.
2. An commercial aircraft is flying at a pressure altitude of 200 hPa, where the
temperature outside is –60°C. The pressure inside the cabin is maintained at 850
hPa.(a) If the outside air were adiabatically compressed to “cabin altitude”, what
would be the temperature inside the cabin? (b) How much heat would need to be
added to (or extracted from) the air drawn into the cabin in J kg-1 to maintain the
cabin at a comfortable temperature of 21°C? (c) How would the answers to (a) and
(b) be different if the cabin altitude were 1000 hPa rather than 850 hPa? (d) If you
solved this problem using the skew T ln p chart, show the equation you could use to
solve for the cabin temperature in (a) analytically. (20 points)
The skew T ln p chart yields almost 50°C.
You could also use T(850) in °K = T(200) (850/200)^(287/1004)
" "
"
"
= 213 (4.25)^0.286
"
"
"
"
= 213 x 1.52 = 322 K = 49°C.
To do the numbers on this one I cheated and used my 50-year old slide rule.
To cool the air from 49 to 21°C, would require removing 1004 x (49 – 21) = 28100 J per
kg of heat from the air. if the airline tried to maintain the pressure at 1000 hPa and the
temperature at 21°C, it would have to cool the air down from
" "
"
"
= 213 (5)^0.286
"
"
"
"
= 213 x 1.585 = 339 K = 66°C
which would require removing 1004 x (66 – 21) = 45180 or almost twice as much air
conditioning. Thatʼs one of the reasons that pressure in the passenger cabin is lowered
as the plane climbs toward cruising altitude.
3. On cold, calm winter days in Fairbanks, Alaska, a shallow temperature inversion often
forms just a few tens of meters above the surface. Surface temperatures often hover
around –40°C while temperatures just above the inversion are 20°C higher. Under
these conditions dense ice fog often forms below the inversion over the city as a
result of the exhaust from motor vehicles and heating systems. If the pressure is 1000
hPa (a) estimate the water vapor mixing ratio at the surface. (b) If local human
activities account for half the water vapor in the air below the inversion, estimate the
relative humidity of the air just above the inversion? (20 points)
At these very low temperatures the saturation mixing ratio is very low. I read 0.13 g / kg
from the chart. The air is saturated, so 0.13 g / kg must be the actual mixing ratio as
well. If we assume that half the water vapor beneath the inversion is from local human
activities and the other half is natural and characteristic of the air mass both below and
above the inversion, then the mixing ratio above the inversion should be 0.065 g / kg
whereas reading from the chart, I get a saturation mixing ratio of about 0. 7 g / kg.
Hence, the relative humidity of the ambient air is only about 10%.
4.The updraft of a heavy thunderstorm remains fixed above a station for 103 s (about 15
min). At cloud base the density of the air is 1 kg m-3, the water vapor mixing ratio is 20 g
kg-1 and the updraft velocity is 1 m s-1. Estimate the depth of the rainfall over this period,
assuming that all the water entering the base of the updraft in the form of water vapor
falls out as rain. (20 points)
The mass flux of water vapor into the updraft in units of kg m-2 s-1 is equal to the air
density times the specific humidity q (in dimensionless units) times the updraft velocity.
It is always allowable to use mixing ratio w as a very close approximation of q. Hence,
the upward mass flux into the cloud is 1 kg m-3 x 0.02 x 1 m s-1 = 0.02 kg m-2 s-1. Over a
period of 1000 s the updraft will deliver 20 kg of water per square meter, all of which is
assumed to fall out as precipitation. If a 1 square meter column of water i m deep has a
mass of 103 kg (i.e., the density of liquid water), then 20 kg of water corresponds to a
depth of 20 mm or 2 cm (almost an inch), which is quite impressive for a rainfall event
lasting only 15 minutes. Weʼll encounter more problems like this one in the cloud
microphysics chapter.
5. Calculate the wavelength of stationary gravity waves induced by the flow of air over
rough terrain under the following conditions. The lapse rate is isothermal, the air
temperature is 0°C, and the wind speed is 30 m s-1. (20 points)
The period of the waves is given by the Brunt Vaisala frequency N. On the whiteboard, I
provided you with the formula N 2 = (gamma sub d minus gamma) x (g/T). Substituting
numerical values, we obtain N 2 = 9.8 x 10-3 x 9.8 / 273 = 3.52 x 10-4 ; N = 0.0188*; and
the period P = 2 pi / N = 334 s. Since the wave is stationary, the wavelength L is simply
the distance traveled by an air parcel passing through the wave during a time interval of
one period; i.e., L = PU = 30 m s-1 x 334 s = 10.0 km.
*When you take square roots in problems like this one, be sure to write the number that youʼre taking the
square root of so that the exponent of the 10 is divisible by 2.