Please show your work on all problems. Check that your results are reasonable.

 1. Using the NCAR web site, calculate the density of air (in kg m^-3) at the Foothills and MESA Labs for the most recent data you can find.

Example - the exact answer will depend on what day you chose.

density = Pressure/RT (where R=287)

Foothills (morning 990917) P= 840 mbar, T=10 C
                (afternoon 990917) P = 848 mbar, T=27 C

Density at Foothills ~ 1.03 kg/m^3 (morning)
                                     0.98 kg/m^3 (afternoon)

2. Using the hypsometric equation, reduce each of the pressure measurements from 1 above to equivalent sea level pressure

Rearrange the hypsometric equation to obtain:

P_o = P_g exp(z/H) where H = RT/g, and you should use the ground temperature (for lack of anything better).

For the example above, Morning T = 283 C, P = 840 hPa
                                        Afternoon T = 300 K, P = 848 hPa

The altitude at Foothills is 1625 m (from web), so z/H = 0.196, so that P_o/P_g = 1.217, or Po = 1022 hPa in morning.
z/H = 0.185, so that Po = 1020 in afternoon.

Note, most of the pressure change at 1625 m from morning to afternoon appears to be related to the change in temperature. As T increased, so did P, but
the reduced P at sea level altitude changed less.

Repeat the same calculation for Mesa.
 

3. A hot air balloon is lifted by the buoyancy force, which can be calculated by

where r _balloon is the density of the air in the balloon, r _env is the density of the displaced environmental air, V is the volume of the balloon, and g is the acceleration of gravity.
 

(a) For a typical hot air balloon with a volume of 2200 m3, what is the maximum total mass (note, this includes the material of the balloon) that can be lifted off the
ground at the Foothills Lab on the morning of January 31, 2003 if the average air temperature inside the balloon is maintained at 100 ^oC?

This example is for 1999 - I include it here to give you an idea of how to go about this problem. The only twist here is that you needed to convert the pressure
on the weekly chart for NCAR to the Foothills Lab pressure, since pressure reduced to sea level is what they plot.

T = 100 + 273 K = 373 K
so density of hot air in the balloon at Foothills lab in the morning, where pressure = 840 mbar is

density = 84000 Pa/287x373 = 0.785 kg/m^3

(Note: since the volume of the balloon is 2200 m^3, the mass of air in the balloon is 0.785 kg/m^3 x 2200 kg = 1727 kg)

Since F=mg, and the buoyancy force is given by the formula above, the mass that can be lifted by the balloon is:

mass = F/g = Volume x (density of ambient air - density of air in balloon) = 2200 m^3 x (1.03 - 0.785) kg/m^3 = 539 kg

This is a reasonable number...accounting for the mass of the balloon, gondola, and a couple of people.

(b) How much more does the pilot of the balloon need to heat the air in balloon to keep this balloon afloat with this maximum payload at the altitude of the Mesa Lab?

Again, this answer is for the 1999 date.

At the altitude of the mesa lab (1885 m) the ambient air density is smaller than at Foothills. From the data posted on the web on Friday Sept. 17, the pressure at the Mesa Lab in the morning was 816 mbar, and temperature was 16 C, so the ambient air density was 0.983 kg/m^3. Thus, to achieve the same value for F/m = 539 kg, the difference in density between ambient air and air inside the balloon must be the same as at the ground (1.03-0.785) = (0.983 - X), so X = 0.738 kg/m^3. From the ideal gas law, this means that the air in the balloon must be heated to 385 K, or 112 degrees C.

(c) Why do hot air balloonists like to fly on cold mornings?

The density of ambient air is greater when the temperature is coldest, so the lift is greater for a given temperature of air inside the balloon.

4. (a) Compute the scale height for an isothermal atmosphere with a temperature of 250 K.

H = RT/g = 287x250/9.81 = 7314 m = 7.31 km

(b) Compare the pressure values in this isothermal atmosphere with the US standard atmosphere for altitudes of 5 km, 11 km, and 25 km.

P(5 km) = 1013 x exp(-5/7.31) = 511 mbar (US Std =540 mbar)
P(11 km) = 225 mbar (US STD = 230 mbar)
P(25 km) = 33 mbar (US STD = 25 mbar)

5. (a) Use the hypsometric equation (with regular temperature) to calculate the heights in meters for the following Denver radiosonde profile. The surface elevation is 1625 m.

 Level     Pressure     Temperature     Water Vapor
                mb                     C                     g/kg

SFC         841                 18.8                     4.82
                700                    3.8                    4.65
                500                -13.9                    1.32
                400                -24.5                    0.07
                300                -40.7                    0.22
 

Using the hypsometric equation deltaZ= -RT/g x ln(P2/P1),

thickness(841-700) = -287x 284.5/9.8 x ln(700/841) = 1529 m
similarly,
thickness(700-500) = 2639 m
thickness(500-400) = 1658 m
thickness(400-300) = 2025 m

So 700 mb -> 3154 m
     500 mb -> 5793 m
     400 mb -> 7451 m
     300 mb -> 9465 m

We can also solve this problem using the hydrostatic balance equation, deltaZ = deltaP/density x g
Again, we should use average temperature and pressure in the layer to calculate the average density, so

deltaZ1 = (84100 - 70000)/[(77050/284x287)x 9.81] = 1520 m
deltaZ2 = (70000 - 50000)/[(60000/268x287)x 9.81] = 2614 m
deltaZ3 = (50000 - 40000)/[(45000/254x287)x 9.81] = 1651 m
deltaZ4 = (40000 - 30000)/[(35000/241x287)x 9.81] = 2014 m

So the heights that correspond to the pressure levels are:

841 mbar = 1625 m
700 mbar = 1625 m + 1520 m = 3145 m
500 mbar = 3145 m + 2614 m = 5759 m
400 mbar = 5759 m + 1651 m = 7410 m
300 mbar = 7410 m + 2014 m = 9424 m

These are very similar to what we got from the hypsometric equation.

(b) What is the 700-500~mb thickness (in m)?

2639 m

(c) Use virtual temperature instead of temperature for calculating the 700 mb height. Is there much difference?

Tv(P) = [1 + watervapor x (29.9/18 - 1)/1000] x T(K)

Tv(841) = [1 + 4.82x(0.661)/1000] x 292.0 = 292.9 K
Tv(700) = 277.8 K (true = 277.0 K)
Tv(500) = 259.5 K (true = 259.3 K)
Tv(400) = 248.7 K (true = 248.7 K)
Tv(300) = 232.5 K (true = 232.5 K)

So for the 700 mbar height, we use an average temperature of 285.4 K instead of 284.5 K, so the 700 mbar hight is 1625 m + 285.4/284.5 x 1520 m = 3150 m. This is very small difference (5 meters), but it can be significant when small pressure differences are responsible for winds (which you'll learn about in 4720).

(You may check your answers using the PAOS Weather Center web site; this sounding was for 0Z September 15, 1999.)
(actual was 700 mbar = 3145 m, 500 mbar = 5800 m, 400 mbar = 7460 m, 300 mbar = 9480 m)

6. (a) Use the PAOS Weather Center sounding plotting web site(http://paos.colorado.edu/~wxp archives section, general sounding archive) to find recent examples of a well mixed boundary layer and a very stable boundary layer. Show the two corresponding Stuve diagrams.

Stable boundary layer
Well mixed boundary layer

(b) Support and explain your selection by quoting a few relevant potential temperatures (THETA) from the text outputs.

(c) Using the text output of the sounding, show that the lapse rate in the well mixed boundary layer (excluding the surface layer) is close to the dry adiabatic lapse rate. Explain why.

(d) Why does a shallow nocturnal inversion not imply, in general, that pollution will be trapped near the surface during the day?

As the surface heats during the day, the inversion will be broken, and air will eventually become unstable, allowing mixing.

7. Commercial aircraft normally cruise near 200 mb where the temperature is typically -60 ^oC.

(a) Calculate the potential temperature of the environmental air.

Theta = T(K) x (1000/P)^0.286 = 213.2 x (1000/200)^.286 = 337.8 K

(b) Calculate the temperature of air if compressed adiabatically to the cabin pressure of 800 mb.

337.8 K = T(at 800 mbar) x (1000/800)^.286, so T(at 800 mbar) = 317 K

You would have arrived at the same answer with the formula T (at 800 mbar) = T (at 200 mbar) x (800/200)^0.286

(c) How much heat per mass of air (J/kg) must be then be added/removed to maintain a cabin temperature of 25 ^oC? During a cross country flight perhaps 2000 kg of air are exchanged in the cabin. Express the total heat added (removed) during a flight in terms of the equivalent mass of water frozen (or ice melted).

To solve this we need the heat capacity of air, and use the formula for constant pressure (we need to cool the air and keep the pressure the same, 800 mbar)

dq = Cp x dT, where dT = 317 K - 298 K

(we've already compressed it adiabatically, i.e. without putting in any heat, to get it to cabin pressure, where its temperature is 317 K, now we need to cool it to 298 K)

Since Cp = 7/2R  for diatomic gases (like air), and R = 287 J/(kg degree),

dq = 1005 J/kg degree x 19 degrees = 19,100 J/kg (or 19.1 kJ/kg air)

For 2000 kg of air, we need a total of 3.82 x 10^7 Joules of heat removed. We could conceivable do this by passing the warm air over ice and melting it, a process that requires 3.34x10^5 J/kg (the "latent heat of fusion"). So we would have to melt 114 kg of water to accomplish this cooling of cabin air. That's the equivalent of about 1 passenger with baggage ($750 to the airlines).

Another way we could achieve this cooling would be to pass the warm air over water and evaporate it. The latent heat of evaporation (condensation) is 2.5x10^6 J/kg, about 7.5 times larger than the heat of fusion, so we'd only need about 114/7.5 kg = 15.2 kg of water to cool the air to a comfortable temperature. This process would also humidify the air...adding about 7.5 kg of water to 2000 kg of air...or it would humidify the air to about 3.75 g/kg....hmm, this is very interesting! This is about the humidity of air near the surface (see problem 4 above). Can you explain why this must be the case?

(5710 questions)

8. Repeat problem 4, only using a surface temperature of 288 K and a lapse rate of -6.5 degrees km-1.

You only have to do part b, sorry I didn't state this specifically.

Use Equation 2.35 in the text, or from the handout from class.

In this case:

(5 km)  ln(P/P_o) = g/(RG) ln(255.5/288), from which you can determine ln(P/Po) = -0.629, or P = 1013.25 x 0.533 = 540 hPa
Note, you have to use 0.0065 deg/m for G because this is MKS units.

(11 km) same analysis, but for T = 216.5 K, you get 226.3 hPa

(25 km) same analysis, but for T = 125.5 K, you get 12.9 hPa

Note that in comparison to STD atmosphere, the error is 0% for 5 km, -1.6% for 11 km, and -48% for 25 km. This is an improvement for low altitudes, but not so good for the stratosphere. This is because the stratosphere is warmer than the upper troposphere, so the temperature doesn't keep dropping. I.e. the STD atmosphere can't use this better formula if we are dealing with the stratosphere. But airplanes usually fly in the troposphere (unless you are in the polar regions or you are a U-2 or SR-71).

9. A typical tropical sounding has the tropopause at 17 km with a pressure of 94 mb and a temperature of 195 K. Calculate the temperature of the parcel compressed adiabatically from the tropopause to sea level using both Poisson's equation and the dry adiabatic lapse rate. Note: the two methods give different results.

Poisson's equation T(1000 mbar) = 195 K x (1000/94)^0.286 = 383 K
Dry adiabatic lapse rate, T(sea level) = 195 K + 10 deg/km x 17 km) = 365 K

(a) Use hydrostatic balance and the ideal gas law to show that Poisson's equation and the dry adiabatic lapse rate will give the same parcel temperature only if the environmental lapse rate is equal to the dry adiabatic lapse rate (neutrally stable). Hint: find an expression for the height of a given pressure level in an atmosphere with a constant lapse rate.
 
To be honest, the hint isn't the best way to do this. The better way is to take the derivative of the Poisson equation for potential temperature of a parcel of air that is undergoing adiabatic expansion, and do some rearranging. Here goes:

Q = T (P/1000)^-R/Cp

take derivative with respect to altitude, set equal to zero (for an adiabatic process, potential temperature is constant)

dQ/dz = 0 = dT/dz (P/1000)^-R/Cp - (RT/PC_p) dP/dz (P/1000)^-R/Cp

0 = [ dT/dz - (RT/PC_p) dP/dz] (P/1000)^-R/Cp = [dT/dz - (RT/PC_p) dP/dz] Q/T

Noting that Q/T cannot be zero, this means that the term in brackets must be zero. What is this term?

dT/dz - (RT/PC_p)dP/dz = 0. To be clear, let's note that T is for a parcel, or write  T_parcel

But note that pressure is determined from the hydrostatic equation, using density based on the ambient temperature profile NOT the temperature of the profile. We can then say the following: dP/dz = - r_ambient g = - Pg/RT_ambient.  We don't need to distinguish the ambient P from the parcel P, because they must be the same (the parcel equilibrates pressure with the surroundings).

So, we can substitute hydrostatic balance for dP/dz, to get

(dT/dz)_parcel - (RT_parcel/PC_p) (Pg/RT_ambient) = 0

or,  ***** (dT/dz)_parcel = (T_parcel/T_ambient) g/C_p  ******

Now, we have our result. The problem asked to show that (dT/dz)_parcel equaled the adiabatic lapse rate (i.e. g/C_p) for adiabatic expansion (i.e. dQ/dz = 0) if and only if the atmospheric temperature lapse rate is also the adiabatic lapse rate. In this case, the only way for (dT/dz)_parcel = g/C_p is if T_ambient = T_parcel explicitly.  This means that the ambient temperature has to decrease at the same rate with altitude as the parcel temperature, and we know the latter is decreasing at the adiabatic lapse rate.

One thing to note from the equation **** **** above is that a parcel of air that is ascending adiabatically in an atmosphere that has a different lapse rate than the adiabatic lapse rate will decrease in temperature at a rate that is slightly different than g/C_p, but that this difference is small (of order a few degress out of 250 or so, or 1%).
 

(b) Carefully examine the derivation of the dry adiabatic lapse rate, and determine which assumption is to blame for getting the wrong parcel temperature.

The derivation of the dry adiabatic lapse rate (see lecture notes, page 5) uses the barometic law (P = Po x exp(-z/H) where H = RT/g with ambient temperature, not the temperature of the parcel!). If instead of ambient temperature we used the parcel's temperature, we wouldn't have this problem, but the pressure of the atmosphere is determined by the bulk of air (i.e. the parcel is too small to alter the bulk properties of the much larger surrounding air).