The Chinook wind is a warm dry wind that comes down from the Rocky Mountains into Alberta, and can turn a winter day into short-sleeve weather in the space of hours. But what heats up the air? It wasn’t that warm on the BC side of the rockies, before it crossed the mountain range. Well, I decided that I was going to calculate it. Let's see if this works.

Let’s say the air comes off the ocean at about 10 degrees Centigrade and 90% relative humidity, which isn't actually typical for a dreary Vancouver winter day—but that's because the Chinook is powered by the Pineapple Express, which is warm, wet air coming inland from Hawaii. It travels inland, raining on Vancouver as it goes, until it hits the coast mountain range, and is forced to rise.

You’ve probably noticed, if you’ve ever changed altitude quickly, that it gets colder the higher you are. Well, that wind from the ocean is going to do exactly that.

To do the math, we’ll look at a small piece of the wind, pretend it stays together to simplify things, and follow it over the mountains.

First up is the ideal gas law, \(PV=nRT\), which relates the pressure (\(P\)), temperature (\(T\)), and volume (\(V\)) of a quantity (\(n\)) of gas.

When our part of the ocean wind reaches land, it is at sea level, so the pressure is 1 atmosphere. We’ll pretend it’s a perfectly standard pressure day, 101.325 kPa - the average air pressure at sea level. And we’ll consider a volume of gas of 1 cubic meter, which is completely arbitrary but we need to pick some starting point. Therefore:

\(P = 101.325\mathrm{kPa}\)

\(V = 1\mathrm{m}^3 = 1000\mathrm{L}\)

\(T = 10\mathrm{C} (283.15\mathrm{K})\)

\(R\) is the ideal gas constant, \(8.314 \mathrm{L.kPa}/\mathrm{mol.K}\).

The only factor in the equation we don’t have now is \(n\), the actual quantity of the gas, in units of moles.

Solving for our starting conditions:

\[n ={PV \over RT} = {(101.325\mathrm{kPa})(1000\mathrm{L})\over (8.314 {\mathrm{L.kPa}\over\mathrm{mol.K}})(283.15\mathrm{K})} = 43\mathrm{mol}\]This equation says nothing about the mass or type of molecules. It’s a funny thing about ideal gases, that the pressure and volume depend on the number of molecules and not on the mass of those molecules, but that’s how it works.

Now, air is a combination of about 78% nitrogen, 21% oxygen, and small quantities of CO_{2}, water vapour, and miscellaneous other things, depending on where it’s been hanging around. That water, despite being only a tiny percentage of the total number of molecules floating around in our cubic meter of air, has a massive effect on it. We’ll get to that later.

Our piece of the wind rises up the side of the mountain. As the altitude increases, the pressure decreases as well as the temperature.

Looking at the ideal gas law above, we know the quantity of gas we’re starting with and the average air pressure at various altitudes is known, but we have two unknowns, temperature and volume. We can’t assume either of those is constant; as air pressure decreases volume increases and vice versa, and the air also gets colder as you go higher—into lower pressure regions.

One assumption we can make is that the gas is adiabatic—no heat is transferred in or out. Since all the air around our piece of wind is doing the same thing and changing in pressure, temperature, and volume at the same time and in roughly the same place, this is a reasonable assumption: in order for heat to flow from one thing to another, a difference in temperature must be present. The equation for adiabatic expansion is \(PV^\gamma = \mathrm{constant}\), where \(\gamma\) is \(7\over5\) for air. We can combine this with the ideal gas law above to hide the volume term (which, really, we’re not interested in) and show the temperature term (which we are interested in), to get \(P^{(1-\gamma)}T^\gamma = \mathrm{constant}\). (Remember, \(R\) is a constant and \(n\) is a defined quantity of air that we are looking at and thus also constant in this case, so they were just rolled into the other constant.)

The constant is dispensed with by simply doing the equation twice for two different conditions. Say, our starting conditions, 1 atm and 10C, and some final conditions, of 0.8 atm (roughly equivalent to 2000m altitude) and an unknown temperature. Now we have only one unknown:

\[P_1^{(1-\gamma)}T_1^\gamma = P_2^{(1-\gamma)}T_2^\gamma\] \[1\mathrm{atm}^{(1-{7\over5})} \times 283.15\mathrm{K}^{7\over5} = 0.8\mathrm{atm}^{(1-{7\over5})} \times T_2^{7\over5}\]Solving for \(T_2\), the temperature is 265.7K, or -7.5C. This is a 17.5C drop from our starting conditions of 10C.

If this is the high point of our mountain range, the air then drops down the slope, we do the equation in reverse, and the air warms up again. This is of course not the whole story; if it were, the air would warm up to a slightly cooler temperature than it started at, because the prairies are well above sea level where they meet the rockies.

This is where we bring water into the equation.

Water in the atmosphere is sometimes measured as %RH (relative humidity), and sometimes as the dew point temperature. %RH is the percentage of water vapour in the air relative to the maximum amount of water vapour the air can hold. The dew point temperature is the temperature at which dew will form given the amount of water vapour in the air. At 100%RH, the dew point temperature is the same as the air temperature; the air cannot hold any more moisture as water vapour.

The actual quantity of water vapour that the air can hold is equivalent to the vapour pressure of water, which is related to the temperature of the air only, according to:

\[P = \mathrm{e}^{(20.386-{5132\over T})}\]where \(T\) is in Kelvin and \(P\) is in millimeters of mercury.

So our starting point, 10C (283.15K) gives a water vapour pressure of 9.6mmHg, or 1.28 kPa (1.26% of 1atm). This means that in our initial \(1\mathrm{m}^3\) of air at 10C and 90% RH, we have 90% of 1.26% of 1atm of water vapour: 1.14% of the cube of air is water vapour. Doesn't sound like much.

If we reverse the equation and use 90% of 9.6mmHg as our water pressure, we can find out at what temperature our cube of air will start to form dew:

\[T = {5132\over{20.386 - \mathrm{ln}P}}\]T = 281.51K (8.4C)

Now let's run the adiabatic expansion equation backwards, and find out the pressure at which our parcel of air starts to form dew.

\[(1\mathrm{atm})^{(1-{7\over5})}\times 283.15\mathrm{K})^{7\over5} = P_2^{(1-{7\over5})}\times 281.51\mathrm{K}^{7\over5}\]This gives a pressure of 0.98 atm (98.3kPa, or about 150m above sea level).

Why is this important?

This is the altitude at which clouds start to form.

From this point on, we can't simply use the adiabatic expansion equation anymore because water vapour is condensing into the visible droplets that make up clouds, and when water vapour condenses, it releases heat. A lot of heat. That heat has to go somewhere, and it goes to warming up (or rather, slowing the cooling of) our parcel of rising and cooling air.

Now, recalling that 1.14% of our parcel of air is water vapour, that means for every kg of air (roughly that cubic metre we started with) we have 0.0114kg of water vapour.

Let's pick a temperature arbitrarily and see how much water vapour condenses: -5C (268.15K), cooling the parcel of air by 13.4C from the dew point calculated above. Using the above vapour pressure equation, this gives us 3.48mmHg, or 0.464 kPa (0.46% of 1atm). This means instead of 0.0114kg of water vapour, we have 0.0046kg of water still as vapour. So, 0.0068kg of water vapour has condensed, releasing its heat at a rate of 2257kJ/kg, for a total of 15.3kJ of energy released. This energy goes both into warming the water and to warming the air.

So if the temperature of our air is -5C after warming, what would the temperature have been without the warming? Once we figure this out, we can use this unwarmed temperature to go back to the adiabatic expansion equation and figure out our altitude.

We're going to warm up three different things by the same amount: 0.0046kg of water vapour, 0.0068kg of liquid water, and 1kg of air. (Actually 9.9886kg, but we'll use 1 since I'm not using enough decimal places for it to make a difference. The amount of water vapour we're heating may not make a difference either; let's find out.) How much energy it takes to warm something up is called its heat capacity. For water vapour this is 1.86 kJ/kg.K, for liquid water it is 4.2kJ/kg.K, and for dry air it is 1.005kJ/kg.K.

\[e = C_p m (T_2-T_1)\]So, we know the total energy available for warming (15.3kJ), we know the heat capacity \(C_p\) for water vapour, liquid water, and dry air, we know the ending temperature \(T_2\), and we know the mass \(m\). We do not know \(T_1\), and we do not know how much energy goes to each of water and air. What we really have is a system of 4 equations and 4 unknowns:

Water vapour: \(e_w = (1.86 {\mathrm{kJ}\over\mathrm{kg.K}})(0.0046\mathrm{kg})(268\mathrm{K} - T_1)\)

Liquid water: \(e_l = (4.2 {\mathrm{kJ}\over\mathrm{kg.K}})(0.0020\mathrm{kg})(268\mathrm{K} - T_1)\)

Air: \(e_a = (1.005{\mathrm{kJ}\over\mathrm{kg.K}})(1\mathrm{kg})(268\mathrm{K} - T_1)\)

Total: \(e_w +e_l + e_a = 15.3\mathrm{kJ}\)

But since the masses of water vs. air, 0.0114kg and 1kg, are so vastly different, we'll start with the assumption that only the air equation is going to be needed:

\[15.3\mathrm{kJ} = (1.005{\mathrm{kJ}\over\mathrm{kg.K}})(1\mathrm{kg})(268\mathrm{K} - T_1)\]Solving for \(T_1\) we get 253.91K or -20.23C, a difference of 15.23C from the actual temperature. To sanity check the assumption that only the mass of air matters for this calculation, let's find the energy to heat the water by 15.23 degrees:

\[e_w = (4.2 {\mathrm{kJ}\over\mathrm{kg.K}}) (0.0068\mathrm{kg}) (15.23\mathrm{K}) = 0.4339\mathrm{kJ}\]As 0.4339kJ is about 2.8% of 15.3kJ, it was reasonable to assume that the water vapour not condensing wouldn't make much difference to the final temperature, at least for the level of precision I'm using. Calculating the energy absorbed by the water and water vapour will refine the calculation somewhat, and slightly lower the overall temperature rise, probably by about 3% or so. As there is less water vapour than liquid water and the heat capacity is about half, it will use about 1% of the energy and can safely be ignored as well.

Now with the adiabatic expansion equation which we've already used a few times, we can find out what altitude we're at. That is, the altitude at which the air would be -20.23C, had there been no condensing water vapour warming it up to -5C, the actual temperature.

\[0.98\mathrm{atm}^{(1-{7\over5})}\times 281.51\mathrm{K}^{7\over5} = P_2^{(1-{7\over5})}\times 252.91\mathrm{K}^{7\over5}\]This gives us a pressure of 0.67atm (68.2kPa, or about 3200m above sea level).

To simplify the calculations for the rest of this exploration, we found here that for a rise of 3050m (remember, we started forming clouds at 150m) we dropped 13.4 degrees C, or 4.4 degrees per 1000m. This means, to reach an altitude of 3500m above sea level, which for the purposes of this calculation I am calling the height of the rockies, (there are peaks higher and passes lower, but as a simplifying assumption it should be reasonable) we move up 3200m from our cloud formation height, losing 14.02C in temperature to bring us to -5.66C.

At this point, all of the water which has condensed and warmed the air is effectively left behind as cloud or rain, and we have much less water vapour to deal with when the air starts going down the other side of the mountain. Not only that, but as water is no longer condensing, we can go back to using simply the adiabatic expansion equation we started with. Then, we found that we had a change of 28.6 degrees C in 3050m, or about 9.38 degrees per 1000m. As the air drops from 3500m (the peak of the mountains) to 1140m (Calgary, right on the edge of the prairies), we drop 2360m, which works out to a 22 degree rise in temperature. From -5.6 at the peak of the mountains, this gives us a balmy 16C - which will easily melt snow, and in the middle of a prairie winter is definitely T-shirt weather.

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