Not true. At 15,000 feet altitude, atmospheric pressure is 8.6 psi. That means only 8.6/14.7 = 58.5% of the atmosphere’s mass is above this elevation.
At the top of Everest, ambient pressure is 4.6 psi; only 31% of the atmosphere is above this elevation.
OK, the standard atmosphere model I have says 20 degrees F, 14.7 psi at sea level.
It also says that at the top of Mount Everest (~29,000 feet), the temperature is -48 degrees F, and the pressure is 4.58 psi. I’ll assume these conditions are accurate, and then conduct the following simulations.
Sim 1:
If I run an adiabatic expansion from sea level conditions to a final pressure of 4.58 psi, I arrive at a temperature of -81 degrees F. This is a surprisingly large discrepancy from the standard atmosphere.
Sim 2:
If I take a parcel of sea-level air; move it to 29K feet, increasing its gravitational potential energy by a calculable amount; and subtract the same amount of thermal energy from the air, I get to -87F, a slightly larger discrepancy.
The problem with Sim 2 is that the air hasn’t been allowed to expand yet. Once you let it expand to the ambient pressure at that altitude, it cools off even further.
Moreover, the reality is that a parcel of air doesn’t propel itself to higher altitudes by virtue of its own thermal energy; it’s pushed/convected there by other air masses, and so its own thermal energy does not have to be compromised in order to obtain altitude.
Either way, I’m surprised that the adiabatic expansion result is so far off. Anyone got any insight? Is my standard atmosphere wrong? Is there conduction/mixing with warmer air?
The air on top of Mount Everest isn’t constantly moved there from sea level.
You sure about that? The US Standard Atmosphere rattling around in my head says 59 deg F at sea level. I can’t be arsed to check on that right now, but it sure as hell isn’t 20.
If you factor in the 40 degree discrepancy in your starting conditions, I bet your calculated temperature at altitude is a lot closer to the measured value.
I mistyped that one, sorry. 20 is the Celcius value; it’s 68F.
The other numbers in my previous post are correct.
Well, as was pointed out, things start to get wacky with extreme elevation. How about running calcs for something closer to earth than 29k feet? (Also, I’m assuming you’re not saying the top of Everest, specifically, is 48F, but average air at that elevation.)
I’m really stunned that we don’t have a clear, agreed-upon answer for this, btw.
I think the OP question is surprisingly hard to answer. The answers relating to hot air rising, and heat capacity, and adiabatic expansion sound skeevy to me, though I am not perfectly sure why. I do understand that there is a global scale circulation of lower wind blowing from the poles toward the equator, hot humid air rising in the tropics, high altitude wind from the equator toward the poles, and recently cooled air falling at the poles. It looks to me like a big convection cell carrying heat away from the earth like a conveyor belt.
I want to guess the important effect here is that there is enough water vapor and maybe carbon dioxide in the air to radiate in the thermal infrared and lose energy to outer space, and that’s the mechanism for heat loss at the top of the system. Solar warming of the ground and especialy the oceans, driving both heating and humidification of the air, is the mechanism for heat gain at the bottom of the system. Overall it would be cold up high because that’s where heat is mostly lost from the system.
But, this isn’t my field, and a few of the other things people have brought up also sound relevant to me.
Joe Frickin Friday, in your sim 2, you moved the entire parcel of air from sea level to 29k feet, but only the air molecules with a large enough vertical velocity to make it up to 29Kft (or to have fallen from 29Kft) should be included.
Looking at the distribution of velocity in a gas, the three coordinates are independent. Taking Z as vertical, you’re left with only particles with abs(V[sub]z[/sub]) >= V[sub]c[/sub], the velocity needed to rise 29Kft. After rising 29Kft, those particles will have a vertical velocity given by V’[sub]z[/sub][sup]2[/sup] = V[sub]z[/sub][sup]2[/sup] - V[sub]c[/sub][sup]2[/sup], and unchanged horizontal velocities. You end up with an identical velocity distribution, just with V’[sub]z[/sub] replacing V[sub]z[/sub], so it has the exact same temperature, but with fewer particles, hence a lower pressure.
So my thought that the vertical velocity accounts for the temperature difference seems fatally flawed. Some questions that maybe you or someone else can follow up on: Does the parcel of air moved up to 29Kft now have the correct pressure for that altitude? If not, and does adiabatic expansion of that parcel then account for the temperature difference?
What? Wait, JFF addresses this point when he says “Moreover, the reality is that a parcel of air doesn’t propel itself to higher altitudes by virtue of its own thermal energy; it’s pushed/convected there by other air masses”. He is of course right, and a little extra thought shows that we are nowhere near the Knudsen regime; the air molecules are primarily interacting with each other and not moving over the scale of the problem by coasting on their thermal velocities.
The standard atmosphere (either US or ISA) is 15 degrees C at sea level, not 20.
You have to include the entire velocity. Most real systems will thermalize fairly quickly, which means that the energy of the particles gets distributed such that the velocities have a thermal distribution. One of the features of a thermal distribution is that it’s isotropic, with all directions being equally likely for the velocities. So even if a particle did give up a little of its vertical velocity to rise to a greater height, it would eventually interact with other particles in such a way as to convert some of the energy in its horizontal motion to vertical. The net effect is that energy from any component of motion can be used to lift the particle.
His Sim 2 was looking at my comment about atoms from higher up gaining energy by falling, with this causing the higher temperatures at lower altitudes and lower temperatures at higher altitude. I was imagining a more-or-less stationary column of air. The particles at lower altitude with small vertical velocity can’t get up to higher altitude (other than by gaining vertical velocity from some other particle that does have enough).
Don’t get hung up on the interactions between particles. You can look at a weakly interacting case, or at a series of small vertical steps, each one with (in my original wrong thinking) a slightly higher temperature at the bottom than the top. Instead, in equilibrium, the temperature is constant top to bottom, and just the pressure varies.
Particles that rise in altitude don’t have to convert energy in horizontal motion to vertical. The subset of those particles that have enough vertical velocity to rise to the higher altitude, or to have come from the higher altitude, will have the correct thermal distribution of vertical velocities at that altitude. The horizontal velocity distribution is already correct.
Napier, I was thinking about your “skeevy” comment, and I think you’re right. If the air at a higher elevation was colder in equilibrium just because it’s higher, you could set up a perpetual motion machine using the temperature difference as a source of power. It has to be colder because of some non-equilibrium condition, like Earth being relatively warm compared to space, like you said in the next paragraph.