Warmer on the top of tall buildings

[Penumbra]

Cecil said that it is cooler on the tops of tall buildings because the air gets cooler the higher up you go. However, inside the building, as hot air rises wouldn't all the hot air from the lower floors rise to the top of the skyscraper and make the upper floors hotter? This effect works on my two story house, wouldn't it work even more so on skyscrapers to the point that the upper floors would require much less heating during winter?

[Saltire]

Oops, you forgot to link to the column. It makes it easier for everyone in the thread to refer to Cecil's words. Here you go:Is it warmer at the top of tall buildings? (http://www.straightdope.com/classics/a1_164.html)

Penumbra, your hypothesis would certainly be true in a skyscraper that had a broken ventilation system. However, they build those buildings specifically to prevent air from circulating by convection.

What struck me about this column was a rare show of humility from the Perfect Master.Originally written by Cecil
There is also the matter of "adiabatic cooling," but God Himself could not render this concept intelligible, so there is little reason to think that I can. Take my word for it.
Come on, Cece, you know God has always been lousy at explaining anything. You're better than Him (by several orders of magnitude).

[SlowMindThinking]

Perhaps Chronos can explain adiabatic cooling?

Not having knowledge, my old physics texts, nor humility, I can make a weak stab at it. In physics, "adiabatic" means a process that occurs without energy loss. So, adiabatic cooling would mean cooling without energy loss. The question is, what is cooling? The air, or the thing measuring the temperature? I'm going to assume the air.

Heat energy manifests itself as random, molecular motion. The only other ways I can picture the atmosphere storing energy is in organized motion, chemical bonds, or excited electronic states. I can't believe the last two count in this case. So, the warmest air on the ground surface rises as it expands; as it expands, it cools. This organized motion (outwards in all direction) is adiabatic cooling?

[toadspittle]

This may be the feeblest Cecil answer I've ever seen. Umm, DUH it's probably colder at the top of the Sears tower ... OUTSIDE. But inside the building envelope is a completely different matter, which Cecil, apparently, left to Saltire to even attempt to explain.

More than 20 years and Cece still hasn't expanded on this thin answer?

[bup]

Cecil...wasn't himself then.

Ted Turner has said the top floor of the CNN building is like an oven.

[Davebear]

Penumbra is closer to being right, on this one (http://www.straightdope.com/classics/a1_164.html), than Cecil is. Perhaps Cecil is getting burnt out, but he totally misunderstood the question, while Penumbra got it. Granted, it was badly phrased, but the point was still pretty clear.

The OP wasn't asking about sitting on the roof; she was talking about the heat rising through the building up to the top floor. As Saltire pointed out, this is not an issue in a building with a working HVAC system, nor is there much variation in temperature between the first and hundredth (or whatever) floors when it is working. But, what if it isn't?

Say the whole system shuts down at 7:00 am on February 29, 2004. The entire building is 70F degrees, give or take. What's the temperature on the top floor at, say, 4:00 pm versus the temp. on the ground floor? How about at 11:00 pm? Or, 7:00 am the next day?

Probably too many variables to be a practical question, but no where near as dumb as Cecil claimed.

[Rube E. Tewesday]

The questioner was going to take suntan lotion to the top of the building instead of going on a tropical vacation. How is it obvious that she was talking about going outside? Anybody can be warmer inside by turing up the thermostat. I think Cecil got it right, FWIW.

[rfs001]

The answer about OUTSIDE air temperature is not entirely accurate either. It is true that in the case of the Sears Tower, the air would be cooler. But for a shorter distance, say, 60-90 feet, on a hot day it could indeed be warmer at the higher altitude. I attended an outdoor wedding at a hillside restaurant in Southern California where, outside the restaurant, it was easily 10 degrees warmer than the air on the highway 70 feet below. All other factors being equal, convection and thermals in the air could indeed make the air warmer at elevation, so long as the distance from the ground was within 200 feet or so.

[ejacobsen]

Even for the outside-the-building case I thought Cecil's response was pretty thin. Anyone who remembers PV = T or even PV = nRT from college physics understands that the atmospheric pressure drop (P) with altitude results in a corresponding temperature (T) drop. My understanding has always been that this is the primary force affecting temperature change with altitude and the 3 degree/thousand feet rule of thumb.

This is also, BTW, the reason airliners have to run air conditioning at altitude despite being in a sub-zero temperature environment.

Just a thot...

[SlowMindThinking]

But not only is the volume increasing, which alone could counter the pressure drop, but the energy flow into the atmosphere changes, hence Cecil's answer. Changes in the energy budget is the reason are regions of the atmosphere which have a much higher temperature than at the Earth's surface. I suspect you'll find that one must make assumptions about energy flux, such as hot air rises adiabatically.

[oldionus]

Cecil avoided the whole subject of adiabatic cooling, but if anyone is interested, there's a reasoanbly comprehensible explanation of adiabatic atmospheric processes at http://daphne.palomar.edu/jthorngren/adiabatic_processes.htm

Oldionus of North Hollywood

[toadspittle]

Wow--adiabatic cooling doesn't seem that tough to explain (though calculating its effects, etc., would be murder): When a mass of air rises, its pressure decreases, causing it to spread out. This expansion cools the air mass, as the energy of the air molecules is being used to drive the expansion.

[SlowMindThinking]

toadspittle, you almost summarize the link correctly, but not quite. As the mass of air rises, the pressure of the surrounding air drops, so the "new air" spreads out, and subsequently cools like you said.

If I remember correctly, it is not that hard to calculate. The assumption, or approximation depending on how you look at it, of no heat loss, relates the volume of an ideal gas to the pressure. (That caculation is, as the text boxes used to say, "nontrivial"). But given the relationship, you can use the ideal gas law to compute the temperature drop. I would think the real bear would be computing the volume as a function of time.

[ejacobsen]

The simple idea for ideal gases is PV = T, or it's more detailed cousin PV = nRT.

P = pressure
V = volume
T = temperature

If the pressure drops and the volume is held constant, the temperature drops. If the volume decreases and the pressure is held constant (tough to do), the temperature drops. If the pressure drops and the volume increases slowly the temperature will still drop.

In the more detailed case n is the number of moles, essentially the molecular count of the gas in the volume V, and R is the universal gas constant. PV = nRT is then the state equation for ideal gases.

Certainly not all gases are ideal, and there is also the assumption that this is for a steady-state system. For the case in question, however, this should be very accurate.

The relationship doesn't require heat to be added or extracted, since T describes the heat content. Adding or extracting heat changes T and the effects on P and V are then described by the equation as well.

This isn't that tough to get one's head around, in my opinion, especially in the most fundamental form of PV = T. Certainly Cecil has gone into far more technical depth on more esoteric topics, so I was disappointed that he missed this simple explanation.

[BondJamesBond]

OK. The original answer may be correct based on abstract, theoretical principles. As one rises in altitude, temprature drops. However, I am sure that one could find easily find localized points where this does not hold true.

A very easy example may be found on the Island of Manhattan. During the afternoon the streets are in shadow while the tops of the buildings receive sunlight. A person standing on the roof of 45 Wall Street may be warmer than somebody standing on street level.

Another example is Buffalo, NY. Cold winds coming off the the lake may cool the first few floors of a building, but higher up, the chilling effects of the wind may be diminished.

I do not mean to diminish the original column, but there may be "exceptions that prove the rule."

[SlowMindThinking]

If the pressure drops and the volume is held constant

A major point of adiabatic cooling is that the volume, for the same number of moles, is not constant. When hot air rises, it expands. (The number of moles within some given volume is changing.) The ideal gas law, in and of itself, is insufficient. Knowing nothing more, it is impossible to state what happens to the temperature, because the volume could increase faster than the pressure decreases. The presumption of adiabatic cooling allows one to relate the pressure to the volume (or the change in n for a given volume) and thus use the ideal gas law to show that the temperature drops. (If I remember correctly, in adiabatic cooling the PV goes like Volume^(1-gamma), where gamma depends on the molecular structure of the substance. In this case, I believe it is 4/3, or maybe 5/3, so as the air expands, the temperature drops slowly.

I also think James Bond has a point. Buildings are known to absorb a significant amount of solar energy, hence the difficulty in cooling them. Since the upper reaches of the tallest buildings are exposed light for longer periods, it is easy to believe there is some localized heating.

[ejacobsen]

I'm not disputing that adiabatic cooling also takes place as warm air rises, or that urban canyons in shadows won't get warmed by the sun as much as higher sunlit roofs, but the most general explanation of why it gets cooler as one goes up in altitude is PV = T, and this was completely missed in Cecil's answer.

Sure, talking about adiabatic cooling is fine for the case where there is warm air rising. That is an important, but not necessarily most general, case. And several here have provided nice synopses of adiabatic cooling, too, so why the uncharacteristically incomplete response from Cecil, even in a 20-year repeat?

I think this is getting beaten to death pretty thoroughly.

[climbersam]

The tops of buildings, especially tall ones,
with lots of mechanical equipment on top, may dissipate enough heat raise the immediate temperature on the rooftop relative to ambient air temperatures. You might be able to suntan comfortably on an otherwise chiily spring day.

[Ether Dragon]

Climbersam hit on my thought. With the various equipment on building tops - heat exchangers, being one, I can easily see skyscraper rooftops being warmer than the weatherman suggests. Also, many commercial rooftops are covered with black, heat absorbing material.

[SlowMindThinking]

First, let me start by stating that I agree with others. There are undoubtedly more effects here than the simple "it gets cold when you climb a mountain". (Where do they vent the output of the cooling system?)


But, with respect to:
the most general explanation of why it gets cooler as one goes up in altitude is PV = T
ejacobsen, I'm afraid you are wrong. The temperature of the atmosphere does not drop uniformly as you climb to the space station. There are layers of the atmosphere that have a very high temperature.

In fact, an object sitting at earth's orbit, and rotating, will have a mean temperature comparable to the earth's - even though the pressure is zero. PV = nRT explains what happens to closed containers as you play with the four variables.

[doug mclure]

I dont know about outside, but in my eight storey buiding without A/C but with good insulation and heating the same in all apartments, it is definately warmer on the seventh floor than the ground floor. It is true of the rooms and particularly of the stairwell. I can only assume that it is a result of warm air rising through the building.

[Evil Captor]

From what I have read about construction work, it's often much more windy at the top of skyscrapers than at ground level. That would account for some cooling ... wind chill effect.

[ejacobsen]

PV = nRT applies to gases, period. The model deviates a bit in dynamic conditions (it's a steady-state model), but it most definitely applies to gas in the atmosphere (i.e., air). Because we have gravity on this planet our atmospheric pressure is higher at lower altitude, so P goes down as one goes up. T goes down accordingly. Very, very simple model.

The statement: "In fact, an object sitting at earth's orbit, and rotating, will have a mean temperature comparable to the earth's - even though the pressure is zero. " doesn't seem to have anything to do with gases, as near as I can tell.

"PV = nRT explains what happens to closed containers as you play with the four variables", isn't correct, as there is no assumption about closed containers. n moles of a gas will behave as described. Whether or not there is a container doesn't affect the basic gas laws.

[Marnie'sDad]

http://www.straightdope.com/classics/a1_164.html
Connie,

Your friend is correct. If it's winter in Chicago and you wish you were lying on the beach in Rio, just do the following...
[list=1] Head for the Sears Tower or any other convenient large building -- with beach chair, suntail oil, bottle of rum, etc.
Especially the bottle of rum.
Proceed to the top floor.
Do whatever you need to in order to gain access to the roof (good luck in these post 9-11 times)
Locate one of the giant exhaust fans you'll find there and set up your chair under or near it.
Close your eyes and imagine real hard that you're at the beach. The rum can be a useful aid here.
Click your heels together three times and say, "There's no place like Rio, There's no place like Rio"
[/list=1] Of course you should realize the unfortunate truth that Rio, like so many other things, probably wouldn't seem as "hot" to you (or me) today as when we were twenty-two years younger :(

[SlowMindThinking]

PV = nRT applies to gases, period

Absolutely! But it is also insufficient in this case. I said a closed container, because then you have a definite volume. What is the volume to which you are applying the ideal gas law? Clearly not the whole atmosphere. In fact, it only applies to volumes small enough so that the terms "pressure" and "temperature" are defined. So you pick small enough parcels of atmosphere of similar volume in a column within the atmosphere. Now, the pressure drops with altitude. This does not necessarily imply that the temperature drops! Why? Well, the one thing you know for sure is that the number of moles in the parcels drops with altitude. You can only determine what happens to the temperature when you have a model for the change in the number of moles.

No matter how you slice it, the ideal gas law is an equation with four unknowns. If you are measuring one, the pressure in your case, then you can only find a second, the temperature, when you have some mechanism for eliminating the other two. (In my example, I picked the volume to be some constant, arbitrary value.) No insistence that the ideal gas law applies to the atmosphere will change this mathematical fact.

To see that it is not all just esoteric, physics mind games, go to this site: http://daac.gsfc.nasa.gov/CAMPAIGN_DOCS/ATM_CHEM/atmospheric_structure.html
A brief quote: The air temperature in the stratosphere remains relatively constant up to an altitude of 25 km. Then it increases gradually to 200-220 degrees Kelvin (K) at the lower boundary of the stratopause (~50 km), Now, I happen to know that the pressure within the stratosphere does not increase with altitude, but drops exponentially, so that it is a fact that the temperature increases as the pressure decreases.

[Dolphin]

The temperature rises in the stratosphere because the expansion is not adiabatic - ozone is absorbing UV light so energy is entering the system.

[Dolphin]

Ah, apologies for the double post. I typed this out earlier but I couldn't get it to post properly. This is my first time posting so I hope you will bear with me if I step on any toes.

Adiabatic (no energy enters or leaves) expansions can easily be explained with PV = nRT. For a given parcel of air (n constant) rising through the atmosphere, pressure (P) decreases exponentially but volume (V) cannot increase exponentially because of the weight of the air above. Hence, temperature (T) must decrease to compensate.

More rigorously from thermodynamics, dU = -PdV for adiabatic expansions (q = 0), so dH = VdP. For an ideal gas dH also equals Cp * dT, so combining the two equations (and substituting V = nRT/P) gives Cp/T * dT = nR/P * dP. Since Cp, T, n, R, and P are all positive, then the sign of dP must equal the sign of dT. If the change in pressure is positive, then the change in temperature must be positive.

This holds true for all adiabatic expansions.

The problem with the stratosphere is that the adiabatic conditions don't hold. Ozone absorbing UV light pumps energy into the system, so the expansion is no longer adiabatic and the equation doesn't hold.

The problem with the interior of a building is that the roof holds up the weight of the air above it. The pressure inside the building does not decrease exponentially (unless someone opens a window) and hence, there is no drop in pressure from the ground floor to the top floor.

[SlowMindThinking]

Welcome aboard Dolphin, nice first post.

I agree with almost everything you said. You are absolutely correct in asserting that you must assume an adiabatic expansion in order to use the ideal gas law. That was a point I was trying to make.

A couple of nits, however. The weight of the air above is not the reason the volume can not increase exponentially; the weight of the air above is the source of the pressure in the first place. So, if the pressure decreases exponentially, the volume could increase exponentially. But, of course, it doesn't, because of the constraints of adiabatic expansion.

Also, the building is either a sealed container, or it isn't. If it isn't, then the pressure within the building will decrease exponentially with altitude. If it is sealed, then it is its own system, and obeying the same laws as the earth, the pressure will again decrease exponentially. It seems to me that some Chicagoans would know if high speed elevators can make your ears pop - due to the sudden pressure drop.

[Chronos]

About the ideal gas law... True, we don't have a reference for V, the volume, but then, we also don't have a reference for N, the number of molecules. Why not re-arrange the equation to P = rkT (where r is the number of molecules per volume), and thereby cancel out our ignorance? Density is a perfectly well-defined quantity, regardless of what volume we choose.

(Note for chemists: I used the logical, physicist's version of the Ideal Gas Equation. If you prefer, you can multiply and then divide the right-hand side by a large, arcane number whose significance is lost to most students and which isn't really relevant here anyway. I prefer my equations simple.)

[SlowMindThinking]

Correct me if I am wrong Chronos, but surely the density is also a function of altitude? And, it is the fact that density does not drop exponentially (uniformly) with altitude, while pressure does, that leads to the temperature dropping with altitude (below the stratosphere).

Dolfin implicity assumed N to be a constant (and so is following the same chunk of air), and explicitly that dQ vanished, so that dU = PdV. This then allows using the ideal gas law.

[The Ryan]

Originally posted by climbersam
The tops of buildings, especially tall ones,
with lots of mechanical equipment on top, may dissipate enough heat raise the immediate temperature on the rooftop relative to ambient air temperatures. You might be able to suntan comfortably on an otherwise chiily spring day.
As far as getting a tan is concerned, the top of the building definitely would be better (although perhaps not significantly). There's slightly less atmosphere for the UV to pass through, and if there's a lot of pollution at ground level, then there's less ozone to pass through as well.