Monday, September 14, 2009

Answer to the Friday Fermi Problem

The result of our Fermi problem: you may need one of these to brave your classes this fall.



On Friday, we posed the following back-to-school-themed Fermi problem:

Assuming you're not in a big lecture hall and the professor shuts the door at the start of class, how long does it take for you and your classmates to deplete the oxygen enough to feel it?


We promised a surprising answer, and here it is. You decide if our back-of-the-envelope calculations are reasonable.

Let's build our classroom first. It's 16 feet wide and long, and 10 feet tall. In handy metric dimensions, that's:
5 meters by 5 meters by 3 meters, or 75 cubic meters.

A cubic meter is 1000 liters, so now we've got 75,000 liters of fresh air.
The oxygen content of air is about 21 percent, and at about 17.5 percent you'll run from the room screaming. To get from fresh and breathable to absolutely stifling, take the difference between 21 percent of 75,000 liters and 17.5 percent of 75,000 liters. That gives us 2,625 liters of oxygen to get through.

How much oxygen does a human consume? It was tough finding a reliable source, but this press release about the 2006 installation of a new oxygen generation system on the International Space Station provides a clue:

During normal operations, it will provide 12 pounds daily; enough to support six crew members.


Aha! So one person needs about 2lb of oxygen a day, or .9 kg. But how many liters is that? Oxygen has a molar mass of 16 grams, so oxygen gas, or O2, has a mass of 32 grams per mole. One mole of gas at standard pressure and temperature takes up 22.4 liters. Now, as my high-school chemistry would say, it's time to hop on the mole-train:
.9 kg x (1000 g/1 kg) x (1 mole O2/32 g O2) x (22.4 L/1 mole O2)


This gives us a daily oxygen intake of 630 liters per person. Let's get a more reasonable rate:

(630 L/day) x (1 day/24 hours) x (1 hour/60 mins)


Now we have the serviceable rate of oxygen consumption of .4375 liters per minute. We're almost there.

Now populate the classroom with 34 students and 1 teacher. The 35 occupants consume 15.3125 liters per minute. Now for the final calculation:

2625 L x (1 minute/ 15.3125 L)


It will take about 171 minutes, or 2 hours and 51 minutes for the room to become unbearably stifling. You can image that you'd start to feel pretty uncomfortable about an hour and a half into the lecture—a good argument for shorter classes.

Posted by scappuccino at 5:29 PM

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26 comments:

  2. You are assuming that the room is hermetically closed. Maybe the gaps around the door alter the result, and it is also customary to have airing systems installed, so the usual answer is that you can be there indefinitely (just taking into account the oxygen and not other physiological needs).

    A much better argument for shorter classes, or classes with breaks, is mental capacity to absorb knowledge.
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  3. Wow, Ignacio. If there was an award for missing the point, you'd win it.
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  4. The better measure to know is that at rest on average we consume about 3.5 ml of O2/kg/min. With an average weight for the 35 people of 80 kg, you would have about 10 L/min of O2 consumption. Where do you get the 17.5% as difficult to maintain? That is still more than enough PO2 for diffusion to occur. As a commenter on another board brought up, the increasing CO2 is the real problem with this (This might be getting off the point of the Fermi problem though). If you truly wanted to measure how much you could sustain you would want to calculate PO2 down to a level that diffusion is just barely still occurring (~40 mm Hg) which would be much lower than 17.5%.
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  5. Anonymous CowardSep 16, 2009 11:14 AM
    To be read tongue in cheek:

    Cheater! Using the web and some NASA resource to answer a Fermi problem? For shame! :)


    Fermi would know that the average human hanging out is roughly a 100 Watt device, and that a chemical bond has a few eV of energy. So to generate 100 W, you need to be burning around 10^20 bonds per second and consuming 10^20 oxygen atoms per second. 1 STP liter of pure gas contains a few times 10^22 molecules (ideal gas constant and all that), so one person burns around 0.01 l of O2 a second.

    To within order-of-magnitude accuracy, this reproduces the NASA number.

    Alternatively, you could estimate O2 consumption (and get a similar answer) by estimating the size of human lungs (a few liters), the rate at which you breathe (a few seconds), the fraction of O2 in the air (around 20%) and take a wild guess at the fraction of that you can absorb (10%?) to get pretty much the same answer.
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  6. Anonymous CowardSep 16, 2009 11:17 AM
    P.S. I forgot to mention: fun problem! Thanks!
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  7. Excellent job tackling the problem without internet-research cheating...as far as what percentage of oxygen we need to be comfortable, I found different answers depending on where I looked. I love the idea of estimating oxygen intake using the volume of the lungs, by the way...

    Also, we welcome Fermi problem ideas, so comment with questions you've been itching to calculate and we might post it on our Friday blog.
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  8. I think there is an important piece of information that is missing for us to make a reasonable calculation on this problem, and that is the elevation of the classroom. Higher elevations have lower pressure, changing the concentration of the room, even though it will have the same volume. I'm assuming that the ISS is pressurized at 1 atmosphere (I don't know for sure though). However, if our classroom is in Denver, Colorado, our students would pass out significantly sooner than if they were in Death Valley, California (or maybe the same time because of the heat).
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  9. Hemoglobin saturation is not linear with oxygen partial pressure and a decrease of atmospheric oxygen content from 21% to 17.5% at sea level is unlikely to have much effect at all. Framing this as a physics problem is amusing, but it it is really an environmental physiology question.
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  10. you didn't take into account the volume of the people in the room...
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  11. > 5 meters by 5 meters by 3 meters

    These dimensions are a near-perfect fit for my living room (it's an old Viennese building, so the rooms are all quite high). So, I know exactly what you're talking about, and I can tell you with confidence: Cramming 35 people into my living room is unrealistic. It would work if you didn't give them proper desks, but it would still be absurdly crammed. Your calculation that after three hours they'd be itching to open the damned windows already seems entirely plausible, it's just that it's very implausible that such a situation would arise in the first place.
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  12. I estimated the size of the classroom that inspired this problem based on my extensive experience estimating the sizes of things (believe it or not, it's a little hobby of mine, and it's much harder than it sounds). I will be there again this afternoon, and will bring both a measuring tape and a camera. You might be right . . . we'll see soon.

    -Buzz
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  13. You cannot reasonably fit 35 students into a 5 by 5 meter room. At 25 square meters of floor space, giving every person a square meter of space (with a reasonably-sized desk) you cannot have more than 25 people in the room, and this would mean that no one can get out (no isles between desks), and the professor must spend the entire class within his 1-meter space. Accounting for isles and whiteboard space in front, the number of people goes down, unless you start to stack them up.
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  14. We'll know for sure when I measure the room later today, but I don't think I agree with you. It's about .75 meters from the back of my chair to the front of my knees. And my chair is about half a meter wide. So, I'm only taking up .375 square meters at the moment. If we bump that up to .5 square meters to give me room to squirm, it still leaves plenty of space to squeeze 35 people into a 5x5 meter room.
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  15. It also doesn't take into account the number of bathroom breaks during that period. Chances are someone is going to have to go, breaking the air lock. Like others said, it's also not realistic based on the classroom specs and the fact that public buildings today are required to have ventilation systems by code. However, it is fun to think about. What if the room has windows and it's a hot sunny day? The greenhouse effect will certainly have a significant impact on the results.
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  16. Re Riordon: unless you plan to take notes on your knees (or keep your laptop on your actual lap), the distance from the back of your chair to the front of your desk will be about a meter, and if you imagine other students sitting less than half a meter away from you on each side, you'd be nudging them with your elbows all the time, so your .5 meter estimate is definitely way too low

    You need space in order to be moderately comfortable. Sure, you could fit 35 people in a small room. Hey, they fit 14 people in a phone booth, didn't they? But that's not realistic if we're talking about a regular classroom. If you try to fit 35 people in a 5x5 room for a physics lecture, you'll likely get injured in a fight caused by too many irritated people jostling each other by accident long before the air runs out.
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  17. So, how many square meters per person do you think should be appropriate? At 1 square meter per person, the classroom must be 5.6 meters on a side. If you prefer that we round up to 6 rather than down to 5 (although 5 is a MUCH nicer number for Fermi calculations),that changes the volume of the room by 100*(36-25)/25=35%. That results in an increase over Scappucino's estimate of about 3 hours to about 4 hours. That's not much of a change, from a Fermi problem point of view. It would still be pretty uncomfortable pretty quick in a room that size.

    Another factor that wasn't included is the fact that there were probably students in the room during the previous hours too. So the oxygen concentration starting point might be a bit low when class begins.
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  18. Anonymous CowardSep 17, 2009 12:32 PM
    Re: the quibbling over numerical factors

    Are you guys kidding? I mean, I suppose the original poster deserves some ribbing for writing down the answer to an approximate calculation to 7 digits of precision (and I agree that 5x5x3m may have a volume approaching an order-of-magnitude smaller than a typical 35-person classroom). But trying to get things to better than a factor of 10 or so is kinda missing the point of what is usually implied by the term "Fermi Problem".
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  19. Good point, AC. Precision is almost the antithesis of this sort of exercise. I went and measured the room and desks in the Math building of the University of Maryland anyway.

    The 5x5x3 (75 square meters) estimate is low, not by too much. The actual room is 7x6x2.4=100 square meters, which is an even better number for Fermi problems. I also packed chairs as tightly as reasonable possible to see how many square meters each student actually needs, the answer, 18 seats easily fit in a 2x5 meter space. That means each student needs about .6 square meters.
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  20. I'll post pics of my seating measurements later.
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  21. Anonymous CowardSep 17, 2009 04:06 PM
    Re: riordon room volume measurements

    OK, I hate myself for saying this, and I am a big jerk, but I can't help it.

    Perhaps you mean 100 cubic meters?

    Best regards,
    A.C.


    P.S. You guys got some low classroom ceilings there.
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  22. Oops, indeed I do mean cubic meters. My bad. That's a much bigger problem than a factor of 1.3 error in the estimation.

    Proper units are critical to Fermi problems, while precision is not.

    Thanks for the correction, AC.
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  23. To be more precise, the room is actually about 100 cubic meters in total volume, rather than 75 cubic meters. But my desk packing experiment shows that each student comfortably fits in about .6 square meters of floor space (I didn't bother trying to estimate a typical student's vertical dimension while seated).
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  24. How about considering if an English grad student was in the class mistakenly? They would suffer a panic attack within the first few minutes and consume a helluva lot of oxygen very rapidly!
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  25. The anwser is wrong. The issue of temperature & altitude have to be addressed.
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  26. This is the funnest physics problem I've ever seen. Maybe it's because I've experienced this before...
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