A physics question



#21

Here is a question I posed to my kids. I'm embarassed to say I'm not quite sure of the correct answer(s) or the reasoning behind them. I've gotten some interesting replies from the kids but I'm not sure any of us are on the right track. I thought I would post it here so we could find out for sure. I did mean to say that all extraneous forces can be ignored (friction, piston mass, etc.).

" Suppose you had a steel cylinder 12" in diameter, 50 miles long with a one square-inch hole through the center from end to end. The cylinder is sealed on one end and in the bore there is a piston which seals perfectly to the inside surface of the bore. You place the cylinder vertically at sea level with the piston completely at the bottom. As you just begin to raise the
piston you measure a force pushing down to be about 14.7 lbs which is the 'weight' of the air column from the top side of the piston to the edge of space. As you begin to raise the piston further what happens to the pulling force required and why?

Now, take the cylinder into space, say to the moon. In this case there is no air pressure at all, essentially a vacuum everywhere. Does this mean the piston can be moved freely from the bottom to anywhere else within the cylinder's length without any real force being required? Are there 'different' kinds of vacuums? Is there something 'more empty' than a perfect vacuum (i.e. 'negative' pressure)?"

As a caveat to the previous question, would a suction cup stick to the outside of the porthole of the Lunar Lander as it sat on the Moon's surface?

Thanks for any thoughts!

LHH

"You are what you don't excrete" (unknown author)


#22

I would say:

As you raise the piston, the force required to move it at any one elevation/height will reduce (in proportion to the amount of air remaining above the piston). For example, here in Flagstaff (elevation almost exactly 7000 feet above sea level), the atmospheric pressure is around 75-80% of sea level, so you would need around 11-12 pounds of "pull." The vacuum inside the tube, as you pull the piston higher, basically does not change - the force on the piston is the difference between air pressure on one side and zero pressure (from the vacuum) on the vacuum. You will be doing a small amount of work as you pull the air up with the piston. That air will spill out the top and once spilled, does not affect the problem any more (except for the exceedingly small affect it has on the total volume of the atmosphere!).

On the Moon, there is no air, so there will be no pressure (difference) across the piston, so zero force is needed to pull the piston (ignoring whatever work needs to be done to actually move the piston, which is zero if you postulate a zero mass piston).

Similarly, the suction cup won't work on the Moon - it works here on Earth thanks to the air pressure pushing on the outside of the cup (versus the lower pressure inside, where you squirted the air from when you pushed the cup against the window.

A vacuum is not "negative" pressure, it is lack of pressure. Hence it can not get more negative - maybe that's the concept that causes any uncertainty.


#23

That's similar to what my son Ben said (he's the son in Flagstaff/Lowell I mentioned before). Somehow it still seems counter-intuitive to me though. Here is my most recent reply:

" I am still in limbo based mostly on practical experiences. My guess at the moment is the force required to withdraw the piston would increase, basically in direct proportion to the surface area of the "empty" cylinder being created below the piston as it rises. Since pressure acts in all directions it would seem that the air pressure required to equalize the vacuum would equal the surface area of the vacuum chamber for any given height. The reason for the heavy steel pipe was to eliminate any forces from
acting in these directions from the outside but I think that's actually irrelevant (assuming a rigid pipe is all that's necessary). Although I have wondered if it might be more as you described my practical experience seems to contradict the idea. If I try to pull up on a bicycle pump with the inlet closed it seems the force required increases the further I go. Seems like I
have experienced similar results with other 'dashpot' objects I have played with over the years. I do think there is a distinct difference between vacuum and pressure though, vacuum being a linear function of surface area and pressure being an exponential function of volume. But maybe I'm wrong about that too!

If the force does increase it would seem to imply the same would happen in space since you would be trying to create a space with less pressure than 0, a 'negative' pressure if you will. The only difference would be the initial force of 14.7 lbs. would be absent. And, if this is the case, it would seem a suction cup would stick to the porthole in a total vacuum for the same
reason but somehow this seems crazy too. So, as I said, I'm still not sure at this point."

BTW, to make this more HP Museum appropriate, it would be fun to have some equations for this that could be run/plotted on the 28C.

Edited: 11 June 2012, 6:45 p.m.


#24

You are confused about vacuum, I think! As I noted earlier, the vacuum is not negative pressure, so you can not get anymore negative by expanding a vacuum chamber.

Air pressure exerts a force (force = pressure * area; do a unit analysis: pressure (in the USofA) is pounds per square inch, which is force (pounds) per area (square inch). Take that time area, and you have force left). If there is no pressure (i.e. a vacuum) there is no force. If there is no force, there is nothing opposing or aiding the motion of the piston seal (or the suction cup).

Your pump perception may be a function of how you pull on the pump handle - as you pull it out, the angles of your hand and arms will change, effectively changing how you apply force with your muscles - this will feel different as you pull!


Quote:
vacuum being a linear function of surface area

Where does that come from?!? I would define the quality of a vacuum based on how much material there is per volume, and a true vacuum would have zero material per (any!) volume. If you transformed a cube with a particular volume into a sphere of the same volume, you would have a change in the internal surface area but no change in volume and hence no change in pressure if there was air inside.


#25

Quote:
You are confused about vacuum, I think!

Yes, probably so, but I think I'm coming around. I now have two valued opinions (yours and Ben's) which agree and do seem to make scientific sense. It's just so odd to think that if I rasied the piston from sea level to a point above the atmosphere there would be no force trying to pull the piston back (assume massless piston).

Fascinating!


#26

Also, remember the perfect gas law: PV=nRT

If n = 0 => perfect vacuum, then PV = 0 , so, if V is non-zero (whatever size!), then P is zero always.

#27

Quote:
Although I have wondered if it might be more as you described my practical experience seems to contradict the idea. If I try to pull up on a bicycle pump with the inlet closed it seems the force required increases the further I go. Seems like I have experienced similar results with other 'dashpot' objects I have played with over the years. I do think there is a distinct difference between vacuum and pressure though, vacuum being a linear function of surface area and pressure being an exponential function of volume. But maybe I'm wrong about that too!

I thought about that too. I think the confusion with this is that you are not starting with an object that has zero air inside. That little bit of residual air in the pump (or whatever) will actually help counter the opposing pressure of the atmosphere outside - this air then gets thinned out as you extend the pump thus causing an increase in force required. If you start off as in the original problem, there is a perfect sealing piston at the bottom against a perfect seal - i.e. no residual air to "thin out", and the problem reduces to Dave's first answer.

#28

Yes, I guess it's all starting to make sense, strange as it seems to me for some reason. Am I correct in assuming that the outside pressure on the vessel (the 'crushing' force on the cylinder in this case) is increasing as more empty interior volume is created by withdrawing the piston? If so this would seem to be a sort of 'force amplifier' if only there were some way to extract the differential? Knowing that the laws of nature are almost always a less-than-zero-sum-game makes this thought seem a little silly too though.

Ahh, the wonders of it all!


#29

Quote:
Am I correct in assuming that the outside pressure on the vessel (the 'crushing' force on the cylinder in this case) is increasing as more empty interior volume is created by withdrawing the piston?
Pressure, no; but force, yes. The force is borne by a longer and longer portion of the cylinder though, so the cylinder doesn't need to get stronger and stronger as the piston is withdrawn.

Edited: 11 June 2012, 10:25 p.m.

#30

Why are you asking this question and what do you hope to gain?

First of all, why all the extraneous details? 12" diameter but only a 1 square inch hole? So the metal is over 5" thick? Do we need to know this?

You did not say that the tube was pumped out. Therefore, I would expect the initial condition to be that it has air in it. The air pressure on both sides is equal and the initial force is ZERO.

You might say, no, the piston is right against the bottom, and there is vacuum there. Then you might get a cold weld

http://en.wikipedia.org/wiki/Cold_welding

and be unable to raise the piston.

As you raise the piston, you are decreasing the pressure in the tube, by the ideal gas law

PV = nKT

V (volume) is increasing, therefore P is decreasing. It is the difference in pressures on the two sides that produces the force.

Anyway, 50 miles up, the pressure is about half of what it is at sea level.

Edited: 12 June 2012, 2:00 p.m.


#31

Quote:
Anyway, 50 miles up, the pressure is about half of what it is at sea level.

No, it's MUCH less, as indicated by this University of Illinois reference. At 50 km, it's about 0.001 of sea level. At 50 miles, you are in the lower ionosphere.

Barring a cold weld, there is a(n expanding) vacuum created at the bottom, since the piston is postulated to completely seal the tube. Since it's a vacuum, there is no pressure at all on that side of the piston, so it's only atmospheric pressure pushing on the exposed side of the piston.

#32

Dave Shaffer is right. Specifically, the force on the massless piston at any elevation is proportional to the air pressure at that elevation. The work required is the integral of F*ds where F is the force on the piston (pressure times area) and s is the distance.

Note that it wouldn't be hard to add gravity and the mass of the piston to the equation either.

Once the piston reaches the top of the atmosphere, there is no suction because the vacuum in the tube is the same as the vacuum in space. To really bend your brain, think about what would happen if you unplugged the bottom of the tube at that point. Air would rush in but the piston would not move. In fact, if anything it would be momentarily pushed up since the only extra force that might be applied would come from below.

Quote:
If I try to pull up on a bicycle pump with the inlet closed it seems the force required increases the further I go.

You aren't starting with a vacuum because the hose is full of air, so as you pull, you create move volume for the fixed amount of air and thus the pressure goes down. It's that decreasing pressure that you're fighting against and feeling.

Quote:
It's just so odd to think that if I raised the piston from sea level to a point above the atmosphere there would be no force trying to pull the piston back (assume massless piston).

It probably seems odd because you had to exert so much force on the way up. Think of the air in the tube as a stack of pennies. As you lift the piston, you're lifting pennies that tumble out the top of the tube. The higher you lift, the fewer pennies there are so it gets easier. When you finally get to the top, there is no pressure on the piston. Your efforts have been expelled lifting a bunch of pennies up and out.

#33

I greatly appreciate all the replies to this somewhat silly question. I think it's all beginning to sink in but there's still something amazing about it all to me, quite a few semesters of Physics way back when not withstanding. Somehow it makes me wonder why drawing a perfect vacuum is seemingly so difficult but I guess it's the detail of trying to extract every last atom or molecule that causes all the trouble. And of course the vessel must be strong enough to withstand the enormous forces on its surface if it's of any size at all.

Thanks all around for some great explanations!

LHH


#34

By the way, realistically, opening the piston more and more will not necessarily cause a greater vacuum. At ultra high vacuum levels, a source of residual gas is hydrogen coming out of the stainless steel itself.

#35

Quote:
And of course the vessel must be strong enough to withstand the enormous forces on its surface if it's of any size at all.

There are no forces greater than (at sea level) approximately 14.7 pounds force per square inch times the area of the piston in square inches (or equivalent units in the unit system of your choice). If the piston area were, say, 1 square inch, even a small child could easily exert the 14.7 lbf required to draw out the piston as far as he wanted.

And as far as differential pressure across the vessel wall goes, it would never exceed 14.7 lbf/inch_squared (psi).


Edited: 13 June 2012, 11:26 p.m.


#36

Yes, but this was intended to imply other situations where vacuum chambers are used here on the Earth's surface. Those 14.7 lb/in^2 add up quickly if the vacuum chamber is larger than a breadbox!

#37

Quote:

To really bend your brain, think about what would happen if you unplugged the bottom of the tube at that point. Air would rush in but the piston would not move. In fact, if anything it would be momentarily pushed up since the only extra force that might be applied would come from below.


This is probably not such a simple situation to consider. There would certainly be a delay in any force, since the tube would have to be large so its length divided by the speed of sound would be a large time (on the order of minutes).

On the other hand, as the "weight" of air builds up in the tube, the pressure at the bottom equalizes, and the air stops rushing in, suggesting the piston will not move.

However, there will still be a force on the piston. The air molecules will have a thermal velocity distribution. This causes our atmosphere to slowly boil off, since some number of molecules in the distribution will have velocity greater than the Earth's escape velocity. However, the air cannot boil off through the piston. This means there will be some (obviously small) pressure up on the piston.

But, ah! The piston is in outer space. So there may be radiation pressure downward on it.

#38

I think part of the difficulty in answering the question stems from the poor wording...I don't mean that unpleasantly; I think that we all get the meaning of the question. I view the problem statement as describing a relative of the [mercury] manometer.

Suggested rewording: A 50 km long massless cylinder has a hole with a 1 cm^2 cross-section bored through the center from end to end. The cylinder is sealed on one end. In the bore there is a massless piston which seals perfectly to the inside walls of the bore, and may moved by pulling it from the open end of the cylinder. You place the cylinder vertically, open end up, with the bottom at sea level and the piston completely at the bottom. The space twixt the piston and the cylinder is evacuated; i.e, there is no gas in the space. What is the pressure on the top of the piston when it is at sea level? At the edge of the troposphere? (Feel free to switch from SI units ;-)

In a sense, this is really a pressure, not a weight, problem, though the difference is really just a point of view. A 1 sq in column of air here on earth exerts a pressure of ~14.7 psia. Note that I didn't just say psi. We usually work in psig. If we report water pressure in our homes is 50 psi, we really mean psig. The distinction is an important one for this question.

I'll jump to this quote for a minute: "Now, take the cylinder into space, say to the moon. In this case there is no air pressure at all, essentially a vacuum everywhere."

This is not true. Interplanetary and interstellar space is filled with plasma, a very dilute "gas" that consists largely of free electrons and free protons. The moon has a very slight atmosphere, and it is actually substantial when compared to interplanetary plasma.

With that on the table, let's return to the problem. The pressure in absolute units that we feel here on Earth is due to the weight of the air above us, in aggregate, pressing down on us due to its gravitational attraction to Earth. If I have a pressure gauge that measures in psia, it would read ~14.7 psia. So let's embed a differential pressure gauge in the piston. One end of the gauge measures the vacuum beneath the piston, and the other measures the pressure of the air above the piston. This is, in effect, an absolute pressure gauge, because we have stated that the volume twixt the piston and the bottom of the sealed cylinder is evacuated --- a perfect vacuum. Important Note: No matter where the piston is in the cylinder, the pressure on the bottom of the piston is always absolute zero!

Now the question to ask is what does the gauge read (in psia) at different altitudes above the surface of the earth? As long as we remain in the troposphere, the pressure is a fairly predictable log formula. Above that, the equations change, but the principle of the differential pressure gauge remains the same. No matter where we are, the gauge will likely read /some/ pressure, however small it may be. The reason that we must differentiate twixt, say, interstellar pressure, and pressure at sea level is that the pressure that we address at sea level is largely due to gravitational attraction, whereas the pressure that we measure in interstellar space is largely influenced by the density and temperature of the plasma.

Quote: "As you just begin to raise the piston you measure a force pushing down to be about 14.7 lbs which is the 'weight' of the air column from the top side of the piston to the edge of space. As you begin to raise the piston further what happens to the pulling force required and why?"

The problem here is conflating pressure with the effort it takes to move an object. The piston is massless, so F = ma = 0. There is no force required to accelerate a massless object. However, if I hook a strain gauge to the top of the piston and anchor the other end at the top of the cylinder, that gauge will show different strains when the piston is suspended at sea level and at an altitude of 100 km. This difference is due to the different differential pressures across the piston at different altitudes.

Question: "...would a suction cup stick to the outside of the porthole of the Lunar Lander as it sat on the Moon's surface?"

Answer: Yes, but the differential pressure holding it there would be very small compared to sea level on Earth.

~R~


#39

Quote:

Question: "...would a suction cup stick to the outside of the porthole of the Lunar Lander as it sat on the Moon's surface?"

Answer: Yes, but the differential pressure holding it there would be very small compared to sea level on Earth.


But in a practical sense, I doubt you could make it work. If went EVA and pushed the suction cup to the side, I bet you'd never get it to stick--even in LEO

Your overall description is excellent. You should be a professor :-)


#40

Thank you! Yes, I think that I should like to teach, for I regard teaching highly, considering the impact of teaching on shaping students' lives.

I had a few minutes so took the discussion a bit further for those who might want to know whence comes weight in the discussion.

Derive the pressure at the bottom of a column of air from the mass of the air in the column.

Ideal Gas Law (r = rho, density):

R = ideal gas constant
k = Boltzmann's constant
m = molecular mass of air

N = # of molecules in a sample of air
n = # of moles in a sample of air
V = volume of a sample of air
r = density of air
P = pressure of air
T = temperature of air (K)
h = altitude above Earth's surface
g = acceleration of gravity

PV = nRT = NkT
P = NkT/V = rkT/m
r = mP/kT

P0 = psia @ sea level

Imagine a slab of area A and thickness dh. In hydrostatic equilibrium, the change in pressure over an infinitesimal change in altitude must oppose the gravitational force on the air in that infinitesimal layer. The change in pressure, dP, from the bottom to the top of the slab, dh, for some air density rho is (using calculus notation and implicit multiplication). Take care to distinguish twixt an expression with variables in it, where, for example, m means mass, and an expression that has units, where, for example, m means meters.

[P(h) - P(h+dh)] A = r g A dh.

r A dh is: the density times the volume of the slab = mass

This leads us to [P(h+dh) - P(h)] / dh = - r g, where [P(h+dh) - P(h)] / dh is the definition of dP/dh.

Substituting for r, we find

dP/dh = - r g = -mgP/kT

This leads us to an expression for pressure as a function of height, assuming some fixed temperature (not too bad an assumption when T is measured in K)

P(h) = P0 exp (-mgh/kT) where mgh is potential energy, and kT is thermal energy (units cancel).

Going back to
dP/dh = - r g, we can rearrange as dP = - r g dh

If we integrate both sides (wrt P on the left and wrt h on the right) from some height h to infinity, we have (pardon trying to do calculus in a text base)
- integ (P(h), P(inf), dP) = integ (h, inf, g r dh).

Because P(inf) = 0 (ideally), P(h) must equal integ (h, inf, g r dh), but this is just the weight of the air in the vertical column of unit cross-sectional area lying above that level. Unit analysis yields

g r h = g m/V h: m/s^2 * kg/m^3 * m = kg/(m s^2). This is the Pascal, the SI unit of pressure, which we commonly know in the US as pound/sq in, or psi, and it is not in units of "weight," which is W = m g, but rather in units of "weight"/area, i.e., N/m^2. I put "weight" in quotes because in US units we use weight specifically for the product m g, but in SI there is the concept of force and mass. Socially, this distinction is still confused, IMO, in a general sort of way because if I am asked how much I "weigh" I can respond with, 74 kg, which is not a measure of /weight/, but of /mass/, or I can respond with 160 lbs, which /is/ a measure of weight (force), and no one wants me to muddy the water with slugs ;-) In SI units, I really "weigh" 726 N, but that just makes me sound fat, and who needs that? ;-) My torque wrench has scales in ft-lbs and N-m, but I expect my tools to be technically correct.

Note that an important part of this discussion is my assumption that the volume enclosed by the closed end of the cylinder and the piston is a perfect vacuum. I think this agrees with the original condition of "with the piston completely at the bottom."

I think that one finds the most interesting discussions on this forum...the original 15C manual and the Advanced manual covered material, and presented it in a manner, that one simply does not see today, and I think that anyone who enjoys that sort of material has an interesting mind.

Yours, &c.

~R~

Edited: 16 June 2012, 7:03 a.m.


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