Cylinder Load: Newsletter Challenge (07/01/08)

I rate this as a good answer but there must be some more complicate, interesting interpretation of the question that we can develop and then discuss.

Sorry - this PC doen't have the leter that follows 'a' and comes in front of 'c' so my sentence construction may seem a little wierd (trying to avoid the letter ' ').

Assuming there is no friction in the system and avoiding the tiny effect of the 'head pressure' resulting from an increase in the height of the gas column on either side, the cylinders will stay in equilirium if the piston on either side is pushed down to any new position. This is due to the fact that when one piston is pushed down, the other one will rise up a sufficient distance to maintain the same volume and pressure within the system. In other words, the fact that the pistons are initially at the same height is something of a red herring - it tells you that the ratio of piston mass to piston area is the same on each side - ut the relative heights of the two pistons is irrelevant.

Now if you add an extra weight to either side, the piston mass to piston area is no longer identical on oth sides. The side added to will drop until it hits the floor of the cylinder while the unweighted piston rises. The pressure and volume of the gas will remain unchanged.

So the gas under the 2+2 kg piston will flow across to the other side (as the 2+2kg) piston fals 0.6 m, and the 4 kg piston wil rise 0.3 m to maintain the same volume and presure.

Pistons now 0.9 m apart (you would get the same result if you aded 2 mg to the 2 kg piston - assuming there is no friction).

I just voted for #44 to be a "good answer" since I have seen that point made nowhere else.

Until I read #44, I had mistaken "pistons in equilibrium" to mean "both pistons at the same height". I realize now that equilibrium here means "not moving". It had not occurred to me (or anyone else apparently), but I now agree that "pistons at the same height" is a "red herring".

Actually, I disagree with #44's statement, "it tells you that the ratio of piston mass to piston area is the same on each side". The pistons being at the same height do not tell us that. (He himself contradicts this point earlier when he rightly pointed out that "the cylinders will stay in equilirium if the piston on either side is pushed down to any new position.") Rather, the implied fact that they are initially "not moving" tells us that they have the same piston mass/area ratios.

Actually, the pistons could move if the system is mechanically disturbed (shaken), but they would stop moving when the disturbance stopped. BTW, the restriction of the "thin pipe" connecting the cylinders dampens such disturbances.

The equilibrium here is actually created by the fact that the gravitational mass force of both pistons is equal to the gas pressure/area force applied to the bottom of the pistons

The initial vertical heights of the pistons is important only for deteremining the "difference" in height after the equilibrium is upset by the added mass. By telling us that "both pistons are at .6 m height", the challenger cleverly tells us what we need to know, the height of both pistons (even though the relative height is irrelevant) while at the same time distracting us with the "red herring". Very Clever indeed.

At anyrate, the answer is the same, a difference of .9 m. ,still a very good point #44.

Hello "Guest". First I am assuming you are the same "guest" that posted #44. BTW, your contributions are good (although you contradict your self at times, don't we all) and you should register so we can follow the responses better. If not, please ignore this post.

You said in #44, "avoiding the tiny effect of the 'head pressure' resulting from an increase in the height of the gas column on either side"...

Strictly speaking, that is what I was doing in post #53 when I replied to your #44, avoiding (or ignoring) the tiny effect of the head pressure (resulting from the mass of the gas).

I like this response the best. Since the cylinder ratio is 1:2, when the smaller pistons falls to the bottom, the other can only rise half that distance.

I agree that we need to assume the pistons are in equilibrium at the initial measurement, but I'm not sure I agree with the rise in temperature. We are already assuming 2kg is enough to overcome starting frictional forces, so basically we are assuming zero friction. If so, then where does the heat come from? Without friction, the pressure of the gas would remain the same throughout the experiment.

What do you mean by "be some increase in temperature due to the losses in the pipe". What "losses"???

Clearly, temperature changes were excluded from the problem, and I was making that explicit. However, in reality the gas under the smaller piston would be heated as soon as the piston starts to fall (look up adiabatic expansion of gases for details). The problem also assumed viscous damping in the small pipe, to allow the large piston to rise without overshooting - that too would convert mechanical energy to heat.

However, if we change the problem slightly we arrive at a more interesting one: what happens if all the lost potential energy is allowed to remain in the gas? For this different challenge, I would reword your question as: how can the lost potential energy be converted into a constant temperature rise in this system?

The only method of conversion to heat that is needed is the heating due to adiabatic compression. However, in the absence of a source of resonance damping this heat gain would be recoverable - so the two pistons would continue bouncing for ever, and the gas temperature would e continuously changing. However, because the resonances are at different frequencies (same masses, different piston volumes), the temperature of the gas on the two sides will generally be different. Thermal conduction will tend to even this out, so eventually all relative movement between the pistons will cease. That is the condition that can be calculated for the adiabatic case.

This is just the "fundamental" source of loss. However, regardless of the source of the loss and the process by which temperatures are equalised: if there are no energy losses from the gas to the environment, the equilibrium condition will be the same.

N.B. that thermal conduction between adiabatically heated regions of gas is also the principal mechanism for acoustic loss in gasses.

Physicist?, thank you for your response. Obviously, I had failed to recognize the thin pipe as a form of resistance.

The neat part of wrenching on cars for 15 years and 3 years on fork lifts is that you get learn a little bit about a whole bunch of stuff. When I first read this problem, I immediately compared it to a hydraulic brake system.

..... "The system is place in vacuum. Each cylinder is covered by a movable piston......"

and ....."Initially the pistons are at same height (0.6 m). ......"

If the system is placed in a vacuum and then covered with the pistons, both being at the same height, then with no gas left in the system (due to the vacuum) each come to rest (having not moved) on the platform 0.6 meters above the reference measuring line.

just a thought.

Yup, I thought of that too - but decided you could push the interpretation of "is covered" to mean "is already covered" - and I'd winged more than enough already...

But... wouldn't this interpretation allow a reasoned answer of anywhere between your zero and 0.6-metres - depending on the reason the pistons were at the same height in the first place? e.g.:
still under gravity, but friction is just sufficient to hold the pistons in place - so smaller piston falls to base position; or
no gravity, but some friction, and placing of additional mass jogs lighter mass, which descends an indeterminate distance.

Of course, we don't actually know that the reference position is the base of the pistons, or indeed that the pistons are not directly connected via a pulley or lever system - but I admit either of those would really push Occam's razor deep into the flesh

There is not a lot of information here that appears to be missing, such as nothing is known about the weight of the gas, so you have to assume the gas weight is insignificant. If you assume the connecting tube attaches to the bottom of the cylinders then when you add weight, all the gas from the one cylinder will be pushed out into the other cylinder. As the cylinders are at a ratio of 2 to 1, the larger cylinder will rise only 0.3 meters.

This is for HTR's evacuated version.

The 0.6-metres was the distance the lighter piston fell...
The other piston didn't move - being held by friction (and zero gas pressure to make any difference to the forces acting on it)

My answer is ZERO. The pistons don't move! The pistons are subject to friction and at rest in both loading cases as follows:

Originally, the smaller side has a force of 2g N acting downwards, and a friction of F also acting downwards, because the pison movedupwards as the vacuum was applied. Total force on right smaller cylinder (2g+F1) N. Similarly the larger cylinder has a force of (4g+F2) N

When the additional weight of 2kg is applied to the smaller side, the piston tends to move downwards reversing the Friction force and resulting in a total force at this side of (4g-F1) N The other, larger, cylinder tends to move upwards and so its friction remains in the same direction as previously.

Using a for before the weight is applied and b for after

Pressure Pa = (2g+F1)/a1 = (4g+F2)/a2 and Pb = (4g-F1)/a1 = (4g+F2)/a2.

Dividing gives Pa/Pb = (2g+F1)/(4g-F1) = 1

From this relationship, it is deduced that the pressure of the gas is unchanged, Pa/Pb = 1. This means rhe volume is unchanged and the position of the pistons can remain unchanged.

For this to be true, then F1 = 1 N and the effect of the added 2 kg is offset by the change in direction of the friction going from +1 N t0 -1 N

If you allow friction, the ratio of cylinder cross-section areas can be anything other than unity. That would mean the answer could be anywhere in a range of zero to somewhat (I think) greater than 0.9-m. If we allow stiction with zero friction, the answer could presumably be anywhere between 0 and (not-quite) 1.2-metres - but perhaps someone will find a way to store some energy in the gas and get an even larger movement? (If so, the question about the largest possible movement becomes worthwhile). Until someone (not me) works that out, it's all getting too variable (at least for me), so perhaps we should look for simplifying assumptions that lead to a meaningful challenge?

Fyz

Hi fyz

I have re-posted my answer under #21 and I hope it clarifies things. I think the .9m is a red-herring and neither piston moves. I look forward to any comments you may have on that posting.

SR

I find this problem a bit confusing? The wording is ''Initially the pistons are at same height (0.6 m). ''. Does this mean, before natural processes brings the system into equilibrium? Eg: the pistons are released at this height and then move to equilibrium, and then the weight of the 2kg piston increased to 4kg?

I can not visualise the pistons at the same height, as described. In my mind I see the two pistons being released as described, and there after the larger piston would descend to the bottom of the cylinder? And the smaller one rise. Then the weight of the smaller piston is increased to 4kg,

So my answer is:-

P(1)V(1) = P(2)V(2)

2 times 0.6a = 4 times xa ( a =area cancels)

1.2 = 4x

x = 0.3.

Regards JD.

I think this is meant to be in equilibrium under gravity, with zero friction, and requires 1:2 ratio of cross-sectional area so that the original weights can be in equilibrium at the same height (that would supposedly be the stable equilibrium if you took the density of the gas into account). I think you then have to assume that the height given is the height above some sort of end-stop. That allows the conditions stated in the challenge to be satisfied and gives a sensible unique solution - that lighter side starts at 0.6m and descends to zero when the weight is added, and the heavier side starts at 0.6m and rises to 0.9m height.

An interesting corollary would be to consider the effect of the (smooth-walled) cylinders not being at the same angle to the vertical? The answer may surprise.

I disagree with the statement:

"Also P_i = 6/( A₁+ A₂)"

I think P_i = 2/A₁= 4/A₂ and P_f = 4/A₂ since piston 1 is resting on the bottom but piston 2 is still held up by the gas inside. The gas from side 1 has moved to side 2 and is presumedly at the same temperature/pressure. Since the total volume of the gas is unchanged, the final height of piston 1 is zero and the final height of piston 2 is 0.9 m.

Sorry,

I failed to look at your revised response.

I agree with you answer.

"...Since the total volume of the gas is unchanged..."

An incompressable gas? What about PV=nRT?

Since A1=2*A2, we can say that the initial pressure, inside of the system, is:

Pi = (4+2)/(2*A2+A2)

and the pressure, after adding the additional 2kg is:

Pf = (4+2+2)/(2*A2+A2)

Assuming that the gas is ideal (PV=nRT), the system is sealed (n==constant), isothermal (t==constant), then Vf = (2/3)*Vi.

Not incompressible - just under the same pressure at the and of the process as it was at the beginning - defined by the 4kg mass and the area of its cylinder; (the 2kg + 2kg masses eventually sit at the base of their cylinder, supported partly by pressure and partly by the base of the cylinder)

Very helpful sketch, thanx

In any case it is Pf = Pi right below the 4kgr weight on the right, as it still needs the same force to keep it in place. The question is whether we should take the specific weight of the gas to be close to zero, or not. If γ=0, then the weight on the right will indeed move by 0.6m/2 = 0.3m.

If on the other hand, γ>0, there is a gradient of pressure as pressure is supposed to increase going towards the bottom of the cylinder. The higher the pressure, the smaller the volume as they are inversely proportional entities according to the formula P = nRVT (I assume that temperature remains stable throughout the process). In this case, we should do some calculus by integrating along the cylinder(s) both before and after, keeping in mind that the number of moles doesn't change, and hopefully find x, as a function of γ and area A2. Eventually, it will be less than 0.3m. I doubt that the quiz assumes γ>0, though, as γ and A2 are not given.

if cylinder with load 2kg = cylinder1 , Cylinder with load 4 kg= Cylinder2. then A₁ = 2A_2. Then volume: V₁= 2V_2. p₁=p₂=p=4kg/A₂=2kg/A1.

Because of gas pressrure P₁=P_2, Piston on Cylinder1 will fall down to the bottom. All the gas volume in cylinder1 will move to cylinder2. there is no Pressure change: P = 4 kg/A₂(do not callculate P using cylinder1, because the piston is support by the bottom of cylinder1)_. The the total volume of the gas is still same. V_t= V₁+V₂=3V₂

V_t= A₂.h_t=3V₂=3A₂ X 0.6m; then h_t=1.8m

the answer is 1.8m

Can I buy shares in your perpetual-motion machine? (It helps to get your relative piston areas the right way around)

Of course ya can. Please enlight me. My friend!

According to your solution, Piston2 rises twice the distance that Piston1 plus its extra load fall. Each side is now the same weight - so the total potential energy has increased as a result of releasing the added load. Do this a few times, and we've gotten ourselves a power source. An investment cheque for US$10Million for the rights to your invention is in the post; the funds in the account on which it was drawn equal the validity of the invention.

Proper Calculation (in wondrous detail)

Equilibrium_Weight2 = 2 x Equilibrium_Weight1

Force = pressure x area

=>

Pressure2 x area2 = 2 x Pressure1 x Area1

But Pressure2 = Pressure1

=>

Area2 = 2 x Area1

But: Volume_Change = Area x Height_Change

=>

Height_Change2 = Height_Change1/2 = 0.6-metres/2

Thanks! You correct. Pressure= Force/Area. so the answer must be 0.9m. Because V1 = 0.5V2.

According to your solution, Piston2 rises twice the distance that Piston1 plus its extra load fall. Each side is now the same weight - so the total potential energy has increased as a result of releasing the added load. Do this a few times, and we've gotten ourselves a power source. An investment cheque for US$10Million for the rights to your invention is in the post; the funds in the account on which it was drawn equal the validity of the invention.

Well then, how about this for good scam potential:

The additional weight added to the 2kg piston does not need to be 2kg. It could be 20 grams (assuming a low density gas is used in the cylinders), and that imbalance will cause the 4kg piston to be lifted .3 meters. (By allowing that 4kg mass to fall back to the start point, we could extract 1.2 meter-kilograms of work.)

Obviously, 1.2 meter-kilograms is a lot more work than that (20 grams x .6m = .012 meter-kilograms) invested in the smaller cylinder. This is only a prototype system, but already we are getting 100 times more energy out than we put in! Sure, there will be frictional losses, but conservatively, we believe that a 50:1 ratio of output to input is realistic. Given this potential, imagine what sort of investment returns there could be for the savvy investor who is not afraid to act now!

If you insist on being even sillier than my original thought: change the mass ratio to 10:1 and the limiting gain is x100. Etc.

My dear former posters,

I'm afraid that all of you think way to complicated. The system of the two zylinders and pistons is first gas filled, and THEN placed in vacuum. No matter how big friction is and also what weigth the pistons have they will under vaccum move upwards as much as they can - both!

I further assume that there is an end stop, which prevents the pistons to jump out. Now we start putting masses on one of them and at a certain mass the one piston with that mass will move down.

If we accept this interpretation of the challenge, we don't necessarily arrive at the answer you propose.
The position of the pistons will simply be low/high enough so that the gas pressure is exactly correct to support them. If the system is assembled at atmospheric pressure, the additional expansion when placed in vacuum will depend on the area of the tubes.
To give a synthetic example:
The two vertical cylinders have cross-sectional areas of 1 sq-cm and 2-sq-cm, respectively, and are filled in an atmosphere with a pressure of 1.g Newton/sq.cm. The cylinders are 1.8-m high, so when the pistons are placed and released they settle to a height of 0.6-m (once the temperature has re-stabilised). They are then moved to vacuum, so the pistons rise to 0.9-m. After the additional 2-kg mass is added to the lighter piston, it falls to the bottom and the other rises to 1.35 metres. So we have a case that gives a difference of 1.35-metres.
I could of course produce any number of such scenarios each giving different answers.

However, the interpretation is somewhat idiosyncratic - if we rely on the order of statements to determine that the tubes are filled before the pistons are placed, we can similarly deduce that the system is also in the vacuum before the pistons are placed, and that the initial 0.6-m height is established after the system is in vacuum. In which case, no gas is left, so the pistons are supported by something other than air pressure. That might be friction? So we don't know whether the 2-kg mass will fall after the addition of the extra mass. But it does mean there is no coupling between the masses.
Therefore, if we fully apply the basis of your interpretation, the answer would be either zero or 0.6 m.

The smallest piston will go to the bottom. The larger piston will rise. The gas pressure is still the same, supporting the larger piston. As the smaller piston drops .6 m, the larger piston rises .3 m. The difference will be 0.9 m.

I would still state that both pistons moved upwards. The reasons therefore are as following:

1. The gas inside the zylinder-piston-tube system has athmospheric pressure.

2. Outside is a much lower pressure, causing a pressure difference of about 1 bar, which is causing a force into direction of lower pressure of 1N per squarecentimeter.

3. The equivalent of 1N on earth is a weight of about 9.81kg (!) which has to be multiplied with the area of each piston.

4. Depending on the cross section and the load of the piston and eventual additonal weight the piston with smaller cross section might move down (toward gravity center= earth)

5. The bigger the surface of the piston the higher ist the force of the gas to move the piston upwards. Since no information at all is given about the size of the pistons there will be certain combinations where any equilibrium is possible.

6. Assume A1=A2=0.407sqcm and M1=M2=4kg then both pistons move half way down, causing an overpressure inside the gas of 2bar. From there a force of 39.24N is produced which just compensates the force of the two masses of the pistons.

7. maybe I made some errors, but probably only a factor of 2 or 0.5... have no more time yet to think more about it... sry

I'm surprised that no one commented on my post #4. I think that it meets all of the conditions. Perhaps the text was too terse to be understood. I am re-stating it here with a sketch which hopefully will help understanding

It includes friction and the fact that the small piston will not move in the band where the friction is changing directions. In essence the system is initially in equilibrium in a friction free environment with a 3kg load. This is realized with an original 2kg load aided by friction or with a 4kg load impeded by friction the friction being the equivalent of a 1kg load (9.81 N).

Note that the diameter ratio, and the amount of friction in the larger side are unimportant, (indeed the larger piston could be fixed in place and the added weight would cause no increase in the pressure, and so no movement of the small piston!), since the direction of the friction force on this side is not changing.

I trust this posting will generate some comments

Was my post #7 invisible? Just in case,I'll say it again: if you include friction/stiction, you can have any answer you choose between 0 and some upper bound - depending on the magnitudes and directions of the initial forces.

I replied to your post #7 separately. You posted it as guest,(although I realized that you had signed it as I replied), and I read it out of context with your post #1. Since it stated that the answer could be any positive value I didn't pay it too much attention.

With regard to posting#1, I agree it is a solution. I find the reference to Losses in the pipe to be a little misleading; if there were any heat loss, the heavier mass would not rise as much. Are you referring to the Joule-Thompson effect, which is an effect of the specific heat not being constant and would depend on which gas was present? Also I don't understand the restriction you give in the PS. Where does that come from?

There are 3 suspicious points in the question.

1) why connect with a thin pipe? To permit heat exchange? To limit the pressure? To confuse and confound us?

2) Why the vacuum? this does not affect your answer, but I took it to mean that the seals were loaded with "downward friction" for the initial condition

3) Why the initial height of 0.6m? Critical to your solution, but meaningless in mine.

It seems to me that there may be multiple answers, just like there was with the "Magnitude of Force" challenge

SR

I commend your approach to looking differently at this problem. On the surface, this appears to be a rather simple if not so clearly written challenge. However, even if not originally intended, we might be able to find a different interpretation of the challenge or evolve a more interesting one of our own.

Thanks,

Jim

When you place the additional mass on the 2-kg, the first thing that happens is that the gas in the smaller cylinder is compressed to twice its original pressure. That will heat it up. There will be further energy dissipation as the gas is forced through the thin pipe. My solution represents the condition of the challenge that the temperature in the larger cylinder is maintained constant - possibly by heat loss from the gas to the connecting pipe (but see also below). If you avoid that heat loss, the 4kg mass will rise further than in the solution given for the challenge. The height in the absence of heat loss from the gas is independent of the initial pressure or temperature.

1) I think the thin pipe is primarily there to limit the velocity of the larger mass so that it doesn't overshoot its final position and add real ambiguity. It may also serve as a heat exchanger as you suggest.
2) I too can see no good reason for the vacuum - so long as the external pressure is constant while the 4kg mass is rising the answer is unchanged. But only for the case of constant temperature in the large cylinder.
3) To make a moderately interesting solution possible??

Regarding the multiple solutions:
a) For this case I can see only one non-void numerically unique solution
b) It turns out that AlphaBeta's alternate answer to the magnitude of force challenge was not quite correct - you need to know the coefficient of friction to find the lower bound on the force; otherwise, you would have to assume you need to accelerate both masses until the smaller one loses contact, after which you can reduce the force to AlphaBeta's solution.

I miss in your drawing the upwards directed forces of the pressurized gas volume. This force is calculated by F=p*A where F is perpendicular to A. The bigger the cross section of the piston is the bigger is also the Force which pushes the piston into direction of lower pressure. The relative pressure difference of 1 Bar means a load of about 10kg per squarecentimeter, which is enormous. Since the expansion of the volume due to the moving pistons is unknown due to the fact of unknown cross section we cannot exactly say where the pistons stop. Probably can we get an equilibrium equation which is dependent on total cross section, with neglecting the volume of the connecting tube.

Sorry, I don't understand your question.

I didn't draw the pressure in but it is uniform throughout the system. See my post #4.

Also a pressure of 1bar is not a high pressure. Industrial use of pressures up to 5000+ bar (500+ MPa) is quite common.

The pressure is uniform in terms of N/cm^2. Which means that the upwards directed force depends on the piston area. I think it is possible to describe the whole set-up with some equations with parameters like area (A1 and A2) and mass (M1 and M2). I assume anything with square and/or cubic terms where the area is squared and the height of movement of the piston mass is in the cubic term.

Personally I'not aware of technical applications with more than let's say 250bar. Probably is it apparatus with hydraulics or heavy presses which have 5000+ bar. I used the expression "high pressure" not to express a high absolut pressure but to remark that vacuum produces a considerable force on e.g. pistons. The force of the vacuum is certainly high enough to overcome the friction, except that there is a really bad seal around the pistons in question. Since it is a more theoretical question I assume that there is no friction at all to consider and thus I wonder why nobody of the posters considered the force of the vacuum in their considerations. I've worked with vacuum equipment in a particle accelerator and there we did have enough problems with the, for those applications certainly unwanted pressure by the surrounding air. In the present theoretical task it is just the opposite: The pistons move out.

Think only of the gas dampers in the car hood or trunk. They operate at probably slightly higher gas pressure than atmospheric and use a relatively small cross section, which IMHO fits well in our actual problem. Those gas dampers produce also something like an equilibrium and the difference to the actual problem is only 1bar in external pressure (assuming the dampers use 2bar filling pressure :-).

My interpretation and solution are:

1-pistons are in place before gas is added, otherwise the question is unanswerable.

2-the system is assumed to be sealed so that when it is placed into a vacuum, no gas escapes around piston. This is a theoretical assumption of course, since this can't be done without some piston friction, which has to be ignored here, otherwise the question is unanswerable.

3-the mass of the gas inside the cylinders has to be ignored since we don't know the volume or type which would be needed to calculate this.

4-"Initially" refers to before the mass is added to the smaller piston to doulbe its mass, but after the system is put into a vacuum. Therefore the expansion of the gas inside the cylinders due to system being in a vacuum has already occured.

5-The .6 m height has to be assumed from the bottoms of the cylinders which also must be assumed at the same level, otherwise the relative volumes cannot be determined.

Since the two pistons are "balanced" by the force applied to them by the gas pressure, the area of the smaller piston must be 1/2 that of the larger piston since the mass is 2 times that of the smaller. Therefore the volume of the smaller cylinder is 1/2 that of the larger.

Now, by adding the extra weight to the smaller cylinder, they are no longer "balanced" and the piston will drop to the bottom where the added mass is supported. The gas pressure is unchanged, since the larger piston still applies 4 kg/area force. The volume and the height of the larger cylinder is now increased by 1/2 to .9 meters. Since the smaller piston is at the bottom or "zero", then the difference is .9 meters.

With the equation of the thermal state of a gas p*V=n*R*T=const. it can not be that the gas pressure remains constant if the volume is reduced by the additional 2kg. Temperature is assumed to remain constant too and we don't have a very quick change of volume where temperature rises quickly (How was that effect called? I forgot...).

I like your statements 1 to 5 because they should have stated in the initial question. :-)

Who was it that said " you are all thinking too complicated?

The added weight only "imbalances" the pressures between cylinders while the smaller piston falls to the bottom causing the gas to flow into the larger cylinder.The added weight is then supported by the bottom of the smaller cylinder.

Otherwise stated: the pressure is maintained constant by the 4kg mass acting over the area of the corresponding piston.

The vacuum in the problem is insignificant as well. Friction is not holding the cylinders in place, only gravity. The pressure inside the cylinders remains constant due to the whole, constant temperature line. This means that the pressure has to be equal to the weight of the cylinders pushing down in order to oppose the force and hold the system in equilibrium. One would think that adding the extra weight to the 2kg piston side will simply cause the pressure to increase, & that the only way the larger piston would move would be if the pressure inside were to be greater than that of the force exterted by gravity's effect on the piston. And if the small piston already had enough downward force to cause the larger piston to hold at the same level, then the pressure is enough to resist the force of gravity on the cylinder. Therefore, once the load is added, the volume will decrease and pressure will increase until the 4 kg cylinder moves, and the system is yet again at equilibrium. I belive that the system will only allow the cylinder to move .15 m however. There is more going on here than is being realize because if the small cylinder is 2 kg and the large is 4, a typical situation would suggest that the smaller cylinder would be higher not even. So I belive there is a factor of 4, not just 2. I am more than likely wrong, but 3 just doesn't seem right to me.

The piston in the smaller cylinder falls to the bottom - a downwards movement of 0.6-m. The other cylinder has twice the area (so it can support the heavier piston with the same pressure), so the volume of air displaced by the falling of the smaller piston only raises the 4-kg piston by 0.3-m. That gives a height difference between the pistons of 0.9-m.

vacuum or not it seems to be like a balance scale. pivoting in the connecting tube.

A few grams either way should tip them fully one way or the other.

Maybe a simple view, but it's mine.

Simple Thinker.

Yes, it is also a good analogy - provided you allow the pivot to be at one third of the distance between the masses.

Different lengths for the arms of the balance will allow the two unequal masses balance each other (no other masses acting), and the situation is equivalent to using the different areas of the cylinders to balance the masses.

We could just say that a mechanical advantage is a mechanical advantage - however applied

I've been thinking about the frictionless answer of .9m (which I agree is an acceptable answer), and considering the energy conservation.

In the final steady state, the 2*2kg weights have fallen .6m adding 2.2*9.81*.6 or 23.54Nm of work on the gas. The 4kg weight has risen .3m extracting 4*9.81*.3or 11.77Nm of work that means that the other 11.77Nm of work have been converted into heat and lost from the gas to keep it at constant temperature. Before the heat can be lost, the gas temperature and therefore volume will increase. If no heat were lost, the heating effect would be sufficient to raise the larger piston to 1.2m.

It takes time for heat to pass through pressure containment boundaries, especially in a vacuum where losses are dependent on radiation only. So a complete answer would seem to be:

"Zero" if you permit friction, regardless of when you measure it, and if friction is absent, "something between 1.2m and .9m depending on when you measure it".

Perhaps this is the reason for the thin pipe! Noe that if the "constant temperature" is the same as the vacuum chamber wall (also assumed to be constant) the steady state value of .9m will be approached asymptotically and we will never get there!

For a non-zero pipe thickness: rather than "not reaching" the final height we would actually overshoot - partly because of the momentum (however small) of the 4-kg piston, and partly because viscous losses in the pipe will heat the system.

I had interpreted a thin pipe to be a thin-walled pipe of non-negligible diameter. with masses moving quickly, and heat loss rate being slow in comparison to the mechanical energy. Your interpretation as a small diameter pipe is probably the correct one.

When I said the final height would not be reached; this is a thermal limitation. In order to finish at .9m and have the same constant temperature as at the .6m level, the system has to reject heat.

In basic thermodynamic terms

Work done on gas = 23.54J

Work done by gas to moving .6m to .9m = 11.77J

Nett work done on gas ΔW= (23.54-11.77)J = 11.77J

Change in internal energy ΔU=0 (constant temperature)

Therefore Heat lost from gas ΔQ=ΔW-ΔU =11.77J

Until this heat has been lost externally, the piston position will be above .9m. when the 2*2kg weights are at 0m as you stated. (Also as you noted, viscous or turbulent heat generated in the pipe will stay in the gas unless it can flow out of the system). To remove this heat, requires a heat sink lower in temperature than the gas "constant temperature". If the external heat sink is "at the constant temperature" the gas temperature and therefore the piston will approach the final position asymptotically.

Any low temperature heat sink used, would of course, have to be removed once the .9m was reached at which point the gas would be at the specified "constant temperature".

That leaves us with a solution that can only be reached by artificial means, or after an infinite time. Fortunately this is something engineers have to get used to. Physicists may not find it so palatable!

This was a double posting by mistake. I have removed the content

SR

The smaller piston will move four times the distance that the larger piston. This assumes that the smaller piston has a diameter that is haft as small as the larger piston.

If the smaller piston is two inches in dia. and the larger piston is 4 inches the distance will decrease on the square of the area.

the smaller piston will go to the bottom of the cylinder.

the smaller piston will only go to the bottom if the gravitational force of its mass (Fg=m*g) (with or without additional 2kg) exceeds the force Fv, which is caused by the vaccum. Since the pressure difference is 1 bar there must be one area for which Fg=Fv. With a smaller area we get Fg>Fv and the piston moves downwards and thus increasing pressure by decreasing volume. With bigger area we get Fg<Fv and the piston is moving upwards and reduces pressure. This happens to both pistons and I am intended to state that there might maybe no equilibrium at all. To stop writing assumptions one should start writing down equations to describe the system. I assume it might give at least a system of minimum 2 equations.

I did have a bit of a problem with this but now accept that the pistons are balanced because of a unit surface pressure. Therefore if we think of the pistons as a single unit, then the height would vary as follows:-

P¹V¹ = P²V².

(2+4)0.6a = (4+4)xa (a= area cancels).

3.6 = 8x

x = 0.45.

Therefore the volume difference after adding 2kg to the smaller piston would be

(0.6 - 0.45) time area. And as the pistons have the ratio of 1:2 and as the smaller piston is doing the work, Then the height difference would be one third greater,

0.15 + 0.15/3 = 0.2.

Answer 0.2

After adding 2 kg to the smaller piston, its new height will be 0.3 m and the new height for the bigger piston will be 0.75 m and the diference is 0.45 m.

Thanks,

Mohammad

I don't know from where the ratio of the piston areas is coming from. The initial question say nothing about this, IMHO.

Marco aka Newton2k1

The area ratio comes from the interperatation (not explicitly stated in the challenge) that the system is initially in equilibrium and with the pistons entirely supported by the gas, and that pressures have been equalised by flow through the pipe. That means that the ratio of cylinder areas is equal to the ratio of the supported masses. But be warned, some of the recent contributions ignore loss of gas through the pipe for the smaller piston, but still manage to assume this has effect on the gas in the larger cylinder...

If the pressure inside the two cylinders remains the same before and after, the amount of gas in the connecting pipe is the same and does not change the results.

Let's see... the challenge is to find a literate person to write challenges for CR4?

The wording of this challenge does present some problems in interpretation.

The system is place in vacuum.

Does this mean the cylinders are in a vacuum before the gas is added, or is the gas added and then the system is placed in a vacuum?

If the cylinders are in the vacuum before the gas is added, then the pistons are sitting at the bottoms of the cylinders, and enough gas is added to raise them to 0.6 meters.

If the cylinders are placed in the vacuum after the gas is added, then the pistons should rise to the tops of the cylinders due to the pressure difference. Then when the extra mass is added to the 2 kg piston, it might not cause any change in position.

When the extra 2 kg mass is gently kept on the smaller piston and released on contact, two things happen. Firstly, the equilibrium is disturbed and the smaller piston must come down to bottom finally. Secondly, an energy equal to 2 x 0.6 x g (PE of the extra mass) is added to the system. Also,it is clear that the cross-section of the larger piston (A2) is twice that of the smaller piston (A1). As the connecting pipe is thin, transfer of gas to the larger cylinder is slow, thus compressing the gas in the smaller cylinder and retarding the movement of piston so that it lands gently on the bottom ( may involve some damped oscillation before that, but I have not checked that!).

Of course, the temperature of the gas will rise but nowhere it is stated that temperature is constant after adding the extra mass - added energy has to go somewhere!

The gas having absorbed this energy gradually fills the larger cylinder as the piston comes to bottom. If, P1,V1 and P2,V2 are initial and final pressure and volume (total ) of the gas, then:

P2 = P1 = P = 4g / A2 and P2V2 = P1V1 + 2 x (0.6) x g (added energy)

V2 = V1 + 2g / P = V1 + A2 /2 = V1 + 0.5 A2

V1 = ( A1 + A2 ) x 0.6 = 0.9 A2 giving a final volume of gas V2 = 1.4 A2

Therefore the larger piston is now at a height of 1.4 m above the smaller piston which is at the bottom and the gas is at higher temperature than before (till it loses energy by radiation, but that it would be doing any way - in long term). Hope this is a good answer.

Sorry for a few mistakes in my #59. Correction below:

Since the larger piston goes up , we have to account for that too. Let h be its final height, the final volume V2 = A2 x h.

Also, P2V2 + 4g (h-0.6) = P1V1 + 1.2g (not 2g as I wrote by mistake), P1=P2=P

Hence, V2 + (4h-2.4) x g/P = V1 + 1.2 g/P ( g/P = A2/4, V1 = 0.9A2, V2 = A2 x h )

2 x A2 x h = 0.6A2 + 0.9A2 + ).3A2 , giving final height of larger piston as 0.9 m only. So, in the end, the gas remains at same temperature as before. Additional P.E.of 2kg at 0.6 m becomes P.E. 4kg mass 0.3 m higher.

This is my third attempt at hurriedly trying to solve the problem after #59 and #60 above and I hope there is no mistake this time;

Total initial energy = P1V1 + 4g (0.6) + 2g (0.6) = P x 0.9 x A2 + 3.6g

Added energy = 2g (0.6) = 1.2g

Final energy = P2V2 + 4gh = P x 0.9 x A2 + 3.6g + 1.2g = P x A2 x h + 4gh = 8gh

4g/A2 x 0.9 x A2 + 4.8g = 8gh

3.6g + 4.8g = 8gh, Therefore, h = 1.05 m and not 0.9 m. The larger piston is therefore 1.05 m above the smaller piston. As the pressure of the gas is unchanged but volume changes (increases), please note that temperature of the gas does change (increases). Comments and corrections welcome.

I don't think so.

The energy is converted to heat, not directly to other potential energy. This is (0.6*4.g-h4.4.g) Joules, where h4 is the distance that the 4kg piston rises. To a first approximation that would be 1.2g Joules as you say. So far, so good. But I don't understand what you have done from there onwards:

For simplicity, I will assume Earth gravity and N2 at standard temperature and pressure - but (other than a small effect due to the weight of the gas itself) this would all cancel out in practice.

The area of the large piston would be about 3.9-sq-cm, so the volume of the gas would be 3.5-c.c. The mass would be about 4.4gm, and the heat capacity about 4.6 Joules/^OC. So the temperature rise of the gas would be about 1.2*9.81/4.6 or 2.6^OC. That represents just under 1% temperature rise, which would raise the higher weight to 0.909 m height (so you can see why I only bothered with first approximations to the heating).
Anyone wanting more precision or to show that neither the initial temperature nor local gravity will affect (to first order) this is welcome to do it for themselves.
However, I would add the comment that the change in height will depend on the nature of the gas, but not on the molecular weight. The calculation was for Nitrogen, a diatomic gas that has five degrees of freedom. On the other hand, if we used a mono-atomic gas such as Helium or Neon with only three degrees of freedom the difference would increase to about 0.912 metres. More complex molecules would result in smaller temperature rises and therefore reduced increases in the height.

Fyz,

I agree with everything you've said, but have one question. Is the value unique for all temperatures? I'm having trouble finding that. I'm assuming a slow isothermal process.

tvp

Do you mean that the temperature does not change at all (the original problem), or that it is uniform throughout the gas?

Considering the case where all the heat generated from the change of potential energy remains in the gas - it doesn't actually matter whether the gas is isothermal or not: for example, twice the temperature rise in half the mass will result in an unchanged total expansion.
And, yes the unique result should be independent of temperature (subject to the gas behaving in a quasi-ideal fashion).

Consider as follows**: the mass of the gas in a specific Volume is inversely proportional to the temperature (at constant pressure). The specific heat is roughly constant*** (exactly so if the gas is ideal). So the temperature rise for a fixed heat input is proportional to the original temperature. As the expansion depends on the proportionate change in temperature, it all cancels out.
In other words, so long as the weight of the gas doesn't disturb the situation, the only variable that will affect the final height is the number of degrees of vibrational freedom of the gas molecule.

**You could write it all as a single equation and see it cancel, but I have this prejudice in favour of looking for pathways...
***Clearly, this will fail over certain temperature ranges where the effective number of degrees of freedom of the molecule changes, the gas decomposes, etc. But that is saying that the gas ceases to be anything like quasi-ideal. Most of our familiar gases behave pretty well in this regard - at least over the sort of temperature ranges we are considering.

Duh! I had a convoluted formula that I should have simplified and temperature canceled. 'Cuse me.

1) I assumed that it is a closed system and their is no energy exchanged with the environment. Radiation loss is assumed negligible over the observation period.

2) Relevant energy in the system is the energy contained in the gas and the potential energy of the pistons.

3) Gas equation PV=nRT is assumed to be applicable. Initial and final temperatures or the actual amount / type of gas don't seem to me to be relevant to the problem.

4) I agree with your observation that the energy is first converted to heat and as the gas is slowly released into larger cylinder, work is done.

5) In accordance with laws of thermodynamics, only part of the energy is converted back, thus raising the temperature.

6) As the temperature rises, volume increases, pressure is constant (as area and piston wt. which it balances do not change ), thus raising the piston higher than 0.9m. As per my calculation in #62, it turns out to be 1.05 m

I agree that it doesn't actually matter what is the temperature distribution, as the total volume increase at constant pressure does not depend on which parts of the gas are heated; calculation based on a specified volume was to minimise the necessary explanation. But (although I tried) I couldn't follow how you got from the heat to the increase in height, which is why I redid it, explaining as well as I could manage (hopefully without overcomplicating things).
As my results did not agree with yours, I'd be grateful if you could do one of the following:
a) find my (more than possible) mistake (you may need to seek clarification of individual steps in my presentation), or
b) explain what you have done in simple steps such as mere mortals can follow.

The way I see it, the route is:
Potential Energy Loss -> Heat -> Temperature Rise in Gas (under constant pressure) -> Thermal Expansion of Gas.
AND I know of no other relevant mechanisms that affect the final height.

For your calculations to work you also have to assume that the cylinders and pistons themselves do not heat up. Assuming frictionless cylinder then the smaller piston with the weight added would acccelerate downwards with the rate of acceleration depending on the size of the pipe connecting the two cylinders. Some of the potential energy would be converted to kinetic energy of the piston which would then be converted to vibration/heat of the piston and the cylinder (as well as the gas) when it hit the bottom (assuming system was closed so no energy was lost due to sound). You would have to know the thermal characteristics of the cylinder and piston to determine how much energy they would store as heat. If they stored a lot (ie. metal piston) then the gas would not heat up as much.

Yes, I have assumed that there is no fricton and that cylinder and piston do not heat up. However, I think that the Kinetic energy would be negligible at the time of piston hitting the bottom (due to the thin connecting tube). Please refer to my reply to Physicist (#75). I have not done the detailed calculations. Please correct me if I am wrong.

Whoops!
The volume would be 350-cc, the mass would be 0.44-g, and the temperature rise 26^OC, or nearly 10% temperature rise. So this first-order approximation wpuld predict a height for the 4-kg mass of 0.99-metre - enough for the initial estimation of the potential energy change to be highly inaccurate; that means we need to work with a fuller equation, to give a total height difference of about 0.96 metre.

I have redid my workings off-line and found an order-of-magnitude mistake (see #70 for correction). But try as I may I can't identify the physical basis for your calculations (maybe it's hidden away in some unstated assumptions?).
That is not to say that I can't get a similar (not-quite-correct) result: if I assume an ideal monomolecular gas, and assume that 1.2-Joules heats the gas, I would get a total height of about 1.04-m for the heavier piston. However, with that increase in height the modification to the final potential energy becomes significant...

This is my attempt to present a complete solution for the quasi-ideal gas with no thermal loss. The form is as similar to devinder's presentation as I could manage.

First, the we may equate the thermal energy stored in a quasi-ideal gas with its pressure and Volume as follows:
Eg = P.V.N/2, where P is the pressure of the gas, V is its volume, and N is the effective number of degrees of freedom of the gas.

The energy of the system (gas plus all masses) when the additional 2-kg mass is first placed is:
E = Eg(1) + 0.6.(2+2+4).g
. = P.V1.N/2 + 4.8.g
Note: Potential energies are referred to the base reference height given in the challenge

The energy of the system when everyhting has settled (all the released potential energy having transferred to the gas) is:
E = Eg(2) + 4.g.h
. = P.V2.N/2 + 4.g.h, where h is the height of the 4kg mass above the reference.

Equating initial and final energies:
P.V1.N/2 +4.8.g = P.V2.N/2 + 4.g.h, or
P.N/2.(V2-V1) = 4.8.g - 4.g.h

Now, V2-V1 = A2.(h-0.9), and P = 4.g/A2
where A2 is the area of the larger cylinder.

Substituting gives:
4.g/A2.N/2.A2.(h-0.9) = 4.8.g - 4.g.h, or
2.g.N.(h-0.9) = 4.8.g - 4.g.h, or
2.h.(N+2) = 4.8+1.8.N
i.e.
h = (2.4+0.9.N)/(N+2)

For a truly ideal gas, N=3, so
h = 1.02 metres

For a day-to-day gas such as Nitrogen at practical temperatures, N=5, so
h = 9.86 metres

N.B. Insanity checks give
1.2-metres for a hypothetical gas with zero specific heat, where there can be no conversion from potential energy to heat in the gas; and
0.9 metres for a gas with infinite specific heat (i.e. the temperature remains constant, as in the original problem)
Clearly, both of these (although non-physical extrema) are as we would expect.

I agree with your result. The difference with my result is as I missed the factor N/2 while doing the calculation. Sorry for that!

Regarding the process of arriving at the final state, the piston first accelerates with a decreasing acceleration to a maximum velocity and then as compression continues, acceleration becomes increasingly negative till the velocity becomes (approx.) zero. During this period, P,V, T and to a negligible extent n in the ideal gas equation all vary. Since the pipe is (very?) thin, this happens fast. Thereafter, slow transfer of gas and slow movement of piston goes on till all the gas is transferred to the larger cylinder at constant pressure. Will writing complete equations and solving the same make any impact on our end result? Further please tell me if I am missing anything here.

I made more than my fair share of mistakes trying to use listed parameters for reasons of explanation!

Regarding your question: once released, the small piston bounces on the gas. If we ignore the non-uniformity of heating due to the weight of the gas, the adiabatic heating will be uniform, so there will be no internal loss. However, there will always be loss of resonance due to the gas at different temperature exiting through the tube and thermal conduction along the tube. So, eventually, the levels will eventually stabilise to an equilibrium.
However, BobD is right that there are other possibilities if the small piston bounces on the bottom of its cylinder. I have assumed the bottom is flat, so the gas compression would have to be infinite before the piston actually touched bottom (the flat base would also guarantee that there was no residual oscillation when the piston reaches the bottom of its path). Otherwise, you would need to assume a pearfectly elastic bounce on the end-stop in the cylinder.

In either of those cases, once we have an equilibrium, the volume of the gas depends only on the stored energy and the applied pressure, so (as per your implementation) there is no need to consider the intermediate process. The only parameter missing (once we assume thermally-isolated gas) is the effect of the weight of the gas itself.

I should also say: if we take the weight of the gas into account for the adiabatic solution, the total volume of the gas and its location become relevant. This is both because the Volume specific heat is dependent on density (and therefore local pressure), and because the temperature will vary with the vertical position of the gas. We could therefore solve meaningfully for "no excess volume"; alternately, we might observe that, if there is a large-to-infinite depth of gas below the cylinders, the situation tends towards that of "constant-temperature-at-the-bottoms-of-the-cylinders".

Another (possibly trivial) note on the effect of using an ideal gas at the no-mass limit**: that would mean infinite thermal conductivity, and also no thermal coupling between the gas and the enclosure. Thus, all the thermal energy would inevitably remain in the gas, and conduction through the tube would rapidly damp the oscillation of the smaller piston. So the "light ideal gas" assumption would properly lead to the solution you proposed.

**QM would cause this to be a superfluid - nevertheless...

The difference in height will be 0.9m. The smaller piston will reach the bottom of the cylinder, i.e. 0.6m lower. The large piston will rise by half this amount making a difference of 0.9m. The pressure in the system remains the same as before.

I agree with uffarndan, plain and simple, with the only the facts stated this is the logical answer......

I am not an engineer but it seems to me that being in a vaccum both pistons will start at the same position, at the top. This is a pressure vs surface area problem. If you double the weight on the small piston making it equal with the other but having half the surface area then it must fall half way down the cylinder. So it will be .3m below the other cylinder due to the same pressure on each but one has a smaller surface area so its supporting pressure is less.

Just a thought