Two-Ended Rocket

The second case will leave the system in a similar condition to the first case, though not quite so extended as in the final condition in the first case.

In both cases, the system will be a bit cooler!

Initially the cylinder is on stable condition (so assyming some sort of a spring support- joining the two ends- may be even the elasticity of the collar working as a spring)

The spring tension (inwards) = the force due to pressure (outwards)

a) The pressure inside the cylinder drops down as the orifice opens.

The spring pressure remaining same, the pressure outwards reduces so the

spring tension (inwards) > new force due to pressure (outwards)

So the telescope moves in

b) The escaping force also tries to move the individual rockets inwards only

am i missing anything?

Hi SB, the idea was not to have any springs, just a relatively free sliding telescopic joint between the two cylinders. We just hold the two ends, put the pressure inside and release them to move freely (for 1st case). For the 2nd and 3rd cases, we open the orifices at the same time as releasing the cylinders.

-J

So the pressure inside the Cylinders is acting on the Supports (ie let us say hand or whatever)

Then we remove the stopper and let us say cap of the orifice(together)

This does complicate the matter a bit.

Now the external pressure tries to expand the Cylinder and the escaping ags tries to contract it.

Of course my guess (wild one ) is that it will expand.

- As it expands the pressure will release and then the flow reduces and the expanding force reduces. Assuming zero friction between the cylinders, and assuming the density of the cylinder material is more than that of air, the expansion continues till pressure equalisation, but then as the pressure differential goes down, flow reduces, and back force reduces.

Interesting, will try to mathematically formulate, but the result seem to be expansion.

Not sure whether the thought is worded - find a bit difficult some time.

With your kind permission Jorrie, let me add my understanding to make the picture clearer:

The joint beween two cylinders are frictionless and leakless. (This is the assumption, though may not be possible practically)

Hi gsuhas, yes you can assume an 'ideal system', if you want to. To make the exercise more interesting, also assume a very long telescope.

-J

Very interesting question:

I am also not sure, as you are not. But logically let me put what I think:

In second case... due to pressure differential, telescope will try to extend. At the same time, due to roket action, of individual sections, it will try to contract. Winning action will depend up on the pressure differential.

In third case, opposite will happen. Due to pressure differential, telescope will try to collapse and due to rocket action, telescope will try to extend.

Actually, we should try it physically to confirm our understandings.

Great question (as usual). Momentum always wins - details to follow tonight (It's morning in the Northern Hemisphere).

There is a point which speaks against what you claim for the 2. situation: the 2 parts of the telescopic cylinder are NOT connected to each other and can slide relatively.

If you consider one of them the jet will generate a force F= dM/dt (M= momentum, t=time) in the direction from the out side to the inside, this force will act on the other part of the pair which is opposite to the considered jet and try to extend in same direction as internal pressure. This is valid for both sides.

In the case 3 the jet acts on the outside mass so that it will not have any effect on the cylinders. The movement will be centripetal for both under out side pressure.

The problem is to consider relative and absolute effects.

<...The problem is to consider relative and absolute effects....>

...and, then, whether it is an underdamped, critically damped or an overdamped system in the second case. If the second case is either of the last two it will oscillate. If it is the first, it won't.

Game on!

Hi PWSlack, yes, I guess the game is on!

Apart from many other assumptions one can make, what about the length of the telescopic sections? The following argument gave me a bad headache.

If the orifices are opened simultaneously, then for a short time (length of the telescope divided by the speed of sound in the gas) each section can be viewed as a separate rocket, being propelled 'inward'. There is no way for the section to immediately 'know' that the other orifice was also opened.

I'll go and take an aspirin or stronger...

-J

each section can be viewed as a separate rocket, being propelled 'inward'.

But can it?
Each section can't be isolated surely...the reaction must act about the centre of gravity of the whole structure, not just the end section.

If the opposite sliding section didn't have a hole and was fixed against a pressure sensor/spring balance or whatever, it would register the force of the rocket, showing that the force is on the whole not just one sliding section.
Del

I read again your comment and you cannot consider only the wall with the orifice as a "rocket" it is from rocket point of view only part of the nozzle and a nozzle does not travel alone.

Hi again nick name, you wrote: "... you cannot consider only the wall with the orifice as a "rocket" it is from rocket point of view only part of the nozzle and a nozzle does not travel alone."

In a practical situation, where lengths are small, you are right - or at least 'close enough for all practical purposes', I would say. If I 'take you to the limit' (over very short time intervals), the decompression in the gas (due to gas exhaust) should start at the two nozzles and move inwards at the speed of sound. This should make the nozzles first move inwards for a short time, not so?

These inward movements should send compression waves down the tubes at the speed of sound in that material. The whole lot should vibrate for a while (transients) and then the usual expansion of the telescope should get under way. Now, make those tubes very long...

-J

If the pressure drops behind the wall then the jet which is proportional to the pressure drop will decrease as intensity so that the impulse force will also decrease.

Hi nick name, you wrote: "... the jet which is proportional to the pressure drop will decrease as intensity so that the impulse force will also decrease."

Do I understand correctly that you reckon that the pressure inside the wall will always win over the impulse, i.e., there will be no inward transient force, even for a short time?

I think I can agree with that. I also think that I can spot a rocket engineer speaking here! (?)

-J

The computation is a bit more complex than the one presented above since the force is proportional to the derivative of the mass. When the pressure decreases the mass flow will also decrease thus influencing the jet reactive force. It has also to be considered if the pressure ratio is over or under critical since the laws are different for volumetric flow and associated mass flow. What I think is that the pressure drop occurs on a very small distance from the orifice edge out-wards so that the pressure on the internal surface of the wall is not affected by the jet locally but can present in case of very long tubes a slight difference with respect to the pressure in the middle (assuming 2 orifices).

If we use the approximation of an under critical pressures ratio with further the assumption of constant specific mass for the fluid then the jet force will be directly proportional to the volumetric flow as Fjet= Q*ρ and Q= K (pin-pout)^0.5 * Aor

The pressure force will be Fp= (A-Aor)*(pin-pout)

Let us make the ratio of the 2 Fjet/Fp= ρ*K*Aor*(pin-pout)^0.5/((A-Aor)*(pin-pout))=

= ρ*K*( Aor/(A-Aor)) / (pin-pout)^0.5 this result is << 1 which means that pressure force > jet force in all situations.

Aor= orifice area A= cross section area of cylinders(internal considered = external for making it simple) ρ= specific mass of fluid

I think you meant presented below. since you answered to wrom one and hence you are above me

Yes I know (see the almost) that's why the further time derivatives I didn't consider (since such a high order differential equation may be for theoritical purpose, but we engineers work on approximation)

I tried to do the derivation, but in the few minute, it became too complex in case of that aspect.

The increase in volume due to the expansion, the resultant drop in pressure, the density of the air at that pressure, just to name a few were inter-related and hence the constants (K) I took were becoming a multi order differentials by themselves.

Left after filling up a page and then just the first lines and written approximately to take care of all these.

But even at the approximates, if I took a random pressure position and calculate the constants, the normal orifice (hole very less in dia wrt the end plate) was showing expansion, and quite a bit of it.

so left with result.

It will be interesting to formulate the whole thing with reverence to the Volume v(t), initial pressure differential p(t) and the few constants - end plate diameter, orifice diameter, orifice constant, ambient temperature, humidity etc.

But then i may get a PHD in theoritical physics for the effort . And I don' want to be known as Dr G (you know we dislike being appreciated)

Good problem to observe practically.

May be some one from us having CFD at disposal may try it and enlighten all of us about the results.

Good answer nick name! I think this settles it for me, thx.

-J

I too think so in second case expanding the case I said in the #20,

The air flow

Q = K √P

P = P2-P1,

K = constant dependant upon with orifice shape, Fluid(density, viscosity etc) and a few other factors.

The thrust due to air flow may be calculated by newton

m₁v₁=m₂v₂ (1: metal cylinder, 2 = fluid - air for you)

The force = the time differential

f= m₁a₁ = d/dt(m₂v₂) V2 = K₂√P

= V₂dm₂/dt + m₂dv₂/dt , dm₂/dt = Q

= QV₂ almost

= K₃P

The nett force outward force

= A.P (A= total cross section area - the orifice area)

so the force acting on the cylinder (in either case) say case (2) outwards

= (A-K₃).P

The net thrust will be of course depend on the (A-K₃) and unless (A) is very small ie the orifice is very large with respect to cross section, the value will be positive

ie the pressure will win over the expansion.

I am not sure of the mathematics at this moment, and nor the values of the constant, but may be substituted

I sincerely do not see where a damping could come. There is no spring to accumulate energy and as the problem was put the parts do slide free in each other.

" ... whether it is an underdamped, critically damped or an overdamped system ..."

"If (it) is either of the last two it will oscillate ..."

Sure you haven't got that the wrong way round? (or am I misreading you?). If anything's overdamped, it won't oscillate.

Oops!

Well done.

Hi nick name, if you look at what I wrote to PWSlack in #13, do you still think the same?

BTW, are you actually from "La Ville-Lumière"?

Fortunately, we do not have to consider the speed of light here!

-J

1- Yes I am at home in "La Ville Lumière". And on greater scale in European Community.

2- I considered in my answer that:

- there is no friction between the two sliding parts.

- even if the tubes are very long and the delay due to the sound limited speed is unelectable the jet effect will affect the internal gas mass and not the part with the orifice, there is from the mechanical (classical) point of view (as far as I know) no reason to have a jet reaction force on the orifice wall. In such a case the jet will generate a "force" wave which will travel till it will reach the opposite wall. During this time the pressure will act on the wall and push out the parts.

You know I may be wrong since no one knows every thing and the more you know the more you know that you know nothing (that's from me)

Taking it to extremes for an illustration of a special case:

1) Pint > Pext

Say Pext ≈ 0, and area of orifices << area of ends of cylinders.

Telescope extends until the two sections separate.

2) Pint < Pext

Say Pint ≈ 0, and again area of orifices << area of ends of cylinders.

Telescope contracts as far as it physically can. Eventually (some time later), pressures equalize with no further movement.

Haven't got as far as thinking it through with larger orifices!

Be gentle coz I a cat, I think in the second case the 'rocket effect' is zero, the rocket forces just cancel eachother and they are acting on the gas or the opposite casing (eg forcing the oppose casing out, rather than the one with the hole in it inwards...??? )
Del<rushes off to hide in cat nest>

Strongly agree.

The momentum is conserved with the equal jets of air in opposite directions.

Cases 2 and 3 are the same (but opposite).

Hang on - isn't this similar to the issue of the design of the shock absorber for the modern road motor carriage suspension system?

• Turn the system 90deg. Let one of the telescopic components be attached to something rigid, like a car body, and the other component be attached to the suspension system.
• Is it necessary to consider the port in one component or a port in both?

Hmmmmmmm........................

As the problem was put there is no rigid connection whith any other mechanical component. The damper has a reaction according to the relative displacement of its ends. Here there free forces generated acting on free bodies. I do not see any similitude from the mechanical point of view.

Jorrie,

Thanks a lot! Now I have a thought experiment I believe I'm not capable of solving!

Specifically because assumptions must be made about friction between cylinders, how fast the nozzles open (not to mention what size they are), and the C_v's of the orfices.

Since each orfice is subject to the same fluid, the properties of such may not come into play, but I can't convince myself that there's not a difference between effects of SF₆ (for example) and helium.

If anyone has expertise in rocket science, I'd be very interested in their analysis.

OK, let me start backwards. Case 3 is fairly straightforward if I make a few simplifying assumptions which, I believe, don't change the question at all. So, here are my assumptions:

The orifices are quite a bit smaller than the end caps, but large enough that edge effects can be neglected (at least at a first cut).

The pressure inside is zero. The pressure outside is, say, 10 bar.

The cylinder is massless.

The orifices are initially closed and can be opened simultaneously.

The two halves slide without friction and are initially pinned together with a pin that can be removed simultaneously with the orifice caps. Otherwise the outside pressure crushes this thing before you start.

1. So, you remove the orifice caps and the pin.

2. The air pressure acts on the air molecules impinging on the end cap at the same time it acts on the air molecules at the opening. Since there is no cylinder mass to retard acceleration, all air molecules on the right accelerate equally to the left. The molecules at the opening never get inside since the orifice moves as fast as they do.

3. The right end cap moves to the left. The left end cap moves to the right.

4. The cylinder eventually bottoms out, still with a vacuum inside.

5. Since there is no cylinder mass, there can be no oscillation.

Now, had any of my assumptions been wrong....

If there were initially air inside, that would slow equally the incoming air and the end cap, but the collapse would now slow exponentially as the inside pressure rose due to volume decrease. In general, this would look like critical damping, although we might arrange it so that it looks overdamped?

If there were friction in the cylinder, the air at the orifice would race ahead of the end caps and we would now get an earlier equilibrium, possibly with a small oscillation. I haven't thought through whether that looks underdamped or not.

If the cylinder had mass, we now get the air at the orifices racing ahead, we get an earlier equilibrium, and we definitely have underdamping.

Hi TVP, a very good classical analysis, but what about the issue that I pointed out to nick name? If I'm right there, then for the 3rd scenario, the end plates should start to move outward for a brief period, because the pressure there should be higher than at the center of the telescope, due to the inrush of the gas. After the transients subside, I agree that it will work as you wrote.

-J

That should not happen. The inrushing molecules should not have a velocity greater than the end cap unless the end cap has a real mass. If that is the case, then this happens: The inrushing molecules have a net momentum only in the -x direction on the right and +x on the left. So, they could not make a "U" turn and apply pressure to the inside of the end cap in a transient. They would, on average, continue in until they collided withe ones from the other side, at which point everything cancels. There should be a surge, not unlike water hammer, at that point, but keep in mind that velocity head must always be no higher than the static head.

This is a tricky problem to think about because our experience usually includes a friction head component, which does change things. But, another way to look at this is as follows. From symmetry, we can argue that we have only mirrors of two cans and that nothing crosses that neutral border. If that is the case, we can place a virtual "bottom" for each can at that neutral point, and the problem reduces to the classic punctured vacuum can which is known to do nothing.

T

"... and the problem reduces to the classic punctured vacuum can which is known to do nothing."

But, doesn't the "punctured vacuum can" reacts exactly like the "Feyman Sprinkler", which turns during the transient when the suction is switched on? In other words, doesn't the can get a short-lived kick?

-J

Yes, you are quite right. Shame on me for accepting a standard explanation without question. The punctured vaccum can will accelerate to the right during the time it takes for the incoming air to reach the bottom. It will then maintain some steady, albeit very small, velocity until drag brings it to a stop.

Case number 3, however, does not do that due to symmetry.

Kudos to you for deflating a long held misconception. You really are pretty darn smart.

In the first case I see a number of things happening, due to differential pressure the cylinders expand, this results in kinetic energy of the moving parts, displacement of the surrounding air, and a drop in temperature as the internal pressure drops? reasonable quickly a dampening effect will bring the system into equilibrium, thereafter the internal temperature will take in heat from the outside and will slowly expand until the temperatures are even.

In the second case the jet of air from either ends of the cylinders reacts with the internal air not the cylinders themselves and the stream of air will act as venturi-es creating a vacuum at both ends, this effect will act on the cylinders. The same effects as mentioned above will occur till the system comes into equilibrium, but in this case as the temperature equalises air will be draw in through the end ports.

The end result is, the first case will expand more than the second case resulting from the temperature drop of the air doing a bit extra work as its temperature stabilises. But I am unsure what effect the vacuum will have on the final result regarding the distance moved apart?

Regards JD.

Case #2: just look at all the forces on one of the cylinders, it must move outward.

Case #3: the opposite.

And now the hard ones. Cases 1 and 2 are simply double acting cylinders except the seals are bad in #2.

It depends on orifice size. Imagine 1" diameter cylinders. With orifices only 0.001" in diameter the telescope will move apart. With orifices of 0.999" the telescope will move closer. At some point in between forces balance.

Forgot to login.

1st. case, the cylinders extend 'till Pint=Pext, then it overshoots and oscilates in spite of the no mass of the cylinder, gas molecules have mass and therefore innertia.

2nd. case, the cylinders extend 'till Pint=Pext, which will happen at a shorter travel due to gas loss, besides, any overshooting will be damped by the holes.

3rd. case, the cylinder couple will collapse, can't expand since Pint<Pext.

In 2nd and 3rd case, the effect of the holes over the inner gas mass will be delayed by the lenght of the cylinders.

Yahlasit

Let's assume the above and that the outer cylinder is infinitely long thus bottoming out or separating is not an option.

1st case. Pistons oscillate away from and then towards each other reducing slightly each cycle due to the inevitable losses, friction, turbulence & ? The force pushing them apart is piston end area X pressure. The forces are equal and opposite and they will accelerate equally. ( note incidentally that this is two equal pistons, not as Jorrie's original drawing which had a larger piston on the right).

2nd case. To put some numbers to this idea, lets say the pistons have an area of 100cm^2 and the gas pressure is 10kg/cm^2. thus the initial force will be 1,000kg.

We then release the system, simultaneously opening a 10cm^2 hole. The force acting on each piston is now 900kg. The acceleration will be slowed accordingly. At this level the situation would almost certainly be over-damped and no oscillation would take place. There would be no "jet" reaction in anyway as the piston has no back to it and thus nothing to react against. If one cap failed to seal but left the two pistons tethered then they would move in the direction of the sealed cap at a rate determined by the pressure differential (in /out) x area of the hole.

If the holes were a pinholes i.e. too small to note against the 1,000kg initial force then oscillation would still take place but in a damped manner and will come to a standstill much quicker as each cycle will loose more gas to atmosphere.

3rd case. is essentially no different to the 2nd. The forces only act on the surfaces that have pressure differentials on opposite sides.

regards

Chas

Hello Jorrie,

hope you are well?

Two ended rocket?.........................Mmmmmm, is this to confuse the 'enemy'?

Take care and interesting question!

bb

The jet reaction will be due to the mass transaction (momentum interaction) it soes not matter whether there is a back or not or does it?

The system is intearcting with its sorrounding and jetting out the fluid so the momentum of that fluid is going to have an applied force. It is like a long straight hose pumping out air, does it have a applied force on the pipe material? I suppose it has

In this case, since the orifice is large (and the pressure on the end plate is zero- no end plate) the applied force is compressive - pure jet force.

So if there is an applied force on one edge of the system, the same will happen on the other edge too. (Put a tee in betwwen to pump in air and let it get out of the two ends of tee)

To see the force in action, bend the two ends and you have a air motor, or a fan or whatever you want to call it

The jet reaction will be due to the mass transaction (momentum interaction) it soes not matter whether there is a back or not or does it?

Quite right...even a cat kno that.
But what does the reaction act upon?
1.The Centre of Gravity of the whole assembly?
2.The nozzle?
3. The remaining gas?
An answer to this would be helpful....
Del

The cat_ch is there It works on the whole system across which the transfer is taking place. But here the system is in two place with a possibility of a sliding motion.

Assuming some sort of a symmetry, from where onwards there is no further transfer across boundar is there (say at the middle point), from there the air rarifies on both sides but except the usual molecular level dispersions, no nett mass transaction is taking place. So say on half of the telescope each the force is inwards, and the result is the telescope trying to slide over each other.

₍Again hunch feeling and looks to be on mathematically viable but then you have to try out the experiment.The air will be too light with reference to normal metals, so beter try to throw something heavier from it _{(what about some sort of canon ball leaving both sides?)}

Jorrie,

I was partly wrong about the single can (as in case 3). If you had only the right side, it would move to the right, then back to the left, then back to rest, sort of a complete triangular wave. There could be no continued drift, conservation of momentum and all that.

Cheers.

But has not some of the mass disappeared off-stage (left or right - lost the thread of which way is which) at some speed? It would therefore take some momentum with it - yes?

(If I've got it wrong, and you're talking Pint<Pext, then the same applies: the mass has increased - and it's come in from one side only).

That was my original thought. Then I put the whole thing inside a big box (mental) and asked whether momentum would be conserved.

But a rocket in a box will still move as far as it can?

Until it collides with Schrödinger's Cat!