Can Air Resist a Piston? A Serious Discussion

[Sailor Al]

TL;DR Summary

I know that sound travels through air at Mach 1.0.
An ideal piston moving in an open ended cylinder will generate a pressure wave of some sort.
I would like to know how this pressure wave impacts the pressure against the piston.
The answer will assist in working out how the shape and orientation of a sail or wing affects the aerodynamic force being generated by the speed of the foil through the air.
Ref youtu.be/dgE9xhIjTOU and paper linked in vid description

 

[A.T.]

Does the required force vary with speed?

The pressure wave is generated by the change in speed. If you move the piston for a long time at constant speed through a very long open ended cylinder, and there is no viscosity or drag at the walls, then the air will move at the same speed as the piston and the force on the piston will go to zero.

 

[Sailor Al]

But here's where I think you're wrong.
At constant speed, since it's a long tube, the piston will be pushing (and pulling) an increasing amount of air. All the air that was in front of of the piston will be moving into the stationary air even further ahead. That air will have to be accelerated to the speed of the piston. Also, the space behind the piston will have to be filled by accelerating the stationary air further behind. Mass x acceleration = force.
I think the force necessary to maintain the piston speed will increase even with a constant velocity piston.

 

[DaveC426913]

I think the force necessary to maintain the piston speed will increase even with a constant velocity piston.

Consider extending that experiment without limit - say a pipe hundreds of miles long that is open at the end.

You are suggesting that it will get harder and harder to push the piston at constant velocity along that pipe. But the inevitable outcome of such a notion is that the stopping force (the pressure), rising without limit as it does, will eventually reach and exceed the force available. Like: you would literally be stopped in your tracks and be unable to push it any farther - first with human strength and, eventually, even with a tractor. You'd be stuck in the tunnel, with your piston, pushed by a tractor, going nowhere ... because of all the air piled up.This is, obviously, an absurd outcome, therefore the premise must be faulty.

 

[A.T.]

All the air that was in front of of the piston will be moving into the stationary air even further ahead. That air will have to be accelerated to the speed of the piston.

I was thinking about a steady state, after a long time at constant speed, where all the air in the tube is moving at the same speed as the piston. With an open tube there would be circulation outside the tube, also at a constant speed.

The speed of the piston affects how long it would take to achieve this steady state. And if it is a significant fraction of the speed of sound, you might not achieve it before you run out of tube.

 

[Sailor Al]

Consider extending that experiment without limit - say a pipe hundreds of miles long that is open at the end.

You are suggesting that it will get harder and harder to push the piston at constant velocity along that pipe. But the inevitable outcome of such a notion is that the stopping force (the pressure), rising without limit as it does, will eventually reach and exceed the force available. Like: you would literally be stopped in your tracks and be unable to push it any farther - first with human strength and, eventually, even with a tractor. You'd be stuck in the tunnel, with your piston, pushed by a tractor, going nowhere ... because of all the air piled up.This is, obviously, an absurd outcome, therefore the premise must be faulty.

Interesting.
On your first point, when I specified a "long" pipe, I meant long, as in really long ... without an end, so you're not really extending the experiment, that is the experiment.
And yes, the conclusion that you've arrived at is exactly what I imagined. As time passes, more air will "pile up" and increase the load - indefinitely... forever. And more and more force would be required to maintain the velocity.
But, please don't shoot the messenger: the premise is not faulty. The conclusion is surprising, but I think it's really worth breaking it down to make sense of it.
I hope, at sometime in the future we'll return to these comments with a wry smile.

 

[Sailor Al]

I was thinking about a steady state, after a long time at constant speed, where all the air in the tube is moving at the same speed as the piston. With an open tube there would be circulation outside the tube, also at a constant speed.

The speed of the piston affects how long it would take to achieve this steady state. And if it is a significant fraction of the speed of sound, you might not achieve it before you run out of tube.

I don't think a steady state ever occurs. As I suggested to @DaveC426913 above, the tube is long, so by definition, you can never get all the air moving.
We never run out of tube.
I would like to break it down: how quickly does the force rise?

 

[256bits]

But here's where I think you're wrong.
At constant speed, since it's a long tube, the piston will be pushing (and pulling) an increasing amount of air. All the air that was in front of of the piston will be moving into the stationary air even further ahead. That air will have to be accelerated to the speed of the piston. Also, the space behind the piston will have to be filled by accelerating the stationary air further behind. Mass x acceleration = force.
I think the force necessary to maintain the piston speed will increase even with a constant velocity piston.

With the piston moving at constant speed, the dm dv/dt is a constant ahead of the piston as well as behind - thus the force is a constant.

At the steady state condition, where there is circulation of air outside the piston, the force drops to zero.
@A.T. has already stated this.

 

[Sailor Al]

I think we need some calculus here.... At time t, the position of the piston, P(t).x = v * t
The position of the pressure front (F(t).x , moving at Mach 1.0 from the initial condition, (Ma) is F(t).x = Ma * t
And some mechanics: we'll need to introduce ρ, the density of air at NTP.
Now we need some calculus to work out the rate of increase in the force required to overcome the acceleration of the air just ahead of the pressure front.
At t + dt what happens to the force?
Here my maths gets a bit hazy, my degree in mathematical physics was 60 years ago.

With the piston moving at constant speed, the dm dv/dt is a constant ahead of the piston as well as behind - thus the force is a constant.

At the steady state condition, where there is circulation of air outside the piston, the force drops to zero.
@A.T. has already stated this.

But, as defined in the problem, the pipe is long, there can be no steady state because the air never gets to circulate outside the cylinder.
And even if the pipe were not long, there would be no steady state because the distance from the piston to the end of the pipe reduces.
I would like to tackle the long pipe condition first, then consider the condition not of a fixed length pipe, but one which extends at different speeds, from zero (fixed) through P.v through to Mach 1.0 . I suspect extending the pipe at Mach 1.0 would yield the same result as a long pipe.

Thanks for the opportunity to expand on this.

 

[jack action]

But, as defined in the problem, the pipe is long, there can be no steady state because the air never gets to circulate outside the cylinder.

That cannot be right. How can you have an open end (no matter how far it is) and no friction, and still have no possibility of air exiting?

I propose another experiment. Imagine a closed tube in a donut shape. When you will accelerate the piston inside, the air will compress, leaving a pressure differential between each side of the piston. But when the piston will reach a constant velocity, the air will "catch up" with the piston and decompress. The result will be air and piston moving together at the same velocity. The air cannot have constantly a velocity of zero at some point while the piston moves at a constant velocity, otherwise, only one revolution would be possible; the piston stopped by some sort of magical force.

Opening the tube to a common atmosphere shouldn't change that. (Assuming no losses when the air enters and leaves the cylinder.)

When you will begin pushing the air with the piston, a pressure wave will go at the speed of sound toward the exit. When it will reach the exit (no matter how long it takes), it will be reflected as a negative pressure wave (vacuum) - still going at the speed of sound - that will assist in pulling the air out of the tube. The opposite goes for the air on the other side of the piston (negative pressure wave created, reflected as a pressure wave at the entry of the tube, helping to push the air inside the tube).

 

[Chestermiller]

Suppose that the tube is short on the right, and the pressure of the right-side air on the piston is always atmospheric. Also suppose that the piston is moving much slower than the speed of sound in the gas. Let Fg represent the force of the air on the left side on the left face of the piston. Let's see your force balance equation on the piston.

 

[Sailor Al]

I think we need some calculus here.... At time t, the position of the piston, P(t).x = v * t
The position of the pressure front (F(t).x , moving at Mach 1.0 from the initial condition, (Ma) is F(t).x = Ma * t
And some mechanics: we'll need to introduce ρ, the density of air at NTP.
Now we need some calculus to work out the rate of increase in the force required to overcome the acceleration of the air just ahead of the pressure front.
At t + dt what happens to the force?
Here my maths gets a bit hazy, my degree in mathematical physics was 60 years ago.

That cannot be right. How can you have an open end (no matter how far it is) and no friction, and still have no possibility of air exiting?

I propose another experiment. Imagine a closed tube in a donut shape. When you will accelerate the piston inside, the air will compress, leaving a pressure differential between each side of the piston. But when the piston will reach a constant velocity, the air will "catch up" with the piston and decompress. The result will be air and piston moving together at the same velocity. The air cannot have constantly a velocity of zero at some point while the piston moves at a constant velocity, otherwise, only one revolution would be possible; the piston stopped by some sort of magical force.

Opening the tube to a common atmosphere shouldn't change that. (Assuming no losses when the air enters and leaves the cylinder.)

When you will begin pushing the air with the piston, a pressure wave will go at the speed of sound toward the exit. When it will reach the exit (no matter how long it takes), it will be reflected as a negative pressure wave (vacuum) - still going at the speed of sound - that will assist in pulling the air out of the tube. The opposite goes for the air on the other side of the piston (negative pressure wave created, reflected as a pressure wave at the entry of the tube, helping to push the air inside the tube).

OK, If you are having trouble with a long pipe, let's limit it to, say 10,000 Km, with the piston starting in the middle - 5000 Km from each end, and confine the experiment to, say one hour.
Let's set Mach 1.0 to 343 m/sec, so in an hour the pressure front will have travelled 3600*343/1000 Km = 1,235 km. It hasn't reached the end of the tube, so the air in the remaining 3,765 km has not been affected by the piston at either end. So there has been no circulation with the outside air.
How do we calculate the force required then, 1 hour after the start , or 2 hours...?
Please, stick with my experiment. It's a bit tricky but I think the results will be rewarding.

 

[Sailor Al]

Suppose that the tube is short on the right, and the pressure of the right-side air on the piston is always atmospheric. Also suppose that the piston is moving much slower than the speed of sound in the gas. Let Fg represent the force of the air on the left side on the left face of the piston. Let's see your force balance equation on the piston.

Sorry, that's a different problem. Please stay with my problem. The tube is long. The pressure front never reaches the end of the tube.

 

[Chestermiller]

Sorry, that's a different problem. Please stay with my problem. The tube is long. The pressure front never reaches the end of the tube.

Then you should be including viscosity also.

 

[jack action]

I think we need some calculus here.... At time t, the position of the piston, P(t).x = v * t
The position of the pressure front (F(t).x , moving at Mach 1.0 from the initial condition, (Ma) is F(t).x = Ma * t
And some mechanics: we'll need to introduce ρ, the density of air at NTP.
Now we need some calculus to work out the rate of increase in the force required to overcome the acceleration of the air just ahead of the pressure front.
At t + dt what happens to the force?
Here my maths gets a bit hazy, my degree in mathematical physics was 60 years ago.

OK, If you are having trouble with a long pipe, let's limit it to, say 10,000 Km, with the piston starting in the middle - 5000 Km from each end, and confine the experiment to, say one hour.
Let's set Mach 1.0 to 343 m/sec, so in an hour the pressure front will have travelled 3600*343/1000 Km = 1,235 km. It hasn't reached the end of the tube, so the air in the remaining 3,765 km has not been affected by the piston at either end. So there has been no circulation with the outside air.
How do we calculate the force required then, 1 hour after the start , or 2 hours...?
Please, stick with my experiment. It's a bit tricky but I think the results will be rewarding.

But you said: "at constant velocity". At constant velocity, there will not be a pressure wave at all as you will be in a steady-state condition.

But if you were in such a long pipe, and the air would pile up as the piston accelerates with no exit close by, I would expect there would be a shock wave created at some point. More precisely, a moving shock wave, one that moves faster than the speed of sound. (The speed depends on the pressure differential across the shockwave.)

 

[A.T.]

But if you were in such a long pipe, and the air would pile up as the piston accelerates with no exit close by, I would expect there would be a shock wave created at some point. More precisely, a moving shock wave, one that moves faster than the speed of sound. (The speed depends on the pressure differential across the shockwave.)

If the piston moves slower than the speed of sound, why would there be a shockwave?

 

[jack action]

If the piston moves slower than the speed of sound, why would there be a shockwave?

At the limit, if the air pressure gets high enough with respect to the exit pressure, the local air velocity must reach the speed of sound somewhere in the pipe (choked flow). That means a pressure wave will move at twice the speed of sound. As the pressure increases, the temperature also increases, thus the speed of sound increases as well.

For sure, at one point, something will give. Again, that is only when accelerating. Once the airflow is accelerated and going at the constant speed of the piston, there are no forces involved, no matter the speed of the piston/airflow.

If you replace the air with water, can you accept that the water will move with the piston? With a compressible fluid, the only difference is the pressure variation along the path during acceleration; but the air will move for sure at one point.

 

[A.T.]

I specified a "long" pipe, I meant long, as in really long ... without an end, so you're not really extending the experiment, that is the experiment.
And yes, the conclusion that you've arrived at is exactly what I imagined. As time passes, more air will "pile up" and increase the load - indefinitely... forever.

The amount of pilled up air moving with the piston is increasing indefinitely, but if the air has no resistance at the walls, then that doesn't increase the force. It seems that if the piston moves at constant velocity, then the rate which the air is pilled up will approach a constant value and so will the force, probably something like: 2v²Aρ.

 

[Chestermiller]

I did a modeling analysis to describe the transient situation that will exist if the piston suddenly starts moving at a constant velocity. The equation for this will be $$\frac{\partial ^2 u}{\partial t^2}=\gamma \frac{RT_0}{M}\left(\frac{1}{1+\frac{du}{dx}}\right)^{\gamma+1}\frac{\partial ^2 u}{\partial x^2}$$where u is the axial displacement of the air, $T_0$ is the initial temperature of the air in the tube, M is the molecular weight, x is the axial coordinate, and M is the molecular weight. For low piston velocities compared to the speed of sound, this reduces to the ordinary wave equation $$\frac{\partial ^2 u}{\partial t^2}=c^2\frac{\partial ^2 u}{\partial x^2}$$where c is the speed of sound: $$c=\sqrt{\frac{\gamma RT_0}{M}}$$Also, for low velocities of the piston compared to the speed of sound, the magnitude of the pressure disturbance relative to the initial pressure is negligible.

 

[Sailor Al]

@Chestermiller Can you break that down a bit please, my maths isn't what it used to be.
1) What is the origin of your initial PDE please?
2) What are you using as the function u(x,t)?
3) Why does it rely on a low piston velocity? That wasn't in my problem definition and , while v < c, I would like the solution to allow for a to be a significant fraction of c, like c/2.
4) does your wave equation yield a solution for my problem? Does it give a function of a) force over time or b) force over piston velocity?

 

[Sailor Al]

At the limit, if the air pressure gets high enough with respect to the exit pressure, the local air velocity must reach the speed of sound somewhere in the pipe (choked flow). That means a pressure wave will move at twice the speed of sound. As the pressure increases, the temperature also increases, thus the speed of sound increases as well.

For sure, at one point, something will give. Again, that is only when accelerating. Once the airflow is accelerated and going at the constant speed of the piston, there are no forces involved, no matter the speed of the piston/airflow.

If you replace the air with water, can you accept that the water will move with the piston? With a compressible fluid, the only difference is the pressure variation along the path during acceleration; but the air will move for sure at one point.

Why must the "local air velocity reach the speed of sound"? The piston is travelling at a constant speed. The air, initially stationary w.r.t. the tube will only ever accelerate to the speed of the piston. Why does it have to go faster? I don't follow, could you elaborate please?

 

[Sailor Al]

The amount of pilled up air moving with the piston is increasing indefinitely, but if the air has no resistance at the walls, then that doesn't increase the force. It seems that if the piston moves at constant velocity, then the rate which the air is pilled up will approach a constant value and so will the force, probably something like: 2v²Aρ.

Agreed, the air has no resistance at the walls, so that doesn't increase the force.
Yes, the piston is moving at a constant velocity. I am not sure what you mean by rate of "pile up" becoming constant. Could you elaborate a bit please?
And I'm looking for a way to get to the bottom of that force equation.

 

[jack action]

Why must the "local air velocity reach the speed of sound"? The piston is travelling at a constant speed. The air, initially stationary w.r.t. the tube will only ever accelerate to the speed of the piston. Why does it have to go faster? I don't follow, could you elaborate please?

If you assume that the pressure builds up on the piston's face (locally) while the exit pressure stays the same (because the air is not moving there) then there is a pressure differential that tends to push the air toward the exit. If it would build high enough (because we assume the flow is somehow restricted to exit), the pressure differential will be enough to create a velocity of Mach 1, i.e. choked flow (most likely a pressure ratio of $\frac{p_{piston}}{p_{exit}} > 1.893$ for air).

Of course, I don't think this will ever happen as @Chestermiller demonstrated (the air will begin moving forward at a much much lower velocity), but it shows that you cannot increase the pressure indefinitely without creating some kind of forward motion at some point.

 

[jbriggs444]

Why must the "local air velocity reach the speed of sound"? The piston is travelling at a constant speed. The air, initially stationary w.r.t. the tube will only ever accelerate to the speed of the piston. Why does it have to go faster? I don't follow, could you elaborate please?

One can make a hand-wavy argument that the scenario is equivalent to the collision of a moving infinite column of air with a stationary piston. If energy is conserved then the result should be an elastic collision with the column of air retreating from the piston at the same speed that it arrived. Shift to the frame in which the air is initially motionless and we have a column of air retreating at twice the speed of the piston.

No speed of sound lurking in this vision.

Of course, the handwave depends on the false assumption that the final state is a column of air in uniform motion. We have an infinite column. Some will be in motion. Some will still be stationary. Forever.

I think that @Chestermiller is on the right track, modelling the situation in terms of an initial disturbance rather than chasing some imagined final steady state.

 

[Sailor Al]

If you assume that the pressure builds up on the piston's face (locally) while the exit pressure stays the same (because the air is not moving there) then there is a pressure differential that tends to push the air toward the exit. If it would build high enough (because we assume the flow is somehow restricted to exit), the pressure differential will be enough to create a velocity of Mach 1, i.e. choked flow (most likely a pressure ratio of $\frac{p_{piston}}{p_{exit}} > 1.893$ for air).

Of course, I don't think this will ever happen as @Chestermiller demonstrated (the air will begin moving forward at a much much lower velocity), but it shows that you cannot increase the pressure indefinitely without creating some kind of forward motion at some point.

I'm not sure why you say the flow is restricted at the exit. It isn't. It's a long cylinder and the exit is always further away than the pressure wave.
I am not familiar with choked flow, and when I followed your link for pressure ratio, I found it related to venturis, so I don't know how that applies to this problem.
You say you don't think this will ever happen, I'm not sure what "this" is.
I am addressing @Chestermiller directly.

 

[Sailor Al]

One can make a hand-wavy argument that the scenario is equivalent to the collision of a moving infinite column of air with a stationary piston. If energy is conserved then the result should be an elastic collision with the column of air retreating from the piston at the same speed that it arrived. Shift to the frame in which the air is initially motionless and we have a column of air retreating at twice the speed of the piston.

No speed of sound lurking in this vision.

Of course, the handwave depends on the false assumption that the final state is a column of air in uniform motion. We have an infinite column. Some will be in motion. Some will still be stationary. Forever.

I think that @Chestermiller is on the right track, modelling the situation in terms of an initial disturbance rather than chasing some imagined final steady state.

Sure there speed of sound is lurking. That's as fast as a pressure wave will move through air. The movement of the piston cannot be detected beyond the the pressure wave. The air ahead of the pressure wave remains stationary and at the ambient pressure. Discounting the speed of sound denies that reality.
And nobody is chasing some imaginary steady state. Certainly not me. From the moment the piston starts moving the system loses equilibrium - the pressure wave advances faster than the piston and I think the pressure in front of the piston increases, but I don't know, and that's what I'm looking for assistance on.
Not making much progress so far.
I suspect @Chestermiller is muddying the waters with his insoluble PDEs.

 

[berkeman]

From the moment the piston starts moving the system loses equilibrium - the pressure wave advances faster than the piston and I think the pressure in front of the piston increases, but I don't know, and that's what I'm looking for assistance on.

A reasonable request.

I suspect @Chestermiller is muddying the waters with his insoluble PDEs.

No, Chet is our local savant when it comes to questions like this. Please watch your attitude.

 

[Sailor Al]

A reasonable request.No, Chet is our local savant when it comes to questions like this. Please watch your attitude.

Which bit offended you. Do you think his PDE's are soluble? (There's a $1M prize on offer if you can show that a solution even exists!)
Muddying the waters? Yes, his response was not an answer to the question. That's muddying the waters in my book.
Are you able to answer the questions I raised?

 

[boneh3ad]

The pressure wave is generated by the change in speed. If you move the piston for a long time at constant speed through a very long open ended cylinder, and there is no viscosity or drag at the walls, then the air will move at the same speed as the piston and the force on the piston will go to zero.

This is not correct.

The piston problem is a classical one in gas dynamics. It will generate a shock wave in front of it that moves to the right faster than the piston (supersonically) and compresses the air ahead of the piston so that it has a higher pressure than ambient. You will get the opposite effect on the back size, instead generating an expansion wave that lowers the pressure below ambient. End result: a net force opposing the piston's motion.

Assuming an infinitely long tube, this force will be maintained forever, even at steady state. For a finite tube, of course the shock will eventually exit the tube and allow some pressure equalization, but that does not appear to be the spirit of the question.

 

[Sailor Al]

This is not correct.

The piston problem is a classical one in gas dynamics. It will generate a shock wave in front of it that moves to the right faster than the piston (supersonically) and compresses the air ahead of the piston so that it has a higher pressure than ambient. You will get the opposite effect on the back size, instead generating an expansion wave that lowers the pressure below ambient. End result: a net force opposing the piston's motion.

Assuming an infinitely long tube, this force will be maintained forever, even at steady state. For a finite tube, of course the shock will eventually exit the tube and allow some pressure equalization, but that does not appear to be the spirit of the question.

Unfortunately I am not a student of gas dynamics, so am looking for advice.
I think your description is spot on. The piston compresses the air in front of it. That requires a force
You suggested the "shock wave" moves supersonically. I don't see how that can be true. Wouldn't it travel at Mach 1.0? Can any pressure wave travel faster than that through air? Please link to a reference that can resolve that.
And yes, absolutely, let's keep the discussion using the infinite tube.
My questions remain:
1) How can can we calculate the force: The piston compresses the air in front of it. That requires a force. It's a one-dimensional mechanics question with a differential equation, but my maths isn't up to it. We know the velocity of the piston v, the speed of sound Ma, the density of air, the relationship PV/T= C. There's no friction or viscosity to contend with. The piston has no mass.
2) Is the force steady, if not, how does it change? On the one hand the piston is pushing (and pulling) more and more air, so I would expect the force to rise, but on the other hand it's only accelerating the new air at the wave front, so maybe it's constant. I just don't know.

 

[boneh3ad]

I am going to use $v$ to represent velocities in the lab reference frame and $u$ to represent velocities in the frame of the compression wave ahead of the piston.

Case 1: Impulsively-started piston
Consider first the impulsively started piston. It is initially at rest, and, at some time $t=0$, it is impulsively accelerated to a velocity of $v_p$ (moving left to right, per the diagram in the original post). No penetration requires that no fluid crosses the face of the piston, so locally at the piston face, the flow is moving with $v=v_p$, but in the far field it is still moving at $u = 0$, creating a velocity discontinuity. It is a classical result that such a change in a 1D duct occurs discontinuously, i.e., as a result of a shock. This shock propagates with a speed $v_s$.

Consider a change in frame of reference such that the coordinate system follows along with the shock, i.e., subtract $v_s$ from everything. Then the velocity upstream of the shock is $u_1=0-v_s=-v_s$ and downstream of the shock is $u_2 = v_p- v_s$. One of the standard methods for determining the Mach number of a shock is to look at the velocity ratio across it (in its reference frame), which follows the relation
$$\dfrac{u_2}{u_1} = \dfrac{2 + (\gamma-1)M_1^2}{(\gamma + 1)M_1^2}$$
or
$$\dfrac{v_p- v_s}{-v_s} = \dfrac{v_s-v_p}{v_s} = 1 - \dfrac{v_p}{v_s} = \dfrac{2 + (\gamma-1)M_s^2}{(\gamma + 1)M_s^2}.$$
From there,
$$\dfrac{v_p}{v_s} = 1 -\dfrac{2 + (\gamma-1)M_s^2}{(\gamma + 1)M_s^2} = \dfrac{(\gamma + 1)M_s^2 - 2 - (\gamma-1)M_s^2}{(\gamma + 1)M_s^2} = \dfrac{2}{\gamma + 1}\left( \dfrac{M_s^2 - 1}{M_s^2} \right)$$
Note that $M_s = v_s/a_1$ where $a_1$ is the speed of sound in the stagnant air in the tube, so you can eliminate $v_s$ above and end up with
$$\boxed{\dfrac{v_p}{a_1} = M_p = \dfrac{2}{\gamma + 1}\left( \dfrac{M_s^2 - 1}{M_s} \right).}$$

Here, $M=u/a$ is the Mach number, $\gamma$ is the ratio of specific heats, and $a$ is the local speed of sound. You can uniquely determine the Mach number of the shock that propagates away from the piston based on the velocity of the piston and the conditions in the tube at the beginning. Indeed, the term $M_p = v_p/a_1$ represents a Mach number for the piston relative to the calm air in the tube (i.e., the lab frame).

You can easily solve the boxed equation for $M_s$ for a given piston Mach number.
$$M_s^2 - \left(\dfrac{\gamma + 1}{2}\right)M_p M_s - 1 = 0.$$
That's a straightforward quadratic equation. If you pick a very low piston Mach number, say, $M_p=0.1$, you still get a supersonic shock Mach number of $M_s=1.06$.

A fundamental truth about shocks is that they increase the temperature, pressure, and density of the air into which the propagate. Thus, the piston face experiences a pressure greater than ambient even if it moves at constant velocity, and a constant force is required to maintain that constant velocity (assuming the tube is sufficiently long).

If it seems odd that the shock is supersonic, that's likely because the shock Mach number is measured based on its speed relative to the medium into which it is propagating (here, the stagnant air). But the shock is propagating through the heated air behind it, which has a higher temperature and, thus, a higher speed of sound. It is still subsonic relative to that medium.

Case 2: Gradually-started piston
If the piston instead starts very slowly, it will not necessarily immediately generate a shock. A sufficiently slow ramp-up would produce a weak compression wave that will travel upstream at the speed of sound. Of course, that weak compression wave is also weakly heating the flow, thus increasing the sound speed behind it. As the piston slowly accelerates, each new wave front propagates into a medium which is slightly hotter and therefore propagates slightly faster than the wave in front of it. Thus, those waves all eventually (and fairly quickly) coalesce into a shock anyway, and the Mach number of that shock is driven by the final piston velocity when it stops accelerating. So if it accelerates to the same $v_p$ as before into the same medium, the shock will have the same $M_s$ and the problem is effectively identical once you hit steady piston motion

Of course I am trying to paint this picture as a series of discrete waves leaving the face of the piston at some finite time interval but the real process is continuous. You could certainly solve the 1D transient Euler equation to get an exact solution to the real problem, which would have a continuous generation of increasingly fast waves, but I don't think you'd learn much more than the simple thought experiment above.

What is the ramp-up time? It's going to be the same order of magnitude as the time it takes to ramp up the piston velocity since the waves are propagating faster than the piston.

Back to the question of force...
If you know the shock Mach number, $M_s$, that uniquely determines the pressure ratio across the shock (for a given gas), $p_2/p_1 = f(M_s,\gamma)$. If you have pressure and area, then you have force.

Some good sources
[1] Liepmann, H. W., Roshko, A.1957. "Elements of gasdynamics." Dover.
[2] Anderson, J. D. 2003. "Modern Compressible Flow with Historical Perspective." McGraw-Hill.
[3] https://www.grc.nasa.gov/www/k-12/airplane/normal.html

 

[Sailor Al]

I am going to use $v$ to represent velocities in the lab reference frame and $u$ to represent velocities in the frame of the compression wave ahead of the piston.

Case 1: Impulsively-started piston
Consider first the impulsively started piston. It is initially at rest, and, at some time $t=0$, it is impulsively accelerated to a velocity of $v_p$ (moving left to right, per the diagram in the original post). No penetration requires that no fluid crosses the face of the piston, so locally at the piston face, the flow is moving with $v=v_p$, but in the far field it is still moving at $u = 0$, creating a velocity discontinuity. It is a classical result that such a change in a 1D duct occurs discontinuously, i.e., as a result of a shock. This shock propagates with a speed $v_s$.

Consider a change in frame of reference such that the coordinate system follows along with the shock, i.e., subtract $v_s$ from everything. Then the velocity upstream of the shock is $u_1=0-v_s=-v_s$ and downstream of the shock is $u_2 = v_p- v_s$. One of the standard methods for determining the Mach number of a shock is to look at the velocity ratio across it (in its reference frame), which follows the relation
$$\dfrac{u_2}{u_1} = \dfrac{2 + (\gamma-1)M_1^2}{(\gamma + 1)M_1^2}$$
or
$$\dfrac{v_p- v_s}{-v_s} = \dfrac{v_s-v_p}{v_s} = 1 - \dfrac{v_p}{v_s} = \dfrac{2 + (\gamma-1)M_s^2}{(\gamma + 1)M_s^2}.$$
From there,
$$\dfrac{v_p}{v_s} = 1 -\dfrac{2 + (\gamma-1)M_s^2}{(\gamma + 1)M_s^2} = \dfrac{(\gamma + 1)M_s^2 - 2 - (\gamma-1)M_s^2}{(\gamma + 1)M_s^2} = \dfrac{2}{\gamma + 1}\left( \dfrac{M_s^2 - 1}{M_s^2} \right)$$
Note that $M_s = v_s/a_1$ where $a_1$ is the speed of sound in the stagnant air in the tube, so you can eliminate $v_s$ above and end up with
$$\boxed{\dfrac{v_p}{a_1} = M_p = \dfrac{2}{\gamma + 1}\left( \dfrac{M_s^2 - 1}{M_s} \right).}$$

Here, $M=u/a$ is the Mach number, $\gamma$ is the ratio of specific heats, and $a$ is the local speed of sound. You can uniquely determine the Mach number of the shock that propagates away from the piston based on the velocity of the piston and the conditions in the tube at the beginning. Indeed, the term $M_p = v_p/a_1$ represents a Mach number for the piston relative to the calm air in the tube (i.e., the lab frame).

You can easily solve the boxed equation for $M_s$ for a given piston Mach number.
$$M_s^2 - \left(\dfrac{\gamma + 1}{2}\right)M_p M_s - 1 = 0.$$
That's a straightforward quadratic equation. If you pick a very low piston Mach number, say, $M_p=0.1$, you still get a supersonic shock Mach number of $M_s=1.06$.

A fundamental truth about shocks is that they increase the temperature, pressure, and density of the air into which the propagate. Thus, the piston face experiences a pressure greater than ambient even if it moves at constant velocity, and a constant force is required to maintain that constant velocity (assuming the tube is sufficiently long).

If it seems odd that the shock is supersonic, that's likely because the shock Mach number is measured based on its speed relative to the medium into which it is propagating (here, the stagnant air). But the shock is propagating through the heated air behind it, which has a higher temperature and, thus, a higher speed of sound. It is still subsonic relative to that medium.

Case 2: Gradually-started piston
If the piston instead starts very slowly, it will not necessarily immediately generate a shock. A sufficiently slow ramp-up would produce a weak compression wave that will travel upstream at the speed of sound. Of course, that weak compression wave is also weakly heating the flow, thus increasing the sound speed behind it. As the piston slowly accelerates, each new wave front propagates into a medium which is slightly hotter and therefore propagates slightly faster than the wave in front of it. Thus, those waves all eventually (and fairly quickly) coalesce into a shock anyway, and the Mach number of that shock is driven by the final piston velocity when it stops accelerating. So if it accelerates to the same $v_p$ as before into the same medium, the shock will have the same $M_s$ and the problem is effectively identical once you hit steady piston motion

Of course I am trying to paint this picture as a series of discrete waves leaving the face of the piston at some finite time interval but the real process is continuous. You could certainly solve the 1D transient Euler equation to get an exact solution to the real problem, which would have a continuous generation of increasingly fast waves, but I don't think you'd learn much more than the simple thought experiment above.

What is the ramp-up time? It's going to be the same order of magnitude as the time it takes to ramp up the piston velocity since the waves are propagating faster than the piston.

Back to the question of force...
If you know the shock Mach number, $M_s$, that uniquely determines the pressure ratio across the shock (for a given gas), $p_2/p_1 = f(M_s,\gamma)$. If you have pressure and area, then you have force.

Some good sources
[1] Liepman, H. W., Roshko, A.1957. "Elements of gasdynamics." Dover.
[2] Anderson, J. D. 2003. "Modern Compressible Flow with Historical Perspective." McGraw-Hill.
[3] https://www.grc.nasa.gov/www/k-12/airplane/normal.html

That's going to take some digesting, many thanks. Case 1 is the one I'm interested in as I am postulating the piston is weightless so can instantaneously accelerate.
I am surprised to find ϒ in the solution. That's a constant for air, right?
Could I trouble you for a worked solution for Case 1 for a piston speed of Mach0.5, piston area 1 m^2 in air at NTP?
And I think you're saying the pressure is constant, correct?
Meanwhile I'll bone on Liepman, Anderson and NASA.

 

[boneh3ad]

Yes, $\gamma$ is generally a constant in most cases (not if you get very rarefied, very hot, etc.). For air, $\gamma=1.4$.

Liepmann and Roshko is a cheap book ($15 on Amazon right now) and NASA is free. Anderson is pricey.

I am not sure I want to just write out your answer for you since this smells like homework. You ought to be able to use the equations I gave, your conditions, and the NASA site to work your way through it.

 

[Sailor Al]

Yes, $\gamma$ is generally a constant in most cases (not if you get very rarefied, very hot, etc.). For air, $\gamma=1.4$.

Liepmann and Roshko is a cheap book ($15 on Amazon right now) and NASA is free. Anderson is pricey.

I am not sure I want to just write out your answer for you since this smells like homework. You ought to be able to use the equations I gave, your conditions, and the NASA site to work your way through it.

I assure you, this is not homework! My BSc. Mathematical Physics was awarded nearly 60 years ago! No, it's in pursuit of my project [Link to unpublished paper redacted by the Mentors].
I have just downloaded both the Anderson and Liepman books from doc.pub.
The NASA paper is totally unintelligible to a neophyte such as me.
I have deep suspicions about Anderson's credibility. You can see why in the literature review on my paper.
I hope Liepman is more credible.
Meanwhile, I really would appreciate a worked example if you could. Many thanks.

 

[boneh3ad]

I assure you, this is not homework! My BSc. Mathematical Physics was awarded nearly 60 years ago! No, it's in pursuit of my project "Origins of the pressure differences around a sail" [Link to unpublished paper redacted by the Mentors].
I have just downloaded both the Anderson and Liepman books from doc.pub.
The NASA paper is totally unintelligible to a neophyte such as me.
I have deep suspicions about Anderson's credibility. You can see why in the literature review on my paper.
I hope Liepman is more credible.
Meanwhile, I really would appreciate a worked example if you could. Many thanks.

Anderson is fine, and if you have Fundamentals handy (as you seem to based on your link) then it has some good chapters on elementary compressible flow.

The discrepancy to which you allude absolutely can be attributed to a typographical error: I have sitting on my desk right now the 5th edition of said book and the arrow points down in both figures you reference. Some publishing staff member likely just got the direction of the arrow wrong when they prettied up the figure for the new edition. All the equations that follow that figure have the signs correct.

Anderson is by no means perfect, but there's no reason to doubt the credibility of the man's textbooks. They're the de facto standards for teaching introductory aerodynamics in the English-speaking world for a reason.