Undergrad Can Air Resist a Piston? A Serious Discussion

[Sailor Al]

Now what happens if you start with stationary air in a long tube and set the piston in motion?
Same. Air hammers will propagate upwind and downwind with the same speed and force as if the air had been moving and piston stationary. The forces on piston will only change if and when either of the air hammers will have reflected from somewhere and travelled back to the piston.

I have no issue with your upstream “suction” wave. It’s just a way of describing a low pressure wave. The downstream wave is a high pressure wave propagating downstream.
What I’m after is a mathematical function that will solve for the pressure on the piston at time t.

 

[boneh3ad]

I have no problem with the pressure wave moving relative to the cylinder. The cylinder is a solid object. In any (inertial) frame of reference, all parts of the cylinder are moving at the same speed.
The air is a compressible fluid - a continuum. All parts of it may be moving at different speeds. The concept of stationary fluid is not so challenging. None of it is moving, no problem. But the speed of a moving, compressible fluid under the influence of a pressure wave is what makes my brain go pop!
Liepman's statement that the fluid flows through the shockwave with speed u1 cannot be allowed to pass unchallenged.

Again, it's a simple change of reference frame. I don't know how else to explain it other than that. A shock can propagate with a velocity past a stationary observer. The frame of reference can also be changed such that the observer moves along with the shock, making the shock appear stationary with the flow passing through it.

Even then, it's actually very easy to create a stationary shock in nature with air flowing through it. Take, for example, any supersonic wind tunnel. Mounting an object in the flow will result in shock waves.

Here is an example of a model aircraft mounted in a supersonic wind tunnel producing oblique and conical shock waves captured via schlieren imaging:

Here is a similar pair of schlieren images of a supersonic inlet in a wind tunnel. This time, the unstarted inlet produces locally normal shocks that are stationary in the lab frame.

 

[boneh3ad]

I have no issue with your suction wave. It’s just a way of describing a low pressure wave. It’s a high pressure wave propagating downstream.
What I’m after is a mathematical function that will solve for the pressure on the piston at time t.

I gave you all the tools you need for that with my post related the piston Mach number and the shock Mach number. You may be able to come up with a closed-form solution by solving that quadratic equation and substituting it into the pressure ratio across a shock of Mach number $M_s$ (available in all of the references I previously gave you).

But aside from all of this, the bigger problem is that this has nothing to do with sail boats or airfoils. Those don't operate in environments with solid walls and no opportunity for air to move around them. And sailboats especially do not generate shock waves.

 

[Sailor Al]

Again, it's a simple change of reference frame. I don't know how else to explain it other than that. A shock can propagate with a velocity past a stationary observer. The frame of reference can also be changed such that the observer moves along with the shock, making the shock appear stationary with the flow passing through it.

A physicist has no problem with the shock wave appearing stationary w.r.t. the cylinder.
But not “the flow passing through it”
As stated, a stationary flow is one thing: all parts of it are stationary. But in the problem context, the flow is a compressible fluid subject to varying pressures. The step from being stationary and moving is not a simple case. When stationary, all parts of the fluid have the same zero velocity. When subject to the pressure of the piston, this homogeneity is destroyed.
I know it’s a bitter pill for an engineer, but for a physicist, it’s a show-stopper.

 

[berkeman]

I know it’s a bitter pill for an engineer, but for a physicist, it’s a show-stopper.

Which are you? I'm not finding your educational background in your Profile page. Clarifying that would help us a lot to try to answer your questions...

 

[Sailor Al]

And sailboats especially do not generate shock waves.

I have not propose a “shock wave”, that has serious supersonic connotations.
But a sailboat certainly does create pressure waves.
Go race a sailboat and experience the pressures generated by nearby boats: wind shadow, lee-bow effect, dirty wind astern and leeward etc. These are pressure variations which move propagate through air and move with the boat. That is a pretty good description of pressure waves.

 

[Sailor Al]

Which are you? I'm not finding your educational background in your Profile page. Clarifying that would help us a lot to try to answer your questions...

My BSc. is in Mathematical Physics, Birmingham UK, 1967. I am a retired but still very active software developer, private pilot and enthusiastic keelboat racer.

 

[berkeman]

My BSc. is in Mathematical Physics, Birmingham UK, 1967. I am a retired but still very active software developer, private pilot and enthusiastic keelboat racer.

I'm not familiar with the Mathematical Physics degree in the UK, but why are you saying that such a fundamental concept is a show-stopper for you with your background? Is there something in the mathematics presented so far in this thread that is problematical?

 

[Sailor Al]

I'm not familiar with the Mathematical Physics degree in the UK, but why are you saying that such a fundamental concept is a show-stopper for you with your background? Is there something in the mathematics presented so far in this thread that is problematical?

Yes, I think I have expressed it pretty clearly in my last reply to @boneh3ad .

 

[boneh3ad]

A physicist has no problem with the shock wave appearing stationary w.r.t. the cylinder.
But not “the flow passing through it”
As stated, a stationary flow is one thing: all parts of it are stationary. But in the problem context, the flow is a compressible fluid subject to varying pressures. The step from being stationary and moving is not a simple case. When stationary, all parts of the fluid have the same zero velocity. When subject to the pressure of the piston, this homogeneity is destroyed.
I know it’s a bitter pill for an engineer, but for a physicist, it’s a show-stopper.

It is a simple case, actually. So simple, that physicists have a name for it: a Galilean transformation.

Not only is this not a contested process (by engineers or physicists) but its accuracy has been confirmed by the successful design of rockets, supersonic airplanes, and engines of all kinds relying on this exact type of analysis with shocks.

 

[boneh3ad]

I have not propose a “shock wave”, that has serious supersonic connotations.
But a sailboat certainly does create pressure waves.
Go race a sailboat and experience the pressures generated by nearby boats: wind shadow, lee-bow effect, dirty wind astern and leeward etc. These are pressure variations which move propagate through air and move with the boat. That is a pretty good description of pressure waves.

And none are well-modeled by the piston-in-tube problem, which, notably, does produce a shock.

 

[snorkack]

I have no issue with your upstream “suction” wave. It’s just a way of describing a low pressure wave. The downstream wave is a high pressure wave propagating downstream.
What I’m after is a mathematical function that will solve for the pressure on the piston at time t.

Were you asking for pressure as a function of t, or of v?

 

[Sailor Al]

Were you asking for pressure as a function of t, or of v?

Well, both, since they both probably affect the result. t>0, 0<v<Mach 1.0.

 

[A.T.]

I gave you all the tools you need for that with my post related the piston Mach number and the shock Mach number. You may be able to come up with a closed-form solution by solving that quadratic equation and substituting it into the pressure ratio across a shock of Mach number $M_s$ (available in all of the references I previously gave you).

I guess you mean the fromula given here?
https://www.grc.nasa.gov/www/k-12/airplane/normal.html

I'm still wondering if there isn't a simpler approach to get the steady state force on the piston, based on momentum conservation: A piston moving at v must continuously accelerate air from 0 to v, and this impulse propagates at the speed u in both directions. This should give you the steady state rate of momentum transfer (force). It assumes that u ≈ const, but would that not be the case if v << Mach 1?

As an analogy: Consider a locomotive coupled to an infinite number of cars in front and back, which starts moving at a constant speed v. The cars are perfectly rigid, but the coupling is such that the impulse propagates at u = 10v. From this you can calculate the average force the locomotive has to supply.

 

[snorkack]

Well, both, since they both probably affect the result. t>0, 0<v<Mach 1.0.

Nope. t actually does not affect the force.

And here is where the difference between free air and tube goes in. Note that the force is independent of t each case.
A sail placed perpendicular to airflow will bring the wind to a halt. Stagnant compressed air forms upstream of the sail.
But since the air is free to flow around the sail, the volume of stagnant air will be modest and constant with time. A small sail will not stop the wind indefinite distance upwind. Air far upwind will be deflected around the sail and the small body of stagnant air upwind of the sail.
Now, there will be a steady flow of wind that arrives at the sail and is stopped. But it will be replaced by equal amount of air that passes around the sail and is accelerated downwind.

The reason the sail experiences any force at all is that although the mass flux of air downwind is identical to the mass flux upwind, the momentum flux is not.
In principle, a fluid decelerated by the obstacle might make a perfect recovery and accelerate to exactly the original speed. The energy for that is there (in the potential energy of compressed stagnant air). But in practice, there is some imperfection in the recovery. Which is why there is any air resistance.

This resistance is independent of t.
In terms of v, the mass of air that hits the sail in unit time is proportional to v, and the momentum of unit mass is also proportional to v. So the aerodynamic forces are generally proportional to v² - but the resistance is not close to it. Because the fraction of imperfection of recovery will depend on flow pattern and thus through Reynolds number on v, in a manner that is complex and cannot be analyzed from first principles.

Now compare the piston/tap closing the long pipeline!

The tap also brings the airflow to stagnation upstream (and downstream).
But since the air cannot get around, unlike sail, the volume of stagnant air will not reach a steady state. It will grow without bound, at the speed of sound (plus a small increment thanks to the warming on the air when stopped).
The volume of stagnant air will not reach a steady state but the force it exercises will. Because the volume of new air stopped in unit time will stay proportional to c.
The force is independent on t. But since the momentum of an unit volume of air is proportional to v (as in previous case), while the amount of air stopped in unit time is proportional to c (and not to v, as in the case of sail), the force on a tap is proportional to v, in contrast to the force on a sail, which was proportional to v²

 

[Sailor Al]

I guess you mean the fromula given here?
https://www.grc.nasa.gov/www/k-12/airplane/normal.html

I'm still wondering if there isn't a simpler approach to get the steady state force on the piston, based on momentum conservation: A piston moving at v must continuously accelerate air from 0 to v, and this impulse propagates at the speed u in both directions. This should give you the steady state rate of momentum transfer (force). It assumes that u ≈ const, but would that not be the case if v << Mach 1?

As an analogy: Consider a locomotive coupled to an infinite number of cars in front and back, which starts moving at a constant speed v. The cars are perfectly rigid, but the coupling is such that the impulse propagates at u = 10v. From this you can calculate the average force the locomotive has to supply.

A big NO to the NASA reference. I think the formulae apply specifically to supersonic flow.

But I like your thinking about the train. I’ll try that out to see what pops out. I’m thinking once we do the basic maths, we may be able to reduce the length of the cars for an integration.

 

[256bits]

A physicist has no problem with the shock wave appearing stationary w.r.t. the cylinder.
But not “the flow passing through it”
As stated, a stationary flow is one thing: all parts of it are stationary. But in the problem context, the flow is a compressible fluid subject to varying pressures. The step from being stationary and moving is not a simple case. When stationary, all parts of the fluid have the same zero velocity. When subject to the pressure of the piston, this homogeneity is destroyed.
I know it’s a bitter pill for an engineer, but for a physicist, it’s a show-stopper.

a pipeline engineer for example does take this into account.
You can look up
https://en.wikipedia.org/wiki/Fanno_flow
https://en.wikipedia.org/wiki/Rayleigh_flow

With fanno flow the temperature of the fluid decreases along the pipe, which would be contrary to expectations. flow will approach M=1 down the pipe.
With rayleigh flow, the loss of heat will cause the flow to not approach M=1.

 

[Sailor Al]

I'm not familiar with the Mathematical Physics degree in the UK, but why are you saying that such a fundamental concept is a show-stopper for you with your background? Is there something in the mathematics presented so far in this thread that is problematical?

In 1967 Birmingham University did have a BSc. Mathematical Physics. Now that I'm back at my desk I have dug out and scanned the certificate attached.

Please feel free to delete after reading.

 

[boneh3ad]

A big NO to the NASA reference. I think the formulae apply specifically to supersonic flow.

But I like your thinking about the train. I’ll try that out to see what pops out. I’m thinking once we do the basic maths, we may be able to reduce the length of the cars for an integration.

The formulae apply to shocks. And they apply here.

 

[A.T.]

As an analogy: Consider a locomotive coupled to an infinite number of cars in front and back, which starts moving at a constant speed v. The cars are perfectly rigid, but the coupling is such that the impulse propagates at u = 10v. From this you can calculate the average force the locomotive has to supply.

But I like your thinking about the train. I’ll try that out to see what pops out. I’m thinking once we do the basic maths, we may be able to reduce the length of the cars for an integration.

If the cars can move without resistance, then you get a constant average force, that corresponds to the rate of total momentum change of the cars:

F = 2vuρ

where:
v : locomotive speed
u : impulse propagation speed = v * (car_length + initial_coupling_length) / coupling_give_distance
ρ : initial linear mass density of the train = car_mass / (car_length + initial_coupling_length)

 

[Sailor Al]

If the cars can move without resistance, then you get a constant average force, that corresponds to the rate of total momentum change of the cars:

F = 2vuρ

where:
v : locomotive speed
u : impulse propagation speed = v * car_length / coupling_give_distance
ρ : initial linear mass density of the train = car_mass / (car_length + initial_coupling_length)

If v = u/2 then coupling_give_distance = car_length/2? Makes sense.
In the impact, the distance between the cars, initially with a value of coupling_give_distance, becomes zero, right?
What happens if the coupling is a spring, initially under no compression or tension? Maybe emulating gas compression?
Could the spring be damped - energy loss -> heat?
Do you know if this approach has been explored elsewhere and shown to be fruitful or shown to diverge seriously from the piston/gas problem?
Is this your own work or from a standard text? This is not a challenge to your work but a request for a source of more information without hassling you.

 

[A.T.]

If v = u/2 then coupling_give_distance = car_length/2? Makes sense.
In the impact, the distance between the cars, initially with a value of coupling_give_distance, becomes zero, right?

I was inconsistent about including the initial_coupling_length in the car_length. I corrected the original post now. I think this is clearer:

F = 2vuρ

where:
v : locomotive speed
u : impulse propagation speed = v * (car_length + initial_coupling_length) / coupling_give_distance
ρ : initial linear mass density of the train = car_mass / (car_length + initial_coupling_length)

So (car_length + initial_coupling_length) is the initial distance from the start of a car to the start of the next car.

coupling_give_distance is the distance a car has to move, before the next car starts to move (or previous car for the back side).

Is this your own work or from a standard text?

This was just the simplest possible model that came to my mind. But whatever happens in the more complex gas piston case, in the steady state it will have to obey F = 2vuρ, to be consistent with momentum conservation.

A difference for gas is that u is not a function of v, if v << Mach 1. But no matter what the relation between u and v in the gas case is: As long as u stays finite, the force F will also stay finite, and not increase without bound.

 

[Sailor Al]

The formulae apply to shocks. And they apply here.

By "shocks" do you mean "shock waves?
According to my Physics text* :
"A shock wave is produced continuously by any object that moves through the air at supersonic speed"
As noted my piston is not supersonic.
And BTW, I think the Wikipedia page is incorrect when it claims, in the opening sentence:
"In physics, a shock wave (also spelled shockwave), or shock, is a type of propagating disturbance that moves faster than the local speed of sound in the medium."
The speed of sound is a medium is the speed that a pressure wave travels in that medium. A shockwave is a pressure wave - it travels at the speed of sound in that medium, not "faster than the local speed of sound in the medium."
So much misinformation!
*Young H. Freedman R. (2018) University Physics with Modern Physics, 15th Edition , Sears & Zemansky

 

[Sailor Al]

@jbriggs444 Why the skeptical emoji? Do you have an alternative proposal? Please don't snipe from the sideline.

 

[Frabjous]

By "shocks" do you mean "shock waves?
According to my Physics text* :
"A shock wave is produced continuously by any object that moves through the air at supersonic speed"
As noted my piston is not supersonic.
And BTW, I think the Wikipedia page is incorrect when it claims, in the opening sentence:
"In physics, a shock wave (also spelled shockwave), or shock, is a type of propagating disturbance that moves faster than the local speed of sound in the medium."
The speed of sound is a medium is the speed that a pressure wave travels in that medium. A shockwave is a pressure wave - it travels at the speed of sound in that medium, not "faster than the local speed of sound in the medium."
So much misinformation!
*Young H. Freedman R. (2018) University Physics with Modern Physics, 15th Edition , Sears & Zemansky

A shock wave travels faster than the sound speed of the medium in front of it (and subsonic relative to the medium behind it). In the infinitesimal limit, it moves at the sound speed which is why there is a lot of similarity between shock and sound wave equations.

 

[Sailor Al]

A shock wave travels faster than the sound speed of the medium in front of it.

Nope. If you position yourself (from a balloon in still air) in front of a supersonic jet plane, you won't hear it till it hits you.

 

[Sailor Al]

@Frabjous Sad emoji because you didn't hear the plane?

 

[Frabjous]

Since other people will be reading this thread…
What you have failed to take into account is that a supersonic plane is moving faster than the speed of sound so your argument is not necessarily valid. There are also detached (bow) shocks which move in front of vehicles.

 

[Sailor Al]

What you have failed to take into account is that a supersonic plane is moving faster than the speed of sound so your argument is not necessarily valid.

What part of "supersonic jet plane" makes you think that? It is moving faster than the speed at which pressure waves will propagate through the medium.

There are also detached (bow) shocks which move in front of vehicles.

Yes, but they only move in front of subsonic vehicles.
Read your physics book: The speed of sound is as fast as a wave will propagate in that medium.
You will get hit by the supersonic jet at the moment you hear it. You won't hear it till it hits you!
I thought this was a serious forum, for discussing real physics.

 

[Sailor Al]

@jbriggs444 Once again a skeptical emoji. Please put up or....