Can Air Resist a Piston? A Serious Discussion

[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....

 

[jbriggs444]

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

It is rude to tell others how to respond.

We know that a flat plate moving through the air broadside can exceed the speed of sound. We know that the disturbance from this flat plate exists upstream of the plate. You might quibble that this is not really "exceeding the speed of sound" because the "speed of sound" is locally elevated due to heating and, perhaps, non-ideality of the fluid. You might quibble that this is not really "exceeding the speed of sound" because there is fluid flow in the direction of motion carrying the disturbance along with it.

But we see no such nuance in your postings. It's just "you can't hear the jet until it hits you". Or "Wikipedia is wrong".

I dislike posting in a thread where experts such as @boneh3ad are already participating because they've forgotten more abound fluid dynamics than I've ever learned.

 

[Sailor Al]

It is rude to tell others how to respond.

Yes, that would be rude. I asked, I didn't tell.

 

[Sailor Al]

We know that a flat plate moving through the air broadside can exceed the speed of sound. We know that the disturbance from this flat plate exists upstream of the plate.

But that's exactly my point, it doesn't.
If the plate is exceeding the speed of sound then the disturbance does not exist upstream of the plate.
If you're upstream of the plate, just like being in front of the supersonic plane, you won't hear the plate till it hits you.
No nuance is needed, it's basic physics.
The speed of sound in a medium is the speed that a disturbance will proceed through the medium.
I have provided the reference above:
Young H. Freedman R. (2018) University Physics with Modern Physics, 15th Edition , Sears & Zemansky

 

[A.T.]

If you're upstream of the plate, just like being in front of the supersonic plane, you won't hear the plate till it hits you.

A supersonic flat plate and other blunt objects will produce a bow shock in front of them, which travels at the same supersonic speed as the object. You will hear that before the objects hits you or passes close to you.

https://en.wikipedia.org/wiki/Bow_shock_(aerodynamics)

 

[Sailor Al]

A supersonic flat plate and other blunt objects will produce a "bow shock-wave" in front of them, which travels at the same supersonic speed as the object. You will hear that before the objects hits you or passes close to you.

That's a pretty picture, but there's nothing in the text that indicates the shock wave is propagating at >Mach 1.0. The caption says the flow is supersonic in the wind tunnel, so that shockwave is clearly propagating slower than the flow. If it were going faster its curve would be convex, not concave as in the pic!
Please, check your physics books. Mach 1.0 is the speed that disturbances propagate through a medium.

 

[A.T.]

That's a pretty picture, but there's nothing in the text that indicates the shock wave is propagating at >Mach 1.0.

The bow shock is in front of the supersonic object, so it will obviously reach you before the object.

The caption says the flow is supersonic in the wind tunnel, so that shockwave is clearly propagating slower than the flow.

No idea what you mean here. Maybe you have problems understanding/accepting Galilean Invariance:

https://en.wikipedia.org/wiki/Galilean_invariance

Relative to the wind tunnel walls the bow shock and object are both stationary, which means that both are moving faster than Mach 1 relative to the air in the wind tunnel.

 

[Sailor Al]

The bow shock is in front of the supersonic object, so it will obviously reach you before the object.

Like I said, there's nothing in the text to support your claim.

Relative to the wind tunnel walls the bow shock and object are both stationary, which means that both are moving faster than Mach 1 relative to the air in the wind tunnel.

Please refer to your physics textbook:
Mach 1.0 is the speed that a disturbance will propagate in a medium.
You will not hear the supersonic jet plane before it hits you.

 

[A.T.]

The bow shock is in front of the supersonic object, so it will obviously reach you before the object.

Like I said, there's nothing in the text to support your claim.

Because it is trivially obvious. How would you hit the object from the front, without first passing through the bow shock in front of it?

You will not hear the supersonic jet plane before it hits you.

We aren't talking about a jet plane with a pointy nose, but flat plate which generates a bow shock in front of it.

Mach 1.0 is the speed that a disturbance will propagate in a medium.

What you are missing is that medium itself is moving at different speeds. A blunt object like a plate or piston will push some air in front of it, and move that air along. So a disturbance traveling at a certain speed relative to that local air can be traveling much faster relative to the surrounding air.

 

[renormalize]

Mach 1.0 is the speed that a disturbance will propagate in a medium.
You will not hear the supersonic jet plane before it hits you.

Your statement is simply wrong.

Please take a look at this recent experimental paper: Measurement of unsteady shock standoff distance around spheres flying at Mach numbers near one, T.Kikuchi & K.Ohtani (2022)
The authors fired plastic, aluminum and steel ball-bearings of diameter $\sim0.8\text{cm}$ through air at speeds spanning Mach 1.0 and observed the shocks produced. For example, here is a photo of a plastic bearing traveling from left to right at Mach 1.03:

The detached bow shock wave clearly leads (stands-off) the projectile by about 2 ball diameters. If you do the math, you find that the shock reaches your ear about $45\mu\text{s}$ before the projectile does. So much for your claim.

Moreover, for all projectile speeds from Mach 0.95 to 1.25, the speed of the shock is equal-to or greater-than that of the projectile:

High-amplitude, non-linear disturbances in air such as shock waves most certainly can travel at supersonic speeds! Perhaps you're the one who needs to refer to research literature?

 

[Sailor Al]

Because it is trivially obvious. How would you hit the object from the front, without first passing through the bow shock in front of it?

Like I said: there's nothing in the text that supports your argument.

What you are missing is that medium itself is moving at different speeds. A blunt object like a plate or piston will push some air in front of it, and move that air along. So a disturbance traveling at a certain speed relative to that local air can be traveling much faster relative to the surrounding air.

But if there's a small region of compressed air in front of the object, then in that compressed region the speed of sound will be higher than the speed of sound in the surrounding air. The shock wave may be travelling faster there but it's still Mach 1.0 in that medium.
From anywhere else in the surrounding air, the shock wave will move at Mach 1.0

 

[Frabjous]

All I can say is, check your physics textbooks.

Looking at a 2012 edition, the couple of pages in Freedman do not address shock velocity. You are reading in your own biases.

 

[A.T.]

But if there's a small region of compressed air in front of the object, then in that compressed region the speed of sound will be higher than the speed of sound in the surrounding air. The shock wave may be travelling faster there but it's still Mach 1.0 in that medium.

The air in front of the blunt object is not just compressed but also moving along with the object, so the speed of the local medium adds to the speed through the local medium (see Galilean Transformation). But whatever the contributions here are, the final result is that the bow shock right in front of the supersonic blunt object moves at the same speed as the object, and stays at a constant distance ahead of it. So it will pass you before the object hits you.

 

[boneh3ad]

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

I've provided you with several references that corroborate my analysis and my claims. They've been supported by 100 years of aerodynamics observations/experiments and have contributed to the designs of aircraft that fly millions of miles per year.

If you want to be a skeptic in the face of a preponderance of evidence, be my guest. I will not be providing further guidance for someone who simply refuses to engage seriously.

 

[berkeman]

I will not be providing further guidance for someone who simply refuses to engage seriously.

And with that, this thread is done.