Pressure felt by a piston, Mach number of speaker making the sound

  • Thread starter PainterGuy
  • Start date
  • #1
PainterGuy
931
68
Hi,

Once someone wrote the following to me in response to my query. My query was somewhat unrelated to it and I don't even have the copy of query anymore.

The pressure felt by the piston at any time is the average component of momentum of a gas molecule that is normal to the piston times the number of molecules striking it per second. The increase in the pressure against the piston when you move it will be the ambient pressure times the mach-number of its speed.

With normal day-to-day sounds -- even the loud ones like a rock concert, the mach-number of object making the sound is a tiny fraction. Say a speaker is making a 60 hz sound and travels 2 cm from crest to trough. Then its max speed is 2pi 60 hz times 1 cm = 377 cm/sec = 3.77 m/sec for a mach number of about 0.01. Such a speaker, assuming it was large would make a very loud sound.


"ambient pressure" - I think it refers to the external pressure applied by an outside agency and this pressure results in compression(or, expansion?) of the gas. So is the total pressure felt by the piston = (piston times the number of molecules striking it per second) + (ambient pressure times the mach-number of its speed)?

"2pi 60 hz times 1 cm" - Where does this formula come from? How is it derived?

Thank you for your help!
 

Answers and Replies

  • #2
36,247
13,301
Ambient pressure is just the pressure of the air everywhere. - including behind the speaker membrane. It cancels out for the net pressure on it.
"2pi 60 hz times 1 cm" - Where does this formula come from? How is it derived?
2 cm total motion means an amplitude of A=1 cm. The position as function of time will be x=A sin(wt) with the angular speed w=2pi f where f is the frequency. The velocity is the derivative of that, the maximal velocity is just A*w and if you plug in everything you get that result.
 
  • #3
PainterGuy
931
68
Thanks a lot!

Ambient pressure is just the pressure of the air everywhere. - including behind the speaker membrane. It cancels out for the net pressure on it.

That part didn't involve a speaker. It was about a piston moving back and forth in a gas container.
 
  • #4
boneh3ad
Science Advisor
Insights Author
Gold Member
3,384
1,125
It honestly just sounds like a bunch of nonsense that someone who doesn't know what they're talking about made up.
 
  • #5
PainterGuy
931
68
It honestly just sounds like a bunch of nonsense that someone who doesn't know what they're talking about made up.

Hi,

Are you referring to the italicized text from post #1? In that case, you are wrong. The guy who wrote that knew a lot about physics, chemistry and mathematics.
 
  • #6
sophiecentaur
Science Advisor
Gold Member
27,840
6,336
It honestly just sounds like a bunch of nonsense
Which bit? The statement about the small changes in pressure for normal levels of sound is quite reasonable. The bit about pressure on a piston, relating to momentum changes for all the molecules hitting it, is the basis of textbook kinetic theory of gases. A bit arm waving, perhaps but it's ' general chat ' and not the sort of stuff that would normally require equations.
Your comment seems a bit harsh and could be applied to a lot of reasonable statements made on PF.
 
  • #7
boneh3ad
Science Advisor
Insights Author
Gold Member
3,384
1,125
Perhaps a bit harsh, and the basic stuff is essentially correct, but then there is this line:

Anonymous said:
The increase in the pressure against the piston when you move it will be the ambient pressure times the mach-number of its speed.

Maybe this is just worded poorly, but this makes zero sense to me. The "mach-number of its speed," otherwise just known as Mach number. This statement equates to ##\Delta p = p M##, which I've never seen anywhere. Typically, pressure change is related to ##M^2## and it's a lot more complicated than ##pM##.

So, perhaps what I should have said to be more correct is that the person who answered has a few basics correct, but I wouldn't trust what they said too much beyond that without sources, because they said some suspect things in that reply.
 
  • #8
sophiecentaur
Science Advisor
Gold Member
27,840
6,336
Perhaps a bit harsh, and the basic stuff is essentially correct, but then there is this line:



Maybe this is just worded poorly, but this makes zero sense to me. The "mach-number of its speed," otherwise just known as Mach number. This statement equates to ##\Delta p = p M##, which I've never seen anywhere. Typically, pressure change is related to ##M^2## and it's a lot more complicated than ##pM##.

So, perhaps what I should have said to be more correct is that the person who answered has a few basics correct, but I wouldn't trust what they said too much beyond that without sources, because they said some suspect things in that reply.
That's a bit more of a friendly response, I think. As for your objection to the M factor, there was an attempt to relate pressure to speed of molecules (rate of collisions with piston) and the speed of sound (also relates to molecular speed). Temperature is proportional to mean velocity squared and the formula:
c_{{{\mathrm  {ideal}}}}={\sqrt  {\gamma \cdot {p \over \rho }}}={\sqrt  {\gamma \cdot R\cdot T \over M}}={\sqrt  {\gamma \cdot k\cdot T \over m}},

implies c is proportional to mean speed. A moving piston (loudspeaker) would add to the mean speed which would cause a corresponding pressure difference. Not a conventional way of looking at it but it's not too crazy. Changes would be adiabatic.
 
  • #9
PainterGuy
931
68
Thank you!

The "mach-number of its speed," otherwise just known as Mach number. This statement equates to ##\Delta p = p M##, which I've never seen anywhere. Typically, pressure change is related to ##M^2## and it's a lot more complicated than ##pM##.

As for your objection to the M factor, there was an attempt to relate pressure to speed of molecules (rate of collisions with piston) and the speed of sound (also relates to molecular speed). Temperature is proportional to mean velocity squared and the formula:
c_{{{\mathrm  {ideal}}}}={\sqrt  {\gamma \cdot {p \over \rho }}}={\sqrt  {\gamma \cdot R\cdot T \over M}}={\sqrt  {\gamma \cdot k\cdot T \over m}},

implies c is proportional to mean speed. A moving piston (loudspeaker) would add to the mean speed which would cause a corresponding pressure difference. Not a conventional way of looking at it but it's not too crazy. Changes would be adiabatic.

Isn't "M" molar mass? I thought that I should confirm it because from @boneh3ad 's post it looked to me that Mach number, M factor and "M" are the same.

Do you mean root-mean-square speed by "mean speed"? Thanks.

1567656919285.png


References:
1: http://www.schoolphysics.co.uk/age16-19/Sound/text/Sound_waves/index.html
2: https://chem.libretexts.org/Bookshe...s_Given_by_the_Maxwell-Boltzmann_Distribution
3: https://www.ck12.org/book/CK-12-Chemistry-Concepts-Intermediate/section/14.9/
 
  • #10
sophiecentaur
Science Advisor
Gold Member
27,840
6,336
Do you mean root-mean-square speed by "mean speed"? Thanks.
If I had used the word "average" then it would have been better.
Also, M was used with two different meanings, which didn't help either. The M in the expression I cut and pasted from Wiki was obvs not Mach number.
Your extended reply was useful (and understandable) and seems to justify the original statement about Mach Number. Afaics, that was a needless (but not altogether unhelpful) complication in the quote that's in the OP.
 
  • #11
PainterGuy
931
68
Thank you!

Also, M was used with two different meanings, which didn't help either.

Yes, in post #7 it was used as Mach number but you used it as the molar mass.

I think it's better to look at the original statement again.

The pressure felt by the piston at any time is the average component of momentum of a gas molecule that is normal to the piston times the number of molecules striking it per second. The increase in the pressure against the piston when you move it will be the ambient pressure times the mach-number of its speed.

I think the statement suggests that when the piston moves total pressure would come from static pressure plus the dynamic pressure which results from its movement. We need to stress that the setup is not that of a cylinder with a piston as shown below because in such a case as the piston moves, it would compress the gas and temperature would be affected too. It's more like a cylinder with one end open, or the piston could be equated with a loudspeaker which can move back and forth in open air.

1567725755840.png


I think that @boneh3ad was referring to the following highlighted equation.

1567726591513.png


I'd say that the person was only trying to make a general statement to convey the idea, or probably he made a typo and could have said 'ambient pressure times squared mach-number of its speed'.


Note to self:
1567728574862.png


References:
1: https://eng-software.com/about-us/p...on-between-total-static-and-dynamic-pressure/
2: https://en.wikipedia.org/wiki/Mach_number#Calculation
 
  • #12
sophiecentaur
Science Advisor
Gold Member
27,840
6,336
he made a typo and could have said 'ambient pressure times squared mach-number of its speed'.
Yes; it's the square of the speed. I the tradition of PF (bringing things to an understandable level when possible) this can be explained by saying that the change in momentum for each collision is proportional to that speed and the rate of collisions is also proportional to speed. So M comes in twice.
The poor originator of that slightly glib statement has certainly had his tail singed in this thread. But I think it's ben sorted out now.
The issue of the risks when cutting and pasting expressions from other sources is also brought out. It's difficult to know the best way round that problem without making threads longer and longer.

It's more like a cylinder with one end open, or the piston could be equated with a loudspeaker which can move back and forth in open air.
I'm not too sure about that. The molecules hitting the piston are not 'aware' of what's happening down at the other end of the cylinder. (Except for an extremely short cylinder.)
 
  • #13
PainterGuy
931
68
Thank you!

this can be explained by saying that the change in momentum for each collision is proportional to that speed and the rate of collisions is also proportional to speed. So M comes in twice.

Just to repeat it: pressure is proportional to change in momentum which in turn is proportional to the speed and rate of collision and the rate of collision also depends upon speed.

I'm not too sure about that. The molecules hitting the piston are not 'aware' of what's happening down at the other end of the cylinder.

I didn't say it properly.

Loudspeaker/piston in open air:
Total pressure = static pressure + dynamic pressure

Static pressure is atmospheric pressure and the movement of loudspeaker wouldn't change this pressure so dynamic pressure, q, only depends upon Mach number, M.

1567817726478.png


Enclosed cylinder with piston:
In this case, static pressure would change as the piston moves therefore it's a variable quantity and so is Mach number, M.

Thanks a lot for your help and time.
 
  • #14
sophiecentaur
Science Advisor
Gold Member
27,840
6,336
dynamic pressure, q, only depends upon Mach number, M.
I don't get that. The piston has an increase in relative velocity to the molecules and it also hits more in a given time as it's going forward. How does that not produce a M2 factor?
 
  • #15
PainterGuy
931
68
How does that not produce a M2 factor?

Thank you.

It does produce M2 factor. I was rather trying to clarify something to myself and it should have been a 'note to self'.

Thanks a lot for helping me to sort this out.
 
  • Like
Likes sophiecentaur

Suggested for: Pressure felt by a piston, Mach number of speaker making the sound

  • Last Post
Replies
7
Views
869
Replies
7
Views
338
  • Last Post
2
Replies
41
Views
544
Replies
10
Views
482
Replies
23
Views
630
Replies
4
Views
447
Replies
3
Views
1K
Replies
8
Views
833
Replies
2
Views
573
Replies
20
Views
376
Top