Science > Physics > Sound pressure, calculating thrust from pressure, theory
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Science > Physics |
| User: |
"kvark" |
| Date: |
19 Dec 2005 02:26:18 AM |
| Object: |
Sound pressure, calculating thrust from pressure, theory |
Hello to all!
I give up... can you help me calculate and understand how to calculate
the thrust from sound?
Let assume we have a speaker, or any source of sounds, 1KW power,
and lets assume that all the sound from that source is being directed
to the receiver, for example, a plate which will reflect the sound...
Now, what impulse will that sound (with 1 KW power) transfer
to the receiver? How many newtons?
How to calculate that thrust force which is being produced by sound
pressure?
Why is that force on the level that we observe it, ie, why sound
gives so low thrust on the objects it is pushing on?
Anyway, how is sound "pushing" the objects, if we know that
particle motion in sound is in both directions, left and right,
ie, consist of compressions and rarifications - in other words,
should they cancel each others effect?
Thank you for answer... if I wasn't clear with my questions,
then please let me know, so I can clarify.
Kvark.
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| User: "Timo Nieminen" |
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| Title: Re: Sound pressure, calculating thrust from pressure, theory |
19 Dec 2005 04:13:55 AM |
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On Mon, 19 Dec 2005, kvark wrote:
Hello to all!
I give up... can you help me calculate and understand how to calculate
the thrust from sound?
Let assume we have a speaker, or any source of sounds, 1KW power,
and lets assume that all the sound from that source is being directed
to the receiver, for example, a plate which will reflect the sound...
Now, what impulse will that sound (with 1 KW power) transfer
to the receiver? How many newtons?
How to calculate that thrust force which is being produced by sound
pressure?
Why is that force on the level that we observe it, ie, why sound
gives so low thrust on the objects it is pushing on?
Anyway, how is sound "pushing" the objects, if we know that
particle motion in sound is in both directions, left and right,
ie, consist of compressions and rarifications - in other words,
should they cancel each others effect?
Waves can make distant things move; they can do work on them, and can
exert force on them. That means that waves, including sound waves, can
carry energy and momentum. In general, the ratio of the momentum flux to
the energy flux (ie power) is the wave speed. Thinking about the Doppler
shift due to reflection from a moving reflector will give the correct
result.
If the sound is absorbed, then the force is:
F = power/(speed of sound).
If the sound is reflected (assuming an incident plane wave, normally
incident, reflected normally), then:
F = 2*P/speed
--
Timo
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| User: "Ian Parker" |
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| Title: Re: Sound pressure, calculating thrust from pressure, theory |
19 Dec 2005 04:38:14 AM |
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But you don't have anything moving at 340m/s. In fact the net thrust
(to first order) will be zero, the compression is balanced by
rarifaction. 1Kw means about 3J/m assuming a speed of 340. Now
atospheric pressure is 100,000N/m^2, this means we are well within the
first order.
Sound is compressional. This is different to light where electric and
magnetic fields are at right angles to the direction of motion. A
photon has a definite mass, a sound wave does not.
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| User: "kvark" |
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| Title: Re: Sound pressure, calculating thrust from pressure, theory |
19 Dec 2005 05:06:02 AM |
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Ian Parker wrote:
===========================================
But you don't have anything moving at 340m/s.
===========================================
The sound is moving at 340 m/s? Particles of the sound
are moving at that speed, correct?
Ian Parker wrote:
==================================================
In fact the net thrust (to first order) will be zero, the compression
is
balanced by rarifaction.
==================================================
This sounds logical, however, some are claiming that sound
will produce thrust on a body it is directed at?
Ian Parker wrote:
==================================================
1Kw means about 3J/m assuming a speed of 340.
Now atmospheric pressure is 100,000N/m^2, this means we are
well within the first order.
==================================================
What difference atmospheric pressure makes?
There is going to be extra pressure on the side where
sound waves are pushing, and that should be enough
to create thrust, right?
Ian Parker wrote:
==================================================
Sound is compressional. This is different to light where electric and
magnetic fields are at right angles to the direction of motion. A
photon has a definite mass, a sound wave does not.
==================================================
But mass of air in the volume of sound wave has a definite mass...
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| User: "Ian Parker" |
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| Title: Re: Sound pressure, calculating thrust from pressure, theory |
19 Dec 2005 05:52:54 AM |
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The sound is moving at 340 m/s? Particles of the sound
are moving at that speed, correct?
No, sometimes when we are talking about quantized sound in a crystal we
talk about a phonon and we have a conservation of "momentum" in the
sense that Sigma (k) Input = Sigma (k) output. However this is
completely different from propagation in a fluid. Have you read
anything on Solid State Physics? Is that the source of confusion?
A compressional wave travels at 340m/s but the individual volumes of
air do not move at that speed.
But mass of air in the volume of sound wave has a definite mass...
Yes, but there no NET movement. There is thrust during the peak and
braking during the trough. You fell yourself vibrated, particularly at
low frequencies but there is no thrust.
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| User: "Jan Panteltje" |
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| Title: Re: Sound pressure, calculating thrust from pressure, theory |
19 Dec 2005 06:08:16 AM |
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On a sunny day (19 Dec 2005 03:52:54 -0800) it happened "Ian Parker"
<ianparker2@gmail.com> wrote in
<1134993173.964798.317280@g49g2000cwa.googlegroups.com>:
The sound is moving at 340 m/s? Particles of the sound
are moving at that speed, correct?
No, sometimes when we are talking about quantized sound in a crystal we
talk about a phonon and we have a conservation of "momentum" in the
sense that Sigma (k) Input = Sigma (k) output. However this is
completely different from propagation in a fluid. Have you read
anything on Solid State Physics? Is that the source of confusion?
A compressional wave travels at 340m/s but the individual volumes of
air do not move at that speed.
But mass of air in the volume of sound wave has a definite mass...
Yes, but there no NET movement. There is thrust during the peak and
braking during the trough. You fell yourself vibrated, particularly at
low frequencies but there is no thrust.
Think for a moment, a big bass speaker.
You move the cone forward, and leave it there.
Now you have a ripple of compressed air travelling.
This will move an object away.
You need no amplifierr, a small (say 4.5 V) battery will do.
In case of a 4 Ohm speaker capable of 1000W.... continued dissipation....
U^2 / 4 = 1000 so U = sqrt(4000) = 63 V DC.
To not damage a 8 Ohm speaker with 4.5 V DC you need one capable of
(4.5 x 4.5) / 8 = 2.5 W
If the speaker box itself is sealed, and the room, then indeed the air
pressure in the room will also increase a bit, because the room volume will
have decreased.
However you cannot move a lot that way, air is a gas, compressible,
maybe a piece of paper or some plastic ball...
Or blow somebodies ear drums out (soliton).
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| User: "Ian Parker" |
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| Title: Re: Sound pressure, calculating thrust from pressure, theory |
20 Dec 2005 04:35:01 AM |
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Or blow somebodies ear drums out (soliton).
I did say zero force to the first order. An atmosphere is 100,000
N/m^2, it will require 100,000 joules to compress it (isothermally) to
1/e of its original size. 1/2 adiabatic is about 100,000J (rough and
ready). We are talking of the order of 10J/m^-3. An explosion travells
(initially) faster than 340m/s and does have positive push. At long
ranges we have 340m/s and no net force.
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| User: "kvark" |
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| Title: Re: Sound pressure, calculating thrust from pressure, theory |
19 Dec 2005 11:12:20 AM |
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Jan... thank you. How to I translate 1000 W to decibels?
If the sound is absorbed, then the force is:
F = power/(speed of sound).
If 1000 w / 340 = ~3N, then, how to express that in decibels?
Also, why sound pressure has not been used as a thrust production?
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| User: "Sam Wormley" |
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| Title: Re: Sound pressure, calculating thrust from pressure, theory |
19 Dec 2005 11:33:46 AM |
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kvark wrote:
If 1000 w / 340 = ~3N, then, how to express that in decibels?
Decibel Scale
http://scienceworld.wolfram.com/physics/DecibelScale.html
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| User: "kvark" |
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| Title: Re: Sound pressure, calculating thrust from pressure, theory |
19 Dec 2005 05:09:55 AM |
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Timo:
==================================
If the sound is absorbed, then the force is:
F = power/(speed of sound).
==================================
Hi Timo... this looks like we should observe a lot of force
exerted on stuff around us... imagine speakers
with 340 watts, sound waves bumping at you...
1 N is not that much, but enough to move you a bit.
Yet, we don't observe such thrust?
Why?
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| User: "Timo Nieminen" |
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| Title: Re: Sound pressure, calculating thrust from pressure, theory |
19 Dec 2005 02:38:02 PM |
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On Mon, 19 Dec 2005, kvark wrote:
Timo:
==================================
If the sound is absorbed, then the force is:
F = power/(speed of sound).
==================================
Hi Timo... this looks like we should observe a lot of force
exerted on stuff around us... imagine speakers
with 340 watts, sound waves bumping at you...
1 N is not that much, but enough to move you a bit.
Yet, we don't observe such thrust?
Why?
We do. I've knocked things off shelves with speakers. I don't know whether
that was sound waves transmitted through the air, or by vibrations through
the floor and wall (ie sound waves transmitted through the various solid
bodies around), but it's still the same thing.
Other than that, various proper experiments have been done. Some of the
interesting ones are acoustic manipulation of microscopic particles, where
small forces can do a lot.
But: what fraction of a sound wave can you absorb/reflect? If the
absorber/reflector gets sound from both sides, the forces from each side
cancel. Finally, most speakers/amps that are supposedly 340W don't do 340W
average output, and 1N is a small force (how heavy is 100g?).
--
Timo
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