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From the golden days of this newsgroup...



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Aeronautics Digest 1.16                                      November 11, 1989 
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[...]

From: leo@riacs.edu (leo)
Subject: Re: Question: humidity vs speed of sound
Date: 8 Nov 89 18:53:31 GMT

rando@skipper.dfrf.nasa.gov (Randy Brumbaugh) wrote:
>How does speed of sound change with relative humidity and/or 
>temperature?
>
Speed of sound is given by
	c = sqrt(gamma*R*T)
where
	gamma = 1.4 for an ideal diatomic gas.  For a more complex
molecule such as water vapor gamma is slightly lower (I beleive people
use 1.3 for steam) but it doesn't vary too much.

	R = gas constant, this is just the universal gas constant
divided by the molucular weight of the gas.  You can just use the
weighted average of the molecular weights of a gas mixture, air
is 80% N2 and 20% O2, throw in the corresponding fraction due
to water vapor and adjust accordingly. (The universal gas constant
is equal to 8.3144 kJ/kg/mol if I remember correctly, you might
want to check the dimensions on that).

	T = temperature.  This is just a measure of the thermal
energy in the gas.  I beleive humidity affects the reading you
get from a thermometer (it improves the heat transfer to the
bulb) but I can't remember what adjustments are typically made.
In any case, remember you want the temperature of the gas and
the effect of the water vapor in the air is handled by the
adjustment to R.

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Aeronautics Digest 2.22                                         March 20, 1990 
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From: awtron@phoenix.Princeton.EDU (Andrew Walter Tron)
Subject: Speed of sound (long)
Date: 10 Mar 90 04:50:19 GMT

In article <SHAFER.90Mar9120634@skipper.dfrf.nasa.gov> shafer@skipper.dfrf.nasa.gov (Mary Shafer (OFV)) writes:
>If, as you claim, the speed of sound is a function of temperature
>(rather than a function of density, as I claim), why does sound move
>faster through water than through air?
>

The reason that sound moves faster through water than through air is
that water is less compressible than air, not that it is less dense.
Let me explain.

If you take a look at the 1-dimensional equations of motion for an
inviscid fluid (by inviscid I mean that the fluid is, roughly speaking,
frictionless), they look like:

d(density)/dt + (velocity)*d(density)/dx + (density)*d(velocity)/dx = 0

which is the differential equation that states that mass is conserved
(i.e there are no 'sources' or 'sinks' of mass in the flow).  The
equation for conservation of momentum looks like:

d(velocity)/dt + velocity*d(velocity)/dx + d(pressure)/dx /density = 0

Now if you define the quantity a^2 = d(pressure)/d(density), you can
transform the second equation to:

d(density)/dt + velocity*d(velocity)/dx + a^2/density * d(density)/dx =
0

Notice that so far I have only assumed that the fluid is in (at most)
1-d flow, and that no viscous (or frictional) forces are present.
Nowhere have I made the assumption that the fluid is a liquid or a gas.

Now if you further assume that the velocity is small with respect to
the quantity 'a', and that the quantity 'a' remains relatively constant
(at some value a0) and that the density only fluctuates by a small
amount, then you could use these assumptions to simplify the equations.
Your equations would then become

d^2(density)/dt^2 - a0^2 d^2(density)/dx^2 = 0

d^2(velocity)/dt^2 - aO^2 d^2(velocity)/dx^2 = 0

These two equations are the 1-dimensional linear wave equations.
The solutions to these equations are signals that propagate at a speed
a0.

So far, I have demonstrated that if your fluid is basically frictionless
(inviscid) and one dimensional, and that if it is relatively still,
small disturbances in density (and velocity) propagate with a speed of
a0.  Thus it is easy to see that a0 is the speed of sound in still
fluid.  The only part of my argument that is contrary to the physics of
the situation is the one-dimensional assumption.  It is possible to
extend this argument to three dimensions, but the math gets more
complicated, and I'm trying to keep this as simple as possible.

So now we've established that the speed of propagation of small
disturbances, called the speed of sound, equals a0, which is the square
root of d(pressure)/d(density).  To evaluate this quantity for a perfect
gas, we assume the perfect gas law (pressure = density * gas constant *
temperature) and we assume that any changes in the gas are such that
entropy does not increase (this is a standard assumption in
gasdynamics).  Using these two relations, we find that 

      d(pressure)/d(density) = square of speed of sound
                             = gamma * gas constant * temperature
     
where gamma is the ratio of specific heats for a gas: this ratio
can vary with temperature but for a wide range of temperatures
(including room temperature) it is 1.4.

Thus for a perfect gas, the speed of sound depends only on the
temperature.  

I can't give you any formula for the speed of sound in water, but
if you take a good look at the quantity d(pressure)/d(density), you can
see it as an 'inverse compressibility factor'.  If the flow process is
such that a moderate increase in pressure results in a substantial
increase in density, then the speed of sound will be relatively small.
This is the case for a perfect gas, which is a good approx. for air.
If a substantial increase in pressure results in only a relatively small
increase in density, then the speed of sound will be large.  The latter
is the case in water -- water can withstand large pressures while
keeping its density relatively constant, and as a result, the speed of
sound in water is larger than the speed of sound in air.


So the fact that the density of water is greater than the density of air
has nothing to do with the fact that the speed of sound in water is
greater than the speed of sound in air.  It's just a coincidence.
Theoretically, one could increase the density of air by a proportional
increase in pressure until the density of that air equals the density of
water.  Since this process is isothermal (by the perfect gas law), the
speed of sound would theoretically be the same.  (However, by this time
the gas would no longer be a perfect gas -- I'm not exactly sure what it
would be).

The above analysis assumed that the fluid was essentially still and that
the disturbances are small.  These assumptions are not necessary, but
they do simplify the math.

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