Understanding the Speed of Sound in Fluids: An Explanation from Thermodynamics

AI Thread Summary
The speed of sound in fluids is defined by the relationship c_s^2 = ∂P/∂ρ, where P is pressure and ρ is density, and this holds true for fluids experiencing small amplitude vibrations. The derivation involves applying the continuity equation in a 1D fluid context, leading to a wave equation that confirms this relationship. The assumption that fluid velocity (v) approaches zero during small amplitude variations is justified because it simplifies the analysis of pressure and velocity changes. In real fluids, the relationship ∂P/∂ρ is not constant and varies with pressure, which limits the applicability of the derived equations to small amplitude waves. Understanding these dynamics is crucial for exploring the behavior of sound waves in various fluid contexts.
Niles
Messages
1,834
Reaction score
0
Hi

The speed of sound in a fluid is defined as
<br /> c_s^2 = \frac{\partial P}{\partial \rho}<br />
where P is the pressure and ρ the density. In my thermodynamics-course this was how we defined the speed of sound in an ideal gas, I have never read the explanation anywhere for, why this relation is also valid in a fluid.

Is the reason that a fluid be approximated as an ideal gas? If not, where does this relation come from?
 
Physics news on Phys.org
This can be derived very generally for any fluid where continuity equation holds. I don't want to think about advection terms in 3D right now, so let's restrict this to a 1D fluid. You can try carrying out all the algebra in 3D yourself. It goes exactly the same. In 1D, continuity is simple enough, ##\frac{\partial \rho}{\partial t} + \frac{\partial}{\partial x}(\rho v) = 0##. Of course, we don't know much about the velocity of the flow from the start, but we know how it changes, so let's differentiate.

\frac{\partial^2 \rho}{\partial t^2} + \frac{\partial}{\partial x}\left(\frac{\partial}{\partial t}(\rho v)\right)

Now, let's look at total time derivative of ##\rho v##. Using chain rule, we arrive at the usual instantaneous change and advection terms.

\frac{d}{dt}(\rho v) = \frac{\partial}{\partial t}(\rho v) + \frac{\partial}{\partial x}(\rho v)\frac{dx}{dt} = \frac{\partial}{\partial t}(\rho v) + v\frac{\partial}{\partial x}(\rho v)

Naturally, I'm interested in that partial time derivative bit, so let me re-arrange things a bit.

\frac{\partial}{\partial t}(\rho v) = \frac{d}{dt}(\rho v) - v\frac{\partial}{\partial x}(\rho v) = \left(\frac{d\rho}{dt}v + \rho\frac{dv}{dt}\right) - \left(v\frac{\partial \rho}{\partial x}v + v\rho \frac{\partial v}{\partial x}\right)

This is probably a good place to linearize. Algebra can be carried on a bit longer to make it a bit more rigorous, but the end result is the same. So long as we are talking about small amplitude vibrations, v is very small, and in fact, we can take it as going to zero. This leaves us with a very simple expression.

\frac{\partial}{\partial t}(\rho v) = \rho \frac{dv}{dt}

This is very fortunate, because ##\rho \frac{dv}{dt}## is the one thing we do know. If I take a small element of fluid of span ##\Delta x##, it experiences a force ##-\Delta P## on it. Taking the limit as ##\Delta x \rightarrow 0##, we get what we are looking for.

\rho \frac{dv}{dt} = -\frac{\partial P}{\partial x} = - \frac{\partial P}{\partial \rho}\frac{\partial \rho}{\partial x}

And now we can put it back into continuity equation.

\frac{\partial^2 \rho}{\partial t^2} - \frac{\partial P}{\partial \rho}\frac{\partial^2 \rho}{\partial x^2} = 0

And that, of course, is just wave equation with ##c^2 = \frac{\partial P}{\partial \rho}##.
 
  • Like
Likes 1 person
K^2 answer is about as detailed as you can get. Just to add one tiny bit to make sure there's no confusion, an ideal gas is a type of fluid.
 
  • Like
Likes 1 person
Hi guys

Thanks a lot for your answers and the detailed derivation. So the relation is something general for fluids that experience small amplitude vibrations.

1) Physically, what does c_s mean? I guess it basically says something about how fast pressure vibrations travel in the media.

2) What is the physics behind the assumption v going to 0, when the amplitude variations are small? v refers to the fluid velocity, how is that related to the amplitude?
 
Last edited:
Niles said:
1) Physically, what does c_s mean? I guess it basically says something about how fast pressure vibrations travel in the media.

It doesn't "say something" about it - that's exactly what the speed of sound is.

2) What is the physics behind the assumption v going to 0, when the amplitude variations are small? v refers to the fluid velocity, how is that related to the amplitude?

You can make K^2's math a bit more explicit by letting the pressure = ##p_0(x,y,z) + p(x,y,z,t)## and the velocity = ##v_0(x,y,z) + v(x,y,z,t)##. ##p_0## and ##v_0## are some arbitrary steady flow conditions which don't change with time. ##p## and ##v## are the time-dependent parts of the pressure and velocity that represent the "sound wave".

If you work through the math, you get to the same result for the speed of sound as before. ##p_0## and ##v_0## drop out of the equations when you differentiate with respect to time.

In a real fluid ##\partial p/\partial \rho## is not constant, it varies with ##p##. As a "sound wave" with a large amplitide travels through the fluid, it changes shape and spreads out in space. As a simple extreme example, of this, the pressure in a real fluid can't be negative. K^2's math only matches the behavior of the real fluid for "small" amplitude waves, where "small" means that ##\partial p/\partial \rho## really is constant.
 
AlephZero said:
In a real fluid ##\partial p/\partial \rho## is not constant, it varies with ##p##. As a "sound wave" with a large amplitide travels through the fluid, it changes shape and spreads out in space. As a simple extreme example, of this, the pressure in a real fluid can't be negative. K^2's math only matches the behavior of the real fluid for "small" amplitude waves, where "small" means that ##\partial p/\partial \rho## really is constant.

And of course, finite amplitude waves are where the excitement really starts. Nonlinear wave equations are fun (no sarcasm intended). :devil:
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top