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Speed of sound in fluid

  1. Dec 16, 2013 #1
    Hi

    The speed of sound in a fluid is defined as
    [tex]
    c_s^2 = \frac{\partial P}{\partial \rho}
    [/tex]
    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?
     
  2. jcsd
  3. Dec 16, 2013 #2

    K^2

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    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 lets 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 lets differentiate.

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

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

    [tex]\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)[/tex]

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

    [tex]\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)[/tex]

    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.

    [tex]\frac{\partial}{\partial t}(\rho v) = \rho \frac{dv}{dt}[/tex]

    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.

    [tex]\rho \frac{dv}{dt} = -\frac{\partial P}{\partial x} = - \frac{\partial P}{\partial \rho}\frac{\partial \rho}{\partial x}[/tex]

    And now we can put it back into continuity equation.

    [tex]\frac{\partial^2 \rho}{\partial t^2} - \frac{\partial P}{\partial \rho}\frac{\partial^2 \rho}{\partial x^2} = 0[/tex]

    And that, of course, is just wave equation with ##c^2 = \frac{\partial P}{\partial \rho}##.
     
  4. Dec 16, 2013 #3

    boneh3ad

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    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.
     
  5. Dec 17, 2013 #4
    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 [itex]c_s[/itex] 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: Dec 17, 2013
  6. Dec 17, 2013 #5

    AlephZero

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

    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.
     
  7. Dec 17, 2013 #6

    boneh3ad

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    And of course, finite amplitude waves are where the excitement really starts. Nonlinear wave equations are fun (no sarcasm intended). :devil:
     
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