Thermodynamics and isentropy in a fluid

Discussion in 'Physics & Math' started by AlphaNumeric, Jan 7, 2011.

  1. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    This is likely something anyone familiar with statistical physics will answer in a snap but I'm too rusty and too unfamiliar with it presently.

    I have a gas which is moving. I know its density, velocity, total energy and at a boundary its temperature. How do I work out its temperature elsewhere given what I know and the fact the system is known to be isentropic? The flow is obtained by some method which includes temperature but I'm not the one doing it and its not information I'm provided. Also I don't know the pressure but if its essential then I could, with a lot of effort, work it out but I'd prefer not to.

    At a glance the Wiki page would lead me to think I can use the 2 relations \(\rho_{1}V_{1} = \rho_{2}V_{2}\) and \(C_{V} = \frac{\gamma R}{\gamma - 1}\) (from here) to reformulate \(T_{2} = T_{1} \left( \frac{V_{1}}{V_{2}} \right)^{\frac{R}{C_{V}}}\) to being in terms of the things I know, ie \(T_{2} = T_{1} \left( \frac{\rho_{2}}{\rho_{1}} \right)^{\gamma-1}\) (which I've just noticed is also given on that page

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    ). Normally I'd just go ahead with that but I would have thought the energy data would be relevant. Also there's the issue that for non-ideal gases \(\gamma\) is temperature dependent itself and if I can't assume T doesn't vary too much then it's less straightforward.

    My alternative idea involves the fact in an ideal gas setup the internal energy is \(U = \xi \frac{1}{2}nRT\) where \(\xi\) relates to gas degrees of freedom, n is the number of particles per unit volume, R is the relevant gas constant and T is the temperature. I can work out n given the density by using the physical properties of the gas's make up. U is the internal energy while I know the total energy. Is the difference E-U due entirely to the kinetic energy of the fluid, which I can compute, ie if I work out U by E-KE and then rearrange to get \(T = \frac{2(E-KE)}{\xi n R}\) will I get the same answer or am I doing something wrong there? If it helps E is constant.

    Should I just stick with the Wiki formula or can I use the energy expression? Both need a roughly ideal gas to be viable but I'm not sure precisely how far to ideal each can tolerate and still be viable.
     
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  3. arfa brane call me arf Valued Senior Member

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    The wiki has
    \( C_P = \frac{\gamma R}{\gamma - 1} \qquad \mbox{and} \qquad C_V = \frac{R}{\gamma - 1} \)

    When you say the gas is "moving", do you mean transported? I don't know a whole lot about it, so can only suggest Fick's law, Fourier's law and a viscous flow \( j_p\; =\; -\eta \frac {dv} {dx} \)
     
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  5. AlphaNumeric Fully ionized Registered Senior Member

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    When I say moving I mean that I have a flow field \(v(x)\) which has physical things like density (and thus momentum) and energy and temperature. I know all but the last one and I want to know how I work out the temperature elsewhere if I know it somewhere, ie an initial condition.

    Along the lines of your suggestion I know that viscosity depends on temperature so if I know it and a boundary condition I can work out the temperature but there's two issues there. Firstly I don't know how the viscosity of the fluid depends on T (different materials depend differently) and secondly the viscosity in the flow I know is taken to be zero, so there's no time dependency even if I worked out the viscosity via something like the Navier-Stokes equations.

    When considering 'per unit volumes' there's definitely the suggestion if density goes up then so does temperature as you're basically packing more stuff into a region and an increase in pressure (which is related to stuff per unit volume) is equivalent to that and the ideal gas law says p and T go similarly for a constant volume. Just seems a little bit weird I don't have any use for the energy values, despite wanting to know about kinetic energy (which I suppose is more momentum related) and temperature!
     
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  7. arfa brane call me arf Valued Senior Member

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    If you know or can calculate the mass of gas molecules, you have:

    \( v_{ave}\; =\; \sqrt {8 k T \over \pi m} \)

    According to my textbook, the three laws are related by:

    \( D\; =\; \frac 1 3 v_{ave} l;\; K\; =\; \frac 1 2 n k v_{ave} l;\; \eta\; =\; \frac 1 3 n m v_{ave} l; \)

    where (coefficents) D is the diffusion, K is the conduction, \eta is the viscosity; l is the mean free path.

    I can't remember if I actually used these formulas for anything, sorry.
     
    Last edited: Jan 8, 2011
  8. arfa brane call me arf Valued Senior Member

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    7,832
    I think what Alphanumeric is trying to do is find the boundary at which either the temperature is zero, or all the coefficients vanish.

    The mean free path approaches zero as the density gets large which means viscosity increases; viscosity is zero where the m.f.p. is infinite--i.e. a molecule has that many degrees of freedom to transport its heat, which is vibration and rotation.

    This is more for thread observers, because Alphanumeric probably knows all that.

    I mean, varying those quantities that can be varied is fairly stat. Average velocity is a statistic though.
     
    Last edited: Jan 9, 2011
  9. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    Although what I'm talking about is a gas I'm not really considering 'gas properties' like mean free path. Instead I'm coming at it from the Navier-Stokes continuum point of view? In addition I know that the fields I have are computed with the assumption of no viscosity so I can't do any procedure which requires it.

    Perhaps an example might be of use. Suppose you're modelling flow in a straight pipe of uniform width and there's no friction between the fluid and the walls nor any heat flow out through the walls. You know the temperature and flow going into the pipe (ie your boundary conditions) and in an ideal case the flow and temperature will remain the same all the way down the pipe. If you allow friction with the walls then the flow will slow down and heat up, turning macroscopic kinetic energy into microscopic kinetic energy. If you allow heat flow through the walls of the pipe then the flow will cool (assuming the outside of the pipe is at a lower temperature).

    For a fluid like water temperature has little effect on its volume but what if the flow is air? A change in temperature can alter the volume of a mole of gas, and thus the pressure and thus the flow itself, quite a lot. Conversely if you know all about the density, flow and energy of the fluid then surely you can work out the temperature everywhere, if given a reference temperature somewhere (ie boundary conditions for temperature). I know that if a 'flow' trapped in a box increases in temperature the pressure goes up by the ideal gas laws but what if you're considering flow through a system?
     
  10. arfa brane call me arf Valued Senior Member

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    7,832
    Perhaps this end of chapter question from the same book is a clue:

    "A fluid has a convection current in the direction of the y axis. The velocity of the convection varies in the x direction.

    What is the direction of
    a) the convection momentum,
    b) the transport of convection momentum,
    c) the transported convection momentum,
    d) the internal stress on the fluid layer?"

    Fluid flow is the diffusion of mass (as particles), the conduction of heat (particle energy), and viscosity (particle collisions or mean free path statistics).
    Hence the 1,2,3 of Fick's law, Fouriers law, and a 'law' of viscous flow. I think you want to formulate something like a viscosity gradient?
     
    Last edited: Jan 9, 2011

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