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Fluid Mechanics: Constant Pressure and Incompressible flow

  1. Jun 2, 2015 #1
    Assuming a flow can be idealised as incompressible, then can you use the constant pressure assumption ?

    I just want to get my understanding clear. My problem is the following.


    Consider a fluid element with volume ##V## and a fixed number of molecules. If the flow is incompressible, then the volume of this element does not change. Right? (otherwise the density would change). Then if the volume does not change, and say you heated it through some process (Joule dissipation, Viscous Dissipation, Conduction etc), then the pressure would have to increase right? So hence, you cannot use the constant pressure assumption and the incompressible assumption at the same time?

    The reason I am asking is because I am looking at some MHD temperature equation derivations. There is one step where the substantial derivative ##DP/DT## disappears, where ##P## is the pressure field. There is no reason given, but if it does, then there must be some assumption that the pressure field is atleast approximately constant in both space and time. But how can this be if the fluid is also considered incompressible?

    Here is the equations that have confused me

    [tex] \rho \frac{Dh}{Dt} = \frac{Dp}{Dt} + \rho \dot{q}_{rad} + \nabla \cdot (k \nabla T) + \frac{1}{\sigma}\mathbf{j}^2+ \mu \nabla^2 \mathbf{V}[/tex]

    [tex]\rho c_p\left(\frac{\partial T}{\partial t} + (\mathbf{V} \cdot \nabla)T\right) = \rho \dot{q}_{rad} + \nabla \cdot (k \nabla T) + \frac{1}{\sigma}\mathbf{j}^2+ \mu \nabla^2 \mathbf{V}[/tex]

    Note that the LHS is just the definition of enthalpy substituted for ##h=c_pT## and the substantial derivative being expanded.
     
  2. jcsd
  3. Jun 2, 2015 #2
    Thermodynamically, for an incompressible fluid dh = CpdT + vdP = CpdT + dP/ρ

    Chet
     
  4. Jun 2, 2015 #3
    .....and that's the key....thanks Chester, so the book used the wrong formula.
     
  5. Jun 2, 2015 #4
    I'm not sure the book used the wrong formula. Did they actually say that dh = Cp dT? Both equations form the book look consistent with the equation I wrote down.

    Chet
     
  6. Jun 2, 2015 #5
    Yes, the book stated dh=cpdT bizarrely
     
  7. Jun 2, 2015 #6
    Go figure. Even the experts make mistakes.

    Chet
     
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