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Mass Conservation and Reynold Transport theorem for Non Uniform flow

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data

    I will honestly be so grateful if someone can explain this to me. I am studying the Reynolds transport theorem, particularly mass conservation. I have read over my notes and I really do not understand how to calculate the mass flow rate through the control surface if it has a non uniform velocity. The equation I was given was:

    [itex]\dot{m}[/itex]i = integralcsi ρVdA

    I don't even understand why this would give the mass flow rate. I think the problem is I don't understand the relevance of the integral. How can integrating with respect to the control surface area give you the mass flow rate?
    Also how can you integrate density x velocity with respect to area?

    Does the mass flow rate have different types of integrals for different types of flow? (e.g. linear, non-linear)

    I am so confused and keep panicking. Please someone help me!


    2. Relevant equations

    [itex]\dot{m}[/itex]i = integralcsi ρVdA
     
  2. jcsd
  3. Apr 30, 2013 #2

    SteamKing

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    What would the integral be if rho and V were constant? What if the control surface were represented by a section of pipe, with a fluid like water flowing through it?
     
  4. Apr 30, 2013 #3
    If rho and V were constant you get mass flow rate = rho x V x A .. which is the equation I use when the velocity is constant to calculate mass flow rate, so that makes sense!

    What would happen if the control surface was a section of pipe with fluid flowing through it? I don't actually know. This is where I lack understanding I think.

    Doing a question on this and there is a linear velocity profile at an inlet.. supposedly the average velocity can be calculated using V= Vmax(y/h).
    This is also what I struggle with: how does the above equation for velocity come about?

    I am useless at this!
     
  5. Apr 30, 2013 #4

    SteamKing

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    Don't you use mass flow rate= rho * A * V when you have water flowing through a pipe? After all, mass in = mass out of the pipe, because the water cannot migrate through the pipe walls.
     
  6. May 1, 2013 #5
    VdA is the volumetric throughput rate through a differential element of area dA, where V is the velocity normal to the area. Once you know the differential volumetric throughput rate, you can get the differential mass throughput rate by multiplying by the density ρ.

    Suppose you have laminar flow in a pipe, and you want to apply your equation. For laminar flow in a pipe, you probably know that the velocity profile is given by
    [tex]V=2\overline{v}(1-(\frac{r}{R})^2)[/tex]

    where [itex]\overline{v}[/itex] is the average axial velocity (the volumetric throughput rate Q divided by the cross sectional area of the pipe)

    Suppose you want to determine the mass throughput rate. In this case, you have to integrate the velocity profile radially, with dA = 2πrdr. The volumetric throughput rate through this annular cross section is given by [itex]2πrVdr[/itex]. Therefore, multiplying by the density, and integrating over the entire cross section gives:
    [tex]\dot{m}=\int_0^R{ρ2πrVdr}=πR^2ρ\overline{v}=ρQ[/tex]
     
  7. May 1, 2013 #6
    This explanation helps me a lot, thank you. It may be a silly question, and I may not need to know the answer, but why do we integrate in the first place? I think I may lack knowledge of the functions of integration.
     
  8. May 1, 2013 #7
    You need to integrate because the fluid velocity may not be constant over the entire surface through which the mass is flowing, and you want to calculate the total mass flow rate through the surface. So you need to break the surface down into smaller surface areas, each of which has a nearly constant velocity through it. By integrating, you add these flow rates up to get the total.
     
  9. May 2, 2013 #8
    Thank you so much! Makes so much more sense :D
     
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