Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Flow through a pipe question

  1. Jan 10, 2014 #1

    O2F

    User Avatar

    I have a flow through a pipe that changes in diameter to become narrower. I can measure the pressure P1 at the wider end where it is applied and I want to calculate the pressure P2 at the narrow end.

    I can't directly apply Bernoulli's equation here because I have no way of measuring the flow rate at either end. I do however know all of the relative dimensions of the pipe (i.e. I know the cross-secional area through which the water flows at P1 and the smaller area at P2), and assuming the water is incompressible the volume flow is constant, I can substitute v2 = (A1/A2)v1. But simplfying leaves P2 in terms of v1: -

    P2 = P1 + 0.5*rho*( V1^2 - ( (A1/A2)*V1 )^2 )

    Is the pressure at the narrow end simply scaled linearly by the ratio of the cross sections? I have a feeling I have solved this problem before and found this to be the case, I just can't see it at the moment.
     
  2. jcsd
  3. Jan 10, 2014 #2

    russ_watters

    User Avatar

    Staff: Mentor

    That equation is definitely not linear.
     
  4. Jan 10, 2014 #3

    O2F

    User Avatar

    Yes, thank you, I can see this.

    I should have kept this simpler: look at the attached drawing, the diameters d1 and d2 and the pressure P1 are all known quantities.... how can I calculate P2?

    Any ideas?
     

    Attached Files:

  5. Jan 10, 2014 #4

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor
    2016 Award

    The value of P2 depends on the detailed geometry ('head loss')- the shape of the contraction and how far downstream you are measuring P2- there is a region directly past the vena contracta where the velocity head is reconverted back into a pressure head and flow is unsteady.

    Using the continuity equation, the head loss 'h' due to the contraction is:

    h = (1/C -1)^2* V2^2/2g, where V2 is the exit velocity well downstream of the contraction, g is gravitational acceleration, and C the 'contraction coefficient'.
     
  6. Jan 10, 2014 #5

    russ_watters

    User Avatar

    Staff: Mentor

    I don't understand: you provided the correct equation in your first post. Have you tried to use it yet?
     
  7. Jan 13, 2014 #6

    O2F

    User Avatar

    Thank you both for your help.

    I'm not sure I completely understand Andy's reply.
    I probably should have said at the start that I am just trying to get an estimate of the pressure at the constriction (the tightest part of the cone), I drew the tube afterwards to simplify things (in reality the cone sits inside a larger reservoir which is at atmospheric pressure).

    I know there is a simpler way to solve this starting with Bernoulli's expression, we are definitely overcomplicating this, I have done it before a long time ago I just can't remember how.

    Thank you for the reply Russ, I can't use the equation in the first post because I cannot measure the flow velocity. I know the input pressure and the diameters at each end of the cone.
     
  8. Jan 13, 2014 #7

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    The pressure difference depends upon the flow rate so you can't ignore it. With no flow, the pressures will all be the same static value (for a level section of pipe).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook