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Help me with Brenouille's equation

  1. Apr 18, 2003 #1
    Hi. Can somebody help me with Brenouille's equation. How is it theoretically derived and what are the effects. I have a project on that. Thank You.
    Last edited by a moderator: Feb 4, 2013
  2. jcsd
  3. Apr 18, 2003 #2
    Brenouille's equation says that the sum of the pressure (P), the kinetic energy per unit volume and the potential energy per unit volume, has teh same value at all points along a streamline.....

    Consider a flow through a nonuniform pipe in a time [del]t....the force on the lower end of the fluid will be P,A, where P, is the pressure at the lower.....the work done at teh lower end by the fluid behind it is.......

    W, = F,[del]x, = P,A,[del]x, = P,V

    , - 1 like P1 = P,

    ; - 2 like P2 = P;

    similarly...the work done on the fluid on the upper portion in time [del]t is....

    W; = -P;A;[del]x; = -P;V

    The assumtion is that the pipe is curved and that the force on the fluid at teh top is opposite to its displacement.....therefore the work done is negative......the net work done by these forces is....

    W = P,V - P;V

    Part of this work goes into changing the kinetic energy of the fluid and part goes into changing its gravitational potential......If m is the mass passing by the pipe in the time interval [del]t, then the change in kinetic energy of the volume of fluid is

    [del]KE = 1/2mv;^2 - 1/2mv,^2

    the change in potential energy is

    [del]PE = mgy; - mgy,

    We can apply the work-energy theorm ......W = [del]KE + [del]PE

    P,V - P;V = 1/2mv;^2 - 1/2mv,^2 + mgy; - mgy,

    density = mass/volume .... [rho] = m/V

    P, - P; = 1/2[rho]v;^2 - 1/2[rho]v,^2 + [rho]gy; - [rho]gy,

    P, + 1/2[rho]v^2 + pgy, = P; = 1/2[rho]v;^2 + [rho]gy;

    P = 1/2[rho]v^2 + pgy = constant
  4. Apr 27, 2003 #3
    Thanks stranger...
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