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Homework Help: Calculating water pressure from bernouliis equation

  1. Jan 18, 2010 #1
    A power station is supplied with water from a reservoir. A pipeline connects the reservoir to the turbine hall
    The flow of water through the pipeline is controlled by a valve which is located exactly 500 metres below the surface of the water in the reservoir. The pipeline is 0.30 m in diameter.

    Calculate the static pressure at the lower end of the pipeline when the valve is in the closed position

    relivant equation

    bernouliis equation

    P + 1/2 rho v^2 + pgh = constant

    my attempt so far -

    for the surface of the reservior
    P1 - atmospheric pressure 1.00x10^5
    rho1 = density of water 1.00x10^3
    g = 9.81
    h1 = 0
    v1 = 0 (water is not moving)

    for the bottom of the pipe
    p2 - unknown
    rho 2 = density of water 1.00x10^3
    g = 9.81
    h2 = 500
    v2 - 0 (water not moving)

    rearrange for p2

    p2 = 1/2 Rho (V1-V2)g(h1-h2)-P1

    this gives

    1/2*1.00x10^3+1.00x10^3*9.81*(0-500)-1.00x10^5

    so p2 = 4904500

    or 4.9x10^6

    can someone let me know if i went wrong anywhere as this doesnt look correct? i dont think the diamter of the pipe is important for this part of the equation,
     
  2. jcsd
  3. Jan 28, 2010 #2
    I've got this question too victoriafello...am currently working on it and will have something, at any rate, posted soon this morning..
     
  4. Jan 28, 2010 #3
    Someone I know worked out the answer to the first question as 4.90e6 N/m² too...so I reckon you may well be right on that one

    I got this of the same person for the second answer:

    2.
    pipe area = πD²/4 = π(0.3)²/4 = 0.0707 m²
    water flow = 5.00 m³/s
    V = water velocity = 5.00/0.0707 = 70.7 m/s
    V²/2g = (70.7)²/2g = 255 m of H2O
    gauge pressure = 500 - 255 = 245 m of H2O => (4.90e6)(245/500) = 2.40e6 N/m² ANS-2
     
    Last edited: Jan 28, 2010
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