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Determining Pipe Pressure of a system that discharges to air

  1. Oct 23, 2016 #1
    1. The problem statement, all variables and given/known data
    A pipe 30 cm in diameter and 420 m long has a wall thickness of 1 cm. The pipe is commercial steel (ks = 0.045 mm) and carries water from a reservoir to an elevation 100 m below the bottom of the reservoir and discharges into the air. A rotary valve is installed at the downstream end. Minor losses include an entry (kL = 0.5) and rotary valve (kL = 10). Calculate the maximum water hammer pressure that can be expected on the valve if it closes in 0.5 sec. Also determine the total pressure the pipeline will be exposed to during the water hammer phenomenon. Assume: K = 2.2 x 109 N/m2, E = 1.9 x 1011 N/m2

    2. Relevant equations
    I am having trouble with determining the water pipe velocity and pipe pressure.
    The Bernoulli:

    $$ \frac{P_1}{2g} + \frac{v_1^2}{2g}+h_1 = \frac{P_2}{2g} + \frac{v_2^2}{2g}+h_2+ \frac{v^2}{2g}(\lambda \cdot L/D + \sum K_L) $$

    3. The attempt at a solution
    I am having a really hard with this problem. I think we need more details about the reservoir depth to solve the problem.

    Applying the Bernoulli equation from the bottom of the reservoir to the exit of the pipe:

    $$ \frac{P_{pipe}}{2g} + \frac{v_{pipe}^2}{2g}+h_1 = \frac{P_{exit}}{2g} + \frac{v_{exit}^2}{2g}+h_2+ \frac{v_{pipe}^2}{2g}(\lambda \cdot L/D + \sum K_L) $$

    Exit pressure is atmospheric, and velocity in pipe and at exit are equal.

    $$\frac{P_{pipe}}{2g} + \frac{v_{pipe}^2}{2g} = \frac{v_{exit}^2}{2g} - 100 + \frac{v_{pipe}^2}{2g}(\lambda \cdot 450/0.3 + 10.5) $$

    So I get this equation with two unknowns velocity ( and lambda depends on v ...) and pipe pressure. Can some one please help - this is so frustrating to solve...
     
    Last edited: Oct 23, 2016
  2. jcsd
  3. Oct 23, 2016 #2
    Shouldn't the velocity in the pipe be the same as the velocity at the exit? Shouldn't the volumetic flow rate be constant? Shouldn't the pressure at the reservoir be atmospheric? Isn't this enough to determine the volumetric flow rate?
     
  4. Oct 23, 2016 #3
    Yes I agree with all that - however only the change in elevation of the bottom of the reservoir to the exit is given ( the 100 m) . So, when we apply the Bernoulli there we'll get 1 equation with 2 unknowns : velocity and pipe pressure. But i don't think it's enough to calculate both velocity and pressure
     
  5. Oct 23, 2016 #4
    If you apply the Bernoulli equation between the reservoir and the pipe exit, the atmospheric pressures cancel.
     
  6. Oct 23, 2016 #5
    Yes I agree, but we don't know what the elevation to the top of the reservoir is. From my understanding the 100 m elevation is from the bottom of the reservoir to the pipe exit, and at the bottom of the reservoir , pressure is not atmospheric
     
  7. Oct 23, 2016 #6
    Ah. I missed that. It seems to me you need to know the depth of the reservoir.
     
  8. Oct 23, 2016 #7
    I guess so, I'm going to email my professor to see if there is a type some where in the question. Thank you for the confirmation Chestermiller!
     
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