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Fluid mechanics of water in a pipe

  1. Apr 20, 2010 #1
    Hello All,

    I've frequented this forum for various purposes and this is the first time I've decided to post! Hopefully I can find the help or conformation I'm looking for. I may not have all the necessary info, but I feel this is basic enough that I should be able to come up with an appropriate answer.

    This is a basic setup where I have water being pumped through steel piping and I'm trying to find the exit pressure and velocity of the water. Here is what I know for sure.

    Pipe Diameter, d = .151in or .003835m
    Pipe Length, l = 72in or 1.8288m
    Pressure from pump, P = 100psi or 689400Pa
    Temp, T = 140°F or 60°C
    Density, ρ = 983.2 kg/m³
    Viscosity, μ = 0.467*10-³ N*s/m²
    Roughness steel, ε = .000025 m

    Now somewhere in here I feel I am going about this wrong, so please bear with me.

    1. To get the velocity of the water at the starting point I use Bernoulli's equation for dynamic pressure, q=1/2*ρ*v². Solving for v, I get v=37.45 m/s.

    2. I calculate my Reynolds Number and relative roughness of the steel pipe, then use the Moody chart to determine the Friction Factor.
    Re = (ρ*v*d)/μ = (983.2*37.45*.003835)/(.467*10-³) = 301,978
    RR = ε/d = (.000025)/(.003835) = .00652
    From the Moody chart: f = .031

    3. Calculating the pressure drop due to the friction in the pipe (I feel this is my problem)
    ΔP= (ρ*v²*f*l)/(2*d) = (983.2*37.45²*.031*1.8288)/(2*.003835) = 10192437.5 Pa

    From there I would use Bernoulli's equation for the flow at the beginning of the pipe, P + 1/2*ρ*v² + ρ*g*h (last part is negligible for me), set equal to an equation for the exit flow which incorporates the friction.

    I feel as though I am making an incorrect assumption (or 2 or 3...) which is leading to such a large drop in pressure. Areas I feel I could have gone astray; calculating the initial velocity using the dynamic pressure equation, assuming that friction will even have an effect here, or even that the pressure of the pump is way too high for such a small diameter pipe (and my calculations are somewhat correct).

    If there is anyone out there that can comment on this and perhaps guide me in the right direction I'd greatly appreciate it.

    Cheers and thanks for reading!
  2. jcsd
  3. Apr 20, 2010 #2

    Andy Resnick

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    Yikes... I was going to use Poiseuille flow, but your Re >>2000, so that's no good.

    The exit is at atmospheric pressure...does the pipe let water freely squirt out, or is it held at some other pressure? That gives you your pressure drop across the length of the pipe.

    100 psi into a 1/10 inch diameter tube... yikes.
  4. Apr 20, 2010 #3
    Yea, I'm way past laminar flow with the current setup. I'm looking into more details right now, but I'm pretty sure there is not a resovoir at the end of the tubing and there is not any kind of nozzle (which would probably shoot off like a bullet anyways).

    After looking at it for a bit I too started to think that the 100 psi was just way too much. Thank you for your reply, it had been some time since I dealt with fluid mechanics and I needed some one else's opinion on this.

    Before I keep digging into this topic, can anyone confirm that the way in which I used the dynamic pressure equation, q = 1/2*ρ*v², was in fact correct? I thought it to be since the only initial pressure in the system is from the pump, but I'd really like to know that I do infact have the basics correct.
  5. Apr 20, 2010 #4
    Your velocity is probably closer to 9 meters per second. Reynolds number is about 36.000. See thumbnail.

    Bob S

    Attached Files:

  6. Apr 27, 2010 #5

    I was wondering how you came to the conclusion that the initial velocity would be around 9 m/s. Assuming everything is correct, pressure drop = 108 psi and pressure of water at the exit = 7.45 psi, I can plug those numbers into the bernoulli equation, which should look like this:

    1/2*ρ*v1² + P1 = 1/2*ρ*v2² + P2

    and with the numbers:

    1/2*983*(9)² + 792810(pascals) = 1/2*983*(v2)² + 51705(pascals)
    solving for v2 = 39.86 m/s²

    I know that a drop in pressure leads to increased velocity typically where the flow path has decreased in area, but I'm not positive that the same applies when the drop in pressure is due to head loss or the friction in the pipe.

    Also, I'm still looking for confirmation on how to calculate an initial velocity with the given data. Need to be sure I have the basic of basics down before I'm satisfied with the other results. Once again, any help is appreciated! Cheers!
  7. Apr 27, 2010 #6
    After going through everything again, I realize I'm completely off base in my last post. That program Bob used gave me my output velocity via the flow rate. If thats the case, then my exit velocity is roughly 0.06 m/s. To me that seems much too low. I am wandering off the path somewhere and making poor assumptions.

    Still interested in how to determine my initial velocity though.
  8. Apr 27, 2010 #7


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    Your initial steps in the OP (steps 1 through 3) are fine. But once you got to step 3 and found your pressure drop is much larger than the actual pressure drop, you need to iterate. Go back and reduce velocity, then recalculate pressure drop. Keep doing that till you home in on the velocity required to produce the pressure drop you know to exist. Once you find the velocity, you can determine flow rate.

    You might also look through the text I've posted online https://www.physicsforums.com/showthread.php?t=234887"called "Pipe-Flo Pro.PDF".

    Note that you don't need to use the Bernoulli equation unless there is a change in elevation or change in pipe diameter. Since you don't mention a change in either of those, the only pressure drop is the permenant, irreversible loss caused by pipe friction and calculated by the Darcy-Weisbach equation. To see how to incorporate velocity or elevation changes, see equation 15 and 16 on page 14 of the Pipe-Flo Pro.PDF file.
    Last edited by a moderator: Apr 25, 2017