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Finding the flow rate through an open-end pipe

  1. Feb 1, 2017 #1
    I have a basic question for my own understanding of things,

    Consider a straight horizontal piece of Ø15 mm pipe (inside), 20 meters in length, with 5 bars static pressure connected to a water main. Trying to find out what the flow rate through the open end of the pipe will be, doesn't seem to be a straight forward operation. The flow rate depends on the head loss, which again depends on the flow rate. So how do I go about estimating the flow rate through the open end?
     
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  3. Feb 1, 2017 #2

    Mech_Engineer

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    For a start, I would treat it as an orifice flow problem:
    • Given an orifice geometry (and therefore a discharge coefficient) and pressure you can calculate a flow rate.
    • Given this initial flow rate you can re-calculate for head loss in the length of pipe and then re-calculate your orifice flow given the slightly reduced pressure at the end of the pipe
    • Iterate towards a solution which balances the pipe's head loss for a particular flow rate, and the flow rate out the end of the pipe.
    It looks like ballpark you're looking at about 12.5 m^3/hr. See here: http://www.tlv.com/global/TI/calculator/water-flow-rate-through-orifice.html

    upload_2017-2-1_9-44-28.png
     
    Last edited: Feb 1, 2017
  4. Feb 1, 2017 #3

    JBA

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    While I agree with the result for a 15 mm orifice, I have a program for piping flow that uses the Moody Chart for friction factor vs Reynolds No. and a iterative process; and, using this program the calculated flow rate for 20 meters of 15 mm copper (smooth) pipe is considerably different. Since my program operates with US units I have had to do several conversions so I will confirm my results; but, to this point the program indicates a flow rate of 13.5 gallons/minute or 3.1 m^3/ hr.

    As a comparison with our US water system with a 65 psi supply pressure through 65.5 feet of .625" I.D. hose (smooth bore) a close to 13.5 gpm flow rate would seem more reasonable.

    As I indicated, I will verify my above calculation later today and report either the verification of the above or any difference or error I find.
     
  5. Feb 1, 2017 #4
    This is much closer to what I expected to find if I only knew how. After all 12,6 m3/h would yield a velocity of almost 20 m/s (66 ft/s), which would probably have the pipe shaking quite a bit.
     
  6. Feb 1, 2017 #5

    Mech_Engineer

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    Good catch, I underestimated the pressure loss due to flow in such a long pipe.

    Breaking out the old Fundamentals of Engineering training book and some MathCAD, I was able to get pretty close to JBA's answer.

    I was able to calculate a flow rate of about 3.54 m^3/hr, at a speed of 5.57 m/s. The PDF shows the calculations, but generally I followed an iterative approach using Bernoulli and Darcy-Weisbach until the pressure drop along the pipe's length and the pressure drop at the pipe exit added up to the original pressure of 5 bar.
     

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    Last edited: Feb 1, 2017
  7. Feb 1, 2017 #6
    I'm playing around with the Swamee-Jain equation together with roughness, viscosity, and Reynold's No., and upon entering the numbers and matching the pressure loss with the initial pressure, I get 3,85 m3/h (17 gpm). Would love to hear what you guys think about it.
     
  8. Feb 1, 2017 #7

    JBA

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    Just for reference, my Excel program is based upon the:"Standard Handbook for Mech Engrs,7th Ed, Baumeister & Marks" and also uses the Darcy method.

    Mech_Engr, I have reviewed your calculations and found that you and I used different values for the Kinematic viscosity of water. I used 1.18E-5 @ 65°F / 18.3°C based upon my available curve; whereas, you used a lower 1.052E-5 (which on my graph would be at 72°F / 22.2°C), whioch would increase my Re value to 7.32E+4 and reduce my ƒD to .033. I used the "Smooth" curve on my Moody Diagram with a Re = 6.54E+4 to obtain a ƒD = .0342 friction factor for my calculation which would account for part of the difference between our values but not all of it. One value that is missing from your calculation is the Vav that you used to obtain what appears to be an initial Re = 8.85E+4 before you started your iterations and why that Re value is not different after all of your iterations to obtain the final ΔP. It should change with each entered Vav during your iterations.

    TSN79, based upon the above, I think we need to know at least the Kinematic Viscosity value and the friction factor you used before offering any response.
     
  9. Feb 1, 2017 #8
    I used the Blasius equation $$f=\frac{0.0791}{Re^{0.25}}$$ to calculate the Fanning friction factor for a smooth pipe at a Reynolds number between 2100 and 100000. The results I got were

    ##\Delta P=3.05\ bars## at ##3.1\ m^3/hr##
    ##\Delta P=3.84\ bars## at ##3.54\ m^3/hr##
    ##\Delta P=5.0\ bars## at ##4.12\ m^3/hr##

    Of course, with the Blasius equation, there is no iteration involved in the calculation.
     
    Last edited: Feb 1, 2017
  10. Feb 1, 2017 #9

    JBA

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    Thanks for the added input. I am going to plot that curve and overlay it on the "Smooth" curve on the Moody chart to see how those two curves compare. I will let you know what I find. I have tried to find some curve equation like that (and even tried to develop one using curve fitting on overlay curves without success), then my program could do that part of the iterations automatically and the iterations can be fully automated by using Excel "Goal Seeking" or a "push the button" VBasic program.
     
  11. Feb 1, 2017 #10
    The comparison you are about to do is already shown in Bird, Stewart, and Lightfoot, Transport Phenomena. As I said, the equation applies over the range of Reynolds numbers from 2100 to 100000. For the present problem, the Reynolds number falls within that range.
     
  12. Feb 1, 2017 #11

    JBA

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    If that is true, then the fact that your flow volume at 5 bar is so much higher than either Mechanical_Engr's or my result concerns me, since we both used the same highly recognized and accepted calculation method and for your values the f would have to be much lower than either of ours.
     
    Last edited: Feb 1, 2017
  13. Feb 1, 2017 #12
    I have a lot of experience with fluid dynamics. That was my thesis area. The Fanning friction factor gives the same information as the Moody chart. The only difference is a factor of 4, because Moody uses L/D and Fanning uses 4L/D. So, if you divide the Moody friction factor by 4, you get the Fanning friction factor. Didn't you know that?
     
  14. Feb 1, 2017 #13

    JBA

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    The only problem is that the calculation we are using does not use the fanning friction factor, it uses the direct f taken from the Moody diagram. You can verify this by referring to a copy of the above "Standard Handbook for Mech Engrs,7th Ed, Baumeister & Marks" Piping Loss Example on pg 3-61 of that edition.

    Like you my background is specializing in gas and liquid flow, I spent the last 20 years of my engineering career in the New Product Development of ASME Section VIII Pressure Vessel Code Pressure Relief Valves, specializing in valve and piping flow analysis and performance certification.
     
  15. Feb 1, 2017 #14

    JBA

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    The result of your equation when multiplied by 4 does in fact give a value of .01979 in line with our required values, so with that modification it may still be usable for our method of calculation and I thank you for that.
     
  16. Feb 1, 2017 #15
    Like I said, the two calculations are equivalent. I'm going to redo the calculation with the Moody chart to compare directly. Why don't you try it with the fanning equation (this is what we ChEs use) to see what you get?
     
  17. Feb 1, 2017 #16
    Using the Moody chart:

    For 4.11 m^3/hr, I get a velocity of 6.46 m/s, and a Reynolds number of 96900 (using 1 centipoise for the viscosity of water at 20 C). At this Reynolds number, for a smooth pipe, I get a Moody friction factor of 0.018 from the Moody chart. The pressure drop is then:

    $$\Delta P=\left(\frac{1}{2}\rho v^2\right)f\frac{L}{D}=\left(\frac{1}{2}(1.0)(626)^2\right)(0.018)\frac{2000}{1.5}=5.16\times 10^6\ dynes/cm^2=5.16\ bars$$
     
  18. Feb 1, 2017 #17

    JBA

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    To be honest I would not know where to start using the fanning equation, I am vaguely familiar with it but have no experience using it.

    Chestermiller, I surrender, I have just finished going through my whole calculation again and with Re = 9.01E+4 and an f = .0182 I am getting 6.477 m/s with a volume flow of 4.254 m^3/hr at a ΔP of 5.0 bar; and, based upon your history I am not really surprised. Thanks for keeping me honest.

    It would appear that the most likely problem with my earlier calculation comes down to the Re value I entered on my Moody Chart, from the .031 f value I used it appears that I slipped an order of magnitude to 9.01E+3 for the Re when using the Moody Chart and that resulted in that higher .031 f value I used and the resulting final errors and I did not catch that error in my quick review earlier today.

    One point I am not clear on is that our method uses Kinematic Viscosity which for the 65°F/18°C I selected is 1.18E-05 ft^2/sec and I don't know how that relates to the 1 cp you used.

    PS I have also plotted the chart values for the "Smooth" curve against the equation X 4 and while they are close there is some variance along the curves. At the same time, since I have never found any table of values for the chart, my only reference for the chart is by reading the chart, so it is hard for me to verify my reading of its values.
     
  19. Feb 1, 2017 #18

    Mech_Engineer

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    It looks to me like the main difference between my numbers above and Chester's are that I applied a discharge coefficient of (Cd = 0.7); this was sort of "left over" from when I was trying to approximate the problem as an orifice flow calculation . It seems it isn't necessary to use a discharge coefficient for this particular flow problem however since there wouldn't be a hydraulic diameter contraction at the pipe exit, so I defer to Chet's expertise!
     
  20. Feb 2, 2017 #19
    I used a dynamic viscosity of 0.01 gm/cm sec and a density of 1.0 gm/cm^3. This gives a kinematic viscosity of 0.01 centistokes (cm^2/sec). That is the same as 1.08E-05 ft^2/sec.
     
  21. Feb 2, 2017 #20

    JBA

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    With regard to the f equation x 4 relative to values taken from the Moody Chart see the below Excel work sheet. I have displayed the graph with a linear horizontal scale and truncated its range for better visibility (for me at least).
     

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