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Calculating Head Losses through a Pipe Network

  1. Dec 27, 2011 #1
    Hey everyone! I'm trying to determine the losses through a pipe network that consists of 254.5 feet of 0.75 inch diameter copper pipe.

    The flow rate is 0.37125 gpm, with a corresponding velocity of 0.26962 fps. The fluid is water at 150 degrees F.

    Based off that information, I have a Reynolds Number of 3655.67

    To calculate my friction factor I have three equations, which I'm trying to determine which one is appropriate:

    1) f = 1.325(ln(0.27(e/D)+5.74(1/Re)^0.9))-2

    this produces a friction factor of 0.22143, with a total head loss (including minor losses from elbows, etc.) of 1.039 feet

    2) f = (1.28*g*K1)/(C1.85*D0.02*(Re*v)0.15)

    using this I get a friction factor of 0.0413, total head loss = 0.2 feet

    3) 1/f0.5 = -0.86*ln((e/3.7D)+(2.51/(Re*f0.5)))

    using #3 it produced an f of 0.9979, total head loss = 4.61 feet

    As you can see these produce vastly different results, I'm just confused as to how to go about calculating the friction factor for this pipe system with my given Reynolds number. For instance equation 1 says it is valid for 8,000,000 > Re > 5000 so I don't believe that is an appropriate head loss.

    Equation 2 doesn't really specify any applicable ranges or Reynolds numbers, just that it incorporates the Hazen-Williams coefficient.

    Equation 3 says it was an empirical equation to represent the Moody diagram for Transition Zone. Since my Re is greater than 2000 but less than 5, thats where I think it falls.

    Any advice is appreciated!!


  2. jcsd
  3. Dec 27, 2011 #2


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    Your flow is in the "critical zone" (4000 > Re > 2000) where the flow can be laminar or it can be turbulent or I suppose some mixture of the two, though generally if it becomes turbulent it should stay turbulent.

    Note in the diagram below, the friction factor in the laminar zone (below Re=2000) is linear and independent of wall roughness. Above about Re=4000, you enter the "transition zone" where the flow is turbulent. When that happens, the friction factor will increase considerably. If your flow is between these two, which yours is, it could be either so you need to use your judgement as to which one you should use. Pick the worst case and ensure the system functions appropriately under those conditions. Note that the "transition zone" is above the velocity where flow is expected to be turbulent (ie: Re > 5000), so your equation 3 gives a relatively high value that may not be realistic.

    PS: There's a good review of friction factors at eng-tips.com here:
    http://www.mathworks.com/matlabcentral/fx_files/7747/1/moody.png [Broken]
    Last edited by a moderator: May 5, 2017
  4. Dec 27, 2011 #3
    Thanks for the reply!

    That site was pretty useful in explaining a few things. How can I have a better understanding of what the real head loss would be? I tried using a few of those equations from that site which were appropriate for my Re:

    Using Serghides Equation: f = 0.04

    Using Churchill's Equation: f = 0.07632

    Using Chen's Equation: 0.041

    On the other hand if it were totally laminar flow (not realistic) f = 0.0175

    Or if I used the colebrook equation: f = 0.9979 (which probably isn't realistic either)

    The maximum head loss produced would be from colebrook equation which would be 4.61 feet.

    It's important to note that given the choices of pumps I have to choose from the manufacturer, the system will still work at this head loss. So I'm not worried about the system not performing. I'm just trying to get a better understanding of what the real head loss would be through the system.

    For instance, say the cutoff for one paticular type of pump would be 3 feet of head loss for this system after applying bernoulli's equation. And if the actual head loss was 4.6 feet it would require a larger, more expensive pump. What would you suggest to do to more accurately predict the actual head loss to ensure your choosing an appropriate pump?
  5. Dec 27, 2011 #4


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    Interesting numbers. Thanks for posting. I agree the colebrook equation value seems unreasonable. I'd love to see a more thorough analysis of friction factors in the critical zone as well as laminar and turbulent zones for that matter, along with a comparison of all the different equations used, but I really don't have time for that. If you'd be interested in doing that, I'm all ears! :smile:

    I'd suggest either using a different size pipe so the flow isn't in the critical zone or use the more expensive pump. The cost of installing a pump is generally fairly significant, so the overall cost of pump plus installation won't go up as much as just the capital cost of the pump alone. You'll be better off being conservative if you can't be accurate.
  6. Dec 27, 2011 #5
    haha, unfortunately I don't have that kind of time either!! :surprised but I agree it would be interesting to see the results to help better predict losses in piping systems.

    Without a doubt its better to be safe if your not confident with the accuracy of the analysis. If 1/2" pipe could be used it would place us in the turbulent zone but this doesn't meet our minimum pipe size requirements. On the other size 1.5" pipe would put us in totally laminar flow, however I feel that it wouldn't be necessary to use that size. We'll just proceed with the .75" pipe, as currently designed.

    Thanks for your help though! It wold be nice to see better analysis of friction factors though!!
  7. Dec 30, 2011 #6
    Use the D'Arcy-Weisbach equation. use it again for all changes of direction, like elbows, where minor lost work occurs
  8. Dec 31, 2011 #7
    What values are you using for e and D? What would you read off the Moody diagram (forgetting the different equations for a moment)?

    When I try e/D = 0.000079 (using e=0.0015 mm & D=0.75 inch=19.05 mm) I get:
    Colebrooke: 0.04107
    Serghides: 0.04107
    Churchill: 0.04182

    my understanding is, such equations are various attempts at matching Moody's chart; they should all give the same answer in their range of applicability (as seen above, they do). Moody's chart represents experimental results (and hence is 'right') -- the various formulas are just fits you can use in spreadsheets or computer programs. If you are solving a single problem, just read the value off the chart.

    This is a different question from 'what's the uncertainty in Moody chart in transition range?' Your design needs to address that.
    Last edited: Dec 31, 2011
  9. Dec 31, 2011 #8
    Yes Indeed Darcy-Weisbach is indeed the equation to use!

    The value of e for copper piping was 0.000005 ft/ft

    straight from the moody diagram: f around 0.04

    The issue comes from the range of uncertainty that exists when fluids are flowing in the transition / critical region. As you can see from some of the previous posts that some equations have been developed to answer what the true f would be, however there are many equations which produce a wide range of results. There really isn't too much knowledge that exists for what the true value of f would be in the critical zone ( 2100 < Re < 3000)

    It's important to note that in either case, because of what is available to choose from pump manufacturers that the design will work with a head loss up to 8 feet, and the maximum head loss any equation produced in the analysis was 4.6 feet, so either way I'm comfortable that the design will work. It was just a matter of looking for additional guidance regarding flows in transition / critical zones.
  10. Dec 31, 2011 #9
    Maybe I was not clear --- check your calc, I think colebrooke yields 0.04 also (ie,your 0.99 is a mistake).
  11. Dec 31, 2011 #10
    colebrook yields 0.00234, however it is only valid at Turbulent Flows. I realize the confusion, earlier i wrote colebrook yields 0.99, that is incorrect. f = 0.99 is an equation that was developed to represent transition flows. However in this case a friction factor of 0.99 probably wouldn't be realistic.
  12. Jan 1, 2012 #11
    Jim, check again. 0.002 is off the chart low; the right value is 0.04.
  13. Jan 1, 2012 #12
    Colebrook yields 0.0234 sorry I added an extra zero, typo.

    Colebrook yields 0.04142 with the original flow rate. I assure you all the calculations are correct. Thank-you. Again colebrook wouldn't be the value I would assign the highest amount of accuracy for transition flows, and therefore will not be used for determining the pump size requirements.
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