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Pressure drop analysis on water pipe system w/pump

  1. May 21, 2015 #1
    Hey everyone. Four days ago I didn't know anything about fluids but I started my internship and got assigned a project that is all about fluids! And I need some help, this topic can get really complicated but I'm trying to take it one step at a time.

    Problem Description:
    There is a pump attached to a tank (really it's two pumps and two tanks but lets keep it simple). The pump can produce 34.5 psi at about 4 gal/min flow rate, and I need to determine the pressure drop at the end of the system. The pipe network is complex, or at least I think so. It is probably 200 feet in total with plenty of bends, elevation changes, fittings(valves and whatnot), hubs where pipes branch off, and so on.

    My Solution:
    I was going to take the system one piece at a time, calculate the pressure drop after each piece, and then use that as my initial pressure for the next piece working my way around. I am going to do this in Excel so I can keep track of the pressure throughout the system. To get a rough estimate I was first going to look at all the bends, straight pieces, and elevation changes and then start adding more details. To do this I am using the Darcy-Weisbach equation but I'm not sure this is correct because my values are coming out wrong when I test it.

    I'm using this currently:
    P1 +rho/2(v1-v2)-[(rho*v2^2)/2]*(f*L/D+kb)-(z2-z1)*rho*g = P2

    P1 - initial pressure
    P2 - final pressure
    Rho - density
    v1 - initial velocity
    v2 - final velocity
    f - friction coefficient
    D = inner pipe diameter
    z1/z2 - heights

    I think one of my problems may be with units, I have a ton of variables in Excel so I do need to straighten that out. But also one of my main concerns is that right now I am using 4 gal/min as my flow rate, and using that to calculate velocity, Reynold's number, and the friction coefficient. But shouldn't flow rate be a variable also? Otherwise the velocity will never change. I know in theory the flow rate and velocity are constant, but wouldn't they be changing in a system with friction?

    If anyone can give me some guidance on this topic I would be really appreciative. I really want to make sure I do a good job with this internship and would love any help I can get.

    Thanks in advance and let me know what questions you may have for me!
  2. jcsd
  3. May 21, 2015 #2


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    In addition to the Bernoulli equation, you should have come across the continuity equation in your study of fluids. Basically, the continuity equation says that the amount of fluid passing by one point in a pipe is equal to the amount of fluid passing by a point in the same pipe further down. In other words, between two points in the same pipe, no fluid is created or destroyed:

    Q1 = Q2

    Since Q = AV, then

    A1V1 = A2V2

    If your pump puts 4 gal. of water per minute into the piping system, then 4 gal. per minute is going to come out of the system somewhere. The velocity of the water will change depending on the pipe size, if more than one size is used in the system.

    You didn't list the units of any of the quantities you are using to calculate the pressure in the system. You didn't explain why your calculations are not coming out the way you expect; the problem could be traced to a mismatch in the units.

    If you care to post your calcs, we can see if the problem is readily apparent.
  4. May 22, 2015 #3
    Hey Steamking, thanks for the reply! Your name is reassuring. Haha. I guess I was really tired when I made this post last night; sorry I didnt include units. Everything is in inches, and although I don't have the numbers or specific units infront of me currently, I will post them in a few hours when I can get at them again.

    Maybe I could start by listing the units that I am using, in inches, with the associated conversation factor I used.

    Also, I did use the Q=VA to calculate the velocity based on my flow rate. So, theoretically, the speed of the water will stay constant for a pipe of the same cross sectional area? I suppose where I was getting caught up was by looking at a coworkers old spreadsheet for an oxygen system. In his case the flow rate and velocity were changing constantly, probably due to many junctions and what not.

    So, do you think it's fine to leave my flow rate as a constant? Maybe I should tackle one thing at a time, but i'm just thinking ahead to when the lines start to split off. I've read that you can treat the flow of water similarly to current in a circuit using Kirchoffs rules, is this correct?

    Sorry that I have so many questions. I always need to get concepts Down fully before I can really understand and feel confident in what i'm doing, and I have a rushed background of studying on this topic.

    Also, I am happy to post the calculations when I am able to look at them again.

    Again, thanks for the help!
  5. May 22, 2015 #4


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    Welcome to PF. It looks to me like you are approaching the task in the right way and have correctly identified the problem of flow (pump performance) being variable, not constant. So here's what I'd do:

    Set up the spreadsheet such that flow is a variable with a single cell reference, so you can change it easily. Then adjust the flow until the flow and resulting pressure match a point on the pump curve.
  6. May 22, 2015 #5
    Hey Russ, thanks for your reply and vote of confidence! I am not quite sure what you mean exactly... wait I may have just Figured it out. Do you mean change the value of flow until the flow and pressure match a point on the flow rate vs pressure Graph associated with the particular pump I have? That sounds like a good idea. Would I do this assuming basically no bends or length in the darcy-weisbach equation? Or Should I just use Q=VA?

    Thanks again!
  7. May 22, 2015 #6


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    You can't just pick up any old pressure drop calculation and assume that it applies to your situation.

    Gases like oxygen don't always behave the same as liquids like water. Gases can become compressed as they flow, which is why the properties like velocity may be changing inside a pipe. Liquids like water can be treated as essentially incompressible, hence their velocity will be a constant as long as the flow rate and the pipe diameter are constant.

    Pipe networks can be analyzed in a fashion similar to analyzing an electrical network, but there is one important difference: unlike an electrical network where the voltage drop is proportional to the current and the resistance in a branch, in a fluid network, the pressure drop is proportional to the fluid velocity squared, so this complicates things a bit. The sum of the flows into a branch must equal the sum of the flows out, so this fact is equivalent to what happens in an electrical circuit.

    Before tackling a complex flow network, I recommend first that you get the bugs worked out of calculating flow thru a single pipe.
  8. May 22, 2015 #7


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    Correct method, but instead of assuming no bends, you should be able to find tables for "equivalent length" of the fittings/elbows.
  9. May 22, 2015 #8
    Ask around at work for a copy of "the Crane book." Officially it is called Crane Tech Paper No. 410, Flow of Fluids through Valves, Fittings, and Pipe. It is an orange book less than 1/2 inch thick and it has nearly everything you need to know. If you spend your entire internship reading and understanding that one book you will have had a very worthwhile summer.

    When I work away from the office I take my calculator and my Crane 410.
  10. May 22, 2015 #9
    Yeah, I came to the realization also that the oxygen system would behave Differently, I shouldnt have gotten so caught up on that aspect of the problem. I need to get my equation working in Excel and as long as there is nothing seriously wrong with the equation I posted I guess the issue may be coming down to units. Im in a conference all day today so I probably wont get back to the calculations until Tuesday. I will try and get the spreadsheet working with a simple system first and then move on. Right now My pressure drop is way too high, which is mainly driven by the velocity sqaured term. I will try and get that worked out soon. Thanks For your help!
  11. May 22, 2015 #10
    Ive had some trouble tracking down kb values For bends, but ill give this a try, this seems like a really good way to simplify things.
  12. May 22, 2015 #11
    Thank you! I will see if I can get my hands.on this Book. I've been looking for something like this.
  13. May 22, 2015 #12
    Ok let me run the units by you guys. Here's the equation:
    P1/P2 - lb/in2
    ρ - lb/in3 (0.0361 for water @ °60 F)
    v1/v2 - in/s
    f - Unitless. Calculated via: 0.316/Re1/4
    Re - Unitless. Calc. via: vD/(kinematic viscosity)
    Kinematic Viscosity - in2/s (0.001515 @ 60 F)
    L - in
    D - in
    z1/z2 - in
    g - 32.2 ft/s (Well I guess I found an error!!)
    Except this doesn't change my calculation because I didn't have any height change. Fooey.
    velocity I got from Q/A which is
    Q - in3/s (15.437 in^3/s for 4 gal/min)
    A - in2

    How does this look to you guys? Anything else I should be looking out for? Thanks again!
  14. May 22, 2015 #13


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    It's not clear what values of L and D you are using for the pipe.
  15. May 22, 2015 #14
    Well the L values will be whatever the length of straight pipe I have. for the D term, its a 0.93 inch inner diameter pipe.
  16. May 22, 2015 #15


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    BTW, g is the acceleration due to gravity, and its value is 32.2 ft/s2 (notice the units), or about 386.4 in/s2.

    My copy of Crane TP-410 lists the ID of a 1" Schedule 80 steel pipe as 0.957 in.

    For 1" clean carbon steel pipe, fT = 0.023, according to Crane, and is constant for fully turbulent flow, which for this size pipe is any Re > 700,000.
    (If you are not familiar with the Moody diagram, this is where it comes in handy). The friction factor for pipe depends on the material, the roughness, and the size.

    The kinematic viscosity of fresh water at 60 °F is 1.20828×10-5 ft2/s according to some tables I have for the properties of fresh and salt water.
    This kinematic viscosity would also be equal to 0.00174 in2/s.
  17. May 22, 2015 #16
    It looks like the values we have for acceleration and kinematic viscosity are the same, I think I just listed the value of kinematic viscosity at 70 F instead of 60. My 60 F value is the same as yours. The f value that I calculated was 0.024 or so, so basically the same as yours. The only thing though is my Reynold's number is no where near 700,000... I think it was 35,000. Also, the OD is 1" and the wall thickness is 0.035" so the inner diameter of this piping is 0.93". I think they're called CRS pipes (corrosion resistant steel). Which is basically stainless steel. I need to get myself a copy of the ellusive Crane TP-410...
  18. May 22, 2015 #17
    Well I found a PDF of the TP-410 (I am going to buy one though) and after about 2 minutes of reading it I think I found my problem. It says the Darcy-Weisbach equation in it's standard form is used to calculate feet of fluid, which you can then convert to inches, and then divide by 27.8 to get psi. I forget what the 27.8 conversion factor is, it's like the amount of inches water can raise per 1 psi or something. So I was assuming that through that calculation I was getting PSI out as an answer, but really it was feet of fluid. I don't have the spreadsheet with me (has to stay at work) otherwise I would test it out. But just by thinking about it out, it still seems like my answer would be off.

    I got a pressure drop of something like 30 psi, or what I thought was psi, so that would be 30 feet of fluid, or 360 inches, and then 13.33 psi or so. That's still way too much for a single 90 degree bend.
  19. May 22, 2015 #18


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    To be clear, I didn't say the Reynolds number for your flow was 700,000, just that for fully turbulent flow inside a 1" pipe, the Reynolds number had to be at least 700,000.

    Using a flow rate of 4 GPM and your velocity of 15.4 fps, the actual Reynolds No. for your flow is about 8200, which brings up another issue: the type of flow regime.

    In fluid flow, there is laminar flow, fully turbulent flow, and a transition zone between the two. In each flow regime, there is a different formula for calculating the friction factor. The friction factor formulas are listed below:


    The absolute roughness values for several different pipe materials are given in the table below:

    This article discusses the various friction factor formulae in more detail:


    The equation for friction factor calculation in the transition zone is known as the Colebrook equation. In its full glory, the friction factor f can be calculated only by using an iterative process. The article above gives several approximate versions of the Colebrook equation which avoid using iteration to calculate f.

    If you are basing your pressure drop calculations on a friction factor which is valid only in the turbulent zone, but you have laminar flow or flow in the transition zone, then your pressure drop calculations will be in error.

    The equation you have been using to calculate the friction factor is based on pipe materials which have no roughness whatsoever, i.e., a perfectly smooth material, which typically doesn't exist. The friction factors obtained from such an equation must therefore be considered approximations to the friction factor inside an actual pipe. In any event, this formula is valid only for flows with Re < 100,000.

    For pipes which are not smooth, once fully turbulent flow develops, the friction factor becomes a constant, independent of the Reynolds number of the flow.
  20. May 22, 2015 #19


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    IDK where this factor of 27.8 come from, but you can convert from feet of water to psi by a very simple calculation.

    1 cubic foot of fresh water weighs 62.4 lbs., which implies that the pressure on the bottom of this 1-foot cube of water is 62.4 lbs. per square foot.

    A head of 1 foot of water is also equivalent to a pressure of 62.4 lbs / ft2, which is equal to
    (62.4 lbs. / ft2) / (144 in2 / ft2) = 0.433 psi.

    To find the pressure in psi given so many feet of water head H, multiply the head value H by 0.433.
  21. May 22, 2015 #20
    Wow, thank you so much for posting all that great information. One thing though, the flow that I calculated in in3/s is 15.4, not the velocity. The velocity was something like 72 in/sec. But after just thinking about my math on that 72 in/sec seems like way too much... maybe thats where part of my error was coming from.

    Anyways, so if you calculate the Re # with 15.4 in^3/sec being the flow rate not the velocity it should be well above the transition region, but also well below the fully turbulent region.That's why I just used the smooth pipe turbulent flow equation. I will look at the TP-410 document and the link you posted tomorrow, I'm super tired. I really do appreciate all the help you have given me on this topic though, i'm really glad I came to this forum for help.
  22. May 23, 2015 #21


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    You are right. I made a mistake in calculating the flow velocity.

    Q = (4 gal/min * 231 in3 / gal) / (60 sec / min.) = 15.4 in3 / sec.

    ID = 0.93 in

    A = π × ID2 / 4 = 0.679 in2

    V = Q / A = (15.4 in3/s) / 0.679 in2 = 22.67 in / sec. = 1.889 fps [EDITED]

    Re = V × ID / ν = 22.67 in / sec × 0.93 in / 0.00174 in2/sec

    Re = 12,117 which is still in the transition zone for flow

    The absolute roughness ε for commercial steel pipe is 0.00015 feet = 0.0018 in

    The relative roughness ε / ID = 0.0018 / 0.93 = 0.0019

    Using the Haaland equation to estimate f

    1/√f = -1.8 log [(0.0019 / 3.7)1.11 + 6.9 / 12,117]

    1/√f = 5.577

    f = 0.0322

    Plugging this value of f into the Colebrook equation to check:

    f = 0.0324

    which should be the friction factor to use to calculate pressure drop.

    Calculating head loss for 100 feet of pipe:

    hL = f (L/ID)×V2 / 2g

    hL = 0.0324 (1200 / 0.93) 1.8892 / (2 ×32.2)

    hL = 2.32 ft of water = 1.0 psi for 4 GPM flow rate thru 100 feet of pipe.
    Last edited: May 23, 2015
  23. May 23, 2015 #22
    Awesome! Thanks man. The only number that was different for me was velocity, but if I do a hand calc. I get the same value as you do, so I will check my velocity calculation. 1 psi for a 100 foot pipe seems very reasonable to me.

    One thing that I still have a question about though is the Reynolds number and flow characterization. I read in a textbook that Re in and around the 2100 value are considered transition, but numbers above 2100 could be considered turbulent. I haven't really looked too hard at a Moody diagram and maybe it says otherwise there, but that is where I was getting my idea of the flow being turbulent and not transitioning. Either way though, it seems like the way you calculated everything is pretty straight forward so I may just stick with that. Thanks a ton!
  24. May 23, 2015 #23


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    There are various Reynolds numbers quoted for the end of laminar flow and the beginning of turbulent flow. Typical Reynolds numbers in the range of 2100-2300 are common. Since turbulence must be studied experimentally, it is difficult to say when laminar flow ceases and fully turbulent flow is developed, since you can't easily do a calculation and say with certainty the flow is fully turbulent here but not there.

    The modified Moody diagrams in Crane TP-410 on pages A-24 and A-25 show how the different zones of flow overlap
  25. May 26, 2015 #24
    Thanks for all the help SteamKing. I think I have the Excel spreadsheet working now after looking over everything you had helped me with. I did some test calculations like the one you posted of 100 feet of steel pipe and got the same result, along with a few other calculations I had done by hand.

    I am having trouble figuring out what some of my Kb values should be though. Like if I have a tee fitting that's not threaded and comes back at an angle (so not a 90 degree tee). Things like that seem to be kind of 'custom' and left out of the tables for standard K values.

    Also, if I am calculating the pressure drop from a pipe that goes straight up, do I calculate the pressure drop from the length of the pipe also? Or just from the height change? I want to make sure i'm not doubling up on the drop.

    Oh and one last question (for now anyways) if I have a set of parallel pipes that come from one pipe, deviate into two paths, and then come back into one pipe, is it correct to say the pressure drop will be equal in both paths even though one path has bends and elevation increase/decrease and the other path is straight?

    Thanks again!
    Last edited: May 26, 2015
  26. May 26, 2015 #25


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    This sounds like a Y-strainer. The pressure drop will depend on what's inside. In large Y-strainers, there is a basket-type filter which collects debris to keep it from flowing downstream. The PD values for these units depend on the strainer used and are usually furnished by the manufacturer.

    Here is a sample of some Y-strainers:


    The PD in this section will include the minor fitting losses (which also includes the friction due to the length of pipe) and the change in elevation.

    In general, the flow going into each branch from the supply line will apportion itself such that the PD in each branch will be the same, so that when the branches rejoin into a single line, you won't see the flow from one branch trying to back up into the other, i.e., you won't see the branches acting as a loop for the flow. This is usually where after adding up the minor losses from the fittings and accounting for the lengths of straight pipe, plus any changes in elevation, you must iterate to find the flow in each branch which gives the same PD, making sure that the combined flow in equals the combined flow out.
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