1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Water feature design - pump mathmatics

  1. Jan 9, 2012 #1
    Hey everyone, as part of a university project working on renewable energy sources in the garden I am performing some simple fluid dynamics analysis to calculate the pump power needed to feature in this garden water feature. I have designed the schematics of the feature and worked out the following:

    Flow rate = 0.0001875m^3/s
    Pipe length = 7.6633m of which 0.357m of length is a curve with a radius of curvature of 0.25m
    Fluid head (vertical distance from starting position to lifted position) = 1.65m
    Reynolds number of around 1500 (as this will keep it laminar which should help keep down energy usage.

    Now as far as i know the only thing i need left for a formula for pump power would be the friction created by the pipes, and i currently dont have a pipe diametre or material, so these are my questions:

    1. I did initially plan to use garden hose but i suspect it may be hard to find the coeffeciant of friction for this so it may be better to use copper pipping as i think there should be some good books in the library on measured values. If anyone knows of any specific books that would be a great help!
    2. What equation would i then use to calculate pump power as i have looked through my fluid dynamics notes and cant locate anything that will help me. Im assuming i may also need to use the friction coeffeciant to calculate a friction factor so if anyone knows of the correct equation that would be very much appreciated!
    3. I know from experiance that friction in bends can be difficult to calculate as its mostly done via experimentation but i have tried to keep the bend as smooth as possible to reduce friction so is there any decent equations or estimations i could use to account for the bend?
    Thanks in advance for any help i can get.
  2. jcsd
  3. Jan 9, 2012 #2
    First, derive a formula for diameter of your pipe as a function of Reynolds number, kinematic viscosity, and your volume flow rate. Kinematic viscosity of water is about 1e-6, so you'll get
    [tex]d = (187.5/R)*(π^2/16)[/tex]
    A Reynolds number of 1500 gives you a diameter of 7.7cm. That would be fire hose as opposed to garden hose -- undoubtedly overkill!

    I think if you start with garden hose (inside diameter ~2cm), and use a Moody diagram to find the friction factor (assume a smooth pipe -- that's good enough for such a small flow rate), you'll find that head loss due to friction will be much less than the 1.65m gravity head. I wouldn't worry about pipe curvature as small as you describe. In that case, pump power needs to be your volume flow rate times the head times a fudge factor to account for motor losses, pump inefficiency, and whatever small friction loses occur in the pipe. A fudge factor of 1.5 is undoubtedly sufficient, but make it 2 to allow for motor deterioration over the life of the garden.
  4. Jan 9, 2012 #3


    User Avatar
    Science Advisor
    Homework Helper

    tommy060289: You could initially try the garden hose. It might be fine. You would use the Bernoulli equation. Post the inside diameter (D) of your garden hose, in mm. Also, describe the inlet and outlet (entrance and exit) geometry of your system. What medium does your water exit into at the outlet, and what is the geometry of the outlet? E.g., is there any contraction at the outlet? Or is it just a straight-cut garden hose, exiting into air? Or is it exiting into water? What is the water reservoir depth at the inlet?

    For a pipe, for Re = 1500, I think the pipe flow friction factor (also sometimes called resistance coefficient) is f = 64/Re, if I recall correctly. However, you must compute Reynolds number (Re) for your garden hose, and you will see Re is not 1500, and the flow is not laminar. Therefore, you will need to look up the pipe friction factor (f), using the pipe relative roughness (e/D) and the Moody chart. At T = 20 C, water absolute (dynamic) viscosity is mu = 0.001 002 Pa*s. Kinematic viscosity is nu = mu/rho = (0.001 002 Pa*s)/(1000 kg/m^3) = 1.002e-6 m^2/s. A good typical value for absolute roughness (e) of typical garden hose is probably e = 0.010 mm.

    By the way, always leave a space between a numeric value and its following unit symbol. E.g., 0.25 m, not 0.25m. There is a template for homework questions. You must list relevant equations yourself, and show your work. And then someone might check your math.
    Last edited: Jan 9, 2012
  5. Jan 10, 2012 #4
    Hey Guys, thanks very much for the help! I think I'm getting there but Im still a little confused.

    To get my pump power Im going to use the equation:

    Pump Power = Volume flow rate * Head

    What head am I actually using to multiply. Would I need the total head (i.e gravity head (negative as rising) - Head loss)

    this I could use the Benoulli's equation with head loss:

    h1 + z1 + (u1^2/2g) = h2 +z2 + (u2^2/2g) + Head loss

    then to calculate Head loss use:

    = Rf * Q^2

    where Q is volumetric flow rate and pipe resistance (Rf) = (32Cf * L)/(g Pi^2 d^5)

    the only this I appear to be missing is the friction coefficient (Cf) which could be found from a moody chart if I have Re and relative roughness

    and Re can be calculated from (density * mean velocity * Diameter) / dynamic viscosity

    Does this look ok?

    now I need to figure out how the bernoulli equation fits on my design. Back to the old note book!

    nvn: for reference the pump is sat in a pool of water about 500mm deep that contains about 50 litres. It then pumps water along the explained path in the garden hose before rising through a small pond at a higher ground level where it will pass through a small pond (will this external pressure have much effect on the pipe as it passes through - I could always shield the pipe to stop the pressure getting to the garden hose) and the water exits into the atmosphere just above the water level
  6. Jan 10, 2012 #5


    User Avatar
    Science Advisor
    Homework Helper

    tommy060289: On the left-hand side of your Bernoulli equation in post 4, you need to add pump-output-power head gain, hp. (You can name this parameter anything you like, if you do not like the name "hp.")

    Head loss due to pipe flow friction is usually called hf. Your equation for hf would almost look correct, except I am not familiar with a 32 in the numerator of Rf. Should 32 in the numerator of Rf instead be 8? If 32 is changed to 8, then your hf would look like what I am familiar with. Are you sure about your 32? Where did it come from? (Or is your Cf equal to one-fourth of the usual friction factor, f?)

    Typically, a dimensioned diagram of your system might be helpful.

    What is the inside diameter (d) of your garden hose, in mm?
    Last edited: Jan 11, 2012
  7. Jan 11, 2012 #6
    hey nun,

    I have attached a system diagram of how I intend to set it out. Sorry its not the greatest diagram you will see all year but hopefully it gives the essential parameters so I could fit it to my Bernoulli equation (with Pump gain added of course - what would this be rated in, would it be the value of the power of the pump?)

    the total height the water is lifted is 2.2m if it is not clear from the diagram (the second bund is 2.1m from datum and pump in 0.25 above datum but then water must pass through 0.35m to clear the water in the second bund. I could remove the 0.35m pipe there and evacuate the water into the bottom of the bund but I expect the pressure would result in needing more power than evacuating into the air.

    Im not sure on the diameter yet as I plan to use an excel spread sheet to demonstrate how different pipe diameters that are available may allow me to reduce power consumption of the pump (the primary purpose of this evaluation) - but for the purpose of this example I think we could say 0.02 m (or 20 mm depending on how you like to work) internal diameter.

    The 32 in my Rf equation came from a equation sheet I have kept with my Fluid dynamics notes, but it is possible that the equation is not 'pure' and like you say may have accounted for Cf = 0.25 f. If I changed in to 8 from 32 would I be in the right area.



    P.S I have changed a couple of dimensions since I started the question so if the diagram doesn't look the same as the original post thats why, this is a finished dimension drawing now

    Attached Files:

    Last edited: Jan 11, 2012
  8. Jan 11, 2012 #7


    User Avatar
    Science Advisor
    Homework Helper

    tommy060289: I think d = 12.5, 16, or 19 mm garden hose is more common. Check your supplier for available sizes. It sounds like you want to initially assume d = 19 mm, right?

    Excellent diagram. I think pump output power is Po = rho*g*Q*hp, if hp is in units of meters, where hp = pump-output-power head gain. Add hp to the left-hand side of your Bernoulli equation.

    Yes, I would currently say, change 32 to 8 in your Rf equation. Double-check to see if I am correct here.

    By the way, always leave a space between a numeric value and its following unit symbol. E.g., 2.2 m, not 2.2m. See the international standard for writing units (ISO 31-0).
  9. Jan 12, 2012 #8
    Ill try different pipe sizes to see what kind of answers I come up.

    with regards calculating the friction in the bend, is there any formula for calculating it as I know a lot is done by experimentation but id really like to theoretically try different pipe bends to justify by design choice.

    with regards to my diagram would I have:

    h1 = not really sure would I specify 0.25 m as I am saying the inlet/outlet is half way up water level (although I am likely to move that slightly
    h2 = 0.32 m
    z1 = 0 m
    z2 = 2.1 m or would it be 2.2 m as the vertical distance from the pump out let to the bottom of the second bund is 1.85 m and then there is another 0.35 m from bund 2 to where the water surfaces

    also, since I want the flow rate to be constant Im assuming u (=Q/A) will be the same either side.

    Once again thanks for all the help
  10. Jan 13, 2012 #9


    User Avatar
    Science Advisor
    Homework Helper

    tommy060289: Yes, h1 = 0.25 m; that is correct. No, h2 = 0; the water in the second reservoir has no effect on the water inside the pipe that passes through the second reservoir. Yes, z2 = 2.2 m. For d = 19 mm, look up pipe bend head loss in a plot or table, for now.
  11. Jan 15, 2012 #10
    Hey nvn, I managed to use the tables to calculate an equivalent length for the pipe bend (it comes out at roughly at 3.5 times a straight pipe length) so thanks very much for the pointer!

    Is h2 = 0 because it is evacuating into the air therefor there is no head of pressure pressing down towards the pump?

    Also, im assuming my u values will be equal to each other as I want the flow rate to be the same from the pump to the exit (and the pipe is the same diameter all the way)

    Also, im calculating the dynamic viscosity in the pipe which is h*Rho*g*d^2/(32L*mean velocity). h would again be 0.25m as that is the head at the pump end?

    Finally, for the friction factor equation Rf = 8 CfL/gPi^2d^5 where I originally thought it was a value of 32 instead of 8, your not getting it mixed with 8 K/gPi^2d^4 are you as I have been looking on my formula sheet and where I had 32 that is listed for pipe resistance and the 8* one is listed for a section change. Im certainly not suggesting your wrong but I'm just a little confused as to which to use as I can't find evidence of either on the internet so I'm assuming they are both derivations of other equations. If you could shed light on this it would be much appreciated!
    Last edited: Jan 15, 2012
  12. Jan 16, 2012 #11


    User Avatar
    Science Advisor
    Homework Helper

    tommy060289: Your pipe bend calculation currently looks grossly in error. If your total pipe length is 7.685 m, and your pipe diameter is d = 19 mm, then I think your equivalent pipe length, including your pipe bend, should be ~7.99 m, which is 1.040 times longer than a straight pipe. Try again.

    h2 = 0 because the water is exiting into air. Flow velocity, u, is constant because pipe area is constant. I did not understand your statement about dynamic viscosity. I am not familiar with that equation.

    I checked your Rf equation again, and it should have an 8 in the numerator, not 32, if Cf is the friction factor (often called f) taken directly from the y axis of the Moody chart. Here is evidence; your form of the hf equation is at the bottom of this Darcy-Weisbach page. I do not know why you have 32.
    Last edited: Jan 16, 2012
  13. Jan 17, 2012 #12
    Hey nvn, what adjuster did you use for the pipe bend. I used the factor 4-8d which is for 90 degree swept bend (i know mine isnt a full 90 degree but i felt it was the likely to be the most accurate) which gave me a total equivilent pipe length 8.639 m - formula obtained from http://www.aquatext.com/tables/frict-wat.htm )

    i did the equations for a 19mm pipe and got a pump output of 72.5W which doesnt seem to bad.
  14. Jan 17, 2012 #13


    User Avatar
    Science Advisor
    Homework Helper

    tommy060289: Your url for the pipe bend does not seem too credible. Head loss due to a 90 deg pipe bend is a function of bend radius (r) and pipe diameter, not just pipe diameter. You have r/d = 0.25/0.019 = 13.16; therefore, I got an additional pipe length, due to the bend, of 16.1*d = 0.3059 m, not 8*d. Therefore, using what you have posted in post 1, the total equivalent length of the pipe bend is 0.357 + 0.3059 = 0.6629 m.

    Your attached diagram says the total pipe length is 7.685 m, but when we compute it ourselves based on what you have posted, the total pipe length is 7.679 m. Therefore, the total equivalent pipe length, including your pipe bend, is Le = 7.679 + 0.3059 = 7.985 m, not 8.639 m.

    There is an inlet head loss of hi = 0.5*ki*(v1^2)/g, where ki = 0.5; and an exit head loss of zero.

    For a 19 mm pipe, using all the data you posted, I currently get a pump output power of Po = 4.055 W. Your Po value is nowhere near my current value. Show your work, if you wish, and then someone might check your math. Also, please see the last paragraph of post 7.
  15. Jan 18, 2012 #14
    Hey nvn, I don't really understand the formula you have you used the calculate the equivalent pipe length. as 0.25/0.019 is 1316, where has the 16.1*d come from?

    Also, for my pipe length, I worked out where the straight pipe section intersects the curve (as it doesn't go a full 90 degrees) and then added a join, is this how you did it? I expect the 0.006 m difference is a rounding error.

    This is my working out, I have currently used my length for the pipe as I didn't want to use yours till I understood it if thats ok. even with the difference it doesn't explain our two large pump differences.

    First I needed a value for friction coefficient (Cf) from the moody chart so I needed a Reynolds Number and Relative roughness. I used relative roughness 0.0000015 and divided my 0.019 to get a value of 0.000079.

    I then used Re = (Roe * u * d)/dyn viscosity to get Re, for u I used 3.467 (u = 0.0009831/Pi*0.0095^2) and for den viscosity I used the formula:

    = (h * Roe * g * d^2) / (32 * L (at value 8.639 m) * u)

    giving dyn as 0.000924 Ns/m^2 giving an Reynolds Number of just over 70,000.

    These two values on a moody chart give a Cf value of 0.02 that I then put into the friction factor equation:

    R = ( 8 * Cf * L (again 8.639) )/(g * Pi^2 * d^5) giving 5,765,629 m^5/s^2

    I then calculated Head loss (Hl) using hl = R * Q^2 to give me a head loss of 5.57 m

    I then used this in my bernoulli equation and rearranged to get:

    hp = 2.2 + 5.57 - 0.25 (the values missing are either 0 or the u values which were both equal and cancelled each other out)

    with hp = 7.52 m

    I did Roe * g * Q * hp to get the 72.5 Watt Pump Power

    I can see that I need to add in the inlet head loss into my total head loss but this would require more pump power.

    How does this method differ from yours?

    Much appreciated

  16. Jan 18, 2012 #15


    User Avatar
    Science Advisor
    Homework Helper

    This is vastly different from what you told us in this thread. Is this a typographic mistake?
  17. Jan 19, 2012 #16
    My apologies nvn, i adjusted the design before i did the arrangement drawing and as such the flow rate Q became 0.0009831 m^3/s. i thought I had posted that but I must have forgot. Everything else is correct so I expect that adjustment shoudl account for the majority of the difference.

    The other thing is the headloss at inlet, did you get the your value for K from a typical values table (ie k value for head loss at inlet pump = 0.5?)
  18. Jan 19, 2012 #17


    User Avatar
    Science Advisor
    Homework Helper

    tommy060289: I used an additional pipe length, due to the bend, of 16.0*d = 0.3040 m, which came from a graph in a fluid mechanics text book.

    I used a total pipe length of Lt = 7.679 m, which was computed from your diagram dimensions, using geometry and trigonometry. I used a total equivalent pipe length of Le = Lt + 16.0*d = 7.983 m.

    Your calculation of dynamic (absolute) viscosity does not make sense to me. Dynamic (absolute) viscosity is a property of a fluid, which you generally look up in a table. For water, it is only a function of temperature, T. It is not a function of the parameters you listed in post 14. I listed dynamic (absolute) viscosity already, in post 3. Your value happens to be close to the correct value at T = 23.5 C. Close enough.

    You must not divide relative roughness (e/d) by diameter; you must divide absolute roughness (e) by diameter, to obtain relative roughness (e/d). I already gave you an absolute roughness (e) for garden hose, in post 3. In post 14, you are using an absolute roughness for drawn copper tubing, which is slightly too smooth, compared to garden hose. If you do not like e = 0.010 mm in post 3, then you can also use e = 0.005 mm for garden hose, if you wish, but not 0.0015 mm. Therefore, e/d = 0.000 526, not 0.000 079.

    Yes, my inlet head loss coefficient, ki, came from a typical values table.

    Putting all of the above together, I got a total head loss of hL = 5.919 m. Therefore, I got, Po = 75.891 W.
  19. Jan 21, 2012 #18
    thank you so much nvn,

    I don't think I could have done this without all your help! It is very much appreciated! Now all I have to do is find a book with the table in for pipe bends so I can reference it and I'm finished with my pump calculations!

    Once again a very big thank you!
  20. Jan 23, 2012 #19


    User Avatar
    Science Advisor
    Homework Helper

    tommy060289: The following equation gives resistance due to a 90 deg bend in circular pipe with fully-developed turbulent flow at bend inlet, where y(x) is dimensionless equivalent additional length due to bend (Leb/d), and x is bend relative radius (r/d), where r = pipe centerline bend radius, d = pipe inside diameter, and Leb = equivalent additional length due to bend. Notice, the first two quantities on each line, below, are boolean quantities, whose value is 1 if true, 0 if false.

    y(x) = (x≥1)(x<5.5)[(1.981e10 + 6.168e8*x)/(1 + 1.518e9*x - 1.853e8*x^2)]
    + (x≥5.5)(x≤20)(-7.722 + 4.4685*x - 0.35279*x^2 + 0.014989*x^3 - 2.6319e-4*x^4).

    E.g., if r = 250 mm, and d = 19 mm, then x = r/d = 13.16. Therefore, Leb/d = y(13.16) = 16.25. Therefore, Leb = 16.25*d = 308.8 mm. The actual length of the bend is Lb = 0.5*pi*r = 392.7 mm. Therefore, the total equivalent length of the 90 deg pipe bend is, Le = Lb + Leb = 392.7 + 308.8 = 701.5 mm.

    tommy060289, you can see I was slightly inaccurate in post 17, where I said Leb = 16.0*d = 0.3040 m. Instead, I should have used Leb = 16.25*d = 0.3088 m, in post 17.
    Last edited: Jan 23, 2012
  21. Jan 24, 2012 #20


    User Avatar
    Science Advisor
    Homework Helper

    tommy060289: y(x) in post 19 is for a 90 deg pipe bend. If pipe bend angle, theta, is not 90 deg, you can multiply y(x) in post 19 by the following equation, y2(theta), per post 3688710, to obtain the resistance for pipe bend angles other than 90 deg, between 0 and 180 deg.

    y2(theta) = 0.03464*theta^3 - 0.28803*theta^2 + 1.0036*theta,
    for 0 ≤ theta ≤ pi radians,​

    where y2 = angle factor, and theta = pipe bend angle, in radians.

    You should find that, at theta = 0.5*pi rad (90 deg), the angle factor (y2) comes to 1.00, and at theta = pi rad (180 deg), the angle factor comes to 1.384.

    E.g., if r = 250 mm, d = 19 mm, and theta = 1.42803 rad (81.82 deg), then y2(1.42803 rad) = 0.94668. Therefore, Leb = y2(1.42803 rad)*[y(13.16)*d] = 0.94668(16.25*d) = 15.384*d = 292.3 mm. Therefore, the total equivalent length of the pipe bend is, Le = Lb + Leb = theta*r + Leb = (1.42803 rad)(250 mm) + Leb = 357.0 + 292.3 = 649.3 mm.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook