1. Not finding help here? Sign up for a free 30min 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!

Hydraulics Basic Problem

  1. Aug 30, 2011 #1

    Femme_physics

    User Avatar
    Gold Member

    I just started hydraulics and this is the first problem I'm trying to solve. Please take that into consideration!

    1. The problem statement, all variables and given/known data

    Given Two Pipes
    Water is 50 degrees Celsius

    http://img31.imageshack.us/img31/9752/74467684.jpg [Broken]

    I wanna see if I understand this stuff correctly.

    YL = The pressure lost through both pipes
    L - Length of pipe
    ID - Inner diameter

    [ I'm not entirely sure what eA and eB mean - I think it says something about the quality of the surface? How smooth it is?]


    3. The attempt at a solution

    So I got these three formulas

    http://img11.imageshack.us/img11/963/yyyyyyj.jpg [Broken]

    I'm looking at my class notes and I got a few things missing.

    Ra and Rb are "Reynolds" number for the pipes, yes? I imagine it's constant.

    And the strange symbol at the bottom of Reynolds formula, that misshapen "V", that is a constant as well, yes?

    I hope it turns out there are only these equations. It seems like a fairly simple problem.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 30, 2011 #2
    Hello,

    I am just an undergraduate studying Civil Engineering/Physics so take this as 'guidance' not complete help if you wish.

    From Understanding Hydraulics, 2nd ed., Hamill

    The equations I find of relevance are:

    Q=V*A
    Flow is equal to Velocity x Area


    Re = V*D/(v)
    Re = Reynolds number
    V = Velocity (m/s)
    D = Diameter (m)
    (v) = 'funny looking v' (which is the greek letter 'nu') This is the kinematic viscosity

    f= 64/Re
    f = Darcy Friction Factor

    hf = (f*L*V^2)/(2*g*D)
    Head loss in a pipe.

    hf = head loss (m)
    f= Darcy Friction Factor (unitless)
    L= Pipe Length (m)
    V= Water Velocity in Pipe (m/s)
    g=acceleration due to gravity (9.81m/s^2)
    D= Pipe Diameter (m)


    hl = (0.5*(V2)^2)/(2*g)

    hl = head loss for sudden contractrion (m)
    V2 = Velocity of the liquid on the output side of the connection (lesser diameter) (m/s)
    g=acceleration due to gravity (9.81m/s^2)



    Initially I would Calculate the velocity of the water on either side of the joint using Q=A*V. Then using the velocity I would then follow the formula through which I have posted above.

    I am not entirely certain on the value of 'nu' but I believe this may be where the 50 degrees celcius may come into play (you may be able to look up a chart for 'nu' values from certain degrees of celcius).

    EDIT: from a website I found that for 50 degrees C 'nu' is equal to 0.553 *10^-6 m^2/s

    The final head loss will equal hf1 + hf2 + hl (head loss in pipe 1 plus head loss in pipe 2 plus head loss by the sudden contraction).

    Once again I am just an undergraduate so feel free to disregard any of this until a 'professional' can confirm or deny. ^^
     
  4. Aug 30, 2011 #3

    Femme_physics

    User Avatar
    Gold Member

    That means I have to find the area first. So I have to use the formula for a cylinder's area to find area I need.

    So area of a cylinder is: 2 x π x r2 + 2 x π x r x h
    So the radius of pipe A = 100mm, the height is 200mm


    The radius of pipe B = 50mm, the height is 100mm (the height is basically diameter, yes?)

    So now plugging for the area I get

    Area A = 188495.556 mm2

    Area B = 47123.89 mm2

    Is this all true so far?
     
  5. Aug 30, 2011 #4
    The area in the equation Q=A*V is referring to the 'cross-sectional area' and so just the area of the circle.

    This is pi*r^2 or pi*d^2/4 which is used more often at the uni I attend for circular cross-sectional area.

    Also be careful to convert the diameters to metres.

    Lastly looking through the information you have been provided the 'f' value or Darcy friction factory need to be calculated differently due to the 'e' values you were provided. Unfortunately I cannot remember a quick
    Formula for this. Maybe someone else can clear this up.

    EDIT: there is an equation for the Darcy friction factor taking into account your 'e'
    Values. Look up the 'Colebrook-White equation' and an estimated equation solving directly for 'f' by 'Moody'.
     
    Last edited: Aug 30, 2011
  6. Aug 30, 2011 #5

    Femme_physics

    User Avatar
    Gold Member


    Wait, I'm confused. I need the CROSS-SECTIONAL area?

    You mean I need the red part in figure (a)? Because I was under the impression I need the red part in figure (b), and that's what they gave me. Isn't that what matters to the flow formula?

    http://img855.imageshack.us/img855/7130/crossssssssssssssssssss.jpg [Broken]



    Should I always use meters in hydraulics?

    Hmm ok. Also, what book would you recommend?

    Thanks, I will :smile:
     
    Last edited by a moderator: May 5, 2017
  7. Aug 30, 2011 #6

    Femme_physics

    User Avatar
    Gold Member

    *bumps* I forgot to add that image (Edited in in the post above)
     
  8. Aug 30, 2011 #7
    Yes, you need the area of figure b. This I'd the cross sectional area. Your 1st reply included the surface area of a cylinder not the area of the circle in figure b.

    I would suggest converting to metres in all hydraulic problems (except maybe some special cases I cannot think of an example)

    Hopefully I have cleared a few things up.
     
  9. Aug 30, 2011 #8

    Femme_physics

    User Avatar
    Gold Member

    Ohhhhhhh you're absolutely right! I see what you mean now :smile: Thanks.

    You most certainly have! I hope you stick around, too!


    So, cross-sectional area of a cylinder, that's just pi x r squared !

    So, that'd equal for pipe A

    Inner diameter = 0.2m

    So, pi x 0.12 = 0.031416m2

    For pipe B

    Inner diameter = 0.1m

    So, pi x 0.052 = 0.007854 m2


    So both QA and QB = 0.0025 m3/s?


    It doesn't seem to make since because Q is volume supply, and the wider diameter clearly has a bigger volume supply. But, I'm only given one of the Q and I'm not told to which pipe it belongs. Confusing!
     
  10. Aug 30, 2011 #9
    There is no loss of water in the transfer of water from pipe 1 to pipe 2. So the volume flowing through one pipe must equal the volume flowing in the other pipe.

    The larger pipe has a larger supply for a constant velocity, but the velocity is not constant the volumetric flow is.

    If you consider the pipes separately. Assume that 0.0025m^3 per second is going into the left end of pipe A. That would mean that 0.0025m^3 is coming out the other end.
    Q in = Q out

    There are no holes in te system just the joint or change in pipe diameter.
    The Q out of the pipe A must now become Q in of pipe B, but as it is narrower the velocity of the water will need to increase.

    Does that help to see why the flow is constant in both pipes?
     
  11. Aug 30, 2011 #10

    Femme_physics

    User Avatar
    Gold Member

    I've handcopied your reply into my notebook and I'm on the way to the bus back home. I'll think about it on the way and will reply soon :smile: Thanks for sticking around!^^
     
  12. Aug 31, 2011 #11

    Femme_physics

    User Avatar
    Gold Member

    But there are different volumes flowing in each pipe, because one pipe is smaller.

    Isn't velocity constant in this case as well?

    Agreed

    What do you mean by "flow is constant"? The fluid's velocity?
     
  13. Aug 31, 2011 #12
    My quoting skills are lacking so I apologise.

    The volumes per second are the same in each pipe.

    If 0.0025m^3 per second is going in the left end then 0.0025m^3 must be coming out the right end. If this was not the case there would either be a build up of fluid in the pipe system somewhere or there would be a hole in the system leaking fluid. As this is the case QA = QB.

    The velocity (average velocity) in pipe A is constant. The velocity (average velocity) in pipe B is constant. But they are not equal to each other.

    By 'flow' I am refering to the volumetric flow per second, so in your case 'Q' which is equal to 0.0025m^3.

    I don't have my textbook on me so I cannot quote the name of the laws but you may need to do some reading on the formula of:

    Qin = Qout

    and how the formula will be applied to your problem.
     
  14. Aug 31, 2011 #13

    Femme_physics

    User Avatar
    Gold Member

    np. Just use quotation marks if you want --> ""

    Or in turn you can just use the actual quote code which is

    http://img196.imageshack.us/img196/7737/quottees.jpg [Broken]

    :smile:

    Oh, I see what you mean. So only a portion of QA flows, whereas all of QB flows, yes?

    Ah yes, that's what I thought :smile: in the smaller pipe - faster!

    Will do.




    SOOOOOO to solve

    For pipe 1


    Q = VA

    0.0025 = 0.031416V
    V = 0.0025/0.031416
    V = 0.0796 [m/s]

    Hmm... are my units correct? meters per second?

    For Pipe 2
    0.0025 = 0.007854V

    V = 0.318 [m/s]

    So I now need Reynolds number so I could later find "f" (my friction coeffecient throughout the pipe)...but I still don't know the kinematic viscosity (which is represented by the greek letter "nu" as you said).

    So I end up with two unknowns here (circled)

    http://img827.imageshack.us/img827/4531/raraaa.jpg [Broken]

    So once again getting a bit stuck. A bit more help, please? :smile:
     
    Last edited by a moderator: May 5, 2017
  15. Aug 31, 2011 #14
    No the whole portion flows in both. So in the smaller pipe the area goes down, but to make the Q constant, the velocity goes up.

    Exactly ^^. Once again Q stays constant, so if A goes down V must go up.


    Your units are correct.
    m^3/s = m/s * m^2 (if that makes sense)

    Seems like an awfully slow speed, but i guess it is what it is.


    The 'nu' value for kinematic viscosity for water can be found on a table if you know the velocity (this is where your 50degrees Celsius will come into play as the viscosity depends on temperature). There should be a table in your text somewhere or try 'Google-ing' kinematic viscosity of water.
     
  16. Aug 31, 2011 #15

    Femme_physics

    User Avatar
    Gold Member

    I fully understand :smile: it's all in the formula Q = AV after all


    Good, and I guess :approve:


    Are you sure that I need to know velocity? Because I see a table graph with kinematic viscosity on the Y axis and Temperature on the X axis and then I see a line that describes how kinematic viscosity changes base on temperature. Such as (just without the numbers):

    http://img847.imageshack.us/img847/5840/kinewaterok.jpg [Broken]

    So, shouldn't it be only temperature that matter?


    At 50 temp' it's our "mu" according to the table is 5.8 (it's hard to see the exact number, so it can turn up pretty inaccurate because it can be anywhere between 5.6 to 5.9). Can I use that?
     
    Last edited by a moderator: May 5, 2017
  17. Aug 31, 2011 #16
    Sorry you are correct I'm not sure why I wrote that bit on velocity. I wrote it from my phone so Iay not have proof read too well, sorry.

    Yes the viscosity only depends on the temperature. Just check the magnitude though it may be *10^-6 or something like that.
     
  18. Sep 1, 2011 #17

    Femme_physics

    User Avatar
    Gold Member

    http://img204.imageshack.us/img204/1996/temp11g.jpg [Broken]

    So its 0.553 x 10-6

    Solving for Reynolds Number


    http://img217.imageshack.us/img217/4820/reyrey.jpg [Broken]


    Since Ra > 2300
    This is turbulent flow.


    Now I can use the formula you gave me to find "f"
    So,

    fA = 64/29868.67 = 0.00214


    And now Rb....

    http://img43.imageshack.us/img43/9986/rbbb.jpg [Broken]

    f = 64/Rb
    fB = 0.0011

    So now I FINALLY can use this

    http://img849.imageshack.us/img849/2505/solvedylt.jpg [Broken]

    WHEW! THat was a long one! Thank you so much lachy^^ :smile:

    2 questions:

    1) Is this all correct?
    2) What are the units of YLT?
     
    Last edited by a moderator: May 5, 2017
  19. Sep 1, 2011 #18
    The Equation for te Darcy friction factor is a 'rough formula' u believe to be more accurate and incorporate your e values look up the Colebrook white equation, this might change your f value.

    Ylt should be in m, I am a little unsure on the formula you used for this. Isn't there a v^2 meant to be in there somewhere?
     
  20. Sep 1, 2011 #19

    Femme_physics

    User Avatar
    Gold Member

    You're right, I just read this:

    from here -> http://sites.google.com/site/excelsolveimpliciteq/

    Which does make sense since they've provided me with values for eA and eB. And I do have turbulent flow if my values are correct.

    But they're not actually giving me the formula. Where can I find it?

    You're right, I forgot to square my V's :smile:

    Thanks!
     
  21. Sep 1, 2011 #20
    http://en.m.wikipedia.org/wiki/Darcy_friction_fzactor_formulae [Broken]

    Look at the equation section ( I realize it is wikipedia but it should stillbe correct). As the 'f' is on both sides of the equation you may need to do some iterations or look for a simplified solution.
     
    Last edited by a moderator: May 5, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Hydraulics Basic Problem
  1. Hydraulic lift problem (Replies: 4)

Loading...