How to Calculate Flow Distribution in Parallel Pipes. Fluid Mechanics

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  • Опубліковано 17 січ 2025

КОМЕНТАРІ • 49

  • @cbh598cbh
    @cbh598cbh 11 місяців тому +3

    Thank you! This helped me a lot. I needed to know this for a case when there was a pressure difference

  • @rezanoormohammadi6914
    @rezanoormohammadi6914 8 місяців тому +2

    It was very interesting, Iranian professor👌

  • @itsfikree
    @itsfikree 7 місяців тому +1

    Wonderful explanation sir 🤩, relieved some pressure 😁

  • @tipsandlearnings9421
    @tipsandlearnings9421 Рік тому +2

    Well done 👍 Sir

  • @DespoinaAnastasopoulou
    @DespoinaAnastasopoulou Рік тому +2

    Thank you Prof. Could please explain what would happen if the two branches wouldn't be connected and had different lengths?

    • @professorbehrang
      @professorbehrang  Рік тому

      In this problem, they do have different lengths but do you mean they are not connected on both ends? Or they are only connected on one end? Can you elaborate?

    • @robertamarinei170
      @robertamarinei170 Рік тому

      @@professorbehrang I have in fact the same curiosity. Say pipes A and B have arbitratry lengths, LA and LB, and they simply have open ends?

    • @professorbehrang
      @professorbehrang  Рік тому

      @@robertamarinei170 Open end would be a different problem. More info on the pressure on the open side is needed.

  • @JamesRogerLackore
    @JamesRogerLackore Рік тому +2

    I have the same question as one of the other commenters. To calculate J, you need to know both FA and FB. You glossed over determining this in the video. Unless I am mistaken determining these values require you to know the velocity in each branch. But we don't know the velocity in each branch because we don't yet know the flow in each branch. Seems like a chicken and egg problem. Can you explain what I am missing?

    • @professorbehrang
      @professorbehrang  Рік тому +1

      No, You are not missing anything. We need to know friction factors to be able to solve this problem. However, there are 2 ways to solve the problem if you don't know the friction factors. Case 1: sometimes you can guess the values of friction factors correctly. For example, if the pipes are really rough and the flow is fully turbulent f values on the moody chart become independent of Reynolds number and can be picked from the Moody chart.
      Case 2: You will need to add 2 equations for friction factor (Colebrook equation) to your set of equations in a code or something and have it iterate until it converges.

    • @onsite2406
      @onsite2406 3 місяці тому

      @@professorbehrang I'm having this same issue, where Qtot = 20 gpm, branches A and B are the same diameter (1") and material, but different lengths. At such a low flow rate, can Fa and Fb be assumed to be equivalent?

    • @professorbehrang
      @professorbehrang  3 місяці тому

      @@onsite2406 f_a and f_b are not necessarily the same. I responded in my last comment. There are 2 cases.

  • @ming91947
    @ming91947 7 місяців тому +1

    hello prof! nice video. can you help me understand again in a case where the 2 pipe in path A and B are made of different material and size will have the same head loss again?

    • @professorbehrang
      @professorbehrang  7 місяців тому

      As long as the process is steady state and then both rejoin again yes!

  • @claudiosanchez4479
    @claudiosanchez4479 3 місяці тому

    Nice video professor! Thank you!
    I just noticed that you used the Darcy-Weisbach for the calculation of head loss of pipes. I would just like to ask, would this still be applicable if I would use the Hazen-Williams equation for head loss?

    • @professorbehrang
      @professorbehrang  3 місяці тому +1

      @@claudiosanchez4479 the head loss in both branches is the same. Whatever valid equation you have you can use. I think the equation you mentioned is an empirical equation used for water systems.

    • @claudiosanchez4479
      @claudiosanchez4479 3 місяці тому +1

      @@professorbehrang I see. Thank you very much professor! :)

  • @sharp_guy2310
    @sharp_guy2310 7 місяців тому

    Good explanation sir. I expect a video from you for same parallel flow with valves of different Cv value installed in 2 pipes. Also valves installed in series too if possible.

  • @aravindkumar-vn7ox
    @aravindkumar-vn7ox Рік тому +1

    Thank you Prof. Thats really a nice video. Just curious to know the gadgets you have used for the video. They are really awesome. If its sharable, what kind of hardware and software you are using for this?

    • @professorbehrang
      @professorbehrang  Рік тому +1

      Thanks for watching. I used Microsoft oneNote with a Microsoft surface for this video.

    • @professorbehrang
      @professorbehrang  Рік тому +1

      For my newest videos I have been using Microsoft Surface Studio

  • @blessingmuningwi7353
    @blessingmuningwi7353 Рік тому +1

    if i add say a turbine in one of these branches is hLA still equal to hLB?

    • @professorbehrang
      @professorbehrang  Рік тому

      If you add a turbine in one of the branches, let’s say branch A, then: h_L_A + h_turbine = h_L_B.

  • @wang-ni8tq
    @wang-ni8tq Рік тому +1

    would this apply if you have 3 or 4 more pipes in parallel?

    • @professorbehrang
      @professorbehrang  Рік тому

      The principle is no matter how many pipes you have in parallel after the whole reaches becomes steady state, the head loss in every pipe becomes the same.

  • @robertamarinei170
    @robertamarinei170 Рік тому

    Thanks for the nice explanation!
    One thing I do not understand is why the pressures (P2) in the two branches are equal when you write the bernoulli equation for the two branches?

    • @professorbehrang
      @professorbehrang  Рік тому

      P2 is the same because there’s only one point that has the pressure of P2.

  • @MathHandsOnwithPython
    @MathHandsOnwithPython 2 роки тому

    Thanks for the nice video Prof. Please prepare a video for the velocity of fluid flow in a curved tube

    • @professorbehrang
      @professorbehrang  2 роки тому +1

      Even if the tubes are curved, it won’t change anything. The flowrate distribution will be exactly the same as this one.

  • @shahin.kimiya
    @shahin.kimiya Рік тому +1

    very well

  • @TP-iv3os
    @TP-iv3os Рік тому

    Professor Behrang,
    I have a 24" line that I need to measure flow. The pipe is liquid full, the pressure varies but is measured and always known, temperature is pretty constant but is also measured and known. The current method being used to measure the flow is an ultrasonic meter which is not working properly due to the fluid causing internal scaling. The scaling has a negative effect on the meter. Taking the line out of service to install an inline meter such as a mag meter or an orifice plate is not an option. I am contemplating hot tapping the pipe so I can install a small 2" lined pipe with a Coriolis meter. My question is can I simply tap and install a 2” line in parallel to the 24” line, and measure the flow in the 2” and be able to determine the total flow in the 24” line?DeDear Professor Behrang,
    I have a liquid-filled 24" pipe that requires flow measurement. The pressure fluctuates but is always known, while the temperature remains relatively constant and is also measured and known. Currently, we are using an ultrasonic meter, but it is not functioning correctly due to internal scaling caused by the fluid, which has a negative impact on the meter. Unfortunately, we cannot shut down the line to install an inline meter such as a mag meter or an orifice plate. Therefore, I am considering hot tapping the pipe to install a small 2" lined pipe with a Coriolis meter. My question is whether I can tap the 24" line and install a 2" line in parallel to it. Can I measure the flow in the 2" line and determine the total flow in the 24" line?
    Thank you.ar Professor Behrang,
    I have a liquid-filled 24" pipe that requires flow measurement. The pressure fluctuates but is always known, while the temperature remains relatively constant and is also measured and known. Currently, we are using an ultrasonic meter, but it is not functioning correctly due to internal scaling caused by the fluid, which has a negative impact on the meter. Unfortunately, we cannot shut down the line to install an inline meter such as a mag meter or an orifice plate. Therefore, I am considering hot tapping the pipe to install a small 2" lined pipe with a Coriolis meter. My question is whether I can tap the 24" line and install a 2" line in parallel to it, then measure the flow in the 2" line and determine the total flow in the 24" line?
    Thank you.

    • @TP-iv3os
      @TP-iv3os Рік тому

      If I measure the flow in the smaller line (Qa) then I should be able to determine the total flow by using the following equation Qa + Qa/J. WOuld this be the correct understanding?

    • @professorbehrang
      @professorbehrang  Рік тому

      That’s true. But when you add the 2” pipe, the flow in the 24” pipe would not be the same as when the 2” pipe didn’t exist.

  • @tipsandlearnings9421
    @tipsandlearnings9421 Рік тому +2

    Also make video in case of more than 2 parallel tubes

    • @professorbehrang
      @professorbehrang  Рік тому

      Good suggestion.

    • @davemorgan8349
      @davemorgan8349 Рік тому +1

      @@professorbehrang I thought that was left as an exercise for the reader ;)

    • @professorbehrang
      @professorbehrang  Рік тому

      @@davemorgan8349 All the head losses will be the same in every branch.

  • @ninara2317
    @ninara2317 Рік тому

    Thanks doctor! How can I contact with you please?

  • @karthickarya2419
    @karthickarya2419 Рік тому +1

    how to know friction in each pipe without velocity

    • @professorbehrang
      @professorbehrang  Рік тому +1

      Good question. In some cases you can use a guess value: For example, if we know that the flow will be fully rough and fully turbulent, the friction factor on moody chart levels off and becomes independent of velocity. That value can be picked from the moody chart.

  • @wang-ni8tq
    @wang-ni8tq Рік тому

    would this be the same if you have 3 or 4 parallel pipes?

    • @professorbehrang
      @professorbehrang  11 місяців тому

      If it is steady state the head loss in every branch becomes the same. But those pipes need to be connected to the same pressure nodes on both ends.

  • @TP-iv3os
    @TP-iv3os Рік тому

    I thought that Q total = Qa + Qa/J.
    Therefore, knowing Qa, I could calculate Q total.

    • @professorbehrang
      @professorbehrang  Рік тому

      That's true but after the process becomes steady state . Adding another branch will change the total steady state solution if the upstream pumping power stays the same. A new branch after becomes steady state will change the total head loss. In other words, without an added branch you have one flowrate in the pipe. Once you add the branch you won't have the same flowrate in the same pipe. You might think it won't matter because you are looking for the total flowrate. But if the pumping power upstream stays the exact same, adding another branch will also change the total steady state flow rate of the system.

  • @swaaaarts4
    @swaaaarts4 Місяць тому

    My professor did not taught this properly