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Suppose that liquid is flowing through a tube with a constant velocity of, say 10 m/s. Within the liquid, there is a constant concentration of some substance, [S], lets say of 20 uM. Then, at some point, the tube splits into two branches. The original tube and the two branches are completely identical in size and volume and other properties.

My two questions are:

  1. What happens to the flow velocity?

    A similar question was asked here: Flow of liquid among branches and I think I can conclude the same for my question: the flow velocity is identical in the original tube and its two branches?

  2. What happens to the concentration of substance? Will it be identical or will it split?

    My gut feeling says it will split, as if there was only 1 particle it can go in just 1 of the tubes, so there will be 1/2 particle in each. However, as it is connected to the flow velocity it might be more complex than this.

I am particularly interested in two cases:

Case I: the tube is a rigid tube

Case II: the tube is a small blood capillary, hence very small and with non-rigid walls. here, flow is non-newtonian.

While writing this I came up with two final questions:

  1. IF anything changes, will it be something simple like a correction factor of 1/2 (because the tube splits in 2) or is it rather a complex mix of the effects of altered velocity and altered concentration?

  2. Are there cases for which would it be justified that nothing changes and total fluid flow (and concentration) would be similar in the original tube and its two branches? enter image description here

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If the diameter of the two tubes after the split is the same as the diameter of the single tube before the split, the flow velocities after the split will be half the original velocity. This is the only way that mass can be conserved. The concentration of particles in the two flows after the split will be the same as the concentration before the split.

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  • $\begingroup$ Your answer sounds very plausible and logic, however, how come it differs from the answer in physics.stackexchange.com/questions/31852/… ? where it states: " A1v1 = A2v2 Where A = area of pipe, v = velocity of flow. Assuming that all pipes are of equal size, they will all have equal flow following the above formula - divide right hand side by number of equal-sized pipes." $\endgroup$ – Phoneheads Nov 7 '18 at 15:48
  • $\begingroup$ You mean this one that they wrote for tubes of unequal size: A1v1 = ( A2v2 + A3v3 + ... + Axvx )? $\endgroup$ – Phoneheads Nov 7 '18 at 16:22
  • $\begingroup$ It differs because that it is incorrect. It should read $A_1v_1=nA_2v_2$. That would make it consistent with their subsequent equation (which is correct). $\endgroup$ – Chet Miller Nov 7 '18 at 16:23
  • $\begingroup$ Yes. That equation is correct both for equal tubes and unequal tubes. $\endgroup$ – Chet Miller Nov 7 '18 at 16:24
  • $\begingroup$ Thank you. Do you happen to have any sources where I can look a bit more into this? Or terms that I can search for? I am not in the field so I have a bit of trouble finding good literature on this topic, while I have the feeling it must be abundant. $\endgroup$ – Phoneheads Nov 8 '18 at 10:54

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