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What happens to the speed of a liquid in a pipe that is one of several branches that a larger diameter pipe splits into?
Here is a question in a practise booklet I want to answer:

"A horizontal tube with diameter $2d$ branches into two horizontal tubes, each with diameter $d$. The speed of liquid flow in the first tube is $v$.
What is the speed of liquid flow in each smaller tube?"
The answers are:
A. v/4
B. v/2
C. 2v
D. 4v

I have not studied physics or water flow, but I know that the volumetric flow rate is supposed to be the same in any cross-sectional area, and that, in order to maintain this, the speed must change. When diameter halves, area is quartered, this is a mathematical fact. Therefore the new flow rate is: (A/4) * V where A is area and V is velocity.
Thus the speed, V, clearly needs to be multiplied by 4 to maintain the same flow rate, to counter the effect of the Area being quartered, so 4V.

However, the answer to this question given by the booklet is 2V. I want to know why. Is it because the larger pipe was split into 2 identical-diameter branches so you divide 4V by 2? If it were split into 3 identical diameter prongs would you divide by 3?

If yes, my question is: Why? How can the speed of the water depend on the quantity of pipes? In order for that to happen, the water would need to "know" how many branches there are ahead of its arrival at that section, so it can share the speed among them, so to speak. If there were just 1 pipe that had half the diameter as the original pipe, would it not be 4V as I originally calculated? And if yes, it proves my point of the water needing to "know" how many branches there are. In other words, as a water molecule flowing through the pipe, as it passes through the smaller diameter pipe, it should not know whether the pipe it is in, is the only pipe with diameter d, or, if there is another pipe of the same diameter, all it knows is the diameter is now d instead of 2d, so it had better speed up (at least according to the rule that flow rate remains the same in any cross sectional area).

Based on this, I would therefore expect the answer to be 4V, regardless of how many branches there are. And that makes sense: If you had a single lane highway that split into 4 lanes later, the cars don't suddenly drive slower just because there are 4 lanes. Or, if a funnel had multiple drainage tubes, water poured into the cup on the top with a much larger diameter would flow out of each drainage tube at the same speed it would as if it had only one drainage tube.

I saw this question and in particular the comment by Joseph saying "The velocity is greater in the smaller pipes (since water needs to move faster through the smaller cross sectional area to constantly deliver the same amount of water)."

This further confirms to me that the answer to the question should be 4V and not 2V, because if it were 2V, it would be less than the calculated 4V, not greater.

Also, when I try to look at it intuitively, things get weirder.

"The velocity is greater in the smaller pipes (since water needs to move faster through the smaller cross sectional area to constantly deliver the same amount of water)".

This bolded statement is a fact, but how does this work intuitively? A pipe is not a live worker that is forced to produce or keep up with a certain demand: why does it need "to deliver the same amount of water"?

I have 2 ways of looking at it intuitively:

  1. If I had a pipe/tube and I pinched it, this decreases its diameter. If anything, as I pinch more and more, this would slow down the flow of water more and more until I pinch it completely shut and then no water flows at all. A real life example is in intravenous drips in hospitals: the tube that drains the fluids that goes into the patient is controlled by a manual dial that the tube is threaded through. Turning the dial one way makes it tighter, tighter means the fluid drips slower and slower, and turning it all the way shut, closes the tube, and no fluid is delivered.

  2. An alternative way to look at it intuitively is to say, as the diameter decreases, but the pressure is constant, the pressure is continually forcing the liquid out so that if the pipe diameter was decreased but pressure is the same, the pressure in the pipe would essentially increase (under the same amount of pressure, smaller things can withstand less pressure than larger things) and the water is forced to move out faster because there is less cross sectional area to escape from the same amount of pressure (to get the same amount of people out of a burning building though a tiny door, each person needs to move faster). The problem with this when applied to the pipe is that water, unlike people, do not know there is only one small door (one smaller diameter pipe).

So I have two approaches to this question: mathematically I would say the new speed in each of the smaller diameter tubes is 4V, but intuitively I have 2 hypotheses that oppose each other, one of them says the speed slows down by a quarter and be v/4 each, and none are the right answer.

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I know that the volumetric flow rate is supposed to be the same in any cross-sectional area, and that, in order to maintain this, the speed must change. When diameter halves, area is quartered, this is a mathematical fact.

Here is your problem. You calculated the area incorrectly. A circular pipe of radius $r$ has a total area $A=\pi r^2$. Two circular pipes of radius $r/2$ have a total area $2 (\pi (r/2)^2)=A/2$. Half the area implies twice the velocity, for an incompressible fluid.

In your body something similar happens. Your capillaries have a much greater total cross sectional area. So blood slows down as it passes from arteries to capillaries and then speeds up as it passes from capillaries to veins.

It is the total cross sectional area that matters. It matters because the fluid is incompressible, which means that the total volumetric flow is constant upstream and downstream. The total volumetric flow is equal to the total cross sectional area times the linear velocity, so as the total cross sectional area changes, the the velocity must change by the inverse.

A lot of your reasoning involved a confusion between “before and after” vs “upstream and downstream”. The volumetric flow is constant upstream and downstream for an incompressible flow. Turning the dial on an IV drip is a before and after comparison. The volumetric flow need not be constant before and after, so none of that reasoning holds.

The other incorrect reasoning was the statement that the fluid doesn’t know how many branches there are. In fact, it does. Not in the sense that the fluid is sentient, but in the sense that such information is communicated throughout the fluid. An incompressible fluid requires an arbitrarily large pressure to change its volume. This information gets passed, both upstream and laterally, as a pressure wave that travels arbitrarily fast throughout the fluid both to communicate the information and to move the fluid as needed to maintain the volumetric flow. Vehicle traffic is not an incompressible flow, and information is not communicated well, so it is not a relevant analogy.

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  • $\begingroup$ Thanks @Dale for this answer, I don't know why Stackexchange doesn't notify me. If it is the total cross sectional area that matters (I suspected the question could be answered by 4v divided by the number of branches, 2, so it is 2V), my question is why? As I said, the water doesn't know how many branches there are and what the total cross sectional area is... having 1 pipe 1/4 the cross sectional area would have the water flow through it anyway. At least intuitively, diverting things that move eg water flow, traffic, into different branches, doesn't automatically change the speed. $\endgroup$ Mar 19, 2023 at 10:52
  • $\begingroup$ @Ihavemanyquestions I have expanded my answer $\endgroup$
    – Dale
    Mar 19, 2023 at 12:32
  • $\begingroup$ Thanks for the updated answer, it is much clearer now except the IV drip example. Why isn't the wider diameter tube "before" the narrowing (due to the dial) not considered upstream while the narrowed section to be downstream? How is it different to if the tube was naturally narrower at the point of the dial, if the dial was not actually there? $\endgroup$ Mar 19, 2023 at 23:18
  • $\begingroup$ That isn’t the issue. The problem is “Turning the dial one way makes it tighter …”. That is a before and after comparison. If you leave the dial at any fixed setting then you see that the linear flow rate is faster downstream in the tubing than upstream in the bag. The ratio of the linear flow rate is indeed the inverse of the area ratio in the tube vs the bag. $\endgroup$
    – Dale
    Mar 19, 2023 at 23:28

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