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Say that we have water flowing horizontally in a pipe of constant diameter. The continuity equation for incompressible fluids, which is a statement of mass conservation, guarantees that the flow rate will be the same throughout the pipe. Given that the diameter is constant it also guarantees that the velocity will be the same throughout.

Now say the pipe reaches a point where it is tilted. The tilted segment leads downhill and reaches another horizontal segment.

Water is now going downhill and so it is accelerating. Let us assume an ideal situation with negligible friction and hydraulic resistance. If the water is accelerating under gravity then velocity must change so I assume the diameter of the water flow itself must shrink and the water will "stretch" in order to maintain continuity. Is this actually the case?

What happens at the bottom when the water flow becomes horizontal again? Does it gradually "bulge up" at the bottom in order to reestablish the horizontal flow of constant diameter in the lower horizontal pipe? Assuming that the conservation of mass holds and continuity is preserved, the horizontal flow and speeds must be the same in both horizontal pipes.

One additional remark on this situation. What if the water were being pumped uphill in a pipe in the same configuration? Is it possible to pump water uphill and maintain a constant velocity in a constant diameter pipe? How can you demonstrate whether it is or isn't from first principles? It seems like water in the pipe would have to slow down when it encounters and uphill segment, but it can't pile up anywhere in the pipe either, and certainly doesn't expand, so it must also preserve continuity somehow I would assume? I feel like I am really missing something.

I apologize if this question seems trivial or silly. My backgrounds are in chemistry and in geology and I could not find any explanations on the internet regarding this. My fluid mechanics, physical hydrology, and hydrogeology textbooks also were of no help.

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  • $\begingroup$ Apply mass continuity - what goes in must come out. $\endgroup$
    – Solar Mike
    Aug 11, 2022 at 19:18
  • $\begingroup$ that's the problem though. How does it apply? My guess is that the water is accelerating in the downslope segment and so it thins out to maintain mass continuity. If you have constant diameter then velocity is the same everywhere, but in this instance you have acceleration downslope in a segment, then at the bottom it becomes horizontal again. What happens downslope and at bottom is what trips me up because I strongly suspect velocity must differ. $\endgroup$
    – MattGeo
    Aug 11, 2022 at 19:24

2 Answers 2

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Vacuum keeps the downhill water from accelerating. You would need to create a void in the fluid for acceleration to happen. Instead, you use the height change in the head loss equation to calculate flow. This is how siphons work.

If the pipe were large (like a drainage culvert) then you would model open channel flow (as Kamran says), which is a different animal, and which I didn't even learn in ME except that it exists. We leave that for the civils to float their concrete canoes in.

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  • $\begingroup$ could you just apply Bernoulli to this? I was sort of originally thinking that there would be no acceleration because as you lose potential energy going downhill by losing elevation you are gaining energy from greater pressure at the bottom, so that increase in pressure holding the fluid back offsets any velocity increase and hence, it's still constant. Is that a correct way to think about it? $\endgroup$
    – MattGeo
    Aug 12, 2022 at 2:13
  • $\begingroup$ @MattGeo there is no acceleration because there isn't. It's just a loss of potential energy. You're really overthinking it. $\endgroup$
    – Tiger Guy
    Aug 12, 2022 at 7:13
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If you have an inclined segment but with the same diameter, the flow stays constant because the flow speed in the entire pipe is constant.

Unless the slope has a drop of more than 10 meters, or there are obstructions in the pipe, then there will be a vacuum, bubbling, foaming, turbulence, and a discontinuity of flow.

The scenario you have in mind lends itself better to open channel flow.

In open channel flow, there is a hydraulic jump similar to what you think after a slope channel, turns horizontal.

Edit

After OP's comment.

Let's say we have a flow in a pipe with length l of which l/4 (horizontal projection) is sloping down by an angle of $\theta $ . Assuming again laminar flow and ignoring friction. and let's say the mass of water in the pipe is 1kg/ (meter length)

  • the mass of water is then $l*kg$ And the acceleration of the flow in the entire length of the pipe with a uniform speed along the entire length of the pipe is. $$\alpha= F/m=F/l.kg$$

$$F= l/4kg*g*sin(\theta)$$

$$\alpha= \frac{g sin(\alpha)l/4}{l}m/s^2=\frac{g sin(\alpha)}{4}m/s^2$$

But we know this initial acceleration will be soon canceled by the friction and there will be a steady flow according to the properties of the pipe.

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  • $\begingroup$ In the downslope segment the water has to stretch and can't actually fill the volume of the pipe though, right? It seems it would have to do so because it is accelerating but mass flow rate must remain constant. $\endgroup$
    – MattGeo
    Aug 11, 2022 at 18:37
  • $\begingroup$ I think you're thinking of flow running on the bottom of a pipe which will be considered open channel flow in most of the situations. $\endgroup$
    – kamran
    Aug 11, 2022 at 18:41
  • $\begingroup$ so how is the flow rate constant and continuity maintained if it is accelerating in the downslope but not in the other segments? velocity would have to be the same for constant diameter pipe to have continuity, this is what really trips me up $\endgroup$
    – MattGeo
    Aug 11, 2022 at 18:45
  • $\begingroup$ it is not accelerated on the downslope. The work of gravity on this segment will impact the entire flow. will modify my answer. please check it. $\endgroup$
    – kamran
    Aug 12, 2022 at 2:42

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