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A steady stream of water falls down a pipe. Assume the flow is incompressible. How does the pressure vary with height in the following stream?

The answer to this problem as stated by my book was that the pressure is same at all points in the stream. I can understand that the pressure would be atmospheric and not vary with height since the liquid is accelerating freely downwards. However, the velocity of water stream is increasing as it falls down, so applying the Bernoulli's theorum there should be variation of pressure. So why is this not the case here?

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    $\begingroup$ if your pipe has the same radius, how can the velocity in the completely file pipe vary with the hight? $\endgroup$
    – trula
    Commented Aug 26, 2021 at 15:34
  • $\begingroup$ Is it not accelerating though? Or do we assume that in "steady state" there is a continuous flow of water, hence it is not accelerating? $\endgroup$
    – marks_404
    Commented Aug 26, 2021 at 16:02
  • $\begingroup$ @trula Also if I replace the pipe system with a tap, would pressure differ in that case? $\endgroup$
    – marks_404
    Commented Aug 26, 2021 at 16:03
  • $\begingroup$ I think you are forgetting about the potential head term (hdg) in while applying Bernoulli's equation as the two points will not be at the same height $\endgroup$
    – Harshal Chaware 3791
    Commented Aug 26, 2021 at 16:14
  • $\begingroup$ @HarshalChaware3791 That really doesn't make a lot of sense. $\endgroup$
    – Gert
    Commented Aug 26, 2021 at 16:39

1 Answer 1

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Assuming no pump at the top of the pipe and no flow restriction (no valve e.g.) at the bottom then the pressure at the top and bottom is identical (atmospheric pressure) and Bernoulli's equation reduces to:

$$\frac12 \rho v_1^2+\rho gh_1=\frac12 \rho v_2^2+\rho gh_2$$

There's no pressure gradient at all.

Assume (for argument's sake) that: $v_1 \approx 0$, then:

$$v_2\approx\sqrt{2 g\Delta h}$$

which is of course the classic formula for free falling objects.

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  • $\begingroup$ I understood what I did wrong, Thanks! $\endgroup$
    – marks_404
    Commented Aug 26, 2021 at 16:53
  • $\begingroup$ You're welcome! $\endgroup$
    – Gert
    Commented Aug 26, 2021 at 16:57
  • $\begingroup$ How do you know the pressure is not different inside the pipe? $\endgroup$
    – user65081
    Commented Aug 26, 2021 at 21:59
  • $\begingroup$ @Wolphramjonny Since as the pressure in the end-points is equal it would have to be a 'parabolic' type of profile? What would cause that? I can't see it... $\endgroup$
    – Gert
    Commented Aug 26, 2021 at 22:35
  • $\begingroup$ @Gert, I dont think it will happen, but I do not think it is that obvious from the math. $\endgroup$
    – user65081
    Commented Aug 26, 2021 at 22:43

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