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There is a classic problem by which a fluid having a uniform velocity profile enters a tube. The pressure at the outlet is atmospheric, and the dimensions of that tube (diameter and length) are known. Pressure and velocity fields can be quite easily be calculated using a CFD tool. By reviewing the results it seems that there is a pressure difference between the inlet and outlet, which is very reasonable since there is a flow resistance to overcome.

However, how the pressure in the inlet is generated? If I wanted to allow this kind of flow, I would need to use a pump generating this pressure.

In the case the tube is just laid in a field of moving fluid - what would happen?

I must miss something...

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Part I (One doesn't take into account viscosity)

One way to generate needed pressure difference will be to use some kind of pump as shown on the first picture:

enter image description here

$$\Delta P = P_{\text{pump}} - P_{0}$$ where $P_{0}$ stands for atmospheric pressure. Note however that in this setup you will run out of fluid after time $\frac{L}{V}$ (unless you modify it)

You can also consider another setup:

enter image description here

$$\Delta P = mgh $$ Here fluid itself will generate pressure difference. However this it may be considered to be constant only within short amount of time. After some time pressure difference will decrease.

Finally you can of course place your pipe into stream of fluid. In this case you will just have to fix it so it doesn't move. In this case this pipe will be somewhat secondary in the sense that it will have nothing to do with generation of pressure difference. It may be produced in external liquid by any of aforementioned methods.

enter image description here

Part II (Laminar flow of viscous incompressible fluid at constant temperature)

Flow produced the same way as in previous case enters pipe while having uniform velocity profile. We imply so called no-slip condition which states that speed of fluid layer near the pipe surface equals zero. Due to presence of viscosity($\approx$ friction between the molecules of fluid) aforementioned layer starts to slow down the one nearest to it and so on. This forces uniform velocity profile to change itself as shown in the picture:

enter image description here

After some time velocity profile reaches certain(parabolic) form and doesn't change afterwards(fully developed velocity profile).
Now let us analyze in details a ring-shaped differential volume element of radius r, thickness $dr$ and length $dx$ oriented coaxially with the pipe, as shown in the picture:

enter image description here

Since speed of every such volume is constant we state that sum of forces acting on it equals zero. From one hand there are $\tau_{r+dr}$ and $\tau_{r}$ - shear stresses due to interaction of the nearest layers of the fluid. Since $\tau$ has radial dependence there appears disbalance in forces acting on a given volume. This generates the same(up to sign) pressure difference in x direction.

$$2 \pi r dr P_x - 2 \pi r dr P_{x+dx} +2 \pi r dr \tau_r - 2 \pi r dr \tau_{r+dr} = 0$$

Once we take limit $dr,dx \to 0$ we obtain:

$$r \frac{dP}{dx}+ \frac{d(r\tau)}{dr} = 0$$

If one takes $\tau = \mu(du/dr)$ where $\mu$ is some constant one can also show that $\frac{dP}{dx} = \text{const}$

PS: Aforementioned explanation is heavily based on the following article. See 8-3 and 8-4.

Hope it helps!

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  • $\begingroup$ Thanks for your answer, Yaroslav. Lets examine the 3rd case you suggested - the pressure seems to be the same in both ends of the pipe. But since it has a resistance, there should be a pressure drop. Where does it comes from in that case (in the first two cases you generated it by two different mechanisms)? $\endgroup$ – Yaniv Ben David Jul 21 '18 at 22:07
  • $\begingroup$ +1 @YanivBenDavid Pressure differs at the two ends of the smaller pipe. Flow in the bigger pipe requires a pressure gradient too, pressure decreasing in the flow direction. $\endgroup$ – Deep Jul 22 '18 at 3:52
  • $\begingroup$ I accept the idea that there is a pressure gradient in the larger pipe. However, the flow resistance of the smaller pipe should be larger, meaning the pressure difference between inlet and outlet would not be sufficient to allow the same flow inside. $\endgroup$ – Yaniv Ben David Jul 22 '18 at 10:50
  • $\begingroup$ @YanivBenDavid I may have overthought your question yet I have added Part II into my answer which deals with viscosity issues. $\endgroup$ – Yaroslav Shustrov Jul 22 '18 at 16:26
  • $\begingroup$ @YaroslavShustrov Again, I appreciate your help but you are still missing my question. You just showed WHY there is a pressure drop in the case of fully developed flow in a pipe. I'm asking WHERE does this pressure in the inlet come from. $\endgroup$ – Yaniv Ben David Jul 22 '18 at 18:46

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