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This seems true when I slightly compress the opening of pipe the speed of water flowing through pipe increases and same amount of water flows out but as I further decrease the opening of pipe. Instead of increasing speed of water to maintain same volume flow rate.the amount of water coming out of it decreases and same volume rate stops. Why is it ?

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  • $\begingroup$ The incompressibility implies that when the outgoing flow decreases and stops, then the flow through the other portions of the pipe also decreases and stops. $\endgroup$
    – stafusa
    Feb 21 '20 at 12:17
  • $\begingroup$ If the water supply delivered a constant flow rate (imagine a positive displacement pump driven at constant speed by an infinitely powerful engine) then the pressure in the supply pipe would increase without bound, and the jet velocity would increase without bound as you decreased the area of the exit orifice toward zero. But, your actual water supply flow rate almost certainly is a decreasing function of pressure, and it almost certainly stops altogether when the pressure reaches some limit. $\endgroup$ Feb 21 '20 at 13:06
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From the law of mass conservation $$\partial_t \rho + \vec \nabla (\rho \vec v) = 0 $$ follows in the case of a static, incompressible fluid $$\vec \nabla \cdot \vec v = 0,$$ implies, when integrated over a volume $dV$ $$\int dV \; \vec \nabla \cdot \vec v = 0,$$ which, by using the Gaussian theorem we can transform from volume, to surface integral $$\int dV \; \vec \nabla \cdot \vec v = \oint d\vec S \cdot \vec v,$$ and then by taking a flow in a pipe with decreasing diameter, choosing a box with 6 sides as surface around the pipe for $dS$, we get $$ 0=\oint d\vec S \cdot \vec v = \sum_{i=1}^6 \Delta S_i\;\, \vec n_i \cdot \vec v$$, where $n_i$ is the normal vector on each side. Now choose the box such that 4 sides are outside the pipe, those vanish trivially, and what remains is $$ 0 = S_1 v_1- S_2 v_2, \;\;\;\;\;\;\; (1)$$ where the minus comes from taking the correct scalar product with the surface normals.
(1) is now true for any portion of the pipe! So if you know $S_1$, $S_2$ and $v_1$, then you can always determine $v_2$.

Now coming finally to your question, what we ignore in this computation is the viscosity, which wold turn up in the Navier-Stokes equation. Viscosity will couple the velocity with the boundary of your pipe, which can stall the flow in the whole pipe, but still at any time (1) will hold.
It is just that when $v_1$ decreases due to effects of viscosity, then $v_2$ has to decrease accordingly.

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Yes, due to incompressibility, the same amount of water flows into and out of any given section of a tube - that is correct. But due to viscosity and a no-slip condition on the tube wall, the pressure of the water drops downstream (if the tube width and flowrate is constant).

If you have a constant pressure potential available (high pressure at the tube inlet, atmospheric at the outlet), the flowrate will adjust with the geometry of the tube. The longer and thinner the tube, the more effect from viscosity (more pressure drop for a given flowrate) and a lesser flow at your constant pressure drop.

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