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Assume I have a straight pipe with diameter D1 and a volumetric flow, Q flowing through it. The inlet pressure is P1 and the outlet pressure is P2.

Now, assume I've added a restriction in the middle of that pipe, which takes the form of another pipe with a smaller diameter D2 and a length L.

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Presumably, the new outlet pressure P'2 would drop due to the increased restriction in the pipe. How can I calculate that drop in outlet pressure P'2?

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  • $\begingroup$ You would still need a pressure gradient across the entire device to even get the fluid to move through the contraction. You can't set a venturi tube like that on a table and suddenly get flow through it. It's always critical to remember that pressure changes cause velocities and not the other way around. $\endgroup$ – tpg2114 Feb 28 '15 at 19:46
  • $\begingroup$ @tpg2114 I agree, but if I place a restriction in a pipe as described in the question, I'm expecting to see a larger pressure drop across it than across the pipe on it's own, and I'm trying to figure out how to calculate that. $\endgroup$ – Amr Bekhit Feb 28 '15 at 19:49
  • $\begingroup$ You said "the overall pressure drop across the whole restriction would be 0" which is only true if there is no pressure gradient within the pipe -- which means there is no flow. The pressure out the outlet of your picture must be lower than the pressure at your inlet no matter what is happening in the middle with the restriction, otherwise you won't get any flow. Therefore, no, it would not be zero. $\endgroup$ – tpg2114 Feb 28 '15 at 19:51
  • $\begingroup$ Ah I see what you mean - yes OK the question needs rephrasing. $\endgroup$ – Amr Bekhit Feb 28 '15 at 19:51
  • $\begingroup$ @tpg2114 OK the question has been reworded. Hopefully it's much clearer and accurate now! $\endgroup$ – Amr Bekhit Feb 28 '15 at 20:01
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In an ideal fluid, assuming the diameter of the pipe after the contraction is the same as the diameter of the pipe before the contraction, $P_2^\prime = P_2$. There is no effect of the contraction downstream from the contraction itself.

If we consider an inviscid, isentropic, incompressible flow, the total pressure in the flow is constant along streamlines (and since all streamlines originate from the same source, it is also constant throughout the flow). This means $P_0 = P + 1/2 \rho U^2 = \text{const}$. We also know that $\rho_1 U_1 A_1 = \rho_2 U_2 A_2$ for any positions 1 and 2 where $\rho$ is the density, $U$ is the velocity and $A$ is the area of the pipe.

So, we know density is the same because it is incompressible. We know $A_{inlet} = A_{outlet}$ because the pipe is the same diameter away from the contraction. Therefore, the velocities are the same $U_1 = U_2$. And since the total pressure is constant, the values of $P_2$ and $P_2^\prime$ must be the same also.

This is probably counter-intuitive. You would expect something to be different. And in a real fluid, it very well likely would be. Turbulence, separation, friction, heating, all kinds of losses will add up to have an influence. But in an ideal world where none of those things happen, there is no effect of the contraction away from the contraction.

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