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I need to know the effect of friction loss on the decrease of pressure in a horizontal duct (such as in an oil pipeline, with a constant diameter and a high length) and therefore, on the decrease of the maximum distance of flow of the fluid flowing in the duct.

So in a duct which length is very big (e.g. 1000 km), with a pump that propels a fluid at a high constant pressure of 10 MPa. Due to friction loss, will the pressure decrease and at some point will become equal to zero, and would the speed would be equal to zero too?

I say this because according to Bernoulli's equation : $vf=\sqrt{\frac{2}{ρ}(Pi-Pf)+vi^2+ρgΔh}$, so if $Pi-Pf>0$ then $vf>0$ and there is a movement ($Δh=0$)

I couldn't find any formulas that actually modelized this phenomenon. Huge thank you for your help.

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Bernoulli's equation is valid along a streamline for inviscid flow - where there isn't any friction or loss of stagnation pressure.

You need to instead look at the loss of stagnation pressure for fully developed turbulent flow (assuming the pipe is very long, and that the fluid is incompressible)

Velocity is constant for an incompressible fluid if the pipe has constant area, because the mass flow rate $\dot{m}=\rho VA$.

So to get the pressure drop, you use the Darcy-Weisbach equation, $\frac{\Delta P}{L}=\frac{1}{2}\frac{\rho V^2}{D}f$, where $D$ is the pipe diameter and $f$ is the friction factor, which is given in charts like the one at https://www.pipeflow.com/pipe-pressure-drop-calculations/pipe-friction-factors

If the pipe is leaving into a big tank, there'll be an additional stagnation pressure loss of $\Delta P=\frac{1}{2}\rho V^2$ for the exit, because the dynamic pressure isn't being recovered, but for very long pipes this becomes comparatively small.

So the friction factor $f$ depends on how rough the pipe walls are - for a pipe with bumps that are about 1% of the diameter, it's about 0.04 for fully turbulent flows (at long distances)

So to calculate the velocity, ignoring the little bits where it's not fully turbulent and the exit loss, you'd get $V=\sqrt{\frac{2\Delta PD}{\rho Lf}}$.

There's no maximum distance - the velocity just decreases with lower pressure and longer length. There'll always be some flow. But at very low velocities, the turbulent assumption breaks down and the maths gets harder.

Also if you're pumping something compressible like a gas, you get all sorts of weirdness where the gas expands as the pressure goes down and it heats up, so velocity increases, up to the speed of sound sometimes, as it goes and the maths gets horrible.

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  • $\begingroup$ Okay, thanks a lot. Just a question to clarify, can I express L with V, DP, D, rho and f this way $L=\frac{2ΔPD}{ρfv^2}$. Sorry for these lame questions, I am no physician. $\endgroup$
    – eengeeneer
    Commented May 23, 2022 at 17:14
  • $\begingroup$ Yes that looks right $\endgroup$
    – sqek
    Commented May 23, 2022 at 17:24
  • $\begingroup$ Okay, just another question please, $ΔP$, is it $Pf-Pi$ or $Pi-Pf$ ? $\endgroup$
    – eengeeneer
    Commented May 23, 2022 at 23:18
  • $\begingroup$ Not sure what you're defining $P_i$ and $P_f$ as. It's the pressure at pipe inlet, minus the pressure at pipe outlet. So the change in pressure from one end of the pipe to the other. So if you have a pump before the pipe, and the outlet is into something open to the air, it's just the outlet pressure of the pump minus atmospheric $\endgroup$
    – sqek
    Commented May 24, 2022 at 10:50
  • $\begingroup$ Oh I meant $P_i$ as in initial pressure (the pressure at the end of the pump) and $P_f$ as the final pressure (the pressure at the end of the pipe). I get it now, thanks a lot. $\endgroup$
    – eengeeneer
    Commented May 24, 2022 at 13:33

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