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If a pipe (of unknown/arbitrary length) with cross section area $A$ and with a continuous supply of a simple fluid with a known pressure $p$ at the opening to the atmosphere, which has a pressure of $p_a$, what would be the rate of flow out of the pipe?

I have looked but not found an answer anywhere, but apologies if this seems trivial or seems to be a duplicate of another question.

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Poiseuille's Law,

$$ Flow=\dfrac{π\cdot r^4 \cdot (P-Po) }{ 8\cdot η\cdot L} cm^3/s $$ And in your case assuming you have a Pipe with the area $Acm^2$,

$$ Flow= A\dfrac{r^2 \cdot (P-Po) }{ 8\cdot η\cdot L} cm^3/s $$

-P = pressure at the entrance, Bar

-P0 = atmosphere pressure, Bar

-eta = viscosity at dyne second/cm2 for water at 20c it is 0.01

-L = length cm.

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  • $\begingroup$ @Kaman. Thanks for your answer. However what will $L$ be the length of? The pipe length is of unknown length before the opening. The only thing known is the pressure at the end (opening). $\endgroup$ Apr 22 '19 at 23:07
  • $\begingroup$ I will edit my answer when I get home. You need to plug in the resistance and head loss. $\endgroup$
    – kamran
    Apr 23 '19 at 0:21
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If you don't know the length L, then you'll have to disregard head losses. Then you can simply apply the Bernoulli equation:

$$ p + \rho *(c^2/2) + \rho*g*z = constant $$

$p$ is the pressure, $\rho$ the density, $c$ the speed, $g$ the gravity, $z$ the height.

Apply this equation in the entry and in the exit and clear $c^2$. After that, multiply $c_2$ by $A_2$ and you'll get the flow rate.

Sorry for the wrong formatting, I still don't know how to format equations properly.

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  • $\begingroup$ Sorted the formatting for you. $\endgroup$
    – am304
    Apr 23 '19 at 7:58
  • $\begingroup$ How did you do it? $\endgroup$
    – user20096
    Apr 23 '19 at 19:00
  • $\begingroup$ Ah, the dollar symbol, thank you. $\endgroup$
    – user20096
    Apr 23 '19 at 19:01

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