# Open pipe and relation between pressure and flow rate

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.

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.

• @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). Apr 22 '19 at 23:07
• I will edit my answer when I get home. You need to plug in the resistance and head loss. Apr 23 '19 at 0:21

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.

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