Calculate volume of water to increase pipe system pressure

Mar 2020
3
0
canada
Hi,

I am hoping someone can help with this fairly simple problem.

A pipe that is closed at both ends is oriented horizontally and is full of water. The static pressure is currently "X" psi. There is a port on the end of the pipe that allows you to pump more fluid into the closed pipe.

What volume of water needs to be injected into the closed pipe to increase its pressure to "Y" psi?

Assume the pipe volume is fixed (it will not increase with pressure)
Assume water is incompressible
 

topsquark

Forum Staff
Apr 2008
3,106
659
On the dance floor, baby!
If you are assuming that water is incompressible then how can you pump more water into a pipe that's full of water? I can't think of how to increase the pressure in the pipe without it being only partly full or compressible.

-Dan
 
Mar 2020
3
0
canada
Hi Dan,

Then we should assume the pipe does expand? The increase in pressure being a result of the stress from the pipe expansion?
 
Mar 2020
3
0
canada
I may be focusing on the wrong concepts for this problem... If we hold the pipe volume as fixed and water as incompressible, the increase in pressure in the systm and the volume to do so would be related to the expansion tank in the system.
 
Apr 2015
1,233
359
Somerset, England
What expansion tank would that be?

Is there anything else you haven't told us?
 
Oct 2017
676
348
Glasgow
If we assume that the pipe is instead open so that water can flow in or out of both sides, then
1. the volume of water inside the pipe is constant; and
2. any water injected is removed at the other end of the pipe.

Then, total pressure = static pressure + dynamic pressure
\(\displaystyle y = x + \frac{1}{2} \rho u^2\)

Volume flow rate:
\(\displaystyle Q = uA\)

\(\displaystyle y = x + \frac{1}{2} \rho \left(\frac{Q}{A}\right)^2\)

Rearrange for Q:
\(\displaystyle \frac{1}{2} \rho \left(\frac{Q}{A}\right)^2 = y-x\)
\(\displaystyle Q = \sqrt{\frac{2A^2 \left(y-x\right)}{\rho}}\)
 
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