# Pressure in the pipe when the pool is draining out

I have a question: the situation is as follows: There is an open pool (in blue) to which a pipe is connected. The pressure above the pool is equal to the air pressure = 1 bar. In normal configuration the valve from the pipe is closed and thus the pressure in point B is equal to the air pressure + ρgH (hydrostatic pressure). There is no flow through the pipe.

In case there is a rupture of the pipe at the height of the valve, water will siphon from the pool via the pipe at a velocity v. My question is how can you calculate the pressure in point A?

I think the pressure in point A is calculated as follows:

Pa+ $$\frac{1}{2}$$ρva²+ρgL=Pb+$$\frac{1}{2}$$ρvb²+ΔPfriction where va=vb and thus:

Pa+ρgL=Pb+ΔPfriction where Pb is equal to the air pressure (1 bar) due to the break.

PA=PB+ΔPfriction - ρgL

ΔPfriction=fD$$\frac{L}{D}$$$$\frac{ρv²}{2}$$

However, my friend does not agree with me and says that Pa is equal to Pb. So who is right?

• The driving force for fluid flow is a pressure difference. That means that you are right. Jun 14, 2019 at 19:27
• P friction should not be used if we use bernoulli's theorem. Because it is conservation of energy. Jul 11, 2019 at 6:38
• Therefore Pa=Pb-pgL. Jul 11, 2019 at 6:40