Flow Rate and PSI of water exiting a pipe

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To determine the flow rate and PSI of water exiting a pipe, the continuity equation Q=Av can be applied, assuming friction losses are ignored. The pressure exerted by a piston of 2000 lbs over an area of 452.16 square inches results in a pressure of 4.42 PSI at the piston. When the pipe narrows to 1 square inch, the water exits at 4.42 PSI, not 2000 PSI. The relationship between the piston speed and the water exit speed can be established through their respective areas. For a comprehensive analysis, Bernoulli's principle should be considered to relate flow rates, heights, and pressures.
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I am trying to figure out the flow rate (gallons per minute) and PSI of the water just as it exits the top of the pipe. Is it possible to figure out these out given the information in the diagram?
 

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If you are ignoring the friction losses, you can use the continuity equation Q=Av= constant to get the flow rate Q and the fact that pressure is transmitted equally through a fluid to get the psi at the end.
 
I don't know the velocity. The piston will only move up as quickly as the water can exit the top. I can't just assign an arbitrary velocity to the piston, can I?


As for the PSI of the water coming out the top of the pipe... I must admit I have a bit of confusion. 2000 lbs. is being applied to a piston with an area of roughly 452.16 square inches... which I guess means the piston is applying 4.42 PSI. But when the pipe narrows at the top to 1 square inch... does that mean the water comes out at 4.42 PSI, or does that mean the water comes out at 2000 PSI?

(yes, I am ignoring the weight of the water and the weight of the piston itself, but I could deduct it and still have the same question with a force a bit lower than 2000 lbs being applied to the water).
 
At the top the water would come out at 4.42 psi.

I am not sure given the information you can find the flow rate at the top.
 
I think you'll want to take a look at Bernoulli's principle.

You can relate the (unknown) speed of the piston to the (unknown) speed of the water exiting at the top via the areas of the piston and pipe opening. Bernoulli's relationship will provide a relationship between the flow rates, heights, and pressures.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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