What is the flow rate in an open-ended pipe with increasing gauge pressure?

[agondouin]

This is not a homework problem but I'm going to try and present it like one.

1. Water is flowing through a horizontal, open-ended pipe, of constant diameter. A pump is mounted upstream and the water flows through the pipe. The pump supplies a gauge pressure of 20 psi and the flow rate is 8 gpm. A larger pump is used to increase the gauge pressure to 80 psi. What is the flow rate?



2. I tried to use Bernoulli's equation: P/rho + 1/2*v² + gz = constant



3. I tried to apply Bernoulli's equation to two points in the pipe, after the pump and at the pipe outlet. P1 is the upstream pressure with the small pump. P2 is the upstream pressure with the big pump, and Pe is the pressure at the pipe exit. I think Pe is atmospheric pressure.


I get: P1-Pe = rho(ve²-v1²), and P2-Pe = rho(ve²-v2²)

Then I tried dividing these two. So I got:
Pgauge 1/Pgauge 2 = (ve²-v1²)/(ve²-v1²)

I'm not sure what to do after this, or if I've done this right so far.I must be missing something. Any help is greatly appreciated

 

[tiny-tim]

welcome to pf!

hi agondouin! welcome to pf!

i think you can asume that v₁ = v₂ = 0

also, gauge pressure = pressure minus atmospheric pressure, so P_e = 0

 

[agondouin]

Thanks Tim!

If I assume that V1 and V2 are both zero, I can simplify and get V2e = V1e * sqrt(P2/P1). If I plug in the numbers from my example, then I get a flow rate of 16 gpm for the larger pump.

It makes sense that the flow should ge greater with a bigger pump. What I don't understand is why the flow is zero after the pump. I thought that conservation of mass meant that the flow rate through the entire pipe had to be constant, so that the flow can't be zero anywhere and still have water coming out. Can you explain why we can set V1 and V2 zero?