Pressure Added to a Flow Loop by a Pump?

[shauntur]

Hi there,

I have set up a flow loop for a uni experiment that will teach other students about the bernoulli equation. The loop will use a few working section that will increase and decrease pressure due to diameter, velocity and height changes.

I first set-up my loop with the pump at the end, pulling the water through the system so that i wouldn't have to worry about any discharge pressures. However i had to reverse this to get the pressure readings in the range of the pressure transducers.

I have been calculating the pressures at a point using:

p = density*gravity*height - 0.5*density*(velocity^2)
where height is the distance below the water level to the transducer.
​

I want to be able match the recorded pressures with some calculations, But now i have the pump at the start i am unsure how this adds to the pressures?
I am able to calculate the flow rate, as the difference p2-p1 will cancel the constant out. If i use the pump curve to find the pressure head at this flow rate, is this the discharge pressure that i can add to the equation above to obtain the actual pressure at each point.

Any help would be appreciated,
Thanks

Shaun

 

[Chestermiller]

Hi there,

I have set up a flow loop for a uni experiment that will teach other students about the bernoulli equation. The loop will use a few working section that will increase and decrease pressure due to diameter, velocity and height changes.

I first set-up my loop with the pump at the end, pulling the water through the system so that i wouldn't have to worry about any discharge pressures. However i had to reverse this to get the pressure readings in the range of the pressure transducers.

I have been calculating the pressures at a point using:

p = density*gravity*height - 0.5*density*(velocity^2)
where height is the distance below the water level to the transducer.
​

I want to be able match the recorded pressures with some calculations, But now i have the pump at the start i am unsure how this adds to the pressures?
I am able to calculate the flow rate, as the difference p2-p1 will cancel the constant out. If i use the pump curve to find the pressure head at this flow rate, is this the discharge pressure that i can add to the equation above to obtain the actual pressure at each point.

Any help would be appreciated,
Thanks

Shaun

Yes. The head is the outlet pressure minus the inlet pressure divided by ρg. Don't forget to include the turbulent frictional pressure drops, if these are significant (which they probably are).

Chet