Liquid outlet pressure concept

[adrich91]

Good evening, everyone,

I'm back here again for further enlightenment (greatly appreciated )

I was reviewing the subject of pumping liquids and I had understood that at the exit of the pipe to the atmosphere (for example a tap or end of open pipe) the pressure was always the atmospheric pressure; However, I have been reviewing a website and it has brought me many doubts and I need someone to help me understand.

I have been misled by the attached drawing (which says that at the tap outlet there is 3 bar g, when I think it would be 0 bar g):



If I pump from a tank that is open to the atmosphere, and I have to circumvent a circuit with a pressure loss X, I understand that I make the calculations saying the flow that I want to move and that at the end of line I have 0 bar g and after that the software (aspen hysys, for example) calculates the power of the pump and the pressure at the discharge of the pump.

If I want more flow, I will have to have more discharge pressure and the pressure drop will be higher, but I will always have atmospheric pressure at the end of the line, right?


Thank you! and sorry for my english!

 

[Latexman]

I would say it like this, just outside the end of the line there will be atmospheric pressure.

 

[LittleInch]

I don't know where that drawing has come from or what it's assumptions are, but it looks very odd to me. Yes there are some losses at the nozzle and the liquid will have kinetic energy which can revert to pressure, e.g if you point the jet at a flat surface, but not 3 bar.

 

[HTURKAK]

If we assume pressure gauge connection parallel to the flow , or say the entrance subject to velocity head, a gauge pressure will develop but will not be 3 kg /cm2.
A simple assumption, say velocity 10 m/ sec, velocity head would be 5 m or pressure 0.5 kg /cm2.

 

[Snickster]

Good evening, everyone,

I'm back here again for further enlightenment (greatly appreciated )

I was reviewing the subject of pumping liquids and I had understood that at the exit of the pipe to the atmosphere (for example a tap or end of open pipe) the pressure was always the atmospheric pressure; However, I have been reviewing a website and it has brought me many doubts and I need someone to help me understand.

I have been misled by the attached drawing (which says that at the tap outlet there is 3 bar g, when I think it would be 0 bar g):

View attachment 6583

If I pump from a tank that is open to the atmosphere, and I have to circumvent a circuit with a pressure loss X, I understand that I make the calculations saying the flow that I want to move and that at the end of line I have 0 bar g and after that the software (aspen hysys, for example) calculates the power of the pump and the pressure at the discharge of the pump.

If I want more flow, I will have to have more discharge pressure and the pressure drop will be higher, but I will always have atmospheric pressure at the end of the line, right?


Thank you! and sorry for my english!

Yes that is correct. The diagram you show makes no sense.