Liquid outlet pressure concept 1

[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.

 

[katmar]

The whole question of entrance and exit pressure losses, resistance coefficients and pressure recovery is generally not well covered in the standard texts. In particular, the texts do not sufficiently emphasize that the pressure changes due to friction and Bernoulli effects must be considered together.

I have written a short article on this question, discussing the theory and also giving examples. It is available as a downloadable PDF. There are also links in the document to a series of excellent videos on YouTube that show real-world experiments illustrating these concepts.

 

[adrich91]

Hello again!

Thank you very much for your comments and contributions!
Now, I have a new question (and thanks! hehe, it's always a good opportunity to learn something new).



What would happen if in the drawing I show you (pumping from an open tank to an open pipe) with the head losses in the circuit and taking into account that it is negative suction, instead of ‘configuring’ the pump for that case, I would say that the pressure at the end of the line would be 3 bar (knowing that it is not, that it is at atmospheric pressure)?

If I were to install this pump (for a final pressure of 3 bar) in my system whose discharge is at atmospheric pressure, would I get more flow? Because I understand that the pump would have been chosen to give me 3 bar at the end of the line when it does not need it, because it is at atmospheric pressure.

I'm trying to see if I can link this with the previous drawing which says that it has 3 bar at the tap outlet.

Is modelling the pump to have 3 bar at the outlet a way of making sure that I will have plenty of pressure in the line?

Thanks again!

 

[LittleInch]

The only way your pressure in the first diagram is 3 bar is if you stick a tube pointing in the opposite direction to flow and measure pressure. This is the dynamic head (sometimes called velocity head) and for water would need a flow velocity of 25m/s. Look up pitot tubes for a better explanation.

It's normally ignored because velocities rarely exceed 5 m/ sec in most systems.

Yes you would get more flow because the static pressure is essentially zero barg.

So your pump needs to add 2 bar plus the negative pressure.

So if this is say -0.5 barg, then your TDH of your pump needs to be 2.5 barg/ 25m.

But you need to tell us all of the data belonging to that first diagram because on its own it doesn't make sense.

 

[adrich91]

The only way your pressure in the first diagram is 3 bar is if you stick a tube pointing in the opposite direction to flow and measure pressure. This is the dynamic head (sometimes called velocity head) and for water would need a flow velocity of 25m/s. Look up pitot tubes for a better explanation.

It's normally ignored because velocities rarely exceed 5 m/ sec in most systems.

Yes you would get more flow because the static pressure is essentially zero barg.

So your pump needs to add 2 bar plus the negative pressure.

So if this is say -0.5 barg, then your TDH of your pump needs to be 2.5 barg/ 25m.

But you need to tell us all of the data belonging to that first diagram because on its own it doesn't make sense.

This is the web where I extracted the image (https://www.iagua.es/blogs/miguel-angel-monge-redondo/fbh-iv-caudal-presion-y-velocidad-agua)

Thanks!

 

[LittleInch]

This is the web where I extracted the image (https://www.iagua.es/blogs/miguel-angel-monge-redondo/fbh-iv-caudal-presion-y-velocidad-agua)

Thanks!

Yup. Total rubbish. Just because someone says something in a blog post doesn't make it true

 

[Snickster]

Hello again!

Thank you very much for your comments and contributions!
Now, I have a new question (and thanks! hehe, it's always a good opportunity to learn something new).

View attachment 6640

What would happen if in the drawing I show you (pumping from an open tank to an open pipe) with the head losses in the circuit and taking into account that it is negative suction, instead of ‘configuring’ the pump for that case, I would say that the pressure at the end of the line would be 3 bar (knowing that it is not, that it is at atmospheric pressure)?

If I were to install this pump (for a final pressure of 3 bar) in my system whose discharge is at atmospheric pressure, would I get more flow? Because I understand that the pump would have been chosen to give me 3 bar at the end of the line when it does not need it, because it is at atmospheric pressure.

I'm trying to see if I can link this with the previous drawing which says that it has 3 bar at the tap outlet.

Is modelling the pump to have 3 bar at the outlet a way of making sure that I will have plenty of pressure in the line?

Thanks again!

If you mistakenly design the pump for 3 bar at the outlet when it is really 0 bar atmospheric pressure then you will have over-sized sized your pump. You will have sized it for a head of 2+3+suction pressure loss (including friction loss and lift height say 0.5 bar as previously indicated by LI) = 5.5 bar total head at design flowrate when you only need 2.5 bar at that flowrate.

The head the system needs to flow a given amount is called the system curve. The system curve plots the required head of the system which includes friction loss, plus velocity head, plus changes in static height per the Bernoulli equation versus flowrate. The system curve is plotted on the pump curve. Where the system curve intersects the pump curve that is where the actual operating point will be.

Look at it this way. If for a given flow your pump is outputting 5.5 bar at a given flowrate but your system pressure drop is only 2.5 bar at that flowrate (per your system curve) then you will get more flow in your system (because you have more available pressure) until the head output of the pump equals the losses in the system at the same flowrate. This is where the system curve crosses the pump curve.

Here is an example of pump and system curve attached. Note that the solid line system curve is the design curve/operating point of pump. If there is actually less losses in the system then calculated the actual operating point/system curve with shift to the right shown by the dashed system curves. If there is more system losses that expected then the system curve/operating point will shift to the left shown by the dashed system curves. Note that as you shift operating points on the pump curve you are changing the horsepower requirements so you may exceed your motor horsepower and damage your motor. Note that the BHP lines are purple on the attached pump curves. Each purple line is a given brake horsepower required. As the flow increases to the right the red pump curve crosses the BHP curve indicating that a higher horsepower is needed as it crosses on BHP curve into the next BHP region.

Also note this is all assuming what you have is a centrifugal pump where pump head varies with flow per the pump curves. If you have a positive displacement pump the flow is always constant since the flow is based on the fact that each revolution of the motor there is a revolution of the pump piston/vane/rotor etc. that pushes out a given volume. For a PD pump the flow is constant but the head output will be whatever the system demands as long as the horsepower is not exceeded.