Pump pressure/outlet pressure

[vinhermes]

Dear All,

Small brain cramp today... This will seem obvious to many of you but it is not clear for me.

So, the components to determine the pump required pressure (for a given flow rate) is:

- static head
- friction (pipe & fitting)
- velocity head (ignored on my case)

But do I need to add the required pressure at the outlet of the discharge pipe? For instance, if this is a fire hydrant and that I want the water to flow at a pressure of say 5 barg, shall I add 5 barg to the pump pressure required by friction and head?

It seems logical but then I am afraid that the pump will only incease the flow rate to match its curve (for a pressure only equal to head + friction, without the 5 barg).

Hope I am clear enough (hard to explain clearly sompething while being confused). Thanks by advance, Vincent

 

[micalbrch]

Yes, you must add the pressure at the end of the line but keep in mind that you need something to create this pressure. The pressure is only the consequence.

 

[vinhermes]

Dear Micalbrch,

Thanks for this but by answering a question, you have raised another one... What do you mean by "something to create this pressure".Is this supposed to be the fitting at the outlet of the pipe? Does this mean that an open pipe alone is not suffcient to build-up the pressure?

Thanks again, vincent

 

[micalbrch]

Correct. If you have an open pipe the pressure at it's end will be atmospheric. Pressure is created by the fluid properties (density, viscosity) and the velocity in combination with the entire pipe system (incl. fittings and valve). Right behind the pump the pressure is at its maximum and at the end of pipe (open outlet to atmosphere) it is equal to atmosphere. If you install an orifice right at the end, you'll be able to increase the pressure at that point.

 

[vinhermes]

OK perfect, thank you

 

[BigInch]

"But do I need to add the required pressure at the outlet of the discharge pipe? For instance, if this is a fire hydrant and that I want the water to flow at a pressure of say 5 barg, shall I add 5 barg to the pump pressure required by friction and head?"

yes.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[Artisi]

remember, if you now run the pump (designed for the 5 bar discharge pressure) without any restriction, ie fire hose or sprinklers etc - the pump will probably run to end of curve and cavitate.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[vinhermes]

OK, all starts to make sense.

So to sum it up, in a classic system, where a pump discharges a fluid at a given static head to a given flow rate, the pressure at the open end of the pipe (no fittings, restrictions, etc.) will be the atmospheric pressure.

In the case of a water monitor, the pump is sized for the required pressure given by the monitor supplier to reach the required dsitace but if one takes the monitor out and runs the system, the flow rate will be huge but the pressure at the monitor flange will still be atmospheric.

What about a tap in my house? What happens in the case where the static head is negative (ie the outlet is below the tank). If there is a pump on this system, will the head pressure be present at the outlet of an open pipe or will it still be atmospheric pressure?

Sorry for this (and any reading recommendations will be apprciated) even if it all takes shape in my head. Being a self learner is great but it has its drawback when things are confusing...

 

[BigInch]

You reduce the required head accordingly, but not due to the tank's elevation. Everything depends on elevation in relation to the pump centerline. If the outlet is below the pump centerline by 5 feet, then you could reduce head by 5 feet than what you would need if the outlet was exactly at the pump's elevation, provided the tank was closed to atmosphere. Otherwise, an open tank, you would have to pump to the highest level you wanted to have in the tank, no matter where there was an outlet on your system, then let the tank feed the outlet by gravity, for example.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[katmar]

From your examples and choice of words it seems that you are looking at a fire-fighting application, and this needs a litle bit more consideration. The 5 barg that you want at the hydrant when it is working is usually called the "residual pressure" by the fire people. It is the pressure that would be measured at the hydrant while it is in operation. This pressure is consumed by the 3 factors you mentioned in your first post - static head, friction and velocity head. In fire fighting system design you absolutely cannot ignore the velocity head. A nozzle exit velocity of 20 m/s gives a velocity head of 2 bar, so if you had 5 bar at the hydrant you only have 3 bar left for static and friction.

Regarding your example of the domestic tap with a pump and a negative static head - the answer was given by Artisi. You are likely to over-run the pump and burn it out.

Katmar Software - Uconeer 3.0

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[TenPenny]

Don't forget that in a house system, you have a pressure tank with a pressurized air chamber, so the pump will cycle on and off, it's not a very good analogy or example in this case.

 

[vinhermes]

Dear All,

Thanks for this.

Katmar, I went through this post from you (at the bottom):

http://www.eng-tips.com/viewthread.cfm?qid=128047

It has clarified everything.

I have developped a small application for pump sizing/pipe losses (not as deep as the one on your website but same approach) and will, from now on, include the velocity head into my calculations.

So Total head:

- Static head
- Friction head
- Velocity head

And in the case where it is not an open pipe outlet (like on a fire system for instance, where the supplier requires "X" bar to be discharged into a fitting), then the residual head (of "X" bar) has to be added.

Thank you all for clarifying this. Have a nice day, Vincent