Piping/Pump + elevation question 1

[ly1phil]

Here’s the piping/pump problem…
I’m leaving out some details in order to simplify the problem in order to explain it.

I have 3 heat exchangers that have cooling water provided from one pump. These heat exchangers are at different elevations, for simplicity say they are at elevation 10ft, 20ft and 30ft above the centerline of the pump discharge. Currently all three heat exchangers discharge into a common discharge line located at elevation 15ft.
If I were to add a section of pipe within the common discharge line but before the pipe exit point (i.e. the point of discharge is still at 15ft) that resembled a upside down U, with the top of the U at elevation 25ft, would the pump see an increase in the system’s pressure if I disregarded frictional losses because they are negligible.
The “loop” would start at the 15ft elevation climb vertically up to the 25ft elevation and back down to the 15ft elevation then to the exit point.

I am unsure as to how much this will affect the system pressure that the pump sees. Will it add 10ft H2O of backpressure to the pump? Or can I neglect the height downstream of the highest point (30ft) and assume it is free flowing at that point? Any ideas would be appreciated. Thanks.

 

[Apakrat]

Or can I neglect the height downstream of the highest point (30ft) and assume it is free flowing at that point?

The 30ft thingy lost me. Believe, You will find the fluid static head pressure at the pump will be at the greatest height above the pump.


At 74th year working on IR-One PhD from UHK - - -

 

[Artisi]

How about a sketch of your installation.
Is it a closed loop , ie recirculating

 

[katmar]

As the water flows up the first leg of the U you will of course get an additional static head of 10'. The question is whether you will get the same pressure recovery (assuming negligible friction) as the water flows down the second leg of the U.

The balance to do is to compare the pressure at the base of the second leg with the height of the U. Say, for example, that the pressure at the base of the second leg was 20' due to friction in the pipe from the U to the discharge point and any static height change. This would mean that the down leg of the U would run full and that you would recover the full 10' of static height that you lost in the up leg.

On the other hand, if the inverted U was near the end of the line and the pressure at the base of the down leg was only 2' it would not be necessary for the downleg to be full to provide this pressure. The pressure at the top of the inverted U could actually go below atmospheric and depending on the temperature of the water you could get boiling and cavitation. As soon as you get boiling you start to lose pressure recovery and the pressure the pump "sees" will increase.

In this sort of installation you will often see a breather pipe or valve (syphon break) above the top of the U to prevent the pressure going below atmospheric (and to prevent the resulting cavitation and vibration). If a breather is installed then there cannot be full pressure recovery.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[chicopee]

theoritically no change in pressure per original post conditions as long as you maintain full flow on the downcomer of the U tube.

 

[ly1phil]

There is a vacuum relief valve being installed at the top of the inverted U, which means that full pressure recovery is not possible. Does that mean that the pump will “see” an extra 10 ft of static head pressure? Even though the highest point in the system is 5 ft above the height of the U?

 

[katmar]

I must admit I had not fully considered the front section of your piping, i.e. around the heat exchangers. The overall situation will depend on how you control the split to the three exchangers and whether you have vacuum relief on the 30 ft line, and also on the frictional losses from the start of the common line to the eventual discharge point. The position of the inverted U relative to the other equipment is also important (see my earlier post). As Artisi said, without a fairly detailed sketch we are guessing.

At a simplistic level, whatever the pressure drop was from the start of the common line to the discharge point, you have now added in additional static head that cannot be fully recovered. It is likely that your pump will see some additional head but without more detailed info nobody can say for sure how much.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[ly1phil]

I understand, and I appreciate all the responses.

Here is a diagram of what is going on.

http://files.engineering.com/getfil...935-9ebd-63b68fc03f91&file=Piping_Diagram.pdf

All three heat exchangers “thrust bearing”, “lower guide”, “turbine bearing” are simply water coils running within an oil tub to cool the various bearings on a hydro turbine generator. The “thrust bearing” (30ft) does not have vacuum relief. The inverted U on the second diagram does have vacuum relief. All three lines going to the 3 heat exchangers have flow setters to adjust flow. It is also safe to assume that flow through the 3 bearings on the second diagram can be adjusted to the flows on the first diagram.
Would it be conservative to assume a 10ft increase in static head based on this diagram?

 

[Artisi]

" Would it be conservative to assume a 10ft increase in static head based on this diagram? " --- Why??

The only difference I see will be an increase in friction head due to the extra pipe-run bought about by the inverted U, this could effect the output of the pump in terms of the flowrate at the new total head. However - you say to neglect friction - so I don't see any problem.

 

[chief]

The obvious thing to do is to arrange your piping , so that the pump discharges to the highest elevation and cascades to the lowest heat exchanger. Otherwise you will short circuit, if you flow the bottom/up arrangement.

Offshore Engineering&Design

 

[ly1phil]

"The only difference I see will be an increase in friction head due to the extra pipe-run bought about by the inverted U, this could effect the output of the pump in terms of the flowrate at the new total head. However - you say to neglect friction - so I don't see any problem. "

This was my initial thought, however I was second guessing myself.

"The obvious thing to do is to arrange your piping , so that the pump discharges to the highest elevation and cascades to the lowest heat exchanger. Otherwise you will short circuit, if you flow the bottom/up arrangement. "

There are flow setter valves to control all the flows through all three heat exchangers, thus short circuiting can be prevented. Cascading the heat exchanger flows is not possible due to requirements of the heat exchangers.

 

[daviwy]

You will loose pressure from the U bend, and friction from the new installed pipework...

I dont think you will make much difference though with the U shape upside down as the head you gain going down the U, you loose going up it in the first place...

so i think you'd end up with less pressure than you initially had...

Please do drop more information or knowledge you have on this subject they will all be taken in and appreciated.


David O.
Mechanical Engineer
Graduate as of: 2007
Oil, Gas & Power

 

[katmar]

There is no easy, definitive answer to this question without doing a complete analysis of all flows and pressures. One thing we can say is that the maximum vacuum that can be pulled by the down leg of the inverted U is 10 ft of Water Column and this means that there will be no boiling unless the water is hotter than 85 deg C (185 deg F). If the water is colder than this you could do away with the vacuum breaker at the top of the inverted U which would increase your chances of getting full static pressure recovery in the down leg, but now you are faced with an analysis of whether there is sufficient flow in the down leg and piping from the down leg to the discharge point to keep the pipe full. This is a whole new can of worms.

Once you have solved the question of whether you get full recovery or not you can determine the pressure change immediately upstream of the inverted U. If the pressure at this point increases does it mean there is an increase in the pressure the pump must deliver? To answer this question you need to investigate the positions (percentage opening) of the flow setting valves on the heat exchangers, and also you need to check whether there is pressure recovery in the down leg from the thrust bearing.

Even though the diagrams look simple, this is not a trivial problem to solve rigorously.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[ly1phil]

Thanks for all the thoughts/inputs.

I'm going to throw another assumption into the equation. If none of the pressure is recovered on the downstream side of the U (due to the vacuum relief valve), does that mean that there would be, at a maximum, an increase of 10ft of static pressure on the pump?

I just need to be appropriately conservative, not necessarily exact.

Thanks

 

[katmar]

I cannot imagine a situation where the impact would be more than 10 ft. I think you would be safe to make that assumption.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[CJKruger]

ly1phil, whether you have pressure recovery in the downhill section after the loop depends on:

- Vapor pres at top of loop (Katmar explained it well).
- How the outlet of the pipe is sealed (is it below a liquid level, or above it in vapor/air).
- Velocity in the downward section of the loop.

Thus, if the outlet is not sealed in liquid, and the velocity is below self venting flow, then you will not recover the head in the downward section and the pipe will probably vibrate (as it sucks in and then blows out air).