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How to calculate NPSHA in a closed pumping Loop?
7

How to calculate NPSHA in a closed pumping Loop?

How to calculate NPSHA in a closed pumping Loop?

(OP)
Hi everybody...

thank you for being a resource for engineers everywhere!  I have a project where we are going to replace a hot water boiler with a cogen driven heat exchanger.  The heat exchanger is going to take the place of the boiler, as the heat generating device.  It will be about 60 feet away, increasing the length of the suction line to the pump from say, 10 feet to 60 or so.

I know how to calc the NPSHA for a open loop like this, but what about a closed loop?  The vapor pressure for water is about 10ft of head at 160F.  The elevation of the piping loop is about 10' above the pump.  Can anyone shed some insight into this for me?

Jack

RE: How to calculate NPSHA in a closed pumping Loop?

3
Pick a point anywhere in the loop.  Take the pressure of that point and convert to equivalent head.  Add, or subtract (if the point is lower than the pump inlet), the elevation difference.  Subtract the head lost at the system's flowrate between the point at which you began to the pump's inlet.  Lastly, subtract the head equivalent of the vapor pressure at the water temperature at the pump's inlet.  What's left is the pump's NPSHA.

http://virtualpipeline.spaces.msn.com

RE: How to calculate NPSHA in a closed pumping Loop?

(OP)
thank you BigInch!  This is what I have done...I was not clear on the technique.  Another question; anywhere on the loop? Seems that I could reduce the NPSHA to it's lowest value by picking, say the point at which is both the highest elevation and the longest run to the pump.  A centrigal hot water pump will go to negative NPSH?  

RE: How to calculate NPSHA in a closed pumping Loop?

Right, anywhere in the loop.  In a closed circuit, sum of heads lost = pump head + initial head.

You have to maintain a minimum (initial) head/pressure in the loop, such that you will not have NPSHR > NPSHA.  So minimum initial pressure is the vapor pressure at the suction flange.  You can use that as a reference point if you like and add or subtract all heads around the loop from there.

Once you've done that, pick any random point in the system and see how it works.  If you randomly pick the pump discharge, then you have that discharge pressure converted to head - head loss at the system flowrate for discharge to suction - elev change from pump discharge flange to high point + elevation change from high point to pump suction - vapor pressure + initial pressure = 0

http://virtualpipeline.spaces.msn.com

RE: How to calculate NPSHA in a closed pumping Loop?

(OP)
thanks BigInch....interesting.  NPSHA on a closed loop is just as important then, as open loop.

RE: How to calculate NPSHA in a closed pumping Loop?

Right, the pump doesn't know if the system is closed or open.  All it wants is its NPSHR.  An example of each,



When the system is open to atmosphere as would be when pumping from a tank open to atmosphere, you get to add 14.7 psia as a credit for NPSHA and subtract the vapor pressure.  

If the system was closed, as when pumping from a pressurized tank, the tank pressure above the fluid is the credit, from that subtract vapor pressure.

Adding the dynamics of the system is easy.  In this example, the operating point on the system curve (flow 0.25 cfs) gives a dynamic loss of 6 ft (OK, just made that up).  The pump operates at that point, so it produces a dH of 6.00 ft.  Just superimpose the dynamics  (red) on the static case.

 

http://virtualpipeline.spaces.msn.com

RE: How to calculate NPSHA in a closed pumping Loop?

From an hydraulics perspective, it doesn't matter where the pump is located in the loop. From a dissolved air removal from the water point of view, it matters very much. This is one of the main reasons why buildings heated with hot water boilers or convertors usually have them located at the top of the structure, along with the circ pumps, expansion tanks, and air separators.

RE: How to calculate NPSHA in a closed pumping Loop?

TBP (Mechanical)Very good point relating to dissolved gases. This seems to be ignored in many designs.

RE: How to calculate NPSHA in a closed pumping Loop?

BigInch - These closed loop systems are a different animal from the "one-way-trip" pump applications, such as boiler feed pumps. In closed loop hot water heating applications, the suction and discharge heads are essentially equal. These systems are like ferris wheels. If the operator has properly distributed his passengers, the weight coming down will equal the weight going up. All the motor (or pump)  has to do is supply a little bit of "oomph" to make it move. Very low heads are involved with these circulating pumps. I have never seen one cavitate, and don't know that's it's even possible. What most people claim is cavitation with pumps in these systems, is, in fact, air locking.

Low head is the name of the game with these heating systems. If you have a high-head pump, you can have high velocities. This will be noisy in applications such as hotels, apartments, offices, etc. It's the same with the air removal. Failure to provide proper air elimination will result in a system sounding like it's full of BB's. You will get endless occupant complaints about system noise.

RE: How to calculate NPSHA in a closed pumping Loop?

(OP)
thanks TBP - I don't think Big Inch was stating anything different.  I was asking what happens when addtional suction piping is added to a closed loop, such as the one in the diagram he posted.  If we were to, say, "extend" the suction piping 60 feet or so to a "new" boiler or other heat generating exchanger, what would happen.  I think I understand the answers to be that we'd have to increase the bladder pressure in the expansion tank to accomidate the increase in suction piping.....I appreciate your comments.  

RE: How to calculate NPSHA in a closed pumping Loop?

Quote (seventhstream):

I think I understand the answers to be that we'd have to increase the bladder pressure in the expansion tank to accomidate the increase in suction piping

Not necessarily. Suppose your existing NPSHa-NPSHr>0, even after adding the suction pipe, you need not increase the bladder pressure.

Do you have any reason why you are placing the HX in the pump suction and not in discharge?

RE: How to calculate NPSHA in a closed pumping Loop?

(OP)
discharge becomes suction at some point right?  If we don't have to increase bladder pressure, (if NPSHa-NPSHr>0)then we won't. (right?)  I was trying to understand what happens when we break into the circuit and "extend" the suction line (see above...).  thanks for your comments.  They help me crystalize the problem....in my mind.

RE: How to calculate NPSHA in a closed pumping Loop?

Quote:

discharge becomes suction at some point right?

Semantically yes and technically no, for a closed loop. The boundary between suction and discharge is where you install your expansion tank (actually, where you should install it).

The static pressure is high at the pump discharge and goes on reducing along the height of upward pipe and becomes minimum at some point (which is the highest point in the loop) and then agian increases along the downward pipe. You will get the best advantage of your tank if you install it at the break point. If you have the heating set up somewhere from the tank to the pump suction flange, the increased vapor pressure decreases the NPSHa. If you have it in the pump discharge (i.e between pump discharge flange and the user point) then there won't be any increment in vapor pressure due to the temperature drop of circulating fluid at user equipment.

RE: How to calculate NPSHA in a closed pumping Loop?

(OP)
..thank you for your comments.  If, as you suggest, we have "heating set up somewhere from the tank to the pump suction flange", won't we need to increase expansion tank pressure?  (Assuming that NPSHa-NPSHr <0?) thanks Quark! Not sure what the incentive is for you to answer these kinds of questions; but it is very much appreciated!

RE: How to calculate NPSHA in a closed pumping Loop?

Come on guys.  

There's no difference between a closed loop system and a run straight configuration, except the closed loop system has a loop... and a closed "vent" valve.  Its still all E = Z + p/g + v^2/2/G - HL.  Where the tanks and heaters go is just a matter of optimization.  In a closed loop system, dissolved air shouldn't be a problem as it should get trapped in an expansion tank, and heaters should not be located before the pump to keep the vapor pressure of the water cooler (and the NPSH higher).  If the last run is small diameter, long and horizontal just before it gets to the pump, you CAN have NPSH problems.  Nothing else to it even it you build them on the Moon.

http://virtualpipeline.spaces.msn.com

RE: How to calculate NPSHA in a closed pumping Loop?

(OP)
Big Inch....not sure who are you directing your comments too and I'm not clear on what you are intending to discuss.  I think we all understand that the loops are the same.  You've made that point earlier. The other folks are trying to address my comment about adding in a future HXT loop.  It appears that this addition may reduce NPSHa and in order to rectify that...one solution might be to increase the pressure in the expansion tank. (that is if we need to...because NPSHr - NPSHa may be adequate.  We dont know at this point.)  So...if we look at your last sentence...and we have NPSH problems after the additional piping has been installed - wouldn't we be able to increase pressure on the expansion tank to alleviate?  thanks again for your comments and willingness to weigh in...

RE: How to calculate NPSHA in a closed pumping Loop?

Right. I should have directed my previous comment to TBP 24 Oct 07 21:57.  BigInch - These closed loop systems are a different animal from the "one-way-trip" pump applications...[I hadn't commented since then].

OK..now.  Yes you're correct. If A is still > R you will not have to increase pressure.  If A < R you must increase pressure somehow.  If you can add compressed air into the tank, the pressures at all points in a closed system will increase by a corresponding amount.

 

http://virtualpipeline.spaces.msn.com

RE: How to calculate NPSHA in a closed pumping Loop?

(OP)
hey thanks again for jumping in.  I do appreciate the help and the comments do help me (force?) me to understand the nuances....I'll see you guys on another topic!

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