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
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?
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RE: How to calculate NPSHA in a closed pumping Loop?
RE: How to calculate NPSHA in a closed pumping Loop?
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
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RE: How to calculate NPSHA in a closed pumping Loop?
RE: How to calculate NPSHA in a closed pumping Loop?
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.
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RE: How to calculate NPSHA in a closed pumping Loop?
RE: How to calculate NPSHA in a closed pumping Loop?
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RE: How to calculate NPSHA in a closed pumping Loop?
RE: How to calculate NPSHA in a closed pumping Loop?
thread407-179090: Closed system and NPSHa
thread378-163255: NPSHA in a closed loop system
RE: How to calculate NPSHA in a closed pumping Loop?
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?
RE: How to calculate NPSHA in a closed pumping Loop?
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?
RE: How to calculate NPSHA in a closed pumping Loop?
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?
RE: How to calculate NPSHA in a closed pumping Loop?
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.
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RE: How to calculate NPSHA in a closed pumping Loop?
RE: How to calculate NPSHA in a closed pumping Loop?
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.
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RE: How to calculate NPSHA in a closed pumping Loop?