Monday, February 15, 2010

Pumps as Turbines (PaT) Motors as Generators


I have been using Pumps as Turbines and motors as generators very successfully for almost four years now. The optimum (maximum power) speed of a PaT is around 1/2 of its no load speed. So to test,run your PaT and generator with the available head but no electrical load or excitation. Then, for maximum power, you should load the system so it runs at 1/2 this No Load speed. So let's assume you have a six pole electric motor running as a generator at 1200 RPM (nominally). You would be looking for a No Load speed of (approximately) 2400 RPM at the generator shaft. If the PaT is direct coupled the only thing that can be tuned is the impeller diameter assuming the head is fixed. Cut down (on a lathe) the impeller diameter to increase the No Load RPM. If the generator is coupled to the PaT by a V belt the shiv ratios can be changed to give the correct RPMs.


How can I estimate roughly the maximum no load speed when impeller diameter and the net head are known?


No Load RPM= (19.1)(SQRT(64 H))/D
Where H= Head in feet, D= impeller diameter in feet (not inches)

So if H = 200 feet and Impeller Diameter is 7 inches we get a No Load RPM = 3699
This is a little high for a 4 pole (1800 RPM) generator, I'd rather be on the low end for better efficiency (less friction due to lower velocities). I would choose an 8" impeller. Or, (even better) go to a 6 pole 1200 RPM (direct coupled) motor, bring the no load speed down to 2400 RPM and calculate the required impeller diameter as below.


More often, you would know the full load normal operating speed of the generator to produce 60Hz from the name plate or the number of poles. Also the head is fixed and known. So you'll want to know what the impeller diameter should be for a direct coupled setup:


Impeller Diameter in inches = 230(SQRT(64 H)/R


Where H = head in feet, R = No Load RPM and SQRT = Square root


So if H = 200' and R = 2400, we get an impeller diameter of 10.8 inches.


These admittedly rough calculations work well with Pelton wheels as well as Francis type runners.


By the way, larger pumps are well designed and optimized to be very efficient. When run in reverse as turbines they perform just like Francis type turbines and are every bit as efficient, assuming you match up head, impeller diameter and operating speed.
The graph above shows how I reverse engineered the pump curves for for operation as a turbine. This analytical approach is not as easy to understand as the empirical approach described above. Note the U shaped > 73 % efficiency region. This is where you want to be operating this particular PaT with a 12 inch impeller, 230 ft of head, and using 750 Gpm with expected output of 25 Hp mechanical energy. BTW, pumps are designated by their outlet X suction inlet X impeller diameter. So this is a 3X4X12 end suction pump with a 30HP, 1800 RPM 3 Phase electric motor.

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Jeff said...

Hey Rob,

Fantastic blog! Now I really need to go down and take a tour!

Nice areal shot of the house too ;-)


Best,
Jeff

Palang Thai said...

I'm looking to do a grid-connected PAT project with a water utility in Thailand. I've had some (good) experiences building PAT off-grid systems on Thai/Burma border (http://palangthai.blogspot.com/2008/02/3-kw-pump-as-turbine-microhydro-at-mae.html).

Our site has 15 meters head and about 93 liters/second flow. Using the "Pumps as Turbines" book by ITDG, I calculate we're looking for a pump optimized for about 73 l/s and 10 meters head. Any ideas of good pumps for this range? Fairly low head, high flow compared to your installation.

Rob said...

Hi Palang,
With a volume of 1500- 2000 GPM at head of 50 feet would require a 6X8 pump.
It will be hard to find a 6X8 with a small enough impeller diameter to get the loaded speed up to 1500 RPM so I would look for a 3 phase motor with 6 poles (1000 RPM for 50Hz) or even 8 poles. Size can be from 10, 15 or 20HP. Not critical as long as it is more watts than you plan to pull out assuming enough water in.

So: with 1000 RPM at full load, the no load speed will be around 2000 RPM. (about 2X)
Substituting in to the impeller diameter formula gives:

230 x SQRT(64x50’) all divided by 2000 RPM = 6.5 inches impeller diameter.
Looks like you might be better off with an 8 pole motor for a No Load speed of 1400 rpm ( 2X normal loaded name plate speed ) giving you an impeller diameter of 9.3 inches.

So my estimate for the pump you want is a 6x8x9.3” .
The motor would be a 10 to 20 HP, 3 phase, 240V 8 pole, 750RPM.

That might be a hard to find motor so you might want to couple with pulleys and belt to match up the speeds, that is to go from 750 RPM to 1500 RPM (4 Pole motor).

Happy Hydro - and contact me directly for more question .
Honders@hvc.rr.com