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.

Estimating your site Hydro Power

HVC.RR
Honderosa Valley Consulting and Renewables Research

One has to remember that actual work done is force delivered times the distance that the force moved (the object) (per unit of time). So if you break a sweat pressing on an immovable wall with 100 lb pressure for several hours, you have done zero work! Because Work=Force X Distance, so 100Lb X zero distance= zero work no matter how long a time you press. But move the wall (or weight) 1 foot with your 100 Lb pressure exerted and you've done 100 ft lbs worth of work. If you do this amount of work in 1 second then you have produced about 1/5th of a horse power (HP) for that second . (550 ft lbs per second = 1 HP) One HP = about 746 watts, so you could have lit up a 100 watt light bulb for about 1.2 seconds with that work you just did. So keep lifting a 100 lb weight 1 ft every second and that is what it takes to light one light bulb!! (Approximately, there are some losses associated with the conversion of the mechanical up and down motion to electrical energy.)

In this photo you are looking at about 6000 Gal/Minute dropping through 200 feet.

Now let's relate that to hydro power.

A gallon of water weighs about 8 Lb. So if you run (drop) 13 Gal/sec (about 100 Lb) through 100 feet of height in 1 second, your total energy would be about 100 lb X 100 ft =10,000 ft lbs/ sec. divide this by 550 ft/lbs per second per HP and we get 18 HP. So at 746 W per HP our gross power in watts = 13,000W. If we sum up all the losses due to friction / heat generated etc we would probably end up with about half of that as usable electrical energy. So we could light up about 70, 100W light bulbs for as long as the water flows at 13 gal/sec or 780 gal / min. (Can you fill 3, 250 gal (standard oil) tanks in 1 minute from your water source and pipe it down hill 100 feet vertical?) Or a 5 gal pail in 1 second? Compare your stream to these kinds of estimates to get an initial approximation of potential.

If I close the valve down in my power house the pressure reads 93 PSI. This is analogous to the pressure exerted on the immovable wall above. No movement (distance = 0) no work. When I open the valve fully the water forces its way through the turbine and out the discharge (draft tube) at a rate of 750 Gal/minute. The inlet pressure drops slightly to 90 psi due to the friction the now moving water encounters against the pipe walls. These are the stats for the smaller of the two turbine / generator setups.