# EnergySmart School Inventors

## Miniature Hydroelectric Power Plant

By MHEPP staff

In the ideal Pelton wheel, a water jet impinges on a bucket which is curved to reverse the direction of the water's flow without reducing the water's speed. In practice, a bucket can't precisely reverse the jet, else the water leaving would crash into the water entering or other buckets, but the reversal is nevertheless adequate. In order to get the maximum power from a jet of water, a motor should be able to reduce the water's speed to zero. If the bucket is moving in the same direction as the water, but half as fast, then when the water is reversed by a bucket, the water's speed will be zero relative to the frame of reference of the laboratory.

Assume the nozzle is 1/8" diameter for a first guess. What is the pressure drop through the nozzle if the flow rate is 2 gpm (7.7 cubic inches per second), and the nozzle is 100% efficient?

v = 7.7/((pi/4)*(1/8)^2) = 627.45 inches/sec

velocity head = v^2/2g = 509.85 inches

equivalent pressure = 509.85 inches * .03611 lb/in^3 = 18.4 psi

A good nozzle will have almost 100% efficiency and so the actual pressure required to deliver 2 gpm through our 1/8" nozzle won't be more than a very few percent higher than this 18.4 psi.

As shown above, jet speed is 627.45 ips for 2 gpm flowing through a 1/8" diameter nozzle. Wheel pitch speed will be half that or 314 ips. We have two wheels; one has a pitch diameter of 4", the other 2". So one has a circumference of 4*pi = 12.57 inches, and the other half that. Shaft speed will then be 314/12.57 = 25 rps = 1498 rpm. The smaller wheel will have twice that shaft speed.

If our wheel speeds are 1500 and 3000 rpm, and we want, say, 10 V out of our generator, then we need servo motors with 150 and 300 rpm per volt, or 6 and 3 volts per 1000 rpm.

Now for wheel or shaft torque. Force is mass flow rate times change in velocity. (This can be derived from the impulse-momentum equation which can be derived from F = m*a.) Change in velocity also includes the sign (direction) of the initial and final velocities.

F = m-dot*dv

m-dot = 2 gpm*specific weight/gravity = 7.202E-4 lb-sec/in

dv is the aforementioned 627.45 in/sec in either lab frame or bucket frame.

So F = m-dot*dv = 7.2E-4 lb-sec/in * 627.45 in/sec = 0.452 lb

If the pitch diameter of the wheel is 4", the torque will be 0.9 in-lb, and if it's 2", 0.45 in-lb.

Power = 2*pi*T*n / 33000 = 2*pi*(.452/12)*3000/33000 = .0215 HP times 746W/HP = 16 watts

Power check:

Hydraulic power = q delta P = 7.7 in^3/sec * 18.4 psi = 141 in-lb/sec divided by 6600 in-lb/sec-HP = .0215 HP

So we have calculated the power coming out of the wheel to be 1/50th horsepower or 16 watts by two independent methods.

What can we say about the general characteristics of a Pelton wheel? If the wheel is not turning, the water will have its direction reversed, ending up traveling at full velocity in the opposite direction. This doubles the dv in the m-dot*dv equation. So if the wheel is not turning, torque is doubled as compared to the torque at optimum wheel speed. If there is no load on the wheel, the wheel will try to go as fast as the water jet, with the result that there will be no change in the water's speed and hence no torque. This top speed is double the optimum speed. We can see that the Pelton wheel is a very forgiving machine, tolerating load abuse without causing disasters such as blowing up or breaking its shafting.

The Pelton wheel is historically important because it is low-tech, and easily made, while having high efficiency. Higher efficiency means that it delivers more power from a given water supply than another low-tech machine.