Microhydro assessment: Estimating power
The
final step of assessing your stream for microhydro is doing a bit of
math to determine the creek's power. I'm simplifying a bit here
because you will lose some power due to friction as the water rubs up
against the inside of your pipe, but this formula is good enough for
estimating whether your creek is worth looking into further.
Power
output (continuous watts) = Flow (gpm) X Head (ft) ÷ 10
If you'd rather have
your estimated energy output in kwh/month so that you can compare it to
your electric bill, continue on to this formula:
Kwh/month
= Power (continuous watts) X 0.72
So, it's finally time to
see if our little creek passes the test. She puts out 20 gpm of
water and has a head of about 3 feet. So:
Power output = 20 gpm X 3 ft ÷ 10 = 6 continous watts
Kwh/month = 6 continuous watts X 0.72 = 4.3 kwh/month
Sadly, our little creek
failed miserably  that would be enough to keep the lights on in our
house, but nothing more. As a rule of thumb, you need either a
large head or a large flow to make microhydro appealing, and our little
creek had neither.
On the other hand, we
have several other possibilities on our property that look more
appealing. If we were willing to pay a lot for a run of the river
system, or to build a big dam, our primary creek would definitely
provide all of our power. On the cheaper side, it's possible that
it would be worth our while to tap energy from the spring that comes
out way up on the hill, although it does stop flowing during dry
weather.
Finally, I'm curious
whether there would be a way to make electricity from the water running
off the barn roof if we installed gutters. I envision using tanks
as a storage system and just letting the water leak out slowly, rather
than buying expensive (and environmentally unfriendly batteries.)
I estimate that nearly 4,000 gallons of water flow off the roof each
month, but I guess that's only 0.09 gpm. Back to the drawing
board....
This post is part of our Microhydro lunchtime series.
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Power from potential energy is mass flow (kg/s) times the accelleration of gravity (m/s^{2}) times height (m). (check for yourself; the product is kg times m^{2} divided by s^{3}, which is Nm/s = J/s = W)
According to the units(1) program, 20 gallons per minute is 1.26 liters per second or kg/s, since the denisty of water is almost 1 kg/l. Three feet is 0.91 meters, and the accelleration of gravity is 9.81 m/s^{2}.
So the total potential power of the creek is 1.269.810.91 = 11.2 W.
Practically though, I think you'd be lucky to extract half of that. An efficient small generator, like the 26pole Schmidt hub generator for a bicycle, which is tuned for low rotational speeds doesn't come higher than 65% efficiency. (The link is in German, but look for "Wirkungsgrad" i.e. efficiency)
You're probably right about the fudge factor. But on more reflection I think I might have been too optimistic.
If 65% is about the best you can realistically get for a small generator at low speed (and I suspect it is; the Schmidt alternator is wellliked among cyclists, a group that doesn't have a large power budget to spend!), coupled with turbulence/friction losses in the turbine and transport losses in the wire or conversion losses in the battery, you might be looking at a much smaller value because the losses are cumulative.
Suppose we put the stream of water at 100% of available power. I'm still doing research on turbine efficiency, but let's say 70% for now for a small/simple one. We now have 1000.7 = 70% of the power left at the shaft. Suppos the bearings of the turbine dissipate 1% of the energy at the shaft. That leaves 100.7.99 = 69.3% available at the generator. Again assuming the generator is 65% efficient, that leaves 100.7.99.65 = 45% of the power coming out of the wires of the generator. Such a small generator will probably not give a voltage high enough to enable you to lay a cable to the house, so you use a rectifier and charging electronics to store the energy in a battery. Assuming an efficiency of 80% for the electronics, we have 100.7.99.65.80 = 36% of the power of the stream loading the battery. Say the battery has a load/discharge efficiency of 75%, which seems a reasonable average, only 100.7.99.65.8*.75 = 27% of the power of the stream will eventually come out of the battery. And since almost all household appliances are AC, you'll need an inverter as well. By the time you're powering appliances, you might have 25% of the original power left.
Of course the flowing water is "free", and in that sense one could make the argument efficiency doesn't matter. The downside is that if you have inefficient conversion, your source and conversion machinery will need to be bigger to meet your demands.