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Microhydro assessment: Estimating power

Our small creekThe 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.  Read all of the entries:





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I've really enjoyed this series about microhydro. I'm definitely going to look at it more. I too was wondering if roof runoff would be an option. Perhaps a small earthbag dam would be feasible and not too expensive though probably labor intensive.
Comment by Andy Fri Mar 5 15:16:43 2010
I'm glad you enjoyed it! Actually, we've been holding onto the plastic burlap bags our catfood and dogfood comes in because they look so useful. I'll bet they'd make great earth bags.
Comment by anna Fri Mar 5 19:19:50 2010

Power from potential energy is mass flow (kg/s) times the accelleration of gravity (m/s2) times height (m). (check for yourself; the product is kg times m2 divided by s3, 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/s2.

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 26-pole 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)

Comment by Roland_Smith Sat Mar 6 09:11:29 2010
I couldn't get the units to match up with those formulas, which were from the Hydropower book. I suspect there is a built in fudge factor to partially deal with loss of efficiency, perhaps? I do notice that your estimate of what we'd actually get out of the creek is in line with his formula.
Comment by anna Sat Mar 6 15:47:02 2010

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 well-liked 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.

Comment by Roland_Smith Sat Mar 6 17:34:23 2010
My brain is fried this weekend, but I'll bet you're right. :-)
Comment by anna Sun Mar 7 08:28:58 2010

One very unique homestead, $1,500 per acre, the opportunity of a lifetime