blog/Microhydro assessment: Estimating power
http://www.waldeneffect.org/blog/Microhydro_assessment:_Estimating_power/
waldeneffectikiwikiSat, 26 Nov 2011 12:34:25 -0500I've really enjoyed this series http://www.waldeneffect.org/blog/Microhydro_assessment:_Estimating_power/comment_1_ebb721440d8f69784a147a38a6d116bc/
http://www.waldeneffect.org/blog/Microhydro_assessment:_Estimating_power/#comment-6950be20544ff56d0cf0a28d65064b40
AndyFri, 05 Mar 2010 15:16:43 -05002011-11-26T17:34:25ZI'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 <a href="http://www.earthbagbuilding.com/faqs/misc.htm#retaining">earthbag dam</a> would be feasible and not too expensive though probably labor intensive.
comment 2http://www.waldeneffect.org/blog/Microhydro_assessment:_Estimating_power/comment_2_7fecfe0b356b8b46da09858dd4d59c94/
http://www.waldeneffect.org/blog/Microhydro_assessment:_Estimating_power/#comment-9a15f5172f8d36cc29ddfa3857bba77f
annaFri, 05 Mar 2010 19:19:50 -05002011-11-26T17:34:25ZI'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.
I think the formula is wronghttp://www.waldeneffect.org/blog/Microhydro_assessment:_Estimating_power/comment_3_d67a8a77d26e071e6788b47ef450b751/
http://www.waldeneffect.org/blog/Microhydro_assessment:_Estimating_power/#comment-92824ddfee7d25762a0393f96b0df0bc
Roland_SmithSat, 06 Mar 2010 09:11:29 -05002011-11-26T17:34:25Z<p>Power from potential energy is mass flow (kg/s) times the accelleration of gravity (m/s<sup>2</sup>) times height (m).
(check for yourself; the product is kg times m<sup>2</sup> divided by s<sup>3</sup>, which is Nm/s = J/s = W)</p>
<p>According to the <a href="http://en.wikipedia.org/wiki/Units_%28software%29">units</a>(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<sup>2</sup>.</p>
<p>So the total potential power of the creek is 1.26<em>9.81</em>0.91 = 11.2 W.</p>
<p>Practically though, I think you'd be lucky to extract half of that. An efficient small generator, like the 26-pole <a href="http://www.nabendynamo.de/produkte/SON_28.html">Schmidt hub generator</a> 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)</p>
comment 4http://www.waldeneffect.org/blog/Microhydro_assessment:_Estimating_power/comment_4_bbd0059d13bc071956cac9c7076334d9/
http://www.waldeneffect.org/blog/Microhydro_assessment:_Estimating_power/#comment-1c24814fef6a332a8f104603e87671bc
annaSat, 06 Mar 2010 15:47:02 -05002011-11-26T17:34:25ZI 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.
My calculation might have been optimistichttp://www.waldeneffect.org/blog/Microhydro_assessment:_Estimating_power/comment_5_2e123db178062d110dfe9f64e251413b/
http://www.waldeneffect.org/blog/Microhydro_assessment:_Estimating_power/#comment-8d0df060443174446f0e71c88f26076e
Roland_SmithSat, 06 Mar 2010 17:34:23 -05002011-11-26T17:34:25Z<p>You're probably right about the fudge factor. But on more reflection I think I might have been too optimistic.</p>
<p>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.</p>
<p>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 100<em>0.7 = 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</em>.7<em>.99 = 69.3% available at the generator. Again assuming the generator is 65% efficient, that leaves 100</em>.7<em>.99</em>.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<em>.7</em>.99<em>.65</em>.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<em>.7</em>.99<em>.65</em>.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 <em>might</em> have 25% of the original power left. <img src="http://www.waldeneffect.org/smileys/sad.png" alt=":-(" /></p>
<p>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.</p>
comment 6http://www.waldeneffect.org/blog/Microhydro_assessment:_Estimating_power/comment_6_8f5f2c19564ec1cc7107645b20a16d90/
http://www.waldeneffect.org/blog/Microhydro_assessment:_Estimating_power/#comment-758e3a72c51c1c83e0c8c1bbfcaf0c85
annaSun, 07 Mar 2010 08:28:58 -05002011-11-26T17:34:25ZMy brain is fried this weekend, but I'll bet you're right. <img src="http://www.waldeneffect.org/smileys/smile.png" alt=":-)" />