<font face="georgia" size="3"><p style="margin:0;padding:0;font-family: georgia; font-size: 12pt; overflow-wrap: break-word;">isn't there a formula like head in feet X gpm divided by 10 = approximate wattage?<br /><br />todd</p>
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<p style="margin:0;padding:0;font-family: georgia; font-size: 12pt; overflow-wrap: break-word;"><br /><br />On Wednesday, July 10, 2019 7:10pm, "Jerry Shafer" <jerrysgarage01@gmail.com> said:<br /><br /></p>
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<div>You need head (pressure) and GPM, clean degree free, no silt, a catch basin to stall the water, big pipe long run, try to keep the rollercoasters out, (rises), and no 90's, that's just to start. Low head waterwheels can produce lots of power to but you need a river of water.
<div>Jerry</div>
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