3 ms·

Sure. With a hydroelectric turbine you can produce electrical power (by black magic as far as I'm concerned) or mechanical power. There is a terrific document

Sure. With a hydroelectric turbine you can produce electrical power (by black magic as far as I'm concerned) or mechanical power. There is a terrific documentary called Ben's Mill on such a mechanical application here: https://www.youtube.com/watch?v=l2KJbRHO76s https://www.youtube.com/watch?v=l2KJbRHO76s That will explain visually how mechanical power can be coached into performing useful work. It's also worth watching just to see the cows reaction to their new water vessel at the end, they are like animals out of Disneyland. Mechanical power when converted to electrical power undergoes a drop in performance (similar to other power conversions, something is lost). There is a fascinating article here on the subject, I quote: "The hydro power installations in use today are actually less efficient than those of earlier centuries" "Direct hydropower is at least twice as economically viable as a hydroelectric operation with the same rate of energy production." http://www.lowtechmagazine.com/2013/08/direct-hydropower.html http://www.lowtechmagazine.com/2013/08/direct-hydropower.htm... Also take a look at this: http://www.lowtechmagazine.com/2013/09/power-from-the-tap-water-motors.html http://www.lowtechmagazine.com/2013/09/power-from-the-tap-wa... That is hacking if I ever saw it! I intend to construct a micro hydroelectric plant myself, so ask away if you've any questions. I love the idea of marrying up old school technology with modern technology. I also have a terrific idea for lighting using heliostats, which combines tech literally thousands of years old with the latest in machine learning.

msandford 10y ago
That's not even close to an explanation. If 1kg of water descends 1m the most power you can hope to extract from that is U = m x g x h or 1 * 1 * 9.8 ~ 10 J (https://en.wikipedia.org/wiki/Potential_energy https://en.wikipedia.org/wiki/Potential_energy) If you can extract 90% of that, it's 9J. If you can double that efficiency, it's 18J. Which would make the "over unity" generator folks on YouTube pretty happy, but the universe generally says you can't really do. So which is it? Is hydro 90% efficient, or can you get double the power output if you want mechanical energy? You can't have both.
- internaut 10y ago
  Hmmm. I'm not claiming to defy physics although I see how it came across that way. That was not my intention. If you read the literature you can see why I was phrasing it like that so perhaps I too was misled. http://www1.agric.gov.ab.ca/$department/deptdocs.nsf/all/eng4431 http://www1.agric.gov.ab.ca/$department/deptdocs.nsf/all/eng... The potential energy harvest from a hydro setup must be 90% of available energy. In practice half is lost due to the reasons outlined in the low tech magazine article I linked to. So the electricity production is lower than if we had the turbine hooked up to a motor. The other possibility is that small scale hydro is of a lower efficiency and that this explains the paradox. That is: the literature is talking of a large scale operation that can achieve 90% efficiency but smaller schemes incur losses that explain why converting a small hydro setup back to providing mechanical energy can double the output.
  - msandford 10y ago
    Assembling and paraphrasing your information: Turning the shaft horsepower from a hydro turbine into electricity, transmitting it through wires, transformers, etc, and turning it back into shaft horsepower through motors, etc is inefficient due to conversion losses. Just using the shaft horsepower directly incurs none of those conversion losses. They can be substantial enough in some situations to approach 50%. This doesn't mean you can harvest more energy than the water can provide, it means that by eliminating losses you can USE more of the energy that the water provides, instead of losing it to heat or noise or whatever other conversion losses. That's an explanation. What you have linked to and hinted at is only the start of one.

msandford 10y ago

That's not even close to an explanation. If 1kg of water descends 1m the most power you can hope to extract from that is U = m x g x h or 1 * 1 * 9.8 ~ 10 J (https://en.wikipedia.org/wiki/Potential_energy https://en.wikipedia.org/wiki/Potential_energy) If you can extract 90% of that, it's 9J. If you can double that efficiency, it's 18J. Which would make the "over unity" generator folks on YouTube pretty happy, but the universe generally says you can't really do. So which is it? Is hydro 90% efficient, or can you get double the power output if you want mechanical energy? You can't have both.

internaut 10y ago
Hmmm. I'm not claiming to defy physics although I see how it came across that way. That was not my intention. If you read the literature you can see why I was phrasing it like that so perhaps I too was misled. http://www1.agric.gov.ab.ca/$department/deptdocs.nsf/all/eng4431 http://www1.agric.gov.ab.ca/$department/deptdocs.nsf/all/eng... The potential energy harvest from a hydro setup must be 90% of available energy. In practice half is lost due to the reasons outlined in the low tech magazine article I linked to. So the electricity production is lower than if we had the turbine hooked up to a motor. The other possibility is that small scale hydro is of a lower efficiency and that this explains the paradox. That is: the literature is talking of a large scale operation that can achieve 90% efficiency but smaller schemes incur losses that explain why converting a small hydro setup back to providing mechanical energy can double the output.
- msandford 10y ago
  Assembling and paraphrasing your information: Turning the shaft horsepower from a hydro turbine into electricity, transmitting it through wires, transformers, etc, and turning it back into shaft horsepower through motors, etc is inefficient due to conversion losses. Just using the shaft horsepower directly incurs none of those conversion losses. They can be substantial enough in some situations to approach 50%. This doesn't mean you can harvest more energy than the water can provide, it means that by eliminating losses you can USE more of the energy that the water provides, instead of losing it to heat or noise or whatever other conversion losses. That's an explanation. What you have linked to and hinted at is only the start of one.