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I have a water channel with a flow of 1.5 m3 per second. What would the potential energy be if I install a waterwheel with fins almost touching the sides and bottom?

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    $\begingroup$ mgh minus some losses, if it's a closed channel (pipe). Trivial math and a little guesswork on the losses. 1/2mv^2 (again minus losses) for an open channel where you're using the kinetic energy of the water. $\endgroup$ – Brian Drummond Sep 17 '16 at 13:29
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    $\begingroup$ You need height. If you have a level channel and put a waterwheel in it and try to extract energy the wheel will stop (or run very slowly) and the wheel will become a dam. $\endgroup$ – Transistor Sep 17 '16 at 13:37
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Flow rate by itself doesn't tell you the potential for extracting power. You need to know the pressure drop, or if this water is open to the atmosphere, the height drop (which is a special case for determining pressure drop).

For example, let's say you have 1.5 m3/s water flow that is dropping 2 m in height. 1.5 m3 of water has a mass of 1500 kg, which on earth weighs 14.7 kN. That times the drop distance of 2 m yields 29.4 kJ/s, which is better expressed as 29.4 kW. Of course any real device won't actually get all of that due to some inevitable inefficiency.

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Mechanical and electrical energy is the product pairs of variables, like torque.angle, force.distance, volts.charge, pressure.volume, and so on. If you only have one of those, it's like the sound of one hand clapping.

Your flow rate is volume per second, so now all you have to do is define the pressure difference available to get the power (energy per second). Keep all your units in SI, and it will just come out correctly (Pascals, m3, watts, Joules).

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