Say you have 20 kg of water moving at a speed of 10m/s, if the KE formula is used, it would have 1000 joules of energy per second. Is this accurate or is there another way of calculating it?
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No its not accurate.
In the case you are describing the 20 kg of water have 1000 joules of energy.
If you had a mass rate of 20 kg/s and the fluid was moving at 10 m/s then you could have claimed that there are available 1000 joules per second.
Transistor
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NMech
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This idea works reasonably well for an impulse turbine (e.g. a Pelton wheel) but it doesn't work at all for a reaction turbine (i.e. something with blades which the water flows through).
The reason is simple when you realize what it is: if you extract all the KE from the water, its velocity is now zero, so how does it move out of the way to let more water through the turbine? For a Pelton wheel, this is possible, because the buckets on the wheel rotate at half the speed of the incoming water, so the water deflecting off the buckets has zero speed relative to the body of the machine and gravity makes it fall out of the way of more water.
Of course practical Pelton wheel turbines also have much higher inlet velocities - e.g. 100m/s not 10m/s.
alephzero
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Usually a first approximation is based on Q * g * h,
Q flow in l/s
g gravity - assumed to be 10 since efficiency will cover difference between 10 and 9.81...
h is height in metres.
Solar Mike
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