- #1
rattis
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How did British engineer John Smeaton measure the efficiency of an overshot waterwheel?
I am doing an experiment with overshot waterwheels to find their efficiency, and found that John Smeaton measured their efficiency at about 65%. However I have no idea how he got this number, and I cannot think of an accurate way of calculating its efficiency.
My latest idea is to use a dynamometer to find the difference in forces, then multiply by the distance traveled (2prn (where n is number of revolutions in t seconds, and r is the radius of the axle)) Then divide it all by t; to get Power. This would be compared with a theoretical power output calculated from gravitational and kinetic energy going in.
Is this the correct way to go about this? Or not?
I am doing an experiment with overshot waterwheels to find their efficiency, and found that John Smeaton measured their efficiency at about 65%. However I have no idea how he got this number, and I cannot think of an accurate way of calculating its efficiency.
My latest idea is to use a dynamometer to find the difference in forces, then multiply by the distance traveled (2prn (where n is number of revolutions in t seconds, and r is the radius of the axle)) Then divide it all by t; to get Power. This would be compared with a theoretical power output calculated from gravitational and kinetic energy going in.
Is this the correct way to go about this? Or not?