hello there,
Does anyone know the formula to caclulate the power in Kw for a horizontal
kaplan turbine:
head of 10M, pipe diam. 0.6M, length 33 M, waterflow 1.2 M3/sec. propeller diam
0.35 M, 3 adjustable blades, alternator 1000 rev/min.
And how can we calculate the waterspeed just before the turbine?
Regards, Peter Dekker, Mt. Kenya
Radu Babau - VARSPEED Hydro <[EMAIL PROTECTED]> wrote:
Hello Jean,
This induction machine (200 kW, 60 Hz, 480 V, 900 rpm) has 8 poles.
The machine torque is proportional with U/f ratio, and since (480/60)=(400/50),
it means that you will be able to get the same torque (2,125 Nm) out of this
machine when connected to European 400 V, 50 Hz grid.
However, 50 Hz on a 8 pole machine means 750 rpm. Since power is torque times
speed, you will only get 166 kW output power.
The rated current and power factor will be the same.
With this motor (iron core) you will not be able to squeeze more power, so
don't waste your money rewinding it or so.
Have fun,
Radu
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