The power is = head*waterflow (in kg/s)*9.81 in your case is
10*1200*9.81 = 120KW .this value must to multiply by effeciency of
turbine. For Kaplan turbine effeciency is about 0.88 to 0.94 max.
The water speed is squareroot(2*head*9.81) in your case is 14m/s.
--- In [email protected], Gpower NGO <[EMAIL PROTECTED]> wrote:
>
>
> 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
>
> [Non-text portions of this message have been removed]
>
>
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