Confirm AC generator output frequency calculation
The synchronous speed for an electric induction motor is determined by the power supply frequency and the number of poles in the motor winding and can be expressed as:

$$ n = f \frac {2}{p} 60 $$

where


n = shaft rotation speed (rev/min, rpm)
f = frequency (Hz, cycles/sec, 1/s)
p = number of poles
60 = number of seconds in a minute


Rearranging for f we get

$$ f = \frac {n \cdot p}{2 \cdot 60} $$


  Based on a ... 12 pole, 2800 RPM ...


$$ f = \frac {2800 \cdot 12}{2 \cdot 60} = 280~Hz$$

Your calculations are correct. Your frequency is rather unusual.


  Will the frequency be suitable for the heating elements designed to be used at 50 Hz?


If the heating elements are just resistive they will behave normally at these low frequencies. If they exhibit significant inductance their impedance to AC will rise with frequency. It's unlikely to be a problem for water heaters as they use straightish elements rather than coils of wire.

Be careful not to run transformers or relay coils from this 280 Hz supply unless you have confirmed that they are suitable.