No argument with the above analysis, but with a diesel engine you can assume that the torque curve is fairly flat and that at least 80 percent of peak torque is available from about 1200 to the power peak, in which case the engine should be able to provide the required 23 HP (17.2kW) at 1800.

Look at the peak torque rating, assume 80 percent of this at 1800 (which is conservative), compute the HP and convert to kW (1HP = 0.746 kW). That's how big an alternator it will drive.

I would recommend direct driving the alternator at 1800 rather than trying to drive it from a belt. That's a lot of power to transfer via belt.

Duke