Okay, let's run the calculations.
Someone should.
570 amps at 24 volts equals 13,680 watts
For ~14kW a genset I'd think you will probably be using at least a 25 HP engine.
I figured a 570 A alternator was probably a Neihoff N1609 or something similar.
http://www.militarysystems-tech.com/files/militarysystems/supplier_docs/N1606-Brochure.pdf
Therefore at 3,500 RPM it'll produce about 500 A at 28 V, e.g. 14,000 watts divided by 746 HP/watt = 18.8 HP, so at an actual efficiency of 0.78 you get 18.8 HP/.78 = 24.1 HP input needed. The 3,500 RPM was picked because it's a realistic RPM for an engine despite that the 570 A isn't actually produced until you exceed 7,500 RPM.
At full rated output of 570 A @ 28 V (15,960 watts or 21.4 HP) you'd need an engine capable of 30.1 HP @ 8,000 RPM, pretty much just as john suggested and basically impractical due to RPMs required.
Note: Power factor correction is likely 80 percent so HP required is probably closer to 23 HP....
Power factor is a question on the load side. That's how a 20 HP/15 kW generator becomes 18.75 kVA @ 0.8 pf. It's the same real work, the apparent power differs with the voltage and current out of phase. Since there was no mention of an inverter or AC power by john this is aside a point that is already aside the point (does that make this an apparent thread?). Power factor is important when sizing AC generators to know the engine's actual capability since it only knows how to develop real power at the shaft. This also why the utility puts power factor correction on your line if your load is out of phase. They're only capable of billing you for real power and not getting paid for the reactive power they had to generate to drive your shop full of motors. So they hang a capacitor bank on your line to make sure they aren't short changed.