The way we are redoing it, it is not 3 systems it will become one large system. Here is what I got from an engineer friend of mine. He is possibly the smartest person I've ever met because he does this type of math in his head for fun.
Power (P) = Volts (V) x Amps (i) x Efficiency (h) x Power factor (f), or i = P/Vhf. The value of f is strongly dependent on load, varying from about 0.4 at 25% power to 0.8 at full power. Your motor is drawing about 50% of full load amps, so it is producing about 0.5 x (0.4/0.8) = 25% of its rated power. You wish to draw 77 amps, or 90% of its full load current, and I'd estimate its power factor at about 0.75 at that load. An affinity law states that power is proportional to the cube of speed, so N2 = N1 x [(77)(0.75)/(44)(0.4)]^(1/3), where N2 is the desired speed and N1 is the current speed.If my assumed power factors are correct, this is about 1857 rpm. Before buying any sheaves, contact the motor manufacturer for a power factor curve, and the fan manufacturer to verify that the fan is rated for this speed.
Any objections to this?
My objection is that besides leaving off a portion of the power formula (the sq. rt. of 3, because it's HIGHLY unlikely that you have a 75HP single phase motor), the entire concept is fraught with assumptions and inaccuracies. And getting a "power factor curve" from the motor mfr. is absolutely pointless. The curve (if one were to exist) would only be able to tell you what the PF will be when you ALREADY know the load on the motor. So if you ALREADY know the load on the motor, why do you need a curve? YOUR problem is, that you do not already know the load on the motor!
What you need is a watt meter. It will take in amps and voltage, and be capable of integrating the phase angle differences in them (cosine theta), which is what we call "power factor", in order to give you the electrical watts consumed (absorbed power). That, times the efficiency, often available as a curve from the motor mfr., will give you a MECHANICAL load on the motor shaft. THAT mechanical load is what will increase at the cube of the speed change by increasing the fan speed.
But before you waste a lot more time on this, let's just pretend for a moment that your friend's guesstimate is correct, that your motor is only 25% loaded. That would mean that your fan is using only just over 18HP. If you increase the speed to 2000RPM from 1250RPM, an increase of 57%, that means the shaft (mechanical) kW required from the motor will be 1.57
3 or 386% of what you are using now. So again, IF you are only using 18HP from a 75HP motor, and you increase the load by 386%, the motor shaft power required by that motor becomes 69HP and it could work.
Now apply logic to this:
If the fan COULD have run with only 18HP, WHY DID THE ORIGINAL DESIGNER USE A 75HP MOTOR???!!!
Honestly, it makes NO sense. CONSERVATIVE design ideals from 20+ years ago might have made an Engineer over design by 20-25%; we used to call that a "fudge factor". But over design by 400%???? No. Something is missing in all of this.
Get / rent yourself a real WATT meter or hire an electrician who has one and knows how to use it. Open up ALL of the dampers, then measure the motor watt use. You will likely find that under full operating conditions, that motor is likely consuming no less than 56HP (25% fudge factor), probably more. So even if it is 56HP, you will need about 135HP from the motor, and your motor is only 75HP.
