Dollars Savings Per Year =
0.746 x HP x [(1/EffA) - (1/EffB)] x hours x $/KWh
Where:
HP = load on the motor
EffA = efficiency of motor A at HP (expressed as decimal)
EffB = efficiency of motor B at HP (expressed as decimal)
EffA < EffB
Hours = hours of operation under load
$/KWh = average cost of electricity per kilowatt hour
The one large dynamic you have done is changed your power factor by raising the voltage . In doing this you have directly effected your motor efficiencys hence the savings.
Mike
