A 20hp supply fan is currently operating 24/7. What is the yearly
motor energy savings if we install a VFD on the supply fan and operate
the fan at 50% speed during unoccupied hours (8pm to 6am). Assume
motor efficiency = 90%

Hi rass27-ga,

  Since your question does not have enough information, Here is the
answer(once you fillin the missing parameters):

Relative motor energy savings= (Energy used at full speed)-(Energy used at half
speed for 10 hrs).
=> (24*365*20*0.746+24*365*20hp*0.746*0.1) -
((14*365*20*0.746+14*365*20*0.746*0.1) +
(10*365*(?hp50%)*0.746+10*365*(?hp50%)*
0.746*(?(100-%eff@50%rpm)/100)))

If you substitute the values for the hp at 50% speed (?hp50%) and the
corresponding efficiency(%eff@50%rpm), you should get the kWh savings for
a year (multiply by the unit kWh charges to get the saved amount).

Regards
ldavinci-ga

The hp is 20 and the efficiency is 90%.

This is a question on a quiz. There is no additional information, is
there an answer?