Answer
$E_{save}=9289.81\frac{kWh}{year}$
$Cost_{save}=1114.7$
$t_{payback}=0.00637 years$
Work Step by Step
$75hp*\frac{0.746kW}{hp}=55.95kW$
The energy save will be:
$E_{save}=\frac{W}{\eta_{1}}t-\frac{W}{\eta_{2}}t=55.95kW*(4368\frac{h}{year}*0.75)*(\frac{1}{0.91}-\frac{1}{0.954})$
$E_{save}=9289.81\frac{kWh}{year}$
The cost save will be:
$Cost_{save}=Cost*E_{save}=0.12\frac{dollars}{kWh}*9289.81\frac{kWh}{year}$
$Cost_{save}=1114.78\frac{dollars}{year}$
To determine the simple payback period first we need to determine the implementation cost, wich is the differential cost betwen the high and standard efficiency motor:
$Cost_{differential}=5520dollars-5449dollars=71dollars$
Then the simple payback period will be:
$t_{payback}=\frac{71dollars}{1114.78\frac{dollars}{year}}=0.00637 years$
