Question 10.1: Determine the cost-effectiveness of replacing a 10 HP motor ......
Determine the cost-effectiveness of replacing a 10 HP motor with an efficiency of 85% with a premium-efficiency motor with a rated full-load efficiency of 91.70%. Assume that
• The cost of electricity is $0.10/kWh
• The differential cost of premium versus standard motor is $300
• The average load factor of the motor 0.80
• The average full-load operating hours of the motor are 5000 \ h/year
To determine the cost-effectiveness of installing a premium-efficiency motor instead of a standard-efficiency motor, a simplified economic analysis is used to estimate the simple payback period. The savings in energy use in kWh for the premium-efficiency motor can be calculated using Equations 10.1 and 10.2:
\Delta P_R=P_M.\left\lgroup\frac{1}{\eta _S}-\frac{1}{\eta _E} \right\rgroup (10.1)
\Delta kWh=\Delta P_R\cdot N_h\cdot LF_M (10.2)
kWh_{saved}=N_h\cdot HP*0.746\cdot LF_M\cdot \left\lgroup\frac{1}{\eta _S}-\frac{1}{\eta _E} \right\rgroupwhere
N_h is the total number of hours per year during which the motor is operating (5000 \ h/year).
HP is the rated motor power output (10 \ HP)
LF_M is the annual average load factor of the motor ( LF_M=0.80)
\eta _S and \eta _E are the efficiencies of the standard and efficient motors, respectively (0.850 and 0.917)
Thus, the energy savings in kWh are calculated as follows:
kWh_{saved}=5000*10*0.746*0.80* \left\lgroup\frac{1}{0.850}-\frac{1}{0.917} \right\rgroup=2565 \ kWh/yearTherefore, the simple payback period, SPP, for using the premium-efficiency instead of the standard-efficiency motor is
SPP=\frac{\$300}{2565 \ kWh*\$0.10/kWh}=1.2 \ years