User:Apc-innovation

Case-I The 6% η improvement of 15 hp motor from 86% to 92%, when it operates at 80% of rated load will influence the kilowatt savings as: Required kW for 86% efficiency = 0.746x15x0.8/0.86 = 10.41 kW. Required kW for 92% efficiency = 0.746x15x0.8/0.92 = 9.73 kW. Thus 92% efficient motor will consume less electrical energy than 86% as: Saved kW = 6.63-6.28 = 0.68 kW If the motor operates in continuous mode, approximate working hours in a year are 6,000 hours and electricity cost is Rs.5. 5 per kWh. The savings from 6% efficiency improvement is:

Calculating Annual Energy Savings Annual savings in Case-I can also be determined from this equation: S = 0.746 x H x L x C x N (100/EB - 100/EA)

Where: S = annual savings, Rs. H = motor rating, hp L = motor loading C = electric energy rate, Rs. /kWh	N = annual operation, hr EA = efficiency of high efficiency motor, % EB = efficiency of standard motor, %

S = 0.746 x 15 x 0.8 x 5.5 x 6000 (100/92 - 100/86) S = Rs. 22,440