Assumptions for this example: You have a 480 volt motor drawing 40 amps and operating 8400 hours/year. The Power Factor is .85. Electricity costs $0.06/kWhr. Also assume that the amp draw is measured before and after the alignment and it is found that there is a 1 amp difference.
- kW =V* A*PF*1.732/1000
- kW =480V* 1A*.85*1.732/1000
- kW =.706 = power savings
- Savings ($) = operating hours * kW savings * kW cost
- Savings ($) = 8400 * .706 * $0.06/kWhr
- Savings ($) = 8400 * .706 * $0.06/kWhr
- Savings ($) = 355.82
If your plant has 80 similar motors, and if only half of them had the same amount of savings, it would total $355.82 * .5 = $14,232!