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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!
The coupling icons next to the data displays show the direction of the misalignment at the coupling and update as you move or shim the motor! You can also zoom in on the coupling to show them in the live display area of the screen.
