Find the horsepower rating top top the nameplate that the motor. Multiply kilowatts (kW) of corresponding horsepower top top the graph by the total variety of hours the engine is used. This figure - kilowatt-hours (kWh) - multiplied by the applicable rate, will offer you the price of operation.
Example:
How much would it expense to operate a 10 h.p. Motor 24 hours per job for three weeks?
10 H.P. 230V 1Ø, 24 hours/day because that 3 weeks.
ANSWER: (assuming the electric rate is $.08)
8.625 kW x 24 hours x 21 days = 4,347 kWh4,347 kWh x $.08 = $347.76
Note:No capacity charge included.
1/6 | .329 | .329 | |
| ¼ | .447 | .447 | |
| 1/3 | .571 | .571 | |
| ½ | .800 | .800 | .568 |
| ¾ | 1.159 | 1.159 | .774 |
| 1 | 1.380 | 1.380 | .999 |
| 1 ½ | 1.794 | 1.794 | 1.335 |
| 2 | 2.180 | 2.180 | 1.893 |
| 3 | 3.167 | 3.167 | 2.868 |
| 5 | 4.701 | 4.478 | |
| 7 ½ | 6.808 | 6.310 | |
| 10 | 8.625 | 8.724 | |
| 15 | 12.269 | ||
| 20 | 16.679 | ||
| 25 | 20.197 | ||
| 30 | 24.858 | ||
| 40 | 33.044 | ||
| 50 | 38.752 | ||
| 60 | 48.078 | ||
| 75 | 60.105 | ||
| 100 | 82.253 |
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