TYPICAL OPERATING COSTS OF FARMSTEAD EQUIPMENT
Grain Aeration & Farm Motors
To calculate electricity used by single phase motors operating aeration fans or other equipment, estimate the hours of operation (with careful consideration of weather conditions and other variables), and apply the following formula:
Motor input watts x hrs. of use/mo. ÷ 1,000 = _______kWh/mo.
Because of design differences and internal electrical losses, motors have different power factors. This means that more input watts are required than are indicated by the motor’s rated horsepower. The formula above, and the input watts in Table A, account for these varying efficiencies. All values assume motors run at 80 to 100 percent of full load, and at rated voltage.
Single Phase Motor Data for Estimating kWh Use |
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Rated Motor Size in HP |
Rated Motor Size in Watts |
Assumed Efficiency Percent |
Input Watts |
1/4 |
187 |
55 |
340 |
1/3 |
248 |
60 |
413 |
1/2 |
383 |
63 |
608 |
3/4 |
560 |
65 |
862 |
1 |
750 |
70 |
1,072 |
1.5 |
1,120 |
75 |
1,493 |
2 |
1,490 |
80 |
1,863 |
3 |
2,240 |
85 |
2,635 |
5 |
3,370 |
85 |
4,388 |
7.5 |
5,600 |
87 |
6,437 |
10 |
7,460 |
88 |
8,477 |
15 |
11,000 |
90 |
12,222 |
Example: (Motor Input watts x 12 hrs/day x 31 days) ÷ 1,000 Multiply kilowatts times 6 cents to get cost. |
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The above information has been provided from the National Food and Energy Council.
Livestock Waterers
Application/Operation |
kWh/Head/Season |
Comments |
Hog waterer, dual trouch |
5-10 2-3 |
1" insulation |
Cattle waterer |
6 5 4 2-3 |
1" insulation 2" insulation, floating covers |
Combination hog & cattle waterer |
10 8 |
1" insulation |
| Data mainly from Iowa and Missouri, during typical winter months. For more northern states add 15 percent for each tier of states to the north. | ||