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
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:
Determine the amount of electricity used during the month of May by a 1/2 hp ventilation fan when it runs an average of 12 hours per day.

(Motor Input watts x 12 hrs/day x 31 days) ÷ 1,000
(608 x 12 x 31) ÷ 1,000 = 226kWh

Multiply kilowatts times 6 cents to get cost.

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
— open lot
— sheltered


5-10
2-3

1" insulation
     

Cattle waterer
— open lot
— sheltered
— inside barn, semi-enclosed
— outside, very well insulated


6
5
4
2-3

1" insulation

2" insulation, floating covers
     

Combination hog & cattle waterer
— open lot
— sheltered

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.