Field Efficiency
Field efficiency is usually used to evaluate the performance of tillage or harvesting machines. It is a comparison of the actual amount of "work" (volume of activity, not Force x Distance) done by a machine compared to what it would do with no lost time or capacity. The maximum rate that a machine can perform is determined by the width of the machine and the speed of travel. When a machine operates with a constant width and travels at a constant speed, it will perform at 100% field efficiency. A machine is capable of operating at 100% field efficiency for short periods of time, but as soon as the speed changes (slow down for turns, etc.), or the width changes (overlap width of the machine to prevent skips), the efficiency drops below 100%. The primary cause of loss efficiency is lost time (unproductive time) and a working width of the machine less than the maximum. Typical field efficiencies for common machines can be found in Appendix IV. This concept is illustrated in more detail in the next section, on capacity.
9.4. Capacity
The term capacity is used to evaluate the productivity of a machine. In agriculture, two types of capacity are commonly used, field capacity and throughput capacity. Field capacity is used to evaluate the productivity of machines used to work the soil, such as plows, cultivators, and drills, sprayers, and harvesting machines. Throughput capacity is used to describe machines that handle or process a product, such as grain augers, balers, forage harvesters, and combines.
An additional concept relating to both types of capacity is the difference between theoretical and actual productivity. If a tillage machine operates at 100% efficiency, it is operating at 100% capacity. This is called the theoretical field capacity. Theoretical field capacity is determined using the width of the machine and the speed of travel. It can be calculated using units cancellation, but an equation is commonly used.
Theoretical field capacity(CT) =
8.25
where Ct = Theoretical field capacity (f); S = Speed of travel (S); W = Width of the machine (ft); 8.25 = Units conversion constant (43,560 f2) ^ (5,280 2^).
This equation can be used as long as the unit used for speed is miles per hour and the unit used for the width of the machine is feet.
Problem: Determine the theoretical capacity for a machine that travels at 5.0 mph and has an operating width of 20.0 ft.
Solution:
When this machine travels at a constant speed and uses a constant width, it has a theoretical capacity of 12 ac/hr.
Effective field capacity is the amount of productivity that actually occurred not what is theoretical possible. Lost capacity is an important concern for the machine operator and/or manager because it represents lost revenues or resources. Usually lost capacity is caused by lost time, time not operating, and operating the machine with less than the maximum working width. Common causes of lost time include:
1. Mechanical breakdowns.
2. Taking time to adjust the machine.
3. Stopping to fill seed hoppers, spray tanks, etc.
4. Slowing down to turn at the end of the row or crossing waterways, etc.
5. Operator rest stops.
The equation for effective capacity is the same as theoretical capacity with a field efficiency added. A range and typical field efficiency values for common machines can be found in Appendix IV. The common equation for effective field capacity is:
8.25
where CE = Effective field capacity (ac/hr); S = Average speed of travel (mph); W = Effective width of the machine (ft); Ef = Field efficiency (decimal form).
Problem: Assume that the operator in the previous problem averages 0.75 hr of lost time per 10.0hr day. What is the effective field capacity?
Solution: The first step is to determine the time efficiency:
f input 10.0 hr
The second step is to determine the effective capacity:
Now the effects of lost productivity are apparent. The theoretical capacity is 12 ac/hr, but because of lost time, the effective capacity is 11 ac/hr.
The concept of effective capacity also can be used to determine the amount of time it would take a machine to cover a field.
Problem: How many hours will it take to cultivate 200.0 acres with a field cultivator that is 24.0 ft wide?
Solution: This is an example of a problem with a hidden intermediate step. Before the hours can be determined, the effective capacity of the machine must be calculated. Note: The effective capacity equation requires three values. Two of these, speed and field efficiency, are not given in the problem. If the actual speed and the field efficiency are unknown, the typical values found in Appendix IV can be used. In Appendix IV, the typical field efficiency for a field cultivator is 85%, and the typical speed is 5.5 mph. With these values, the effective capacity can be calculated:
Once the effective capacity is known, the time required to work the field can be calculated. Using units cancellation:
14 ac
If the average field speed is 5.5 mph and the operator can maintain an 85% field efficiency, it will take 14 hr to cultivate the 200 acre field.
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