Metric Problems

The principles of animal waste handling treatment and disposal are the same in SI units. The only differences in the problems are the units that are used.

Problem: Determine the amount of solid waste from 91 kg finishing pigs that can be applied to a coarse loamy soil with apH of 6.5 that will be used to raise 5.0 cubic meters per hectare of wheat. A soil test reveals 35.3 kg of nitrogen and 15.0 kg of phosphorus per hectare available for the plants.

Solution: The first step is to use Appendix VI to determine the amount of nitrogen and phosphorus produced by the animals. Note: SI values in Appendix VI are direct conversions.

Nitrogen = 0.0305 kg/day Phosphorus = 0.0204 kg/day

The next step is to determine the pounds of nutrients per pound of waste.

kg of nitrogen 0.0305 kgn 1 day kg of nitrogen

kg of waste 1 day 5.9 kgw kg of waste kg of phosphorous 0.0204 kgp 1 day kg of phosphorous

kg of weight 1 day 5.9 kgw kg of waste

The next step is to determine the net amount of nutrients needed by the crop. This is determined by subtracting the nutrients available in the soil from the nutrients used by the crop, Appendix VII.

ha ha ha ha

The last step is to determine the kilograms of waste that can be applied per hectare. This is determined by multiplying the net nutrients used by the crop by the kilograms of nutrients per kilogram of waste.

kg waste 1 kg waste 104.7 kg N kg waste

Nitrogen waste: —-=--- x -— = 20134.61 ... or 20,000 —-

6 ha 0.0052 kg N ha ha kg waste 1 kg waste 40 kg P kg waste

Phosphorus waste: —-=--- x -— = 11764.7 ... or 12,000 —-

ha 0.0034 kg P ha ha kg waste 1 kg 448 kg P kg waste

Phosphorus soil limits: —-=--- x -— = 131764.7 ... or 130, 000 —-

ha 0.0034 kg P ha ha

The least amount (kg/ha) is 12,000 for the phosphorus in the waste. This sets the limits of 12,000 kg of waste per hectare per year that can be applied.

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