Handling Solid Animal Waste

Waste containing 20% or more solids or with a moisture content of 50% MWB or less is considered to be solid waste. Proper handling of solid waste inside buildings requires solid floors that can be bedded or drained. The waste is collected and usually is not treated except by the natural processes, as it is stored. The preferred method for waste disposal is mimicking the natural system—spread it on the land. If not done correctly, disposal of solid animal waste on the land can reduce plant production, produce offensive odors, and contribute to the contamination of ground and surface water. TheNRCS1 has developed standards for the application of solid animal waste on land. These standards are based on soil type, slope, and distance from surface and ground water. The general standards are:

1. No application to frozen or snow covered soil when slope is greater than 5% unless special provisions are made to control runoff.

2. No application to cropland that exceeds soil loss tolerances.

3. No application on any cropland with slopes greater than 15%.

4. No application within 200 ft of wells, sinkholes, or surface water.

5. Liquid cannot be applied to soils with less than 10 inches of at least moderately permeable soil.

6. No application on organic soils with seasonal water table within 1 ft of the surface.

7. No application on flood plains where flooding occurs more frequent than once in 10 years.

8. Application can occur on flood plains if the injection or immediate incorporation method is used.

9. No intentional application on established waterway or any area where there many be concentrated water flow.

10. No more than 25% of the surface may be covered.

Failure to follow these standards can cause contamination of the ground or surface water and the possibility of fines or litigation.

Note: Different computer programs are available to help producers calculate the amount of livestock waste that can be applied. The manual process explained in the following sections shows the steps and types of information required to use one of the computer programs.

For accurate results, the first step in determining the amount of solid animal waste that can be applied to the soil is to analyze both the solid waste and the soil for the amounts of nitrogen and phosphorous they contain. If the waste has not been tested, Appendix VI may be used to estimate the nutritional content. These values are estimates because the animal's ration, the type and quantity of bedding used, the amount of liquid added, the type of housing and manure handling system, and the storage system all affect the nutrient content of the animal waste.

1 Waste Utilization, Code 633, Natural Resources Conservation Service, 2002.

When the nutritional content of the soil is unknown, it must be estimated before Appendixes VI (Solid Animal Waste Production and Characteristics), VII (Nutrient Utilized by Crops), and VIII (Maximum Annual Application Rates for Phosphates Based on Soil Family) can be used to determine the amount of waste that may be applied.

The primary governing principle for the application of solid animal waste is that no more nutrients can be applied in the soil than what will be removed from the field in crops or animal weight. It is unlikely that the amount of each nutrient in the waste can be matched with the needs of the plants; so the nutrient with the most restrictive amount determines the maximum amount of waste that can be applied. Note: This method assumes that more than one application per year will be used to distribute the total amount of waste. The steps in calculating the amount of waste that can be applied are:

1. Determine the amount of nutrients produced per day, Appendix VI.

2. Calculate the pounds of nutrients per pound of waste.

3. Determine the amount of nutrients used for the expected crop, Appendix VII.

4. Determine the maximum amount of phosphorous for the soil type, Appendix VIII.

5. Calculate the pounds of waste that can be applied.

Problem: Determine the amount of solid waste from 1,000 lb dairy cows that can be applied to a coarse loamy soil with a pH of 6.5 that will be used to raise 60 bushels per acre of wheat. Note: In this example we will assume that there are no nutrients in the soil.

Solution: In this type of a problem a table is useful for determining the answer, for example Table 21.1. First, determine the amount of nutrients in the waste and the maximum amounts of phosphorous and nitrogen that can be applied. This is accomplished by selecting the correct values from Appendixes VI, VII, and VIII, and converting them to pounds of nutrient per pound of waste. From Appendix VI the nutrients produced per day are:

day lb

Each cow produces 82 lb of waste per day, Appendix VI. The pounds of nutrient per pound of waste can be calculated using units cancellation:

lb of nitrogen 0.14 lb N 1 day lb N

lb of waste 1 day 82 lb waste lb waste lb of phosphorus 0.27 lb P 1 day lb P

lb of waste 1 day 82 lb waste lb waste

TABLE 21.1. Solution to problem.

Nutrient

Soil limits (lb/ac)

Net nutrients used by crop (lb/ac)

Pounds of nutrient per pound of waste (lb/lb)

Application rate of waste (lb/ac/yr)

Nitrogen

125

1.7 x 10-3

7,400*

Phosphorous

50

3.3 x 10-3

15,000

Phosphorous

400

3.3 x 10-3

120,000

125 lb N 1 lb waste

125 lb N 1 lb waste

50 lb of P 1 lb waste

Application rate = - x - = 15151.51... or 15,000 lb

400 lb P 1 lb waste

Application rate = - x - = 121212.12... or 120,000 lb

The next step is to use Appendix VII to determine the nutrients used by the crop:

Nitrogen = 125 lb/ac Phosphorous = 50 lb/ac

The last information needed to find the solution is the maximum amount of phosphorous that can be applied to the soil based on soil type. Consult Appendix VIII for this information:

Phosphorous = 400 lb/ac

The solution is found by calculating the application rate (lb/ac/yr) of waste for each nutrient. The maximum amount of waste that can be applied is determined by the smallest of three values: the nitrogen used by the plant, the phosphorous used by the plant, and the phosphorous that can be applied to the soil. The results are shown in Table 21.1.

Table 21.1 shows that nitrogen is the most restricted nutrient (7,400 lb/ac/yr). Therefore, the maximum amount of waste from the dairy cows that can be annually applied is 7,400 lb per acre per year.

In the previous sample problem, the nutrients available in the soil were not considered. The actual amount of waste than can be applied is the difference between the nutrients used by the crop and the amount of nutrients already available in the soil.

Another aspect of solid waste management is determining how many acres are required to dispose of the waste being produced.

Problem: How many acres will be required to distribute all of the waste from a 50-cow dairy? Use the cows in the previous problem.

Solution: The answer to this question can be found by using the 7,400 lb of waste per acre per year from the previous problem, additional information from Appendix VI, and units cancellation. Appendix VI shows that each 1,000 lb dairy cow produces 82.0 lb of waste per day. Then:

7,400 lb waste cow - day

1496500

7,400

For the 50-cow dairy herd, a minimum of 202 acres are required to dispose of the animal waste each year. It is important to remember that in the original problem we assumed that the soil contained no nitrogen or phosphorous. Any nitrogen or phosphorous in the soil reduces the amount of waste that may be applied and increases the number of acres required.

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