Reviewed October 1993
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A primary need and concern for most confinement livestock producers is managing manure so that groundwater and surface water are protected and regulatory requirements are fulfilled. This objective is usually accomplished by applying manure to the land in such a manner that the potential polluting nutrients (N, P, K and organic matter) are used by the soil-plant complex and are not allowed to enter the groundwater/surface water infrastructure.
Manure should be viewed as a fertilizer resource and managed similarly to commercial fertilizer in the fertility program. The occasional practice of meeting fertility requirements with commercial fertilizer, then applying manure in addition "for good measure," can easily lead to adverse impacts on water quality. In general, Missouri waste application regulations are based on the rate of nitrogen application. With this scenario, the phosphorus and potash applied may greatly exceed crop needs. Therefore, optimum use of plant nutrients may necessitate applying less nitrogen from waste than the crop needs and buying supplemental nitrogen to balance crop needs.
Applying phosphorus to fields with a Bray 1-P test level exceeding 800 pounds per acre may aggravate surface water quality problems
It is highly recommended that a representative sample of dairy waste be analyzed for nutrient values immediately prior to spreading, in addition to soil tests, before determining the land application rate. The purpose of this publication, however, is to provide guidance for application of waste without the benefit of a lab analysis or a soil test. Other publications in this series address application of dairy waste with other scenarios.
Unlike commercial fertilizers, manure is a highly variable substance, even within a given animal specie, and variations of 50 to 100 percent among test samples are not unusual. Other management considerations peculiar to livestock operations, such as lagoon pumping in the fall to provide storage during winter and spring months, or manure storage-tank emptying at whatever intervals are required to prevent overflow, dictate different management than commercial fertilizer that can just be "ordered and spread."
If a laboratory analysis is not available, average values of manure nutrients in similar waste management systems, as reported in the literature, must be used. Table 1 lists values for dairy waste.
Table 1
Average nutrient levels in dairy waste.1
| Waste type | Total nitrogen | Organic nitrogen | Ammonia nitrogen | ||
|---|---|---|---|---|---|
| Solid2 with bedding (pounds per ton) | 9 | 4 | 5 | 4 | 10 |
| Solid3 without bedding (pounds per ton) | 9 | 5 | 4 | 4 | 10 |
| Lagoon (pounds per acre-inch) | 69 | 23 | 46 | 79 | 144 |
| Liquid (slurry; pounds per acre-inch) | 26 | 16 | 10 | 14 | 26 |
In contrast to commercial fertilizer, manure has the potential for nutrients (primarily nitrogen in the form of ammonia) to be lost to the atmosphere after field spreading. Table 2 shows the loss of ammonia nitrogen before incorporation. Table 3 lists the percent of available organic nitrogen available with time. Table 4 gives the percent of various nutrients available in the growing season after application. Table 5 provides a basis for estimating the expected nitrogen release from soil organic matter for major annual crops in lieu of a soil test. Table 6 lists nitrogen credits for crops following various legumes.
Table 2
Manure ammonia-nitrogen loss by days until incorporated into the soil (unavailable portion is lost to the atmosphere)
| Days until incorporation | Percent of ammonia-N available for crops |
|---|---|
| 0 to 2 | 80 |
| 2 to 4 | 60 |
| 4 to 7 | 40 |
| 7 | 20 |
Table 3
Manure organic nitrogen available by year
| Manure applied | Percent of organic-N available during current year |
|---|---|
| Current year | 40 to 60 |
| 1 year ago | 10 |
| 2 years ago | 5 |
| 3 years ago | 5 |
Table 4
Other minerals and micronutrients available in manure
| Nutrient | Percent available in growing season |
|---|---|
| P | 80 |
| K | 100 |
| S, Mn, Cu, Zn | 80 |
| Ca, Mg | 100 |
Table 5
Expected N release from soil organic matter for major annual crops when a current soil test is not available (assumes a cation exchange capacity from 10.1 to 18.0 meq per 100 grams and organic matter content less than or equal to 2 percent (no credit given for N released with perennial crops)
| Summer annuals | 40 pounds N per acre |
| Winter annuals | 20 pounds N per acre |
Table 6
Nitrogen supplied by legumes for succeeding crops (optimum)
| Legume crop | Nitrogen added (pounds N per acre) first year after |
|---|---|
| Alfalfa | |
| 80 to 100 percent stand | 120 to 140 |
| 40 to 60 percent stand | 40 to 60 |
| less than 50 percent | 0 to 20 |
| Sweet clover (green manure) | 100 to 120 |
| Red clover (pure stand) | 40 to 60 |
| Soybeans (add about 1 pound per bushel) | 15 to 60 |
If soil tests are not available for guidance on nutrient application rates, a standard rate of 100 pounds of N per acre per year might be used. This application rate would conform to the regulatory guideline for sizing soil-plant filters under the conservative management approach. This publication, however, details a procedure for estimating the amount of manure to apply to satisfy the projected crop needs for nitrogen, which may exceed the 100 pounds per acre allowed under the conservative management approach. However, one may wish to use this worksheet with 100 pounds of N per acre applied (conservative approach) to see what happens with P and K. A blank "Manure fertility worksheet" is included for actual applications.
Note
This approach cannot be used (to apply more than 100 pounds of N per acre per year) if the Department of Natural Resources has issued a letter of approval based on the "conservative approach" of applying not more than 100 pounds of nitrogen per year regardless of the crop and the production level of the crop.
A fescue hayfield (soil-plant filter) is available for receiving dairy waste. No laboratory analysis of the manure to be applied is available and no soil tests have been performed on the soil-plant filter area. A yield goal of 3 tons of hay per acre is assumed. Given this information, how many inches of lagoon effluent, how many gallons per acre of slurry, and how many tons per acre of solid manure can be applied?
Since no soil test is available, the nitrogen requirement for fescue production found in Table 7 will be used and manure will be applied to supply adequate nitrogen for the desired yield goal. For fescue (a perennial), no credit is given for nitrogen release from soil organic matter or from a previous legume crop.
Table 7
Nitrogen, phosphate and potash removal from soil by various crops
| Crop | Pounds removed per unit production1 | |||
|---|---|---|---|---|
| Units | N | P2O5 | K2O | |
| Corn, grain | bushel | 1.0 | 0.4 | 0.3 |
| Corn, stover | ton | 20.6 | 7.5 | 37.2 |
| Corn, silage | ton | 7.4 | 2.9 | 8.9 |
| Soybeans, grain2 | bushel | 3.4 | 1.0 | 1.5 |
| Soybeans, residue2 | ton | 15.0 | 6.5 | 15.8 |
| Wheat, grain | bushel | 1.3 | 0.5 | 0.3 |
| Wheat, straw | ton | 13.0 | 3.6 | 24.6 |
| Oats, grain | bushel | 0.7 | 0.3 | 0.2 |
| Oats, straw | ton | 12.4 | 4.6 | 32.9 |
| Barley, grain | bushel | 1.0 | 0.4 | 0.3 |
| Barley, straw | ton | 13.5 | 4.7 | 31.0 |
| Sorghum, grain | bushel | 1.1 | 0.4 | 0.3 |
| Sorghum, silage | ton | 7.0 | 2.6 | 10.0 |
| Rye, grain | bushel | 1.0 | 0.5 | 0.3 |
| Rye, straw | ton | 10.0 | 6.0 | 16.9 |
| Alfalfa2 | ton | 49.0 | 11.0 | 50.0 |
| Reed canarygrass | ton | 60.0 | 13.4 | 49.0 |
| Orchardgrass | ton | 50.0 | 16.6 | 62.5 |
| Bromegrass | ton | 33.2 | 13.2 | 50.8 |
| Tall fescue | ton | 55.0 | 18.6 | 52.9 |
| Bluegrass | ton | 25.8 | 18.3 | 60.0 |
| Clover-grass2 | ton | 41.0 | 13.3 | 38.9 |
| Timothy | ton | 37.5 | 13.8 | 62.5 |
| Sorghum-Sudan grass | ton | 39.9 | 15.3 | 55.9 |
From Table 7, for a yield goal of 3 tons per acre per year, we calculate the following nutrient removal by fescue hay:
55 pounds of N per ton x 3 tons per acre = 165 pounds of N per acre
18.6 pounds of P5 per acre
52.9 pounds of K2O per acre
Since no laboratory analysis of the manure is available, the average values from Table 1 will be used. Assume that the waste applied as solid or liquid will not be incorporated into the soil, therefore the loss of ammonia-nitrogen will be 80 percent. Assume that the waste applied as lagoon effluent will be incorporated into the soil within two days after application (by infiltration into the soil), therefore the loss of ammonia-nitrogen will be only 20 percent.
For the application of solid manure with no bedding, complete the "Solid dairy manure worksheet" below to determine the proper application rate. Assume the soil-plant filter area has not received manure from any source the past three years. See Table 1 for average nutrients per unit of manure applied.
For the application of liquid manure (slurry) with no bedding, complete the "Liquid dairy manure worksheet" below to determine the proper application rate. Assume the soil-plant filter area received 3,800 gallons of liquid dairy manure per acre two years ago. See Table 1 for average nutrients per unit of manure applied.
For the application of waste from a lagoon, complete the "Lagoon effluent worksheet" below to determine the proper application rate. Assume the soil-plant filter area has received 1.45 inches of dairy lagoon effluent each of the past six years. See Table 1 for average nutrients per unit of manure applied.
Crop ________
Yield ________
N, pounds per acre ________
P5, pounds per acre ________
K2O, pounds per acre ________
4-N per acre-inch
4-N/K-gal (K-gal = 1,000 gallons)
4-N per ton
(Percent available from Table 2)
________ x ________ = __________
Note
K-gal = 1,000 gallons
Lagoon
pounds N per acre-inch x percent available = ________pounds N per acre-inch
Slurry
pounds N/K-gal x percent available = ________pounds N/K-gal
Solid
pounds N per ton x percent available = ________pounds N per ton
(Percent available from Table 3)
________ x ________ = __________
Note
K-gal=1,000 gallons
Lagoon
inches x pounds N per acre-inch. x percent available = pounds N per acre
Slurry
K-gal per acre x pounds N/K-gal x percent available = pounds N per acre
Solid
tons per acre x pounds N per ton x percent available = pounds N per acre
(Percent available from Table 3)
| 1 year ago: | ________ x | ________ x | ________ | = __________ |
| 2 years ago: | ________ x | ________ x | ________ | = __________ |
| 3 years ago: | ________ x | ________ x | ________ | = __________ |
| Total | = __________ | |||
| (crop N requirement, line 1) - (residual N, line 4) - (N from O.M., Table 5) - (N from legumes, Table 6) (available NH4-N, line 2)+(available organic fraction, line 3) |
= application rate |
| (______) - (______) - (______) - (______) (_____) + (______) |
= ____________pounds N per acre |
Lagoon
inches x pounds P per acre-inch x percent available = ______pounds P per acreSlurry
K-gal per acre x pounds P/K-gal x percent available = ______pounds P per acreSolid
tons per acre x pounds P per ton x percent available = ______pounds P per acre(Percent available from Table 4)
________ x ________ x ________ = __________ pounds P per acre
pounds P per acre x 2.27 = pounds P5 per acre
Note
Do not perform the conversion from P to P5.
________ x 2.27 = ________ pounds P5 per acre
Lagoon
inches x pounds K per acre-inch x percent available = ______pounds K per acreSlurry
K-gal per acre x pounds K/K-gal x percent available = ______pounds K per acreSolid
tons per acre x pounds K per ton x percent available = ______pounds K per acre(Percent available from Table 4)
________ x ________ x ________ = __________ pounds K per acre
pounds K per acre x 1.2 = pounds K2O per acre
Note
Do not perform the conversion from K to K2O.
________ x 1.2 = pounds K2O per acre
Crop Fescue
Yield 3 tons per acre
N, pounds per acre 165
P2O5, pounds per acre 56
K2O, pounds per acre 159
pounds NH4-N per ton x percent available = pounds NH4-N per ton
(Percent available in Table 2)4 pounds per ton x 0.2 available = 0.8 pounds per ton
pounds N per ton x percent available = pounds N per ton
(Percent available first year from Table 3)5 pounds per ton x 0.5 available = 2.5 pounds per ton
______(crop N requirement) - (residual N) - (N, OM) - (N, leg)______
(available NH4-N) + (available organic fraction)= application rate 165 - 0 - 0 - 0
0.8 + 2.5= 50 tons per acre
tons per acre x pounds P2O5 per ton x percent available = pounds P2O5 per acre
(P2O5 per ton from Table 1 = 4; percent available from Table 4)
50 tons per acre x 4 pounds per ton x 0.8 = 160 pounds per acre
Note
160 pounds per acre of P2O5 is applied
versus 56 pounds per acre removed by drop.
tons per acre x pounds K2O per ton x percent available = pounds K2O per acre
(K2O per ton from Table 1 = 10; percent available from Table 4)
50 tons per acre x 10 pounds per ton x 1.0 = 500 pounds per acre
Note
500 pounds per acre of K2O is applied versus 159
pounds per acre removed by crop.
Crop Fescue
Yield 3 tons per acre
N, pounds per acre 165
P2O5, pounds per acre 56
K2O, pounds per acre 159
pounds NH4-N/K-gal x percent available = pounds NH4-N/K-gal
(Percent available from Table 2)
10 pounds per K-gal x 0.2 available = 2 pounds per K-gal
Note
K-gal = 1,000 gallons, e.g. 5 K-gal = 5,000 gallons
pounds N per K-gal x percent available = pounds N per K-gal
(Percent available first year from Table 3)
16 pounds per K-gal x 0.5 available = 8 pounds per K-gal
Number of K-gal per acre x pounds N per K-gal x percent available = pounds N per acre
(Percent available from Table 3)2 years ago: 3.8 K-gal x 16 pounds per K-gal x 0.05 = 3.0 pounds
______(crop N requirement) - (residual N) - (N, OM) - (N, leg)______
(available NH4-N) + (available organic fraction)= application rate 165 - 3 - 0 - 0
2 + 8= 16.2 K-gal per acre = 16,200 gallons per acre
Number of (K-gal per acre) x pounds P2O5 per K-gal x percent available = pounds P2O5 per acre
(P2O5 per K-gal from Table 1 = 14; percent available from Table 4)
16.2 (K-gal per acre) x 14 pounds per K-gal x 0.8 = 181.4 pounds per acre
Note
181.4 pounds per acre of P2O5 is applied
versus 56 pounds per acre removed by crop.
Number of (K-gal per acre) x pounds K2O per K-gal x percent available = pounds K2O per acre
(K2O per K-gal from Table 1 = 26; percent available from Table 4)
16.2 (K-gal per acre) x 26 pounds per K-gal x 1.0 = 421.2 pounds per acre
Note
421.2 pounds per acre of K2O is applied versus
159 pounds per acre removed by crop.
Crop Fescue
Yield 3 tons per acre
N, pounds per acre 165
P2O5, pounds per acre 56
K2O, pounds per acre 159
pounds NH4-N per acre-inch x percent available = pounds NH4-N per acre-inch
(Percent available from Table 2)
46 pounds per acre-inch x 0.8 available = 36.8 pounds per acre-inch
pounds N per acre-inch x percent available = pounds N per acre-inch
(Percent available first year from Table 3)
23 pounds per acre-inch x 0.5 available = 11.5 pounds per acre-inch
inches x pounds N per acre-inch x percent available = pounds N per acre
(Percent available from Table 3)
1 year ago: 1.45 inches x 23 pounds per acre-inch x 0.10 = 3.3 pounds per acre 2 years ago: 1.45 inches x 23 pounds per acre-inch x 0.05 = 1.7 pounds per acre 3 years ago: 1.45 inches x 23 pounds per acre-inch x 0.05 = 1.7 pounds per acre Total = 6.7 pounds per acre
______(crop N requirement) - (residual N) - (N, OM) - (N, leg)______
(available NH4-N) + (available organic fraction)= application rate 165 - 6.7 - 0 - 0
36.8 + 11.5= 3.28 inches
Number of inches applied x pounds P2O5 per acre-inch x percent available = pounds P2O5 per acre
(P2O5 per acre-inch from Table 1 = 79, percent available from Table 4)
3.28 inches x 79 pounds per acre-inch x 0.8 = 207.3 pounds per acre
Note
207.3 pounds per acre of P5 is applied versus 56
pounds per acre removed by crop.
Number of inches applied x pounds K2O per acre-inch x percent available = pounds K2O per acre
(K2O per acre-inch from Table 1 = 144; percent available from Table 4)
3.28 inches x 144 pounds per acre-inch x 1.0 = 472.3 pounds per acre
Note
472.3 pounds per acre of K2O is applied versus 159 pounds per acre removed by crop.
WQ309, reviewed October 1993