When sludge is applied to meet nitrogen needs, we often over apply phosphorus. Therefore, application rates may need to be reduced to meet phosphorus needs. This example assumes no residual organic nitrogen source.
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Example: We will apply 5 dry tons sludge/A since this is often the legal limit.
N crop needs = 145 lb/A (corn 125 bu/A) 145 lb/A ÷ available N (14.24) lb/ton (from our example) If the field has had previous applications of organic waste residual nitrogen should be calculated. |
1 year ago we applied 20 lb/A organic nitrogen (anaerobically digested). From Table 4.
20 lb/acre organic nitrogen x 0.20 mineralization rate = 4 lb/A organic nitrogen mineralized. 1 year ago 20 - 4 = 16 lb/acre organic nitrogen X 0.10 mineralization rate = 1.6 lb mineralized organic nitrogen.
Remaining organic N = 1.6 lb/A
2 years ago we applied biosolids that were 15 lb per ton organic nitrogen (aerobically digested) at rate of 4 tons/acre.
The amount of residual nitrogen =
15 lb/ton x 4 tons/acre = 60 lb organic N/A
60 lb/A organic N x 0.30 mineralization rate = 18 lb/A mineralized organic N.
60 - 18 = 42 lb/A organic N X 0.15 = 6.3 lb/A mineralized organic N.
42 - 6.3 = 35.7 lb/A organic N X 0.08 mineralization rate = 2.8 lb/A mineralized organic N
Total PAN from residual organic N = 1.6 (1 year ago) + 2.8 (2 years ago) = 4.4 lb/A PAN from residual organic nitrogen.
Using our analyzed sludge 14.24 lb/ton available N and crop nutrient needs of 145 lb N our application rate of biosolids should be:
145 lb N/A - 4.4 lb residual N = 9.87 tons/acre to meet N needs.
14.24 lb N/ton in biosolids
In this case, the amount of PAN from residual organic nitrogen is insignificant.
The phosphorus needs of the plants are most likely to be met first when applying biosolids.
P crop needs according to soil test = 19.80 lb P/A Biosolids contains 0.568
lb P/ton (from our analysis)
If we apply 5.47 tons of sludge x 0.568 lb/ton = 3.12 lb P applied at this rate.
Potassium levels in sludge are low.
K crop needs = 58.1 lb K/acre Sludge contains ? (not listed in our analysis) (Sludge usually contains 7 lb K/ton)
If our sludge is 15 percent solids then we can convert dry ton application
rate to wet ton application rate utilizing the following equation.
dry tons/A x 100 = wet tons/A percent dry solids
Example: Total percent solids 15 Max sludge rate = 5 ton ÷ 15 percent solids = 33.3 wet tons/acre
For more specific information on spreader calibration, refer to Ohio State University Extension fact sheet, AEX-707.
Injecting sludge will reduce the percent nitrogen lost to 1-5 percent for solids and liquids. Surface spreading of solids will lose 15-30 percent N in 4 days and liquids will lose 10-25 percent N in 4 days. Injected sludge will lose 0-2 percent nitrogen in 4 days.