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Practical review of fertilizer management problems Fundamentals of Nutrient Management December 16-17, 2009 West Virginia University Extension Service T.C. Griggs Division of Plant & Soil Sciences, WVU Review of N2 fixation • Plant nitrogen (N) requirement can be met via uptake of NH4+ or NO3- from soil. • Atmospheric N2 fixed into NH3 (then amino acids ) via symbiotic relationship with rhizobia: Rhizobium spp. Mesorhizobium spp. Bradyrhizobium spp. Ensifer (prev. Sinorhizobium) spp. • Nitrogenase, the N2-fixing enzyme synthesized in nodules, reduces N2 that diffuses in to NH3, which is converted to amino acids or amides for transport. Legume Alfalfa Birdsfoot trefoil Crownvetch Cicer milkvetch Crimson clover Hairy vetch Kura clover Lentil Red clover Soybean Subterranean clover Sweetclover White clover N2 fixed (lb N/ac-yr) 70-200 44-150 98 140 57 99 17-158 149-168 60-200 20-200 52-163 120 115-180 Legume nodules Effective nodules have a pinkish, rather than whitish, interior Nodules are easily stripped away from roots during careless sampling Birdsfoot trefoil Red clover Match legume species with correct inoculum species (e.g., types ‘A’, ‘B’, ‘K’, etc. from EMD Crop BioScience, formerly Nitragin, Inc.) N2 fixation • Nitrogenase is inactivated by O2. • To keep O2 away from nitrogenase, the O2carrier leghemoglobin synthesized in nodules captures O2 before it reaches nitrogenase, then releases it for respiration to drive N2-fixation. • Leghemoglobin with captured O2 imparts pink color to healthy nodules - a good check of nodule effectiveness (effective vs. ineffective nodules). N2 fixation and transfer to grass depend on: • N transfer to grass is primarily from nodule sloughing, legume leaf litter and plant death, and grazing animal urine and manure. • Site conditions including adequate P, S, B, and pH > 6.0-6.5. • Legume density and health. • Leaf area for growth (defoliation management). Residual nitrogen contributions (‘credits’) from legumes Species and density Lb available N/ac* Alfalfa, 25-49% of stand 80 Red clover, 25-49% of stand 70 Soybeans harvested for grain 1 per bu/ac harvested *Lb/ac = pound(s)/acre (43,560 ft2) WVDA, 2009 Example annual nutrient uptakes by forage crops. If crops are fed on-farm, some nutrients recycle back to soil in manure and urine. If crops are exported off-farm, nutrients must be replaced by mineralization of soil organic matter, nutrient release from soil parent material, or fertilizer and manure applications. Species DM yield N P K lb/ac Alfalfa 16,000 450 35 400 Orchardgrass 12,000 300 45 310 Silage corn 24,000 240 45 250 Note higher N-use efficiency (lb DM production/lb N) for silage corn (C4 grass) than for orchardgrass (C3 grass). Feed values expressed as elements and fertilizer values expressed as oxides. Crude protein (CP) = N x 6.25 (17% CP = 2.7% N) E. Rayburn (ed.). 2006. Forage production for pasture-based livestock production. NRAES-172. Natural Resource, Agriculture, and Engineering Service, Cooperative Extension, Ithaca, NY. Fertilizer mixtures • Simple (‘straight’) fertilizer: single nutrient, e.g. urea, ammonium nitrate, or triple superphosphate. • Compound or mixed fertilizer: contains at least 2 of the 3 primary macronutrients N, P, and K, e.g., diammonium phosphate (DAP). • Complete fertilizer: contains the 3 primary macronutrients N, P, and K, e.g., 10-20-20. Definition of a ‘unit’ of fertilizer • Variable interpretations; check with dealer/ applicator/grower to be sure what is meant. • Typically, 1 unit is equivalent to 1 lb of plant nutrient, e.g., 1 unit of N = 1 lb of N. • Example: spreading urea (46-0-0) at 300 lb bulk material/ac delivers 138 lb N/ac or 138 ‘units’ of N/ac (300 x 0.46 = 138). • Sticking to lb of nutrient applied is probably safer terminology. Fertilizer grade and P and K interconversions • Soil test results are usually in elemental form for N, P, and K (ppm or lb/ac of available element). • Fertilizer grading and application recommendations are usually in terms of oxide forms for P and K. • P2O5 and K2O are remnants from when contents of minerals were expressed as the oxides formed upon heating. • Grade: Percentages by weight of total N - available P2O5 - soluble K2O (and S, if a fourth). • Nutrient budgets and calculations often concern elemental, rather than oxide forms, e.g., N, P, and K rather than N, P2O5, and K2O. Why do we do this to ourselves? P and K interconversions • (No interconversions necessary for N) • Phosphorus (element vs oxide): P x 2.29 = P2O5 P2O5 x 0.44 = P • Potassium (element vs oxide): K x 1.2 = K2O K2O x 0.83 = K Common fertilizer sources, grades, and prices (09/25/09), delivered locally Grade* (% N-available P2O5-soluble K2O) Price** ($/ton material) Urea 46-0-0 449 NH4NO3 (ammonium nitrate) 34-0-0 Not available (NH4)2SO4 (ammonium sulfate) 21-0-0 (+ 24 S) 339 DAP (diammonium phosphate) 18-46-0 519 Triple super phosphate 0-46-0 529 KCl (muriate of potash) 0-0-60 950 19-19-19 19-19-19 609 20-10-10 20-10-10 449 Material *Nutrient concentrations can vary slightly among sources of a material. See label. **Add approx. $8.50/ac to spread Note differing costs of P from DAP vs triple superphosphate (what are they?) Fertilizer application rates - 1 • Fertilizer rate recommendations are typically given in lb/ac of N, P2O5, K2O, and S. • To convert recommendations to lb/ac of fertilizer material: lb nutrient recommended/ac x 100 % nutrient in fertilizer material = lb fertilizer material needed/ac Example 1: To supply 120 lb N/ac using ammonium nitrate (34-0-0): 120 x 100 = 353 lb/ac of 34-0-0 34 Mahler, 2002 Fertilizer application rates - 2 Example 2: To supply 120 lb N/ac using urea (45-0-0) and 40 lb P2O5/ac using triple superphosphate (0-44-0): For N, 120 x 100 = 267 lb/ac of 45-0-0 45 For P2O5, 40 x 100 = 91 lb/ac of 0-44-0 44 Mahler, 2002 Fertilizer application rates - 3 Example 3: To supply 120 lb N/ac using urea (45-0-0) and DAP (16-48-0), and 40 lb P2O5/ac using DAP: Start with the amount of DAP needed to meet the P2O5 requirement: 40 x 100 = 83 lb/ac of 16-48-0. 48 Then calculate N supplied by rate of DAP application: 83 lb x 0.16 = 13 lb N/ac Lastly, calculate remaining N requirement to be met by urea: 120 – 13 = 107 lb N/ac from urea 107 x 100 = 238 lb/ac of 45-0-0 45 Example problems - 1 Q. How much 34-0-0 is needed to apply 30 lb of N? A. 88 lb (30 ÷ 0.34) of 34-0-0. Q. If 15-8-10 is used to apply 45 lb of N, how much P2O5 and K2O is also be applied? A. 300 lb (45 ÷ 0.15) of 15-8-10 is used to apply 45 lb of N. Therefore, 24 lb of P2O5 (300 X 0.08) and 30 lb of K2O (300 X 0.10) are also applied. (McCauley et al. 2009) Example problems - 2 Q. How much urea (46-0-0) is needed to apply 135 lb N/ac to 30 ac? A. First, calculate how much urea is needed to provide 135 lb of N/ac. This is 293.5 lb (135 ÷ 0.46). The total urea needed for 30 ac is then 8,804 lb (293.5 X 30) or 4.4 tons (2,000 lb/ton). (McCauley et al. 2009) Example problems - 3 Q. How much N, P, and K are in a 50 lb bag of 16-6-12? Recall that the numbers in the fertilizer grade are percentages by weight and can be expressed as decimal fractions (i.e., 6% = 0.06). A. Since a conversion factor is not required for N, N content = 0.16 x 50 lb = 8 lb of N The conversion factor for P2O5 is 0.44, so P content = 0.06 x 0.44 x 50 lb = 1.3 lb of P The conversion factor for K2O is 0.83, so K content = 0.12 x 0.83 x 50 lb = 5 lb of K (McCauley et al. 2009) Problem 1 How many lb of N are in one ton of 34-0-0? Problem 2 A producer has applied 200 lb/ac of ammonium sulfate (21-0-0-24). How much N and S were applied/ac? lb N/ac: lb S/ac: Problem 3 How many lb of K are removed from the soil by 5 tons of alfalfa hay, assuming the hay has 2.5% K in the dry matter (DM)? Assume that air-dry hay is approx. 85% DM, i.e., 15% H2O (‘moisture’) Total lb air-dry hay: Total lb hay DM: Total lb K removed: Problem 4 How much fertilizer N, P2O5, and K2O will be needed to replace the N, P, and K removed in alfalfa hay yielding 6 tons dry matter (DM)/ac annually (from 4 harvests)? The DM contains 4% N, 0.3% P, and 3% K Total DM yield, lb/ac: Total N, P, and K removals, lb/ac: Total P2O5 and K2O removals, lb/ac: Does N need to be replaced? P2O5 and K2O replacement required, lb/ac: Problem 5 If a fertilizer spreader applies 10 lb of material to a 300-square foot area, approximately how many tons would it apply over an acre (43,560 ft2/ac)? Problem 6 If a fertilizer dealer mixes 1000 lb each of 45-0-0, 0-45-0, and 0-0-60, approximately what analysis of fertilizer has the dealer made? If a fertilizer dealer mixes 1000 lb each of 45-0-0, 11-52-0 (MAP), and 0-0-60, approximately what analysis of fertilizer has the dealer made? Fertilizing a 40-ac field for corn - 1 A farmer will fertilize a 40-acre field for corn. Nutrient requirements for the crop are: 120 lb N/ac, 150 lb P2O5/ac, and 180 lb K2O/ac. Commercial fertilizers that are available are urea, diammonium phosphate (DAP), and muriate of potash (KCl). A. To meet these nutrient requirements, the amount of DAP (tons) to be applied to the whole field is: Total field P2O5 requirement: DAP requirement: Fertilizing a 40-ac field for corn - 2 B. The urea (tons) that will be added to the blend to meet N requirements for the entire field is: Total field N requirement: Less N being provided by DAP: Urea requirement: C. The muriate of potash (tons) that will be added to the blend to meet K requirements for the entire field is: Total field K2O requirement: KCl requirement: Fertilizing a 40-ac field for corn - 3 D. What rate of blended material (complete fertilizer) will be applied/ac to meet crop requirements? DAP: Urea: KCl: Fertilizing a 40-ac corn crop following grass-legume hay Based on soil test results, recommended nutrient application rates for a corn crop are: 150 lb N/ac, 60 lb P2O5 /ac, and 140 lb K2O/ac. However, this corn is following a hay crop that was a grass and red clover mixture. Red clover constituted about 40% of the crop stand. How much urea will be needed to meet the crop N requirement? Nutrient availability in a field Soil test results show that available P in a field was 180 lb/acre. A corn crop that was harvested from this field removed 20 lb P/acre. How much plant-available P was left in the field after the crop harvest? What information do we need to answer this? P fixation by soil: Available P released from parent material: Available P from mineralization of organic matter including manure: Problem 7 Urease activity is greatest: [ ] a. Below 50o F [ ] b. Between 50o F and 100o F [ ] c. Under dry soil conditions [ ] d. Below pH 6.5 (McCauley et al. 2009) Additional resources • D.B. Beegle. Undated. Comparing fertilizer materials. Agronomy Facts 6. Penn State Univ. Dept. of Crop and Soil Sciences Cooperative Extension, http://cropsoil.psu.edu/extension/facts/agfacts6.cfm • G.D. Binford. 2006. Commercial fertilizers. Chap. 8 in The midAtlantic nutrient management handbook. MAWP 06-02. MidAtlantic Regional Water Program, http://www.mawaterquality.org/publications/pubs/manhcompl ete.pdf • R.L. Mahler. 2002. Fertilizer primer. Terminology, calculations, and application. CIS 863. Univ. of Idaho Extension, http://www.cals.uidaho.edu/edComm//pdf/CIS/CIS0863.pdf • A. McCauley, C. Jones, and J. Jacobsen. 2009. Commercial fertilizers and soil amendments. Nutrient Management Module No. 10. 4449-10. Montana State Univ. Extension, http://msuextension.org/publications/AgandNaturalResources/ 4449/4449_10.pdf Add next time • Pre-sidedress N (NO3?) test. • Fall stalk N test (for lux. consumption NO3?); D. Beegle? • Cap soybean N credit at 40 lb? (Craig Yohn may have been the person who mentioned this). Also pH-dependent: •Al (toxic, like Fe) •bacteria •fungi •mineralization of OM •earthworms (optimum for all of above is pH 6-7) •pH scale is logarithmic; pH 5 and 7 are each 10x different from pH 6 Grass response to N rates Typical cool-season forage grass N rates range up to 150-200 lb N/ac-yr. Forage production responds to higher N rates, but above 180 lb N/ac there is increased risk of NO3 leaching to groundwater Growth increment = 10-50 lb forage DM/lb N: Lower efficiency for C3 and at higher application rates Higher efficiency for C4 and at lower application rates For WV cool-season grasses, assume 20-30 lb forage DM production/lb N applied (how does the value of 25 lb forage DM compare to cost of 1 lb N?) Interaction of N x S; alleviating soil S deficiency improves efficiency of N response; same concept holds for other nutrients Tall fescue response to inorganic and organic N sources E. Rayburn (ed.). 2006. Forage production for pasture-based livestock production. NRAES-172. Natural Resource, Agriculture, and Engineering Service, Cooperative Extension, Ithaca, NY. Example of timothy response to N rates