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 N 2 fixation Plant nitrogen (N) requirement can be met via uptake of NH 4 + or NO 3 - from soil. Atmospheric N 2 fixed into NH 3 (then amino acids ) via symbiotic relationship with rhizobia: Rhizobium spp. Mesorhizobium spp. Bradyrhizobium spp. Ensifer (prev. Sinorhizobium) spp. Nitrogenase, the N 2 -fixing enzyme synthesized in nodules, reduces N 2 that diffuses in to NH 3, which is converted to amino acids or amides for transport.

LegumeN 2 fixed (lb N/ac-yr) Alfalfa Birdsfoot trefoil Crownvetch Cicer milkvetch Crimson clover Hairy vetch Kura clover Lentil Red clover Soybean Subterranean clover Sweetclover White clover

Birdsfoot trefoil Red clover Legume nodules Effective nodules have a pinkish, rather than whitish, interior Nodules are easily stripped away from roots during careless sampling

Match legume species with correct inoculum species (e.g., types ‘A’, ‘B’, ‘K’, etc. from EMD Crop BioScience, formerly Nitragin, Inc.)

N 2 fixation Nitrogenase is inactivated by O 2. To keep O 2 away from nitrogenase, the O 2 - carrier leghemoglobin synthesized in nodules captures O 2 before it reaches nitrogenase, then releases it for respiration to drive N 2 -fixation. Leghemoglobin with captured O 2 imparts pink color to healthy nodules - a good check of nodule effectiveness (effective vs. ineffective nodules).

N 2 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 > Legume density and health. Leaf area for growth (defoliation management).

Residual nitrogen contributions (‘credits’) from legumes Species and densityLb available N/ac* Alfalfa, 25-49% of stand80 Red clover, 25-49% of stand70 Soybeans harvested for grain1 per bu/ac harvested WVDA, 2009 *Lb/ac = pound(s)/acre (43,560 ft 2 )

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. SpeciesDM yieldNPK lb/ac Alfalfa16, Orchardgrass12, Silage corn24, Note higher N-use efficiency (lb DM production/lb N) for silage corn (C 4 grass) than for orchardgrass (C 3 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.) 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.,

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. P 2 O 5 and K 2 O are remnants from when contents of minerals were expressed as the oxides formed upon heating. Grade: Percentages by weight of total N - available P 2 O 5 - soluble K 2 O (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, P 2 O 5, and K 2 O. Why do we do this to ourselves?

P and K interconversions (No interconversions necessary for N) Phosphorus (element vs oxide): P x 2.29 = P 2 O 5 P 2 O 5 x 0.44 = P Potassium (element vs oxide): K x 1.2 = K 2 O K 2 O x 0.83 = K

Common fertilizer sources, grades, and prices (09/25/09), delivered locally Material Grade* (% N-available P 2 O 5 -soluble K 2 O) Price** ($/ton material) Urea NH 4 NO 3 (ammonium nitrate)34-0-0Not available (NH 4 ) 2 SO 4 (ammonium sulfate) (+ 24 S)339 DAP (diammonium phosphate) Triple super phosphate KCl (muriate of potash) *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, P 2 O 5, K 2 O, 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 Mahler, 2002

Fertilizer application rates - 2 Example 2: To supply 120 lb N/ac using urea (45-0-0) and 40 lb P 2 O 5 /ac using triple superphosphate (0-44-0): For N, 120 x 100 = 267 lb/ac of For P 2 O 5, 40 x 100 = 91 lb/ac of Mahler, 2002

Fertilizer application rates - 3 Example 3: To supply 120 lb N/ac using urea (45-0-0) and DAP ( ), and 40 lb P 2 O 5 /ac using DAP: Start with the amount of DAP needed to meet the P 2 O 5 requirement: 40 x 100 = 83 lb/ac of 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

Example problems - 1 Q. How much is needed to apply 30 lb of N? A. 88 lb (30 ÷ 0.34) of Q. If is used to apply 45 lb of N, how much P 2 O 5 and K 2 O is also be applied? A. 300 lb (45 ÷ 0.15) of is used to apply 45 lb of N. Therefore, 24 lb of P 2 O 5 (300 X 0.08) and 30 lb of K 2 O (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 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 ? 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 P 2 O 5 is 0.44, so P content = 0.06 x 0.44 x 50 lb = 1.3 lb of P The conversion factor for K 2 O 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 ?

Problem 2 A producer has applied 200 lb/ac of ammonium sulfate ( ). 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% H 2 O (‘moisture’) Total lb air-dry hay: Total lb hay DM: Total lb K removed:

Problem 4 How much fertilizer N, P 2 O 5, and K 2 O 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 P 2 O 5 and K 2 O removals, lb/ac: Does N need to be replaced? P 2 O 5 and K 2 O 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 ft 2 /ac)?

Problem 6 If a fertilizer dealer mixes 1000 lb each of , , and , approximately what analysis of fertilizer has the dealer made? If a fertilizer dealer mixes 1000 lb each of , (MAP), and , 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 P 2 O 5 /ac, and 180 lb K 2 O/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 P 2 O 5 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 K 2 O 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 P 2 O 5 /ac, and 140 lb K 2 O/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 50 o F [ ] b. Between 50 o F and 100 o 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, G.D. Binford Commercial fertilizers. Chap. 8 in The mid- Atlantic nutrient management handbook. MAWP Mid- Atlantic Regional Water Program, ete.pdf ete.pdf R.L. Mahler Fertilizer primer. Terminology, calculations, and application. CIS 863. Univ. of Idaho Extension, A. McCauley, C. Jones, and J. Jacobsen Commercial fertilizers and soil amendments. Nutrient Management Module No Montana State Univ. Extension, /4449_10.pdf /4449_10.pdf

Add next time Pre-sidedress N (NO 3 ?) test. Fall stalk N test (for lux. consumption NO 3 ?); 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 lb N/ac-yr. Forage production responds to higher N rates, but above 180 lb N/ac there is increased risk of NO 3 leaching to groundwater Growth increment = lb forage DM/lb N: Lower efficiency for C 3 and at higher application rates Higher efficiency for C 4 and at lower application rates For WV cool-season grasses, assume 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.) 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