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P and K interconversions

Part 2: Answers to practical 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. P and K interconversions. (No interconversions necessary for N)

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P and K interconversions

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  1. Part 2: Answers to practical fertilizer management problemsFundamentals of Nutrient ManagementDecember 16-17, 2009 West Virginia University Extension ServiceT.C. GriggsDivision of Plant & Soil Sciences, WVU

  2. 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

  3. Common fertilizer sources, grades, and prices (09/25/09), delivered locally Note differing costs of P from DAP vs triple superphosphate (what are they?)

  4. 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

  5. Problem 1 How many lb of N are in one ton of 34-0-0? 1 ton = 2000 lb 34-0-0 2000 lb x 100 or 2000 lb 34 0.34 (my preference)= 680 lb N INCORRECT!! REDO!!

  6. 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: 200 lb x 0.21 = 42 lb N/ac lb S/ac: 200 x 0.24 = 48 lb S/ac (a high rate!)

  7. 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: 5 tons x 2000 lb/ton = 10000 lbTotal lb hay DM: 10000 lb x 0.85 DM concentration = 8500 lb DMTotal lb K removed: 8500 lb DM x 0.025 K = 212 lb K removed

  8. 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: 6 tons DM x 2000 lb/ton = 12000 lb DM Total N, P, and K removals, lb/ac: 12000 x 0.04 = 480 lb N 12000 x 0.003 = 36 lb P 12000 x 0.03 K = 360 lb KTotal P2O5 and K2O removals, lb/ac: 36 lb P x 2.29 = 82 lb P2O5 360 lb K x 1.2 = 432 lb K2O Does N need to be replaced? Not by fertilizer; rely instead on N2 fixation by properly-nodulated alfalfa at correct soil pH (> 6.5). Problem 4

  9. 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)? 300 = 0.0069 43560 10 lb material = 1452 lb/ac 0.0069 ac 1452 lb = 0.73 ton material/ac 2000 lb/ton

  10. 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?1000 lb 45-0-0 = 450 lb N 450 lb N in 3000 lb blend = 15% N 1000 lb 0-45-0 = 450 lb P2O5 450 lb P2O5 “ “ = 15% P2O5 1000 lb 0-0-60 = 600 lb K2O 600 lb K2O “ “ = 20% K2O (15-15-20) 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? Same approach as above, but 1000 lb 11-52-0 contains 110 lb N as well as 520 lb P2O5, so total N in 3000 lb blend = 450 lb from 45-0-0 + 110 lb N from MAP = 560 lb = 18.7% N in blend (19-17-20).

  11. 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 (45-0-0), diammonium phosphate (DAP, 16-48-0), and muriate of potash (KCl, 0-0-60). A. To meet these nutrient requirements, the amount of DAP (tons) to be applied to the whole field is:Total field P2O5 requirement: 150 lb x 40 ac = 6000 lb P2O5 DAP requirement: 6000 lb = 12500 lb = 6.25 tons DAP. 0.48*Note DAP also provides 12500 lb x 0.16 N = 2000 lb N

  12. 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: 120 lb x 40 ac = 4800 lb NLess N being provided by DAP: 4800 – 2000 lb from DAP = 2800 lbUrea requirement: 2800 lb N = 6222 lb urea = 3.11 tons 0.45 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: 180 lb/ac x 40 ac = 7200 lb K2O KCl requirement: 7200 = 12000 lb KCl = 6.0 tons 0.60

  13. 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: 6.25 tons (= 2000 lb N + 6000 lb P2O5) Urea: 3.11 tons (= 2800 lb N) KCl: 6.0 tons (= 7200 lb K2O)

  14. Residual nitrogen contributions (‘credits’) from legumes *Lb/ac = pound(s)/acre (43,560 ft2) WVDA, 2009

  15. 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?Using previous table, assign available N credit of 70 lb/ac, so urea requirement will be to meet 150-70 = 80 lb N/ac(= 178 lb 45-0-0)

  16. 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? I have no simple answer to this, other than listing the information that we need to provide to answer this, some or all of which you no doubt covered in your workshop:P fixation by soil:Available P released from parent material:Available P from mineralization of organic matter including manure:Other?

  17. Problem 7 Urease activity is greatest: [ ] a. Below 50o F [x] b. Between 50o F and 100o F[ ] c. Under dry soil conditions[ ] d. Below pH 6.5Urease is the enzyme excreted by soil microbes that converts urea to ammonium ion (NH4+). Its activity level is temperature-, pH-, and moisture-dependent (see McCauley et al. 2009). (McCauley et al. 2009)

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