1. 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

2. 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

3. 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) Urea46-0-0449 NH 4 NO 3 (ammonium nitrate)34-0-0Not available (NH 4 ) 2 SO 4 (ammonium sulfate)21-0-0 (+ 24 S)339 DAP (diammonium phosphate)18-46-0519 Triple super phosphate0-46-0529 KCl (muriate of potash)0-0-60950 19-19-19 609 20-10-10 449 *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?)

4. 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 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 100or 2000 lb 34 0.34 (my preference) = 680 lb NINCORRECT!! 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% H 2 O (‘moisture’). Total lb air-dry hay: 5 tons x 2000 lb/ton = 10000 lb Total lb hay DM: 10000 lb x 0.85 DM concentration = 8500 lb DM Total lb K removed: 8500 lb DM x 0.025 K = 212 lb K removed

8. 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: 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 K Total P 2 O 5 and K 2 O removals, lb/ac:36 lb P x 2.29 = 82 lb P 2 O 5 360 lb K x 1.2 = 432 lb K 2 O Does N need to be replaced?Not by fertilizer; rely instead on N 2 fixation by properly-nodulated alfalfa at correct soil pH (> 6.5).

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 ft 2 /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 N450 lb N in 3000 lb blend = 15% N 1000 lb 0-45-0 = 450 lb P 2 O 5 450 lb P 2 O 5 ““ = 15% P 2 O 5 1000 lb 0-0-60 = 600 lb K 2 O 600 lb K 2 O““ = 20% K 2 O (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 P 2 O 5, 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 P 2 O 5 /ac, and 180 lb K 2 O/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 P 2 O 5 requirement: 150 lb x 40 ac = 6000 lb P 2 O 5 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 N Less N being provided by DAP: 4800 – 2000 lb from DAP = 2800 lb Urea 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 K 2 O requirement: 180 lb/ac x 40 ac = 7200 lb K 2 O 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 P 2 O 5 ) Urea: 3.11 tons (= 2800 lb N) KCl: 6.0 tons (= 7200 lb K 2 O)

14. 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 )

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 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? 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 50 o F [x] b. Between 50 o F and 100 o F [ ] c. Under dry soil conditions [ ] d. Below pH 6.5 Urease is the enzyme excreted by soil microbes that converts urea to ammonium ion (NH 4 +). Its activity level is temperature-, pH-, and moisture-dependent (see McCauley et al. 2009). (McCauley et al. 2009)