Lab 8 Applications and Problem Solving

Transcription

1 Lab 8 Applications and Problem Solving OBJECTIVES In this lab you will apply some of the concepts learned during Unit 2 (Soil Chemistry) to solving practical problems such as fertilizer calculations. INTRODUCTION Fertilizer recommendations are based on the results of soil tests. Soil test results do not specifically tell how much fertilizer is to be added. Rather, soil tests report the relative levels of particular nutrients such as P, K, Ca, and Mg using terms like: very low, low, medium, high and very high (Table 1 and Table 2). Specific amounts of fertilizer needed are then determined using the test level terms mentioned above, the type of crop to be grown, geographic location of the crop, and the relative yield desired. It is often best to fertilize to a soil test level of "medium" to "high" because any less gives poor yields and any more gives no return on the fertilizer invested. Further, extremely high soil test levels may result in certain micronutrient (Zn and Cu) deficiencies. TABLE 1. Interpretation of Soil Test Results Obtained by Mehlich III Soil Extraction Method. Pounds/Acre P K Ca Mg Very low Low Medium High Very High TABLE 2. Soil Test Interpretations and Recommendations Based on Soil Test Results Soil Test Relative Yield Recommendations Rating W/O Nutrient Very low < 50 % Large applications for soil building purposes Low % Annual application to produce maximum response & increase soil fertility Medium % Normal application to produce maximum yield High 100 % Small applications to maintain soil level. Amount suggested may be doubled & applied in alternate years

2 Unlike P, K, Ca, and Mg, recommendations for N applications do not depend on soil test levels. Nitrogen recommendations are the same for a given crop regardless of soil conditions. In fact, nitrogen fertility of soil is not reported by standard soil tests. Nitrogen recommendations also differ from P and K recommendations in that nitrogen content of fertilizer is simply the percentage of nitrogen the fertilizer contains. For P, it is the amount of phosphorous the fertilizer contains as P 2 O 5 and for K it is the amount of potassium contained as K 2 O. The practice of reporting P and K as oxides is an old practice that has not been changed. The consequence of using this old system is that the amount of P 2 O 5 and K 2 O necessary to meet the fertilizer requirement is greater than the amount that would be needed if fertilizers were reported as just P and K. Therefore, conversion factors are needed. Conversion Factors: % P 2 O 5 = % P x 2.29 % P = % P 2 O 5 x 0.44 % K 2 0 = % K x 1.20 % K = % K 2 0 x 0.83 All fertilizer packaging must contain three numbers. The first number always refers to the %N, the second number refers to the amount of P reported as %P 2 O 5, and the third number refers to amount of K as %K 2 O. The combination of these percentages, N-P 2 O 5 -K 2 O, make up what is known as the fertilizer grade. So, a fertilizer having a grade of contains 15% N, 3% P 2 O 5, and 4% K 2 O or, 15% N, 1.32% P, and 3.32% K. Other Fertilizer Terms Fertilizer label: legally required notice on all fertilizer containers. Includes: fertilizer grade name and address of manufacturer net weight specialty fertilizers Fertilizer grade: minimum guaranteed nutrient content given as % total nitrogen % available phosphoric acid % water soluble potash Available phosphoric acid is the phosphorus soluble in water plus that soluble in ammonium citrate; expressed as P 2 O 5 Water soluble potash: that potassium soluble in water expressed as K 2 O. The amount of nutrient required must be converted to the amount of the particular fertilizer grade to be used. More than one fertilizer grade may be required to supply the amount of nutrient required.

3 Conversion: [ Amount of fertilizer of a specific grade ] = = [ Amount of nutrient required ] [ Fertilizer grade for that nutrient ] x 100 Examples: 1. Nutrient Requirements 110 lb/acre N = 110 lb/acre N 24 lb P /acre / (0.44) = 55 lb/acre P 2 O 5 45 lb K /acre / (0.83) = 55 lb/acre K 2 O Fertilizer Grade Available = Fertilizer Required to Supply: (Note: always start with lowest nutrient requirement) P 2 O 5 lb/acre fertilizer = (55/5) x 100 = 1100 lb/a of fertilizer K 2 O lb/acre fertilizer = (55/5) x 100 = 1100 lb/a of fertilizer (in this case, one fertilizer supplies enough of both P and K) Calculate other nutrients supplied by the 1100 lb of fertilizer Amount of N supplied = amount of fertilizer x grade for N 100 = 1100 lb/acre x 10/100 = 110 lbs N Coincidentally, N supplied by adding 1100 lbs/acre of fertilizer is also adequate. 2. Nutrient Requirements 180 lb/acre N = 180 lb/acre N 35 lb/acre P (/.44) = 80 lb/acre P 2 O 5 67 lb/acre K (/.83) = 80 lb/acre K 2 O Fertilizer Grade Available = , Fertilizer Required to Supply: (Note: start with lowest nutrient requirement) P 2 O 5 lb/acre fertilizer = (80/10) x 100 = 800 lb/a fertilizer K 2 O lb/acre fertilizer = (80/10) x 100 = 800 lb/a fertilizer Calculate other nutrient supplied by the 800 lb of fertilizer: Amount of N supplied = amount of fertilizer x grade for N/100 = 800 lb/a fertilizer x 10/100 = 80 lb/a N Additional N required = Total required - Amount supplied = = 100 lb/a additional N needed Fertilizer required to supply 100 lb/a N N lb/acre fertilizer = (100/33) x 100 = 303 lb/a fertilizer

4 3. Nutrient Requirements 220 lb/a N 183 lb/a P 2 O lb/a K 2 O Fertilizer Grades Available = , , , Fertilizer Required to Supply: (Note: start with lowest nutrient requirement) 144 lb/a of K 2 O lb/acre fertilizer = (144/10) x 100 = 1440 lb/a fertilizer Calculate other nutrient supplied by the 1440 lb/a of fertilizer: 1440 lb/a of fertilizer contains 144 lb/a of P 2 O 5 and 144 lb/a of N Thus, 183 lb/a 144 lb/a = 39 lb/a of P 2 O 5 need to be applied 220 lb/a 144 lb/a = 76 lb/a of N need to be applied Calculation of additional fertilizer application: P 2 O 5 lb/acre fertilizer = (39/45) x 100 = 86.7 lb/a fertilizer N lb/acre fertilizer = (76/33) x 100 = lb/a fertilizer Nutrients in Manures As nutrient sources, animal wastes have two problems. First, nutrient composition and moisture are variable. Secondly, nutrients are released at variable rates. In New Jersey, horse manure is abundant (Rutgers NJAES Cooperative Extension, publication E307). Lime requirements Lime requirement depends on the active and residual acidity that need to be neutralized to raised the ph to the target (desired) ph. The target ph is a function of the type of crop (Rutgers NJAES Cooperative Extension, FS903, FS902 and FS904) Total acidity can be determined by titration or estimated by soil ph and texture (a surrogate for cation exchange capacity).

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