Fertilizer Calibration Objective: The objective of this lesson is to help you to learn to convert a fertilizer recommendation to the required amounts per unit area. These calculations are on a hectare basis or for a given size of an experimental plot. Goal: After completion of this lesson, you should: Introduction Know the fertilizer materials and the expression of its chemical analysis and grade. (% nutrient content). Be able to calculate the needed quantities of fertilizers using combinations of single and double nutrient fertilizers to apply a recommended fertilizer rate. Have learned the procedure to compute the required quantities of fertilizers for a given experimental plot or row. Have had practice in solving the problems on fertilizer calibration. A fertilizer is any substance containing plant nutrients that are usually added to soil to supplement the required plant nutrients. Chemical fertilizers may be natural or synthetic. Natural inorganic fertilizers would include material such as the Chilean nitrate, rock phosphate, etc. Synthetic fertilizers are manufactured products, such as urea, ammonium sulfate, ammonium phosphate, single super phosphate, etc. Synthetic fertilizers are available in various grades and analyses. Fertilizer grade or analysis is the minimum guaranteed plant nutrient content as percentage of total nitrogen, available phosphorus, and water-soluble potassium. A complete fertilizer has all three primary fertilizer ingredients ( N, P, and K ) as part of its formulation. For example, 14-15-14 is a complete fertilizer. This fertilizer has 14% N, 15% P, and 14% K. Fertilizers may be classified according to the nutrient elements present.
Fertilizers may be: Single nutrient fertilizers or straight fertilizers: fertilizer containing only ONE nutrient element. Double nutrient fertilizers: those that contain TWO nutrient elements. Complete fertilizers or complex fertilizers: materials contain the elements N, P, and K. Types of fertilizers Now let us look at a few common fertilizer materials and their grades. Table: Different fertilizers and their grades. Fertilizer Grade (%) N P * K * Single nutrient fertilizers: Ammonium sulphate 20 0 0 Urea 46 0 0 Calcium Ammonium nitrate 28 0 0 Single super phosphate 0 07 0 Triple super phosphate 0 20 0 Potassium sulphate 0 0 40 Potassium chloride 0 0 48 Double fertilizers: Ammonium phosphate 11 23 0 Diammonium phosphate 18 20 0 Complete fertilizers: Sampurna ** 19 08 16 Vijay complex ** 17 07 14 IFFCO grade I ** 12 11 22 *In most of the commercial fertilizers P is expressed as P 2 O 5 and K as K 2 O. To convert P 2 O 5 to P, multiply the value by 0.437 and to convert K 2 O to K, multiply the value by 0.830. In this table we have given nutrient contents in elemental form N, P, and K. **Examples of complex fertilizers available in India.
Computation of single nutrient fertilizers The nutrient fertilizer grade indicates that everything in a bag of fertilizer is not plant nutrient. Much of the material in the bag is made up of carriers and filler materials. For example, ammonium sulfate contains 20% N. That means, 100 kg of ammonium sulfate will contain 20 kg of N and the remaining 80 kg is sulfate and filler material. With this background about fertilizer material, we need to calculate the amounts of each nutrient source to meet the fertilizer recommendations. Fertilizer recommendations are expressed in Kilograms of nitrogen (N), phosphorus (P), and potassium (K) per ha. Fertilizers available in the market also contain carrier and filler materials in addition to N, P, or K. Hence, there is need to compute the amount of fertilizer required to supply the recommended rate of nutrients. For example, if 90 kg of N is recommended for sorghum, we must convert this recommendation to kg of urea or some other N fertilizer to be applied. Let us study this calculation procedure. This section gives the calculations to convert a fertilizer recommendation into amounts of required fertilizer materials using combinations of single nutrient fertilizers. Suppose a fertilizer rate of 90-21-30 is recommended for sorghum. This means that in 1 ha, 90 kg of N, 21 kg of P, and 30 kg of K should be applied. How do we translate this recommendation into kg per ha of the commercial fertilizers which are available? To solve for the mass of the fertilizer material, we can use the following formula: mass of nutrient recommended mass of fertilizer = --------------------------------- analysis of fertilizer (%) Using the above formula, find the amounts of ammonium sulfate (20% N), single super phosphate (7% P), and potassium chloride (48% K) needed to meet the recommended rate of 90-21-30. Recommended rate = 90-21-30 Fertilizers available: Ammonium sulfate (20% N), Single super phosphate (7% P), Potassium chloride (48% K).
Using the formula: mass of nutrient recommended mass of fertilizer = ---------------------------------- analysis of fertilizer (%) 90 Ammonium sulfate = ------------ = 450 kg (20/100) 21 Single super phosphate = ------------ = 300 kg (7/100) 30 Potassium chloride = ------------ = 62.5 kg (48/100) Computation of double-nutrient fertilizers per hectare Now let us look at the calculation procedures, for the use of double nutrient fertilizers or complex fertilizers to supply the recommended rate. One should realize that double nutrient fertilizers or complex fertilizers contain more than one nutrient and these nutrients are inseparable. The amount of nitrogen will usually be higher than the phosphorous in fertilizer recommendations; e.g., 90N-17P-30K. Hence, in converting the nutrient rate to the amount of double nutrient fertilizer or complex fertilizer, we should begin by calculating the nutrient required in the smallest amount. Suppose if we start with nitrogen, then the quantity of phosphorous would exceed the required amount. With this understanding, let us review these calculation procedures. A farmer has diammonium phosphate (18-20-0) and urea (46% N) to fertilize his pearl millet. How much of these fertilizers are required to supply the recommended rate of 100N-17P-0K? We must start with phosphate. If we do that; 17 Diammonium phosphate (kg) to supply 17 P = ------------ = 85 (20/100)
Diammonium phosphate contains nitrogen also. Hence, we should calculate the amount of N in 85 kg of diammonium phosphate. 100 kg of diammonium phosphate contains 18 kg of nitrogen. Then, 85 kg of diammonium phosphate will have = (18 X 85)/100 = 15.3 kg N. The recommendation calls for 100 kg N. 85 kg of diammonium phosphate supplies 15.3 kg of N. Hence, the balance of nitrogen needed = 100-15.3 = 84.7 kg The balance of 84.7 kg of N can be from urea. Hence: 84.7 kg of Urea to supply 84.7 kg of N = ------------ = 184 (46/100) Therefore: To meet the recommended rate of 100-17-0, we need to supply: 85 kg of diammonium phosphate, and 184 kg of urea. Computation of fertilizers per plot or rows So far, the calculations were for one hectare (ha) because the fertilizer recommendations are on a ha basis. But, in research, the experimental plots are usually confined to a few sq m and the fertilizers are to be applied to each row. Hence, there is need to calculate the required amount of fertilizers for small plots as well as for a row, based on the fertilizer recommendation per ha. Let us review these calculation procedures. Recommended rate = 60-17-0 Fertilizers available: Urea (46% N), Single super phosphate (7% P). Experimental plot size = 5 m x 3 m Rows per plot = 4 Row spacing = 75 cm
Calculate the quantity of urea and single super phosphate required: 1. per plot 2. per row Recommended rate = 60-17-0 Fertilizers available: Single super phosphate (7% P), Urea (46% N) Experimental plot size = 5 m x 3 m Rows per plot = 4 Row spacing = 75 cm First, calculate the amount of urea required per ha. 60 Urea per ha = ------------ = 130.43 kg (46/100) Secondly, calculate the amount required per sq m in g. 130.43 x 1000g Urea per sq m = ----------------------- = 13.04 g 10000 sq m Based on the amount of urea required per sq m, calculate the amount needed per plot and per row. Now calculate the amount of urea needed per plot of 5 x 3m=15 sq m. Urea per plot = 13.04 x 15 = 195.65 g Calculate the amount of urea needed per row of 0.75 x 5m=3.75 sq m. Urea per row = 13.04 x 3.75 = 48.91 g Similarly, the amount of single super phosphate required per plot and per row could be calculated. Recommended rate = 60-17-0 Fertilizers available: Single super phosphate (7% P), Urea (46% N) Experimental plot size = 5 m x 3 m Rows per plot = 4 Row spacing = 75 cm First, calculate the amount of single super phosphate (SSP) required per ha.
17 SSP per ha = ------------- = 242.8 kg (7/100) Secondly, calculate the amount required per sq m in g. 242.8 kg X 1000 g SSP per sq m = ------------------------- = 24.28 g 10000 sq m Based on the amount of SSP required per sq m, calculate the amount needed per plot and per row. Single super phosphate (SSP) required per sq m = 24.28 g Calculate the amount of SSP needed per plot of 5 x 3 m =15 sq m. SSP per plot = 24.28 x 15 = 364.2 g Calculate the amount of SSP needed per row of 0.75 x 5m=3.75 sq m. SSP per row = 24.28 x 3.75 = 91.0 g Thus, the amount of urea and single super phosphate required per plot and per row are calculated. Solving Problems on single nutrient fertilizers Solving Problems on double nutrient fertilizers Solving problems on per row and per plot