13CropSci_fertilizer-calculation 98.pptx

It helps you maintain and balance the potential of fertilizer applications in field

1. Fertilizer Calculation

2. Fertilizer CALCULATION FORMULA Amount of fertilizer Material (Kg)= Recommended Rate (kg nutrient/ha) Nutrient in Fertilizer Material % x Area (ha)

3. Single fertilizer combination Sample Problem: Calculate the amount of fertilizer materials for 1 ha at the recommended rate of 120-90-60 using the following fertilizers: 46-0-0 0-20-0 0-0-60 Solution:

4. Single fertilizer combination Sample Problem: Calculate the amount of fertilizer materials for 1 ha at the recommended rate of 120-90-60 using the following fertilizers: 46-0-0 0-20-0 0-0-60 Solution: Amount of fertilizer Material (Kg)= x Area (ha)

5. Amount of 0-0-60= Amount of 0-20-0= Amount of 46-0-0= 60 kg K ₂O 0.60 X 1ha =100 kg 90 kg P ₂O₅ 0.20 X 1ha =450 kg 120 kg N 0.46 X 1ha =260.87 kg

6. Combination of incomplete and single fertilizer Sample Problem: How much 21-0-0, 0-0-60 and 18-46-0 will you apply if you are to fertilized an area of ¾ hectare if the recommended rate of fertilizer is 120-90-60 per hectare. Solution:

7. Combination of incomplete and single fertilizer Sample Problem: How much 21-0-0, 0-0-60 and 18-46-0 will you apply if you are to fertilized an area of ¾ hectare if the recommended rate of fertilizer is 120-90-60 per hectare. Solution: First you have to convert ¾ hectare into decimal number Recommended Rate (kg nutrient/ha) Nutrient in Fertilizer Material % Amount of fertilizer Material (Kg)= x Area (ha) 3 divided by 4 = 0.75 ha

8. Amount of 0-0-60= Amount of 18-46-0= Calculate the amount of Nitrogen present in 18-46-0 Calculate the remaining Nitrogen requirement by subtracting the Nitrogen present in 18-46-0 with the N recommended rate. 120kg-26kg= 94kg 60 kg K ₂O 0.60 X 0.75ha =75 kg 90 kg P ₂O₅ 0.46 X 0.75ha =146.74kg 146.74kg x .18 N = 26.41 kg N

9. Amount of 21-0-0 = Answer: You have to apply 75 kg 0-0-60, 146.74 kg 18-46-0 and 335.71 kg 21-0-0 to suffice the 120-90-60 fertilizer requirement of ¾ hectare. 94 kg N 0.21 X 0.75ha = 335.71kg

10. COMBINATION OF COMPLETE AND SINGLE fERTILIZER Sample Problem: You are tasked to apply fertilizer on a one hectare land using 14-14-14, 46-0-0 and 0-20-0. How much of each fertilizer material will you apply if the recommend rate is 90-80-60. Solution:

11. COMBINATION OF COMPLETE AND SINGLE fERTILIZER Sample Problem: You are tasked to apply fertilizer on a one hectare land using 14-14-14, 46-0-0 and 0-20-0. How much of each fertilizer material will you apply if the recommend rate is 90-80-60. Solution: Amount of fertilizer Material (Kg)= Tips: You have to identify the lowest recommended rate and solve it first. In the case 90-80-60, 60 which is K ₂O₅ is the lowest. Recommended Rate (kg nutrient/ha) Nutrient in Fertilizer Material % x Area (ha)

12. Amount of 14-14-14= You have to identify the Nitrogen and P ₂O₅ content of 14-14-14 hence it is a complete fertilizer. After identifying the Nitrogen and P ₂O₅ of 14-14-14, Subtract it to its respective elements to get the remaining fertilizer requirement. Amount of Nitrogen present in 14-14-14=Amount of 14-14-14 applied x Nitrogen grade of 14-14-14 =428.57kg x .14 = 60 kg Amount of P ₂O₅ present in 14-14-14=Amount of 14-14-14 applied x P ₂O₅ grade of 14-14-14. =428.57kg x .14 = 60 kg 60 kg K ₂O 0.14 X 1ha =428.57 kg

13. Amount of 0-20-0= = = 100kg 80 kg P ₂O₅ – 60 kg P ₂O₅ in 14-14-14 0.20 X 1ha 20 kg P ₂O₅ 0.20 X 1ha

14. Amount of 46-0-0 = = = 65.22 kg Therefore, you have to apply 65.22 kg 46-0-0, 100 kg 0-20-0 and 428.57 kg 14-14-14 to satisfy the recommended rate 90-80-60 on one hectare land. 90 kg N – 60 kg N in 14-14-14 0.46 30 kg N 0.46 X 1ha X 1ha

15. Combination of complete, incomplete and single fertilizer Sample Problem: How much 14-14-14, 16-20-0 and 46-0-0 are needed to satisfy the fertilizer requirement of a one hectare land if the recommended rate is 100-50-30. (NPK) Solution:

16. Combination of complete, incomplete and single fertilizer Sample Problem: How much 14-14-14, 16-20-0 and 46-0-0 are needed to satisfy the fertilizer requirement of a one hectare land if the recommended rate is 100-50-30. Solution: Amount of fertilizer Material (Kg)= Amount of 14-14-14= Amount of P ₂O₅ in 14-14-14 is 30 kg Amount of N in 14-14-14 IS 30 kg Recommended Rate (kg nutrient/ha) Nutrient in Fertilizer Material % x Area (ha) 30 kg K ₂O 0.14 X 1ha =214.29 kg

17. Amount of 16-20-0= = = 100kg Amount of Nitrogen present in 16-20-0=Amount of 16-20-0 applied x Nitrogen grade of 16-20-0 =100kg x .16 = 16 kg 50 kg P ₂O₅ – 30 kg P ₂O₅ in 14-14-14 0.20 X 1ha 20 kg P ₂O₅ 0.20

18. Amount of 46-0-0= = = 117.39 kg Therefore, you have to apply 117.39 kg 46-0-0, 100 kg 16-20-0 and 214.29kg 14-14-14 100 kg- 30kg N 14-14-14 - 16 kg N 16-20-0 0.46 54 kg N 0.46

19. Finding Recommended rate Sample Problem: Mang Jose was able to satisfy the recommended rate of fertilizer by applying 4 bags of 46-0-0, 5 bags 0-20-0 and 2.5 bags of 0-0-60. What is the recommended rate being followed by Mang Jose ?

20. COMBINATION OF COMPLETE AND SINGLE fERTILIZER Sample Problem: You are tasked to apply fertilizer on a one hectare land using 14-14-14, 46-0-0 and 0-20-0. How much of each fertilizer material will you apply if the recommend rate is 90-80-60. Solution: Amount of fertilizer Material (Kg)= Tips: You have to identify the lowest recommended rate and solve it first. In the case 90-80-60, 60 which is K ₂O₅ is the lowest. Recommended Rate (kg nutrient/ha) Nutrient in Fertilizer Material % x Area (ha)

21. 4 bags 46-0-0 x 50 kg/bag= 200 kg 5 bags 0-20-0 x 50 kg/bag = 250 kg 2.5 bags 0-0-60 x 50 kg/bags= 125 kg RR= 200 kg x .46 N= 92 kg N = 250 kg x .20 P ₂O₅ = 50 kg P ₂O₅ = 125 kg x .60 K ₂O = 75 kg K ₂O Therefore, the recommended rate being followed is 92-50-75.

22. DETERMINING THE NUTRIENT CONTENTOF Fertilizer Materials 3 bags of Urea (46-0-0) 3 x 50 x .46= 69 kg N 4 bags 16-20-0 4 x 50 x .16= 32 kg N 4 x 50 x .20= 40 kg P ₂O₅

23. Determining the cost per kg nitrogen Example: Determine the cost per kg N of 21-0-0 which have a price of Php 800.00 per bag. Note that 1 bag is 50 kg. Step 1: Determine the Nitrogen present in one bag of fertilizer material. Nitrogen present= 50kg/bag × .21N = 10.5 kgN/ bag
Step 2: Divide the price per bag with kgN present/ bag to get the cost per kg nitrogen. Cost/kg N = Php 800/ bag ÷ 10.5 kg N/bag = Php 76.19/ kg N

24. 2. Mang Pedro borrowed 6 bags of (21-0-0) from Mang Juan. Mang Pedro returned 6 bags (45-0-0). Was the deal fair. Assume (21-0-0) @ Php 1150 and (45-0-0) @ Php 1550. Solution: 50 × .21= 10.5 1150÷ 10.5= Php 109.52/kgN
50 × .45= 22.5 1550÷ 22.5= Php 68.89/kgN
In terms of total price: 6 × 1150= Php 6,900.00 4 × 1550= Php 6,200.00 Total Nitrogen Content: 6 × 10.5= 63kgN 4 × 22.5= 90kgN

25. If 6 bags (21-0-0), 8 bags (0-20-0) and 2 bags (0-0-60) were applied by a farmer to one hectare of sitao, how much nutrients had been applied? 6bags × 50kg/bag= 300kg× .21N=63kgN 8bags × 50kg/bag= 400kg × .20P2O5=80kgP2O5 2bags x 50kg/bag= 100kg×.60K2O=60kgK2O Therefore, 63kgN, 80kg P2O5, and 60kg K2O were applied.

26. Determining the cheapest fertilizer combination to use Fertilizer Materials Price per bag (Php) 14-14-14 1,400 16-20-0 1,500 46-0-0 1,125 0-0-60 2,100 21-0-0 800

27. Combination 1 Amount of 14-14-14 =