5 A-- 1 ha of rice field required 170 kg nitrogen, 130 kg phosphorus and 95kg potash. The 5 nutrient given by Urea, Diamonium phosphate and murrate of potash. Calculate the amount of urea, Diamonium phosphate and murrate of potash without using conversion factor.
a - Sure, let's calculate the amount of Urea, Diammonium Phosphate (DAP), and Muriate of Potash (MOP) required without using conversion factors.
Given:
- 1 ha of rice field requires:
- 170 kg nitrogen (N)
- 130 kg phosphorus (P)
- 95 kg potash (K)
Nutrient Content (per 100 kg):
- Urea (N): 46 kg N
- Diammonium phosphate (DAP): 18 kg P₂O₅ (phosphorus)
- Muriate of Potash (MOP): 60 kg K₂O (potash)
### Calculations:
1. **Urea (N):**
- Since Urea contains 46% nitrogen (N) per 100 kg, we can calculate the amount of Urea required for 170 kg of N.
- Amount of Urea = (170 kg N) / (46 kg N per 100 kg Urea)
- Amount of Urea = 170 / 0.46 ≈ 369.57 kg
2. **Diammonium Phosphate (DAP) (P₂O₅):**
- Since DAP contains 18% phosphorus (P₂O₅) per 100 kg, we can calculate the amount of DAP required for 130 kg of P₂O₅.
- Amount of DAP = (130 kg P₂O₅) / (18 kg P₂O₅ per 100 kg DAP)
- Amount of DAP = 130 / 0.18 ≈ 722.22 kg
3. **Muriate of Potash (MOP) (K₂O):**
- Since MOP contains 60% potash (K₂O) per 100 kg, we can calculate the amount of MOP required for 95 kg of K₂O.
- Amount of MOP = (95 kg K₂O) / (60 kg K₂O per 100 kg MOP)
- Amount of MOP = 95 / 0.60 ≈ 158.33 kg
### Summary:
- Amount of Urea required: Approximately 369.57 kg
- Amount of Diammonium Phosphate (DAP) required: Approximately 722.22 kg
- Amount of Muriate of Potash (MOP) required: Approximately 158.33 kg
BA - farmer using NPK fertilizer in his field, their grade is 13:10:27. The required dose of 5 nitrogen is 72 kg, phosphorus is 75kg and potash is 53kg. Calculate the remaining amount of fertilizer if they using Urea, Double super phosphate and murrate of potash.
A - To calculate the remaining amount of NPK fertilizer required after applying Urea, Double Super Phosphate (DSP), and Muriate of Potash (MOP), we first determine the nutrient content in each fertilizer and then subtract the amounts used from the required doses.
Given:
- NPK fertilizer grade: 13:10:27
- Required doses:
- Nitrogen (N): 72 kg
- Phosphorus (P): 75 kg
- Potash (K): 53 kg
Nutrient Content (per 100 kg):
- Urea (N): 46 kg N
- Double Super Phosphate (DSP) (P): 20 kg P₂O₅ (phosphorus)
- Muriate of Potash (MOP) (K): 60 kg K₂O (potash)
### Calculations:
1. **Amount of Urea (N):**
- Amount of Urea required = 72 kg N (required) / 46 kg N per 100 kg Urea
- Amount of Urea required = 72 / 0.46 ≈ 156.52 kg
2. **Amount of Double Super Phosphate (DSP) (P):**
- Amount of DSP required = 75 kg P (required) / 20 kg P per 100 kg DSP
- Amount of DSP required = 75 / 0.20 = 375 kg
3. **Amount of Muriate of Potash (MOP) (K):**
- Amount of MOP required = 53 kg K (required) / 60 kg K per 100 kg MOP
- Amount of MOP required = 53 / 0.60 ≈ 88.33 kg
### Remaining Amount of NPK Fertilizer:
- NPK fertilizer grade: 13:10:27
- Amount of NPK fertilizer used:
- Nitrogen (N): 156.52 kg
- Phosphorus (P): 375 kg
- Potash (K): 88.33 kg
Remaining amount of NPK fertilizer = Total required - Amount used
- Remaining nitrogen (N) = 72 kg - 156.52 kg = -84.52 kg (Negative means excess applied)
- Remaining phosphorus (P) = 75 kg - 375 kg = -300 kg (Negative means excess applied)
- Remaining potash (K) = 53 kg - 88.33 kg = -35.33 kg (Negative means excess applied)
