With the popularization of soil testing, formula fertilization and the development of compound fertilizers, the majority of farmers and agrochemical service personnel are required to master the fertilizer compounding technology and the related methods of fertilizer application. The basic principle of different fertilizers mix has been briefly introduced to readers. Here, at the request of the readers, experts are required to further explain the simple calculation methods related to fertilizer compounding through several examples so that they can be used in practical applications. learn by analogy.
Nutrients and Fertilizer Fertilizer Conversion
Most of the general fertilization recommendations are based on nutrient measurement and often need to be converted into specific fertilizer application amounts. Conversion method see (1):
Nutrient application rate
Divided by fertilizer nutrient content
Multiply fertilizer with nutrient content
Fertilizer application amount
(1) The nutrient content of the fertilizer can be found in the fertilizer bag surface or fertilizer manual.
Application example: If winter wheat has nitrogen (N) 6 kg/mu, if urea (including N46%) is used, how much fertilizer is needed?
A: According to formula (1), then
Urea application amount = nutrient application amount Urea nitrogen content = 60.46 = 13 (kg/mu)
Fertilizing ternary compound fertilizer with simple fertilizer
Application example: Mr. Cui of Handan City, Hebei Province, asked: The base fertilizer formula for winter wheat is: N7 kilograms of Mushi, P2O56 kilograms, and K2O3 kilograms. Existing "16-20-8" compound fertilizer, how many kilograms need to be applied? If nutrient ratios are inconsistent with the expert's recommendations, how to use single fertilizer urea (containing N46%) and potassium chloride (K2O60%) to regulate?
A: Because of the long-lasting effect of phosphate fertilizer, the amount of fertilizer is relatively stable; if the amount of nitrogen applied to the basal fertilizer is inappropriate, it can also be adjusted by top dressing. Therefore, the application amount of fertilizer should be determined by the amount of phosphorus applied and the phosphorus content of the fertilizer (if topdressing compound fertilizer is applied). The amount of nitrogen can also be used to determine the amount of compound fertilizer). If the nitrogen and potassium nutrients are insufficient, then use a single fertilizer supplement. So get:
Compound fertilizer application amount = 620% = 30 kg
When the compound fertilizer is applied, N=3016%=4.8 kg K2O applied when applying the compound fertilizer=308%=2.4 kg The amount of urea to be supplemented=(7-4.8)46%=4.8 kg Need to supplement the application of chlorination Potassium amount = (3-2.4) 60% = 1 kg
Finally, the basic fertilizer application plan was: 30 kg of compound fertilizer in Mushi, 4.8 kg of urea, and 1 kg of potassium chloride.
Diluted compound fertilizer with binary compound fertilizer
Application example: Ms. Liu from Fujin City, Heilongjiang Province asked: Here, soybeans are applied with sulfur-based compound fertilizers with nitrogen, phosphorus, and potassium nutrients (%) of 12-22-11, 350 kg per hectare. Application, the effect is very good. Only 15-15-15 compound fertilizers are available in the hands. Can it be formulated with diammonium phosphate (containing N18%, P2O546%) and potassium sulfate (including K2O50%)?
A: The problem involves two types of compound fertilizers. The method of establishing a binary equation group can be used to solve the nitrogen and phosphorus levels simultaneously. Based on this, the following formula is obtained and used to calculate the 15-15-15 compound fertilizer and diammonium application amount respectively. The amount of potassium fertilizer can be solved according to the formula (1).
Assume that 15-15-15 after the deployment of the amount of x, its nitrogen content, phosphorus content were a1, a2; diammonium application rate is y, its nitrogen content, phosphorus content were b1, b2; known Or you can directly calculate the total amount of nitrogen and phosphorus nutrients respectively applied for m1 and m2. Then, the solving formulae for x and y are Formula (2) and Formula (3):
x=(b2m1-b1m2)(a1b2-a2b1)(2)
y=(m1-a1x)b1, or y=(m2-a2x)b2(3)
For this example, the 15-15-15 compound fertilizer has a nitrogen content of a1=0.15, a phosphorus content of a2=0.15, a diammonium nitrogen content of b1=0.18, a phosphorus content of b2=0.46, and a compound known as 12-22-11 per hectare. The total nitrogen applied to the fertilizer was m1 = 3500.12 = 42, and the total phosphorus applied m2 = 3500.22 = 77. Substituting Formula (2), the application amount of 15-15-15 compound fertilizer can be obtained:
x=(b2m1-b1m2)(a1b2-a2b1)
= (0.1542-0.1877) (0.150.46-0.150.18) = 130 (kg)
Substituting formula (3) can obtain the diammonium application amount:
y=(m1-a1x)=(42-0.15130)0.18=125 (kg) or
y=(m2-a2x)b2=(77-0.15130)0.46=125(kg)
For the application amount of potassium sulfate, the amount of 350 kg of 12-22-11 compound fertilizer applied should be calculated first: 3500.11 = 38.5, and then the application amount of potassium sulfate (K2O50%): 38.50.5 = 77 (kg).
So far, if the application of 350 kg of 12-22-11 compound fertilizer per hectare is changed to 15-15-15 compound fertilizer, diammonium and potassium sulfate, the compounding scheme is: 15-15-15 per hectare. 130 kg of compound fertilizer, 125 kg of diammonium and 77 kg of potassium sulfate.
Compared with the original fertilizer, the compounded fertilizer has the same amount of nutrients and equal fertilizer effect. This point has been proved by scientific experiments and production practices, but in practice it is necessary to look at both the fertilizer cost and the fertilizer supply situation. If it is a large-scale planting, it should also consider the ease of application and mixed labor.
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