Special thread on compound fertilizer formulation issues
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Formula for High Tower to produce 30-0-5: Nitrogen: 1000*30%/46%/98% = 665 kilograms; Potassium: 1000*5%/53% = 94 kilograms; Others: 1000-665-94 = 241 kilograms. I would appreciate some advice from everyone. Communication leads to progress. Don’t stick to old ways.
Your formula should indicate whether it is for producing thio-based or chloro-based products. Also, it’s clear that 46% of the ingredients are urea, but what is the composition of the 53% remaining? Potassium sulfate? I haven’t heard of potassium sulfate with a 53% concentration; and even if it does exist, it seems like it would be somewhat wasteful as well, given its high cost. As for potassium chloride, those with a 57% concentration are less expensive. In any case, it’s important to understand the differences in cost when it comes to fertilizers containing chlorine or sulfur. . . . . . .
Both the moisture content of the raw materials and that of the final product need to be taken into account
Thank you to the two of you for participating. I work in the compound fertilizer industry, and there are many things I don’t understand. At the workplace, there’s no one to talk to either; everyone regards technical skills as something extremely valuable, fearing that others will take their jobs. How is it possible to make progress under such circumstances? It’s difficult to learn something; one has to figure things out trial and error. Thank you once again for the support from the moderator and the other two above. My recipe should use potassium chloride. Is this the correct way to do it? How should the moisture content of the raw materials be considered? Please explain
53 potassium is probably potassium chloride on a wet basis, right? What does the rest of that include? We have used ammonium chloride to reduce nutrients in the production of chlorinated compounds. I’m not sure whether your system is of the high-tower type or the drum type; it’s quite difficult to work with 66.5% urea in a drum-type system
66.5% urea – that’s a bit too high; use some other nitrogen fertilizer to reduce it
There’s no need to lower it; using a high tower works well. It just seems like your formula is based on calculations for dry materials
It’s clear that this is a high-tower fertilizer; we produce large quantities of this type of formula. The poster is correct in his calculations, and the other items likely refer to concrete fertilizer additives.
You need to calculate the formula using the method based on the product standard moisture content
Excuse me, upstairs. . If the moisture content of the raw material is very low. It can be ignored. . Can’t we just use the dry material algorithm?
Moisture content should be taken into consideration; for example, the composition of the raw materials used in the production of compound fertilizers is shown in the table, with the moisture content of the compound fertilizers set at 1.0%. Table: Composition of raw materials for compound fertilizers/%
Raw material N P2O5 K2O H2O
Ammonium oxalate 34.6 0 0 2.0
Ammonium phosphate 11 44 0 1.2
Ammonium sulfate 21 0 0 2.1
Potassium chloride 0 0 60 1.5
Filler 0 0 0 8.0
(1) When no filler is used, the amount of raw materials required to produce 100 t of high-efficiency fertilizer with a nutrient ratio of 15 – 15 – 15 is as follows. Since no filler is used, E = 0. Substituting these values into equations (1)–(4) yields the following system of equations:
34.6% (1 – 2%)A1 + 11% (1 – 1.2%)A2 + 21% (1 – 2.1%)A3 = 100 × 15% (1′)
44% (1 – 1.2%)A2 = 100 × 15% (2′)
60% (1 – 1.5%)D = 100 × 15% (3′)
A1 (1 – 2%) + A2 (1 – 1.2%) + A3 (1 – 2.1%) + D (1 – 1.5%) = 100 × (1 – 1%) (4′)
After calculation, the values are: A1 = 21.53 t, A2 = 34.5 t, A3 = 19.22 t, and D = 25.38 t. That is, to produce 100 t of 15-15-15 high-efficiency fertilizer, 21.53 t of monoammonium phosphate, 34.5 t of ammonium phosphate, 19.22 t of sulfuric acid, and 25.38 t of potassium chloride are required as raw materials. (2) Since sulfuric acid AN is not used, the amounts of raw materials required to produce 100 tons of a compound fertilizer specific for corn with a nutrient ratio of 14–10–11 are as follows: A3 = 0. By substituting these values into equations (1) to (4), the following system of equations is obtained: 34.6% (1 – 2%) × A1 + 11% (1 – 1.2%) × A2 = 100 × 14% (1″); 44% (1 – 1.2%) × A2 = 100 × 10% (2″); 60% (1 – 1.5%) × D = 100 × 11% (3″); E (1 – 8.0%) + A1 (1 – 2%) + A2 (1 – 1.2%) + D (1 – 1.5%) = 100 × (1 – 1.0%) (4″). The calculated values are: A1 = 33.92 tons, A2 = 23.00 tons, D = 18.61 tons, and E = 26.85 tons.
Raw material N P2O5 K2O H2O
Ammonium oxalate 34.6 0 0 2.0
Ammonium phosphate 11 44 0 1.2
Ammonium sulfate 21 0 0 2.1
Potassium chloride 0 0 60 1.5
Filler 0 0 0 8.0
(1) When no filler is used, the amount of raw materials required to produce 100 t of high-efficiency fertilizer with a nutrient ratio of 15 – 15 – 15 is as follows. Since no filler is used, E = 0. Substituting these values into equations (1)–(4) yields the following system of equations:
34.6% (1 – 2%)A1 + 11% (1 – 1.2%)A2 + 21% (1 – 2.1%)A3 = 100 × 15% (1′)
44% (1 – 1.2%)A2 = 100 × 15% (2′)
60% (1 – 1.5%)D = 100 × 15% (3′)
A1 (1 – 2%) + A2 (1 – 1.2%) + A3 (1 – 2.1%) + D (1 – 1.5%) = 100 × (1 – 1%) (4′)
After calculation, the values are: A1 = 21.53 t, A2 = 34.5 t, A3 = 19.22 t, and D = 25.38 t. That is, to produce 100 t of 15-15-15 high-efficiency fertilizer, 21.53 t of monoammonium phosphate, 34.5 t of ammonium phosphate, 19.22 t of sulfuric acid, and 25.38 t of potassium chloride are required as raw materials. (2) Since sulfuric acid AN is not used, the amounts of raw materials required to produce 100 tons of a compound fertilizer specific for corn with a nutrient ratio of 14–10–11 are as follows: A3 = 0. By substituting these values into equations (1) to (4), the following system of equations is obtained: 34.6% (1 – 2%) × A1 + 11% (1 – 1.2%) × A2 = 100 × 14% (1″); 44% (1 – 1.2%) × A2 = 100 × 10% (2″); 60% (1 – 1.5%) × D = 100 × 11% (3″); E (1 – 8.0%) + A1 (1 – 2%) + A2 (1 – 1.2%) + D (1 – 1.5%) = 100 × (1 – 1.0%) (4″). The calculated values are: A1 = 33.92 tons, A2 = 23.00 tons, D = 18.61 tons, and E = 26.85 tons.