If the concentration of an element in an agricultural input is known (mg kg-1), and the application rate is known (kg ha-1), how is the final concentration in the treated soil determined?
The final concentration in the treated soil can be determined by lab sample. and depth of the soil, Volume and weight can be consider.
If you are using certain amount of given fertilizer for your experiment then you must have added certain dose lets say x% and then you might have use different doses for other treatment like 75% of x, 50% of x and so on. If this is the case you can easily calculate required amount of dose for your given field area.Just convert the required area into m2 (suppose for one acre you got 4,000m2). At the same time convert the experimental plot or pot into m2. Then figure out the most responsive fertilizer % which you want to use in real field. Lets say if area is 1m2 for the experiment unit and you get good response when you used 200gm of fertilizer ( suppose having concentration of (10%, 10%,10% NPK).You can get your required amount 200 gm *4,000m2.
If amount of applied fertilizer = 2% ,it means adding 2 ton of fertilizer to 100 ton soil or 100000 kg soil .
Therefore for one acre = 2 * (1.6 million /100000) =32 ton per acre.
Thank you everyone for your answers, especially Prem Baboo who went above and beyond my expectations.
Hi, we did not calculate the concentrazione at the end if the crop.
Fertilizer rate to be applied (kg/ha) = Nutrient need by the crop - nutrient available in the soil.
Yihenew,
How does one determine the nutrient need of a crop and nutrient available in a soil? For example, how much potassium is needed by a crop in your region and how much potassium is available in soils in your region?
For example:
- If your available input is Urea (46 % N). Thus, 100 kg of Urea contains 46 kg of nitrogen,
- Plot area (or Experimental unit area) is 4m x 3m (= 12 m2), and
- The application rate is 45 kg N per ha.
If 100 kg Urea = 46 kg N
??? Urea = 45 kg N
The Amount of Urea needed to supply the 45 kg N per hectare = (45/46)*100 = 97.83 kg of Urea per ha. = 97.83 kg of Urea for 1ha (10000 m2) of land.
- NB: 1 hectare = 10000 m2
So, if you need 97.83 kg of Urea for one hectare (10000 m2) of land, how much Urea will be needed for an experimental plot of 12 m2?
If 1 ha (10000m2) = 97.83 kg Urea
12 m2 = ??? Urea
Thus, for the 12 m2 plot, the amount of Urea needed will be
= (12/10000)*97.83 = 0. 1174 kg of Urea = 117.4 g of Urea per plot.
This is just an example.
It can be calculated as follows:
Amount of fertilizer kg/ha = kg/ha nutrient ÷ % nutrient in fertiliser x 100.
For example,
If you need 20 units (kg) /ha of phosphorus (P) and you plan to use single superphosphate with 8.8% P. You can reverse this calculation to work out how much of a nutrient you are applying.
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