Hello friends, my name is Kashinath and I got bumper yield from my Onion crop this season. Do you know why? Actually I used recommended dose of fertilizers as per my lands soil testing report. It’s so easy that you can also do it and get more yields with less cost on fertilizers. Do you want to learn this? this? If yes then I will shar share e m ex erie erienc nce e with with ou -----
Before cultivation of Onion, I did soil testing and found that in my 1 Hector land amount of Nitrogen (N) is less than required, Amount of Phosphorus (P) is medium and Amount of Potash (K) is high than required quantity. Then I learnt how to calculate r equired fertilizer dose for onion crop with simple mathematical calculations.
Recomanded dose of Onion crop as per Mahatma Phule Kr ushi Vidyapeeth (MPKV), Rahuri is 100:50:50 Kg / he. This means we have to supply 100 Kg of N , 50 Kg of P and 50 Kg of K to onion crop for 1 Hector area. Now as you know this you can supply by adding straight and/or mixed fertilizers like Urea / DAP / SSP/ MOP etc to
I have used only straight fertilizers for my crop so let’s learn that first
So if we want to give 100:50:50 Kg / he recommended dose to onion o nion crop let’s learn how to calculate how much Kg of Urea / SSP and MOP will be required for this. Know as you know Urea contains 46 % N , SSP contains 16 % P and MOP contains 58 % K in it. So this means for supplying 1 Kg of N by using Urea we have to use 2.17 Kg urea in field and you can calculate this simply formula as – as –
Step --1 Kg of fertilizer required for one Kg nutrient nutrient = 100/ Content (%) of nutrient nutrient in fertilizer Let’s Let’s see in case of Urea = 100/46
= 2.17 ------------------- this we will call as factor.
By the same way we can calculate factor for SSP and MOP. (Factor of SSP = 6.25, MOP = 1.72.) 1 .72.)
Step-- 2 Now we have to only multiply this factor w ith required nutrient quantity for each nutrient as that of N/P/K. So required quantity of fertilizer will be --Urea = 100 * 2.17 = 217 Kg / He. SSP = 50* 6.15 = 307.5 Kg/ He. MOP= 50* 1.72= 86 Kg /He. This means we will required 217 Kg of Urea , 307 Kg of SSP and 86 Kg of MOP for onion crop on 1 Hector area (2.5 Acre). As urea leach down with water this is also recommended that it should be given in 2 or 3 split doses.
Step -- 3 Now as we have done soil testing and found that Nitrogen is low, phosphorus is medium and potash is high in our soil. We need to increase incr ease nitrogen dose by 25 % and lower potash dose by 25 % whereas phosphorus dose will remain as it is. So final required dose as per soil test report will be –
Urea = 217 +25 % =271 Kg/He.
SSP = 307 Kg/ He.
MOP = 86 – 25 % = 65 Kg / He.
But friends, sometimes you don’t get straight fertilizers in market and mix fertilizers are also cost effective. Then let’s see how we can use mix fertilizers for same crop and area. ar ea. Now let’s assume that we have decided to use Di-ammonium phosphate (DAP) which contains 18 % Nitrogen (N) +46 % phosphorus (P) and MOP which contains 58 % K for our onion crop.
Dose of Onion crop – crop – 100:50:50 Kg / He. Available fertilizers – fertilizers – DAP, MOP and Urea.
Step 1 – Calculate factors as shown in above formula – formula – Kg of fertilizer required for one Kg nutrient = 100/ Content (%) of nutrient in fertilizer Factor of N in DAP = 100/18 = 5.55. Factor of P in DAP = 100/46 = 2.17. Factor of K in MOP = 100/ 58 = 1.72. Factor of N in Urea = 100/ 46 = 2.17.
Step –2 Now, first we have to calculate dose for highest nutrient content in given fertilizer. As that of in case DAP, Phosphorus (P) content is more (46%) than Nitrogen (N) content (18%), so we will calculate re quirement of Phosphorus (P) first in DAP – DAP – Phosphorus required through DAP is 50 Kg so, DAP for P = 50*2.17 = 108.5 Kg / He.
Step – 3 Now, as we have fulfilled Phosphorus (P) requirement, let’s check how much nitrogen we have given as this is is Mix fertilizer – fertilizer – So , IF
100 Kg of DAP content 18 Kg N (i.e 18%) , how much N will be there in in 108.5 Kg DAP. 108.5*18 = 1953/100= 19.53 Kg let’s say 20 Kg N requirement is fulfilled fulfilled through DAP.
Step--4 Remaining Nitrogen(N) Nitrogen(N) requirement requirement = 100 – 100 – 20 = 80 Kg / He These we can fulfill by using any straight fer tilizer like Urea – Urea – Remaining N = 80*2.17 (Factor of urea) = 173.6 Kg / He.
Step – 5 Potash (K) requirement = 50*1.72 (Factor of MOP) = 86 Kg / He. So by this we can sa y that, for Onion crop recommended fertilizer dose will be fulfilled by giving – giving –
108.5 Kg – DAP 173.6 Kg – Urea 86 Kg – MOP Step – 6 Now , as per our soil test report we have to reduce or increase calculated dose by 25 % ( if nutrient content in soil is high dose will be reduced by 25 % and if nutrient content if low it will be increase by 25 % ). So actual fertilizer dose will be – be –
108.5 Kg / He – DAP (As Phosphorus Phosphorus (P) content c ontent in soil testing report id medium, so it will not change). 217 Kg / He – Urea ( As Nitrogen(N) content in soil testing testing report is less , dose increased by 25 %). 65 Kg / He – MOP (As Potash (K) content in soil testing report is high, dose reduced by 25 %).
So friends, this is how you can also calculate required fertilizer for any crop and land area to get more yield from your crop and save lot o f many on cost of fertilizers. Now by the same method, you can calculate fertilizer dose for any crop on the basis of your soil test report.
