Fertilizer recommendations for optimum crop production

The nutrients needed by plants are mostly taken from the soil. If the supply of nutrients in the soil is abundant, crops will grow well and produce high yields. However, if one of the nutrients needed is in short supply, plant growth is limited and crop yields are reduced.

The nutrients needed by plants are mostly taken from the soil. If the supply of nutrients in the soil is abundant, crops will grow well and produce high yields. However, if one of the nutrients needed is in short supply, plant growth is limited and crop yields are reduced. Nutrient soil test and fertilizers application is important to supplement nutrients needed by crops.  More and better quality food and cash crops can be produced and the fertility of soils, which have been over-exploited, can also be restored.

Soil test laboratories provide a critical step in management decisions of nutrients for optimum crop production. A Good Nutrient Soil Test is able to predict nutrient application rate for responsive soil.

How to interpret fertilizer recommendation:

Example 1

After soil test, a farmer is advised that a maize crop will require 80 kg N, and 60 kg P2O5 to achieve 7 tons per Ha.

The available fertilizers in the market are: DAP (18:46:0) and CAN (26:0:0).

The farmer comes to you for assistance and ask “How much of each fertilizer will be required for optimum maize yields?”

Method:

Step 1:

4R Plant Nutrient Management – Rate, Source, Time & Place.

Apply the basal fertilizer first i.e. DAP to boost root growth (P2O5)

Determine how many kg DAP are required to supply 60kg P2O5

100 kg DAP supplies 46 kg P2O5

Y                                  60 kg P2O5 per Ha

 

Y             =             60 kg P2O5 /ha x 100 kg DAP

46 kg P2O5

                                                                       

Y             =   60 kg P2O5 /ha x 100 kg DAP     =  60 x 100 kg DAP per Ha

46 kg P2O5                                               46

=             130.4 kg DAP per Ha

 

Step 2: Determine how much nitrogen is supplied by 130.4 kg DAP

 

 

100 kg DAP supplies 18 kg N

 

130.4 kg DAP per Ha          Y kg N

 

Y =          130.4 kg DAP / ha x 18 kg N

100 kg DAP

 

Y = 23.47 kg N /ha

 

Step 3: Determine how much additional nitrogen required from CAN

80 N kg/ha required – 23.47 kg N per Ha supplied by DAP = 56.53 kg N from CAN

 

100 kg CAN                                         26 kg N

Z kg CAN                                              56.53 kg N

 

Z =          56.53 kg N/ha x 100 kg CAN per Ha

——————————————————–

                        26 kg N per Ha

 

Z =          217.4 kg CAN per Ha.

 

Conclusion:

Based on the above example where we are assuming our soil test recommend  that a maize crop will require 80 kg N, and 60 kg P2O5 to achieve 7 tons per Ha. We can conclude that:

  • How much of each fertilizer is required for optimum maize yields = 130 kg DAP and 4 (50kg bags of CAN)
  • To arrive at bags Divide Kg of Each by Capacity of Bag e.g. 50 for 50 kg bag
  • How much per plant: Convert kgs of fertilizer into grams then divide by plant population

 

Example 2

A maize crop requires that, there is 80kg of Nitrogen(N), and 60kg of Phosphorus(P) to achieve 7 tons per Ha

Only Triple super phosphate (TSP) (0:46:0) fertilizer, CAN (26:0:0) and urea (46:0:0) are available.

How much of each fertilizer is required for optimum maize yields.

Step 1: Determine how many kg of P2O5 is equivalent to P2

Atomic weight of P2O5 is [2 x 31.0g (P) =62g] + [5 x 16.0g (O) =80]

                                Total =62.0 +80 =142g

                                62g P in 142g P2O5

Step 2: Calculate the P2O5 equivalent to 60kg P by calculating the conversion factor

                If 62g P in  142g P2O5, conversion factor for P2O5 into P is:

                                142/62 =2.29

(Implying you can get P by dividing with this factor or get P2O5 by multiplying with this factor)

Step 3: How much P2O5 will supply 60kg P

                142g P2O5                                                            62g P                    

                X                                                                       60kg P

X = 60 kg P x 142kg P2O5

                62kg P

 X = 137.4 kg P2O5

Step 4: How much TSP will supply 137.4 kg P2O5?

100 kg TSP                           46 kg P2O5

Y kg TSP                                137.4 kg P2O5

Y = 137.4 kg P2OX 100 kg TSP

46 kg P2O5

Y = 298.7 kg TSP

 

Step 5: How much CAN is required to supply 80kg N?

100 kg CAN                                         26 kg N

Z kg CAN                                              80 kg N

Z = 80 kg N/ha x 100 kg CAN

                26 kg N

Z = 307.7 kg CAN per Ha

 

Assignment: if the farmer decides to use Urea instead of CAN, How much urea will be required to supply 80kg N?

Recommended nutrient rate 80 kg N/ha; N % of Urea is 46%. How many kg of Urea do you need in a hectare?

80 kg N/ha x 100 kg Urea

                46 kg N

 = 173.9 kg Urea per Ha  

Last updated on Friday, August 25, 2023 at 4:20 pm

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