 Soil test recommendations are given in lb/acre or kg/ha of nutrients. To determine the
fertilizer rate for a particular nutrient, multiply the rate of the desired nutrient by
100 and divide by the percentage of the nutrient in the fertilizer.
Example 1
Recommended rate of N is 80 lb./ac.
Using 46-0-0, the rate of fertilizer required is:
(80 x 100) / 46 = 174 lb./acre
Example 2
Recommended rate of P2O5 is 40 lb/acre.
Using 11-52-0, the rate of fertilizer required is:
(40 x 100) / 52 = 77 lb/acre
77 lb./acre of 11-52-0 would also supply (11/100) x 77 = 8.5 lb/acre of N.
Example 3
Recommended rate of K2O is 15 lb/acre.
Using 0-0-60, the rate of fertilizer required is:
(15 x 100) / 60 = 25 lb/acre.
Converting fertilizer prices into price per unit of nutrient
The cost of a fertilizer is related to its plant nutrient content. If a nitrogen
fertilizer such as 34-0-0 is being purchased, the cost should be about three-quarters that
of 46-0-0. When buying fertilizer, one should compare prices on the basis of cost per
pound of "actual" nutrient, not the price per tonne of fertilizer material.
Example 1
If urea (46-0-0) costs $367/tonne, the cost per pound of nitrogen (N) is calculated as
follows:
Nitrogen in one tonne (1,000 kg or 2,204 lb.) of 46-0-0 (containing 46% N):
(46/100) x 2,204 = 1,014 lb.
Cost per lb. of N is:
$367/1,014 = $0.362
Example 2
(Based on 11-521-0 at $391/tonne)
In order to calculate the cost of phosphate in 11-52-0, the value of nitrogen must
first be subtracted.
Nitrogen in one tonne (1,000 kg or 2,204 lb.) of 11-52-0 is (11/100) x 2,204 = 242 lbs.
The value of nitrogen is 242 x $0.362 = $88 (from example 1, which calculated the value of
N to be $0..362/lb)
Cost of phosphate per tonne is $391 - $88 = $303.
Phosphate in one tonne (1,000 kg or 2,204 lb.) of 11-52-0 is:
(52/100) x 2,204 = 1,146 lb.
Cost per lb. of P2O5 is :$303/1,146 = $0.264
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