Page 112 - Calculating Agriculture Cover 20191124 STUDENT - A

9-28 Land, Soils and Fertilizers CH 9]

Example F. How many lbs. of each are needed; Ammoniated Superphosphate (3-20-0) and
Ammonium Nitrate (36% N), to provide the nutrients in a ton of 17-7-0 fertilizer?
Note
N P K _________ # Ammoniated Superphosphate (AS)
AS: 3 – 20 – 0 ________________ # Ammonium Nitrate (AN)
AN: 36 – 0 – 0
K : 0 – 0 – 0 To build an equation that resolves our question of how much of each is required, it is important to
1 ton: 17 – 7 – 0 set up simultaneous equations; one each for Nitrogen, Phosphorus, and then one for Potassium;
though Potassium in each is 0 (zero). In this problem, note that you are adding together separate
elements of Nitrogen (N) and Phosphorus (P), and Potassium (K) is not present (has a value of 0 )
and will be ending with a container of Nitrogen and Phosphorus mixed. As such, the parenthesis
( ) for separate N and P are combined ( = )in a single parenthesis ( ) for these elements.
Question: What is being mixed? = Question: What will you end with?
Separately Combined
( N ) + ( P ) = ( N + P ) Nutrient
( AS ) + ( AN ) = ( AS + AN ) Formulation

Next, you attribute the percentage strength of each factor with the equivalent decimal value. For the nitrogen
element, what is mixed is 3% Ammoniated Superphosphate (AS) and (+) 36% Ammonium Nitrate (AN) which is
mixed and ( = ) combined as 17% Fertilizer ( AS + AN ) that will weigh 2000 pounds. This equation becomes:

(AS + AN) Formulation with element strength (%)
N —> (0.03) (AS) + (0.36) (AN) = (0.17) ( 2000 ) and the weight. NOTE: The arithmetic
operation between the parenthesis is
multiplication.

For the phosphorus element, what is mixed is 20% Ammoniated Superphosphate (AS) and (+) 0% Ammonium
Nitrate (AN), as the AN contains ZERO Phosphate, and this will be mixed ( = ) in, and combined as 7%
Fertilizer ( AS + AN ) that will weigh 2000 pounds. This equation becomes:

(AS + AN) Formulation with element strength (%)
P —> (0.20) (AS) + (0.00) (AN) = (0.07) ( 2000 ) and the weight. NOTE: The arithmetic
operation between the parenthesis is
multiplication.

In this equation, the phosphorus element (AN) will equal zero ( 0.00 x AN = 0; zero times any number
equals zero ) leaving only one unknown, AS. So now you solve for Ammoniated Superphosphate, then
insert this solution value calculated for AS into the nitrogen formula and then solve for AN...

Thus:

P —> (0.20) (AS) + (0.0) (AN) = (0.07) (2000) Formulation with element strength (%)
(0.20) (AS) + 0.0 = (0.07) (2000) and the weight. NOTE: The arithmetic
(0.20) (AS) = 140 operation between the parenthesis is
(1/0.20) (0.20) (AS) = (140) (1/0.20) multiplication.
AS = 140 / 0.20
AS = 700# # Ammoniated Superphosphate

Substituting 700# for AS in the nitrogen equation we have:

N —> (0.03) (AS) + (0.36) (AN) = (0.17) (2000)
Formulation with element strength (%)
and now solve for AN: and the weight. NOTE: The arithmetic
(0.03) (700) + (0.36) (AN) = (0.17 ) (2000) operation between the parenthesis is
multiplication.
21 + (0.36) (AN) = 340#
21 - 21 + (0.36) (AN) = 340 - 21
(0.36) (AN) = 319
(1/0.36) (0.36) (AN) = (319) (1/0.36)
AN = 319/0.36
AN = 886.11 # Ammonium Nitrate

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