Page 111 - Calculating Agriculture Cover 20191124 STUDENT - A

CH 9] Calculating Agriculture 9-27

FERTILIZER

As plants use nutrients, the soil must be replenished by means of fertilization. Fertilizers are spoken of
in terms of these three plant foods N-P-K. A 10-12-8 fertilizer means that 10% of the fertilizer is nitrogen
(N), 12% of the fertilizer is phosphoric acid (P 20 5), and 8% of the fertilizer is potash (K 20). These nutrients
are provided by materials commonly called plant food carriers. Some of the carriers are: ammonium
nitrate, super-phosphate, muriate of potash, and ammonium sulphate.
None of these carriers have 100% nutrient content, and some of them contain more than one of the
three major nutrients. Hence, the quantity of these carriers to be used in mixing fertilizer becomes an issue
of consideration and arithmetic calculation.

Example A: The three main plant nutrients in commercial fertilizers are nitrogen, phosphate and
potash. The amounts of each vary to meet the needs of a particular soil and crop. An 8-10-
4 fertilizer contains (by weight) 8% nitrogen (N), 10% available phosphate (P 20 5), and 4%
available potash (K 20). Determine the number of lbs. of nitrogen in 1,000 lbs. of 8-10-4
fertilizer; also determine the lbs. of Phosphorus (P), and lbs. of potassium (K).
Solution: N — 1000 x 0.08 = 80# Nitrogen
P — 1000 x 0.10 = 100# Phosphate
K — 1000 x 0.04 = 40# Potassium

Example B: How many lbs. of plant nutrients are in a ton of 15-12-4 fertilizer?
_________ nutrients

Solution: Calculate the pounds of each and then add them together. (Add the %’s of each
together and then multiply.)

2000#/ton x (0.15 + 0.12 + 0.04) =
2000#/ton x (0.31) = 620# nutrients

Example C: How many lbs. of potash are furnished by 40# of Sulphate of Potash (60% K 20)?
_____________ # K
Solution: (40#) x (0.60) = 24 #

Example D: A material containing a plant nutrient is called a carrier. Ammonium Sulphate is a
carrier often used as a source of nitrogen. Compute the number of lbs. of Ammonium
Sulphate containing 21% nitrogen is needed to provide 80# of nitrogen.

____________ # Ammonium Sulphate
Solution: (0.21) (AS) = 80# Nitrogen

1 1 Multiply 0.21 through by its reciprocal to maintain the
—— x (0.21) (AS) = 80# N x —— equality of the equation. This will isolate AS on one
0.21 0.21 side of the equation for the solution.

80#
AS = ——
0.21

AS = 380.95#

Example E: Compute the number of lbs. of Treble Superphosphate (45% P 20 5) that will
supply the phosphate in 800# of 6-10-4 fertilizer.
________ # Treble Superphosphate

(0.45) (TS) = (0.10) (800 #) Multiply 0.45 through by its reciprocal to maintain the

(0.10) (800 #) equality of the equation. This will isolate TS on one
TS = —————— side of the equation for the solution.
0.45 9

TS = 177.78 #

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