Horticulture Principles and Practices

4.4.8 GREENHOUSE FERTILIZATION
Most greenhouse production occurs in containers, which restricts the amount of soil or
growth medium from which a plant can obtain nutrition. Further, greenhouse media used
in greenhouses for plant cultivation are often artificial and have little or no nutrition to
support plants. As a result, fertilization is critical in greenhouse production. Fertilization
programs should include minor or trace elements in addition to major elements. Liquid
fertilizers, applied through irrigation water, are important in greenhouse fertilization.
The subject is discussed further in Chapter 12.
4.4.9 FERTILIZER CALCULATIONS
Following a soil test, recommendations are made as to the type and amounts of fertilizer
the producer may apply. An amount of a certain nutrient may be purchased. However, the
weight of fertilizer applied depends on the source, since fertilizers come in all kinds of
grades. The three common calculations involving dry fertilizer formulations are (a) nutrient
percentage, (b) amount or weight of source (commercial) fertilizer to apply, and (c) the
amounts or weights of component materials to use in preparing a bulk of mixed fertilizer.
1. Nutrient percentage
Problem:
Solution:
2. Simple fertilizer mixture
Problem:
Solution:
What is the percentage of nitrogen in the fertilizer urea?
Urea has a formula of (NH 2 ) 2 CO, and molecular weight
of 60.056 g Molecular weight of nitrogen (N 2 ) = 28.014
Percentage of nitrogen = [28.014/60.056] × 100 =
46.6 or 46% approximately (round down to nearest
whole number).
Given ammonium nitrate (34-0-0) and treble superphosphate
(0-45-0), prepare 1,000 kg of fertilizer of
grade 15-10-0.
Final mixture will contain 150 kg of nitrogen (i.e.,
15% of 1,000 kg).
It will also contain 100 kg P 2 O 5 (i.e., 10% of
1,000 kg).
Amount of ammonium nitrate needed (p.s., it contains
34 kg of N per 100 kg): [100 × 150]/34
= 441 kg of 34-0-0
Similarly, for phosphorus (contains 45 kg of P 2 O 5 per
100 kg)[100 × 100]/45
= 222 kg of 0-45-0
Total nutrients = 441 + 222 = 663 kg (leaving 337
balance of the desired 1,000 kg).
126 Chapter 4 Plant Growth Environment
The balance is satisfied by adding a filler (inert material) or lime. The procedure is
the same for compounding a mixture of three components (i.e., N-P-K).
What if calculated proportions of nutrient components add up to more than total
desired weight? In this example, what if the two amounts exceeded 1,000 kg? The
component amounts cannot exceed the total desired weight. If that happens, the desired
grade should be lowered or a source with higher analysis (e.g., for nitrogen, use urea with
46% N) should be used.
3. Amounts (weights) of sources (fertilizers) to apply
Problem: It has been recommended that a producer apply 80 kg of
nitrogen and 40 kg of phosphorus per acre to his field.