How to prepare the stock solution for fertigation

How to prepare the starting solution for fertigation

In proportional fertigation, concentrated nutrient solutions are prepared in several tanks. Solutions injected in the irrigation water in suitable proportions. These concentrated solutions is known as the "mother solutions".

Fertigation is not enough to know what amount of fertilizer should be applied. Also other factors must be taken into account in the preparation of the solutions. The main factors are:

• Compatible fertilizers.

• Number of storage tanks.

• The solubility of the fertilizers.

• The injection rate (or injection time).

• The type of used fertilizer.

• Use chelates.

• The interaction of fertilizers with water (endothermic reactions, reactions with elements present in water).

Compatibility of fertilizers

Some fertilizer interact to form insoluble compounds and precipitate. Precipitated nutrients are blocked and not available for plants. Another unfavorable effect of this is a clogging in the irrigation equipment.

Characteristics of the main fertilizers used in fertigation

For example, do not mix fertilizer, containing calcium, with fertilizers containing sulfates or phosphates. Incompatible fertilizers must be dissolved in different tanks.

Compatibility-soluble fertilizer used in fertigation

C – compatible,
– The solubility of reduced N – incompatible.
Most
common reactions of incompatibility with precipitation are:

How to determine the number of required
tanks (the number of stock solutions)?
The type of used fertilizers and their compatibility determine the minimum number
necessary masterbatch solutions. The quality of irrigation water and nutrients
available in the soil affect the number of storage tanks, as they
determine which fertilizer to use.

If the source water contains essential nutrients, such as calcium, sulfur and
magnesium, in sufficient concentrations, will not need to use
fertilizers containing these elements in the nutrition program. As a rule,
the use of fertilizers containing calcium, magnesium or sulphur requires
use from 2 to 4 storage tanks due to restrictions
compatibility.

For example, suppose needs to be used following fertilizers: potassium nitrate, calcium nitrate, monoammonium phosphate (MAP) and sulfate of magnesium.
In this case necessary, at least three tanks. Calcium nitrate is incompatible
with both MAP and magnesium sulfate is incompatible with MAP.
A possible distribution is as follows:
Tank 1: MAP.
Tank 2: calcium nitrate + potassium nitrate.
Tank 3: the magnesium Sulfate.

The solubility of fertilizers
Solubility of a fertilizer is defined as the maximum amount of fertiliser that can be completely dissolved in a given volume of water. Exceeding the maximum
amount will result in precipitation of the fertilizers in the irrigation system and
can be a very serious problem. Solubility is expressed in units
weight/volume of water.
For example: grams/liter or pounds/Gallon.
The solubility of each of the fertilizers depends on the temperature of the water in which it dissolves.
The solubility of most fertilizers increases with increasing temperature. This
way, at lower temperatures, the mother liquors should be more
diluted. At higher temperatures the mother liquors can be
more concentrated.
Common ion effect - solubility of a fertilizer is also dependent on other fertilizers that are dissolved in the mother solution. When the fertilizer dissolves in the storage tank with other fertilizers both contain a common ion, the solubility of both fertilizers
reduced. For example, potassium nitrate and potassium sulfate are compatible
and can be dissolved in the same storage tank. However,
as both contain potassium, its solubility is reduced when mixing.
Injection rate
The injection efficiency is defined as the ratio between the volumes of fluid fertilizers injected into irrigation water. Thus, it is expressed in units of volume/volume.
Example: liters/m3, gallon/100 gallon or % (percent).
It is possible to calculate the following ratio: dosage of fertilizer input / throughput
the ability of an irrigation system. Where the dosage input and throughput
the capacity expressed in units of volume / time.
For example, if the injector has a suction capacity 200 l/h throughput
the ability of the system 40 m3/h, injection rate as follows:
200L/h / 40m3/HR = 5 l/m3. This result can also be expressed as 0.5% or
in the ratio of 1:200.
The minimum injection rate depends on the solubility of fertilizers and the crop needs nutrients. Needs of crops for nutrients determine
the amount of fertilizer that will be applied on the plantation. Solubility
fertilizer determines the maximum quantity that can be dissolved in
the tank. If, for example, the solubility of the fertilizer is 100 g/l, and
the required concentration of this fertilizer in the irrigation water is 500 g/m3
the minimum injection rate will be as follows:
500 (g/m3) / 100 (g/l) = 5 l/m3.
At a lower injection rate require a larger amount of dissolved
fertilizer in the tank, in order to achieve the same concentration,
equal to 500g / m3 in irrigation water.
500 (g/m3) / 100 (g/l) = 5 l/m3.
Suppose that the injection ratio 4 l/m3. 4 l/m3 = 500 (g/m3) / X (g/l)
X = 500 (g/m3) / 4 (l/m3) = 125
g/l, which exceeds the solubility of this fertilizer.
To convert the injection rate during the injection use the following
equation:
Time of injection (min.) = (F X D X PI) / DI,
where F = Bandwidth of the irrigation system (m3/h);
D = Duration of irrigation (min);
PI = injection rate (l/m3);
DI = Suction capacity of the injector (l/h).
The need for nutrients of some crops

Note: needs must be distributed according to phenological phase of the crop development.
Technical characteristics of the fertilizers used in fertigation