Author: Mr. Guy Sela, CEO of SMART! Fertilizer Management software and an international expert in plant nutrition and irrigation.
In proportional fertigation, concentrated nutrient solutions are prepared in a number of tanks.
The solutions are then injected to the irrigation water at adequate ratios. These concentrated solutions are known as "stock solutions" or "mother solutions".
The advantages of this method are that it reduces the number of times that the nutrient solution has to be prepared, and therefore the grower saves time and labor. It also gives more flexibility in adjusting the salinity of the final nutrient solution and the ratios between nutrients, by enabling the grower to inject different ratios from each stock tank.
In fertigation, knowing the fertilizer amounts that have to be applied is not enough.
Other factors must be taken in consideration when preparing fertilizer stock solutions.
The main ones are:
Some fertilizer materials interact to form insoluble compounds and precipitates.
The precipitates tie up the nutrients and make them unavailable to the plant and cause clogging in the irrigation equipment.
For example, fertilizers containing calcium must not be mixed with fertilizers containing sulfates or phosphates.
Incompatible fertilizers must be separated and dissolved in different tanks.
The types of fertilizers that are being used and their compatibility determine the minimum number of tanks that is required.
The quality of the irrigation water and the nutrients available in the soil affect the number of stock tanks as well, since they determine which fertilizers should be used.
If the source water contains essential nutrients, such as sulfur, calcium and magnesium, at sufficient concentrations, fertilizer containing these elements may not be needed for the fertilizer recipe.
Usually, using fertilizers that contain calcium, magnesium or sulfur require using 2-4 stock tanks, due to incompatibility limitations.
For example, assume that the fertilizers that have to be used are Potassium nitrate, Calcium nitrate, MAP and Magnesium sulfate.
In this case, a minimum of three tanks is required. Calcium nitrate is incompatible with both MAP and Magnesium sulfate, and Magnesium sulfate is incompatible with MAP. A possible distribution is:
Tank 1 – MAP, Tank 2 – Calcium nitrate + potassium nitrate, Tank 3 – Magnesium sulfate.
The solubility of a fertilizer is determined as its maximum amount that can be fully dissolved in a determined volume of water. Exceeding this maximum amount will result in precipitation of the fertilizers in the irrigation system and can be a very serious problem.
The solubility is expressed in units of weight/volume of water. For example: grams/liter or lb/gallon.
The solubility of each fertilizer is dependent on the temperature of the water in which it is being dissolved. The solubility of most fertilizers increases with the temperature. Therefore, at lower temperatures, the fertilizer stock solutions must be more diluted. At higher temperatures, more concentrated stock solutions can be prepared.
|Fertilizer / Temperature (C˚)||5||10||20||25||30||40|
|MAP (Mono Ammonium Phosphate)||250||295||374||410||464||567|
|MKP (Mono Potassium Phosphate)||110||180||230||250||300||340|
The common ion effect - The solubility is also dependent on other fertilizers in the stock solution. If a certain fertilizer is being dissolved in the same stock tank with another fertilizer that contains a common ion, the solubility of both fertilizers is reduced.
For example, Potassium nitrate and Potassium sulfate are compatible, and can be dissolved in the same stock tank. However, since both contain potassium, their solubility is reduced when mixed together.
The injection ratio is defined as the ratio between the volumes of the fertilizer solution injected and the irrigation water. Therefore, it has units of volume/volume. For example: L/m3, gallon/100 gallon or % (percent).
It can be calculated by the ratio: Injection rate / Irrigation flow rate. Where flow rates of the injection and the irrigation are expressed in units of volume/time. For example, if the injector has a discharge of 200 L/hr and the irrigation flow rate is 40 m3/hr, then the injection ratio is:
200 L/hr / 40 m3/hr = 5 L/m3. This result can also be expressed as 0.5% or a ratio of 1:200.
The minimum injection ratio required is dependent on the solubility of the fertilizers and on the nutrient requirements of the crop. The nutrient requirement of the crop determines the amount of fertilizer to be applied to the field. The solubility of the fertilizer determines the maximum amount that can be dissolved in the tank. If, for example, the solubility of a certain fertilizer is 100 g/L and the required concentration of this fertilizer in the irrigation water is 500 g/m3, the minimum injection ratio will be:
500 g/m3/ 100 g/L = 5 l/m3.
A lower injection ratio requires dissolving a higher amount of fertilizer in the tank, in order to reach the same concentration of 500 g/m3 in the irrigation water:
Injection ratio 4 L/m3 = 500 g/m3 / x g/l
X = 500 g/m3 / 4 L/m3 = 125 g/m3, which exceeds the solubility of the fertilizer.
To convert Injection Ratio to necessary injection time or vice versa, use the following equation:
Injection time (min.) = (F X D X IR) / IFR
F = Irrigation flow rate (m3/hr)
D = Irrigation duration (min)
IR = Injection ratio (L/m3)
IFR = Injector Flow Rate (L/hr)