In proportional hydroponics, concentrated nutrient solutions—also known as stock solutions or mother solutions—are prepared in a number of tanks. These solutions are then injected to the system’s irrigation water at adequate ratios.
The main advantage of utilizing this method is that it reduces the number of times that a nutrient solution has to be prepared; 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, with the grower injecting different amounts from each stock tank.
However, in hydroponics, 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 the compatibility of fertilizers, number of stock tanks, solubility of fertilizers, injection ratio or injection time, types of fertilizers to be used, use of chelates and interaction of fertilizers with water (endothermic reactions, reactions with elements present in water, ect.).
In this article, we will concentrate on the first four factors.
Some fertilizer materials are non-compatible—for example, fertilizers containing calcium must not be mixed with fertilizers containing sulfates or phosphates. Often, when incompatible fertilizers interact, they’ll form insoluble compounds and precipitates. These precipitates tie up the nutrients and make them unavailable to the plant and cause clogging in the irrigation equipment. In order to avoid this, incompatible fertilizers must be separated and dissolved in different tanks.
Determining the number of tanks required
The types of fertilizers 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 also affect the number of stock tanks, since they determine which fertilizers should be used.
That is, if the source water or soil contains essential nutrients—such as sulfur, calcium and magnesium—at sufficient concentrations, fertilizers containing these elements might not be needed for the recipe.
For example, two to four stock tanks would be required when using fertilizers that contain calcium, magnesium or sulfur due to incompatibility limitations. Assume that the fertilizers you have to use are potassium nitrate, calcium nitrate, monoammonium phosphate (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 in this situation is: Tank one—MAP, Tank two—calcium nitrate and potassium nitrate, and Tank three—magnesium sulfate.
The solubility of fertilizers and its effect on hydroponic system design
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 serious problem.
The solubility is expressed in units of weight/volume of water (for example: lb./gal.). The solubility of each fertilizer is also dependent on the temperature of the water in which it is 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.
The solubility is also dependent on other fertilizers in the stock solution due to the common ion effect. 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.
Injection ratio or injection time
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).
The injection ratio can be expressed through this formula: Injection rate ÷ irrigation discharge (flow), where the injection rate and the irrigation discharge are expressed in units of volume/time.
For example, if the injector has a capacity of 200 L/hr and the irrigation discharge 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. In order to reach the same concentration of 500 g/m3 in the irrigation water, a lower injection ratio requires dissolving a larger amount of fertilizer in the tank: 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, where F is the irrigation flow rate (m3/hr), D is the irrigation duration (min), IR is the injection ratio (L/m3) and IFR is the injector flow rate (L/hr).