# hydroponic nutrient solution for vegetables and fruits formula indoor outdoor

Important conversions:
Parts per million = 1 milligram/liter of water.

1 gallon = 3,785 liters.

2.2 pounds = 1 kilogram.

How to calculate the nutritional needs of plants:
Calculating the nutritional needs of plants depends on several factors, including:

1 – The final desired concentration of the element, in parts per million.

2 – Analyzing the water to find out what it contains of essential nutrients for the plant, so what is the final required concentration of the element, but always remember that this element is part of the compound.

3- What is the percentage of the element in the final compound (the nutrient solution)

4- What is the size of the fertilizer injection tank?

5- What is the dilution factor when injected?

Note: Fertilizer calculations are not completed before completing the calculation of the amount of fertilizers for all components.

## The composition of the nutrient solution:

• Element N = 189 parts per million.
• Element P = 39 parts per million.
• Element K = 341 parts per million.
• Elemental Ca = 170 parts per million.
• Mg element = 48 parts per million.

*Fe = 2.0 parts per million.

• Mn = 0.55 ppm.
• Zn = 0.33 parts per million.

The element Cn = 0.05 parts per million.

• Element B = 0.28 parts per million.

Mo element = 0.05 parts per million.

Calculation of calcium is usually done (how to determine the proportion of calcium in a nutrient solution):

• Final required concentration of calcium = 170 parts per million.

Calcium is not added in a single form, but it is added in the form of calcium nitrate Ca (NO2).

Note that the percentage of calcium in calcium nitrate = 19%.

Thus, the required amount of calcium nitrate to get 170 parts per million 0.19/170 = 894.73 parts per million or mg / liter of water, or we need 0.8947 grams of calcium nitrate (nutrient solution).

But if the volume of the concentrated nutrient solution tank is 100 liters and it is intended to make a nutrient solution, it is diluted 100 times and multiplied by 100 x 100.

Therefore, the required amount of calcium is:

0.8974 g/L x 100 x 100 = 8947 g = 8,947 kg of calcium nitrate fertilizer.

• Nitrogen concentration of calcium nitrate = 15% and therefore 894.73 mg/L x 0.15 = 134 mg/L or 134 parts per million nitrogen.

Therefore, the remaining concentration is 189 – 134 = 55 parts per million.

So is it possible to use a compound fertilizer such as 5-5-50?

Compound fertilizers can be used by following the following steps.

It is preferable to calculate the total amount of phosphorous required from the compound fertilizer, where the concentration of the phosphorous element is the lowest concentration of the major elements.

The percentage of phosphorous in the compound fertilizer used is 5%

Calculate the amount of compound fertilizer required to give the required phosphorous concentration in the nutrient solution 34/0.05 = 680 mg/liter = 6800 grams of compound fertilizer to make 100 liters of concentrated solution.

What is the percentage of nitrogen present in this amount of compound fertilizer = 680 x 0.05 = 34 mg / liter.

So the amount of residual nitrogen = 55 – 34 = 21 parts per million.

What is the percentage of potassium in this amount of fertilizer (50% potassium concentration) = 680 x 0.50 = 340 mg / liter.

Accordingly, all the required amount of phosphorous and potassium have been added from the compound fertilizer, and the remaining 21 parts per million of nitrogen can be added from other sources.

Where the remaining amount can be added using 33% ammonium nitrate fertilizer.

Through the following: 21 x 0.33 = 63.63 mg of ammonium nitrate fertilizer = 636 g to make 100 liters of concentrated solution.

The rest of the minor elements are calculated in this way, taking into account.

1- The concentration of the required element in the nutrient solution, expressed in parts per million.

2- The percentage of the fertilizer component in the compound used.

3- The size of the tank.

4- Dilution rate.

5- Dilution ratio of the compound solution during the different growth stages of plants