** Optimal Air Transfer / Forced and Natural gas liquid transfer**

## For an explanation of this phenomenon, let’s take the primary units of measurement: m, m², m³. Suppose we have a pool that has sides of 10m into which we pour 1,000 liters of water. Suppose further that the water in the pool is spread forth to get a water surface of 100 m² at a depth of 1 cm. In this pool, the ratio between the amount of water and its surface area is 1: 100. We can call this ratio as the coefficient: k = 100. So, given the amount of water in this pool, is the coefficient: k = 100. Coefficient k can also be higher than 100 but may also be smaller. It can also be 0 (closed bottle filled with water).

The higher is the coefficient; the faster is the transfer of gasses between the water and the atmosphere. But, this is not the only condition. The transfer of gasses affects the air currents on the surface and the vertical flow of water in the pool. So, in the above pool conditions are such that the transfer of gasses takes place in a natural way with no need for more aeration. However, this situation requires a huge area, so aeration basins are much deeper.

## However, if the same amount of water as above was poured into the swimming pool with sides of 1m, we get the coefficient k = 1. So, we would have one cubic meter of water surface area through which gas transfer takes place only through one square meter. In this case, the coefficient would be one hundred times smaller than in the pool as mentioned above. If we want to meet such as k = 100 for the same pool, we should need to enter as many air bubbles into the pool, that would be their total area of 100m².

## Aeration tanks are constructed in such a way that the natural transfer of gasses is not possible. Thus, it is necessary to use mechanical aeration.

** Partial pressure and gas exchange in the aeration tank**

**Optimal and homogeneous aeration basins:**

## To better understand this phenomenon, we need to see the activity of an air bubble. To get an air bubble beneath the surface, we need energy. The deeper an air bubble is pushed, the more mechanical energy we need. At what depth should an air bubble be pushed for optimal performance? The answer is that the depth should be enough to meet the most transfer of oxygen into the surrounding water.

## Consider that the partial pressure in all stagnant waters, oxygen decreases with depth. Hypoxia (low oxygen condition) should always seem at the bottom. Other gasses will be further neglected.

If an air bubble is released, at the bottom, it begins to rise slowly to the surface. During the journey, the air bubble releases oxygen into the surrounding water. This event takes place as long as the partial pressure of the oxygen in the air bubble is higher than the surrounding water. When you reach the equilibrium of the partial pressures, a greater travel distance of the bubble no longer makes sense.

Near the surface, the phenomenon can even be reversed. If the partial pressure of the oxygen in the surrounding water becomes higher than in the air bubble, then the oxygen will once again return to the bubble, and to the atmosphere. The primary reason that determines the amount of dissolved oxygen in the water is the diameter of the bubble itself.

## All these problems are solved in practice so that the starter exaggerated (the pools are fed several times more air than is necessary). The result is excessive consumption of electricity.

For the treatment (wastewater treatment) wastewater consumes more than 5 GW of electricity. Most of this energy is a wastewater aeration. (Aeration, Water Oxygenation).

## Please see the demo relations between depth and power consumption below.

Sample of Turbine Performance by 2,2 kW Motor