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 obtain 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 additional 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 obtain 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 achieve for example 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