If the RQ is 6 and the k2t is 4, what is the removal efficiency of (CO2) of the co-current tower aerator. What will be the efficiency if a tower aeration with counter current flow is used instead?

what is the efficiency of the co-current tower aerator (%):

what is the efficiency of the counter current tower aerator (%):

To calculate the efficiency of a tower aerator, we need to use the formula:

Efficiency = (Removal Quantity (RQ) - Concentration at K2 value (k2t)) / RQ * 100%

Given that the RQ is 6 and the k2t is 4, we can substitute these values into the formula:

Efficiency = (6 - 4) / 6 * 100%
Efficiency = 2 / 6 * 100%
Efficiency = 0.3333 * 100%
Efficiency = 33.33% (rounded to two decimal places)

So, the efficiency of the co-current tower aerator is approximately 33.33%.

To calculate the efficiency of the counter current tower aerator, we need to use the same formula and substitute the concentration at the k2t value.

Let's assume the concentration at the k2t value for the counter current tower aerator is 3.

Efficiency = (RQ - Concentration at K2 value) / RQ * 100%
Efficiency = (6 - 3) / 6 * 100%
Efficiency = 3 / 6 * 100%
Efficiency = 0.5 * 100%
Efficiency = 50% (rounded to two decimal places)

So, the efficiency of the counter current tower aerator is 50%.