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?

The efficiency of the co-current tower aerator (%): ____

The efficiency of the counter current tower aerator (%): _____

To calculate the efficiency of a tower aerator, we need the removal efficiency of a particular component and the knowledge of the RQ (Removal Quantity) and k2t (Contact Time Exponent).

To calculate the removal efficiency (%), we use the formula:

Efficiency = (1 - e^(-k2t)) * 100

Given that RQ = 6 and k2t = 4, we can calculate the efficiency for the co-current tower aerator using the formula provided above.

Efficiency of the co-current tower aerator:
Efficiency = (1 - e^(-4)) * 100
Efficiency ≈ (1 - 0.0183) * 100
Efficiency ≈ 98.17%

Now, let's find the efficiency for the counter current tower aerator. In this case, the counter current flow allows for greater contact time between the gas and the liquid, which typically results in higher efficiency.

To calculate the efficiency of the counter current tower aerator, we use the same formula as before, but with a different value for k2t:

Efficiency = (1 - e^(-k2t_counter_current)) * 100

However, since the value of k2t for the counter current tower aerator is not provided in the question, we cannot calculate the efficiency without this information. If you have the value of k2t for the counter current tower aerator, please provide that, and I will be able to calculate the efficiency for you.