## Proposed Model of the American Society of Civil Engineers

In this advanced kind of evaluation we should consider that the molar fraction of oxygen and the air flow rate QG will change as a result of both the consumption of O2 and the uptake of CO2 as well as the change of inert gases. The OTE can be written, while introducing the air flow rate QG into Eq. (5.55):

QG,in cin with:

the O2 concentration in the air flowing in and out of the reactor, Eq. (5.56) gives: OTE = 1 - Qc,out Po2 yO2,out (5.59)

QG,in Po2yo with the molar fraction:

inert,in i=1

where N°2 is the number of moles of O2, NCO2 is the number of moles of CO2 and Ninert is the number of moles of inert components i.

A first approximation is obtained for neglecting the change in the flow rate Qg (QG in = QG out) respectively:

yO2,in

If we want to achieve greater precision, we must consider that the air flow rate QG changes during the rise of the bubbles. This change is caused by:

• the decrease in oxygen,

• the increase in carbon dioxide,

• the increase in water vapor and all other inert components inside the air bubble.

In the following model, we assume that the water vapor and the concentration of all other inert components do not change. But there is a further problem: it is seldom possible to measure the flow rate of the influent air. Therefore, the air flow rate is replaced by the nitrogen mass flow rate QN2,in pN2, which must be corrected with Eq. (5.62) (Redmon et al. 1983):

QG,in P°2 y°2,in = QN2,in Pn -¡"J— — (5.62)

Mn2 NN2

where QN2in pN2 is the mass flow rate of influent nitrogen (g N2 h 1),-is the

ratio of the molar masses oxygen/nitrogen in influent air (-) and-

is the ratio of the moles oxygen/nitrogen in the influent air (-).

mol N2

Introducing Eq. (5.62), reflecting whether "in" or "out" into Eq. (5.59) gives:

 M02 /No2 Mn2 lNN2, M02 [No2 Mn2 with: ^N^/in (Nn2/(NO2 + NN2 + NcO2))in yN2,in f NoA (No2/(No2 + Nn2 + NcoJ)out yO2, as well as: 2,in 5.4 Oxygen Transfer Rate, Energy Consumption and Efficiency in Large-scale Plants 1103 Equation (5.64) can be transformed into: If CO2 is removed before the O2 concentration is measured (e.g. in NaOH), from Eq. (5.67) we obtain: Finally, the oxygen transfer rate can be calculated based on Eq. (5.37) by considering the changing gas flow rate and by using: Qn2 ,in PN2 QN2 ,in PN2 M02/MN QN2 ,in PN2 MO2/MN yO2, yO2, yN2,in yN2,out y02,in y02,out 2,out yC02,in If CO, is removed, we obtain: QN2 ,in PN2 MO2/MN y02,in(1 y02,out) y02,out(1 y02,in) (1-y02,in)(1-y02,out) With Eq. (5.54a) or Eq. (5.54b), the overall mass transfer coefficient KLa can be calculated from OTR. M02 N02 MN2 NN2 MN2 NN2,out out