WWTP - Post Aeration Air Requirement Based on Fick's Law

[RJSH]

I am trying to understand the calculations of air requirement for Post Aeration Tank based on Fick's Law that I found. Can someone decodify how the followings were calculated based on the data provided?
Interfacial Area Between Gas and Liquid A/V
Gas Transfer Coefficient KG
Time Gas is in Contact with Liquid T

Post Aeration Process- Diffused Aeration
The following will allow the evaluation of a Diffused Air Type Post Aeration Unit. The procedure is based
upon FICKS Law and uses an assumed value for gas transfer coefficient (KG) based upon a chosen
wastewater temperature. Ficks law is defined as:
CT= CS+(CO-CS)e-KG(A/V)T
Where
CT = DO CONC. OF EFFLUENT AT TIME T
CS = DO SOLUBILITY AT WASTEWATER TEMP.
CO = DO CONC. OF INFLUENT, USUALLY ASSUME 0.00
KG= GAS TRANSFER COEFFICIENT
A/V INTERFACIAL AREA BETWEEN GAS AND LIQUID
T= TIME GAS IS IN CONTACT WITH LIQUID

ENTER FACILITY DESIGN FLOW (GPD)= 15000
ENTER WASTEWATER TEMPERATURE ( °C)= 25
ENTER BUBBLE DIAMETER (mm)= 20
(IF UNKNOWN USE 1.6 mm)
ENTER BUBBLE RISE VELOCITY (ft/sec)= 0.8
(IF UNKNOWN USE 0.8 ft/sec)
ENTER DEPTH OF DIFFUSER (ft)= 3.5
ENTER INFLUENT DO CONC (mg/l)= 0.00
ENTER AIR INPUT (CFM) TO POST AIR TANK= 10

RESULTS

1. THE DO SATURATION CONCENTRATION (mg/l) FOR THE
CHOSEN WASTEWATER TEMPERATURE IS EQUAL TO= 8.38
2. THE VALUE OF A/V (ft-1) FOR THE CHOSEN BUBBLE
DIAMETER AND BUBBLE RISE VELOCITY IS = 656.610
3.THE GAS TRANSFER COEFFICIENT, KG (ft/hr)
KG=32.2x1.018(T-20) IS EQUAL TO 1.15
4. THE PREDICTED EFFLUENT DO CONC (mg/l) IS = 7.54

DIFFUSER DEPTH (FT)= 3.5
CFM INPUT =10
DO SAT (mg/l)= 8.38
GAS TRANS. COEF.= 1.15
EFFLEUNT DO (mg/l)= 7.54

BUBBLE VOLUME= 1.48E-04 cft
BUBBLE AREA= 1.35E-02 sft
NO. OF BUBBLES= 6490 bubbles/gal
BUBBLE RISE TIME= 3.04E-03 hours

 

[RJSH]

CT= CS+(CO-CS)e[sup]-KG(A/V)T[/sup]
KG=32.2x1.018[sup](T-20)[/sup]