I have tried to figure this out but I was under the impression that you must be
ID: 1920943 • Letter: I
Question
I have tried to figure this out but I was under the impression that you must be given the MLSS measurement in order to find the F/M ratio. Please help!
the BOD5 of a liquid flowing from the primary clarifier is 120,000 g/L at a flow rate of 180 m3/hr. If the concentration of suspended solids in the liquid is 2 g per liter, and if the dimensions of the aeration tank are 66.6 feet x 32.8 feet ft x 9.8 feet, then _____:
(a) what is the food-to-micro-organism ratio in the aeration tank
(b) please comment on the performance of the aeration tank evaluated above
Explanation / Answer
a) Yo= BOD Effluent = 120,000 g/l
Q= Flow rate = 180 m3/hr
Xt = Concentration of suspended solids = 2 g/l = 2000 mg/l
V = Volume of aeration tank = 66.6X32.8x9.8 = 21407.9 ft3
F = Mass of BOD removal
= Q.Yo
= 180 x 120000
= 21600000/1000
= 21600 Kg
M = Mass of MLSS
= V.Xt
= 21407.9 x 2000
= 42815800/1000
= 42815.8 Kg
F/M = 21600/42815.8
= 0.5 Kg
b) F/M ratio is the main controlling factor. if the F/M ratio is low, BOD removal in the tank will be high. however F/M ratio can be altered by varying MLSS concentration in the tank