Answer:
a) 0.488 days ≈ 11.72 hour , 8542.38 m^3
b) 2.05 kg/m^3.day
c) 0.37 day^-1
d)
i) volume of wasted solid = 783 m^3/day
ii) mass of wasted solid = 7830 kg/day
e) 1.12
Explanation:
Given data :
Flow ( Q ) = 17500 m^3/day
BOD 5 ( influent ) = 1000 mg/L
BOD 5 ( efficient ) = 120 mg/L
mean cell residence time ( ∅ c ) = 6-day
MLSS concentration ( X ) = 5500 mg/L
a) Determine the hydraulic retention time and volume of the activated sludge
i) Hydraulic retention time ( OH ) we will use the relation below
X = [tex]\frac{OcY(sin-s)}{OH(1 + Kdθc)}[/tex]
5500 = [tex]\frac{6*0.6(1000-120)}{OH(1 + 0.03*6)}[/tex] . therefore OH = 0.488 days ≈ 11.72 hour
ii) volume of the activated sludge using the relation below
OH = V / Q
V = OH * Q
= 0.488 * 17500
= 8542.38 m^3
b) Determine the volumetric loading rate in kg BOD,/m- day to the reactor
loading rate = ( Q * Sin ) / V
= ( 17500 * 1000 ) / 8542.38
= 2.05 kg/m^3.day
c) Determine The F/M ratio in the reactor
F/M ratio : F/M = ( Q .sin ) / VX
= ( 17500 * 1000 ) / ( 8542.37 * 5500)
= 0.37 day^-1
d) Determine The mass and volume of solids wasted each day, at an underflow solids concentration
i) volume of wasted solid = 783 m^3/day
ii) mass of wasted solid = 7830 kg/day
attached below is the remaining part of the solution
e) Sludge recirculation ratio
= 1.12
attached below is the remaining part of the solution
