Please do all sections 2. (Learning objective #3, #7) In an activated sludge sys
ID: 1844762 • Letter: P
Question
Please do all sections 2. (Learning objective #3, #7) In an activated sludge system, 1 MGD is leaving the primary clarifier with a BOD concentration of 130 ppm and a VSS of 200 ppm. The effluent flow is 0.95 MGD with a VSS of 20 ppm and a BOD concentration of 5 ppm. Qr/Q = 1,MLVSS = 2000 ppm. If the residence time in the primary clarifier was 4.5 hours, calculate the BOD and SS coming into the primary clarifier. What is the VSS in the recycle stream?(3791 ppm) c. How many lbs/day of sludge exit the system?(1581 lb/day) What is the efficiency of the activated sludge system for BOD removal?Explanation / Answer
Given:
Effluent from the primary clarifier = 1 MGD
BOD Concentration = 130 ppm
VSS = 200 ppm
Effluent flow from the aeration tank = 0.95 MGD
VSS = 20 ppm
BOD Concentration = 5 ppm
Qr/Q = 1
MLVSS = 2000 ppm
Solution:
a) If the residence time in the primary clarifier was 4.5 hours, calculate the BOD and SS coming into the primary clarifier;
First we should find the volume of the tank, since the residence time is provided it can be easily found with the formula given below
Hydraulic retention time = 24 * Volume of the primary clarifier / inflow
4.5 = 24 * Volume of the primary clarifier / 1 MGD
Volume of the primary clarifier = 0.1875 MG
To find the Volumetric BOD Loading (VBL) the following formula is used,
VBL = 8.34 (Q * S) / V
Where VBL = Volumetric BOD Loading (lb BOD/1000 ft3 * day)
Using this formula the
Q = Influent flow (MGD)
S = Influent BOD concentration (mg/L)
V = Volume of the clarifier
VBL = 8.34 ( 1 MGD * 129.85 mg/L) / 187500
VBL = 0.0057 lb BOD/1000 ft3 * day