8/18/2019 Lectut CE 202 PDfF Tutorial 7

1/1

 

CE 202: Wastewater Engineering, Spring 2014-15

Problem Set # 7

1. A conventional completely mixed activated sludge process is operated with a mean cell

residence time of 10 day. The reactor volume is 7500 cum. The system is operated with MLSS

concentration of 3000 mg/L. Determine a) the rate of sludge production per day, b) The sludgewasting flow rate. The recycle line has MLSS concentration of 10,000 mg/L.

2. An activated sludge secondary treatment unit is to treat an inflow Q = 0.5 m3/s with an inflow

BOD concentration So = 150 mg/L so that the outflow from the unit has a BOD concentration S =

30 mg/L. The MLVSS of the plant is to be maintained at X = 2000 mg/L, and the concentration of

the solids in the flow leaving the secondary clarifier is X s = 10,000 mg/L. Determine the bacterial

growth rate(µ), the sludge age (θc), the aeration tank volume (V), the sludge return flow Q r, the

sludge volume outflow Q s, and the F/M ratio. Assume that Y = 0.5 mg VSS/mg BOD, µm = 2.5/day,

and Ks = 100 mg/L.

3. 

An activated sludge process is to treat 3500 cum/d of wastewater with 200 mg/L of BOD 5. The

process loading rate is 0.3 kg BOD5/(kg of MLSS.day). The hydraulic retention time is 6 hours.Find out the value of MLVSS in aeration tank.

4. 

Design an activated sludge process with the following data:

a)  Population served = 50,000

b)  Average sewage flow rate= 150 lpcd

c)  BOD of raw sewage = 250 mg/L

d) 

Efficiency required=85%.

Assume any other data that are required for the design.

5. 

Design a completely mixed Activated sludge process to treat 0.25 cum/sec of wastewater with

250 mg/L BOD5. The effluent should have 30 mg/L of BOD5. The inlet microbe concentration is

negligible. Concentration of return sludge MLSS is 10,000 mg/L. MLVSS to be kept in the reactoris 3500 mg/L. MCRT is 10 days. Find out the volume of air required per hour considering that

there is 21% oxygen in air and oxygen transfer efficiency of the aeration equipment is 10

percent. Keep a safety factor of 2 while finding out the design capacity of the air-blowers.

6.  An activated sludge unit is being used to treat 1500 cum/day of wastewater containing a BOD5 

concentration of 250 mg/L. Measured parameter for this particular unit are as follows: Y= 0.6,

m=2.0 per day, kd=0.1 per day, Ks= 4 mg/L. Make the typical assumptions for activated sludge

process modeling. (a) Calculate the mean cell residence time and volume of the reactor

(aeration basin) required for 99% removal of BOD for a system with no sludge recycling b)

Repeat the same design considering that the system recycles 50 percent of the cells in percent of

the influent flow. Assume that the secondary settling tank works perfectly.7.  Wastewater from a primary treatment tank is to undergo secondary treatment in an activated

sludge unit. The following data and operational parameters are specified:

a)  HRT= 6hr; b)MCRT = 10 days; c)m= 3 per day d)kd= 0.05 /day e)Ks=30 g/cum; f) Y=0.4 and g)

S0=150 mg/L

Two identical reactors are designed to operate in parallel and to process a combined wastewater

flow rate of 0.5 cum/sec. Assume the units operate ideally.

1) 

Find out the volume of the reactor

2)  What is the rate at which the cell mass will be wasted from the units in steady-state

operation (in tons per day)