Department of Civil Engineering-I.I.T.

Delhi
CEL 212: Environmental Engineering
Sample Wastewater Questions

Biological Processes
Courtesy: Dr. Arvind K. Nema

Example 1. Design of the aeration basis based on solids retention time. You are provided the
following information about a municipal wastewater treatment plant. This plant will use the traditional
activated sludge process. Population = 150,000 people, flow rate of 33.75x106 L/day (equals
225L/person/day) and influent BOD5 concentration of 444 mg/L (note this is high strength wastewater).
Assume that the regulatory agency enforces an effluent standard of BOD5 = 20 mg/L and suspended
solids standard of 20 mg/L in the treated wastewater. A wastewater sample is collected from the
biological reactor and is found to contain a suspended solids concentration of 4,300 mg/L. The suspended
solids concentration in the secondary sludge is 15,000 mg/L and the concentration in the secondary
sludge is 5,000 mg/L. The concentration of suspended solids in the plant influent is 200 mg/L and that
which leaves the primary clarifier is 100 mg/L. The microorganisms in the activated sludge process can
convert 100 grams of BOD5 into 55 grams of biomass. They have a maximum growth rate of 0.1/day, a
first-order death rate constant of 0.05/day, and they reach ½ of their maximum growth rate when the
BOD5 concentration is 10 mg/L. The mean cell retention time of the solids is 4 days and sludge is
processed on the belt filter press every 5 days.

A. What is the design volume of the aeration basin?

Solution: Assuming that 30% o f the plant influent BOD5 is removed during primary sedimentation, this
means that So = 444 mg/L × 0.70 = 310 mg/L. Thus,

1/SRT = [(Qo × Y)/(V × X)) × (So-S)] – kd

1 / 4 days=[33.75x106 L/day (0.55 gm SS/gm BOD5) /V 4,300 mg SS/L (310 mg/L – 20 mg/L)] –
0.05/day

Solve for V = 5,000,000 Liters

B. What is the plant’s aeration period?

Solution: The plant’s aeration period is the number of hours that the wastewater is aerated during the
activated sludge process. This equals the hydraulic detention time of the biological reactor.

HRT= V/Q = 5,000,000 L / 33.75 x 106 L/day = 0.15 days = 3.6 hours

C. How many kg of primary and secondary dry solids need to be processed daily from the treatment
plant?

Solution: The amount of solids processed from the primary sedimentation tanks equals the difference in
suspended solids concentrations measured across the sedimentation tanks multiplied by the plant flow
rate.

33.75x106 L/day (200 mg SS/L – 100 mgSS/L) × kg/1,00,000 mg = 2,275 kg primary solids per day

We have not provided with the concentration difference of suspended solids across the secondary
sedimentation tanks so we can determine the amount of secondary solids produced daily in the same

1
manner that we used for primary solids. However, careful examination of the expression of solids
retention time shows that that the term QwXw equals the answer.

4 days = V × X / Qw × Xw = 5,000,000 L (4,300 mg SS/L) / Qw × Xw

Solve for QwXw which equals 5,400 kg secondary dry solids per day.

D. Determine the F/M ratio (in units of lbs BOD5/lb MLSS-day) using data provided in the above
example problem.

Solution.

By definition,
F/M = Q So / X V = [33.75x106 L/day x 310 mg/L) / [4,300 mgSS/L × 5,000,000 L]
= 0.48 lbs BOD5/lb MLSS-day

Various reactor configurations are available, each with its own set of advantages and disadvantages. The
two basic types are plug flow (PF) and completely mixed flow (CMF) reactors. PF reactors offer a higher
treatment efficiency than CMF reactors, but are less able to handle spikes in the BOD load. Other
modifications of the process are based on the manner in which waste and oxygen are introduced to the
system.
++++++++++

2
Problem 1: Design a complete mix activated sludge process to treat 0.25 m3/s of wastewater with BOD5
of 250 mg/L. The effluent is to have BOD5 of 20mg/L or less. Assume the temperature is 20oC and the
following conditions are applicable.
- The influent and effluent microorganism concentrations are negligible. Food and microorganisms
are completely mixed in the aeration basin. Wastewater contains adequate nitrogen, phosphorus
and other trace nutrients for biological growth.
- Ratio of MLVSS to MLSS is 0.8. (MLSS and MLVSS: Mixed liquor suspended solids (represents
total solids) and mixed liquor volatile suspended solids (represents biological solids).
- MLVSS concentration in the reactor = 3500 mg/L; Return sludge concentration = 10,000 mg/L;
- Design mean cell residence time (θC) is 10 days
- Effluent contains 22 mg/L of biosolids, of which 65% is biodegradable
- Kinetic coefficients: Ks = 50 mg/L; µm = 5.0 d-1; kd = 0.06 d-1 and Y = 0.50
Suggested Steps:

1. Estimate the concentration of soluble BOD5 in the effluent (eff)

BOD5 in the eff = Soluble BOD5 in the eff + BOD5 in the eff suspended solids
BOD5 = 0.68 x BODL
2. Determine the treatment efficiency

3. Compute the reactor volume

K s (1 + k d θ c ) θ Y (S o − S ) V VX
S= X = c θ= θc =
θ c (µ m − k d ) − 1 θ (1 + k d θ c ) Q Qw X r
S = BOD concentration in activated sludge (S = BOD5 allowed – BOD5 in SS) [mg/L]
X = microorganism concentration in activated sludge [mg/L of MLVSS]
Xr = microorganism concentration in recycle [mg/L of VSS]
SO = influent BOD [mg/L]
θc = mean cell resident time in the aeration tank [d]
θ = hydraulic detention time [d]
V = aeration tank volume [m3]
Q = flow rate [m3/d]
Qw = flow rate of waste sludge [m3/d]
µm = maximum specific substrate utilization rate [d-1]
KS = half-maximum rate concentration [mg/L]
-1
kd = endogenous-decay rate coefficient [d ]
Y = yield coefficient [mg/L MLVSS/mg/L]
4. Food to Microorganism Ratio:
To keep the microorganisms efficient, the Food to Microorganism Ratio (F/M) must be
keep low (around 0.10 to 1.0 mg/L-d).
F QS o
= To achieve a low F/M ratio, use a low sludge wasting rate (Qw) creating a
M VX
long cell detention time (θc)

5. Waste Sludge Production:

Excess sludge is produced during the activated sludge process that must be treated and
disposed of. To estimate the excess sludge production, use the following equation:

3
 Y
Px =
1 + k dθ c
(
Q (S o − S ) 10 − 3 kg g )
Px = waste activated sludge produced [kg VSS/d]
6. Oxygen Requirements:
Activated sludge uses large volumes of oxygen in the production of sludge and the
consumption of BOD. However, oxygen is produced during cell formation by moving
from right to left per Equation 5-44:
C5H7O2N + 5O2 <-----> 5CO2 + 2H2O + NH3 + energy
The ratio of oxygen usage to cell formation is 5(32)/113 = 1.42.
Subtracting cell formation from the oxygen consumed in the reduction of BOD (SO-S), the
oxygen requirements of activated sludge can be estimated:

M O2 =
(
Q(S o − S ) 10−3 kg g )
− 1.42Px ; f = conversion factor to convert BOD5 to BODL.
f
Calculate air requirement

4
Q1. An anaerobic reactor, operated at 35oC, treats wastewater with a flow of 2000m3/d and a
biological soluble COD (bsCOD) concentration of 500g/m3. At 90% bsCOD removal and a biomass
synthesis yield of 0.04 g Volatile Suspended Solids/ g bsCOD used, estimate the amount of methane
produced in m3/d.
Step 1: Prepare a steady state mass balance for COD to determine the amount of influent COD
converted to methane.
CODin = CODeff + CODVSS(biomass) + CODmethane
CODin = COD concentration in influent x Wastewater Inflow
CODeff = COD concentration in effluent x Wastewater Inflow
CODVSS = 1.42 g COD/g VSS x Yield coeff g VSS/g COD x Efficiency of system
CODmethane= ?
Step 2: Determine the volume of gas occupied by 1 mole of gas at 35oC.
Step 3: The CH4 equivalent of COD converted under anaerobic conditions =
(L/ mole)/ (64 g COD/ mole CH4)
Step 4: CH4 production = CODmethane x CH4 equivalent of COD converted

Solution:
CODin = COD concentration in influent x Wastewater Inflow = ( 500g/m3) *(2000m3/d)=1000 Kg/d
Biological soluble COD removal = 90%
CODeff = 0.10* 1000 Kg/d = 100 Kg/d
CODused = 0.90* 1000 Kg/d = 900 Kg/d

Biomass synthesis yield = 0.04 g Volatile Suspended Solids/ g bsCOD used

Biomass synthesis yield = (900*1000g/day)*(0.04 g VSS)=36000 g/day
COD corresponding to biomass = (36000 g VSS/day)*(1.42 g COD/g VSS)=51.120 Kgday

CODin = CODeff + CODVSS(biomass) + CODmethane

CODmethane = CODin - {CODeff+CODVSS(biomass)}
CODmethane = (1000 Kg/d)- {100 Kg/d+ 51.120} =848.88 Kg/d (answer)
CODmethane = 848.88 Kg/d (answer)

As 64 g COD comes from 1mole of CH4. So 848.88 Kg/d COD would result from
= (1 mole CH4 /64)*(848.88 Kg/d) = 13.26 mole CH4/day

Now we know that for STP conditions (i.e., 273 K and 1 atm pressure, 1 mole gas occupies 22.4 Liter
volume). Now we need to calculate volume of methane gas at 35°C (i.e., 308K) and 1 atm pressure:

V/T=constant (as P is constant here)

[22.4/273] = [V2/308] ==> V2=25.27 Liters/mole (this is true for methane gas also)

Now as biological process is producing 13.26 moles methane gas/day, it indicates that 335.1 liters/day (or
0.335 m3/day).

5
Q2. The water content of solids slurry (WW sludge) is reduced from 98 to 95 %. What is the
percent reduction in volume assuming that solids contain 70% organic matter of specific gravity
1.0 and 30% mineral matters of specific gravity 2.0? What is the specific gravity of 98 and 95%
slurry?
Solution:
w1=98% (i.e., 2% solids)
w2=95% (i.e., 5% solids) (this solid content is increased due to dewatering process)

Wsolids/(Ssolids*ρw) = Wmineral/(Smineral *ρw) + Worganic matter/(Sorganic matter* ρw)

Here S: specific gravity
W: weight
ρw = density of water (i.e., 1 Kg/liter)

Case 1:
Water content =98% (i.e., 2% solids);
Organic matter content =70% =0.7 Wsolids
Mineral matter =30% =0.3 Wsolids

ρw = 1 Kg/liter
Smineral=specific density of mineral = 2
Sorganic matter=specific density of organic matter =1

For determining specific gravity of all solids (Ssolids):

Wsolids/(Ssolids) = 0.3Wsolids/(2) +0.7 Wsolids/(1)
1/Ssolids = 0.15 +0.7 = 0.85
=> Ssolids =1/0.85 =1.18 (this is specific density of solids)

Now if specific density of water = 1 and water content =98% (i.e., solid content =2%), we can calculate
specific gravity of sludge =

1/Ssludge = (0.98)/1+(0.02)/1.18 =0.9969

Ssludge = 1/0.9969 =1.003 (or Density of sludge = 1.003*1=1.003 Kg/L)

Density of sludge = Mass of sludge /Volume of sludge

=>Volume of sludge = Mass of sludge/Density of sludge
For 1 Kg mass of sludge, volume of sludge = 1/(solid content*density of sludge)
= 1/(0.02*1.003) = 49.85 liters (initial volume of sludge per Kg mass of sludge)

After solid waste treatment, water content is 95% (i.e., 5% solids).

V1=49.85 liters/Kg

Assuming specific gravity of sludge remains same (i.e., Ssludge = 1.003) or Density =1.003 Kg/L
New volume of sludge after reduction in water content per Kg of sludge
=V2= 1/(0.05*1.003) = 19.94 liters
So volume reduction = [1-(19.94/49.85)] *100 =60%

6
Q3. Assume that the data given in Q2 belongs to primary sludge. What will be the volume of
digested slurry in case 60% of the volatile solids are destroyed and water content is reduced to
90%?
Solution:
Say 60% volatile solids (or organic matter) is destroyed. Initially 70% solid was organic matter and now
remaining organic matter = (70%)*(0.40) = 28% (or mineral content =1-0.28=0.72 (i.e., 72%). Revised
water content =90% (and thus solid content =10%).

1/(Ssludge) = 0.72/(2) +0.28/(1) =0.64 Ssludge =1/0.64 =1.56

Density of sludge = 1.56*1=1.56 Kg/L

Volume of sludge = Mass of sludge/Density of sludge

For 1 Kg mass of sludge, volume of sludge = 1/(0.10*1.56) =6.41 liters

Q4. What are three stages of anaerobic digestion process? Discuss their importance and BOD
requirements at different stages.
Hint: See Lecture Notes.

Q5. Some substances can be toxic to bacterial growth. How can you incorporate this effect in
determining specific biomass growth rate constant?
Solution:
Some substance can impart toxicity to bacteria and thus it might affect their growth. For incorporating
this effect, microbial decay constant due to toxic substance (i.e., kdecay,substance) need to be determined.
Then this aspect needs to be incorporated in the overall biomass rate equation.
Roverall=Rlag+Rgrowth+Rstationary+Rendogenous decay+Rsubstance decay

Q6. Why do we keep SRT greater than HRT and how does it affect plant’s performance?
Solution:
Keeping SRT> HRT results in long retention of biological solids (i.e., biomass) in aeration tank where
aerobic biological processes occur. This increased retention of biological solids results in higher removal
of organic matter compared to SRT=HRT situation (i.e., no recirculation case).

Q7.Comment on performance of chlorination unit placed before and after aeration tank.
Solution:
Placing chlorination unit before aeration tank can kill bacteria however; it would not reduce
microorganisms and pathogen in wastewater effluent which we might discharge to river. In this case, the
discharge stream might have pathogens unsuitable for discharging in river water.

Q8. Calculate BOD and alkalinity requirements for removing 14 mg NH4+-N during nitrification
process? Also explain the need for maintaining alkalinity in the nitrification reactor.
Solution:

Q9. A completely mixed activated sludge plant is to treat 10000 m3/d industrial wastewater (BOD5
= 1200 mg/L; BOD effluent = 100 mg/L prior to discharge). For 5-day SRT, 5000 mg/L MLSS is
required (Y= 0.7 kg/kg and kd=0.03/day). Calculate following:
(i) The volume of the reactor
(ii) The mass and volume of solids wasted each day
(iii)The sludge recirculation ratio.
(iv) To reduce BOD effluent to 30 mg/L would increase SRT a good idea? Discuss.
7
Q10. Write overall oxidation and cell synthesis reactions during nitrification process and calculate
amount of BOD required?
Hint: See solution of Q8.

Q11. Calculate total carbonaceous and nitrogen oxygen demand of a water sample that contains 5
mg/L organic compounds having chemical formula C6H6N2O2. Assume that nitrogen is converted
to ammonia and then to nitrate.
Hint: See notes. This question has been discussed in class and asked in HW and Exam.

Q12. Discuss different stages of anaerobic digestion process in the order they occur.
Hint: See notes

Q13. Name three disinfection kinetics models generally used to model disinfection process. Can CT
concept be applied to all three kinetic models? Why or why not?
Hint:
For part 1: See notes.
For part 2 Ct concept is applied only for first order kinetic model (i.e., for Chick’s Law when n=1 in the
Watson’s Law).

Q14. Comment on difference between domestic and hospital solid waste management (SWM)
treatment methodologies.
Solution:
domestic SWM hospital SWM
more treatment is required for inert more treatment is required for sharps, infectious
and biodegradable materials microorganisms and unused drugs, etc.
generally, segregration, screening, generally segregation, followed by incineration is done to
biological processes are used and kill pathogens; biological process and advanced oxidation
then land fill disposal and methan processes such as ozonation, UV radiation and hydrogen
gas capturing is done. peroxide usage for properly treating unused drug before
doing land filling and methane gas capturing if its feasible.

8
Q15. Look at the following information for a completely mixed biological reactor.
Influent water information:
Parameter Influent water Effluent water
3
Flow rate 50 m /d 50 m3/d
Biomass 0 mg VSS/L ??
Substrate 95 mg BOD5/L ??
Tank volume 200 m3
VSS: volatile suspended solids
Calculate following:
1. effluent BOD5
2. biodegradable organic matter removal efficiency of tank
3. biomass leaving the tank
Solution:
Here HRT =θ= 200m3/(50m3/d)= 4 days
X in aerobic reactor =?
S0=95 mg BOD5/L
Sfinal=?

1/θ =[(µ mS)/(KS+S)]-kd

so: determine expression for S.

S=[ KS (1+ θ kd)]/[( θµ m -1- θ kd)]

Given KS=60 mg BOD5/L (this is average value)

kd=0.06/day (this is average value)
µ m =3/day (this is average value)

Solve for S now. It is coming out to be: Sfinal= 6.9 mg BOD5/L

S removal efficiency =(1-6.9/95)100% = 92.74%

For determining X coming out of the system:

Do a mass balance for substrate first.
0=QS0-QSf+VrS here rS is rate of substrate utilization.
so rS = -Q(S0-S)/V

And we also know that rS = -kXS/[ KS +S]

so {-Q(S0-S)/V} = -kXS/[ KS +S]
X = [ (KS +S)(S0-S)(Q/V)]/(kS)
here Q/V = 1/HRT
solve for X. X=43 mg VSS/L

9
Q16. Discuss removal efficiencies of different units for different contaminants in wastewater
treatment plant with following schematic:
Influent water  Primary settling tank  Biological aeration  Secondary settling-  Effluent
water:
Parameter Influent water After settling (and Effluent
influent to aeration tank) water
BOD 200 mg/L 130 mg/L 30 mg/L
suspended solids 240 mg/L 120 mg/L 30 mg/L
phosphorous 7 mg/L 6 mg/L 5 mg/L
nitrogen 35 mg/L 30 mg/L 26 mg/L

Solution:
Removal efficiency table
Parameter After settling after aeration remarks
tank
BOD =(200- =(130- more removal is observed in aeration
130)/200=35% 30)/130=76.9% tank than settling basin
suspended solids =50% 75% more removal is observed in aeration
tank than settling basin
phosphorous 14.29% 16.67% more removal is observed in aeration
tank than settling basin
nitrogen 14.29% 13.33% comparable removal is observed in
settling tank and in aeration tank
Now for every contaminant, remaining concentration values need to be compared with receiving body
standard (say for river) to determine if further water treatment is required and which parameters need to
be removed and then one needs to decides about another unit process in the given treatment train.

Q17. Visit your (1) Dining hall and (2) Hostel. Identify types of solid wastes produced in terms of
biodegradable and non-biodegradable components and list two methods of their proposal disposal in
environment. Would you like to treat them in wastewater treatment plant and/or dispose in landfills?
You may form a group of 5 students. Submit a two-page report as a class participation activity.

10