SOLUTIONS OF HOMEWORK PROBLEMS ON ACTIVATED SLUDGE PROCESS

1. 1
Department of Civil Engineering-I.I.T. Delhi
CEL 795: Water and Wastewater Treatment Processes
1st
Semester 2011-2012
HW 8 Solution and Additional Questions (Solution: Nipun)
Q1. Assuming that the endogenous coefficient (kd) can be neglected, develop expressions for
determining substrate and cell concentration as a function of time for a batch reactor. If the initial
concentration of substrate and cell is 100 and 200 mg/L, respectively, calculate the amount of
substrate remaining after 1 h. If the endogenous coefficient is equal to 0.04/day, estimate the error
made be neglecting this factor. Assume: k=2/h; Ks = 80 mg/L; Y=0.4 mg/mg. [4+4=8 points]

2. 2
S=8.98 mg/L
%Error = 100*[(8.98-8)/8.98] = 12.25%

3. 3
Q2. An anaerobic digester is designed to remove 85% of BOD5 of an industrial organic waste with an
ultimate BOD =2000 mg/L. If (θc) = 10 days, estimate the amount of sludge to be wasted daily and the
quantity of gas produced each day. Assume that the flow =0.1 million gallons/day; Y=0.1;
kd=0.01/day. [3+3=6 points]

4. 4

5. 5
Q3. In the activated sludge reactor, bacterial oxidation and synthesis can be described using Eq. (1)
and its endogenous respiration can be described using Eq. (2). Calculate amounts of oxygen required
for oxidation and synthesis of 3500 mg/L MLVSS (i.e., bacterial concentration)? [5 points]
Oxidation and synthesis (in presence of bacteria):
COHNS (organic matter) +O2 + nutrients
CO2 +NH3 + C5H7NO2 (new cells) + other end products (1)
Endogenous respiration (in presence of bacteria):
C5H7NO2 (cells) + 5O2
5CO2 +2H2O+ NH3 + energy …… ……………(2)
Solution (3)

6. 6
Q4. An organic waste with 250 mg/L BOD5 (S0) needs to be treated using a completely-mixed
activated sludge process. The effluent BOD is to =10 mg/L (Seff). Assume that the temperature is
20°C, the flow rate is 5 million gallons/day and following information are available:
(i) Influent volatile suspended solids to reactor are negligible.
(ii) Return sludge concentration = 10000 mg/L of suspended solids = 8000 volatile suspended
solids
(iii)Mixed-liquor volatile suspended solids (MLVSS)=X=3500 mg/L
(iv) Mean cell-residence time (θc)= 10 days
(v) Microbial parameters: kd =0.06/day; Y=0.65 mg cells/mg BOD5 utilized.
(vi) Effluent has 20 mg/L biological solids (here 80% is volatile and 65% is biodegradable).
Assume that the biodegradable biological solids can be converted from ultimate BOD demand
to a BOD5 demand using the factor 0.68 [i.e., BOD K value =0.1/day (base 10)].
Calculate (i) Biomass production rate; (ii) Observed yield (i.e., Y/ [1+kdθc]); (iii) Food to
microorganisms ratio (i.e., S0/ [Xθ]). [2×3=6 points]
Solution 4

7. 7

8. 8
Q5. During nitrification process, calculate amounts of oxygen required and alkalinity required to
completely convert 10 mg/L ammonia to nitrate using balanced equations? [5 points]

9. 9
Additional Questions-Part 2 (Courtesy: Dr. Arvind K. Nema)
QA5 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?
QA6. Assuming that the data given in QA5 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%?
QA7 An anaerobic reactor, operated at 35o
C, 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.
Solution
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 35o
C.
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

10. 10
QA8: 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 20o
C 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
Solution:
Step1: 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
Step 2: Determine the treatment efficiency
Step 3. Compute the reactor volume
( )
( ) 1
1
−
−
+
=
d
m
c
c
d
s
k
k
K
S
µ
θ
θ ( )
( )
c
d
o
c
k
S
S
Y
X
θ
θ
θ
+
−
=
1 Q
V
=
θ
r
w
c
X
Q
VX
=
θ
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]
kd = endogenous-decay rate coefficient [d
-1
]
Y = yield coefficient [mg/L MLVSS/mg/L]

11. 11
Step 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).
VX
QS
M
F o
= To achieve a low F/M ratio, use a low sludge wasting rate (Qw) creating a long cell
detention time (θc)
Step 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:
( )( )
g
kg
o
c
d
x S
S
Q
k
Y
P 3
10
1
−
−
+
=
θ
Px = waste activated sludge produced [kg VSS/d]
Step 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:
( )( )
x
g
kg
o
O P
f
S
S
Q
M 42
.
1
10 3
2
−
−
=
−
; f = conversion factor to convert BOD5 to BODL.
Calculate air requirement