CE6331: Biological Processes
Homework 3 Solutions
Problem 1
An aerobic biochemical process uses a CSTR w/o recycle. The feed characteristics are as
follows: Influent substrate concentration, So = 200 mg/L; half- velocity coefficient, Ks = 50
mg/L; maximum, specific substrate utilization rate, k = 5 g/g- day; yield coefficient, Y = 0.5
g Xa/g substrate consumed; microorganism decay coefficient, b = 0.10 day-1.
1. The process goal is the production of active biomass. Determine the hydraulic
retention time that the reactor should be operated in order to maximize the reactor
active biomass concentration (Xa). How does this retention time compare with the
minimum solids retention time that the system can theoretically be operated?
Calculate the solids retention time safety factor.
2. Based on the solids retention time for maximum Xa calculated above, estimate the
active biomass concentration (Xa) and the substrate removal efficiency (E, %).
SOLUTION
Part 1
𝑋! =
𝑆=
When
/0#
/+"
"($% ! &%'
(eq 1)
()(*+"
()*+"
𝐾, "-+ &(()*+
"
")
(eq 2)
𝑑𝑋!
$𝑑𝜃 = 0, then Xa = maximum, thus
"
"($% ! &%'
/
= /+ %
()*+"
"
&
(eq 3)
Substituting for equation 2 in equation 3 and taking derivative of a reciprocal you get
/0#
/+"
&"% ! *
= (()*+
Setting
"
"1 ("-&*)
%
+ [("-&*)+
)$
$
" &(]
(eq 4)
𝑑𝑋!
$𝑑𝜃 = 0 and substituting values for Y, S0, b, k, Ks, solve for 𝜃"
"
𝜃" = 1.6 days
The minimum theoretical 𝜃" that the system can be operated at can be calculated as
1 )% !
%
𝜃4567 = % !("-&*)&*1
%
(eq 5)
𝜃"#$% = 0.52 days
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The actual retention time for maximum biomass production is higher than the
minimum retention time to operate this reactor.
Safety factor =
)𝜃8567 *
965
𝜃"
' #$% &$#
%𝜃" &
(
= "-&*
(eq 6)
[𝜃4567 ]965 = 0.417 days
i.e., Safety factor = 1.6/0.417 = 3.84 days.
Part 2
Using equation 1 for a retention time of 1.6 days
Xa = 77.4 mg/L
Using equation 2 for retention time of 1.6 days
S = 20.4 mg/L
And the removal efficiency (%) is 89.8%
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Problem 2
An activated sludge plant (CSTR w/ recycle) has the following characteristics: Influent
substrate concentration, So = 400 mg/L BODL (all soluble); influent VSS, Xvo = 0; effluent
substrate concentration, S = 10 mg/L BODL; flow rate, Q = 1000 m3/day; aeration basin
volume, V = 200 m3; reactor MLVSS, Xv = 4,280 mg/L; average, steady-state oxygen
uptake rate, Oxygen uptake rate (OUR) = 0.3 mg O2/L-min. The biomass viability (i.e., v =
Xa/Xv) was measured using the oxygen uptake rate (OUR) technique and found equal to
0.926. An aliquot of mixed liquor was placed in a small reactor, aerated without any
exogenous substrate addition (i.e., starvation conditions) and it’s OUR was measured over
time resulting in the following data:
Time (days)
OUR (mg O2/L-min)
0
1
2
3
4
5
6
8
10
0.3 0.274 0.243 0.222 0.203 0.182 0.165 0.135 0.110
By use of the above data, the microorganism decay coefficient (b, day-1) can be estimated if
you assume that a constant relationship exists between Xa and OUR, i.e., OUR = a Xa, where
a = proportionality constant (g O2/g Xa - min).
1. Estimate b (1/d) by using the above data (Use regression).
2. Calculate the solids retention time for this plant.
3. Calculate the net biodegradable fraction of the viable biomass (fd).
SOLUTION
'(!
')
Part 1
= 𝑏𝑋! (starvation contion)
(eq 1)
Integrating equation 1 from t=0 to t=t
𝑋! = 𝑋!* 𝑒 +,)
(eq 2)
Since, OUR= aXa
𝑂𝑈𝑅 = 𝑂𝑈𝑅* 𝑒 +,)
(eq 3)
thus, the slope of ln(OUR/OUR0) vs t will give us -b
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0.2
Ln (OUR/OUR0)
0
-0.2
y = -0.1002x + 0.0017
R² = 0.9997
-0.4
-0.6
-0.8
-1
0
2
4
6
8
10
12
Time (days)
Therefore, b = 0.1 days
Part 2
Xa = vXv
Xa = 0.926 (4280 mg/l) = 3963.3 mg/l
-$.#!// $% /1/)2#
𝜃" = ,$.#!// 3!/)2' 45.# /1/)2#
Biomass in system = Xa * V = 3963.3 mg/l*200m3*1000 l/m3 * 1kg/106mg= 792 kg
biomass.
All biomass produced should be removed to maintain SRT,
Thus, biomass removed = Y*Q*(S0-S)
Assuming Y for heterotrophic biomass = 0.5 g VSS/g BODL
Biomass removed = 0.5 g VSS/g BODL* 1000 m3/day * 1000 l/m3 *(400 mg BODL/l –
10 mg BODL /l)* 0.001mg BODL/gBODL* 1kg/103g = 195 kg biomass removed/day
𝟕𝟗𝟐 𝒌𝒈
𝜽𝒄 = 𝟏𝟗𝟓 𝒌𝒈/𝒅 = 4.06 days.
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𝑋@ = 𝑋! (1 + (1 − 𝑓' )𝑏𝜃"
Xa = vXv
Solving for fd
fd = 0.8
Part 3
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Problem 3
A "three-sludge" biological treatment system (i.e., three suspended-growth reactors in series,
each with its own clarifier) is to be designed to remove BOD and nitrogen from a municipal
wastewater. Assume each process is "reactor wasted".
Stage I: CSTR with recycle for BOD removal (a high-rate, sloppy system) has already been
designed. The following data apply to the effluent from stage I:
Q = 1.82 x 107 L/day
Xa (active organic-carbon users) = 37.4 mg/L Xi (inert biomass) = 12.6 mg/L
S (carbonaceous BOD) = 6.3 mg/L
Alkalinity = 139 mg/L as CaCO3
NH4+-N = 26.8 mg/L.
Stage II: PF with recycle (for nitrification and further BOD removal). Ignore ammonia
production and alkalinity change due to the carbonaceous BOD removal in the second stage.
Preliminary: The equations for the MLVSS in AS systems as a function of various
parameters need to be modified if carbonaceous and nitrogenous oxidations are occurring in
the same reactor. Also, modification is required wherever there is an appreciable active
organism concentration in the influent to a reactor. Both modifications are required for Stage
II, since the problem specifies that Stage I is a high-rate, sloppy system (i.e., additional BOD
removal will take place in Stage II and, with a high-rate system for Stage I, the degree of
clarification in Clarifier I is rather poor, resulting in significant active carbon-users in the
influent stream to Stage II).
𝜃" (𝑋!* + 𝑌(𝑆* − 𝑆))
𝑋! =
(1 + 𝑏𝜃" )
𝜃
𝜃" 𝑌A (𝑁* − 𝑁)
𝑋A =
𝜃 (1 + 𝑏A 𝜃" )
𝜃" *
𝑋@ = 𝑋$ + 𝑋! (1 + 0.2𝑏𝜃" ) + 𝑋A (1 + 0.2𝑏A 𝜃" )
𝜃
Where
𝑋! and 𝑋!* = active, carbon oxidizing bacteria in influent and reactor, respectively
𝑋A = active nitrifier concentration in bioreactor
𝑋$* = inert biomass influent
𝑋@ = total biomass in reactor
For Stage II, assume: SRT = 15 days; mean MLVSS =2000 mg/L; Y=0.448gVSS/gBODL;
b=0.10day-1; YN =0.34gVSS/gN; bN =0.05day-1
Specify:
1. Reactor volume and hydraulic retention time (assume effluent BODL and
ammonia from Stage II are negligibly small from this PF reactor);
2. Xa and XN concentration;
3. Oxygen requirements for both heterotrophs and autotrophs (kg O2/d);
4. Effluent NO3-N concentration;
5. Alkalinity consumed. If addition is required, specify dose and form to maintain
a pH = 7. In any event, specify effluent alkalinity for pH = 7 (assume this is an
open system and at equilibrium with air at a PCO2 = 10-3.5 atm);
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6. Excess VSS handled (kg VSS/d, assuming no significant effluent VSS] and
waste flow rate (Qw).
SOLUTION
Part 1
𝜃"
𝑋@ = 𝑋$* + 𝑋! (1 + 0.2𝑏𝜃" ) + 𝑋A (1 + 0.2𝑏A 𝜃" )
𝜃
𝑋@ =
𝜃" * 𝜃" (𝑋!* + 𝑌(𝑆* − 𝑆))
𝜃" 𝑌A (𝑁* − 𝑁)
(1 + 0.2𝑏𝜃" ) +
(1 + 0.2𝑏A 𝜃" )
𝑋$ +
(1 + 𝑏𝜃" )
𝜃
𝜃
𝜃 (1 + 𝑏A 𝜃" )
𝑋@ =
𝜃" * (𝑋!* + 𝑌(𝑆* − 𝑆))
𝑌A (𝑁* − 𝑁)
(1 + 0.2𝑏𝜃" ) +
(1 + 0.2𝑏A 𝜃" )]
[𝑋$ +
(1 + 𝑏𝜃" )
(1 + 𝑏A 𝜃" )
𝜃
assume effluent BODL and ammonia from Stage II are negligibly small from this PF
reactor, S=0, N=0 and rearrange to ger
𝜃" * (𝑋!* + 𝑌(𝑆* ))
𝑌A (𝑁* )
(1 + 0.2𝑏𝜃" ) +
(1 + 0.2𝑏A 𝜃" )]
𝜃=
[𝑋$ +
(1 + 𝑏𝜃" )
(1 + 𝑏A 𝜃" )
𝑋@
Substituting all known values, 𝜽 = 7.1 hours
Volume = 5.39 x106 L
Part 2:
𝑿𝒂 =
𝑿𝑵 =
𝜽𝒄 (𝑿𝟎
𝒂 )𝒀(𝑺𝟎 &𝑺))
(𝟏)𝒃𝜽𝒄 )
𝜽
𝜽𝒄 𝒀𝑵 (𝑵𝟎 &𝑵)
𝜽 (𝟏)𝒃𝑵 𝜽𝒄 )
= 815 mg/L
= 264 mg/L
Part 3:
E.GHI
MOUR BCD = Q(S* − S) C1 − EJK "#$
(1 + 0.2bMCD θN )F = 76.7 kg/day
L
"#$ %
E.QRI&'$
MOUR OPD = Q(N* − N) C1 − EJK
&'$ L%
(1 + 0.2bOPD θN )F = 2013 kg/day.
Part 4:
From class notes, overall stoichiometry of nitrification.
• 0.13 moles ammonium oxidation produced 0.125 moles nitrate
• Complete oxidation of 26.8 mg-N/l of ammonium would thus result in
25.77 mg-N/L as nitrate
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Part 5:
For 1 mg ammonium consumed, 7.14 mg of alkalinity as CaCO3 is consumed (see
class notes on nitrification)
+
+
Alkalinity consumed = 26.8 mg NH4 -N /l *7.14 mg alkalinity/1mg NH4 -N = 191.35
mg as CaCO3/L.
The influent alkalinity is 139 mg/L as CaCO3. Thus, alkalinity will need to be added.
The desired alkalinity at pH 7 in an open system can be calculated as
Desired Alkalinity = [HCO3-] + 2[CO32-] + [OH-] – [H+]
=
𝐾B 𝐾C( 𝑃8DE
𝐾B 𝐾C( 𝐾CE 𝑃8DE
+
2
∗
+ [𝑂𝐻& ] + [H) ]
[𝐻) ]E
[𝐻) ]
PCO2 =10-3.5 atm
KH = 10-1.5 mol/L.atm
[OH-] = 10-7
[H+] = 10-7
K 𝛼1 = 10-6.37
K 𝛼2 = 10-10.32
Desired alkalinity = 4.27 x 10-5 eq/l = 2.55 mg CaCO3/l
Influent alkalinity – Alkalinity consumed + added alkalinity = 2.55 mg CaCO3/L
Therefore, added alkalinity = 2.55 + 191.35- 139 = 54.9 ml CaCO3/L
Part 6:
Excess VSS = Xv*V/θc = 2000 x 5.39x106/15 = 719 kg/day
Qw = V/θc = 5.39x106/15 = 3.59*105 L/day.
Stage III: PF with recycle (for denitrification, using methanol). Assume: SRT 5 days;
MLVSS = 2000 mg/L; [D.O.]o = 0; [NO2-]o = 0 and a covered tank.
Specify:
1. Influent flow rate;
2. Methanol dose;
3. Reactor volume and hydraulic retention time;
4. Concentration of active denitrifier population (XD);
5. Effluent alkalinity and excess VSS produced (kg VSS/d).
SOLUTION
Part 1
Influent flow rate = Stage 2 effluent – Stage 2 waste = 1.82 x 107 L/day - 3.59*105
L/day = 1.78x107 L/day
Influent flow rate = 1.78x107 L/day
Part 2
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From class notes (methanol use stoichiometry), 0.1667 moles of methanol required
for 0.1561 moles of nitrate.
Effluent nitrate concentration for stage 2 is 25.77 mg-N/L = 1.84 mmoles/L.
Methanol dose required = 1.84*0.1667/0.1561 * 32.04 mg/mmol = 62.9 mg/L
Part 3
From class notes (see table) 0.49 gVSS produced g NO3--N consumed when
methanol is electron donor
Assuming all nitrate is consumed,
The excess VSS produced each day will be
= 25.77 mg NO3—N/L * 1.78x107 L/day* 0.49 = 225.77 kg/day
The excess VSS must be remove daily to maintain SRT
i.e., SRT = XvV/QwXv
therefore V = SRT * QwXv / Xv = 5 days * 225.77 kg/day/2000 mg/l*10-6 kg/mg
V = 564.4 *103 liters.
HRT = V/Q = 564.4 *103 L/1.78x107 L/day = 45 minutes
Part 4
For this you will need to estimate yield on methanol.
From class notes (see table) – net yield when methanol is electron donor is 0.135
gVSS/g OD. The COD of methanol is 1.5 g O2/g methanol. Therefore, the yield per
gram of methanol = 0.2 g VSS/g methanol
𝑋/ =
+" (0*! )"(%! &%))
+
(()*+" )
F
= (GF/(EG∗JK))
K)K.E(JE.M&K))
(()K.KF∗F)
= 1610 mg/l
Part 5
From class notes (see table), 3.57 grams of alkalinity (as CaCO3) is produced per
gram of NO3--N consumed.
Thus effluent alkalinity = (input alkalinity to stage 3)+ (alkalinity produced)
= (3 )+ (25.77*3.57) = 94.9 mg/L
From class notes (see table) 0.49 gVSS produced g NO3--N consumed when
methanol is electron donor
Assuming all nitrate is consumed,
The excess VSS produced each day will be
= 25.77 mg NO3—N/L * 1.78x107 L/day* 0.49 = 225.77 kg/day
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