Chapter 16
Problem 16.1. Data
Kinetic Expressions – Batch Growth, 2 Competing Organisms
iii
dX X i EC, AV
EC EC AV AV
EC AV
dt Y Y
Assumption – Since
sO
KS
, growth rate is not limited by glucose. Therefore,
S
max
i di max i di
S
ik k i EC, AV
KS
= − − =
+
So eq. (1) can be integrated to (3)
it
i iO
X X e i EC, AV
==
To determine the final cell mass, first determine the duration of the fermentation
by substituting eq. (3) into eq. (2) and collecting terms.
tt
EC AV
OO
EC AV
EC AV
EC AV
dS X e X e
dt Y Y
− = +
Integrating
t
t
EC AV
ECo AVo
O
EC AV
XX
S S e e
YY
− = +
Iterate to find
ex
t (S 0)=
Tex = 4.06 hr
Substituting into eq. (3) XEC= 1.4 g dw/l XAV =0.76 g dw/l
Therefore, the ratio
AV EC
X / X 0.54=
107
Problem 16.2.
Data
Steady State Material Balances for a Chemostat with 2 Competing Organisms; B
is adherent.
AA A A
dX V FX X V 0
dt
= − + =
O A B B
AB
YY
Am
S
B 1 B B
B
a
Substituting eq. (5) into eq. (3), and solving for XB
( )
at
B dB 1 B aB Bm 1 B B
( k )J X k X J X X 0
− + − =
XB = 0 (trivial solution) or
at
B dB 1 aB Bm
B
aB 1
( k )J k X
X 0.70 g/l
kJ
−+
==
Substituting the non-trivial XB into eq. (5),
at 5 2
B
X 7.0 10 g/cm .
−
=
Solving eq. (4) for XA
at
BA
A O B B
BA
Y
X D(S S) (X X a) 1.8 g/l
Y
= − − − =
Problem 16.3.
Steady State Material Balances for a Chemostat with 2 Organisms and
Commensalism
A A A
dX V SX
B
P/A
S X /P p
dt K S Y K P
++
Solve eq. (1) for D (F/V) as a function of S (5)
A
s
S
DKS
=+
Solve eq. (2) for P as a function of S and D (6)
B
p
P
DKP
=+
(7)
P
B
DK
PD
=−
Solve eq. (3) for XA as a function of S, D and P
A
P/A
A
OA
S X /S P/S
Y
S1
(S S)F VX
K S Y Y
− = +
+
(8)
A
A
OO
AP/A
P/A
X /S P/S
X /S P/S
(S S)D (S S)
XY
1
Y
1
DYY
YY
−−
==
+
+
Solve eq. (4) for XB as a function of S, D, P and XA
B
AB
P/A A B
S X /P p
SP
1
FP Y X V X V
K S Y K P
=−
++
B
P/A A B
X /P
1
DP Y DX DX
Y
=−
(9)
B
B P/A A X /P
X (Y X P)Y=−
Choose S and use eq (5) to compute D, eq (7) to compute (P), eq. (8) to compute
XA, and eq. (9) to compute XB.
S (g/L) D (1/hr) P (g/L) Xa (g/L) Xb (g/L)
1 10.0000 0.172745 0.441973 0.00000 0.00000
8 3.00000 0.157895 0.358566 0.954545 1.72981
Graphing this data
Since organism B relies on organism A to produce its substrate, P, washout of
organism A, and thus the cessation of P production, results in washout of
organism B as well. In examining eq. (8), notice the direct dependence of XB on
XA.
Problem 16.4. (a, b, & c.)
m
net d
cs
S
1k
KS
= = −
+
111
c
1.5d 30 mg/l
10.07d 0.0347d
400 mg/l 30 mg/l
−−−
= − =
+
Qc = 28.8 d (unusually long – typically less than 15 d)
M
X/S C 0
dC
Y (S S) (0.5)(28.8)(300 30)
V/F (1 k ) X (1 0.07.28.8) 5000
0.258d
− −
==
+ +
=
76
V 0.258d 2 10 l/d 5.16 10 l= =
r
r
X /X
C
X /X
rr
V/F (1 )
0.258 28.8 (1 0.5 0.5 )
X / X 2.98 X 29.5 5 g/l 14.9 g/l
= + −
= + −
= = =
d)
H
θ V/F 0.258 d or 6.2 hour==
e) O2 requirement:
7
96
F S (300 30)mg/l 2 10 l/d
5.4 10 mg/d or 5.4 10 g/d
= −
=
Since S is measured as BOD
O2 consumption.
Problem 16.5. a) need
C
to get
net
which is needed for S
(1 ) F 0X (1 )F X= − 0e (x ), F+ + 0
r
X (16.32)
r
r
r 0.4 0.1
X/X 0.357
1 1 0.4
X /X 2.8
++
= = =
++
=
( )
netVX 1 F X
=−
( )
er
r
net
1C
FX (16.35)
0.1 500 l / h
FX2.8
X
V 1500l
0.0932h θ 10.7 h
−
+
==
= =
1
m
net d
S
111
Sk 0.0932h
KS
1h S 0.05h 0.0932h
10 mg/l S
−
−−−
= − =
+
−=
+
S = 1.67 mg/l
b)
M
X/S C O
dC
Y F(S S)
X from eq.16.39
V(1 k )
−
=+
1
(0.5)(10.7)(500)(1000 1.67)
X 1163 mg/l
1500 (1 0.05h 16.7h)
1.16 g/l
−
−
==
+ +
=
c)
Problem 16.6 .
F = 6 m3/h , COD0 = S0 = 1000 mg/l , CODe = S = 30 mg/l , k = 5 d-1 ,
KS = 125 mg/l , Y = 0.35 gX/gS , b = 0.03 d-1 , SVI = 125 ml/g , V = 50 m3
r = 0.362
Sludge Production rate
PrX = F(1+r)X –r F Xr = 6(1.362)(2.887) – 0.362(6) (10) = 1.872 kgX/h = 44.93 kgX/d
b. r = 0.3 , S = 30 mg/L , cm = 0.655 d , c = 3.24 d
Problem 16.7.
F = 10 m3/h , S0 = 2000 mg/L , k = 6 d-1 , Ks = 150 mg/L , Y = 0.4 gX/gS
b = 0.02 d-1 , SVI = 100 ml/g , r= 0.4 , c = 4 cm
r
X /X 2.8 X 2.8 (1.16 g/l) 3.26 g/l
S S 1.67 mg/l
r
r
= = =
==
Problem 16.8.
Aerated Lagoon: Activated sludge without sludge recycle
F = 5 m3/h , S0 = 500 mg/L , S = 30 mg/l , k= 4 d-1 , Ks = 100 mg/l , b = 0
b. Monod Kinetics:
c. Vmonod > V1st order , H monod > H,1st order
Problem 16.9.
11
1
11
0.2h 20 mg/l 0.01h
80 mg/l 20 mg/l
133.3h
0.03h
−−
−
−
+
==
b)
C
V F (1 X /X)
r
= + −
(eqn. 16.40)
note Xr/X = 3 based on “sedimentation unit concentrates biomass by a factor of
3”
50000 l 10,000 l/h 33.3h (1 3)
0.15 (1 2 )
0.425
= + −
=−
=
Problem 16.10. Since
net C
1=
eqn. 16.30 is
Cr
1/ XV FX (1 )FX
+ = +
which can be rearranged to:
Cr
1/ D(1 X /X)
= + −
s d c
c m d
K (1 k )
S( ) 1k
+
=−−
(Eqn. 16.29a)
So:
s r d
m d r
1
1 1 1
K (D(1 X / X)) k
Sk b(1 X / X )
0.2g e/l(0.1h (1 0.7 0.7 2) 0.05)
S0.5h 0.05h 0.1h (1 0.7 0.7 2)
S 0.038 g/l
−
− − −
+ − +
=− − + −
+ − +
=− − + −
=
Problem 16.11.
M
Cd
ss
ss C
C m d
S
1/θk
K X S
K X(1 kd )
S( k ) 1
=−
+
+
=−−
Eqn 16.39 still applies:
M
X/S C O
dC
M
X/S C ss d C
O
d C C m d
31
3 1 1 1
YF
X (S S) or
V(1 k )
Y F K X(1 k )
X (S )
V(1 k ) ( k ) 1
(0.5)(120 h)(400 m /h) 0.029 BOD / g MLVSS X (1 0.005h 120h)
X 800 mg / l
3200 m (1 0.005h 120 h) 120 h(0.2h 0.005h ) 1
X 4.688 80
−
− − −
=−
+
+
=−
+ − −
+
=−
+ − −
=0.032X
0 mg/l 22.4
X 3725 mg/l
−
=
0.02 (3725) (1 .6)
S 5.32 mg/l (still 20mg/l)
22.4
+
= =
C
33
7
VX
Sludge Production Rate
(3200 m )(3725 mg/l)(1000 l/m )
120 h
9.93 10 mg/h
99.3 kg/h
=
=
=
=
Problem 16.12.
Data
F = 103 l/hr Soin=500 mg/l Soout=10 mg/l
2
52
O
D 2 10 cm /s
−
=
rm=20 mg S/l hr KS = 200 mg S/l
From eq 12.43 – A Balance on a differential volume of the reactor can be
written
o m o
So
dS r S
dz K S
+
Assume the reaction is first order,
OO
S O S
SS
K S K
+
. Eq. (1) can be simplified and
integrated
OmO
S
dS r LaA
FS
dz K
−=
OO
O oin
S S z H
oin mm
O
S S z 0
O O S S
Sr LaA r LaA
1d S ln dz H
S S K F K F
==
==
= = − =
Solving for H (2)
S oin
mO
K F S
H ln
r LaA S
=
Determining
from eq. 16.48 and 16.46
0.0118
=
4
2.27 10−
=
1.00
=
Substituting into eq. (2) H = 3.9 m
Problem 16.13.
a. Si = (S0 +R S) / (1 +R) , 500 = (1500 + 50 R)/ (1+R) , R = 2.22
Problem 16. 14
a. RBC tank is completely mixed S = 0.05 S0 = 40 mg/l
b. RBC tank behaves like a plug flow reactor (PFR)
Problem 16.15.
Two RBCs in series. E1 = E2 , 1- S1/S0 = 1- S2/S0
S1 = (1500×50)1/2 = 274 mg/l = 0.274 kg/m3