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Chapter 9
Problem 9.1
V 1000 l=
S
K 1.5 g/l=
0
S 10 g/l=
1
m0.4 h−
=
X/S
Y 0.5 g/g=
F 100 l/h=
a)
1
F 100 l/h
D 0.1h
V 1000 l
−
= = =
At steady state:
1m
S
S
D 0.1h KS
−
= = = +
11
S
0.1h 0.4 h 1.5 g/l S
−−
=
+
Solving for S:
S 0.5 g/l=
X/S 0
X Y (S S) 0.5 g/g (10 0.5 g/l) 4.75 g/l= − = − =
1
DX 0.1h (4.75 g/l) 0.475 g / (l h)
−
= =
b) Cell Recycle Flow Rate = 10 l/h
recycle reactor
X 5 X=
Recycle Flow Rate 10 l/h 0.1
Feed Flow Rate 100 l/h
= = =
recycle
reactor
X
C5
X
==
-1
[1+α (1-C)]D
[1+0.1(1-5)] (0.1h )
=
=
(equation 9.9)
-1
0.06 h=
(Note: μ < D)
As above:
11
S
0.06 h 0.4 h 1.5 g/l S
−−
=
+
Solving for S:
S 0.265 g/l=
1
X/S 0 1
D 0.1h
X Y (S S) 0.5 g/g (10 0.265 g/l)
μ 0.06 h
8.11g/l
−
−
= − = −
=
1
DX 0.1h (8.11g/l) 0.811g / (l h)
−
= =
Problem 9.2
-1
m
μ 0.3 h=
S
K 0.1g/l=
X/S
Y 0.4 g/g=
a)
1
1
1
F 100 l/h
D 0.2 h μ
V 500 l
−
= = = =
, at steady state
11
11
m
S 1 1
SS
0.2 h μ 0.3 h
K S 0.1g/l S
−−
= =
++
2 P/S 1 2
Problem 9.3
F = 100 l/h V = 900, 600, 300 l
1. Calculate dilution rates for the tanks:
1
F
−
4. Determine X1, as follows: X1 corresponds to the intersection point between the line with a
slope of D = 0.111 h–1 (through the origin) and the dx/dt versus X curve, i.e.
5. Determine P1, from X and P versus t plots, i.e.
6. Step off the subsequent stages on
dP
dt
versus P curve:
Equations for the subsequent stages :
22
0.167 (P 0.09) 0.378 0.619 P− = −
or
2
P 0.5 g/l=
33
0.333 (P 0.50) 0.378 0.619 P− = −
or
3
P 0.572 g/l=
7. Repeat steps 3–6 for the remaining combinations. The results are summarized in table
blow.
Since the dilution rate for the 300 l tank is 0.333 h–1 which is larger than μm, washout of
biomass will occur if the 300 l vessel is the first vessel in the series. Two combinations are
55
Problem 9.4. Two CSTRs in series
a. 1st reactor: H1 = V1/F = (S0 –S1)/ kS1
Problem 9.5.
a.
μR
m = 0.693/ td = 0.139 h-1
V1 = 8.13 L
2 =
m S2/(Ks + S2) = 0.139(1.6)/(1+1.6) = 0.0855 h-1
e. qP1 = D1 (P1 –P0)/X1 = 0.123(16)/3.2 = 0.615 gP/gX-h
Problem 9.6
0
V 500 l=
0
S 300 g/l=
1
m
μ 0.2 h−
=
F 50 l/h=
0
X 20 g/l=
S
K 0.5 g/l=
X/S
Y 0.3 g/g=
m
μ D 0.2 0.05 h
−
−−
c)
t0 0 X/S 0
t
X X V FY S t
20 g/l (500 l) 50 l/h (0.3 g/g)(300 g/l)(10 h)
X 55,000 g
=+
=+
=
m
X 55,000 g
[X ] 55.0 g/l X
V 1000 l
t
t= = = =
d)
P
q 0.05 g / (g h)=
and P0 = 0.1 g/l
( )
0 0 0
Pm
1
P V V Dt
P q X t
V V 2
0.1g/l (500 l) g g 500 l 1
0.05 55 (0.05 h )(10 h) (10 h)
1000 l g h l 1000 l
equati
2
on 9.44
−
= + +
= + +
P 20.7 g/l=
Problem 9.7.
P
D 0.5 cm=
3
m
r 100 mg / (cm h)=
3
S
K 10 mg / cm=
mm
SL
SS
r S r
r S k S (At low S)
K S K
= =
+
23
a 25 cm cm=
2
A 100 cm=
62
S
D 10 cm S
−
=
Calculating effectiveness factor, n :
3
πP
m S 6
P m P m
3
P S P S S S S
31
6 2 1 3 1
D
r / K
V r D r
AD πD D K 6 D K
0.5 cm 100 mg cm h
6 10 cm S (10 mg cm )(3600 S h )
4.392
1 1 1
η 0.210
tanh 3 3
−−
− − − −
= = =
=
=
= − =
Steady state substrate balance over element dz:
mP
SP
rV
FdS η S a A dz 0
KA
− − =
Problem 9.8.
Db = 10 cm X0 = 45 g/l
0
X X (1 0.005z ) 45 (1 0.005z ) g/l= − = −
a) Substrate balance over element dz
S0
FdS qS X A dz
FdS q A X (1 0.005z) dz
−=
− = −
Integrating from (0, S0) to (H, S):
0
SH
S0
S0
2
0 S 0
F dS q A X (1-0.005z) dz
F (S S ) q A X (H 0.0025 H )
−=
− − = −
Solving for S :
22
S0
0b
1 1 1 22
13
qX π
S S D (H 0.0025 H )
F4
2 g g h (45 g l ) πl
160 g/l (10 cm) [200 0.0025 (200) ] cm
5 l h 4 1000 cm
S 160 141.37 g/l 18.6 g/l
− − −
−
= − −
= − −
= − =
3
1
1
4 1000 cm
V 15.71l
F 5 l/h
D 0.318 h
V 15.71l
DP 0.318 h (67.9 g/l) 21.6 g / (l h)
−
−
−
=
= = =
= =
Problem 9.9.
a.)
μ
=
μ
m S/(Ks +S) = 0.2(0.5)/(0.5 +0.5 ) = 0.1 h-1
Problem 9.10.
a. Chemostat without cell recycle
b. Chemostat with cell recycle. r = 0.4, c = 3 , = 0.1
S = 1.5 g/l , 1/ c = uN = ug –kd = 0.108 h-1 , c = 9.26 h
Problem 9.11.
L = 0.5 mm S0 = 2000 mg/l
mm
S
SS
r S r
rS
K S K
=
+
(true at low values of S)
Steady state substrate balance over element dz :
m
S
r
FdS S η L a A dz
K
− =
Integrating :
0,e
0
SH
m
S0
S
0,e m
0S
rη L a A
dS
F dz
SK
Srη L a A
F ln H
SK
−=
−=
or
S0
m 0,e
3 1 3 1
22
33
K F S
H ln
rη L a A S
25 mg cm (2 l h )(1000 cm l ) 2000
ln
mg cm π100
50 (0.7)(0.05 cm)(2.5 ) (10 cm)
cm h cm 4
H 436 cm 4.4 m
− − −
=
=
==
Problem 9.12. Similar to example problem 9.4. Based on mass balance over a differential
height and integrating over the length of the column
P
η q X A H
Problem 9.13. See Attached Figure. The minimization process requires trial and error
solution. Typically volumes are minimized by operating at a dilution rate slightly
below the maximum in the dx/dt us X curve for the first vessel
The reactor order is Vfirst = 5, 180 L , Vsecond = 19, 580 L
Problem 9.14. a)
net
net
net
d (total mass) / dt
μtotal mass
d(VX) / dt
μVX
X dV / dt V dx / dt
μVX VX
=
=
=+
but X is constant ∴ dX/dt = 0
net 1 VdV dt =
b) at t = 5 h
5
00
V V dV dt
V 100 L 20 L h 5 h
V 200 L
=+
= +
=
1
net
20 L h 0.1h
200 L
−
= =
Problem 9.15
S
m
K D (1 α αC)
+−
200D (1.6 0.6(2))
−
0.4D
0.5 0.25 0.4D
=−
1
1
F
D 0.208 h V
100 L / h
V 481L
0.208 h
−
−
==
==
b.
01
1
y(S S ) 0.4 (5000 100)
X 4900 mg/L 4.9 g/L
1α αC [1.6 0.6(2)]
−−
= = = =
+ − −
r1
X c X 2 (4900) 9800 mg/L 9.8 g/L= = = =
c.
ST
H,ST
V
θ 2 h
(1 ) F
==
+
;
ST
V 2 (1.6)100 320 L==
Biomass balance around the sedimentation tank is :
1 2 1
1 2 1
(1 α)F X F X αF X
(1 α) X X αC X
+ = +
+ = +
2
2
X 1.6 (4900) 0.6 (2) (4900)
X 1960 mg/L 1.96 g/L
=−
==
Problem 9.16.
For DP = 4 mm, r4 = 200 mg/L h
a.
0
4
4
0
7
7
0
Rate with dif Limit
ηRate without dif Limit
r
ηr
r200
η 0.67
r 300
r100
η 0.33
r 300
=
=
= = =
= = =
b.
m0
S0
rS
rηKS
=+
or
S
m m 0
K
1 1 1
rηr ηr S
=+
A plot of
1r
versus
0
1S
yields
3m3
m
11
0.5 10 (intercept);r
ηr (0.67) (0.5)10
−
−
= =
S
m
K1.14 (slope)
ηr =
m
m
r 2985 mg/L h
r 3000 mg/L h
=
S
S
K 0.67(2985)(1.14)
K 2280 mg/L
=
=
Problem 9.17.
KS = 0.2 kg/m3 , S0 = 2 kg/m3 , Se = 0.03 kg/m3 , X = 10 g/m3
Substrate Balance on “dƶ” yields
S
S
kXS
QdS η r dV η LaA dz
KS
− = − = − +
H 8.8 m.
=
Problem 9.18.
Packed column biofilm reactor with effluent recycle
b. Substitute the numbers into the design eqn.
c. af1 V1 = af2 V2 V2 = (3.14/4) Do22 H2 = (3.14/4) (1.5)2 (3) = 5.3 m3
