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S
m
D
K1
SD
1
+
−
=
+
−
−
S
D
K
D (1 )
1
+
+ −
−
33
0 S X S 0 m X S 0 S X S m
43
2
0 S X S X S S m 22
0 m X S
2
2
0 m X S 0 S X S m 0 S X S
2
0 m X S
(S K ) Y S Y (S K ) Y
O = D D
(1 ) 1 1
(S K ) Y (4 ) Y K D S Y
(1 ) (1 )
3 S Y 3 (S K ) Y (S K ) Y (6 )
(1 ) (1 ) (1 )
S Y (S
+ − +
+ − +
− − −
+
− + −
− −
+ +
− − +
− − −
− − 0 S X S
0 S m X S
X S S m 2
0 m X S 0 m X S
2 2 3
0 m X S 0 S X S m 0 S X S
2
2
0 m X S 0 m X S 0 S m X S
(S K ) Y
K ) Y (1 )
2 Y K
2S Y D 2 S Y
(1 )
3 S Y 3 (S K ) Y 4 (S K ) Y
(1 ) (1 ) (1 )
+ S Y (1 ) 2 S Y 2 (S K ) Y
2 (S
+
+ + +
−
+ − + −
−
+ +
− − + +
− − −
− − − + +
+0 S X S 2
0 m X S 0 m X S
23
X S S m 0 m X S
22
0 m X S
34
0 S X S m 0 S X S 2
0 m X S
2
22
0 m X S 0 S m X S
K ) Y S Y (1 ) 2S Y
(1 )
Y K S Y
SY
(1 ) (1 )
(S K ) Y (S K ) Y S Y (1 )
(1 ) (1 )
(S
S Y (S K ) Y
+
− − + −
−
− + − −
− −
+ +
− + + − −
− −
− − + +
2
0 S X S
K ) Y
(1 )
+
−
Sm
S
2
X S X S 0
2
m
m
S
2
X S 0
2
m
K
1D
K
D1
d(DX) 1 1
D Y 2DY S D
dD D+
D
1
1
D
K
(1 ) 1
D Y S D
(D )
1
−
−
+
+ −
− −
= + −
+
+
−
−
−
−
+
− −
−−
+
+
−
−
dD =
m
32
0S
X S X S S m m X S 0 S X S 0 m
2
0S
X S X S S m m X S 0 m X S 0 S
22
0S
X S 0 X S m X S 0
2
m X S 0 S
(S K )
D D Y D Y K Y (S K ) Y S
(1 )
(S K )
4 Y D Y K Y (1 )S 3 Y (S K )
(1 )
(S K )
3 Y S 5 Y 2 Y (1 )S
(1 )
2 Y (S K )
+
= + − − + − +
−
+
+ + − + − − +
−
+
− + + − −
−
− + 23
0S
X S 0 m X S
(S K )
2 Y S 2 Y (1 )
+
− +
−
-1
0.0823 hr
=
Problem 6.12.
-1
opt
D 0.482 h=
[EtOH] 0.803g L=
X 10.6 g cells L=
22
2
X O X O
X DX
OUR 20.4 go L hr
YY
= = =
OUR > OTR = 10 gO2/L·hr
Therefore, the reactor cannot be operated at Dopt.
Problem 6.13
t
0
0
0
XS
X X e
(X X )
SS
Y
=
−
= − +
t(hr) X(g/L) Sglc(g/L) Slac(g/L)
0 0.10 2 3
Problem 6.14
//
1 1 m
D
Ap M
X S X S
YY
=+
0
S 500 mg/L=
1
D(h )
−
( )
H
θ 1 D h=
/
0
X
(S S)
Ap
XS
Y=−
Ap
X/S
1Y
0.05 20 0.33 3.03
A plot of
Ap
X/S
1Y
vs
1D
yields :
M
X/S
1 1.5 (intercept)
Y=
/0.67 gX/gS
M
XS
Y=
and
m 0.08 gS/gX h (slope)=
M1
d
d
MX/S
X/S
k
M , k m (0.67)(0.08) 0.054 h
Y
Y
−
= = = =
mm
d d d
SS
S
d m m
SS
D k k ; D k
K S K S
K
1 1 1
or D k S
= − = − + =
++
=+
+
d
Dk+
( )
d
1 D k+
1S
A plot of
( )
d
1 D k+
vs
1S
10.8 (intercept)
m
S
K 1.25 (135) 169 mg / L
==
Problem 6.15.
c. Anaerobic treatment
Dopt = µm (1- (Ks/(Ks + S0))1/2 ) = 1/HRTopt
Problem 6.16.
(a) Determine X,
2
o
q
at steady state:
at steady state
M
0
X/S
D, so X (S S)
Y
= = −
M
X/S 0.45 g X gS
Y=
0
S 2g l=
S 0.1g l=
X 0.45 gX/gS (2g/l 0.1 g/l)
X 0.855 gX/l
=−
=
2
2
o
XO
X
OUR q X Y
==
2
2
0o
XO
Sq Y
=
22
1
o o 2
2
0.28 h
q q 1.12 gO gX hr
0.25 gX gO
−
= =
(b)
L
Ka
to overcome oxygen transfer limitation when oxygen transfer is rate-limiting,
growth= oxygen transfer
2
*
LL
XO
Xk a (C C )
Y
=−
1
L
2
3
2
L
(0.28 h )(0.855 g l) k O (8 mg/l 2 mg/l)
0.25 gX gO
gO
0.9576 k a (6 10 g/l)
ml
−
−
=−
=
1
L
k a 159.6 hr−
=
Problem 6.17.
a) From cell balance
Problem 6.18
H
5.29kcal gS
Y 0.076 gS kcal
=
39
H,S
H,S
Y
15.26 kcal g Substrate
Y
=
Problem 6.19.
p
DP q X (S.S. product balance)
=−
DP X X (Luedeking-Piret Eqn. 6.20)= +
DP 0.394 mg P / L h
=−
Extra Credit-S should not go directly to product it should go to X first & then
X→P
Problem 6.20
V V net
RR
V dt V
m,rep
V
dV
S,rep
[S]
dn k D n
dt K [S]
= − −
+
(This could be integrated if [S], Kd, and D were not f(t))
b)
m,rep
V
d
S,rep
[S]
dn 0 k D
dt K [S]
= = +
+
Solving for [S]:
d
d S,rep
n,rep-K -D
(k D)K
[S] ()
+
=
c)
d
d
R R d V n
dn
V V K n F (From a balance on dead cells)
dt =−
d
d V d
dn k n Dn
dt =−
d
dn 0 k n D
Problem 6.21.
1
0
S 5 g / L; D 0.2 h−
==
0.005S
e 0.2; S 322 mg / L
−
==
n
SD 0.2
0.25 S 16 0.19685
S 311 mg / L
=+
=
Problem 6.22. Since
d net growth
k 0, D→ = =
m2
S1
2
2S
I
mm
I
SD
K S S K
4D K
K
S ( D) ( D)
2D K
=
++
= − − −
b) There are 3 possible steady states: the washout steady – state and the 2
2
2S
4D K
Problem 6.23.
Sd
md
K (D k )
SDk
+
= − −
M
X S 0 n
d P X S P/S
D
X Y (S S) D k q Y Y
=−
++
P
P
qX
DP q X P D
= =
For
m d 0
D ( k ), S S , X 0, P 0 (wash out) − = = = −
a)
b) Maximize productivity of product(DP).
d(DP) dX
( )
( )
( )
XS 2
M
md
d p X S P S
M
d P X S P S
Sd
0M
md d P X S P S
dD ( D k )
D+k q Y Y
D k q Y Y D
K (D k )
SDk D k q Y Y
− −
+
+ − −
+
+−
− − ++
setting this to zero brings us, after some algebra, to:
( )
( ) ( )
( )
2Sm
S0
M
d P X/S P/S
M
P X/S P/S S m
d S m
m S m 0
MM
d P X/S P/S d P X/S P/S
d 0 d S d 0
2
m S d m d 0 m 0 S
a
b
K
D K S
k q Y Y
q Y Y K
kK
D K 2 S
k q Y Y k q Y Y
2k S 2k K k S
( K k 2 k S S K
−
++
+
−
+ − − −
+ +
+ + +
+ − − + − 22
d d 0
c
k k S ) 0−=
a 5.00
b 2.06
c 0.196
=
=−
=
md
11
sensible answer not a sensible answer since for
D ( k ) there is a wash out
D 0.149 [h ] or D 0.263[h ]
−−
− −
= =
Maximize productivity of biomass (DX):
set
d(DX) 0
dD =
It is a considerable time saving to use a computer to solve the latter
equation and find
1
D 0.165 [h ]
−
=
.
