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20
Chapter 6
Problem 6.1.
Volume of a single cell: Vc = (3.14/4) (1))2 (2) = 1.57 um3/cell = 1.57 x10-12 cm3/cell
Problem 6.2.
a.
max
1 = 0.693/td1 = 0.693/ (2 h) = 0.346 h-1
max
Problem 6.3 a.
max
net 0
1 dX
μ = , when is constant ln X ln X μt
X dt → − =
1
max
0.296 hr−
=
21
b.
App
x/s
Y
App
X/S
ΔX (6.2 0.2) g cells/L g calls
Y = = 0.65
substrate
ΔS g substrate
(0 9.23) g L
− − − =
−
c.
d
t
d0
0
X
ln μt, t occurs when X 2X
X
==
d d d
ln 2
ln 2 t , t , t 2.34 hr= = =
d.
S
K
m
S
[s]
K [S]
= +
S 0.077 g / L=
dX g
= 0.123
dt L hr
1
0.020 hr−
=
-plot μ vs.S
-
S
K [S]=
at
max
1
2
=
s
K 0.45 g/L=
23
e. at t = 10 hr,
1
net max 0.296 hr−
= =
Problem 6.4. a.
m
S
1
S
1 dx
x dt H
K 1 S
K
+
= =
++
X dt
2) rearrange,
S
mm
K
1 1 1
S
=+
plot
1
vs.
1
S
,
S
m
K
slope
=
m
1
intercept =
-for pH data,
+
S
SS
1
K
K K (H )
K
=+
plot
S
K
vs. H+
S1
slope K /K=
, intercept=KS
b.
+
S1
mm
K (1 H /K )
1 1 1
S
+
=+
Problem 6.5.
E/RT
k = Ae−
25
Problem 6.6.
c. Monod eqn. dX/dt =
max
S / (Ks + S) X
Problem 6.7.
6 variables:
*
1
S
, S1,P, M, T, V
S1*= g substrate external to cell [S1*]= g/L reactor
S1= g substrate internal to cell [S1]= g/L reactor
trans
rate of transport g
Kcell area area hr
==
KS*=saturation constant for permease, g/L
KIS=inhibition constant for permease, g substrate/g biomass
k1. k2=rates of conversion, g formed/hrg biomass
K1S, K2S, K2P = saturation constants, g/g biomass
(i) Balance on S1*
*
*
*
trans 1 IS
1
*
1
1
SIS
K A[S ] K
dS -T
[S ]
dt K +[S ] K+
[T]
=
(ii) Balance on S1
*
*
trans 1 IS
1 1 1 2 1
*
1
1 1S 1 2S 1 2P
SIS
loss to formation of P lo
transport in
K A[S ] K
dS K [S ] [T] K [S ] [T]
[M] [P] [T] [M]
=[S ]
dt K +[S ] K +[S ] [T] [T] K +[S ] [T] K +[P] [T] [T]
K+
[T]
−−
ss to formation of M
T
(iii) Balance on P
1 1 2 1
1S 1 2S 1 2P
formation of P
k [S ] [T] k [S ] [T]
dP [M] [P] [T] [M]
=T
dt K +[S ] [T] [T] K +[S ] [T] K +[P] [T] [T]
−
(iv) Balance on M
21
2S 1 2P
k [S ] [T]
dM [P] [T] [M]
=T
dT K +[S ] [T] K +[P] [T] [T]
(v) Balance on T
1
T = S + P + M
(vi) volume
dV 1 dT
=
dt P dt
Problem 6.8. a.
2
22
2
O
XO O Used for
maintenance
O Used
for growth
μX
OUR= m X
Y+
2
O2
m [=] gO g dry weight
2
X/O 2
Y [ ] g dry weight gO=
X [ ] g dry weight L=
1 dX ,
X dt
=
substitute into above equation
2
2
O
XO
1 dX
OUR + m X
Y dt
=
To estimate
dX
dt
: In exponential growth phase O2 for growth dominates,
2
XO
1 dX
OUR Y dt
To estimate X: During stationary phase cells aren't growing and O2 is needed for maintenance
only
2
O
OUR m X
b. plot log X vs. t
exponential growth between t = 3 hr & t = 10 hr
-1
max 0.325 hr=
μX
2
X O 2
during stationary phase, t = 18 hr to t = 21 hr
2
O
OUR m X=
( )
OUR g L hr
( )
X g L
( )
2
-1
O
m hr
2
2
O
go
m 0.0600 gdw hr
=
Problem 6.9. a.
-1
max
d
ln 2 0.289 hr
t
= =
max max
11
D = D
22
=
S
X S 0 X S 0 S
m
KD
X = Y S = Y (S K )
μ -D
X =16.9 g dry weight/L
−−
b.
max max
D= 0.8D = 0.8μ
S
S
m
KD
S = 4K = 5.2g L
μD
=
−
c.
-1
max max
D=μ 0.289 hr=
d.
S
X/S 0 max
m
KD
DX DY S w/D = 0.8μ
μD
=−
−
g dry weight
DX 3.49 L hr
=
Problem 6.10.
m
d
S
μS
Dμk
K +S
= = −
S
d
d m m
K
1 1 1 , first must find K
D+K μ S μ
=+
max
d X S S
K Y m=
S
app max app
X S X S X S
m
1 1 1 ΔS 1
+ , plot vs.
Y Y D Y X D
==
c.
max -1
XS
Y = (intercept)
= 0.491 gdw gS
d.
S
S
m slope
gS
m 0.0578 gdw hr
=
=
max -1
d X S S
K Y m 0.0284 hr==
11
b.
-1
max
-1
max
(intercept)
0.95 hr
=
=
a.
Sm
S
K = (slope) μ
K = 0.141 mg S mL
Problem 6.11. Steady State Balances around Chemostat
p
q
0
P/X
VX
X: FX FX V X 0
Y
− + − =
with
0
F
X 0 & D
V
==
p
p
PX
q
D+ , with q
Y
= = +
pX
pX
pX
pX
DY
, let Y
1Y
Y
+
= =
−
=
0
XS
VX
S: FS FS 0
Y
− − =
X S 0
(1 )
X Y D(S S) (D )
−
=−
+
