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Chapter 7
Problem 7.1. Total nitrogen in cells =(Nitrogen content of cells)×(Final cell
concentration)×(Culture volume)=(0.12 g nitrogen/g cells) (30 g cell/liter) (1000 liters) = 3.6
kg nitrogen.
Problem 7.2. a. C6H12O6 = glucose, MW = 180; C6H10NO3 = yeast, MW = 144;
(NH4)2SO4=ammonium sulfate, MW 132.
b. Yield coefficients:
Yx/s= (50g yeast/liter) / (130 g glucose/liter) = 0.385 g yeast/g glucose.
7.3) a. Biomass yield coefficient
b. Product yield coefficients:
c. β coefficient determined by nitrogen balance.
Problem 7.4.
2 5 2 3 1.704 0.149 0.408 2 2
C H OH a O b NH c CH N O d CO e H O+ + = + +
.
RQ d/a 0.66==
Solve 5 equations for 5 unknowns: a, b, c, d, and e.
2
Problem 7.5.
2
C H COOH a O b NH c C H NO d H O e CO
5 5 7
6 2 3 2 2
+ + → + +
e 4.5
=
6 5 2 3 5 7 2 2 2
C H COOH 5O 0.5NH 0.5C H NO 2H O 4.5CO+ + → + +
b.
2
x/s
x/ 2o
0.5 (60 7 14 32) 56.5
Y 0.463gX / gS
(72 6 12 32) 122
0.5 (60 7 14 32) 56.5
Y 0.353
51/
(32) 60 gX gO
+ + +
= = =
+ + +
+ + +
= = =
c.
'b
's
20
5(4) 7(1) 1( 3) 2( 2) 20; 4
5
30
7(4) 6(1) 2( 2) 30 ; 4.28
7
b
S
= + + − + − = = =
= + + − = = =
Problem 7.6.
3 6 3 2 3 5 7 2 2 2
C H O a O b NH c C H NO d H O e CO+ + → + +
'
20
Problem 7.8
C6H4CH3OH + a O2 + bNH3 ----→ c C5H7NO2 + d CO2 + e H2O
a. Elemental balances
b. Yx/s = 1.1(113)/108 = 1.15 gX/gS
Problem 7.9.
47
Yx/s = 119.2c /180 = 0.1
Solution: a = 0.15, b = 0.03, c= 0.15, d = 2.24, e=0.67, f = 1.265
C6H12O6 +0.15 NH3 +0.03 H3PO4 →0.15C5H7NO2P0.2 +2.24CH3COOH +0.67CO2 +1.265H2
b.
γ,
s = 24 ,
γ,
b = 21,
γ,
Ac = 8 ,
γ,
H2 = 2
Problem 7.10.
b. Yx/s = 0.227(119.2)/60 = 0.45 gX/gHAc
Problem 7.11.
a. Consider 1 g dw cell mass
mc = 0.45 g, mo = 0.25, mH = 0.05 g, ms = 0.005 g, mN = 0.12 g , mP = 0.025 g, mmin = 0.1 g
C5 H6.65 N1.14 O2.1 P0.1 S0.02
b. C4H8O2.2N1.2P0.08 S0.02 Consider 1 mol cell mass.
Problem 7.12.
c. Rc : 1/5 CO2 +1/20 NH4+ + 1/20 HCO3- + H+ + e- --> 1/20 C5H7NO2 + 9/20 H2O
Problem 7.13.
a. Rd : 1/8 NH4+ + 3/8 H2O ---> 1/8 NO3- + 5/4 H+ + e- ∆ G0 = 35.11 kJ/eq e-
Ra : 1/4 O2 + H+ + e- ---> 1/2 H2O ∆ G0 = -78.72 kJ/eq e-
Problem 7.14. ∆G0 (kJ/eq e)
a. Rd : 1/8 CH3COO- + 3/8 H2O → 1/8 CO2 + 1/8 HCO3- + H+ + e- - 27.40
c. R = ee Re + es Rs = 0.9 Re + 0.1 Rs
R = 0.9 (1/8 CH3COO- +1/8 H2O) + 0.1 (1/8CH3COO- + 1/20NH4+ + 3/40CO2) →
Problem 7.15.
Methane formation from glucose, acetate,ethanol and H2 as electron donors CO2 as
electron acceptor.
Glucose: From Fig 7.1. ∆G0 = 18 kJ/eq e-
