Test problems 1-5
1. If -1 S = -6(X/4) O2 # Yx4 = 0?
need Y to write
stoichiometry for oxygen
Growth ONLY
Stoichiometry
true growth yield - can be measured exactly
R = Rd - feRa - fsRc
fs = YH (COD basis)
fc = 1 - YH
R = Rd - (1 - YH)Ra - YH Rc
2a. YH = 0.6 mgX-COD / mgS-COD = fs
MOLAR
Ss = C10H19O3N
Rd = 1/50 C10H19O3N + 9 H2O -> 9/50 O2 + 1/50 NH4+ + 1/50 HCO3- + 1/50 H+
- 0.6 Rc = 0.12 CO2 + 0.03 HCO3- + 0.03 NH4+ + 0.04 H2 + 0.6 -> 0.03 C5H7NO2 + 0.21 H2O
- 0.4 Ra = 0.1 O2 + 0.4 H+ + 0.4 e- -> 0.24 H2O
0.02 C10H19O3N + 0.01 NH4+ + 0.15 O2 + 0.01 HCO3- -> 0.11 H2O + 0.03 C5H7NO2 + 0.06 CO2
x50
C10H19O3N + 0.5 NH4+ + 5 O2 + 0.5 HCO3- -> 5.5 H2O + 1.5 C5H7NO2 + 3 O2
LHS RHS
C 10.5 10.5
N 1.5 1.5
O 17.5 17.5
H 21.5 21.5
charge 0 0
Σν reactant = Σν product
3. mw C10H19O3N = 201
ψ1 = 1, ψ2 = 0.5(18)/201 = 0.045, ψ3 = 5(32)/201 = 0.796
ψ4 = 0.5(61)/201 = 0.152, ψ5 = 5.8(18)/201 = 0.493, ψ6 = 1.5(13)/201 = 0.843
ψ7 = 3(44)/201 = 0.657
LHS: 1.993, RHS: 1.993
C10H19O3N + 0.015 NH4+ + 0.796 O2 + 0.152 HCO3- -> --
-> 0.493 H2O + 0.843 C5H7NO2 + 0.657 CO2
4. C10H19O3N : 1.99 gCOD / g OM
C10H19O3N-COD + 0.796(-1)/1.99 O2 -> 0.843(1.42)/1.99 C5H7NO2
C10H19O3N-COD + (-0.40) O2 -> 0.60 C5H7NO2
5. Monod: μ = μ̂ Ss / (Ks + Ss)
Andrews (inhibition) μ = μ̂ Ss / (Ks + Ss + Ss²/KI)
rg = μXB = rs / (1/YH) = ro / (-(1-YH)/YH) = ro / (-(1-YH)/YH)
μ̂
max. inhibited growth rate
μ* = μ̂ / (2(Ks/KI)^0.5 + 1)
S* = (KsKI)^0.5
μ̂
μ*z
μ*z
Monod and KI -> ∞
Andrews
KI1
KI2
Ks - same in both models but no longer gives 1/2 max rate in Andrews
K12 < K1
if KI = Ss
μ̂ = μ̂ KI / (Ks/KI + KI + KI) = μ̂ / (2 + Ks/KI) < μ̂/2 unless KI >> Ks