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Sef Heijnen, Department of Biotechnology, Faculty of Applied Sciences
Learning about the microorganism: q-‐rates and chemostat
From measurements to process rates
Fin cs,in cx,in
FN,out yN2,out yO2,out yCO2,out
Fout cs,out cx,out cp,out
FN,in yN2,in yO2,in yCO2,in
RO [mol O2/h] RC [mol CO2/h]
Rs [mol s/h] Rx [mol x/h] Rp [mol p/h] Measurements
& balances
VL cs cx cp
VL cs cx cp
Rs [mol s / h] Rx [mol x / h] Rp [mol p / h] RO [mol O2 / h] RC [mol CO2 / h]
qp qCO2
qs qO2
µ (= qx)
From process rates (R) to biomass specific rates (q)
Biomass specific: q-‐rates
Biomass specific rates
Biomass specific: q-‐rates Process rates
Biomass specific: q-‐rates
Total amount of biomass
Why are q-‐rates important?
Q-‐rates quanIfy the performance of the organisms
Drawing conclusions only from product concentraQons is dangerous!
Why drawing conclusions from concentraIons is dangerous: Product yield (exp 1)
Broth (steady state)
Fin = 0.1 m3/h cs,in = 300 mol glucose/m3
cp,in = 0 mol product/m3
Product yield (exp 1)
pH control acid/base FpH = 0.02 m3/h
Fout = 0.12 m3/h cs,out = 150 mol glucose/m3
cp,out = 100 mol product/m3
mol p/mol s = Rp/Rs = qp/qs = 12 / (30 -‐18) = 1.0 mol p/mol s
VL = 1.00 m3
cx = 1000 mol x/m3
mol p/mol s = (cp,out-‐cp,in)/(cs,in-‐cs,out) = (100 -‐ 0)/(300 -‐150) = 0.667 mol p/mol s
Why drawing conclusions from concentraIons is dangerous: Product inhibiIon (exp 2)
Product inhibiIon (exp 2)
Fin = 0.1 m3/h cs,in = 300 mol glucose/m3
cp,in = 100 mol product/m3
pH control acid/base FpH = 0.02 m3/h
Fout = 0.12 m3/h cs,out = 150 mol glucose/m3
cp,out= 183.3 mol product/m3
qp (exp 2) = (0.12 · 183.3 – 0.10 · 100) / (1.00 · 1000) = 0.012 mol p/h
cp,out = 100 (exp1) + 100 = 200 mol p/m3 à inhibition?
qp (exp 1) = (0.12 ·∙ 100 – 0.10 · 0) / (1.00 · 1000) = 0.012 mol p/h
Broth (steady state)
VL = 1.00 m3
cx = 1000 mol x/m3
Compare microorganisms
• Make balances
• Do measurements
• Calculate q-‐values for two different organisms • Draw your conclusions
• Do not compare concentraQons
• Do experiments
Behaviour of microorganisms
• µ determines the behaviour of the microorganisms
• Forcing the organisms to have different µ values : how?
• Run fermenter in chemostat mode
• Growth rate µ can be controlled in chemostat Other q-‐rates will change when we change µ
Fin [m3 / h] cx,in [mol x / m3 ]
Fout [m3 / h] cx,out [mol x / m3]
How to control μ in a chemostat: Biomass balance
VL [m3] cx [mol x/m3]
SeXng up the biomass balance for chemostat
AccumulaQon ProducQon RX(t)
Inflow Ouglow
Steady state
0
20
40
60
80
100
120
0 2 4 6 8 10 12 14 t (h)
Cx (gCDW)
Cs (mM)
Cp (mM)
V (L)
Steady state, Dynamic
cx [g CDW]
cs [mM]
cp [mM]
VL [L]
but Ri ≠ 0
SeXng up the steady state biomass balance for chemostat
VL and cx are constant in Qme
/h
cx = cx,out
Fout = Fin
AssumpIons
Olen true Olen not true
Always measure cx, cx,out, Fin and Fout
Ideal broth ouglow: cx,out = cx
Biomass concentraQon in the inflow zero
Chemostat with ideal broth ou[low and no biomass in the feed
à We control µ!
0 cx
Obtaining other q-‐rates:
Substrate balance: (for qs)
Product balance: (for qp)
The remainder you can do yourselves J Other balances:
Wrap up: Learning about the microorganism: q-‐rates and chemostat
q-‐rates are key performance indicators • Are calculated from proper compound balances and
measurements • Can be controlled in chemostat by us
Never judge microorganism performance based on concentraIons • Use q-‐rates!
See you in the next unit!
