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Sagot :
Answer:
The rate of cell metabolism is limited by mass transfer since the value of maximum cell concentration obtained (38 g/l) is lower than 50 g l-1, the value planed.
Explanation:
Data
kLa = 0.17/s
Solubility of oxygen = 8 × 10^-3 kg / m^3
The maximum specific oxygen uptake rate = 4 mmol O2 / g h.
Concentration of oxygen = 0.5 × 10^-3 kg/ m^3
**The maximum cell density = 50 g/l
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The calculated maximum cell concentration:
xmax= kLa · CAL*/ qo
CAL* is the solubility of oxygen in the broth and qo is the specific oxygen uptake rate
Replacing the data given
xmax= ( 0.17/s ) · (8 × 10^-3 kg / m^3) / 4 mmol O2 / g h
4 mmol O2 / g h to kg O2/ g s
[tex]4 \frac{mmol}{gh} \frac{1 gmol}{1000mmol}\frac{1h}{3600s}\frac{32 g}{gmol} \frac{1Kg}{1000g}[/tex]
= 3.56 x 10^-3 kg O2/ g s
So then,
xmax= ( 0.17/s ) · (8 × 10^-3 kg / m^3) / 3.56 x 10^-3 o kg O2/ g s
xmax= 3. 8 x 10^4 g/ m^3 = 38 g/l
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