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aev [14]
2 years ago
11

Serratia marcescens bacteria are used for the production of threonine. The maximum specific oxygen uptake rate of S. marcescens

in batch culture is 4 mmol O2 g-1 h-1. It is planned to operate the fermenter to achieve a maximum cell density of 50 g l-1. At the fermentation temperature and pressure, the solubility of oxygen in the culture liquid is 8 × 10^-3 kg m^-3. The concentration of oxygen in the culture liquid is 0.5 × 10^-3 kg m^-3. At a particular stirrer speed, kLa is 0.17 s^-1.
Required:
Under these conditions, will the rate of cell metabolism be limited by mass transfer?
Engineering
1 answer:
Sati [7]2 years ago
5 0

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

<u>kLa</u> = 0.17/s

<u>Solubility of oxygen</u> =  8 × 10^-3 kg / m^3

<u>The maximum specific oxygen uptake rate </u>= 4 mmol O2 / g h.

<u>Concentration of oxygen</u> =  0.5 × 10^-3 kg/ m^3

<u>**The maximum cell density</u> = 50 g/l

___________________

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

4 \frac{mmol}{gh} \frac{1 gmol}{1000mmol}\frac{1h}{3600s}\frac{32 g}{gmol} \frac{1Kg}{1000g}

= 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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Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can b
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Answer:

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

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Explanation:

Given that;

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5

h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

conditions for plates of length L = 0.1 m and 1 m

Now

Average heat transfer coefficient is expressed as;

h⁻ = 1/L ₀∫^L hxdx

so for Laminar flow

h_lam(x)= 1.74 . Kx^-0.5  W/m^1.5

from the expression

h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5   dx

= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L

= 1.74k/L = [ (x^0.5)/0.5)]⁰^L

= 1.74K × L^0.5 / L × 0.5

h⁻_lam= 3.48KL^-0.5

For turbulent flow

h_turb(x)= 3.98. Kx^-0.2 W/m^1.8

form the expression

1/L ₀∫^L 3.98 . Kx^-0.2   dx

= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L

= (3.98K/L) × (L^0.8 / 0.8)

h⁻_turb = 4.975KL^-0.2

Now at L = 0.1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(0.1)^-0.5  W/m^1.5

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(1)^-0.5  W/m^1.5

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2

h⁻_turb = 4.975K   W/m^1.8

Therefore

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

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