Answer:
Explanation:
Given that:
The dilution rate D = 0.28 h⁻¹
The concentration of lactose in the feed ![S_o = 2.0 \ g/L](https://tex.z-dn.net/?f=S_o%20%3D%202.0%20%5C%20g%2FL)
The effluent S = 0.10 g/L
Also;
![Y_{X/S} = 0.45 g\ X/g \ S , \\ \\ Y_{X/O2 }= 0.25 g \ X/g \ O2, \\ \\](https://tex.z-dn.net/?f=Y_%7BX%2FS%7D%20%3D%200.45%20g%5C%20%20X%2Fg%20%20%5C%20S%20%2C%20%5C%5C%20%5C%5C%20%20Y_%7BX%2FO2%20%20%7D%3D%200.25%20g%20%20%5C%20X%2Fg%20%20%5C%20O2%2C%20%20%5C%5C%20%5C%5C)
Saturation C* = 8 mg/l
To calculate the steady-state biomass, we use the formula:
![X = Y_{X/S}(S_o-S_e) \\ \\ X = 0.45(2.0 -0.10) \ g/L \\ \\ X= 0.45 (1.9) \ g/L \\ \\ X = 0.855\ g/L \\ \\ X = 855 \ mg/L](https://tex.z-dn.net/?f=X%20%3D%20Y_%7BX%2FS%7D%28S_o-S_e%29%20%5C%5C%20%5C%5C%20X%20%3D%200.45%282.0%20-0.10%29%20%5C%20g%2FL%20%5C%5C%20%5C%5C%20X%3D%200.45%20%281.9%29%20%5C%20g%2FL%20%5C%5C%20%5C%5C%20%20X%20%3D%200.855%5C%20g%2FL%20%5C%5C%20%5C%5C%20%20X%20%3D%20855%20%5C%20mg%2FL)
The biomass is 0.855 g/L
For a steady-state condition, the oxygen uptake rate can be illustrated by using the formula:
![q_{o_2}X =\dfrac{\mu_X}{Y_{X/O_2}}](https://tex.z-dn.net/?f=q_%7Bo_2%7DX%20%3D%5Cdfrac%7B%5Cmu_X%7D%7BY_%7BX%2FO_2%7D%7D)
where;
dilution rate (D)
Thus, the steady-state can be expressed as:
![q_{o_2}X =\dfrac{D}{Y_{X/O_2}}](https://tex.z-dn.net/?f=q_%7Bo_2%7DX%20%3D%5Cdfrac%7BD%7D%7BY_%7BX%2FO_2%7D%7D)
![q_{o_2}X =\dfrac{0.28}{0.25}](https://tex.z-dn.net/?f=q_%7Bo_2%7DX%20%3D%5Cdfrac%7B0.28%7D%7B0.25%7D)
![q_{o_2}X =1.12 \ h^{-1}](https://tex.z-dn.net/?f=q_%7Bo_2%7DX%20%3D1.12%20%5C%20h%5E%7B-1%7D)
The specific rate of oxygen consumption ![q_{o_2}X =1.12 \ h^{-1}](https://tex.z-dn.net/?f=q_%7Bo_2%7DX%20%3D1.12%20%5C%20h%5E%7B-1%7D)
b)
In the fermentation medium, if the desired DO concentration
= 2 mg/L
Here, the oxygen transfer is regarded as the rate-limiting step.
As such, the oxygen transfer rate(OTR) is equivalent to the oxygen uptake rate.
In this scenario, let's determine the oxygen transfer coefficient
by using the formula:
![OTR = K_{La}(C^* - C_L)](https://tex.z-dn.net/?f=OTR%20%3D%20%20K_%7BLa%7D%28C%5E%2A%20-%20C_L%29)
where;
= coefficient of oxygen transfer
C* = saturation
Since ![OTR = q_{O_2}X](https://tex.z-dn.net/?f=OTR%20%3D%20q_%7BO_2%7DX)
![q_{o2}X = K_{La}(C^*-C_L) \\\\ (1.12 )(855) = K_{La}(8-2) \\ \\ 957.6 = K_La (6) \\ \\ K_{La}= \dfrac{957.6}{6} \\ \\ K_{La} = 159.6 \ h^{-1}](https://tex.z-dn.net/?f=q_%7Bo2%7DX%20%3D%20K_%7BLa%7D%28C%5E%2A-C_L%29%20%5C%5C%5C%5C%20%281.12%20%29%28855%29%20%3D%20K_%7BLa%7D%288-2%29%20%5C%5C%20%5C%5C%20%20957.6%20%3D%20K_La%20%286%29%20%5C%5C%20%5C%5C%20K_%7BLa%7D%3D%20%5Cdfrac%7B957.6%7D%7B6%7D%20%5C%5C%20%5C%5C%20K_%7BLa%7D%20%3D%20159.6%20%5C%20h%5E%7B-1%7D)
Thus, the oxygen transfer coefficient
= 159.6 h⁻¹