Answer:
a) Cell concentration when the dilution rate is one-half of the maximum is 0.598g cell/L
b) the substrate concentration when the dilution rate is 0.8 [tex]D_{max}[/tex] is 5.2g/l
c) the maximum dilution rate is : 0.41 h⁻¹
d) the cell productivity at 0.8 [tex]D_{max}[/tex] is 2.40g cell/L
Explanation:
Given data :
Mass doubling time of Pseudomonas sp. = 2.4 hr
Saturation constant = 1.3 g/L
Cell yield on acetate = 0.46g cell/g acetate
We are to find;
a. Cell concentration when the dilution rate is one-half of the maximum.
Here, cell yield =amount of cell produced / amount of substrate consumed.
[S] at 0.5D max is determined using the Monod's equation.
Using the formula :
[tex]D = \frac {D_{max}[S] }{ks+[S]}[/tex]
, where D is the dilution rate,
[S] is substrate concentration; &
Ks is the saturation constant.
By replacing the values, we get :
[tex]0.5 = \frac{S}{1.3+[S]}[/tex]
[tex]\\\0.65=0.5[S][/tex]
[S]=1.3g/L
The cell concentration at 0.5Dmax= cell yield x substrate consumed at 0.5Dmax.
=0.46×1.3
= 0.598g cell/L
b)
Substrate concentration when the dilution rate is 0.8 [tex]D_{max}[/tex] is calculated as:
[tex]D = \frac {D_{max}[S] }{ks+[S]}[/tex]
0.8=[S]/1.3+[S]
1.04+0.8[S]=[S]
[S]= 5.2g/L
Therefore , the substrate concentration when the dilution rate is 0.8 [tex]D_{max}[/tex] is 5.2g/l
c)
Maximum dilution rate is calculated using the expression [tex]D_{max} = \frac{1}{time}[/tex]
=1/2.4
=0.41 h⁻¹
So, the maximum dilution rate is : 0.41 h⁻¹
d)
The cell productivity at 0.8 [tex]D_{max}[/tex] can be calculated by multiplying the amount of the cell yield with the amount of the substrate consumed at 0.8[tex]D_{max}[/tex]
Cell yield = [tex]\frac {cell \ productivity \ at \ 0.8Dmax}{amount \ of \ substrate\ consumed \at\ 0.8 \D_{max}}[/tex]
Cell productivity at 0.8 [tex]D_{max}[/tex] = 0.46 × 5.2
=2.40g cell/L
Therefore, the cell productivity at 0.8 [tex]D_{max}[/tex] is 2.40g cell/L
