Homework Answers
We will assume that the given cell growth is following Monod equation and as there is no endogenous material so the Kd=0, μ=μnet,
μ=μmaxCs/Ks+Cs
As the system is a chemostat system μg=D,
Therefore D=μmaxCs/Ks+Cs
1/D=Ks+Cs/μmaxCs
1/D=(Ks/μmax)(1/Cs)+(1/μmax)
Now try to plot the graph of 1/D against 1/Cs,
|
1/D hr |
1/Cs g-1L |
|
11.905 |
18.519 |
|
10 |
12.658 |
|
6.250 |
7.246 |
|
5.051 |
5.376 |
|
4.132 |
4.425 |
1/μmax=y-intercept=2
μmax=0.500 hr-1
Ks/μmax=slope=11.8-4.6/17.2-4.6=0.5714 gL-1.hr
Ks=0.5714 gL-1.hrx0.5 hr-1=0.2857 gL-1
rx=μg Cx= (μmaxCs/Ks+Cs)Cx
rx=(0.500Cs/0.2857+Cs)Cx
(B) To find the rate equation for the ethanol formation, assume its a growth associated product
(1/Cx)(dP/dt)=Yp/xμgrp=dP/dt=Yp/xμgCx
rp=Yp/x(μmaxCs/Ks+Cs)Cs
Now from the slope of graph of Cp against Cx, obtain Yp/x
From the graph we can see Yp/x=deltaP/deltax=slope
Yp/x=8.2-2.5/2.2-0.6=3.5625 g P.(gX)-1
rp= 3.5625(0.500Cs/0.2857+Cs)Cx
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