Anthony J. Wavrin Week 4
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Figures
Standard Conditions
Parameter |
Parameter Value
|
q |
0.15
|
un |
120
|
r |
1.0
|
K |
5
|
v |
0.5
|
uc |
60
|
Changed q
Parameter |
Parameter Value
|
q |
0.30
|
un |
120
|
r |
1.0
|
K |
5
|
v |
0.5
|
uc |
60
|
Changed un
Parameter |
Parameter Value
|
q |
0.15
|
un |
240
|
r |
1.0
|
K |
5
|
v |
0.5
|
uc |
60
|
Changed r
Parameter |
Parameter Value
|
q |
0.15
|
un |
120
|
r |
2.0
|
K |
5
|
v |
0.5
|
uc |
60
|
Changed K
Parameter |
Parameter Value
|
q |
0.15
|
un |
120
|
r |
1.0
|
K |
10
|
v |
0.5
|
uc |
60
|
Changed v
Parameter |
Parameter Value
|
q |
0.15
|
un |
120
|
r |
1.0
|
K |
5
|
v |
1.0
|
uc |
60
|
Changed uc
Parameter |
Parameter Value
|
q |
0.15
|
un |
120
|
r |
1.0
|
K |
5
|
v |
0.5
|
uc |
120
|
Questions
- Find steady states by replacing the d/dt’s with 0. This will involve a bit of algebra.
- How do the steady states depend on the feed rates? This will also involve a bit of algebra.
- This question is dependent on where the feed rate constants are in the steady states that were supposed to be derived in the first question. Since I was unable to successfully derive those states, I will be using Paul Magnano's derived steady states to answer this question. For the steady state of nitrogen, it is clear that as the feed rates increase, so will the steady state. This is because the feed rate is positive and in the numerator. For the steady state of carbon, it is dependent on both the feed rate of nitrogen and carbon, thus while it will clearly effect the steady state, it is not possible to predict trends as it depends on two feed rates.
- Suppose we replace the product term, how might that affect the steady state? Or the dynamics?
- I do not believe that this substituting this will affect the steady state or the dynamics. This is because all that is being done is combining two constants into one combined constant. Thus, it should not have any bearing on the outcome in comparison with the one we used.
Week 4 Assignment
Links