Using deviation variables, derive the response when the inlet concentration increase in a step by 1 kg m-3
A continuous stirred tank reactor is used to break down a chemical in an irreversible first-order reaction. The reactor has a volume of 10 m3 and nominal inlet and outlet concentrations of the chemical of 5 kg m-3 and 0.1 kg m-3 respectively. The flowrate through the reactor is perfectly controlled at 0.5 m3 h -1 .
a) Derive a dynamic model for the reactor.
b) By considering steady state, calculate the reaction rate constant.
c) Using deviation variables, derive the response when the inlet concentration increase in a step by 1 kg m-3
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SHOW WORKING ON HOW TO REACH ANSWER FOR PART C
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![1. Most managed this question well:
a) Derive a dynamic model for the reactor.
The model is the same as the one we found during the lecture.
b) By considering steady state, calculate the reaction rate constant.
Set the accumulation term to zero for steady-state. Sub in the steady-state
values of concentration that were provided and rearrange for k. Pay attention
that the units of k are consistent with the order of reaction.
V
0 ==20 h k =
F
c* (t)
=
1 Cin - Css
0 Css
c) Using deviation variables, derive the response when the inlet concentration
increase in a step by 1 kg m-³.
1
=
Again the solution is the one we obtained during the lecture, although in this case
with specific numerical values for the time constant and steady-state gain.
50
4.9
0.1 × 20
= 2.45 h-¹
[1 - exp(-2.5t)]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fde55ad6b-27eb-44c5-bca1-afdb32f2dfc7%2F39e52fa2-688a-4363-b193-6c65365e428f%2Fvxrowvs_processed.png&w=3840&q=75)
![Deviation Variables
• We use the following notation for these:
CA* (t)= C₁ (t) - CASS
CA,in (t)=CA,in (t) - CA, inss
• At the nominal steady state (at t=0), both of these
are zero
Substituting into our model rate equation we relate
changes in CA* (t) to changes in CA,in* (t)
• Often we deal with a step-change in the input
Deviation Variables
From previously:
Substitute in the deviation variables:
d(C₁* + CA₂)
dt
CA (t)=CA (t) - Cass
CA,in' (t)= CA,in (t) - CA,inss
T
dC₁
dt
dc
dt
dC₁
KCAIR - CA
dCA
dt
Deviation Variables
=KCA,in - CA
2 = K (Cain" + Gina) - CA + Cris
In fact, for a linear system an equivalent cancellation is always
possible, so we can take deviation variables directly
dt
= [²4/1/
T
0
=KCAin - CA
CA"(t)= KCA,in (1-expl
= KCA.in - CA
xp (--))
0=(residence time)
1
K = 1+k@
T=
1=1+kQ²
Cancel since: CA = KCA,ings
0 ==(residence time)
1
1+k@
T=
0
1+k@'
K=
Thus, we have a solution that deals only with the changes in the
input and output of the system](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fde55ad6b-27eb-44c5-bca1-afdb32f2dfc7%2F39e52fa2-688a-4363-b193-6c65365e428f%2Fqgwcyid_processed.png&w=3840&q=75)
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