2. Exercise 0
• Assume dynamic evolution
of volume
• Perform a degree of freedom
Analysis
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3. Course Questions
1. What is the difference between process
design and control?
2. What is the difference between feedback and
feedforward control?
3. What is the difference between manipulated
controlled and disturbance variables?
4. What is the reason behind a LaPlace model?
5. What is the reason for employing TFs?
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4. Course Answers
1. Dynamics: Represents the key equations of
the process and its response to changes in
inputs . Control: Deals with the manipulation
of the studied process to suit certain control
objectives: Speed, response, etc
2. Feedback, measures change in output
variable and changes input accordingly.
Feedforward: manipulates input
independently of output
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5. Course Answers
3. Controlled Variables: Variables that need to
be at their pre-defined set point.
Manipulated Variables: Input variables that
are changed so at to render controlled
variables equal to their set point.
Disturbance variables : Input variables that
are not manipulated but direct impact the
value of controlled variables. 𝑤1, 𝑤2, 𝑤,
𝑥1, 𝑥2
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6. Course Answers
4. The LaPlace Model is used in order to
simplify the resolution of Differential
Equations encountered in chemical process
dynamics and control
5. The TF model is employed in order to
standardize the study of dynamic processes,
namely because many systems exhibit similar
behaviors.
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7. Degrees of Freedom Analysis
Structured Approach
1. List all quantities in the model that are known constants (or
parameters that can be specified) on the basis of equipment
dimensions, known physical properties, and so on.
2. Determine the number of equations NE and the number of
process variables, NV. Note that time t is not considered to be
a process variable, because it is neither an input nor an
output.
3. Calculate the number of degrees of freedom, 𝑁 𝐹 (Nv-NE)
4. Identify the NE output variables (include dependent variables
in the ODEs) obtained by solving the process model.
5. Identify the 𝑁 𝐹 input variables that must be specified as
either disturbance variables (DVs) or manipulated variables
(MVs). 7Rami Bechara3/29/2019
8. Equations and variables
• First we need to set the equations
• Global balance
• 𝜌
𝑑𝑉
𝑑𝑡
= 𝑤1 + 𝑤2 − 𝑤
• Component Balance
• 𝜌
𝑑𝑉𝑥
𝑑𝑡
= 𝑤1 𝑥1 + 𝑤2 𝑥2 − 𝑤𝑥
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10. Exercise 1
• y(0) and dy/dt (0) both equal to zero
• What are the functions of the time (e.g.,
e−t/τ) in the solution for each of the following
cases?
• (a) u(t) = 2e−2t.
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14. Exercise 2 Jacketed Vessel
• Both the tank contents and the jacket contents are well mixed and have
significant thermal capacitances.
• Imagine direct heat transfer between the terms
• Do a DOF analysis on the system
• Derive transfer function
𝑻
𝑸 𝑱
• Presentation and
assumptions
• The volume of liquid in the
tank V and the volume of
coolant in the jacket VJ
remain constant. Volumetric
flow rate qF, TF, Ti, TJ are
constant, but qJ varies with
time. Heat losses from the
jacketed vessel are negligible.
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19. Exercise 3
• Liquid storage system
• Normal operating conditions are q1 = 10 ft3∕min, q2 =
5 ft3∕min, h = 4 ft.
• The tank is 6 ft in diameter, and the density of each
stream is 60 lb/ft3.
• Assume that the exit flow rate q is related to height h
as 𝑞 = 𝑐 𝑣 ℎ
• Suppose that a change in q1 occurs.
• The system and the change are highlighted afterwards
• (a) What is the TF relating H′ to Q′1?
• (b) Derive an expression for h(t) for this input change.
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24. Linearization method
• Suppose a nonlinear dynamic model derived
from first principles y output
u input
• Linear approximation of this equation can be
obtained by using a Taylor series expansion
and truncating after the first-order terms.
• Final expression
Deviation Variables
24Rami Bechara3/29/2019
28. Exercise 5
• A step change from 15 to 31 psi in actual
pressure results in the measured response
from a pressure-indicating element shown
below
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29. Question
• Assuming second-order dynamics, calculate all
important parameters and write an approximate
transfer function in the form
• where R′ is the instrument output deviation
(mm), P′ is the actual pressure deviation (psi).
• (b) Write an equivalent differential equation
model in term of actual (not deviation) variables
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30. Step Response characteristics of a
2nd -order underdamped process
• Important terms
• X-axis: time relate terms
• Rise Time (tr) is the time the process output
takes to first reach the new steady-state
value.
• Time to First Peak. (tp) is the time required
for the
• output to reach its first maximum value.
• Settling Time. ts is the time required for the
process output to reach and remain inside a
band whose width is equal to ±5% of the
total change in y (±1% sometimes used).
30
• Period of Oscillation. P time between 2 successive peaks or valleys of the response.
• Y-axis terms
• Overshoot. OS = a/b (% overshoot is 100 a/b).
• Decay Ratio.DR = c/a (where c is the height of the second peak).
• True for the step response of any underdamped process.
• If no overshoot: rise time definition: time to go from 10% to 90% of steady-state
response Rami Bechara3/29/2019
31. Expressions of terms for
2nd order underdamped processes
31
• Expressions for terms in case of 2nd order underdamped
process:
• For an underdamped second-order transfer function,
Equations and figures can be used to obtain estimates
of ζ and τ based on step response characteristicsRami Bechara3/29/2019
34. Exercise 6
• A process consists of an
integrating element
operating in parallel
with a 1st-order element
• (a) What is the order of the overall transfer
function, G(s) =Y(s)/U(s)?
• (b) What is the gain of G(s)?
• (c) What are the poles of G(s)?Where are they
located in the complex s-plane?
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35. Questions
• (d) What are the zeros of G(s)? Where are they
located? Under what condition(s) will one or
more of the zeros be located in the right-half s-
plane?
• (e) For what conditions, will this process exhibit
both a negative gain and a right-half plane zero?
• (f) For any input change, what functions of time
(response modes) will be included in the
response, y(t)?
• (g) Is the output bounded for any bounded input
change, for example, u(t) = M?
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36. Solution
• (a) 𝐺 =
𝐾1
𝑠
+
𝐾2
𝜏𝑠+1
=
𝐾1 𝜏𝑠+1 +𝐾2 𝑠
𝑠(𝜏𝑠+1)
• G=
𝐾1 𝜏+𝐾2 𝑠+𝐾1
𝑠(𝜏𝑠+1)
= 𝐾1
𝜏+
𝐾2
𝐾1
𝑠+1
𝑠(𝜏𝑠+1)
• G= 𝐾1
𝜏 𝑎 𝑠+1
𝑠(𝜏𝑠+1)
; 𝜏 𝑎 = 𝜏 +
𝐾2
𝐾1
• Gain: 𝐾1
• Poles of G: 0 and −
1
𝜏
located on the origin and
real axis (left –half)
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37. Solution
• There is one zero: −
1
𝜏 𝑎
• Right half 𝜏 𝑎<0 𝜏 +
𝐾2
𝐾1
<0
𝐾2
𝐾1
<- 𝜏
• (-) Gain 𝐾1<0, right-half plane zero 𝐾2>-𝐾1 𝜏
• 𝜏 𝑎, and 𝜏 will appear
• Bounded: 𝑈 =
𝑀
𝑠
, 𝑌 = 𝐾1 𝑀
𝜏 𝑎 𝑠+1
𝑠2
(𝜏𝑠+1)
• No need for solving, use the Final Value theorem
• Lim y (t∞)= lim
𝐾1 𝑀
𝑠
(s0), it is unbounded
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38. Exerice 7
• Derive an approximate First order plus time
delay model for this function
• Make use of Skogestad’s rule
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39. Skogestad’s “Half Rule”
Approximation
• Approximation method for higher-order models that
contain multiple time constants.
• He approximates the largest neglected time constant in the
denominator in the following manner.
• One-half of its value is added to the existing time delay (if
any), and the other half is added to the smallest retained
time constant.
• Time constants smaller than the largest neglected time
constant are approximated as time delays, along with plane
zero according to previous equations.
• The motivation for this “half rule” is to derive approximate
low-order models more appropriate for control system
design.
39Rami Bechara3/29/2019
40. Exercise 8
• Given the table below, Develop a transfer
function model of the second order plus time
delay form to a response of step input of
amplitude M=4
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41. Second order functions
Approximating 𝜏 and 𝜁
• Smith’s method requires the
times (with apparent time
delay removed) at which the
normalized response reaches
20% and 60%, respectively
• Using the figure, the ratio of
t20/t60 gives the value of ζ.
• An estimate of τ can be
obtained from the plot of t60/τ
vs. t20/t60.
41Rami Bechara3/29/2019
42. • 𝜃 defined as the time delay, thus intuitively,
𝜃 = 2𝑠, KM is the last term: K=3/4
• 𝑡20 = 4 − 𝜃 = 2s, 𝑡60 = 7 − 𝜃 = 5𝑠
• t20/t60=2/5=0.4
• t60/τ=2, τ=2.5
• 𝜁 = 1.05
• 𝐺 =
0.75𝑒−2𝑠
6.25𝑠2+5.25𝑠+1
𝜃
0%
20%
40%
60%
80%
100%
120%
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
NormalizedResponsey
Time t
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43. • 𝐺 =
0.75𝑒−2𝑠
6.25𝑠2+5.25𝑠+1
, b1=-0.54, b2=-0.29
• 𝐺 =
0.75𝑒−2𝑠
(1.82𝑠+1)(3.42𝑠+1)
→ 𝑌 =
3𝑒−2𝑠
𝑠(1.25𝑠+1)(5𝑠+1)
• 𝑦 = (1 +
1
3.75
(5𝑒−
𝑡
5 − 1.25 𝑒−
𝑡
1.25)) ∗ S t − 2
• Did not work very well
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44. Linear Regression
• Also does not work very well
0
0.5
1
1.5
2
2.5
3
3.5
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Responey
Time t
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