Solution to problem 4.2
a) A first-order transfer function has the form:
G ( s) = K s +1
Where K is the steady-state gain, and is the time constant. In our case, the gain is:
K =5
b) The time constant is:
= 10
c) Remember the Final Value Theorem (FVT), f

Solution to problem 4.5
We have two coupled differential equations relating two outputs ( y1 , y2 ) with two inputs ( u1 , u2 ):
2 dy1 = 2 y1 3 y2 + 2u1 dt dy2 = 4 y1 6 y2 + 2u1 + 4u2 dt (1) (2)
The objective of the exercise is to obtain the four transfer

28150. Introduction to process control 6. Feedback controllers
Learning objectives
At the end of this lesson you should be able to: Explain advantages and disadvantages of different feedback controllers Explain the effect of PID controller parameters on

Solution to problem 3.5
Before writing T(s), we first need to write down an expression for T(t). To this purpose we need to decompose the temperature profile. The temperature profile can be written as:
T (t ) = 20 S (t ) +
55 55 t S (t ) (t 30 ) S (t 30 )

Solution to problem 3.4
a) Before writing F(s), we first need to write down an expression for f(t). To this purpose we need to decompose the function. The function can be written as:
f (t ) = 5 S (t ) 4 S (t 2 ) 1 S (t 6 )
In the above equation S(t) repre

Laplace transform: Extra homework problems
1) Assume that the following system of two differential equations is given:
dx1 = 2.4048 x1 + 7u dt dx2 = 0.8333 x1 2.2381x2 1.1170u dt
Furthermore y = x2. The initial conditions are x1(0) = x2(0) = 0. a) Demonst

Learning objectives 28150. Introduction to process control 8. PID controller design
At the end of this lesson you should be able to: Design a feedback controller for a given open-loop system Tune the controller parameters for a closed-loop system with fe

Learning objectives 28150. Introduction to process control 7. Dynamics and stability of closed-loop systems
At the end of this lesson you should be able to: Derive closed-loop transfer functions based on a given block diagram Investigate the stability of

Learning objectives 28150. Introduction to process control 9. Control system design based on frequency response analysis (2)
At the end of this lesson you should be able to: Investigate the stability of a closed-loop system using the Bode stability crite

Learning objectives 28150. Introduction to process control 9. Frequency response analysis (1)
At the end of this lesson you should be able to: Derive and plot the frequency response of a transfer function using Bode diagrams or Nyquist diagram
Krist V. G

Solution to problem 3.6
a) X (s ) =
s (s + 1) (s + 2)(s + 3)(s + 4)
We apply partial fraction expansion:
X (s ) =
s (s + 1) = 1+ 2+ 3 (s + 2)(s + 3)(s + 4) s + 2 s + 3 s + 4
We use Heaviside expansion to find values for 1, 2 and 3 Multiplication by (s+2)

Solution to problem 3.8
a) The Laplace transform of
& + 3 x + 2 x = 2 e d & x
0
t
can be obtained as follows: For the right hand side of the above expression, we know that L f t * dt * =
0
t
()
1 F (s ) s
And thus we can write that:
t 1 1 L e d = L e

Solution to problem 11.7
a) Servo problem: The servo problem relates the output with the setpoint, there is no disturbance, i.e. D = 0 .
Using the general procedure to solve the closed-loop system (starting from Y and going backwards) we will find ourselv

Solution to problem 11.6
One of the limitations of proportional-only control is that a steady-state error (or offset) occurs after a set-point change or a sustained disturbance. Section 11.3.1 in Seborg et al. 2004 demonstrates that this offset will occur

Solution to problem 6.6
The first step is to write the overall transfer function G ( s ) in standard form before starting the analysis. Since the two processes (integrating element and first-order element) are operating in parallel, the overall transfer f

Solution to problem 6.2
a) The standard form is:
G ( s) = K ( a s + 1) e s ( 1s + 1)( 2 s + 1)
2( s + 0.5) 5 s e in order to get terms of the form ( s + 1) , ( s + 2)(2 s + 1)
Thus we need to rearrange G ( s) =
as follows:
G(s) = 2
0.5(2 s + 1) e5 s 2(0.5

Solution to problem 5.14
a) 1) From the initial and final steady states we can obtain the gain, knowing that the step change is from 15 psi to 31 psi. 2) From the overshoot we can obtain the damping coefficient . 3) From the period (once the damping coeff

Solution to problem 5.6
a) The overall gain of G is G(s=0) (recall the Final Value Theorem). K = K1 K 2 b) We just need to check the response at t = 1 + 2 = 5 + 3 = 8 :
8 8 y (8) 5e 5 3e 3 = 1 KM 53
= 0.599 0.632
Therefore, the equivalent time constant

Solution to problem 5.2
a) For a step change in input of magnitude M: y (t ) = KM 1 e
(
t
) + y(0)
where M = 1.5 1 = 0.5 Kw
We need to find both K and , and we have information about the initial conditions and two time points. One way of obtaining these p

Solution to problem 4.17 (Corrected formulation!)
System of 2 coupled differential equations:
dX S = max X D X = f1 dt KS + S dS S = max X + D (S f S ) = f 2 dt YX / S K S + S
Input: D States: X, S Disturbance: Sf Remember that the following applies for l

Solution to problem 4.15
From example 4.4, the system equations are:
dh '1 1 1 = q 'i h '1 dt A1 A1R1 dh '2 1 1 h '1 h '2 = dt A2 R1 A2 R2 , , q '1 = q '2 = 1 h '1 R1 1 h '2 R2
In this system, the modeled variables are h '1 and h '2 (liquid heights in tan

Solution to problem 3.9
a) As explained in section 3.4.1 in Seborg et al. 2004, the Final Value Theorem can only be applied if the lim [ sX ( s ) ] exists for all p with Re( p ) 0 . This limit does not exist when pi
s p
cancels the denominator of a fracti

28150. Introduction to process control 5. Second and higher order systems
Learning objectives
At the end of this lesson you should be able to: Explain the response of basic process transfer functions (first-order, second-order, integrating process) to st

28150. Introduction to proces control 5. First-order systems
Learning objectives
At the end of this lesson you should be able to: Explain the response of first-order systems to standard inputs
Krist V. Gernaey
28150
21 September 2009
Outline
Standard pr

5. ICAS Tutorial Basics
5.3. Use of Utility Toolbox (Calculation of SLE & VLE phase diagrams)
5.3.1 Compound selection & property model selection (general - all problems)
I. Draw a stream and then select compounds by clicking on the compounds button.
II.

C28420: Lecture 2 Tutorial - Use of ProCAMD and Database for solvent selection The exercises during the tutorial will cover searching for solvents through the database search engine in ICAS and through generation of candidates through ProCAMD in ICAS. 1.

SOLVENT BASED SEPARATION Solvent Selection
Rafiqul Gani, Peter M. Harper & Martin Hostrup CAPEC, Department of Chemical Engineering, Technical University of Denmark, DK-2800 Lyngby, Denmark.
The objective of this article is to highlight the important issu