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Chapter 2
ERT 210/4
Process Control & Dynamics
DYNAMIC BEHAVIOR OF
PROCESSES :
Theoretical Models of Chemical
Processes
MISS. RAHIMAH BINTI OTHMAN
(Email: [email protected])
1
2
Chapter 2
Chapter 2
Course Outcome 1 (CO1)
Ability to derive and develop theoretical model of chemical
processes, dynamic behavior of first and second order processes
chemical and dynamic response characteristics of more complicated
processes.
1. Introduction Concepts
•
•
Introduction to Process Control:
Theoretical Models of Chemical Processes
Apply and derive unsteady-state models (dynamic model)
of chemical processes from physical and chemical principles.
2. Laplace Transform
3. Transfer Function Models
4. Dynamic Behavior of First-order and Second-order
Processes
5. Dynamic Response Characteristics of More Complicated
Processes
3
Chapter 2
Process Dynamics
a) Refers to unsteady-state or transient behavior.
b) Steady-state vs. unsteady-state behavior;
i.Steady state: variables do not change with time
c) Continuous processes : Examples of transient behavior:
i. Start up & shutdown
ii. Grade changes
iii. Major disturbance;
e.g: refinery during stormy or hurricane conditions
iv. Equipment or instrument failure (e.g., pump failure)
d) Batch processes
i. Inherently unsteady-state operation
ii. Example: Batch reactor
1.Composition changes with time
2.Other variables such as temperature could be
constant.
4
Process Control
Chapter 2
1. Objective:
Enables the process to be maintained at the desired operation
conditions, safely and efficiently, while satisfying environmental
and product quality requirements.
2. Control Terminology:
Controlled variables (CVs)
- these are the variables which quantify the performance or quality of
the final product, which are also called output variables (Set point).
Manipulated variables (MVs)
- these input variables are adjusted dynamically to keep the
controlled variables at their set-points.
Disturbance variables (DVs)
- these are also called "load" variables and represent input variables
that can cause the controlled variables to deviate from their
respective set points (Cannot be manipulated).
5
Development of Dynamic Models
Chapter 2
Illustrative Example: A Blending Process
Figure 2.1. Stirred-tank blending process.
An unsteady-state mass balance for the blending system:
rate of accumulation   rate of   rate of 




 of mass in the tank  mass in  mass out 
(2-1)
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or
d Vρ 
dt
 w1  w2  w
(2-2)
Chapter 2
where w1, w2, and w are mass flow rates.
The unsteady-state component balance is:
d Vρx 
dt
 w1x1  w2 x2  wx
(2-3)
The corresponding steady-state model was derived in Ch. 1 (cf.
Eqs. 1-1 and 1-2).
0  w1  w2  w
(2-4)
0  w1x1  w2 x2  wx
(2-5)
7
Conservation Laws
Theoretical models of chemical processes are based on
conservation laws.
Chapter 2
Conservation of Mass
 rate of mass  rate of mass  rate of mass 



 (2-6)
in
out
accumulation  
 

Conservation of Component i
rate of component i  rate of component i 



accumulation
in

 

rate of component i  rate of component i 



out
produced

 

(2-7)
8
Conservation of Energy
Chapter 2
The general law of energy conservation is also called the First
Law of Thermodynamics. It can be expressed as:
rate of energy  rate of energy in  rate of energy out 




accumulation
by
convection
by
convection

 
 

net rate of work
net rate of heat addition  


 

  to the system from   performed on the system 

  by the surroundings 
the
surroundings

 

(2-8)
The total energy of a thermodynamic system, Utot, is the sum of its
internal energy, kinetic energy, and potential energy:
U tot  U int  U KE  U PE
(2-9)
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Chapter 2
For the processes and examples considered in this book, it
is appropriate to make two assumptions:
1. Changes in potential energy and kinetic energy can be neglected
because they are small in comparison with changes in internal
energy.
2. The net rate of work can be neglected because it is small compared
to the rates of heat transfer and convection.
For these reasonable assumptions, the energy balance in
Eq. 2-8 can be written as;


dU int
  wH  Q
dt
(2-10)
  denotes the difference
between outlet and inlet
conditions of the flowing
the system
streams; therefore
H  enthalpy per unit mass
-Δ wH = rate of enthalpy of the inlet
w  mass flow rate
stream(s) - the enthalpy
Q  rate of heat transfer to the system
of the outlet stream(s)
U int  the internal energy of


10
The analogous equation for molar quantities is,


Chapter 2
dU int
  wH  Q
(2-11)
dt
where H is the enthalpy per mole and w is the molar flow rate.
In order to derive dynamic models of processes from the general
energy balances in Eqs. 2-10 and 2-11, expressions for Uint and Ĥ
or H are required, which can be derived from thermodynamics.
The Blending Process Revisited
For constant  , Eqs. 2-2 and 2-3 become:
dV

 w1  w2  w
dt
 d Vx 
 w1x1  w2 x2  wx
dt
(2-12)
(2-13)
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Equation 2-13 can be simplified by expanding the accumulation
term using the “chain rule” for differentiation of a product:

d Vx 
dt
 V
dx
dV
 x
dt
dt
(2-14)
Chapter 2
Substitution of (2-14) into (2-13) gives:
V
dx
dV
 x
 w1x1  w2 x2  wx
dt
dt
(2-15)
Substitution of the mass balance in (2-12) for  dV/dt in (2-15)
gives: dx
V
 x  w1  w2  w   w1x1  w2 x2  wx
(2-16)
dt
After canceling common terms and rearranging (2-12) and (2-16),
a more convenient model form is obtained:
dV 1
  w1  w2  w 
dt

w2
dx w1

x

x

1 
 x2  x 
dt V 
V
(2-17)
(2-18)
12
Chapter 2
Stirred-Tank Heating Process
Figure 2.2 Stirred-tank heating process with constant holdup, V.
13
Stirred-Tank Heating Process (cont’d.)
Case 1: Constant Holdup, V
Chapter 2
Assumptions:
1. Perfect mixing; thus, the exit temperature T is also the
temperature of the tank contents.
2. The liquid holdup V is constant because the inlet and outlet
flow rates are equal.
3. The density  and heat capacity C of the liquid are assumed to
be constant. Thus, their temperature dependence is neglected.
4. Heat losses are negligible.
14
Model Development
Chapter 2
1. For a pure liquid at low or moderate pressures, the internal
energy is approximately equal to the enthalpy, Uint  H , and
H depends only on temperature.
2. Consequently, in the subsequent development, we assume
that Uint = H and Uˆ int  Hˆ where the caret (^) means per unit
mass. As shown in Appendix B, a differential change in
temperature, dT, produces a corresponding change in the
internal energy per unit mass,dUˆ int ,
dUˆ int  dHˆ  CdT
(2-29)
where C is the constant pressure heat capacity (assumed to be
constant). The total internal energy of the liquid in the tank is:
Uint  VUˆ int
(2-30)
15
Model Development (cont’d.)
Chapter 2
An expression for the rate of internal energy accumulation can be
derived from Eqs. (2-29) and (2-30):
dU int
dT
 VC
dt
dt
(2-31)
Note that this term appears in the general energy balance of
Eq. 2-10.
Suppose that the liquid in the tank is at a temperature T and has an
enthalpy, Ĥ . Integrating Eq. 2-29 from a reference temperature
Tref to T gives,

Hˆ  Hˆ ref  C T  Tref

(2-32)
where Hˆ ref is the value of Ĥ at Tref. Without loss of generality, we
assume that Hˆ ref  0 (see Appendix B). Thus, (2-32) can be
written as:

Hˆ  C T  Tref

(2-33)
16
Model Development (cont’d.)
For the inlet stream

Chapter 2
Hˆ i  C Ti  Tref

(2-34)
Substituting (2-33) and (2-34) into the convection term of (2-10)
gives:






 wHˆ  w C Ti  Tref   w C T  Tref 




(2-35)
Finally, substitution of (2-31) and (2-35) into (2-10)
dT
V C
 wC Ti  T   Q
dt
(2-36)
17
Chapter 2
Degrees of Freedom Analysis for the Stirred-Tank
Model:
3 parameters:
V , ,C
4 variables:
T , Ti , w, Q
1 equation:
Eq. 2-36
Thus the degrees of freedom are NF = 4 – 1 = 3. The process
variables are classified as:
1 output variable:
T
3 input variables:
Ti, w, Q
For temperature control purposes, it is reasonable to classify the
three inputs as:
2 disturbance variables:
Ti, w
1 manipulated variable:
Q
18
Stirred-Tank Heating Process (cont’d.)
Case 2: Variable Holdup, V
Chapter 2
Tank holdup vary with time, based on four assumptions;
Assumptions:
1. Perfect mixing; thus, the exit temperature T is also the
temperature of the tank contents.
2. The liquid holdup V is constant because the inlet and outlet
flow rates are equal.
3. The density  and heat capacity C of the liquid are assumed to
be constant. Thus, their temperature dependence is neglected.
4. Heat losses are negligible.
19
Model Development
The overall mass balance;
Chapter 2
d (V )
 wi  w
dt
(2-37)
The energy balance for the current stirred-tank heating system
can be derived from Eq. 2-10 in analogy with the derivation of
Eq. 2-36. We again assume that Uint = H for the liquid in the tank.
Thus, for constant ;
d (U int ) dH
d ( VHˆ )
d (VHˆ )



dt
dt
dt
dt
(2-38)
From the definition of and  ( wHˆ ) Eq. 2-33 and 2-34, it follows
that;
(2-39)
 ( wHˆ )  wi Hˆ i  wHˆ  wiC (Ti  Tref )  wC (T  Tref )
20
Model Development (cont’d.)
Chapter 2
where wi and w are the mass flow rates of the inlet and outlet
streams, respectively. Substituting (2-38) and (2-39) into (2-10)
gives;
d (VHˆ )

 wiC (Ti  Tref )  wC (T  Tref )  Q
dt
(2-40)
Next we simplify the dynamic model. Because ρ is constant,
Eq (2-37) can be written as;
(2-41)

dV
 wi  w
dt
The ‘chain rule’ can be applied to expand the left side of (2-40)
for constant C and ;
d (VHˆ )
dHˆ
dV
ˆ

 V
 H
dt
dt
dt
(2-42)
21
Model Development (cont’d.)
Chapter 2
From Eq. 2-29 or 2-33, it follows that dHˆ /dt  CdT/dt .
Substituting this expression and Eq. 2-33 and 2-41 into Eq. 2-42
gives;
d (VHˆ )
dT

 C (T  Tref )( wi  w)  CV
dt
dt
(2-43)
Substituting (2-43) into (2-40) and rearranging give;
dT
C (T  Tref )( wi  w)  CV
 wiC (Ti  Tref )  wC (T  Tref )  Q
dt
(2-44)
Rearranging (2-41) and (2-44) provides a simpler form for the
dynamic model;
dV
  ( wi  w)
dt
dT
w
Q
 i (Ti  T ) 
dt V
CV
(2-45)
(2-46)
22
Chapter 2
Liquid Storage Systems
Figure 2.5 A liquid-level storage process.
23
Model Development
Chapter 2
A typical liquid storage process is shown in Fig. 2.5, where qi and
q are volumetric flow rates. A mass balance yields:
d ( V )
 qi  q
dt
(2-53)
Assume that liquid density  is constant and the tank is
cylinderical with cross-sectional area, A. Then the volume of
liquid in the tank can be expressed as V = Ah, h is the liquid level
(or head). Thus, (2.53) becomes;
A
dh
 qi  q
dt
(2-54)
24
Chapter 2
Continuous Stirred Tank Reactor (CSTR)
Fig. 2.6. Schematic diagram of a CSTR.
25
Chapter 2
CSTR: Model Development
Assumptions:
1. Single, irreversible reaction, A → B.
2. Perfect mixing.
3. The liquid volume V is kept constant by an overflow line.
4. The mass densities of the feed and product streams are equal
and constant. They are denoted by .
5. Heat losses are negligible.
6. The reaction rate for the disappearance of A, r, is given by,
r = kcA
(2-62)
where r  =  moles of A reacted per unit time, per unit volume, cA
is the concentration of A (moles per unit volume), and k is the rate
constant (units of reciprocal time).
7. The rate constant has an Arrhenius temperature dependence:
k = k0 exp(-E/RT )
(2-63)
where k0 is the frequency factor, E is the activation energy,
and R is the the gas constant.
26
CSTR: Model Development (continued)
• Unsteady-state mass balance
Chapter 2
(2-64)
Because  and V are constant,
(2-65)
• Unsteady-state component balance
(2-66)
27
Assumptions for the Unsteady-state Energy Balance:
8.
9.
Chapter 2
10.
11.
12.
28
CSTR Model: Some Extensions
• How would the dynamic model change for:
Chapter 2
1. Multiple reactions (e.g., A → B → C) ?
2. Different kinetics, e.g., 2nd order reaction?
3. Significant thermal capacity of the coolant liquid?
4. Liquid volume V is not constant (e.g., no overflow line)?
5. Heat losses are not negligible?
6. Perfect mixing cannot be assumed (e.g., for a very
viscous liquid)?
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