Transcript Slayt 1

Process Control
CHAPTER II
PROCESS DYNAMICS AND
MATHEMATICAL MODELING
Dynamic Models are also referred as unsteady
state models.
These models can be used for;
1. Improve understanding of the process
2. Train plant operating personnel
3. Develop a control strategy for a new
process
4. Optimize process operating conditions.
1.
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Theoretical Models are developed using the
principles of chemistry, physics and biology.
Advantages:
they provide physical insight into process
behavior.
they are applicable over a wide range of
conditions.
Disadvantages:
they tend to be time consuming and expensive to
develop.
theoretical models of complex processes include
some model parameters that are not readily
available.
2. Empirical Models are obtained by fitting
experimental data.
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Advantages:
 easier to develop
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Disadvantages:
 do not extrapolate well
 the range of data is usually quite small
compared to the full range of operating
conditions.
3. Semi-Empirical Models are a combination
of theoretical and empirical models where
the numerical values of parameters in a
theoretical model are calculated from
experimental data.
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Advantages:
 they incorporate theoretical knowledge
 they can be extrapolated over a large range
of conditions
 they require less development of effort than
theoretical models.
A Systematical Approach for
Modeling
1.
2.
3.
State the modeling objectives and the end
use of the model. Then determine the
required levels of model detail and model
accuracy.
Draw a schematic diagram of the process
and label all process variables.
List all of the assumptions involved in
developing the model. The model should
not be no more complicated than necessary
to meet the modeling objectives.
4. Determine if spatial variations are important.
If so, a partial differential equation model will
be required.
5. Write appropriate conservation equations
(mass, component, energy etc)
6. Introduce constitutive equations, which are
equilibrium relations and other algebraic
equations
7. Perform a degrees of freedom analysis to
ensure that the model equations can be
solved.
8. Simplify the model.
output = f (inputs)
This model form is convenient for computer
simulation and subsequent analysis.
9. Classify inputs as disturbance variables or as
manipulated variables.
Degrees of freedom
To simulate a process, model equations should
be solvable set of relations. In order for a
model to have a unique solution, number of
degrees of freedom should be zero.
Number of
degrees of
freedom
Number of
process
variables
Number of
independent
equations
For Degrees of Freedom Analysis
1.
List all quantities in the model that are known
constants (or parameters).
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 a process input nor a process output.
3.
Calculate the number of degrees of freedom.
4.
Identify the NE output variables that will be
obtained by solving the process model.
5.
Identify the NF input variables that must be
specified as either disturbance variables or
manipulated variables, in order to utilize the NF
degrees of freedom.
Summary
if DOF=0 The system is exactly specified
if DOF<0 The system is over specified, and in
general no solution to model exists. The likely cause
is either (1) improperly designating a variable(s) as a
parameter or external variable or (2) including an
extra, dependent equation(s) in the model.
if DOF>0 The system is underspecified, and an
infinite number of solutions to the model exists. The
likely cause is either (1) improperly designating a
parameter or external variable as a variable or (2)
not including in the model all equations that
determine the system’s behavior.
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Example
Consider a continuous stirred tank blending system
where two input systems are blended to produce an
outlet stream that has the desired composition.
Stream 1 is a mixture of two species A and B. We
assume that its mass flow rate is constant but the
mass fraction of A (x1) varies with time. Stream 2
consists of pure A and thus x2=1. The mass flow rate
of stream 2 (w2) can be manipulated using a control
valve. The mass fraction of A in the exit stream is
denoted by x and the desired value by xsp.
X1, w1
X2, w2
X, w
Control Question: Suppose that inlet concentration x1 varies with
time. How can we ensure that the outlet composition x remains at
or near its desired value.

Method 1.
Measure x and adjust w2.
 if x is high, w2 should be reduced
 if x is low, w2 should be increased
X2, w2
X1, w1
X, w
AT
(Analyzer-Transmitter)
(Feedback Control)
Method 2
Measure x1 and adjust w2.
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AT
X2, w2
X1, w1
X, w
(Feed forward Control)
Design Question: If the nominal value of x1 is x1,s
what nominal flow rate w2 is required to produce the
desired outlet concentration xsp.
With a st-st material balance,
w1+ w2 –w = 0
( overall balance)
w1x1,s + w2x2,s – wxsp = 0 (component A balance)
w1x1,s+w2(1.0)-(w1+w2)xsp=0

w2  w1
xsp  x1,s
1  xsp
X1, w1
X2, w2
X, w
Consider a more general
version of the blending
system where stream 2
is not pure and volume of
the tank may vary with
time.
(! Not an overflow system
any more but a draining
system)

Objective is again to keep x at the desired value
A unsteady state mass balance gives;
rate of accumulation of
mass in the tank
rate of
rate of
mass in
mass out
The mass of liquid in the tank can be
expressed as product of the liquid and the
density.
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1.
2.
system: liquid in the tank
assumptions:
tank is well mixed
density of liquid is not changing with composition
change
Total material balance;
{rate of mass in} - {rate of mass out} = {rate of accumulation of mass}
The rate expression in real form is,
w1t  w2 t  wt  Vt t  Vt
Dividing by Δt and taking limit as Δt →0 gives
d
w1  w2  w 
(V )
dt
Component balance;
d
w1 x1  w2 x2  wx  (Vx)
dt
Considering the constant density assumption
equations become;
w1  w2  w  
dV
dt
d (Vx )
dx
dV
w1 x1  w2 x2  wx  
 V
 x
dt
dt
dt
Equation 1
Equation 2
Replacing Equation (1) into Equation (2) gives;
dx w1 ( x1  x) w2 ( x2  x )


dt
V
V
dV
1
 ( w1  w2  w)
dt

With these two equations system behavior is
mathematically defined.
Degrees of Freedom Analysis:
Parameter(s): ρ
variables: V, x1, w1, x2, w2, x, w
equations: (dV/dt and dx/dt)
D.O.F = 7-2 = 5
outputs: V, x
inputs: x1, w1, x2, w2, w
disturbances: x1, w1, x2
manipulated variables: w2, w (x vs w2 and V vs w as
the control structures)
Example
Goal: The dynamic response of
temperature of the liquid in the
tank is to be determined.
Fi, Ti
System: The liquid in the tank.
Assumptions:
1. tank is well mixed
F, T
2. Physical properties of the
system are not changing during
the process.
Total mass balance:
Fi t  Ft  Aht  t  Aht
d
Fi   F 
( Ah)
dt
dh
A
 Fi  F
dt
with constant ρ and cross-sectional area A
flow rates are given in units of volumetric flow rates
Total energy balance:
Fi c p (Ti  Tref )t  Fc p (T  Tref )t  Qt
 Ahcp (T  Tref )t  t  Ahcp (T  Tref )t
Fi c p (Ti  Tref )  Fc p (T  Tref )  Q 
d
Ahcp (T  Tref )
dt
with constant ρ, Cp and assuming Tref =0 gives
Q
d (hT )
dT
dh
FiTi  FT 
A
 Ah
 AT
c p
dt
dt
dt
dh
A
 Fi  F
dt
dT
Q
Ah
 Fi (Ti  T ) 
dt
c p
D.O.F Analysis
Parameter(s): ρ,cp
variables: V, T
equations: (dh/dt and dT/dt)
D.O.F = 2-2 = 0
outputs: V, T
inputs: Fi, Ti, Fst
disturbances: Fi, Ti,
manipulated variables: no control structures
Example
Consider the typical liquid
storage process shown in the
figure, where qi and q are
volumetric flow rates.
qi
h
q
Assuming constant density and
cross sectional area A a mass
balance gives:
dh
A
 qi  q
dt
There are three important variations in the
liquid storage processes:
1. The inlet or outlet flow rates might be
constant. In that case the exit flow rate is
independent of the liquid level over a wide
range of conditions. Consequently qin=qout
at the steady state conditions.
2.
The tank exit line may function simply
as a resistance to flow from the tank or
it may contain a valve that provides
significant resistance to flow at a single
point. In the simplest case, the flow may
be assumed to be linearly related to the
driving force, the liquid level.
q 
A
1
h
Rv
dh
1
 qi 
h
dt
Rv
3.
A more realistic expression for flow rate q can
be obtained when a fixed valve has been
placed in the exit line and turbulent flow can be
assumed. The driving force for flow through the
valve is the pressure drop ΔP, ΔP=P-Pa where
P is pressure at the bottom of the tank and Pa
is pressure at the end of the exit line.
F  Cv *
P  Pa

Cv* is the valve constant
Example
Fi, Ti, CAi
Consider the Continuous Stirred Tank
Reactor (CSTR) in which a simple liquid
phase irreversible chemical reaction
takes place.
F, T, CA
A
B
r=kCA
r : rate of reaction
k : reaction rate constant,
k=k0exp(-E/RT)
Cooling
medium
CA : molar concentration
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1.
2.
3.
4.
system: liquid in the tank
assumptions;
CSTR is perfectly mixed
Mass densities of feed and product are
equal and constant
Liquid volume is kept constant by an
overflow line.
The thermal capacitances of the coolant
and the cooling coil wall are negligible
compared to the thermal capacitance of the
liquid.
5.
6.
7.
8.
Coolant temperature is constant. (change in
the tank is negligible)
Rate of heat transfer to coolant is given by,
Q=UA(Tc-T), where U,A are parameters.
Heat of mixing is negligible compared to the
heat of reaction.
Shaft work and heat losses are negligible.

Total mass balance;
Fi  F 
d
( V )
dt
With constant ρ and V, Fi=F
 Component balance for species A (in molar units);
dC A
FC Ai  FC A  VkC A  V
dt

Energy balance;
dT
wC p (Ti  T )  (H R )VkC A  UA (Tc  T )  VC p
dt