Download Q Degrees of Freedom Analysis for the Stirred-Tank Model

Chapter 2
Mathematical Modeling of
Chemical Processes
Mathematical Model (Eykhoff, 1974)
“a representation of the essential aspects of an existing
system (or a system to be constructed) which
represents knowledge of that system in a usable form”
Everything should be made as simple as possible, but
no simpler.
General Modeling Principles
• The model equations are at best an approximation to the real
process.
Chapter 2
• Adage: “All models are wrong, but some are useful.”
• Modeling inherently involves a compromise between model
accuracy and complexity on one hand, and the cost and effort
required to develop the model, on the other hand.
• Process modeling is both an art and a science. Creativity is
required to make simplifying assumptions that result in an
appropriate model.
• Dynamic models of chemical processes consist of ordinary
differential equations (ODE) and/or partial differential equations
(PDE), plus related algebraic equations.
Chapter 2
Table 2.1. A Systematic Approach for
Developing Dynamic Models
1. State the modeling objectives and the end use of the model.
They determine the required levels of model detail and model
accuracy.
2. Draw a schematic diagram of the process and label all process
variables.
3. List all of the assumptions that are involved in developing the
model. Try for parsimony; the model should be no more
complicated than necessary to meet the modeling objectives.
4. Determine whether spatial variations of process variables are
important. If so, a partial differential equation model will be
required.
5. Write appropriate conservation equations (mass, component,
energy, and so forth).
Table 2.1. (continued)
Chapter 2
6. Introduce equilibrium relations and other algebraic
equations (from thermodynamics, transport phenomena,
chemical kinetics, equipment geometry, etc.).
7. Perform a degrees of freedom analysis (Section 2.3) to
ensure that the model equations can be solved.
8. Simplify the model. It is often possible to arrange the
equations so that the dependent variables (outputs) appear
on the left side and the independent variables (inputs)
appear on the right side. This model form is convenient
for computer simulation and subsequent analysis.
9. Classify inputs as disturbance variables or as manipulated
variables.
Chapter 2
Modeling Approaches
 Physical/chemical (fundamental, global)
• Model structure by theoretical analysis
 Material/energy balances
 Heat, mass, and momentum transfer
 Thermodynamics, chemical kinetics
 Physical property relationships
• Model complexity must be determined
(assumptions)
•
Can be computationally expensive (not realtime)
•
May be expensive/time-consuming to obtain
•
Good for extrapolation, scale-up
•
Does not require experimental data to obtain
(data required for validation and fitting)
• 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 



in
 accumulation  

rate of component i  rate of component i 



out
produced

 

(2-7)
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 
 the surroundings
  by the surroundings 

 

(2-8)
The total energy of a thermodynamic system, Utot, is the sum of its
internal energy, kinetic energy, and potential energy:
Utot  U int  U KE  U PE
(2-9)
 Black box (empirical)
• Large number of unknown parameters
• Can be obtained quickly (e.g., linear regression)
Chapter 2
• Model structure is subjective
• Dangerous to extrapolate
 Semi-empirical
• Compromise of first two approaches
• Model structure may be simpler
• Typically 2 to 10 physical parameters estimated
(nonlinear regression)
• Good versatility, can be extrapolated
• Can be run in real-time
• linear regression
y  c0  c1 x  c2 x 2
• nonlinear regression

Chapter 2
y  K 1  e t /

• number of parameters affects accuracy of model,
but confidence limits on the parameters fitted must
be evaluated
• objective function for data fitting – minimize sum of
squares of errors between data points and model
predictions (use optimization code to fit
parameters)
• nonlinear models such as neural nets are
becoming popular (automatic modeling)
Chapter 2
Number of
births (West
Germany)
Number of sightings of storks
Uses of Mathematical Modeling
•
to improve understanding of the process
•
to optimize process design/operating conditions
•
to design a control strategy for the process
•
to train operating personnel
•Development of Dynamic Models
Chapter 2
•Illustrative Example: A 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)
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)
Chapter 2
The Blending Process Revisited
For constant  , Eqs. 2-2 and 2-3 become:
dV

 w1  w2  w
dt
 d Vx 
dt
 w1x1  w2 x2  wx
(2-12)
(2-13)
Equation 2-13 can be simplified by expanding the accumulation
term using the “chain rule” for differentiation of a product:
dx
dV
 x
(2-14)
dt
dt
dt  d Vx 
 w1x1  w2 x2  wx (2-13)
Substitution of (2-14) into (2-13) gives: dt
dx
dV
V   x
 w1x1  w2 x2  wx
dV (2-15)

 w1  w2  w (2-1
dt
dt
dt
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 
(2-17)
dt 
w2
dx w1

(2-18)
 x1  x    x2  x 
dt V 
V

Chapter 2
d Vx 
 V
Chapter 2
Chapter 2
Stirred-Tank Heating Process
Figure 2.3 Stirred-tank heating process with constant holdup, V.
Stirred-Tank Heating Process (cont’d.)
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.
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.
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 
 the surroundings
  by the surroundings 

 

(2-8)
Chapter 2
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
U int  the internal energy of
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)


Chapter 2
Model Development - I
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. Consequently, in the subsequent
development, we assume that Uint = H and Uˆ int  Hˆ where the
caret (^) means per unit mass. 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)
Chapter 2
Model Development - II
An expression for the rate of internal energy accumulation can be
Uint  VUˆ int (2-30)
derived from Eqs. (2-29) and (2-30):
dU int
dUˆ int


V
dU int
dT
dt
dt
 VC
(2-31)
dt
dt
dUˆ int  dHˆ  CdT (2-29)
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
dUˆ int  dHˆ  CdT
(2-29)
Tref to T gives,
Hˆ  Hˆ  C T  T
(2-32)

ref
ref

where Hˆ ref is the value of Ĥ at Tref. Without loss of generality, we
assume that. Hˆ ref  0 Thus, (2-32) can be written as:

Hˆ  C T  Tref

(2-33)
Model Development - III

For the inlet stream

Hˆ i  C Ti  Tref
Chapter 2

Hˆ  C T  Tref (2-33)

(2-34)
Substituting (2-33) and (2-34) into the convection term of (2-10)
dU int
gives:
   wH   Q (2-10)
dt
 




 wHˆ  w C Ti  Tref   w C T  Tref 




(2-35)
Finally, substitution of (2-31) and (2-35) into (2-10)
V C
dT
 wC Ti  T   Q
dt
dU int
dT
 VC
dt
dt
(2-36)
(2-31)
Define deviation variables (from set point)
y  T T
T is desired operating point
Chapter 2
u  ws  ws
V dy
w dt
ws (T ) from steady state
 y 
note when
H v
u
wC
dy
0
dt
note that
H v
V
 K p and
 1
wC
w
y  K pu
dy
1   y  K pu
dt
General linear ordinary differential equation solution: sum of exponential(s)
Suppose u  1 (unit step response)
t
 

y (t )  K p 1  e 1 




Chapter 2
In fact, after a time interval equal to the process time constant (t=τ), the process
response is still only 63.2% complete. Theoretically, the process output never
reaches the new steady-state value except as t→ ∞. It does approximate the final
steady-state value when t≈5τ.
Chapter 2
Process Dynamics
Process control is inherently concerned with unsteady
state behavior (i.e., "transient response", "process
dynamics")
Stirred tank heater: assume a "lag" between heating
element temperature Te, and process fluid temp, T.
Chapter 2
heat transfer limitation = heA(Te – T)
Energy
Tank:
Chest:
At s.s.
Suppose that we
have a significant
thermal
balances
capacitance
and
dT
that the electrical
wCTi +h e A(Te -T)-wCT=mC
heating
rate
Q
dt
directly affects the
dTe
temperature of the
Q - h e A(Te - T) = m e C e
element rather than
dt
the liquid contents.
dT
dT
 0, e  0 Where m C is product of mass of the metal in
e e
dt
dt
the heating system
Specify Q  calc. T, Te
2 first order equations  1 second order equation in T
Relate T to Q (Te is an intermediate variable)
y=T-T
u =Q-Q
Ti fixed
Chapter 2
mm e C e d 2 y  m e C e m e C e m  dy
1






y

u
2


wh e A e dt
wC
w  dt
wC
 h eAe
Rv
Note Ce  0 yields 1st order ODE (simpler model)
(2.56)
Rv: line resistance
*Solve by substituting
2.56 into 2.54
1
q=
h - - - -(2.56)
Rv
A
dh
1
 qi 
h
dt
Rv
(2 - 57)
Chapter 2
If
q = Cv P - Pa
ΔP=P-Pa
Pa : ambient pressure (2 - 59) P  p  ρgh (2  60)
dh
A  qi  Cv* ρgh  qi  Cv h
dt
nonlinear ODE
(2-61)
Chapter 2
Table 2.2. Degrees of Freedom Analysis
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, etc.
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, NF = NV - NE.
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.
Degrees of Freedom Analysis for the Stirred-Tank
Model:
V , ,C
Chapter 2
3 parameters:
4 variables:
T , Ti , w, Q
1 equation:
V C
Eq. 2-36
dT
 wC Ti  T   Q
dt
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
Example 2.2
• Analyze the degree of freedom for the
blending model of d Vρx   w1x1  w2 x2  wx
dt
for special condition where volume V
constant.
(2-3)
Solution
• 2 parameters: V, ρ
• 4 variables (Nv=4): x, x1, w1, w2
d Vρx 
• 1 equation(NE=1) dt  w1x1  w2 x2  wx
• Degree of freedom NF=4-1=3
So we must identify three input variables that
can be specified as function of time in order
for the equation to have a unique soluation.
(2-3)
• consequently, we have
• 1 output: x
• 3 input: x1, w1, w2
Three degree of freedom can be utilized by
specifying the input as
2 disturbance variable: x1, w1
1 manipulated variable: w2
Example 2.3
• Analyze the degree of freedom of the
blending system model in
dV 1
  w1  w2  w 
dt 
w2
dx w1

 x1  x    x2  x 
dt V 
V
(2-17)
(2-18)
Is this set of equation linear or nonlinear,
according to the usual working definition
Solution
• Volume is variable rather than a constant
parameter
• 1 parameter: ρ
• 7 variables (Nv=7): V, x, x1,x2, w,w1,w2
• 2 equations: NE= dVdt  1  w  w  w
(2-17)
1
2
w
dx w1

 x1  x   2  x2  x 
dt V 
V
(2-18)
• NF= 7-2= 5
• 2 outputs: V and x
• 5 inputs: w,w1,w2,x1,x2
Because the two outputs are the only variables
to be determined in solving the system of
two equations, no degree of freedom is left.
So the system of equations is exactly
specified and hence solvable.
Utilization of the degree of freedom:
The five inputs are classified as either
disturbance variables or manipulated
variables.
3 disturbance variables: w1,x1,x2
2 manipulated variables: w,w2
For example, w could be used to control V
and w2 to control x