1. CHAPTER II PROCESS DYNAMICS AND MATHEMATICAL MODELING
Process Control
CHAPTER II PROCESS DYNAMICS AND MATHEMATICAL MODELING

2. Dynamic Models are also referred as unsteady state models.
These models can be used for;
Improve understanding of the process
Train plant operating personnel
Develop a control strategy for a new process
Optimize process operating conditions.

3. 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.

4. 2. Empirical Models are obtained by fitting experimental data.
Advantages:
easier to develop
Disadvantages:
do not extrapolate well
the range of data is usually quite small compared to the full range of operating conditions.

5. 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.
Advantages:
they incorporate theoretical knowledge
they can be extrapolated over a large range of conditions
they require less development of effort than theoretical models.

6. A Systematical Approach for Modeling
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.

7. 4. Determine if spatial variations are important
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. 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.

9. 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

10. For Degrees of Freedom Analysis
List all quantities in the model that are known constants (or parameters).
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.
Calculate the number of degrees of freedom.
Identify the NE output variables that will be obtained by solving the process model.
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.

11. 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.

12. 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.

13. 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.

14. (Feedback Control) Method 1. Measure x and adjust w2.
if x is high, w2 should be reduced
if x is low, w2 should be increased
(Feedback Control)
X2, w2
X1, w1
X, w AT
(Analyzer-Transmitter)

15. (Feed forward Control)
Method 2
Measure x1 and adjust w2.
(Feed forward Control) AT
X2, w2
X1, w1
X, w

16. 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 = ( overall balance)
w1x1,s + w2x2,s – wxsp = (component A balance)
w1x1,s+w2(1.0)-(w1+w2)xsp=0

17. (! Not an overflow system any more but a draining system)
X2, w2
X1, w1
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)
X, w
Objective is again to keep x at the desired value

18. A unsteady state mass balance gives;
The mass of liquid in the tank can be expressed as product of the liquid and the density.
rate of
mass in
rate of
mass out
rate of accumulation of mass in the tank

19. 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}

20. The rate expression in real form is,
Dividing by Δt and taking limit as Δt →0 gives

21. Considering the constant density assumption equations become;
Component balance;
Considering the constant density assumption equations become;
Equation 1
Equation 2

22. Replacing Equation (1) into Equation (2) gives;
With these two equations system behavior is mathematically defined.

23. 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)

24. Example
Goal: The dynamic response of temperature of the liquid in the tank is to be determined.
System: The liquid in the tank.
Assumptions:
tank is well mixed
Physical properties of the system are not changing during the process.
Fi, Ti
F, T

25. Total mass balance:
with constant ρ and cross-sectional area A
flow rates are given in units of volumetric flow rates

26. Total energy balance:
with constant ρ, Cp and assuming Tref =0 gives

27. 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

28. Example
Consider the typical liquid storage process shown in the figure, where qi and q are volumetric flow rates.
Assuming constant density and cross sectional area A a mass balance gives: qi h q

29. There are three important variations in the liquid storage processes:
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.

30. 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.

31. 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.
Cv* is the valve constant

32. Example
Fi, Ti, CAi
Consider the Continuous Stirred Tank Reactor (CSTR) in which a simple liquid phase irreversible chemical reaction takes place.
A B
r=kCA
r : rate of reaction
k : reaction rate constant,
k=k0exp(-E/RT)
CA : molar concentration
F, T, CA
Cooling medium

33. 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.

34. 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.

35. Total mass balance;
With constant ρ and V, Fi=F
Component balance for species A (in molar units);
Energy balance;