CHAPTER IIPROCESS DYNAMICS AND

MATHEMATICAL MODELING

Process Control

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. 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.

• 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.

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.

A Systematical Approach for Modeling

1. State the modeling objectives and the end use of the model. Then 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 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 Analysis1. 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.

Summaryif DOF=0 The system is exactly specifiedif 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.

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

(Feedback Control)

X1, w1

X2, w2

X, wAT

(Analyzer-Transmitter)

Method 2

Measure x1 and adjust w2.

(Feed forward Control)

X1, w1

X2, w2

X, w

AT

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

sp

ssp

x

xxww

1

,112

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)

X1, w1X2, w2

X, w

Objective is again to keep x at the desired value

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 accumulation of mass in the tank

rate of

mass in

rate of

mass out

system: liquid in the tank assumptions:

1. tank is well mixed

2. 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,

Dividing by Δt and taking limit as Δt →0 gives

ttt VVtwtwtw 21

)(21 Vdt

dwww

Component balance;

Considering the constant density assumption equations become;

)(2211 xVdt

dwxxwxw

dt

dVx

dt

dxV

dt

Vxdwxxwxw

dt

dVwww

)(2211

21 Equation 1

Equation 2

Replacing Equation (1) into Equation (2) gives;

With these two equations system behavior is mathematically defined.

)(1

)()(

21

2211

wwwdt

dV

V

xxw

V

xxw

dt

dx

Degrees of Freedom Analysis:

Parameter(s): ρ

variables: V, x1, w1, x2, w2, x, wequations: (dV/dt and dx/dt)D.O.F = 7-2 = 5outputs: 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

F, T

Fi, Ti

Goal: The dynamic response of temperature of the liquid in the tank is to be determined.

System: The liquid in the tank.

Assumptions:

1. tank is well mixed

2. Physical properties of the system are not changing during the process.

Total mass balance:

with constant ρ and cross-sectional area A

flow rates are given in units of volumetric flow rates

FFdt

dhA

Ahdt

dFF

AhAhtFtF

i

i

ttti

)(

Total energy balance:

with constant ρ, Cp and assuming Tref =0 gives

)()()(

)()(

)()(

refprefprefipi

trefpttrefp

refprefipi

TTAhcdt

dQTTcFTTcF

TTAhcTTAhc

tQtTTcFtTTcF

pii

i

pii

c

QTTF

dt

dTAh

FFdt

dhA

dt

dhAT

dt

dTAh

dt

hTdA

c

QFTTF

)(

)(

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

h

qi

q

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:

qqdt

dhA i

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.

hR

qdt

dhA

hR

q

vi

v

1

1

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.

Cv* is the valve constant

a

v

PPCF

*

Example

F, T, CA

Fi, Ti, CAi

Cooling medium

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

system: liquid in the tank assumptions;1. CSTR is perfectly mixed2. Mass densities of feed and product are

equal and constant3. Liquid volume is kept constant by an

overflow line.4. The thermal capacitances of the coolant

and the cooling coil wall are negligible compared to the thermal capacitance of the liquid.

5. Coolant temperature is constant. (change in the tank is negligible)

6. Rate of heat transfer to coolant is given by, Q=UA(Tc-T), where U,A are parameters.

7. Heat of mixing is negligible compared to the heat of reaction.

8. Shaft work and heat losses are negligible.

Total mass balance;

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

Energy balance;

)( Vdt

dFFi

dt

dCVVkCFCFC A

AAAi

dt

dTCVTTUAVkCHTTwC pcARip )()()(