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CHAPTER II PROCESS DYNAMICS AND MATHEMATICAL

MODELING

Process Control

- 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.

- 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

- 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
- 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.

- Summary
- if DOF0 The system is exactly specified
- if DOFlt0 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 DOFgt0 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 systems 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 x21. 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)

X2, w2

X1, w1

X, w

AT

(Analyzer-Transmitter)

- Method 2
- Measure x1 and adjust w2.
- (Feed forward Control)

AT

X2, w2

X1, w1

X, w

- 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,sw2(1.0)-(w1w2)xsp0

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

- 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

- 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,
- Dividing by ?t and taking limit as ?t ?0 gives

- Component balance
- Considering the constant density assumption

equations become

Equation 1

Equation 2

- Replacing Equation (1) into Equation (2) gives
- With these two equations system behavior is

mathematically defined.

- Degrees of Freedom Analysis
- Parameter(s) ?
- 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. - 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

- Total mass balance
- with constant ? and cross-sectional area A
- flow rates are given in units of volumetric flow

rates

- Total energy balance
- with constant ?, Cp and assuming Tref 0 gives

- D.O.F Analysis
- Parameter(s) ?,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. Assuming constant density and cross

sectional area A a mass balance gives

qi

h

q

- 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 qinqout at the steady state

conditions.

- 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.

- 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 ?P, ?PP-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

Example

Fi, Ti, CAi

Consider the Continuous Stirred Tank Reactor

(CSTR) in which a simple liquid phase

irreversible chemical reaction takes place. A

B rkCA r rate of reaction k reaction

rate constant, kk0exp(-E/RT) CA molar

concentration

F, T, CA

Cooling medium

- 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.

- Coolant temperature is constant. (change in the

tank is negligible) - Rate of heat transfer to coolant is given by,

QUA(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
- With constant ? and V, FiF
- Component balance for species A (in molar units)
- Energy balance