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ERT 210/4Process Control & Dynamics

DYNAMIC BEHAVIOR DYNAMIC BEHAVIOR OF PROCESSES :OF PROCESSES :Theoretical Models of Theoretical Models of Chemical ProcessesChemical Processes

MISS. RAHIMAH BINTI OTHMAN(Email: rahimah@unimap.edu.my)

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Course Outcome 1 (CO1)Ability to derive and develop theoretical model of chemical processes, dynamic behavior of first and second order processes chemical and dynamic response characteristics of more complicated processes.1. Introduction Concepts

• Introduction to Process Control:• Theoretical Models of Chemical Processes

Apply and derive unsteady-state models (dynamic model) of chemical processes from physical and chemical principles.

2. Laplace Transform

3. Transfer Function Models

4. Dynamic Behavior of First-order and Second-order Processes

5. Dynamic Response Characteristics of More Complicated Processes

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Process Dynamics

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a) Refers to unsteady-state or transient behavior.b) Steady-state vs. unsteady-state behavior;

i.Steady state: variables do not change with timec) Continuous processes : Examples of transient behavior:

i. Start up & shutdownii. Grade changesiii. Major disturbance; e.g: refinery during stormy or hurricane conditionsiv. Equipment or instrument failure (e.g., pump failure)

d) Batch processesi. Inherently unsteady-state operationii. Example: Batch reactor

1.Composition changes with time2.Other variables such as temperature could be constant.

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Process Control1. Objective:Enables the process to be maintained at the desired operation conditions, safely and efficiently, while satisfying environmental and product quality requirements.

2. Control Terminology:Controlled variables (CVs) - these are the variables which quantify the performance or quality of

the final product, which are also called output variables (Set point).

Manipulated variables (MVs) - these input variables are adjusted dynamically to keep the

controlled variables at their set-points.

Disturbance variables (DVs) - these are also called "load" variables and represent input variables

that can cause the controlled variables to deviate from their respective set points (Cannot be manipulated).

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Development of Dynamic ModelsIllustrative Example: A Blending Process

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An unsteady-state mass balance for the blending system:

rate of accumulation rate of rate of(2-1)

of mass in the tank mass in mass out

Figure 2.1. Stirred-tank blending process.

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The unsteady-state component balance is:

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1 1 2 2

ρ(2-3)

d V xw x w x wx

dt

The corresponding steady-state model was derived in Ch. 1 (cf. Eqs. 1-1 and 1-2).

1 2

1 1 2 2

0 (2-4)

0 (2-5)

w w w

w x w x wx

or

where w1, w2, and w are mass flow rates.

1 2

ρ(2-2)

d Vw w w

dt

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Conservation Laws

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Theoretical models of chemical processes are based on conservation laws.

Conservation of Mass

rate of mass rate of mass rate of mass(2-6)

accumulation in out

Conservation of Component i

rate of component i rate of component i

accumulation in

rate of component i rate of component i(2-7)

out produced

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Conservation of Energy

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 heat addition net rate of work

to the system from performed on the sys

the surroundings

tem (2-8)

by the surroundings

The total energy of a thermodynamic system, Utot, is the sum of its internal energy, kinetic energy, and potential energy:

int (2-9)tot KE PEU U U U

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For the processes and examples considered in this book, itis 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.

For these reasonable assumptions, the energy balance inEq. 2-8 can be written as;

int (2-10)dU

wH Qdt

int the internal energy of

the system

enthalpy per unit mass

mass flow rate

rate of heat transfer to the system

U

H

w

Q

denotes the differencebetween outlet and inletconditions of the flowingstreams; therefore

-Δ wH = rate of enthalpy of the inlet

stream(s) - the enthalpyof the outlet stream(s)

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The analogous equation for molar quantities is,

int (2-11)dU

wH Qdt

where is the enthalpy per mole and is the molar flow rate. H w

In order to derive dynamic models of processes from the general energy balances in Eqs. 2-10 and 2-11, expressions for Uint and or are required, which can be derived from thermodynamics.

H

The Blending Process Revisited

For constant , Eqs. 2-2 and 2-3 become:

1 2 (2-12)dV

w w wdt

1 1 2 2 (2-13)

d Vxw x w x wx

dt

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Equation 2-13 can be simplified by expanding the accumulation term using the “chain rule” for differentiation of a product:

(2-14)

d Vx dx dVV x

dt dt dt

Substitution of (2-14) into (2-13) gives:

1 1 2 2 (2-15)dx dV

V x w x w x wxdt dt

Substitution of the mass balance in (2-12) for in (2-15) gives:

/dV dt

1 2 1 1 2 2 (2-16)dx

V x w w w w x w x wxdt

After canceling common terms and rearranging (2-12) and (2-16), a more convenient model form is obtained:

1 2

1 21 2

1(2-17)

(2-18)

dVw w w

dt

w wdxx x x x

dt V V

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Stirred-Tank Heating Process

Figure 2.2 Stirred-tank heating process with constant holdup, V.

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Stirred-Tank Heating Process (cont’d.)

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.

Case 1: Constant Holdup, V

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1. For a pure liquid at low or moderate pressures, the internal energy is approximately equal to the enthalpy, Uint , and H depends only on temperature.

2. Consequently, in the subsequent development, we assume that Uint = H and where the caret (^) means per unit mass. As shown in Appendix B, a differential change in temperature, dT, produces a corresponding change in the internal energy per unit mass,

H

ˆ ˆintU H

ˆ ,intdU

intˆ ˆ (2-29)dU dH CdT

where C is the constant pressure heat capacity (assumed to be constant). The total internal energy of the liquid in the tank is:

int intˆ (2-30)U VU

Model Development

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An expression for the rate of internal energy accumulation can be derived from Eqs. (2-29) and (2-30):

int (2-31)dU dT

VCdt dt

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, . Integrating Eq. 2-29 from a reference temperature Tref to T gives,

H

ˆ ˆ (2-32)ref refH H C T T

where is the value of at Tref. Without loss of generality, we assume that (see Appendix B). Thus, (2-32) can be written as:

ˆrefH H

ˆ 0refH

ˆ (2-33)refH C T T

Model Development (cont’d.)

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ˆ (2-34)i i refH C T T

Substituting (2-33) and (2-34) into the convection term of (2-10) gives:

ˆ (2-35)i ref refwH w C T T w C T T

Finally, substitution of (2-31) and (2-35) into (2-10)

(2-36)idT

V C wC T T Qdt

Model Development (cont’d.)For the inlet stream

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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

Degrees of Freedom Analysis for the Stirred-Tank Model:

3 parameters:

4 variables:

1 equation: Eq. 2-36

, ,V C

, , ,iT T w Q

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Stirred-Tank Heating Process (cont’d.)

Case 2: Variable Holdup, V

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.

Tank holdup vary with time, based on four assumptions;

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The overall mass balance;

Model Development

wwdt

Vdi

)( (2-37)

The energy balance for the current stirred-tank heating system can be derived from Eq. 2-10 in analogy with the derivation of Eq. 2-36. We again assume that Uint = H for the liquid in the tank. Thus, for constant ;

dt

HVd

dt

HVd

dt

dH

dt

Ud )ˆ()ˆ()( int (2-38)

From the definition of and Eq. 2-33 and 2-34, it follows that;

)ˆ( Hw

)()(ˆˆ)ˆ( refrefiiii TTwCTTCwHwHwHw (2-39)

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QTTwCTTCwdt

HVdrefrefii )()(

)ˆ(

wwdt

dVi

dt

dVH

dt

HdV

dt

HVd ˆˆ)ˆ(

where wi and w are the mass flow rates of the inlet and outlet streams, respectively. Substituting (2-38) and (2-39) into (2-10) gives;

(2-40)

Model Development (cont’d.)

Next we simplify the dynamic model. Because ρ is constant, Eq (2-37) can be written as;

(2-41)

The ‘chain rule’ can be applied to expand the left side of (2-40) for constant C and ;

(2-42)

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QTTwCTTCwdt

dTCVwwTTC refrefiiiref )()())((

)( wwdt

dVi

CV

QTT

V

w

dt

dTi

i

)(

dt

dTCVwwTTC

dt

HVdiref ))((

)ˆ(

dtCdTdtHd //ˆ From Eq. 2-29 or 2-33, it follows that . Substituting this expression and Eq. 2-33 and 2-41 into Eq. 2-42 gives;

Model Development (cont’d.)

(2-43)

Substituting (2-43) into (2-40) and rearranging give;

(2-44)

Rearranging (2-41) and (2-44) provides a simpler form for the dynamic model;

(2-45)

(2-46)

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Liquid Storage Systems

Figure 2.5 A liquid-level storage process.

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A typical liquid storage process is shown in Fig. 2.5, where qi and q are volumetric flow rates. A mass balance yields:

Model Development

qqdt

Vdi

)(

qqdt

dhA i

(2-53)

Assume that liquid density is constant and the tank is cylinderical with cross-sectional area, A. Then the volume of liquid in the tank can be expressed as V = Ah, h is the liquid level (or head). Thus, (2.53) becomes;

(2-54)

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Continuous Stirred Tank Reactor (CSTR)

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Fig. 2.6. Schematic diagram of a CSTR.

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CSTR: Model Development

A

A

(2-62)

where = moles of A reacted per unit time, per unit volume, is the concentration of A (mol

r = kc

r ces per unit volume), and is the rate

constant (units of reciprocal time).

7. The rate constant has an Arrhenius temperature dependence:

exp(- ) 0

k

k = k E/RT (2-63)

where is the frequency factor, is the activation energy,

and is the the gas constant.0

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Assumptions:1. Single, irreversible reaction, A → B.2. Perfect mixing.3. The liquid volume V is kept constant by an overflow line.4. The mass densities of the feed and product streams are equal

and constant. They are denoted by .5. Heat losses are negligible.6. The reaction rate for the disappearance of A, r, is given by,

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• Unsteady-state component balance

• Unsteady-state mass balance

Because and V are constant,

CSTR: Model Development (continued)

(2-64)

(2-66)

(2-65)

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Assumptions for the Unsteady-state Energy Balance:

8.

9.

10.

11.

12.

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CSTR Model: Some Extensions

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1. Multiple reactions (e.g., A → B → C) ?

2. Different kinetics, e.g., 2nd order reaction?

3. Significant thermal capacity of the coolant liquid?

4. Liquid volume V is not constant (e.g., no overflow line)?

5. Heat losses are not negligible?

6. Perfect mixing cannot be assumed (e.g., for a very viscous liquid)?

• How would the dynamic model change for: