Mathematical Modeling of Mathematical Modeling of Chemical ProcessesChemical Processes

Mathematical ModelMathematical Model

“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.

to improve understanding of the process

to optimize process design/operating conditions

to design a control strategy for the process

to train operating personnel

Uses of Mathematical ModelingUses of Mathematical Modeling

General Modeling PrinciplesGeneral Modeling Principles

The model equations are at best an approximation to the real process.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.

A Systematic Approach for Developing A Systematic Approach for Developing Dynamic ModelsDynamic 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).

A Systematic Approach for Developing A Systematic Approach for Developing Dynamic ModelsDynamic Models

6. Introduce equilibrium relations and other algebraic equations (from thermodynamics, transport phenomena, chemical kinetics, equipment geometry, etc.).

7. Perform a degrees of freedom analysis 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.

Conservation LawsConservation Laws

Theoretical models of chemical processes are based on conservation laws.

Conservation of Mass

Conservation of Component i

Conservation of Energy

The general law of energy conservation is also called the First Law of Thermodynamics. It can be expressed as:

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

ExampleExample

Simple tank Problem

Degrees of Freedom AnalysisDegrees of Freedom Analysis1. 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.

Ch

apte

r 2

Stirred-Tank Heating ProcessStirred-Tank Heating Process

Stirred-tank heating process with constant holdup, V.

Ch

apte

r 2

Stirred-Tank Heating Process (cont’d.)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.

Ch

apte

r 2

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 Degrees of Freedom Analysis for the Stirred-Tank Model:Model:

3 parameters:

4 variables:

1 equation:

, ,V C

, , ,iT T w Q

Ch

apte

r 2

Degrees of Freedom AnalysisDegrees of Freedom Analysis

Degrees of Freedom AnalysisDegrees of Freedom Analysis

System comprises of only 2 chemical species A and B

Can write only 2 independent mass balanceswrite for species A and species B

write overall balance & one component balance (either for species A or B)

Degrees of Freedom AnalysisDegrees of Freedom Analysis

Degrees of Freedom AnalysisDegrees of Freedom Analysis

Focus on the control volume (A ∆z) over the time interval t to t + ∆t

Degrees of Freedom AnalysisDegrees of Freedom Analysis

Dimensional AnalysisDimensional Analysis

A conceptual tool often applied to understand physical situations involving a mix of different kinds of physical quantities. It is routinely used by physical scientists and engineers to check the plausibility of derived equations. Only like dimensioned quantities may be added, subtracted, compared, or equated. When unlike dimensioned quantities appear opposite of the "+" or "−" or "=" sign, that physical equation is not plausible, which might prompt one to correct errors before proceeding to use it. When like dimensioned quantities or unlike dimensioned quantities are multiplied or divided, their dimensions are likewise multiplied or divided.

Dimensional AnalysisDimensional Analysis

Dimensions of a physical quantity is associated with symbols, such as M, L, T which represent mass, length and timeAssume to determine the power required to drive a house fan. Torque is chosen as the dependent variable and the following are known physical variables

Fan diameter (d)Fan design (R)Air density ()Rotative speed (n)

DimensionsDimensions

QuantityQuantity SymbolSymbol MM LL TT

Torque 1 2 -2

Fan diameter D 0 1 0

Fan design R 0 0 0

Air density 1 -3 0

Relative speed n 0 0 -1

Dividing torque by density gives

/ divided by D5n2 gives:1

25

nD

SymbolSymbol MM LL TT

/ 0 5 -2

D 0 1 0

n 0 0 -1

Final analysis

The torque for a given design R is proportional to the dimensionless product

0,25

R

nDf

25nD

Buckingham Buckingham theorem theorem

every physically meaningful equation involving n variables can be equivalently rewritten as an equation of n – m dimensionless parameters, where m is the number of fundamental dimensions used it provides a method for computing these dimensionless parameters from the given variables, even if the form of the equation is still unknown

Buckingham Buckingham theorem theorem

In mathematical terms, if we have a physically meaningful equation such as

where the qi  are the n  physical variables, and they are expressed in terms of k  independent physical units, then the above equation can be restated as

where the πi are dimensionless parameters constructed from the qi  by p = n − k  equations of the form

where the exponents mi  are constants.

ExampleExample

If a moving fluid meets an object, it exerts a force on the object, according to a complicated (and not completely understood) law. We might suppose that the variables involved under some conditions to be the speed, density and viscosity of the fluid, the size of the body (expressed in terms of its frontal area A), and the drag force.

ExampleExample

Buckingham theorem states that there will be two such groups

Development of Dynamic ModelsDevelopment of Dynamic ModelsIllustrative Example: A Blending Process

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

The unsteady-state component balance is:

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

The Blending Process RevisitedThe 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

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