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Mathematical Modeling of Chemical Processes. Mathematical Model “a representation of the essential aspects of an existing system (or a system to be constructed)

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Presentation on theme: "Mathematical Modeling of Chemical Processes. Mathematical Model “a representation of the essential aspects of an existing system (or a system to be constructed)"— Presentation transcript:

1 Mathematical Modeling of Chemical Processes

2 Mathematical 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.

3 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 Modeling

4 General 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.

5 A Systematic Approach for Developing Dynamic 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).

6 A Systematic Approach for Developing Dynamic 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.

7 Conservation Laws Theoretical models of chemical processes are based on conservation laws. Conservation of Mass Conservation of Component i

8 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, U tot, is the sum of its internal energy, kinetic energy, and potential energy:

9 Example Simple tank Problem

10 Degrees of Freedom Analysis 1.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 N E and the number of process variables, N V. 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, N F = N V - N E. 4.Identify the N E output variables that will be obtained by solving the process model. 5.Identify the N F input variables that must be specified as either disturbance variables or manipulated variables, in order to utilize the N F degrees of freedom. Chapter 2

11 Stirred-Tank Heating Process Stirred-tank heating process with constant holdup, V. Chapter 2

12 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. Chapter 2

13 Thus the degrees of freedom are N F = 4 – 1 = 3. The process variables are classified as: 1 output variable:T 3 input variables:T i, w, Q For temperature control purposes, it is reasonable to classify the three inputs as: 2 disturbance variables:T i, w 1 manipulated variable:Q Degrees of Freedom Analysis for the Stirred-Tank Model: 3 parameters: 4 variables: 1 equation: Chapter 2

14 Degrees of Freedom Analysis

15 System comprises of only 2 chemical species A and B Can write only 2 independent mass balances write for species A and species B write overall balance & one component balance (either for species A or B)

16 Degrees of Freedom Analysis

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

18 Degrees of Freedom Analysis

19 Dimensional 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.

20 Dimensional Analysis Dimensions of a physical quantity is associated with symbols, such as M, L, T which represent mass, length and time Assume 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)

21 Dimensions QuantitySymbolMLT Torque  12-2 Fan diameterD010 Fan designR000 Air density  1-30 Relative speedn00

22 Dividing torque by density gives  /  divided by D 5 n 2 gives:SymbolMLT // 05-2 D010 n00

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

24 Buckingham  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

25 Buckingham  theorem In mathematical terms, if we have a physically meaningful equation such as where the q i 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 π i are dimensionless parameters constructed from the q i by p = n − k equations of the form where the exponents mi are constants.

26 Example 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.

27 Example Buckingham  theorem states that there will be two such groups

28 Development of Dynamic Models Illustrative Example: A Blending Process An unsteady-state mass balance for the blending system:

29 The unsteady-state component balance is: The corresponding steady-state model was derived in Ch. 1 (cf. Eqs. 1-1 and 1-2). or where w 1, w 2, and w are mass flow rates.

30 The Blending Process Revisited For constant, Eqs. 2-2 and 2-3 become:

31 Equation 2-13 can be simplified by expanding the accumulation term using the “chain rule” for differentiation of a product: Substitution of (2-14) into (2-13) gives: Substitution of the mass balance in (2-12) for in (2-15) gives: After canceling common terms and rearranging (2-12) and (2-16), a more convenient model form is obtained:

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