CHAPTER 3 : MATHEMATICAL MODELLING PRINCIPLES When I complete this - PowerPoint PPT Presentation

CHAPTER 3 : MATHEMATICAL MODELLING PRINCIPLES When I complete this chapter, I want to be able to do the following. • Formulate dynamic models based on fundamental balances • Solve simple first-order linear dynamic models • Determine how key aspects of dynamics depend on process design and operation

CHAPTER 3 : MATHEMATICAL MODELLING PRINCIPLES Outline of the lesson. • Reasons why we need dynamic models • Six (6) - step modelling procedure • Many examples - mixing tank - CSTR - draining tank • General conclusions about models • Workshop

WHY WE NEED DYNAMIC MODELS Do the Bus and bicycle have different dynamics? • Which can make a U-turn in 1.5 meter? • Which responds better when it hits s bump? Dynamic performance depends more on the vehicle than the driver! The process dynamics are more important than the computer control!

WHY WE NEED DYNAMIC MODELS Feed material is delivered periodically, but the process requires a continuous feed flow. How large should should the tank volume be? Periodic Delivery flow Continuous Feed to process Time We must provide process flexibility for good dynamic performance!

WHY WE NEED DYNAMIC MODELS The cooling water pumps have failed. How long do we have until the exothermic reactor runs away? Temperature F Dangerous T L A time

WHY WE NEED DYNAMIC MODELS The cooling water pumps have failed. How long do we have until the exothermic reactor runs away? Temperature F Dangerous T L A time Process dynamics are important for safety!

WHY WE DEVELOP MATHEMATICAL MODELS? Input change, Effect on e.g., step in Process output coolant flow rate variable T A • How far? How does the process • How fast influence the response? • “Shape”

WHY WE DEVELOP MATHEMATICAL MODELS? Input change, Effect on e.g., step in Process output coolant flow rate variable T A • How far? Math models How does the help us answer process • How fast these questions! influence the response? • “Shape”

SIX-STEP MODELLING PROCEDURE 1. Define Goals We apply this procedure 2. Prepare • to many physical systems information • overall material balance 3. Formulate • component material balance the model • energy balances 4. Determine the solution 5. Analyze Results T A 6. Validate the model

SIX-STEP MODELLING PROCEDURE 1. Define Goals • What decision? 2. Prepare • What variable? information • Location T 3. Formulate A the model 4. Determine Examples of variable selection the solution → liquid level total mass in liquid 5. Analyze → pressure total moles in vapor Results → temperature energy balance 6. Validate the model concentration → component mass

SIX-STEP MODELLING PROCEDURE 1. Define Goals Key property • Sketch process of a “system”? 2. Prepare • Collect data information • State 3. Formulate assumptions the model • Define system 4. Determine the solution 5. Analyze T Results A 6. Validate the model

SIX-STEP MODELLING PROCEDURE 1. Define Goals Key property • Sketch process of a “system”? 2. Prepare • Collect data information • State 3. Formulate assumptions the model • Define system 4. Determine Variable(s) are the the solution same for any location within 5. Analyze the system! T Results A 6. Validate the model

SIX-STEP MODELLING PROCEDURE CONSERVATION BALANCES 1. Define Goals Overall Material 2. Prepare { } { } { } = − Accumulati on of mass mass in mass out information Component Material 3. Formulate the model       Accumulati on of component component = −       component mass mass in mass out       4. Determine   generation of the solution +   component mass   Energy* 5. Analyze Results  Accumulati on  { } { } = + + − + + H PE KE H PE KE   in out + + U PE KE   6. Validate the + Q - W s model * Assumes that the system volume does not change

SIX-STEP MODELLING PROCEDURE 1. Define Goals • What type of equations do we use first? 2. Prepare Conservation balances for key variable information • How many equations do we need? 3. Formulate Degrees of freedom = NV - NE = 0 the model • What after conservation balances? 4. Determine the solution Constitutive Not equations, e.g., 5. Analyze fundamental, Results based on Q = h A ( ∆ T) empirical data 6. Validate the r A = k 0 e -E/RT model

SIX-STEP MODELLING PROCEDURE Our dynamic models will involve 1. Define Goals differential (and algebraic) equations 2. Prepare because of the accumulation terms. information dC = − − A V F ( C C ) VkC 3. Formulate A 0 A A dt the model With initial conditions 4. Determine C A = 3.2 kg-mole/m 3 at t = 0 the solution 5. Analyze And some change to an input Results variable, the “forcing function”, e.g., 6. Validate the C A0 = f(t) = 2.1 t (ramp function) model

SIX-STEP MODELLING PROCEDURE 1. Define Goals We will solve simple models analytically to provide excellent relationship between 2. Prepare process and dynamic response, e.g., information − τ t / = + ∆ − C ( t ) C ( t ) ( C ) K ( 1 e ) 3. Formulate A A A 0 = t 0 the model for t 0 f 4. Determine Many results will have the same the solution form! We want to know how the 5. Analyze process influences K and τ , e.g., Results F V 6. Validate the = τ = K + + F kV F Vk model

SIX-STEP MODELLING PROCEDURE We will solve complex models 1. Define Goals numerically, e.g., 2. Prepare dC information = − − 2 A V F ( C C ) VkC A 0 A A dt 3. Formulate the model Using a difference approximation for the derivative, we can derive the 4. Determine Euler method. the solution   2 − − F ( C C ) VkC 5. Analyze = + ∆ A 0 A A C C ( t )   A A V − n n 1   Results − n 1 6. Validate the Other methods include Runge-Kutta model and Adams.

SIX-STEP MODELLING PROCEDURE 1. Define Goals • Check results for correctness - sign and shape as expected 2. Prepare - obeys assumptions information - negligible numerical errors 3. Formulate • Plot results the model • Evaluate sensitivity & accuracy 4. Determine the solution 5. Analyze Results 6. Validate the • Compare with empirical data model

MODELLING EXAMPLE 1. MIXING TANK Textbook Example 3.1: The mixing tank in the figure has been operating for a long time with a feed concentration of 0.925 kg-mole/m 3 . The feed composition experiences a step to 1.85 kg-mole/m 3 . All other variables are constant. Determine the dynamic response. F C A0 C A V (We’ll solve this in class.)

Let’s understand this response, because we will see it over and over! Output is smooth, monotonic curve 1.8 1.6 Maximum tank concentration At steady state slope at ≈ 63% of steady-state ∆ C A 1.4 “t=0” ∆ C A = K ∆ C A0 1.2 1 τ 0.8 0 20 40 60 80 100 120 time Output changes immediately 2 inlet concentration 1.5 ∆ C A0 Step in inlet variable 1 0.5 0 20 40 60 80 100 120 time

MODELLING EXAMPLE 2. CSTR The isothermal, CSTR in the figure has been operating for a long time with a feed concentration of 0.925 kg-mole/m 3 . The feed composition experiences a step to 1.85 kg- mole/m 3 . All other variables are constant. Determine the dynamic response of C A . Same parameters as textbook Example 3.2 F C A0 → A B C A V − = r kC A A (We’ll solve this in class.)

MODELLING EXAMPLE 2. CSTR Annotate with key features similar to Example 1 1 reactor conc. of A (mol/m3) 0.8 Which is faster, mixer or CSTR? 0.6 Always? 0.4 0 50 100 150 time (min) 2 inlet conc. of A (mol/m3) 1.5 1 0.5 0 50 100 150 time (min)

MODELLING EXAMPLE 2. TWO CSTRs Two isothermal CSTRs are initially at steady state and experience a step change to the feed composition to the first tank. Formulate the model for C A2 . Be especially careful when defining the system! F C A0 C A1 V1 C A2 V2 → A B − = r kC A A (We’ll solve this in class.)

SIX-STEP MODELLING PROCEDURE 1. Define Goals We can solve only a few models analytically - those that are linear 2. Prepare (except for a few exceptions). information We could solve numerically. 3. Formulate the model We want to gain the INSIGHT from learning how K (s-s gain) and τ ’s 4. Determine (time constants) depend on the the solution process design and operation. 5. Analyze Results Therefore, we linearize the models, even though we will not achieve an 6. Validate the exact solution! model

LINEARIZATION Expand in Taylor Series and retain only constant and linear terms. We have an approximation. This is the only variable 2 dF 1 d F 2 = + − + − + F ( x ) F ( x ) ( x x ) ( x x ) R s s s 2 dx 2 ! dx x x s s Remember that these terms are constant because they are evaluated at x s We define the deviation variable: x’ = (x - x s )