Introduction Process Simulation Classification of the models Black

Introduction Process Simulation

Classification of the models Black box – white box Black box – know nothing about process in apparatus, only dependences between inputs and outputs are established. Practical realisation of Black box is the neural network White box – process mechanism is well <? ? > known and described by system of equations

Classification of the models Deterministic – Stochastic Deterministic – for one given set of inputs only one set of outputs is calculated with probability equal 1. Stochastic – random phenomenon affects on process course (e. g. weather), output set is given as distribution of random variables

Classification of the models Microscopic- macroscopic Microscopic – includes part of process or apparatus Macroscopic – includes whole process or apparatus

Elements of the model Balance dependences 1. Based upon basic nature laws of of conservation of of mass energy atoms number electric charge, etc. Balance equation (for mass): (overall and for specific component without reaction) Input – Output = Accumulation or (for specific component if chemical reactions presents) Input – Output +Source = Accumulation

Elements of the model Constitutive equations 2. Newton eq. – for viscous friction Fourier eq. – for heat conduction Fick eq. – for mass diffusion

Elements of the model 3. 4. 5. Phase equilibrium equations – important for mass transfer Physical properties equations – for calculation parameters as functions of temperature, pressure and concentrations. Geometrical dependences – involve influence of apparatus geometry on transfer coefficients – convectional streams.

Structure of the simulation model Structure corresponds to type of model equations Structure depends on: Type of object work: Continuous, steady running Periodic, unsteady running Distribution of parameters in space Equal in every point of apparatus – aggregated parameters (butch reactor with ideal mixing) Parameters are space dependent– displaced parameters

Structure of the model Steady state Unsteady state Aggregated parameters Algebraic eq. Ordinary differential eq. Displaced parameters Differential eq. 1. Ordinary for 1 dimensional case 2. Partial for 2&3 dimensional case (without time derivative, usually elliptic) Partial differential eq. (with time derivative, usually parabolic)

Process simulation the act of representing some aspects of the industry process (in the real world) by numbers or symbols (in the virtual world) which may be manipulated to facilitate their study.

Process simulation (steady state) Flowsheeting problem Specification (design) problem Optimization problem Synthesis problem by Rafiqul Gani

Flowsheeting problem Given: All of the input information All of the operating condition All of the equipment parameters To calculate: All of the outputs INPUT FLOWSHEET SCHEME OPERATING CONDITIONS PRODUCTS EQUIPMENT PARAMETERS

R. Gani

Specyfication problem INPUT Given: OPERATING CONDITIONS Some input & some output information Some operating condition Some equipment parameters To calculate: FLOWSHEET SCHEME Undefined inputs&outputs Undefined operating condition Undefined equipment parameters PRODUCTS EQUIPMENT PARAMETERS

Specyfication problem NOTE: degree of freedom is the same as in flowsheeting problem.

Given: feed composition and flowrates, target product composition Assume value to be guessed: D, Qr Solve the flowsheeting problem Is target product composition satisfied ? STOP Find: product flowrates, heating duties Adjust D, Qr

Process optimisation the act of finding the best solution (minimize capital costs, energy. . . maximize yield) to manage the process (by changing some parameters, not apparatus)

Given: feed composition and flowrates, target product composition Assume value to be guessed: D, Qr Solve the flowsheeting problem Is target product composition satisfied AND =min. STOP Find: product flowrate, heating duty Adjust D, Qr

Process synthesis/design problem the act of creation of a new process. Given: inputs (some feeding streams can be added/changed latter) Outputs (some byproducts may be unknown) To find: Flowsheet (topology) equipment parameters operations conditions

Process synthesis/design problem INPUT flowsheet undefined OUTPUT

Given: feed composition and flowrates, target product composition Assume value to be guessed: D, Qr, N, NF, R/D etc. Solve the flowsheeting problem Is target product composition satisfied AND =min. STOP Find: product flowrate, heating duty, column param. etc. Adjust D, Qr As well as N, NF, R/D etc.

Process simulation - why? COSTS Material – easy to measure Time – could be estimated Risc – hard to measure and estimate

Software for process simulation Universal software: Worksheets – Excel, Calc (Open Office) Mathematical software – Math. CAD, Matlab Specialized software – process simulators. Equipped with: Data base of apparatus models Data base of components and mixtures properties Solver engine User friendly interface

Software process simulators (flawsheeting programs) Started in early 70’ At the beginning dedicated to special processes Progress toward universality Some actual process simulators: 1. 2. 3. 4. 5. ASPEN Tech /HYSYS Chem. CAD PRO/II Pro. Sim Design II for Windows

Chemical plant system The apparatus set connected with material and energy streams. Most contemporary systems are complex, i. e. consists of many apparatus and streams. Simulations can be use during: Investigation works – new technology Project step – new plants (technology exists), Runtime problem identification/solving – existing systems (technology and plant exists)

Chemical plant system characteristic parameters can be specified for every system separately according to: 1. 2. Material streams Apparatus

Apparatus-streams separation Assumption: All processes (chemical reaction, heat exchange etc. ) taking places in the apparatus and streams are in the chemical and thermodynamical equilibrium state. Why separate? It’s make calculations easier

Streams parameters Flow rate (mass, volume, mol per time unit) Composition (mass, volume, molar fraction) Temperature Pressure Vapor fraction Enthalpy

Streams degrees of freedom DFs=NC+2 e. g. : NC=2 -> DFs=4 Assumed: F 1, F 2, T, P Calculated: • enthalpy • vapor fraction

Apparatus parameters & DF Characteristics for each apparatus type. E. g. heat exchanger : Heat exchange area, A [m 2] Overall heat-transfer coefficient, U (k) [Wm -2 K-1] Log Mean Temperature Difference, LMTD [K] degrees of freedom are unique to equipment type

Calculation subject Number of equations of mass and energy balance for entire system Can be solved in two ways:

Types of balance calculation Overall balance (without use of apparatus mathematical model) Detailed balance on the base of apparatus model

Overall balance Apparatus is considered as a black box Needs more stream data User could not be informed about if the process is physically possible to realize.

Overall balance – Example 3 2 Countercurrent, tube-shell heat exchanger Given three streams data: 1, 2, 3 hence parameters of stream 4 can be easily calculated from the balance equation. 1 4 DF=5 There is possibility that calculated temp. of stream 4 can be higher then inlet temp. of heating medium (stream 1).

Overall balance – Example 3, m. A 2 1, m. B 4 Given: 1. m. A=10 kg/s 2. m. B=20 kg/s 3. t 1= 70°C 4. t 2=40°C 5. t 3=20°C cp. A=cp. B=idem

Apparatus model involved Process is being described with use of modeling equations (differential, dimensionless etc. ) Only physically acceptable processes taking place Less stream data required (smaller DF number) Heat exchange example: given data for two streams, the others can be calculated from a balance and heat exchange model equations

Loops and cut streams Loops occur when: some products are returned and mixed with input streams when output stream heating (cooling) inputs some input (also internal) data are undefined To solve: one stream inside the loop has to be cut (tear stream) initial parameters of cut stream have to be defined Calculations have to be repeated until cut streams parameters are converted.

Loops and cut streams

Simulation of system with heat exchanger using Math. CAD

I. Problem definition Simulate system consists of: Shell-tube heat exchanger, four pipes and two valves on output pipes. Parameters of input streams are given as well as pipes, heat exchanger geometry and valves resistance coefficients. Component 1 and 2 are water. Pipe flow is adiabatic. Find such a valves resistance to satisfy condition: both streams output pressures equal 1 bar.

II. Flawsheet 5 s 6 s 7 2 1 s 1 4 3 s 2 s 3 s 4 s 8 s 5 7 6 s 9 s 10

Numerical data: Stream s 1 Ps 1 =200 k. Pa, ts 1 = 85°C, f 1 s 1 = 10000 kg/h Stream s 6 Ps 6 =200 k. Pa, ts 6 = 20°C, f 2 s 6 = 10000 kg/h

Equipment parameters: 1. L 1=7 m d 1=0, 025 m 2. L 2=5 m d 2=0, 16 m, s=0, 0016 m, n=31. . . 3. L 3=6 m, d 3=0, 05 m 4. z 4=50 5. L 5=7 m d 5=0, 05 m 6. L 6=10 m, d 6=0, 05 m 7. z 7=40

III. Stream summary table Uknown: Ts 2, Ts 3, Ts 4, Ts 5, Ts 7, Ts 8, Ts 9, Ts 10, Ps 2, Ps 3, Ps 4, Ps 5, Ps 7, Ps 8, Ps 9, Ps 10, f 1 s 2, f 1 s 3, f 1 s 4, f 1 s 5, f 2 s 7, f 2 s 8, f 2 s 9, f 2 s 10 number of unknown variables: 26 WE NEED 26 INDEPENDENT EQUATIONS.

Equations from equipment information f 1 s 2= f 1 s 1 f 1 s 7= f 1 s 6 f 1 s 3= f 1 s 2 f 1 s 8= f 1 s 7 f 1 s 4= f 1 s 3 f 1 s 9= f 1 s 8 f 1 s 5= f 1 s 4 f 1 s 10= f 1 s 9 14 equations. Still do define 2614=12 equations

Heat balance equations New variable: Q Still to define: 12+1 -2=11 equations

Heat exchange equations New variables: k, DTm: number of equations to find 11+2 -2=11

Heat exchange equations Two new variables: a. T and a. S number of equations to find: 11+2 -1=12

Heat exchange equations Three new variables: Nu. T, Nu. S, deq, number of equations to find: 12+3 -3=12

Heat exchange equations

Heat exchange equations Two new variables Re. T and Re. S, number of equations to find: 12+2 -4=10

Pressure drop

Pressure drop Two new variables Re 1 and l 1, number of equations to find: 10+2 -3=9

Pressure drop One new variables and l 2 T, number of equations to find: 9+1 -3=7

Pressure drop Two new variables Re 3 and l 3, number of equations to find: 7+2 -3=6

Pressure drop Number of equations to find: 6 -1=5

Pressure drop Two new variables Re 5 and l 5, number of equations to find: 6+2 -3=4

Pressure drop One new variables and l 2 S, number of equations to find: 4+1 -3=2

Pressure drop Two new variables Re 6 and l 6, number of equations to find: 2+2 -3=1

Pressure drop Number of equations to find: 1 -1=0 !!!!!!!

Agents parameters Temperatures are not constant Liquid properties are functions of temperature • Density • Viscosity • Thermal conductivity • Specyfic heat cp • Prandtl number Pr

Agents parameters Data are usually published in the tables

Agents parameters Data in tables are difficult to use Solution: Approximate discrete data by the continuous functions.

Approximation Approximating function Polynomial Approximation target: find optimal parameters of approximating function Approximation type Mean-square – sum of square of differences between discrete (from tables) and calculated values is minimum.

Polynomial approximation

The end as of yet.