process simulation introduction. model, what for? a good model of the apparatus is needed for:...
Post on 20-Dec-2015
219 views
TRANSCRIPT
The model It is a representation of real object
(apparatus or part of the apparatus) It let us foresee behavior of physical objects
without experiments in the real world. However, usually the basis of the models are
experiments Approximation of constants/coefficients in model
equations Physical parameters of chemicals, etc.
Classification of the models Physical – mathematical
Physical – one physical quantity replaced with another (easier to measure) or use of scaled down objects (cars, planes in wind tunnels)
Mathematical - representation by the use of equations.
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 are found 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
1. Balance dependences Based upon basic nature laws
of conservation of mass of conservation of energy of conservation of electric charge, etc.
Balance equation: Input – Output = Accumulation
Elements of the model
2. Constitutive equations – apply to unconvectional streams
Newton eq. – for viscous friction Fourier eq. – for heat conduction Fick eq. – for mass diffusion
dt
dx
dt
xdm
2
2
S
STdt
Q
2
2
xD
t
Elements of the model3. Phase equilibrium equations – important
for mass transfer
4. Physical properties equations – for calculation parameters as functions of temperature, pressure and concentrations.
5. Geometrical dependences – involve influence of apparatus geometry on transfer coefficients – convectional streams.
Structure of the simulation model 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 vary with space – 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 Design (specification) problem Optimization problem Synthesis problem
by Rafiqul Gani
Flowsheeting problem Given:
All input information All operating condition All equipment parameters
To calculate: All outputs FLOWSHEET
SCHEMEINPUT
OPERATING CONDITIONS
EQUIPMENT PARAMETERS
PRODUCTS
Specyfication problem
Given: Some input&output information Some operating condition Some equipment parameters
To calculate: Rest of inputs&outputs Rest of operating condition Rest of equipment parameters
FLOWSHEETSCHEME
INPUT
OPERATING CONDITIONS
EQUIPMENT PARAMETERS
PRODUCTS
To gues: D, Qr
Solve the flowsheetingproblem
STOP
Is target product composition satisfied
?
Adjust D, Qr
Given: feed composition and flowrates, target product composition
Find: product flowrates, heating duties
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)
To gues: D, Qr
Solve the flowsheetingproblem
STOP
Is target product composition satisfied
AND =min.
Adjust D, Qr
Given: feed composition and flowrates, target product composition
Find: product flowrate, heating dutie
Process synthesis/design problem the act of creation of a new process. Given:
inputs (some feeding streams can be added/changed)
Outputs (some byproducts may be unknown)
To find: flowsheet equipment parameters operations conditions
To gues: D, Qr
Solve the flowsheetingproblem
STOP
Is target product composition satisfied
AND =min.
Adjust D, Qr
As well asN, NF, R/D etc.
Given: feed composition and flowrates, target product composition
Find: product flowrate, heating dutie
Process synthesis/design problem
Separation method & equipment
methanolwater
methanol
water
Methods: distillation, membrane separation, flash, extraction
Equipment: how many apparatus are needed what is apparatus design and conditions
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 – MathCAD, Matlab
Specialized software – flowsheeting programs. Equipped with:
Data base of apparatus models Data base of components 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. ASPEN One 2. HYSIM3. ChemCAD4. PRO/II5. ProMax
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 according to separately:
1. Material streams2. Apparatus
Apparatus-streams separation Why separate?
It’s make calculations easier Assumption:
All processes (chemical reaction, heat exchange etc.) taking places in the apparatus and streams are in the chemical and thermodynamical equilibrium state.
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=4Assumed: F1, F2, T, PCalculated:
•enthalpy•vapor fraction
Apparatus parameters Characteristics for each apparatus
type. E.g. heat exchanger :
Heat exchange area, A [m2] Overall heat-transfer coefficient, U (k)
[Wm-2K-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 apparatus mathematical model use)
Detailed balance on the base of apparatus model
Overall balance Apparatus is treated 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
Countercurrent, tube-shell heat exchangerGiven three streams data: 1, 2, 3 hence parameters of stream 4 can be easily calculated from the balance equation. 1
2
4
3
There is possibility that calculated temp. of stream 4 can be higher then inlet temp. of heating medium (stream 1).
DF=5
2134 ttcmttcm pBBpAA
Overall balance – ExampleGiven: 1. mA=10kg/s2. mB=20kg/s3. t1= 70°C4. t2=40°C5. t3=20°CcpA=cpB=f(t)
1, mB
2
4
3, mA
2134 ttm
mtt
A
B
Ct 80407010
20204
2134 ttcmttcm pBBpAA
Apparatus model involved Process is being described with use of
modeling equations (differential, dimensionless etc.)
Only physically possible 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 initial parameters of cut stream has to be
defined Calculations has to be repeated until cut
streams parameters are converted.
I.Problem definitionSimulate 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 1bar.
Stream s1
Ps1 =200kPa, ts1 = 85°C, f1s1 = 1000kg/h
Stream s6
Ps6 =200kPa, ts6 = 20°C, f2s6 = 1000kg/h
Numerical data:
Equipment parameters:1. L1=7m d1=0,025m
2. L2=5m d2=0,16m, s=0,0016m, n=31...
3. L3=6m, d3=0,025m
4. 4=50
5. L5=7m d5=0,025m
6. L6=10m, d6=0,025m
7. 7=40
III. Stream summary table
s1 s2 s3 s4 s5 s6 s7 s8 s9 s10
f1 f1s1 X X X X
f2 f2s6 X X X X
T Ts1 X X X X Ts6 X X X X
P Ps1 X x x X Ps6 X X X X
Uknown:Ts2, Ts3, Ts4, Ts5, Ts7, Ts8, Ts9, Ts10, Ps2, Ps3, Ps4, Ps7, Ps8, Ps9, f1s2,
f1s3, f1s4, f1s5, f2s7, f2s8, f2s9, f2s10
number of unknown variables: 24WE NEED 24 INDEPENDENT EQUATIONS.
f1s2= f1s1 f1s7= f1s6
f1s3= f1s2 f1s8= f1s7
f1s4= f1s3 f1s9= f1s8
f1s5= f1s4 f1s10= f1s9
12 TT
34 TT
45 TT
67 TT
89 TT
910 TT
Equations from equipment information
14 equations. Still do define 24-14=10 equations
Heat balance equations
8
7
2
3
T
T
pS
T
T
pT
QdTcm
QdTcm
QTTcf
QTTcf
pSs
pTs
786
321
2
1
New variable: QStill to define: 10+1-2=9 equations
Heat exchange equations
mm TkFQ
New variables: k, Tm: number of equations to find 9+2-2=9
101
65
10165
lnss
ss
ssssm
TT
TTTTTT
T
Heat exchange equations
Two new variables: T and S number of equations to find: 9+2-1=10
SstT
sk
11
1
Heat exchange equations
Three new variables: NuT, NuS, deq,
number of equations to find: 10+3-3=10
2d
Nu TTT
eq.d
Nu SSS
22
22
22
ndD
ndDdeq
Heat exchange equations
10000Re2300RelnRelnRelnReln
lnlnlnexp
5,2300Re62,1PrRe62,1
5,2300Re5,0PrRe5,0
10000RePrRe023,0
3/1
3/1
2
2
4,08,0
TTLTTuTLT
TuTLTLT
TTTT
TTHEX
HEXTT
TT
T
NuNuNu
GzGzl
d
GzGzl
d
Nu
T
s
T
T d
fwd
2
12 14Re
Heat exchange equations
10000Re2300RelnRelnRelnReln
lnlnlnexp
5,2300Re62,1PrRe62,1
5,2300Re5,0PrRe5,0
10000RePrRe023,0
3/1
3/1
4,08,0
SSLSTuSLS
TuSLSLS
SSHEX
HEXSS
SSHEX
HEXTS
SS
S
NuNuNu
GzGzl
d
GzGzl
d
Nu
Two new variables ReT and ReS,
number of equations to find: 10+2-4=8
SCSA
eqs
S
eqS F
dfwd
.6. 2
Re
Pressure drop
Ps1-Ps2=P1
Ps2-Ps3=P2T
Ps3-Ps4=P3
Ps4-Ps5=P4
Ps6-Ps7=P5
Ps7-Ps8=P2S
Ps8-Ps9=P6
Ps9-Ps10=P7
Eight new variables: P1, P2T, P3, P4, P5,
P2S, P6, P7, number of equations to find: 8+8-
8=8
Pressure drop
1
4
21
1
2
1
116
2 d
l
d
f
d
lwP s
1
237,0
525,01
Re
221,00032,0
10Re2300,Re
3164,0
2300Re,Re
64
1
11
14Re
d
f s
Two new variables Re1 and 1, number of equations to find: 8+2-3=7
Pressure drop
Two new variables Re2t and 2T, number of equations to find: 7+2-3=6
THEX
HEXs
T
T d
l
dn
f
d
lwP
2
4
21
2
2
2
116
2
T
T
2237,0
525,02
Re
221,00032,0
10Re2300,Re
3164,0
2300Re,Re
64
T
sT nd
f
2
12
14Re
Pressure drop
Two new variables Re3 and 3, number of equations to find: 6+2-3=5
3
4
21
3
2
3
116
2 d
l
d
f
d
lwP s
3237,0
525,03
Re
221,00032,0
10Re2300,Re
3164,0
2300Re,Re
64
3
13
14Re
d
f s
Pressure drop
Two new variables Re5 and 5,
number of equations to find: 4+2-3=3
5
4
22
5
2
5
216
2 d
l
d
f
d
lwP s
5237,0
525,05
Re
221,00032,0
10Re2300,Re
3164,0
2300Re,Re
64
5
25
24Re
d
f s
Pressure drop
Two new variables Re2S and 2S, number of equations to find: 3+2-3=2
SeqCSA
s
SeqS d
l
F
f
d
lwP
2.
22
2
2.
22
2
216
2
S
S
2237,0
525,02
Re
221,00032,0
10Re2300,Re
3164,0
2300Re,Re
64
Seq
sS d
f
2.
22
24Re
Pressure drop
Two new variables Re6 and 6,
number of equations to find: 2+2-3=1
6
4
22
6
2
6
216
2 d
l
d
f
d
lwP s
5237,0
525,06
Re
221,00032,0
10Re2300,Re
3164,0
2300Re,Re
64
5
26
24Re
d
f s
Agents parameters
Temperatures are not constant Liquid properties are functions of temperature
•Specyfic heat cp
•Viscosity
•Density
•Thermal conductivity
•Prandtl number Pr
Agents parametersData are usually published in the tables
t cp Pr
0,00 999,80 17,89 0,551 4237 13,7610,00 999,60 13,04 0,575 4212 9,5520,00 998,20 10,00 0,599 4203 7,0230,00 995,60 8,014 0,618 4199 5,4540,00 992,20 6,531 0,634 4199 4,3350,00 988,00 5,495 0,648 4199 3,5660,00 983,20 4,709 0,659 4203 3,0070,00 977,70 4,059 0,668 4211 2,5680,00 971,80 3,559 0,675 4216 2,2290,00 965,30 3,147 0,68 4224 1,95
100,00 958,30 2,822 0,683 4229 1,75
Data in tables are difficult to use
Solution:
Approximate discrete data by the continuous functions.
Agents parameters
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 y = 1,540E-05x3 - 5,895E-03x2 + 2,041E-02x +
9,999E+02
950
960
970
980
990
1000
1010
0 20 40 60 80 100
t [°C]
[k
g/m
3 ]