selected differential system examples from lectures
DESCRIPTION
Selected Differential System Examples from Lectures. w i. V = Ah. w o. Liquid Storage Tank. Standing assumptions Constant liquid density r Constant cross-sectional area A Other possible assumptions Steady-state operation Outlet flow rate w 0 known function of liquid level h. - PowerPoint PPT PresentationTRANSCRIPT
Selected Differential System Examples from Lectures
Liquid Storage Tank
Standing assumptions» Constant liquid density » Constant cross-sectional area A
Other possible assumptions» Steady-state operation» Outlet flow rate w0 known function of liquid level h
V = Ah
wi
wo
Mass Balance
Mass balance on tank
Steady-state operation: Valve characteristics
Linear ODE model
Nonlinear ODE model
0
out
0
inonaccumulati
)( wwdtdhAww
dtAhd
ii
ii wwww 000
hCwhCw vovo NonlinearLinear
0)0( hhhCwdtdhA vi
0)0( hhhCwdtdhA vi
Stirred Tank Chemical Reactor
Assumptions» Pure reactant A in feed stream» Perfect mixing» Constant liquid volume» Constant physical properties (, k)» Isothermal operation
A
k
kCrBA
Overall mass balance
Component balance
qqqqdt
Vdii 0)(
0)0()(
)(
AAAAAiA
AAAiiA
CCkCCCVq
dtdC
VkCqCCqdtVCd
Plug-Flow Chemical Reactor
Assumptions» Pure reactant A in feed stream» Perfect plug flow» Steady-state operation» Isothermal operation» Constant physical properties (, k)
A
k
kCrBA
z
qi, CAi qo, CAoCA(z) z
Plug-Flow Chemical Reactor cont.
Overall mass balance
qqqdzdq
zqq
i
zzz
z
zzz
0
0
out Massin Mass
0
0)()(lim
0)()(
z
qi, CAi qo, CAoCA(z) z
Component balance
AiAAA
AA
AzzAzA
z
AzzAzA
CCkCdz
dCAq
kCdz
dCAq
kCzCC
Aq
zAkCqCqC
)0(0
0
0)()(lim
0)()(
0
consumedA outA inA
Continuous Biochemical Reactor
Fresh Media Feed (substrates)
Exit Gas Flow
Agitator
Exit Liquid Flow(cells & products)
Cell Growth Modeling Specific growth rate
Yield coefficients» Biomass/substrate: YX/S = -X/S» Product/substrate: YP/S = -P/S» Product/biomass: YP/X = P/X» Assumed to be constant
Substrate limited growth
» S = concentration of rate limiting substrate» Ks = saturation constant» m = maximum specific growth rate (achieved when S >> Ks)
(g/L)ion concentrat biomass 1 X
dtdX
X
SKSS
S
m
)(
Continuous Bioreactor Model
Assumptions Sterile feed Constant volume Perfect mixing Constant temperature and pH Single rate limiting nutrient Constant yields Negligible cell death
Product formation rates» Empirically related to specific growth rate» Growth associated products: q = YP/X» Nongrowth associated products: q = » Mixed growth associated products: q = YP/X
Mass Balance Equations Cell mass
» VR = reactor volume» F = volumetric flow rate» D = F/VR = dilution rate
Product
Substrate
» S0 = feed concentration of rate limiting substrate
XDXdtdXXVFX
dtdXV RR
qXDPdtdPqXVFP
dtdPV RR
XY
SSDdtdSXV
YFSFS
dtdSV
SXR
SXR
/0
/0
1)(1
Exothermic CSTR
Scalar representation
Vector representation
00 )0()exp()( AAAAAfA CCCRTEkCC
Vq
dtdC
00 )0()()/exp()()( TT
VCTTUA
CCRTEkHTT
Vq
dtdT
p
c
p
Af
0
0
0
0
)0()(
)()/exp()()(
)/exp()()(
TC
dtd
TTVC
UAC
CRTEkHTTVq
CRTEkCCVq
TC
A
cpp
Af
AAAfA
yyfy
yfy
)(/exp)( 0
TTUAQRTEkTk
BA
c
k
Isothermal Batch Reactor
CSTR model: A B C
Eigenvalue analysis: k1 = 1, k2 = 2
Linear ODE solution:
0)0(,10)0(211 BABAB
AA CCCkCk
dtdCCk
dtdC
11
110
2
2101
)()(
)(
)2(2
)1(1 xx
A
yydtdy
tCtC
tyB
A
tttt
B
A ecececectCtC
t
11
10
)()(
)( 22
1)2(
2)1(
121 xxy
Isothermal Batch Reactor cont.
Linear ODE solution:
Apply initial conditions:
Formulate matrix problem:
Solution:
tt
B
A ecectCtC
t
11
10
)()(
)( 22
1y
010
11
10
)0()0(
)0( 21 ccCC
B
Ay
1010
010
1110
2
1
2
1
cc
cc
tt
ttt
B
A
eee
eetCtC
101010
11
1010
10)()(
22
Isothermal CSTR
Nonlinear ODE model
Find steady-state point (q = 2, V = 2, Caf = 2, k = 0.5)
)(2)(2 2AAAAf
Ak CfkCCCVq
dtdCBA
12
31)1)(2(
)2)(1)(4(11
02
0)5.0)(2()2(222)()(
2
2
22
A
AA
AAAAAfA
C
CC
CCCkCCVqCf
Isothermal CSTR cont.
Linearize about steady-state point:
This linear ODE is an approximation to the original nonlinear ODE
'''
''
3])5.0)(2)(2([
0
)(
AAAAA
ACA
AA
CCCCdt
dC
CCfCf
dtdC
A
Continuous Bioreactor Cell mass balance
Product mass balance
Substrate mass balance
XDXdtdX
qXDPdtdP
XY
SSDdtdS
SX
/
01)(
Steady-State Solutions Simplified model equations
Steady-state equations
Two steady-state points
),()(1)(
)(),()(
2/
0
1
SXfXSY
SSDdtdS
SKSSSXfXSDX
dtdX
SX
S
m
0)(1)(
)(0)(
/0
XSY
SSD
SKSSXSXD
SX
S
m
0:Washout
)()(:Trivial-Non
0
0/
XSS
SSYXD
DKSDS SXm
S
Model Linearization Biomass concentration equation
Substrate concentration equation
Linear model structure:
SSKSX
SKXXDS
SSSfXX
XfSXf
dtXd
S
m
S
m
SXSX
2
,
1
,
11
zero),(
SDSKSX
SKX
YX
SKS
Y
SSSfXX
XfSXf
dtSd
S
m
S
m
SXS
m
SX
SXSX
2//
,
2
,
22
11zero
),(
SaXadtSd
SaXadtXd
2221
1211
Non-Trivial Steady State
Parameter values» KS = 1.2 g/L, m
= 0.48 h-1, YX/S = 0.4 g/g
» D = 0.15 h-1, S0 = 20 g/L
Steady-state concentrations
Linear model coefficients (units h-1)
529.31375.01
472.10
2/
22/
21
21211
DSKSX
SKX
Ya
SKS
Ya
SKSX
SKXaa
S
m
S
m
SXS
m
SX
S
m
S
m
g/L 78.7)(g/L 545.0 0/
SSYXD
DKS SXm
S
Stability Analysis Matrix representation
Eigenvalues (units h-1)
Conclusion» Non-trivial steady state is asymptotically stable» Result holds locally near the steady state
Axxdtdx
SX
x
529.3375.0
472.10
365.3164.0529.3375.0472.1
11
IA
Washout Steady State Steady state: Linear model coefficients (units h-1)
Eigenvalues (units h)
Conclusion» Washout steady state is unstable» Suggests that non-trivial steady state is globally stable
15.01132.11
0303.0
2
maxmax
/22
max
/21
12
max
11
DSKSX
SKX
Ya
SKS
Ya
aDSKSa
SSSXSSX
S
g/L 0g/L 20 XSS i
15.0303.015.0132.1
0303.011
IA
Gaussian Quadrature Example Analytical solution
Variable transformation
Approximate solution
Approximation error = 4x10-3%
067545.2125
1
5
121
21
xx edxe
32)(223
21
txxab
baxt
066691.21)533346.10(2
533346.10)55555.0()88889.0()55555.0(
22
5
1
1
1
77459.0077459.0
1
1
1
1
)32(5
1
21
23
23
23
23
23
21
21
dxe
eeedte
dtedtedxe
x
t
ttx
Plug-Flow Reactor Example
Ai
N
NAA
AiAnAnA
nAnAnA
AiAAA
CqzkA
zCLC
CzCzCqzkA
zC
zkCz
zCzCAq
CCkCdz
dCAq
11)()(
)()(1
1)(
0)()()(
)0(0
01
11
A
k
kCrBA
z
qi, CAi qo, CAoCA(z) z
0 L
Plug-Flow Reactor Example cont. Analytical solution
Numerical solution
Convergence formula
Convergence of numerical solution
Ai
N
Ai
N
A CNqkAL
CqzkA
LC
1
11
1)(
L
qkACLC AiA exp)(
LqkACC
NqLkA AiAi
N
Nexp
11lim
aN
Ne
Na
1
1lim
Matlab Example
Isothermal CSTR model
Model parameters: q = 2, V = 2, Caf = 2, k = 0.5
Initial condition: CA(0) = 2 Backward Euler formula
Algorithm parameters: h = 0.01, N = 200
)(2)(2 2AAAAf
Ak CfkCCCVq
dtdCBA
)(2)( ,,2
,,,1, nAnAnAnAAfnAnA ChfCkCCCVqhCC
Matlab Implementation: iso_cstr_euler.m
h = 0.01;
N = 200;
Cao = 2;
q = 2;
V = 2;
Caf = 2;
k = 0.5;
t(1) = 0;
Ca(1) = Cao;
for i=1:N
t(i+1) = t(i)+h;
f = q/V*(Caf-Ca(i))-2*k*Ca(i)^2;
Ca(i+1)= Ca(i)+h*f;
end
plot(t,Ca)
ylabel('Ca (g/L)')
xlabel('Time (min)')
axis([0,2,0.75,2.25])
Euler Solution
>> iso_cstr_euler
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0.8
1
1.2
1.4
1.6
1.8
2
2.2
CA
(g/L
)
Time (min)
Solution with Matlab Function
function f = iso_cstr(x)
Cao = 2;
q = 2;
V = 2;
Caf = 2;
k = 0.5;
Ca = x(1);
f(1) = q/V*(Caf-Ca)-2*k*Ca^2;
>> xss = fsolve(@iso_cstr,2)
xss = 1.0000
>> df = @(t,x) iso_cstr(x);
>> [t,x] = ode23(df,[0,2],2);
>> plot(t,x)
>> ylabel('Ca (g/L)')
>> xlabel('Time (min)')
>> axis([0,2,0.75,2.25])
Matlab Function Solution
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0.8
1
1.2
1.4
1.6
1.8
2
2.2
Ca
(g/L
)
Time (min)
EulerMatlab
CSTR Example Van de Vusse reaction
CSTR model
Forward Euler
233
2211
3
21
2 Ar
BArr
CkrDACkrCkrCBA
0221
012
31
)0(),(
)0(),(2)(
BBBABABB
AABAAAAAiA
CCCCfCkCkCVq
dtdC
CCCCfCkCkCCVq
dtdC
00,,2,1,,,,2,1,
00,2
,3,1,,,,1,1,
),(
2)(),(
BBnBnAnBnBnBnAnBnB
AAnAnAnAAinAnBnAnAnA
CCCkCkCVqhCCChfCC
CCCkCkCCVqhCCChfCC
Stiff System Example
CSTR model: A B C
Homogeneous system:
Eigenvalue analysis: q/V = 1, k1 = 1, k2 = 200
BABB
AAAiA CkCkC
Vq
dtdCCkCC
Vq
dtdC
211)(
BBBAAA CtCtCCtCtC )()()()( ''
'2
'1
''
'1
''
BABB
AAA CkCkC
Vq
dtdCCkC
Vq
dtdC
2012
''201102'
)()(
)('
21
'
'
yydtdy
tCtC
tyB
A A
Explicit Solution Forward Euler
First iterative equation
Second iterative equation
)201(201
)2(2
',
',
',
'1,
'''
',
',
'1,
''
nBnAnBnBBAB
nAnAnAAA
CChCCCCdt
dC
ChCCCdt
dC
unstable andy oscillator1stablebut y oscillator1
behaved well0)21()21()2(
21
21
'0,
',
',
',
',
'1,
hh
hChCChChCC A
nnAnAnAnAnA
unstable andy oscillatorstablebut y oscillator
behaved well0)2011()21()201(
2012
2012
2011
2011
'0,
'0,
',
',
',
',
'1,
hh
hChhChCCChCC B
nA
nnBnBnAnBnB
Implicit Solution Backward Euler
First iterative equation
Second iterative equation
)201(201
)2(2
'1,
'1,
',
'1,
'''
'1,
',
'1,
''
nBnAnBnBBAB
nAnAnAAA
CChCCCCdt
dC
ChCCCdt
dC
behaved well0)21(
1)2( '0,
',
'1,
',
'1,
h
Ch
CChCC AnnAnAnAnA
behaved well0)2011(
1)21(
1)201( '0,
'0,1
',
'1,
'1,
',
'1,
h
Ch
hCh
CCChCC BnAnnBnBnAnBnB
Matlab Solution
function f = stiff_cstr(x)Cai = 2;qV = 1;k1 = 1;k2 = 200;Ca = x(1);Cb = x(2);f(1) = qV*(Cai-Ca)-k1*Ca;f(2) = -qV*Cb+k1*Ca-k2*Cb;f = f';
>> xo = fsolve(@stiff_cstr,[1 1])xo = 1.0000 0.0050>> df = @(t,x) stiff_cstr(x);>> [t,x] = ode23(df,[0,2],[2 0]);>> [ts,xs] = ode23s(df,[0,2],[2 0]);>> size(t)ans = 173 1>> size(ts)ans = 30 1
Matlab Solution cont.>> subplot(2,1,1)>> plot(t,x(:,1))>> holdCurrent plot held>> plot(ts,xs(:,1),'r')>> ylabel('Ca (g/L)')>> ylabel('Ca (g/L)')>> xlabel('Time (min)')>> legend('ode23','ode23s')>> subplot(2,1,2)>> plot(t,x(:,2))>> holdCurrent plot held>> plot(ts,xs(:,2),'r')>> ylabel('Cb (g/L)')>> xlabel('Time (min)')>> legend('ode23','ode23s')
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 21
1.5
2
Ca
(g/L
)
Time (min)
ode23ode23s
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.005
0.01C
b (g
/L)
Time (min)
ode23ode23s