ORDINARY DIFFERENTIAL ORDINARY DIFFERENTIAL EQUATIONSEQUATIONS
(ODE)(ODE)
Differential EquationsDifferential Equations
Heat transfer
Mass transfer
Conservation of momentum, thermal energy or mass
dz
TCd
A
q pz)(
dz
dCDJ AABAZ
Rzt
2
2
(4.1)
(4.2)
(4.3)
ODEODE
PDEPDE
ODEODE
Definition
Example
A 3rd order differential equation for = (t)
Solution
independent
dependent
0)(),...,('),(, )( ttttf n (4.4)
4'''''' 2 tet (4.5)
kQdt
dQ (4.6)
tcetQ kt ,)( (4.7)
Important IssuesImportant Issues
1. Existence of a solution
2. Uniqueness of the solution
3. How to determine a solution
Linear Equation (1)Linear Equation (1)
1. Rewrite 4.9
2. Determine
where (t) is called an integrating factorintegrating factor
)()()()(...)()( 0)0(
1'
2)1(
1)( tgtatatatata n
nn
n (4.8)
)()()(')( 012 tgtatata (4.9)
0)(;)(
)()(
)(
)(' 2
2
0
2
1
tata
tatg
ta
ta for all t (4.10)
dt
ta
tat
)(
)(exp)(
2
1 (4.11)
Linear Equation (2)Linear Equation (2)
3. Multiply both sides of equation 4.10 by (t)
Observe that the left-hand side of eqn 4.12 can be written as
or
)(tdt
d
)()(
)()()(
)(
)('
2
0
2
1 tta
tatgt
ta
ta
(4.12)
)()(
)(exp
)(
)(
)(
)(exp'
2
1
2
1
2
1 tdt
ddt
ta
ta
ta
tadt
ta
ta
(4.13)
Linear Equation (3)Linear Equation (3)
Equation (4.12) can be rephrase as:
4. Integrate both sides of Equation (4.14) with respect to the independent variable:
dt
ta
ta
ta
tatgdt
ta
ta
dt
d
)(
)(exp
)(
)()(
)(
)(exp
2
1
2
0
2
1
cdtdtta
ta
ta
tatgdt
ta
ta
)(
)(exp
)(
)()(
)(
)(exp
2
1
2
0
2
1
(4.14)
dt
ta
tacdtdt
ta
ta
ta
tatgdt
ta
tat
)(
)(exp
)(
)(exp
)(
)()(
)(
)(exp)(
2
1
2
1
2
0
2
1 (4.15)
where cc is the constant of integrationconstant of integration
Example 1Example 1
Water containing 0.5 kg of salt per liter is poured into a tank at a rate of 2 l/min, and the well-stirred mixture leaves at the same rate. After 10 minutes, the process is stopped and fresh water is poured into the tank at a rate of 2 l/min, with the new mixture leaving at 2 l/min. Determine the amount (kg) of salt in the tank at the end of 20 minutes if there were 100 liters of pure water initially in the tank.
2 l/min
2 l/min, CA (l/min)CA
½ kg salt/l
SolutionSolution
Example 2Example 2
Consider a tank with a 500 l capacity that initially contains 200 l of water with 100 kg of salt in solution. Water containing 1 kg of salt/l is entering at a rate of 3 l/min, and the mixture is allowed to flow out of the tank at a rate of 2 l/min. Determine the amount (kg) of salt in the tank at any time prior to the instant when the solution begins to overflow. Determine the concentration (kg/l) of salt in the tank when it is at the point of overflowing. Compare this concentration with the theoretical limiting concentration if the tank had infinite capacity.
SolutionSolution
THEOREMTHEOREM
If the functions p and g are continuous on an open interval < x < containing the point x = x0, then there exists a unique function y = (x) that satisfies the differential equation
y’ + p(x)y = g(x)
for < x < , and that also satisfies the initial condition
y(x0) = y0
where y0 is an arbitrary prescribed initial value.
Higher ODE Reduces to 1Higher ODE Reduces to 1stst Order Order
2
2( ) ( )
Define , we have
( ) ( )
d y dyq x r x
dx dxdy
zdx
dyz
dxdz
r x q x zdx
2
1 2 3
12
23
231 3 2 1
'''( ) ( ) ''( ) 2( '( )) ( ) 0
Define , ', '', we have
2( )
y x y x y x y x y x
y y y y y y
dyy
dxdy
ydxdy
y y y ydx
In general, it is sufficient to solve first-order ordinary differential equations of the form
1( , , , ), 1, 2, ,ii N
dyf x y y i N
dx
Nonlinear equations can be reduced to linear ones by a substitution. Example:
y’ + p(x)y = q(x)yn
and if n 0,1 then
(x) = y1-n(x)
reduces the above equation to a linear equation.
(4.16)
(4.17)
Example 3Example 3
Suppose that in a certain autocatalytic chemical reaction a compound A reacts to form a compound B. Further, suppose that the initial concentration of A is CA0 and that CB(t) is the concentration of B at time t. Then CA0 – CB (t) is the concentration of A at time t. Determine CB(t) if CB(0) = CB0.
Solution Solution
NONLINEAR ORDINARY NONLINEAR ORDINARY DIFFERENTIAL EQUATIONSDIFFERENTIAL EQUATIONS
NONLINEAR EQUATIONSNONLINEAR EQUATIONS
Rewrite as
0),(),( dt
dtNtM
),( tf
dt
d
If M is a function of t only, and N is a function of only, then
0)()( dt
dNtM
dttMdN )()( Separable
NONLINEAR EQUATIONSNONLINEAR EQUATIONS
Consider
)( 0 BABB CCkC
dt
dC
0)0( BB CC
kdtCCC
dC
BAB
B )( 0
subject to
Then, it is separable and results in:
(4.16)
Simplifying left-hand side; 1st consider the fraction
BABBAB CCCCCC
00 )(
1
1)( 0 BBA CCC
1:0 0 AB CC
0
1
AC
where and are constants to be determined. Then:
If we put
then
(4.17)
Rewrite equation 4.17
1: 00 AAB CCC If we put
0
1
AC
then
BA
A
B
A
BAB CC
C
C
C
CCC
0
00
0
11
)(
1
And equation 4.16 becomes
which integrates to:
kdtdCCCCC BBABA
00
111
)exp(1
1
0
0
ktmCC
C AC
BA
B
where m1 is an arbitrary constant to be determined with the given initial condition. @ t = 0, CB = CB0, then
1
1
0
0
mCC
C AC
BA
B
0000
00
)exp()()(
BABA
ABB CtkCCC
CCtC
Example of Problem SetupExample of Problem Setup
Consider the continuous extraction of benzoic acid from a mixture of benzoic acid and toluene, using water as the extracting solvent. Both streams are fed into a tank where they are stirred efficiently and the mixture is then pumped into a second tank where it is allowed to settle into two layers. The upper organic phase and the lower aqueous phase are removed separately, and the problem is to determine what proportion of the acid has passed into the solvent phase.
Example (cont…)Example (cont…)
List of assumptions1. Combine the two tanks into a single stage2. Express stream-flow rates on solute-free basis3. Steady flowrate for each phase4. Toluene and water are immiscible5. Feed concentration is constant6. Mixing is efficient, the two streams leaving the stage are in
equilibrium with each other given by y = mx7. Composition stream leaving is the same with the composition
in the stage8. The stage initially contains V1 liter toluene, V2 liter water and
no benzoic acid
Problem 1Problem 1
Consider an engine that generates heat at a rate of 8,530 Btu/min. Suppose this engine is cooled with air, and the air in the engine housing is circulated rapidly enough so that the air temperature can be assumed uniform and is the same as that of the outlet air. The air is fed to the housing at 6lb-mole/min and 65oF. Also, an average of 0.20 lb-mole of air is contained within the engine housing and its temperature variation can be neglected. If heat is lost from the housing to its surroundings at a rate of Q(Btu/min) = 33.0(T-65oF) and the engine is started with the inside air temperature equal to 65oF.
1. Derive a differential equation for the variation of the outlet temperature with time.
2. Calculate the steady state air temperature if the engine runs continuously for indefinite period of time, using Cv = 5.00 Btu/lb-mole oF.
Problem 2Problem 2
A liquid-phase chemical reaction with stoichiometry A B takes place in a semi-batch reactor. The rate of consumption of A per unit volume of the reactor is given by the first order rate expression
rA (mol/liter.s) = kCA
where CA (mol/liter) is the reactant concentration. The tank is initially empty. At time t=0, a solution containing A at a concentration CA0(mol/liter) is fed to the tank at a steady rate (liters/s). Develop differential balances on the total volume of the tank contents, V, and on the moles of A in the tank, nA .
Solving ODEs using Numerical MethodsSolving ODEs using Numerical Methods
Initial Value Problem (IVP)y’’ = -yx
y(0) = 2, y’(0) = 1
Boundary Value Problem (BVP)y” = -yx
y(0) = 2, y’(1) = 1
General ProcedureGeneral Procedure
Re-write the dy and dx terms as Δy and Δx and multiply by Δx
Literally doing this is Euler’s method
Niyyyxfdx
xdyNi
i ,...,1),...,,()(
21 equations for
xyxfyy
yxfx
xy
yxfdx
xdy
iiii
),(
),()(
),()(
1
Tank mixing problemTank mixing problem
tccV
Vcc
ccV
V
dt
dc
iinii
in
)(
)(
1tank
tank
Mixing tankMixing tank
t Error Et
at t=600
300 1.4
150 0.61
100 0.39
50 0.19
30 0.11
15 0.055
10 0.036
5 0.018
3 0.011
Improvements to Euler’s MethodImprovements to Euler’s Method
EulerEuler
Heun’s methodHeun’s method (predictor-corrector)
Procedure
calc yi+1 with Euler (predictor)
calc slope at yi+1
calc average slope
use this slope to calc new yi+1 (corrector)
xyxfyy
yxfdx
xdy
iiii
),(
),()(
1
Heun exampleHeun example
xyy
yx
y
xyxfyy
yx
y
x
eyeyxy
.xyydx
dy
ii
ii
x
1
1
1
0
10
),(
)5.0(
1
36.15.0)1()(
105.0)0(
location updated
slope average
location predicted at slope :corrector
location predicted
slope (Euler) predictor
Δ withNumerical
so Analytical
at Solve
0
0.5
1
1.5
0 1x
y
Midpoint MethodMidpoint Method
Use Euler to calculate a midpoint location
evaluate slope y’ at the midpoint
use that to calculate full step location
0
0.5
1
1.5
0 1x
y
xyxfyy
xyxfyy
yxfy
iiii
iiii
),(2
),(
),(
2/12/11
2/1
Runge-KuttaRunge-Kutta
slope) a (aka function increment is
form General
integral the evaluate to quadrature Gaussian use Or
between locations at evaluated f of values on based
integral for fit l)(polynomia order higher use Could
hyy
hxx
dxyxfxyhxyyxfdx
dy
ii
hx
x
1
),()()(),(
R-K – General formR-K – General form
)...,(
),(
),(
),(
,,,
11,122,111,11
22212123
11112
1
21
2211
1
hkqhkqhkqyhpxfk
hkqhkqyhpxfk
hkqyhpxfk
yxfk
aaa
where
kakaka
hyy
nnnnninin
ii
ii
ii
n
nn
ii
constants
:as Write
form General
R-K – 1st Order FormR-K – 1st Order Form
hyxfyy
yxfk
a
ka
hyy
iiii
ii
ii
),(1
),(
constant 1
where
form General
1
1
1
11
1
R-K – 2R-K – 2ndnd Order Form Order Form
11212
21
11121
11112
1
22111
2
1
1
constants
),(
),(
qapa
aa
,q,p,aa
hkqyhpxfk
yxfk
hkakayy
ii
ii
ii
y(x)
xi xi+1 x
RK2 – OptionsRK2 – Options
0
0
2
1;1
),(;),(
1
2
1121221
111121
22111
a
a
qapaaa
hkqyhpxfkyxfk
hkakayy
iiii
ii
y(x)
xi xi+1
x
RK2 – OptionsRK2 – Options
3
2
2
1
2
1;1
),(;),(
2
2
1121221
111121
22111
a
a
qapaaa
hkqyhpxfkyxfk
hkakayy
iiii
ii
y(x)
xi xi+1 x
y(x)
xi xi+1 x
R-K – 2R-K – 2ndnd Order Form Order Form
constants
)(),(
),(
1121
212
1
22111
,q,p,aa
hOy
fqfh
x
fphfhqkyphxfk
yxfk
hkakayy
ii
ii
ii
RK – 3rd Order FormRK – 3rd Order Form
)2,(
)2
1,
2
1(
),(
46
1
213
12
1
3211
hkhkyhxfk
hkyhxfk
yxfk
hkkkyy
ii
ii
ii
ii
y(x)
xi xi+1 x
RK – 4th OrderRK – 4th Order
),(
)2
1,
2
1(
)2
1,
2
1(
),(
226
1
33
23
12
1
43211
hkyhxfk
hkyhxfk
hkyhxfk
yxfk
hkkkkyy
ii
ii
ii
ii
ii
y(x)
xi xi+1 x
Example Example yy΄́=x+y, y(0)=0=x+y, y(0)=0
x yo k1=fik2=f(x+h/
2,y+h/2k1)k3=f(x+h/
2,y+h/2k2)k4=f(x+h,y
+hk3)yn=yo+1/6(k1+2k2+2k
3+k4)h
0 0 0 0.1 0.11 0.222 0.02140
0.20.021
4 0.221 0.344 0.356 0.493 0.0918
0.4 0.092 0.492 0.641 0.656 0.823 0.2221
0.6 0.222 0.822 1.004 1.023 1.227 0.4255
0.8 0.426 1.226 1.448 1.470 1.720 0.718
1 0.718 1.718 1.990 2.017 2.322 1.120
1.2 1.120 2.320 2.652 2.685 3.057 1.655
1.4 1.655 3.055 3.461 3.501 3.955 2.353
1.6 2.353 3.953 4.448 4.498 5.052 3.250
1.8 3.250 5.050 5.654 5.715 6.393 4.389
0
0.5
1
1.5
2
2.5
3
3.5
0 0.5 1 1.5 2
x
y
Euler
analytical
RK4
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