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Differential equation
• An equation relating a dependent variable to one or more independent variables by means of its differential coefficients with respect to the independent variables is called a “differential equation”.
xeydx
dy
dx
yd x cos44)( 23
3
Ordinary differential equation --------only one independent variable involved: x
)(2
2
2
2
2
2
z
T
y
T
x
Tk
TC p
Partial differential equation ---------------more than one independent variable involved: x, y, z,
Order and degree
• The order of a differential equation is equal to the order of the highest differential coefficient that it contains.
• The degree of a differential equation is the highest power of the highest order differential coefficient that the equation contains after it has been rationalized.
xeydx
dy
dx
yd x cos44)( 23
3
3rd order O.D.E.
1st degree O.D.E.
Linear or non-linear
• Differential equations are said to be non-linear if any products exist between the dependent variable and its derivatives, or between the derivatives themselves.
xeydx
dy
dx
yd x cos44)( 23
3
Product between two derivatives ---- non-linear
xydx
dycos4 2
Product between the dependent variable themselves ---- non-linear
First order differential equations
• No general method of solutions of 1st O.D.E.s because of their different degrees of complexity.
• Possible to classify them as:– exact equations– equations in which the variables can be separated– homogenous equations– equations solvable by an integrating factor
Exact equations
• Exact? 0),(),( dyyxNdxyxM
dFdyy
Fdx
x
F
General solution: F (x,y) = C
For example
0)2(cossin3 dx
dyyxxyx
Separable-variables equations
• In the most simple first order differential equations, the independent variable and its differential can be separated from the dependent variable and its differential by the equality sign, using nothing more than the normal processes of elementary algebra.
For example
xdx
dyy sin
Homogeneous equations
• Homogeneous/nearly homogeneous?• A differential equation of the type,
• Such an equation can be solved by making the substitution u = y/x and thereafter integrating the transformed equation.
x
yf
dx
dy
is termed a homogeneous differential equationof the first order.
Homogeneous equation example
• Liquid benzene is to be chlorinated batchwise by sparging chlorine gas into a reaction kettle containing the benzene. If the reactor contains such an efficient agitator that all the chlorine which enters the reactor undergoes chemical reaction, and only the hydrogen chloride gas liberated escapes from the vessel, estimate how much chlorine must be added to give the maximum yield of monochlorbenzene. The reaction is assumed to take place isothermally at 55 C when the ratios of the specific reaction rate constants are:
k1 = 8 k2 ; k2 = 30 k3
C6H6+Cl2 C6H5Cl +HClC6H5Cl+Cl2 C6H4Cl2 + HClC6H4Cl2 + Cl2 C6H3Cl3 + HCl
Take a basis of 1 mole of benzene fed to the reactor and introducethe following variables to represent the stage of system at time ,
p = moles of chlorine presentq = moles of benzene presentr = moles of monochlorbenzene presents = moles of dichlorbenzene presentt = moles of trichlorbenzene present
Then q + r + s + t = 1and the total amount of chlorine consumed is: y = r + 2s + 3tFrom the material balances : in - out = accumulation
d
dtVpsk
d
dsVpskprk
d
drVprkpqk
d
dqVpqk
3
32
21
10
1)(1
2 q
r
k
k
dq
dr
u = r/q
Equations solved by integrating factor
• There exists a factor by which the equation can be multiplied so that the one side becomes a complete differential equation. The factor is called “the integrating factor”.
QPydx
dy where P and Q are functions of x only
Assuming the integrating factor R is a function of x only, then
)(Rydx
d
dx
dRy
dx
dyR
RQyRPdx
dyR
PdxR exp is the integrating factor
Example
Solve
2
3exp
24 x
ydx
dyxy
Let z = 1/y3
4
3
ydy
dz
dx
dy
ydx
dz4
3
2
3exp33
2x
dx
dzxz
integral factor
2
3exp3exp
2xxdx
32
3exp
2
3exp3
22
x
dx
dzxxz
32
3exp
2
xz
dx
dCx
xz
3
2
3exp
2
Cxx
y
3
2
3exp
1 2
3
Summary of 1st O.D.E.
• First order linear differential equations occasionally arise in chemical engineering problems in the field of heat transfer, momentum transfer and mass transfer.
First O.D.E. in heat transfer
An elevated horizontal cylindrical tank 1 m diameter and 2 m long is insulated withasbestos lagging of thickness l = 4 cm, and is employed as a maturing vessel for abatch chemical process. Liquid at 95 C is charged into the tank and allowed tomature over 5 days. If the data below applies, calculated the final temperature of theliquid and give a plot of the liquid temperature as a function of time.
Liquid film coefficient of heat transfer (h1) = 150 W/m2CThermal conductivity of asbestos (k) = 0.2 W/m CSurface coefficient of heat transfer by convection and radiation (h2) = 10 W/m2CDensity of liquid () = 103 kg/m3
Heat capacity of liquid (s) = 2500 J/kgCAtmospheric temperature at time of charging = 20 CAtmospheric temperature (t) t = 10 + 10 cos (/12)Time in hours ()Heat loss through supports is negligible. The thermal capacity of the lagging can be ignored.
T
Area of tank (A) = ( x 1 x 2) + 2 ( 1 / 4 x 12 ) = 2.5 m2
Tw
Ts
Rate of heat loss by liquid = h1 A (T - Tw)Rate of heat loss through lagging = kA/l (Tw - Ts)Rate of heat loss from the exposed surface of the lagging = h2 A (Ts - t)
t
At steady state, the three rates are equal:
)()()( 21 tTAhTTl
kATTAh ssww tTTs 674.0326.0
Considering the thermal equilibrium of the liquid,
input rate - output rate = accumulation rate
d
dTsVtTAh s )(0 2
)12/cos(235.0235.00235.0
Td
dT
B.C. = 0 , T = 95
Second O.D.E.
• Purpose: reduce to 1st O.D.E.• Likely to be reduced equations:
– Non-linear• Equations where the dependent variable does not occur explicitly.
• Equations where the independent variable does not occur explicitly.
• Homogeneous equations.
– Linear• The coefficients in the equation are constant
• The coefficients are functions of the independent variable.
Non-linear 2nd O.D.E. - Equations where the dependent variables does not occur explicitly
• They are solved by differentiation followed by the p substitution.
• When the p substitution is made in this case, the second derivative of y is replaced by the first derivative of p thus eliminating y completely and producing a first O.D.E. in p and x.
Solve axdx
dyx
dx
yd
2
2
Let dx
dyp 2
2
dx
yd
dx
dpand therefore
axxpdx
dp
2
2
1exp xintegral factor
222
2
1
2
1
2
1xxx
axexpeedx
dp
22
2
1
2
1
)(xx
axepedx
d
Bx
Aerfaxy
dxxCaxy
2
2
1exp 2
error function
Non-linear 2nd O.D.E. - Equations where the independent variables does not occur explicitly
• They are solved by differentiation followed by the p substitution.
• When the p substitution is made in this case, the second derivative of y is replaced as
Letdx
dyp
dy
dpp
dx
dy
dy
dp
dx
dp
dx
yd
2
2
Solve 22
2
)(1dx
dy
dx
ydy
Let dx
dyp and therefore
21 pdy
dpyp
Separating the variables
dy
dpp
dx
yd
2
2
dyy
dpp
p 1
12
)1ln(2
1lnln 2 pay
)1( 22 yadx
dyp
bayax
ya
dyx
)(sinh1
)1(1
22
Non-linear 2nd O.D.E.- Homogeneous equations
• The homogeneous 1st O.D.E. was in the form:• The corresponding dimensionless group containing the 2nd differential
coefficient is• In general, the dimensionless group containing the nth coefficient is• The second order homogenous differential equation can be expressed in a
form analogous to , viz.
x
yf
dx
dy
2
2
dx
ydx
n
nn
dx
ydx 1
x
yf
dx
dy
dx
dy
x
yf
dx
ydx ,
2
2 Assuming u = y/x
dx
duxuf
dx
udx ,
2
22
Assuming x = et
dt
duuf
dt
du
dt
ud,
2
2
If in this form, called homogeneous 2nd ODE
Solve2
222
222
dx
dyxy
dx
ydyx
Dividing by 2xy
2
2
2
2
1
2
1
dx
dy
y
x
x
y
dx
ydx
homogeneous
2
2
2
2
dt
du
dt
udu
dx
dy
x
yf
dx
ydx ,
2
2
Let uxy 2
22
22 22
dx
dux
dx
duux
dx
udux
Let tex
dt
dup
22 pdu
dpup
2)ln( CxBxy
Axy
Singular solution
General solution
A graphite electrode 15 cm in diameter passes through a furnace wall into a watercooler which takes the form of a water sleeve. The length of the electrode betweenthe outside of the furnace wall and its entry into the cooling jacket is 30 cm; and asa safety precaution the electrode in insulated thermally and electrically in this section,so that the outside furnace temperature of the insulation does not exceed 50 C.If the lagging is of uniform thickness and the mean overall coefficient of heat transferfrom the electrode to the surrounding atmosphere is taken to be 1.7 W/C m2 of surface of electrode; and the temperature of the electrode just outside the furnace is1500 C, estimate the duty of the water cooler if the temperature of the electrode atthe entrance to the cooler is to be 150 C.The following additional information is given.
Surrounding temperature = 20 CThermal conductivity of graphite kT = k0 - T = 152.6 - 0.056 T W/m CThe temperature of the electrode may be assumed uniform at any cross-section.
x
T
T0
x
T
T0
The sectional area of the electrode A = 1/4 x 0.152 = 0.0177 m2
A heat balance over the length of electrode x at distance x from the furnace is
input - output = accumulation
0)()( 0
xTTDUx
dx
dTAk
dx
d
dx
dTAk
dx
dTAk TTT
where U = overall heat transfer coefficient from the electrode to the surroundingsD = electrode diameter
xTTA
DUx
dx
dTk
dx
dT )( 0
0)()( 00
TT
dx
dTTk
dx
d
0)()( 0
2
2
2
0
TT
dx
dT
dx
TdTk
dx
dTp
2
2
dx
Td
dT
dpp
0)()( 02
0 TTpdT
dppTk
0)()( 02
0 TTpdT
dppTk
zp 2 )( 0TTy
022])[( 0 yzdy
dzyTk
Integrating factor 200
00
2exp yTk
yTk
dy
])(32))(([
)(3
02
00
0
TTTTTkC
dTTkx
Linear differential equations
• They are frequently encountered in most chemical engineering fields of study, ranging from heat, mass, and momentum transfer to applied chemical reaction kinetics.
• The general linear differential equation of the nth order having constant coefficients may be written:
)(... 11
1
10 xyPdx
dyP
dx
ydP
dx
ydP nnn
n
n
n
where (x) is any function of x.
2nd order linear differential equationsThe general equation can be expressed in the form
)(2
2
xRydx
dyQ
dx
ydP
where P,Q, and R are constant coefficients
Let the dependent variable y be replaced by the sum of the two new variables: y = u + v
Therefore
)(2
2
2
2
xRvdx
dvQ
dx
vdPRu
dx
duQ
dx
udP
If v is a particular solution of the original differential equation
02
2
Ru
dx
duQ
dx
udP
The general solution of the linear differential equation will be the sum of a “complementary function” and a “particular solution”.
purpose
The complementary function
02
2
Rydx
dyQ
dx
ydP
Let the solution assumed to be: mxmeAy mx
mmeAdx
dy mx
m emAdx
yd 22
2
0)( 2 RQmPmeA mxm
auxiliary equation (characteristic equation)
Unequal rootsEqual rootsReal rootsComplex roots
Unequal roots to auxiliary equation
• Let the roots of the auxiliary equation be distinct and of values m1 and m2. Therefore, the solutions of the auxiliary equation are:
• The most general solution will be
• If m1 and m2 are complex it is customary to replace the complex
exponential functions with their equivalent trigonometric forms.
xmeAy 11 xmeAy 2
2
xmxm eAeAy 2121
Equal roots to auxiliary equation
• Let the roots of the auxiliary equation equal and of value m1 = m2 = m. Therefore, the solution of the auxiliary equation is: mxAey
mxmx mVedx
dVe
dx
dymxVey Let
mxmxmx Vemdx
dVme
dx
Vde
dx
yd 22
2
2
2
2
where V is a function of x 02
2
Rydx
dyQ
dx
ydP
02
2
dx
VdDCxV
mxedCxy )(
Particular integrals
• Two methods will be introduced to obtain the particular solution of a second linear O.D.E.– The method of undetermined coefficients
• confined to linear equations with constant coefficients and particular form of (x)
– The method of inverse operators• general applicability
)(2
2
xRydx
dyQ
dx
ydP
Method of undetermined coefficients
• When (x) is constant, say C, a particular integral of equation is
• When (x) is a polynomial of the form where all the coefficients are constants. The form of a particular integral is
• When (x) is of the form Terx, where T and r are constants. The form of a particular integral is
)(2
2
xRydx
dyQ
dx
ydP
RCy /
nn xaxaxaa ...2
210
nn xxxy ...2
210
rxey
Method of undetermined coefficients
• When (x) is of the form G sin nx + H cos nx, where G and H are constants, the form of a particular solution is
• Modified procedure when a term in the particular integral duplicates a term in the complementary function.
)(2
2
xRydx
dyQ
dx
ydP
nxMnxLy cossin
Solve 32
2
8444 xxydx
dy
dx
yd
32 sxrxqxpy 232 sxrxq
dx
dy
sxrdx
yd62
2
2
3322 84)(4)32(4)62( xxsxrxqxpsxrxqsxr
Equating coefficients of equal powers of x
84
0124
4486
0442
s
sr
qrs
pqr
32 26107 xxxy p
0442 mmmauxiliary equation
xc eBxAy 2)(
pcgeneral yyy
Method of inverse operators
• Sometimes, it is convenient to refer to the symbol “D” as the differential operator:
n
nn
dx
ydyD
dx
ydyDDyD
dx
dyDy
...
)(2
22
But, 2
2)(
dx
dyDy
ydx
dy
dx
yd23
2
2
yDDyDDyDyyD )2)(1()23(23 22
The differential operator D can be treated as an ordinary algebraicquantity with certain limitations.
(1) The distribution law:A(B+C) = AB + ACwhich applies to the differential operator D
(2) The commutative law:AB = BAwhich does not in general apply to the differential operator DDxy xDy(D+1)(D+2)y = (D+2)(D+1)y
(3) The associative law:(AB)C = A(BC)which does not in general apply to the differential operator DD(Dy) = (DD)yD(xy) = (Dx)y + x(Dy)
The basic laws of algebra thus apply to the pure operators, but therelative order of operators and variables must be maintained.
Differential operator to exponentials
pxpx
pxnpxn
pxpx
epfeDf
epeD
peDe
)()(
...
pxpx eppeDD )23()23( 22
ypDfeyeDf
ypDeyeD
ypDeyeD
ypDeyDeDyeyeD
pxpx
npxpxn
pxpx
pxpxpxpx
)())((
)()(
...
)()(
)()(22
More convenient!
Differential operator to trigonometrical functions
ipxnipxnipxnn eipeDeDpxD )Im(ImIm)(sin
pxpppxD
pxppxD
pxpppxD
pxppxD
pxipxe
nn
nn
nn
nn
ipx
sin)()(cos
cos)()(cos
cos)()(sin
sin)()(sin
sincos
212
22
212
22
where “Im” represents the imaginary part of the function which follows it.
The inverse operator
The operator D signifies differentiation, i.e.
)()( xfdxxfD )()( 1 xfDdxxf
•D-1 is the “inverse operator” and is an “intergrating” operator.•It can be treated as an algebraic quantity in exactly the same manner as D
Solve xeydx
dy 24
differential operator
xeyD 2)4(
xeD
y 2
)4(
1
1...])4
1()
4
1()
4
1(1[
4
1 322 DDDey x
binomial expansion
=2
xey 2
2
1
pxpx epfeDf )()(
xeD
y 2
)41
1(4
1
2p
xey 2
)42(
1
...])2
1()
2
1()
2
1(1[
4
1 322 xey
xeD
y 2
)4(
1
pxpx epfeDf )()(
2p
xey 2
)42(
1
pxpx epf
eDf )(
1
)(
1
如果 f(p) = 0, 使用因次分析
pxn
px eDpD
eDf )()(
1
)(
1
n
pxpx
pDp
ee
Df )(
1
)()(
1
pxpx epfeDf )()(
n
pxpx
Dp
ee
Df
1
)()(
1
非 0 的部分
ypDfeyeDf pxpx )()(
y = 1, p = 0, 即將 D-p 換為 D
npx
px Dp
ee
Df
)()(
1
integration
!)()(
1
n
x
p
ee
Df
npxpx
Solvexxey
dx
dy
dx
yd 42
2
6168
differential operator
xxeyDyDD 422 6)4()168(
01682 mm
xc eBxAy 4)(
xp xe
Dy 4
2)4(
6
246 Dxey xp
f(p) = 0
)()( pDfeeDf pxpx
integration
!26
24 x
xey xp x
p exy 433
y = yc + yp
Solve23
2
2
346 xxydx
dy
dx
yd
differential operator
232 34)2)(3()6( xxyDDyDD
062 mm
xxc BeAey 23
)34()2)(3(
1 23 xxDD
y p
expanding each term by binomial theorem
y = yc + yp)34(
)2(
1
)3(
1
5
1 23 xxDD
y p
)34(...16842
1...
812793
1
5
1 233232
xxDDDDDD
y p
...01296
2413
216
)624(7
36
612
6
34 223
xxxxx
y p
O.D.E in Chemical Engineering
• A tubular reactor of length L and 1 m2 in cross section is employed to carry out a first order chemical reaction in which a material A is converted to a product B,
• The specific reaction rate constant is k s-1. If the feed rate is u m3/s, the feed concentration of A is Co, and the diffusivity of A is assumed to be constant at D m2/s. Determine the concentration of A as a function of length along the reactor. It is assumed that there is no volume change during the reaction, and that steady state conditions are established.
A B
uC0
x
L
x
uC
A material balance can be taken over the element of length x at a distance x fom the inlet
The concentraion will vary in the entry sectiondue to diffusion, but will not vary in the sectionfollowing the reactor. (Wehner and Wilhelm, 1956)
x x+x
Bulk flow of A
Diffusion of A
uC
xdx
dCD
dx
d
dx
dCD
dx
dCD
xdx
dCuuC
Input - Output + Generation = Accumulation
0
xkCx
dx
dCD
dx
d
dx
dCDx
dx
dCuuC
dx
dCDuC
分開兩個 sectionCe
0
xkCx
dx
dCD
dx
d
dx
dCDx
dx
dCuuC
dx
dCDuC
dividing by x
0
kC
dx
dCD
dx
d
dx
dCu
rearranging
02
2
kCdx
dCu
dx
CdD
auxillary function
02 kumDm
a
D
uxBa
D
uxAC 1
2exp1
2exp
In the entry section0
2
2
dx
Cdu
dx
CdD
auxillary function
02 umDm
D
uxC exp
2/41 ukDa
a
D
uxBa
D
uxAC 1
2exp1
2exp
D
uxC exp
B. C.
0
0
dx
dCLx
dx
Cd
dx
dCx
B. C.
CCx
CCx
0
0
xL
D
uaaxL
D
uaa
D
ux
KC
C
2exp1
2exp1
2exp
2
0
DuLaaDuLaaK 2/exp12/exp1 22
if diffusion is neglected (D0)
u
kx
C
CCexp1
0
0
The continuous hydrolysis of tallow in a spray column 連續牛油水解
1.017 kg/s of a tallow fat mixed with 0.286 kg/s of high pressure hot water is fed intothe base of a spray column operated at a temperature 232 C and a pressure of4.14 MN/m2. 0.519kg/s of water at the same temperature and pressure is sprayedinto the top of the column and descends in the form of droplets through the rising fatphase. Glycerine is generated in the fat phase by the hydrolysis reaction and is extractedby the descending water so that 0.701 kg/s of final extract containing 12.16% glycerineis withdrawn continuously from the column base. Simultaneously 1.121 kg/s of fattyacid raffinate containing 0.24% glycerine leaves the top of the column.If the effective height of the column is 2.2 m and the diameter 0.66 m, the glycerineequivalent in the entering tallow 8.53% and the distribution ratio of glycerine betweenthe water and the fat phase at the column temperature and pressure is 10.32, estimatethe concentration of glycerine in each phase as a function of column height. Also findout what fraction of the tower height is required principally for the chemical reaction.The hydrolysis reaction is pseudo first order and the specific reaction rate constant is0.0028 s-1.
Glycerin, 甘油
Tallow fat Hot water G kg/s
Extract
RaffinateL kg/s L kg/s
xH
zH
x0
z0
y0
G kg/syH
h
xz
x+xz+z y+y
y
h
x = weight fraction of glycerine in raffinatey = weight fraction of glycerine in extracty*= weight fraction of glycerine in extract in equilibrium with xz = weight fraction of hydrolysable fat in raffinate
Consider the changes occurring in the element of column of height h:
Glycerine transferred from fat to water phase, hyyKaS )*(
S: sectional area of towera: interfacial area per volume of towerK: overall mass transter coefficient
Rate of destruction of fat by hydrolysis, hSzk
A glycerine balance over the element h is:
hyyKaSw
hSzkh
dh
dxxLLx
*
Rate of production of glycerine by hydrolysis, whSzk /k: specific reaction rate constant: mass of fat per unit volume of column (730 kg/m3)w: kg fat per kg glycetine
A glycerine balance between the element and the base of the tower is:
00
w
Lz
w
LzLx
L kg/sxH
zH
x0
z0
y0
G kg/syH
h
xz
x+xz+z y+y
y
h
The glycerine equilibrium between the phases is:
mxy *
hyyKaSGyhdh
dyyG *
in the fat phase
in the extract phase
00 GyGy
in the fat phase
in the extract phase
02
2
00
2
dh
dy
L
KaSm
dh
yd
dh
dy
G
KaS
dh
dyy
G
KaS
L
Skyy
L
mG
w
mz
LG
KaSk
L
mGr
L
Skp
)1( rG
KaSq
w
mzry
r
pqpqy
dh
dyqp
dh
yd 002
2
1)( 2nd O.D.E. with constant coefficients
Complementary functionParticular solution
0)(2 pqmqpm Constant at the right hand side, yp = C/R
)exp()exp( qhBphAyc pqw
mzry
r
pqy p /
10
0
w
mzry
rqhBphAy 0
01
1)exp()exp(
B.C.
0,
0,0
yHh
xhWe don’t really want x here!
Apply the equations two slides earlier (replace y* with mx)
0,
,0 0
yHh
yyh
dh
dy
q
rymx
1
We don’t know y0, either
)exp()exp(1
qhqBphpAq
rymx
0,0 yyh
pHqHqhpHpHqHpHqH reveeveeerw
mzry
rrevey
)()(
1
1)( 0
0
q
prpqv
Substitute y0 in terms of other variables
qH
pHqHqH
qH
pHpH
er
revee
er
vee
vrw
mzy
)(0
730
66.04
0028.0
32.10
165.0286.0701.0
276.01216.0701.0~
403.02
519.0286.0~
069.12
121.1017.1~
2
0
S
k
m
y
G
L
Simultaneous differential equations
• These are groups of differential equations containing more than one dependent variable but only one independent variable.
• In these equations, all the derivatives of the different dependent variables are with respect to the one independent variable.
Our purpose: Use algebraic elemination of the variables until only one differential equation relating two of the variables remains.
Elimination of variableIndependent variable or dependent variables?
),(
),(
2
1
yxfdt
dy
yxfdt
dx
Elimination of independent variable
),(
),(
2
1
yxf
yxf
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較少用
Elimination of one or more dependent variables
It involves with equations of high order andit would be better to make use of matrices
Solving differential equations simultaneously usingmatrices will be introduced later in the term
Elimination of dependent variablesSolve
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1.25 kg/s of sulphuric acid (heat capacity 1500 J/kg C) is to be cooled in a two-stage counter-current cooler of the following type. Hot acid at 174 C is fed to a tank where it is well stirred incontact with cooling coils. The continuous discharge from this tank at 88 C flows to a secondstirred tank and leaves at 45C. Cooling water at 20 C flows into the coil of the second tank andthence to the coil of the first tank. The water is at 80 C as it leaves the coil of the hot acid tank.To what temperatures would the contents of each tank rise if due to trouble in the supply, the cooling water suddenly stopped for 1h?
On restoration of the water supply, water is put on the system at the rate of 1.25 kg/s. Calculatethe acid discharge temperature after 1 h. The capacity of each tank is 4500 kg of acid and the overall coefficient of heat transfer in the hot tank is 1150 W/m2 C and in the colder tank750 W/m2 C. These constants may be assumed constant.
Example of simultaneous O.D.E.s
1.25 kg/s
0.96 kg/s
88 C
45 C
174 C
20 C40 C80 C
Steady state calculation: Heat capacity of water 4200 J/kg C
)2080(4200)45174(150025.1 waterF skgFwater /96.0
)20(420096.0)4588(150025.1 middleT CTmiddle40
Heat transfer area A1 Heat transfer area A2
32.22
)4088()8088(
ln
)4088()8088(
1150)88174(150025.1 1
T
and
TA
Tank 1 Tank 2
Note: 和單操課本不同
44.68
)4088()80174(
ln
)4088()80174(
T
When water fails for 1 hour, heat balance for tank 1 and tank 2:
dt
dTVCMCTMCT 1
10
dt
dTVCMCTMCT 2
21
Tank 1
Tank 2
M: mass flow rate of acidC: heat capacity of acidV: mass capacity of tankTi: temperature of tank i
dt
dTTT
dt
dTTT
221
110
B.C.t = 0, T1 = 88 teT 861741
t = 1, T1=142.4 C
dt
dTTe t 2
286174 integral factor, et tetT )12986(1742
t = 1, T2 = 94.9 C
B.C.t = 0, T2 = 45
When water supply restores after 1 hour, heat balance for tank 1 and tank 2:
dt
dTVCMCTtWCMCTtWC ww
22213
Tank 1
Tank 2
W: mass flow rate of waterCw: heat capacity of watert1: temperature of water leaving tank 1t2: temperature of water leaving tank 2t3: temperature of water entering tank 2
dt
dTVCMCTtWCMCTtWC ww
11102
1 2T0 T1 T2
t3t2t1
Heat transfer rate equations for the two tanks:
)ln()ln(
)()()(
2111
21111121 tTtT
tTtTAUttWCw
)ln()ln(
)()()(
3222
32222232 tTtT
tTtTAUttWCw
4 equations have to besolved simultaneously
dt
dTVCMCTtWCMCTtWC ww
22213
dt
dTVCMCTtWCMCTtWC ww
11102
)ln()ln(
)()()(
2111
21111121 tTtT
tTtTAUttWCw
)ln()ln(
)()()(
3222
32222232 tTtT
tTtTAUttWCw
觀察各 dependent variable 出現次數 , 發現 t1 出現次數最少,先消去! (i.e. t1= *** 代入 )
211 )1( ttT
322 )1( ttT
wWC
AU
e11
wWC
AU
e22
再由出現次數次少的 t2 消去
…..
30975.708.6 22
22
2
Tdt
dT
dt
Td
代入各數值
…..
B.C. t = 0, T2 = 94.9 C 同時整理 T1
基本上, 1st order O.D.E. 應該都解的出來,方法不外乎:
Check exactSeparate variableshomogenous equations, u = y/xequations solvable by an integrating factor
2nd order 以上的 O.D.E Non-linear O.D.E.
Linear O.D.E. 缺 x 的 O.D.E., reduced to 1st O.D.E.缺 y 的 O.D.E., reduced to 1st O.D.E. homogeneous 的 O.D.E., u = y/x
我們會解的部分
General solution = complementary solution + particular solution
我們會解的部分
constant coefficientsThe method of undetermined coefficientsThe method of inverse operators
尋找特殊解的方法variable coefficient?用數列解, next course