化工應用數學 授課教師: 郭修伯 助理教授 lecture 3...
TRANSCRIPT
化工應用數學
授課教師: 郭修伯 助理教授
Lecture 3 應用數學方程式表達物理現象
建立數學模式 The conservation laws
– material balance– heat balance– enery balance
Rate equations– the relationship between flow rate and driving force
in the field of fluid flow– heat transfer– diffusion of matter
建立數學模式
The conservation laws– material balance– heat balance– enery balance
(rate of) input - (rate of) output = (rate of) accumulation
範例說明
A single-stage mixer settler is to be used for the continuous extractionof benzoic acid from toluene, using water as the extracting solvent.The two streams are fed into a tank A where they are stirred vigorously,and the mixture is then pumped into tank B where it is allowed to settleinto two layers. The upper toluene layer and the lower water layer areremoved separately, and the problem is to find what proportion of thebenzoic acid has passed into the solvent phase.
watertoluene
+benzoic acid
toluene+
benzoic acid water+
benzoic acid
簡化(理想化)
S m3/s tolueney kg/m3 benzoic acid
R m3/s toluenex kg/m3 benzoic acid
S m3/s water
R m3/s toluenec kg/m3 benzoic acid
Rate equation for the extraction efficiency : y = mx
Material Balance : Input of benzoic acid = output of benzoic acid
Rc = Rx +SySame method can be applied to multi-stages.
隨時間變化
Funtion of time
非穩定狀態 (unsteady state)
In unsteady state problems, time enters as a variable and someproperties of the system become functions of time.
Similar to the previous example, but now assuming that the mixer isso efficient that the compositions of the two liquid streams are inequilibrium at all times. A stream leaving the stage is of the samecomposition as that phase in the stage. The state of the system at a general time t, wher x and y are now functions of time.
S m3/s tolueney kg/m3 benzoic acid
R m3/s toluenex kg/m3 benzoic acid
S m3/s water
R m3/s toluenec kg/m3 benzoic acid
V1, x
V2, y
Material balance on benzoic acid
S m3/s tolueney kg/m3 benzoic acid
R m3/s toluenex kg/m3 benzoic acid
S m3/s water
R m3/s toluenec kg/m3 benzoic acid
V1, x
V2, y
Input - output = accumulation
dt
dxV
dt
dxVSyRxRc 21)(
單位時間的變化
CmVV
t
mSR
xmSRRc
21
])(ln[
t = 0, x = 0
tmVV
mSR
mSR
Rcx
21
exp1
Mathematical Models
Salt accumulation in a stirred tank
t = 0Tank contains 2 m3 of water
Q: Determine the salt concentration in the tankwhen the tank contains 4 m3 of brine
Brineconcentration 20 kg/m3
feed rate 0.02 m3/s
Flow0.01 m3/s
建立數學模式 V and x are function of time t During t:
– balance of brine
– balance of salt
Brineconcentration 20 kg/m3
feed rate 0.02 m3/s
Brine0.01 m3/s
V m3
x kg/m3
tdt
dVtt 01.002.0
Vxtdt
dxxt
dt
dVVtxt ))((01.02002.0
解數學方程式 Solve
x = 20 - 20 (1 + 0.005 t)-2
V = 2 + 0.01 t
0)0(
2)0(
01.04.0
01.0
x
V
xdt
dxV
dt
dVx
dt
dV
Mathematical Models
MixingPure water3 l/min
Mixture2 l/min
Mixture3 l/min
Mixture4 l/min
Mixture1 l/min
Tank 1 Tank 2
t = 0Tank 1 contains 150 g of chlorine dissolved in 20 l waterTank 2 contains 50 g of chlorine dissolved in 10 l water
Q: Determine the amount of chlorine in each tank at any time t > 0
建立數學模式 Let xi(t) represents the number of grams of chlorine in tank i at
time t. Tank 1: x1’(t) = (rate in) - (rate out)
Tank 2: x2’(t) = (rate in) - (rate out)
Mathematical model:
x1’(t) = 3 * 0 + 3 * x2/10 - 2 * x1/20 - 4 * x1/20
Pure water3 l/min
Mixture2 l/min
Mixture3 l/min
Mixture4 l/min
Mixture1 l/min
Tank 1 Tank 2
x2’(t) = 4 * x1/20 - 3 * x2/10 - 1 * x2/10
50)0(
150)0(5
2
5
110
3
10
3
2
1
212
211
x
x
xxdt
dx
xxdt
dx
解數學方程式 How to solve?
Using Matrices X’ = AX ; X(0) = X0 where
– x1(t)=120e-t/10+30e-3t/5
– x2(t)=80e-t/10-30e-3t/5
50
150
5
2
5
110
3
10
30XandA
50)0(
150)0(5
2
5
110
3
10
3
2
1
212
211
x
x
xxdt
dx
xxdt
dx
Mathematical Models
Mass-Spring System– Suppose that the upper weight is
pulled down one unit and the lower weight is raised one unit, then both weights are released from rest simultaneously at time t = 0.
Q: Determine the positions of the weights relative totheir equilibruim positions at any time t > 0
k1=6
k3=3
k2=2
m1=1
m2=1
y2
y1
建立數學模式 Equation of motion weight 1: weight 2: Mathematical model:
m1 y1”(t) = - k1 y1 +k2 (y2 - y1)
0)0()0(
1)0(
1)0(
52
28
21
2
1
212
211
yy
y
y
yyy
yyyk1=6
k3=3
k2=2
m1=1
m2=1
y2
y1
m2 y2”(t) = - k2 (y2 - y1) - k3 y2
解數學方程式 How to solve?
– y1(t)=-1/5 cos (2t) + 6/5 cos (3t)
– y2(t)=-2/5 cos (2t) - 3/5 cos (3t)
0)0()0(
1)0(
1)0(
52
28
21
2
1
212
211
yy
y
y
yyy
yyy
隨位置變化
Funciotn of position
Mathematical Models
Radial heat transfer through a cylindrical conductor
Temperature at a is To
Temperature at b is T1
Q: Determine the temperature distributionas a function of r at steady state
r
r +dr
a
b
建立數學模式 Considering the element with thickness r Assuming the heat flow rate per unit area = Q
Radial heat flux
A homogeneous second order O.D.E.
))((22 rdr
dQQrrrQ
dr
dTkQ
where k is the thermal conductivity
02
2
dr
dT
dr
Tdr
解數學方程式 Solve
1
0
2
2
)(
)(
0
TbT
TaTdr
dT
dr
Tdr
)lnln
lnln)(()( 010 ab
arTTTrT
流場 (Flow systems) - Eulerian
The analysis of a flow system may proceed from either of two different points of view:– Eulerian method
the analyst takes a position fixed in space and a small volume element likewise fixed in space
the laws of conservation of mass, energy, etc., are applied to this stationary system
In a steady-state condition:– the object of the analysis is to determine the properties of t
he fluid as a function of position.
流場 (Flow systems) - Lagrangian
– the analyst takes a position astride a small volume element which moves with the fluid.
– In a steady state condition: the objective of the analysis is to determine the properties of the fluid
comprising the moving volume element as a function of time which has elapsed since the volume element first entered the system.
The properties of the fluid are determined solely by the elapsed time (i.e. the difference between the absolute time at which the element is examined and the absolute time at which the element entered the system).
– In a steady state condition: both the elapsed time and the absolute time affect the properties of t
he fluid comprising the element.
Eulerian 範例
A fluid is flowing at a steady state. Let x denote the distance from theentrance to an arbitrary position measured along the centre line in thedirection of flow. Let Vx denote the velocity of the fluid in the x direction, A denote the area normal to the x direction, and denote thefluid density at point x.Apply the law of conservation of mass to an infinitesimal element of volume fixed in space and of length dx.
xdx
, A, Vx +d, A+dA, Vx+dVx
xdx
, A, Vx +d, A+dA, Vx+dVx
If Vx and are essentially constant across the area A,The rate of input of mass is: wAVx
The rate of mass output is:
dxdx
dww
dxAVdx
dAV
dVVdAAd
xx
xx
]][][[
Rate of input - rate of output = rate of accumulation0
0)( dwAVd x Equation of continuity
Lagrangian 範例Consider a similar system. An infinitesimal volume element whichmoves with the fluid through the flow system.Let denote the elapsed time : = t -t0
where t is the absolute time at which the element is observed andt0 is the absolute time at which the element entered the system.At elapsed time , the volume of the element is Aa, the density is ,and the velocity of the element relative to the stationary wall is Vx.Apply the law of conservation of mass to the volume element.
xa
, A, Vx
2
a
dx
dVV x
x
2
a
dx
dVV x
x
0)( aAdt
d
xa
, A, Vx
2
a
dx
dVV x
x
2
a
dx
dVV x
x
t integralconstmaA
x
The elapsed time : 00tt
V
dxx
x
The difference between the relative velocity of the forward face and the relativevelocity of the trailing face is the change rate of the length of the element:
adx
dV
dt
ad x
)(
Mass balance of the element at steady-state
xVdt
dx
x
x
V
dV
a
ad
)(
0)( aAd
0)( xAVd
和 Eulerian結果一樣
獨立參數 (independent variable)
These are quantities describing the system which can be varied by choice during a paticular experiment independently of one another.
Examples:– time– coordinates
非獨立參數 (dependent variable)
These are properties of the system which change when the independent variables are altered in value. There is no direct control over a dependent variable during an experiment.
The relationship between independent and depend variables is one cause and effect; the independent variable measures the cause and the depend variable measures the effect of a particular action.
Examples:
– temperature
– concentration
– efficiency
變數 (Parameter)
It consists mainly of the charateristics properties of the apparatus and the physical properties of the materials.
It contains all properties which remain constant during an individual experiment. However, a different constant value can be taken by a property during different experiments.
Examples
– overall dimensions of the apparatus
– flow rate
– heat transfer coefficient
– thermal conductivity
– density
– initial or boundary values of the depent variables
各符號之間的關係 A dependent variable is usually differentiated with
respect to an independent variable, and occasionally with respect to a parameter.
When a single independent variable is involved in the problem, it gives rise to ordinary differential equations.
When more than one independent variable is needed to describe a system, the usual result is a partial differential equation.
邊界條件 (Boundary conditions)
There is usually a restriction on the range of values which the independent variable can take and this range describes the scope of the problem.
Special conditions are placed on the dependent variable at these end points of the range of the independent varible. These are natually called “boundary conditions”.
常見的邊界條件
熱傳 (heat transfer)– Boundary at a fixed temperature, T = T0.
– Constant hear flow rate through the boundary, dT/dx = A.
– Boundary thermally insulated, dT/dx = 0.
– Boundary cools to the surroundings through a film resistance described by a heat transfer coefficient, k dT/dx = h (T-T0).
k is the thermal conductivity; h is the heat transfer coefficient; and T0 is the temperature of the surrendings.
邊界值與起始值 (Boundary value and initial value)
Specifying conditions on a solution and its derivative at the ends of an interval (boundary value problem) is quite different from specifying the value of a solution and its derivative at a given point (initial value problem).
Boundary value problems usually do not have unique solutions, and it is this lack of uniqueness which makes certain boundary value problems important in solving P.D.E. of physics and engineering.
心得如何建立一個應用數學問題?
– 由「假設」,將問題簡化。– 確定所要探討的目標,找出「非獨立參數」。例如溫度、濃度等。– 找出「獨立參數」,使得非獨立參數可經由獨立參數表示。例如位置、
時間等。– 找出可將「獨立參數」 及「非獨立參數」的關係經由數學式表示出的
「變數」 。例如氣體流速、熱傳係數等。– 選定一個「特殊點」,應用「非獨立參數」來描述該系統的狀態。– 增加微量「非獨立參數」。應用泰勒展開式來表示該微量增加後,該
系統的狀態。– 應用「守恆定律或速率方程式」來顯示增加的微量。– 將增加的微量取極限值,建立該模型方程式。– 將邊界條件確定。