化工应用数学

1、化工應用數學化工應用數學授課教師：授課教師： 郭修伯郭修伯 助理教授助理教授Lecture 3應用數學方程式表達物理現象建立建立數學模式數學模式lThe conservation laws material balance heat balance enery balancelRate equations the relationship between flow rate and driving force in the field of fluid flow heat transfer diffusion of matter建立建立數學模式數學模式lThe conservation law

2、s material balance heat balance enery balancel(rate of) input - (rate of) output = (rate of) accumulation範例說明範例說明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 the

3、y 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+benz

4、oic acidtoluene+benzoic acidwater+benzoic acid簡化（理想化）簡化（理想化）S m3/s tolueney kg/m3 benzoic acidR m3/s toluenex kg/m3 benzoic acidS m3/s waterR m3/s toluenec kg/m3 benzoic acidRate equation for the extraction efficiency : y = mxMaterial Balance : Input of benzoic acid = output of benzoic acidRc = Rx +

5、SySame method can be applied to multi-stages.隨時間變化隨時間變化Funtion of time非穩定狀態非穩定狀態 (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 co

6、mpositions 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 acidR m3/s toluenex kg/m3 benzoic ac

7、idS m3/s waterR m3/s toluenec kg/m3 benzoic acidV1, xV2, yMaterial balance on benzoic acidS m3/s tolueney kg/m3 benzoic acidR m3/s toluenex kg/m3 benzoic acidS m3/s waterR m3/s toluenec kg/m3 benzoic acidV1, xV2, yInput - output = accumulationdtdxVdtdxVSyRxRc21)(單位時間的變化CmVVtmSRxmSRRc21)(lnt = 0, x =

8、 0tmVVmSRmSRRcx21exp1Mathematical ModelslSalt accumulation in a stirred tankt = 0Tank contains 2 m3 of waterQ: Determine the salt concentration in the tankwhen the tank contains 4 m3 of brineBrineconcentration 20 kg/m3feed rate 0.02 m3/sFlow0.01 m3/s建立數學模式建立數學模式lV and x are function of time tlDuring

9、 t: balance of brine balance of saltBrineconcentration 20 kg/m3feed rate 0.02 m3/sBrine0.01 m3/sV m3x kg/m3tdtdVtt01.002.0VxtdtdxxtdtdVVtxt)(01.02002.0解數學方程式解數學方程式lSolvelx = 20 - 20 (1 + 0.005 t)-2lV = 2 + 0.01 t0)0(2)0(01.04.001.0 xVxdtdxVdtdVxdtdVMathematical ModelslMixingPure water3 l/minMixture2

10、 l/minMixture3 l/minMixture4 l/minMixture1 l/minTank 1Tank 2t = 0Tank 1 contains 150 g of chlorine dissolved in 20 l waterTank 2 contains 50 g of chlorine dissolved in 10 l waterQ: Determine the amount of chlorine in each tank at any time t 0建立數學模式建立數學模式lLet xi(t) represents the number of grams of c

11、hlorine in tank i at time t. lTank 1: x1(t) = (rate in) - (rate out)lTank 2: x2(t) = (rate in) - (rate out)lMathematical model: x1(t) = 3 * 0 + 3 * x2/10 - 2 * x1/20 - 4 * x1/20 Pure water3 l/minMixture2 l/minMixture3 l/minMixture4 l/minMixture1 l/minTank 1Tank 2x2(t) = 4 * x1/20 - 3 * x2/10 - 1 * x

12、2/10 50)0(150)0(525110310321212211xxxxdtdxxxdtdx解數學方程式解數學方程式lHow to solve?lUsing MatriceslX = AX ; X(0) = X0 where x1(t)=120e-t/10+30e-3t/5 x2(t)=80e-t/10-30e-3t/55015052511031030XandA50)0(150)0(525110310321212211xxxxdtdxxxdtdxMathematical ModelslMass-Spring System Suppose that the upper weight is p

13、ulled 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 0k1=6k3=3k2=2m1=1m2=1y2y1建立數學模式建立數學模式lEquation of motionlweight 1: lweight

14、 2: lMathematical model: m1 y1”(t) = - k1 y1 +k2 (y2 - y1) 0)0()0(1)0(1)0(52282121212211yyyyyyyyyyk1=6k3=3k2=2m1=1m2=1y2y1m2 y2”(t) = - k2 (y2 - y1) - k3 y2 解數學方程式解數學方程式lHow 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(52282121212211yyyyyyyyyy隨位置變化隨位置

15、變化Funciotn of positionMathematical ModelslRadial heat transfer through a cylindrical conductorTemperature at a is ToTemperature at b is T1Q: Determine the temperature distributionas a function of r at steady staterr +drab建立數學模式建立數學模式lConsidering the element with thickness rlAssuming the heat flow ra

16、te per unit area = QlRadial heat fluxlA homogeneous second order O.D.E.)(22rdrdQQrrrQdrdTkQwhere k is the thermal conductivity022drdTdrTdr解數學方程式解數學方程式lSolve1022)()(0TbTTaTdrdTdrTdr)lnlnlnln)()(010abarTTTrT流場流場 (Flow systems) - EulerianlThe analysis of a flow system may proceed from either of two dif

17、ferent points of view: Eulerian methodlthe analyst takes a position fixed in space and a small volume element likewise fixed in spacelthe laws of conservation of mass, energy, etc., are applied to this stationary systemlIn a steady-state condition: the object of the analysis is to determine the prop

18、erties of the fluid as a function of position.流場流場 (Flow systems) - Lagrangian the analyst takes a position astride a small volume element which moves with the fluid. In a steady state condition:lthe objective of the analysis is to determine the properties of the fluid comprising the moving volume e

19、lement as a function of time which has elapsed since the volume element first entered the system.lThe 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 t

20、he system). In a steady state condition:lboth the elapsed time and the absolute time affect the properties of the fluid comprising the element.Eulerian 範例範例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 thed

21、irection 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+dVxxdx,

22、 A, Vx+d, A+dA, Vx+dVxIf Vx and are essentially constant across the area A,The rate of input of mass is:wAVxThe rate of mass output is: dxdxdwwdxAVdxdAVdVVdAAdxxxxRate of input - rate of output = rate of accumulation00)( dwAVdxEquation of continuityLagrangian 範例範例Consider a similar system. An infini

23、tesimal volume element whichmoves with the fluid through the flow system.Let denote the elapsed time : = t -t0where 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 densit

24、y 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, Vx2adxdVVxx2adxdVVxx0)(aAdtdxa, A, Vx2adxdVVxx2adxdVVxxt integralconstmaAxThe elapsed time : 00ttVdxxxThe difference between the relative velocity of the for

25、ward face and the relativevelocity of the trailing face is the change rate of the length of the element:adxdVdtadx)(Mass balance of the element at steady-state xVdtdxxxVdVaad)(0)(aAd0)(xAVd和Eulerian結果一樣獨立參數獨立參數 (independent variable)lThese are quantities describing the system which can be varied by

26、choice during a paticular experiment independently of one another.lExamples: time coordinates非獨立參數非獨立參數 (dependent variable)lThese 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.

27、lThe 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.lExamples: temperature concentration efficiency變數變數 (Parameter)lIt consists mainly of the charateristics

28、properties of the apparatus and the physical properties of the materials.lIt contains all properties which remain constant during an individual experiment. However, a different constant value can be taken by a property during different experiments.lExamples overall dimensions of the apparatus flow r

29、ate heat transfer coefficient thermal conductivity density initial or boundary values of the depent variables各符號之間的關係各符號之間的關係lA dependent variable is usually differentiated with respect to an independent variable, and occasionally with respect to a parameter.lWhen a single independent variable is in

30、volved in the problem, it gives rise to ordinary differential equations.lWhen more than one independent variable is needed to describe a system, the usual result is a partial differential equation.邊界條件邊界條件 (Boundary conditions)lThere is usually a restriction on the range of values which the independ

31、ent variable can take and this range describes the scope of the problem. lSpecial conditions are placed on the dependent variable at these end points of the range of the independent varible. These are natually called “boundary conditions”.常見的邊界條件常見的邊界條件l熱傳熱傳 (heat transfer) Boundary at a fixed temperature, T = T0. Constant hear flow rate throu