化工應用數學

1. 化工應用數學 Mathematical Statement of the Problem 授課教師： 林佳璋

2. 建立數學模式 Mass and heat balance -mass balance applying to each component -heat balance Rate equations -the transfer of heat across all phase interfaces -the mass transfer of each individual component across phase interfaces -each of the individual chemical reactions occurring in the various phases. Equilibriums -thermodynamic equilibrium -chemical equilibrium (rate of) input - (rate of) output = (rate of) accumulation

3. 溶劑萃取-Single Stage A single-stage mixer settler is to be used for the continuous extraction of 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 settle into two layers. The upper toluene layer and the lower water layer are removed separately, and the problem is to find what proportion of the benzoic acid has passed into the solvent phase. toluene + benzoic acid water toluene + benzoic acid water + benzoic acid

4. R m3/s toluene c kg/m3 benzoic acid R m3/s toluene x kg/m3 benzoic acid S m3/s water y kg/m3 benzoic acid S m3/s water 溶劑萃取-Single Stage Equilibrium equation for the extraction : y = mx Mass Balance : Input of benzoic acid = output of benzoic acid Rc = Rx +Sy Same method can be applied to multi-stages.

5. 溶劑萃取-Single Stage Proportion of benzoic acid extracted is If S=12R, m=1/8, and c=1.0; then x=0.4, y=0.05, and the proportion of benzoic acid extracted is 60% 無因次群

6. 溶劑萃取-Two Stages R R R x2 c x1 Stage 1 Stage 2 S S S 0 y1 y2 Equilibrium equation for the extraction : Mass Balance : Input of benzoic acid = output of benzoic acid

7. 溶劑萃取-Two Stages If S=12R, m=1/8, and c=1.0; then x2=0.21, y1=0.066, and the proportion of benzoic acid extracted is 79%

8. 溶劑萃取-N Stages y1 c R S Stage n Stage 1 mass balance x1 equilibrium y2 Stage 2 y3 x2 xn-1 yn Stage n yn+1 xn xN-2 yN-1 利用finite difference求得xn Stage N-1 yN xN-1 Stage N xN 0 S R

9. R m3/s toluene c kg/m3 benzoic acid R m3/s toluene x kg/m3 benzoic acid S m3/s water y kg/m3 benzoic acid S m3/s water 溶劑萃取-Time Change In unsteady state problems, time enters as a variable and some properties of the system become functions of time. Similar to the previous example, but now assuming that the mixer is so efficient that the compositions of the two liquid streams are in equilibrium at all times. A stream leaving the stage is of the same composition as that phase in the stage. The state of the system at a general time t, where x and y are now functions of time. V1, x V2, y

10. 溶劑萃取-Time Change

11. 溶劑萃取-Time Change During a time interval t Input of acid = Output of acid = Accumulation of acid = input - output = accumulation

12. 溶劑萃取-Time Change t0 t = 0, x = 0 t

13. 蒸餾問題 A distillation apparatus consisting of a boiler B with a constant level device C fed with the condenser cooling water. The steam is condensed in A and collected in the receiver D. Some of the latent heat of evaporation is returned to the boiler by preheating the feed. Denoting the condenser feed rate by F kg/s and the temperature by T0 C, the exit water temperature by T C, the excess water over-flow rate by W kg/s and the distillation rate by G kg/s, the performance of the apparatus can be investigated. F F A T T0 G T G D W C T B

14. 蒸餾問題 Input of water to the still (kg/s) = F Output of water from the still (kg/s) = W Output of steam from the still (kg/s) = G F=W+G Heat input (J/s) = H+FCvT Heat output (J/s) = GCv*100+GL+WCvT H+GCvT=G(Cv*100+L) H= G(L+Cv(100-T)) Heat supplied to the boiler at a rate H J/s, latent heat of evaporation of water is L J/kg Heat gained by cooling water (J/s) = FCv(T-T0) Heat lost by condensing steam (J/s) = G(L+Cv(100-T)) FCv(T-T0) = G(L+Cv(100-T)) FCv(T-T0) = H T=T0+H/CvF

15. 蒸餾問題 H=G(100Cv+L-CvT0-H/F) T=T0+H/CvF≤100 T=100 G F F

16. Brine concentration 20 kg/m3 feed rate 0.02 m3/s Flow 0.01 m3/s 攪拌混合問題 A tank contains 2 m3 of water. A stream of brine containing 20 kg/m3 of salt is fed into the tank at a rate of 0.02 m3/s. Liquid flows from the tank at a rate of 0.01 m3/s. If the tank is well agitated, what is the salt concentration in the tank when the tank contains 4 m3 of brine? t = 0 Tank contains 2 m3 of water

17. 攪拌混合問題 Simple Treatment liquid accumulates in the tank at a rate of (0.02-0.01)=0.01 m3/s increase in tank contents in 4-2=2 m3 the operation will last for 2/0.01=200 s During 200 s, 4 m3 of brine enter carrying 80 kg of salt and 2 m3 of brine leave Input of salt during 200 s (kg)=80 Output of salt during 200 s (kg)=2(X/2) Accumulation of salt in the tank (kg)=4X 假設濃度和時間線性增加 80-X=4X X=16 kg/m3

18. 攪拌混合問題 Detailed Treatment

19. 攪拌混合問題 During a time interval t Input of brine = Output of brine = Accumulation of brine= input - output = accumulation

20. 攪拌混合問題 During a time interval t Input of salt = Output of salt = Accumulation of salt = input - output = accumulation

21. 攪拌混合問題 t = 0, V = 2 t = 0, x = 0 t = 200, x = 15 kg/m3

22. r +dr r a b 熱傳問題 Two concentric cylindrical metallic shells are separated by a solid material. If the two metal surfaces are maintained at different constant temperatures, what is the steady temperature distribution within the separating material? Temperature at a is To Temperature at b is T1

23. 熱傳問題 Considering the element of the thickness, r Heat Input to inner surface = Heat output from outer surface = Accumulation of heat = 0

24. 熱傳問題 input - output = accumulation r0

25. 熱傳問題 r = a, T = T0 r = b, T = T1

26. 熱傳問題 A closed kettle of total surface area A m2 is heated through this surface by condensing steam at temperature Ts C. The kettle is charged with M kg of liquid of specific heat C J/kgC at a temperature of T0 C. If the process is controlled by a heat transfer coefficient h W/m2C, how does the temperature of the liquid vary with time? Considering a time interval, t Heat input (J) = Heat output (J) = 0 Accumulation of heat =

27. 熱傳問題 input - output = accumulation t = 0, T = T0

28. 流體力學問題 A fluid is flowing at a steady state. Let x denote the distance from the entrance to an arbitrary position measured along the centre line in the direction 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 the fluid density at point x. Apply the law of conservation of mass to an infinitesimal element of volume fixed in space and of length dx. , A, Vx +d, A+dA, Vx+dVx x dx

29. 流體力學問題 If Vxand  are essentially constant across the area A, The rate of mass input is: The rate of mass output is: 0 rate of input - rate of output = rate of accumulation Equation of continuity

30. , A, Vx x a 流體力學問題 Consider a similar system. An infinitesimal volume element which moves 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 and t0 is the absolute time at which the element entered the system. At elapsed time , the volume of the element is Aa, 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.

31. 流體力學問題 Mass balance of the element at steady-state t integral The elapsed time : The difference between the relative velocity of the forward face and the relative velocity of the trailing face is the change rate of the length of the element:

32. 反應問題 A countercurrent packed absorption tower is to be used for carrying out the liquid-phase reaction A+BC L(XA+dXA) G(YA+dYA) This reaction is irreversible, and the reaction rate may be expressed as follows: dZ GYA LXA Z The rate of transfer of A from the gas phase to the liquid phase is proportional to B and C are nonvolatile and never appear in the gas phase

33. 反應問題 For component A Input rate = Output rate = H: moles of inert solvent held up by packing per unit of tower volume Accumulation = 0

34. 反應問題 For component A in gas phase Input rate = Output rate = Accumulation = 0

35. 反應問題

36. 反應問題 P=kHS, R=KgaS

37. 參數(variable) 獨立參數(independent variable) These are quantities describing the system which can be varied by choice during a particular 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. Examples: temperature, concentration, efficiency The relationship between independent and depend variables is one cause and effect; the independent variable measures the cause and the dependent variable measures the effect of a particular action.

38. 變數(Parameter) Itconsists mainly of the characteristic 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, the correct term for them is “parameters”. Examples: overall dimensions of the apparatus, flow rate, heat transfer coefficient, thermal conductivity, density, initial or boundary values of the dependent variables independent variable : t dependent variables : x, y Parameter : m, R, S, c, V1,V2

39. 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 variable. These are naturally called “boundary conditions”. 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, kdT/dx = h(T-T0). k is the thermal conductivity; h is the heat transfer coefficient; and T0 is the temperature of the surroundings.

40. 如何建立應用數學問題？ 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。