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1、化工應用數學http:/ equationAn 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”.Ordinary differential equation -only one independent variable involved: xPartial
2、 differential equation -more than one independent variable involved: x, y, z, http:/ and degreeThe 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 different
3、ial coefficient that the equation contains after it has been rationalized.3rd order O.D.E.1st degree O.D.E.http:/ or non-linearDifferential equations are said to be non-linear if any products exist between the dependent variable and its derivatives, or between the derivatives themselves.Product betw
4、een two derivatives - non-linearProduct between the dependent variable themselves - non-linearhttp:/ order differential equationsNo general method of solutions of 1st O.D.E.s because of their different degrees of complexity.Possible to classify them as:exact equationsequations in which the variables
5、 can be separatedhomogenous equationsequations solvable by an integrating factorhttp:/ equationsExact?General solution: F (x,y) = CFor examplehttp:/ equationsIn the most simple first order differential equations, the independent variable and its differential can be separated from the dependent varia
6、ble and its differential by the equality sign, using nothing more than the normal processes of elementary algebra.For examplehttp:/ equationsHomogeneous/nearly homogeneous?A differential equation of the type,Such an equation can be solved by making the substitution u = y/x and thereafter integrating
7、 the transformed equation.is termed a homogeneous differential equationof the first order.http:/ equation exampleLiquid 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 chlor
8、ine 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 sp
9、ecific reaction rate constants are: k1 = 8 k2 ; k2 = 30 k3C6H6+Cl2 C6H5Cl +HClC6H5Cl+Cl2 C6H4Cl2 + HClC6H4Cl2 + Cl2 C6H3Cl3 + HClhttp:/ 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 = mole
10、s of benzene presentr = moles of monochlorbenzene presents = moles of dichlorbenzene presentt = moles of trichlorbenzene presentThen q + r + s + t = 1and the total amount of chlorine consumed is: y = r + 2s + 3tFrom the material balances : in - out = accumulationu = r/qhttp:/ solved by integrating f
11、actorThere 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”.where P and Q are functions of x onlyAssuming the integrating factor R is a function of x only, thenis the integrating facto
12、rhttp:/ z = 1/y3integral factorhttp:/ 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.http:/ O.D.E. in heat transfer An elevated horizontal cylindrical tank 1 m diameter and
13、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 th
14、e 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/m3Heat capacity of liquid (s)= 2500 J/kgCAtmos
15、pheric 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.http:/ of tank (A) = ( x 1 x 2) + 2 ( 1 / 4 x 12 ) = 2.5 m2TwTsRate of heat loss by liquid = h1 A
16、(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)tAt steady state, the three rates are equal:Considering the thermal equilibrium of the liquid,input rate - output rate = accumulation rateB.C. = 0 , T = 95http:/ O.D.E.P
17、urpose: reduce to 1st O.D.E.Likely to be reduced equations:Non-linearEquations where the dependent variable does not occur explicitly.Equations where the independent variable does not occur explicitly.Homogeneous equations.LinearThe coefficients in the equation are constantThe coefficients are funct
18、ions of the independent variable.http:/ 2nd O.D.E. - Equations where the dependent variables does not occur explicitlyThey 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 th
19、us eliminating y completely and producing a first O.D.E. in p and x.http:/ thereforeintegral factorerror functionhttp:/ 2nd O.D.E. - Equations where the independent variables does not occur explicitlyThey are solved by differentiation followed by the p substitution.When the p substitution is made in
20、 this case, the second derivative of y is replaced as Lethttp:/ thereforeSeparating the variableshttp:/ 2nd O.D.E.- Homogeneous equationsThe homogeneous 1st O.D.E. was in the form:The corresponding dimensionless group containing the 2nd differential coefficient is In general, the dimensionless group
21、 containing the nth coefficient isThe second order homogenous differential equation can be expressed in a form analogous to , viz.Assuming u = y/xAssuming x = etIf in this form, called homogeneous 2nd ODEhttp:/ by 2xyhomogeneousLetLetSingular solutionGeneral solutionhttp:/ graphite electrode 15 cm i
22、n 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 th
23、is 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
24、 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.
25、056 T W/m CThe temperature of the electrode may be assumed uniform at any cross-section.xTT0http:/ sectional area of the electrode A = 1/4 x 0.152 = 0.0177 m2A heat balance over the length of electrode x at distance x from the furnace isinput - output = accumulationwhereU = overall heat transfer coe
26、fficient from the electrode to the surroundingsD = electrode diameterhttp:/ factorhttp:/ differential equationsThey 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 differe
27、ntial equation of the nth order having constant coefficients may be written:where (x) is any function of x.http:/ order linear differential equationsThe general equation can be expressed in the formwhere P,Q, and R are constant coefficientsLet the dependent variable y be replaced by the sum of the t
28、wo new variables: y = u + vThereforeIf v is a particular solution of the original differential equationThe general solution of the linear differential equation will be the sum of a “complementary function” and a “particular solution”.purposehttp:/ complementary functionLet the solution assumed to be
29、:auxiliary equation (characteristic equation)Unequal rootsEqual rootsReal rootsComplex rootshttp:/ roots to auxiliary equationLet 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 beIf m1 and
30、 m2 are complex it is customary to replace the complex exponential functions with their equivalent trigonometric forms. http:/ functionhttp:/ roots to auxiliary equationLet the roots of the auxiliary equation equal and of value m1 = m2 = m. Therefore, the solution of the auxiliary equation is:Letwhe
31、re V is a function of xhttp:/ functionhttp:/ functionhttp:/ integralsTwo methods will be introduced to obtain the particular solution of a second linear O.D.E.The method of undetermined coefficientsconfined to linear equations with constant coefficients and particular form of (x)The method of invers
32、e operatorsgeneral applicabilityhttp:/ of undetermined coefficientsWhen (x) is constant, say C, a particular integral of equation isWhen (x) is a polynomial of the form where all the coefficients are constants. The form of a particular integral isWhen (x) is of the form Terx, where T and r are const
33、ants. The form of a particular integral ishttp:/ of undetermined coefficientsWhen (x) is of the form G sin nx + H cos nx, where G and H are constants, the form of a particular solution isModified procedure when a term in the particular integral duplicates a term in the complementary function.http:/
34、coefficients of equal powers of xauxiliary equationhttp:/ of inverse operatorsSometimes, it is convenient to refer to the symbol “D” as the differential operator:But, http:/ differential operator D can be treated as an ordinary algebraicquantity with certain limitations.(1) The distribution law:A(B+
35、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) =
36、 (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.http:/ operator to exponentialsMore convenient!http:/ operator to trigonometrical functionswhere “Im” represents the imaginary part of the function which foll
37、ows it.http:/ inverse operatorThe operator D signifies differentiation, i.e.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 Dhttp:/ operatorbinomial expansion=2http:/ f(p) = 0, 使用因次分析非0的部分y = 1, p = 0, 即將 D-p換為
38、Dintegrationhttp:/ operatorf(p) = 0integrationy = yc + yphttp:/ operatorexpanding each term by binomial theoremy = yc + yphttp:/ in Chemical EngineeringA 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
39、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
40、change during the reaction, and that steady state conditions are established.A Bhttp:/ material balance can be taken over the element of length x at a distance x fom the inletThe concentraion will vary in the entry sectiondue to diffusion, but will not vary in the sectionfollowing the reactor. (Wehn
41、er and Wilhelm, 1956)xx+xBulk flow of ADiffusion of AInput - Output + Generation = Accumulation分開兩個sectionCehttp:/ by xrearrangingauxillary functionIn the entry sectionauxillary functionhttp:/ C. B. C. if diffusion is neglected (D0)http:/ continuous hydrolysis of tallow in a spray column連續牛油水解1.017
42、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 t
43、hrough 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 conta
44、ining 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,
45、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http:/
46、 fatHot water G kg/sExtractRaffinateL kg/sL kg/sxHzHx0z0y0G kg/syHhxzx+xz+zy+yyhx = 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 raffinatehttp:/ the chan
47、ges occurring in the element of column of height h:Glycerine transferred from fat to water phase,S: sectional area of towera: interfacial area per volume of towerK: overall mass transter coefficientRate of destruction of fat by hydrolysis,A glycerine balance over the element h is:Rate of production
48、of glycerine by hydrolysis,k: specific reaction rate constant: mass of fat per unit volume of column (730 kg/m3)w: kg fat per kg glycetineA glycerine balance between the element and the base of the tower is:L kg/sxHzHx0z0y0G kg/syHhxzx+xz+zy+yyhThe glycerine equilibrium between the phases is:in the
49、fat phasein the extract phasein the fat phasein the extract phasehttp:/ O.D.E. with constant coefficientsComplementary functionParticular solutionConstant at the right hand side, yp = C/Rhttp:/ dont really want x here!Apply the equations two slides earlier (replace y* with mx)We dont know y0, either
50、Substitute y0 in terms of other variableshttp:/ differential equationsThese 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 inde
51、pendent variable.Our purpose: Use algebraic elemination of the variables until only one differential equation relating two of the variables remains.http:/ of variableIndependent variable or dependent variables?Elimination of independent variable較少用Elimination of one or more dependent variablesIt inv
52、olves with equations of high order andit would be better to make use of matricesSolving differential equations simultaneously usingmatrices will be introduced later in the term http:/ of dependent variablesSolvehttp:/ Ey = yc + yphttp:/ kg/s of sulphuric acid (heat capacity 1500 J/kg C) is to be coo
53、led 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
54、 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
55、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 si
56、multaneous O.D.E.shttp:/ kg/s0.96 kg/s88 C45 C174 C20 C40 C80 CSteady state calculation:Heat capacity of water 4200 J/kg CHeat transfer area A1Heat transfer area A2Tank 1Tank 2Note: 和單操課本不同http:/ water fails for 1 hour, heat balance for tank 1 and tank 2:Tank 1Tank 2M: mass flow rate of acidC: heat
57、capacity of acidV: mass capacity of tankTi: temperature of tank iB.C.t = 0, T1 = 88t = 1, T1=142.4 Cintegral factor, ett = 1, T2 = 94.9 CB.C.t = 0, T2 = 45http:/ water supply restores after 1 hour, heat balance for tank 1 and tank 2:Tank 1Tank 2W: mass flow rate of waterCw: heat capacity of watert1:
58、 temperature of water leaving tank 1t2: temperature of water leaving tank 2t3: temperature of water entering tank 212T0T1T2t3t2t1Heat transfer rate equations for the two tanks:4 equations have to besolved simultaneouslyhttp:/ variable 出現次數, 發現 t1 出現次數最少，先消去！(i.e. t1= *代入)再由出現次數次少的 t2 消去.代入各數值.B.C. t
59、 = 0, T2 = 94.9 C 同時整理T1http:/ 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 factor2nd order 以上的O.D.ENon-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 coursehttp:/
