化工應用數學http:/www.docin.com/sundae_mengDifferential 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 differential equation -more than one independent variable involved: x, y, z, http:/www.docin.com/sundae_mengOrder 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 differential coefficient that the equation contains after it has been rationalized.3rd order O.D.E.1st degree O.D.E.http:/www.docin.com/sundae_mengLinear 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 between two derivatives - non-linearProduct between the dependent variable themselves - non-linearhttp:/www.docin.com/sundae_mengFirst 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 can be separatedhomogenous equationsequations solvable by an integrating factorhttp:/www.docin.com/sundae_mengExact equationsExact?General solution: F (x,y) = CFor examplehttp:/www.docin.com/sundae_mengSeparable-variables equationsIn 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 examplehttp:/www.docin.com/sundae_mengHomogeneous 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 the transformed equation.is termed a homogeneous differential equationof the first order.http:/www.docin.com/sundae_mengHomogeneous 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 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 k3C6H6+Cl2 C6H5Cl +HClC6H5Cl+Cl2 C6H4Cl2 + HClC6H4Cl2 + Cl2 C6H3Cl3 + HClhttp:/www.docin.com/sundae_mengTake 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 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:/www.docin.com/sundae_mengEquations solved by integrating factorThere 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 factorhttp:/www.docin.com/sundae_mengExampleSolveLet z = 1/y3integral factorhttp:/www.docin.com/sundae_mengSummary 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:/www.docin.com/sundae_mengFirst 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/m3Heat 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.http:/www.docin.com/sundae_mengTArea of tank (A) = ( x 1 x 2) + 2 ( 1 / 4 x 12 ) = 2.5 m2TwTsRate 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)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:/www.docin.com/sundae_mengSecond O.D.E.Purpose: 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 functions of the independent variable.http:/www.docin.com/sundae_mengNon-linear 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 thus eliminating y completely and producing a first O.D.E. in p and x.http:/www.docin.com/sundae_mengSolveLetand thereforeintegral factorerror functionhttp:/www.docin.com/sundae_mengNon-linear 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 this case, the second derivative of y is replaced as Lethttp:/www.docin.com/sundae_mengSolveLetand thereforeSeparating the variableshttp:/www.docin.com/sundae_mengNon-linear 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 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:/www.docin.com/sundae_mengSolveDividing by 2xyhomogeneousLetLetSingular solutionGeneral solutionhttp:/www.docin.com/sundae_mengA 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.xTT0http:/www.docin.com/sundae_mengxTT0The 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 coefficient from the electrode to the surroundingsD = electrode diameterhttp:/www.docin.com/sundae_mengIntegrating factorhttp:/www.docin.com/sundae_mengLinear 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 differential equation of the nth order having constant coefficients may be written:where (x) is any function of x.http:/www.docin.com/sundae_meng2nd 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 two 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:/www.docin.com/sundae_mengThe complementary functionLet the solution assumed to be:auxiliary equation (characteristic equation)Unequal rootsEqual rootsReal rootsComplex rootshttp:/www.docin.com/sundae_mengUnequal 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 m2 are complex it is customary to replace the complex exponential functions with their equivalent trigonometric forms. http:/www.docin.com/sundae_mengSolveauxiliary functionhttp:/www.docin.com/sundae_mengEqual 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:Letwhere V is a function of xhttp:/www.docin.com/sundae_mengSolveauxiliary functionhttp:/www.docin.com/sundae_mengSolveauxiliary functionhttp:/www.docin.com/sundae_mengParticular 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 inverse operatorsgeneral applicabilityhttp:/www.docin.com/sundae_mengMethod 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 constants. The form of a particular integral ishttp:/www.docin.com/sundae_mengMethod 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:/www.docin.com/sundae_mengSolveEquating coefficients of equal powers of xauxiliary equationhttp:/www.docin.com/sundae_mengMethod of inverse operatorsSometimes, it is convenient to refer to the symbol “D” as the differential operator:But, http:/www.docin.com/sundae_mengThe 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.http:/www.docin.com/sundae_mengDifferential operator to exponentialsMore convenient!http:/www.docin.com/sundae_mengDifferential operator to trigonometrical functionswhere “Im” represents the imaginary part of the function which follows it.http:/www.docin.com/sundae_mengThe 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:/www.docin.com/sundae_mengSolvedifferential operatorbinomial expansion=2http:/www.docin.com/sundae_meng如果 f(p) = 0, 使用因次分析非0的部分y = 1, p = 0, 即將 D-p換為 Dintegrationhttp:/www.docin.com/sundae_mengSolvedifferential operatorf(p) = 0integrationy = yc + yphttp:/www.docin.com/sundae_mengSolvedifferential operatorexpanding each term by binomial theoremy = yc + yphttp:/www.docin.com/sundae_mengO.D.E 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 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 Bhttp:/www.docin.com/sundae_menguC0xLxuCA 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. (Wehner and Wilhelm, 1956)xx+xBulk flow of ADiffusion of AInput - Output + Generation = Accumulation分開兩個sectionCehttp:/www.docin.com/sundae_mengdividing by xrearrangingauxillary functionIn the entry sectionauxillary functionhttp:/www.docin.com/sundae_mengB. C. B. C. if diffusion is neglected (D0)http:/www.docin.com/sundae_mengThe 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http:/www.docin.com/sundae_mengTallow 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:/www.docin.com/sundae_mengConsider the changes 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 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 fat phasein the extract phasein the fat phasein the extract phasehttp:/www.docin.com/sundae_meng2nd O.D.E. with constant coefficientsComplementary functionParticular solutionConstant at the right hand side, yp = C/Rhttp:/www.docin.com/sundae_mengB.C.We dont really want x here!Apply the equations two slides earlier (replace y* with mx)We dont know y0, eitherSubstitute y0 in terms of other variableshttp:/www.docin.com/sundae_menghttp:/www.docin.com/sundae_mengSimultaneous 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 independent variable.Our purpose: Use algebraic elemination of the variables until only one differential equation relating two of the variables remains.http:/www.docin.com/sundae_mengElimination of variableIndependent variable or dependent variables?Elimination of independent variable較少用Elimination of one or more dependent variablesIt involves 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:/www.docin.com/sundae_mengElimination of dependent variablesSolvehttp:/www.docin.com/sundae_meng= Ey = yc + yphttp:/www.docin.com/sundae_meng1.25 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.shttp:/www.docin.com/sundae_meng1.25 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:/www.docin.com/sundae_mengWhen water fails for 1 hour, heat balance for tank 1 and tank 2:Tank 1Tank 2M: mass flow rate of acidC: heat 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:/www.docin.com/sundae_mengWhen 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: 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:/www.docin.com/sundae_meng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)再由出現次數次少的 t2 消去.代入各數值.B.C. t = 0, T2 = 94.9 C 同時整理T1http:/www.docin.com/sundae_meng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 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:/www.docin.com/sundae_meng