化工應用數學 授課教師： 郭修伯 Lecture 4 Ordinary Differential Equations Differential equation • An equation relating a dependent variable to one or more independent variables by means.
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化工應用數學 授課教師： 郭修伯 Lecture 4 Ordinary Differential Equations Differential equation • An 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”. d 3 y dy 2 x ( ) 4 y 4 e cos x 3 dx dx Ordinary differential equation -------only one independent variable involved: x T 2T 2T 2T Partial differential equation --------------C p k ( 2 2 2 ) more than one independent variable involved: x, y, z, x y z Order and degree • The 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. 3 d y dy 2 x ( ) 4 y 4 e cos x 3 dx dx 3rd order O.D.E. 1st degree O.D.E. Linear or non-linear • Differential equations are said to be nonlinear if any products exist between the dependent variable and its derivatives, or between the derivatives themselves. d 3 y dy 2 x ( ) 4 y 4 e cos x 3 dx dx Product between two derivatives ---- non-linear dy 4 y 2 cos x dx Product between the dependent variable themselves ---- non-linear First order differential equations • No general method of solutions of 1st O.D.E.s because of their different degrees of complexity. • Possible to classify them as: – exact equations – equations in which the variables can be separated – homogenous equations – equations solvable by an integrating factor Exact equations • Exact? M ( x, y)dx N ( x, y)dy 0 F F dx dy dF x y General solution: F (x,y) = C For example dy x y sin x (cos x 2 y ) 0 dx 3 Separable-variables equations • In 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 example dy y sin x dx Homogeneous equations • Homogeneous/nearly homogeneous? • A differential equation of the type, dy y f dx x is termed a homogeneous differential equation of the first order. • Such an equation can be solved by making the substitution u = y/x and thereafter integrating the transformed equation. Homogeneous equation example • Liquid 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 k3 C6H6+Cl2 C6H5Cl +HCl C6H5Cl+Cl2 C6H4Cl2 + HCl C6H4Cl2 + Cl2 C6H3Cl3 + HCl Take a basis of 1 mole of benzene fed to the reactor and introduce the following variables to represent the stage of system at time , p = moles of chlorine present q = moles of benzene present r = moles of monochlorbenzene present s = moles of dichlorbenzene present t = moles of trichlorbenzene present Then q + r + s + t = 1 and the total amount of chlorine consumed is: y = r + 2s + 3t From the material balances : in - out = accumulation 0 k1 pq V dq d dr d ds k 2 pr k 3 ps V d dt k 3 ps V d k1 pq k 2 pr V u = r/q dr k2 r ( ) 1 dq k1 q Equations solved by integrating factor • There 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”. dy Py Q dx where P and Q are functions of x only Assuming the integrating factor R is a function of x only, then dy R yRP RQ dx dy dR d R y (Ry) dx dx dx R exp Pdx is the integrating factor Example 2 dy 3 x 4 Solve xy y exp dx 2 Let z = 1/y3 dz 3 4 dy y dz 3 dy 4 dx y dx 3x 2 dz 3xz 3 exp dx 2 integral factor 3x 2 1 3x C exp y3 2 3x 2 exp 3xdx exp 2 3x 2 dz 3x 2 exp 3 3xz exp 2 dx 2 3x 2 z exp 2 3 x C 3x 2 d z exp 3 dx 2 Summary 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. First O.D.E. in heat transfer An elevated horizontal cylindrical tank 1 m diameter and 2 m long is insulated with asbestos lagging of thickness l = 4 cm, and is employed as a maturing vessel for a batch chemical process. Liquid at 95 C is charged into the tank and allowed to mature over 5 days. If the data below applies, calculated the final temperature of the liquid and give a plot of the liquid temperature as a function of time. Liquid film coefficient of heat transfer (h1) = 150 W/m2C Thermal conductivity of asbestos (k) = 0.2 W/m C Surface coefficient of heat transfer by convection and radiation (h2) = 10 W/m2C Density of liquid () = 103 kg/m3 Heat capacity of liquid (s) = 2500 J/kgC Atmospheric temperature at time of charging = 20 C Atmospheric 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. Area of tank (A) = ( x 1 x 2) + 2 ( 1 / 4 x 12 ) = 2.5 m2 Rate 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) At steady state, the three rates are equal: kA (Tw Ts ) h2 A(Ts t ) l h1 A(T Tw ) Ts 0.326T 0.674t Considering the thermal equilibrium of the liquid, input rate - output rate = accumulation rate t Ts T Tw 0 h2 A(Ts t ) Vs dT d dT 0.0235T 0.235 0.235 cos( / 12) d B.C. = 0 , T = 95 Second O.D.E. • Purpose: reduce to 1st O.D.E. • Likely to be reduced equations: – Non-linear • Equations where the dependent variable does not occur explicitly. • Equations where the independent variable does not occur explicitly. • Homogeneous equations. – Linear • The coefficients in the equation are constant • The coefficients are functions of the independent variable. Non-linear 2nd O.D.E. - Equations where the dependent variables does not occur explicitly • They 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. Solve Let d2y dy x ax 2 dx dx dy p dx and therefore dp d 2 y dx dx2 dp xp ax dx integral factor 1 1 exp x 2 2 1 1 x2 x2 dp 2 x 2 e xpe2 axe2 dx 1 y ax C exp x 2 dx 2 x y ax Aerf B 2 error function d ( pe dx 1 2 x 2 1 2 x 2 ) axe Non-linear 2nd O.D.E. - Equations where the independent variables does not occur explicitly • They 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 Let p dy dx d 2 y dp dp dy dp 2 p dx dx dy dx dy Solve Let d2y dy 2 y 2 1 ( ) dx dx dy p dx and therefore d2y dp p dx2 dy dp yp 1 p 2 dy Separating the variables p dy ( a 2 y 2 1) dx p 1 dp dy 2 p 1 y x ln y ln a 1 ln( p 2 1) 2 dy (a 2 y 2 1) x 1 sinh1 (ay) b a Non-linear 2nd O.D.E.- Homogeneous equations dy y f dx x • The homogeneous 1st O.D.E. was in the form: • The corresponding dimensionless group containing 2 the 2nd differential coefficient is x d y2 dx • In general, the dimensionless group containing the n n 1 d y th x n coefficient is dxn • The second order homogenous differential equation dy y f can be expressed in a form analogous to dx x , viz. Assuming u = y/x Assuming x = et 2 du d y y dy 2 d u x f u , x x f , 2 2 dx dx dx x dx d 2u du du f u, If in this form, called homogeneous 2nd ODE dt 2 dt dt 2 Solve 2 2 d y 2 2 dy 2x y 2 y x dx dx 2 Dividing by 2xy d 2 y 1 y 1 x dy x 2 dx 2 x 2 y dx homogeneous Let x 2 d2y y dy f , 2 dx x dx y Ax y ux 2 d u du 2 2 du 2ux 2ux x 2 dx dx dx Let 2u y x( B ln x C ) 2 2 General solution x et d u du 2 dt dt 2 Singular solution 2 p du dt dp 2up p2 du A graphite electrode 15 cm in diameter passes through a furnace wall into a water cooler which takes the form of a water sleeve. The length of the electrode between the outside of the furnace wall and its entry into the cooling jacket is 30 cm; and as a 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 transfer from 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 is 1500 C, estimate the duty of the water cooler if the temperature of the electrode at the entrance to the cooler is to be 150 C. The following additional information is given. Surrounding temperature = 20 C Thermal conductivity of graphite kT = k0 - T = 152.6 - 0.056 T W/m C The temperature of the electrode may be assumed uniform at any cross-section. T0 T x The sectional area of the electrode A = 1/4 x 0.152 = 0.0177 m2 A heat balance over the length of electrode x at distance x from the furnace is input - output = accumulation dT dT d dT (kT A )x DU (T T0 )x 0 kT A kT A dx dx dx dx where U = overall heat transfer coefficient from the electrode to the surroundings D = electrode diameter d dT DU k x (T T0 )x T dx dx A d dT (k0 T ) (T T0 ) 0 dx dx 2 d 2T dT (k0 T ) 2 (T T0 ) 0 dx dx T0 T x p (k0 T ) p dT dx p dp d 2T dT dx2 dp p 2 (T T0 ) 0 dT (k0 T ) p dp p 2 (T T0 ) 0 dT p2 z [(k0 T ) y ] y (T T0 ) dz 2z 2 y 0 dy Integrating factor exp x 2dy k0 T0 y 2 k0 T0 y (k0 T )dT [C (k0 T )(T T0 ) 2 2 3 (T T0 )3 ] Linear differential equations • They 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: dny d n1 y dy P0 n P1 n1 ... Pn1 Pn y ( x) dx dx dx where (x) is any function of x. 2nd order linear differential equations The general equation can be expressed in the form d2y dy P Q Ry ( x) dx2 dx where P,Q, and R are constant coefficients Let the dependent variable y be replaced by the sum of the two new variables: y = u + v Therefore d 2u d 2v du dv P Q Ru P Q Rv ( x) 2 2 dx dx dx dx If v is a particular solution of the original differential equation d 2u du P Q Ru 0 2 dx dx The general solution of the linear differential equation will be the sum of a “complementary function” and a “particular solution”. purpose The complementary function d2y dy P Q Ry 0 2 dx dx Let the solution assumed to be: y Ame mx dy Am me mx dx Amemx ( Pm2 Qm R) 0 auxiliary equation (characteristic equation) Unequal roots Equal roots Real roots Complex roots d2y 2 mx A m e m 2 dx Unequal roots to auxiliary equation • Let the roots of the auxiliary equation be distinct and of values m1 and m2. Therefore, the solutions of the auxiliary equation are: y A1em1x y A2em2 x • The most general solution will be y A1em1x A2em2 x • If m1 and m2 are complex it is customary to replace the complex exponential functions with their equivalent trigonometric forms. Solve d2y dy 5 6y 0 2 dx dx auxiliary function m2 5m 6 0 m1 2 m2 3 y Ae2 x Be3 x Equal roots to auxiliary equation • Let the roots of the auxiliary equation equal and of value m1 = m2 = m. Therefore, the solution of the auxiliary equation is: y Aemx Let y Ve mx dy mx dV e mVe mx dx dx where V is a function of x 2 d2y d V mx mx dV 2 mx e 2 m e m Ve dx2 dx2 dx d2y dy P Q Ry 0 2 dx dx y (Cx d )emx d 2V 0 2 dx V Cx D Solve d2y dy 6 9y 0 2 dx dx auxiliary function m2 6m 9 0 m1 m2 3 y ( A Bx)e3 x Solve d2y dy 4 5y 0 2 dx dx auxiliary function m2 4m 5 0 m 2i ( 2i ) x y Ae ( 2i ) x Be y e ( E cos x F sin x) 2x Particular integrals • Two methods will be introduced to obtain the particular solution of a second linear O.D.E. – The method of undetermined coefficients • confined to linear equations with constant coefficients and particular form of (x) – The method of inverse operators • general applicability d2y dy P Q Ry ( x) 2 dx dx Method of undetermined coefficients P d2y dy Q Ry ( x) 2 dx dx • When (x) is constant, say C, a particular integral of equation is y C/R • When (x) is a polynomial of the form a0 a1x a2 x2 ... an xn where all the coefficients are constants. The form of a particular integral is y 0 1 x 2 x 2 ... n x n • When (x) is of the form Terx, where T and r are constants. The form of a particular integral is y e rx Method of undetermined coefficients P d2y dy Q Ry ( x) 2 dx dx • When (x) is of the form G sin nx + H cos nx, where G and H are constants, the form of a particular solution is y L sin nx M cosnx • Modified procedure when a term in the particular integral duplicates a term in the complementary function. Solve d2y dy 4 4 y 4 x 8x3 2 dx dx auxiliary equation m2 4m 4m 0 y p qx rx 2 sx 3 d2y 2r 6sx dx2 dy q 2rx 3sx 2 dx (2r 6sx) 4(q 2rx 3sx 2 ) 4( p qx rx 2 sx3 ) 4x 8x3 Equating coefficients of equal powers of x 2r 4 q 4 p 0 6s 8r 4q 4 4r 12s 0 4s 8 y p 7 10x 6x2 2x3 yc ( A Bx)e2 x ygeneral yc y p Method of inverse operators • Sometimes, it is convenient to refer to the symbol “D” as the differential operator: Dy dy dx d2y D ( Dy) D y d x2 ... 2 But, dy ( Dy) dx 2 2 dny D y d xn n d2y dy 3 2y 2 dx dx D2 y 3Dy 2 y (D2 3D 2) y (D 1)(D 2) y The differential operator D can be treated as an ordinary algebraic quantity with certain limitations. (1) The distribution law: A(B+C) = AB + AC which applies to the differential operator D (2) The commutative law: AB = BA which does not in general apply to the differential operator D Dxy 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 D D(Dy) = (DD)y D(xy) = (Dx)y + x(Dy) The basic laws of algebra thus apply to the pure operators, but the relative order of operators and variables must be maintained. Differential operator to exponentials De px pe px ... ( D2 3D 2)e px ( p 2 3 p 2)e px D n e px p n e px f ( D )e px f ( p )e px D( ye px ) e px Dy yDe px e px ( D p) y D 2 ( ye px ) e px ( D p) 2 y ... D n ( ye px ) e px ( D p) n y f ( D)( ye ) e f ( D p) y px px More convenient! Differential operator to trigonometrical functions Dn (sin px) Dn Imeipx Im Dneipx Im(ip)n eipx where “Im” represents the imaginary part of the function which follows it. eipx cos px i sin px D 2 n (sin px) ( p 2 ) n sin px D 2 n 1 (sin px) ( p 2 ) n p cos px D 2 n (cos px) ( p 2 ) n cos px D 2 n 1 (cos px) ( p 2 ) n p sin px The inverse operator The operator D signifies differentiation, i.e. D f ( x) dx f ( x) f ( x)dx D 1 f ( x) •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 D Solve dy 4 y e2 x dx differential operator ( D 4) y e2 x 1 y e2 x ( D 4) y f ( D)e px f ( p)e px p2 1 e2 x 1 4(1 D) 4 binomial expansion y 1 2x 1 1 1 e [1 ( D ) ( D ) 2 ( D ) 3 ...]1 4 4 4 4 =2 1 1 1 1 y e 2 x [1 ( ) ( ) 2 ( ) 3 ...] 4 2 2 2 1 y e2 x 2 y 1 e2 x ( 2 4) 1 y e2 x ( D 4) f ( D)e px f ( p)e px p2 y 1 e2 x ( 2 4) 1 1 px e e px f ( D) f ( p) 如果 f(p) = 0, 使用因次分析 1 1 px e px e f ( D) ( D p ) n ( D) 非0的部分 f ( D)e px f ( p)e px px 1 e 1 e px f ( D) ( p) ( D p) n 1 e px x n px e f ( D) ( p) n! f ( D) ye px e px f ( D p) y y = 1, p = 0, 即將 D-p換為 D integration 1 e px 1 px e f ( D) ( p) D n 1 e px px e D n f ( D) ( p) Solve d2y dy 4x 8 16 y 6 xe dx2 dx m2 8m 16 0 differential operator ( D2 8D 16) y ( D 4)2 y 6xe4 x yp 6 4x xe ( D 4) 2 yc ( A Bx)e4 x y = yc + yp f(p) = 0 f ( D)e px e px f ( D p) y p 6xe4 x D2 integration y p 6 xe4 x x2 2! y p 3x3e4 x Solve d 2 y dy 3 2 6 y 4 x 3 x dx2 dx m2 m 6 0 differential operator ( D2 D 6) y ( D 3)(D 2) y 4x3 3x 2 yp yc Ae3x Be2 x 1 (4 x 3 3x 2 ) ( D 3)(D 2) 1 1 1 3 2 yp ( 4 x 3 x ) 5 (3 D) (2 D) y = yc + yp expanding each term by binomial theorem 1 D D 2 D3 1 1 D D 2 D 3 y p ... ... (4 x 3 3x 2 ) 5 3 9 27 81 8 16 2 4 4 x 3 3x 2 12x 2 6 x 7(24x 6) 13 24 yp 0... 6 36 216 1296 O.D.E in Chemical Engineering • A 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, A 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. L u C0 The concentraion will vary in the entry section due to diffusion, but will not vary in the section following the reactor. (Wehner and Wilhelm, 1956) u C Ce x 分開兩個section x x Bulk flow of A uC Diffusion of A dC D dx x+x dC x dx dC d dC D D x dx dx dx uC u A material balance can be taken over the element of length x at a distance x fom the inlet Input - Output + Generation = Accumulation dC dC dC d dC uC D dx uC u dx x D dx dx D dx x kCx 0 dC dC dC d dC uC D uC u x D D x kCx 0 dx dx dx dx dx dividing by x dC d dC u D kC 0 dx dx dx rearranging d 2C dC D 2 u kC 0 dx dx In the entry section auxillary function d 2C dC D 2 u 0 dx dx auxillary function Dm2 um k 0 Dm2 um 0 ux 1 a B exp ux 1 a C A exp 2D 2D ux C exp D a 1 4kD / u 2 ux C exp D ux 1 a B exp ux 1 a C A exp 2D 2D B. C. B. C. dC dC x0 dx dx dC xL 0 dx x C C0 x0 C C C 2 ux ua L x a 1exp ua L x exp a 1exp C0 K 2 D 2D 2D K a 1 expuLa / 2D a 1 exp uLa / 2D 2 if diffusion is neglected (D0) 2 C0 C kx 1 exp C0 u The continuous hydrolysis of tallow in a spray column 連續牛油水解 Glycerin, 甘油 1.017 kg/s of a tallow fat mixed with 0.286 kg/s of high pressure hot water is fed into the base of a spray column operated at a temperature 232 C and a pressure of 4.14 MN/m2. 0.519kg/s of water at the same temperature and pressure is sprayed into the top of the column and descends in the form of droplets through the rising fat phase. Glycerine is generated in the fat phase by the hydrolysis reaction and is extracted by the descending water so that 0.701 kg/s of final extract containing 12.16% glycerine is withdrawn continuously from the column base. Simultaneously 1.121 kg/s of fatty acid 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 glycerine equivalent in the entering tallow 8.53% and the distribution ratio of glycerine between the water and the fat phase at the column temperature and pressure is 10.32, estimate the concentration of glycerine in each phase as a function of column height. Also find out 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 is 0.0028 s-1. Raffinate L kg/s L kg/s xH zH x+x z+z h Tallow fat x z G kg/s yH y+y h y Hot water G kg/s Extract x0 z0 y0 x = weight fraction of glycerine in raffinate y = weight fraction of glycerine in extract y*= weight fraction of glycerine in extract in equilibrium with x z = weight fraction of hydrolysable fat in raffinate Consider the changes occurring in the element of column of height h: Glycerine transferred from fat to water phase, KaS( y * y)h Rate of destruction of fat by hydrolysis, kSzh Rate of production of glycerine by hydrolysis, kSzh / w S: sectional area of tower k: specific reaction rate constant a: interfacial area per volume of tower : mass of fat per unit volume of column (730 kg/m3) K: overall mass transter coefficient w: kg fat per kg glycetine A glycerine balance over the element h is: dx kSzh Lx L x h KaS y * y h dh in the fat phase w dy G y h Gy KaS y y *h dh L kg/s xH zH G kg/s yH in the extract phase A glycerine balance between the element and the base of the tower is: Lz Lz0 in the fat phase Lx 0 w w in the extract phase h Gy Gy 0 x+x z+z y+y h x z y 0 The glycerine equilibrium between the phases is: y* m x x0 z0 y0 kS 2 Ka m z0 m G dy KaS dy d 2 y KaSm dy kS KaS y y0 y 0 2 LG w L L G dh G dh dh L dh p k S L r mG L q KaS (r 1) G m z0 d2y dy pq ( p q ) pqy ry 0 dh2 dh r 1 w Complementary function m2 ( p q)m pq 0 yc A exp( ph) B exp(qh) 2nd O.D.E. with constant coefficients Particular solution Constant at the right hand side, yp = C/R yp m z0 pq ry0 / pq r 1 w y A exp( ph) B exp(qh) m z0 1 ry0 r 1 w B.C. We don’t know y0, either h 0, x 0 h H, y 0 We don’t really want x here! h 0, y y0 h H, y 0 Apply the equations two slides earlier (replace y* with mx) mx y mx y r 1 pAexp( ph) qB exp(qh) q y(veqH re pH ) q rp p v q r 1 dy q dh mz 1 qH pH (e pH v)e qh veqH re pH ry0 0 (r e )e r 1 w Substitute y0 in terms of other variables h 0, y y0 m z0 pH e pH v qH ve qH re pH e y e qH qH w(r v) r e r e 1.017 1.121 1.069 2 0.286 0.519 G~ 0.403 2 0.701 0.1216 0.276 y0 ~ 0.165 0.701 0.286 m 10.32 L~ k 0.0028 S 0.662 4 730 Simultaneous differential equations • These 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. Elimination of variable Independent variable or dependent variables? dx f1 ( x , y ) dt dy f 2 ( x, y ) dt Elimination of independent variable dx f1 ( x, y ) dy f 2 ( x, y ) 較少用 Elimination of one or more dependent variables It involves with equations of high order and it would be better to make use of matrices Solving differential equations simultaneously using matrices will be introduced later in the term Elimination of dependent variables Solve ( D 2 D 6) y ( D 2 6 D 9 ) z 0 and ( D 2 3D 10) y ( D 2 3D 2) z 0 ( D 3)(D 2) y ( D 3) 2 z 0 and ( D 2)(D 5) y ( D 2)(D 1) z 0 ( D 5) ( D 3) ( D 3)(D 2)(D 5) y ( D 5)(D 3) 2 z 0 and ( D 3)(D 2)(D 5) y ( D 3)(D 2)(D 1) z 0 ( D 5)(D 3) 2 z ( D 3)(D 2)(D 1) z 0 ( D 3)(11D 13) z 0 ( D 3) ( D 2 8 D 15) ( D 2 3D 2) z 0 ( D 3)(11D 13) z 0 z Ae 13 x 11 Be3 x ( D2 D 6) y ( D2 6D 9) z 0 13 2 6 13 13 x ( D D 6) y 9 Ae 11 11 11 13 x 1 yp 2 Ee 11 ( D D 6) 2 =E yp 3 x yc He Je 2x f ( D)e px f ( p)e px p 13 11 1 13 13 (( ) 2 ( ) 6) 11 11 13 y = yc + yp 121 11 x yp Ee 700 Ee 13 x 11 Example of simultaneous O.D.E.s 1.25 kg/s of sulphuric acid (heat capacity 1500 J/kg C) is to be cooled in a two-stage countercurrent cooler of the following type. Hot acid at 174 C is fed to a tank where it is well stirred in contact with cooling coils. The continuous discharge from this tank at 88 C flows to a second stirred tank and leaves at 45C. Cooling water at 20 C flows into the coil of the second tank and thence 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. Calculate the 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 tank 750 W/m2 C. These constants may be assumed constant. Tank 1 Tank 2 80 C 40 C 20 C 0.96 kg/s 88 C 1.25 kg/s 45 C 174 C Heat transfer area A1 Heat transfer area A2 Steady state calculation: Heat capacity of water 4200 J/kg C 1.251500 (174 45) Fwater 4200 (80 20) Fwater 0.96kg / s 1.251500 (88 45) 0.96 4200 (Tmiddle 20) Tmiddle 40 C 1.251500 (174 88) 1150 A1 T and (88 80) (88 40) T 22.32 (88 80) ln (88 40) Note: 和單操課本不同 T (174 80) (88 40) 68.44 (174 80) ln (88 40) When water fails for 1 hour, heat balance for tank 1 and tank 2: Tank 1 Tank 2 dT1 dt dT MCT1 MCT 2 VC 2 dt MCT0 MCT1 VC d T1 dt d T2 dt T0 T1 T1 T2 B.C. t = 0, T1 = 88 M: mass flow rate of acid C: heat capacity of acid V: mass capacity of tank Ti: temperature of tank i T1 174 86et t = 1, T1=142.4 C dT2 integral factor, et T2 174 (86t 129)et 174 86e T2 dt t B.C. t = 0, T2 = 45 t = 1, T2 = 94.9 C When water supply restores after 1 hour, heat balance for tank 1 and tank 2: Tank 1 Tank 2 WCwt2 MCT0 WCwt1 MCT1 VC dT1 dt WCwt3 MCT1 WCwt2 MCT 2 VC dT2 dt t1 T0 t2 1 T1 t3 2 T2 W: mass flow rate of water Cw: heat capacity of water t1: temperature of water leaving tank 1 t2: temperature of water leaving tank 2 t3: temperature of water entering tank 2 Heat transfer rate equations for the two tanks: (T1 t1 ) (T1 t2 ) WC w (t1 t2 ) U1 A1 ln(T1 t1 ) ln(T1 t 2 ) (T2 t2 ) (T2 t3 ) WC w (t2 t3 ) U 2 A2 ln(T2 t2 ) ln(T2 t3 ) 4 equations have to be solved simultaneously WCwt2 MCT0 WCwt1 MCT1 VC dT1 dt WCwt3 MCT1 WCwt2 MCT 2 VC dT2 dt (T1 t1 ) (T1 t2 ) WC w (t1 t2 ) U1 A1 ln(T1 t1 ) ln(T1 t 2 ) (T2 t2 ) (T2 t3 ) WC w (t2 t3 ) U 2 A2 ln(T2 t2 ) ln(T2 t3 ) e e U1 A1 WCw T1 (1 ) t1 t2 U 2 A2 WC w T2 (1 ) t2 t3 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 ….. d 2T2 dT2 6.08 7.75T2 309 2 dt dt B.C. t = 0, T2 = 94.9 C 同時整理T1 基本上，1st order O.D.E. 應該都解的出來，方法不外乎： Check exact Separate variables homogenous equations, u = y/x equations solvable by an integrating factor 2nd order 以上的O.D.E Non-linear O.D.E. 我們會解的部分 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 General solution = complementary solution + particular solution variable coefficient? 用數列解，next course 我們會解的部分 constant coefficients 尋找特殊解的方法 The method of undetermined coefficients The method of inverse operators