Cap4 lec5

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Cap4 lec5

  1. 1. UNIVERSIDAD NACIONAL ABIERTA Y A DISTANCIA - UNAD. ESPECIALIZACIÓN EN INGENIERÍA DE PROCESOS DE ALIMENTOS Y BIOMATERIALES MÉTODOS MATEMÁTICOS SEGUNDA UNIDAD: FORMULACIÓN INTEGRAL Y DIFERENCIAL CAPÍTULO CUATRO: FUNCIONES ESPECIALES. LECCIÓN DOCE. SERIES DE FOURIER.Un caso especial y muy utilizado de la expansión de funciones ortogonales en series son las “series de Fourier”.En un intervalo simétrico –L,L, la representación de las series de Fourier de una función f(x) se define como: f § nSx · f § nSx · f( x) ¦ A n cos¨ ¸ ¦ B n sen¨ ¸ n 0 © L ¹ n 0 © L ¹Donde n es un entero positivo.
  2. 2. Si la ecuación presentada se multiplica por cos nSx L dx y el resultado se integra entre –L y L, se pueden
  3. 3. determinar los coeficientes An. De manera similar, utilizando sen nSx L dx se obtienen lo coeficientes Bn: 1 L 1 L nSx A0 ³ f( x)dx ; An ³ f( x) cos dx 2L L L L L 1 L nSx B0 0 ; Bn ³ f( x )sen dx L L LAunque la expansión en series de Fourier de la función f(x) dependa solo de términos de seno o de coseno, o deambos, si depende de si la función es regular, impar o ninguna de las dos.Una función regular, es una donde f(x)=f(-x). x 2 , cos nx, xsenmx u un número puro son funciones regulares ( opares). Si una función es regular: 1L 2L nSx A0 ³ f( x)dx ; An ³ f( x) cos dx ; B0 0 L0 L0 LEntonces se puede decir que: nSx § nSx · f( x)sen¨ f( x)sen ¸ L © L ¹Consecuentemente, en la evaluación de los coeficientes An, la integral desde –L hasta 0 se suma a la integral de 0a L, mientras que en la evaluación de los términos Bn estas integrales se cancelan. Esto se puede mostrar demanera formal a continuación:Sea f(x) una función regular (o par), entonces: 1 L 1 ªL 0 º A0 ³ f( x )dx « ³ f( x )dx ³ f( x )dx » 2L ¬ 0 2 L L L ¼ 1 ªL L º A0 « ³ f( x )dx ³ f( x )dx» 2L ¬ 0 0 ¼ 1L A0 ³ f( x )dx L0 Los coeficientes An y Bn se obtienen de la misma manera. 1
  4. 4. UNIVERSIDAD NACIONAL ABIERTA Y A DISTANCIA - UNAD. ESPECIALIZACIÓN EN INGENIERÍA DE PROCESOS DE ALIMENTOS Y BIOMATERIALES MÉTODOS MATEMÁTICOSUna función impar, es una en la que f(x)=-f(-x). x , x 3 , sen( nx), x cos(nx) son funciones impares.Si f(x) es impar: A0 0 An 0 2L nSx Bn ³ f( x )sen dx L0 LSi solo el intervalo 0 a L es de interés, f(x) puede expandirse solo en series senoidales o solo en seriescosenoidales, y esto puede realizarse aunque la función sea par o impar. Cuando este procedimiento se realiza laexpansión en términos de se senos genera una función impar que, en general, no representará a f(x) fuera delintervalo o a L. y la expresión en términos de cosenos, genera una función par que de igual manera norepresenta a f(x) fuera del intervalo.Cuando el intervalo de despliegue de las series incluye al infinito, y si la función es ortogonal, se puede llegauaren el despliegue a uno de los siguientes casos:* Integral del seno de Fourier: 2 ªf f º F( x) « ³ senax
  5. 5. ³ F( v)sen(av)dvda» S ¬0 0xf 0 ¼* Integral de coseno de Fourier: 2 ªf f º F( x) « ³ cosax
  6. 6. ³ F( v) cos(av)dvda» S ¬0 0xf 0 ¼* Integral de Fourier-Bessel: ff F( x ) ³ ³ avF( v )J p (ax )dvda 0xf ; p ! 1 00* Integral completa de Fourier: 1 f f F( x ) ³ ³ F( v ) cos(a( v x )dvda fxf 2 S f fEn la carpeta del capítulo cuatro ud. podrá encontrar un ejemplo de solución utilizando series de Fourier,extraído de BRDKEY y HERSHEY, Transport Phenomena. 2
  7. 7. 654 OF TRANSPORT PHENOMENAIt is common practice to abbreviate Eqs. (13.33) and (13.34) as (x, 0) = (0, = 0 (13.35)Similarly, it is desirable to transform the mass transfer equation, Eq.using some convenient variable such as = (13.36)where again the transformation is useful only if the boundary conditions arenot a function of time. For one-dimensional transient mass transfer, Eq.(13.11) becomes (13.37)The boundary conditions in terms of are almost identical to those in Eq.(13.35); the variable is simply substituted forW.2.1 Fourier Series SolutionThe solution of partial differential equations using Fourier series is usuallygiven in an advanced mathematics course at most universities. Hence, in thissection only a typical Fourier series solution to Eq. (13.32) will be given. Thereader is referred to the several excellent texts devoted to a more completetreatment Ml, M4, A Fourier series may be defined as + + sin (13.38)where the function f(x) is represented in terms of two periodic infinite series,as shown in Eq. (13.38). If the function f(x) is assumed to be periodic, withperiod’ 2L as shown in Fig. 13.6, then it is easily shown that (13.39) (13.40) (13.41)A function is periodic if + for all
  8. 8. UNSTEADY-STATE TRANSPORT The power of Fourier series arises from the fact that any function may beassumed periodic, even if it is not, by assuming that the length of the period isthe region of interest. For clarification, let us consider the following boundaryconditions: 0) = =0 =0 (13.35)These are the simplest possible; since the transformed temperature is zero ateither end, these boundary conditions are termed “homogeneous”. Obviously,those are not in themselves periodic; yet they may be considered as a periodicfunction of period as shown in Fig. 13.7. Physically the boundaryconditions in Fig. 13.7 have no meaning for less than zero or greater than but the mathematical assumption of periodicity allows a Fourier seriessolution, as will be shown later. Many functions likely to be encountered in our engineering problems canbe expanded in a Fourier series. There are some mathematical restrictions,known as the Dirichlet conditions that cannot be violated. A function isFourier expandable if in the interval 0 2L the following are true:1. The function f(x) is single-valued.2. The function f(x) never becomes infinite.3. The function f(x) has a finite number of maxima and minima.4. The function f(x) has a finite number of discontinuities.A practical concern that does limit the utility of Fourier series solutions is that -4L -2L 0 2L 4L (13.35) as a periodic function.
  9. 9. 656 APPLICATIONS OF TRANSPORT PHENOMENAthe integrals in Eqs. (13.39) to (13.41) must be analytic, since and arecoefficients in an infinite series. Also, not all boundary conditions areamenable to Fourier series solution C3, C6, Ml, M4, The first step in the Fourier series solution of a partial differentialequation is to assume that the solution is a product of two quantities: = (13.42)where is a function of only and is a function of only. This assumptioncan be justified only in that it leads to a solution satisfying the partialdifferential equation and its boundary conditions. The assumed solution, Eq.(13.42) is substituted into the partial differential equation of interest, Eq.(13.32). Since is a function of only, it follows that (13.43)Similarly (13.44)Substituting the above into Eq. the result isThe variables in Eq. (13.45) are separated as follows X ” (13.46)It is argued that, if is varied there is no effect on the term since is not a function of x. Thus, the term must be independentof x. A similar argument about varying and its effect on the term leadsto the conclusion that each side of Eq. (13.46) must be equal to a constant.This constant shall be designated as to facilitate later forms of thesolution: (13.47)The constant in Eq. (13.47) must be negative in order to avoid anexponential solution that would be inconsistent with the boundary conditions Equation (13.47) may be decomposed into two ordinary differentialequations: (13.48) (13.49)
  10. 10. UNSTEADY-STATE TRANSPORT 657The solutions to Eqs. (13.48) and (13.49) are assumed to be = exp( (13.50) (13.51)Appropriate boundary conditions for a direct Fourier series solution must bethe simplest possible, i.e., homogeneous: 0) = =0 =0 (13.35)At this stage in the solution, there are four constants and to bedetermined from the boundary conditions. Using the boundary condition atthe point = 0, Eq. (13.42) becomes = =0 (13.52)Since for the nontrivial case T(t) cannot be zero for all it follows that Eq.(13.52) is true only if =0 (13.53)Substituting the results of Eq. (13.53) into Eq. Eq. (13.51) becomes = + (13.54)Since the sine of zero is zero and the cosine of zero is one, then by Eq. (13.54) must be zero. Next, the boundary condition at the other end is applied, andby similar reasoning =0= sin (13.55)Equation (13.55) equals zero only if is zero or if sin is zero or if bothare zero. However, if is zero, then is zero and our assumed solution, Eq.(13.42) is trivial. Therefore sin =0 (13.56)The sine of an arbitrary angle is zero if = etc. Hence 1, 2, 3, . . (13.57)From Eq. the constant is found to be (13.58) The solution as determined so far is substituted from Eqs.(13.55) and (13.58) into Eq. (13.42): = (13.59)
  11. 11. APPLICATIONS OF TRANSPORT PHENOMENAwhere each and every j, as j goes from 1 to is a solution to the originalpartial differential equation, Eq. (13.32). Thus, the total solution is the sumover j of all possible solutions, since Eq. (13.32) is a linear partial differentialequation: (13.60)where the product has been replaced by The remaining boundary condition at time zero, Eq. is used toevaluate Applying that condition to Eq. (13.60): 0) (13.61)Since the exponential of zero is one, Eq. (13.61) reduces to = (13.62)A comparison of Eq. (13.62) with the Fourier series of Eq. (13.38) shows that must be zero for all j and (13.63)By substituting Eq. (13.63) into Eq. a complete solution is nowavailable. Note that a Fourier series solution is possible as long as theintegration in Eq. (13.63) can be performed. For the case of constantintegration of Eq. (13.63) yields (-cos 0)] = [for odd j] (13.64)The simplification of Eq. (13.64) resulted from the following reasoning. Thecosine of zero is one. The cosine of equals -1 for odd j and for even j.Hence -(cos jn) + 1 = 0 if j is even -(cos +1=2 if j is odd (13.65)Note that a single value of from Eq. (13.64) cannot possibly satisfy theboundary condition of Eq. (13.33). Hence, it is argued that only the sum fromone to infinity of all possible will satisfy the boundary condition, since Eq.(13.32) [or Eq. is a linear partial differential equation. Combining results, the final solution to Eq. (13.32) as determined by the
  12. 12. UNSTEADY-STATE TRANSPORT 659method of Fourier series isEquation (13.66) expresses in terms of an infinite series the dimensionlesstemperature for any and when the boundary conditions of Eq. (13.35)are valid. However, the engineer will usually be interested in the temperaturefor a particular and c or for a series of and combinations. In general aFourier series such as Eq. (13.38) or Eq. (13.66) converges very slowly,especially for small Often thousands of terms are required in order toevaluate to the required accuracy. In fact, evaluation of Eq. (13.66) on adigital computer requires almost as much effort as a direct numerical method(to be discussed subsequently), not including the lengthy steps required to getEq. (13.66). A mass transfer problem with nonhomogeneous boundary conditions issolved in Example 13.2. Example A 3-in. schedule 40 pipe is 3 ft long and contains helium at 26.03 atm and 317.2 K as shown in Fig. 13.8. The ends of the pipe are initially capped by removable partitions. At time zero, the partitions are removed, and across each end of the pipe flows a stream of air plus helium at the same temperature and pressure. On the left end, the stream is 90 percent air and 10 percent He (by volume) and on the right 80 percent air and 20percent He. It may be assumed that the flow effectively maintains the helium concentration constant at the ends. If isothermal conditions are maintained and there are no end effects associated with the air flowing past the pipe, calculate the composition profile (to four decimal places) after 1.2 h at space increments of Use Fourier series. The value of is 0.7652 x Answer. First, note that mass transfer in Fig. 13.8 occurs in the z direction only; there is no transport in either the or e-directions. Since the previous equations (derived with heat transfer as the example) are in terms of the direction, the . solution to this problem arbitrarily use the direction as the direction of mass transfer. FIGURE 90% air, 10% He 80% air, 20% He Transient diffusion of helium atm, 44°C 1 atm, 44°C in a pipe.
  13. 13. OF TRANSPORT PHENOMENA Concentration is related to partial pressure by Eq. (2.37): = (2.37)where is the gas constant (0.082057 atm from Table C.l). Thepartial pressure of species A is defined by Eq. (2.38): PA = (2.38)where is the mole fraction of A and is the total pressure. Example 2.7 illustrated the method of converting partial pressures intoconcentrations. Initially, the partial pressure of helium in the tube equals the totalpressure, 26.03 atm. When the partitions are removed, the ofhelium at the ends of the pipe are = = 2.603 atm = = = atmInserting these into Eq. the concentrations are = = x = 1.0 kmol = x = 0.1 kmol (ii) = x = 0.2 kmolFollowing the nomenclature of Eq. these are 0) = 1.0 kmol = 0.1 kmol (iii) = 0.2 kmolwhere the total length of the pipe is or 3 Although does not equalC it is still convenient to transform to 0,: = 0) = = 0.9 = = = t) = = 0.1 (vii)where the definition of 0, is arbitrarily based on for ease of notation,and have also been introduced in the above equations. Using the abovetransformation, Eq. (13.37) still applies: (13.37)where the total length of the pipe is or 3 ft. The boundary conditions in Eqs. (iv) through (vii) do not allow a solutionby Fourier series because the condition in Eq. (vii) does not equal zero. Thislimitation is easily circumvented by expressing the total solution as the sum of theunsteady-state (or transient or particular) solution 0, and the steady-statesolution 0, = + (viii)
  14. 14. TRANSPORT 661The steady-state solution is obtained from Fick’s law, Eq. (2.4). Forequimolar counter diffusion, it is easily shown that the ratio must beconstant at steady-state. If the ratio is constant, then the ratiois also constant [cf. Eq. Since two points uniquely determine the equationfor a straight line, the equation for is =0+ = = = The boundary conditions for the transient solution are found by combiningthe last two equations. First, Eq. (viii) is solved for t); then the conditionsin Eqs. (v) through (vii) are inserted into the resulting equation: 0) = 0) = 0.9 = t) = t) =0 0=0 = = 0. =0The solution to Eq. (13.37) subject to the above boundary conditions follows thederivation of Eq. (13.66) from the inception, Eq. up through Eq. which after replacing w i t h D i s (xiii) boundary condition of Eq. (x) is used to evaluate the remaining constant, = = 0.9 [A comparison of Eq. (xiv) with the Fourier series of Eq. (13.38) shows thatmust be zero for all and 2LEquation (xv) equals the sum of two integrals:The first integral is identical to that in Eq. (13.64): [for oddThe second integral may be located in a standard table of integrals (xviii) a
  15. 15. APPLICATIONS OF TRANSPORT PHENOMENAwhere equals Substituting, the second integral isThe first term in the above is ( x x )since cos = for even values of j and -1 for odd values of j. The second termis zero since sin equals zero for all values of j: sin 0 = sin 0=0 The value of is the combination of Eqs. (xvii) and (xx). The finalexpression is obtained by the summation of all between 1 and infinity, which isthen included in Eq. (xiii): (xxii) ,.As a rule, these infinite series converge slowly; hence, it may be advantageous tocombine the two infinite series: (xxiii)Note that if Eq. (xxiii) reduces to Eq. (13.66). Inserting Eqs. (ix) and(xxiii) into Eq. (viii) yields the final expression for the concentration or for asa function of time and distance: (xxiv)where D= A computer program to evaluate Eq. (xxiv) at the and of interest isgiven in Fig. 13.9. Equation (xxiv) is a converging infinite series, in which thesigns of the terms alternate in an irregular pattern. The best procedure is tocombine all terms of like sign, then test the magnitude to see if that combinationis less than the accuracy desired. In the case of a simple alternating convergingseries, the truncation error is smaller in absolute value than the first termneglected and is of the same sign. The results are given in Table 13.2. The lastcolumn is an indication of how many terms are required for the infinite series toconverge.

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