Cap4 lec5

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UNIVERSIDAD NACIONAL ABIERTA Y A DISTANCIA - UNAD. ESPECIALIZACIÓN EN INGENIERÍA DE PROCESOS DE ALIMENTOS Y BIOMATERIALES MÉTODOS MATEMÁTICOS 1 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: 0 n n 0 n n L x n sen B L x n cos A ) x ( f Donde n es un entero positivo. Si la ecuación presentada se multiplica por dx L x n cos y el resultado se integra entre –L y L, se pueden determinar los coeficientes An. De manera similar, utilizando dx L x n sen se obtienen lo coeficientes Bn: L L n L L 0 dx L x n cos ) x ( f L 1 A ; dx ) x ( f L 2 1 A L L n 0 dx L x n sen ) x ( f L 1 B ; 0 B Aunque la expansión en series de Fourier de la función f(x) dependa solo de términos de seno o de coseno, o de ambos, si depende de si la función es regular, impar o ninguna de las dos. Una función regular, es una donde f(x)=f(-x). u un número puro son funciones regulares ( o pares). Si una función es regular: xsenmx , nx cos , x 2 0 B ; dx L x n cos ) x ( f L 2 A ; dx ) x ( f L 1 A 0 L 0 n L 0 0 Entonces se puede decir que: L x n sen ) x ( f L x n sen ) x ( f Consecuentemente, en la evaluación de los coeficientes An, la integral desde –L hasta 0 se suma a la integral de 0 a L, mientras que en la evaluación de los términos Bn estas integrales se cancelan. Esto se puede mostrar de manera formal a continuación: Sea f(x) una función regular (o par), entonces: L 0 0 L 0 L 0 0 0 L L 0 L L 0 dx ) x ( f L 1 A dx ) x ( f dx ) x ( f L 2 1 A dx ) x ( f dx ) x ( f L 2 1 dx ) x ( f L 2 1 A Los coeficientes An y Bn se obtienen de la misma manera.

Transcript of Cap4 lec5

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UNIVERSIDAD NACIONAL ABIERTA Y A DISTANCIA - UNAD.ESPECIALIZACIÓN EN INGENIERÍA DE PROCESOS DE ALIMENTOS Y BIOMATERIALES

MÉTODOS MATEMÁTICOS

1

SEGUNDA UNIDAD: FORMULACIÓN INTEGRAL Y DIFERENCIALCAPÍTULO 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:

� ���

���

� ���

���

��

� 0nn

0nn L

xnsenBL

xncosA)x(f

Donde n es un entero positivo.Si la ecuación presentada se multiplica por �dxL

xncos y el resultado se integra entre –L y L, se pueden

determinar los coeficientes An. De manera similar, utilizando �dxLxnsen se obtienen lo coeficientes Bn:

�����

L

Ln

L

L0 dx

Lxncos)x(f

L1A;dx)x(f

L21A

���

L

Ln0 dx

Lxnsen)x(f

L1B;0B

Aunque la expansión en series de Fourier de la función f(x) dependa solo de términos de seno o de coseno, o de ambos, si depende de si la función es regular, impar o ninguna de las dos.

Una función regular, es una donde f(x)=f(-x). u un número puro son funciones regulares ( o pares). Si una función es regular:

xsenmx,nxcos,x2

0B;dxL

xncos)x(fL2A;dx)x(f

L1A 0

L

0n

L

00 ��

���

Entonces se puede decir que:

���

��� ����

L

xnsen)x(fL

xnsen)x(f

Consecuentemente, 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:

��

��

���

�����

��

���

������

��

L

00

L

0

L

00

0

L

L

0

L

L0

dx)x(fL1A

dx)x(fdx)x(fL21A

dx)x(fdx)x(fL21dx)x(f

L21A

Los coeficientes An y Bn se obtienen de la misma manera.

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UNIVERSIDAD NACIONAL ABIERTA Y A DISTANCIA - UNAD.ESPECIALIZACIÓN EN INGENIERÍA DE PROCESOS DE ALIMENTOS Y BIOMATERIALES

MÉTODOS MATEMÁTICOS

2

Una función impar, es una en la que f(x)=-f(-x). son funciones impares.)nxcos(x),nx(sen,x,x 3

Si f(x) es impar:

��

L

0n

n

0

dxL

xnsen)x(fL2B

0A0A

Si 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:

� �����

���

���

��x0dvda)av(sen)v(Faxsen2)x(F

00

* Integral de coseno de Fourier:

� �����

���

���

��x0dvda)avcos()v(Faxcos2)x(F

00

* Integral de Fourier-Bessel:

1p;x0dvda)ax(J)v(avF)x(F0 0

p ������ ����

* Integral completa de Fourier:

������ � �

��

��

��xdvda)xv(acos()v(F

21)x(F

En la carpeta del capítulo cuatro ud. podrá encontrar un ejemplo de solución utilizando series de Fourier,extraído de BRDKEY y HERSHEY, Transport Phenomena.

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654 OF TRANSPORT PHENOMENA

It 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 for

W.2.1 Fourier Series Solution

The 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

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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.

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656 APPLICATIONS OF TRANSPORT PHENOMENA

the 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

Similarly

Substituting the above into Eq. the result is

(13.43)

(13.44)

The 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 independent

of 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)

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The 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, and

by 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)

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where 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 to

evaluate 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 constant integration 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

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method of Fourier series is

Equation (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 at26.03 atm and 317.2 K as shown in Fig. 13.8. The ends of the pipe areinitially capped by removable partitions. At time zero, the partitions areremoved, and across each end of the pipe flows a stream of air plus helium at thesame temperature and pressure. On the left end, the stream is 90 percent air and10 percent He (by volume) and on the right 80 percent air and 20percent He. Itmay be assumed that the flow effectively maintains the helium concentrationconstant at the ends. If isothermal conditions are maintained and there are noend effects associated with the air flowing past the pipe, calculate the compositionprofile (to four decimal places) after 1.2 h at space increments of UseFourier 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, thesolution to this problem arbitrarily use the direction as the direction of masstransfer.

.

FIGURE90% air, 10% He 80% air, 20% He

atm, 44°C 1 atm, 44°CTransient diffusion of heliumin a pipe.

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OF TRANSPORT PHENOMENA

Concent ra t ion i s re la ted to par t ia l pressure by Eq. (2 .37) :

= (2.37)

where is the gas constant (0.082057 atm from Table C.l). Thepart ia l pressure of species A is def ined by Eq. (2 .38) :

PA = (2.38)where is the mole fract ion of A and i s the to ta l p ressure .

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 ofhel ium at the ends of the pipe are

= = 2 .603 a tm

= = = atm

Inser t ing these in to Eq . the concent ra t ions are

= = x = 1.0 kmol = x = 0.1 kmol = x = 0.2 kmol

Fol lowing the nomenclature of Eq. these are

0) = 1.0 kmol = 0.1 kmol = 0.2 kmol

(ii)

(iii)

where 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 abovetransformat ion, Eq. (13 .37) s t i l l appl ies :

(13.37)

where the total length of the pipe is or 3 ft.The boundary cond i t ions in Eqs . ( iv ) th rough (v i i ) do no t a l low a so lu t ion

by Four ie r ser ies because the condi t ion in Eq. (v i i ) 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)

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TRANSPORT 661

The steady-state solution is obtained from Fick’s law, Eq. (2.4). Forequimolar counter d i f fus ion , i t i s eas i ly shown tha t the ra t io must beconstant a t s teady-s ta te . I f the ra t io i s cons tan t , then the ra t io is also constant [cf. Eq. Since two points uniquely determine the equationfor a s t ra ight l ine , the equat ion for i s

= 0 + = = =

The boundary condi t ions for the t rans ien t so lu t ion are found by combiningthe las t two equa t ions . F i r s t , Eq . (v i i i ) i s so lved for t ) ; t hen the cond i t i onsin Eqs. (v) through (vii) are inserted into the resulting equation:

0) = 0) = 0.9 = t) = t) = 0 0 = 0

= = 0. = 0

The 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 af ter replacing w i t h D i s

(xiii)

boundary condi t ion of Eq. (x) i s used to evaluate the remaining cons tant ,

= = 0.9 [

A comparison of Eq. (xiv) with the Fourier series of Eq. (13.38) shows that must be zero for all and

2L

Equat ion (xv) equals the sum of two in tegra ls :

The first integral is identical to that in Eq. (13.64):

[for odd

The second in tegral may be located in a s tandard table of in tegrals

a(xviii)

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APPLICATIONS OF TRANSPORT PHENOMENA

where equals Substituting, the second integral is

The f i rs t term in the above is

( x x )

s ince 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):

( x x i i ), .

As a rule, these infinite series converge slowly; hence, it may be advantageous tocombine the two inf in i te ser ies :

(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 t ime and distance:

where(xxiv)

D =

A computer program to evaluate Eq. (xxiv) at the and of interest isgiven in F ig . 13 .9 . Equat ion (xxiv) i s a converging inf in i te ser ies , in which thesigns of the terms alternate in an irregular pattern. The best procedure is tocombine a l l terms of l ike s ign, then tes t the magni tude to see i f that combinat ionis less than the accuracy des i red . In the case of a s imple a l ternat ing convergingseries, the truncation error is smaller in absolute value than the first termneglected and is of the same s ign. The resul ts are given in Table 13.2. The las tcolumn is an indicat ion of how many terms are required for the inf ini te ser ies toconverge.