Transport Phenomena in Materials Processing

Abstract

Before studying Electromagnetic Processing of Materials, let us see how the transport phenomena of mass, heat and momentum can be formulated in mathematical expressions. Later we will learn that a magnetic field can be described as a transport phenomenon where the governing equation of the magnetic field is the same type equation as those appearing in the transport phenomena of mass, heat and momentum, i.e., the partial differential equation of parabolic type.

Appendixes

1.1 Appendix A: Methods for Solving Eq. 2.37

• (1) The method of solving the partial differential equation (2.37) by transforming it to an ordinary differential equation.

  Let us assume c′ is a function of only the non-dimensional variable η $\ \equiv z/2\sqrt {{Dt}}$

  $$ c\prime = f(\eta ) $$

  (2.103)

  Then,

  $$ \partial \eta /\partial t = - \left( {z/4\sqrt {{Dt}} } \right)\,\,\,{t^{{ - 1}}} = - (\eta /2t) $$

  (2.104)

  Thus,

  $$ \partial c\prime/\partial t = (df/d\eta )(d\eta /dt) = - (\eta /2t)(df/d\eta ) $$

  (2.105)

  On the other hand,

  $$ \partial \eta /\partial t = 1/2\sqrt {{Dt}} $$

  (2.106)

  is valid, so that

  $$ {\partial^2}c\prime/\partial {z^2} = \frac{\partial }{{\partial z}}\,(\partial f/\partial z) = (\partial \eta /\partial z)\frac{d}{{d\eta }}\left( {\frac{{\partial \eta }}{{\partial z}}\frac{{df}}{{d\eta }}} \right) = \frac{1}{{4Dt}}\frac{{{d^2}f}}{{d{\eta^2}}} $$

  (2.107)

  By substituting Eqs. 2.105 and 2.107 into (2.37) and rearranging, the following equation is obtained.

  $$ {d^2}f/d{\eta^2} + 2\eta \,\,(df/d\eta ) = 0 $$

  (2.108)

  The boundary conditions of Eqs. 2.38 and 2.39 and the initial condition of Eq. 2.40 are written as follows:

  The boundary conditions are

  $$ f = {c_s} - {c_0}\quad \quad {\text{at}}\quad \eta = 0 $$

  (2.109)
  $$ f = 0\quad \quad {\text{at}}\quad \eta = \infty $$

  (2.110)

  The initial condition is

  $$ f = 0\quad \quad {\text{at}}\,\,\,\,\eta = \infty $$

  (2.111)

  Equation 2.108 is an ordinary differential equation so that it is easily integrated twice to get a solution as Eq. 2.112.

  $$ f = {c_1}\,\,\,\,\int_0^{\eta } {\mathit{exp}\,( - {\eta^2})\,\,\,d\eta + } {c_2} $$

  (2.112)

  The integral constants of c ₁ and c ₂ are determined by use of Eqs. 2.109, 2.110, and 2.111.

  $$ {c_1} = - 2\,\,({c_s} - {c_b})/\sqrt {\pi }, \,\,\,\,{c_2} = ({c_s} - {c_b}) $$

  (2.113)

  Then, the solution is obtained as Eq. 2.114.

  $$ c\prime = f = ({c_s} - {c_b})\,\left\{ {1 - \mathit{erf}\,\,\left( {z/2\sqrt {{Dt}} } \right)} \right\} $$

  (2.114)

  The key point in this method is to find the dimensionless variable η. The finding of η is not derived by chance, but can be done from the dimensionless procedure given in Sect. 3.

• (2) The method to solve the partial differential equation (2.37) by use of the Laplace Transform.

  The Laplace transform $\mathcal{L}[c\prime\,\,(z,t)] \equiv C\,\,(z,s)$ of Eq. 2.37 with respect to t is given as:

  $$ sC\,\,(z,\,s) - C\,(z,\,0) = D\,\,{d^2}C\,\,(z,\,\,s)/d\,{z^2} $$

  (2.115)

  From Eq. 2.40,

  $$ \mathcal{L}[c\prime\,\,(z,\,\,0)] \equiv C\,\,(z,\,\,0) = 0 $$

  (2.116)

  so that Eq. 2.115 is reduced to Eq. 2.117.

  $$ {d^2}C\,\,(z,\,\,s)/d\,{z^2} - (s/D)\,\,C\,\,(z,\,s) = 0 $$

  (2.117)

  The solution of this second-order ordinary differential equation is given as

  Eq. 2.118.

  $$ C\,\,(z,\,\,s) = A\,\,\mathit{exp}\,\left\{ {\sqrt {{s/D}} \,\,\,\,z} \right\} + B\,\,\mathit{exp}\left\{ {\sqrt {{ - (s/D)}} \,\,\,\,z} \right\} $$

  (2.118)

  The Laplace transform of the boundary conditions are given as:

  $$ C\,\,(0,\,\,\,s) = ({c_s} - {c_b})/s $$

  (2.119)
  $$ C\,\,(\infty, \,\,s) = 0 $$

  (2.120)

  By substituting Eq. 2.119 into Eq. 2.118, the following relation is obtained.

  $$ ({c_s} - {c_b})/s = A + B $$

  (2.121)

  From Eqs. 2.118, 2.120, and 2.121, A and B are obtained as:

  $$ A = 0, $$

  (2.122)
  $$ B = ({c_s} - {c_b})/s $$

  (2.123)

  Then, C (z,s) is given as:

  $$ C(z,s) = \left\{ {({c_s} - {c_b})/s} \right\} \exp\left\{ { - \sqrt {{(s/D)}} \,\,\,z} \right\} {=} \left\{ {({c_s} - {c_b})/s} \right\} \exp \left\{ { - ( {z/\sqrt {D} } )\sqrt {s} } \right\} $$

  (2.124)

  By use of the inverse Laplace transform defined as ${\mathcal{L}^{{ - 1}}}\{ (1/s)\, \mathit{exp}(- a\sqrt {s} ) =\break \mathit{erfc}\,(a/2\sqrt {t} )$, Eq. 2.41 is obtained as follows:

  $$ C\,\,(z,\,\,t) = ({c_s} - {c_b})\,\mathit{erfc}\left\{ {\left( {c/\sqrt {D} } \right)/\left( {2\sqrt {t} } \right)} \right\} = ({c_s} - {c_b})\left\{ {1 {-} {\text{erf}}\,\,\,( {z/2\sqrt {{Dt}} } )} \right\} $$

  (2.41)

1.2 Appendix B: Derivation of Eq. 2.54

Let us put c = c _s  + c ₂. Then, c ₂ should be satisfied by the governing equation (2.125) under the boundary and initial conditions of (2.126), (2.127), and (2.128), which are modified from Eqs. 2.51, 2.52, and 2.53 as follows:

$$ \partial {c_2}/\partial t = D\,\,{\partial^2}{c_2}/\partial {z^2}, $$

(2.125)

$$ \partial {c_2}/\partial z\,\,{|_{{z = 0}}} = 0 $$

(2.126)

$$ {c_2}\,\,{|_{{z = l}}} = 0, $$

(2.127)

$$ {c_2}\,\,{|_{{t = 0}}} = {c_b} - {c_S} $$

(2.128)

Putting c ₂(z, t) = Z(z) · T(t) and substituting it into Eq. 2.125, then the following equation is obtained.

$$ (1/T)(dT/dt) = (D/Z)({d^2}T/d{z^2}), $$

(2.129)

where T (t) and Z (z) can be obtained by solving the following equations, respectively.

$$ dT/dt - DkT = 0 $$

(2.130)

$$ {d^2}Z/d{z^2} - kZ = 0 $$

(2.131)

From Eq. 2.130, T = C exp (Dkt) is the solution.

On the other hand, the solution of Z = az + b when k = 0 is obtained from (2.131). However, it is found that this solution is not suitable since Z = 0 is from a = b = 0 which are obtained from Eqs. 2.126 and 2.127. If we put k = p ² where k is positive and p is a real number, the solution of Z = A exp (pz) + B exp (−pz) is obtained. This solution is also not suitable since A = B = 0 is from Eqs. 2.126 and 2.127. Then, let k be negative, that is k = −q ² . The solution is obtained as Z = A cos (qz) + B sin (qz). From (2.126), dZ/dz = Bq = 0 is so that B = 0 yields. Then, the solution is Z = A cos (qz). From Eq. 2.127, the condition of A cos (ql) = 0 is obtained. That is, q should be satisfied in the following condition.

$$ \mathit{cos} (ql) = 0 $$

(2.132)

That is, the condition which q should be satisfied is as follows:

$$ q = (2m + 1)\,\pi /2l,\,\,(m\,\,{\text{is}}\,\,{\text{integers}}) $$

(2.133)

Then, c2 is obtained as Eq. 2.134.

$$ {c_2} = {A_m}\,\,\mathit{exp}\,\,( - D{q^2}t)\,\, \cdot \,\,\cos \,\,(qz), $$

(2.134)

where A _m is a function of m.

As Eq. 2.134 holds for all of integers m, c ₂ is given as:

$$ {c_2} = \sum_{{m = 0}}^{\infty } {{A_m}} \,\,\mathit{exp}\,\,\left\{ { - D{{(2m + 1)}^2}{\pi^2}t/4{l^2}} \right\}\,\,\, \cdot \,\,\cos \,\,\left\{ {(2m + 1)\pi /2l} \right\}z $$

(2.135)

Finally, the function of A _m is determined to satisfy the initial condition of Eq. 2.128.

$$ {c_2}\,\,{|_{{t = 0}}} = \sum_{{m = 0}}^{\infty } {{A_m}} \cdot \cos \,\,\left\{ {(2m + 1)\pi z/2l} \right\} = {c_b} - {c_s},\,\,|z|\,\,\,\, < \,\,\,l $$

(2.136)

As the cosine is the even function, a Fourier series transform is applied over 0 < z < l. That is,

$$ \begin{aligned}[b] {A_m} & = (2/l)\,\,\int_0^l {({c_b} - {c_s})\,\, \cdot \,\,\cos } \left\{ {(2m + 1)\pi z/2l} \right\}dz \\ & = \left\{ {2\,({c_b} - {c_s})/l} \right\}\,\int_0^l {\cos \,\left\{ {(2m + 1)\pi z/2l} \right\}dz} \\ & = \left\{ {4\,({c_b} - {c_s})/(2m + 1)\pi } \right\}{( - 1)^m} \end{aligned} $$

(2.137)

Then, c ₂ is obtained as Eq. 2.138.

$$ \begin{aligned}[b] {c_2} & = \sum_{{m = 0}}^{\infty } \left\{ {4\,\,({c_b} - {c_s})/(2m + 1)\pi } \right\}\,\,{{( - 1)}^m}\\ &\quad \cdot \mathit{exp}\left\{ { - D\,\,{{(2m + 1)}^2}{\pi^2}t/4{l^2}} \right\} \times \cos \,\left\{ {(2m + 1)\pi z/2l} \right\} \end{aligned} $$

(2.138)

The final solution is c = c _s  + c ₂, so that

$$ \begin{aligned}[b] c & = {c_s} + 4\left\{ {({c_b} - {c_s})/\pi } \right\}\,\sum_{{m = 0}}^{\infty } {\left\{ {1/(2m + 1)} \right\}\,\,{{( - 1)}^m}} \\ & \quad \times \mathit{exp}\left\{ { - D\,{{(2m + 1)}^2}{\pi^2}t/4{l^2}} \right\}\,\, \cdot \,\,\cos \,\,\left\{ {(2m + 1)\pi z/2l} \right\} \end{aligned} $$

(2.54)

1.3 Appendix C: Derivation of Eq. 2.57

By putting c = c ₂ + c _b , Eq. 2.33 and Eqs.2.51, 2.52, and 2.53 become as Eq. 2.139 and Eqs. 2.140, 2.141, and 2.142, respectively.

$$ \partial {c_2}/\partial t = D\,\,{\partial^2}{c_2}/\partial {z^2} $$

(2.139)

$$ \partial {c_2}/\partial z\,\,{|_{{z = 0}}} = 0 $$

(2.140)

$$ {c_2}\,\,{|_{{z = \pm l}}} = {c_s} - {c_b} $$

(2.141)

$$ {c_2}\,\,{|_{{t = 0}}} = 0 $$

(2.142)

The Laplace transform $\mathcal{L}[{c_2}(z,t)] \equiv {C_2}(z,s)$ of Eq. 2.139 with respect to t is given as Eq. 2.143.

$$ s\,\,{C_2}(z,\,\,s) - {C_2}(z,\,\,0) = D\,\,{d^2}{C_2}(z,\,\,s)/d\,\,{z^2} $$

(2.143)

Also, the Laplace transforms of Eqs. 2.140, 2.141, and 2.142 are given as:

$$ d\,\,{C_2}(0,\,\,s)/dz = 0 $$

(2.144)

$$ {C_2}(x,\,\,s)\,\,{|_{{z = \pm l}}} = ({C_s} - {C_b})/s $$

(2.145)

$$ {C_2}(z,\,\,0) = 0 $$

(2.146)

The general solution of Eq. 2.143 is obtained as follows:

$$ {C_2}(z,\,\,s) = A\,\,\mathit{exp}\left( { - \sqrt {{s/D}} \,\,z} \right) + B \mathit{exp}\left( {\sqrt {{s/D}} \,\,z} \right) $$

(2.147)

From the boundary conditions of Eqs. 2.144 and 2.145, A = B and the following relation are obtained.

$$ {C_2}(l,\,\,s) = A\left\{ {\mathit{exp}\left( { - \sqrt {{s/D}} \,\,\,l} \right) + \mathit{exp}\left( {\sqrt {{s/D}} \,\,l} \right)} \right\} = ({c_s} - {c_b})/s $$

(2.148)

Thus,

$$ A = B = ({c_s} - {c_b})/\left[ {s\left\{ {\mathit{exp}\left( { - \sqrt {{s/D}} \,\,l} \right) + \mathit{exp}\left( {\sqrt {{s/D}} \,\,l} \right)} \right\}} \right] $$

(2.149)

Substituting Eq. 2.149 into Eq. 2.147 yields Eq. 2.150.

$$ \begin{aligned}[b] {C_2}(z,\,\,s) & = ({c_s} - {c_b})\left\{ {\exp \left( { - \sqrt {{s/D}} \,\,z} \right) + \exp \left( {\sqrt {{s/D}} \,\,\,z} \right)} \right\}\Big/\\ &\left[ {s\left\{ {\exp \,\,\left( { - \sqrt {{s/D}} \,\,l} \right) + exp\,\,\left( {\sqrt {{s/D}} \,\,l} \right)} \right\}} \right] \\ &= \left\{ {({c_s} - {c_b})/D} \right\}\,\, \cdot \,\,\left[ {\cosh \,\,\left( {\sqrt {{s/D}} \,\,\,z} \right)/\left\{ {(s/D)\cosh \,\,\left( {\sqrt {{s/D}} \,\,\,l} \right)} \right\}} \right] \end{aligned} $$

(2.150)

The Taylor expansion of (1 + e ^x)^–1 is given as follows:

$$ {(1 + {e^x})^{{ - 1}}} = 1 - {e^x} + {e^{{2x}}} - {e^{{3x}}} + \cdot \cdot \cdot \cdot + {( - 1)^n}{e^{{nx}}}, $$

(2.151)

so that Eq. 2.150 can be written as:

$$ \begin{aligned}[b] {C_2}(z,\,\,s) & = \left\{ {({c_s} - {c_b})/s} \right\}\,\, \cdot \,\,\left( {{e^{{ - \sqrt {{s/D}} \,\,z}}} + {e^{{\sqrt {{s/D}} \,\,z}}}} \right)\,\, \cdot \,\,\sum_{{n = 0}}^{\infty } {{{( - 1)}^n}{e^{{ - (2n + 1)\sqrt {{s/D}} \,\,l}}}} \\ & = \left\{ {({c_s} - {c_b})/s} \right\}\,\, \cdot \,\,\sum_{{n = 0}}^{\infty } {{{( - 1)}^n}\,\,} {e^{{ - \sqrt {{s/D}} \,\,z}}}{e^{{ - (2n + 1)\sqrt {{s/D}} \,\,l}}} \\ & \quad + \left\{ {({c_s} - {c_b})/s} \right\} \cdot \sum_{{n = 0}}^{\infty } {{{( - 1)}^n}} {e^{{\sqrt {{s/D}} \,z}}}{e^{{ - (2n + 1)\sqrt {{s/D}} \,\,l}}} \\ & = \left\{ {({c_s} - {c_b})/s} \right\} \cdot \,\,\sum_{{n = 0}}^{\infty } {{{( - 1)}^n}} {e^{{ - \sqrt {{s/D}} \left\{ {(2n + 1)l + z} \right\}}}} \\ & \quad + \left\{ {({c_s} - {c_b})/s} \right\} \cdot \sum_{{n = 0}}^{\infty } {{{( - 1)}^n}} {e^{{ - \sqrt {{s/D}} \left\{ {(2n + 1)l - z} \right\}}}} \end{aligned}\raisetag{17pt} $$

(2.152)

The inverse Laplace transformation gives the following relation as:

$$ {\mathcal{L}^{{ - 1}}}\left\{ {(1/s)\,\, \cdot \,\,\exp \,\,\left( { - a\sqrt {s} } \right)} \right\} = \mathit{erfc}\,\,\left( {a/2\sqrt {t} } \right) $$

(2.153)

By using the above relation, the inverse Laplace transformation of Eq. 2.152 is written as:

$$ \begin{aligned}[b] {c_2} & = ({c_s} - {c_b})\,\,\sum_{{n = 0}}^{\infty } {{{( - 1)}^n}\,\,\mathit{erfc}\left\{ {\frac{{(2n + 1)l + z}}{{2\sqrt {{Dt}} }}} \right\}} \\ & \quad + ({c_s} - {c_b})\,\,\sum_{{n = 0}}^{\infty } {{{( - 1)}^n}} \mathit{erfc}\left\{ {\frac{{(2n + 1)l - z}}{{2\sqrt {{Dt}} }}} \right\} \end{aligned} $$

(2.154)

Finally, the solution c is obtained from c = c ₂ + c _b as follows:

$$ \begin{aligned}[b] c & = {c_b} + ({c_s} - {c_b})\left[ \sum_{{n = 0}}^{\infty } {{( - 1)}^n} \mathit{erfc}\left\{ {\frac{{(2n + 1)l + z}}{{2\sqrt {{Dt}} }}} \right\} \right.\\ &\qquad\qquad\qquad\qquad \left. + \sum_{{n = 0}}^{\infty } {{{( - 1)}^n} \mathit{erfc}\left\{ {\frac{{(2n + 1)l - z}}{{2\sqrt {{Dt}} }}} \right\}}\right] \end{aligned} $$

(2.57)

1.4 Appendix D: Derivation of Eq. 2.65

By putting θ = c – c _s , Eqs. 2.61, 2.62, 2.63, and 2.64 are written as follows:

$$ \partial \theta /\partial t = D\,\,\left\{ {{\partial^2}\theta /\partial {r^2} + (1/r) \cdot (\partial \theta /\partial r)} \right\} $$

(2.155)

$$ \partial \theta /\partial r = 0,\,\,\,\,\,\,{\text{at}}\,\,\,r = 0,\,\,\,\,\,t \geqslant 0 $$

(2.156)

$$ \theta = 0,\quad \quad {\text{at}}\quad r = R,\,\,\,\,\,\,t \geqslant 0 $$

(2.157)

$$ \theta = {c_b} - {c_s}\,\,\,\,\,\,\,{\text{at}}\,\,\,\,\,t = 0,\,\,\,\,0 \leqslant r \leqslant R $$

(2.158)

We may attempt to find a solution of Eq. 2.155 by putting θ(r, t) = X(r) · T(t), where the X and T are functions of r and t, respectively. Equations 2.159 and 2.160 are obtained as follows:

$$ X\prime\prime + (1/r)X\prime - kX = 0 $$

(2.159)

$$ T\prime - kDT = 0, $$

(2.160)

where k is constant.

From Eqs. 2.156 and 2.157, the following boundary conditions are obtained.

$$ X\prime(0) = 0 $$

(2.161)

$$ X(R) = 0 $$

(2.162)

In the case of k = 0, the solution in the ordinary differential equation of Eq. 2.159 is X = C ₁ In(r) + C ₂. In this case, from the above boundary conditions, C ₁ = C ₂ = 0 is obtained. Then, the solution is X = 0. That is, it is understood that k = 0 does not provide a suitable solution. In the case of k > 0, the solution of Eq. 2.159 is $X = {C_3}{I_0}(\sqrt {k} r) + {C_4}{K_0}(\sqrt {k} r)$, where I ₀ and K ₀ are the modified Bessel functions of the first and second kinds of order zero, respectively. From the boundary condition of Eq. 2.161 with a finite value and the conditions of $I_0^{\prime}(0) = {I_1}(0) = 0$ and $K_0^{\prime}(0) = - {K_1} = - \infty$, so that C ₄ = 0 is obtained, and from the boundary conditions of Eq. 2.162 and the condition of I (0)| _r=0 ≠ 0, C ₃ = 0 is. Thus, the condition of k > 0 is also an unsuitable choice since X = 0 is. Finally, in the case of k < 0, the solution of Eq. 2.159 is $X = {C_5}{J_0}(\sqrt {{ - k}} r) + {C_6}{Y_0}(\sqrt {{ - k}} r)$, where J ₀ and Y ₀ are Bessel functions of the first and second kinds of order zero, respectively. From the boundary condition of Eq. 2.161 and the conditions of $J_0^{\prime}(0) = \mathit{finite}$ and $Y_0^{\prime}(0) = \infty$, so that C ₆ = 0 is obtained, and from the boundary condition of Eq. 2.162, ${C_5}{J_0}(\sqrt {{ - {k_n}}} R) = 0$ must be held, where k _n should be the n-th positive root of ${J_0}(\sqrt {{ - {k_n}}} R) = 0$. Then, X is given as follows:

$$ X = \sum_{{n = 1}}^{\infty } {{J_0}} ({\lambda_n}r),\,\,\,\,\,\,{\lambda_n} \equiv \sqrt {{ - {k_n}}} $$

(2.163)

The other hand, from Eq. 2.160, the function of T is given as:

$$ T = {C_7}\,\,\,\mathit{exp}\,\,\left( { - \lambda_n^2Dt} \right) $$

(2.164)

Then, θ(r, t) is obtained as Eq. 2.165.

$$ \theta \,\,(r,\,t) = \sum_{{n = 1}}^{\infty } {{C_n}} {J_0}({\lambda_n}r)\,\,\,\mathit{exp}\,\,\,\left( { - \lambda_n^2Dt} \right) $$

(2.165)

From the initial condition of (2.158), the following condition is obtained.

$$ \theta \,\,(r,\,\,0) = \sum_{{n = 1}}^{\infty } {{C_n}} {J_0}({\lambda_n}r) = {c_b} - {c_s} $$

(2.166)

In order to determine the values of ${C_n},\smallint_0^Rr{J_0}({\lambda_{{n\,}}}r)dr$ is multiplied on both sides of Eq. 2.166 and integrated from 0 to R. By using the following characteristics of the orthogonal functions Eqs. 2.167, 2.168, and 2.169, the values of C _n are obtained as Eq. 2.170.

$$ \int_0^R {r{J_0}} ({\lambda_n}r)\,{J_0}\,({\lambda_m}r)\,dr = 0,\,\,\,{\text{n}} \ne {\text{m}} $$

(2.167)

$$ \int_0^R {r{{\left\{ {{J_0}({\lambda_n}r)} \right\}}^2}} dr = (1/2){R^2}J_1^2({\lambda_n}R) $$

(2.168)

$$ \int_0^R {r{J_0}({\lambda_n}r)\,} dr = R{J_1}({\lambda_n}R)/{\lambda_n}, $$

(2.169)

where J ₁(x) is the Bessel function of the first order,

$$ {C_n} = ({c_b} - {c_s})\int_0^R {r{J_0}({\lambda_n}r)\,} dr\bigg/\!\int_0^R {r{{\left\{ {{J_0}({\lambda_n}r)} \right\}}^2}dr = 2({c_b} - {c_s})} /R{\lambda_n}{J_1}({\lambda_n}R) $$

(2.170)

Substituting Eq. 2.170 into Eq. 2.165 and using the relation of θ = c – c _s yields Eq. 2.65.

$$ c = \left\{ {2\,\,({c_b} - {c_s})/R} \right\}\,\,\sum_{{n = 1}}^{\infty } {\left\{ {{J_0}({\lambda_n}r)/{\lambda_n}{J_1}({\lambda_n}R)} \right\}\,\, \cdot \,\,\mathit{exp}\left( { - D\lambda_n^2t} \right)} + {c_s} $$

(2.65)

1.5 Appendix E: Derivation of Eq. 2.72

By setting $\eta (z,\,\,y) \equiv (c - {c_s})/({c_b} - {c_s}),y \equiv r/R$, and $z \equiv Dt/{R^2}$ to reduce to non-dimensional form, Eq. 2.68 is expressed as Eq. 2.171.

$$ \partial \eta /\partial z = {\partial^2}\eta /\partial {y^2} + (2/y) \cdot (\partial \eta /\partial y),\,\,\,(0 < z < \infty, \,\,\,0 < y < 1) $$

(2.171)

From the initial and boundary conditions, η (0, y) = 1, η (z, 1) = 0, η (z, 0)= $\partial \eta /\partial y{|_{{y = 0}}}$= 0 are obtained. The right-hand side of Eq. 2.171 is modified by multiplying y as follows:

$$ \begin{aligned}[b] y({\partial^2}\eta {/}\partial {y^2}) + 2 \cdot (\partial \eta /\partial y) & = y({\partial^2}\eta /\partial {y^2}) + (\partial \eta /\partial y) + (\partial \eta /\partial y) \\ & = \partial \left\{ {y(\partial \eta /\partial y) + \eta } \right\}/\partial y = {\partial^2}(\eta y)/\partial {y^2} \end{aligned} $$

(2.172)

And multiplying y on the left-hand side of Eq. 2.171 yields Eq. 2.173.

$$ y(\partial \eta /\partial z) = \partial \eta y/\partial z $$

(2.173)

That is, by introducing $\varphi (z,y) \equiv \eta y$, Eq. 2.171 can be expressed as Eq. 2.174.

$$ \partial \varphi /\partial z = {\partial^2}\varphi /\partial {y^2} $$

(2.174)

Then, the initial and boundary conditions become as follows:

$$ \varphi (0,y) = y,\,\,\,\varphi (z,1) = 0,\,\,\,\varphi (z,0) = 0 $$

We may attempt to find a solution of Eq. 2.174 by putting $\varphi = Z(z) \cdot Y(y)$, where Z and Y are functions of z and y, respectively. Substitution of $\varphi = Z \cdot Y$ in Eq. 2.174 yields Eq. 2.175.

$$ (1/Z) \cdot (dZ/dz) = (1/Y) \cdot ({d^2}Y/d{y^2}) = - {\alpha^2},\,\,\,(\alpha \,\,{\text{is}}\,\,{\text{a}}\,\,{\text{real}}\,\,{\text{number}}) $$

(2.175)

Then, we get the two ordinary differential equations as follows:

$$ dZ/dz + {\alpha^2}Z = 0 $$

(2.176)

$$ {d^2}Y/d{y^2} + {\alpha^2}Y = 0 $$

(2.177)

The corresponding general solutions are shown as follows:

$$ Z = {C_1}\,\,\mathit{exp}\,\,( - {\alpha^2}z) $$

(2.178)

$$ Y = {C_2}\,\,\cos \,\,(\alpha y) + {C_3}\sin \,\,(\alpha y) $$

(2.179)

From the boundary conditions, the following conditions are obtained.

$$ {C_2} = 0,\quad \quad {\text{from}}\,\,\varphi \,\,(z,\,\,0) = 0 $$

(2.180)

$$ \mathit{sin}\,\,\alpha = 0,\,\,\,{\text{from}}\,\,\varphi \,\,(z,\,\,1) = 0 $$

(2.181)

From Eq. 2.181, α should be given as:

$$ \alpha = n\pi, \,\,\,\,\,\,\,\,(n\,\,{\text{expresses}}\,\,{\text{integer}}\,\,{\text{numbers}}) $$

(2.182)

Thus, the general solution of φ is given as:

$$ \varphi = \sum_{{n = 1}}^{\infty } {{C_n}\mathit{exp}\,\,( - {n^2}{\pi^2}z)\,\, \cdot \,\,\sin \,\,{{(n\pi y)}^{\alpha }}} $$

(2.183)

C _n is determined to satisfy the condition of $\varphi (0,y) = y$ by using the Fourier sine transform.

$$ \varphi = \sum_{{n = 1}}^{\infty } {{C_n}} \sin \,\,(n\pi y) = y $$

(2.184)

$$ {\text{Thus}},\,\,\,{C_n} = 2\,\,\int_0^1 {\xi {\rm si} \,n(n\pi \xi )} \,\,d\xi = (2/n\pi ) \cdot {( - 1)^{{n + 1}}} $$

(2.185)

Then, the solution of φ is obtained as:

$$ \varphi = (2/\pi )\,\,\sum_{{n = 1}}^{\infty } {\left\{ {{{( - 1)}^{{n + 1}}}/n} \right\}} \cdot \mathit{exp}\,\,( - {n^2}{\pi^2}z) \cdot \sin \,\,(n\pi y) $$

(2.186)

Finally φ(z, y) is reduced to c (r, t) and then Eq. 2.72 is obtained.

$$ c = {c_s} + \left\{ {2R\,\,({c_b} - {c_s})/\pi r} \right\}\sum_{{n = 1}}^{\infty } {\left\{ {{{( - 1)}^{{n + 1}}}/n} \right\} \mathit{exp}\,( - {n^2}{\pi^2}Dt/{R^2})\sin (n\pi r/R)} $$

(2.72)