ow> T m the melting temperature, and R is the reflectivity at the front surface.

One obtains Cattaneo equation in dimensionless form as:

$\frac{\partial \phi }{\partial \tau }+2\phi =-\nabla \theta ,\text{ }\tau =\frac{t}{{t}_{k}}$ (10)

the energy conservation equation in the dimensionless scale attains the form:

$\frac{\partial \theta }{\partial \tau }=-\nabla \phi$ (11)

and the hyperbolic heat conduction equation (HHCE) in the following dimensionless form is:

$\frac{{\partial }^{2}\theta }{\partial {\tau }^{2}}+2\frac{\partial \theta }{\partial \tau }=\frac{{\partial }^{2}\theta }{\partial {X}^{2}}$ (12)

Equation (12) is subjected to the following dimensionless initial and boundary conditions:

$\text{at}\text{\hspace{0.17em}}\tau =0,\text{ }\theta \left(X,0\right)=0$ (13)

$\text{at}\text{\hspace{0.17em}}X={X}_{d},\text{ }\frac{\partial \theta }{\partial X}\left({X}_{d},\tau \right)=0$ (14)

$\text{at}\text{\hspace{0.17em}}X=0,\text{ }\frac{\partial \theta }{\partial X}\left(0,\tau \right)=-\left[2\phi \left(0,\tau \right)+\frac{\partial \phi }{\partial \tau }\left(0,\tau \right)\right]$ (15)

Equation (14) indicates that the rear surface of the target is insulated.

Laplace integral transform technique is applied to solve Equation (12). Taking Laplace transform on Equation (12) w.r.t. the time t one gets:

$\frac{{\partial }^{2}}{\partial {X}^{2}}\stackrel{¯}{\theta }\left(X,s\right)-\left({s}^{2}+2s\right)\stackrel{¯}{\theta }\left(X,s\right)=0$ (16)

The solution of equation (16) is as follows:

$\stackrel{¯}{\theta }\left(X,s\right)=A{\text{e}}^{\sqrt{s\left(s+2\right)}X}+B{\text{e}}^{-\sqrt{s\left(s+2\right)}X}$ (17)

at X = 0 one gets:

${\frac{\partial \stackrel{¯}{\theta }}{\partial X}|}_{X=0}=A\sqrt{s\left(s+2\right)}-B\sqrt{s\left(s+2\right)}$ (18)

And at $X={X}_{d}$ one gets:

${\frac{\partial \stackrel{¯}{\theta }}{\partial X}|}_{X={X}_{d}}=A\sqrt{s\left(s+2\right)}\text{ }{\text{e}}^{\sqrt{s\left(s+2\right)}{X}_{d}}-B\sqrt{s\left(s+2\right)}\text{ }{\text{e}}^{-\sqrt{s\left(s+2\right)}{X}_{d}}=0$

Thus: $A{\text{e}}^{\sqrt{s\left(s+2\right)}{X}_{d}}-B{\text{e}}^{-\sqrt{s\left(s+2\right)}{X}_{d}}=0$ (19)

Taking Laplace transform to the boundary condition Equation (15) one gets:

$\frac{\partial \stackrel{¯}{\theta }}{\partial X}\left(X,s\right)=-\left[2\stackrel{¯}{\phi }\left(X,s\right)+\left\{s\stackrel{¯}{\phi }\left(X,s\right)\phi \left(X,0\right)\right\}\right]$ ,

At the front surface X = 0 this equation can be rewritten as:

${\frac{\partial \stackrel{¯}{\theta }}{\partial X}\left(0,s\right)|}_{X=0}=-2\stackrel{¯}{\phi }\left(0,s\right)-s\stackrel{¯}{\phi }\left(0,s\right)$ , Where $\phi \left(X,0\right)=0$

$\therefore \text{ }{\frac{\partial \stackrel{¯}{\theta }}{\partial X}\left(0,s\right)|}_{X=0}=-\left(s+2\right)\stackrel{¯}{\phi }\left(0,s\right)$ (20)

Comparing the two Equations (18), (20) A and B can be obtained as follows:

$A\sqrt{s\left(s+2\right)}-B\sqrt{s\left(s+2\right)}=-\left(s+2\right)\stackrel{¯}{\phi }\left(0,s\right)$

$\left(A-B\right)=-\frac{\sqrt{s\left(s+2\right)}}{s}\stackrel{¯}{\phi }\left(0,s\right)$ or: $\left(A-B\right)=-\sqrt{\frac{s+2}{s}}\text{ }\stackrel{¯}{\phi }\left(0,s\right)$

Multiplying both sides by ${\text{e}}^{\sqrt{s\left(s+2\right)}{X}_{d}}$ then by ${\text{e}}^{-\sqrt{s\left(s+2\right)}{X}_{d}}$ respectively one gets:

$A{\text{e}}^{\sqrt{s\left(s+2\right)}{X}_{d}}-B{\text{e}}^{\sqrt{s\left(s+2\right)}{X}_{d}}=-\stackrel{¯}{\phi }\left(0,s\right)\sqrt{\frac{s+2}{s}}\text{ }{\text{e}}^{\sqrt{s\left(s+2\right)}{X}_{d}}$ (21)

$A{\text{e}}^{-\sqrt{s\left(s+2\right)}{X}_{d}}-B{\text{e}}^{-\sqrt{s\left(s+2\right)}{X}_{d}}=-\stackrel{¯}{\phi }\left(0,s\right)\sqrt{\frac{s+2}{s}}\text{ }{\text{e}}^{-\sqrt{s\left(s+2\right)}{X}_{d}}$ (22)

Subtracting (19), (21) one gets:

$B\left[{\text{e}}^{+\sqrt{s\left(s+2\right)}{X}_{d}}-{\text{e}}^{-\sqrt{s\left(s+2\right)}{X}_{d}}\right]=\stackrel{¯}{\phi }\left(0,s\right)\sqrt{\frac{s+2}{s}}\text{ }{\text{e}}^{+\sqrt{s\left(s+2\right)}{X}_{d}}$

Thus:

$B=\frac{\stackrel{¯}{\phi }\left(0,s\right)\sqrt{\frac{s+2}{s}}\text{ }{\text{e}}^{+\sqrt{s\left(s+2\right)}{X}_{d}}}{2\mathrm{sinh}\sqrt{s\left(s+2\right)}{X}_{d}}$

Also from (19), (22) one gets:

$A\left[{\text{e}}^{+\sqrt{s\left(s+2\right)}{X}_{d}}-{\text{e}}^{-\sqrt{s\left(s+2\right)}{X}_{d}}\right]=\stackrel{¯}{\phi }\left(0,s\right)\sqrt{\frac{s+2}{s}}\text{ }{\text{e}}^{+\sqrt{s\left(s+2\right)}{X}_{d}}$

Thus:

$A=\frac{\stackrel{¯}{\phi }\left(0,s\right)\sqrt{\frac{s+2}{s}\text{ }}{\text{e}}^{-\sqrt{s\left(s+2\right)}{X}_{d}}}{2\mathrm{sinh}\sqrt{s\left(s+2\right)}{X}_{d}}$

Substituting in Equation (17) one gets:

$\stackrel{¯}{\vartheta }\left(X,s\right)=\frac{\stackrel{¯}{\phi }\left(0,s\right)\sqrt{\frac{s+2}{s}}\left[{\text{e}}^{+\sqrt{s\left(s+2\right)}\left({X}_{d}-X\right)}+{\text{e}}^{-\sqrt{s\left(s+2\right)}\left({X}_{d}-X\right)}\right]}{{\text{e}}^{\sqrt{s\left(s+2\right)}{X}_{d}}\left[1-{\text{e}}^{-2\sqrt{s\left(s+2\right)}{X}_{d}}\right]}$

Considering that: $\frac{1}{1-a}={\sum }_{n=0}^{\infty }{a}^{n},|a|<1$ 

Thus:

$\frac{1}{\left[1-{\text{e}}^{-2\sqrt{s\left(s+2\right)}{X}_{d}}\right]}={\sum }_{n=0}{\text{e}}^{-2n\sqrt{s\left(s+2\right)}{X}_{d}}$

This gives:

$\stackrel{¯}{\vartheta }\left(X,s\right)=\stackrel{¯}{\phi }\left(0,s\right)\sqrt{\frac{s+2}{s}}\left[{\sum }_{n=0}^{\infty }{\text{e}}^{-\left(X+2n{X}_{d}\right)\sqrt{s\left(s+2\right)}}+{\sum }_{n=0}^{\infty }{\text{e}}^{-\left(2\left(1+n\right){X}_{d}-X\right)\sqrt{s\left(s+2\right)}}\right]$

Neglecting the last term with respect to the first one, one gets:

$\stackrel{¯}{\vartheta }\left(X,s\right)={\sum }_{n=0}^{\infty }\stackrel{¯}{\phi }\left(0,s\right)\frac{s+2}{\sqrt{s\left(s+2\right)}}{\text{e}}^{-\left(X+2n{X}_{d}\right)\sqrt{s\left(s+2\right)}}$

This equation can be rewritten in the form:

$\begin{array}{c}\stackrel{¯}{\vartheta }\left(X,s\right)={\sum }_{n=0}^{\infty }\left[\stackrel{¯}{\phi }\left(0,s\right)\frac{s}{\sqrt{s\left(s+2\right)}}{\text{e}}^{-\left(X+2n{X}_{d}\right)\sqrt{s\left(s+2\right)}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\stackrel{¯}{\phi }\left(0,s\right)\frac{2}{\sqrt{s\left(s+2\right)}}{\text{e}}^{-\left(X+2n{X}_{d}\right)\sqrt{s\left(s+2\right)}}\right]\end{array}$

Let, $f\left(s\right)=\frac{1}{\sqrt{s\left(s+2\right)}}{\text{e}}^{-\left(X+2n{X}_{d}\right)\sqrt{s\left(s+2\right)}}$ we get:

$\begin{array}{l}\stackrel{¯}{\vartheta }\left(X,s\right)=\underset{n=0}{\overset{\infty }{\sum }}\left[\stackrel{¯}{\phi }\left(0,s\right)\left\{sf\left(s\right)\right\}+2\stackrel{¯}{\phi }\left(0,s\right)f\left(s\right)\right]\\ \text{Put}:\\ \text{ }\text{ }2\stackrel{¯}{\phi }\left(0,s\right)f\left(s\right)={M}_{1},\text{ }\stackrel{¯}{\phi }\left(0,s\right)\left\{sf\left(s\right)\right\}={M}_{2}\end{array}$ (23)

From Equation (9):

${L}^{-1}\left\{\stackrel{¯}{\phi }\left(s\right)\right\}=\frac{q\left(t\right)\left(1-R\right)}{W\rho {c}_{p}\left({T}_{m}-{T}_{0}\right)}$ (24)

${L}^{-1}\left\{f\left(s\right)\right\}={L}^{-1}\left\{\frac{1}{\sqrt{s\left(s+2\right)}}{\text{e}}^{-\left(X+2n{X}_{d}\right)\sqrt{s\left(s+2\right)}}\right\}$ (25)

We have: 

${L}^{-1}\frac{{\text{e}}^{-c\left[{\left(s+a\right)}^{\frac{1}{2}}{\left(s+b\right)}^{\frac{1}{2}}\right]}}{{\left(s+a\right)}^{\frac{1}{2}}{\left(s+b\right)}^{\frac{1}{2}}}={\text{e}}^{-\left(\frac{a+b}{2}\right)\text{\hspace{0.17em}}\tau }{I}_{0}\left[\left(\frac{a-b}{2}\right){\left({\tau }^{2}-{c}^{2}\right)}^{\frac{1}{2}}\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}\tau >c,\text{\hspace{0.17em}}c>0$

Put: $c=X+2n{X}_{d},\text{\hspace{0.17em}}b=0,\text{\hspace{0.17em}}a=2,\text{\hspace{0.17em}}\tau =\frac{t}{2{t}_{k}}$ one gets:

${L}^{-1}\frac{{\text{e}}^{-\left(X+2n{X}_{d}\right)\sqrt{s\left(s+2\right)}}}{\sqrt{s\left(s+2\right)}}={\text{e}}^{-\tau }{I}_{0}\sqrt{{\tau }^{2}-{c}^{2}}$ (26)

The modified Bessel function ${I}_{\eta }\left(x\right)$ is written in the following form: 

${I}_{\eta }\left(x\right)=\underset{m\text{\hspace{0.17em}}=1}{\overset{\infty }{\sum }}\frac{{\left(\frac{x}{2}\right)}^{\left(\eta +2m\right)}}{m!}\Gamma \left(\eta +m+1\right)$ (27)

Comparing (25), (26) (at $\eta =0$ ) one gets:

${L}^{-1}\left\{{f}_{s}\right\}={\text{e}}^{-\tau }\underset{m=1}{\overset{\infty }{\sum }}\frac{{\left[{\tau }^{2}-{\left(X+2n{X}_{d}\right)}^{2}\right]}^{m}}{{2}^{2m}m!}\Gamma \left(m+1\right)$ (28)

Thus the inverse in time domain is as follows:

${L}^{-1}\left\{{f}_{s}\right\}={\text{e}}^{-\frac{t}{2{t}_{k}}}\underset{m=1}{\overset{\infty }{\sum }}\frac{{\left[{\left(\frac{t}{2{t}_{k}}\right)}^{2}-{\left(\frac{W}{2\alpha }\right)}^{2}{\left(x+2n{x}_{d}\right)}^{2}\right]}^{m}}{{2}^{2m}m!}\Gamma \left(m+1\right)$ (29)

where: $\Gamma \left(m+1\right)=m!$

${L}^{-1}\left\{\stackrel{¯}{\phi }\left(s\right)\right\}=\frac{q\left(t\right)}{W\rho {c}_{p}\left({T}_{m}-{T}_{0}\right)}$ (30)

The convolution theorem:

${L}^{-1}\left[{f}_{1}\left(s\right){f}_{2}\left(s\right)\right]=\underset{0}{\overset{t}{\int }}{F}_{1}\left(t-u\right){F}_{s}\left(u\right)\text{d}u$ (31)

Thus

$\begin{array}{c}{L}^{-1}\left[{M}_{1}\right]={L}^{-1}\left[2\phi \left(s\right)f\left(s\right)\right]\\ =2\underset{0}{\overset{t}{\int }}\frac{q\left(t-u\right)}{W\rho {c}_{p}\left({T}_{m}-{T}_{0}\right)}{\text{e}}^{-\frac{u}{2{t}_{k}}}\underset{m=1}{\overset{\infty }{\sum }}\frac{{\left[{\left(\frac{u}{2{t}_{k}}\right)}^{2}-{\left(\frac{W}{2\alpha }\right)}^{2}{\left(x+2n{x}_{d}\right)}^{2}\right]}^{m}}{{2}^{2m}}\end{array}$ (32)

Also we have:

$L\left[\frac{\text{d}}{\text{d}t}F\left(t\right)\right]=sf\left(s\right)-f\left(0\right)$ ,

For such a case one has $f\left(0\right)=0$ , thus:

$\begin{array}{c}{L}^{-1}\left[{M}_{2}\right]={L}^{-1}\left[\stackrel{¯}{\phi }sf\left(s\right)\right]=\underset{0}{\overset{t}{\int }}\phi \left(t-u\right)\frac{\text{d}}{\text{d}u}f\left(u\right)\text{d}u\\ =\underset{0}{\overset{t}{\int }}\frac{Aq\left(t-u\right)}{W\rho {c}_{p}\left({T}_{m}-{T}_{0}\right)}\frac{\text{d}}{\text{d}u}\left\{{\text{e}}^{-\frac{u}{2{t}_{k}}}\underset{m=1}{\overset{\infty }{\sum }}\frac{{\left\{{\left(\frac{u}{2{t}_{k}}\right)}^{2}-\frac{{W}^{2}}{4{\alpha }^{2}}{\left(x+2n{x}_{d}\right)}^{2}\right\}}^{m}}{{2}^{2\text{\hspace{0.17em}}m\text{\hspace{0.17em}}\text{\hspace{0.17em}}}}\right\}\text{ }\text{ }\text{d}u\end{array}$ (33)

${L}^{-1}\left[\stackrel{¯}{\vartheta }\left(X,s\right)\right]={L}^{-1}\left[{M}_{1}\right]+{L}^{-1}\left[{M}_{2}\right]$

$\begin{array}{l}\therefore \vartheta \left(x,t\right)\\ =\underset{n=1}{\overset{\infty }{\sum }}\left[\left(2-\frac{1}{2{t}_{k}}\right)\underset{0}{\overset{t}{\int }}\frac{Aq\left(t-u\right)}{W\rho {c}_{p}\left({T}_{m}-{T}_{0}\right)}{\text{e}}^{-\frac{u}{2{t}_{k}}}\underset{m=1}{\overset{\infty }{\sum }}\frac{{\left\{{\left(\frac{u}{2{t}_{k}}\right)}^{2}-\frac{{W}^{2}}{4{\alpha }^{2}}{\left(x+2n{x}_{d}\right)}^{2}\right\}}^{m}}{{2}^{2m}}\text{ }\text{ }\text{d}u\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{0}{\overset{t}{\int }}\frac{Aq\left(t-u\right)}{W\rho {c}_{p}\left({T}_{m}-{T}_{0}\right)}\frac{1}{2{t}_{k}^{2}}{\text{e}}^{-\frac{u}{2{t}_{k}}}\underset{m=1}{\overset{\infty }{\sum }}\frac{m{\left\{{\left(\frac{u}{2{t}_{k}}\right)}^{2}-\frac{{W}^{2}}{4{\alpha }^{2}}{\left(x+2n{x}_{d}\right)}^{2}\right\}}^{m-1}}{{2}^{2m}}\text{ }\text{ }u\text{d}u\right]\end{array}$ (34)

where: $q\left(t-u\right)={q}_{\mathrm{max}}{\text{e}}^{-{\left(\frac{\left(t-u\right)-{t}_{0}}{\gamma }\right)}^{2}}$ , ${q}_{\mathrm{max}}$ , W/m2 is the laser maximum power density

At x = 0 we get:

$\begin{array}{l}\vartheta \left(0,t\right)\\ =\underset{n=1}{\overset{\infty }{\sum }}\left[\left(2-\frac{1}{2{t}_{k}}\right)\underset{0}{\overset{t}{\int }}\frac{Aq\left(t-u\right)}{W\rho {c}_{p}\left({T}_{m}-{T}_{0}\right)}\text{ }{\text{e}}^{-\frac{u}{2{t}_{k}}}\underset{m=1}{\overset{\infty }{\sum }}\frac{{\left\{{\left(\frac{u}{2{t}_{k}}\right)}^{2}-\frac{{W}^{2}}{4{\alpha }^{2}}{\left(2n{x}_{d}\right)}^{2}\right\}}^{m}}{{2}^{2\text{\hspace{0.17em}}m\text{\hspace{0.17em}}\text{\hspace{0.17em}}}}\text{ }\text{d}u\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{0}{\overset{t}{\int }}\frac{Aq\left(t-u\right)}{W\rho {c}_{p}\left({T}_{m}-{T}_{0}\right)}\frac{1}{2{t}_{k}^{2}}{\text{e}}^{-\frac{u}{2{t}_{k}}}\underset{m=1}{\overset{\infty }{\sum }}\frac{m{\left\{{\left(\frac{u}{2{t}_{k}}\right)}^{2}-\frac{{W}^{2}}{4{\alpha }^{2}}{\left(2n{x}_{d}\right)}^{2}\right\}}^{m-1}}{{2}^{2\text{\hspace{0.17em}}m\text{\hspace{0.17em}}\text{\hspace{0.17em}}}}\text{ }u\text{d}u\right]\end{array}$ (35)

3. Computations

The obtained thermal profile is computed for different laser pulses with different maximum power densities as follows:

${q}_{\mathrm{max}}=\text{2E7},\text{1}.\text{5E7},\text{1E7},0.\text{8E7},0.\text{7E7},0.\text{6E7},0.\text{5E7},0.\text{4E7},0.\text{3E7},0.\text{2E7},\text{W}/{\text{m}}^{\text{2}}.$

The laser pulse shape is taken of Gaussian form as: $q\left(t\right)={q}_{\mathrm{max}}{\text{e}}^{-{\left(\frac{t-{t}_{0}}{\gamma }\right)}^{2}}$ , γ is

the full width at half maximum of the suggested pulse, t0 in seconds is the time required for $q\left(t\right)$ to reach the maximum value ${q}_{\mathrm{max}}$ .

The other laser pulse parameters are as follows: t0 = 6 μsec, γ = 6 μ sec. pulse duration = 12 μsec, the slab thickness = 300 μm, the absorption coefficient A = 0.67, tk = 1 μsec.

The physical and thermal properties of the silver selenide slab material   are given in Table 1.

The temperature of the irradiated front surface $\vartheta \left(0,t\right)$ for the different pulses is computed and are illustrated graphically in Figure 1 and Figure 2 and are tabulated in Table 2 for m = 1, n = 1, and in Table 3 for m = 1, n = 1.2.

The critical time tph required to start phase transition, and the critical time tm required to initiate melting at the front surface are obtained for the considered pulses and are tabulated in Table 4 for m = 1, n = 1, and in Table 5 for m = 1, n = 1.2 and are illustrated graphically in Figure 3 and Figure 4.

4. Conclusions

As a result of the study the following conclusions can be made:

1) The temperature of the irradiated surface does depend linearly on the maximum power density qmax of the laser pulse and also on the absorption coefficient (1 − R).

2) The dependence on the slab thickness and the pulse parameters γ & t0 are no longer linear.

3) The dependence of the tph and tm on qmax is not linear.

Table 1. The physical and thermal properties of the silver selenide slab material   .

Table 2. The temperature of the front surface as a function of the qmax, W/m2 and exposure time, (m = 1, n = 1).

Table 3. The temperature of the front surface as a function of the qmax, W/m2 and exposure time, (m = 1, n = 1, 2)

Table 4. The critical time tph required to start phase transition, and the critical time tm required to initiate melting at the front surface are obtained for the considered pulses for m = 1, n = 1.

Table 5. The critical time tph required to start phase transition, and the critical time tm required to initiate melting at the front surface are obtained for the considered pulses for m = 1, n = 1, 2.

Figure 1. The temperature of the irradiated front surface as a function of the laser exposure time “t”. qmax = 0.2 × 107 W/m2, m = 1 and n = 1.

Figure 2. The temperature of the irradiated front surface as a function of the laser exposure time “t”. qmax = 0.2 × 107 W/m2, m = 1 and n = 1.

Figure 3. The critical time tph required to start phase transition as a function of the maximum power density of the laser pulse for m = 1 and n = 1. The other parameters are kept constant.

Figure 4. The critical time tph required to start phase transition and the critical time tm required to initiate melting as a function of the maximum power density of the laser pulse for m = 1 and n = 1, 2. The other parameters are kept constant.

4) Successive pulses may be required to initiate phase transition or melting.

5) For the summation over “m” and “n” one gets acceptable values for θ(0, t) only for m = 1.

But for m > 1 one gets negative values.

Increasing values of θ(0, t) are obtained with increasing n values.

6) The extension of the present technique to other materials makes it possible to specify the optimum operation conditions to attain and maintain a certain phase for specific medical and technological applications.

Cite this paper
EL-Adawi, M. and Shalaby, S. (2018) Pulsed Laser Heating of a Finite Silver Selenide Slab Using (HHCE) Model. Applied Mathematics, 9, 355-368. doi: 10.4236/am.2018.94027.
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