Heat conduction problem for a half-space medium containing a penny-shaped crack

Abstract

This paper deals with the analytical investigation of the steady-state heat conduction within a semi-infinite medium embedding an insulating penny-shaped crack and subjected to a disk heat source of an arbitrary heat flux distribution. This axisymmetric heat conduction problem is formulated as a mixed boundary value problem, which is then converted to dual integral equations using the Hankel integral transform technique. A solution approach based on the Gegenbauer addition formula is employed to directly reduce the resulting integral equations into an infinite set of simultaneous equations. Closed-form expressions for temperature, heat flux, heat flux intensity factor near the crack tip, and constriction resistance are subsequently derived. Moreover, four practical configurations of heat flux conditions are further investigated and discussed. Sets of dimensionless plots are provided to show the effect of the size and depth of the crack on the thermal fields, heat flux intensity factor, and constriction resistance. Additionally, the validity of the obtained results is tested by comparison with numerical simulations and shows a good agreement.

Appendices

Derivation of the integral formula (41)

From [45, Eq. (6.522 11)], it is easy to find that

$$\begin{aligned} \int _{0}^{\infty }\xi \,M_n(\xi )\,J_0(\xi \,\rho )\,\textrm{d}\xi = {\left\{ \begin{array}{ll} \displaystyle {\frac{2\,T_{2\,n+1}(\rho /c)}{\pi \,\rho \,\sqrt{c^2-\rho ^2}}},&{} \quad 0<\rho <c,\\ 0, &{} \quad \rho >c, \end{array}\right. } \quad n=0,1,2,\ldots \end{aligned}$$

(A1)

where $T_n$ is Chebyshev’s polynomials of the first kind of degree n defined by [45, Eq. (8.940 1)] $T_n(x)=\cos (n\,\arccos \,x)$. Taking the difference between the $(n+1)\textrm{th}$ and $n\textrm{th}$ equations and using the following recursion formulas [45, Eqs. (8.941 1 and 4)]

$$\begin{aligned}&T_{n-1}(x)=2\,x\,T_n(x)-T_{n+1}(x),\end{aligned}$$

(A2)

$$\begin{aligned}&\left( 1-x^2\right) U_{n-1}(x)=x\,T_n(x)-T_{n+1}(x), \end{aligned}$$

(A3)

yield

$$\begin{aligned} \int _{0}^{\infty }N_n(\xi )\,J_0(\xi \,\rho )\,\textrm{d}\xi = {\left\{ \begin{array}{ll} \displaystyle {\frac{4\,\sqrt{c^2-\rho ^2}}{\pi \,\rho \,c^2}\,U_{2\,n+1}(\rho /c)},&{}\quad 0<\rho <c,\\ 0, &{} \quad \rho >c, \end{array}\right. }\quad n=0, 1, 2,\ldots \end{aligned}$$

(A4)

Asymptotic expansion formulas of $\scriptstyle {N_n}$ and $\scriptstyle {Y_m}$

For large values of $\xi$, the asymptotic expansion of the Bessel function $J_{\nu }$ reads [45, Eq. (8.451 1), $\nu =1$]

$$\begin{aligned}&J_\nu (\xi )\!=\!\sqrt{\dfrac{2}{\pi \,\xi }}\left[ \cos \!\left( \xi -\left( 2\,\nu +1\right) \dfrac{\pi }{4}\right) - \dfrac{4\,\nu ^2-1}{8\,\xi }\,\sin \!\left( \xi -\left( 2\,\nu +1\right) \dfrac{\pi }{4}\right) +O\!\left( \dfrac{1}{\xi ^2}\right) \right] . \end{aligned}$$

(B1)

As $M_n(\xi )=J_{n+1/2}\left( \xi \,c/2\right) \,J_{-n-1/2}(\xi \,c/2)$ and $X_m(\xi )=J^2_m\left( \xi \,c/2\right)$, then, we use Eq. (B1) along with appropriate trigonometric identities stated below

$$\begin{aligned}&\cos (x)\,\cos (y)=\left[ \cos (x+y)+\cos (x-y)\right] /2,\end{aligned}$$

(B2)

$$\begin{aligned}&\cos (x)\,\sin (y)=\left[ \sin (x+y)-\sin (x-y)\right] /2,\end{aligned}$$

(B3)

$$\begin{aligned}&\cos (x+y)=\cos (x)\,\cos (y)-\sin (x)\,\sin (y),\end{aligned}$$

(B4)

$$\begin{aligned}&\sin (x+y)=\cos (x)\,\sin (y)+\sin (x)\,\cos (y), \end{aligned}$$

(B5)

to get the following asymptotic expansions of $M_n$ and $X_m$

$$\begin{aligned}&M_n(\xi )=\dfrac{2}{\pi \,\xi \,c}\left[ \sin (\xi \,c)+\dfrac{2\,n\left( n+1\right) \cos (\xi \,c)}{\xi \,c}\right] +O\!\left( \dfrac{1}{\xi ^3}\right) , \end{aligned}$$

(B6)

$$\begin{aligned}&X_m(\xi )=\dfrac{2}{\pi \,\xi \,c}\left[ 1+(-1)^m\,\sin (\xi \,c) +\dfrac{(-1)^m\left( 4\,m^2-1\right) \cos (\xi \,c)}{2\,\xi \,c}\right] +O\!\left( \dfrac{1}{\xi ^3}\right) . \end{aligned}$$

(B7)

Substituting Eqs. (B6) and (B7) in the equations $N_n(\xi )=\xi \left[ M_n(\xi )-M_{n+1}(\xi )\right]$ and $Y_m(\xi )=\xi \left[ X_m(\xi )-X_{m+2}(\xi )\right]$, respectively, yields

$$\begin{aligned}&N_n(\xi )=-\dfrac{8\left( n+1\right) \cos (\xi \,c)}{\pi \,\xi \,c^2}+O\!\left( \dfrac{1}{\xi ^2}\right) , \end{aligned}$$

(B8)

$$\begin{aligned}&Y_m(\xi )=-\dfrac{16\,(-1)^m\left( m+1\right) \cos (\xi \,c)}{\pi \,\xi \,c^2}+O\!\left( \dfrac{1}{\xi ^2}\right) . \end{aligned}$$

(B9)

Calculation of $\scriptstyle {\varphi _H}$ and $\scriptstyle {\int _{0}^{\infty } \xi ^{-1}\,\varphi _H(\xi )\,J_1(\xi )\,\textrm{d}\xi }$

In Table 8, we provide final expressions of $\varphi _H$ and $\int _{0}^{\infty } \xi ^{1}\,\varphi _H(\xi )\,J_1(\xi )\,\textrm{d}\xi$ corresponding to each case of heat flux loading.

Table 8 Expressions of $\varphi _H$ and $\int _{0}^{\infty } \xi ^{-1}\,\varphi _H(\xi )\,J_1(\xi )\,\textrm{d}\xi$ for different configurations of heat flux loading

Evaluation of $\scriptstyle {I}$ and $\scriptstyle {\partial I/\partial \zeta }$

1.1 D.1 Case (a): uniform loading

In this case, we have $I(\rho ,\zeta )=\int _{0}^{\infty } \xi ^{-1}\,\textrm{e}^{-\xi \left( \zeta +d\right) }\,J_0(\xi \,\rho )\,J_1(\xi )\,\textrm{d}\xi$.

For $\zeta =-d$, using [45, Eqs. (6.574 1 and 2)], it yields

$$\begin{aligned} I(\rho ,-d)=\int _{0}^{\infty } \xi ^{-1}\,J_0(\xi \,\rho )\,J_1(\xi )\,\textrm{d}\xi = {\left\{ \begin{array}{ll} \,F\!\left( \dfrac{1}{2},\,-\dfrac{1}{2};\,1;\,\rho ^2\right) , \quad &{} \quad \rho \le 1, \\ \,\dfrac{1}{2\,\rho }F\!\left( \dfrac{1}{2},\,\dfrac{1}{2};\,2;\,\rho ^{-2}\right) , \quad &{} \quad \rho >1, \end{array}\right. } \end{aligned}$$

(D1)

where F denotes the Gauss hypergeometric function defined by [45, Eq. (9.14)]

$$\begin{aligned}&F\Big (\alpha ,\,\beta ;\,\gamma ;\,x\Big ) =\sum _{n=0}^{\infty }\frac{\left( \alpha \right) _{n} \left( \beta \right) _{n}}{\left( \gamma \right) _{n}}\,\frac{x^{n}}{n!},\quad \text {with} \quad \left( \alpha \right) _{n}=\frac{\Gamma (\alpha +n)}{\Gamma (\alpha )}, \qquad \left| x\right| <1. \end{aligned}$$

(D2)

The relation (D1) can be simplified by expressing the Gauss hypergeometric function in terms of the complete elliptic integrals of the first and second kinds [45, p. 396]

$$\begin{aligned}&K(y)=\int _{0}^{\pi /2} \dfrac{\textrm{d}x}{\sqrt{1-y^2\,\sin ^2x}},\end{aligned}$$

(D3)

$$\begin{aligned}&E(y)=\int _{0}^{\pi /2} \sqrt{1-y^2\,\sin ^2x}\,\textrm{d}x. \end{aligned}$$

(D4)

In view of [45, Eqs. (9.137 13) and (9.131 1)], we have

$$\begin{aligned}&\dfrac{\rho }{2}\,F\!\left( \dfrac{1}{2},\,\dfrac{1}{2};\,2;\,\rho ^2\right) =\dfrac{1-\rho ^2}{\rho }\left[ F\!\left( \dfrac{3}{2},\,\dfrac{1}{2};\,1;\,\rho ^2\right) -F\!\left( \dfrac{1}{2},\,\dfrac{1}{2};\,1;\,\rho ^2\right) \right] ,\end{aligned}$$

(D5)

$$\begin{aligned}&\left( 1-\rho ^2\right) F\!\left( \dfrac{3}{2},\,\dfrac{1}{2};\,1;\,\rho ^2\right) =F\!\left( -\dfrac{1}{2},\,\dfrac{1}{2};\,1;\,\rho ^2\right) . \end{aligned}$$

(D6)

Then, knowing from [45, Eqs. (8.114 1) and (8.113 1)] that

$$\begin{aligned}&F\!\left( \dfrac{1}{2},\,-\dfrac{1}{2};\,1;\,\rho ^2\right) =F\!\left( -\dfrac{1}{2},\,\dfrac{1}{2};\,1;\,\rho ^2\right) = \dfrac{2}{\pi }\,E(\rho ),\end{aligned}$$

(D7)

$$\begin{aligned}&F\!\left( \dfrac{1}{2},\,\dfrac{1}{2};\,1;\,\rho ^2\right) =\dfrac{2}{\pi }\,K(\rho ), \end{aligned}$$

(D8)

one obtains

$$\begin{aligned} I(\rho ,-d)= {\left\{ \begin{array}{ll} \,\dfrac{2}{\pi }\,E(\rho ), \quad &{} \quad \rho \le 1, \\ \,\dfrac{2\,\rho }{\pi }\left[ E\!\left( 1/\rho \right) - \left( 1-1/\rho ^2\right) K\!\left( 1/\rho \right) \right] , \quad &{} \quad \rho >1. \end{array}\right. } \end{aligned}$$

(D9)

For $\rho =0$, we use [45, Eq. (6.623 3)] to get

$$\begin{aligned} I(0,\zeta )=\int _{0}^{\infty } \xi ^{-1}\,\textrm{e}^{-\xi \left( \zeta +d\right) }\,J_1(\xi )\,\textrm{d}\xi = -\left( \zeta +d\right) +\sqrt{\left( \zeta +d\right) ^2+1}. \end{aligned}$$

(D10)

Differentiating the previous equation with respect of $\zeta$, we obtain

$$\begin{aligned} \dfrac{\partial I}{\partial \zeta }(0,\zeta ) = \dfrac{\zeta +d}{\sqrt{\left( \zeta +d\right) ^2+1}}-1. \end{aligned}$$

(D11)

For $\zeta >-d$ and $\rho >0$, I and $\partial I/\partial \zeta$ may be evaluated numerically.

1.2 D.2 Case (b): triangular loading

In this case, I and $\partial I/\partial \zeta$ are calculated numerically.

1.3 D.3 Case (c): semi-elliptic loading

In this case, we have $I(\rho ,\zeta )=\dfrac{3\,\sqrt{\pi }}{2\,\sqrt{2}}\displaystyle {\int _{0}^{\infty } \xi ^{-3/2}\,\textrm{e}^{-\xi \left( \zeta +d\right) }\,J_0(\xi \,\rho )\,J_{3/2}(\xi )\,\textrm{d}\xi }$.

For $\zeta =-d$, using [45, Eqs. (6.574 1 and 2)], it yields

$$\begin{aligned} I(\rho ,-d)&=\dfrac{3\,\sqrt{\pi }}{2\,\sqrt{2}}\int _{0}^{\infty } \xi ^{-3/2}\,J_0(\xi \,\rho )\,J_{3/2}(\xi )\,\textrm{d}\xi \nonumber \\&= {\left\{ \begin{array}{ll} \dfrac{3\,\pi }{8}F\!\left( \dfrac{1}{2},\,-1;\,1;\,\rho ^2\right) ,&{}\quad \rho \le 1,\\ \dfrac{1}{2\,\rho }F\!\left( \dfrac{1}{2},\,\dfrac{1}{2};\,\dfrac{5}{2};\,\rho ^{-2}\right) , &{}\quad \rho >1. \end{array}\right. } \end{aligned}$$

(D12)

For $\zeta >-d$, I and its derivative with respect to $\zeta$ are calculated numerically.

1.4 D.4 Case (d): loading involving singularity

In this case, we have $I(\rho ,\zeta )=\displaystyle {\dfrac{1}{2}\int _{0}^{\infty } \xi ^{-1}\,\textrm{e}^{-\xi \left( \zeta +d\right) }\,\sin (\xi )\,J_0(\xi \,\rho )\,\textrm{d}\xi }$.

For $\zeta =-d$, using [45, Eq. (6.693 6)] and taking in to account that $\textrm{arccosec}\,\rho =\arcsin (1/\rho )$, it yields

$$\begin{aligned} I(\rho ,-d)=\dfrac{1}{2}\int _{0}^{\infty } \xi ^{-1}\,\sin (\xi )\,J_0(\xi \,\rho )\,\textrm{d}\xi = {\left\{ \begin{array}{ll} \dfrac{\pi }{4}, &{} \quad \rho \le 1,\\ \dfrac{\arcsin (1/\rho )}{2},&{} \quad \rho >1. \end{array}\right. } \end{aligned}$$

(D13)

For $\zeta >-d$, we use [45, Eq. (6.752 1)], to get

$$\begin{aligned} I(\rho ,\zeta )=\dfrac{1}{2}\arcsin \left( \dfrac{2}{\kappa ^-+\kappa ^+}\right) , \end{aligned}$$

(D14)

where $\kappa ^\pm =\sqrt{\left( \rho \pm 1\right) ^2+\left( \zeta +d\right) ^2}$.

By differentiating Eq. (D14) with respect to $\zeta$, we obtain

$$\begin{aligned} \dfrac{\partial I}{\partial \zeta }(\rho ,\zeta ) = -\dfrac{\zeta +d}{\kappa ^-\,\kappa ^+\sqrt{2\left[ \rho ^2+\left( \zeta +d\right) ^2+\kappa ^-\,\kappa ^+-1\right] }}. \end{aligned}$$

(D15)

It is worth mentioning that, for $\zeta >1-d$, we can alternatively evaluate I and $\partial I/\partial \zeta$ in the first and third case by using [45, Eq. (6.626 1)].

Half-space medium subjected to a circular heat spot

The boundary conditions of a half-space heated by an arbitrary heat flux are given by Eqs. (11) and (16), where the superscripts (1) and (2) has been omitted here since the medium is continuous. Using the Hankel transform technique, the temperature expression satisfying the governing partial differential equation of this problem can be obtained as

$$\begin{aligned}&T(\rho ,\zeta )=\int _{0}^{\infty } A(\xi )\,\textrm{e}^{-\xi \,\zeta }\,J_0(\xi \,\rho )\,\textrm{d}\xi , \end{aligned}$$

(E1)

where A is an unknown function to be determined. Thereafter, the temperature derivative with respect to $\zeta$ is

$$\begin{aligned}&\frac{\partial T}{\partial \zeta }(\rho ,\zeta )=-\int _{0}^{\infty } \xi \,A(\xi )\,\textrm{e}^{-\xi \,\zeta }\,J_0(\xi \,\rho )\,\textrm{d}\xi . \end{aligned}$$

(E2)

Using the boundary condition (11) and Hankel transform theorem, we obtain

$$\begin{aligned}&A(\xi )=\varphi _H(\xi )\,\textrm{e}^{-\xi \,d}. \end{aligned}$$

(E3)

Substitution Eq. (E3) into Eq. (E1) leads to the following explicit formulas of the scaled temperature

$$\begin{aligned}&T(\rho ,\zeta )=I(\rho ,\zeta ), \end{aligned}$$

(E4)

where I is given by Eq. (38).