Miscellaneous

The evaluation of the integrals to derive the PKN governing equation

The average velocity is:

(1)\[\bar u_x = \frac{\partial p}{\partial x} \frac{2ab}{\mu \pi (a^2+b^2)} [\int_0^{a} \int_0^{b\sqrt{1-(\frac{z}{a})^2}} (\frac{z^2}{a^2}+\frac{y^2}{b^2})dydz - \frac{\pi}{4}ab]\]

And we are looking for the integral:

(2)\[I = \int_0^{a} \int_0^{b\sqrt{1-(\frac{z}{a})^2}} (\frac{z^2}{a^2}+\frac{y^2}{b^2})dydz\]

The procedure is as follows:

(3)\[\begin{split}\begin{equation} \begin{split} I& = \int_0^a \int_0^{b\sqrt{1-(\frac{z}{a})^2}} (\frac{z^2}{a^2}+\frac{y^2}{b^2})dydz \\ & = \int_0^a ( \frac{z^2}{a^2}y + \frac{y^3}{3b^2} ) \vert_0^{ b\sqrt{ 1-(\frac{z}{a})^2} }dz \\ & = \int_0^a \lbrace \frac{z^2}{a^2} b \sqrt{1-(\frac{z}{a})^2} +\frac{b}{3}[1-(\frac{z}{a})^2]^{\frac{3}{2}}\rbrace dz \\ & = \frac{b}{a^3}\int_0^a z^2 \sqrt{a^2-z^2}dz + \frac{b}{3a^3}\int_0^a(a^2-z^2)^{\frac{3}{2}}dz \\ & = \frac{b}{a^3}I_1 + \frac{b}{3a^3}I_2 \\ & = \frac{b}{a^3}(I_1 + \frac{I_2}{3}) \end{split} \end{equation}\end{split}\]

And:

  • (4)\[\begin{split}\begin{equation} \begin{split} I_1& = \int_0^a z^2 \sqrt{a^2-z^2}dz \\ & = \frac{z}{8}(2z^2-a^2)\sqrt{a^2-z^2} +\frac{a^4}{8} \textrm{arcsin} \frac{z}{a} \vert_0^a \\ & = \frac{a^4}{8} \textrm{arcsin}(1) \\ & = \frac{a^4}{8} \times \frac{\pi}{2} \\ & = \frac{\pi a^4}{16} \end{split} \end{equation}\end{split}\]
  • (5)\[\begin{split}\begin{equation} \begin{split} I_1& = \int_0^a(a^2-z^2)^{\frac{3}{2}}dz \\ & = \frac{z}{8}(5a^2-2z^2)\sqrt{a^2-z^2} +\frac{3a^4}{8} \textrm{arcsin} \frac{z}{a} \vert_0^a \\ & = \frac{3a^4}{8} \textrm{arcsin}(1) \\ & = \frac{3a^4}{8} \times \frac{\pi}{2} \\ & = \frac{3\pi a^4}{16} \end{split} \end{equation}\end{split}\]

Substitute Eqn (4) and Eqn (5) into Eqn (3):

(6)\[\begin{split}\begin{equation} \begin{split} I& = \frac{b}{a^3}(I_1 + \frac{I_2}{3}) \\ & = \frac{b}{a^3}( \frac{\pi a^4}{16} + \frac{1}{3} \times \frac{3\pi a^4}{16} ) \\ & = \frac{\pi ab}{8} \end{split} \end{equation}\end{split}\]

Substitute Eqn (6) into Eqn (1):

(7)\[\begin{split}\begin{equation} \begin{split} \bar u_x& = \frac{\partial p}{\partial x} \frac{2ab}{\mu \pi (a^2+b^2)} (I- \frac{\pi}{4}ab) \\ & = -\frac{\partial p}{\partial x} \frac{a^2b^2}{4\mu (a^2+b^2)} \end{split} \end{equation}\end{split}\]

Obtaining \(\bar{T}\)

The governing equation with the boundary condition and initial condition is:

(8)\[\begin{split}\left\{ \begin{align} & \frac{d^2 \bar{T}}{dr^2} + \frac{1}{r}\frac{d \bar{T}}{dr} - q^2 \bar{T} = -\frac{f(r)}{\kappa}, \ \ \ \ r>r_w \\ & \bar{T}=\frac{T_0}{s}, \ \ \ \ r=r_w \\ & \bar{T}=0, \ \ \ \ \ \ \ \ t=0 \\ \end{align} \right.\end{split}\]

Because only \(K_0(qr)\) is involved in the solution, the solution should be in the form of \(c_1K_0(qr)\) with \(c_1\) an arbitrary constant.

With the boundary condition, \(K_0(qr)\) is reduced to a constant \(K_0(qr_w)\) . Thus, a term that equals to \(K_0(qr_w)\) must be included in the denominator of \(c_1\) to cancel itself so that the value \(\frac{T_0}{s}\) can be obtained. In this way, \(c_1\) may be written as \(c_1 = \frac{T_0}{sK_0(qr_w)}\) . Therefore, the solution is:

(9)\[\bar{T}=\frac{T_0K_0(qr)}{sK_0(qr_w)}\]

Similar procedures can be used for the other two boundary conditions.

Asymptotic expansion of the temperature solution

If \(\bar{f}(\lambda)\) has a branch point at the origin, a series of equation is available:

(10)\[f(t) = \frac{1}{2\pi i} \int_{\gamma-i\infty}^{\gamma+i\infty} e^{\lambda t} \bar{f}(\lambda)d\lambda = \frac{1}{2\pi i} \int_{\infty}^0 e^{\lambda t} \bar{f}(\lambda)d\lambda\]
(11)\[\frac{1}{2\pi i} \int_{\infty}^0 e^{\lambda t} \lambda^{-\alpha-1} d\lambda =\frac{t^\alpha}{\Gamma(1+\alpha)}, \ \ \ \ \alpha \in \mathbb{R}\]
(12)\[\frac{1}{2\pi i} \int_{\infty}^0 e^{\lambda t} \lambda^n d\lambda =0, \ \ \ \ \ n \in \mathbb{N^0}\]
(13)\[\frac{1}{2\pi i} \int_{\infty}^0 e^{\lambda t} \frac{\ln (k\lambda)} {\lambda} d\lambda = -\ln{ \frac{ct}{m} }\]
(14)\[\frac{1}{2\pi i} \int_{\infty}^0 e^{\lambda t} \ln (k\lambda) d\lambda = -\frac{1}{t}\]
(15)\[\frac{1}{2\pi i} \int_{\infty}^0 e^{\lambda t} \ln^2 (k\lambda) d\lambda = -\frac{2}{t} \ln{ \frac{ct}{m} }\]

In addition, the modified nth order Bessel function of second kind can be expanded as:

(16)\[K_0(z) = -\ln(\frac{zc}{2})I_0(z) + \sum_{p=1}^\infty \sum_{r=1}^p r^{-1} \frac{ (\frac{z}{2})^{2p} }{p!}\]

and

(17)\[\begin{split}\begin{equation} \begin{split} K_n(z)&= (-1)^{n+1} \ln (\frac{zc}{2}) + \frac{1}{2}(-1)^n \sum_{r=0}^\infty \frac{ (\frac{z}{2})^{n+2r} }{r!(n+r)!} [ \sum_{p=1}^{n+r}m^{-1} + \sum_{p=1}^r m^{-1}] \\ & \ \ \ \ + \frac{1}{2} \sum_{r=0}^{n-1}(-1)^r( \frac{z}{2} )^{-n+2r} \frac{ (n-r-1)! }{r!} , \ \ \ \ n > 0 \\ \end{split} \end{equation}\end{split}\]

where:

  • \(\ln c = 0.5722\) is known as Euler’s constant;
  • \(m\) is real and positive;
  • \(p \in \mathbb{N}\) .

And the temperature solution is:

(18)\[\bar{T} = \frac{q_0''K_0(qr)}{k s q K_1(qr_w)}\]

Apply Eqn (10) on Eqn (18) with writing \(\mu = \sqrt{ \frac{\lambda}{\kappa} }\):

(19)\[T=\frac{q_0''}{2\pi k i} \int_{-\infty}^{0+} \frac{e^{\lambda t}K_0(qr)} {\lambda \mu K_1(qr_w)}d\lambda\]

Then the application of Eqn (11) through Eqn (17) on Eqn (19) yields:

(20)\[\begin{split}\begin{equation} \begin{split} T&= \frac{q_0''r_w}{2k} \lbrace \ln\frac{4\kappa t}{cr^2} + \frac{r_w^2}{\kappa t}\ln \frac{4\kappa t}{cr^2} + \frac{1}{4\kappa t}[r_w^2-r^2-2r_w^2 \ln \frac{r_w}{r}+.. .]\rbrace \\ &= \frac{q_0''r_w}{2k}\ln\frac{4\kappa t}{cr^2} + O(\frac{r_w^2}{\kappa t}) \end{split} \end{equation}\end{split}\]

Eqn (20) is suitable for large \(t_D\) defined as \(t_D = \frac{\kappa t}{r_w^2}\) as the higher order terms vanish with large \(t_D\) . When \(t_D\) becomes smaller, say \(t_D \le 0.02\) , the asymptotic expansions of the Bessel functions are needed. User may refer Chapter 13.3 of Ref.1 [1] for details.

[1]Carslaw and Jaeger: Conduction of Heat in Solids, 2nd edition. Oxford University Press, Oxford, UK (1959)