博文

热穿透深度(thermal penetration depth)的推导

已有 8818 次阅读 2013-7-11 22:05 |系统分类:科研笔记| thermal, depth, penetration, 热穿透深度, 瞬态导热

Periodic heating on the surface of a semi-infinite body induces an exponentially decaying temperature profile with a thermal penetration depth

$L_{p}=\sqrt{k/C\pi f}$

which identifies the depth normal to the surface at which the temperature amplitude is $e^{-1}$ of its surface amplitude. This concept is commonly adopted in time domain thermoreflectance (TDTR) and other transient thermoreflectance techniques, and is derived as follows:

Let's consider the one-dimensional transient heat conduction problem in a semi-infinite body. The governing equation and corresponding initial and boundary conditions are as follows:

$.frac{{{.partial ^2}T(x,t)}}{{.partial {x^2}}} = .frac{1}{.alpha }.frac{{.partial T(x,t)}}{{.partial t}}.quad .quad x > 0,.quad t > 0 .qquad .qquad(1)$ $T = {T_i} + f(t) = {T_i} + A.cos (.omega t - .beta ),{.rm{ }}x = 0,.quad t > 0 .qquad .qquad(23)$ $T(x,t) = {T_i}.quad .quad x .to .infty ,.quad t > 0 .qquad .qquad(3)$

$T(x,t) = {T_i}.quad .quad x > 0,.quad t = 0 .qquad .qquad(4)$

where A is the amplitude of oscillation, ω is the angular frequency, and β is the phase delay.

Introducing excess temperature $.vartheta = T - {T_i}$ , the governing equation and corresponding boundary and initial conditions become

$.frac{{{.partial ^2}.vartheta (x,t)}}{{.partial {x^2}}} = .frac{1}{.alpha }.frac{{.partial .vartheta (x,t)}}{{.partial t}}.quad .quad x > 0,.quad t > 0 .qquad .qquad(24)$ $.vartheta (x,t) = f(t) = A.cos (.omega t - .beta ).quad .quad x = 0,.quad t > 0 .qquad .qquad(25)$ $.vartheta (x,t) = 0.quad .quad x .to .infty ,.quad t > 0 .qquad .qquad(26)$

$.vartheta (x,t) = 0.quad .quad x > 0,.quad t = 0 .qquad .qquad(27)$

Duhamel’s theorem can be used to handle the time-dependent boundary condition. Instead of solving eqs. (24) – (27) directly, we will start with a simpler auxiliary problem which is similar to eqs. (24) - (27), only that in eq. (25) t is treated as a parameter $τ$ , rather than time, as defined below:

$.frac{{{.partial ^2}.Phi (x,t)}}{{.partial {x^2}}} = .frac{1}{.alpha }.frac{{.partial .Phi (x,t)}}{{.partial t}}.quad .quad x > 0,.quad t > 0 .qquad .qquad(28)$ $.Phi (x,t) = f(.tau ) = A.cos (.omega .tau - .beta ).quad .quad x = 0,.quad t > 0 .qquad .qquad(29)$ $.Phi (x,t) = 0.quad .quad x .to .infty ,.quad t > 0 .qquad .qquad(30)$

$.Phi (x,t) = 0.quad .quad x > 0,.quad t = 0 .qquad .qquad(31)$

Duhamel’s theorem (Ozisik, 1993) stated that the solution to the original problem is related to the solution of the auxiliary problem by

$.vartheta (x,t) = .frac{.partial }{{.partial t}}.int_{.tau = 0}^t {.Phi (x,t - .tau ,.tau )d.tau } .qquad .qquad(32)$

which can rewritten using Leibniz’s rule

$.vartheta (x,t) = .int_{.tau = 0}^t {.frac{.partial }{{.partial t}}.Phi (x,t - .tau ,.tau )d.tau } + .Phi {.left. {(x,t - .tau ,.tau )} .right|_{.tau = t}} .qquad .qquad(33)$

The second term on the right hand side is

$.Phi {.left. {(x,t - .tau ,.tau )} .right|_{.tau = t}} = .Phi (x,0,.tau ) = 0 .qquad .qquad(34)$

therefore, eq. (33) becomes

$.vartheta (x,t) = .int_{.tau = 0}^t {.frac{.partial }{{.partial t}}.Phi (x,t - .tau ,.tau )d.tau } .qquad .qquad(35)$

The solution of the auxiliary problem can be expressed as

$.Phi (x,t,.tau ) = f(.tau ).left[ {1 - {.rm{erf}}.left( {.frac{x}{{.sqrt {4.alpha t} }}} .right)} .right] = .frac{{2f(.tau )}}{{.sqrt .pi }}.int_{x/.sqrt {4.alpha t} }^.infty {{e^{ - {.eta ^2}}}d.eta } .qquad .qquad(36)$

The partial derivative appearing in eq. (35) can be evaluated as

$.frac{.partial }{{.partial t}}.Phi (x,t - .tau ,.tau ) = f(.tau ).frac{x}{{.sqrt {4.pi .alpha } {{(t - .tau )}^{3/2}}}}.exp .left[ { - .frac{{{x^2}}}{{4.alpha (t - .tau )}}} .right] .qquad .qquad(37)$

Substituting eq. (37) into eq. (35), the solution of the original problem becomes

$.vartheta (x,t) = .frac{x}{{.sqrt {4.pi .alpha } }}.int_{.tau = 0}^t {.frac{{f(.tau )}}{{{{(t - .tau )}^{3/2}}}}.exp .left[ { - .frac{{{x^2}}}{{4.alpha (t - .tau )}}} .right]d.tau } .qquad .qquad(38)$

Introducing a new independent variable

$.xi = .frac{x}{{.sqrt {4.alpha (t - .tau )} }}$

eq. (38) becomes

$.vartheta (x,t) = .frac{2}{{.sqrt .pi }}.int_{x/.sqrt {4.alpha t} }^.infty {f.left( {t - .frac{{{x^2}}}{{4.alpha {.xi ^2}}}} .right).exp ( - {.xi ^2})d.xi } .qquad .qquad(39)$

For the periodic boundary condition specified in eq. (25), we have

$.vartheta (x,t) = .frac{{2A}}{{.sqrt .pi }}.int_{x/.sqrt {4.alpha t} }^.infty {.cos .left[ {.omega .left( {t - .frac{{{x^2}}}{{4.alpha {.xi ^2}}}} .right) - .beta } .right].exp ( - {.xi ^2})d.xi } .qquad .qquad(40)$

which can be rewritten as

$.begin{array}{l} .vartheta (x,t) = .frac{{2A}}{{.sqrt .pi }}.int_0^.infty {.cos .left[ {.omega .left( {t - .frac{{{x^2}}}{{4.alpha {.xi ^2}}}} .right) - .beta } .right].exp ( - {.xi ^2})d.xi } .. {.rm{ }} - .frac{{2A}}{{.sqrt .pi }}.int_0^{x/.sqrt {4.alpha t} } {.cos .left[ {.omega .left( {t - .frac{{{x^2}}}{{4.alpha {.xi ^2}}}} .right) - .beta } .right].exp ( - {.xi ^2})d.xi } .. .end{array} .qquad .qquad(41)$

Evaluating the first integral on the right hand side of eq. (41) yields

$.begin{array}{l} .vartheta (x,t) = A.exp .left[ { - x{{.left( {.frac{.omega }{{2.alpha }}} .right)}^{1/2}}} .right].cos .left[ {.omega t - x{{.left( {.frac{.omega }{{2.alpha }}} .right)}^{1/2}} - .beta } .right] .. {.rm{ }} - .frac{{2A}}{{.sqrt .pi }}.int_0^{x/.sqrt {4.alpha t} } {.cos .left[ {.omega .left( {t - .frac{{{x^2}}}{{4.alpha {.xi ^2}}}} .right) - .beta } .right].exp ( - {.xi ^2})d.xi } .. .end{array} .qquad .qquad(42)$

It can be seen that as $t .to .infty$ , the second term will become zero and the first term represents the steady oscillation.

${.vartheta _s}(x,t) = A.exp .left[ { - x{{.left( {.frac{.omega }{{2.alpha }}} .right)}^{1/2}}} .right].cos .left[ {.omega t - x{{.left( {.frac{.omega }{{2.alpha }}} .right)}^{1/2}} - .beta } .right] .qquad .qquad(43)$

where $A.exp .left[ { - x{{.left( {.omega /(2.alpha )} .right)}^{1/2}}} .right]$ represents the amplitude of oscillation at point x. It is apparent that the amplitude of the oscillation decreases as $x$ increases. The location at which the oscillation amplitude drops to $e^{-1}$ of its surface amplitude is at