扩展热阻(1)-矩形通道偏心热源的扩展热阻模型

电热输运

矩形偏心热源的扩展热阻模型

Muzychka Y S, Culham J R, Yovanovich M M. Thermal spreading resistance of eccentric heat sources on rectangular flux channels[J]. J. Electron. Packag., 2003, 125(2): 178-185.

扩展热阻

当高温热源和低温热沉面积相等时，此时温度场呈一维线性分布，此时的热阻为： \[ \begin{aligned} R_{1D} &= \frac{T_h - T_c}{Q}\\ Q &= kA\frac{T_h - T_c}{\Delta x} \end{aligned} \] 于是 \[ R_{1D} = \frac{\Delta x}{kA} \] 两界面间的一维热阻只取决于层厚度、层间热导率以及层面积。

当热量从一个面积较小的热点流向宽阔区域时，此时热流线分布不再是一维的，在靠近热点的区域内，热量流线有横线传播的部分，此时温度分布是多维的；在远离热点的区域内，热流量流线基本上又变成均匀的，温度分布又近似变成一维的。这种情况下，体系的总热阻会大于一维热阻，热阻多出来的部分被称之为扩展热阻。因此，总热阻可以认为由一维热阻和扩展热阻两部分组成： \[ R_{t} = R_{1D} + R_s \] 其中一维热阻可以用热流、热源平面的平均温度（\(\overline{T_{z=0}}\)）和远离热源的热沉温度（\(T_{z \rightarrow \infty}\)）来定义： \[ R_{1D} = \frac{\overline{T_{z=0}} - T_{z \rightarrow \infty} }{Q} \] 总热阻可以用热源平均温度（\(\overline{T_{s}}\)）来定义： \[ R_t = \frac{\overline{T_{s}} - T_{z \rightarrow \infty }}{Q} \] 于是扩展热阻的一般定义式为： \[ R_s = R_t - R_{1D} = \frac{\overline{T_{s}} - \overline{T_{z = 0 }}}{Q} \]

在电子器件的总热阻中，扩展热阻占了绝大部分。

问题陈述

考虑下面这个问题：

系统的控制方程为拉普拉斯方程： \[ \nabla^{2} T=\frac{\partial^{2} T}{\partial x^{2}}+\frac{\partial^{2} T}{\partial y^{2}}+\frac{\partial^{2} T}{\partial z^{2}}=0 \] 在顶部存在热源的区域，满足等热流边界条件： \[ \left.\frac{\partial T}{\partial z}\right|_{z=0}=-\frac{\left(Q / A_{s}\right)}{k_{1}} \] 其中\(A_s = cd\)，在热源区域外，满足： \[ \left.\frac{\partial T}{\partial z}\right|_{z=0}=0 \] 底部满足对流换热边界条件： \[ \left.\frac{\partial T}{\partial z}\right|_{z=t_{1}}=-\frac{h}{k_{1}}\left[T\left(x, y, t_{1}\right)-T_{f}\right] \] 侧边界为绝热边界条件： \[ \begin{aligned} &\left.\frac{\partial T}{\partial x}\right|_{x=0, a}=0 \\ &\left.\frac{\partial T}{\partial y}\right|_{y=0, b}=0 \end{aligned} \]

通解

首先采用分离变量法求拉普拉斯方程通解，拉普拉斯方程的一般形式为： \[ \nabla^{2} T=\frac{\partial^{2} T}{\partial x^{2}}+\frac{\partial^{2} T}{\partial y^{2}}+\frac{\partial^{2} T}{\partial z^{2}}=0 \] 假设通解的形式为： \[ \theta(x,y,z) = X(x)Y(y)Z(z) = T(x,y,z) - T_f \] 代入控制方程后，可以得到： \[ \begin{aligned} \nabla^{2} T &=\frac{\partial^{2} T}{\partial x^{2}}+\frac{\partial^{2} T}{\partial y^{2}}+\frac{\partial^{2} T}{\partial z^{2}} \\ &= Y(y)Z(z)\frac{\partial^2 X}{\partial x^2} + X(x)Z(z)\frac{\partial^2 Y}{\partial y^2} + X(x)Y(y)\frac{\partial^2 Z}{\partial z^2} \\ &= \frac{1}{X}\frac{\partial^2 X}{\partial x^2} + \frac{1}{Y}\frac{\partial^2 Y}{\partial y^2} + \frac{1}{Z}\frac{\partial^2 Z}{\partial z^2} \\ &=0\\ \end{aligned} \] 即： \[ - \frac{1}{Z}\frac{\partial^2 Z}{\partial z^2} = \frac{1}{Y}\frac{\partial^2 Y}{\partial y^2} + \frac{1}{X}\frac{\partial^2 X}{\partial x^2} \] 式子左边是\(z\)的函数，式子右边分别是\(x\)和\(y\)的函数，可以假设： \[ \begin{aligned} \frac{1}{Z}\frac{\partial^2 Z}{\partial z^2} &= \lambda^2 + \delta^2 \\ \frac{1}{X}\frac{\partial^2 X}{\partial x^2} &= - \lambda^2 \\ \frac{1}{Y}\frac{\partial^2 Y}{\partial y^2} &= - \delta^2 \\ \end{aligned} \] 这是二阶常系数线性常微分方程的标准形式，

于是\(Z,X,Y\)的特征方程的特征方程分别为： \[ \begin{aligned} Z^2 - (\lambda^2 + \delta^2) &= 0 \\ X^2 + \lambda^2 &= 0 \\ Y^2 + \delta^2 &= 0 \\ \\ \end{aligned} \] 特征根分别为： $$ \[\begin{aligned} Z&: \frac{\pm \sqrt{4(\lambda^2 + \delta^2)}}{2} = \pm \sqrt{\lambda^2 + \delta^2} = \pm \beta\\ X&: \frac{\pm \sqrt{-4\lambda^2}}{2} = \pm i \lambda \\ Y&: \frac{\pm \sqrt{-4\delta^2}}{2} = \pm i \delta \\ \end{aligned}\]

$$ 特征根取的值不同，微分方程通解的形式也会有所不同

  1. \(\lambda < 0, or \, \sigma<0\)时得到的解无意义。

  2. 当\(\lambda > 0, \sigma > 0\)时，通解的表达式为：

\[ \begin{aligned} X &= C_0 \sin(\lambda x) + C_1 \cos(\lambda x)\\ Y &= C_2 \sin(\delta y) + C_3 \cos(\delta y)\\ Z &= C_4 e^{ \beta z } + C_5e^{ -\beta z }\\ \end{aligned} \]

代入边界条件：

\(x = 0, x = a\)时满足 \[ \frac{\partial X}{\partial x} = \lambda \left(C_0 \cos(\lambda x) - C_1 \sin(\lambda x)\right) = 0 \] 于是： \[ X = C_1 \cos(\lambda x), \lambda = m\pi/a, m=1,2,\ldots \]

\(y = 0, y = b\)时满足 \[ \frac{\partial Y}{\partial y} = \lambda \left(C_2 \cos(\delta y) - C_3 \sin(\delta y)\right) = 0 \] 于是： \[ \begin{aligned} Y &= C_3 \cos(\delta y), \delta = n\pi/b , n=1,2,\ldots\\ Z &= C_4 e^{ \beta z } + C_5e^{ -\beta z } \end{aligned} \]

  1. 当\(m = 0, n \neq 0\)，此时\(\beta = \sigma\)：

\[ \begin{aligned} X &= C_0 + C_1 x\\ Y &= C_2 \sin(\delta y) + C_3 \cos(\delta y)\\ Z &= C_4 e^{ \beta z } + C_5e^{ -\beta z }\\ \end{aligned} \]

代入边界条件：

\(x = 0, x = a\)时满足 \[ \frac{\partial X}{\partial x} = C_1= 0 \] 于是： \[ X = C_0 \]

\(y = 0, y = b\)时满足 \[ \frac{\partial Y}{\partial y} = \lambda \left(C_2 \cos(\delta y) - C_3 \sin(\delta y)\right) = 0 \] 于是： \[ \begin{aligned} Y &= C_3 \cos(\delta_n y), \delta = n\pi/b , n=1,2,\ldots\\ Z &= C_4 e^{ \beta z } + C_5e^{ -\beta z } \end{aligned} \]

  1. 当\(m \neq 0, n = 0\)时，此时\(\beta = \lambda\)，和上面同理，此时：

\[ \begin{aligned} Y &= C_3 \\ X &= C_1 \cos(\lambda x), \lambda = m\pi/b ,m=1,2,\ldots\\ Z &= C_4 e^{ \beta z } + C_5e^{ -\beta z } \end{aligned} \]

  1. 当\(m = 0, n = 0\)时，\(\beta = 0\)，此时：

\[ \begin{aligned} X &= C_0 + C_1 x \\ Y &= C_2 + C_3 y \\ Z &= C_4 + C_5 z \end{aligned} \]

代入边界条件： \[ \begin{aligned} X &= C_0 \\ Y &= C_2 \\ Z &= C_4 + C_5 z \end{aligned} \] 根据叠加原理，可以得到方程的通解为： \[ \begin{aligned} \theta(x, y, z)=& A_{0}+B_{0} z\\ &+\sum_{m=1}^{\infty} \cos (\lambda x)\left[A_{1} \cosh (\lambda z)+B_{1} \sinh (\lambda z)\right] \\ &+\sum_{n=1}^{\infty} \cos (\delta y)\left[A_{2} \cosh (\delta z)+B_{2} \sinh (\delta z)\right] \\ &+\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \cos (\lambda x) \cos (\delta y)\left[A_{3} \cosh (\beta z)\right.\left.+B_{3} \sinh (\beta z)\right] \end{aligned} \]

其中\(\lambda = m\pi/a, \sigma = n\pi/b, \beta = \sqrt{\lambda^2 + \sigma^2}\)，双曲函数\(\cosh\)和\(\sinh\)是两项指数相加的另一种表示形式， \[ \begin{aligned} \sinh(x) &= \frac{e^x - e^{-x}}{2} \\ \cosh(x) &= \frac{e^x + e^{-x}}{2} \\ \end{aligned} \] 通解由四部分组成，第一部分是线性部分的贡献，剩下三项是扩展部分的贡献，当\(z=0\)平面的热流均匀分布时，扩展部分的三项会消失。之后通过代入\(z\)平面的边界条件，可以解出各个系数的具体表达式。由于通解是线性叠加出来的，因此各个子项都应满足边界条件。代入热沉平面\(z=t_1\)的边界条件： \[ \left.\frac{\partial \theta}{\partial z}\right|_{z=t_{1}}=-\left.\frac{h}{k_{1}}\theta\right|_{z=t_1} \]

\[ \left[ A_{3} \sinh (\beta z)+B_{3} \cosh (\beta z) \right]\beta = - \frac{h}{k_1}\left[A_{3} \cosh (\beta z)\right.\left.+B_{3} \sinh (\beta z)\right] \]

于是\(B_i\)和\(A_i\)的关系可以整理为： \[ B_i = -\phi(\zeta)A_i,\quad i = 1,2,3,\ldots \] 其中： \[ \phi(\zeta)=\frac{\zeta \sinh \left(\zeta t_{1}\right)+h / k_{1} \cosh \left(\zeta t_{1}\right)}{\zeta \cosh \left(\zeta t_{1}\right)+h / k_{1} \sinh \left(\zeta t_{1}\right)} \] 其中\(\zeta\)分别代表\(\lambda, \sigma,\beta\)

对于理想热沉，底边界为等温边界条件，\(h\rightarrow +\infty\)，此时\(\phi(\zeta)\)退化为 \[ \left. \phi(\zeta) \right|_{h\rightarrow +\infty} = \coth (\zeta t_1) \] 最终通解中的系数可以通过对热源平面的边界条件做傅里叶展开得到，先求温度梯度表达式： \[ \begin{aligned} \frac{\partial \theta(x, y, z)}{\partial z}=& B_0\\ &+\lambda \sum_{m=1}^{\infty} \cos (\lambda x)\left[A_{1} \sinh (\lambda z)+B_{1} \cosh (\lambda z)\right] \\ &+\sigma \sum_{n=1}^{\infty} \cos (\delta y)\left[A_{2} \sinh (\delta z)+B_{2} \cosh (\delta z)\right] \\ &+\beta \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \cos (\lambda x) \cos (\delta y)\left[A_{3} \sinh (\beta z)\right.\left.+B_{3} \cosh (\beta z)\right] \end{aligned} \] 于是 \[ \begin{aligned} \frac{\partial \theta(x, y, z)}{\partial z}\Big|_{z=0}=& B_0\\ &+\lambda \sum_{m=1}^{\infty} \cos (\lambda x)B_{1} \\ &+\sigma \sum_{n=1}^{\infty} \cos (\delta y)B_{2} \\ &+\beta \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \cos (\lambda x) \cos (\delta y)B_3 \end{aligned} \] 边界条件\(f(x,y)\)为： \[ f(x,y) = \left\{ \begin{array}{l} \dfrac{\partial T}{\partial z} = -\dfrac{\left(Q / cd\right)}{k_{1}}, \quad X_c - \dfrac{c}{2} < x < X_c + \dfrac{c}{2}, \quad Y_c - \dfrac{d}{2} < y < Y_c + \dfrac{d}{2} \\ \dfrac{\partial T}{\partial z} = 0, \quad others \end{array} \right. \] 作二维非周期傅里叶级数展开： $$ \[\begin{aligned} f(x, y) =& \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \left[ c_{mn} \sin \left(\frac{m \pi x}{a}\right) \sin \left(\frac{n \pi y}{b}\right) \right. \\ &\left. + d_{mn} \cos \left(\frac{m \pi x}{a}\right) \cos \left(\frac{n \pi y}{b}\right)\right] \\ =& d_{00} + \sum_{m=1}^{\infty} d_{m0} \cos(\frac{m\pi x}{a}) + \sum_{n=1}^{\infty} d_{0n} \cos(\frac{n\pi y}{b}) \\ &+ \sum_{m,n=1}^{\infty}\left[ c_{mn} \sin \left(\frac{m \pi x}{a}\right) \sin \left(\frac{n \pi y}{b}\right) + d_{mn} \cos \left(\frac{m \pi x}{a}\right) \cos \left(\frac{n \pi y}{b}\right)\right] \end{aligned}\]

$$

\[ \begin{aligned} c_{mn} &= \frac{\kappa}{ab} \int_{0}^{b} \int_{0}^{a} f(x, y) \sin \frac{m \pi x}{a} \sin \frac{n\pi y}{b} d x d y \\ d_{mn} &= \frac{\kappa}{ab} \int_{0}^{b} \int_{0}^{a} f(x, y) \cos \frac{m \pi x}{a} \cos \frac{n\pi y}{b} d x d y \end{aligned} \]

\[ \begin{aligned} \text { Where }\kappa &=1 \text { if } n=0 \text { and } m=0 \\ &=2 \text { if } n=0 \text { or } m=0 \\ &=4 \text { if } n>0 \text { and } m>0 \end{aligned} \]

对比\(\dfrac{\partial \theta(x, y, z)}{\partial z}\Big|_{z=0}\)中的各系数，得到： \[ B_0 = - \frac{Q}{abk_1} \]

\[ \begin{aligned} B_1 &= d_{m0} / \lambda \\ &= \frac{2}{ab\lambda} \int_{0}^{b} \int_{0}^{a} f(x, y) \cos \frac{m \pi x}{a} d x d y \\ &= -\dfrac{2\left(Q / cd\right)}{ab\lambda k_{1}} \int_{X_c - \frac{c}{2}}^{X_c + \frac{c}{2}} \int_{Y_c - \frac{d}{2}}^{Y_c + \frac{d}{2}} \cos(\lambda x)dxdy \\ &= -\dfrac{2Q}{abc\lambda k_{1}} \int_{X_c - \frac{c}{2}}^{X_c + \frac{c}{2}} \cos(\lambda x)dx \\ &= \dfrac{2Q}{\lambda^2 abck_{1}} \left[ \sin(\lambda (X_c - c/2)) - \sin(\lambda (X_c + c/2)) \right] \end{aligned} \]

\[ \begin{aligned} B_2 &= d_{0n} / \sigma \\ &= \frac{2}{ab\sigma} \int_{0}^{b} \int_{0}^{a} f(x, y) \cos \frac{n\pi y}{b} d x d y \\ &= -\dfrac{2\left(Q / cd\right)}{ab\sigma k_{1}} \int_{X_c - \frac{c}{2}}^{X_c + \frac{c}{2}} \int_{Y_c - \frac{d}{2}}^{Y_c + \frac{d}{2}} \cos(\sigma y)dxdy \\ &= -\dfrac{2Q}{abd\sigma k_{1}} \int_{Y_c - \frac{d}{2}}^{Y_c + \frac{d}{2}} \cos(\sigma y)dy \\ &= \dfrac{2Q}{\sigma^2 abdk_{1}} \left[ \sin(\sigma (Y_c - d/2)) - \sin(\sigma (Y_c + d/2)) \right] \end{aligned} \]

\[ \begin{aligned} B_3 &= d_{mn} / \beta \\ &= \frac{4}{ab\beta} \int_{0}^{b} \int_{0}^{a} f(x, y) \cos \frac{m \pi x}{a} \cos \frac{n\pi y}{b} d x d y \\ &= -\dfrac{4\left(Q / cd\right)}{ab\beta k_{1}} \int_{X_c - \frac{c}{2}}^{X_c + \frac{c}{2}} \int_{Y_c - \frac{d}{2}}^{Y_c + \frac{d}{2}} \cos(\lambda x)\cos(\sigma y)dxdy \\ &=\frac{16 Q \cos \left(\lambda X_{c}\right) \sin \left(\frac{1}{2} \lambda c\right) \cos \left(\delta Y_{c}\right) \sin \left(\frac{1}{2} \delta d\right)}{a b c d k_{1} \beta \lambda \delta} \end{aligned} \]

对照一下通解中的系数： \[ \begin{aligned} \theta(x, y, z)=& A_{0}+B_{0} z\\ &+\sum_{m=1}^{\infty} \cos (\lambda x)\left[A_{1} \cosh (\lambda z)+B_{1} \sinh (\lambda z)\right] \\ &+\sum_{n=1}^{\infty} \cos (\delta y)\left[A_{2} \cosh (\delta z)+B_{2} \sinh (\delta z)\right] \\ &+\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \cos (\lambda x) \cos (\delta y)\left[A_{3} \cosh (\beta z)\right.\left.+B_{3} \sinh (\beta z)\right] \end{aligned} \] 其中， \[ B_i = -\phi(\zeta)A_i,\quad i = 1,2,3 \]

\[ \phi(\zeta)=\frac{\zeta \sinh \left(\zeta t_{1}\right)+h / k_{1} \cosh \left(\zeta t_{1}\right)}{\zeta \cosh \left(\zeta t_{1}\right)+h / k_{1} \sinh \left(\zeta t_{1}\right)} \]

其中\(\zeta\)分别代表\(\lambda, \sigma,\beta\)

于是， \[ A_{1}=\frac{4 Q \cos \left(\lambda_{m} X_{c}\right) \sin \left(\lambda_{m}(c / 2)\right)}{a b c k_{1} \lambda_{m}^{2} \phi\left(\lambda_{m}\right)} \]

\[ A_{2}=\frac{4 Q \cos \left(\delta_{n} Y_{c}\right) \sin \left(\delta_{n}(d / 2)\right)}{a b d k_{1} \delta_{n}^{2} \phi\left(\delta_{n}\right)} \]

\[ A_{3}=\frac{16 Q \cos \left(\lambda_{m} X_{c}\right) \sin \left((1 / 2) \lambda_{m} c\right) \cos \left(\delta_{n} Y_{c}\right) \sin \left((1 / 2) \delta_{n} d\right)}{a b c d k_{1} \beta_{\mathrm{mn}} \lambda_{m} \delta_{n} \phi\left(\beta_{\mathrm{mn}}\right)} \]

线性部分的解为： \[ A_0 = \frac{Q}{ab}\left( \frac{t_1}{k_1} + \frac{1}{h} \right) \] 热源平面的温度分布为： \[ \begin{aligned} \theta(x, y, 0)=& A_{0}+\sum_{m=1}^{\infty} A_{1} \cos \left(\lambda_{m} x\right)+\sum_{n=1}^{\infty} A_{2} \cos \left(\delta_{n} y\right) \\ &+\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} A_{3} \cos \left(\lambda_{m} x\right) \cos \left(\delta_{n} y\right) \end{aligned} \] 通过对热源区域积分，可以得到： \[ \begin{aligned} \bar{\theta}_{s} =& \frac{1}{cd} \int_{X_c - \frac{c}{2}}^{X_c + \frac{c}{2}} \int_{Y_c - \frac{d}{2}}^{Y_c + \frac{d}{2}} \theta(x,y,0) dxdy \\ =& \bar{\theta_{1D}}+2 \sum_{m=1}^{\infty} A_{1} \frac{\cos \left(\lambda_{m} X_{c}\right) \sin \left((1 / 2) \lambda_{m} c\right)}{\lambda_{m} c} \\ &+2 \sum_{n=1}^{\infty} A_{2} \frac{\cos \left(\delta_{n} Y_{c}\right) \sin \left((1 / 2) \delta_{n} d\right)}{\delta_{n} d} \\ &+4 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} A_{3} \frac{\cos \left(\delta_{n} Y_{c}\right) \sin \left((1 / 2) \delta_{n} d\right) \cos \left(\lambda_{m} X_{c}\right) \sin \left((1 / 2) \lambda_{m} c\right)}{\lambda_{m} c \delta_{n} d} \end{aligned} \] 其中 \[ \bar{\theta}_{1 D}=\frac{Q}{a b}\left(\frac{t_{1}}{k_{1}}+\frac{1}{h}\right) \] 总热阻的表达式为： \[ R_t = \frac{\bar{\theta_s}}{Q} = R_{1D} + R_s \] 一维热阻的表达式为： \[ R_{1D} = \frac{1}{ab}( \frac{t_1}{k_1} + \frac{1}{h_s}) \] 扩展热阻 \[ \begin{aligned} &R_{s}=\frac{2}{Q} \sum_{m=1}^{\infty} A_{m} \frac{\cos \left(\lambda_{m} X_{c}\right) \sin \left((1 / 2) \lambda_{m} c\right)}{\lambda_{m} c} \\ &+\frac{2}{Q} \sum_{n=1}^{\infty} A_{n} \frac{\cos \left(\delta_{n} Y_{c}\right) \sin \left((1 / 2) \delta_{n} d\right)}{\delta_{n} d} \\ &+\frac{4}{Q} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} A_{\mathrm{mn}} \frac{\cos \left(\delta_{n} Y_{c}\right) \sin \left((1 / 2) \delta_{n} d\right) \cos \left(\lambda_{m} X_{c}\right) \sin \left((1 / 2) \lambda_{m} c\right)}{\lambda_{m} c \delta_{n} d} \end{aligned} \]