扩展热阻(1)-矩形通道偏心热源的扩展热阻模型
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
\(\lambda < 0, or \, \sigma<0\)时得到的解无意义。
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: \[ \begin{aligned} Y &= C_3 \cos(\delta y), \delta = n\pi/b , n=1,2,\ldots\\ Z &= C_4 e^{ \beta z } + C_5e^{ -\beta z } \end{aligned} \]
- 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: \[ \begin{aligned} Y &= C_3 \cos(\delta_n y), \delta = n\pi/b , n=1,2,\ldots\\ Z &= C_4 e^{ \beta z } + C_5e^{ -\beta z } \end{aligned} \]
- 当\(m \neq 0, n = 0\)时,此时\(\beta = \lambda\),和上面同理,此时:
\[ \begin{aligned} Y &= C_3 \\ X &= C_1 \cos(\lambda x), \lambda = m\pi/b ,m=1,2,\ldots\\ Z &= C_4 e^{ \beta z } + C_5e^{ -\beta z } \end{aligned} \]
- 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 \[ \begin{aligned} &R_{s}=\frac{2}{Q} \sum_{m=1}^{\infty} A_{m} \frac{\cos \left(\lambda_{m} X_{c}\right) \sin \left((1 / 2) \lambda_{m} c\right)}{\lambda_{m} c} \\ &+\frac{2}{Q} \sum_{n=1}^{\infty} A_{n} \frac{\cos \left(\delta_{n} Y_{c}\right) \sin \left((1 / 2) \delta_{n} d\right)}{\delta_{n} d} \\ &+\frac{4}{Q} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} A_{\mathrm{mn}} \frac{\cos \left(\delta_{n} Y_{c}\right) \sin \left((1 / 2) \delta_{n} d\right) \cos \left(\lambda_{m} X_{c}\right) \sin \left((1 / 2) \lambda_{m} c\right)}{\lambda_{m} c \delta_{n} d} \end{aligned} \]