锂离子电池组的三维电化学-热耦合仿真分析

锂离子电池组的三维电化学-热耦合仿真分析

韦雪晴, 邓海鹏, 周宇, 王冰川

Three-dimensional electrochemical-thermal coupling model of a lithium-ion battery module

Xueqing WEI, Haipeng DENG, Yu ZHOU, Bingchuan WANG

表1 电化学控制方程和能量方程

Table 1 Electrochemical governing equations and energy equations

方程原理	控制方程	边界条件
固相颗粒中的锂离子扩散	$\frac{\partial c_{s}}{\partial t} = \frac{1}{r^{2}} \frac{\partial}{\partial r} (D_{s} r^{2} \frac{\partial c_{s}}{\partial r})$	$\frac{\partial c_{s}}{\partial r} \|\begin{matrix} r = 0 \end{matrix} = 0,$ $- F D_{s} \frac{\partial c_{s}}{\partial r} \|\begin{matrix} r = r_{n} \end{matrix} = j_{n}, c_{s} \|\begin{matrix} t = 0 \end{matrix} = c_{s, 0}$
固相电子输运	$i_{s} = - σ_{s}^{e f f} \nabla ϕ_{s}$	$i_{s} \|\begin{matrix} x = 0 \end{matrix} = i_{s} \|\begin{matrix} x = l_{p} + l_{s} + l_{n} \end{matrix} = I$ ， $i_{s} \|\begin{matrix} x = l_{p} \end{matrix} = i_{s} \|\begin{matrix} x = l_{p} + l_{s} \end{matrix} = 0$
液相锂离子输运	$i_{l} = - σ_{l}^{e f f} \nabla ϕ_{l} + \frac{2 R T σ_{l}^{e f f}}{F} (1 - t_{0}) (1 + \frac{\partial l n f_{\pm}}{\partial c_{l}}) \nabla l n c_{l}$	$i_{l} \|\begin{matrix} x = 0 \end{matrix} = i_{l} \|\begin{matrix} x = l_{p} + l_{s} + l_{n} \end{matrix} = 0$ ， $i_{l} \|\begin{matrix} x = l_{p} \end{matrix} = i_{l} \|\begin{matrix} x = l_{p} + l_{s} \end{matrix} = I$
物质守恒	$ε_{l} \frac{\partial c_{l}}{\partial t} = \nabla ∙ (D_{l}^{e f f} \nabla c_{l}) + (1 - t_{0}) \frac{\nabla ∙ i_{l}}{F}$	$i_{l} \|\begin{matrix} x = 0 \end{matrix} = i_{l} \|\begin{matrix} x = l_{p} + l_{s} + l_{n} \end{matrix} = 0$ ， $i_{l} \|\begin{matrix} x = l_{p} \end{matrix} = i_{l} \|\begin{matrix} x = l_{p} + l_{s} \end{matrix} = I,$ $c_{l} \|\begin{matrix} t = 0 \end{matrix} = c_{l, 0}$
电荷守恒	$\nabla ∙ i_{l} + \nabla ∙ i_{s} = 0$ ， $\nabla ∙ i_{s} = - α_{p} j_{n}$	$i_{l} \|\begin{matrix} x = 0 \end{matrix} = i_{l} \|\begin{matrix} x = l_{p} + l_{s} + l_{n} \end{matrix} = 0,$ $i_{l} \|\begin{matrix} x = l_{p} \end{matrix} = i_{l} \|\begin{matrix} x = l_{p} + l_{s} \end{matrix} = I,$ $c_{l} \|\begin{matrix} t = 0 \end{matrix} = c_{l, 0}$
界面电化学反应	$j_{n} = j_{0} \{e x p (\frac{α_{a} F}{R T} η) - e x p (- \frac{α_{c} F}{R T} η)\}$ ， $j_{0} = F k_{i} {(\frac{c_{l}}{c_{l, r e f}})}^{α_{a}} {c_{s, s u r f}}^{α_{c}} {(c_{s, m a x} - c_{s, s u r f})}^{α_{a}}$ ， $η = ϕ_{s} - ϕ_{l} - U_{e}$	—
电芯能量传输方程	$ρ c_{p} \frac{\partial T}{\partial t} - \nabla ∙ (k \nabla T) = q_{o h m} + q_{a c t} + q_{r e v} + q_{t a b}$ =-i_s· $\nabla ϕ_{s}$ i_l - $i_{l} ∙ \nabla ϕ_{l} + α_{p} j_{n} η + α_{p} j_{n} T \frac{\partial U_{e}}{\partial T}$ + $q_{t a b}$	$k \frac{\partial T}{\partial x} \|\begin{matrix} x = 0 \end{matrix} = k \frac{\partial T}{\partial x} \|\begin{matrix} x = l_{p} + l_{s} + l_{n} \end{matrix} = - h (T - T_{a})$
极耳产热	$q_{t a b} = \frac{I^{2} R_{t a b}}{V_{t a b}} = I^{2} \frac{1}{σ_{t a b}} \frac{h_{t a b}}{A_{t a b}} \frac{1}{h_{t a b} A_{t a b}} = \frac{1}{σ_{t a b}} {(\frac{I}{A_{t a b}})}^{2}$	—