我编译此代码时出现错误

Question

问题描述

这段代码有什么问题？我遇到了错误。请帮帮我。

梅威瑟：

\documentclass[12pt]{article}
\usepackage[a4paper,top=0.6 in,bottom=0.6 in,left=0.6 in,right=0.6 in]{geometry}
\usepackage{amsmath}
\begin{document}
\large
\begin{equation}
\begin{align*}
 \lambda&=\dfrac{Gr_{x}}{R_{e}^2}   &   \widetilde{N}&=\dfrac{Gr_{x}^{\ast}}{Gr_{x}} \\[3ex]
 Gr_{x}&=\dfrac{g\beta_{T}\left(T_{w}-T_{\infty}\right)x^{3}}{\nu^{2}}       &   Gr_{x}^{\ast}&=\dfrac{g\beta_{C}\left(C_{w}-C_{\infty}\right)x^{3}}{\nu^{2}}\\[3ex]
 M^{2}&=\dfrac{\sigma\,B_{0}^{2}}{a\,\rho\, x^{n-1}}   &   Nr&=\dfrac{16\sigma^{\ast}}{3k^{\ast}}\,\dfrac{T^{3}_{\infty}}{k_{2}} \\[3ex]
 \dfrac{1}{Pr}&=\dfrac{k_{2}}{\rho c_{p}}       &   A&=\dfrac{b}{a}\\[3ex]
S_{c}&=\dfrac{\nu}{D_{B}}   &   \dfrac{1}{k_{1}}&=\dfrac{\mu \phi}{k'}\\[3ex]
 S_{r}&=\dfrac{D_{m}\,k_{T}\,\left(T_{w}-T_{\infty}\right)}{\alpha_{m}\,C_{p}\,C_{s}\,\left(C_{w}-C_{\infty}\right)}      &   D_{f}&=\dfrac{D_{m}\,k_{T}\,\left(T_{w}-T_{\infty}\right)}{\alpha_{m}\,C_{p}\,C_{s}\,\left(C_{w}-C_{\infty}\right)}
\end{align*} 
\end{equation}
\end{document}

Answer 1

这应该有效：

\documentclass[12pt]{article}
\usepackage[a4paper,top=0.6 in,bottom=0.6 in,left=0.6 in,right=0.6 in]{geometry}
\usepackage{amsmath}
\begin{document}
\large
\begin{align}
 \lambda&=\dfrac{Gr_{x}}{R_{e}^2}   &   \widetilde{N}&=\dfrac{Gr_{x}^{\ast}}{Gr_{x}} \\[3ex]
 Gr_{x}&=\dfrac{g\beta_{T}\left(T_{w}-T_{\infty}\right)x^{3}}{\nu^{2}}       &   Gr_{x}^{\ast}&=\dfrac{g\beta_{C}\left(C_{w}-C_{\infty}\right)x^{3}}{\nu^{2}}\\[3ex]
 M^{2}&=\dfrac{\sigma\,B_{0}^{2}}{a\,\rho\, x^{n-1}}   &   Nr&=\dfrac{16\sigma^{\ast}}{3k^{\ast}}\,\dfrac{T^{3}_{\infty}}{k_{2}} \\[3ex]
 \dfrac{1}{Pr}&=\dfrac{k_{2}}{\rho c_{p}}       &   A&=\dfrac{b}{a}\\[3ex]
S_{c}&=\dfrac{\nu}{D_{B}}   &   \dfrac{1}{k_{1}}&=\dfrac{\mu \phi}{k'}\\[3ex]
 S_{r}&=\dfrac{D_{m}\,k_{T}\,\left(T_{w}-T_{\infty}\right)}{\alpha_{m}\,C_{p}\,C_{s}\,\left(C_{w}-C_{\infty}\right)}      &   D_{f}&=\dfrac{D_{m}\,k_{T}\,\left(T_{w}-T_{\infty}\right)}{\alpha_{m}\,C_{p}\,C_{s}\,\left(C_{w}-C_{\infty}\right)}
\end{align} 
\end{document}

删除\begin{equation}、\end{equation}环境并更改align*为align可解决所有问题。

如果你希望它们单独垂直：

\documentclass[12pt]{article}
\usepackage[a4paper,top=0.6 in,bottom=0.6 in,left=0.6 in,right=0.6 in]{geometry}
\usepackage{amsmath}
\begin{document}
\large

\begin{align}
 \lambda=\dfrac{Gr_{x}}{R_{e}^2}
\end{align}

\begin{align}
 \widetilde{N}=\dfrac{Gr_{x}^{\ast}}{Gr_{x}}
\end{align}

\begin{align}
 Gr_{x}=\dfrac{g\beta_{T}\left(T_{w}-T_{\infty}\right)x^{3}}{\nu^{2}}
\end{align}

\begin{align}
 Gr_{x}^{\ast}=\dfrac{g\beta_{C}\left(C_{w}-C_{\infty}\right)x^{3}}{\nu^{2}}
\end{align}

\begin{align}
 M^{2}=\dfrac{\sigma\,B_{0}^{2}}{a\,\rho\, x^{n-1}}
\end{align}

\begin{align}
 Nr=\dfrac{16\sigma^{\ast}}{3k^{\ast}}\,\dfrac{T^{3}_{\infty}}{k_{2}}
\end{align}

\begin{align}
 \dfrac{1}{Pr}=\dfrac{k_{2}}{\rho c_{p}}
\end{align}

\begin{align}
 A=\dfrac{b}{a}
\end{align}

\begin{align}
 S_{c}=\dfrac{\nu}{D_{B}}
\end{align}

\begin{align}
 \dfrac{1}{k_{1}}=\dfrac{\mu \phi}{k'}
\end{align}

\begin{align}
 S_{r}=\dfrac{D_{m}\,k_{T}\,\left(T_{w}-T_{\infty}\right)}{\alpha_{m}\,C_{p}\,C_{s}\,\left(C_{w}-C_{\infty}\right)}     
\end{align}

\begin{align}
 D_{f}=\dfrac{D_{m}\,k_{T}\,\left(T_{w}-T_{\infty}\right)}{\alpha_{m}\,C_{p}\,C_{s}\,\left(C_{w}-C_{\infty}\right)}
\end{align} 
\end{document}