ویکی‌جزوه/دانشکده:فنی و مهندسی/انتقال حرارت/رسانش پایای دو بعدی

${\displaystyle \rho c{\frac {\partial T}{\partial t}}=k({\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}})}$

${\displaystyle {\frac {\partial T}{\partial t}}=0\to {\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}}=0}$

${\displaystyle T(x,y)=?}$

${\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}}=0\\&T(x,0)=T0\\&T(x,H)=T0+A\sin(\pi x/L)\\&T(0,y)=T0\\&T(L,y)=T0\\\end{aligned}}}$

${\displaystyle \theta =T-T0}$

${\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}\theta }{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}\theta }{\partial {{y}^{2}}}}=0\\&\theta (x,0)=0\\&\theta (x,H)=A\sin(\pi x/L)\\&\theta (0,y)=0\\&\theta (L,y)=0\\\end{aligned}}}$

${\displaystyle \theta =X(x).Y(y)}$

${\displaystyle {\frac {{X}^{''}}{X}}=-{\frac {{Y}^{''}}{Y}}=-{{\lambda }^{2}}}$

${\displaystyle {\begin{aligned}&{{X}^{''}}+{{\lambda }^{2}}x=0\\&X(0)=X(L)=0\\\end{aligned}}}$

${\displaystyle {\begin{aligned}&X(x)=a\sin(\lambda x)+b\cos(\lambda x)\\&X(0)=0\to b=0\\&X(L)=0\to X(L)={{a}_{n}}\sin({{\lambda }_{n}}L)=0\to {{\lambda }_{n}}={\frac {n\pi }{L}}\\\end{aligned}}}$

${\displaystyle {{X}_{n}}(x)=a\sin({{\lambda }_{n}}x)}$

${\displaystyle {\begin{aligned}&{{Y}_{n}}^{''}-{{\lambda }_{n}}^{2}{{Y}_{n}}=0\\&{{Y}_{n}}(y)={{c}_{n}}\sinh({{\lambda }_{n}}y)+{{d}_{n}}\cosh({{\lambda }_{n}}y)\\&{{y}_{n}}(0)=0\to {{d}_{n}}=0\to {{Y}_{n}}(y)={{c}_{n}}\sinh({{\lambda }_{n}}y)\\\end{aligned}}}$

${\displaystyle {\begin{aligned}&{{\theta }_{n}}(x,y)={{X}_{n}}(x).{{Y}_{n}}(y)\to \theta (x,y)=\sum \limits _{n=1}^{\infty }{{e}_{n}}\sin({\frac {n\pi x}{L}}).\sinh({\frac {n\pi y}{L}})\\&\theta (x,H)=A\sin({\frac {\pi x}{L}})\to \theta (x,H)=\sum \limits _{n=1}^{\infty }{{e}_{n}}\sin({\frac {n\pi x}{L}}).\sinh({\frac {n\pi H}{L}})=A\sin({\frac {\pi x}{L}})\\&T(x,y)=T0+A\sin({\frac {\pi x}{L}}).{\frac {\sinh({\frac {\pi y}{L}})}{\sinh({\frac {\pi H}{L}})}}\\\end{aligned}}}$

${\displaystyle T(x,y)=?}$

${\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}}=0\\&T(x,0)=T0\\&T(x,H)=T0+A\sin(\pi x/L)\\&T(0,y)=T0\\&T(L,y)=T0+y(H-y)\\\end{aligned}}}$

${\displaystyle \theta =T-T0}$

${\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}\theta }{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}\theta }{\partial {{y}^{2}}}}=0\\&\theta (x,0)=0\\&\theta (x,H)=A\sin(\pi x/L)\\&\theta (0,y)=0\\&\theta (L,y)=y(H-y)\\\end{aligned}}}$

${\displaystyle \theta (x,y)={{\theta }_{1}}(x,y)+{{\theta }_{2}}(x,y)}$

${\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}{{\theta }_{1}}}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}{{\theta }_{1}}}{\partial {{y}^{2}}}}=0\\&{{\theta }_{1}}(x,0)={{\theta }_{1}}(0,y)={{\theta }_{1}}(L,y)=0\\&{{\theta }_{1}}(x,H)=A\sin({\frac {\pi x}{L}})\\\end{aligned}}}$

${\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}{{\theta }_{2}}}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}{{\theta }_{2}}}{\partial {{y}^{2}}}}=0\\&{{\theta }_{2}}(x,0)={{\theta }_{2}}(x,H)={{\theta }_{2}}(0,y)=0\\&{{\theta }_{2}}(L,y)=y(H-y)\\\end{aligned}}}$

${\displaystyle \theta ={{\theta }_{1}}+{{\theta }_{2}}}$

${\displaystyle \rho c{\frac {\partial T}{\partial t}}=k({\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}})}$

${\displaystyle {\frac {\partial T}{\partial t}}=0\to {\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}}=0}$

${\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}}=0\\&T(x,0)=T0\\&T(x,H)=T0+T1\sin(\pi x/L)\\&T(0,y)=T0\\&T(L,y)=T0+T1\sin(\pi x/L)\\\end{aligned}}}$

${\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}T_{1}}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T_{1}}{\partial {{y}^{2}}}}=0\\&T_{1}(x,0)=0\\&T_{1}(x,H)=0\\&T_{1}(0,y)=0\\&T_{1}(L,y)={\acute {T}}sin({\frac {\pi y}{H}}\\\end{aligned}}}$

${\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}T_{2}}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T_{2}}{\partial {{y}^{2}}}}=0\\&T(x,0)=T0\\&T(x,H)=T0+T1\sin(\pi x/L)\\&T(0,y)=T0\\&T(L,y)=T0\\\end{aligned}}}$

${\displaystyle T(x,y)=T_{1}(x,y)+T_{2}(x,y)}$