Heat Transfer|L3_1D Steady Heat conduction_2

最新推荐文章于 2020-03-06 11:43:29 发布

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最新推荐文章于 2020-03-06 11:43:29 发布 · 515 阅读

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CATALOG

Steady Problem without Heat Source
Critical Thickness（临界厚度）
Steady Problem with Heat Source

Steady Problem without Heat Source

Conduction through a spherical wall (shell)

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The governing condition is
$\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(\lambda r^{2} \frac{\partial T}{\partial r}\right)=0$
and the BC is
$\begin{aligned} &r=r_{1} \quad T=T_{1}\\ &r=r_{2} \quad T=T_{2} \end{aligned}$

from the g function, we have
$\frac{\partial}{\partial r}\left(\lambda r^{2} \frac{\partial T}{\partial r}\right)=0$
$\begin{aligned} \Rightarrow \lambda r^{2} \frac{\partial T}{\partial r}=C_{1} \Rightarrow \frac{\partial T}{\partial r} &=\frac{C_{1}}{\lambda r^{2}} \\ T =-\frac{C_{1}}{\lambda r}+C_{2} \end{aligned}$

so we have

$\begin{array}{l} T_{1}=-\frac{c_{1}}{\lambda r_{1}}+c_{2} \\ T_{2}=-\frac{c_{1}}{N_{2}}+c_{2} \end{array}$
and
$\begin{aligned} \quad C_{1} &=-\frac{T_{1}-T_{2}}{\frac{1}{\frac{1}{r}-\frac{1}{r_{2}}}} r \\ C_{2} &=T_{1}-\frac{T_{1}-T_{2}}{\frac{1}{r_{1}}-\frac{1}{r_{2}}} \cdot \frac{1}{\lambda r_{1}} \\ &=T_{2}-\frac{T_{1}-T_{2}}{\frac{1}{r_{1}}-\frac{1}{r_{2}}} \cdot \frac{1}{\lambda r_{2}} \end{aligned}$
so we have the function
$T=T_{2}+\left(T_{1}-T_{2}\right) \frac{1 / r-1 / r_{2}}{1 / r_{1}-1 / r_{2}}$
so we have the Heat flution
$q=-\lambda \frac{d T}{d r}=\frac{\lambda}{r^{2}} \frac{T_{1}-T_{2}}{1 / r_{1}-1 / r_{2}}$

the Heat transfer rate is
$\Phi=q A=4 \pi r^{2} \frac{\lambda}{r^{2}} \frac{T_{1}-T_{2}}{1 / r_{1}-1 / r_{2}}=4 \pi \lambda \frac{T_{1}-T_{2}}{1 / r_{1}-1 / r_{2}}$
and the Transfer rate
$R=\frac{T_{1}-T_{2}}{\Phi}=\frac{1}{4 \pi \lambda}\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)$

Conduction through a composite（复合的） spherical(球形) wall (shell)

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Thermal resistance:
$R_{t}=R_{1}+R_{2}+R_{3}=\frac{1}{4 \pi \lambda_{1}}\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)+\frac{1}{4 \pi \lambda_{2}}\left(\frac{1}{r_{2}}-\frac{1}{r_{3}}\right)+\frac{1}{4 \pi \lambda_{3}}\left(\frac{1}{r_{3}}-\frac{1}{r_{4}}\right)$