计算传热学第一次大作业

1 Taylor级数展开法

1.1 网格划分

Taylor级数展开发的网格划分使用外点法

1.2 离散方程表达式

$\rho u\frac{d\phi}{d x}=\Gamma \frac{d^{2}\phi}{dx^{2}}\tag{1}$ρudxdϕ​=Γdx2d2ϕ​(1)
将$\phi$ϕ分别在$i$i+1点和$i$i-1点在$i$i点展开
$\phi(i+1) = \phi(i)+\left.\frac{d\phi}{dx}\right|_{i}\delta x+\left.\frac{d^{2}\phi}{dx^{2}}\right|_{i}\frac{\delta x^{2}}{2!}+…… \tag{2}$ϕ(i+1)=ϕ(i)+dxdϕ​∣∣∣∣​i​δx+dx2d2ϕ​∣∣∣∣​i​2!δx2​+……(2)
$\phi(i-1) = \phi(i)-\left.\frac{d\phi}{dx}\right|_{i}\delta x+\left.\frac{d^{2}\phi}{dx^{2}}\right|_{i}\frac{\delta x^{2}}{2!}+…… \tag{3}$ϕ(i−1)=ϕ(i)−dxdϕ​∣∣∣∣​i​δx+dx2d2ϕ​∣∣∣∣​i​2!δx2​+……(3)
联立(2)、(3)式得
$\left.\frac{d\phi}{dx}\right|_{i}=\frac{\phi_{i+1}-\phi_{i-1}}{2\delta x},o(\delta x^{2})\tag{4}$dxdϕ​∣∣∣∣​i​=2δxϕi+1​−ϕi−1​​,o(δx2)(4)
同样的，联立(2)、(3)式得
$\left.\frac{d^{2}\phi}{dx^{2}}\right|_{i}=\frac{\phi_{i+1}-2\phi_{i}+\phi_{i-1}}{\delta x^{2}},o(\delta x^{2})\tag{5}$dx2d2ϕ​∣∣∣∣​i​=δx2ϕi+1​−2ϕi​+ϕi−1​​,o(δx2)(5)
将(4)、(5)式带入(1)式
$\rho u \frac{\phi_{i+1}-\phi_{i-1}}{2\delta x}=\Gamma \frac{\phi_{i+1}-2\phi_{i}+\phi_{i-1}}{\delta x^{2}}\tag{6}$ρu2δxϕi+1​−ϕi−1​​=Γδx2ϕi+1​−2ϕi​+ϕi−1​​(6)
化简得
$4\Gamma \phi_{i}=(\rho u \delta x+2\Gamma)\phi_{i-1}+(2\Gamma-\rho u \delta x)\phi_{i+1},o(\delta x^{2})\tag{7}$4Γϕi​=(ρuδx+2Γ)ϕi−1​+(2Γ−ρuδx)ϕi+1​,o(δx2)(7)

2 控制容积积分法

2.1 网格划分

控制容积积分法使用内节点法划分网格

2.2 离散方程表达式

$\rho u\frac{d\phi}{d x}=\Gamma \frac{d^{2}\phi}{dx^{2}}\tag{8}$ρudxdϕ​=Γdx2d2ϕ​(8)
将方程在空间内积分
$\int_{w}^{e}\rho u\frac{d\phi}{d x}dx=\int_{w}^{e}\Gamma \frac{d^{2}\phi}{dx^{2}}dx\tag{9}$∫we​ρudxdϕ​dx=∫we​Γdx2d2ϕ​dx(9)
其中
$\int_{w}^{e}\frac{d\phi}{d x}dx=\left.\phi\right|_{e}-\left.\phi\right|_{w}\tag{10}$∫we​dxdϕ​dx=ϕ∣e​−ϕ∣w​(10)
假设$\phi$ϕ对空间呈分段线性变化
$\left.\phi\right|_{e}-\left.\phi\right|_{w}=\frac{\phi_{E}+\phi_{P}}{2}-\frac{\phi_{P}+\phi_{W}}{2}=\frac{\phi_{E}-\phi_{W}}{2}\tag{11}$ϕ∣e​−ϕ∣w​=2ϕE​+ϕP​​−2ϕP​+ϕW​​=2ϕE​−ϕW​​(11)
对于扩散项
$\int_{w}^{e}\frac{d^{2}\phi}{dx^{2}}dx=\left.\frac{d\phi}{dx}\right|_{e}-\left.\frac{d\phi}{dx}\right|_{w}\tag{12}$∫we​dx2d2ϕ​dx=dxdϕ​∣∣∣∣​e​−dxdϕ​∣∣∣∣​w​(12)
假设$\frac{d\phi}{dx}$dxdϕ​在空间呈分段线性变化
$\left.\frac{d\phi}{dx}\right|_{e}-\left.\frac{d\phi}{dx}\right|_{w}=\frac{\phi_{E}-\phi_{P}}{(\delta x)_{e}}-\frac{\phi_{P}-\phi_{W}}{(\delta x)_{w}}\tag{13}$dxdϕ​∣∣∣∣​e​−dxdϕ​∣∣∣∣​w​=(δx)e​ϕE​−ϕP​​−(δx)w​ϕP​−ϕW​​(13)
若使用均分网格
$~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{\phi_{E}-\phi_{P}}{\Delta x}-\frac{\phi_{P}-\phi_{W}}{\Delta x}\tag{14}$                         =ΔxϕE​−ϕP​​−ΔxϕP​−ϕW​​(14)
将式(11)、(14)带入(8)
$\rho u (\frac{\phi_{E}+\phi_{P}}{2}-\frac{\phi_{P}+\phi_{W}}{2})=\Gamma (\frac{\phi_{E}-\phi_{P}}{\Delta x}-\frac{\phi_{P}-\phi_{W}}{\Delta x})\tag{15}$ρu(2ϕE​+ϕP​​−2ϕP​+ϕW​​)=Γ(ΔxϕE​−ϕP​​−ΔxϕP​−ϕW​​)(15)
整理得
$4\Gamma \phi_{P}=(2\Gamma-\rho u \Delta x)\phi_{E}+(2\Gamma+\rho u \Delta x)\phi_{W}\tag{16}$4ΓϕP​=(2Γ−ρuΔx)ϕE​+(2Γ+ρuΔx)ϕW​(16)

3. Gauss-Seidel迭代法

原问题的代数方程
$4\Gamma \phi_{i}=(\rho u \delta x+2\Gamma)\phi_{i-1}+(2\Gamma-\rho u \delta x)\phi_{i+1}$4Γϕi​=(ρuδx+2Γ)ϕi−1​+(2Γ−ρuδx)ϕi+1​
令$\alpha = \frac{\rho u}{\Gamma}$α=Γρu​则
$4 \phi_{i}=(\alpha\delta x+2)\phi_{i-1}+(2-\alpha \delta x)\phi_{i+1}\tag{17}$4ϕi​=(αδx+2)ϕi−1​+(2−αδx)ϕi+1​(17)
假设有$N$N个格点，则该方程得Guass-Seidel迭代格式
$\left\{ \begin{aligned} \phi_{1}^{k+1} & = &\frac{1}{4}[(\alpha\delta x+2)\phi_{0}+(2-\alpha \delta x)\phi_{2}^{k}] \\ \phi_{2}^{k+1} & = & \frac{1}{4}[(\alpha\delta x+2)\phi_{1}^{k+1}+(2-\alpha \delta x)\phi_{3}^{k}] \\ \vdots\\ \phi_{N}^{k+1} & = & \frac{1}{4}[(\alpha\delta x+2)\phi_{N-1}^{k+1}+(2-\alpha \delta x)\phi_{N+1}] \end{aligned} \right.$⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​ϕ1k+1​ϕ2k+1​⋮ϕNk+1​​===​41​[(αδx+2)ϕ0​+(2−αδx)ϕ2k​]41​[(αδx+2)ϕ1k+1​+(2−αδx)ϕ3k​]41​[(αδx+2)ϕN−1k+1​+(2−αδx)ϕN+1​]​
写为矩阵形式：
$\begin{pmatrix} -4 & 2-\alpha \delta x & {} & {} & {} \\ 2+\alpha \delta x & {} & {} & {} & {} \\ {} & {} & \ddots & {} & {} \\ {} & {} & {} & {} & 2-\alpha \delta x \\ {} & {} & {} & 2+\alpha \delta x & -4 \\ \end{pmatrix} \begin{pmatrix} {{\phi }_{1}} \\ {{\phi }_{2}} \\ {{\phi }_{3}} \\ \vdots \\ {{\phi }_{N}} \\ \end{pmatrix} \doteq \begin{pmatrix} -(2+\alpha \delta x){{\phi }_{0}} \\ 0 \\ 0 \\ \vdots \\ -(2-\alpha \delta x){{\phi }_{N+1}} \\ \end{pmatrix}\tag{18}$⎝⎜⎜⎜⎜⎛​−42+αδx​2−αδx​⋱​2+αδx​2−αδx−4​⎠⎟⎟⎟⎟⎞​⎝⎜⎜⎜⎜⎜⎛​ϕ1​ϕ2​ϕ3​⋮ϕN​​⎠⎟⎟⎟⎟⎟⎞​≐⎝⎜⎜⎜⎜⎜⎛​−(2+αδx)ϕ0​00⋮−(2−αδx)ϕN+1​​⎠⎟⎟⎟⎟⎟⎞​(18)
解得方程在$\alpha=(-5,-1,0,1,5)$α=(−5,−1,0,1,5)时的曲线

4. TDMA方法

对于任意一个三对角阵
$A_{i}T_{i}=B_{i}T_{i+1}+C_{i}T_{i-1}+D_{i}\tag{19}$Ai​Ti​=Bi​Ti+1​+Ci​Ti−1​+Di​(19)
我们都可以将其转化至
$T_{i-1}=P_{i-1}T_{i}+Q_{i-1}\tag{20}$Ti−1​=Pi−1​Ti​+Qi−1​(20)
联立(19)、(20)可得
$T_{i}-C_{i}P_{i}T_{i}=B_{i}T_{i+1}+D_{i}+C_{i}Q_{i-1}\tag{21}$Ti​−Ci​Pi​Ti​=Bi​Ti+1​+Di​+Ci​Qi−1​(21)
整理得
$T_{i}=\frac{B_{i}}{A_{i}-C_{i}P_{i-1}}T_{i+1}+\frac{D_{i}+C_{i}Q_{i-1}}{A_{i}-C_{i}P_{i-1}}\tag{22}$Ti​=Ai​−Ci​Pi−1​Bi​​Ti+1​+Ai​−Ci​Pi−1​Di​+Ci​Qi−1​​(22)
观察(20)、(22)式，可以发现
$P_{i}=\frac{B_{i}}{A_{i}-C_{i}P_{i-1}}; Q_{i}=\frac{D_{i}+C_{i}Q_{i-1}}{A_{i}-C_{i}P_{i-1}}\tag{23}$Pi​=Ai​−Ci​Pi−1​Bi​​;Qi​=Ai​−Ci​Pi−1​Di​+Ci​Qi−1​​(23)
结合代数形式(16)及矩阵形式(18)、(19)与(23)可以得到有$N$N个格点的各个系数向量
$\begin{aligned} A & = & (4~~~4~~~4~~~4~~~4~~~4~~~4~~~4~~~4~~~~~\cdots~~~~4)^{T} \\ B & = & (2-\alpha \Delta x~~2-\alpha \Delta x~~2-\alpha \Delta x~~\cdots~~0)^{T} \\ C & = & (0~~2+\alpha \Delta x~~2-\alpha \Delta x~~2-\alpha \Delta x~~\cdots)^{T} \\ D & = & ((2+\Delta x)\phi_{0}~~0~~0~~\cdots~~(2-\Delta x)\phi_{N+1})^{T} \end{aligned}$ABCD​====​(4   4   4   4   4   4   4   4   4     ⋯    4)T(2−αΔx  2−αΔx  2−αΔx  ⋯  0)T(0  2+αΔx  2−αΔx  2−αΔx  ⋯)T((2+Δx)ϕ0​  0  0  ⋯  (2−Δx)ϕN+1​)T​

解得方程在$\alpha=(-5,-1,0,1,5)$α=(−5,−1,0,1,5)时的曲线

5. 网格独立化考核

网格数量对数值计算结果有着重要的影响。当网格足够细密以至于再进一步加密网格已对数值计算结果基本上没有影响时所得到的数值解称为网格独立解。

为了避免浪费计算资源，在精度足够高的同时，减少网格数，这里使用TDMA方法计算在不同网格划分情况下$x$x=0.7处的值，进行网格独立化考核。

显然，当格点数大于80时，再增加格点数时$\phi$ϕ(0.7)的值其变化已经不明显了，故我们认为格点数达到80时，此时的解为网格独立解。