User:Egm6321.f10.team2.oztekin/Newwwwwwww

=Problem 6.10:Heat problem with new boundary condition=

Given
Governing equation was derived in meeting 32-4.
 * $$\displaystyle Q_{1}$$:Heat flow into w throgh $$\displaystyle \partial w$$
 * $$\displaystyle Q_{2}$$:Heat generated in w by heat source.
 * $$\displaystyle Q_{3}$$:Heat in w due to change in term u.


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$$  \displaystyle Q_{1}=-\int_{\partial w}\mathbf{q}\mathbf{n}d(\partial w)=\int_{w}div\mathbf{q}dw $$     (10.1)
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$$  \displaystyle Q_{2}=\int_{w}f(\mathbf{x},t)dw $$     (10.2)
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$$  \displaystyle Q_{3}=\int_{w}\rho c\frac{\partial u}{\partial t}dw $$     (10.3)
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$$  \displaystyle Q_{1}+Q_{2}=Q_{3} $$     (10.4)
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$$  \displaystyle \mathbf{q}=-\mathbf{K}grad(u) $$     (10.5)
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$$  \displaystyle div(\mathbf{K}grad(u))+f=\rho c\frac{\partial u}{\partial t} $$ (10.6)
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Boundry Conditions

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$$  \displaystyle u(\mathbf{x},t)]_{\Gamma _{g}}=g(\mathbf{x},t) $$     (10.7)
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$$  \displaystyle (q.n)]_{\Gamma _{h}}=h(\mathbf{x},t) $$     (10.8)
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$$  \displaystyle (q.n)]_{\Gamma _{H}}=H(u-u_{\infty }) $$     (10.9)
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WRF

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$$  \displaystyle \int_{\Omega }w[div(Kgrad(u))+f-\rho c\frac{\partial u}{\partial t}]d\Omega =0 $$     (10.10)
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$$  \displaystyle\begin{align} K=  & \int_{\Omega }wdiv(Kgrad(u))d\Omega =\frac{\partial }{\partial x_{i}}(K_{ij}\frac{\partial u}{\partial x_{j}})\\ & \int_{\Omega }n_{i}wK_{ij}\frac{\partial u}{\partial x_{j}}d(\partial\Omega)-\int_{\Omega }\frac{\partial w}{\partial x_{i}}K_{ij}\frac{\partial u}{\partial x_{j}}d(\Omega) \end{align} $$     (10.11)
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$$  \displaystyle K=-\int_{\Gamma _{g}}wnqd\Gamma _{g}-\int_{\Gamma _{h}}wnqd\Gamma _{h}-\int_{\Gamma _{H}}wnqd\Gamma _{H}+\int_{\Omega }div(w).K.div(u)d\Omega $$     (10.12)
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We deliberately chose w i.e. i vanishes on essential boundry. So the first term vanishes too. In the first term we have unknown flux that we can not deal with right now. But other fluxes are defined.We can write our continuous form of components as:


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$$  \displaystyle \tilde{m}(w,u)=\int_{\Omega}w\rho c\frac{\partial u}{\partial t}d\Omega $$     (10.13)
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$$  \displaystyle \tilde{k} (w,u)=\int_{\Omega}div(w)Kdiv(u)d\Omega $$     (10.14)
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$$  \displaystyle \tilde{f}(w,u)=\int_{\Omega}wfd\Omega-\int_{\Gamma_{h}}whd\Gamma_{h}-\int_{\Gamma_{H}}wH(u-u_{\infty })d\Gamma_{H} $$     (10.15)
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!Note this time f depeneds on both w and u.

Discrete WF

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$$  \displaystyle w^{app}=\sum_{I=1}^{n}N(x,y)_{I}w_{I} $$     (10.16)
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$$  \displaystyle u^{app}=\sum_{J=1}^{n}N(x,y)_{J}d_{J} $$     (10.17)
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$$  \displaystyle div(u)=div(N(x,y)_{I})d_{I}=B(x,y)_{I}d_{I} $$     (10.18)
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$$  \displaystyle \tilde{m}(w,u)+\tilde{k}(w,u)=\tilde{f}(w,u) $$     (10.19) Now lets substitute everything in the equation above.
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$$  \displaystyle \sum_{I=1}^{n}w^{T}\left [\sum_{J=1}^{n}\left \{ \int_{\Omega }\rho cN_{I}N_{J}d_{I}^{s}d\Omega+\int_{\Omega}B_{I}KB_{J}d_{I}d\Omega -\int_{\Omega}N_{I}fd\Omega+\int_{\Gamma_{h}}N_{I}hd\Gamma_{h}+\int_{\Omega}N_{I}N_{J}(\Gamma_{H})H(d(\Gamma_{H})-d_{\infty })\right \} \right ]=0 $$     (10.20)
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$$  \displaystyle \tilde{\mathbf{d}}=\begin{bmatrix} \bar{de}\\ df
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\end{bmatrix} $$     (10.21)
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$$  \displaystyle \tilde{\mathbf{w}}=\begin{bmatrix} 0\\ wf
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\end{bmatrix} $$     (10.22)
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$$  \displaystyle \tilde{\mathbf{K}}=\begin{bmatrix} K_{E} &K_{EF} \\ K_{EF}^{T} & K_{F} \end{bmatrix} $$     (10.23)
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$$  \displaystyle\begin{align} & K_{E}=\int_{\Omega }B_{1}.K.B_{1}d\Omega&\\ & K_{EF}=\int_{\Omega }B_{1}.K.B_{J}d\Omega& j=2,n\\ & K_{FF}=\int_{\Omega }B_{I}.K.B_{J}d\Omega& i,j=2,n
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\end{align} $$     (10.24)
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$$  \displaystyle \tilde{\mathbf{M}}=\begin{bmatrix} M_{E} &M_{EF} \\ M_{EF}^{T} & M_{F} \end{bmatrix} $$     (10.25)
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$$  \displaystyle\begin{align} & M_{E}=\int_{\Omega }N_{1}\rho cN_{1}d\Omega&\\ & M_{EF}=\int_{\Omega }N_{1}\rho cN_{J}d\Omega& j=2,n\\ & M_{FF}=\int_{\Omega }N_{I}\rho cN_{J}d\Omega& i,j=2,n
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\end{align} $$     (10.26)
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