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Uchanganuzi wa upitishaji joto usio steady

Sehemu hii inaonyesha time discretization na iterative solution ya uchanganuzi wa upitishaji joto katika solid kwa Finite Element Method. Kwa governing equation na boundary conditions katika continuum, rejelea Mlinganyo wa upitishaji joto.

Mlinganyo uliodiskretishwa (mwanzo)

Tukidiskretisha mlinganyo wa upitishaji joto (Mlinganyo wa upitishaji joto (gov_he_main)) kwa Galerkin method, tunapata

\[\begin{equation} K T + M \frac{\partial T}{\partial t} = F \label{eq:2.4.8} \end{equation}\]

ambapo,

\[\begin{equation} K = \int\left( k_x \frac{\partial N^T}{\partial x}\frac{\partial N}{\partial x} + k_y \frac{\partial N^T}{\partial y}\frac{\partial N}{\partial y} + k_z \frac{\partial N^T}{\partial z}\frac{\partial N}{\partial z} \right) dV + \int hc N^T N ds + \int hr N^T N ds \label{eq:2.4.9} \end{equation}\]
\[\begin{equation} M = \int \rho c N^T N dV \label{eq:2.4.10} \end{equation}\]
\[\begin{equation} F = \int Q N^T dV - \int q_s N^T dS + \int{hc} T c N^T dS + \int{hcTr} ({T+Tr}) ({T^2 + T r^2}) N^T dS \label{eq:2.4.11} \end{equation}\]
\[\begin{equation} N = (N^1, N^2, \ldots, Ni) \label{eq:2.4.12} \end{equation}\]

Hapa \(K\), \(M\), \(F\), na \(N\) ni thermal-conduction matrix (ikijumuisha convection na radiation boundary contributions), mass matrix, thermal-load vector, na shape-function matrix, mtawalia. Ufafanuzi wa alama za properties (\(\rho\), \(c\), \(k_x, k_y, k_z\), \(Q\), \(hc\), \(hr\), n.k.) unafuata Mlinganyo wa upitishaji joto.

Time discretization na iterative solution

Mlinganyo \(\eqref{eq:2.4.8}\) ni nonlinear na unsteady. Sasa tudiskretishe muda kwa backward Euler method. Ikiwa temperature katika \(t=t_0\) inajulikana, temperature katika \(t=t_0+\Delta t\) huhesabiwa kwa mlinganyo ufuatao.

\[\begin{equation} K_{t=t_0+\Delta t} T_{t=t_0+\Delta t} + M_{t=t_0+\Delta t} \frac{T_{t=t_0+\Delta t} - T_{t=t_0}}{\Delta t} = F_{t=t_0+\Delta t} \label{eq:2.4.13} \end{equation}\]

Kwa equation \(\eqref{eq:2.4.13}\), tuchukulie temperature vector \(T_{t=t_0+\Delta t}^{(i)}\) inayolitimiza kwa ukadiriaji. na tuiboreshe ili kupata solution sahihi zaidi \(T_{t=t_0+\Delta t}^{(i)+1}\).

Kwa hiyo, kwanza tunaandika temperature vector kama ifuatavyo.

\[\begin{equation} T_{t=t_0+\Delta t}= T_{t=t_0+\Delta t}^{(i)} + \Delta T_{t=t_0+\Delta t}^{(i)} \label{eq:2.4.14} \end{equation}\]

Bidhaa ya thermal-conduction matrix na temperature vector, mass matrix, na quantities nyingine hukadiriwa kama ifuatavyo.

\[\begin{equation} K_{t=t_0+\Delta t} T_{t=t_0+\Delta t} = K_{t=t_0+\Delta t}^{(i)} T_{t=t_0+\Delta t}^{(i)} + \frac{\partial \big(K_{t=t_0+\Delta t}^{(i)} T_{t=t_0+\Delta t}^{(i)}\big) } {\partial T_{t=t_0+\Delta t}^{(i)} } \Delta T_{t=t_0+\Delta t}^{(i)} \label{eq:2.4.15} \end{equation}\]
\[\begin{equation} M_{t=t_0+\Delta t} = M_{t=t_0+\Delta t}^{(i)} + \frac{\partial M_{t=t_0+\Delta t}^{(i)}}{\partial T_{t=t_0+\Delta t}^{(i)}} \Delta T_{t=t_0+\Delta t}^{(i)} \label{eq:2.4.16} \end{equation}\]

Tukibadilisha milinganyo \(\eqref{eq:2.4.14}\), \(\eqref{eq:2.4.15}\), na \(\eqref{eq:2.4.16}\) katika \(\eqref{eq:2.4.13}\) na kuacha terms za order ya pili na zaidi, tunapata

\[\begin{equation} \bigg(\frac{M_{t=t_0+\Delta t}^{(i)}}{\Delta t} + \frac {\partial M_{t=t_0+\Delta t}^{(i)} } { \partial T_{t=t_0+\Delta t}^{(i)} } \frac{T_{t=t_0+\Delta t}^{(i)} - T_{t=t_0}}{\Delta t} + \frac{\partial \big(K_{t=t_0+\Delta t}^{(i)} T_{t=t_0+\Delta t}^{(i)}\big)} {\partial T_{t=t_0+\Delta t}^{(i)}} \bigg) \Delta T_{t=t_0+\Delta t}^{(i)} \\\ = F_{t=t_0+\Delta t} - M_{t=t_0+\Delta t}^{(i)} \frac{T_{t=t_0+\Delta t}^{(i)} - T_{t=t_0}}{\Delta t} - K_{t=t_0+\Delta t}^{(i)} T_{t=t_0+\Delta t}^{(i)} \label{eq:2.4.17} \end{equation}\]

Zaidi ya hayo, coefficient matrix ya upande wa kushoto hukadiriwa kwa mlinganyo ufuatao.

\[\begin{equation} K^{(i)} = \frac{M_{t=t_0+\Delta t}^{(i)}}{\Delta t} + \frac{\partial \big( K_{t=t_0+\Delta t}^{(i)} T_{t=t_0+\Delta t}^{(i)} \big)}{\partial T^{(i)}_{t=t_0+\Delta t}} = \frac{M_{t=t_0+\Delta t}^{(i)}}{\Delta t} + K_{T_{t=t_0+\Delta t}}^{(i)} \label{eq:2.4.18} \end{equation}\]

Hapa \(K_{T_{t=t_0+\Delta t}}^{(i)}\) ni tangent matrix.

Hatimaye, temperature katika \(t=t_0+\Delta t\) inaweza kuhesabiwa kwa iterative calculation kwa kutumia mlinganyo ufuatao.

\[\begin{equation} K^{(i)} \Delta T_{t=t_0+\Delta t}^{(i)} = F_{t=t_0+\Delta t} - M_{t=t_0+\Delta t}^{(i)} \frac{T_{t=t_0+\Delta t}^{(i)} - T_{t=t_0}}{\Delta t} - K^{(i)} T_{t=t_0+\Delta t}^{(i)} \label{eq:2.4.19} \end{equation}\]

Hasa kwa steady-state analysis, iterative calculation hutumia mlinganyo ufuatao.

\[ K_T^{(i)} \Delta T_{t=\infty}^{(i)} = F_{t=\infty} - K_T^{(i)} \Delta T_{t=\infty}^{(i)} \]
\[\begin{equation} T_{t=\infty}^{(i+1)} = T_{t=\infty}^{(i)} + \Delta{T}_{t=\infty}^{(i)} \label{eq:2.4.20} \end{equation}\]

Katika unsteady analysis, kwa kuwa implicit method hutumika kwa time discretization, uchaguzi wa time increment \(\Delta t\) kwa kawaida hauna stability restriction ya ukubwa. Hata hivyo, ikiwa \(\Delta t\) ni kubwa mno, idadi ya iterations zinazohitajika kwa convergence huongezeka. Kwa ujumla, time increment \(\Delta t\) ikiwa kubwa mno, idadi ya iterations huongezeka. Katika utekelezaji, ukubwa wa residual vector hufuatiliwa; ikiwa convergence ni polepole \(\Delta t\) hupunguzwa, na ikiwa iterations ni chache \(\Delta t\) huongezwa kwa automatic increment control (kwa maelezo, rejelea Step control).

Vipengee vinavyohusiana

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