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Uchanganuzi wa upitishaji joto wa mpito

Mbinu ya udiskretishaji wa muda na utatuzi wa iteresheni kwa uchanganuzi wa upitishaji joto katika yabisi kwa kutumia mbinu ya elementi finiti inaelezwa hapa. Kwa milinganyo tawala na masharti ya mpaka katika kontinuamu, rejelea Mlinganyo wa upitishaji joto.

Mlinganyo uliodiskretishwa (sehemu ya kuanzia)

Mlinganyo wa upitishaji joto (mlinganyo wa upitishaji joto (gov_he_main)) ukidiskretishwa kwa mbinu ya Galerkin, hupatikana

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

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, kwa mpangilio, matriki ya upitishaji joto (ikijumuisha vipashio vya konveksheni na mionzi vya mpaka), matriki ya masi, vekta ya mzigo wa joto, na matriki ya funksi za umbo. Ufafanuzi wa alama za sifa za nyenzo (\(\rho\), \(c\), \(k_x, k_y, k_z\), \(Q\), \(hc\), \(hr\), n.k.) unafuata Mlinganyo wa upitishaji joto.

Udiskretishaji wa muda na utatuzi wa iteresheni

Mlinganyo \(\eqref{eq:2.4.8}\) ni mlinganyo usio wa mstari na usiotulia. Tudiskretishe muda kwa mbinu ya Euler ya nyuma. Ikiwa halijoto wakati wa \(t=t_0\) inajulikana, halijoto wakati wa \(t=t_0+\Delta t\) huhesabiwa kwa mlinganyo ufuatao.

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

Tuchukulie vekta ya halijoto \(T_{t=t_0+\Delta t}^{(i)}\) inayokidhi mlinganyo \(\eqref{eq:2.4.13}\) kwa takriban, kisha tuiboreshe ili kupata suluhisho lenye usahihi zaidi \(T_{t=t_0+\Delta t}^{(i)+1}\).

Kwanza, vekta ya halijoto huandikwa kama ifuatavyo.

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

Zao la matriki ya upitishaji joto na vekta ya halijoto, pamoja na matriki ya masi, hukadiriwa kama ifuatavyo.

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

Milinganyo \(\eqref{eq:2.4.14}\), \(\eqref{eq:2.4.15}\), na \(\eqref{eq:2.4.16}\) ikiingizwa katika \(\eqref{eq:2.4.13}\), na vipashio vya daraja la pili na zaidi vikipuuzwa, hupatikana

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

Zaidi ya hayo, matriki ya mgawo 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 matriki ya ugumu wa tanjenti.

Hatimaye, halijoto wakati wa \(t=t_0+\Delta t\) inaweza kuhesabiwa kwa kurudia hesabu 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 katika uchanganuzi wa hali thabiti, hesabu ya iteresheni hufanywa kwa 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 uchanganuzi usiotulia, udiskretishaji wa muda hutumia mbinu fiche, kwa hiyo uchaguzi wa ukubwa wa nyongeza ya muda \(\Delta t\) kwa kawaida hauwekewi kizuizi. Hata hivyo, \(\Delta t\) ikiwa kubwa mno, idadi ya iteresheni zinazohitajika kufikia konverjensi huongezeka. Kwa ujumla, nyongeza ya muda \(\Delta t\) ikiwa kubwa mno, idadi ya iteresheni huongezeka. Katika utekelezaji, ukubwa wa vekta ya masalio hufuatiliwa; konverjensi ikiwa polepole, \(\Delta t\) hupunguzwa, na idadi ya iteresheni ikiwa ndogo, \(\Delta t\) huongezwa kwa udhibiti wa nyongeza otomatiki (kwa maelezo, rejelea Udhibiti wa hatua).

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