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The governing differential equation for steady state one dimensional conduction heat transfer with internal heat generation is given

The governing differential equation for steady state one dimensional conduction heat transfer with internal heat generation is given by $\frac{d}{dx}[K \frac{dT}{dx}]=q \hspace{1cm} \text{for} \hspace{1cm} 0 \leq x \leq 1$ were develop the finite element termulation of linear element. use R-R method, mapped over general element.


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Given D.E. $ \frac{d}{dx}[K \frac{dT}{dx}] = q $ for 0 $\leq$ x $\leq$ i.e. $ \frac{d}{dx}[K \frac{dT}{dx}] - q = 0 $

In local co-ordinates, replace $dx$ by $d \bar{x}$

$ \frac{d}{d \bar{x}}[K \frac{dT}{d \bar{x}}] - q = 0 $

With local boundary conditions as,

$ K \frac{dT}{d \bar{x}}|_{\bar{x}=0} = - \frac{Q_1}{A} $

$ K \frac{dT}{d \bar{x}}|_{\bar{x}=he} = \frac{Q_2}{A} $

Since the solution is approximate, residue is given by,

$ \frac{d}{d \bar{x}}[K \frac{dT}{d \bar{x}}] - q = R $

Weighted integral form,

$ \int_0^1 w_iR \,\, dx = 0 $

$ \int_0^{he} w_i (\frac{d}{d \bar{x}}[K \frac{dT}{d \bar{x}}] - q ) \,\, d \bar{x} = 0 \\ \int_0^{he} w_i (\frac{d}{d \bar{x}}[K \frac{dT}{d \bar{x}}]) \,\, d \bar{x} - \int_0^{he}w_iq \,\,d \bar{x} = 0 \hspace{0.5cm} ....(1) $

Weak formulation

Consider equation, $ \int_0^{he} w_i (\frac{d}{d \bar{x}}[K \frac{dT}{d \bar{x}}]) \,\, d \bar{x} $

Integrating by parts, we get,

$ [w_i(K \frac{dT}{d \bar{x}})]_0^{he} - \int_0^{he} \frac{dw_i}{d \bar{x}}[w_i(K \frac{dT}{d \bar{x}})] \,\, d \bar{x} $

Substituting above value in eq(1), we get,

$ [w_i(K \frac{dT}{d \bar{x}})]_0^{he} - \int_0^{he} \frac{dw_i}{d \bar{x}}[w_i(K \frac{dT}{d \bar{x}})] \,\, d \bar{x} - \int_0^{he} w_iq \,\, d \bar{x} = 0 \hspace{0.5cm} ....(2) $

Let the approximate solution be,

$ T = T_1 \phi_1 + T_2 \phi_2 $

where $ \phi_1 = 1 - \frac{\bar{x}}{he}; \hspace{0.5cm} \phi_2 = \frac{\bar{x}}{he} $

Rayleigh-Ritz method, $ w_i = \phi_1 $

For i = 1, $ w_1 = 1 - \frac{\bar{x}}{he}; \hspace{0.5cm} \frac{dw_1}{d \bar{x}} = - \frac{1}{he} $

Substituting in eq(2), we get,

$ \frac{Q_1}{A} - K \int_0^{he} ( \frac{-1}{he}) (- \frac{T_1}{he} + \frac{T_2}{he}) \,\, d \bar{x} - \int_0^{he} (1 - \frac{\bar{x}}{he}).q \,\, d \bar{x} = 0 \\ \frac{Q_1}{A} = \frac{-K}{he}[\frac{-T_1 \bar{x}}{he} + \frac{T_2 \bar{x}}{he}]_0^{he} - q[\bar{x} - \frac{(\bar{x})^2}{2he}]_0^{he} \\ \frac{Q_1}{A} = \frac{-K}{he}[-T_1 + T_2] - q[\frac{he}{2}] $

or

$ \frac{Q_1}{A} = \frac{K}{he}[T_1 - T_2] - q \frac{he}{2} \\ \frac{K}{he} [T_1 - T_2] = \frac{Q_1}{A} + q \frac{he}{2} \hspace{0.5cm} ...(3) $

For i = 2, $ w_2 = \frac{\bar{x}}{he}, \hspace{0.5cm} \frac{dw_2}{d \bar{x}} = \frac{1}{he} $

Substituting in eq(2), we get,

$ \frac{Q_2}{A} - K \int_0^{he} \frac{1}{he}( \frac{-T_1}{he} + \frac{T_2}{he}) d \bar{x} - \int_0^{he} \frac{ \bar{x}}{he}q \,\, d \bar{x} \\ \frac{Q_2}{A} = \frac{K}{he} [ \frac{-T_1 \bar{x}}{he} + \frac{T_2 \bar{x}}{he}]_0^{he} - \frac{q}{he}[ \frac{ ( \bar{x})^2}{2}]_0^{he} \\ \frac{Q_2}{A} = \frac{K}{he}[-T_1 + T_2] - \frac{q(he)}{2} $

or

$ \frac{K}{he}[-T_1 + T_2] = \frac{Q_2}{A} +\frac{q(he)}{2} \hspace{0.5cm} ....(4) $

Putting eq (3) and (4) in matrix form, we get,

$ \frac{K}{he} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{Bmatrix} T_1 \\ T_2 \end{Bmatrix} = \begin{Bmatrix} \frac{Q_1}{A} \\ \frac{Q_2}{A} \end{Bmatrix} + \frac{q(he)}{2} \begin{bmatrix} 1 \\ 1 \end{bmatrix} $

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