A - Steel

B - Insulation

$K_A = 55 \times 10^{-3} \space \space w/mm^{\circ} c $

$K_B = 0.1 \times 10^{-3} \space \space w/mm^{\circ} c $

(i) To find nodes and elements

(ii) Element matrix equation

$$ \frac{KA}{h_e} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{Bmatrix} \theta_1 \\ \theta_2 \end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2 \end{Bmatrix} $$

For element 1

K = 55 x 10$^{-3}$ W/mm$^\circ$ C; A = 1mm$^2$; h$_e$ = 2mm

$ \frac{KA}{h_e} = \frac{55 \times 10^{-3} \times 1}{2} = 27.5 \times 10^{-3} $

$$ \therefore 27.5 \times 10^{-3} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{Bmatrix} \theta_1 \\ \theta_2 \end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2 \end{Bmatrix} $$

$$ 10^{-3} \begin{bmatrix} 27.5 & -27.5 \\ -27.5 & 27.5 \end{bmatrix} \begin{Bmatrix} \theta_1 \\ \theta_2 \end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2 \end{Bmatrix} $$

For element 2

K = 0.1 x 10$^{-3}$ W/mm$^\circ$ C; A = 1mm$^2$; h$_e$ = 4mm

$ \frac{KA}{h_e} = \frac{0.1 \times 10^{-3} \times 1}{4} = 0.025 \times 10^{-3} $

$$ 10^{-3} \begin{bmatrix} 0.025 & -0.025 \\ -0.025 & 0.025 \end{bmatrix} \begin{Bmatrix} \theta_2 \\ \theta_3 \end{Bmatrix} = \begin{Bmatrix} Q_2 \\ Q_3 \end{Bmatrix} $$

For element 3

K = 55 x 10$^{-3}$ W/mm$^\circ$ C; A = 1mm$^2$; h$_e$ = 2mm

$ \frac{KA}{h_e} = \frac{55 \times 10^{-3} \times 1}{2} = 27.5 \times 10^{-3} $

$$ 10^{-3} \begin{bmatrix} 27.5 & -27.5 \\ -27.5 & 27.5 \end{bmatrix} \begin{Bmatrix} \theta_3 \\ \theta_4 \end{Bmatrix} = \begin{Bmatrix} Q_3 \\ Q_4 \end{Bmatrix} $$

(iii) Global matrix equation

$$ 10^{-3} \begin{bmatrix} 27.5 & -27.5 & 0 & 0 \\ -27.5 & 27.525 & -0.025 & 0 \\ 0 & -0.025 & 27.525 & 27.5 \\ 0 & 0 & -27.5 & 27.5 \end{bmatrix} \begin{Bmatrix} \theta_1 \\ \theta_2 \\ \theta_3 \\ \theta_4 \end{Bmatrix} = \begin{Bmatrix} Q_1 \\ Q_2 \\ Q_3 \\ Q_4 \end{Bmatrix} $$

(iv) Imposing boundary condition

$ \theta_1 = 0^\circ C \hspace{0.5cm} \theta_4 = 20^\circ C \\ Q_2 = Q_3 = 0 \,\, (For \,\, balancing) $

$$ 10^{-3} \begin{bmatrix} 27.5 & -27.5 & 0 & 0 \\ -27.5 & 27.525 & -0.025 & 0 \\ 0 & -0.025 & 27.525 & 27.5 \\ 0 & 0 & -27.5 & 27.5 \end{bmatrix} \begin{Bmatrix} 0 \\ \theta_2 \\ \theta_3 \\ 20 \end{Bmatrix} = \begin{Bmatrix} Q_1 \\ 0 \\ 0 \\ Q_4 \end{Bmatrix} $$

Frame the equation,

$ 10^{-3}(-27.5 \theta_2) = Q_1 \hspace{0.25cm} ...(1) \\ 10^{-3}(27.525 \theta_2 - 0.025 \theta_3) = 0 \hspace{0.25cm} ...(2) \\ 10^{-3}(-0.025 \theta_2 + 27.525 \theta_3 - 27.5 \times 20) = 0 \hspace{0.25cm} ...(3) \\ 10^{-3}(-27.5 \theta_3 + 27.5 \times 20) = Q_4 \hspace{0.25cm} ...(4) $

By solving (2) and (3)

$ \theta_2 = 0.0182^\circ C \\ \theta_3 = 19.982^\circ C $

Substituting above values in equation (1) and (4), we get

Q$_1$ = -0.5005 x 10$^{-3}$ watt

Q$_4$ = 0.495 x 10$^{-4}$ watt

Now,

$ \sum Q = Q_1 + Q_2 + Q_3 + Q_4 \\ = -0.5005 \times 10^{-3} + 0 + 0 + 0.495 \times 10^{-3} \\ = -0.010 \times 10^{-3} $

The answer is equivalent to 0. Hence verified.