Element matrix equation,
$\begin{bmatrix}K\end{bmatrix}^e=\frac{1}{R_e}\begin{bmatrix} 1&-1\\ -1&1 \end{bmatrix},\hspace{3cm}$ where, $R_e=\frac{128\mu h_e}{\pi d_e^4}$

$For\,\, element \,\,1,\,\,d=15mm \,\,,\,\, h_e=7.5m$
$R_e=\frac{128*8*10^{-4}*7.5}{\pi\,(0.015)^4}=4828878.87$
$\begin{bmatrix} K \end{bmatrix}^1=10^{-7}\begin{bmatrix} 2.07&-2.07\\ -2.07&2.07 \end{bmatrix}$

$For\,\, element \,\,2,\,\,d=20mm \,\,,\,\, h_e=10m$
$R_e=\frac{128*8*10^{-4}*10}{\pi\,(0.02)^4}=2037183.27$
$\begin{bmatrix} K \end{bmatrix}^2=10^{-7}\begin{bmatrix} 4.91&-4.91\\ -4.91&4.91 \end{bmatrix}$

$For\,\, element \,\,3,\,\,d=12.5mm \,\,,\,\, h_e=10m$
$R_e=\frac{128*8*10^{-4}*10}{\pi\,(0.0125)^4}=13350884.29$
$\begin{bmatrix} K \end{bmatrix}^3=10^{-7}\begin{bmatrix} 0.75&-0.75\\ -0.75&0.75 \end{bmatrix}$

$For\,\, element \,\,4,\,\,d=7.5mm \,\,,\,\, h_e=7.5m$
$R_e=\frac{128*8*10^{-4}*7.5}{\pi\,(0.0075)^4}=24446199.26$
$\begin{bmatrix} K \end{bmatrix}^4=10^{-7}\begin{bmatrix} 0.41&-0.41\\ -0.41&0.41 \end{bmatrix}$

c) Element Stiffness matrix
$\begin{bmatrix} K \end{bmatrix}= 10^{-7}\begin{bmatrix} 2.07+4.91&-2.07&-4.91&0\\-2.07&2.07+0.75&0&-0.75\\-4.91&0&4.91+0.41&-0.41\\0&-0.75&-0.41&0.75+0.41 \end{bmatrix}$

d) Global Stiffness Matrix Equation,
$\hspace{2cm} \begin{bmatrix} K \end{bmatrix}\begin{Bmatrix} P \end{Bmatrix}^e=\begin{Bmatrix} Q \end{Bmatrix}$
$\therefore 10^{-7}\begin{bmatrix} 6.98&-2.07&-4.91&0\\-2.07&2.82&0&-0.75\\-4.91&0&5.32&-0.41\\0&-0.75&-0.41&1.16 \end{bmatrix}\begin{Bmatrix} P_1\\ P_2 \\ P_3 \\ P_4 \end{Bmatrix}=\begin{Bmatrix} Q_1 \\ Q_2\\Q_3\\ Q_4 \end{Bmatrix}$

e) Imposing boundary condition,
$Q_1=0.16*10^{-3}\,m^3/sec\,,\, \,P_4=0 \,\,,\,\,Q_3=0$
$Q_2=0$

f) Frame the equation,
$10^{-7}(6.98\,P_1-2.07\,P_2-4.91\,P_3)=0.16*10^{-3}\hspace{4cm}$ -(1)
$10^{-7}(-2.07\,P_1+2.82\,P_2)=0\hspace{7cm}$ -(2)
$10^{-7}(-4.91\,P_1+5.32\,P_3)=0\hspace{7.3cm}$ -(3)
$10^{-7}(-0.75\,P_2-0.41\,P_3)=Q_4\hspace{7cm}$ -(4)

Solving equation (2) and (3) we get,
$\therefore \,\,\,P_1=1722.4\,N/m^2\hspace{2cm}P_2=1264.32\,N/m^2\hspace{2cm}P_3=1589.662\,N/m^2$
Substituting in equation (4), we get,
$Q_4=-1600$

Flow rate in pipe,
$Q_1=\frac{P_1-P_2}{R_1}=\frac{1722.4-1264.32}{4828878.87}*1000$
$\therefore \, Q_1=0.09486\,lit/sec$

$Q_2=\frac{P_1-P_3}{R_2}=\frac{1722.4-1589.662}{2037183.27}*1000$
$\therefore \, Q_2=0.06516\,lit/sec$

$Q_3=\frac{P_2-P_4}{R_3}=\frac{1264.32-0}{13350884.29}*1000$
$\therefore \, Q_3=0.0947\,lit/sec$

$Q_4=\frac{P_3-P_4}{R_4}=\frac{1589.662-0}{24446199.26}*1000$
$\therefore \, Q_4=0.06502\,lit/sec$

Check,
$Q_1+Q_2 \approx 0.16002\,lit/sec$
$Q_3+Q_4 \approx 0.15972\,lit/sec$

Also,
$Q_1 \approx Q_3$
$Q_2 \approx Q_4 \hspace{5cm}$ - Hence Verified.