The nodal co-ordinate of the triangular element for ground water simulation is shown in Fig. The nodal values of hydraulic head $(\phi)$ at nodes are (3,5, 2.2, 4.4) respectively. Find the value of hydraulic head at point P.

Given,
$x_1,y_1=1,2$
$x_2,y_2=4,0.5$
$x_3,y_3=3,4$

Now,
$\alpha_1=x_2y_3-x_3y_2=4*4-3*0.5=14.5$
$\alpha_2=x_3y_1-x_1y_3=3*2-1*4=2$
$\alpha_3=x_1y_2-x_2y_1=1*0.5-4*2=-7.5$

$\beta_1=y_2-y_3=0.5-4=-3.5$
$\beta_2=y_3-y_1=4-2=2$
$\beta_3=y_1-y_2=2-0.5=0.5$

$\gamma_1=-(x_2-x_3)=-(4-3)=-1$
$\gamma_2=-(x_3-x_1)=-(3-1)=-2$
$\gamma_3=-(x_1-x_2)=-(1-4)=3$

$\therefore 2A= \begin{vmatrix} 1& x_1&&y_1\\1&x_2&y_2\\1&x_3\\y_3 \end{vmatrix}=\begin{vmatrix} 1&1&2\\1&4&0.5\\1&3&4 \end{vmatrix}$

$\hspace{1cm}=1(16-1.5)-1(4-0.5)+2(3-4)=9 $

$N_1=\frac{\alpha_1+\beta_1x+\gamma_1y}{2A}=\frac{14.5-3.5x-y}{9}$
$N_2=\frac{\alpha_2+\beta_2x+\gamma_2y}{2A}=\frac{2+2x-2y}{9}$
$N_3=\frac{\alpha_3+\beta_3x+\gamma_3y}{2A}=\frac{-7.5+1.5x+3y}{9}$

At, (2.5,2.5),
$N_1=\frac{3.25}{9}\hspace{2cm}N_2=\frac{2}{9}\hspace{2cm}N_3=\frac{3.75}{9}$

Now,
$N_1+N_2+N_3=\frac{9}{9}=1$

Now,
$\phi(2.5,2.5)=N_1\phi_1+N_2\phi_2+N_3\phi_3$

$\therefore \phi_1=3.5 \hspace{2cm}\phi_2=2.2\hspace{2cm}\phi_3=4.4 $

$\phi=\frac{3.25}{9}*3.5+\frac{2}{9}*2.2+\frac{3.75}{9}*4.4$

$\therefore \phi(2.5,2.5)=3.586$