Hydraulic head and discharges from four grids

The subscripts in the data are partly unclear. They are read as the four neighbours of the central grid $(i,j)$: $h_{i+1,j}=50$ m, $h_{i-1,j}=75$ m (along $x$) and $h_{i,j+1}=95$ m, $h_{i,j-1}=10$ m (along $y$).

Step 1: Finite difference equation

For steady flow without withdrawal, $K_{xx}\dfrac{\partial^2h}{\partial x^2}+K_{yy}\dfrac{\partial^2h}{\partial y^2}=0$:

$$K_{xx}\frac{h_{i+1,j}-2h_{i,j}+h_{i-1,j}}{\Delta x^2}+K_{yy}\frac{h_{i,j+1}-2h_{i,j}+h_{i,j-1}}{\Delta y^2}=0$$ $$h_{i,j}=\frac{\dfrac{K_{xx}}{\Delta x^2}\left(h_{i+1,j}+h_{i-1,j}\right)+\dfrac{K_{yy}}{\Delta y^2}\left(h_{i,j+1}+h_{i,j-1}\right)}{2\left(\dfrac{K_{xx}}{\Delta x^2}+\dfrac{K_{yy}}{\Delta y^2}\right)}$$

Step 2: Coefficients

$$\frac{K_{xx}}{\Delta x^2}=\frac{0.95\times10^{-5}}{100^2}=9.500\times10^{-10},\qquad \frac{K_{yy}}{\Delta y^2}=\frac{0.99\times10^{-5}}{80^2}=1.547\times10^{-9}$$

Step 3: Hydraulic head at the central grid

$$h_{i,j}=\frac{9.500\times10^{-10}(50+75)+1.547\times10^{-9}(95+10)}{2(9.500\times10^{-10}+1.547\times10^{-9})}=\frac{2.812\times10^{-7}}{4.994\times10^{-9}}=56.30\ \text{m}$$

Step 4: Discharges into the central grid (per unit thickness)

Flow from a neighbour into the central grid (Darcy's law, per unit thickness): $Q=K\dfrac{h_{nb}-h_{i,j}}{\Delta}\times(\text{face width})$. The face width is $\Delta y=80$ m for the $x$-neighbours and $\Delta x=100$ m for the $y$-neighbours.

A negative value means flow out of the central grid towards that neighbour. Check: the four discharges add up to zero (continuity satisfied). Multiply by the saturated thickness to get m$^3$/s.

Answer: $h_{i,j}=56.30$ m. Inflows: from $(i-1,j)$ $1.421\times10^{-4}$, from $(i,j+1)$ $4.789\times10^{-4}$ m$^3$/s per m; outflows to $(i+1,j)$ $4.792\times10^{-5}$ and to $(i,j-1)$ $5.730\times10^{-4}$ m$^3$/s per m.