Darcy fluxes around a central grid

Darcy's law for a confined (transmissivity) aquifer gives the flux per unit width across the face between two grids as

$$q=T\,\frac{\phi_{\text{from}}-\phi_{\text{to}}}{\Delta}$$

where $\Delta$ is the grid spacing in the direction of flow. Here each flux is measured out of the central grid in the direction of its arrow (positive = leaving the central grid, negative = entering it). The face width is $\Delta Y$ for $x$-direction fluxes and $\Delta X$ for $y$-direction fluxes.

              j-1  (qB, up)
               ^
  i-1 (qC) <-- (i,j) --> (qA) i+1
               v
              j+1  (qD, down)

Fluxes ($T_x=0.013$, $T_y=0.015$ m$^2$/s; $\Delta X=20$ m, $\Delta Y=25$ m)

$$q_A=T_x\frac{\phi_{i,j}-\phi_{i+1,j}}{\Delta X}=0.013\,\frac{13-14}{20}=-6.500\times10^{-4}\ \text{m}^2/\text{s}$$ $$q_B=T_y\frac{\phi_{i,j}-\phi_{i,j-1}}{\Delta Y}=0.015\,\frac{13-13.5}{25}=-3.000\times10^{-4}\ \text{m}^2/\text{s}$$ $$q_C=T_x\frac{\phi_{i,j}-\phi_{i-1,j}}{\Delta X}=0.013\,\frac{13-12}{20}=6.500\times10^{-4}\ \text{m}^2/\text{s}$$ $$q_D=T_y\frac{\phi_{i,j}-\phi_{i,j+1}}{\Delta Y}=0.015\,\frac{13-11}{25}=1.200\times10^{-3}\ \text{m}^2/\text{s}$$

Flow through each face (flux $\times$ face width)

Answer: $q_A=-6.500\times10^{-4}$, $q_B=-3.000\times10^{-4}$, $q_C=6.500\times10^{-4}$, $q_D=1.200\times10^{-3}$ m$^2$/s. Negative signs mean water flows into the central grid from the A and B sides; water leaves through C and D.

Check: the net outflow is $\Sigma Q=0.018$ m$^3$/s $\ne0$, so the given potentials are not a steady-state set. For exact steady state the central potential would be $\phi_{i,j}=12.681$ m (from $\sum$ fluxes $=0$).