---
title: "Darcy fluxes around a central grid"
subject: "Computational Techniques in Civil Engineering"
chapter: 7
key: "semester_8/computational_techniques_in_civil_engineering:darcy-flux-central-grid"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/darcy-flux-central-grid
site: BCT Engineering
---

# Darcy fluxes around a central grid

Computational Techniques in Civil Engineering · Chapter 7: Simulation of Ground water flow

- Asked in: 2073 Magh, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/darcy-flux-central-grid
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-7

## Question

The figure below shows a central grid surrounded by four grids for simulating two dimensional groundwater flow under steady state condition. Values of potential function ($\phi$) are given below: $\phi_{i-1,j} = 12$, $\phi_{i+1,j} = 14$, $\phi_{i,j} = 13$, $\phi_{i,j-1} = 13.5$, $\phi_{i,j+1} = 11$. Transmissivity in X-direction $= 0.013$ m$^2$/s for all grids, transmissivity in Y-direction $= 0.015$ m$^2$/s for all grids. Taking $\Delta X = 20$ m and $\Delta Y = 25$ m, compute Darcy fluxes $q_A$, $q_B$, $q_C$ and $q_D$ from the finite difference equation in terms of $\phi$.

[Figure: central grid $(i, j)$ with $q_A$ to the right (towards $i+1$), $q_B$ upward (towards $j-1$), $q_C$ to the left (towards $i-1$) and $q_D$ downward (towards $j+1$); x to the right, y downward]

## Short answer

Darcy fluxes are qA = –6.500×10⁻⁴, qB = –3.000×10⁻⁴, qC = 6.500×10⁻⁴, qD = 1.200×10⁻³ m²/s. Negative signs indicate flow entering the central grid from the A and B directions.

## Answer

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.

```text
              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)

| Face | Direction | $q$ (m$^2$/s) | Face width (m) | $Q$ (m$^3$/s) |
|---|---|---|---|---|
| A | to $i+1$ | $-6.500\times10^{-4}$ | 25 | -0.01625 |
| B | to $j-1$ | $-3.000\times10^{-4}$ | 20 | -0.00600 |
| C | to $i-1$ | $6.500\times10^{-4}$ | 25 | 0.01625 |
| D | to $j+1$ | $1.200\times10^{-3}$ | 20 | 0.02400 |

**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 $=0$).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the flux out of the central grid in the x-direction computed?
   A. q = T (phi_i,j - phi_nb)/dX, using the x-direction transmissivity
   B. q = T (phi_nb - phi_i,j)/dY
   C. q = T (phi_i,j + phi_nb)/dX
   D. q = T dX (phi_i,j - phi_nb)
   Answer: A. Flux per unit width is T times head drop divided by spacing in the flow direction. (Key point: Darcy flux formula)

2. What is q_C = Tx (13 - 12)/20?
   A. -6.5e-4 m^2/s
   B. 1.2e-3 m^2/s
   C. 6.5e-4 m^2/s
   D. 0.05 m^2/s
   Answer: C. 0.013 x 1/20 = 6.5e-4 m^2/s, leaving the central grid towards i-1. (Key point: Value of q_C)

3. The net outflow of the grid is 0.018 m^3/s, not zero. What does this show?
   A. The fluxes were computed with wrong signs
   B. T is wrong
   C. The given potentials are not a steady-state set
   D. Flow is exactly steady
   Answer: C. For steady state the sum of flows must be zero; since it is not, the potentials are not steady. (Key point: Steady-state check)

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