---
title: "Hydraulic head and discharges from four grids"
subject: "Computational Techniques in Civil Engineering"
chapter: 7
key: "semester_8/computational_techniques_in_civil_engineering:hydraulic-head-four-grids"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/hydraulic-head-four-grids
site: BCT Engineering
---

# Hydraulic head and discharges from four grids

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

- Asked in: 2078 Kartik, 5+1 marks
- Video: [Groundwater: hydraulic gradient in nested piezometers](https://www.youtube.com/watch?v=hGxpUj4qQb8) — Tom Gleeson
- 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/hydraulic-head-four-grids
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-7

## Question

The final values of hydraulic head at four grid points are: $h_{i+1}^{j} = 50$ m, $h_i^{j-1} = 10$ m, $h_i^{j+1} = 95$ m and $h_{i-1}^{j} = 75$ m [subscripts partly smudged in scan]. The hydraulic conductivities are: $k_{xx} = 0.95\times10^{-5}$ m/s and $k_{yy} = 0.99\times10^{-5}$ m/s. Assuming steady flow with no withdrawal, determine the hydraulic head at $h_i^j$ and the discharges in it from the other four grids. Take $\Delta x = 100$ m and $\Delta y = 80$ m.

## Short answer

The central hydraulic head is 56.30 m. Inflows per unit thickness are 1.421×10⁻⁴ m³/s from the left and 4.789×10⁻⁴ m³/s from above; outflows are 4.792×10⁻⁵ m³/s to the right and 5.730×10⁻⁴ m³/s downward.

## Answer

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.

| Neighbour | $h$ (m) | $h-h_{i,j}$ (m) | Darcy velocity (m/s) | Discharge in (m$^3$/s per m) |
|---|---|---|---|---|
| $(i+1,j)$ | 50 | -6.3048 | $-5.990\times10^{-7}$ | $-4.792\times10^{-5}$ |
| $(i-1,j)$ | 75 | 18.6952 | $1.776\times10^{-6}$ | $1.421\times10^{-4}$ |
| $(i,j+1)$ | 95 | 38.6952 | $4.789\times10^{-6}$ | $4.789\times10^{-4}$ |
| $(i,j-1)$ | 10 | -46.3048 | $-5.730\times10^{-6}$ | $-5.730\times10^{-4}$ |

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.

## Check yourself

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

1. Which expression gives the central head for steady flow with Kxx and Kyy and no withdrawal?
   A. Simple average of the four neighbours regardless of K and spacing
   B. Weighted average with weights Kxx/dx^2 for x-neighbours and Kyy/dy^2 for y-neighbours
   C. Sum of neighbours divided by 2
   D. Weighted with weights Kxx dx and Kyy dy
   Answer: B. Substituting central differences into Kxx h_xx + Kyy h_yy = 0 gives the weighted average using K/Delta^2. (Key point: Finite difference equation)

2. How do the four discharges at the central grid relate once h is found?
   A. Inflows are all positive
   B. They sum to zero because there is steady flow with no withdrawal
   C. Their sum equals the withdrawal rate
   D. They sum to the largest discharge
   Answer: B. Steady state without sources means net flow into the grid is zero. (Key point: Continuity check)

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