---
title: "Flow lines cut equipotentials at right angles"
subject: "Soil Mechanics"
chapter: 7
key: "semester_4/soil_mechanics:prove-flow-equipotential-orthogonal"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/prove-flow-equipotential-orthogonal
site: BCT Engineering
---

# Flow lines cut equipotentials at right angles

Soil Mechanics · Chapter 7: Seepage through Soils

- Asked in: 2075 Bhadra, 4 marks; 2073 Magh, 5 marks; 2074 Bhadra, 6 marks (asked 3 times)
- Video: [Orthogonality of Stream Lines & Equi - Potential Lines  | Fluid Kinematics | FMM #engineering #2025](https://www.youtube.com/watch?v=IxnAyiQVLtI) — Impact Academy Official
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/prove-flow-equipotential-orthogonal
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-7

## Question

Prove that flow lines intersect the equipotential lines at right angles.

## Short answer

For isotropic soil, velocity potential φ = -k h, so equipotential line slope is -v_x/v_z and flow line slope is v_z/v_x. Their product is -1, proving they intersect at right angles.

## Answer

#### Statement

At every point of a flow net for steady 2D seepage, the flow line and the equipotential line passing through the point are perpendicular.

#### Proof

Let the head be $h(x,z)$. By Darcy's law (isotropic soil), the velocity components are

$$
v_x = -k\frac{\partial h}{\partial x}, \qquad v_z = -k\frac{\partial h}{\partial z}
$$

Define the velocity potential $\phi = -k\,h$ so $v_x = \dfrac{\partial\phi}{\partial x}$, $v_z = \dfrac{\partial\phi}{\partial z}$. Equipotential lines are $\phi = $ constant.

**Slope of the equipotential line:** along it $d\phi = 0$:

$$
\frac{\partial\phi}{\partial x}dx + \frac{\partial\phi}{\partial z}dz = 0 \;\Rightarrow\; v_x\,dx + v_z\,dz = 0 \;\Rightarrow\; \left(\frac{dz}{dx}\right)_{\phi} = -\frac{v_x}{v_z}
$$

**Slope of the flow line:** the velocity vector is tangent to the flow line, so

$$
\left(\frac{dz}{dx}\right)_{\psi} = \frac{v_z}{v_x}
$$

**Product of the slopes:**

$$
\left(-\frac{v_x}{v_z}\right) \times \left(\frac{v_z}{v_x}\right) = -1
$$

Since the product of the slopes is $-1$, the two lines are at right angles. $\blacksquare$

**Physical reasoning:** along an equipotential line there is no head difference, hence no driving gradient and no flow component along it. The velocity (flow direction) is therefore wholly normal to it.

## Check yourself

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

1. In the proof, what is the velocity potential defined as, so that v_x = dphi/dx and v_z = dphi/dz?
   A. phi = k h
   B. phi = -k h
   C. phi = h / k
   D. phi = -h / k
   Answer: B. Darcy's law gives v = -k dh/dx, so defining phi = -k h makes the velocity components equal the partial derivatives of phi. (Key point: Velocity potential)

2. Along an equipotential line d(phi) = 0. What slope dz/dx does this give?
   A. +v_z / v_x
   B. +v_x / v_z
   C. -v_x / v_z
   D. -v_z / v_x
   Answer: C. v_x dx + v_z dz = 0 gives dz/dx = -v_x / v_z. (Key point: Slope of equipotential)

3. Why is the slope of the flow line equal to v_z / v_x?
   A. The velocity vector is tangent to the flow line
   B. The velocity vector is normal to the flow line
   C. Head is constant along a flow line
   D. Darcy's law fixes the slope of every line to k
   Answer: A. A flow line is by definition tangent to the velocity vector, so its slope is the ratio of the velocity components. (Key point: Slope of flow line)

4. What does the product of the two slopes equal, and what does it show?
   A. +1, the lines are parallel
   B. 0, the lines are horizontal
   C. -1, the lines are perpendicular
   D. k, the lines are at the Darcy angle
   Answer: C. (-v_x/v_z)(v_z/v_x) = -1, which is the condition for perpendicular lines. (Key point: Product of slopes)

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