Flow lines cut equipotentials at right angles

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.