Superelevation of the Water Surface in a Channel Bend

Water forced round a bend does not turn by itself. Its own inertia pushes it toward the outside bank, and the surface tilts across the channel until the slope of the water balances that push. The tilt is the answer here: $\Delta d$ is how much higher the water stands at the outer bank than at the inner bank. The ratio $V^{2}/R_c$ measures how sharply the flow has to be turned, dividing by $g$ turns it into a slope, and multiplying by the surface width $T$ gives the rise across the whole channel.

Use the radius to the centreline for $R_c$, and give $T$ and $R_c$ in the same length unit. The outer bank sits half of $\Delta d$ above the water level at the centreline and the inner bank half of it below, so a designer adds half the difference to the normal depth to find the depth at the outside of the bend, and sets freeboard and lining height above that. The same effect puts a lining in a bend under more shear than one in a straight reach. Gravity is standard, 9.80665 m/s² (32.174 ft/s²); the manuals round it to 9.81 and 32.2. This is the water surface in a curved channel, not the banking of a road.

The relation is published in the Federal Highway Administration's urban drainage manual (HEC-22, fourth edition, equation 6.6) and in its companion Introduction to Highway Hydraulics (HDS 4, equation 4.33), which writes the surface width as the outer radius less the inner radius. It holds for subcritical flow only, so check that the Froude number is below 1: in supercritical flow a bend raises standing waves and oblique jumps that this relation does not describe. HDS 4 adds that the equation assumes a uniform velocity and uniform curvature, so it gives a value somewhat lower than a natural channel shows, generally within 10 percent, and that for a lined canal with strong curvature and large velocities its more detailed equation 4.34 should be used instead.

HDS 4 works it as Example Problem 4.7. With $V = 3.4$ m/s, an outer radius of 12 m, a centreline radius of 10 m and an inner radius of 8 m, the width is $T = 12 - 8 = 4$ m. The page gives $11.56 \times 4 / (9.80665 \times 10) = 0.47152$ m, which agrees with the manual's printed 0.47 m to 0.01 m. In English units, with $V = 11.15$ ft/s, $T = 13.12$ ft and $R_c = 32.81$ ft, the manual prints 1.54 ft.