Why does the water rise on the outside of a bend? #

When water flows around a horizontal bend, the water surface rises against the outer wall and falls at the inner wall. We call this superelevation. We can picture centrifugal force pushing the water outward; in the derivation, we will look at the inward pressure force needed to turn the flow.

The handbooks give different-looking equations for this rise. So which one is right? First we need to check what each equation measures: the full difference across the channel, or just the rise from the centreline to the outer wall.

In this article we will derive both quantities, compare the FHWA and USACE formulas, and then consider what we can calculate for a sump whose inlet and outlet meet at an angle.

The calculation needs a free surface. It can apply to an open channel or a partially full conduit. A gravity pipe can become full during a storm, so we have to check the flow condition at the time being considered. In a completely full pipe, curvature produces a pressure difference; there is no internal water surface to tilt.

What are we calculating? #

Let \eta denote water-surface elevation above a common datum. The subscripts i , c , and o mean the inner wall, centreline, and outer wall respectively. Then

\displaystyle \Delta h_{io}=\eta_o-\eta_i, \qquad \Delta h_o=\eta_o-\eta_c.

The first quantity is the full inner-to-outer difference; the second is the outer-wall rise above the centreline surface. For a narrow, symmetric channel, we will obtain \Delta h_o\approx\Delta h_{io}/2 . It is easy to miss that factor of two when you are rushing through a calculation.

Deriving the surface slope #

Geometry and assumptions #

We start with a smooth circular bend and a rectangular channel. In a cross-section through the bend, define:

Symbol	Meaning
R	Horizontal radius to the channel centreline
B	Water-surface width, equal to the clear width for this rectangular section
n	Distance across the channel, measured outward from the centreline
r=R+n	Radius of the streamline at that position
z	Vertical elevation above the datum
\eta(n)	Water-surface elevation at that position
u(n)	Speed along the curved streamline, treated as uniform over depth
p , \rho , g	Pressure, water density, and gravitational acceleration; g=9.81\,\mathrm{m/s^2}

The inner wall is at n=-B/2 and the outer wall at n=B/2 . Their radii are therefore r_i=R-B/2 and r_o=R+B/2 . We require R>B/2 so that the inner radius is positive. This bend radius is different from the hydraulic radius, which is flow area divided by wetted perimeter.

For the basic derivation, we assume steady flow of constant density, negligible vertical acceleration, and streamlines that follow concentric circles. We also neglect flow across the channel and the transverse effects of shear stress. These assumptions describe water that has settled into a turn. At an abrupt bend entrance, waves and cross-channel motion may prevent that balance from developing.

Pressure at a fixed elevation #

With negligible vertical acceleration, pressure increases with depth according to

\displaystyle \frac{\partial p}{\partial z}=-\rho g.

At the free surface, pressure is atmospheric, p_{atm} . Integrating down from the surface to elevation z gives

\displaystyle p(n,z)-p_{atm}=\int_{\eta(n)}^z-\rho g\,dz’=\rho g[\eta(n)-z].

Here z’ is just the integration variable. Now compare pressures across the channel at the same elevation. Since z and atmospheric pressure are fixed,

\displaystyle \frac{\partial p}{\partial n}=\rho g\frac{d\eta}{dn}.

So a higher water surface on the outside means a higher pressure there at the same elevation below the surface.

The force needed to turn the water #

Water moving at speed u around a circle of radius r accelerates inward at u^2/r . We defined the positive n direction outward, so the acceleration in that direction, a_n , is negative. The pressure force per unit mass is also directed from high pressure to low pressure:

\displaystyle a_n=-\frac{u(n)^2}{R+n}=-\frac{1}{\rho}\frac{\partial p}{\partial n}.

Substitute the pressure gradient from above and cancel the minus signs:

\displaystyle \frac{u(n)^2}{R+n}=g\frac{d\eta}{dn}.

Thus,

\displaystyle \boxed{\frac{d\eta}{dn}=\frac{u(n)^2}{g(R+n)}}.

The slope is positive: water level increases towards the outside. The higher outer pressure supplies the inward force that turns the water.

From surface slope to water-level difference #

Integrating across the width #

To find the full difference, integrate the slope from the inner wall to the outer wall:

\displaystyle \eta_o-\eta_i=\int_{-B/2}^{B/2}\frac{d\eta}{dn}\,dn=\int_{-B/2}^{B/2}\frac{u(n)^2}{g(R+n)}\,dn.

We cannot evaluate this without knowing how velocity changes across the channel. For the usual engineering approximation, we replace u(n) by the mean velocity V=Q/A , where Q is discharge and A is the wetted cross-sectional area. Then V can be taken outside the integral:

\displaystyle \Delta h_{io}=\frac{V^2}{g}\int_{-B/2}^{B/2}\frac{dn}{R+n}.

An antiderivative of 1/(R+n) is \ln[(R+n)/R] . Evaluating it at the two walls gives

\displaystyle \Delta h_{io}=\frac{V^2}{g}\left[\ln\left(\frac{R+B/2}{R}\right)-\ln\left(\frac{R-B/2}{R}\right)\right].

Using \ln a-\ln b=\ln(a/b) , we obtain

\displaystyle \boxed{\Delta h_{io}=\frac{V^2}{g}\ln\left(\frac{R+B/2}{R-B/2}\right)}.

Here \ln means the natural logarithm. This result is exact within our uniform-velocity, concentric-flow assumptions. It does not include velocity changes caused by the bend.

For the rise from the centreline to the outer wall, the lower integration limit is zero:

\displaystyle \Delta h_o=\frac{V^2}{g}\int_0^{B/2}\frac{dn}{R+n}=\frac{V^2}{g}\ln\left(1+\frac{B}{2R}\right).

Likewise, the fall from the centreline to the inner wall is

\displaystyle \eta_c-\eta_i=-\frac{V^2}{g}\ln\left(1-\frac{B}{2R}\right).

These two amounts are slightly different. The surface is curved, so the centreline elevation is not exactly halfway between the two wall elevations. Also, the slope equation gives only differences in elevation. We still need a known water level or a separate hydraulic calculation to place the whole surface above the datum.

The narrow-channel approximation #

If the width is small compared with the bend radius, R+n\approx R across the section. The slope becomes nearly constant:

\displaystyle \frac{d\eta}{dn}\approx\frac{V^2}{gR}.

Multiplying that slope by the full width, or by half the width, gives

\displaystyle \boxed{\Delta h_{io}\approx\frac{V^2B}{gR}}, \qquad \boxed{\Delta h_o\approx\frac{V^2B}{2gR}}.

That is where the factor of two comes from. We are measuring the same nearly straight surface over two different distances.

We can also see what the approximation omits. Put x=B/(2R) . Expanding the two logarithms,

\displaystyle \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots, \qquad \ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots.

Subtracting cancels the even powers:

\displaystyle \ln\left(\frac{1+x}{1-x}\right)=2x+\frac{2x^3}{3}+\cdots=\frac{B}{R}+\frac{1}{12}\left(\frac{B}{R}\right)^3+\cdots.

Keeping only B/R gives the compact full-width formula. For the outer half, \ln(1+x)\approx x gives the compact outer-wall formula.

A numerical example #

Take a rectangular channel with B=1\,\mathrm m , R=3\,\mathrm m , and V=2\,\mathrm{m/s} . The compact formulas give

\displaystyle \Delta h_{io}\approx\frac{2^2\times1}{9.81\times3}=0.136\,\mathrm m, \qquad \Delta h_o\approx0.068\,\mathrm m.

The logarithmic formulas give \Delta h_{io}=0.137\,\mathrm m and \Delta h_o=0.063\,\mathrm m . Notice that the full-width approximation is slightly low, while the outer-wall approximation is slightly high. An error quoted for one cannot be transferred to the other.

Where Froude number enters #

The hydraulic depth is D=A/B . The Froude number compares mean flow speed with the speed of a small shallow-water wave:

\displaystyle Fr=\frac{V}{\sqrt{gD}}.

Flow is subcritical when Fr<1 and supercritical when Fr>1 ; the older handbook terms are tranquil and rapid. At Fr=1 , flow is critical.

For a rectangular section of reference depth y , A=By and therefore D=y . Dividing the full-width formula by y gives

\displaystyle \frac{\Delta h_{io}}{y}\approx\frac{V^2}{gy}\frac{B}{R}=Fr^2\frac{B}{R}.

Thus, a higher Froude number or a sharper bend produces a larger rise relative to the flow depth. For a nonrectangular section, use D=A/B when calculating Fr .

Why do the handbook formulas differ? #

FHWA and USACE #

FHWA HEC-22, Section 6.1.5, gives the subcritical bend equation as

\displaystyle \Delta d=\frac{V^2T}{gR_c}.

Its T is our water-surface width B , R_c is our centreline radius R , and \Delta d is the full inner-to-outer difference. Half is assigned above the centreline and half below. See HEC-22, Fourth Edition.

USACE gives the outer-wall rise as

\displaystyle \Delta h_o=C\frac{V^2W}{gR},

where W is the width at the centreline water level and C depends on the flow and curve treatment. Its Table 2-4 gives:

Flow	Curve treatment	Rectangular C	Trapezoidal C
Subcritical	Simple circular	0.5	0.5
Supercritical	Simple circular	1.0	1.0
Supercritical	Spiral transitions	0.5	1.0
Supercritical	Spiral transitions and banked invert	0.5	Not tabulated

For subcritical flow, C=0.5 agrees with our half-width result. The other coefficients include standing-wave effects for the listed configurations. A spiral changes curvature gradually; a banked invert also tilts the channel bed across the bend. See USACE EM 1110-2-1601, Section 2-5 and Table 2-4.

So the factor-of-two difference between the subcritical formulas comes from the quantity being measured. Differences between flow regimes also involve waves and curve treatment; they cannot all be explained by halving the width.

How sharp can the bend be? #

The logarithmic formula requires R/B>0.5 . At R/B=0.5 , the inner radius becomes zero and the integral diverges. This is a mathematical restriction, not a design recommendation.

To check the narrow-channel approximation, let \lambda=R/B . After cancelling the common factor V^2/g , the exact full-width factor is L=\ln[(\lambda+0.5)/(\lambda-0.5)] and the approximate factor is a=1/\lambda . The percentage underestimate is 100(L-a)/L :

R/B	Compact full-width result below the logarithmic result
10	0.08%
5	0.33%
3	0.93%
2	2.12%
1	8.98%

Both results assume the same ideal flow. A small difference between them says nothing about the error caused by separation, circulation across and through the depth of the channel, or bend-entry waves.

USACE suggests R/B\ge3 for subcritical channel design to reduce circulation and velocity distortion. For rectangular rapid-flow curves with spiral transitions, it gives

\displaystyle R_{min}=\frac{4V^2B}{gy}, \qquad \frac{R_{min}}{B}=4\frac{V^2}{gy}=4Fr^2.

These recommendations apply to their respective channel and transition arrangements. They do not establish a universal validity limit for every bend or sump. See USACE EM 1110-2-1601, Section 2-5, Limiting curvature.

For direct use of the simple subcritical formula, we also need a smooth turn, approximately hydrostatic pressure, flow that follows the bend, and enough bend length for the transverse slope to develop. Strong separation, a plunging jet, or a rapidly changing flow can matter more than the value of R/B . The calculated rise also leaves the project’s freeboard requirement to be checked separately.

What can we calculate for a sump? #

A sump is usually an abrupt chamber. Its inlet and outlet lines tell us the change in direction, but they do not tell us the radius followed by the water. At their sharp intersection, ordinary smooth-curve curvature is undefined. Substituting a zero radius into the bend equation does not solve that problem.

However, the sump footprint gives us a length scale. We can fit a circular arc between the inlet and outlet directions and use its radius for a preliminary comparison. Our derivation does not establish that the water will follow that arc. We use the construction because it is available from the plan geometry and cheap to calculate.

We will call the fitted radius R_{ref} , the selected flow width B_{ref} , and the selected mean velocity V_{ref} . The subscript means these are the inputs to the sump reference calculation. It does not imply that a velocity field has been calculated inside the chamber.

Constructing the tangent arc #

In plan, let S be the usable sump footprint, excluding solid obstructions. Let P_i and P_o be the inlet and outlet connection midpoints. Draw the inlet axis in the direction of incoming flow and the outlet axis in the direction of outgoing flow. The change between those directions is the deflection angle \theta .

For 0<\theta<180^\circ , extend the axes to their intersection PI . For the construction below, PI must lie ahead of the inlet connection and behind the outlet connection in the respective flow directions. Let T_i and T_o be the available distances from PI to the two connections along those axes. In the diagram, C is the circle centre and T is the distance from PI to either tangent point.

To derive the tangent length, join the circle centre C to PI and to either tangent point. The radius is perpendicular to the tangent, so this makes a right triangle. The angle between the two rays from PI towards the connections is 180^\circ-\theta . The line from PI to C bisects it, giving

\displaystyle \tan\left(\frac{180^\circ-\theta}{2}\right)=\frac{R}{T}.

Using \tan(90^\circ-\theta/2)=1/\tan(\theta/2) ,

\displaystyle T=R\tan(\theta/2), \qquad R=\frac{T}{\tan(\theta/2)}.

If T_i=T_o , the arc can touch both axes at the connection midpoints. If they differ, the shorter distance limits the radius:

\displaystyle R_{candidate}=\frac{\min(T_i,T_o)}{\tan(\theta/2)}.

The longer side then has a straight segment between its connection and the arc. For a 90-degree turn, \tan45^\circ=1 , so the candidate radius equals the shorter tangent distance. For example, perpendicular axes meeting at the centre of a 400 mm square sump, with each connection midpoint 200 mm away, give R_{candidate}=200\,\mathrm{mm} .

We must still check that the arc and connecting straight segments lie inside S . Define R_{ref} as the largest positive radius at or below R_{candidate} that satisfies those checks. In an irregular footprint, making an arc smaller does not necessarily make it fit: an obstruction may exclude some radii and allow others. Search the permitted range and report no radius if none fits.

This checks the centreline path only. A fitted arc does not establish that a whole flow strip of width B_{ref} fits beside it. Keep that limitation when interpreting the result.

For collinear straight-through connections, there is no bend contribution in this idealisation. Parallel offset connections, a 180-degree return, or axes that fail the direction check need a different path construction. Do not force them through the tangent formula or insert a small radius just to obtain a number.

Choosing width and velocity #

Use a stated reference section on the selected inlet-to-outlet path. Obtain its discharge Q_{ref} , flow area A_{ref} , and water-surface width at the same time. Then

\displaystyle V_{ref}=\frac{|Q_{ref}|}{A_{ref}}, \qquad D_{ref}=\frac{A_{ref}}{B_{ref}}, \qquad Fr_{ref}=\frac{V_{ref}}{\sqrt{gD_{ref}}}.

For a rectangular channel, the top width is its clear width. For a trapezoid with bottom width b , side slope m horizontal to one vertical, and depth y , it is B_{ref}=b+2my . For a partly full circular pipe of diameter d_p , the horizontal water-surface chord is

\displaystyle B_{ref}=2\sqrt{y(d_p-y)}, \qquad 0<y<d_p.

The chord follows from a circle of radius d_p/2 : its half-width squared is (d_p/2)^2-(y-d_p/2)^2=y(d_p-y) . Thus the pipe diameter equals the top width only at half depth.

If only a geometric width is available, report which width was substituted. Calculate the reference section’s Froude number using its hydraulic area and top width, rather than silently substituting the chamber width into that calculation. If the inlet and outlet sections differ, identify which section supplies each estimate; separate section-based estimates are possible, but they are not bounds on the true sump rise.

The sump reference calculation #

For this preliminary method, we use the subcritical coefficient C_{ref}=0.5 :

\displaystyle \boxed{\Delta h_{ref}=\frac{V_{ref}^2B_{ref}}{2gR_{ref}}}.

Apply it only when the relevant flow has a free surface, the reference section is subcritical, and one turning path can be identified. A subcritical connected section alone does not prove that the chamber flow satisfies the bend assumptions. Withhold the estimate if a plunging jet, conflicting inflows, or an unresolved reversal makes the chosen path unsuitable.

Once those conditions are checked, use the radius ratio as follows:

Radius ratio	Treatment
R_{ref}/B_{ref}\le0.5	No estimate: the assumed inner radius is nonpositive.
0.5<R_{ref}/B_{ref}<3	Return the preliminary rise with a warning that the fitted turn is sharper than the USACE subcritical channel recommendation.
R_{ref}/B_{ref}\ge3	Return the preliminary rise. The ratio alone does not verify flow inside the chamber.

This is a proposed sump-screening convention. The handbook coefficient was derived for a smooth channel; it has not been calibrated here for a sump. Selecting the largest radius also gives the smallest rise among the permitted arcs when width and velocity are held fixed. We therefore cannot use a low result to establish that overtopping will not occur.

Comparing the reference level with the rim #

A one-dimensional network model reports one hydraulic grade line (HGL) elevation at a node. Denote it by H_{node} . To form an outside-level estimate, we have to make one further assumption: use that node HGL as the reference centreline water level. With both elevations on the same datum,

\displaystyle Z_{outside,ref}(t)=H_{node}(t)+\Delta h_{ref}(t),

where t is time. A single node level cannot tell us the water-level distribution within the sump, so this addition remains part of the reference method.

Calculate all inputs from the same timestep, then find the highest reference outside level among the times for which the method applies. Do not add the highest HGL from one time to a rise calculated from the highest velocity at another time. If the flow path changes, reassess its radius and reference section too.

Compare the result with the rim elevation, allowing for the required freeboard. Also report periods when the method could not be applied: the highest valid estimate may miss the governing condition during surcharge or rapid flow. If the estimate approaches the containment level, or the design decision depends on a reliable local maximum, investigate the chamber flow in more detail.

How does a 2D model calculate the rise? #

A full shallow-water model solves for depth and horizontal velocity throughout the channel or sump. Here, “full” means that the momentum equations retain acceleration as water changes speed or direction. A diffusion-wave model omits those acceleration terms. HEC-RAS specifically identifies bend superelevation as a reason to use its full shallow-water equations. See its equation-selection guidance.

We can connect those equations to our derivation without introducing the whole numerical scheme. Let h be depth, z_b bed elevation, and \eta=z_b+h water-surface elevation. Let \mathbf u be the depth-averaged horizontal velocity vector, and \mathbf f the acceleration due to friction and any modelled stresses. In wet regions, the momentum equation can be written as

\displaystyle \frac{\partial\mathbf u}{\partial t}+(\mathbf u\cdot\nabla)\mathbf u=-g\nabla\eta+\mathbf f.

The first term is the change in velocity with time at a fixed location. The second is the change experienced by water as it travels through a velocity field; this includes turning. The symbol \nabla denotes horizontal spatial derivatives, so \nabla\eta is the water-surface slope vector.

In a steady circular turn, project the equation onto our outward direction. Neglecting the transverse stress contribution gives

\displaystyle -\frac{u^2}{r}=-g\frac{d\eta}{dn}, \qquad \frac{d\eta}{dn}=\frac{u^2}{gr}.

This is the same balance we derived earlier. In the numerical model, wall boundary conditions and pressure forces turn the flow. Water and momentum pass between computational cells, and the computed depths develop a transverse slope. The mesh must resolve that variation; its cell directions do not themselves exert a turning force.

For a useful result, refine the mesh across the turn and reduce the timestep until the reported inner and outer levels change by less than the project’s tolerance. Use the actual inflow and downstream conditions, and compare levels at stated locations. A simulation that completes without error has only shown that it ran.

Depth averaging still assumes hydrostatic pressure and removes the vertical velocity structure. For plunging jets, splash, air entrainment, or nonhydrostatic wall impact, consider a validated three-dimensional free-surface model or a physical model. Check the model against measurements where available, as well as conservation and sensitivity to mesh, timestep, and boundary conditions.

What if velocity varies across the bend? #

The earlier integral allows other velocity distributions. Two ideal examples show how the surface shape changes: solid-body rotation, where the whole cross-section turns together like a rigid disk, and the free vortex, where water trades speed for radius the way it does spiralling into a drain. They are examples of the mathematics; choosing either for a real bend requires evidence about its velocity distribution.

In plan, inside the actual curved conduit, the two assumptions look like this. The arrow lengths are not illustrative: they are drawn to scale from the R=3\,\mathrm m , B=1\,\mathrm m , V=2\,\mathrm{m/s} example used earlier, with \omega=V/R=0.667\,\mathrm{s^{-1}} and K_v=VR=6\,\mathrm{m^2/s} chosen so both idealisations pass through V at the centreline.

The arrows are velocity, not water level, drawn at the same scale in both panels so their lengths can be compared directly: longer means faster. Solid-body rotation carries the slowest water, 1.67 m/s, against the inner wall and the fastest, 2.33 m/s, against the outer wall, like a rigid disk turning about a point outside the channel. The free vortex reverses that pattern, 2.40 m/s at the inner wall down to 1.71 m/s at the outer wall, the way water speeds up as it draws in toward the centre of a drain. Both are constructed to equal V=2\,\mathrm{m/s} at the centreline, so the two panels only disagree about how speed varies away from it. Each panel shares the same inner wall, centreline and outer wall as the cross-section further up this article; only the assumed velocity pattern across the width changes.

Because r=R+n , we have dr=dn and

\displaystyle \frac{d\eta}{dr}=\frac{u(r)^2}{gr}.

For solid-body rotation, all water rotates at the same angular speed \omega , so u(r)=\omega r : water near the inside moves slowest, water near the outside moves fastest, and neighbouring streamlines do not slide past one another. Let C_0 be the integration constant fixing the vertical position of the surface. Substitution and integration give

\displaystyle \frac{d\eta}{dr}=\frac{\omega^2r}{g}, \qquad \eta(r)=C_0+\frac{\omega^2}{g}\int r\,dr=C_0+\frac{\omega^2r^2}{2g}.

The surface is parabolic. Evaluating it at the two walls and subtracting cancels the constant C_0 :

\displaystyle \eta_o-\eta_i=\left(C_0+\frac{\omega^2r_o^2}{2g}\right)-\left(C_0+\frac{\omega^2r_i^2}{2g}\right),

\displaystyle \Delta h_{io}=\frac{\omega^2}{2g}(r_o^2-r_i^2).

For an ideal free vortex, picture water spiralling into a drain instead: fast and tight near the centre, slow and wide further out. The product of speed and radius is constant: u(r)r=K_v . Thus u(r)=K_v/r , with K_v in \mathrm{m^2/s} , and

\displaystyle \frac{d\eta}{dr}=\frac{K_v^2}{gr^3}, \qquad \eta(r)=C_0+\frac{K_v^2}{g}\int r^{-3}\,dr=C_0-\frac{K_v^2}{2gr^2}.

Subtracting the inner level from the outer again cancels C_0 :

\displaystyle \eta_o-\eta_i=\left(C_0-\frac{K_v^2}{2gr_o^2}\right)-\left(C_0-\frac{K_v^2}{2gr_i^2}\right),

\displaystyle \Delta h_{io}=\frac{K_v^2}{2g}\left(\frac{1}{r_i^2}-\frac{1}{r_o^2}\right).

The same R=3\,\mathrm m , B=1\,\mathrm m , V=2\,\mathrm{m/s} example, plotted exactly rather than sketched, shows both the assumed velocity and the resulting surface it produces:

The left panel is drawn to scale from the numbers above: solid-body rotation’s velocity line is exactly straight because u=\omega r is linear in r , and the free vortex curve is the actual u=K_v/r hyperbola, not a sketch of one. The right panel plots the exact \eta(r)-\eta(R) formulas for all three cases over the same width. Read together, they show why the three overall rises end up close (0.136 to 0.144 m for this example, all within about 6% of each other): calibrating every case to the same centreline velocity V keeps their totals similar, even though the shape across the width, flattest-to-steepest for solid-body, steepest-to-flattest for the free vortex, differs clearly. Neither vortex example supplies an upper or lower bound for an ordinary channel bend; they show how the total rise can stay roughly fixed while its distribution across the width changes.

A real bend’s velocity field rarely matches any of the three exactly. Flow entering a sharp turn behaves more like the free vortex close to the inside wall, before secondary currents and boundary friction redistribute momentum across the section; well into a long, gradual bend, the uniform-velocity assumption used throughout this article is usually closer to the measured result. Where a design depends on the fine shape of the surface rather than the depth-averaged rise, the two-dimensional model described above, not this comparison, is the tool to use.

Conclusion #

In this article we explained the physical picture behind superelevation, derived the surface-slope equation from first principles and worked it through for the ordinary channel bend, a sump, and the two-dimensional shallow-water case, and connected the resulting terms back to how FHWA HEC-22 and USACE report them.