superelevation - NPTELnptel.ac.in/courses/105106114/pdfs/Unit36/36_4.pdf · Superelevation is defined as the difference in elevation of water surface between inside and outside wall

Hydraulics Prof. B.S. Thandaveswara

Indian Institute of Technology Madras

36.4 Superelevation Superelevation is defined as the difference in elevation of water surface between inside

and outside wall of the bend at the same section.

0 1∆y=y y− (1)

This is similar to the road banking in curves. The centrifugal force acting on the fluid

particles, will throw the particle away from the centre in radial direction, creating

centripetal lift.

Superelevation in other words means the greater depth near the concave bank than

near convex bank of a bend. This phenomenon was first observed by Ripley in 1872,

while he was surveying Red River in Loussiana for the removal of the great raft

obstructing the stream.

36.4.1 Transverse water surface slope in bends Gockinga was first to derive the following formula for determining the difference in

elevation of the water surface on opposite sides of channel bends.

2y=0 235V log 1-

rx. ⎛ ⎞

⎜ ⎟⎝ ⎠

in which V is the velocity in -1ms , x is the distance at which y is to be determined, r is

the radius of the river bend. But above equation is found to fit a particular stream to

which it was designed. Also he found that the transverse slope was twice greater than

the longitudinal slope. He showed that the increased depth in bend is caused by the

helicoidal flow induced by centrifugal force. Fargue while conducting studies on scour in

meandering devised the formula and called " Fargue's law of greatest depth " in 1908.

But unfortunately it was found to be applicable to the stream Garonne at Barsac only.

3 21C =0.03H -0.23H +0.78H-0.76

in which 1C is the reciprocal of radius of curvature in kilometer and H is the lowest water

depth at the deepest point of the channel in meter.

Mitchell also derived another equation applicable to Delawave river, Philadelphia.



2 2 2 2

2 21100 289 110033 2

r1100 1100x xy x

⎛ ⎞ ⎛ ⎞− −= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Ripley in 1926 arrived at the formulae based on the field observations, having their own

limitations.

2 2

1 12 24 17.52 4D 1 D 1

rT Tx xy x

⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2 21

1 2 226.28P4 4P 1 1

rT Tx xy x

⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

In the above equations, parabolic sections are assumed whose focal distances are 2

1

TD

and 2

1

TP

respectively and with the origin at a point on the axis at a distance of 1D and 1P

from apex respectively.

O

90Vmax

v

ydydr

W1

CENTRIFUGAL FORCE

WEIGHT OF FLUID

r

r i

O 0r

=

MECHANICAL MODEL OF THE HELICOIDAL FLOW AS PROPOSEDBY GRASHOF

The above two equations when combined yield a simplified form in FPS units. Figure

represents the general profile for equation given below.

2

20

17 526 35D 0 437 0 433 1rT

x . xy . . .⎛ ⎞⎛ ⎞⎜ ⎟= − − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠



r0

T/2T/2

origin

Y AXIS

0 0

1.445D

PROFILE FOR EQUATION

Grashof was the first to try an analytical solution for superelevation. He obtained

equation by applying Newton's second law of motion to every streamline and integrating

the equation of motion. Equation gives a logarithmic profile. Referring to figure given

below one may write

Centrifugal force = 2max1

c

VWg r

2max1

2c max

1 c

VWg r V

W grdydr

= =

Assuming the boundary conditions near inside wall of the bend and integrating above

equation reduces to the form

2max 0

i

V r∆y=2.3 logg r

in which ir and 0r are the inner and outer radii of the bend respectively.

Woodward in 1920 assumed the velocity to be zero at banks and to have a maximum

value at the centre of the bend and the velocity distribution varying in between



according to parabolic curve. Using Newton's second law of motion he obtained the

following equation for superelevation.

23 22 c c c cmax

c

r r r 2r + b20∆y = V -16 + 4 -1 ln3 b b b 2r - b

⎡ ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎢ ⎥⎣ ⎦

Shukry obtained the following equation for maximum superelevation based on free-

vortex flow and principle of specific energy.

( )2

2 2max 0 i2 2

0 i

C∆y = r - r2g r r

The Euler equation of motion.

( ) sρa 0p zs

γ∂+ + =

∂

( )p + γz = γh Since

( ) hp + γz = γ s n∂ ∂

−∂ ∂

2

0H y z2vg

= + +

( )2

h h = y + z2vg

= + ∵

yisovels

r0

ri

n

H0z

2g

V___2

ri

r0

X

X

Section xx

r

datum

2dhdn gr

v=



differentiating above equation

2

dh v 0dn g n

v+ 0gr g n

v

v v

∂+ =

∂

∂=

∂

v 0r nv ∂+ =∂

Thus v decreases and h increases from inner boundary to outer boundary.

The equation can be rewritten as r 0vv r∂ ∂

+ = .

Therefore it can be shown as = constantv r which is in the free vortex condition. Consider a rectangular channel bend, Discharge per unit width q yv= .

cq = yr

or y q=r c

a constant.

r

H y

2g

Outer

Total Energy Line

Inner Wall 0Wall

0

ri

z

y0

V___2

Modification to the bed profile to obtain the horizontal water surfacein a bend

Subcritical flow F < 1

Supercriticalflow F > 1

dy y=dr r

2dy dz 1= -dr dr 1-F

z decreases from inner wall to outer wall for subcritical flow (as shown in the above figure).

( ) ( )2 2dz dy y= - 1-F - 1-Fdr dr r

=

and

( )2dh dz dy dy dy= - 1-Fdr dr dr dr dr

+ = +

22 2dy y-1+F 1 = F

dr r grv⎡ ⎤+ =⎣ ⎦



Transverse bed profile. 220

0 0

2

0 2

H =z+y y2g 2g

rq CH =z+ +C 2gr

vv+ = +

The above equation gives resonably good result as long as the angle of the circular

bend in plan is greater than 90 a correction factor was suggested by Shukry for

circulation constant C, assuming it to vary linearly from 0 to 90 .

mAf

r Vvr = CU =C + 1 90 90 Cθ θ⎡ ⎤⎛ ⎞−⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

mA1 X 1

1

Vv =w U =w + 1r 90 90 rw

θ θ⎡ ⎤⎛ ⎞−⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

However, by applying Newton's 2nd law of motion based on one dimensional analysis

i.e., all the filamental velocities in the bend are equal to the mean velocity mbV and that

of all the streamlines having the same radii of curvature cr , an equation can be obtained

for a rectangular channel namely

2mb

maxc

V 2b∆y =2g r

⎛ ⎞⎜ ⎟⎝ ⎠

For the channels other than the rectangular channel, the bed width (b) can be replaced

by the water surface width (T) then

2mb

maxc

V 2T∆y =2g r

⎛ ⎞⎜ ⎟⎝ ⎠

The above equation is only a first approximation and gives transverse profile as the

straightline . This assumes that the rise and the drop of the water surface level from the

normal level is equal on either side of center line of bend.

As a better approximation, Ippen and Drinker obtained an equation for superelevation.

The derivation of equation is based on the assumptions of free vortex or irrotational flow

with the uniform specific head over the cross section, and the mean depth in bend being

equal to the mean approaching flow depth.



2

2c

2c

V 2T 1∆y=2g r T1

4r

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠

The bends in nature will not have the symmetry due to entrance conditions, length of

curvature and boundary resistance. Hence above equation will not give accurate result.

If the forced vortex condition exists with constant stream cross section and constant

average specific energy, then equation for superelevation assumes the form

2

2c

2c

V 2T 1∆y=2g r T1+

12r

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

The above equation is applicable to a smooth rectangular boundary with circular bend

with the flowing fluid being ideal.

Better results can be obtained by combining the effects of the free and forced vortex

conditions simultaneously. The minimum angle of bend is to be 90 for applying the

above equation in combination. For the smaller angles the difference in computed

values from the above equation becomes larger than the actual ones.

However, for a rectangular channel, circular bend with , applying the free vortex

formula, velocity at inside wall of the bend becomes as thus a depth of should exist at

the boundary (i.e. at ), which is physically impossible.

Muramoto obtained an equation for superelevation based on equations of motion.

1vr = C

20 2

1

g S r CU = + 3C r

Then

( ) 0

i

r2 22 4 1 20 0 22 2

11r

C + Cg S r 2S C r∆y = + 3C36 C 2gr

⎡ ⎤⎢ ⎥−⎢ ⎥⎣ ⎦

in which 1C and 2C are circulation constants obtained, after integration. The special

feature of above equation lies in including the effect of bed slope on superelevation.



Thus it can be observed from the above discussion that superelevation in a bend is a

function of shape of the cross section, Reynolds number, approach flow, slope of the

bed, Froude number, crθ180 b

, and boundary resistance. The superelevation is also

affected by the presence of secondary currents and separation.

However, before applying to field situations, the validity of the transverse flow profile

equation has to be justified. The observations in the field have shown the occurrence of

troughs near the concave profile in the bend. Further, these troughs have been

observed during rising and as well as falling stages of the channels. The channels of

smaller width have exhibited accumulation of debris instead of troughs.

36.4.2 Superelevation and transverse profile

Normalised super elevation max∆y/∆y was correlated with normalised bend angle 0θ/θ

for all the three bends, for two different Reynolds numbers.

From Figure, it may be observed that two peaks of superelevation occur at 0θ/θ 0 17.=

and 0θ/θ 0 67.= for all the cases except for bend B1 for eR = 42, 280 . The influence of

Reynolds number on the trend of the variation of superelevation at various section of

the bend is insignificant in all the three bends.



1.0

0.34 0.67 1.00SECTION B1

SECTION B2

SECTION B3

1.0

1.0

0.34 0.67 1.00

0.34 0.67 1.00

Variation of normalised Super elevationwith normalised angle forReynolds number 42280

0

y___ymax

θ/θ0

y___ymax

y___ymax



0.34 0.67 1.00

0.34 0.67 1.00

0.34 0.67

1.0

1.0

1.0

1.000

SECTION B1

SECTION B2

SECTION B3

Variation of normalised Super elevationwith normalised angle forReynolds number 101,700

y___ymax

y___ymax

y___ymax

θ/θ0

The maximum value of superelevation was normalised with the approaching velocity

head 2

V /2g and correlated with the Froude number. The equations of these lines is in

the form max2

∆y =m log F + CV /2g

in which 'm' is the slope of the line and 'C' a constant.

max2

∆yBend 1: = 1.19 log F + 0.20V /2g

−

max2

∆yBend 2 : = 5.33 log F 0.49V /2g

−



max2

∆yBend 3: = 1.96 log F 0.06V /

2g

−

View of SET-UP



Flow Condtions D/S of B1 The greatest difference in elevation between the longitudinal profiles at outer and inner

walls is maximum at the 30 section of the 180 bend.



112

1

2

3

4

56

78

910

11

12

Wheel volve

Inlet pipe

Adjustment valveEntrance chamber

Transition

Leading Channel

Rectangular notch

Stilling chamber

Masonry honey comb

Main channel

Tail control

Transition

1 236 9

11

Point gauge

56

18072G.L 230

100

Experimental Set - Up Scale = 1: 100 (all dimensions in cm)

1

2

4

5

6

7

8

9

10

11

100 200100 100100

P.G.

468 468

Bend I314

782

50

1960

Bend II440

2400

870

50

280

3270

rc = 300

900

737

P.G. 44315 50

1800

rc = 140

AA

200 100

LAYOUT PLAN

Bed slope 1000

N



Section and points of measurementb) points of measurement

70

5013

G.L15

100Section AA

Inner Outer

-22.5 -12.5 0 12.5 22.50.4

00

15 3045

6075

90

2 m

rc

rc 1 m Station B

A

00

0

0

0

0

200

A

00

30

6090

0

00120150

180

0

0

0

Station C

rc = 30000

1530

4560

7590

0

0

0

00

0

1 m

13

1 m

All dimensions are in cm



0

1.0

2.0

3.0

4.0

5.0

6.5

- 0.25- 0.31- 0.60- 1.05log F

Bend 2

Bend 3

Bend 1

Variation of with log F

y______

V_2

2g/

y______

V_2

2g/



Comparison of Observed and Theoretical Superelevation

experimental

theoretical

24

23

bed in cm0 12.5 25.0 37.5 50.0

ObservedTheoretical : eqnExperimental : eqn

Inner wall Outer wall

Longitudinal Water Surface Profile

Bed slope 1:1000

Bed slope

1.2

1.0

0.8

1.2

1.0

0.8

STN ASTN B STN C

STN D

STN ASTN B STN C

STN D

B1

Q = 26.1 lps

B2 B3

B1 B2 B3

Re 42280 F = 0.49

Q = 71.9 lps Re 101760 F = 0.55 Scalex axis 1:100y axis 5:1Outer wallInner wall

y/yA

y/yA



Inner wall Outer wallCL1.10

1.111.12

1.20 1.15

1.121.051.00 0.90

0.80

0.95 0.95

1.18 1.151.15

1.051.00

0.90

1.20

1.151.10 1.05

0.81

0.90

0.85 0.98

1.05

1.15

1.25

CL

CLCL

Inner wall Outer wall

Inner wall Outer wallInner wall Outer wall

STATION A STATION B

STATION DSTATION C

0.98

1.00

0.90

ISOVELS in open channel bend [Normalised with V ] Q = 26.1 lps, F = 0.18, R = 36050emax



Inner wall Outer wallCL

STATION A STATION B

STATION DSTATION C


Inner wall Outer wallCLInner wall Outer wallCL

0.85

1.45 1.22

1.25 1.000.85

1.301.251.151.10

1.151.30

1.251.101.0

0.77

1.30

1.27

0.92

1.001.001.08

0.950.75

1.301.251.23

1.20

0.95 1.000.80

1.30

ISOVELS in open channel bend [Normalised with V ] Q = 71.9 lps, F = 0.44, R = 95420emax Inner wall Outer wallCL

STATION A STATION B

STATION DSTATION CISOVELS in open channel bend [Normalised with V ] Q = 83.5 lps, F = 0.41, R = 103460emax

1.30


Inner wall Outer wallCL Inner wall Outer wallCL

1.251.151.10

0.77

1.251.30

1.151.101.00

0.93 0.78

1.20

1.15

1.10

1.081.000.83

1.151.051.00

0.950.80



Reference 1. Ippen A.T and Drinker P.A., "Boundary shear stresses in curved trapezoidal

channels" , Proc. ASCE. journal Hydraulic Diivision HY5, Part - I, Volume 88, p 3273, pp

- 143 - 179, September 1962, and discussion by Shukry A, Proc. ASCE. journal

Hydraulic Division, pp 333, May 1963.

2. Henderson F.M. , "Open Channel Hydraulics", Mac Millan Company Limited 1966.

3. Thandaveswara B.S, "Characteristics of flow around a 90° open channel bend", M.Sc

Engineering Thesis, Department of Civil and Hydraulic Engineering, Indian Institute of

Science, Bangalore - 12, May 1969.

superelevation - NPTELnptel.ac.in/courses/105106114/pdfs/Unit36/36_4.pdf · Superelevation is defined as the difference in elevation of water surface between inside and outside wall

Documents