channel cross sections

7/25/2019 Channel Cross Sections

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Lecture 16

Channel Cross Sections

I. Channel Cross Section Parameters

Common cross-sectional shapes are rectangular, trapezoidal and circular Geometrically, rectangular cross-sections are just special cases of trapezoidal

sections

Circular cross sections are hydraulically more efficient than other cross-sectional shapes, but they are only used for small channel sizes

Circular cross sections are usually made of precast concrete mixes, and elevatedabove the ground

Tunnels designed for open-channel flow are sometimes built with special crosssections (e.g. horseshoe sections)

The standard horseshoe cross section has a semicircular top portion, and anintersection of three larger circles for the lower portion it can be considered a

modification of a circular section A semi-circular channel cross section is the best shape for an open-channel,including open-channel flow in tunnels, but the horseshoe shape has been usedin dozens of tunnels to allow for greater floor width, thereby facilitating thepassage of equipment through the tunnel

Standard horseshoe cross section (bold curves): diameter of upper section

is half of the diameter of the three larger circles

BIE 5300/6300 Lectures 171 Gary P. Merkley


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Trapezoidal cross sections can be symmetrical or non-symmetrical

For trapezoidal cross sections, the inverse side slope (H:V) is usuallybetween zero and 2.0

Common inverse side slopes are zero (rectangular section), 1.0 and 1.5 There are tradeoffs between low and high values of side slope:

1. canals with low inverse side slopes occupy less land area2. high inverse side slopes are more stable and may require

less maintenance3. high inverse side slopes are safer, if animals or people could

fall into the canal, because it is easier to climb out

4. rectangular cross sections can be simpler to build, when linedwith concrete (especially for small cross sections)

5. it may be easier to build and install structures and transitionsfor rectangular sections

6. medium-range side slopes correspond to greater hydraulicefficiency

Compound sections are not uncommon For example, a combination of a trapezoidal and rectangular section:

Gary P. Merkley 172 BIE 5300/6300 Lectures


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II. Freeboard Recommendations

Freeboard means the extra depth of a canal section, above the water surface for100% flow rate capacity, usually for uniform-flow conditions

A freeboard value should be added to the maximum expected depth to allow for:

1. deviations between design and construction (these areubiquitous, only varying in magnitude from place to place)

2. post-construction, non-uniform land settlement3. operational flexibility (including operator mistakes)4. accommodate transient flow conditions5. provide a more conservative design (in terms of flow

capacity)6. increase in hydraulic roughness due to lining deterioration,

weed growth, and for other reasons7. wind loading8. other reasons

Thus, with freeboard, under maximum flow conditions (full supply level, FSL),canal overflow is not impending

If the canal starts to overflow, enormous erosive damage can occur in just a fewminutes

According to Murphys Law, these things usually happen about 3:00 am, when noone is around. Then, everyone finds out at about 6:30 am after it has beenspilling for hours.

Many reaches of canal in many countries are routinely operated with virtually nofreeboard, and disasters often occur



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A canal which is overtopping the banks and spilling water.

A canal with impending spillage (zero freeboard).



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Traditional wisdom says that the minimum design freeboard for small canals is1 ft (0.3048 m), so this would be the minimum for most any canal, but for verysmall canals it would certainly be excessive

For canals with flow rates up to 3,000 cfs (85 m3/s), the freeboard should be upto 4 ft (1.2 m), and for intermediate flow rates, the freeboard should be 1 ft plus25% of the maximum expected water depth

However, the design freeboard can extend above the lined portion of a concrete-lined canal, and it often does (the berm of a lined canal is almost always higherthan the top of the lining)

Also, considerable judgment is required to know what the required freeboardmight actually be, including knowledge of the operational modes in the canal(these will help determine the need for freeboard)

Special analysis should go into the determination of required freeboard for canalswith capacities exceeding 3,000 cfs, but such analysis can also be included forany size canal

The analysis should also include economic criteria, because on large and or longcanals, a difference of a few centimeters in freeboard is likely to mean adifference of millions of dollars in construction costs

Nevertheless, the USBR published data on freeboard guidelines for up to 20,000cfs (560 m3/s) capacities, and these are found in the plots below

To put things in perspective, note that very few irrigation canals exceed acapacity of 100 m

3/s; most main canals have a capacity of less than 20 m

3/s



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III. Trapezoidal Cross Section

Symmetrical section:

( )A h b mh= + (16)

T b 2mh= + (17)

2pW b 2h m 1= + + (18)



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2w h m 1= + (19)

2 2

T bm

4w (T b)

=

(20)

2

ch 2hm

h b2A 3

= +

(21)

2h 4hmh b

2A 3

= +

(22)

where h is the depth from the bottom (or invert) of the cross section up to the

centroid of the cross-sectional area; and hcis the depth from the water surfacedown to the area centroid:

ch h h= (23)

IV. Nonsymmetr ical Trapezoidal Cross Section

[ ]1 2A h b 0.5(m m )h= + + (24)

( )1 2T b h m m= + + (25)

2 2p 1W b h m 1 m 1

= + + + + 2 (26)

2 21 2w h m 1 or h m 1= + + (27)



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( )2

c 1 2

h hh b m m

2A 3

= + + (28)

( )

2

1 2h 2hh b m m2A 3 = + + (29)

V. Circular Cross Section

D

hc

hh

T

In the following, angle is in radians

1 2h2cos 1D

=

(30)

( )2D

A sin8

= (31)

or,

2 2 1 h rA (h r) 2hr h r sin2r

= + + (32)



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where r = D/2

22h

T D sin D 1 12 D

= = (33)

p DW2= (34)

Dh 1 cos

2 2

= (35)

( )3 / 2

2D 2

h hD h2 3A

= (36)

ch h h= (37)

Circular Channel with D = 1.0

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Depth

Beta,

T,

A,

Wp

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

Hydraulicradius&depthtocentroid

beta

Top Width

Area

Wetted Perimeter

Hydraulic Radius

Depth to Centroid

Nondimensional curves of circular cross-section geometry.



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VI. Standard Horseshoe Cross Section

rr/2

h1

h2

h3

The following is for a standard horseshoe cross section

Divide the depth into three segments Note that h1+ h2+ h3= r (see the above figure) Determine h1by solving for the intersection of two of the circles

1

1 7h r 1

4

+=

(38)

then,

2 1

r

h h2= (39)

and,

3

rh

2=

(40)

In the following, all angles are in radians



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Top width (at water surface):

For 0 < h h1:

2h

T 2r 1 1 r

= (41)

For h1< h r/2:

2 2

2 2r r r r T r h r h2 2 2 2

= +

(42)

or,

=

2

2 rT 2 r h r 2

(43)

For r/2 < h < r:

22h

T r 1 1r

=

(44)

and T = 0 when h = 0 or h = r

Cross-sectional area:

For 0 h h1:

( ) 2 1h r

A (h r) h 2r h r sin2r

= + + (45)

For h1< h r/2:

[ ] = + +

2

2 1 1 2 a b

11A r cot( ) cot( ) A A

4A (46)

where A1is the cross-sectional area corresponding to h = h1; cot(1) is thecotangent of 1, equal to 1/tan(1); and,



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2 2

2

1

2

r h

h

= (47)

2

2

2

r

r h2

rh

2

=

(48)

( ) = 11 1tan (49)

( ) = 12 2tan (50)

22

2

a

2

1 rA r h

2 2

=

r (51)

and,2

2 2

b 2

1

1 rA r h

2

= (52)

Note that h2= r/2-h1

Note that 1and Abare constants for a given value of r Note that 2= /2 and Aa= 0 when h = r/2

Another way to calculate this (h1< h r/2) area is by integration:

2 2

1 1

y y 2 21 1y y

rA 2 xdy A 2 r y dy A

2

= + = +

+ (53)

which yields the following expression:

2

1

y2 2 2 1

1y

yA r y y r y r sin A

r

= + + +

(54)

where y2and y1are the integration limits:



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2r

y h2

= (55)

11

r Cy

2= (56)

where,

11 7

C 12

+=

(57)

and,

211 1

2C C

C 1 1 sin2 4

=

1C

2 (58)

Finally, applying the integration limits:

22 1 2

2 12h r r r

A r C sin h r r h A2r 2 2

= + +

(59)

where A1is the area corresponding to h = h1(Eq. 45)

Equation 59 is preferred over Eq. 46 because it is simpler and yields the sameresult for h1< h r/2

For r/2 < h r:

( )2

12

r r 2h rA h h r h sin

2 4 r

= + + A (60)

where A2is the area corresponding to h = r/2 (Eq. 59)

Depth to area centroid:

For 0 h h1:

( )3

1 2h(2r h)r hh sin 1 hr 2hA 2 r 3A

= +

23r (61)



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hwhere A is as calculated by Eq. 45; and is the depth measured from the areacentroid to the bottom of the cross section

For h1< h r/2, the moment of area with respect to x is:

( )

( )2

1

2 2x

2 2x

y2 3 / 2

2 2x

y

M yx dy y r 2 r y

M r ydy 2 y r y dy

ry 2M r y

2 3

= = +

= +

=

dy

(62)

where y1and y2are integration limits, exactly as defined above for cross-

sectional area. Applying the integration limits:

3 / 22 23 2

x 3r 2r rM r C h r h2 32 2

= (63)

where,3 / 2

2 21 1

3C C2

C 18 3 4

= +

(64)

where C3is a constant; and C1is as defined in Eq. 57

The value of Mxwill be negative because it is calculated based on coordinateorigins at h = r/2, so the depth to centroid for a given depth, h, must be shiftedupward by the amount r/2:

xx

x

Mrh

2 A= + (65)

which will be a positive value, with Axbeing the cross-sectional areacorresponding to the same integration limits, y1and y2:

22 1 2

x 22h r r r

A r C sin h r r h2r 2 2

= +

(66)

and C2is also as previously defined



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h The composite value of must account for the calculations up to h = h1, so fordepths from h1to r/2, the following area-weighted relationship is used to obtainthe exact depth to the area centroid:

x

x 1x

x 1

Mr

1A A h

2 Ah

A A

+ +

=+

(67)

where A1and h 1are the values corresponding to h = h1(Eqs. 45 and 61)

For r/2 < h r, the moment of area with respect to x is:

[ ]3

3 / 2x

r 2M h(r h)

12 3= (68)

The cross-sectional area from r/2 up to some h value is:

21

xr 2hr

A h(r h) sinh4 r2

r = + (69)

which is Eq. 60 minus the A2term

The composite value of h must account for the calculations up to h = r/2, so fordepths from r/2 to r, the following area-weighted relationship is used to obtain theexact depth to the area centroid:

xx 2

x

x 2

Mr2A A h

2 Ah

A A

+ +

=+

(70)

where A2and h 2are the values corresponding to h = r/2 (Eqs. 59 and 67)

Wetted perimeter:

For 0 h h1:

1p

hW 2r cos 1

r

=

(71)

For h1< h r/2:



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1 1 1p p

r 2h CW 2r cos cos W

2r 2

=

1+ (72)

where C1is as defined in Eq. 57; and Wp1is the wetted perimeter correspondingto h = h1(Eq. 71)

Note that the term cos-1(-C1/2) is a constant, based on h1

For r/2 < h r:

1

p p

2hW r cos 1 W

r 2

= +

2 (73)

where Wp2is the wetted perimeter corresponding to h = r/2 (Eq. 72)

Standard Horseshoe Cross Section

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Depth, h/r

T/r,

A/r2,andh

bar/r

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Wp

/r

Top width

Area

Centroid depth

Wetted perimeter

Nondimensional geometric values in a standard horseshoe cross section.

The increase in area with the standard horseshoe cross section (compared to acircular section with a diameter of r) is only about 5.6% for a full section



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VII. Efficient Canal Sections

Sometimes it is useful to apply an efficient cross section to maximizechannel capacity for a given bed slope and roughness

However, other considerations such as side slope stability, safety, and liningmaterial may be more important

Comparisons between the most efficient section and other sections show thatrelative changes in the section often do not affect the capacity significantly capacity is much more sensitive to changes in roughness and bed slope

VIII. Most Efficient Trapezoidal Canal Section

How to calculate the most efficient trapezoidal cross section? Minimize the wetted perimeter with respect to cross-sectional area of flow (or

depth), or maximize the hydraulic radius (R = A/Wp)

Express the wetted perimeter as a function of A, m, and h, where h is depth

Keep area, A, as a constant, otherwise you will get Wp= 0 for the mostefficient section Differentiate Wpwith respect to depth, h, and set it equal to zero For a symmetrical trapezoidal cross section:

A h(b mh)= + (1)

2pW b 2h m 1= + + (2)

1. Write the wetted perimeter in terms of A, h, and m (get rid of b by combining

Eqs. 1 and 2):A

b mh

= h (3)

2p

AW mh 2h m

h1= + + (4)

2. Differentiate Wpwith respect to h (A and m constant) and equate to zero (tominimize Wpfor a given area):

p 22

W A m 2 m 1 0h h

= + + =

(5)



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3. Solve Eq. 5 for A

2 2A h 2 m 1 m = +

(6)

4. For R = A/Wp, use Eq. 6 to obtain

2 2

2

h 2 m 1 m

Rb 2h m 1

+ =

+ + (7)

5. Now, manipulate Eq. 7

2 2

2

bh 2h m 1 bh mhR

b 2h m 1

+ + =

+ +

2

(8)

( )2

2

h b 2h m 1 h b mh

Rb 2h m 1

+ + + =

+ + (9)

p

p

hW AR h

WR

= = (10)

6. Therefore, h = 2R, or, hR

2= (11)

You could also directly manipulate Eq. 6 to get the same result: 2A = hW p

7. For the most efficient rectangular section,

p

A bh hR

W b 2h= = =

+ 2 (12)

which results in b = 2h (bed width twice the maximum flow depth).

8. For the most efficient trapezoidal sectionwe will get Wp= T + b, where T is thetop width of flow (b+2mh), which for a symmetrical trapezoid means that thelength of each side slope (for depth h) is T/2. It also means that b = T/2, and thiscorresponds to half of a regular six-sided polygon, or a hexagon. The interior

angle of a hexagon is 120 degrees, so m = 1/tan(60) = 0.577.



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IX. Parabolic Canal Section

Suppose you have a parabolic channel section Define half of a symmetrical parabolic section as:

2

h Kx= (13)

The cross-sectional area of flow (for half of the section) is:

3 / 2h h

0 0

h 2hA x dh dh

K 3 K= = = (14)

The wetted perimeter (again, half of the section) is:

)( ) ( ) (2 2

px 0

W lim f x x f x x

= + + (15)

or,

p

1/ 22

Wdf

1 ddx

=

+ x (16)where f = Kx2. This derivation can be described graphically as follows:



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where the curve is broken up (discretized) into successive linear segments...

)( ) ( ) (2 2

s f x x f x x + + (17)

For y = f(x) = Kx2

, 22 2df 4K x

dx

= (18)

Then,h /K

2 2p

0

W 4K x= + 1 dx (19)

After integration (using integration tables), the wetted perimeter for half of theparabolic section is:

p 2 2

h h 1 1 h h 1W K ln 2K

K K 4K K K4K 4K

= + + + +

(20)

which of course is a function of both K (curvature) and depth (h)

An analysis of the hydraulic radius for such a parabolic section shows that thehydraulic radius decreases monotonically as K increases from an infinitesimally

small value, so there is no most efficient value of K Chow (1959) has some equations (exact and approximate) for various channel

section shapes, including the parabola defined in this case

References & Bibliography

Davis, C.V. and K.E. Sorensen (eds.). 1969. Handbook of applied hydraulics. McGraw-Hill Book

Company, New York, N.Y.

Hu, W.W. 1973. Hydraulic elements for USBR standard horseshoe tunnel. J. of the Transportation

Engrg. Div., ASCE, 99(4): 973-980.

Hu, W.W. 1980. Water surface profile for horseshoe tunnel. Transportation Engrg. Journal, ASCE,

106(2): 133-139.

Labye, Y., M.A, Olsen, A. Galand, and N. Tsiourtis. 1988. Design and optimization of irrigation

distribution networks. FAO Irrigation and Drainage Paper 44, Rome, Italy. 247 pp.

USBR. 1963. Linings for irrigation canals. U.S. Government Printing Office, Washington, D.C. 149 pp.


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