lecture 16 channel cross sections

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 elevated

above the ground • Tunnels designed for open-channel flow are sometimes built with special cross

sections (e.g. “horseshoe” sections) • The standard horseshoe cross section has a semicircular top portion, and an

intersection 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 used in dozens of tunnels to allow for greater floor width, thereby facilitating the passage 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

• Trapezoidal cross sections can be symmetrical or non-symmetrical

• For trapezoidal cross sections, the inverse side slope (H:V) is usually between 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 area 2. high inverse side slopes are more stable and may require

less maintenance 3. 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 lined

with concrete (especially for small cross sections) 5. it may be easier to build and install structures and transitions

for rectangular sections 6. medium-range side slopes correspond to greater hydraulic

efficiency

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

Gary P. Merkley 172 BIE 5300/6300 Lectures

II. Freeboard Recommendations

• Freeboard means the extra depth of a canal section, above the water surface for 100% 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 are ubiquitous, only varying in magnitude from place to place)

2. post-construction, non-uniform land settlement 3. operational flexibility (including operator mistakes) 4. accommodate transient flow conditions 5. provide a more conservative design (in terms of flow

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

weed growth, and for other reasons 7. wind loading 8. 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 few

minutes • According to Murphy’s Law, these things usually happen about 3:00 am, when no

one is around. Then, everyone finds out at about 6:30 am after it has been spilling for hours.

• Many reaches of canal in many countries are routinely operated with virtually no freeboard, and disasters often occur


A canal which is overtopping the banks and spilling water.

A canal with impending spillage (zero freeboard).


• Traditional wisdom says that the minimum design freeboard for “small canals” is 1 ft (0.3048 m), so this would be the minimum for most any canal, but for very small canals it would certainly be excessive

• For canals with flow rates up to 3,000 cfs (85 m3/s), the freeboard should be up to 4 ft (1.2 m), and for intermediate flow rates, the freeboard should be 1 ft plus 25% 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 higher than the top of the lining)

• Also, considerable judgment is required to know what the required freeboard might 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 canals with capacities exceeding 3,000 cfs, but such analysis can also be included for any size canal

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

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

• To put things in perspective, note that very few irrigation canals exceed a capacity of 100 m3/s; most main canals have a capacity of less than 20 m3/s


III. Trapezoidal Cross Section

Symmetrical section:

( )A h b mh= + (16)

T b 2mh= + (17)

2

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


2w h m 1= + (19)

2 2

T bm4w (T b)

−=

− − (20)

2

ch 2hmh 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 hc is the depth from the water surface down to the area centroid: ch h h= − (23)

IV. Nonsymmetrical 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)


( )2

c 1 2h hh b m m2A 3

⎡ ⎤= + +⎢ ⎥⎣ ⎦ (28)

( )2

1 2h 2hh b m m2A 3

⎡ ⎤= + +⎢ ⎥⎣ ⎦ (29)

V. Circular Cross Section

D

hc

hh

Tβ

• In the following, angle β is in radians

1 2h2cos 1D

− ⎛ ⎞β = −⎜ ⎟⎝ ⎠

(30)

( )2DA sin

8= β − β (31)

or,

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

− π⎡ ⎤−⎛ ⎞= − − + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ (32)


where r = D/2

22hT D sin D 1 1

2 Dβ⎛ ⎞ ⎛ ⎞= = − −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

(33)

p

DW2β

= (34)

Dh 1 cos2 2

β⎛ ⎞= −⎜ ⎟⎝ ⎠ (35)

( )3 / 22D 2h 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

Bet

a, T

, A, W

p

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

Hyd

raul

ic ra

dius

& d

epth

to c

entr

oid

betaTop WidthAreaWetted PerimeterHydraulic RadiusDepth to Centroid

Nondimensional curves of circular cross-section geometry.


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 h1 by solving for the intersection of two of the circles

11 7h r 1

4⎡ ⎤⎛ ⎞+

= −⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ (38)

then,

2 1

rh h2

= − (39)

and,

3

rh2

= (40)

• In the following, all angles are in radians


Top width (at water surface):

• For 0 < h ≤ h1:

2hT 2r 1 1

r⎛ ⎞= − −⎜ ⎟⎝ ⎠

(41)

• For h1 < h ≤ r/2:

2 2

2 2r r r rT r h r h2 2 2 2

⎛ ⎞ ⎛⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜= − − − − − − − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝

⎞⎟⎟⎠

(42)

or,

⎛ ⎞= − − −⎜ ⎟⎝ ⎠

22 rT 2 r h r

2 (43)

• For r/2 < h < r:

22hT r 1 1

r⎛ ⎞= − −⎜ ⎟⎝ ⎠

(44)

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

Cross-sectional area:

• For 0 ≤ h ≤ h1:

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

− π⎡ ⎤−⎛ ⎞= − − + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ (45)

• For h1 < h ≤ r/2:

[ ]⎛ ⎞= α − α − α − α − + +⎜ ⎟⎝ ⎠

22 1 1 2 a b

11A r cot ( ) cot ( ) A A

4A (46)

where A1 is the cross-sectional area corresponding to h = h1; cot(α1) is the cotangent of α1, equal to 1/tan(α1); and,


2 22

12

r hh−

ϕ = (47)

22

2

rr h2

r h2

⎛ ⎞− −⎜ ⎟⎝ ⎠ϕ =−

(48)

( )−α = ϕ11 1tan (49)

( )−α = ϕ12 2tan (50)

22

2a

2

1 rA r h2 2

⎛ ⎞⎛ ⎞⎜ ⎟= − − −⎜ ⎟ϕ ⎜ ⎟⎝ ⎠⎝ ⎠

r (51)

and,

2

2 2b 2

1

1 rA r h2

⎛ ⎞= − −⎜ ⎟ϕ ⎝ ⎠ (52)

• Note that h2 = r/2-h1 • Note that α1 and Ab are constants for a given value of r • Note that α2 = π/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 A2

⎛ ⎞= + = − + −⎜ ⎟⎝ ⎠∫ ∫ + (53)

which yields the following expression:

2

1

y2 2 2 1

1y

yA r y y r y r sin Ar

−⎡ ⎤⎛ ⎞= − + − + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ (54)

where y2 and y1 are the integration limits:


2ry h2

= − (55)

11

r Cy2

= (56)

where,

11 7C 1

2⎛ ⎞+

= − ⎜ ⎟⎜ ⎟⎝ ⎠

(57)

and,

2

11 12

C CC 1 1 sin2 4

−⎛ ⎞ ⎛ ⎞⎜ ⎟= − − − ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

1C2

(58)

Finally, applying the integration limits:

2

2 1 22 1

2h r r rA r C sin h r r h A2r 2 2

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

(59)

where A1 is the area corresponding to h = h1 (Eq. 45)

• Equation 59 is preferred over Eq. 46 because it is simpler and yields the same

result for h1 < h ≤ r/2

• For r/2 < h ≤ r:

( )2

12

r r 2h rA h h r h sin2 4 r

−⎛ ⎞ −⎛ ⎞= − − + +⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠A (60)

where A2 is 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 3 A

− −⎡ ⎤π ⎛ ⎞= − − − − +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦23r (61)


hwhere A is as calculated by Eq. 45; and is the depth measured from the area centroid 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 / 22 2x

y

M yx dy y r 2 r y

M r ydy 2 y r y dy

ry 2M r y2 3

⎛ ⎞= = − + −⎜ ⎟⎝ ⎠

= − + −

⎡ ⎤= − − −⎢ ⎥⎢ ⎥⎣ ⎦

∫ ∫

∫ ∫

dy

(62)

where y1 and y2 are 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 / 22 21 1

3C C2C 18 3 4

⎛ ⎞= + −⎜ ⎟⎜ ⎟

⎝ ⎠ (64)

where C3 is a constant; and C1 is as defined in Eq. 57

• The value of Mx will be negative because it is calculated based on coordinate origins at h = r/2, so the depth to centroid for a given depth, h, must be shifted upward by the amount r/2:

xx

x

Mrh2 A

= + (65)

which will be a positive value, with Ax being the cross-sectional area corresponding to the same integration limits, y1 and y2:

2

2 1 2x 2

2h r r rA r C sin h r r h2r 2 2

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

(66)

and C2 is also as previously defined


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

xx 1

x

x 1

Mr1A A h

2 Ah

A A

⎛ ⎞+ +⎜ ⎟

⎝ ⎠=+

(67)

where A1 and h 1 are 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:

2

1x

r 2hrA h(r h) sinh4 r2

− r−⎛ ⎞⎛ ⎞= − +−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ (69)

which is Eq. 60 minus the A2 term

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

xx 2

x

x 2

Mr2A A h

2 Ah

A A

⎛ ⎞+ +⎜ ⎟

⎝ ⎠=+

(70)

where A2 and h 2 are the values corresponding to h = r/2 (Eqs. 59 and 67)

Wetted perimeter:

• For 0 ≤ h ≤ h1:

1p

hW 2r cos 1r

− ⎛ ⎞= −⎜ ⎟⎝ ⎠

(71)

• For h1 < h ≤ r/2:


1 1 1p p

r 2h CW 2r cos cos W2r 2

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

1+ (72)

where C1 is as defined in Eq. 57; and Wp1 is the wetted perimeter corresponding to h = h1 (Eq. 71)

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

• For r/2 < h ≤ r:

1p p

2hW r cos 1 Wr 2

−⎡ ⎤π⎛ ⎞= − − +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦2 (73)

where Wp2 is 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 , a

nd h

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 widthAreaCentroid depthWetted perimeter

Nondimensional geometric values in a standard horseshoe cross section.

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


VII. “Efficient” Canal Sections

• Sometimes it is useful to apply an “efficient” cross section to maximize channel capacity for a given bed slope and roughness

• However, other considerations such as side slope stability, safety, and lining material may be more important

• Comparisons between the most efficient section and other sections show that relative 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 most

efficient section • Differentiate Wp with 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):

Ab mh

= − h (3)

2p

AW mh 2h mh

1= − + + (4)

2. Differentiate Wp with respect to h (A and m constant) and equate to zero (to minimize Wp for a given area):

p 22

W A m 2 m 1 0h h

∂ −= − + + =

∂ (5)


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 mR

b 2h m 1

⎛ ⎞+ −⎜ ⎟⎝ ⎠=+ +

(7)

5. Now, manipulate Eq. 7

2 2

2

bh 2h m 1 bh mhRb 2h m 1

+ + − −=

+ +

2 (8)

( )2

2

h b 2h m 1 h b mhR

b 2h m 1

⎛ ⎞+ + − +⎜ ⎟⎝ ⎠=

+ + (9)

p

p

hW AR h

WR

−= = − (10)

6. Therefore, h = 2R, or,

hR2

= (11)

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

7. For the most efficient rectangular section,

p

A bh hRW b 2h

= = =+ 2

(12)

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

8. For the most efficient trapezoidal section we will get Wp = T + b, where T is the top width of flow (b+2mh), which for a symmetrical trapezoid means that the length of each side slope (for depth h) is T/2. It also means that b = T/2, and this corresponds 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.


IX. Parabolic Canal Section

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

2h Kx= (13)

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

3 / 2h h

0 0h 2hA x dh dhK 3 K

= = =∫ ∫ (14)

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

)( ) ( ) (2 2p

x 0W lim f x x f x x

∆ →⎡ ⎤= + ∆ − +⎣ ⎦∑ ∆ (15)

or,

p

1/ 22W df 1 d

dx

⎛ ⎞⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

+∫ x (16)

where f = Kx2. This derivation can be described graphically as follows:


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

)( ) ( ) (2 2s 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 the

parabolic section is:

p 2 2h h 1 1 h h 1W K ln 2KK 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 the

hydraulic 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.


lecture 16 channel cross sections

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