Lecture 16
Channel Cross Sections
I. Channel Cross Section Parameters
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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 crosssectional 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
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Gary P. Merkley
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Trapezoidal cross sections can be symmetrical or non-symmetrical
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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:
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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
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Compound sections are not uncommon
For example, a combination of a trapezoidal and rectangular section:
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II. Freeboard Recommendations
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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
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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
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Gary P. Merkley
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” 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 concretelined 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
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III. Trapezoidal Cross Section
Symmetrical section:
Gary P. Merkley
A = h ( b + mh )
(16)
T = b + 2mh
(17)
Wp = b + 2h m2 + 1
(18)
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w = h m2 + 1
m=
(19)
T −b
4w 2 − (T − b)2
(20)
h2 ⎛
2hm ⎞
⎜b +
⎟
2A ⎝
3 ⎠
(21)
h2 ⎛
4hm ⎞
h=
⎜b +
⎟
2A ⎝
3 ⎠
(22)
hc =
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:
hc = h − h
(23)
IV. Nonsymmetrical Trapezoidal Cross Section
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A = h [b + 0.5(m1 + m2 )h]
(24)
T = b + h ( m1 + m2 )
(25)
Wp = b + h ⎛⎜ m12 + 1 + m22 + 1 ⎞⎟
⎝
⎠
(26)
w = h m12 + 1 or h m22 + 1
(27)
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Gary P. Merkley
h2 ⎡
h
⎤
hc =
b + ( m1 + m2 ) ⎥
⎢
2A ⎣
3
⎦
(28)
h2 ⎡
2h
⎤
+
b
m1 + m2 ) ⎥
(
⎢
2A ⎣
3
⎦
(29)
h=
V. Circular Cross Section
D
T
β
hc
h
•
h
In the following, angle β is in radians
⎛ 2h ⎞
β = 2cos −1 ⎜ 1 − ⎟
D⎠
⎝
(30)
D2
A=
(β − sinβ )
8
(31)
or,
⎡
h−r ⎞
A = (h − r) 2hr − h2 + r 2 ⎢sin−1 ⎛⎜
⎟+
r
⎝
⎠
⎣
Gary P. Merkley
178
π⎤
2 ⎥⎦
(32)
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where r = D/2
⎛β⎞
2h ⎞
T = D sin ⎜ ⎟ = D 1 − ⎛⎜ 1 −
⎟
D⎠
⎝
⎝2⎠
Wp =
(33)
βD
2
(34)
D⎛
β⎞
⎜ 1 − cos ⎟
2⎝
2⎠
h=
h=
2
(35)
D 2
−
(hD − h2 )3 / 2
2 3A
(36)
hc = h − h
(37)
Circular Channel with D = 1.0
7.0
0.50
beta
Top Width
Area
Wetted Perimeter
Hydraulic Radius
Depth to Centroid
Beta, T, A, Wp
5.0
0.45
0.40
0.35
0.30
4.0
0.25
3.0
0.20
0.15
2.0
0.10
Hydraulic radius & depth to centroid
6.0
1.0
0.05
0.0
0.00
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Depth
Nondimensional curves of circular cross-section geometry.
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Gary P. Merkley
VI. Standard Horseshoe Cross Section
r
r/2
h3
h2
h1
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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
⎡ ⎛ 1 + 7 ⎞⎤
h1 = r ⎢1 − ⎜⎜
⎟⎟ ⎥
⎢⎣ ⎝ 4 ⎠ ⎥⎦
(38)
then,
h2 =
r
− h1
2
(39)
r
2
(40)
and,
h3 =
•
In the following, all angles are in radians
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Top width (at water surface):
•
For 0 < h ≤ h1:
h
T = 2r 1 − ⎛⎜ 1 − ⎞⎟
r⎠
⎝
•
2
(41)
For h1 < h ≤ r/2:
2
2
⎛
⎞ ⎛
r
r
r
r⎞
⎛
⎞
⎛
⎞
2
2
T = ⎜ r − ⎜h − ⎟ − ⎟ − ⎜ − r − ⎜h − ⎟ + ⎟
2⎠
2⎟ ⎜
2⎠
2⎟
⎜
⎝
⎝
⎝
⎠ ⎝
⎠
(42)
or,
2
r⎞
⎛
T = 2 r − ⎜h − ⎟ − r
2⎠
⎝
2
•
(43)
For r/2 < h < r:
2h ⎞
T = r 1 − ⎜⎛ 1 −
⎟
r ⎠
⎝
2
(44)
and T = 0 when h = 0 or h = r
Cross-sectional area:
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For 0 ≤ h ≤ h1:
⎡
h−r ⎞
A = (h − r) h ( 2r − h ) + r 2 ⎢sin−1 ⎛⎜
⎟+
⎝ r ⎠
⎣
•
π⎤
2 ⎥⎦
(45)
For h1 < h ≤ r/2:
1
⎛
⎞
A = r 2 ⎜ α 2 − α1 − [cot (α1 ) − cot (α 2 )] ⎟ − A a + A b + A1
4
⎝
⎠
(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,
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Gary P. Merkley
r 2 − h22
ϕ1 =
(47)
h2
⎛r
⎞
r − ⎜ − h⎟
⎝2
⎠
ϕ2 =
r
−h
2
2
2
(48)
α1 = tan−1 ( ϕ1 )
(49)
α 2 = tan−1 ( ϕ2 )
(50)
2
1 ⎛⎜ 2 ⎛ r
r ⎞⎟
⎞
Aa =
r − ⎜ − h⎟ −
2⎟
ϕ2 ⎜
⎝2
⎠
⎝
⎠
2
(51)
and,
1⎛
r⎞
Ab = ⎜ r 2 − h22 − ⎟
2⎠
ϕ1 ⎝
2
(52)
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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:
A = 2∫
y2
y1
x dy + A1 = 2 ∫
y2 ⎛
y1
r
2
2⎞
⎜ − + r − y ⎟ dy + A1
⎝ 2
⎠
(53)
which yields the following expression:
y
2
⎡
−1 ⎛ y ⎞ ⎤
2
2
2
A = ⎢ −r y + y r − y + r sin ⎜ ⎟ ⎥ + A1
⎝ r ⎠ ⎦ y1
⎣
(54)
where y2 and y1 are the integration limits:
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y2 = h −
y1 =
r
2
(55)
r C1
2
(56)
where,
⎛ 1+ 7 ⎞
C1 = 1 − ⎜⎜
⎟⎟
2
⎝
⎠
(57)
and,
C1 ⎛⎜
C12
C2 =
1− 1−
2 ⎜
4
⎝
⎞
⎟ − sin−1 ⎛⎜ C1 ⎞⎟
⎟
⎝ 2 ⎠
⎠
(58)
Finally, applying the integration limits:
2⎞
⎛
r ⎞⎜
r⎞ ⎟
⎛
2 ⎛
A = r ⎢C2 + sin ⎜
⎟ ⎥ − ⎜ h − ⎟ ⎜ r − r − ⎜ h − ⎟ ⎟ + A1
2r
2⎠
2⎠
⎝
⎝
⎠⎦ ⎝
⎣
⎝
⎠
2⎡
−1 ⎛ 2h − r ⎞ ⎤
(59)
where A1 is the area corresponding to h = h1 (Eq. 45)
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Equation 59 is preferred over Eq. 46 because it is simpler and yields the same
result for h1 < h ≤ r/2
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For r/2 < h ≤ r:
r⎞
r2
⎛
2h − r ⎞
A = ⎜ h − ⎟ h ( r − h ) + sin−1 ⎛⎜
⎟ + A2
2⎠
4
⎝ r ⎠
⎝
(60)
where A2 is the area corresponding to h = r/2 (Eq. 59)
Depth to area centroid:
•
For 0 ≤ h ≤ h1:
r3
h=
A
⎡π
h ⎞⎤
−1 ⎛
⎢ 2 − sin ⎜ 1 − r ⎟ ⎥ −
⎝
⎠⎦
⎣
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(
h(2r − h)
hr − 2h2 + 3r 2
3A
183
)
(61)
Gary P. Merkley
where A is as calculated by Eq. 45; and h is the depth measured from the area
centroid to the bottom of the cross section
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For h1 < h ≤ r/2, the moment of area with respect to x is:
Mx = ∫ ( yx ) dy = ∫ y ⎛⎜ −r + 2 r 2 − y 2 ⎞⎟ dy
⎝
⎠
Mx = −r ∫ y dy + 2∫ y r 2 − y 2 dy
⎡ ry 2 2 2
− r − y2
Mx = ⎢ −
⎢⎣ 2 3
(
3/2⎤
)
(62)
y2
⎥
⎥⎦ y
1
where y1 and y2 are integration limits, exactly as defined above for crosssectional area. Applying the integration limits:
2
2⎤
r⎛
2⎡
r⎞
Mx = r C3 − ⎜ h − ⎟ − ⎢r 2 − ⎜⎛ h − r ⎟⎞ ⎥
2⎝
3⎣
2⎠
⎝
2⎠ ⎦
3/2
3
(63)
where,
C12 2 ⎛ C12 ⎞
+ ⎜1 −
C3 =
⎟
8 3 ⎜⎝
4 ⎟⎠
3/2
(64)
where C3 is a constant; and C1 is as defined in Eq. 57
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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:
hx =
r Mx
+
2 Ax
(65)
which will be a positive value, with Ax being the cross-sectional area
corresponding to the same integration limits, y1 and y2:
2⎞
⎛
r ⎞⎜
r⎞ ⎟
⎛
2 ⎛
A x = r ⎢C2 + sin ⎜
⎟⎥ − ⎜ h − ⎟ ⎜ r − r − ⎜ h − ⎟ ⎟
2⎠
2⎠
⎝
⎝ 2r ⎠ ⎦ ⎝
⎣
⎝
⎠
2⎡
−1 ⎛ 2h − r ⎞ ⎤
(66)
and C2 is also as previously defined
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The composite value of h 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:
⎛r M ⎞
A x ⎜ + x ⎟ + A1h1
⎝ 2 Ax ⎠
h=
A x + A1
(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:
r3 2
3/2
Mx =
− [h(r − h)]
12 3
•
(68)
The cross-sectional area from r/2 up to some h value is:
2
r
⎛ 2h − r ⎞
r
A x = ⎜⎛ h − ⎟⎞ h(r − h) + sin−1 ⎜
⎟
4
⎝
2⎠
⎝ 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:
⎛r M ⎞
A x ⎜ + x ⎟ + A 2h2
⎝ 2 Ax ⎠
h=
A x + A2
(70)
where A2 and h 2 are the values corresponding to h = r/2 (Eqs. 59 and 67)
Wetted perimeter:
•
For 0 ≤ h ≤ h1:
⎛ h⎞
Wp = 2r cos−1 ⎜ 1 − ⎟
r⎠
⎝
•
(71)
For h1 < h ≤ r/2:
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Gary P. Merkley
⎡
⎛ r − 2h ⎞
−1 ⎛ C1 ⎞ ⎤
Wp = 2r ⎢cos−1 ⎜
cos
−
⎟
⎜ − ⎟ ⎥ + Wp1
⎝ 2 ⎠⎦
⎝ 2r ⎠
⎣
(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:
⎡
2h ⎞ π ⎤
⎛
Wp = r ⎢cos−1 ⎜ 1 −
⎟ − 2 ⎥ + Wp 2
r
⎝
⎠
⎣
⎦
(73)
where Wp2 is the wetted perimeter corresponding to h = r/2 (Eq. 72)
Standard Horseshoe Cross Section
1.0
5.0
0.9
4.5
0.8
4.0
Top width
Area
Centroid depth
Wetted perimeter
0.6
3.5
3.0
2.5
0.4
2.0
0.3
1.5
0.2
1.0
0.1
0.5
2
0.5
0.0
Wp/r
T/r, A/r , and hbar/r
0.7
0.0
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
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
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VII. “Efficient” Canal Sections
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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
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•
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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)
Wp = b + 2h m2 + 1
(2)
1. Write the wetted perimeter in terms of A, h, and m (get rid of b by combining
Eqs. 1 and 2):
b=
Wp =
A
− mh
h
(3)
A
− mh + 2h m2 + 1
h
(4)
2. Differentiate Wp with respect to h (A and m constant) and equate to zero (to
minimize Wp for a given area):
∂Wp
∂h
BIE 5300/6300 Lectures
=
−A
2
h
− m + 2 m2 + 1 = 0
187
(5)
Gary P. Merkley
3. Solve Eq. 5 for A
A = h2 ⎛⎜ 2 m2 + 1 − m ⎞⎟
⎝
⎠
(6)
4. For R = A/Wp, use Eq. 6 to obtain
h2 ⎛⎜ 2 m2 + 1 − m ⎞⎟
⎠
R= ⎝
b + 2h m2 + 1
(7)
5. Now, manipulate Eq. 7
R=
bh + 2h2 m2 + 1 − bh − mh2
(8)
2
b + 2h m + 1
h ⎛⎜ b + 2h m2 + 1 ⎞⎟ − h ( b + mh )
⎠
R= ⎝
b + 2h m2 + 1
R=
hWp − A
Wp
(9)
= h−R
(10)
6. Therefore, h = 2R, or,
R=
h
2
(11)
You could also directly manipulate Eq. 6 to get the same result: 2A = hWp
7. For the most efficient rectangular section,
R=
A
bh
h
=
=
Wp 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.
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IX. Parabolic Canal Section
•
•
Suppose you have a parabolic channel section…
Define half of a symmetrical parabolic section as:
h = Kx 2
•
The cross-sectional area of flow (for half of the section) is:
h
h
0
0
A = ∫ x dh = ∫
•
(13)
h
2h3 / 2
dh =
K
3 K
(14)
The wetted perimeter (again, half of the section) is:
∑
∆x →0
Wp = lim
2
⎡⎣ f ( x + ∆x ) − f ( x ) ⎤⎦ + ( ∆x )
2
(15)
or,
Wp =
∫
1/ 2
2
⎛
⎞
⎜ ⎛ df ⎞ + 1⎟
⎜ ⎜⎝ dx ⎟⎠
⎟
⎝
⎠
dx
(16)
where f = Kx2. This derivation can be described graphically as follows:
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Gary P. Merkley
where the curve is broken up (discretized) into successive linear segments...
2
∆s ≈ ⎡⎣ f ( x + ∆x ) − f ( x ) ⎤⎦ + ( ∆x )
•
2
(17)
For y = f(x) = Kx2,
2
⎛ df ⎞
2 2
⎜ ⎟ = 4K x
⎝ dx ⎠
•
(18)
Then,
h/K
Wp =
∫
4K 2 x 2 + 1 dx
(19)
0
•
After integration (using integration tables), the wetted perimeter for half of the
parabolic section is:
Wp = K
h⎛h
1 ⎞
1 ⎡ ⎛ h
h
1
+
+
+
+
ln
2K
⎢
⎜
K ⎜⎝ K 4K 2 ⎟⎠ 4K ⎢⎣ ⎜⎝ K
K 4K 2
⎞⎤
⎟⎟ ⎥
⎠ ⎥⎦
(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.
Gary P. Merkley
190
BIE 5300/6300 Lectures