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Chapter
2.1.
2.2.
2.3.
2.4.
2.5.
UNIFORM FLOW
Introduction
Basic equations in uniform open-channel flow
Most economical cross-section
Channel with compound cross-section
Permissible velocity against erosion and sedimentation
Summary
The chapter on uniform flow in open channels is basic knowledge required for all
hydraulics students. In this chapter, we shall assume the flow to be uniform, unless
specified otherwise. This chapter guides students how to determine the rate of discharge,
the depth of flow, and the velocity. The slope of the bed and the cross-sectional area
remain constant over the given length of the channel under the uniform-flow conditions.
The same holds for the computation of the most economical cross section when designing
the channel. The concept of permissible velocity against erosion and sedimentation is
introduced.
Key words
Uniform flow; most economical cross-section; discharge; velocity; erosion; sedimentation
2.1.
INTRODUCTION
2.1.1. Definition
Uniform flow relates to a flow condition over a certain length or reach of a stream
and can occur only during steady flow conditions. Uniform flow may be also defined as the
flow occurring in a channel in which equilibrium has been reached between gravitational
force and shear force. Many irrigation and drainage canals and other artificial channels are
designed to carry water at uniform depth and cross section all along their lengths. Natural
channels as rivers and creeks are seldom of uniform shape. The design discharge is set by
considerations of acceptable risk and frequency analysis, whereas the channel slope and
the cross-sectional shape are determined by topography, and soil and land conditions.
Uniform equilibrium open-channel flows are characterized by a constant depth and a
constant mean flow velocity:
h
V
0
and
0
(2-1)
s
s
where s is the coordinate in the flow direction, h the flow depth and V the flow velocity.
Uniform equilibrium open-channel flows are commonly called “uniform flows” or “normal
flows”.
Note: The velocity distribution in fully-developed turbulent open channel flows is given
approximately by Prandtl’s power law (Fig. 2.1):
V
 y N
 
(2-2)
Vmax  h 
where the exponent 1/N varies from ¼ down to ½ depending on the boundary friction and
the cross-section shape. The most commonly-used power law formulae are the one-sixth
1
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power (1/6) and the one-seventh power (1/7) formulas. It should be noted that the velocity
in open-channel flow is assumed constant over the entire cross-section.
Vmax
V
velocity
distribution
h
v
y
Fig. 2.1. Velocity distribution profile in turbulent flow
Such flow conditions are represented schematically in Fig. 2.2. Considering Bernoulli’s
theorem of the conservation of energy, between cross-sections 1 and 2, leads to the
expression:


p
V2 
p
V2 
(2-3)
E1   z1  1  1 1   E 2  h L   z 2  2   2 2   h L

2g 

2g 


where 1 and 2 are the Corriolis-coefficients corresponding to the velocities V1 and V2,
respectively. They are also called the kinetic-energy correction coefficients.  is equal to
or larger than 1 but rarely exceeds 1.1. (Li and Hager, 1991). For a uniform velocity
distribution,  = 1.
The slope of the energy gradient line S is equal to the bed slope i of the channel, or:
h
S= i L
(2-4)
L


energy-gradient line
hL
V12
2g
V 22
2g
S
E1
p
h1  1
g
hydraulic-gradient line
h2 
i
z1
E2
z2
L

p2
g

Datum
Fig. 2.2. Energy and hydraulic gradient in uniform-flow channel
If the flow is uniform, the cross sections at points 1 and 2 must be constant. Consequently,
the velocity and the depth of flow must also remain constant, or:
V1 = V2
and h1 = h2
(2-5)
The flow resistance in an open channel is more difficult to quantify. The importance of the
resistance coefficient goes beyond its use in channel design for uniform flow.
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2.1.2. Momentum analysis
Consider a control volume of length L in uniform flow, as shown in Fig. 2.3.
L
FP1
A
Wsin
h
FP2 = FP1
oPL

W = gAL
P
Fig. 2.3. Force balance in uniform flow
By definition, the hydrostatic forces, Fp1 and Fp2, are equal and opposite. In addition, the
mean velocity is invariant in the flow direction, so that the change in momentum flux is
zero. Thus, the momentum equation reduces to a balance between the gravity force
component in the flow direction and the resisting shear force:
A L sin = o P L
(2-6)
in which  = g = specific weight of the fluid, A = cross-sectional area of flow, o = mean
boundary shear stress, and P = wetted perimeter of the boundary on which the shear stress
acts. If Eq. (2-6) is divided by PL, the hydraulic radius R = A/P appears as an intrinsic
variable. Physically, Eq. (2-6) represents the ratio of flow volume to boundary surface
area, or shear stress to unit weight, in the flow direction. Eq. (2-6) can be written as:
o = R sin
  RS
(2-7)
if we replace sin with S = tan for small values of . Furthermore, if we solve Eq. (2-7)
for the bed slope, which equals the slope of the energy grade line, hL/L, and express the
shear stress in terms of the friction factor f for uniform pipe flow according DarcyWeisbach:
o
V
2

f
8
(2-8)
we have the Darcy-Weisbach equation (for uniform pipe flow):
i=S=
h f  o f ..V 2
f V2
.



L R
8 R
4R 2g
(2-9)
from which it is evident that the appropriate length scale, when applied to open-channel
flow, is 4R. It seems reasonable to use 4R as the length scale in the Reynolds-number and
the relative roughness as well. Before applying uniform flow formulas to the design of
open channels, the background of Chezy’s as well as Manning’s formulas for steady,
uniform in open channels are presented in the next section.
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2.2. Basic equations in uniform open-channel flow
2.2.1. Chezy’s formula
Consider an open channel of uniform cross-section and bed slope as shown in Fig. 2.4:
L
Q
Let
and
L
A
V
P
f
i
=
=
=
=
=
=
i
VA
Q
Fig.2.4. Sloping bed of a channel
length of the channel;
cross-sectional area of flow;
velocity of water;
wetted perimeter of the cross-section;
friction coefficient according to Darcy-Weisbach;
uniform slope of the bed.
It has been experimentally found, that the total frictional resistance along the length L of
the channel, follows the law:
f
Frictional resistance =    contact area  (velocity)2
8
f
=    P.L  Vn
(2-10)
8
The exponent n has been experimentally found to be nearly equal to 2. But for all practical
purposes, its value is taken to be 2. Therefore,
Frictional resistance =
f
   P.L  V2
8
(2-11)
Since the water moves over a distance V in 1 second, therefore, the work done in
overcoming the friction reads as:
Frictional resistance  distance V in 1 second =
f
f
PLV2V =  PL V3 (2-12)
8
8
The weight of the water, W, in the channel over a length of L is:
W
= .A.L
(2-13)
This water “falls” vertically down over a distance V.i in 1 second, so
Loss of potential energy = Weight of water  Height
= .A.L.V.i
We know that work done in overcoming friction = Loss of potential energy
f
i.e.
   P.L  V3
= .A.L.V.i
8
(2-14)
(2-15)
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
V2
=8
or
V
=
where C =
 .A.i
f . .P
8
A

i
 .f
P
(2-16)
A
8
8g
is known as Chezy’s coefficeint and R =
as hydraulic radius.

 .f
f
P
The discharge of flow then is
Q = A  V = AC Ri
(2-17)
Note: Unlike the Darcy-Weisbach coefficient f , which is dimensionless, the Chezy
coefficient C has the dimension, [L1/2T-1], as mentioned in Chapter 1. Chezy’s coefficient
C depends on the mean velocity V, the hydraulic radius R, the kinematic viscosity  and
the relative roughness. There is experimental evidence that the value of the resistance
coefficient does vary with the shape of the channel and therefore with R and possibly also
with the bed slope i, which for uniform flow will be equal to the slope of the energy-head
line io, yielding a relationship for the velocity of the form:
V = K. Rx.ioy
where K, x and y are constants.
(2-18)
Example 2.1: A rectangular channel is 4 m deep and 6 m wide. Find the discharge through
the channel, when it runs full. Take the slope of the bed as 1:1000 and Chezy’s coefficient
as 50 m1/2s-1.
Solution:
Given: Depth h = 4 m, Width b = 6 m,
h
Bed slope i = 1/1000 = 0.001,
Chezy’s coefficient C = 50 m1/2s-1
b
Q = ? (m3/s)
A=hb
P = b + 2h
Area of the rectangular channel:
Perimeter of the rectangular channel:
Hydraulic radius of the flow:
Discharge through the channel:
= 24 m2
= 14 m
A
= 1.71 m
P
Q = AC Ri = 49.62 m3s-1
R =
Ans.
Example 2.2 : Water is flowing at the rate of 8.5 m3s-1 in an earthen trapezoidal channel
with a bed width 9 m, a water depth 1.2 m and side slope 2:1. Calculate the bed slope, if
the value of C in Chezy’s formula be 49.5 m1/2s-1.
Solution:
Given: Discharge Q = 8,5 m3/s, Bed width b = 9 m,
Depth h = 1.2 m, Side slope m = 2,
h
Chezy’s coefficient C = 49.5 m1/2s-1,
Bed slope = ?
Surface width of the trapezoidal channel
B
2
h/2
h
B = b + 2( )
2
b
1
= 10.2 m
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Area of the trapezoidal channel:
A =
 b  B h
2
 
Wetted perimeter:
Hydraulic radius:
Now using the relation:

2
P = b  2  h 2  h 2 


A
R =
P
Q
= AC Ri
Q2
i
=
R(AC)2
= 11.52 m2
= 11.68 m
= 0.986 m
=
1
4440
Ans.
2.2.2. Manning’s formula
Manning, after carrying out a series of experiments, deduced the following relation
for the value of C in Chezy’s formula:
1
1
C
= R 6
(2-19)
n
where n is the Manning constant in metric units, n = [m-1/3s]. n is expressing the channel’s
relative roughness properties and values are given in Table 2.1 Now we see that the
velocity:
1
1
1
1
1
1
V
= C Ri =  R 6  Ri =  R 6  R 2  i 2
n
n
2
1
1

V
=  R 3 i 2
(2-20)
n
2
1
1
Now, the discharge is: Q = AV = A   R 3  i 2
(2-21)
n
Table 2.1: Values of Manning coefficient n [m-1/3s]
Wetted perimeter
n
Wetted perimeter
n
A. Natural channel
Clean and straight
Sluggish with deep pools
Major rivers
B. Flood plain
Pasture, farmland
Light brush
Heavy brush
Trees
C. Excavated earth channels
Clean
Gravelly
Weedy
Stony, cobbles
0.030
0.040
0.035
0.035
0.050
0.075
0.150
0.022
0.025
0.030
0.035
D. Artificially lined channel
Glass
Brass
Steel, smooth
Steel, painted
Steel, riveted
Cast iron
Concrete, finished
Concrete, unfinished
Planned wood
Clay tile
Brickwork
Asphalt
Corrugated metal
Rubble masonry
0.010
0.011
0.012
0.014
0.015
0.013
0.012
0.014
0.012
0.014
0.015
0.016
0.022
0.025
For a more detailed description, we can take the value of Manning’s n from the Table at
the end of this chapter.
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Example 2.3: An earthen trapezoidal channel with a 3 m wide base and side slopes 1:1
carries water with a depth of 1 m. The bed slope is 1/1600. Estimate the discharge. Take
the value of n in Manning’s formula as 0.04 m-1/3s.
Solution:
Given:
Base width b = 3 m, Side slope = 1:1,
Water depth h = 1 m,
Bed slope 1/1600,
Manning’s coefficient n = 0.04 m-1/3s
Discharge Q = ? (m3/s)
Surface width of the trapezoidal channel
Area of the trapezoidal channel:
Wetted perimeter:
Hydraulic radius:
Now using the relation:
B
h
1
h
b
1
B = b + 2h = 5 m
 b  B
A=
= 4 m2
h
2
P = b  2 h2  h2
A
R =
P
2
1
1
Q = A  R 3 i 2
n
= 5.828 m
= 0.686 m
= 1.94 m3/s
Ans.
Example 2.4 : Water at the rate of 0.1 m3/s flows through a vitrified sewer with a diameter
of 1 m with the sewer pipe half full. Find the slope of the water surface, if Manning’s n is
0.013 m-1/3s.
Solution:
Given: Discharge Q = 0.1 m3/s,
Diameter of pipe D = 1 m,
Manning’s n = 0.013 m-1/3s
Sewer slope i = ?
Area of the flow:
A=
Wetted perimeter:
P =
Hydraulic radius:
Using Manning’s formula:
 Water surface slope:
1   D2 
2

 = 0.393 m
2 4 
D
2
A
R=
P
D
= 1.57 m
= 0.25 m
2
1
1
Q = A  R 3 i 2
n
 Qn 
1
i =
=
2 
 A  R 3  1430
2
Ans.
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2.2.3. Discussion of factors affecting f and n
The dependence of f on the relative roughness in open channel flow is not as well
known as in pipe flow, because it is difficult to assign an equivalent sand-grain roughness
to the large values of the absolute roughness height typically found in open channels. The
dependence of flow resistance on the cross-sectional shape occurs as a result of changes of
both the channel hydraulic radius, R, and the cross-sectional distribution of velocity and
shear.
There is no substitute for experience in the selection of Manning’s n for natural channels.
Table 2.2. (at the end of this chapter) from Ven Te Chow (1959) gives an idea of the
variability to be expected in Manning‘s n.
2.3. MOST ECONOMICAL CROSS-SECTION
2.3.1. Concept
A typical uniform flow problem in the design of an artificial canal is the
economical proportioning of the cross-section. A canal, having a given Manning
coefficient n and a slope i, is to carry a certain discharge Q, and the designer’s aim is to
minimize the cross-sectional area. Clearly, if A is to be a minimum, the velocity V is to be
a maximum. The Chezy and Manning formulas indicate, therefore, that the hydraulic
radius R = A/P must be a maximum. It can be shown that the problem is equivalent to that
of minimizing P for a given constant value of A. This concept has a practical application in
estimating the cost for a canal excavation and /or lining.
From economic considerations of minimizing the flow cross-sectional area for a given
design discharge, a theoretically optimum cross-section will be introduced.
2.3.2. Conditions for maximum discharge
The conditions for maximum discharge for the following cross-sections will be
dealt with: (a). Rectangular cross-section, and (b). Trapezoidal cross-section.
(a). Channel with rectangular cross-section
Consider a channel of rectangular cross-section
as shown in Fig. 2.5.
Let
h
b = breadth of the channel, and
h = depth of the channel.
Area of flow:
A=b h b=
Discharge:
Q=AV
b
A
h
(2-20) Fig.2.5. A rectangular channel
= AC Ri  AC
A
i
P
(2-22)
Keeping A, C and i constant in the above equation, the discharge will be maximum when
A/P is maximum or the wetted perimeter P is minimum. Or in other words, when:
dP
0
(2-23)
dh
A
We know that
P = b + 2h =  2h
(2-24)
h
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Differentiating the above equation with respect to h and equating to zero yields:
dP
A
 2 20
dh
h
2

A = 2h = b  h
or
b = 2h
i.e. the breadth is two times the depth
In this case, the hydraulic radius is:
A
bh
2h  h
2h 2 h
R 



P b  2h 2h  2h 4h 2
(2-25)
(2-26)
(2-27)
Hence, we can say that for the maximum discharge or the maximum velocity, these two
conditions (i.e. b = 2h and R = h/2) should be used for solving the problem of optimizing
channels of rectangular cross-section.
1.00
0.95
Q
Q max 0.90
h
b
A = bh = constant
0.85
0
1
2
3
4
Fig. 2.6. Experimental relationship between
5
b
h
Q
b
and
h
Qmax
As can be seen in Fig. 2.6, the maximum represented by this optimal configuration is a
b
rather weak one. For example, for aspect ratios, , between 1 and 4, the flow rate is within
h
96% of the maximum flow rate obtained with the same area and by b/h = 2.
Example 2.5.: Find the most economical cross-section of a rectangular channel to carry
0.3 m3/s of water, when the bed slope is 1/1000. Assume Chezy’s C = 60 m-1/3s-1.
Solution:
Given: Discharge Q = 0.3 m3/s, Bed slope i = 1/1000, Chezy coefficient C = 60 m-1/3s-1
Breadth of channel b = ? (m) and depth of the channel h = ? (m)
We know that for the most economical rectangular section:
b = 2h
Area:
A = b  h = 2h  h = 2h2
and hydraulic radius: R = h/2
= 0.5 h
Using the relation:
Q = AC Ri
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and squaring both sides yields:
0.09 m5  7.2h 5
or
h 5  0.0125  h  5 0.0125 m
Depth of the channel: h = 0.416 m
And breadth:
b = 2d = 0.832 m
Ans.
Ans.
(a). Channel with trapezoidal cross-section
Consider a channel of trapezoidal cross-section ABCD as shown in Fig. 2.7.
B = b + 2nh
D
C

B nh
1
n
A
b
h
Fig.2.7. A trapezoidal channel
Let
b = breadth of the channel at the bottom,
h = depth of the channel, and
1
= side slope (i.e. 1 vertical to n horizontal)
n
Area of flow:
or
Discharge:
A = h(b + nh)
A
A
= b + nh  b =
- nh
h
h
Q = A  V  AC Ri  AC
(2-28)
A
i
P
(2-29)
Keeping A, C and i constant in the above equation, the discharge will be maximum, when
A/P is maximum or the wetted perimeter P is minimum. Or in other words:
dP
0
dh
We know that:
P = b  2 n 2 h 2  h 2  b  2h n 2  1
(2-30)
Substituting the value of b from equation (2-28) yields:
A
P=
(2-31)
 nh  2h n 2  1
h
Differentiating the above equation with respect to d and equating to zero results into:
dP
A
  2  n  2 n2  1   0
(2-32)
dh
h
A
 n  2 n2 1
h2
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h(b  nh)
 n  2 n2 1
2
h
[A = h(b + nh)]
b  nh
h
 n    2 n2 1
h
h
b  2nh
 h n2 1
2
or
(2-33)
We see that b + 2nh = B is the top width of the channel and h n 2  1 is the length of the
sloping side,
i.e.
the length of the sloping side is equal to half the top width
In this case, the hydraulic radius:
A
h(b  nh)
h(b  nh)
h(b  nh) h
R=




2
P b  2h n  1 b  (b  2nh) 2(b  nh) 2
(2-34)
Hence, we can say that for the maximum discharge or the maximum velocity, these two
b  2nh
h
conditions (i.e.
 h n 2  1 and/or R = ) should be used for solving the problems
2
2
in the case of channels of trapezoidal cross-section.
Example 2.6: A canal of trapezoidal cross-section has to be excavated through hard clay at
the least cost. Determine the dimensions of the channel for a discharge equal to 14 m3/s, a
side slope for hard clay n = 1:1, a bed slope 1:2500 and Manning’s n = 0.02 m-1/3s.
Solution:
Given: Discharge Q = 14 m3/s;
Bed slope i = 1/2500;
Side slope n = 1:1
Manning’s n = 0.02 m-1/3s
Breath b = ? (m)
Depth h = ? (m)
h
1
h
b
1
We know that for the least cost:
half of the top width = length of sloping side
b  2nh
2
 h n 1
2
with n =1  b  2h  2h 2  2.83h
b = 0.83 h
Area of flow: A = h(b + nh) = 1.83 h2
Using Manning’s formula:
2
1
1
Q = A  R 3 i 2
n
yields:
And
h 3  12.142 m8/3
h  2.55 m
Ans.
b = 0.83 h = 2.12 m
Ans.
8
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Note: The semi-circular section (the semi-circle having its center in the surface) is the best
hydraulic section. The best hydraulic cross-section for other shapes can be drawn as
presented in Fig. 2.8.
D
R
D
2
R
D
2
R
D
2
R
2
60
90
trapezoidal
triangular channel
channel
Fig. 2.8. Cross-sections of maximum flow rate: i.e. “optimum design”
circular channel
rectangular channel
Students should try to proof the conditions for circular and triangular channels for the best
hydraulic cross-section based on the below relationship:
Q 1 2 1
1 A 1
V     .R 3 .i 2  .   .i 2
A n
n P
2
3
 A   Q.n = constant

 P  i 
5
2
3
3
1
(2-25)
2
Table 2.2.: The “best” design summary for several channel cross-sections
Cross-section
Optimum
width B
Optimum
wetted perimeter P
 D2
8
D
2
D
D2
2
2D
2D
3 2
D
4
Semi-circle
D
Rectangular
Trapezoidal
3
Triangular
Optimum
cross-section A
2D
D2
2
Optimum
hydraulic radius R
D
4
D
4
D
4
3D
D
4
2D
2.3.3. Problems of uniform-flow computation
The computation of uniform flow may be performed by the use of two equations:
the continuity equation and a uniform-flow formula. When the Manning formula is used as
the uniform-flow formula, the computation will involve the following six variables:
(1) the normal discharge Q
(2) the normal depth
h
(3) the channel slope
i
(4) the mean velocity of flow V
(5) the coefficient of roughness
(6) the geometric elements that
depend on the shape of the
channel section, such as
n
A,R, etc, …
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When any four of the above six variables are given, the remaining two unknowns can be
determined by the two equations. The following represents some types of problems of
uniform flow computation:
A. to compute the normal discharge:
In practical applications, this computation is required for the determination of the
capacity of a given channel or for the construction of a sysnthetic rating curve of
the channel.
B. to determine the velocity of the flow:
This computation has many applications. For example, it is often required for the
study of scouring and silting effects in a given channel.
C. to compute the normal depth:
This computation is required for the determination of the stage of flow in a given
channel.
D. to determine the channel roughness:
This computation is used to ascertain the roughness coefficient in a given channel;
the coefficient thus determined may be used in other similar channels.
E. to compute the channel slope:
This computation is required for adjusting the slope of a given channel.
F. to determine the dimensions of the channel section:
This computation is required mainly for design purposes.
Table 2.3. lists the known and unknown variables involved in each of the six types of
problems mentioned above.
Table 2.3: Problems of uniform-flow computation
Type of
problem
Discharge
Q
Velocity
V
Depth
d
Roughness
n
Slope
i
Geometric
elements
A
B
C
D
E
F
?




?
-


?






?






?






?
Notes:
 The known variables are indicated by the check mark () and the unknown required
in the problems by the question mark (?). The unknown variable(s) that can be
determined from the known variables is(are) indicated by a dash (-).
 Table 2.3. does not include all types of problems. By varying combinations of
various known and unknown variables, more types of problems can be formed. In
design problems, the use of the best hydraulic section and of empirical rules is
generally introduced and thus new types of problems are created.
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2. 4.
CHANNEL WITH COMPOUND CROSS-SECTION
A compound channel consists of a main channel, which carries the base flow
(frequently running off up to bank-full conditions), and a floodplain on one or both sides
that carries over-bank flow during the time of flooding as sketched in Fig. 2.9. The
compound cross-section of a channel may be composed of several distinct subsections with
each subsection different in roughness form the others.
canal bank
high water level
dyke
low water level
(1): A1, P1, n1
(2): A2, P2, n2
(3): A3, P3, n3
Fig.2.9. Over-bank flow in a compound channel
The roughness of the side channels will be different (generally rougher) from that of the
main channel and the method of analysis is to consider the total discharge to be the sum of
the component discharges computed by the Manning equation. The mean velocity for the
whole channel section is equal to the total discharge divided by the total water area. The
classical method of computation of discharge, as presented by Chow in 1959, consisted in
subdividing the composite cross-section into sub-areas with vertical interfaces in which the
shear stresses are neglected. The discharge for each sub-area is calculated by assuming a
common friction slope i for the whole channel. Thus in the channel, as shown in Fig. 2.9,
assuming that the bed slope is the same for the three sub-areas, it holds:
AR 3
Q   Qi  i  i i
ni
i 1
i 1
m
m
2
(2-26)
The division of the channel by these artificial vertical boundaries assumes implicity that
the shear stress on these interfaces is relatively small with respect to the shear stress acting
on the wetted perimeter of the channel.
Note: It has been found by more recent experimentation that this hypothesis is incorrect
and that it leads to a considerable over-estimation of the discharge in the compound
channel.
Example 2.7: Water flows along a drainage canal having the properties shown in the figure
below. If the bottom slope i = 1/500=0.002, estimate the discharge.
3m
0.6 m
n1 = 0.020 (1)
2m
(2)
n2 = 0.015
3m
n3 = 0.030
0.8 m
(3)
[ni] = m-1/3s
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Solution:
We divide the cross-section into three subsections as is indicated in the figure and write the
discharge as Q = Q1 + Q2 + Q3, where for each section, it holds:
Qi =AiVi = A i 
1 2 3 12
Ri i
ni
The appropriate values of Ai, Pi, Ri and ni are listed in the table below:
i
(1)
(2)*
(3)
Ai (m2)
1.8
2.8
1.8
Pi (m)
3.6
3.6
3.6
Ri (m)
0.500
0.778
0.500
ni (m-1/3s)
0.020
0.015
0.020
Note that the imaginary portions of the wetted perimeter between the sections (denoted by
the dashed lines in the figure) are not implemented in Pi. That is, for section (2):
So that
A2 = 2  (0.8 + 0.6) m2
P2 = {2 + 2(0.8)} m
A
2.8
R2  2 
m
P2 3.6
= 2.8 m2
= 3.6 m
= 0.778 m
Thus the total discharge is:
Q = Q1 + Q2 + Q3 =
2
3
2
2

3
3
1
1.8
0.500
2.8
0.778
1.8
0.500










2
Q =  0.002 



0.020
0.015
0.030


 m3s-1


Q = 11.275 m3/s
Ans.
If the entire channel cross-section were considered as one flow area, then :
A = A1 + A2 + A3
= 6.4 m2
P = P1 + P2 + P3
= 10.8 m
A
Then
R=
= 0.593 m
P
The total discharge can be written as
1
1 2 1
Q=
A  R 3i 2
n eff
n
where neff is the effective value of n for the whole compound channel.
With Q = 11.275 m3/s, as determined above, the value of neff is found to be:
A R 3 i 2
n eff 
= 0.01681 m-1/3s
Q
As expected, the effective roughness (Manning’s n) is between the minimum (n2 = 0.015
m-1/3s) and maximum (n3 = 0.030 m-1/3s) values for the individual subsections.
2
1
----------------------------------------------------------------------------------------------------------------------------------Chapter 2: UNIFORM FLOW
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2.5. PERMISSIBLE VELOCITY AGAINST EROSION AND SEDIMENTATION
The excavation and lining cost of open channels or conduits varies with their size.
With respect to water-resources-system economics, erosion of and sedimentation in
channels are problems in hydraulic engineering. Erosion and sedimentation must be
predicted because they can change the bed slope, the channel width and therefore the flow
conditions.
So, if the available slope permits, the cost of the initial construction may be reduced by
using the highest velocity. However, if the velocity is becoming too high, the channel may
be damaged or destroyed by erosion. This must be avoided by limiting the velocities
according to the boundary materials. For clear water in hard-surfaced water conductors, the
limiting velocity is beyond practical requirements. Velocities above 10 m/s for clear water
in concrete channels have been found to do no harm. If the water carries abrasive material,
damage may occur at lower velocities. No definite relation has been established between
the nature of abrasive materials, the material of channel bank and bed, and a permissible
velocity.
In unlined earthen channels, the limiting velocity involves many factors. Generally, a fine
soil is eroded more easily than a coarse one, but the effect of the grain size may be
obscured by the presence or absence of a cementing or binding material. The tendency to
erode is reduced by seasoning. Groundwater conditions can exert an important influence.
Seepage out of the channel, particularly if the water is turbid, tends to toughen the banks,
whereas infiltration reduces the resistance to erosion. Erosion can be reduced or avoided by
designing for low velocities.
If the water carries an appreciable amount of silt in suspension, too low a velocity will
cause the canal to fill up until its capacity is impaired. It is necessary to choose a velocity
that will keep the silt in motion but that will not erode the bank or bottom of the canal. The
margin of permissible velocities depends on the amount and nature of the silt in the water,
the nature of the bank material, the size and shape of the canal, and many other factors.
The silt content of most turbid water varies with the season, as does also the demand for
water and the resultant velocity of the flow.
The determination of non-scouring, non-silting velocities for earthen canals has attracted
the attention of many investigators over a long period of time, and a considerable mass of
data and formulas have been accumulated. However, for preliminary purposes, and for
design in many cases, use may be made of the approximate values purposed by Fortie and
Scobey, in 1926, as shown in Table 2.4. Where the silt is important, it is better to make the
slope a little too steep rather than a little too flat. A gradient that proves to be too steep can
be controlled by checks. In hard-surfaced channels, silting is easily controlled if fall for
scouring velocity is available.
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OPEN CHANNEL HYDRAULICS FOR ENGINEERS
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Table 2.4: Permissible canal velocities (Fortier and Scobey, 1926)
Original material excavated for canal
(1)
Fine sand (non-colloidal)
Sandy loam (non-colloidal)
Sandy loam (non-colloidal)
Alluvial silts when non-colloidal
Ordinary firm loam
Volcanic ash
Fine gravel
Stiff clay (very colloidal)
Graded, loam to cobbles, when non-colloidal
Alluvial silts when colloidal
Graded, loam to cobbles, when colloidal
Coarse gravel (non-colloidal)
Cobbles and shingles
Shales and hardpans
Velocity, m/s, after aging, of canal carrying:
Clear
Water
Water transporting
water, no
transporting
non-colloidal silts,
detritus
colloidal silts
sands, gravels, or
rock fragments
(2)
(3)
(4)
0.46
0.76
0.46
0.53
0.76
0.61
0.61
0.91
0.61
0.61
1.07
0.61
0.76
1.07
0.69
0.76
1.07
0.61
0.76
1.52
1.14
1.14
1.52
0.91
1.14
1.52
1.52
1.14
1.52
0.91
1.22
1.68
1.52
1.22
1.83
1.98
1.52
1.68
1.98
1.83
1.98
1.52
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OPEN CHANNEL HYDRAULICS FOR ENGINEERS
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Table 2.5: Value of Manning’s Roughness Coefficient n [m-1/3s] (Ven Te Chow, 1973)
Type of channel and description
A. Closed Conduits Flowing Party Full
A.1. Metal
a. Brass, smooth
b. Steel
1. Lock bar and welded
2. Riveted and spiral
c. Cast iron
1. Coated
2. Uncoated
d. Wrought iron
1. Black
2. Galvanized
e. Corrugated metal
1. Subdrain
2. Storm drain
A.2. Nonmetal
a. Lucite
b. Glass
c. Cement
1. Neat surface
2. Mortar
d. Concrete
1. Culvert, straight and free of debris
2. Culvert with bends, connections, and some
debris
3. Finished
4. Sewer with manholes, inlet, etc., straight
5. Unfinished, steel form
6. Unfinished, smooth wood form
7. Unfinished, rough wood form
e. Wood
1. Stave
2. Laminated, treated
f. Clay
1. Common drainage tile
2. Vitrified sewer
3. Vitrified sewer with manholes, inlet, etc.
4. Vitrified subdrain with open joint
g. Brickwork
1. Glazed
2. Lined with cement mortar
h. Sanitary sewers coated with sewage slimes, with
bends and connections
i. Paved invert, sewer, smooth bottom
j. Rubble masonry, cemented
Minimum
Normal
Maximum
0.009
0.010
0.013
0.010
0.013
0.012
0.016
0.014
0.017
0.010
0.011
0.013
0.014
0.014
0.016
0.012
0.013
0.014
0.016
0.015
0.017
0.017
0.021
0.019
0.024
0.021
0.030
0.008
0.009
0.009
0.010
0.010
0.013
0.010
0.011
0.011
0.013
0.013
0.015
0.010
0.011
0.011
0.013
0.013
0.014
0.011
0.013
0.012
0.012
0.015
0.012
0.015
0.013
0.014
0.017
0.014
0.017
0.014
0.016
0.020
0.010
0.015
0.012
0.017
0.014
0.020
0.011
0.011
0.013
0.014
0.013
0.014
0.015
0.016
0.017
0.017
0.017
0.018
0.011
0.012
0.012
0.013
0.015
0.013
0.015
0.017
0.016
0.016
0.018
0.019
0.025
0.020
0.030
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OPEN CHANNEL HYDRAULICS FOR ENGINEERS
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Type of channel and description
B. Lined or built-up channels
B.1. Metal
a.
Smooth steel surface
1. Unpainted
2. Painted
b. Corrugated
B.2. Nonmetal
a.
Cement
1. Neat surface
2. Mortar
b. Wood
3. Planed, untreated
4. Planed, creosoted
5. Unplanted
6. Plank with battens
7. Lined with roofing paper
c. Concrete
1. Trowel finish
2. Float finish
3. Finished, with gravel on bottom
4. Unfinished
5. Gunite, good section
6. Gunite, navy section
7. On good excavated rock
8. On irregular excavated rock
d. Concrete bottom float finished with sides of
1. Dressed stone in mortar
2. Random stone in mortar
3. Cement rubble masonry, plastered
4. Cement rubble masonry
5. Dry rubble or riprap
e. Gravel bottom with sides of
1. Formed concrete
2. Random stone in mortar
3. Dry rubble or riprap
f. Brick
1. Glazed
2. In cement mortar
g. Masonry
1. Cemented rubble
2. Dry rubble
h. Dressed ashlar
i. Asphalt
1. Smooth
2. Rough
j. Vegetal lining
Minimum
Normal
Maximum
0.011
0.012
0.021
0.012
0.013
0.025
0.014
0.017
0.030
0.010
0.011
0.011
0.013
0.013
0.015
0.010
0.011
0.011
0.012
0.010
0.012
0.012
0.013
0.015
0.014
0.014
0.015
0.015
0.018
0.017
0.011
0.013
0.015
0.014
0.016
0.018
0.017
0.022
0.013
0.015
0.017
0.017
0.019
0.022
0.020
0.027
0.015
0.016
0.020
0.020
0.023
0.025
0.015
0.017
0.016
0.020
0.020
0.017
0.020
0.020
0.025
0.030
0.020
0.024
0.024
0.030
0.035
0.017
0.020
0.023
0.020
0.023
0.033
0.025
0.026
0.036
0.011
0.012
0.013
0.015
0.015
0.018
0.017
0.023
0.013
0.025
0.032
0.015
0.030
0.035
0.017
0.013
0.016
0.030
0.013
0.016
….
0.500
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OPEN CHANNEL HYDRAULICS FOR ENGINEERS
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Type of channel and description
C. Excavated or dredged
a. Earth, straight and uniform
1. Clean, recently completed
2. Clean, after weathering
3. Gravel, uniform section, clean
4. With short grass, few weeds
b. Earth, winding and sluggish
1. No vegetation
2. Grass, some weeds
3. Dense weeds or aquatic plants in deep channels
4. Earth bottom and rubble sides
5. Stony bottom and weedy banks
6. Cobble bottom and clean sides
c. Dragline-excavated or dredged
1. No vegetation
2. Light brush on banks
d. Rock cuts
1. Smooth and uniform
2. Jagged and irregular
e. Channels not maintained, weeds and brush uncut
1. Dense weeds, high as flow depth
2. Clean bottom, brush on sides
3. Same, highest stage of flow
4. Dense brush, high stage
D. Natural streams
D.1. Minor stream (top width at flood stage < 100 ft)
a. Streams on plain
1. Clean, straight, full stage, no rifts or deep pools
2. Same as above, but no more stones and weeds
3. Clean, winding, some pools and shoals
4. Same as above, but some weeds and stones
5. Same as above, lower stages, more ineffective
slopes and sections
6. Same as 4, but more stones
7. Sluggish reaches, weedy, deep pools
8. Very weedy reaches, deep pools, or floodways
with heavy stand of timber and underbrush
b. Mountain stream, no vegetation in channel, banks
usually steep, trees and brush along banks
submerged at high stages
1. Bottom: gravels, cobbles, and few boulders
2. Bottom: cobbles with large boulders
Minimum
Normal
Maximum
0.016
0.018
0.022
0.022
0.018
0.022
0.025
0.027
0.020
0.025
0.030
0.033
0.023
0.025
0.030
0.028
0.025
0.030
0.025
0.030
0.035
0.030
0.035
0.040
0.030
0.033
0.040
0.035
0.040
0.050
0.025
0.035
0.028
0.050
0.033
0.060
0.025
0.035
0.035
0.040
0.040
0.050
0.050
0.040
0.045
0.080
0.080
0.050
0.070
0.100
0.120
0.080
0.110
0.140
0.025
0.030
0.033
0.035
0.040
0.030
0.035
0.040
0.045
0.048
0.033
0.040
0.045
0.050
0.055
0.045
0.050
0.075
0.050
0.070
0.100
0.060
0.080
0.150
0.030
0.040
0.040
0.050
0.050
0.070
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OPEN CHANNEL HYDRAULICS FOR ENGINEERS
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Type of channel and description
D. Natural streams
D.2. Flood plains
a. Pasture, no brush
1. Short grass
2. High grass
b. Cultivated areas
1. No crop
2. Mature row crops
3. Mature field crops
c. Brush
1. Scattered brush, heavy weeds
2. Light brush and trees, in winter
3. Light brush and trees, in summer
4. Medium to dense brush, in winter
5. Medium to dense brush, in summer
d. Trees
1. Dense willows, summer, straight
2. Cleared land with tree stumps, no sprouts
3. Same as above, but with heavy growth of sprouts
4. Heavy stand of timber, a few down trees, little
undergrowth, flood stage below branches
5. Same as above, but with flood stage reaching
branches
D.3. Major streams (top width at flood stage > 100 ft). The n
value is less than that for minor streams of similar
description, because banks offer less effective resistance.
a. Regular section with no boulders or brush
b. Irregular and rough section
Minimum
Normal
Maximum
0.025
0.030
0.030
0.035
0.035
0.050
0.020
0.025
0.030
0.030
0.035
0.040
0.040
0.045
0.050
0.035
0.035
0.040
0.045
0.070
0.050
0.050
0.060
0.070
0.100
0.070
0.060
0.080
0.110
0.160
0.110
0.030
0.050
0.080
0.150
0.040
0.060
0.100
0.200
0.050
0.080
0.120
0.100
0.120
0.160
0.025
0.035
….
…
0.060
0.100
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45