NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 33
Dr. Niranjan Sahoo
IIT-Guwahati
DYMAMICS OF FLUID FLOW
UNIFORM DEPTH OPEN CHANNEL FLOW
dy
0 , then it is treated as
dx
In an open channel, if the depth of flow is constant
uniform depth. In some cases, the uniform depth flow can be accomplished by
adjusting the bottom slope so that it precisely equals to the slope of the energy line.
Physically, the loss of potential energy of the fluid as it flows downhill is exactly
balanced by the dissipation of energy through viscous effects.
Consider the flow in an open channel as shown in Fig. 1. The cross-sectional area is
constant in shape and size. If the cross-sectional area is A and the wetted perimeter
(i.e. length of perimeter of the cross section in contact with fluid) is P , then a new
parameter may be defined as hydraulic radius i.e.
Rh
Free surface
A
P
Free surface
Flow in
(Q)
a
(1)
Flow out
(Q)
Section at a-a
Flow area (A)
Wetted
perimeter (P)
Fig. 1: Concept of hydraulic radius.
Since the fluid must adhere to the solid surface, the actual velocity distribution is not
uniform. The velocity is maximum on the free surface and becomes zero on the wetted
perimeter where the wall shear stress w is developed.
Chezy and Manning Equations
The fundamental relations used to determine the uniform flow rate in open channel are
semi-empirical and are governed by Chezy and Manning equations. Consider a control
volume flow with volume flow rate of Q and weight W in an open channel as shown in
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 33
Dr. Niranjan Sahoo
IIT-Guwahati
Fig. 2. Since it is uniform depth flow, so it can be shown from continuity equation that
V1 V2 . Then, x -momentum equation for the control volume can be written as,
Q V2 V1 0
(1)
or, F1 F2 w Pl W sin 0
l
Control surface
= 0
F1
y2 = y1
V2 = V1
F2
wPl
(1)
(2)
Fig. 2: Control volume flow in an open channel.
As the flow is at uniform depth, so the hydrostatic pressure forces across either end of the
control volume balance each other i.e. F1 F2 . Thus, Eq. (1) becomes,
W sin Al S0
Rh S0
Pl
Pl
(2)
where is the specific weight of the fluid and sin tan S0 (since S0 = 1) . More
often the open channel flows are turbulent rather that laminar. So, Reynolds number is
quite large and for such flows, the wall shear stress w is proportional to dynamic
pressure and may be written as
w K
V 2
2
(3)
where K is a constant that depends on the roughness of the pipe. Now, equating Eqs. (2)
and (3), we get,
V C Rh S0
(4)
This equation is known as Chezy equation in which C is called Chezy coefficient. Its
value is determined from experiment and also, it has the unit of
(m)1 2
.
s
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 33
Dr. Niranjan Sahoo
IIT-Guwahati
This equation is modified by Manning by incorporating dependence of hydraulic
radius on the bottom slope and is given by
V
Rh2 3 S01 2
n
(5)
In this equation, the parameter n is the Manning resistance coefficient. Its value depends
on the surface material of the channels wetted perimeter and is obtained from
experiment. It is also not dimensionless and has the unit of
13
. The typical values of
n are given in Table 1.
Table 1: Values for Manning coefficient n (Ref. 1; Table 10.1)
Wetted perimeter
Natural channels
Floodplains
Excavated earth channel
Artificially lined channel
n
0.03 - 0.45
0.035-0.15
0.022-0.035
0.01-0.025
In open channel flows, sometimes it is necessary to determine the best possible
hydraulic cross-section (i.e. minimum area) for a given flow rate Q , slope S0 and
roughness coefficient n . The flow rate can be written as,
23
A
A S01 2
P
Q
n
(6)
which can be rearranged as,
3
nQ 5
A 1 2 P 2 5
S0
(7)
For a channel with given flow rate the quantity in the parentheses is constant which
means the channel with minimum A is also with minimum P .
GRADUALLY AND RAPIDLY VARIED OPEN CHANNEL FLOW
The depth of the open channel flow varies (either increases or decreases) in the flow
direction depending upon the bottom slope and energy line slope. Physically, the
3
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 33
Dr. Niranjan Sahoo
IIT-Guwahati
difference between component of weight and shear forces in the direction of flow
produces a change in fluid momentum. Thus, there is a change in velocity and
consequently a change in depth. The shape of the surface y y x can be calculated by
solving the governing equation obtained form combination of Manning equation and
energy equation. The result will be a nonlinear differential equation, which is beyond the
scope of this syllabus. However, some physical interpretation of gradually varied flows
can be made from the following equation;
dy S f S0
dx 1 Fr2
(8)
S f S0
dy
0 , the factor
2 becomes a non-zero quantity, which is essentially the
dx
1 Fr
For
gradually or rapidly varying flow. Now, the sign of
dy
i.e. whether the flow depth
dx
increases or decreases with distance along the channel depends on both numerator and
denominator of Eq. (8). The sign of denominator depends on whether the flow is subcritical or super-critical. In fact, for a given channel, there exists a critical slope
S0 S0c
and a corresponding critical depth
y yc
that leads to Fr 1 under
conditions of uniform flow.
The character of a gradually varied flow is classified in terms of actual channel slope
S0
compared to that of slope required for producing uniform flow S0c . They may be,
Mild slope with S0 S0c (i.e. the flow would be sub-critical Fr 1 , if it were of
uniform depth)
Steep slope with S0 S0c (the flow would be super-critical Fr 1 , if it were of
uniform depth)
Horizontal slope with S0 0
Adverse slope, S0 0 (i.e. flow uphill)
Thus, it may be inferred from the above that the determination of whether the flow is
sub-critical or super-critical depends solely on whether S0 S0c or S0 S0c respectively.
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 33
Dr. Niranjan Sahoo
IIT-Guwahati
For gradually varying flows, the conditions for sub-critical and super-critical flows are
met by S0 and depth of flow. For example, with S0 S0c , it is possible to have either
Fr 1 or Fr 1 depending upon the depth of flow i.e. y yc or y yc respectively.
A rapidly varied flow in an open channel is characterized by
dy
: 1 i.e. the flow
dx
depth changes occur over a relatively short distance. One such example of a rapidly
varied flow is hydraulic jump in which flow changes from a relatively shallow, highspeed condition into a relatively deep, low-speed condition within a horizontal distance
of just a few channel depths. The mathematical analysis of hydraulic jump will be
discussed in the subsequent lectures. Many open channel flow-measuring devices are
based on the principle associated with rapidly varied flows. These devices include broadcrested weirs, sharp-crested weirs and sluice gate.
Example 1
Water flows in an open channel of trapezoidal cross section with a velocity of 0.9m/s at a
rate 14m3/s. The bed slope and side slopes are 1:2500 and 1:1 respectively. Find the depth
and bottom width of the channel. Take Chezys constant as 40.4.
Solution
Given that,
Discharge, Q 14m3 /s ; Velocity, V 0.9m/s ; Bed slope, S0
Area of the flow, A
1
; Side slope, N 1
2500
Q
15.57m 2
V
2
V 1
By Chezys formula, V C Rh S0 Rh
C S0
0.9
40.4
2500 1.24m
The area of the flow and wetted perimeter for a trapezoidal section is given by,
A b Ny y 15.57m 2
P b 2 y N 2 1 b 2.83 y
By definition of hydraulic radius,
Rh
A
15.57
1.24
P b 2.83 y
b 2.83 y 12.55
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 33
Dr. Niranjan Sahoo
IIT-Guwahati
Thus,
A b Ny y b y y 15.57 12.55 2.83 y y y 15.57
12.55 y 1.83 y 2 15.57
y 2 6.86 y 8.5 0
Solving the above quadratic equation, we get, y 1.63m ; Thus, b 7.92 m .
Hence, the required depth and bottom width of the channel are 1.63m and 7.92m
respectively.
Example 2
A trapezoidal channel with base width 2m and side slope of 1:2 carries water with a depth
of 1m. The bed slope is 1 in 625. Calculate the discharge and average shear stress at the
channel boundary. Take Manning coefficient as n 0.03 .
Solution
Given that,
b 2m; y 1m; N 1/ 2 ; S0
1
; Side slope, N 1
625
The area of the flow and wetted perimeter for a trapezoidal section is given by,
A b Ny y 2.5m 2
P b 2 y N 2 1 4.24m
By definition of hydraulic radius,
Rh
A 2.5
0.59
P 4.24
Using Mannings formula,
2
R 2 3 S 1 2 0.59 3 1 625
V h 0
n
0.03
0.5
0.95m/s
Discharge, Q A V 2.375m 3 /s
Average shear stress at the channel boundary,
gRh S0 1000 9.81 0.59 1/ 625 9.26 N m 2
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 33
Dr. Niranjan Sahoo
IIT-Guwahati
Example 3
Water flows in channel (cross-section of the shape of isosceles triangle) of bed width a
and sides making an angle 450 with the bed. Determine the relation between depth of flow
d and the bed width a for the condition of: (i) maximum velocity; (ii) maximum
discharge. Use Mannings formula.
Solution
Water flow in a channel of isosceles triangular cross-section is shown in the following
figure for which depth of flow and bed width are d and a , respectively.
(a/2)
d
45 deg.
45 deg.
(i)
By Manning equation,
Rh2 3 S01 2
n
The area of the flow and wetted perimeter for the above cross-section is given by,
a 2d a
d A
EF JH
d a d d ; a 2d
d
2
2
d d
d P
P EF FJ EH a 2 2d ; 2 2
d d
A
For a given bed slope, the velocity will be maximum when,
d A P
d Rh
d
0;
0; P
d d
dd
d
Thus by substitution,
A
d P
0
A
d
d d
2.83d 2 2ad a 2 0
Solving the above quadratic equation,
d 0.34a
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 33
Dr. Niranjan Sahoo
IIT-Guwahati
1
ARh2 3 S01 2 1 A5 3 2
(ii) Discharge, Q AV
.
2 S0
n
n P
1
For a given bed slope, the discharge will be maximum when,
d A5 P 2
dd
Substituting
0; 5P d
A
d A
0
2 A
d d
d d
d P
d A
and , we get,
d d
d d
22.63d 2 1.5ad 5a 2 0
d 0.44a
Solving the above quadratic equation,
Example 4
A rectangular channel of 5m width and 1.2m deep has a slope of 1 in 1000 and is lined
with rubber for which the Mannings coefficient is 0.017. It is desired to increase the
discharge to a maximum by changing the section so that the channel has same amount of
lining. Find the new dimensions and probable increase in discharge.
Solution
Using Mannings formula, the discharge through the channel is given by,
Q1 AV
.
Here, S0
ARh2 3 S01 2
n
1
A
; A 5 1.2 6m 2 ; P 5 2 1.2 7.4m; Rh 0.81
1000
P
3
Substituting the values, Q1 9.7 m s
Let b and y be the width and depth of the flow for the new section of the channel. In
order to have the same amount of lining,
P b 2 y 7.4
For the discharge to be maximum in a rectangular channel, it can be proved that
b 2y
Solving above two equation, y 1.85m; b 3.7m
So, the area of cross-section and hydraulic radius for new channel cross-section becomes,
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 33
Dr. Niranjan Sahoo
IIT-Guwahati
A 6.845m 2 ; Rh
A
0.925m
P
By Mannings formula, new discharge, Q2 AV
.
Percentage increase in discharge
ARh2 3 S01 2
12.1m3 s
n
Q2 Q 1
100 24.45%
Q1
EXERCISES
1. Design a concrete lined channel (trapezoidal cross section) to carry a discharge of
480m3/s at a velocity of 2.3m/s. The bed slope and the side slope of the channel are
1:4000 and 1:1 respectively. Take Mannings roughness coefficient for lining as 0.015.
2. A canal of trapezoidal cross-section has a bed width of 8m and bed slope of 1:4000. If
the depth of flow is 2.4m and side slopes are 1:3, then determine the average flow
velocity and discharge in the channel. Also, compute the average shear stress at the
channel boundary.
3. A trapezoidal canal is to carry 50m3/s of water with a mean velocity of 0.6m/s. One
side of the canal is vertical where as the other side has a slope of 1:3. Find the minimum
hydraulic slope if the Mannings coefficient is 0.015.
4. Water flows in a channel as shown in the following figure. Find the discharge if
Chezys constant is 60 and bed slope is 1 in 1000.
1.4m
0.7m
0.7m
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 33
Dr. Niranjan Sahoo
IIT-Guwahati
5. Water flows in canal with bed slope of 1:2300. The cross-section of the canal is shown
in the following figure. Estimate the discharge when the depth of water is 2.5m. Assume
Chezys constant as 40.
0.6m
2.5m
15m
140m
Side slope = 1 vertical to 2 horizontal
6. A concrete lined circular channel of 0.6m diameter has a bed slope of 1: 500. Find the
depth of flow when the discharge is 0.3m 3/s. Also, determine the velocity and flow rate
for conditions of: (i) maximum velocity; (ii) maximum discharge. Assume Chezys
constant as 40.
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