Engineering Hydrology & Hydraulics

(CEC 505 )
M.Tech. 1st Semester
Monsoon 2021-22

Department of Civil Engineering

Indian Institute of Technology (Indian School
of Mines), Dhanbad
 Outline

Hydraulics

• Introduction: Principles and Classification of open channel flow, Governing

equations of channel flows, Energy and momentum equations.

• Uniform and critical flow computations: Energy depth relationships,

critical and normal depths, Hydraulically efficient channel sections.

• Non uniform flow: Gradually and Rapidly varied flow, basic concepts of
spatially varied flow. Applications with Case studies

2
Introduction
• An open channel is a conduit in which a liquid flow with a free surface.

Classification of flow in channels

1. Steady flow and unsteady flow

2. Uniform flow and non-uniform flow
3. Laminar flow and turbulent flow, and
4. Sub-critical, critical and super critical flow

Steady flow and Unsteady flow: Flow characteristics such as depth of flow,
velocity of flow, rate of flow at any point in open channel flow do not change
with respect to time, the flow is said to be steady flow.

3
 Introduction

If at any point in open channel flow, the velocity of flow, depth of flow or
rate of flow changes with respect to time, the flow is said to be unsteady
flow.

Uniform flow and Non-uniform flow:

If for a given length of the channel, the velocity of flow, depth of flow, slope
of the channel and cross-section remain constant, the flow is said to be
uniform. And, if for a given length of the channel, the velocity of flow, depth
of flow etc., do not remain constant, the flow is said to be non-uniform flow.
4
Non-uniform flow in open channels is known as varied flow, which is classified
into two types as:
(i) Rapidly varied flow (RVF) and
(ii) Gradually varied flow (GVF)
(i) Rapidly varied flow (RVF): It is defined as that flow in which depth of
flow changes abruptly over a small length of the channel. There is any
obstruction in the path of flow of water, the level of water rises above the
obstruction and then falls and again rises over a small length of channel. Thus,
the depth of flow changes rapidly over a short length of the channel. For this
short length of the channel the flow is called rapidly varied flow (RVF).
5
Introduction
(ii) Gradually varied flow (GVF): If
the depth of flow in a channel change
gradually over a long length of the
channel, the flow is said to be gradually
varied flow and is denoted by GVF. Fig: Uniform and non-uniform flow

• Laminar flow and Turbulent flow: 𝜌𝑉𝑅

𝑅𝑒 =
𝜇
If the Reynold number (Re) is less
than 500 or 600, the flow is called where V = Mean velocity of water
laminar flow. If the Reynold number
R= Hydraulic radius or Hydraulic mean depth
(Re) is more than 2000, the flow is
known as turbulent flow. If Re lies = Cross-section area of flow normal to the
between 500 to 2000, the flow is direction of flow/wetted perimeter
considered to be in transition state. 𝜌 and 𝜇 = Density and viscosity of water.

6
• Sub-critical, Critical and Super critical flow: If the Froude number (Fe) is
less than 1 (Fe < 1), the flow is called critical flow. Sub-critical flow is also
called tranquil or streaming flow. The flow is called critical if Fe=1, and if
Fe>1, the flow is called super critical or shooting or rapid or torrential.

𝑉
• The Froude number is 𝐹𝑒 =
𝑔𝐷

• where, V=mean velocity of water

• D = Hydraulic depth of channel and is equal to the ratio of wetted area to the
top width of channel = A/T, where T= Top width of channel.

7
Discharge through open channel by Chezy’s formula:
• Consider uniform flow of water in
a channel as shown in Fig. As the
flow is uniform, it means the
velocity, depth of flow and area of
flow will be constant for a given
length of the channel.

• Consider sections 1-1 and 2-2.

Fig: Uniform flow in open channel.
Let L = Length of channel,
V = Mean velocity of flow of water,
A = Area of flow of water,
P = Wetted perimeter of the cross-section,
I = Slope of the bed,
f =Friction resistance per unit velocity per
unit area.
8
• The weight of water between sections 1-1 and 2-2.

W=Specific weight of water x volume of water = w x A x L

Component of W along direction of flow = W x sin i = wAL sin i (i)

Frictional resistance against motion of water = f x surface area x (velocity)𝑛

The value of n is found experimentally equal to 2 and surface area = P x L

∴ Frictional resistance against motion = f x P x L x (V)2 (ii)

• The forces acting on the water between sections 1-1 and 2-2 are:

9
1. Component of weight of water along the direction of flow
2. Friction resistance against flow of water,
3. Pressure force at section 1-1,
4. Pressure force at section 2-2,
As the depths of water at the sections 1-1 and 2-2 are the same, the pressure forces
on these two sections are same and acting in the opposite direction.
Hence, they cancel each other. In case of uniform flow, the velocity of flow is
constant for the given length of the channel.
Hence there is no acceleration acting on the water. Hence the resultant force acting
in the direction of flow must be zero.

10
Resolving all forces in the direction of flow, it gives
𝑤𝐴𝐿 sin 𝑖 − 𝑓 × 𝑃 × 𝐿 × 𝑉 2 = 0
𝑤𝐴𝐿 sin 𝑖 = 𝑓 × 𝑃 × 𝐿 × 𝑉 2
𝑤𝐴𝐿𝑠𝑖𝑛 𝑖 𝑤 𝐴
𝑉2 = = × × sin 𝑖
𝑓×𝑃×𝐿 𝑓 𝑃

𝑤 𝐴
𝑉= × × sin 𝑖 …(iii)
𝑓 𝑃

𝐴
= 𝑚 = ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑚𝑒𝑎𝑛 𝑑𝑒𝑝𝑡ℎ 𝑜𝑟 ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑟𝑎𝑑𝑖𝑢𝑠
𝑃

𝑤
= 𝐶 = 𝐶ℎ𝑒𝑧𝑦 ′ 𝑠𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑓

𝐴 𝑤
Substituting the values of and in equation (iii), V = C 𝑚 sin 𝑖
𝑃 𝑓

For small values of i, sin i ≃ tan i ≈ i ∴ V = C 𝑚𝑖

∴ Discharge, Q = Area x Velocity = A x V
= A x C 𝑚𝑖 11
Que. 1. Find the discharge through a trapezoidal channel of width 8m and side slope of 1
horizontal to 3 vertical. The depth of flow of water is 2.4 m and value of Chezy’s constant, C =
50. The slope of the bed of the channel is given 1 in 4000.

12
 Empirical formulae for the value of Chezy’s constant
Equation (A x C 𝒎𝒊) is known as Chezy’s formulae after the name of French Engineer, Antoine Chezy,
who developed this formula in 1975. ‘C’ is Chezy’s constant, which is not a dimensionless co-efficient.
The dimension of C is

𝑉 𝐿/𝑇 𝐿/𝑇 𝐿 𝐿
= = = = =
𝑚𝑖 𝐴 𝐿2 𝑇 𝐿𝑖 𝑇
𝑖 𝑖
𝑃 𝐿

= 𝐿1/2 𝑇 −1 {i is dimensionless}

Hence the value of C depends upon the system of units. The following are the empirical formulae,

After the name of their inventors, used to determine the value of C:

157.6
1. Bazin formula (In MKS units): 𝐶 = 𝐾 , where K=bazin’s constant and depends upon the
1.81+
𝑚
roughness of the surface of channel. m=hydraulic mean depth or hydraulic radius
13
2. Ganguillet-Kutter Formula. The value of C is given in MKS unit as

0.00155 1
23 + +
𝐶= 𝑖 𝑁
0.00155 𝑁
1 + (23 + )
𝑖 𝑚

Where N = Roughness coefficient which is known as Kutter’s constant.

i=Slope of the bed, m= Hydraulic mean depth.

3. Manning’s Formula. The value of C according to this formula is given by

1 1/6
𝐶= 𝑚
𝑁

Where m = Hydraulic mean depth

N = Manning’s constant which is having same value as Kutter’s constant for the normal range
of slope and hydraulic mean depth.
14
 Que. 2. Find the discharge through a rectangular channel of width 2 m, having a bed slope of 4
in 8000. The depth of flow is 1.5 m and take the value of N in Manning’s formula as 0.012.

Using Manning’s formula, given by

1 1/6 1
𝐶= 𝑚 = 0.61/6 = 76.54
𝑛 0.012
1
Discharge, Q is given by equation A x C 𝑚𝑖 = 3 x 76.54 0.6 × 𝑚2 /𝑠 = 3.977 𝑚3 /𝑠
2000

15
 MOST ECONOMICAL SECTION OF CHANNELS
• When the cost of construction of the channel is minimum, the channel is said to be
most economical. But the cost of construction of channel depends upon the excavation
and the lining. To keep the cost down or minimum, the wetted perimeter for a given
discharge should be minimum. This condition is utilized for determining the
dimensions of a economical sections of different form of channels.

• Most economical section is also called the best section or most efficient section as
the discharge passing through a most economical section of channel for a given cross-
section area (A), slope of the bed (i) and a resistance co-efficient, is maximum.

𝐴×𝑖 𝐴
• Discharge (Q) is given by 𝑄 = 𝐴𝐶 𝑚𝑖 = 𝐴𝐶 since, 𝑚 =
𝑃 𝑃

• For a given A, i and resistance coefficient C, the above equation is written as

16
 1
𝑄=𝐾 𝑃
, where 𝐾 = 𝐴𝐶 𝐴𝑖 = constant

Hence the discharge, Q will be maximum, when the wetted perimeter P is

minimum.

Following shapes of the channels are considered to be most economical channel:

a) Rectangular channel

b) Trapezoidal channel

c) Circular channel.

Most economical Rectangular Channel:

The condition for most economical section is that for a given area, the wetted
perimeter should be minimum.
17
Rectangular channel
A rectangular channel is shown in figure.

Let, b = width of channel

d = depth of the flow

Area of flow, A=b x d (i) Figure: Rectangular channel

Wetted perimeter, P = d + b + d = b + 2d (ii)

From equation (i)

b = A/d

Substituting the value of b in (ii),

𝐴
P = b + 2d = + 2𝑑
𝑑
(iii)
18
For most economical section, P should be minimum for a given area.

𝑑𝑃
or, =0
𝑑(𝑑)

Differentiating the equation (iii) with respect to d and equating the same to zero, it gives
𝑑𝑃 𝐴 𝐴
+ 2𝑑 = 0 𝑜𝑟 − 2 + 2 = 0 𝑜𝑟 𝐴 = 2𝑑 2
𝑑(𝑑) 𝑑 𝑑

But 𝐴 = 𝑏 × 𝑑, ∴ 𝑏 × 𝑑 = 2𝑑 2 𝑜𝑟 𝑏 = 2𝑑

𝐴 𝑏×𝑑
Now hydraulic mean depth, 𝑚 = = (since A = bd, P = b+2d)
𝑃 𝑏+2𝑑

2𝑑 × 𝑑 𝑑
=
2𝑑 + 2𝑑 2

Either b = 2d means width is two times depth of flow.

𝑑
or m =
2
means hydraulic mean depth is half the depth of flow.
19
Q. A rectangular channel carries water at the rate of 400 litres/s when bed slope is 1 in 2000. Find
the most economical dimensions of the channel if C = 50.
Solution. Given: Q = 400 litres/s = 0.4 m3/s
1
Bed slope, 𝑖 = , Chezy’s constant, C = 50
2000

For the rectangular channel to be most economical,

(i)Width, b = 2d
𝑑
(ii)Hydraulic mean depth, 𝑚 =
2

Area of flow 𝐴 = 𝑏 × 𝑑 = 2𝑑 × 𝑑 = 2𝑑 2
Using discharge equation, 𝑄 = 𝐴𝐶 𝑚𝑖

𝑑 1 1
0.4 =2𝑑 2 × 50 × × = 2 × 50 × 𝑑 5/2 = 1.581 𝑑5/2
2 2 2×2000

5/2
0.4 2
𝑑 = = 0.253𝑑 = (.253)5 = 0.577 𝑚
1.581
𝑏 = 2𝑑 = 2 × .577 = 𝟏. 𝟏𝟓𝟒 𝑚
20
MOST ECONOMICAL TRAPEZOIDAL CHANNEL
• The trapezoidal section of a channel
will be most economical, when its
wetted perimeter is minimum. Consider
a trapezoidal section of channel as
shown in figure.

Let, b = width of channel Figure: Trapezoidal section

d = depth of the flow

𝜃 =angle made by the sides with

horizontal.

(1) The side slope is given as 1 vertical

to n horizontal.

21
22
 𝐴
2
+ 𝑛 = 2 𝑛2 + 1
𝑑
Substituting the value of A from equation (i) in the above equation,
𝑏 + 𝑛𝑑 𝑑 2 + 1 𝑜𝑟
𝑏 + 𝑛𝑑
2
+ 𝑛 = 2 𝑛 + 𝑛 = 2 𝑛2 + 1
𝑑 𝑑
𝑏 + 𝑛𝑑 + 𝑛𝑑 𝑏 + 2𝑛𝑑 𝑏 + 2𝑛𝑑
= = 2 𝑛2 + 1 𝑜𝑟 = 𝑑 𝑛2 + 1
𝑑 𝑑 2
𝑏+2𝑛𝑑
But from Figure, = Half of top width
2

And 𝑑 𝑛2 + 1 = 𝐶𝐷 = one of the sloping side

Half of the top width = one of the sloping side
(ii) Hydraulic mean depth
𝐴
Hydraulic mean depth, 𝑚 =
𝑃

Value of A from (i), 𝐴 = (𝑏 + 𝑛𝑑) × 𝑑

Value of P from (iia), 𝑃 = 𝑏 + 2𝑑 𝑛2 + 1 = 𝑏 + (𝑏 + 2𝑛𝑑) (from equation, 𝑏 + 2𝑛𝑑 = 2𝑑 𝑛2 + 1)
2𝑏 + 2𝑛𝑑 = 2(𝑏 + 𝑛𝑑) 23
 𝐴 𝑏+𝑛𝑑 𝑑 𝑑
Hydraulic mean depth, 𝑚 = = =
𝑃 2(𝑏+𝑛𝑑) 2

Hence for a trapezoidal section to be most economical hydraulic mean depth must be equal
to half the depth of flow.

Q . A trapezoidal channel has side slopes of 3 horizontal to 4 vertical and slope of its bed is 1
in 2000. Determine the optimum dimensions of the channel, if it is to carry water at 0.5 m3/s.
Take Chezy’s constant as 80.

24
For most economical
section,

25
26
Q. For a trapezoidal channel with bottom width 40 m and side slopes 2H : 1 V, Manning’s
N is 0.015 and bottom slope is 0.0002. If it carries 60 m3/s discharge, determine the normal
depth.

27
28
Flow through Circular Channel
The flow of a liquid through a circular pipe, when the
level of liquid in the pipe is below the top of the pipe is
classified as an open channel flow. The rate of flow
through circular channel is determined from the depth
of flow and angle subtended by the liquid surface at the
centre of the circular channel. Figure: Circular Channel

Figure shows a circular channel through which water is

flowing.

• Let d = depth of water,

• 2𝜃 = angle subtended by water surface AB at the

centre in radians,

• R = radius of the channel,

29
Then the wetted perimeter and wetted area is determined as:

2𝜋𝑅
Wetted perimeter, 𝑃 = × 2𝜃 = 2𝑅𝜃
2𝜋

Wetted area, A = Area ADBA

= Area of sector OADBO – Area of ∆𝐴𝐵𝑂

𝜋𝑅 2 𝐴𝐵×𝐶𝑂 2𝐵𝐶×𝐶𝑂
= × 2𝜃 − = 𝑅2 𝜃 − (since AB = 2BC)
2𝜋 2 2

2×𝑅 sin 𝜃 ×𝑅 cos 𝜃

𝑅2𝜃 − (since BC = R sin 𝜃, 𝐶𝑂 = 𝑅 cos 𝜃)
2

2
𝑅 2 × 2 sin 𝜃 cos 𝜃 2
𝑅 2 sin 2𝜃
𝑅 𝜃− =𝑅 𝜃− (2 sin 𝜃 cos 𝜃 = sin 2𝜃)
2 2
sin 2𝜃
A= 𝑅 2 (𝜃 − )
2

sin 2 𝜃
𝐴 𝑅 2 (𝜃− ) 𝑅 sin 2𝜃
2
Then hydraulic mean depth, 𝑚 = = = (𝜃 − )
𝑃 2𝑅𝜃 2𝜃 2

And discharge, Q is given by 𝑄 = 𝐴𝐶 𝑚𝑖

30
Q. Find the discharge through a circular pipe of diameter 3 m, if the depth of water in the pipe
is 1 m and the pipe is laid at a slope of 1 in 1000. Take the value of Chezy’s constant as 70.
Solution. Given:

𝐷 3
Dia. of pipe, D = 3.0; Radius, 𝑅 = 2
= 2 = 1.50 𝑚

Depth of water in pipe, d=1.0 m

1
Bed slope, 𝑖 = 1000, Chezy’s constant, C = 70

We have OC = OD – CD = R – 1

= 1.5 – 1 = 0.5

AO = R = 1.5 m
31
32
Most Economical Circular Section:
• For most economical section the discharge for a constant cross-sectional
area, slope of bed and resistance coefficient, is maximum. But in case of
circular channels, the area of flow can not be maintained constant. With
the change of depth of flow in a circular channel of any radius, the wetted
area and wetted perimeter changes. Thus, in case of circular channels, for
most economical section, two separate conditions are obtained.

1. Condition for maximum velocity, and

2. Condition for maximum discharge.

1.Condition for maximum velocity for Circular section:

33
• Figure shows a circular channel through which water is
flowing.

• Let d = depth of water,

• 2𝜃 angle subtended at the centre by water surface,

• R = radius of channel and

• i =slope of the bed

the velocity of flow according to Chezy’s formula is given

𝐴 𝐴
𝑣 = 𝐶 𝑚𝑖 = 𝐶 𝑃
𝑖 (since, 𝑚 = 𝑃 )

34
The velocity of flow through a circular channel will be maximum when the hydraulic
mean depth m or A/P is maximum for a given value of C and i. In case of circular pipe,
the variable is 𝜃 only.

Hence for maximum value of A/P we have the condition.

𝐴
𝑑(𝑃)
=0 (i)
𝑑𝜃

Where A and P both are functions of 𝜃.

The value of wetted area, A is given by

sin 2𝜃
𝐴 = 𝑅 2 (𝜃 − 2
) (ii)

The value of wetted perimeter, P is given by 𝑃 = 2𝑅𝜃 (iii)

Differentiating equation (i) with respect to 𝜃

35
36
The depth of flow for maximum velocity,

d =OD – OC = R – R cos 𝜃
= 𝑅[1 − cos 𝜃] = 𝑅[1 − cos 128° 45′] = 𝑅[1 − cos(180° − 51° 15′ )]
= 𝑅[1 − (− cos 51° 15′)] = 𝑅[1 + cos 51° 15′]

𝐷
= 𝑅 1 + 0.62 = 1.62 𝑅 = 1.62 × 2 = 0.81 𝐷

where D=diameter of the circular channel.

Thus for maximum velocity of flow, the depth of water in the circular channel
should be equal to 0.81 times the diameter of the channel.

Hydraulic mean depth for maximum velocity is

37
 𝐷
= 0.611 × = 0.3055 𝐷 = 0.3 𝐷
2

38
Thus, for maximum velocity, the hydraulic mean depth is equal to 0.3 times
the diameter of circular channel.

2.Condition for Maximum Discharge for circular section

The discharge through a channel is given by

𝐴 𝐴
𝑄 = 𝐴𝐶 𝑚𝑖 = 𝐴𝐶 𝑖 since, 𝑚 =
𝑃 𝑃

𝐴3
=𝐶 𝑖
𝑃

39
 𝐴3
The discharge will be maximum for constant values of C and i, when is
𝑃
𝐴3 𝑑 𝐴3
maximum. will be maximum when = 0.
𝑃 𝑑𝜃 𝑃
Differentiating this equation with respect to 𝜃 and equation the same to zero,
𝑑𝐴 𝑑𝑃
𝑃 × 3𝐴2 − 𝐴3
𝑑𝜃 𝑑𝜃 = 0 𝑜𝑟 3𝑃𝐴2 𝑑𝐴 − 𝐴3 𝑑𝑃 = 0
𝑃2 𝑑𝜃 𝑑𝜃
𝑑𝐴 𝑑𝑃
Dividing by 𝐴2 , 3𝑃 𝑑𝜃 − 𝐴 𝑑𝜃 = 0 …(i)
𝑃 = 2𝑅𝜃
𝑑𝑃
= 2𝑅
𝑑𝜃
sin 2𝜃
𝐴 = 𝑅2 (𝜃 − )
2
40
41
Depth of flow for maximum discharge
d=OD-OC=R-R cosθ
= 𝑅 1 − cos 𝜃 = 𝑅[1 − cos 154°]
= 𝑅[1 − cos(180° − 26°)] = 𝑅[1 + cos 26°] = 1.898 𝑅
𝐷
= 1.898 × = 0.948 𝐷 ≃ 0.95 𝐷
2
where D = Diameter of the circular channel.
Thus, for maximum discharge through a circular channel the depth of flow
is equal to 0.95 times its diameter.

42
 Equation of Continuity

The continuity equation is a statement of the law of conservation of matter.

Steady Flow: In a steady flow the volumetric rate of flow (discharge in 𝑚3 /𝑠) past
various section must be the same.

In a varied flow, if Q = discharge, V = mean velocity and A = area of cross-section

Q = VA = 𝑉1 𝐴1 = 𝑉2 𝐴2 = …….

Energy Equation

In the one-dimensional analysis of steady open channel flow, the energy equation in
the form of the Bernoulli equation is used. As per this equation, the total energy at a
downstream section differs from the total energy at the upstream section by an
amount equal to the loss of energy between the sections.
43
Figure shows a steady varied fl ow
in a channel. If the effect of the
curvature on the pressure
distribution is neglected, the total
energy head (in N.m/newton of fl
uid) at any point A at a depth d
below the water surface is

𝑉2
𝐻 = 𝑍𝐴 + 𝑑 𝑐𝑜𝑠𝜃 + 𝛼 2𝑔 (1) Fig. Definition sketch for the energy
equation

44
This total energy will be constant for all values of d from zero to y at a normal
section through point A (i.e. Section OAB), where y = depth of fl ow measured
normal to the bed. Thus, the total energy at any section whose bed is at an
elevation Z above the datum is

𝐻 = 𝑍 + 𝑦 cos 𝜃 + 𝛼𝑉 2 /2𝑔 (2)

In figure the total energy at a point on the bed is plotted along the vertical
through that point. The elevation of energy line on the line 1-1 represents the
total energy at any point on the normal section through point 1. The total
energies at normal sections through 1 and 2 are therefore
𝑉12
𝐻1 = 𝑍1 + 𝑦1 cos 𝜃 + 𝛼1
2𝑔
𝑉22
𝐻2 = 𝑍2 + 𝑦2 cos 𝜃 + 𝛼2
2𝑔

45
 The term (Z + y cos θ) = h represents the elevation of the hydraulic grade
line above the datum. If the slope of the channel θ is small, cos θ ≈ 1.0, the
normal section is practically the same as the vertical section and the total
energy at any section can be written as
𝐻 = 𝑍 + 𝑦 + 𝛼𝑉2 /2𝑔

Since most of the channels in practice happen to have small values of θ (θ

< 10°), the term cos θ is usually neglected.

Due to energy losses between Sections 1 and 2, the energy head H 1 will be
larger than H2 and H1 − H2 = hL= head loss. Normally, the head loss (hL )
can be considered to be made up of frictional losses (h f ) and eddy or form
loss (he ) such that hL = hf + he . For prismatic channels, he = 0.

46
 MOMENTUM EQUATION

• Steady Flow Momentum is a vector

quantity. The momentum equation
commonly used in most of the open channel
flow problems is the linear-momentum
equation.
• This equation states that the algebraic sum
of all external forces, acting in a given
direction on a fluid mass equals the time
rate of change of linear momentum of the
fluid mass in the direction.
• In a steady flow the rate of change of Figure: Definition sketch for the momentum
equation
momentum in a given direction will be
equal to the net flux of momentum in that
direction.

47
Figure shows a control volume (a volume fixed in space) bounded by Sections 1 and 2, the boundary
and a surface lying above the free surface. The various forces acting on the control volume in the
longitudinal direction are as follows:

i) Pressure forces acting on the control surfaces, F1 and F2 .

ii) Tangential force on the bed, F3 ,

iii) Body force, i.e., the component of the weight of the fluid in the longitudinal direction, F4.

By the linear-momentum equation in the longitudinal direction for a steady-flow discharge of Q,

𝐹1 − 𝐹2 − 𝐹3 + 𝐹4 = 𝑀2 − 𝑀1

In which 𝑀1 = 𝛽1 𝜌𝑄𝑉1 = momentum flux entering the control volume, 𝑀2 = 𝛽2 𝜌𝑄𝑉2 = momentum flux
leaving the control volume.

48
Unsteady Flow: The momentum equation would then state that in an unsteady flow the algebraic sum
of all external forces in a given direction on a fluid mass equals the net change of the linear-momentum
flux of the fluid mass in that direction plus the time rate of increase of momentum in that direction
within the control volume.

Specific Force: The steady-state momentum equation (Eq. 1.45) takes a simple form if the tangential
force F3 and body force F4 are both zero. In that case
𝐹1 − 𝐹2 = 𝑀2 − 𝑀1
𝐹1 − 𝑀1 = 𝐹2 − 𝑀2

1
Denoting (F + M) = 𝑃𝑠
𝛾

(𝑃𝑠 )1 = (𝑃𝑠 )2

The term 𝑃𝑠 is known as the specific force and represents the sum of the pressure force and momentum
flux per unit weight of the fluid at a section. Equation states that the specific force is constant in a
horizontal, frictionless channel.

49
Non uniform flow through open
channels

50
 Introduction
• A flow is called uniform if the velocity of flow, depth of flow, slope of the bed of
the channel and area of cross-section remain constant for a given length of the
channel.

• If the velocity of flow, depth of flow, area of cross-section and slope of the bed
channel do not remain constant for a given length of channel, the flow is known
as non-uniform.

• Non-uniform flow is further divided into rapidly varied flow (RVF) and
Gradually varied flow (GVL) depending upon the change of depth of flow over
the length of the channel.

• If the depth of flow changes abruptly over a small length of the channel, the flow
is said as rapidly varied flow. If the depth of flow in a channel changes gradually
over a long length of channel, the flow is said gradually varied flow.
51
Specific Energy and Specific energy curve
• The total energy of a flowing liquid per
unit weight is given by,

𝑽𝟐
• Total energy = 𝒛 + 𝒉 + 𝟐𝒈

where z = Height of the bottom of channel Fig. 1. Specific energy

above datum,

h=Depth of liquid and V = mean velocity

of flow.

• If the channel bottom is taken as the

𝑽𝟐
datum as shown in figure, then the total 𝑬 = 𝒉 + 𝟐𝒈 (1)
energy per unit weight of liquid will be,
52
Equation (1) is called as specific energy. Hence specific energy of a flowing liquid
is defined as energy per unit weight of the liquid with respect to the bottom of
the channel.

Specific energy curve

It is defined as the curve which shows the variation of specific energy with
depth of flow.

From equation (1), the specific energy of a flowing liquid

𝑽𝟐
𝑬=𝒉+ = 𝑬𝒑 + 𝑬𝒌
𝟐𝒈

where 𝐸𝑝 = potential energy of flow = h

𝑽𝟐
𝐸𝑘 = Kinetic energy of flow = 𝟐𝒈 53
Consider a rectangular channel in which a steady but non-uniform flow is taking
place.

Let, Q = discharge through channel, b=width of the channel, h = depth of flow

and q = discharge per unit width.

𝑸 𝑸
Then 𝒒 = 𝒘𝒊𝒅𝒕𝒉 = = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (since, Q and b are constant)
𝒃

𝑸 𝑸 𝒒 𝑄
Velocity of flow, 𝑽 = 𝑨
= 𝒃×𝒉 = 𝒉 (𝑠𝑖𝑛𝑐𝑒, 𝑏 = 𝑞)

Substituting the values of V in equation (1)

𝒒𝟐
𝑬=𝒉+ = 𝑬𝒑 + 𝑬 𝒌 (2)
𝟐𝒈𝒉𝟐

Equation (2) gives the variation of specific energy (E) with depth of flow (h).
54
• For a given discharge, for different values of depth of flow, the corresponding
values of E can be obtained.

• Then a graph between specific energy (along X-X axis) and depth of flow, h
(along Y-Y axis) can be obtained.

Fig 2. Specific energy curve

55
• The specific energy curve will also be obtained by first drawing a curve for
potential energy (i.e 𝐸𝑝 = ℎ,) which will be staright line passing through the
origin, making an angle of 45° with X-axis as shown in Fig. 2.

𝒒𝟐
• The drawing another curve for kinetic energy (i.e., 𝑬𝒌 = 𝒐𝒓 𝑬𝒌 =
𝟐𝒈𝒉𝟐
𝑲 𝑞2
, 𝑤ℎ𝑒𝑟𝑒 𝐾 = 2𝑔 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) which will be parabola as shown in Fig. 2.
𝒉𝟐

• By combining two curve (potential and Kinetic energy curve) we obtain specific
energy curve.

• In Fig. 2 ACB denotes the specific energy curve.

56
Critical depth (𝒉𝒄 ):
• Critical depth is defined as the depth of flow of water at which the specific
energy is minimum. It is represented by ℎ𝑐 .

• In Fig., curve ACB is a specific energy curve and point C corresponds to the
minimum specific energy. The depth of flow of water at C is known as critical
depth.

• The expression for critical depth is obtained by differentiating the specific energy
equation (ii) with respect to depth of flow and equating the same to zero.

𝑑𝐸 𝒒𝟐
= 0, where 𝑬 = 𝒉 + from eqn. (2)
𝑑ℎ 𝟐𝒈𝒉𝟐

57
𝑑 𝑞2 𝑞2 −2 𝑞2
ℎ+ = 0 𝑜𝑟 1 + = 0 (Since, is constant)
𝑑ℎ 2𝑔ℎ2 2𝑔 ℎ3 2𝑔

𝑞2 𝑞2 3
𝑞2
1 − 3 = 0 𝑜𝑟 1 = 3 𝑜𝑟 ℎ =
𝑔ℎ 𝑔ℎ 𝑔
1/3
𝑞2
ℎ=
𝑔

It is mentioned earlier that when specific energy is minimum, depth is

critical and it is denoted by ℎ𝑐 . Hence critical depth is
1/3
𝑞2
Critical depth, ℎ𝑐 =
𝑔

58
Critical Velocity (𝑽𝒄 )
The velocity of flow at the critical depth is known as critical velocity. It is
represented by 𝑉𝑐 . The mathematical expression for critical velocity is obtained
from equation as

𝟏/𝟑
𝒒𝟐
𝒉𝒄 = (3)
𝒈

𝑞2
Taking cube to both sides, it gives ℎ𝑐 3 = 𝑜𝑟 𝑔ℎ𝑐 3 = 𝑞 2 (i)
𝑔

𝑄
But we know, q = Discharge per unit width = 𝑏
𝐴×𝑉 𝑏×ℎ×𝑉
𝑞= = = ℎ × 𝑉 = ℎ𝑐 × 𝑉𝑐
𝑏 𝑏
59
 Substituting the value of q in eqn.(i)
𝑔ℎ𝑐 3 = ℎ𝑐 × 𝑉𝑐 2

𝑔ℎ𝑐 3 = ℎ𝑐 2 × 𝑉𝑐 2 𝑜𝑟 𝑔ℎ𝑐 = 𝑉𝑐2 [Dividing by ℎ𝑐 2 ]

Critical Velocity, 𝑽𝒄 = 𝒈 × 𝒉𝒄 (4)

60
Minimum Specific energy in terms of critical depth:
𝑞2
Equation (2) is the specific energy equation. 𝐸 = ℎ +
2𝑔ℎ2

When specific energy is minimum, depth of flow is critical and hence equation
becomes as

𝑞2
𝐸𝑚𝑖𝑛 = ℎ𝑐 + (ii)
2𝑔ℎ𝑐 2

1/3
𝑞2 𝑞2
But from equation (3), ℎ𝑐 = 𝑜𝑟 ℎ𝑐3 =
𝑔 𝑔

𝑞2
Substituting the value of = ℎ𝑐3 in equation (ii),
𝑔

ℎ𝑐3 ℎ𝑐 𝟑𝒉𝒄
𝑬𝒎𝒊𝒏 = ℎ𝑐 + = ℎ𝑐 + =
2ℎ𝑐2 2 𝟐
(5)
61
Que. The discharge of water through a rectangular channel of width 8 m, is 15 m3/s
when depth of flow of water is 1.2 m. Calculate: (a) Specific energy of the flowing
water, (b) critical depth and critical velocity, (c) Value of minimum specific energy.

Solution. Given,

Discharge, Q = 15 m3/s,

width, b = 8 m,

Depth, h = 1.2m

𝑄 15
Discharge per unit width, q = = = 1.875 𝑚2 /𝑠
𝑏 8

𝑄 15 15
Velocity of flow, 𝑉 = = 𝑏×ℎ = 8×1.2 = 1.5625 𝑚/𝑠
𝐴

62
(a) Specific energy (E) is from eqn. (1)
𝑽𝟐 1.56252
𝑬=𝒉+ = 1.2 + = 1.20 + 0.124 = 1.324 𝑚.
𝟐𝒈 8 × 9.81
(b) Critical depth (ℎ𝑐 ) is from eqn. (3)
𝟏/𝟑 1/3
𝒒𝟐 1.8752
𝒉𝒄 = = = 0.71 𝑚
𝒈 9.81

Critical velocity (𝑉𝑐 ) is from eqn. (4)

𝑽𝒄 = 𝒈 × 𝒉𝒄 = 9.81 × 0.71 = 2.639 𝑚/𝑠

(c) Minimum specific energy (𝐸𝑚𝑖𝑛 ) is from eqn. (5)

𝟑𝒉𝒄 3×0.71
𝑬𝒎𝒊𝒏 = = = 1.065 𝑚.
𝟐 2
63
Critical flow:
The flow at which the specific energy is minimum.

Equation (4) gives the relation for critical velocity in terms of critical depth as

𝑉𝑐 𝑉𝑐
𝑉𝑐 = 𝑔 × ℎ𝑐 𝑜𝑟 =1 where, = Froude number
𝑔ℎ𝑐 𝑔ℎ𝑐

Froude number, 𝐹𝑒 = 1.0 for critical flow.

Streaming flow or sub-critical flow or Tranquil flow:

When the depth of flow in a channel is greater than the critical depth (ℎ𝑐 ), the
flow is said to be sub-critical flow or streaming flow or tranquil flow. The Froude
number is less than one i.e., 𝐹𝑒 < 1.
64
Super critical flow or shooting flow or torrential flow:

• when the depth of flow in a channel is less than the critical depth (ℎ𝑐 ), the
flow is called as super critical flow or shooting flow or torrential flow. The
Froude number is greater than one i.e., 𝐹𝑒 > 1.

Alternate depths:

• In the specific energy curve (shown in figure) the point C corresponds to the
minimum specific energy and the depth of flow at C is called critical depth.

• There are two depths one greater than critical depth and other smaller than
the critical depth. For a given specific energy these two depths are called as
alternate depths. These depths are shown in Fig. 2 as ℎ1 and ℎ2 or the depths
corresponding to points G and H.

65
Condition for maximum discharge for a given value of
specific energy.
Specific energy (E) at any section of a channel is given by equation (1)

𝑉2 𝑄 𝑄
𝐸 =ℎ+ , where 𝑉 = =
2𝑔 𝐴 𝑏×ℎ

𝑄2 1 𝑄2
𝐸 =ℎ+ 2 × =ℎ+
𝑏 × ℎ2 2𝑔 2𝑔𝑏 2 ℎ2
𝑄 2 = 𝐸 − ℎ 2𝑔𝑏 2 ℎ2 𝑜𝑟 𝑄 = 𝐸 − ℎ 2𝑔𝑏 2 ℎ2 = 𝑏 2𝑔(𝐸ℎ2 − ℎ3 )

For maximum discharge, Q the expression (𝐸ℎ2 − ℎ3 ) should be maximum.

𝑑
or 𝑑ℎ 𝐸ℎ2 − ℎ3 = 0 𝑜𝑟 2𝐸ℎ − 3ℎ2 = 0 (since, E is constant)

66
or 2𝐸 − 3ℎ = 0 (dividing by h)

𝟐
𝒉 = 𝟑𝑬 (6)

3ℎ
𝐸= (i)
2

But from eqn. (5), specific energy is minimum when it is equal to 3/2 times the
value of depth of critical flow.

In eqn. (i), the specific energy (E) is equal to 3/2 times the depth of flow.

Thus equation (i) represents the minimum specific energy and h is the critical
depth.

67
68
Que. The specific energy for a 5 m wide rectangular channel is to be 4 Nm/N. If the
rate of flow of water through the channel is 20 m3/s, determine the alternate depths of
flow.

Solution: given, width of channel, b=5 m, specific energy, E=4Nm/N=4m,

Discharge, Q = 20 m3/s

The specific energy (E) is from eqn. (1) as

𝑽𝟐 𝑄 𝑄 20 4
𝑬=𝒉+ 𝟐𝒈
, where 𝑉 = 𝐴
= 𝑏×ℎ = 5×ℎ = ℎ

𝑉2 4 2 1 8
Specific energy, 𝐸 = ℎ + =ℎ+ × 2𝑔 = ℎ + 𝑔×ℎ2
2𝑔 ℎ

But, E = 4.0

69
Equating the two values of E,

8 0.8155
4 = ℎ + 9.81×ℎ2 = ℎ + ℎ2
4ℎ2 = ℎ3 + 0.8155 𝑜𝑟 ℎ3 − 4ℎ2 + 0.8155 = 0

This is a cubic equation, so solving by trial and error.

h = 3.93 m and 0.48 m.

70
Que. The specific energy for a 3 m wide channel is to be 3 kg-m/kg. What would be the
maximum possible discharge?

Solution. Given:

width of channel, b=3 m,

specific energy, E=3 kg-m/kg = 3 m

For the given value of specific energy, the discharge will be maximum, when depth
of flow is critical. Hence, from eqn. (6) for maximum discharge.

2 2
ℎ𝑐 = ℎ = 𝐸 = × 3.0 = 2.0 𝑚
3 3

Maximum discharge, 𝑄𝑚𝑎𝑥 is given by

71
 𝑄𝑚𝑎𝑥 = 𝐴𝑟𝑒𝑎 × 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑏 × 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑓𝑙𝑜𝑤 × 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦

= (𝑏 × ℎ𝑐 ) × 𝑉𝑐 since, at critical depth, velocity will be critical

where 𝑉𝑐 is critical velocity and it is expressed by (4)

𝑽𝒄 = 𝒈 × 𝒉𝒄 = 9.81 × 2.0 = 4.4249 𝑚/𝑠
𝑄𝑚𝑎𝑥 = 𝑏 × ℎ𝑐 × 𝑉𝑐 = 3 × 2 × 4.4249 = 26.576 𝑚3 /𝑠

72
Trapezoidal Channel Problems:

Q. Calculate the bottom width of a channel required to carry a discharge

of 15 m3/s as a critical flow at a depth of 1.2 m, if the channel section is
trapezoidal with side slope of 1.5 H: 1 V.

Solution: The solution in this case is by trial-and-error.

73
Q. Find the critical depth for a specific energy head of 1.5 m in the
trapezoidal channel, B = 2.0 m and m = 1.0.

Sol.: For Trapezoidal Channel:

𝑄2 𝐴3𝑐
At critical flow, =
𝑔 𝑇𝑐

Solving by trial-and-error, yc =1.095 m.

74
 Further References
Text Books:
1. Subramanya, K. (2015), 4th edition, “Flow in Open Channels”, Tata McGraw Hill.
2. Bansal, R. K. A textbook of fluid mechanics and hydraulic machines:(in SI units). LAXMI
Publications, Ltd., 2005.
Reference Books:
1. Srivastava, R. (2014), 1st edition, “Flow through Open Channels”, Oxford University
Press.

2. Rajput, R. K. A textbook of fluid mechanics and hydraulic machines. S. Chand Publishing,

2015.

75
Engineering Hydrology & Hydraulics
(CEC 505 )
M.Tech. 1st Semester
Monsoon 2021-22

Department of Civil Engineering

Indian Institute of Technology (Indian School
of Mines), Dhanbad
 Gradually Varied Flow (GVF)
Equation of Gradually Varied Flow:
• Following are the assumptions applied for GVF:

a) The bed slope of the channel is small.

b) The flow is steady and hence discharge Q is constant.

c) Accelerative effect is negligible and hence hydrostatic pressure distribution prevails

over channel cross-section.

d) The energy correction factor, ∝ is unity.

e) The roughness coefficient is constant for the length of the channel and it does not
depend on the depth of flow.

f) The equations such as Chezy’s equation, manning’s equation are applicable for finding
the slope of energy line (these eqns. are also applicable for uniform flow).

g) The channel is prismatic. 2

• A steady non-uniform flow in a prismatic channel
with gradual changes in its water surface elevation is
known as Gradually Varied Flow (GVF).
• The backwater produced by a dam or weir across a
river and the drawdown produced at a sudden drop in
a channel are few typical examples of GVF.
• In GVF, the velocity varies along the channel and
consequently the bed slope, water surface slope, and
energy slope will all differ from each other.
Consider a rectangular channel having GVF is shown Figure: Gradually varied flow.
in Figure.

Let Z=height of bottom of channel above datum,

V=mean velocity, h=depth of channel, b=width of
channel, 𝒊𝒃 =bed slope, 𝒊𝒆 =energy line slope, and Q=
discharge through channel.

3
The energy equation at any section is given by Bernoulli’s equation

𝑽𝟐
H= 𝒁 + 𝒉 + (i)
𝟐𝒈

Differentiating this equation with respect to x, where x is measured along the

bottom of the channel in the direction of flow.

𝑑𝐻 𝑑𝑍 𝑑ℎ 𝑑 𝑉2
𝑑𝑥
= 𝑑𝑥
+ 𝑑𝑥
+ 𝑑𝑥 2𝑔
(ii)

𝑑 𝑉2 𝑑 𝑄2 𝑄 𝑄
= (since, 𝑉 = 𝐴 = 𝑏×ℎ)
𝑑𝑥 2𝑔 𝑑𝑥 𝐴2 ×2𝑔

𝑑 𝑄2 𝑄2 𝑑 1
= =𝑏2 ×2𝑔 𝑑𝑥 ℎ2 (since, Q, b and g are constant)
𝑑𝑥 𝑏2 ℎ2 ×2𝑔

4
 𝑄2 𝑑 1 𝑑ℎ 𝑄2 −2 𝑑ℎ −2𝑄2 𝑑ℎ
= 𝑏2 ×2𝑔 𝑑ℎ ℎ2 𝑑𝑥
= 𝑏2 ×2𝑔 ℎ3 𝑑𝑥
= 𝑏2 ×2𝑔ℎ3 𝑑𝑥

𝑄2 𝑑ℎ 𝑉 2 𝑑ℎ
=− 𝑏2 ℎ2×𝑔ℎ 𝑑𝑥 = − 𝑔ℎ 𝑑𝑥 (since, Q=AV)

𝑑 𝑉2
Substituting the value of in equation (ii),
𝑑𝑥 2𝑔

𝑑𝐻 𝑑𝑍 𝑑ℎ 𝑉 2 𝑑ℎ 𝑑𝑍 𝑑ℎ 𝑉2
= + − = + 1− (iii)
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑔ℎ 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑔ℎ

𝑑𝐻
=slope of energy line = −𝑖𝑒
𝑑𝑥

5
𝑑𝑍
=slope of the bed of the channel = −𝑖𝑏
𝑑𝑥

-ve sign with 𝑖𝑒 and 𝑖𝑏 is taken as with the increase of x, the value of E and Z decreases.

𝑑𝐻 𝑑𝑍
Substituting the value of and in equation (iii)
𝑑𝑥 𝑑𝑥
Chezy’s formula, 𝑉 = 𝐶 𝑚 × 𝑖𝑒
1
𝑑ℎ 𝑉2 Manning’s equation, 𝑉 = 𝑁 𝑚2/3 𝑖𝑏 1/2
𝑖𝑏 − 𝑖𝑒 = 1−
𝑑𝑥 𝑔ℎ 𝐴 𝐴
Hydraulic mean depth, 𝑚 = 𝑃 = 𝑏+2ℎ for
rectangular channel.

𝒅𝒉 𝒊𝒃 − 𝒊𝒆 𝑑ℎ
= Slope of free water surface =
𝑑𝑥
𝒅𝒙 𝑽𝟐
𝟏−
𝒈𝒉
𝒊𝒃 −𝒊𝒆 𝑽 Dynamic Equation of Gradually Varied Flow
= [since, 𝑭𝒆 = ]
𝟏−(𝑭𝒆 )𝟐 𝒈𝒉

6
 𝑑ℎ
• As h is the depth of flow and x is the distance measured along the bottom of the channel, hence
𝑑𝑥
represents the variation of the water depth along the bottom of the channel.

• This is also called the slope of the free water surface.

𝑑ℎ
(i) When = 0, ℎ (depth of water) is constant, it means that free surface of water is parallel to the bed
𝑑𝑥
of the channel.

𝑑ℎ 𝑑ℎ
(ii) When > 0 𝑜𝑟 is positive, it means the depth of water increases in the direction of flow. The
𝑑𝑥 𝑑𝑥
profile of the water is called back water curve.

𝑑ℎ 𝑑ℎ
(iii) When < 0 𝑜𝑟 is negative, it means that the depth of water decreases in the direction of flow.
𝑑𝑥 𝑑𝑥
The profile of the water is called drop down curve.

7
 Expression for the Length of Back water curve :

• Consider the flow of water through a channel in which depth of water is rising. Let
the two sections 1-1 and 2-2 are at such a distance that the distance between them
represents the length of back water curve.
Let ℎ1 =depth of flow at section 1-1,

𝑉1 =velocity of flow at section 1-1,

ℎ2 =depth of flow at section 2-2,

𝑉2 =velocity of flow at section 2-2,

𝑖𝑏 =bed slope,

𝑖𝑒 =energy line slope, and

L
Figure: Length of back water curve 8
L= length of back water curve.

Applying Bernoulli’s equation at sections 1-1 and 2-2,

𝑉12 𝑉22
𝑍1 + ℎ1 + 2𝑔
= 𝑍2 + ℎ2 + 2𝑔
+ ℎ𝐿 (i)

where ℎ𝐿 = Loss of energy due to friction = 𝑖𝑒 × 𝐿

Taking datum line passing through the bed of the channel at section 2-2. Then
𝑍2 = 0

𝑉12 𝑉22
∴ eqn. (i) becomes as 𝑍1 + ℎ1 + 2𝑔
= 𝑍2 + ℎ2 + 2𝑔
+ 𝑖𝑒 × 𝐿

𝑍1 = 𝑖𝑏 × 𝐿
9
 𝑉12 𝑉22
𝑖𝑏 𝐿 + ℎ1 + = ℎ2 + + 𝑖𝑒 × 𝐿
2𝑔 2𝑔

𝑉2
2 1𝑉2
or 𝑖𝑏 × 𝐿 − 𝑖𝑒 × 𝐿 = (ℎ2 + 2𝑔 ) − (ℎ1 + 2𝑔 )

𝑽𝟐 𝑽𝟐
𝑳 𝒊𝒃 − 𝒊𝒆 = 𝑬𝟐 − 𝑬𝟏 , where 𝑬𝟐 = (𝒉𝟐 + 𝟐𝒈𝟐 ), 𝑬𝟏 = (𝒉𝟏 + 𝟐𝒈𝟏 )

𝐸2 −𝐸1
𝐿= . This equation is used to calculate the Length of back water curve.
𝑖𝑏 −𝑖𝑒

𝑖𝑒 is calculated either by Manning’s equation or by Chezy’s formula.

For calculating 𝑖𝑒 , mean values of velocity, depth of flow, hydraulic mean depth
etc. are used between sections 1-1 and 2-2.

10
Q. Find the slope of the free water surface in a rectangular channel of width 20 m, having depth of
flow 5 m. The discharge through the channel is 50 m3/s. The bed of the channel is having a slope of
1 in 4000. Take the value of Chezy’s constant C = 60.

11
12
Q. Determine the length of the back water curve caused by an afflux of 2 m in a rectangular
channel of width 40 m and depth 2.5 m. The slope of the bed is given as 1 in 11000. Take
Manning’s N = 0.03.

13
14
15
Length of the back
water curve, L

16
 References
Text Books:
1. Subramanya, K. (2015), 4th edition, “Flow in Open Channels”, Tata McGraw Hill.
2. Bansal, R. K. A textbook of Fluid Mechanics and Hydraulic Machines:(in SI units). Laxmi
Publications, Ltd., 2005.

Reference Books:
1. Srivastava, R. (2014), 1st edition, “Flow through Open Channels”, Oxford University
Press.

2. Rajput, R. K. A textbook of fluid mechanics and hydraulic machines. S. Chand Publishing,

2015.

17
Engineering Hydrology & Hydraulics
(CEC 505 )
M.Tech. 1st Semester
Monsoon 2021-22

Department of Civil Engineering

Indian Institute of Technology (Indian School
of Mines), Dhanbad
 Gradually Varied Flow (GVF)
• CLASSIFICATION OF FLOW PROFILES
In a given channel, y 0 and yc are two fixed depths if Q, n and S0 are fixed. Also,
there are three possible relations between y 0 and yc as (i) y0 > yc, (ii) y0 < yc and
(iii) y0 = yc. Further, there are two cases where y 0 does not exist, i.e., when (a) the
channel bed is horizontal, (S0 = 0), (b) when the channel has an adverse slope, (S 0
is –ve). Based on the above, the channels are classified into five categories as
indicated in the table.
• For each of the five categories of channels, lines representing the critical depth
and normal depth (if it exists) can be drawn in the longitudinal section.

• These would divide the whole flow space into three regions as:

• Region 1: Space above the top most line

2
Region 2: Space between top line and the next lower line
Region 3: Space between the second line and the bed

Table 1 Classification of Channels

3
Fig. Regions of flow profiles
4
Depending upon the channel category and region of flow, the water surface profiles
will have characteristic shapes. Whether a given GVF profile will have an
increasing or decreasing water depth in the direction of flow will depend upon the
term dy/dx being positive or negative.
𝑑𝑦
It can be seen that, 𝑑𝑥 is positive
(i) If the numerator > 0 and the denominator > 0
(ii) if the numerator < 0 and the denominator < 0
𝑑𝑦
i.e. 𝑑𝑥 is positive if (i) (i) K > K0 and Z > Zc or
(ii) K < K0 and Z > Zc
For channels of the first kind, K is a single-valued function of y, and hence
𝑑𝑦
> 0 if (i) y > y0 and y > yc or
𝑑𝑥
5
(ii) y < y0 and y < yc
𝑑𝑦
Similarly, < 0 if (i) yc > y > y0 or
𝑑𝑥

(ii) y0 > y > yc

Further, to assist in the determination of flow profiles in various regions, the behaviour of
dy/dx at certain key depths is given
𝑑𝑦
1.As y → 𝑦0, → 0, i.e. the water surface approaches the normal depth line
𝑑𝑥
asymptotically.
𝑑𝑦
2.As y → 𝑦𝑐, → ∞, i.e. the water surface meets the critical depth line vertically.
𝑑𝑥
𝑑𝑦
3. y → ∞ → 𝑆0, , i.e. the water surface meets a very large depth as a horizontal
𝑑𝑥
asymptote.
Based on this information, the various possible gradually varied flow profiles are grouped
into twelve types. In reality the GVF profiles, especially M1 , M2 and H2 profiles, are very
flat.
6
Table 2. Types of GVF Profiles

7
Fig. Various GVF Profiles

8
Fig. Various GVF profiles
9
 Rapidly Varied Flow
• Hydraulic jump first described by imagination of Leonardo da Vinci.
• The Italian engineer Bidone (1818) is credited with the first experimental investigation of
this phenomenon. Since then considerable research effort has gone into the study of this
subject.
• A hydraulic jump primarily serves as an energy dissipator to dissipate the excess energy
of flowing water downstream of hydraulic structures, such as spillways and sluice gates.
• Some of the other uses are: (a) efficient operation of flow-measurement flumes, (b)
mixing of chemicals, (c) to aid intense mixing and gas transfer in chemical processes, (d)
in the desalination of sea water, and (e) in the aeration of streams which are polluted by
bio-degradable wastes.
• A hydraulic jump occurs when a supercritical stream meets a subcritical stream of
sufficient depth. The supercritical stream jumps up to meet its alternate depth. While
doing so it generates considerable disturbances in the form of large-scale eddies and a
reverse flow roller with the result that the jump falls short of its alternate depth.
10
 Fig. 1 Definition sketch of a hydraulic jump

• Figure 1 is a schematic sketch of a typical hydraulic jump in a horizontal channel.

11
• Section 1, where the incoming supercritical stream undergoes an abrupt rise in the depth
forming the commencement of the jump, is called the toe of the jump.
• The jump proper consists of a steep change in the water-surface elevation with a reverse
flow roller on the major part.
• The roller entrains considerable quantity of air and the surface has white, frothy and
choppy appearance. The jump, while essentially steady, will normally oscillate about a
mean position in the longitudinal direction and the surface will be uneven.
• Section 2, which lies beyond the roller and with an essentially level water surface is
called the end of the jump and the distance between Sections 1 and 2 is the length of the
jump, Lj.
• The initial depth of the supercritical stream is y 1 and y2 is the final depth, after the jump,
of the subcritical stream.
• As indicated earlier, y1 will be smaller than the depth alternate to y 2. The two depths y1.
and y2 at the ends of the jump are called sequent depths.

12
• Due to high turbulence and shear action of the roller, there is
considerable loss of energy in the jump between Sections 1 and 2.
• In view of the high energy loss, the nature of which is difficult to
estimate, the energy equation cannot be applied to Sections 1 and 2 to
relate the various flow parameters.
• In such situations, the use of the momentum equation with suitable
assumptions is advocated.

13
The momentum equation formulation for the jump

• The channel is
inclined to the
horizontal at an
angle θ. Sections
1 and 2 refer to
the beginning and
end of the jump
respectively.
Fig. 2 Definition sketch for the general momentum equation

14
• A control volume enclosing the jump as shown by dashed lines in the figure, is
selected. The flow is considered to be steady. Applying the linear momentum equation
in the longitudinal direction to the control volume,

P1 −P2 −Fs+ W sin θ = M2 − M1 (1)

where P1 = pressure force at the control surface at Section 1 = 𝛾𝐴1 𝑦1 cos 𝜃 by assuming
hydrostatic pressure distribution, where 𝑦1 = depth of the centroid of the area below the
water surface.

P2 = pressure force at the control surface at Section 2 = 𝛾𝐴2 𝑦2 cos 𝜃 if hydrostatic

pressure distribution is assumed.

Fs = shear force on the control surface adjacent to the channel boundary.

W sin θ = longitudinal component of the weight of water contained in the control

volume
15
M2 = momentum flux in the longitudinal direction going out through the
control surface = 𝛽2 𝜌𝑄𝑉2 .

M1 = momentum flux in the longitudinal direction going in through the control

surface =𝛽1 𝜌𝑄𝑉1 .

• The hydraulic jump is a rapidly-varied flow phenomenon and the length of

the jump is relatively small compared to GVF profiles. Thus, frictional force
Fs is usually neglected as it is of secondary importance. Alternatively, for
smaller values of θ, (W sin θ−Fs) can be considered to be very small and
hence is neglected. For a horizontal channel, θ = 0 and W sinθ = 0.

16
HYDRAULIC JUMP IN A HORIZONTAL RECTANGULAR CHANNEL

(a) Sequent Depth Ratio: Consider a horizontal, frictionless and rectangular channel.
Considering unit width of the channel, the momentum equation, Eq. 1, can be
written in the form
(2)

(3)
17
• On non-dimensionalising,

(3.1)

(4)

This equation which relates the ratio of the sequent depths (y2 /y1) to the initial Froude
number F1 in a horizontal, frictionless, rectangular channel is known as the Belanger
momentum equation. For high values of F1, say F1 > 8.0, Eq. 4 can be approximated for
purposes of quick estimation of the sequent depth ratio as
y2 / y1 ≈ 1.41F1 (4.1)

18
Equation 4 can also be expressed in terms of 𝐹2 = 𝑉2 / 𝑔𝑦2 = the subcritical Froude
number on the downstream of the jump as

𝑦1 1
= −1 + 1 + 8𝐹22 (5)
𝑦2 2

(b) Energy Loss: The energy loss EL in the jump is obtained by the energy equation
applied to Sections 1 and 2 as

EL = E1 − E2

19
Substituting for q2 /g from Eq. 3 and simplifying

(𝑦2 −𝑦1 )3
𝐸𝐿 = (6)
4𝑦1 𝑦2

(6.1)

𝐸𝐿 𝐸𝐿 𝐸1
The relative energy loss = /
𝐸1 𝑦1 𝑦1

20
• Substituting for ( y2 /y1 ) from Eq. 4 and simplifying,

(7)

• Equation 7 gives the fraction of the initial energy lost in the hydraulic jump. The
variation of EL / E1 with F1 is shown in Fig. 3 which highlights the enormous energy
dissipating characteristic of the jump. At F1 = 5, about 50 per cent of the initial energy
in the supercritical stream is lost and at F1 = 20, EL / E1 is about 86 per cent. Figure 3
also serves as a yardstick for comparing the efficiencies of other types of jumps and
energy-dissipating devices.

21
Fig. 3 Relative energy loss in a jump

22
Classifications of Jumps:

• The hydraulic jumps in horizontal rectangular channels are classified into

five categories based on the Froude number F1 of the supercritical flow, as
follows:

• (i)Undular Jump 1.0 < F1 ≤ 1.7 The water surface is undulating with a very
small ripple on the surface. The sequent-depth ratio is very small and EL / E1
is practically zero.

23
• (ii)Weak Jump 1.7 < F1 ≤ 2.5 The surface roller
makes its appearance at F 1 ≈ 1.7 and gradually
increases in intensity towards the end of this
range, i.e. F1 ≈ 2.5. The energy dissipation is
very small, is E L/ E1 about 5 per cent at F1 = 1.7
and 18 per cent at F1 = 2.5.

• (iii) Oscillating Jump 2.5 < F 1 ≤ 4.5 This

category of jump is characterised by an
instability of the high-velocity flow in the jump
which oscillates in a random manner between the
bed and the surface. These oscillations produce
large surface waves that travel considerable
distances downstream.

Special care is needed to suppress the waves in stilling basins having this kind of jump.
Energy dissipation is moderate in this range; EL / E1 = 45 per cent at F1 = 4.5 24
• (iv)‘Steady’ Jump 4.5 < F 1 ≤ 9.0 In this range of
Froude numbers, the jump is well-established, the
roller and jump action is fully developed to cause
appreciable energy loss. The relative energy loss
EL / E1 ranges from 45 per cent to 70 per cent in
this, class of jump. The ‘steady jump’ is least
sensitive in terms of the toe-position to small
fluctuations in the tailwater elevation.

• (v) Strong or Choppy Jump F1 >9.0 In this class

of jump the water surface is very rough and
choppy. The water surface downstream of the
jump is also rough and wavy. The sequent-depth
ratio is large and the energy dissipation is very
efficient with EL / E1 values greater than 70 per
cent.
25
• It is of course obvious that the above classification is based on a purely subjective
consideration of certain gross physical characteristics.

• An interesting and useful relationship involving F 1 and a non-dimensional parameter

made up of EL and q is obtained as below:

(8)

26
Fig. Hydraulic jumps at different
Froude numbers (K.Subramanya
Book) [Note: The flow is right to
left]

27
 Characteristics of Jump in a Rectangular Channel
Length of the Jump: The length of the jump Lj is an
important parameter affecting the size of a stilling
basin in which the jump is used.
It is usual to take the length of the jump as the
horizontal distance between the toe of the jump to a
section where the water surface levels off after
reaching the maximum depth.
Fig. Definition sketch of a hydraulic jump
Because the water-surface profile is very flat
towards the end of the jump, large personal errors
are introduced in the determination of the length Lj .

28
Experimentally, it is found that Lj / y2
=f (F1). The variation of Lj / y2 with F1
obtained is shown in Figure.

This curve is usually recommended for

general use. It is evident from Figure
that while Lj / y2 depends on F1 for
small values of the inlet Froude number,
at higher values (i.e. F1 > 5.0) the Fig. Length of the hydraulic jump on a horizontal floor
relative jump length Lj / y2 is practically
constant beyond a Froude number value
of 6.1.

Lj = 6.9 (y2 − y1) (9)

Lj = 5 to 7 (y2 − y1) 29
• Pressure Distribution: The pressures at the toe
of the jump and at the end of the jump follow
hydrostatic pressure distribution. However,
inside the body of the jump, the strong
curvatures of the streamlines cause the pressures
to deviate from the hydrostatic distribution.

• Water-Surface Profile: A knowledge of the

surface profile of the jump is useful in the Fig. Definition sketch for the jump
efficient design of side walls and the floor of a profile (6)
stilling basin. Consider the coordinate system
shown in Figure. The coordinates of the profile
are (x, h) with the boundary condition that at x =
0, h = 0, and at x = Lj , h = (y2 –y1).

30
• Velocity profile: When the supercritical stream at
the toe enters the jump body, it undergoes shearing
action at the top as well as at the solid boundaries.

• The top surface of the high velocity flow will have

high relative velocities with respect to the fluid
mass that overlays it.

• The intense shear at the surface generates a free

shear layer which entrains the fluid from the
overlying mass of fluid. The boundary shear at the
bed causes a retardation of the velocity in a
boundary layer.
Fig.: Velocity distribution in a jump
• velocity distribution in a section at a distance x
from the toe is shown in Fig.
31
• Computations related to hydraulic jumps in rectangular channels are
assumed to be simple and following relations are used

(i) Continuity equation

(ii) Momentum equation for sequent depths and

(iii) Energy equation for energy loss in the jump.

• The basic variables can be discharge intensity q; sequent depths y1 and y2 ;

and energy loss EL .

32
 Que. In a hydraulic jump occurring in a rectangular channel of 3.0-m,
width, the discharge is 7.8 m3/s and the depth before the jump is 0.28 m. Estimate
(i) sequent depth, and (ii) the energy loss in the jump

Solution:

The sequent
depth ratio is
given by Eq.

(ii) The energy loss EL is given by Eq.

33
Que. A rectangular channel carrying a supercritical stream is to be provided with a hydraulic
jump type of energy dissipater. It is desired to have an energy loss of 5.0 m in the hydraulic jump
when the inlet Froude number is 8.5. What are the sequent depths of this jump?
Solution Given F1 = 8.5 and EL = 5.0 m
By eqn. (4) and (6.1)

(6.1)

34
Que. A hydraulic jump takes place in a rectangular channel with sequent depths of 0.25 m
and 1.50 m at the beginning and end of the jump respectively. Estimate the (i) discharge per
unit width of the channel and (ii) energy loss.
Solution: By Eqn.

The energy loss EL is given by Eq. (6).

35
Que. A spillway discharges a flood flow at a rate of 7.75 m3/s per metre width. At the
downstream horizontal apron the depth of flow was found to be 0.50 m. What tailwater depth is
needed to form a hydraulic jump? If a jump is formed, find its (a) type, (b) length, (c) head
loss, (d) energy loss as a percentage of the initial energy.
Solution:

Sequent-depth By Eq. (4)

36
37
USE OF THE JUMP AS AN ENERGY DISSIPATOR
• The high energy loss that occurs in a hydraulic jump has led to its adoption as a
part of the energy-dissipator system below a hydraulic structure.
• The downstream portion of the hydraulic structure where the energy dissipation is
deliberately allowed to occur so that the outgoing stream can safely be conducted
to the channel below is known as a stilling basin.
• It is a fully-paved channel section and may have additional appurtenances, such
as baffle blocks and sills to aid in the efficient performance over a wide range of
operating conditions.
• Stilling basins are so designed that not only a good jump with high energy-
dissipation characteristics is formed within the basin but it is also stable.
• For economic considerations the basin must be as small as practicable.

38
• Designing a stilling basin for a
given hydraulic structure involves
considerations of parameters
peculiar to the location of the
structure in addition to the
mechanics of flow.

• Model studies are usually resorted

to arrive at an efficient design.

• The US Bureau of Reclamation has

developed a series of type designs
which is represented in below
figure. Fig. USBR-type III stilling basin

39
 ENERGY DISSIPATOR

End Sill Baffle blocks Chute blocks

40
• This stilling basin is recommended for F1 > 4.5 and V1<18 m/s. Note the chute blocks to
assist in splitting and aerating of flow; baffle blocks which offer additional resistance to
flow; and the end sill which helps the outgoing stream to be lifted up into a trajectory so
that the basin end is not subjected to scouring action.

• The effect of these appurtenances is to shorten the stilling basin length to 2.7 y 2 as
against 6.1 y2 required for a free unaided hydraulic jump. Also, the minimum tailwater
depth required is 0.83 y2 as against y2 for an unaided jump.

41
Bucket type Energy Dissipator
42
43
44
 References
Text Books:
1. Subramanya, K. (2015), 4th edition, “Flow in Open Channels”, Tata McGraw Hill.
2. Bansal, R. K. A textbook of fluid mechanics and hydraulic machines:(in SI units). LAXMI
Publications, Ltd., 2005.
Reference Books:
1. Srivastava, R. (2014), 1st edition, “Flow through Open Channels”, Oxford University
Press.

2. Rajput, R. K. A textbook of fluid mechanics and hydraulic machines. S. Chand Publishing,

2015.

45
Engineering Hydrology & Hydraulics
(CEC 505 )
M.Tech. 1st Semester
Monsoon 2021-22

Department of Civil Engineering

Indian Institute of Technology (Indian School
of Mines), Dhanbad
Spatially Varied Flow
• A steady spatially varied flow represents a gradually-varied flow with non-
uniform discharge.
• The discharge in the channel varies along the length of the channel due to
lateral addition or withdrawal.
• Spatially varied flow (SVF) classified into two categories:
(i) SVF with increasing discharge and
(ii) SVF with decreasing discharge.

2
 SVF WITH INCREASING DISCHARGE
• SVF with increasing discharge finds considerable
practical applications.

• Flows in side-channel spillway, wash-water troughs in

filter plants, roof gutters, highway gutters are some of
the typical instances.

• Figure shows a typical side-channel spillway causing

an SVF in the channel below it.

• The lateral flow enters the channel normal to the Fig. Lateral spillway channel flow
channel-flow direction causing considerable
turbulence.

• It is difficult to assess the net energy imparted to the

flow and as such the energy equation is not of much
use in developing the equation of motion.

3
 Differential Equation of SVF with Increasing Discharges
In applying the momentum equation, the following assumptions are made:
1. The pressure distribution is assumed to be hydrostatic. This amounts to assuming the
water-surface curvatures to be moderate. The regions of high curvature, if any, must be
delineated and excluded from the analysis.
2. The one-dimensional method of analysis is adopted. The momentum correction factor β
is used to adequately represent the effect of non-uniformity of velocity distribution.
3. The frictional losses in SVF are assumed to be adequately represented by a uniform flow
resistance equation, such as Manning’s formula.
4. The effect of air entrainment on forces involved in the momentum equation is neglected.
5. It is assumed that the lateral flow does not contribute any momentum in the longitudinal
direction.
6. The flow is considered to be steady.
7. The channel is prismatic and is of small slope. 4
• Consider a control volume formed by two
Sections 1 and 2, distance Δx apart as
shown in Figure. Applying the momentum
equation in the longitudinal x direction.

M2 − M1 = P1 − P2 + W sin θ − Ff (1)

ΔM = − ΔP + W sin θ − Ff (1.1)

in which M = momentum flux = βρQ 2 /A

P = pressure force = 𝛾𝐴𝑦ത

Fig. Definition sketch of SVF with lateral inflow

where 𝑦ത = depth of the centre of gravity of the flow cross-section from the water surface, W
sin θ = component of the weight of the control volume in the x direction and F f = frictional
force = γASf Δx.
5
 Dividing Eq. (1.1) by Δx and taking limits as Δx → 0,
𝑑𝑀 𝑑𝑃
=− + 𝛾𝐴𝑆0 − 𝛾𝐴𝑆𝑓 (2)
𝑑𝑥 𝑑𝑥

By taking moments of the areas about the new water surface after a small change dy
in depth (shown in figure),

6
 Fig. Definition sketch

Eq. (2) simplifies to

(3)

7
• Unlike GVF, SVF with lateral inflow has not received extensive attention and
as such the detailed classification and analysis of a general-flow situation are
not available in literature. By assuming zero friction and β = 1.0.
• Hence, flow is classified into the following categories

• Type A The flow is subcritical throughout the channel and the Froude number
increases continuously in the downstream direction.

• Type B The flow is subcritical throughout but the Froude number will first
increase, reach a maximum value less than unity and then decrease.

8
• Type C The flow is subcritical initially, passes through a critical section to
become supercritical in the downstream portions of the channel and then
terminates in a jump due to downstream control such as a submerged outlet.

• Type D The same as Type C, but the jump is not formed in the channel. The
outlet is free.

• These four types of flow can be determined by a study of the transitional

profile and the critical-depth line along with the downstream end conditions.
• In general, Type C and D situations can occur in side spillway channel design
and Type A and B can occur in wash water-trough and gutter-design problems.

9
 SVF WITH DECREASING DISCHARGE
• SVF with decreasing discharges occurs in a variety of field situations, typical
examples being side weirs, bottom racks and siphon tube irrigation systems.
• The abstraction of water from a canal by using the above means is normally
achieved in such a manner as to cause minimum obstruction and with
consequent little energy losses in the parent channel.
• It is usual to assume that energy loss due to diversion of water is zero and the
energy equation is used to derive the basic equation of motion.

10
11
 Differential Equation for SVF with Decreasing Discharge
• The following assumptions are made:

1. The pressure distribution is hydrostatic

2. The one-dimensional method of analysis is used (the energy-correction factor α is

used to adequately represent the non-uniformity of velocity distribution).

3. The friction losses are adequately represented by Manning’s formula.

4. Withdrawal of water does not affect the energy content per unit mass of water in
the channel

5. The flow is steady

6. The channel is prismatic and is of small slope.

12
Consider the total energy at a Section x,
𝑉2
𝐻 = 𝑍 + 𝑦 +∝ (4)
2𝑔

Differentiating this with respect to x

(5)

13
 (6)

Equation (6) is the basic differential equation governing the motion of SVF
with decreasing discharges. Note the difference between Eq. (6) and Eq. (3).
When q * = 0, Eq. (6) will be the same as the differential equation of GVF, Eq.

GVF eqn.

14
15
 References
Text Books:
1. Subramanya, K. (2015), 4th edition, “Flow in Open Channels”, Tata McGraw Hill.

Reference Books:
1. Srivastava, R. (2014), 1st edition, “Flow through Open Channels”, Oxford University
Press.

16