BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

Basic concepts of free
surface flows
 Open Channel

An open channel may be defined

as a passage in which liquid
flows with its upper surface
exposed to atmosphere. In open
channels the flow is due to
gravity, thus the flow conditions
are greatly influenced by the
slope of the channel.
Comparison between Open Channel and Pipe Flow
Comparison between Open Channel and Pipe Flow
Types of Channels
The various types of channels are:
1. Natural channel : It is the one which has irregular sections of varying shapes, developed in a
natural way. Examples: Rivers, streams etc.
2. Artificial channel: It is the one which is built artificially for carrying water for various purposes.
They have the cross-sections with regular geometrical shapes (which usually remain same
throughout the length of the channel). Examples: Rectangular channel, trapezoidal channel,
parabolic channel etc.
3. Open channel: A channel without any cover at the top is known as an open channel. Examples:
Irrigation canals, rivers, streams, flumes and water falls.
4. Covered or closed channels : The channel having a cover at the top is known as a covered or
closed channel. Examples: Partly filled conduits carrying public water supply such as sewerage
lines, under ground drains, tunnels etc. not running full of water.
Types of Channels
5. Prismatic channel: A channel with constant bed slope and the same cross-section along its length is known
as a prismatic channel. Key Features
• Uniform channel cross-section throughout the length.
• Constant bottom slope.
• Prismatic channels can be triangular, rectangular, parabolic, trapezoidal or circular.
• Artificial channels are usually prismatic channels.
The prismatic channels can be further subdivided as:
(i) Exponential channel: It is the one in which area of cross-section of flow is directly proportional to any
power of depth of flow in channel. Examples: Rectangular, triangular and parabolic channels.
(ii) Non-exponential channel: Trapezoidal and circular channels are non-exponential channels.

6. Non-prismatic Channels : If the cross-section of a channel is non-uniform or the bottom slope is not
constant, it is called Non-Prismatic Channels. Key Features
• Non-uniform channel cross-section
• Varying bottom slope.
• Natural channels are usually non-prismatic channel
TYPES OF FLOW IN CHANNELS
The flow in channels is classified into the following types, depending upon the change
in the depth of flow with respect to space and time:
1. Steady flow and unsteady flow
2. Uniform flow and non-uniform (or varied) flow
3. Laminar flow and turbulent flow
4. Subcritical flow, critical flow and supercritical flow.
DEFINITIONS
1. Depth of flow (y) : It is the vertical distance of the lowest point of a channel section (bed of the channel)
from the free surface.
2. Depth of flow section: It is the depth of flow normal to the bed of the channel. d = y cos θ 90º
where, θ = The angle which the channel bed makes with the horizontal.
Since the slopes of the channels are very small, cos θ ≈1 and d ≈ y. The depth of flow and depth of flow
section are assumed equal, unless mentioned otherwise
3. Top width (T): It is the width of the channel section at the free surface (i.e. the width of the liquid
surface exposed to the atmospheric pressure.
4. Wetted area (A): It is the cross-sectional area of the flow section of the channel.
5. Wetted perimeter (P): It is the length of the channel boundary in contact with the flowing water at any
section.
6. Hydraulic radius (R): It is ratio of the cross-sectional area of flow to wetted perimeter. It is also called
hydraulic mean depth. i.e. R = A/P
7. Hydraulic depth (D): It is the ratio of the wetted area A to the top width T. i.e. D = A/T
 Velocity and Pressure distribution,
Mass, energy and momentum principle for
prismatic and non-prismatic channels critical,

https://www.youtube.com/watch?v=g086qmOsTDY
 Velocity and Pressure distribution
Velocity Distribution
• The presence of corners and boundaries in an
open channel causes the velocity vectors of the
flow to have components not only in the
longitudinal and lateral direction but also in
normal direction to the flow.
• Figure shows isovels (contours of equal velocity)
of v for a natural channel.
• The velocity v is zero at the solid boundaries and
gradually increases with distance from the
boundary.
• The maximum velocity of the cross-section
occurs at a certain distance below the free surface.
 Velocity Distribution
• The max velocity occurs at a point little below from the free
surface. This is due to the presence of secondary currents.
• The max velocity occurs at a distance 0.05 y to 0.25 y from the
free surface, (where y is the depth of flow)
• Deeper & narrower the channel, more deep the point of max
velocity be.
• The average vel occurs at a distance 0.6y from the free liquid
surface.
𝑉 0.2 +𝑉 0.8
• Velocity average =
2
• This velocity is slightly less than the surface velocity (about
0.8 to 0.85 times).
OCF – 1D (Approach) : Velocity Profile

Discharge through elemental area da having velocity v :

𝑑𝑄 = 𝑣 𝑑𝐴

So, Total discharge : 𝑄 = ‫𝐴𝑑 𝑣 ׯ‬

Considering average velocity = V and area = A,
𝑄 = 𝑉𝐴
Equating Q,
𝑉𝐴 = ර 𝑣 𝑑𝐴
1
Therefore Average Velocity: 𝑉 = 𝐴 ‫𝐴𝑑 𝑣 ׯ‬
 OCF – 1D (Approach) : Kinetic Energy
Mass of Liquid Passing through area 𝑑𝐴 = 𝜌 × 𝑣 × 𝑑𝐴
𝑣2 𝑚𝑣 2
K.E. Through Area 𝑑𝐴 = 𝜌 × 𝑣 × 𝑑𝐴 × ⇒ 𝐾. 𝐸. =
2 2
1
K.E. Through section for variable velocity = 2 ‫ 𝑣 𝜌 ׯ‬3 𝑑𝐴
If area is A and Avg Velocity is V
𝐴
K.E. = 𝜌𝑉 3 2

𝐾𝐸 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑎𝑐𝑡𝑢𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

K.E. Correction Factor : ∝= 𝐾𝐸 𝐵𝑎𝑠𝑒𝑑 𝑜𝑛 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦

1 3 𝑑𝐴
‫ׯ‬ 𝜌𝑣 1 𝑣3
𝛼= 2 = ර 3 𝑑𝐴
1 3𝐴 𝐴 𝑉
𝜌 𝑉
2
OCF – 1D (Approach) : Momentum
Momentum of Liquid passing through area 𝑑𝐴 = 𝜌 𝑣 𝑑𝐴 × 𝑣
Momentum of Liquid passing through the full section M = ‫ 𝑣 𝜌 ׯ‬2 𝑑𝐴

If Area = A and Avg Velocity = V

Momentum : 𝑀 = 𝜌 𝑉 2 𝐴

Moment Correction Factor :

1 𝑣 2
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑎𝑐𝑡𝑢𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝛽= :𝛽 = ර 𝑑𝐴
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝐵𝑎𝑠𝑒𝑑 𝑜𝑛 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝐴 𝑉
If the velocity distribution is uniform, 𝛼 = 𝛽 = 1

𝛼 & 𝛽 increase in non-uniformity of velocity but for any variation of velocity

𝛼>𝛽>1
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

OPEN CHANNEL FORMULAE FOR UNIFORM FLOW
For uniform flow in open channels, the following formulae will be
discussed:

1. Chezy’s formula
2. Manning’s formula.
1.Chezy’s formula
Chezy's equation applies to open channel flow under several key assumptions:

• Steady Flow: The flow velocity and depth remain constant over time at any given location in the
channel. There's no acceleration or deceleration.

• Uniform Flow: The water depth and flow characteristics (like velocity) are the same across the entire
width of the channel at any given location.

• Prismatic Channel: The channel has a consistent cross-sectional shape throughout its length. It doesn't
widen or narrow.

• Hydrostatic Pressure Distribution: The pressure within the flowing water varies only with depth,
following the principles of hydrostatic pressure.

• Small Bed Slope: The slope of the channel bottom is relatively small. This allows simplifying
calculations by assuming the sine of the slope angle is approximately equal to the tangent of the slope
angle.
Chezy’s Formula
Substituting the Manning Formula in chezy Contsant

1 1 1 1
𝑉 = × 𝑅 × 𝑅 × 𝑆2
6 2
𝑁

1 2 1
𝑉 = 𝑅3 𝑆 2
𝑁
Question 1: Find the rate of flow and
conveyance for a rectangular channel
7.5 m wide for uniform flow at a depth
of 2.25 m. The channel is having bed
slope as 1 in 1000. Take Chezy’s
constant C = 55. Also state whether the
flow is tranquil or rapid.
PYQ _(2014-15): A rectangular channel has width of 2.5 m and slope of 1: 400,
Find depth of flow of the discharge is 10 cume/sec. Use Chezy's formula and take
C = 50 (10 Marks)
PYQ _(2014-15): flows at a uniform depth of 2 m in trapezoidal channel having bottom width 6m,
side slope 2 H: 1V of it is to carry a discharge of 65 m³/sec. compute the bottom slope required to be
provided. Take manning's coefficient n = 0.025. (10 Marks)
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

MOST ECONOMICAL SECTION OF A CHANNEL
The most economical section (also called the best section or most efficient section)
is one which gives the maximum discharge for a given amount of excavation.
• From continuity equation it is evident that discharge is maximum when velocity
is maximum, the area of cross section of channel remaining constant.
• From Chezy’s formula and Manning’s formula it can be seen that for a given
value of slope and surface roughness the velocity of flow will be maximum if
A
hydraulic radius R = is maximum.
P
• Further the area being constant hydraulic radius is maximum if the wetted
perimeter is minimum; this condition is used to determine the dimensions of
economical sections of different forms of channels.
• The best form of channel which complies with this condition is one which has a
semi-circular cross-section.
Most Economical Rectangular Channel Section
Problem 1: A rectangular channel is to be dug in the rocky portion of a soil. Find its most economical
cross-section if it is to convey 12 m3 /s of water with an average velocity of 3 m/s. Take Chezy’s
constant C = 50.
Problem 2 : Determine the most economical section of a rectangular channel carrying water at the rate of 0.5 m3
/s; the bed slope of the channels being 1 in 2000. Take Chezy’s constant C = 50
Most Economical Triangular Channel Section
Substituting the value of θ from eqn. (16.18) in the above eqn., we get:
Most Economical Trapezoidal Channel Section
PYQ 14-15 : Design a concrete lined channel to carry a discharge of 500 m³ at a slope of 1:4000. The side
sloper of channel may be taken as 1 H: 1 V. The meanings coefficient for the lining is 0.014. Assume
permissible velocity in the section as 2.5 m/sec. (10 Marks)
Most Economical Trapezoidal Channel Section
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

PYQ 22-23 : A trapezoidal canal is to carry 45 m3/s with a mean velocity of 0.6 m/s. One side of canal is
vertical and the other has a slope of 2 horizontal to 1 vertical. Find the minimum hydraulic slope, if Manning’s
N = 0.013.. (10 Marks)
Most Economical Circular Channel Section
(i) Condition for maximum velocity:
Y≈ 0.81 d
& 2θ = 257.57°
where, d = Diameter of the circular channel.
Thus, maximum velocity occurs when the depth of flow is 0.81 times the
diameter of the circular channel.
R ≈ 0.6086r ≈ 0.305d
Thus, for maximum mean velocity in a channel of circular section hydraulic
radius equals 0.305 times the channel diameter.

(ii) Condition for maximum discharge:

Y≈ 0.95d
& 𝟐𝛉=308°
where, d is the 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.
R ≈ 0.573r ≈ 0.29d
Thus for maximum discharge through a circular channel, the hydraulic radius
equals 0.29 times channel diameter
Problem : A concrete lined circular channel of 3.6 m diameter has a bed slope of 1 in 600. Determine the velocity and flow
rate for the conditions of:
(i) Maximum velocity, and
(ii) Discharge.
(Take Chezy’s constant, C = 50.)
PYQ 17-18 : An open channel to be made of concrete is to be designed to carry 1.5m3/s at a slope of
0.00085. Find the most efficient cross section for
(a) Rectangular section
(b) Trapezoidal section
(c) Semicircular section. (7 Marks)
PYQ 17-18 : An open channel to be made of concrete is to be designed to carry 1.5m3/s at a slope of
0.00085. Find the most efficient cross section for
(a) Rectangular section
(b) Trapezoidal section
(c) Semicircular section. (7 Marks)
PYQ 17-18 : An open channel to be made of concrete is to be designed to carry 1.5m3/s at a slope of
0.00085. Find the most efficient cross section for
(a) Rectangular section
(b) Trapezoidal section
(c) Semicircular section. (7 Marks)
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

Gradually Varied Flow

With Derivation
1. differential equation of GVF
2. governing equation of GVF
 Introduction

Gradually varied flow (G.V.F.) is one in which the depth changes gradually over a long
distance. In a rapidly varied flow, the change in depth takes place in a short distance.
 Causes For gradually varied flow ( GVF)

❖ The change in the shape and size of the channel

cross-section,

❖ The change in slope of the channel,

❖ The presence of obstruction (e.g., weir etc.), and

❖ The change in frictional forces at the boundaries.

 Equation of Gradually Varied Flow
Assumption
1. the channel is a prismatic (a channel with constant section and alignment).

2. The bed slope is small.

3. The flow is steady and hence discharge is constant.

4. The pressure distribution over the channel section is hydrostatic i.e. stream lines are practically straight and
parallel.

5. The energy correction factor (α) is unity.

6. The roughness co-efficient is constant for the length of the channel and it does not depend on the depth of flow.

7. The Chezy and Manning correlations are equally applicable to gradually varied flow for determining the slope
of energy line.
 Equation of Gradually Varied Flow

𝑆0 ≅𝑖
𝑆𝑓 ≅𝑗=

According to Bernoulli’s equation, the energy equation at any section is given by:

𝑉2
𝐸 =𝑧+𝑦+
2𝑔

Taking the bottom of the channel on the X-axis and the vertically upwards direction measured
from the channel bottom, as the Y-axis, differentiation of eqn. (i), with respect to x yields:
 Equation of Gradually Varied Flow
According to Bernoulli’s equation, the energy equation at
any section is given by:

𝑉2
𝐸 =𝑧+𝑦+
2𝑔

Taking the bottom of the channel on the X-axis and the

vertically upwards direction measured from the channel
bottom, as the Y-axis, differentiation of eqn. (i), with
respect to x yields:

𝑑𝐸 𝑑𝑧 𝑑𝑦 𝑑 𝑉 2
= + +
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 2𝑔
 Equation of Gradually Varied Flow
𝑑𝐸 𝑑𝑧 𝑑𝑦 𝑑 𝑉 2
= + +
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 2𝑔

𝑑𝐸
1. 𝑑𝑥
Represents the energy slope. Since the total energy of flow always decreases in the direction of
motion.. Denoting it by 𝑆𝑓 , we have
𝑑𝐸
= −𝑆𝑓
𝑑𝑥
𝑑𝑍
2. 𝑑𝑥
denotes the bottom slope. Denoting it by as 𝑆0 , we have
𝑑𝑍
= −𝑆0
𝑑𝑥
𝑑𝑦
3. 𝑑𝑥
represents the water surface slope relative to the bottom of the channel. is also called the slope of the
free water surface.
 dy
1) When = 0: y is constant (or depth of water above the bottom of channel is constant); it means that
dx
free water surface is parallel to the channel bed.

dy dy
2) When > 0 or is + ve ∶It indicates that the depth of water increases in the direction of flow,
dx dx
the profile of water so obtained is called back water curve.

dy dy
3) When < 0 or is − ve : It indicates that the depth of water decreases in the direction of flow.
dx dx
The profile of water so obtained is known as drop down curve.
Back water curve &
Afflux
𝐄𝟐 − 𝐄𝟏
𝐥=
𝐒𝟎 − 𝐒𝐟
With Derivation
1. Length of back water curve
 Back Water Curve and Afflux
In an open channel when the flow is uniform, the flow has constant depth at all the sections
and the surface of the free water lies parallel to bed of the channel. But when an obstruction
like a dam, weir etc. comes across the channel width the water level rises and it has maximum
depth from the bed at some section (Fig. 16.34). If y1 is the depth of water at the point, where
the water starts rising up and y2 is the maximum height of rising water from the bed, then this
increase in depth (i.e. y2 – y1) is known as ‘afflux’ and the curved surface of the liquid with
its concavity upwards, is known as ‘back water curve’.
 Length of Back Water Curve

y1 = Depth of flow at section 1-1,

V1 = Velocity of flow at section 1-1,
y2 = Depth of flow at section 2-2,
V2 = Velocity of flow at section 2-2,
S0 = Bed slope,
Sf = Energy line slope, and
l = Length of back water curve.

𝐸2 − 𝐸1
𝑙=
𝑆0 − 𝑆𝑓
Gvf मे ज़ोन कैसे बाटे जाते ैं
 Classification of flow profile (gvf)
In a given channel, 𝒚𝟎 and 𝒚𝒄 are two fixed depths if 𝑸, 𝒏 and 𝑺𝟎 are fixed
1. 𝑦0 > 𝑦𝑐 𝐀𝐜𝐭𝐮𝐚𝐥 𝐝𝐞𝐩𝐭𝐡 𝐲 → real depth at any section in the Channel.

𝐍𝐨𝐫𝐦𝐚𝐥 𝐃𝐞𝐩𝐭𝐡 𝐲𝟎 , 𝐲𝐧 → normal depth is the depth of flow that occur

2. 𝑦0 < 𝑦𝑐 if the flow was 𝐮𝐧𝐢𝐟𝐨𝐫𝐦 and 𝐬𝐭𝐞𝐚𝐝𝐲 and it is usually predicted using
the 𝐦𝐚𝐧𝐧𝐢𝐧𝐠 ′ 𝐬 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧

3. 𝑦0 = 𝑦𝑐 𝐂𝐫𝐢𝐭𝐢𝐜𝐚𝐥 𝐝𝐞𝐩𝐭𝐡 𝐲𝐜 → specific energy will be minimum.

𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑡𝑤𝑜 𝑐𝑎𝑠𝑒𝑠 𝑤ℎ𝑒𝑟𝑒 𝑦0 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡

a) 𝑡ℎ𝑒 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 𝑏𝑒𝑑 𝑖𝑠 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑆0 = 0

b) 𝑤ℎ𝑒𝑛 𝑡ℎ𝑒 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 ℎ𝑎𝑠 𝑎𝑑𝑣𝑒𝑟𝑠𝑒 𝑠𝑙𝑜𝑝𝑒 𝑆0 𝑖𝑠 − 𝑣𝑒

Source-Flow in Open Channels - K. Subramanya - Google Books

Classification of flow profile (gvf)

Source-Flow in Open Channels - K. Subramanya - Google Books

 Region of flow profile (gvf)
Region 1: Space Above The top Most line
Region 2: Space Between Top line and the next lower line
Region 3: Space between The second Line and the bed

𝑆0 = +𝑣𝑒

Source-Flow in Open Channels - K. Subramanya - Google Books

 Region of flow profile (gvf)
Region 1: Space Above The top Most line
Region 2: Space Between Top line and the next lower line
Region 3: Space between The second Line and the bed

𝑆0 = +𝑣𝑒

𝑆0 = −𝑣𝑒
𝑆0 = 0
Source-Flow in Open Channels - K. Subramanya - Google Books
Classification of g.v.f.
surface profile
𝑀1
𝑀2

𝑀3
Types of GVF profile- steep slope
𝑆1

𝑆2

𝑆3
Types of GVF profile- critical slope

𝐶1

𝐶3
Types of GVF profile- horizontal bed

𝐻2

𝐻3
Types of GVF profile-adverse slope

𝐴2

𝐴3
Control Section Or
Point
 Control section ( or point)
1. A control section is defined as a section in which a fixed relationship exists between the
discharge and depth of flow.
2. Any channel section at which there is a unique relationship between the flow depth and
discharge is referred to as a control.
3. Weirs, spillways, sluice gate are some typical example of structure which give rise to
control section.
4. The critical depth is also a critical point.

Source-Flow in Open Channels - K. Subramanya - Google Books

Control section ( or point)

Source-Flow in Open Channels - K. Subramanya - Google Books

Control section ( or point)

Source-Flow in Open Channels - K. Subramanya - Google Books

Break in grad
Open channel flow
 Break in grade
1. Simple situations of a series combination of two channel sections with differing bed slopes are considered.
2. A break in grade from a mild channel to a milder channel

1. It is necessary to first draw the critical-depth line (CDL) and the normal-depth line (NDL) for both slopes.
2. Since yc does not depend upon the slope for a taken Q = discharge, the CDL is at a constant height above
the channel bed in both slopes.
3. The normal depth y01 for the mild slope is lower than that of the milder slope (y02).

In this case, y02 acts as a control, similar to the weir or spillway case and an M1 backwater curve is produced in the
mild slope channel.

Source-Flow in Open Channels - K. Subramanya - Google Books

Source-Flow in Open Channels - K. Subramanya - Google Books
Source-Flow in Open Channels - K. Subramanya - Google Books
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

Problem: In a rectangular channel of width 24 m and depth of flow 6 m, the rate of flow of water is 86.4 m3
/s. If the bed slope of the channel is 1 in 4000 find the slope of the free water surface. Take Chezy’s constant C
= 60.
Problem: A wide channel laid to a slope of 1 in 1000 carries a discharge of 3.5 m3 /s per metre width at a depth of 1.6
m. Find out the value of Chezy’s constant C. Consider the flow to be uniform.
If the actual depth varies from 1.5 m at an upstream location to 1.7 m at a location 300 m downstream or in other words
the flow is gradually varied, what will be the value of Chezy’s coefficient
Problem: A weir is installed across a rectangular open channel thereby raising the flow depth from 1.5 m in a
normal flow to 2.5 m at the weir. The width of the channel is 10 m and it is laid to a slope of 1 in 10000. Find
an approximate length of the backwater curve considering the average velocity, average depth and average
slope midway between the two sections. Take the value of Manning’s rugosity co-efficient equal to 0.02.
PYQ 2022-23: A rectangular channel with bottom width of 4 m and a bottom slope of 0.0008 has a
discharge of 1.5 m3 /s. In a gradually varied flow in this channel, the depth at a certain location is found to
be 0.30 m. Assuming n = 0.016, determine & sketch the GVF profile.
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

Introduction

Source- @R. K. Rajput

 hydraulic jump or Standing Wave

❖
visualisation
Source- @practical_Engineering
 Use of hydraulic jump

❖
Source- @internet
 Important formula
✓ 𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑗𝑢𝑚𝑝 𝐻𝑗 = 𝑦2 − 𝑦1
✓ 𝐿𝑒𝑛𝑔ℎ𝑡 𝑜𝑓 𝐽𝑢𝑚𝑝 𝐿𝑗 = 6.9 𝑦2 − 𝑦1
✓ 𝑃𝑜𝑤𝑒𝑟 𝐿𝑜𝑠𝑠 = 𝑤𝑄 ∆𝐸
𝑊ℎ𝑒𝑟𝑒
𝑤 = 𝛾 = 𝜌𝑔 = 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡
𝑄 = 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒
𝑦2 −𝑦1 3
∆𝐸 = (ℎ𝑓 ) = 𝐸1 − 𝐸2 =
4𝑦1 𝑦2

Source- @R. K. Rajput

Types of hydraulic jump

Source- @internet
 Types of hydraulic jump

Source- @internet
 Application of Hydraulic jump
❖

❖
 Location of hydraulic jump

❖
 Effect of Hydraulic Jump
❖

❖
 Analysis of hydraulic jump
The following assumptions are made in the analysis of hydraulic jump:
1. Loss of head due to friction at the walls and channel bed is negligible.
2. The flow is uniform and the pressure distribution is hydrostatic before and after the jump.
3. The channel is horizontal or it has a very small slope. The weight component in the direction of flow is
neglected.
4. The momentum correction factor (β) is unity.
Height of hydraulic jump (𝐻𝑗 ):
Height of hydraulic jump (𝐻𝑗 ):
Height of hydraulic jump (𝐻𝑗 ):
Height of hydraulic jump (𝐻𝑗 ):
 Length of hydraulic jump (𝐿𝑗 ).

Length of hydraulic jump represents that short distance over which

the jump occurs For rectangular channels with horizontal floor,
length of a jump has been found to vary between 5 to 7 times the
height of the jump.
𝐿𝑗 = 5 𝑡𝑜 7 𝐻𝑗
Loss of energy due to hydraulic jump
Loss of energy due to hydraulic jump
Hydraulic jump
Numerical
hydraulic jump Problem-1
energy dissipators

&
Types of energy
dissipators
Tihari dam AT BHAGIRATHI RIVER (UTTARAKHAND)
 Energy dissipator

Grand Coulee dam, (Columbia River)

in the United States of America. 𝐷𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 → 28320 𝑚3 /𝑠

𝐻𝑒𝑎𝑑 𝑤𝑎𝑡𝑒𝑟 𝑙𝑒𝑣𝑒𝑙 → 393.8 𝑚

𝑡𝑎𝑖𝑙 𝑤𝑎𝑡𝑒𝑟 𝑙𝑒𝑣𝑒𝑙 → 308.23 𝑚

𝐸𝑛𝑒𝑟𝑔𝑦 𝑎𝑡 𝑡𝑜𝑒 = 𝜌𝑔𝑄𝐻

𝐸𝑛𝑒𝑟𝑔𝑦 𝑎𝑡 𝑡𝑜𝑒 = 23𝐺𝑊

where H is the difference between the

headwater and tailwater levels
 Types of energy dissipators

1) Stilling basins

2) Flip buckets

3) Roller buckets
ski-jump dissipator
 Stilling basins

➢
Stilling basins
Flip buckets (ski-jump dissipator)

20° − 30°
 Some dam using Flip buckets

Gandhi Sagar Dam

( Madhya Pradesh)
Hirakund dam RanaPratap Sagar Dam
(Odisa) (Rajasthan)
Gandhi Sagar Dam ( Madhya Pradesh)
Roller buckets
Roller buckets
 Some dam using Roller buckets

Dhorai dam
(at Sabarmati river)
HIRAKUND dam AT MAHANADI RIVER (ODISHA)
Surge and celerity

&
+ve surge and
–ve surge
 Surge and celerity

1. The sudden changes of flow in open channel results in the increase or decrease of flow
depth is called the "SURGE" in open channel.

2. This could take place when there is a breaking of dams due to earthquake or regulating the
hydropower sluice gates.

3. Results in positive and negative surges in downstream river channel or in downstream tail
channel of hydropower projects.

4. This phenomenon also governs when there would be flood (unsteady flow) during
monsoon period in natural river channels.

5. The flood wave which generates during the positive or negative surges is called the
celerity (wave velocity) of the flood in unsteady flow situation
Opening of sluice gate
 Upstream negative surge
𝑪𝒓𝒆𝒔𝒕 ⇒ 𝑪𝟏 = 𝒈𝒚𝟏 − 𝑽𝟏

𝑻𝒓𝒐𝒖𝒈𝒉 ⇒ 𝑪𝟐 = 𝟑 𝒈𝒚𝟐 − 𝟐 𝒈𝒚𝟏 − 𝑽𝟏

Downstream positive surge

1
𝑔𝑦1 2
𝑐= 𝑦1 + 𝑦2 + 𝑉2
2𝑦2
closing of sluice gate
Upstream positive surge

1
𝑔𝑦2 𝑦2 + 𝑦1 2
𝐶= − 𝑉1
2𝑦1
Downstream negative surge
Typical situations in which unsteady flows occur
1. Surges in power canals or tunnels produced by starting or stopping of turbines or
due to the opening or closing of the turbine gates to meet the load changes.

2. Surges in upstream or downstream channels produced by starting or stopping of

pumps and opening or closing of control gates.

3. Waves generated by landslides and avalanches in rivers, channels, reservoirs, and

lakes.

4. Waves in lakes, reservoirs, estuaries, bays, inlets, and oceans produced by wind
storms, cyclones, and earthquakes.

5. Circulation in lakes and reservoirs produced by wind or by temperature and density

gradients.
Surface water wave
 deep-water wave or a short wave
L
1. In deep water wave ⇒ 𝑑 >
2

2. The speed of deep-water waves depends on the wavelength of the waves.

3. deep-water waves show dispersion

𝑔𝐿
𝑉=
2𝜋

the velocity is independent of the depth of the tank.

Particle near the surface moves in circular path
 Shallow -water wave or a long wave
L
1. In shallow water wave 𝑑 < .
20

2. The speed of deep-water waves depends on depth.

3. shallow-water waves show no dispersion.

𝑉= 𝑔𝑑

the velocity is independent of the Wavelength.

Particle near the surface moves in the form of ellipse
summary
 dispersion
In fluid dynamics, dispersion of water waves
generally refers to frequency dispersion which
means that waves of different wavelengths travel
at different phase speeds. Water waves, in this
context, are waves propagating on the water
surface, with gravity and surface tension as the
restoring forces.
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

RECIPROCATING PUMPS
• The reciprocating pump is a positive displacement pump as it sucks and raises the liquid
by actually displacing it with a piston/plunger that executes a reciprocating motion in a
closely fitting cylinder. The amount of liquid pumped is equal to the volume displaced
by the piston

• The total efficiency of a reciprocating pump is about 10 to 20% higher than a comparable
centrifugal pump.

• The reciprocating pump is best suited for relatively small capacities and high heads. This
type of pump is very common in oil drilling operations

• Reciprocating pumps for industrial uses have almost become obsolete owing to their high
capital cost as well as maintenance cost as compared to that of centrifugal pumps.
This type of pump is very common in oil drilling operations. The
reciprocating pump is generally employed for:
(i) Light oil pumping,
(ii) Feeding small boilers condensate return
(iii) Pneumatic pressure systems
CLASSIFICATION OF RECIPROCATING PUMPS
1. According to the water being in contact with piston:
(i) Single-acting pump ...water is in contact with one side of the piston
(ii) Double-acting pump ...water is in contact with both sides of the piston.
2. According to number of cylinders:
• Single cylinder pump
• Double cylinder pump (or two throw pump)
• Triple cylinder pump (or three throw pump)
• Duplex double-acting pump (or four throw pump)
• Quintuplex pump or (five throw pump).
In general the reciprocating pumps having more than one cylinder are known as multi-cylinder pumps.
MAIN COMPONENTS AND WORKING OF A
RECIPROCATING PUMP
The main parts of a reciprocating pump are:
1. Cylinder
2. Piston
3. Suction valve
4. Delivery valve
5. Suction pipe
6. Delivery pipe
7. Crank and connecting rod
Working of a single-acting reciprocating pump:
Working of a single-acting reciprocating pump:
DISCHARGE, WORK DONE AND POWER REQUIRED TO DRIVE
RECIPROCATING PUMP
CO-EFFICIENT OF DISCHARGE AND SLIP OF RECIPROCATING
PUMP
Problem 1: A single-acting reciprocating pump, running at 50 r.p.m. delivers 0.00736 m3 /s of water. The diameter of the
piston is 200 mm and stroke length 300 mm. The suction and delivery heads are 3.5 m and 11.5 m respectively.
Determine:
(i) Theoretical discharge, (ii) Co-efficient of discharge
(iii) Percentage slip of the pump (iv)Power required to run the pump
Problem 2: A single-acting reciprocating pump operating at 120 r.p.m. has a piston diameter of 200 mm and
stroke of 300 mm. The suction and delivery heads are 4 m and 20 m, respectively. If the efficiency of both
suction and delivery strokes is 75 percent, determine the power required by the pump.
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

 Centrifugal pumps
A centrifugal pump is a mechanical device designed to move a fluid by means of the transfer of
rotational energy from one or more driven rotors, called impellers. Fluid enters the rapidly rotating
impeller along its axis and is cast out by centrifugal force along its circumference through the
impeller’s vane tips. The action of the impeller increases the fluid’s velocity and pressure and also
directs it towards the pump outlet. .

https://www.michael-smith-engineers.co.uk/resources/useful-info/centrifugal-pumps
Classification of centrifugal pumps
ADVANTAGES OF CENTRIFUGAL PUMP OVER DISPLACEMENT
(RECIPROCATING) PUMP
1. The cost of a centrifugal pump is less as it has fewer parts.
2. Installation and maintenance are easier and cheaper.
3. Its discharging capacity is much greater than that of a reciprocating pump.
4. It is compact and has smaller size and weight for the same capacity and energy transfer.
5. Its performance characteristics are superior.
6. It can be employed for lifting highly viscous liquid such as paper pulp, muddy and sewage
water, oil, sugar molasses etc
7. It can be operated at very high speeds without any danger of separation and cavitation.
8. It can be directly coupled to an electric motor or an oil engine.
9. The torque on the power source is uniform, the output from the pump is also uniform
COMPONENT PARTS OF A CENTRIFUGAL PUMP
A centrifugal pump consists of the following main components:
1. Impeller 2. Casing 3. Suction pipe 4. Delivery pipe

https://www.gibbonsgroup.co.uk/media/blog/5-golden-rules-of-centrifugal-pump-maintenance/
 Impeller
An impeller is a wheel (or rotor) with a series of backward curved vanes (or blades). It is mounted on a
shaft which is usually coupled to an electric motor
(i) Shrouded or closed impeller: In this type of impeller vanes are provided with metal cover plates or
shrouds on both the sides. It provides better guidance for the liquid and has a high efficiency. It is
employed when the liquid to be pumped is pure and relatively free from debris
(ii) Semi-open impeller. A semi-open impeller is one in which vanes have only the base plate and no crown
plate. This impeller can be used even if the liquids contain some debris.
(iii)Open impeller. : the vanes have neither the crown plate nor the base plate i.e. the vanes are open on
both sides. Such impellers are employed for pumping liquids which contain suspended solid matter
(e.g. sewage, paper pulp, water containing sand or grit)
Closed impeller Semi-open impeller Open impeller

https://www.castlepumps.com/info-hub/pump-impellers-the-types-their-impact/
 Casing
The casing is an airtight chamber surrounding the pump impeller. It contains suction and discharge
arrangements, supporting for bearings, and facilitates to house the rotor assembly. It has provision to fix
stuffing box and house packing materials which prevent external leakage. The essential purposes of the
casing are:
(i) To guide water to and from the impeller,
(ii) To partially convert the kinetic energy into pressure energy

1. Volute casing: This type of casing gradually increases the flow area from the impeller outlet to the
delivery pipe, reducing flow velocity and converting kinetic energy into pressure energy.
2. Vortex casing: This casing includes a circular chamber between the impeller and the volute chamber,
known as a vortex or whirlpool chamber. It converts some kinetic energy into pressure energy before the
volute chamber further increases the pressure energy, making the pump more efficient than a simple volute
pump.
3. Casing with guide blades: This casing surrounds the impeller with guide blades or vanes mounted on a
ring, known as a diffuser. The liquid passes through passages with increasing area, reducing velocity and
converting kinetic energy into pressure energy. These pumps have high efficiency but are more expensive
and less versatile across a range of operating conditions compared to volute pumps.
WORKING OF A CENTRIFUGAL PUMP
WORK DONE BY THE IMPELLER (OR CENTRIFUGAL PUMP)
ON LIQUID
Assumptions:
1. Liquid enters the impeller eye in radial direction, the whirl component 𝑉𝑤1 of the inlet absolute velocity
(V1) is zero and the flow component 𝑉𝑓1 equals the absolute velocity itself (i.e. 𝑉𝑓1 = 𝑉1 ); α = 90°.
2. No energy loss in the impeller due to friction and eddy formation.
3. No loss due to shock at entry.
4. There is uniform velocity distribution in the narrow passages formed between two adjacent vanes
 Velocity Triangles

V2, Vw2, Vr2, Vf2, β and φ are the corresponding values at outlet.
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

The heads of a centrifugal pump
The heads of a centrifugal pump are as follows:

(1)Suction head (2) delivery head (3)Static head (4) Manometric head
1. Suction head (hs) : It is vertical distance between level of sump and eye of an impeller. It is also called suction lift.
2. Delivery head (hd): It is the vertical distance between eye of an impeller and the level at which water is delivered.
3. Static head (Hs): It is sum of suction head and delivery head. It is given by Hs = (hs+ hd)
4. Manometric head (Hm): The head against which the centrifugal Pump has to work
CHARACTERISTICS OF CENTRIFUGAL PUMPS
Or PERFORMANCE CURVE
CHARACTERISTICS OF CENTRIFUGAL PUMPS/PERFORMANCE CURVE
Ordinarily a centrifugal pump is worked under its maximum efficiency conditions. However, when the
pump is run at conditions different from the design conditions, it performs differently. Therefore, to
predict the behavior of the pump under varying conditions of speeds, heads, discharges or powers, tests
are usually conducted. The results obtained from these tests are plotted in from of characteristic curves;
these curves delineate useful information about the performance of a pump in its installation

The following four types of characteristic curves are

usually prepared for centrifugal pumps:
1. Main characteristic curves,
2. Operating characteristic curves,
3. Constant efficiency or Muschel curves, and
4. Constant head and constant discharge curves.
3. Constant efficiency or Muschel curves:
The constant efficiency curves (also called iso-efficiency
curves), depict the performance of a pump over its entire range
of operations. These curves are obtained from main
characteristic curves as follows:
• For a given efficiency, the values of discharges are obtained
from Fig. (3.25) (c). These points are projected on the head
(Hmano) v/s discharge (Q) for that speed in Fig. 3.25 (a).
• Similarly, for another value of efficiency and speed, the
points are obtained and projected.
• The points corresponding to one efficiency are joined.
• The curves so obtained are the constant efficiency or
Muschel curves.
• The curve/ line of maximum efficiency (or best
performance) is obtained when the peak points of various
iso-efficiency curves are joined.
The constant efficiency curves help to locate the regions where
the pump would operate with maximum efficiency.
4. Constant head and constant discharge curves:
The performance of a variable speed pump for which the speed constantly varies can be determined by
these curves. When the pump has a variable speed, the plots between Q and N, and Hmano and N may be
obtained. In the first case Hmano is kept constant and in the second case, Q is kept constant. The curves are
shown in Fig. 3.28.
 MULTI-STAGE
CENTRIFUGAL PUMPS
MULTI-STAGE CENTRIFUGAL PUMPS

A multi-stage centrifugal pump is one which has two or more identical impellers mounted on the same shaft
or on different shafts. The important functions performed by a multi-stage pump are:

1. To produce heads greater than that permissible with a single impeller, ‘Discharge remaining constant’.
The task can be achieved by ‘Series arrangement’ where in the impellers are mounted on the same shaft
and enclosed in the same casing.
2. To discharge a large quantity of liquid, ‘Head remaining same’. This task is accomplished by ‘Parallel
Arrangement’ wherein impellers are mounted on separate shafts
1. Pumps in Series
The series arrangement is employed for delivering a relatively
small quantity of liquid against very high heads.

The advantages of multi–stage pumps– impellers in series over

single-stage pumps are as follows:
1. Less loss due to friction.
2. Reduced stresses.
3. Small slip leakage.
4. The number of stages may be so chosen that the pump speed
suits the driving motor speed.
5. By proper arrangement of impellers a thrust can be
eliminated.
6. Owing to lower specific speed of individual impellers a
higher suction lift is possible.
2. Pumps in Parallel
• Used when a large quantity of liquid is needed to be
pumped against a small head
• Impossible for a single pump to accomplish this task
• Two or more pumps are employed in parallel
• Each pump works separately to lift liquid from a
common sump
• Pumps deliver liquid to a common collecting pipe
• Liquid is carried to the required height through this
pipe
• This arrangement is called "pumps in parallel" since
each pump delivers liquid against the same head
• If the discharge capacity of one pump is Q and there
are 𝑛 identical pumps in parallel, then the total
discharge is Qtotal =nQ
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

NET POSITIVE SUCTION HEAD (NPSH)
The difference between the net inlet head and the head
corresponding to the vapor pressure of the liquid”

NPSH = Suction Pressure- Vapor Pressure

Understanding of NPSH
Analysis of NPSH
• NPSHa: The Net Positive Suction Head Available at the pump impeller inlet. It is a value that expresses
the absolute pressure acting on a liquid as it enters the pump. It is a measure of the pressure that stands
between the liquid in its current state and the formation of vapor bubbles (boiling).

• NPSHr: The Net Positive Suction Head Required by the pump to operate without experiencing damaging
cavitation and a dramatic reduction in pumping production. It is a value that expresses the minimum
absolute pressure that must be acting on a liquid as it enters the pump impeller to avoid excessive
cavitation and degradation of pump performance
 Cavitation
Cavitation in a centrifugal pump is a nasty phenomenon that can wreak havoc on your equipment. It
happens when the pressure of the liquid flowing through the pump dips below its vapor pressure. Let's
break down what that means and how it causes trouble.

1. Pressure Drop, Vapor Bubbles: As the liquid

encounters the impeller, its pressure decreases due to
the increase in velocity. If this pressure falls below the
vapor pressure of the liquid, tiny vapor bubbles start to
form around the impeller.
2. Implosion! Shockwaves!: As the liquid moves further
through the pump and the pressure increases again,
these vapor bubbles collapse with a mighty implosion.
This implosion creates high-energy shock waves that
can damage the pump components.
Causes of Cavitation:
There are a couple of main reasons why cavitation occurs:
1. Low Net Positive Suction Head Available (NPSHa): This refers to the pressure available at the pump
inlet. If the NPSHa is lower than the pump's Net Positive Suction Head Required (NPSHr), cavitation is
likely to happen. In simpler terms, the pump isn't getting enough "push" from the incoming liquid.
2. Defects or Design Issues: A blocked suction pipe, a badly designed system with too much friction, or even
a pump running outside its optimal operating range can all contribute to cavitation.

Effects of Cavitation:
The constant implosion of bubbles due to cavitation is like a tiny
jackhammer going off inside your pump. Here's what it can lead to:
• Erosion and Damage: The shock waves can pit and erode the
impeller, housing, and other pump components. This can lead to
premature failure of the pump.
• Performance Issues: Cavitation can reduce the pump's
efficiency and flow rate. You may not be getting the performance
you expect from your pump.
• Noise and Vibration: Cavitation can cause the pump to vibrate
excessively and make a rattling or knocking sound, like marbles
being pumped through the system.
PRIMING OF A CENTRIFUGAL PUMP
Priming a centrifugal pump involves filling the suction pipe, pump casing, and part of the delivery pipe
with the liquid to be pumped, removing any air, gas, or vapor. This step is crucial because if the pump is
not primed, air pockets can form in the impeller, causing flow disruptions and potential damage from dry
running.
1. Small Pumps: Primed by pouring liquid into a funnel, with an air-vent valve open to release air until
all air is removed.
2. Large Pumps: Primed using a vacuum pump or ejector to evacuate air, drawing liquid up the suction
pipe.
3. Self-Priming Pumps: Designed with special arrangements that automatically fill the pump with liquid,
eliminating the need for manual priming.
SELECTION OF PUMPS
• The main criteria of the selection of the type of pump are values of discharge (Q), head (H) and speed (N).
From these values the specific speed of the pump is calculated and subsequently the type of the pump can be
decided.
• When the specific speed is low and it is possible to increase the pump speed, it is better to use multi-stage
pump; the number of stages are decided on the basis of the head and the type of the pump to be used.

The type of impeller is another aspect of pump selection:

PYQ 2014-15: A centrifugal pump delivers water against a net need of 14.5 m and design speed of 1000 rpm. The vanes
are curved back to an angle of 30° with the periphery. The impeller dia is 300 mm and outlet width 500 mm. Determine
the discharge of pumping manometric efficiency is 95%.
PYQ 2018-19: A Centrifugal pump having outer diameter equal to two times the inner diameter and running at 1000 rpm
works against a total head of 40m. The velocity of the flow through the impeller is constant and is equal to 2.5m/s. The
vanes are set back at an angle of 40° at outlet. If the outer diameter of the impeller is 500mm and width at outlet is 50mm,
determine
i. Vane angle at inlet ii. Work done by impeller on water per second iii. Manometric efficiency.
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

HYDRAULIC TURBINES
A hydraulic turbine is a prime mover (a machine which uses
the raw energy of a substance and converts it into mechanical
energy) that uses the energy of flowing water and converts it
into the mechanical energy (in the form of rotation of the
runner).

Hydro (water) power is a conventional renewable source of

energy which is clean, free from pollution and generally has a
good environment effect. However, the following factors are
major obstacles in the utilisation of hydropower resources:
(i) Large investments,
(ii) Long gestation period, and
(iii)Increased cost of power transmission.
CLASSIFICATION OF HYDRAULIC TURBINES
Comparison between Impulse
and Reaction Turbines
 Aspects Impulse turbine Reaction turbine
Conversion of fluid The available fluid energy is converted The energy of the fluid is partly transformed into K.E.
energy into K.E. by a nozzle. before it (fluid) enters the runner of the turbine.
Changes in pressure The pressure remains same After entering the runner with an excess pressure, water
and velocity (atmospheric)throughout the action of undergoes changes both in velocity and pressure while
water on the runner passing through the runner.
Admittance of Water may be allowed to enter a part or Water is admitted over the circumference of the wheel.
water over the whole of the wheel circumference.
wheel
Extent to which the The wheel/turbine does not run full and Water completely fills all the passages between the blades
water fills the air has a free access to the buckets. and while flowing between inlet and outlet sections does
wheel/ turbine work on the blades.
Installation of unit Always installed above the tail race. Unit may be installed above or below the tail race, use of a
No draft tube is used. draft tube is made.
Flow regulation By means of a needle valve fitted into By means of a guide-vane assembly.
the nozzle.
Relative velocity of Either remaining constant or reduces Due to continuous drop in pressure during flow through
water slightly due to friction. the blade, the relative velocity increases.
Construction and working of Pelton Wheel/Turbine
 Work done per second per unit
weight of water in Pelton turbine
Design Aspects of Pelton wheel
5. Jet ratio (m). It is defined as the ratio of the pitch diameter (D) of the Pelton wheel to the
diameter of the jet (d). It is denoted by ‘m’ and is given as :
m = D/d (lies between 11 and 16 for maximum hydraulic efficiency) ...(2.14)
Normally, the jet ratio is adopted as 12 in practice.
Definitions of Heads and Efficiencies
Problem: A Pelton wheel is to be designed for the following specifications :
Power (brake or shaft) ... 9560 kW Head ... 350 metres
Speed ... 750 r.p.m. Overall efficiency ... 85%
Jet diameter ... not to exceed 1/6 th of the wheel diameter
Determine the following :
(i) The wheel diameter, (ii) Diameter of the jet, and
(iii) The number of jets required. Take Cv = 0·985, Speed ratio = 0.45.
Construction and working of Pelton Wheel/Turbine
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

REACTION TURBINES
• In reaction turbines, the runner utilizes both potential and kinetic energies.
• As the water flows through the stationary parts of the turbine, a portion of its pressure energy is
transformed to kinetic energy and
• when the water flows through the moving parts, there is a change both in pressure and in the direction and
velocity of flow of water.
• As the water gives up its energy to the runner, both its pressure and absolute velocity get reduced.
• The water which acts on the runner blades is under a pressure above atmospheric and the runner passages
are always completely filled with water

Important reaction turbines are Francis, Kaplan and Propeller

Francis Turbine
Main components of Francis Turbine
1. Penstock : It is a large size conduit which conveys water from the upstream of the
dam/reservoir to the turbine runner.
2. Spiral/scroll casing: It constitutes a closed passage whose cross-sectional area gradually
decreases along the flow direction, area is maximum at inlet and nearly zero at exit.
3. Guide vanes/wicket gates: These vanes direct the water onto the runner at an angle
appropriate to the design. The motion to them is given by means of a hand wheel or
automatically by a governor.
4. Governing mechanism: It changes the position of the guide blades/vanes to affect a
variation in water flow rate, when the load conditions on the turbine change.
5. Runner and runner blades: The driving force on the runner is both due to impulse and
reaction effects; — The number of runner blades usually varies between 16 to 24.
6. Draft tube: It is a gradually expanding tube which discharges water, passing through the
runner, to the tail race.
Work done and efficiency of Francis turbine
Work done by the runner

This discharge in this case is radial. For radial discharge, the

absolute velocity at exit is radial.
Working proportions of a Francis turbine
Advantages and disadvantages of a Francis turbine over a Pelton wheel
Advantages
Disadvantages/Drawbacks
1. In Francis turbine the variation in the
1. Water which is not clean can cause very rapid
operating head can be more easily
wear in high head Francis turbine.
controlled.
2. In Francis turbine the ratio of maximum and
2. The overhaul and inspection is much more
minimum operating heads can be even two.
difficult comparatively,
3. The operating head can be utilized even
when the variation in the tail water level is
3. Cavitation is an ever-present danger.
relatively large when compared to the total
head.
4. The water hammer effect is more troublesome
4. The mechanical efficiency of Pelton wheel
with Francis turbine.
decreases faster with wear than Francis
turbine.
5. If Francis turbine is run below 50 percent
5. The size of the runner, generator and power
head for a long period it will not only lose its
house required is small and economical if
efficiency but also the cavitation danger will
the Francis turbine is used instead of Pelton
become more serious.
wheel for same power generation
Problem : An inward flow reaction turbine has an external diameter of 1 m and its breadth at inlet is
250 mm. If the velocity of flow at inlet is 2 m/s, find weight of water passing through the turbine per
second. Assume 10 per cent of the area of flow is blocked by blade thickness. If the speed of the runner
is 210 r.p.m. and guide blades make an angle of 10° to the wheel tangent, draw the inlet velocity triangle
and find :
(i) The runner vane angle at inlet,
(ii) The velocity of wheel at inlet,
(iii)The absolute velocity of water leaving the guide vanes, and
(iv) The relative velocity of water entering the runner blade. Roorkee University
 Propeller turbine

The need to utilize low heads where large volume of

water is available makes it essential to provide a large
flow area and to run the machine at very low speeds.
The propeller turbine is a reaction turbine used for
heads between 4 m and 80 m. It is purely axial-flow
device providing the largest possible flow area that
will utilize a large volume of water and still obtain
flow velocities which are not too large.
Kaplan turbine
A propeller turbine is quite suitable when the load on
the turbine remains constant. At part load its
efficiency is very low; since the blades are fixed, the
water enters with shock (at part load) and eddies are
formed which reduce the efficiency. This defect of the
propeller turbine is removed in Kaplan turbine. In a
Kaplan turbine the runner blades are adjustable and
can be rotated about pivots fixed to the boss of the
runner. The blades are adjusted automatically by
servomechanism so that at all loads the flow enters
them without shock. Thus, a high efficiency is
maintained even at part load. The servomotor cylinder
is usually accommodated in the hub. Figs. 2.37 and
2.38 show the Kaplan turbine runner and Kaplan
turbine (schematic diagram) respectively
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

 UNIT QUANTITIES
• A turbine operates most efficient at its design point, at a particular combination of
head, discharge, speed and power output. but in actual practice hardly any turbine
operates at its designed parameters.
• In order to predict the behaviour of a turbine working under varying conditions of
• head, speed, and power, recourse has been made to the concept of unit.
• The unit quantities give the speed, discharge and power for a particular turbine
under a head of 1m assuming the same efficiency.
The following are the three important unit quantities.
1. Unit speed
2. Unit power
3. Unit discharge
MODEL RELATIONSHIP
Problem : A hydro-turbine is required to give 25 MW at 50 m head and 90 r.p.m. runner speed. The
laboratory facilities available permit testing of 20 kW model at 5 m head. What should be the model runner
speed and model to prototype scale ratio? [UPTU]
 P P P P
⇒ Power Coefficent = = = 3 = 3
N 3 D5 ND 3 D2 H × D2 H 2 D2
Cavitation
Cavitation is when bubbles form, grow, and collapse in a flowing liquid because of a drop in pressure.
When the pressure matches the liquid's vapor pressure, vapor bubbles form. These bubbles collapse with
great force, which can damage surfaces by causing tiny indentations, known as pitting.

In reaction turbines the cavitation may occur at the runner exit or the draft tube inlet where the pressure
is negative. The hydraulic machinery is affected by the cavitation in the following three ways :

1. Roughening of the surface takes place due to loss of material caused by pitting.
2. Vibration of parts is caused due to irregular collapse of cavities.
3. The actual volume of liquid flowing through the machine is reduced (since the volume of cavities is
many times more than the volume of water from which they are formed) causing sudden drop in
output and efficiency.
Cavitation factor.
Prof. Dietrich Thoma of Munich (Germany) suggested a cavitation factor (sigma) to determine the zone
where turbine can work without being affected from cavitation. The critical value of cavitation factor (σc)
is given by,
 BCE403 : CIVIL ENGINEERING

Hydraulic Engineering &

Machine

Explained By- Poorn Prakash

PERFORMANCE CHARACTERISTICS OF HYDRAULIC TURBINE
The turbines are normally designed for specific values of head, speed, discharge, power and
efficiency (known as the designed conditions). But oftenly turbines may be required to operate under
conditions different from those for which these have been designed. Thus, to know about their exact
behavior under varying conditions it becomes necessary to conduct tests either on the actual turbines
at the site or on their small scale models in a research laboratory. The results so obtained are usually
represented graphically and the curves obtained are known as “Characteristic curves”. These curves
are usually plotted in terms of unit quantities (for sake of convenience). The characteristic curves are
of the following types :
1. Main or constant head characteristic curves.
2. Operating or constant speed characteristic curves.
3. Constant efficiency or iso-efficiency or Muschel
curves.
1. Main or Constant Head Characteristic Curves
• Head and gate opening are maintained constant.
• Speed is varied by allowing a variable quantity of water to flow through the inlet opening.
• The brake power (P) is then measured mechanically by means of a dynamometer.
• The overall efficiency and unit quantities are then calculated by using the basic data; these are then plotted
against unit speed as abscissa

The discharge Qu for a Pelton wheel depends only upon the gate opening and is independent of Nu; the
curves for Qu are horizontal.
The curves between Qu and Nu for a Francis turbine are falling curves. This is due to the fact that a
centrifugal head develops which acts outwards and opposes the external head causing flow,
eventually decreasing the discharge as the speed increases.
The curves between Qu and Nu for a Kaplan turbine are rising curves; the discharge increases with
the increase in speed.

The maximum efficiency for a Pelton wheel usually occurs at the same speed for all gate openings;
this speed usually corresponds to a speed ratio of 0.45 . However, the maximum efficiency for a
reaction turbine usually occurs at different speeds for different gate openings.
2. Operating or Constant Speed Characteristic Curves
• As the percentage full load increases 𝜂0 also increases (In other words, at reduced loads 𝜂0 is also less).
• At 100 per cent , full load 𝜂0 is near about the maximum efficiency in all cases.
• The Kaplan, the Deriaz and the Pelton wheel maintain a high efficiency over a longer range of part
load as compared with either the Francis or the fixed blade propeller turbine.
• The maximum overall efficiency of all the turbines is almost the same (about 85%).
(b) Overall efficiency (η0) and output (shaft) power (P) v/s discharge (Q) curves:

Fig. 2.57 shows overall efficiency (𝜂0 ) and shaft

power (P) v/s discharge curves. 𝑄𝑚𝑖𝑛 is the minimum
discharge required to set the turbine runner into
motion from its state of rest. These curves yield the
following information :

• Shaft power or output power (P) is a straight line,

since P ∝ Q if H (head) is constant.
• 𝜂0 v/s discharge (Q) graph is curvilinear and 𝜂0
increases with Q and remains nearly constant
beyond a particular value of discharge.
3. Constant efficiency or iso-efficiency or Muschel curves

Refer to Fig. 2.58. As η–N curve is of parabolic nature, there exits two speeds for one
value of efficiency except for maximum efficiency which occurs at one speed only.
Corresponding to these values of speeds there are also two values of discharge for each
value of efficiency (Q-N curve). Hence on Q-N curve we can plot two points for each
value of efficiency and one point for maximum efficiency. By adopting this procedure
for different gate openings or heads we can get number of Q-N curves and we can plot
on them efficiency points (as described above). The points denoting the same efficiency
can now be joined to get constant iso-efficiency curves or Muschel curves (The German
word ‘Muschel’ means shell, indicating shape of curve). The diagram showing these
curves is also called Hill diagram ( since it looks like top view of a hill). In actual
practice unit speed and unit discharge are taken along the co-ordinate axes.
• The curve for the best performance is obtained by joining the
peak points of the various efficiency curves.
• The constant efficiency curves are helpful for determining
the zone of constant efficiency and for predicting the
performance of the turbine at various efficiencies.
Problem: A 1/5 scale model of a centrifugal pump absorbs 20 kW when pumping against a test head of 8 m
at its best speed of 400 r.p.m. If the actual pump works against 32 m head, find the speed and power
required for the actual pump. Determine also the quantities of water discharged by the two pumps. [UPSC]
MODEL RELATIONSHIP