HYDRAULIC 1
CVE 303
BASICS OF FLUID FLOW

Types of flow – The type of flow depends on the manner in which the particles
unite, the particles group themselves in a variety of ways e.g. regular or irregular.
The type of flow is identified by the Reynolds number.
R 
vL

Flow of an IDEAL FLUID – No Viscousity
Flow of a REAL FLUID – Viscousity included





Laminar (low velocity, motion in layers) &
Turbulent Flow ( high velocity, chaotic motion)
Steady flow (conditions at any point remain constant, may differ from point to point)
Uniform Flow ( velocity is the same at any given point in the fluid)
Flow lines : Path lines, Stream lines and Streak lines

Straight line streamlines of moving particle- One dimensional flow (Flow in pipes)
Curve – Two dimensional flow (Flow over a spillway)
Streamlines represented in space – Three dimensional (flow in a river bed)

Motion of Fluid Particles




Lagrangian method
Eulerian method
Equation of continuity of a liquid flow: If an incompressible liquid us
is continuously flowing through a pipe or a channel (whose cross-sectional area
may or may not be constant) the quantity of liquid passing per second is the
same at all sections.
Q1  Q2  Q3
a1v1  a2 v2  a3v3
Questions
 Water is flowing through a tapered pipe having end
diameter of 150m and 50m respectively. Find the discharge
at the larger end and velocity head at the smaller end if the
velocity of water at the larger end is 2m/s.


A circular pipe of 250 mm diameter carries an oil of specific
gravity 0.8 at the rate of 120 litres/s and under a pressure
of 20kPa. Calculate the total energy in meters at a point
which is 3m above the datum line.
A horizontal pipe 100 m long uniformly tapers from 300
mm diameter to 200 mm diameter. What is the pressure
head at the smaller end, if the pressure at the larger end is
100 KPa and the pipe is discharging 50 litres of water per
second. (5 marks
BERNOULLI’S EQUATION AND ITS
APPLICATIONS
 Introduction
Energy of liquid in motion
Potential: Energy a liquid posses by virtue of its position
 Kinetic: Energy posessed by a liquid by virtue of its motion
 Pressure head of a liquid particle in motion


Total Energy : This is the sum of
a liquids Potential, Kinetics and
Pressure energy
v2
p
E  Z 

2g
w

Total Head of a liquid 2in motion
v
p
H Z 

2g
w
BERNOULLI’S EQUATION (CONTD.)

Definition: For a perfect incompressible fluid, flowing in
a continuous stream, the total energy of a particle remains
the same, while the particles moves from one point to
another.

Limitations of Bernoulli’s equation

Practical application of Bernoulli’s equation

Venturimeter ( Convergent cone, throat, divergent cone)

Discharge through a venturimeter
Q  c.a2 .v2 

a12  a22
Inclined venturimeter
Orificemeter
 Pitot tube

ca1a2
2 gh
Questions




A pipe 500 m long has a slope of 1 in 100 and tapers from 1 m
diameter at the higher end to 0.5 m at the lower end. The
quantity of water flowing is 900 l/sec. If the pressure at the higher
end is 70KPa. Find the pressure at the lower end.
A pipe AB branches into two pipes C and D. The pipe has a
diameter of 0.45 m at A, 0.3 m at B, 0.2 m at C and 0.15 m at D.
Find the discharge at A if the velocity of water at A is 2 m/s also
find the velocities at B and D if the velocity at C is 4 m/s.
A venturimeter with a 150 mm diameter at inlet and 100 mm at
throat is laid with its axis horizontal and is used for measuring
the flow of oil specific gravity 0.9. The oil mercury differential
manometer shows a guage difference of 200 mm. Assume
coefficient of meter as 0.98. Calculate the discharge in litres per
minute
A venturimeter has 400 mm diameter at the main and 150
mm at the throat. If the difference of pressure is 250 mm of
mercury and the metre coefficient is 0.97, calculate the
discharge of oil (specific gravity = 0.75) through the
venturimeter
OPEN CHANNEL FLOW

General Definition: An

Types of open channel flow




open channel is a passage through which the water flows under the
force of gravity and atmospheric pressure e.g. canals, sewers, aqauduct. It is also known as freesurface flow or gravity flow. The flow here is not due to pressure as in the case of pipe flow. Most
open channel flow are turbulent flow and froude number is the relevant parameter here
Natural streams and rivers
Artificial canals, flumes
Sewers, tunnels, partially full pipelines
Gutters
Applications of artificial channels

Water power development
Irrigation
Water supply
Drainage
Flood control

Reynolds no. for pipes

Reynolds no. for open channels

Wetted perimeter: Length of cross-sectional border in contact with water.

Slopes in open channel flow




IR pipe 
VD

IRchannel 
friction acts.
VRn

; Rn 
A
P
Length where
OPEN CHANNEL FLOW (CONTD.)

Uniform flow through open channels

Chezy’s formula
V  C Rn So
Values of Chezy’s constant in the formula for
discharge in open channel
 Mannings formula

V
1 2 3 12
Rn S o
n
Basin’s formula
 Kutter’s formula


Geometric properties of cross-sections

Trapezoidal channels

Triangular channels

Rectangular channels
Exercises
 A trapezoidal channel 3.5 m wide at the bottom
has side and bed slopes of 1:1 and 1:1000
respectively. Using manning’s formula, find the
discharge through the channel, if the depth of
water is 0.5 m. Take N = 0.03.
 A non symmetrical trapezoidal cross-section with
b = 1.2 m, m1 = 2 and m2 = 1, So = 0.00009, Q =
120 m3/s and y = 0.6 m, find the manning’s
roughness factor n.
 Unnatural stream’s cross-section can be
approximated by a parabolic shape with B = 4m
and a depth y = 2 m. If the stream is laid on a
slope of 0.001 m/m and n = 0.028, determine the
discharge.
SPECIFIC ENERGY

Introduction
v2
q3
E  y  y
2g
2 gy 2
Specific energy diagram
Critical conditions

E = Emin i.e.

Critical depth

Critical velocity





dE
0
dy
q2
yc  3
g

q
 gyc
yc
Subcritical: y1 > yc tranquil, upper stage flow V < Vc
Supercritical: y1 < yc rapid, lower stage flow V >Vc
Froude’s number
F

vc 
v
gy
Critical depth in non rectangular channels
Occurrence of critical flow

Critical energy
yc 
2
Ec  2 Emin
3
3
qmax  gyc3

Occurrence of Critical Flow

Change from mild to steep slope in a channel
Entrance from a reservoir into a steep slope channel
Free fall from mild slope channel
Free fall from steep slope channel

Humps and Contractions



Questions



A rectangular channel 3.25 m wide discharges 2600 litres of water per second.
What is the critical depth and critical velocity?
A channel of rectangular section 8 m wide is discharging water at the
rate of 12 m3/s with an average velocity of 1.2 m/s. Find the type of
flow?
A rectangular channel 3 m wide carries 4 m3/s of water in subcritical
uniform flow at a depth of 1.2 m, a frictionless hump is to be installed
across the bed. Find the critical hump height?
HYDRAULIC JUMP

Introduction : Local non-uniform flow phenomenon –
supercritical flow going into subcritical. The shooting flow is an
unstable type of flow and does not continue on the downstream
side, the flow transform itself to the streaming flow by increasing
its depth. The rise in water level which occurs during the
transformation of the unstable shooting flow to the stable
streaming flow is called hydraulic jump.
Depth relations
 Energy losses

hij
3

y2  y1 


v2  
v2 
H1  H 2   d1  1    d 2  2 
2g  
2g 

Types of jumps
 Stilling basins

4 y1 y2
Questions



A horizontal rectangular channel of constant breadth has a
sluice opening from the bed upwards. When the sluice is
partially opened water issues at 6 m/s with a depth of 600
mm. Determine the loss of head per KN of water.
A discharge of 1000 l/s flows along a rectangular channel
1.5 m wide. If a standing wave is to be formed at a point
where the upstream depth is 180 mm. What would be the
rise in water level?
Water flows at the rate of 1 m3/s along a channel of
rectangular section of 1.6 m width. If a standing wave
occurs at a point where upstream depth is 250 mm. Find
the rise in water level after the hydraulic jump. Also find
the loss of head in the standing wave.