Manometers, Pitot tube and Pumps & Turbines Final.

MANOMETERS & PITOT TUBE
By: Dr. Ezzat El-Sayed G. SALEH

What is Civil Engineering?
A civil engineering is profession in which the
respective professional is responsible to change the
natural and artificial environment in such a way it in
best accommodates the basic needs of the human. For
example, to build a sustainable and safe living place, to
build a good communication system and to use the
best of the environment without hampering (
‫عاقة‬‫إ‬
) or
causing disturbance in nature that can affect human
life.

Who is a Civil Engineer?
A civil engineer is a person who practices the laws of
civil engineering and implies the law and rules to the
real world.
How is the yearly gross salary of a Civil Engineer? For
your kind information the salary amount varies from
region to region and from institutions to institutions.
According to a Civil Engineer, the median salary is
found to be around $ 60000 per annual “in USA”.

What are the major branches of Civil Engineering?
There are five major branches of civil engineering. They
are:
Structural Engineering
Water Resources Engineering
Transportation Engineering
Geotechnical Engineering
Environmental Engineering.

Which branch is more demandable?
Each and every branch of civil engineering have good
demand because every branch has its own field to
work with. Choosing the best branch depends mainly
on your choice and taste. Its solely your personal
decision.
How to get enrolled in Civil Engineering Jobs?
To get civil engineering jobs you must to have a B. Sc.
or diploma degree in civil engineering.

John Smeaton, the "father of
civil engineering"

A multi-level stack interchange, buildings,
houses, and park ”Shanghai, China”.

The engineering of this roundabout, attempts to make traffic flow free-moving
“Bristol, England”

Philadelphia City Hall in
the United States is
still the world's tallest
masonry load
bearing structure.

Chichen Itza was a large pre-Columbian city in Mexico built by the Maya
people of the Post Classic. The northeast column temple also covers
a channel that funnels all the rainwater from the complex some 40 meters
(130 ft) away to a rejollada, a former cenote.

The Akashi Kaikyo Bridge, currently the world's longest suspension span
“Japan”.

A Roman aqueduct [built circa 19 BC] near
Pont du Grand, “France”

Oosterscheldekerina. A storm surge barrier “Netherlands”.

A standard Brunton Geo “Compass”, still used commonly today by
geographers, geologists and surveyors for field-based measurements

Civil engineering students using a Theodolite

Optical Theodolite
Robotic Total Station

RTK GPS surveying Station: uses
one static antenna and one
roving antenna

What are Manometers?
Manometers are devices used for measuring the
pressure at a point in a fluid, by balancing the column
of fluid by the same or another column of fluid. Types
of Manometers are classified as:
 Simple manometers,
 Piezometer U-tube manometer

What is a Simple Manometers:
A simple manometer is one which consists of a glass
tube, whose one end is connected to a point where
pressure is to be measured and the other end is open
to atmosphere. Piezometer is one of the simplest forms
of manometers.

Fig.1. In its simplest form the manometer is a U-tube about half filled with
liquid. With both ends of the tube open, the liquid is at the same height
in each leg.
Fig. 2. When positive pressure is applied to one leg, the liquid is forced
down in that leg and up in the other. The difference in height, "h,"
which is the sum of the readings above and below zero, indicates the
pressure.
Fig. 3. When a vacuum is applied to one leg, the liquid rises in that leg and
falls in the other. The difference in height, "h," which is the sum of the
readings above and below zero, indicates the amount of vacuum.

The manometer principle is most easily demonstrated in the U-type
manometer illustrated in Figure 4. Here, with both legs of the instrument open
to atmosphere or subjected to the same pressure, gravity forces the surfaces
of the liquid to be at exactly the same level or reference zero.
Fig. 4

There are three types of pressure measurement:
Positive pressure or gauge pressure are those greater than atmospheric;
Negative pressures or Vacuums are pressures less than atmospheric;
Differential pressure is the difference between two pressures.
Fig. 5
Fig. 6
U tube-
Manometer

As illustrated in Figure 7, if one leg of the
manometer is increased many times in
area to that of the other, the volume of
fluid displaced will represent very little
change of height in the smaller area leg.
This condition results in an ideal
arrangement whereby it is necessary to
read only one convenient scale adjacent
to a single indicating tube rather than two
in the U-type. The larger area leg is
called the "well".
Fig. 7

Inclined Manometer
Many applications require accurate measurement of low pressure.
In these applications the manometer is arranged with the
indicating tube inclined, as in Figure 8, therefore providing an
expanded scale. This arrangement can allow 12" of scale length to
represent 1" of vertical liquid height.
Fig. 8

PA= P2 = Patm. + ρmg h3
PA= PB
PB= 𝑃𝑎𝑖𝑟 + ρoilg h2 + ρoilg h1

PC= PD
PD= PF = Patm. + ρoil g h3 + ρmg h2
Pc= (𝑃𝑎𝑖𝑟= 𝑃𝐵) + ρwg h1

What is an Orifice?
An orifice is a small opening provided on the side or
bottom of a tank, through which a fluid is flowing. The
opening can be of any shape or cross-section, like
rectangular, triangular or circular. The orifices may
discharge fluid into the atmosphere or from one tank
to another.

Orifices are classified based on the size of the orifice and the head
of fluid above the orifice as:
• Small orifice, and
• Large orifice.
Depending on the shape of the upstream edge of the orifices, they
are classified as:
• Sharp-edged orifice and
• Bell-mouthed orifice.
They are also classified based on the nature of the discharge as:
• Partially submerged or drowned orifice,
• Fully submerged or drowned orifice,
• Free discharging orifices.

Classification of Orifices
1. The orifices are classified as Small or Large orifice depending upon
the size of the orifice and head of liquid from the center of orifice.
 If the head of the liquid from the center of the orifice is more
than five times the depth of the orifice, the orifice is called small
orifice.
 And if the head of the liquid from the center of the orifice is less
than five times the depth of the orifice, the orifice is known as
large orifice.

2. On the basis of their cross-sectional areas they are classified as:
 Circular orifice,
 Triangular orifice,
 Rectangular, and
 Square orifice.

Orifices are classified based on the shape or the cross-
sectional area as:
• Rectangular orifice,
• Circular orifice,
• Triangular orifice, and
• Square orifice

Discharge Through A Partially
Drowned Orifice
If the outlet side of the
is partly under water, it is
known as a partly drowned
or partly submerged orifice
as shown in the figure.
The discharge through a partially drowned orifice is obtained
by treating the lower portion as a drowned orifice and the
upper portion as an orifice running free, and then by adding
the two discharges thus obtained.
Partially drowned Orifice

Discharge Through A Wholly Drowned Orifice
When the outlet side of an orifice is beneath the surface of liquid it is
known as a wholly submerged orifice as shown in the figure. In such
orifices, the coefficient of contraction is equal to one.
If depth of the drowned orifice (d) is given instead of H1 and H2, then in
such cases the discharge through the wholly drowned orifice is:
𝑄 = 𝐶𝑑 B (𝐻2-𝐻1) 2 𝑔 ℎ
Wholly drowned
Orifice

Spiral propeller connected to a streamlined housing,
held by a hand

Schematic of the discharge measurement and
bathymetric survey field method.

Vortex flow is defined as the flow of a fluid along a
curved path or the flow of a rotating mass of fluid is
known as Vortex flow. The following are the two types
of Vortex flow:
Free Vortex Flow and
Forced Vortex Flow.

Effect of abruptly valve closer

What is Pitot tube:
Pitot Tube is a device used for measuring the velocity
of flow at any point in a pipe or a channel. The Pitot
tube was invented by the French Engineer Henri Pitot.
The principle of Pitot tube is based on the Bernoulli's
equation, where each term can be interpreted as a
form of pressure.

Pitot Tube
It is used to measure flow rate at a particular point in a
flowing fluid. Such devices are frequently used as air speed
indicators in aircraft
The Pitot tube consists of a tube with a small opening at the
measuring end. This small hole faces the flowing fluid,
When the fluid contacts the Pitot tube, the fluid velocity is
zero and the pressure is maximum.
This small hole or impact opening, provide the higher
pressure for pressure measurement.
The purpose of the static ports is to sense the true
static pressure of the free airstream.

Note: the gray circles outlined in dotted red boxes are Pitot-static ports,
which, combined with an aircraft's Pitot tubes

With no flow, the pressure at
both ports is the same and the
level in each “leg” of the U-tube
is the same.
Flow Sensors are typically a Form of Pitot Tube.
 

The true static
pressure hole
Impact Opening with
higher pressure hole
Pitot Static tube

The Pitot tube is used to find the Velocity of a fluid. The parameters
for the model include density of fluid (L) , density of manometer
liquid (i), differential head (h).

Laminar flow: slow, organized, parallel to vessel
walls, parabolic profile.
Vortex flow: swirling, often countercurrent eddies
seen at bifurcations and after stenosis
Turbulent flow: disorganized, random, with high
velocities.

Types of Flow
Laminar Flow
Low velocity
Particles flow in straight
lines
No Mixing
Rare in water system
Turbulent Flow
High velocity
Particles move in irregular
paths
Mixing
Most common type of flow.

1. Classification of Turbines
2. Selection of Turbines
3. Design of Turbines - Pelton, Francis, Kaplan
4. Draft Tube
5. Surge Tanks
6. Governing of Turbines
7. Unit Speed, Unit Discharge, Unit Power
8. Characteristic Curves of Hydraulic Turbines
9. Similitude or Model Analysis
10. Cavitations
Topics to be Discussed

Pumps and Turbines
Pumps and turbines: Fluid machines.
Pumps: Add energy to the fluid – they do work on the
fluid.
Turbines: Extract energy from the fluid – the fluid does
work on them.

The Dam: is the key component of the hydroelectric power plant.
This raises the water level that creates the Reservoir behind it
which will provide the falling water (kinetic energy) to ultimately
drive the Turbines. The higher the water the higher the potential
energy.
There are intake and control gates within the dam that control the
flow of the water.
The Penstock: is a long pipe that brings the water to the turbines.
Water Turbines: The falling water coming from the reservoir spins
the water turbines that then rotates the shaft within the generators.

The Generator: is where the power is actually produced and works
just like any other generator whether it is a power plant utilizing
fossil fuels, wind, tidal, or solar. The generator converts the
mechanical energy from the water turbines into electricity.
Transmission Lines: bring the electricity generated at the
hydroelectric power plant to the grid to be used by homes and
businesses.

Application range
Pelton Turbines
Heads:
 from 50 m to 1500 m outputs from 100 kW to 300 MW
 runner diameters from 400 mm to 4000 mm
Types:
Horizontal arrangement
 they are used for outputs up to 5 MW, one nozzle-, two nozzle-,
and three nozzle types
Vertical arrangement
 they are used for outputs from 5 up to 300MW, two to six
nozzle types
Type series :
 covers usable head and discharge range of hydraulic machine
prototypes

Low Head Turbine
Types:
Bulb turbines
Horizontal axial flow turbines equipped with a diagonal distributor
and a straight cone draft tube enable to achieve high discharge
thus also output at low head of hydro power plants the direct
driven generator is located in a bulb like casing braced with
columns and ribs
Pit turbines:
They are similar to the BULB turbines high speed generator is
driven by a speed increaser located in the turbine pit these
turbines make even very low head power plants cost effective.

Low Head Turbine
S-type tubular turbines:
horizontal S- type turbines equipped with diagonal distributor and
S – type draft tube the shaft is conducted through the elbow to
the generator draft tubes can be designed with one or two
Vertical axial flow turbines:
Vertical axial flow turbines equipped with diagonal distributor are
arranged with the generator located over the turbine the shaft is
conducted through the turbine inlet elbow to the generator.
Draft tube: can be of a straight type or with one elbow.

A view from the bottom of an operating Pelton wheel illustrating the
splitting and turning of the water jet in the bucket. The water jet
enters from the left, and the Pelton wheel is turning to the right.

Application range
Francis Turbines
Heads:
 from 10 m to 800 m
 outputs from 200 kW to 300 MW
 runner diameters up to 7000 mm
Francis turbines are offered for wide range of heads and
discharges wide range of the specific speed enables
choosing a type most suitable for particular project
parameters

Types:
Vertical arrangement
 In the case of higher output, vertical arrangement is offered as a
more suitable solution.
 A steel spiral case of circular cross sections is connected with a
circular penstock.
 The turbine draft tube is designed with an elbow.
Horizontal arrangement
 Horizontal arrangement is suitable for power plants with lower
output.

The runner of a Francis mixed-flow turbine used at a hydroelectric power
station. There are 17 runner blades of outer diameter 6.19 m. The turbine
rotates at 100 rpm and produces 194 MW of power at a volume flow rate of
375 m3/s from a net head of 54.9 m

Application range
Low Head Turbine
Heads: from 1.5 m to 30 m
Outputs: from 100 kW to 20 MW
Runner diameters: up to 6 000 mm

Application range
Kaplan Turbines
mm
Heads: from 5 m to 75 m
Outputs: from 200 kW to 100 MW
Runner diameters: up to 9 500 mm

Types
 They are supplied in vertical or horizontal arrangement,
 They are supplied in a spiral case or in a flume arrangement.

Kaplan Turbines:
The turbine is double regulated (adjustable runner blades and
adjustable guide vanes)
Advantage:
 Is the wide regulation range having very flat efficiency curve
at great variations of heads and discharges
 This type enables effective operation from zero to maximum
turbine output.

Examples of positive- displacement pumps:
a) flexible-tube peristaltic pump,
b) Three-lobe rotary pump,
c) Gear pump, and
d) Double screw pump.

a) Flexible-tube peristaltic pump,

A sketch of four phases of half cycle of a simple rotary pump with
two lobes on each rotor. The two lobes rotors are synchronized
by an external gear box so as to rotate at the same angular
speed, but in opposite directions. The tope rotor turns clockwise
and the bottom rotor turns counter clockwise (see the above
diagram), sucking in fluid from the left and discharging it to the
right.

 Delivery heads from 20 m to 800 m,
 Inputs from 5 MW to 150 MW,
 Impeller diameters up to 2500 mm.
Large Pumps
Application Range

Storage pumps
 the storage pumps are single- or multistage centrifugal pumps
 they are offered for pumped storage power plants
 pump is fitted with a standard spiral case
Cooling pumps
 diagonal pumps for cooling systems of power plants
 the diagonal pump impeller is equipped with fixed or adjustable blades
 this type enables achieving a larger range of delivery heads.
Irrigation pumps
 centrifugal pumps for higher delivery heads
 axial pumps for smaller delivery heads
 the impeller of the axial pump is equipped with fixed or adjustable
blades
Large Pumps “Types”

Screw pumps are classified as positive-displacement pumps.
These pumps are used for pumping irrigation water, drainage water,
and waste water.
They are based on Archimedes screw principle in which a revolving
shaft fitted with helical blades rotates in inclined trough pushing water
up the trough,
Two advantages of screw pump over a centrifugal pump are:
1. they can pump large solids without clogging, and
2. They operate at a constant speed over a wide range of flows at
relatively good efficiencies.

Archimedes screw pump
Flow rate up to 6000 l/s, Efficiency up to 86%, with a diameter of up to 4 m,
Head up to 12 m and installation angle: 30 to 40°.
:

Centrifugal pump has two main components:
(1) Impeller
(2) Stationary Casing,
Housing, or Volute

There are many different types of pumps (positive-displacement
pumps, turbine pumps, horizontal centrifugal pump, and vertical
pumps).
The centrifugal pump are the most commonly used types used in
water distributing system pump, (because of their low cost, simplicity,
and reliability in the range of flows and heads encountered).
As a result, the discussion on pumps in this course is restricted to
centrifugal pumps.
A centrifugal pump is any pump in which fluid is emerged by a rotating
impeller.

Centrifugal pumps are classified into three groups according to the
manner in which the fluid moves through the pump,
Radial flow pumps: displace the flow radially in the pumps,
Axial flow pumps or propeller pumps: displace the flow axially in the
pump,
Mixed-flow pumps: displace the flow radially and axially in the pump,

The casing shape is designed to reduce the velocity as
the fluid leaves the impeller, and this decrease in kinetic
energy is converted into an increase in pressure,
The volute-shaped casing, with its increase area in the
direction of flow, is used to produce an essentially
uniform velocity distribution as the fluid moves around the
casing into the discharge opening.
A stationary casing, housing, or volute enclosing the
impeller.

Schematic diagram of
basic elements of
Centrifugal Pump
Schematic diagram of Axial Flow
Elbow Pump “Horizontal”
Schematic diagram of Axial Flow
Elbow Pump “Vertical”

As the impeller rotates, fluid is sucked in through the eye
of the casing and flows radially outward.
Energy is added to the fluid by the rotating blades, and
both pressure and absolute velocity are increased as the
fluid lows from the eye to the periphery of the blades.
Operation of the Centrifugal Pump

™
Simple stage pump: Only one impeller is mounted on the
shaft.
Multistage pump: Several impellers are mounted on the
same shaft.
 The flow rate is the same through all stages.
 Each stage develops an additional pressure rise.
 For a very large discharge pressure.
Stages of the Centrifugal Pump

Multistage pumps are pumps with more than one impeller (stage).
The stages are in series that the discharge of the first stage (first
impeller)discharge directly into the suction side or second stage
(second impeller), etc.
The impellers are on a single shaft and are enclosed in a single pump
housing.

What are pump characteristics?
Pump characteristic means the characteristic curves of a
pump. Characteristic curves of centrifugal pumps are
defined as those curves which are plotted from the results
of a number of tests on the centrifugal pump. These
curves are necessary to predict the behavior and
performance of the pump when the pump is working under
different flow rate, head and speed.

Pump Performance Characteristics
Rise Head Curve: the head curve continuously rises as the
flow rate decreases.
Falling Head Curve: H-Q curves initially rise as Q is
decreased from the design value and then fall with a
continued decrease in Q.
Shutoff Head: the head developed by the pump at zero
discharge. It represents the rise in pressure head across the
pump with the discharge valve closed.
™
Best Efficiency Points (BEP): the points on the various curves
corresponding to the maximum efficiency.

As the discharge is increased from zero the brake
horsepower increases, with a subsequent fall as the
maximum discharge is approached.
The efficiency is a function of the flow rate and reaches a
maximum value at some particular value of the flow rate,
commonly referred to as the normal or design flow rate or
capacity for the pump.
The performance curves are very important to the
engineer responsible for the selection of pumps for a
particular flow system.
Pump Performance Characteristics

Define separation pressure and separation pressure
head.
Due to cavitation process the continuous flow of fluid will
get affected and separation takes place. The pressure at
which separation takes place is known as separation
pressure and the head corresponding to separation
pressure is called separation pressure head. For water the
limiting value of separation pressure head is,
hsep = - 7.8 m (Gauge pressure)
hsep = 10.3 – 7.8 = 2.5 m (Absolute pressure)

What is an air vessel?
An air vessel is a closed chamber containing compressed
air in the top portion and liquid at the bottom of the
chamber. At the base of the chamber there is an opening
through which the liquid may flow into the vessel or out
from the vessel.
When the liquid enters the air vessel, the air gets
compressed further and when the liquid flows out of the
vessel, the air will expand into the chamber.

What is the purpose of an air vessel fitted in the
pump?
To obtain a continuous supply of liquid at a uniform rate.
To save a considerable amount of work in overcoming
the frictional resistance in the suction and delivery pipes,
and
To run the pump at a high speed with out separation.

Define Cavitation.
If the pressure in the cylinder is below the vapor
pressure, the dissolved gases will be liberated from the
liquid and air bubbles are formed. This process is
termed as cavitation.

II. Show that the normal depth in a triangular channel of side slope
Z:1 is given by:
𝑦𝑛 = 1.19
𝑄 ∙ 𝑛
𝑆𝑜
3/8
∙
𝑍2
+ 1
𝑍5
1/8
I. The longitudinal slope of the channel bed is 0.001. Compute the
average shear stress in N/m2 on the boundary. Also, compute
the value of Manning’s n.
(𝜏𝑜 = 9.134 𝑃𝑎; 𝑛 = 0.025)
1.5 m
3.0 m
30o

I. In a flow through a rectangular channel for a certain discharge the
Froude numbers corresponding to type equation. If the two alternate
depths are 𝐹𝑟,1 and 𝐹𝑟,2 , show that:
𝐹𝑟,2
𝐹𝑟,1
2 3
=
2 + 𝐹𝑟,2
2
2 + 𝐹𝑟,1
2

II. Water flows in a trapezoidal channel at a rate of 300 ft3/s. The
channel has a bottom width of 10 ft and side slopes of 1:1. If a
hydraulic jump is forced to occur where the upstream depth is 1 ft,
 what will be the downstream depth and velocity?
 What are the values of Fr,1 & Fr,2
1
1
1 ft
10 ft

The section properties
Section “1”
Cross sectional area 𝐴 = 𝑦 𝐵 + 𝑍𝑦 = 1.0 10 + 1 × 1 = 11 𝑚2
Wetted perimeter 𝑃 = 𝐵 + 2𝑦 𝑍2+1 = 10 + 2 × 1.0 12+1 = 12.83 𝑚
Top width 𝑇 = 𝐵 + 2𝑍𝑦 10 + 2 × 1 × 1 = 12 𝑚
Mean Hydraulic depth 𝑦ℎ = 𝐴 𝑇 = 11 12.0 = 0.92 𝑚
Mean velocity 𝑉1 = 𝑄 𝐴 = 300 11 = 27.27 𝑚/𝑠
The location of the centroid of the area “A” can be obtained by taking
moments of the sub-areas about the water surface and expressed as
𝑦1 =
𝐵 𝑦2 2 + 𝑍 𝑦3 3
𝐴 1
=
10 × 12 2 + 1 × 13 3
11
= 0.485 𝑚
𝑦 =
𝐴𝑖 ∙ 𝑦𝑖
𝐴𝑖
=
𝐵 𝑦2 2 + 𝑍 𝑦3 3
𝐴

For the downstream depth – Section “2”
𝑦2 =
𝐵 𝑦2 2 + 𝑦3 3
𝐴 2
=
10 × 𝑦2
2
2 + 𝑦2
3
3
𝑦2 (10 + 𝑦2)
=
5 𝑦2
2
+ 𝑦2
3
3
𝑦2 (10 + 𝑦2)
The total force at section “1” equals that at section “2”
𝜌 𝑔 ℎ1 ∙ 𝐴 + 𝜌 𝑄 𝑉1 = 𝜌 𝑔 ℎ2 ∙ 𝐴 + 𝜌 𝑄 𝑉2
or
ℎ1 ∙ 𝐴1 +
𝑄2
𝑔 𝐴1
= ℎ2 ∙ 𝐴2 +
𝑄2
𝑔 𝐴2
Substituting the data given earlier “ h = y for a vertical body“,

0.485× 11 +
3002
32.2×11
=
5 𝑦2
2+ 𝑦2
2 3
𝑦2 (10+𝑦2)
× 𝑦2 10 + 𝑦2 +
3002
32.2× 𝑦2 10+𝑦2
259.43 =
5 𝑦2
2+ 𝑦2
3 3
𝑦2 (10+𝑦2)
× 𝑦2 10 + 𝑦2 +
3002
32.2× 𝑦2 10+𝑦2
Solving by trial and error, y2 = 5.75 ft

Froude numbers at both sections are;
𝐹𝑟,1 =
𝑉1
𝑔 𝑦ℎ
=
𝑉1
𝑔 𝐴1 𝑇1
=
27.27
32.2 × 11 /12
= 5.02
Similarly, for the downstream depth “y2”
𝑉2 =
𝑄
𝐴2
=
300
𝐴2
=
300
90.56
= 3.31 𝑓𝑡
where,
𝐴2 = 𝑦 𝐵 + 𝑍𝑦 2
= 5.75 10 + 5.75 = 90.56 𝑓𝑡2
and,
𝑇2 = 𝐵 + 2 𝑍𝑦 2 = 10 + 2 × 1 × 5.75 = 21.50 𝑓𝑡
𝐹𝑟,2 =
𝑉2
𝑔 𝑦ℎ
=
𝑉2
𝑔 𝐴2 𝑇2
=
3.31
32.2 × 90.56 /21.5
= 0.284

An overflow spillway has its crest at elevation 136.0 m and a
horizontal apron at an elevation of 102.0 m on the downstream
side.
Estimate the tail water elevation required to form a hydraulic
jump when the elevation of the total energy line just upstream
of the spillway crest is 138.0 m. “Assume Cd for the spillway =
0.735 and neglect energy loss due to flow over the spillway”.

EL. 138.00 m
EL. 136.00 m
EGL
V2
2/ 2g
y2
∆ ES
y1 EL. 102.00 m

The discharge per unit width of the spillway “q” is
𝑞 =
2
3
𝐶𝑑 ∙ 2 𝑔 𝐻3 2
, where H = 138.0 – 136.0 = 2 m
∴ 𝑞 =
2
3
𝐶𝑑 ∙ 2 𝑔 𝐻3 2
=
2
3
× 0.735 × 2 × 9.81 × 2 3 2
= 6.14 𝑚3 𝑠 𝑝𝑒𝑟 𝑚 𝑤𝑖𝑑𝑒
Taking the horizontal bed of the spillway as a datum,
𝐸𝑆,1 = 138.0 − 102.0 = 36.0 𝑚
Since,
𝐸𝑆,1 = 𝑦1 +
𝑉1
2
2 𝑔
= 𝑦1 +
𝑞2
2 𝑔 𝑦1
2
Substituting the values of this condition gives,
36 = 𝑦1 +
6.142
2 × 9.81 × 𝑦1
2

Solving for y1 by trial and error method
𝑦1 = 0.232 𝑚
With a Froude number
𝐹𝑟,1 =
𝑉1
𝑔 𝑦1
=
𝑞
𝑔 𝑦1
3
=
6.14
9.81 × 0.2323
= 17.54
The sequent depth y2,
𝑦2 = 0.5 𝑦1 ∙ −1 + 1 + 8𝐹𝑟,1
2
= 0.5 × 0.232 ∙ −1 + 1 + 8 × 17.542 = 5.64 m
The tail water elevation required to form a hydraulic jump
𝑦2 = 5.64 + 102.0 = 107.64 𝑚

The head loss through the jump,
ℎ𝐿 =
𝑦2 − 𝑦1
3
4 𝑦1 ∙ 𝑦2
=
5.64 − 0.232 3
4 × 0.232 × 5.46
= 30.22 𝑚
The power loss through the jump = 𝜌 𝑔 𝑄 ℎ𝐿
= (1000 × 9.81 × 6.14 × 30.22) 1000 = 1820 𝑘𝑁. 𝑚/𝑠

A river consists of a rectangular channel of width 5 m. At one point
the piers of a simple beam bridge cause a local narrowing to width
3 m. The bottom of the bridge deck is 1.7 m above the bed of the
river.
(a) When the river flow is 11 m3 /s the constriction of the channel
by the bridge causes a subcritical to supercritical flow transition.
Calculate the depths of water upstream, downstream and under
the center of the bridge, stating any assumptions made.
(b) At the flow rate above, a hydraulic jump occurs a short distance
downstream of the bridge. Find the depth of flow immediately
downstream of the hydraulic jump.
(c) Show that if the river flow is 22 m3/s then the flow passage
beneath the bridge will be completely choked.

An artificial channel has a V-shaped cross-section with semi angle
40°. Depths are measured from the bottom of the vee.
(a) If the flow rate is 16 m3/ s, find the critical depth.
(b) A hydraulic jump is observed to occur at this flow rate. If the
depth on one side of the jump is 1.85 m, find the depth on the
other side of the jump.
Note: you would be well-advised to work from first principles,
not invalid formulae.
y

Solve the following problem and select the correct or the closet
numerical answer, one to each question. Then circle your answer.
(Note: any select answer without proving will not be considered)
A rectangular open channel of width B = 10 ft, laid on a slope So, carries a
discharge Q = 40 ft3/s in a gradually varied flow. The Manning’s coefficient is
n = 0.020.
(a) 0.22 (b) 0.022 (c) 0.0022 (d) 0.00022
II. Determine the critical flow depth ycr:
(a) 1.20 ft (b) 1.00 ft (c) 0.80 ft (d) 0.6 ft
I. if a uniform flow with depth “yn = 2.56 ft” were to occur in this channel,
determine the bed slope So:

(a) 1000 ft (b) 3000 ft (c) 2000 ft (d) 4000 ft
IV. Determine the distance ∆ 𝑋 from the section of depth of flow “y1 = 3.0 ft”
to that section where the depth of flow is “y2 = 3.3 ft. Using the direct
step method with a single step, the closest value of ∆ 𝑋 :
(a) M-1 (b) S-1 (c) M-2 (d) other
III. At a section of the channel the depth of flow is “y1 = 3.0 ft” and at
another section the depth of flow is “y2 = 3.3 ft. What type of gradually
varied flow curve would exist between the two sections described above:

Given:
B = 10 ft, n = 0.020. Q = 40 ft3/s
I - Req.: So
Section properties:
Area of flow = 𝐵 ∙ 𝑦 = 10 × 2.56 = 25.6 𝑓𝑡2
Wetted perimeter P =𝐵 + 2𝑦 = 10 + 2 × 2.56 = 15.12 𝑓𝑡2
Hydraulic radius Rh =
𝐴
𝑃
=
25.6
15.12
= 1.69 𝑓𝑡
Applying Manning’s equation
𝑄 =
1.486
𝑛
× 𝑅ℎ
2 3
× 𝑆𝑜
1 2
× 𝐴 or 𝑆𝑜 =
𝑄∙𝑛
1.486∙𝑅ℎ
2 3
×𝐴
2
Substituting the given data, gives
𝑆𝑜 =
40×0.020
1.486×1.692 3×25.60
2
= 0.00022
(d)

II - Req.: ycr
The unit discharge q = 𝑄 𝐵 = 40 10 = 4 𝑓𝑡2
/𝑓𝑡 𝑤𝑖𝑑𝑒
The critical depth 𝑦𝑐𝑟 =
3
𝑞2 32.2 =
3
42 32.2 = 0.79 𝑓𝑡 ≅ 0.80 𝑓𝑡
(a)
(c)
III - Req.: Type of G.V.F curve would exist between the two
sections
Since 𝑦𝑛 = 2.56 𝑓𝑡 & 𝑦𝑐𝑟 = 0.80 𝑓𝑡 ∴ The slope is Mild “M”
𝑦1 = 3.0 𝑓𝑡, 𝑦2 = 3.3 𝑓𝑡 𝑎𝑛𝑑 𝑏𝑜𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒𝑚 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒𝑟 𝑡ℎ𝑎𝑛 𝑦𝑛
𝑦𝑛 = 2.56 𝑓𝑡 & 𝑦𝑐𝑟 = 0.80 𝑓𝑡 ∴ The slope is Mild “M”
𝑦1 = 3.0 𝑓𝑡, 𝑦2 = 3.3 𝑓𝑡 𝑎𝑛𝑑 𝑏𝑜𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒𝑚 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒𝑟 𝑡ℎ𝑎𝑛 𝑦𝑛
The gradually varied flow curve is M-1

B 10 ft & n = 0.020 and q = 4 ft3/s
𝑦1 = 3.0 𝑓𝑡 𝑦2 = 3.3 𝑓𝑡
𝑉1 = 𝑞 𝑦1 = 4 3 = 1.33 𝑓𝑡 𝑉2 = 𝑞 𝑦2 = 4 3.3 = 1.21 𝑓𝑡
𝑉1
2
2𝑔 = 1.332
2 × 32.2 = 0.0275𝑓𝑡 𝑉2
2
2𝑔 = 1.212
2 × 32.21 = 0.0227𝑓𝑡
𝐸𝑆,1
= 𝑦1 + 𝑉1
2
2𝑔 = 3 + 0.0275 = 3.0275𝑓𝑡
𝐸𝑆,2
= 𝑦2 + 𝑉2
2
2𝑔 = 3.3 + 0.022 = 3.322𝑓𝑡
∆ 𝐸𝑆 = 𝐸𝑆2 − 𝐸𝑆,1 = 3.3220 − 3.0275 = 0.2945 𝑓𝑡
Section properties:
𝐴1 = 10 × 3 = 30𝑓𝑡2 𝐴2 = 10 × 3.3 = 33𝑓𝑡2
𝑃1 = 10 + 3 × 2 = 16 𝑓𝑡 𝑃2 = 10 + 3.3 × 2 = 16.60 𝑓𝑡
𝑅ℎ,1 = 𝐴1 𝑃1 = 30 16 = 1.875 𝑓𝑡 𝑅ℎ,2 = 𝐴2 𝑃2 = 33 16.6 = 1.980 𝑓𝑡
𝑅ℎ 𝑎𝑣𝑔. = 𝑅ℎ,1 + 𝑅ℎ,2 2 = 1.875 + 1.980 2 = 1.9275
𝑉 𝑎𝑣𝑔. = 𝑉1 + 𝑉2 2 = 1.33 + 1.21 2 = 1.27 ft/s

B 10 ft & n = 0.020 and q = 4 ft3/s
Applying Beroulli′s equation
𝑉
𝑎𝑣𝑔. =
1.486
𝑛
× 𝑅ℎ
2 3
𝑎𝑣𝑔.
× 𝑆𝑜
1 2
𝑎𝑣𝑔.
or
𝑉
𝑎𝑣𝑔. × 𝑛
1.486 × 𝑅ℎ
2 3
𝑎𝑣𝑔.
2 3
2
= 𝑆𝑜
1 2
𝑎𝑣𝑔.
𝑆𝑜 𝑎𝑣𝑔. =
1.27 × 0.02
1.486 × 1.9275 2 3
2
= 1.218 × 10−4
∆𝑥 =
∆ 𝐸𝑆
𝑆𝑜 − 𝑆𝐸 𝑎𝑣𝑔.
=
0.2945
0.00022 − 0.0001218
≅ 3000 𝑓𝑡 (b)

A penstock 1 m in diameter and 10 km long carries water from a
reservoir to an impulse turbine. If the turbine is 85% efficient,
 what power can be produced by the system if the upstream
reservoir elevation is 650 m above the turbine jet and the jet
diameter is 16.0 cm?
“Assume that f = 0.016 and neglect head losses in the nozzle”
 What should the diameter of the turbine wheel be if it is to have
an angular speed of 360 rpm?
Assume ideal conditions for the bucket design (Vbucket = 0.5 Vj)

A 35 MW generator is to operate by a double Pelton wheel turbines.
The effective head is 350 m at the base of the nozzle.
Find:
The size of jet, mean diameter of the runner and specific
speed of wheel.
Assume Pelton wheel efficiency 84%, velocity coefficient of nozzle
0.96, jet ratio 12, and speed ratio 0.45.
Given:
 The generator is fed by two Pelton turbines,
 The total power 35 MW,
 The effective head = 350 m,
 Pelton wheel efficiency = 84% & Cv = 0.96 & speed ratio
= 0.45 and.

Req. :
The size of jet, mean diameter of the runner and specific speed
of wheel.
Solution:
Power developed by each turbine
= 35 × 106
2 = 17. 50 × 106
𝑊
In order to determine required discharge for each turbine,
 = 0.84 =
𝑂𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
𝐼𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
=
17 500
𝑊𝐻𝑃
=
17.50×106
𝜌 𝑔 𝑄 𝐻𝐸
=
17.50×106
1000×9.81×𝑄×350
∴ 𝑄 =
17. 50 × 106
1000 × 9.81 × 350 × 0.84
= 6.07 𝑚3
/𝑠

Applying continuity equation,
𝑉
𝑗 = 𝑄 𝐴𝑗 = 6.07
𝜋 × 𝑑𝑗
2
4
… … … … … … . . (1)
and 𝑉
𝑗 = 𝐶𝑉 2 𝑔 𝐻𝐸 = 0.96 ∙ 2 × 9.81 × 350 = 79.55 𝑚/𝑠
Substituting in Eq. “1”
𝑑𝑗 =
4 × 6.07
𝜋 × 79.55
0.5
= 0.312 𝑚
Since, jet ratio = 12
 The diameter of the wheel D
= 10 𝑑𝑗= 12 × 0.312 = 3.74𝑚
 with a velocity 𝑉𝐵 = 𝜑 2 𝑔 𝐻𝐸
= 0.45 × 2 × 9.81 × 350 = 37.29 𝑚/𝑠

The velocity 𝑉𝐵 =
𝜋 𝐷 𝑁
60
∴ 𝑁 = 60 𝑉𝐵 (𝜋 ∙ 𝐷) = 60 × 37.74 𝜋 × 3.12 ≅ 193 𝑟𝑝𝑚
and the specific speed
𝑁𝑆 =
𝑁 𝑃
𝐻5 4 =
193 17 500
3505 4 ≅16.9

Vertical Turbine Pumps
Design Features

Daytime: Water flows downhill
through turbines, producing
electricity
Nighttime: Water pumped uphill to
reservoir for tomorrow’s use.

Reservoir
Pipeline Turbine
H
H L
H E
Q

Straight Type
4o
Simple-elbow Type

Dia.“d”
h =2 to 5 d
Elbow Type withVaryingCross-Section

Turbines
Impulse
Pelton Turbine
Reaction
Francis Turbine Kaplan Turbine
and propeller

1. According to type of energy at Inlet
a) Impulse Turbine - Pelton Wheel Turbine
Req.: (High Head and Low Flow Rate)
b) Reaction Turbine - Francis, Kaplan Turbine
Req. : (Low Head and High Flow Rate)
Classification of Turbines

2. According to direction of flow through runner
a) Tangential Flow Turbine - Pelton Wheel
b) Radial Flow Turbine - Francis Turbine
c) Axial Flow Turbine - Kaplan Turbine
d) Mixed Flow Turbine - Modern Francis Turbine

3. According to Head at Inlet of turbine
a) High Head Turbine - Pelton Wheel Turbine
b) Medium Head Turbine - Francis Turbine
c) Low Head Turbine - Kaplan Turbine
4. According to Specific Speed of Turbine
a) Low Specific Speed Turbine - Pelton Wheel
b) Medium Specific Speed Turbine -Francis Turbine
c) High Specific Speed Turbine - Kaplan Turbine

Classification According to Specific Speed
Type of Turbine Type of runner Specific Speed
Pelton Wheel
Slow
Normal
Fast
10 To 20
20 To 28
28 to 35
Francis
Slow
Normal
Fast
60 to 120
120 to 180
180 to 300
Kaplan -- 300 to 1000

5. According to Disposition of Turbine Shaft
a) Horizontal Shaft - Pelton Wheel
b) Vertical Shaft- Francis Turbine & Kaplan Turbines

Problems:
1. A Pelton wheel has a mean bucket speed of 10 m/s with a jet of water
flowing at the rate of 700 L/s under a head of 30 m. The buckets
deflect the jet through an angle of 160°. Calculate the power given by
water to the runner and the hydraulic efficiency of the turbine. Assume
the coefficient of nozzle as 0.98.
2. A Pelton wheel has to develop 13230 kW under a net head of 800 m
while running at a speed of 600 rpm. If the coefficient of Jet CV = 0.97,
speed ratio is 0.46 and the ratio of the Jet diameter is 1/16 of
wheel diameter. Calculate
i) Pitch circle diameter
ii) the diameter of jet
iii) the quantity of water supplied to the wheel

3. Design a Pelton wheel for a head of 80m and speed of 300 rpm. The
Pelton wheel develops 110 kW. Take coefficient of velocity = 0.98,
speed ratio= 0.48 and overall efficiency = 80%.
4. A double jet Pelton wheel develops 895 MkW with an overall
efficiency of 82% under a head of 60 m. The speed ratio = 0.46, jet
ratio = 12 and the nozzle coefficient = 0.97. Find the jet diameter,
wheel diameter and wheel speed in rpm.

Problems:
1. A reaction turbine works at 450 rpm under a head of 120 m. Its diameter
at inlet is 1.2 m and the flow area is 0.4 m2 . The angle made by the
absolute and relative velocities at inlet are 20º and 60º respectively with
the tangential velocity. Determine
(i) the discharge through the turbine
(ii) power developed (iii) efficiency.
Assume radial discharge at outlet.
2. A Francis turbine has inlet wheel diameter of 2 m and outlet diameter of
1.2 m. The runner runs at 250 rpm and water flows at 8 m3/S. The
blades have a constant width of 200 mm. If the vanes are radial at inlet
and the discharge is radially outwards at exit, make calculations for the
angle of guide vane at inlet and blade angle at outlet.

Problems:
1. A Kaplan turbine develops 9000 kW under a net head of 7.5 m.
Overall efficiency of the wheel is 86% The speed ratio based on
outer diameter is 2.2 and the flow ratio is 0.66. Diameter of the
boss is 0.35 times the external diameter of the wheel. Determine
the diameter of the runner and the specific speed of the runner.
2. A Kaplan turbine working under a head of 25 m develops 16,000
kW shaft power. The outer diameter of the runner is 4 m and hub
diameter is 2 m. The guide blade angle is 35˚. The hydraulic and
overall efficiency are 90% and 85% respectively. If the velocity of
whirl is zero at outlet, determine runner vane angles at inlet and
outlet and speed of turbine.

Turbine Application Chart
A representation of the range of turbines

Kaplan Turbine
Francis Turbine
Pelton Wheel Turbine

Draft Tube
The water after working on the turbine, imparts its energy to the vanes
and runner, there by reducing its pressure less than that of atmospheric
pressure. As the water flows from higher pressure to lower pressure, It
can not come out of the turbine and hence a divergent tube is connected
to the end of the turbine.

Draft Tube
Draft tube: is a divergent tube one end of which is connected to the
outlet of the turbine and other end is immersed well
below the tailrace (Water level).
The major function of the draft tube is:
 To increase the pressure from the inlet to outlet of the draft tube as it
flows through it, and hence increase it more than atmospheric
pressure.
 The other function is to safely discharge the water that has worked
on the turbine to tailrace.

Draft Tube
Surge tank (or surge chamber):
is a device introduced within a hydropower water conveyance system
having a rather long pressure conduit to absorb the excess pressure
rise in case of a sudden valve closure.
The surge tank is located between the almost horizontal or slightly
inclined conduit and steeply sloping penstock and is designed as a
chamber excavated in the mountain.
It also acts as a small storage from which water may be supplied in
case of a sudden valve opening of the turbine.
In case of a sudden opening of turbine valve, there are chances of
penstock collapse due to a negative pressure generation, if there is no
surge tank.

Pumped Storage Power Plant, Carinthia, Austria

Governing means Speed Regulation.
Governing system or governor is the main controller of the hydraulic
turbine. The governor varies the water flow through the turbine to
control its speed or power output.
1. Impulse Turbine
a) Spear Regulation
b) Deflector Regulation
c) Combined
2. Reaction Turbine

The gate is a mobile system of hydraulic barrier, which is placed on a
natural watercourse or on a channel to adjust the outflow and thus the
flow. It works by changing the area of free cross section which can be
crossed by the water.

Detail of spear valve used on impulse
turbine

𝑁𝑆 =
𝑁 ∙ 𝑄
𝐻3 4
NS referred to rated point of
turbine speed N (rpm)
Q: turbine discharge “m3/s”,
H: net head “m”.

Principle of a Centrifugal Pump

Water is flowing at a critical depth of 1.4 m on a channel of cross-section shown
below. Estimate:
The average shear boundary
The discharge, and
Specific energy in the channel.
(𝑄 = 11.50 𝑚3 𝑠 ; 𝐸𝑆 = 2.59 𝑚)
0.5
1.0
1.4 m
If y1 and y2 are alternate depths in a triangular channel of side slope Z horizontal
: 1 vertical; show that:
4 𝑦1
4
∙ 𝑦2
4
𝑦2
+ 𝑦2
𝑦 + 𝑦
= 𝑦𝑐𝑟
5

show that for a horizontal wide rectangular channel, by using Manning’s formula,
the gradually varied flow profiles are given by:
𝑦𝑐𝑟
4 3
𝑛2 𝑔
3
4
𝑦
𝑦𝑐𝑟
4 3
−
3
13
𝑦
𝑦𝑐𝑟
13 3
+ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝑋
If “Es” is the specific energy in a triangular channel of side slopes Z : 1, Show
that:
𝐸𝑆
𝑦1
=
4 + 3 + 2 + 
1 + 2 ∙ 1 + 
where: = 𝑦2 𝑦1

Analysis Arranging dissimilar pumps in series can create problems because the
volume flow rate through each pump must be the same, but the overall pressure
rise is equal to the pressure rise of one pump plus that of the other. If the pumps
have widely different performance curves, the smaller pump may be forced to
operate beyond its free delivery flow rate, whereupon it acts like a head loss,
reducing the total volume flow rate. Arranging dissimilar pumps in parallel can
create problems because the overall pressure rise must be the same, but the
net volume flow rate is the sum of that through each branch. If the pumps are
not sized properly, the smaller pump may not be able to handle the large head
imposed on it, and the flow in its branch could actually be reversed; this would
inadvertently reduce the overall pressure rise. In either case, the power supplied
to the smaller pump would be wasted. Discussion If the pumps are not
significantly dissimilar, a series or parallel arrangement of the pumps might be
wise.
We are to explain why dissimilar pumps should not be
arranged in series or in parallel.

𝑄 𝑜𝑟 𝑉.
𝐻 Shutoff head
Free delivery
Operating point
System curve
Performance curve
We are to label several items on the provided plot.
labeled are:
The available net head, corresponding to the pump performance curve,
The required net head, corresponding to the system curve.
The intersection of these two curves is the operating point of the pump.

Since both free surfaces are open to the atmosphere, the pressure term
vanishes. Since both V1 and V2 are negligibly small at the free surface (the tanks
are large), the second term on the right also vanishes. The elevation difference
(Z2 – Z1) does not change, and so the only term in Eq. 1 that is changed by
closing the valve is the irreversible head loss term.

𝑄 𝑜𝑟 𝑉.
𝐻
Original operating point
Original
system curve
Performance curve
(does not change)
We know that the minor loss associated with a valve increases significantly as
the valve is closed. Thus, the system curve (the curve of Hrequired versus Q )
increases more rapidly with volume flow rate (has a larger slope) when the valve
is partially closed. A sketch of H versus Q is plotted, and the new operating point
is labeled. Because of the higher system curve, the operating point moves to a
lower value of volume flow rate, as indicated on the figure. i.e., the volume flow
rate decreases.
New operating point

‫إلفياضاانت‬ ‫إء‬‫ر‬‫ج‬ ‫بريو‬ ‫يف‬ ‫إرع‬‫و‬‫إلش‬‫و‬ ‫إملنازل‬ ‫إق‬‫ر‬‫غ‬‫إ‬
«
2017
»

r1
r2
V1
Velocity diagram at the
inlet of Francis Turbine

V1
inlet of Francis Turbine
exit of Francis Turbine

This Second Edition of Hydraulic Machines is devoted to:
The operating principles,
Design, and performance characteristics of hydraulic machines
used in:
 electric power plants,
 municipal facilities,
 construction works,
 hydraulic engineering,
 industry, and agriculture. The student will learn:
How to select hydraulic turbines, pumps, and reversible pump-
turbines,
Analyze their efficiency, and maintain them for peak performance.

This notes-book emphasizes:
 The types and construction of the machinery, especially the
mechanical aspects of their operation, including head, discharge,
power, efficiency, cavitation factors, reliability, and maintenance.
 Performance characteristics and recommendations for their use
are provided,
Numerical examples promote a better understanding of the
methods and relationships discussed, and excellent technical
drawings help illustrate the design of components and workings

Manometers, Pitot tube and Pumps & Turbines Final.

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