design-of-turbomachinery-2021.pdf

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Design OF Turbomachinery 2021
Mechnical Engineerin (Takoradi Technical University)
Studocu is not sponsored or endorsed by any college or university
Design OF Turbomachinery 2021
Mechnical Engineerin (Takoradi Technical University)
Downloaded by Jordan Moses (mosesjordan24@gmail.com)
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DESIGN OF TURBOMACHINERY
1.0 Introduction to turbomachinery and Basic Principles
1.1 What is a turbomachine?
It is a system in which the stored energy in fluid is converted to mechanical energy or
vice versa (mechanical energy converted to stored energy in fluid). Turbomachines are
Fluid Machines
If the stored energy in fluid is converted to mechanical energy such that the flowing fluid
runs/turn a shaft, the device is referred to as Energy Producing (or Energy Extracting
device). E.g. Turbine.
If the device converts mechanical energy to stored energy in the fluid, it is an Energy
Absorbing device (or Energy Adding device). E.g. Pump, compressor, blower, fan.
What are the stored energy in fluid and mechanical energy?
Stored Energy in Fluid Fluid at Rest Potential Energy
Intermolecular energy
Fluid in Motion Potential Energy
Intermolecular Energy
Kinetic Energy
Pressure Energy (Flow work)
Mechanical energy is transferred through rotating shaft against a load.
Some applications of turbomachinery
Turbomachine
Stored energy
in fluid
Mechanical
energy
Turbomachine
Stored energy
in fluid
Mechanical
energy
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Fig. 1.1 Some applications of turbomachine
1.2 Classification of Turbomachines
Turbomachines are classified based on:
• Fuel Used
• Principle of operation
• Direction of energy transferred
Table 1: Turbomachines can be classified in three different ways as shown below;
Bases of
Classification
Class Type of Machine
Fluid Used Liquid Those that work on incompressible fluids (water,
oil). Eg. Pumps and hydraulic turbines,
Limitation for this type is cavitation.
Gas Those that work on compressible fluid (Air, Gas)
Eg. Gas turbine, steam turbine, fan, blowers.
Limitation for this type is if the fluid velocity exceeds
the speed of sound, fluid shock waves might occur.
Principles of
Operation
Dynamic
Action
Machines with rotation vanes and continuous
motion of fluid. Rotodynamic machines
Static Action Piston and cylinder arrangement. Positive
displacement machine
Direction of
Energy Transfer
𝐸𝑠 → 𝐸𝑚
Stored energy of the fluid is given to the machine.
i.e. those that absorb power from the fluid.
Eg. Turbines
𝐸𝑚 → 𝐸𝑠
Mechanical energy transferred to the fluid. i.e. those
that deliver power to the fluid.
Eg. Pumps, compressors, fans, blowers
1,2,1 Machines classification based on Direction of Energy Transfer
Machines classified under Direction of Energy Transfer are referred to as Energy
Extraction and Energy Adding devices.
1.2.1.1 Energy Extracting Device
Energy extracting device refers to machines that are powered by the stored energy in
the fluid. They are called Turbines.
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Below are some types of turbines
Fig 1.3 Types of turbines
A Simple Turbine
The figures below show a simple turbine. The exploded view shows the nose cone,
stator, casing, shaft casing and cover.
The stator is made up of stationary blades fixed to the casing. The rotor is made up of
rotating blades mounted of the shaft.
Fig. 1.2a A Simple Turbine Fig. 1.2b An Exploded View
Fig 1.2c Fluid flow through the stator and rotor
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The basic mechanism of operation of the simple turbine is as follows;
a) The fluid flows directly into the device in an axial direction ( parallel to the axis of
rotation of the shaft)
b) The stator blades (nozzle) turns the flow so that it is lined up with the rotor
(turbine) blade
c) The rotor blades turn the flow back towards the axial direction, thereby causing
the rotor to rotate along its axis, producing shaft work output in the process.
Gas Turbine
Basically air flows through the compressor (intake) that brings it to high pressure. The
highly pressurized air is combusted to produce high-temperature high-pressure gas,
which enters the turbine, and expands down to the exhaust pressure. This produces
shaft work used to drive the compressor. The energy of the exhaust gas which was not
used by the compressor comes out as exhaust gases that can be used to to produce
thrust for airplanes, generate electricity, etc.
Fig 1.4a A typical Gas Turbine
Steam Turbine
Energy production is accomplished by extracting energy from superheated high
pressure steam.
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Fig 1.5 An electricity power station
1.2.1.2 Energy Adding Devices
Energy Adding Devices transfer mechanical energy to fluids. Machine using liquid is
basically called pump. But if gases are involved, three different terms are in use,
depending on the pressure rise achieved. If the pressure rise is very small (a few inches
of water), a gas pump is called a fan; up to 1 atm, it is usually called a blower; and above
1 atm it is commonly termed a compressor.
Classification of Pumps
There are two basic types of pumps: positive-displacement and roto-dynamic or
momentum change pumps
Fig 1.3.1 Pump Classification
Centrifugal pumps are the most commonly used turbomachinery device which are used
to raise pressure or induce flow in control-volume. They are radial flow devices.
Basic mechanism of operation of a simple centrifugal pump with single suction semi-
open impeller
a) The impeller is driven by an external motor
b) As the impeller rotates, it sucks fluid from the outside through the eye of impeller
c) The rotating impeller transfers energy to the fluid
d) The fluid is pushed radially along the impeller to the casing
e) Casing collects the moving fluid and it is pumped out of the discharge nozzle
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Fig 1.3.2 Operation of centrifugal pump
1.2.2 Machines classification based on Principle of Operation
In industrial practices, the most common way of classifying turbomachines is based on
their principles of operation. The popular ones are the rotodynamic machines – where
there is a continuous motion of fluid as well as a part of the machine called the rotor.
Rotodynamic machines are classified as Axial or Radial Flow Machines.
1.2.2.1 Axial Flow Machine
Axial Compressor
Mode of Operation of Axial Compressor (Fig 3.1.1a). for this type, the fluid enters the
rotor, gains energy and exits the machine in a direction parallel to the axis of rotation of
the rotor. (Mechanical energy converted to stored energy of the fluid)
Description and how they operate;
➢ An axial compressor consists of a row of rotor blades, fixed on a disc mounted on
a shaft, followed by a row of stator blades. The working fluid traverses through
these without significant change in radius as it flows in a direction parallel to the
axis of rotation.
➢ The energy level of the fluid flowing through the machine is increased by the
action of the rotor blades, which exert a torque on the fluid supplied by an
external source.
➢ An axial compressor is a relatively low pressure ratio turbomachine with higher
mass flow rate as compared to a centrifugal compressor.
➢ The flow stream lines passing through the blades are nearly parallel to the shaft
axis.
➢ Flow enters axially and discharges almost axially.
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➢ The blade passages diverge from inlet to exit, and hence the flow decelerates.
➢ Due to density variation from inlet to exit, the compressor end walls have flare
➢
Fig 3.1.1a Axial Compressor Rotor
Axial Flow Turbine
Mode of Operation of Axial Turbine. For this type, the fluid enters the rotor, transfers its
stored energy to the rotor and exits the machine in a direction parallel to the axis of
rotation of the rotor. (Stored energy in fluid converted to mechanical energy)
➢ The kinetic energy of fluid is converted to mechanical power by its impulse or
reaction with a series of blades arranged around the circumference of a wheel or
cylinder.
➢ Stationary blades / guide blades act as nozzles and they convert fluid pressure
into kinetic energy. The following rotating blades convert kinetic energy into
useful work.
➢ Axial turbines have low pressure drop per stage and higher mass flow rate
compared to radial turbines.
➢ The flow stream lines through the bladings are nearly parallel to the shaft axis.
➢ Flow enters axially and discharges almost axially.
➢ The blade passages converge from inlet to exit, and hence the flow accelerates.
➢ Blade profile is thicker at the inlet and thinner at the exit.
➢ Due to density variation from inlet to exit, the turbine end walls have flare with
flow area increasing from inlet to exit
1.2.2.2 Radial Flow Machines
Radial Flow Compressor
Mode of Operation of Centrifugal Compressor. The fluid enters the machine axially and
exits radially.
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➢ The flow enters impeller axially through an inlet duct. The impeller may be
preceded by a row of inlet guide vanes.
➢ The impeller, through its blades, imparts velocity and pressure to the gas, which
flows in radial direction.
➢ The rise in pressure takes place due to the centrifugal action of the impeller and
diverging passages of the downstream diffuser and/or volute and diverging
passages of the downstream diffuser and / or volute.
➢ Vaned or vaneless diffuser with volute are provided to convert kinetic energy at
impeller exit into static pressure at compressor discharge.
➢ Centrifugal compressors are used to produce large pressure ratios.
➢ A single stage centrifugal compressor may have typical pressure ratio of about
4:1. Some test compressors are designed for pressure ratio up to 8:1.
➢ Centrifugal compressors are suitable for low specific speed, high pressure ratio
per stage and low mass flow rate applications.
➢ Based on application, the centrifugal compressors can be either single stage or
multistage type
Fig. 3.2.1a Axial flow and centrifugal (radial) flow machines
Radial Flow Turbine
Mode of Operation of Radial Turbine (Fig. 3.2.1b)
➢ Flow enters the impeller radially and exits axially. These machines are termed
as inward flow turbines.
➢ A radial turbine stage consists of volute, nozzle guide vanes and impeller.
➢ High pressure gas passes through the volute and / or nozzle guide vanes,
increasing its kinetic energy. The high velocity gas transfers its energy to the
impeller shaft by flowing radially inward through the impeller.
➢ The nozzles with adjustable vanes provide highest efficiency.
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➢ Radial turbines employ a relatively higher pressure drop per stage with low mass
flow rate.
➢ The specific speed and power range of the radial turbines are low.
➢ Since rotors / impellers are made of single piece construction, they are
mechanically strong and are more reliable
Fig 3.2.1b Radial turbine
1,2,3 Machines classification based on Fluid Used
Gas Turbine
Mode of Operation of Gas Turbine (Fig 3.3.1)
➢ Gas turbine unit mainly comprises compressor module, turbine module,
combustor and many auxiliary components.
➢ Gas turbines find wide application as aeroengines and in power generation.
➢ In power application, all the power developed by the turbine is used to drive the
compressor, generator and the auxiliary systems of the power plant.
➢ In aeroengines, the turbine develops power only to drive the compressor and the
remaining energy of the combustion gas is used to generate thrust for aircraft
propulsion.
➢ Gas turbines are available in a range of sizes from micro scale to very large
units.
➢ Gas turbine units have high power to weight ratio, small frontal area and high
efficiency
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Fig 3.3.1 Axial compressors and axial turbine
Hydraulic Turbines
Hydraulic turbines are devices for extracting energy from a reservoir of fluid collected
above sea level. The fluid at height is passed through series of pipes to a turbine which
discharges into a river. Used for electricity generation. There are three major types;
Pelton, Kaplan and Francis turbine
Pelton Turbine
➢ The Pelton turbine is a tangential flow impulse turbine.
➢ It is most efficient in high head applications.
➢ Pelton turbines in power plants operate with net heads ranging from 656 to 4,921
ft (200 to 1,500 m).
Fig 3.4.1 Pelton turbine
Francis Turbine
➢ The Francis turbine is a reaction turbine, which means that the working fluid
changes pressure as it moves through the turbine, giving up its energy.
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➢ The inlet is spiral shaped. Guide vanes direct the water tangentially to the runner.
The radial flow acts on the runner vanes, causing it to spin.
➢ The guide vanes (or wicket gates) are adjustable in order to allow efficient turbine
operation for a range of water flow conditions.
➢ Power plants operate with net heads ranging from 66 to 2,461 ft (20 to 750 m).
Fig 3.4.2 The Francis turbine
Kaplan Turbine
➢ The Kaplan turbine is a propeller-type water turbine that has adjustable blades.
➢ The Kaplan turbine was an evolution of the Francis turbine. Its invention allowed
efficient power production in low head applications that was not possible with
Francis turbines.
➢ Kaplan turbines are now widely used throughout the world in high flow low head
power production.
➢ Power plants operate with net heads ranging from 33 to 230 ft (10to70m).
Fig 3.4.3 The Kaplan turbine
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2.0 RELATIVE MOTION
The relative velocity VBA is the velocity of an object or observer B in the rest frame of another object
or observer A
2.1 Two-Dimensional Motion
Consider a propeller aircraft travelling in the direction shown with wind direction as
shown in fig 2.2
Fig. 2.2 Velocity Triangle for a Flying Aircraft
Let the desired direction (propeller) be W and the wind direction be U. As the wind
blows the aircraft, it deviates from its original path as shown. The deviation is the
velocity V of the aircraft resulting from the effect of the wind, as shown in the velocity
triangle. Mathematically, it can be expressed as:
𝑉
⃗ = 𝑈
⃗
⃗ + 𝑊
⃗⃗⃗
• To add two vectors U and V graphically: place them nose to tail and the result is
given by movement from the tail of the first to the nose of the second.
𝑈
⃗
⃗ 𝑊
⃗⃗⃗
𝑈
⃗
⃗ 𝑉
⃗
𝑊
⃗⃗⃗ 𝑉
⃗ = 𝑈
⃗
⃗ + 𝑊
⃗⃗⃗
Fig. 2.2.1 Adding Two Vectors Graphically
• To subtract two vectors V-U graphically, reverse the direction of U and then U to
the tail of V.
𝑈
⃗
⃗ 𝑊
⃗⃗⃗ 𝑊
⃗⃗⃗ = 𝑉
⃗ − 𝑈
⃗
⃗
𝑉
⃗ Fig 2.2.2 Subtracting Two Vectors Graphically
NB: A velocity triangle can only be drawn when there is a relative motion involved.
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2.3 Application of Velocity Triangle to Turbomachinery
Take a turbine that consists of a stator and a rotor as shown below. The three stations
shown are
Station 1 – incoming flow with velocity strikes the stator (which is stationary). Here there
is no relative motion, but the fluid is deflected towards the rotor as shown. Meaning no
velocity triangle can be drawn at this point.
Station 2 – flow (known) leaves the stator and enters the rotor with velocity𝑉
⃗ . The rotor
is moving with a tangential velocity, ωr (ω is the rotational speed of the rotor blades).
Here, there is a relative velocity between the velocities of flow into the rotor and the
rotor blade. Meaning a velocity triangle can be drawn here. Fig 2.3a. To draw the
velocity triangle at station 2;
i. Draw the known velocity 𝑉2
⃗⃗⃗ entering the rotor (after being deflected by rotor)
ii. Draw the rotor velocity 𝑈2
⃗⃗⃗⃗ ,
iii. Appropriately join nose to tail of the above two velocity vectors and close with the
relative velocity vector𝑊2
⃗⃗⃗⃗⃗ .
iv. Always check that 𝑉
⃗ = 𝑈
⃗
⃗ + 𝑊
⃗⃗⃗ .
Station 3 – here, the flow exits the rotor. A velocity triangle can be drawn for the moving
rotor blades and the exiting flow. Fig 2.3b. Drawing the velocity triangle at station 3;
i. Draw the known velocity 𝑊3
⃗⃗⃗⃗⃗ exiting the rotor
ii. Draw the rotor velocity 𝑈3
⃗⃗⃗⃗ , (tangential velocity of rotor)
iii. Appropriately join nose to tail of the two velocity vectors above and close with the
relative velocity vector 𝑉3
⃗⃗⃗ .
1
Stator 𝑈2
⃗⃗⃗⃗
2 𝑉2
⃗⃗⃗ 𝑊2
⃗⃗⃗⃗⃗ 𝑉2
⃗⃗⃗
𝑈
⃗⃗⃗⃗⃗⃗
Rotor 𝑉3
⃗⃗⃗ 𝑊3
⃗⃗⃗⃗⃗
3 𝑊3
⃗⃗⃗⃗⃗ 𝑈3
⃗⃗⃗⃗
Fig 2.3.1 Velocity Triangle for s Turbine Stage
Example 1 An office desk fan rotates at 200 rpm and has a diameter of 30 cm. Air
enters the fan at 3 m/s, parallel to the axis of rotation. Calculate the relative velocity at
the tip of the fan.
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Components of Velocity Triangle as Applied to Turbomachinery
Now, let’s single out one blade and analyze the velocity triangles at its inlet and outlet.
Cartesian coordinates used here are; x – for axial, r – for radial, and θ – for tangential.
Thus axial components will have a subscript x and tangential components will have θ as
subscript.
Consider a single vane as shown in Fig 2.3.2a. Let V1 be the absolute velocity of fluid at
the inlet and V2 be that of outlet. For a stationary vane, V1 and V2 move tangentially to
the blade in order to have a smooth operation (i.e. avoid frictional losses). Both V1 and
V2 have horizontal component called whirl velocity (Vw1 and Vw2) and vertical component
called flow velocity (Vf1 and Vf2) as shown in Fig 2.3.
𝑉𝑤2
⃗⃗⃗⃗⃗⃗ 𝑈2
⃗⃗⃗⃗ 𝑉𝑤2
⃗⃗⃗⃗⃗⃗
𝑉2
⃗⃗⃗ 𝑉2
⃗⃗⃗ 𝑉𝑓2
⃗⃗⃗⃗⃗ 𝑉2
⃗⃗⃗ 𝑉𝑓2
⃗⃗⃗⃗⃗
𝑊2
⃗⃗⃗⃗⃗
U
𝑉1
⃗⃗⃗ 𝑉1
⃗⃗⃗ 𝑉𝑓1
⃗⃗⃗⃗⃗
𝑉1
⃗⃗⃗ 𝑊1
⃗⃗⃗⃗⃗ 𝑉𝑓1
⃗⃗⃗⃗⃗
𝑉𝑤1
⃗⃗⃗⃗⃗⃗
𝑉𝑤1
⃗⃗⃗⃗⃗⃗
Fig 2.3.2a Stationary Vane Fig 2.3.2b Moving Vane
But in most situations, there is a continuous motion of fluid and the rotor of the machine,
resulting in a continuous relative velocity between the flow and the rotor. Assuming the
rotor has velocity U, as shown in Fig 2.3.2b,
There are angles that can be defined as follows:
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3.0 Fundamental Laws and Equations Used in Turbomachinery
3.1 Continuity Equation (Conservation of Mass)
Under steady state conditions, the mass of fluid entering the machine must be equal to
the mass of the fluid leaving the machine. Thus;
𝑚𝑖𝑛
̇ Machine 𝑚𝑜𝑢𝑡
̇
Fig 3.1 Conservation of mass
𝑡ℎ𝑒 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒𝑠: 𝑚𝑖𝑛
̇ = 𝑚𝑜𝑢𝑡
̇ 𝑖𝑓 𝑚
̇ = 𝜌𝐴𝑉, 𝑡ℎ𝑒𝑛
𝜌𝑖𝑛𝐴𝑖𝑛𝑉𝑖𝑛 = 𝜌𝑜𝑢𝑡𝐴𝑜𝑢𝑡𝑉𝑜𝑢𝑡 … … … … … … . . 3.1
Example 3.1.1
For an air compressor with equal inlet and outlet areas, and with both density and
pressure increasing, determine how the average speed at the outlet compares to the
average speed at the inlet.
Example 3.1.2
For a water pump, determine how the average speed at the outlet compares to the
average speed at the inlet.
3.1.1 Application to Axial Flow Machines
Fig 3.2 Flow through axial machine
𝐿𝑒𝑡 𝑟ℎ = 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 ℎ𝑢𝑏, 𝑟𝑡 = 𝑡𝑖𝑝 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑙𝑎𝑑𝑒, 𝐴 = 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑓𝑜𝑟 𝑓𝑙𝑜𝑤
𝐴 = 𝜋(𝑟𝑡
2
− 𝑟ℎ
2) … … … … … … … … … … … … 3.1.
𝐿𝑒𝑡 𝑟𝑚 = 𝑚𝑒𝑎𝑛 𝑟𝑎𝑑𝑖𝑢𝑠, 𝑎𝑛𝑑 𝑏 = 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑙𝑎𝑑𝑒, 𝑡ℎ𝑒𝑛
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𝑟𝑚 =
𝑟𝑡 + 𝑟ℎ
2
⇒ 2𝑟𝑚 = 𝑟𝑡 + 𝑟ℎ 𝑎𝑛𝑑 𝑏 = 𝑟𝑡 − 𝑟ℎ
𝑏𝑢𝑡 𝑟𝑡
2
− 𝑟ℎ
2
= (𝑟𝑡 + 𝑟ℎ)(𝑟𝑡 − 𝑟ℎ) = 2𝑟𝑚𝑏
𝑇ℎ𝑒𝑛 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑓𝑙𝑜𝑤, 𝐴 = 2𝜋𝑟𝑚𝑏 … … … … … … … 3.2
3.1.2 Application to radial Machines
Fig 3.3 Meridional view of radial impeller
The flow is perpendicular to the cross-sectional area
𝑇ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎 𝐴 = 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 × 𝑏𝑙𝑎𝑑𝑒 𝑤𝑖𝑑𝑡ℎ
𝐴 = 2𝜋𝑟𝑏 𝑆𝑜 𝑓𝑜𝑟 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦,
𝜌1𝐴1𝑉1 = 𝜌2𝐴2𝑉2 ⇒ 𝜌12𝜋𝑟1𝑏1𝑉1 = 𝜌22𝜋𝑟2𝑏2𝑉2
For incompressible fluids with constant radial velocities (V1 = V2), b2 < b1
3.3 Conservation of Momentum
Consider energy transfer from stored energy in fluid to mechanical energy.
Fig 3.4 A jet impinging on a moving vane
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Analyzing the interaction of energy in the fluid flow along the moving curved vane,
From the momentum theorem under steady state,
𝐹
𝑥 = 𝑚
̇ (−𝑉𝑟2𝐶𝑜𝑠∅ − 𝑉𝑟1𝐶𝑜𝑠𝜃)
𝐹
𝑥 = −𝑚
̇ (𝑉𝑤1 + 𝑉𝑤2)
Note that Fx is negative because of the direction chosen. To calculate the power, use
the magnitude of the force. Thus the Power output is
𝑃 = |𝐹
𝑥||𝑈|
∴ 𝑃 = 𝑚
̇ (𝑉𝑤1 + 𝑉𝑤2)𝑈 − − − − − − − −3.3
3.3 Basic Equation of Energy Transfer in Rotodynamic Machines
Assume 1. Steady state flow
2. Flow is uniform over any cross-sectional area
For a jet entering the rotor with velocity V1 at radius r1, and exits with velocity V2 at r2, ,
Applying Newton’s second law here; Torque = change in angular momentum with time
At inlet, the moment of momentum per mass
= 𝑉𝑤1𝑟1
At outlet, the moment of momentum per mass
= 𝑉𝑤2𝑟2
The change in moment of momentum per mass
= 𝑉𝑤2𝑟2 − 𝑉𝑤1𝑟1
The net angular momentum
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= 𝑚
̇ (𝑉𝑤2𝑟2 − 𝑉𝑤1𝑟1)
According to the angular momentum theory;
Rate of change in angular momentum = Torque
𝑖. 𝑒 𝑇𝑜𝑟𝑞𝑢𝑒 𝑇 =
𝑚2V𝑤2𝑟2
𝑡
−
𝑚1V𝑤1𝑟1
𝑡
=
∆𝑚𝑉
𝑤𝑟
𝑡
Where Vw = Tangential velocity component of the fluid
r = the radial position of the fluid
𝐹𝑜𝑟 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦, 𝑚1 =
̇ 𝑚2
̇ ⟹ 𝑇 = 𝑚
̇ (V𝑤2𝑟2 − V𝑤1𝑟1)
By definition, Power P is torque multiplied by angular velocity
𝑖. 𝑒. 𝑃𝑜𝑤𝑒𝑟 𝑃 = 𝑇ω = 𝑚
̇ ω(V𝑤2𝑟2 − V𝑤1𝑟1)
𝑜𝑟 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑊 = 𝐸 = 𝑚ω(V𝑤2𝑟2 − V𝑤2𝑟1)
The work done (Energy) by a unit mass of fluid is:
𝑊 =
𝐸
𝑚
= (V𝑤2𝜔𝑟2 − V𝑤1𝜔𝑟1)
𝐵𝑢𝑡 𝜔𝑟1 = 𝑈1 𝑎𝑛𝑑 𝜔𝑟2 = 𝑈2
𝑊 =
𝐸
𝑚
= (V𝑤2𝑈2 − V𝑤1𝑈1) − − − − − − − −3.4
Equation 3.4 is called the Euler Turbomachinery Equation, applicable to all kinds of
turbomachines.
Generally, the Euler Equation for energy transfer between fluid and turbomachines can
be written in the forms:
𝐸 = 𝑚(V𝑤2𝑈2 − V𝑤1𝑈1) 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝐸𝑛𝑒𝑟𝑔𝑦 … … … … … … … … … … … . . … … .3.5𝑎
𝐸
𝑚
= (V𝑤2𝑈2 − V𝑤1𝑈1) 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝐸𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑚𝑎𝑠𝑠 … … … … … … … 3.5𝑏
𝐸
𝑚𝑔
= 𝐻 =
1
𝑔
(V𝑤2𝑈2 − V𝑤1𝑈1) 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝐸𝑛𝑒𝑟𝑔𝑦 𝐻𝑒𝑎𝑑 (𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡) … … … .3.5𝑐
NB
1. The Euler equation gives positive for compressor and negative value for turbines
(indicating the direction of energy flow).
V𝑤1𝑈1 > V𝑤2𝑈2 , − 𝑣𝑒 𝐸. 𝑀𝑒𝑎𝑛𝑠 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑠 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑒𝑑 𝑓𝑟𝑜𝑚 𝑓𝑙𝑢𝑖𝑑 𝑡𝑜 𝑟𝑜𝑡𝑜𝑟. 𝑇𝑢𝑟𝑏𝑖𝑛𝑒
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V𝑤1𝑈1 < V𝑤2𝑈2, +𝑣𝑒 𝐸. 𝑀𝑒𝑎𝑛𝑠 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑠 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑒𝑑 𝑓𝑟𝑜𝑚 𝑟𝑜𝑡𝑜𝑟 𝑡𝑜 𝑓𝑙𝑢𝑖𝑑. 𝑃𝑢𝑚𝑝
2. The work input or output is determined by the flow turning ( that is the change in
angular momentum)
3. Since angular momentum is a product of mass, radius and tangential velocity
that changes in radius, radial machines tend to deliver or absorb much more
power per stage than a corresponding axial machine.
Example 3.1.1
A turbine state with a rotational speed of 3000 rpm is to be designed with an
absolute inlet angle of 60o and an absolute exit angle of -60o at a mean radius of 0.4
m. The machine is to be designed for a constant axial velocity of 450 m/s. estimate
the specific work from this stage.
3.4 Another Form of the Energy Equation in Turbomachinery
Figure 3.5 Velocity diagram at inlet and outlet of an outward flow machine
Considering the velocity triangle at the inlet;
𝑉𝑟1
2
= 𝑉1
2
+ 𝑈1
2
− 2𝑈1𝑉1𝐶𝑜𝑠𝛼1 𝑏𝑢𝑡 𝑉1𝐶𝑜𝑠𝛼1 = 𝑉𝑤1
𝑉𝑟1
2
= 𝑉1
2
+ 𝑉𝑤1
2
− 2𝑈1𝑉𝑤1
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𝑉𝑤1𝑈1 =
1
2
(𝑉1
2
+ 𝑈1
2
− 𝑉𝑟1
2 ) … … … … … … … … … … … … … … 3.6𝑎
For the outlet velocity triangle;
𝑉𝑟2
2
= 𝑉2
2
+ 𝑈2
2
− 2𝑈2𝑉2𝐶𝑜𝑠𝛼2 𝑏𝑢𝑡 𝑉2𝐶𝑜𝑠𝛼2 = 𝑈2 − 𝑉𝑤2
𝑉𝑟2
2
= 𝑉2
2
+ 𝑈𝑤2
2
− 2𝑈2𝑉𝑤2
𝑉𝑤2𝑈2 =
1
2
(𝑉2
2
+ 𝑈2
2
− 𝑉𝑟2
2 ) … … … … … … … … … … … … … … 3.6𝑏
Putting equations 3.6a and 3.6b into equation 3.5c gives the energy head H as;
𝐻 =
1
2𝑔
{(𝑉2
2
− 𝑉1
2) + (𝑈2
2
− 𝑈1
2) + (𝑉𝑟1
2
− 𝑉𝑟2
2
)} … … … … … … … … 3.7
Here there are three distinct components of energy transfer:
(𝑉2
2
− 𝑉1
2) – This represents change in dynamic head (change in absolute velocities) of
the fluid. It is the energy transferred due to fluid motion.
(𝑈2
2
− 𝑈1
2) – This represents change due to the rotational motion of rotor from one location
to another. It is the change in centrifugal head and also a component of change in static
head.
(𝑉𝑟1
2
− 𝑉𝑟2
2
) – This represents change in pressure head and also a component of change
in static head.
With this concept, types of fluid machines can be identified:
Figure 3.6 Axial and Radial flow machines
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1. Axial flow machines;
The main direction of flow is along the axis of rotation, meaning the radius of flow
at the inlet is the same as that of the outlet (i.e. no flow in the radial direction). It
implies that:
𝑟1 = 𝑟2, 𝑎𝑛𝑑 𝜔𝑟1 = 𝜔𝑟2 , ∴ 𝑈1 = 𝑈2.
Meaning there is no change in centrifugal head because the rotor velocities at
the inlet and outlet are equal.
Also the blades are curved and arranged in such a way that the relative velocities
at the inlet and outlet are equal, meaning there is no change in pressure head
across the blade.
2. Radial flow machines
Here the radial location of the fluid changes
Figure 3.7 Inward flow machine
For inward flow (turbines) the inlet is at a higher radial location than the outlet; -
𝑟1 > 𝑟2, 𝑎𝑛𝑑 𝜔𝑟1 > 𝜔𝑟2 , ∴ 𝑈1 > 𝑈2
i.e centrifugal head is released by the fluid to the rotor as fluid flows inward. This
happens in turbine
Figure 3.8 Outward flow machine
For outward flow; 𝑈1 < 𝑈2 i.e pressure is released as fluid flows outward. Thus
fluid gains static head. Eg Pumps and centrifugal head is gained by the fluid. Eg.
Pumps, compressor.
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3.5 Energy Transfers – Impulse and Reaction Machines,
Making reference to equation 3.7, it can be seen that the energy head H has two
main components – the change in dynamic head (i.e. change in absolute
velocities) and the change due to static pressure (i.e. changes in radial position
and pressure).
𝐻 =
1
2𝑔
{(𝑉2
2
− 𝑉1
2) + [(𝑼𝟐
𝟐
− 𝑼𝟏
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)]}
These two characterize the degree of reaction of turbomachines. By definition;
𝐷𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛, 𝑅 =
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑠𝑡𝑎𝑡𝑖𝑐 ℎ𝑒𝑎𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑜𝑡𝑜𝑟
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑
i.e. what fraction of the total head is taken by the static head
𝑅 =
1
2𝑔 {(𝑈2
2
− 𝑈1
2) + (𝑉𝑟1
2
− 𝑉𝑟2
2
}
𝐻
This leads to the definition of two important types of fluid machines;
1. Impulse Machines: 𝑅 = 0 ⟹ 𝑈1 = 𝑈𝑢 𝑎𝑛𝑑 𝑉𝑟1 = 𝑉𝑟2
I.e there is no change in static head of the machine. Here the only change in the
flow is the dynamic head. The static pressure along the flow through rotor is
relatively constant (may be opened to atmosphere). These are mostly axial flow
machines. Eg Pelton wheel
2. Reaction Machines: 0 < 𝑅 < 1
Here, both the dynamic and static heads change along the flow (i.e. change in
radial location of flow from inlet to outlet, change in pressure due to varying area
of passage of the fluid, etc.).
Figure 3.9 Impulse and reaction machines
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Degree of reaction tells about the efficiency of the turbomachine. It is also used in
selecting the proper machine to perform particular works.
3.6 Efficiencies of Fluid Machines
Definition: 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑜𝑓 𝐹𝑙𝑢𝑖𝑑 𝑀𝑎𝑐ℎ𝑖𝑛𝑒 𝜂
𝑈𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 (𝑤𝑜𝑟𝑘 𝑜𝑢𝑡𝑝𝑢𝑡)
𝐸𝑛𝑒𝑟𝑔𝑦 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑
There are basically two types:
1. Hydraulic Efficiency 𝜂ℎ - Concerns energy transferred between fluid and rotor
2. Overall Efficiency 𝜂𝑜 – Concerns the energy transfer between the fluid and the
shaft.
The difference between the two is the energy lost in the mechanical transmission.
Take a turbine for example;
Es = stored energy of the fluid
S & R = Stator + Rotor
W = work done by fluid
Ws = shaft work (work delivered at the coupling of the shaft)
Es S & R W Mech Shaft Ws
Transmission
𝐻𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝜂ℎ =
𝑊
𝐸𝑠
𝑎𝑛𝑑 𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝜂𝑜 =
𝑊
𝑠
𝐸𝑠
𝜂𝑜
𝜂ℎ
=
𝑊
𝑠
𝑊
= 𝜂𝑚 𝑤ℎ𝑖𝑐 𝑖𝑠 𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 (𝑡𝑎𝑘𝑒𝑠 𝑐𝑎𝑟𝑒 𝑜𝑓 𝑙𝑜𝑠𝑒𝑠 𝑤𝑖𝑡ℎ𝑖𝑛 𝑚𝑒𝑐ℎ 𝑠𝑦𝑠𝑡𝑒𝑚)
Thus
𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝜂𝑜 = 𝜂ℎ × 𝜂𝑚
For Pump and Compressor
Em = Mechanical losses due to bearings, couples, friction, etc.
Es = Stored energy of fluid/ useful energy delivered to fluid
E = Energy received by rotor
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Em Mechanical E S & R Es
Transmission
𝐻𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝜂ℎ =
𝐸𝑠
𝐸
𝑎𝑛𝑑 𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝜂𝑜 =
𝐸𝑠
𝐸𝑚
𝜂𝑜
𝜂ℎ
=
𝐸
𝐸𝑚
= 𝜂𝑚 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 (𝑡𝑎𝑘𝑒𝑠 𝑐𝑎𝑟𝑒 𝑜𝑓 𝑙𝑜𝑠𝑒𝑠 𝑖𝑛 𝑚𝑒𝑐ℎ 𝑠𝑦𝑠𝑡𝑒𝑚)
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4.0 PRINCIPLES OF PHYSICAL SIMILARITIES AS APPLIED TO
TURBOMACHINERY
Solutions to engineering problems are determined mostly by experiments due to
i. Economic constraints,
ii. Time limitations
iii. Ease of investigations.
But
• How can results obtained from the lab be directly applied to real designs
successfully?
• If the performance of the system is governed by a large number of operating
parameters, then large number of experiments are required to find the influence of
each operating parameter on the performance of the system.
The principle of physical similarity answers these challenges by making it possible to
apply results from the lab experiments under altered state of conditions, and also perform
a lesser number of tests with variations of lesser independent parameters to predict the
influence of large number of operating parameters on the performance of the system. It
saves time and money!
In similarities, comparison is made of all relevant dimensionless parameters that have the
same corresponding values for models test and prototypes. Basically, there are three
types of similarities considered:
1. Geometric similarities – a model and prototype are geometrically similar if and only
if all the body dimensions in all three coordinates have the same linear scale ratios.
Thus the model must be the reduced size of the prototype. All lengths ratios must
be the same. But angles and flow directions are preserved.
2. Kinetic similarity – all velocity ratios must be the same for corresponding points.
3. Dynamic similarity – all force ratios acting on corresponding mass element in the
model and prototype must be the same.
For example if a process is governed by a number of variables, the process can be
expressed as a functional relationship of n variables. If m is the number of basic
dimensions – mostly called repeating variables (usually m=3 which must be identified),
then by Buckingham Pi-Theorem, the can be n-m number of π-terms. i.e
𝑓(𝑎, 𝑏, 𝑐, … … 𝑛) = 0
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝜋 − 𝑡𝑒𝑟𝑚𝑠 = 𝑛 − 𝑚. 𝑡ℎ𝑢𝑠 𝑓(𝜋1, 𝜋2, 𝜋3 … … … … 𝜋𝑛−𝑚) = 0
All π-terms represent the condition/criteria of similarities, and they are dimensionless.
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Eg. Consider a pipe flow problem, where the difference in pressure ΔP over the length of
the pipe L is a function of flow velocity V, diameter of the pipe D, density of the fluid ρ and
viscosity of the fluid µ.
ΔP
L
Fig 4.1 Flow through a pipe
𝑖. 𝑒.
∆𝑃
𝐿
= 𝑓(𝑉, 𝐷, 𝜌, 𝜇)
The pipe problem can be described by the function:
𝐹 (
∆𝑃
𝐿
, 𝑉, 𝐷, 𝜌, 𝜇) = 0
Applying dimensional analysis, n=5 and m=3 (V,D and ρ)
𝑛 − 𝑚 = 5 − 3 = 2 𝑖𝑚𝑝𝑙𝑦𝑖𝑛𝑔 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 2 𝜋 − 𝑡𝑒𝑟𝑚𝑠
𝐹(𝜋1, 𝜋2) = 0
For π1
𝜋1 =
∆𝑃
𝑙
, 𝑉𝑎
𝜌𝑏
𝐷𝑐
In dimensional form;
𝑀0
𝐿0
𝑇0
= (
𝑀
𝐿2𝑇2
)(
𝐿
𝑇
)𝑎
(
𝑀
𝐿3
)𝑏
𝐿𝑐
Equating exponents;
0 = 1 + 𝑏 ⟹ 𝑏 = −1
0 = −2 − 𝑎 ⟹ 𝑎 = −2
0 = −2 + 𝑎 − 3𝑏 + 𝑐 ⟹ 𝑐 = 1
𝜋1 =
∆𝑃
𝐿
×
𝐷
𝑉2𝜌
=
∆𝑃
𝜌𝑉2
×
𝐷
𝑙
For π2
𝜋2 = 𝜇𝑉𝑎
𝜌𝑏
𝐷𝑐
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In dimensional form;
𝑀0
𝐿0
𝑇0
= (
𝑀
𝐿𝑇
)(
𝐿
𝑇
)𝑎
(
𝑀
𝐿3
)𝑏
𝐿𝑐
Equation exponents;
0 = 1 + 𝑏 ⟹ 𝑏 = −1
0 = −1 − 𝑎 ⟹ 𝑎 = −1
0 = −1 + 𝑎 − 3𝑏 + 𝑐 ⟹ 𝑐 = 1
𝜋2 =
𝜇
𝜌𝑉𝐷
Hence the pipe problem can be described as;
𝑓(𝜋1, 𝜋2) = 0 𝑖𝑠 𝑓 (
∆𝑃
𝜌𝑉2
×
𝐷
𝑙
,
𝜇
𝜌𝑉𝐷
) = 0
So instead of 5 variables describing the pipe flow problem, there are now 2 dimensionless
variables. So if the lab experiments are performed based on these two dimensionless
parameters, the results can be replicated into the real design by maintaining π1 and π2
parameters.
Again it can be shown that
∆𝑃𝐷
𝜌𝑉2𝑙
= 𝑓 (
𝜇
𝜌𝑉𝐷
) 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑎 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠ℎ𝑖𝑝 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑝𝑖𝑝𝑒 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛, 𝑓, 𝑎𝑛𝑑 𝑅𝑒𝑦𝑛𝑜𝑙𝑑′
𝑠 𝑁𝑜, 𝑅𝑒
This is a typical case in Moody’s chart for lamina flow.
Applying this principle of similarities to a fluid machine, the physical variables describing
the problems of fluid machines are;
D = a characteristic dimension of fluid machine (e.g. the rotor diameter)
Q = discharge/volumetric flow rate through the machine
N = rotational speed
gH = Head across the machine
ρ = fluid density
µ = fluid viscosity
E = coefficient of elaticity
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P = power transformed between the rotor
Then the fluid machine problem can be described by the function;
𝑓(𝐷, 𝑄, 𝑁, 𝑔𝐻, 𝜌, 𝜇, 𝐸, 𝑃) = 0
Here, n = 8. m = 3 so there will be 5 π-terms
𝜋1 =
𝑄
𝑁𝐷3
, 𝜋2 =
𝑔𝐻
𝑁2𝐷2
, 𝜋3 =
𝜌𝑁𝐷2
𝜇
, 𝜋4 =
𝑃
𝜌𝑁3𝐷5
, 𝜋5 =
𝐸
𝜌𝑁2𝐷2
So in any fluid machine design, these dimensionless π-terms should be maintained.
Physical significance of the π-terms:
𝜋1 =
𝑄
𝑁𝐷3
=
𝑄 𝐷2
⁄
𝑁𝐷
∝
𝑎 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑓𝑙𝑢𝑖𝑑 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑉
𝑎 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑟𝑜𝑡𝑜𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑈
𝜋2 =
𝑔𝐻
𝑁2𝐷2
;
𝜋2
𝜋1
1 =
𝑔𝐻
𝑁2𝐷2
× (
𝑁𝐷
𝑄 𝐷2
⁄
)2
=
𝑔𝐻
(
𝑄
𝐷2)2
∝
𝑇𝑜𝑡𝑎𝑙 𝑓𝑙𝑢𝑖𝑑 𝑒𝑛𝑒𝑟𝑔𝑦
𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑
𝜋3 =
𝜌𝑁𝐷2
𝜇
𝑡ℎ𝑖𝑠 𝑖𝑠 𝑎 𝑐ℎ𝑎𝑟𝑎𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑅𝑒𝑦𝑛𝑜𝑙𝑑′
𝑠𝑛𝑢𝑚𝑏𝑒𝑟 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑟𝑜𝑡𝑜𝑟 𝑠𝑝𝑒𝑒𝑑
𝜋3𝜋1 =
𝜌𝑁𝐷2
𝜇
𝑄
𝑁𝐷3
=
𝜌(𝑄 𝐷2
⁄ )𝐷
𝜇
𝑎 𝑐ℎ𝑎𝑟𝑎𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑅𝑒𝑦𝑛𝑜𝑙𝑑′
𝑠 𝑁𝑜 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑓𝑙𝑜𝑤 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝜋4
𝜋1𝜋2
=
𝑃
𝜌𝑁3𝐷5
𝑁𝐷
𝑄 𝐷2
⁄
𝑁2
𝐷2
𝑔𝐻
=
𝑃
𝜌𝑄𝑔𝐻
∝
𝑃𝑜𝑤𝑒𝑟 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑓𝑙𝑢𝑖𝑑 𝑎𝑛𝑑 𝑟𝑜𝑡𝑜𝑟
𝑇𝑜𝑡𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒
𝑖. 𝑒. 𝑓𝑜𝑟 𝑡𝑢𝑟𝑏𝑖𝑛𝑒𝑠,
𝜋4
𝜋1𝜋2
= 𝜂ℎ 𝑡ℎ𝑒 ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
𝑖. 𝑒. 𝑓𝑜𝑟 𝑝𝑢𝑚𝑝/𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟,
𝜋4
𝜋1𝜋2
=
1
𝜂ℎ
𝜋1
√𝜋5
=
𝑁𝐷
√𝐸 𝜌
⁄
𝑄 𝐷2
⁄
𝑁𝐷
=
𝑄 𝐷2
⁄
√𝐸 𝜌
⁄
∝
𝑉
𝑎
=
𝐹𝑙𝑢𝑖𝑑 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝐴𝑐𝑜𝑢𝑠𝑡𝑖𝑐 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦(𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑜𝑢𝑛𝑑 𝑖𝑛 𝑓𝑙𝑢𝑖𝑑)
𝑉
𝑎
⟶ 𝑀𝑎𝑐ℎ 𝑛𝑢𝑚𝑏𝑒𝑟
After identifying the physical significance of the π-terms, certain parameters which are
not relevant in fluid machines can be eliminated;
lOMoARcPSD|21319694

29
𝜋5 =
𝐸
𝜌𝑁2𝐷2
𝑖𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑖𝑠 𝑖𝑛𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑙𝑒, 𝑡ℎ𝑖𝑠 𝑡𝑒𝑟𝑚 𝑖𝑠 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑑
𝜋3 =
𝜌𝑁𝐷2
𝜇
𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑑 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑡ℎ𝑒 𝑒𝑓𝑓𝑒𝑐𝑡 𝑜𝑓 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝑖𝑠 𝑛𝑒𝑔𝑙𝑒𝑔𝑖𝑏𝑙𝑒 𝑖𝑠 𝑓𝑙𝑢𝑖𝑑 𝑚𝑎𝑐ℎ𝑖𝑛𝑒𝑠
Therefore the relevant dimensionless parameters are;
𝐹 (
𝑄
𝑁𝐷3
,
𝑔𝐻
𝑁2𝐷2
,
𝑃
𝜌𝑁3𝐷5
) = 0 − − − − − − − 4.1
These 3 terms clearly define the problem of fluid machine handling incompressible fluids.
5.0 The Concept Of Specific Speed In Fluid Machines
The flow in fluid machines (handling incompressible fluid) can be described by
𝐹 (
𝑄
𝑁𝐷3
,
𝑔𝐻
𝑁2𝐷2
,
𝑃
𝜌𝑁3𝐷5
) = 0
Fluid machines are of different shapes and sizes with varying geometrical dimensions.
Machines with the same physical similarities belong to what is called Homologous series
– thus the machines may vary in geometrical dimensions and operate under different
working conditions but have the same shape.
The performance of fluid machines is characterized by the speed, N, power, P, head H
and flow rate Q, depending on the type of machine.
For turbines, the performance is characterized by NPH
For pumps, it is characterized by NQH
When fluid machines are designed, these three parameters that are specified. Therefore
it is more appropriate to determine dimensionless parameters (called dimensionless
specific speed) expressed in terms of these performance parameters.
For turbine, π2 and π4 are manipulated such that the dimensionless specific speed;
𝜋4
1
2
⁄
𝜋2
5
4
⁄
=
(𝑃 𝜌𝑁3
𝐷5
⁄ )
1
2
⁄
(𝑔𝐻 𝑁2𝐷2
⁄ )
5
4
⁄
=
𝑁𝑃1 2
⁄
𝜌1 2
⁄ (𝑔𝐻)5 4
⁄
= 𝐾𝑆𝑇
− − − − − −5.1
For pumps, manipulating π1 and π2
lOMoARcPSD|21319694

30
𝜋1
1
2
⁄
𝜋2
3
4
⁄
=
(𝑄 𝑁𝐷3
⁄ )
1
2
⁄
(𝑔𝐻 𝑁2𝐷2
⁄ )
3
4
⁄
=
𝑁𝑄1 2
⁄
(𝑔𝐻)3 4
⁄
= 𝐾𝑆𝑃
− − − − − − − 5.2
Therefore the two terms 𝐾𝑆𝑇
and 𝐾𝑆𝑃
represent similarity conditions in fluid machines,
and help in determining which choice of machine is most suitable for a given operating
conditions. The parameters can be arranged in such a way to enable the machine to run
at maximum efficiency. i.e. for a particular fluid machine design, there are unique set of
values of NPHQ which give unique values of 𝐾𝑆𝑇
and 𝐾𝑆𝑃
for maximum efficiency.
Practical Concept in Design,
For example, if a turbine for a power generation plant designed is to be selected, the key
parameters to determine will be the rotational speed N, power P to be generated and
head H to be developed (or available). The 𝐾𝑆𝑇
will be calculated from these known
parameters. This 𝐾𝑆𝑇
will then be used to find the class of turbines (from different
Homologous series) that will give maximum efficiency for this operation, and the choice
machine based on this 𝐾𝑆𝑇
will be the most suitable turbine for the design.
Also since g is a constant and density ρ is constant for incompressible fluids,
𝑺𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝑺𝒑𝒆𝒆𝒅 𝒇𝒐𝒓 𝒕𝒖𝒓𝒃𝒊𝒏𝒆 𝑵𝑺𝑻
=
𝑵𝑷𝟏 𝟐
⁄
𝑯𝟓 𝟒
⁄
− − − − − − − 𝟓. 𝟑
And
𝒇𝒐𝒓 𝒑𝒖𝒎𝒑𝒔 𝑺𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝑺𝒑𝒆𝒆𝒅 𝑵𝑺𝑷
=
𝑵𝑸𝟏 𝟐
⁄
𝑯𝟑 𝟒
⁄
− − − − − − − 𝟓. 𝟒
These are dimensional specific speed relation that contain only NPHQ.
6. HYDRAULIC TURBINES (IMPULSE TURBINE)
6.1 Analysis of Force on Impulse Hydraulic Turbine and Power Generation
Impulse Hydraulic Turbine, also known as Pelton turbine or Pelton wheel.
It consists of spoon-shaped buckets mounted on the periphery of a disc (rotor). Water jet
at high velocity (nozzle converts high pressure into high velocity) strikes the bucket
tangentially, transferring energy to the bucket and causes it to rotate about its axis.
lOMoARcPSD|21319694

31
Side view of section of bucket
Fig 6.1 An impulse hydraulic turbine – the Pelton Wheel
As the jet strikes each bucket tangentially, it glides smoothly along the bucket such that
the inlet flow velocity (V1) of the jet and the outlet flow velocity (V2) are at the same radial
location – meaning U1 = U2. For the maximum change in momentum (max force) to be
exerted on bucket, the relative velocities (velocity of jet relative to velocity of bucket)
should be 180o. But in practice, the deflection is designed at 165o in order to prevent the
je from hitting the back of the following buckets during rotation.
𝑈𝑠𝑢𝑎𝑙𝑙𝑦, 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑢𝑐𝑘𝑒𝑡 > 15
𝑎𝑛𝑑 𝑎 𝑚𝑎𝑐ℎ𝑖𝑛𝑒 𝑚𝑎𝑦 ℎ𝑎𝑣𝑒 2 − 6 𝑛𝑜𝑧𝑧𝑙𝑒𝑠.
Consider a Pelton turbine shown in figure 4.2 above;
From the velocity diagrams at inlet, 𝑈 = 𝑈1; 𝑉1 = 𝑉𝑤1 𝑎𝑛𝑑 𝑉1 = 𝑉𝑟1 + 𝑈
𝑉𝑟1 = 𝑉1 − 𝑈
From the velocity diagrams at inlet, 𝑈 = 𝑈2; 𝑉𝑤2 = −(𝑉𝑟2𝑐𝑜𝑠∅ − 𝑈)
From Euler’s equation, the energy per unit mass
𝐸
𝑚
= (𝑉𝑤2𝑈2 − 𝑉𝑤1𝑈1) = [−𝑉𝑟2𝑐𝑜𝑠∅ + 𝑈 − (𝑉𝑟1 + 𝑈)] 𝑈
𝐸
𝑚
= −(𝑉𝑟1 + 𝑉𝑟2𝑐𝑜𝑠∅)𝑈 − − − − − − − −6.1
NB here, 𝑉𝑟1 ≠ 𝑉𝑟2 due to frictional losses between the fluid and the rotor. But the fluid
enters and exits the buckets at atmospheric pressure.
lOMoARcPSD|21319694

32
𝑆𝑜, 𝑉2 = 𝐾𝑉1 𝑤ℎ𝑒𝑟𝑒 𝐾 < 1, 𝐾 𝑖𝑠 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑎𝑐𝑡𝑜𝑟 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑓𝑙𝑢𝑖𝑑 𝑎𝑛𝑑 𝑟𝑜𝑡𝑜𝑟
𝐸
𝑚
= −𝑉𝑟1(1 + 𝐾𝑐𝑜𝑠∅)𝑈, 𝑏𝑢𝑡 𝑉𝑟1 = 𝑉1 − 𝑈 ⟹
𝐸
𝑚
= −(𝑉1 − 𝑈)(1 + 𝐾𝑐𝑜𝑠∅)𝑈
Therefore, the rate at which energy is given to the rotor is
𝐸̇ = 𝑚
̇ (𝑉1 − 𝑈)(1 + 𝐾𝑐𝑜𝑠∅)𝑈 ∴ 𝐸̇ = 𝜌𝑄(𝑉1 − 𝑈)(1 + 𝐾𝑐𝑜𝑠∅)𝑈 − − − −6.2
Define a bucket efficiency, such that 𝜂𝑤 =
𝐸𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑏𝑢𝑐𝑘𝑒𝑡
𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑
𝜂𝑤 =
𝜌𝑄(𝑉1 − 𝑈)(1 + 𝐾𝑐𝑜𝑠∅)𝑈
𝜌𝑄𝑉1
2
/2
𝜂𝑤 = 2(1 + 𝐾𝑐𝑜𝑠∅) (1 −
𝑈
𝑉1
)
𝑈
𝑉1
− − − − − 6.3
This efficiency shows how the effective the bucket can convert the kinetic energy it
receives from the fluid to develop power required.
𝜂𝑤 𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐾, 𝜙 𝑎𝑛𝑑
𝑈
𝑉1
. 𝐹𝑜𝑟 𝑎 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑐ℎ𝑖𝑛𝑒, 𝐾 𝑎𝑛𝑑 𝜙 𝑎𝑟𝑒 𝑓𝑖𝑥𝑒𝑑
So a graph of 𝜂𝑤 𝑣𝑟𝑠
𝑈
𝑉1
;
Fig 6.2
𝑇ℎ𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑜𝑖𝑛𝑡 𝑜𝑐𝑐𝑢𝑟𝑒𝑠 𝑎
𝑑𝜂𝑤
𝑑 (
𝑈
𝑉1
)
= 0 ⟹ 𝜂𝑤𝑚𝑎𝑥 𝑜𝑐𝑐𝑢𝑟𝑒𝑠 𝑎𝑡
𝑈
𝑉1
=
1
2
𝐴𝑡 𝑈 = 0. 𝜂𝑤 = 0 𝑎𝑛𝑑 𝑎𝑡 𝑈 = 𝑉1, 𝜂𝑤 = 0
In reality, the actual power developed by the turbine depends on friction (fluid to buckets
friction, mechanical friction – bearings, couplings, etc.). This loss increases with
lOMoARcPSD|21319694

33
increasing speed. This means that in reality, the max efficiency is lower than the
theoretical efficiency and 𝜂𝑤 = 0 𝑎𝑡 𝑈/𝑉1 < 1 . Hence the graph looks like;
Fig 6.3 Wheel efficiency of an impulse turbine
The overall efficiency
𝐿𝑒𝑡 𝑃 = 𝑠ℎ𝑎𝑓𝑡 𝑝𝑜𝑤𝑒𝑟 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑, 𝐻 = 𝑔𝑟𝑜𝑠𝑠 ℎ𝑒𝑎𝑑, ℎ𝑓 = ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 𝑖𝑛 𝑝𝑒𝑛𝑠𝑡𝑜𝑐𝑘
𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑, 𝑄 = 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒, 𝐻 − ℎ𝑓 = 𝑛𝑒𝑡 ℎ𝑒𝑎𝑑
𝜂0 =
𝑆ℎ𝑎𝑓𝑡 𝑝𝑜𝑤𝑒𝑟 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑, 𝑃
𝐼𝑚𝑝𝑢𝑡 𝑒𝑛𝑒𝑟𝑔𝑦
=
𝑃
𝜌𝑄𝑔(𝐻 − ℎ𝑓)
− − − − − 6.4
The kinetic energy at the nozzle can be determined by applying Torricelli’s formular for
and ideal flow through a nozzle;
𝑉1 = 𝐶𝑣√2𝑔𝐻 ⟹ 𝑔𝐻 =
𝑉1
2
2𝐶𝑣
2
− − − − − − − −6.5
𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑚𝑎𝑠𝑠 =
𝑉1
2
2
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34
6.2 Specific Speed of an Impulse Turbine
𝐹𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 5.3, 𝑁𝑆𝑇
=
𝑁𝑃1 2
⁄
𝐻5 4
⁄
𝑏𝑢𝑡 𝑃 = 𝜌𝑔𝑄𝐻𝜂ℎ 𝑎𝑛𝑑 𝑔𝐻 =
𝑉1
2
2𝐶𝑣
2
, 𝑄 =
𝜋
4
𝑑2
𝑉1 𝑎𝑛𝑑 𝑁 =
𝑈
𝜋𝐷
𝑁𝑆𝑇
=
(
𝑈
𝜋𝐷
) [𝜌 (
𝜋
4
𝑑2
𝑉1 ×
𝑉1
2
2𝐶𝑣
2) 𝜂ℎ]
1/2
(
𝑉1
2
2𝑔𝐶𝑣
2)
5/4
𝑁𝑆𝑇
=
𝑔5/4
21/4𝜋1/2
𝐶𝑣
3/2
(
𝑈
𝑉1
) (
𝑑
𝐷
) 𝜂ℎ
1/2
𝜌1/2
For a turbine using water with no appreciable change in temperature, density is
constant. Also Cv and g are constant. Hence (
𝑈
𝑉1
) 𝑎𝑛𝑑 (
𝑑
𝐷
) decide the value of the
specific speed, 𝑁𝑆𝑇
, for the turbine.
The actual max efficiency of an impulse turbine is obtained at (
𝑈
𝑉1
) = 0.46, 𝜂ℎ =
0.8 𝑡𝑜 0.9 and 𝐶𝑣 ≅ 0.97. With these, the specific speed;
𝑁𝑆𝑇
≅ 105 (
𝑑
𝐷
)
Usually, Pelton wheel has 𝑁𝑆𝑇
= 4 ⟷ 14 𝑎𝑛𝑑
𝐷
𝑑
= 6 ⟷ 26
For optimum operation,
𝐷
𝑑
= 14 ⟷ 16
6.3 Reaction Type Hydraulic Turbine – The Francis Turbine
A reaction machine is a type of machine in which there are changes in both pressure
energy and kinetic energy as the fluid flows through the rotor. A moving fluid through the
penstock possesses both pressure and kinetic energies. The basic difference between
an impulse and reaction machines is:
Impulse Turbine Reaction Turbine
As the fluid exits the nozzle (stator) of the
impulse machine, the high pressure
energy of the fluid is reduced to ambient
pressure (the fluid is expanded), resulting
Here the stator converts part of the
pressure energy into kinetic energy. The
fluid approaching the rotor possesses both
pressure and kinetic energies. Work done
lOMoARcPSD|21319694

35
in a high velocity jet. Here, the pressure
energy is totally exploited in the form of
kinetic energy, and therefore in the rotor,
work done results from the change in
kinetic energy (change in absolute
velocities) only. Pressure of fluid remains
almost constant (ambient pressure) in and
out of the rotor.
by the rotor is as a result of changes in
both pressure and kinetic energies (i.e.
changes in the absolute velocities and
pressure energy of the fluid take place as
the fluid flows in and out of the rotor).
Free jet engages one bucket (blade) at a
time with no change in pressure as the
fluid engages the blade.
Runner is completely enclosed and the
fluid completely fills the entire passage of
the runner, so that the pressure in the rotor
is always above atmospheric pressure,
and goes on changing while the fluid flows
through the runner.
A typical example of a reaction machine is the Francis turbine (Fig 8.2.1).
Spiral/Scroll/Volute Casing: Fluid, at high pressure and high kinetic energy enters the
spiral/scroll/volute casing which has been designed to ensure that approximately equal
amount of fluid enter the guide vane (wicket gate) at the same velocity.
Guide Vane/Wicket Gate (Stator): They 1. Partially convert pressure energy into kinetic
energy as the fluid flows, 2. Direct the fluid in a proper manner such that the angle at
which the fluid leaves it matches with the inlet angle of the runner so as to reduce frictional
losses and avoid shocks, 3. Impart angular or tangential momentum to the fluid as it
deflects the fluid. The guide vanes are pivoted in such a way that they can be turned to
900 so they can stop the flow. i.e. they can be used to start or stop the machine.
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36
Runner (Rotor): this extracts the energy (head) of the fluid in the form of useful work.
Here, both pressure and kinetic energies change as the fluid flows in and out of the
runner.
Draft Tube: this enables some exit kinetic energy to be recovered. Thus it reduces the
kinetic energy at the runner exit to a lesser value so that the fluid is discharged out of the
tube at a very low velocity.
6.4 Analysis Approach
Figure 8.2.2 shows a typical set-up of a Francis turbine. The upstream reservoir is set at
a height H0 (gross head) above the tailrace, with the turbine positioned Z meters above
the tailrace. As fluid flows from the reservoir through the penstock, it enters the turbine at
section 1 (entrance of volute casing), with hf as the loss due to friction in the penstock.
Thus the fluid energy at the entrance of the turbine is
𝐻1 = 𝐻0 − ℎ𝑓 (𝑖. 𝑒 𝑛𝑒𝑡 ℎ𝑒𝑎𝑑 𝑎𝑡 𝑒𝑛𝑡𝑟𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡𝑢𝑟𝑏𝑖𝑛𝑒)
At section 1, the fluid has high pressure and kinetic energy as it enters the spiral vane.
Section 1-2: As the fluid flows through the turbine, work is developed in the rotor. The
fluid exits the turbine through the draft tube to the tailrace where the flow is fully developed
at atmospheric pressure. As shown,
energy lost in the draft tube =
𝑉3
2
2𝑔
Therefor the net head producing work is
𝐻 = 𝐻1 − 𝐻2.
Neglecting friction within the Draft tube, 𝐻2 = 𝐻3 . 𝑎𝑛𝑑 𝐻 = 𝐻1 − 𝐻2 = 𝐻1 − 𝐻3
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Figure 8.2.2 A typical set-up of a Francis turbine
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38
6.5 Analysis of Force on Runner of Francis Turbine
A sectional view of a typical Francis turbine runner blade is shown below with velocities
triangle at both inlet and outlet. The runner is designed such that water enters the blade
with both radial and tangential velocities (caused by the guide vanes), but as it exits the
blade, the tangential velocity diminishes. Hence the flow exits with radial or axial velocity.
In designing, the flow velocities at inlet and outlet must be equal, and also equal to axial
velocity at inlet. 𝑖. 𝑒. 𝑉𝑓1 = 𝑉𝑓2 = 𝑉𝑎1
From the velocity triangle at inlet,
𝑉𝑤1 = 𝑉𝑓1𝑐𝑜𝑡𝛼1
𝑈1 = 𝑉𝑤1 − 𝑉𝑓1 cot(180 − 𝛽1) = 𝑉𝑓1𝑐𝑜𝑡𝛼1 + 𝑉𝑓1 cot 𝛽1 = 𝑉𝑓1(𝑐𝑜𝑡𝛼1 + cot 𝛽1)
The runner is designed such that the tangential velocity at outlet, 𝑉𝑤2 = 0 (for max work
done)
Power developed by shaft:
𝐸
𝑚
= −(𝑉𝑤2𝑈2 − 𝑉𝑤1𝑈1) = 𝑉𝑤1𝑈1
𝐸
𝑚
= 𝑉𝑓1𝑐𝑜𝑡𝛼1(𝑉𝑓1𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1) = 𝑉𝑓1
2
𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)
This is the energy produced per unit mass as the water flows through the runner.
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39
What is the hydraulic efficiency, ηh?
𝜂ℎ =
𝐻𝑒𝑎𝑑 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑
𝐻𝑒𝑎𝑑 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒
=
𝐸 𝑚
⁄
𝐸 𝑚
⁄ + 𝑉2
2
2
⁄
𝜂ℎ =
𝐸 𝑚
⁄
𝐸 𝑚
⁄ + 𝑉𝑓1
2
2
⁄
=
𝑉𝑓1
2
𝑉𝑓1
2
𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1) + 𝑉𝑓1
2
2
⁄
𝜂ℎ =
2𝑉𝑓1
2
𝑉𝑓1
2
[2𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1) + 1]
𝜂ℎ =
2𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)
2𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1) + 1
=
1 − 1 + 2𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)
1 + 2𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)
∴ 𝜂ℎ = 1 −
1
1 + 2𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)
NB: If friction is neglected, the hydraulic efficiency =runner efficiency
Degree of Reaction
𝑅 =
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑡𝑎𝑡𝑖𝑐 ℎ𝑒𝑎𝑑 𝑖𝑛 𝑟𝑜𝑡𝑜𝑟
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑
𝑅 =
𝐸 𝑚
⁄ − (
𝑉1
2
− 𝑉2
2
2
)
𝐸 𝑚
⁄
=
𝐸 𝑚
⁄ − (𝑉1
2
− 𝑉2
2)
2 𝐸 𝑚
⁄
= 1 −
(𝑉1
2
− 𝑉2
2)
2 𝐸 𝑚
⁄
But 𝑉1
2
= 𝑉𝑤1
2
+ 𝑉𝑓1
2
= 𝑉𝑓1
2
(𝑐𝑜𝑡2
𝛼1 + 1)
⟹ 𝑅 = 1 −
𝑉𝑓1
2
𝑐𝑜𝑡2
𝛼1
2𝑉𝑓1
2
∴ 𝑅 = 1 −
𝑐𝑜𝑡𝛼1
2(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)
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40
Specific Speed, NST
𝑁𝑆𝑇
=
𝑁𝑃1 2
⁄
𝐻5 4
⁄
P= input power 𝑃 = 𝜂ℎ𝜌𝑔𝑄𝐻
H = head available 𝐻 = 𝐸 𝑚
⁄ + 𝑉2
2
2
⁄
N= rotor speed rev/min 𝑁 =
𝑈1
𝜋𝐷1
𝑃 = 𝜂ℎ𝜌𝑔𝑄𝐻
𝑁𝑆𝑇
=
𝑁(𝜂ℎ𝜌𝑔𝑄𝐻)1 2
⁄
𝐻5 4
⁄
=
𝑁(𝜂ℎ𝜌𝑔𝑄)1 2
⁄
𝐻3 4
⁄
But 𝐻 = 𝐸 𝑚
⁄ + 𝑉2
2
2
⁄ = 𝑉𝑓1
2
𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1) +
𝑉𝑓1
2
2
𝐻 =
1
2
𝑉𝑓1
2
[1 + 2𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)]
Also
𝑁 =
𝑈1
𝜋𝐷1
=
𝑉𝑓1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)
𝜋𝐷1
𝑁𝑆𝑇
=
𝜋𝐷1
(𝜂ℎ𝜌𝑔𝑄𝐻)1 2
⁄
{
1
2
𝑉𝑓1
2
[1 + 2𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)]}−5 4
⁄
𝑁𝑆𝑇
= (𝜂ℎ𝜌𝑔𝑄)1 2
⁄
𝜋𝐷1
{
1
2
𝑉𝑓1
2
[1 + 2𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)]}−3 4
⁄
𝑁𝑆𝑇
=
23 4
⁄
𝜋𝐷1
(𝜂ℎ𝜌𝑔𝑄)1 2
⁄
(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)
1
𝑉
𝑓1
−1/2
[1 + 2𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)]−3 4
⁄
∴ 𝑁𝑆𝑇
=
23 4
⁄
𝜋𝐷1
(𝜂ℎ𝜌𝑔𝑄)1 2
⁄
𝑉
𝑓1
−1/2
(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)[1 + 2𝑐𝑜𝑡𝛼1(𝑐𝑜𝑡𝛼1 + 𝑐𝑜𝑡𝛽1)]−3 4
⁄
So a relationship is established between specific speed NST and the blade angles.
Typical values of blade angles are between:
𝛼 = 10ᵒ 𝑡𝑜 40ᵒ 𝛽 = 45ᵒ 𝑡𝑜 120ᵒ
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41
𝐵𝑙𝑎𝑑𝑒 𝑤𝑖𝑑𝑡ℎ
𝐼𝑛𝑙𝑒𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
=
𝑏
𝐷
=
1
20
𝑡𝑜
2
3
𝑡ℎ𝑖𝑠 𝑖𝑠 𝑖𝑚𝑝𝑜𝑟𝑡𝑎𝑛𝑡 𝑖𝑛 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑖𝑛𝑔 𝑓𝑙𝑜𝑤 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑉𝑓1
The combination of these values give a wide range of specific speeds 𝑁𝑆𝑇
= 40 𝑡𝑜 400
At specific speed 𝑁𝑆𝑇
> 400, the head H under which the turbine operates reduces,
meaning to achieve the required speed under the reducing head, the runner design must
be modified in order to take a relatively large amount of flow to be able to achieve the
required work output. This challenge is addressed using the Kaplan turbine.
The Kaplan machine is suitable for low head operations and specific speeds 𝑁𝑆𝑇
> 400.
The shape of the runner blades should be such that it will allow maximum flow through it
and the flow should be axial, in the direction parallel to the axis of rotation. Kaplan turbine
is an axial machine and is also called Propeller turbine.
8.5 Draft Tube
This is a diffuser that allows recovery of some of the kinetic energy exiting the runner.
That is, it minimizes energy loss at the outlet of the runner. It is a diverging duct attached
to the outlet of the runner.
How it works:
Using the Bernoulli’s equation with the inlet of the drat tube attached to the exit of the
runner and the outlet exposed to ambient conditions in the tailrace, the pressure at the
inlet becomes the outlet pressure of the runner, and the outlet pressure of the draft tube
assumes atmospheric pressure with a decreasing velocity at the outlet. This means that
the inlet velocity at the inlet of the tube will be greater than its outlet velocity,
corresponding to a pressure lower that atmospheric pressure at the inlet of the tube. So
the tube’s inlet pressure becomes suction pressure (which effectively becomes the exit
pressure of the runner). This results in an increased in the effective head of the runner.
From Bernoulli’s equation:
𝑃3
𝜌
+
𝑉3
2
2
+ 𝑔𝑍3 =
𝑃4
𝜌
+
𝑉4
2
2
+ 𝑍4
But 𝑉3 >> 𝑉4 𝑎𝑛𝑑 𝑃4 = 𝑃
𝑎 𝑎𝑛𝑑 𝑍4 = 0 (𝑑𝑎𝑡𝑢𝑚 𝑖𝑛 𝑡𝑎𝑖𝑙𝑟𝑎𝑐𝑒)
So we have
𝑃3
𝜌
+
𝑉3
2
2
+ 𝑔𝑍3 =
𝑃𝑎
𝜌
+ 0 + 0
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42
𝑃3
𝜌
=
𝑃𝑎
𝜌
− (
𝑉3
2
2
+ 𝑔𝑍3)
It is clear here that the inlet pressure of the tube is less than atmospheric pressure since
Z3 and V3 are positive integers.
That is to increase the work done by the turbine (increase effective head) the pressure at
the runner exit must be low. But caution must be taken so that
𝑃3 < 𝑃
𝑣
Pv = vapour pressure of water at a given operating temperature.
If 𝑃4 < 𝑃
𝑣 , the water will begin to boil and bubbles will start forming even at low ambient
temperatures. These bubbles might collapse on solid surfaces (usually on runner blades)
causing pits or cavities on the surface. This phenomenon is known as Cavitation.
Re-writing the Bernoulli equation above, there is a minimum pressure 𝑃3 = 𝑃𝑚𝑖𝑛 = 𝑃
𝑣
𝑉3
2
2𝑔
=
𝑃𝑎
𝜌𝑔
− (
𝑃𝑚𝑖𝑛
𝜌𝑔
+ 𝑍3)
𝑉3
2
2𝑔
= 𝜎𝑐𝐻 =
𝑃𝑎 − 𝑃𝑚𝑖𝑛
𝜌𝑔
− 𝑍3
Where 𝜎𝑐 = 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑐𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟
𝜎𝑐 =
𝑃𝑎/𝜌𝑔 − 𝑃𝑚𝑖𝑛/𝜌𝑔
𝐻
− 𝑍3
To avoid cavitation, a design criteria known as Thoma parameter σ is defined as
𝜎 =
𝑃 − 𝑃
𝑣
𝐻
− 𝑍
Where P = pressure of the fluid
Pv = vapour pressure of the fluid
Z = elevation of the outlet of the runner above the tailrace
From
𝜎 =
𝑃 − 𝑃
𝑣
△ 𝐻
− 𝑍
Cavitation can be alleviated by reducing Z
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43
𝑍𝑚𝑎𝑥 = 𝑃 − 𝑃𝑣 − 𝜎𝑐 △ 𝐻
And 𝜎 > 𝜎𝑐
9 ANALYSIS OF PUMPS
Pump is a machine where the mechanical energy of the machine is converted into stored
energy of the fluid. They work on the principles of fluid dynamics (i.e. continuous motion
of fluid through a machine results in energy transfer between the fluid and the machine).
Just like turbines, pumps are classified as (1) Axial flow pumps – where the flow direction
is parallel to the axis of rotation of the rotor (i.e. the inlet and outlet are at the same radial
location from the axis of rotation), (2) Radial flow pumps – where the fluid gains centrifugal
head as it flows through the rotor blades outwards (eg. Centrifugal pumps).
Centrifugal pump is the direct inverse of the Francis turbine (radially inward flow machine)
Fig 9.1 shows a diagram of a general pumping system.
Fig 9.1 A general pumping system
Consider a pump that takes fluid from a lower reservoir (sump) to a higher reservoir. The
suction pipe conveys fluid from the sump to the inlet of the pump. The fluid in the sump
is opened to the atmosphere. So in order for the fluid to flow up, the pressure in the pipe
must be lower than the atmospheric pressure (i.e. suction) so that the fluid can be
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44
“sucked” up. As the fluid flows through the pump, it gains energy from the pump and
comes out of the pump. It then flows through the delivery pipe to the upper reservoir,
where the fluid is brought to rest (stored). Usually, the suction pipe diameter is bigger
than the delivery pipe diameter (they can be equal).
At any point in the flow from the suction pipe through to the end of the delivery pipe, the
total energy of the fluid comprises pressure energy, kinetic energy and potential energy.
The fluid in the upper reservoir is at rest, so its total energy head is
𝐻𝑆 + 𝐻𝐴
HS = Static head (difference in total energy of liquid in the upper and lower reservoirs). It
is the difference in elevation between the upper and lower reservoirs, where both are
opened to the atmosphere.
Analysis of Fluid Flow from Sump to Upper Reservoir (Fig 9.1 a and b)
A – B: Consider point A to be the fluid surface at sump. As it flows through the strainer
and enters the suction pipe to point B (almost same as A) the total energy of the fluid
drops a little, which accounts for hin (loss at the inlet to the suction pipe).
B – C: as the fluid flows from B to C (outlet of suction pipe; inlet of pump) there is a head
loss hf1 in the suction pipe due to friction and pipe fittings and bends (major and minor
losses).
C – D: the fluid enters the pump, where it gains energy from the rotor and stator of the
pump. There is an increase in head at point D.
D – E: as the fluid flows through the delivery pipe, there is a loss hf2 (major and minor
losses)
E – F: as the fluid exits the delivery pipe into the reservoir at F, all its kinetic energy is lost
(converted to intermolecular energy) as the fluid comes to rest. So the total energy of the
fluid at this point is 𝐻𝑆 + 𝐻𝐴 Let;
𝑇𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝑎𝑡 𝑖𝑛𝑙𝑒𝑡 𝑡𝑜 𝑝𝑢𝑚𝑝 𝐻1 =
𝑃1
𝜌𝑔
+
𝑉1
2
2𝑔
+ 𝑍1
𝑇𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝑎𝑡 𝑜𝑢𝑡𝑙𝑒𝑡 𝑡𝑜 𝑝𝑢𝑚𝑝 𝐻2 =
𝑃2
𝜌𝑔
+
𝑉2
2
2𝑔
+ 𝑍2
𝐻𝑒𝑎𝑑 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑏𝑦 𝑝𝑢𝑚𝑝 𝐻 = 𝐻2 − 𝐻1 = (
𝑃2
𝜌𝑔
+
𝑉2
2
2𝑔
+ 𝑍2) − (
𝑃1
𝜌𝑔
+
𝑉1
2
2𝑔
+ 𝑍1) − − − −9.1
But for a pump, 𝑍1 ≈ 𝑍2 𝑎𝑛𝑑 𝑉1 ≈ 𝑉2 ℎ𝑒𝑛𝑐𝑒 𝐻 = 𝐻2 − 𝐻1 = (
𝑃2−𝑃1
𝜌𝑔
) = 𝑀𝑎𝑛𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝐻𝑒𝑎𝑑
This can be approximately determined by pressure gauges at the inlet and outlet of the
pump.
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Applying Bernoulli’s equation from A – C;
0 + 0 + 𝐻𝐴 =
𝑃1
𝜌𝑔
+
𝑉1
2
2𝑔
+ 𝑍1 + ℎ𝑖𝑛 + ℎ𝑓1 𝑃𝑜𝑖𝑛𝑡 𝐴 𝑖𝑠 𝑑𝑎𝑡𝑢𝑚, 𝑎𝑛𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟
Applying Bernoulli’s equation from D – F;
𝑃2
𝜌𝑔
+
𝑉2
2
2𝑔
+ 𝑍2 = 0 + 0 + 𝐻𝐴 + 𝐻𝑆 + ℎ𝑓2 + ℎ𝑒𝑥𝑖𝑡
Substituting for HA;
𝐻 = 𝐻𝑠 + ℎ𝑖𝑛 + ℎ𝑓1 + ℎ𝑓2 + ℎ𝑒𝑥𝑖𝑡
𝐻 = 𝐻𝑠 + ∑ ℎ𝑙𝑜𝑠𝑠𝑒𝑠
𝑖. 𝑒. 𝑇𝑜𝑡𝑎𝑙 𝐻𝑒𝑎𝑑 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑏𝑦 𝑝𝑢𝑚𝑝 = 𝑆𝑡𝑎𝑡𝑖𝑐 𝐻𝑒𝑎𝑑 + 𝑆𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑙𝑜𝑠𝑠𝑒𝑠
Also;
𝑇𝑜𝑡𝑎𝑙 𝐻𝑒𝑎𝑑 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑏𝑦 𝑝𝑢𝑚𝑝 = 𝑇𝑜𝑡𝑎𝑙 𝐻𝑒𝑎𝑑 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑓𝑙𝑢𝑖𝑑
= 𝑇𝑜𝑡𝑎𝑙 𝐻𝑒𝑎𝑑 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑎𝑡 𝑜𝑢𝑡𝑙𝑒𝑡 𝑜𝑓 𝑝𝑢𝑚𝑝 − 𝑇𝑜𝑡𝑎𝑙 𝐻𝑒𝑎𝑑 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑎𝑡 𝑖𝑛𝑙𝑒𝑡 𝑜𝑓 𝑝𝑢𝑚𝑝
9.1 Description of a Centrifugal Pump
Fig 9.1.1 (a) (b)
Generally, the centrifugal pump has 3 main components;-
1. The Rotor blade / impeller
2. Stator / guide vane / diffuser vanes
3. Volute casing
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An inlet pipe is connected to the eye (suction eye) of the impeller, through which fluid
(water) is drawn into the rotor axially.
The rotor consists of a metal disc, keyed to a shaft powered by a motor. It has a number
of curved blades attached to the surface of the disc. The blade forms passages for the
water flow. The blades are positioned in a diverging manner (i.e. the cross-sectional area
increases outwardly), and water flows in a radially outward manner.
As water flows through the blades,
i. Its relative velocity decreases as the cross-sectional area increases, and its static
pressure increases
ii. Since it is a radially outward flow the water increases its centrifugal head (as radius
increases outwardly from the axis of rotation) in addition to its static head.
NB: shrouded blades are sometimes used to prevent the flow from leaking from one blade
to another.
As water flows through the blades, it exits the impeller and passes through the guide vane
(stator) positioned just at the outlet of the impeller. Here, part of the kinetic energy
(velocity) of the water is gradually converted to pressure energy.
From the stator, the water enters the volute casing, which has a spiral shape. The volute
casing also converts part of the kinetic energy to pressure energy. The water flows to the
delivery end of the pump at high pressure. The energy gained is mainly pressure energy.
Some pumps do not have guide vanes. In such cases, the volute casing does the
conversion of part of the KE into pressure energy.
9.2 Flow and Energy Transfer in a Centrifugal Pump
Consider the velocity diagram in Fig 9.2.1 below.
Fig. 9.2.1
Inlet
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Water enters the eye in such a way that it has a negligible tangential (whirling) velocity
component (this is to avoid shocks and reduce friction at inlet). The blade angle design is
such that water is drawn in purely axially. So Vw1 ≈ 0
Outlet
Water exits the impeller with both radial and tangential velocity components as shown.
From the Euler’s equation,
𝐻𝑒𝑎𝑑 𝑖𝑚𝑝𝑎𝑟𝑡𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑖𝑚𝑝𝑒𝑙𝑙𝑒𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 =
𝑉𝑤2𝑈2 − 𝑉𝑤1𝑈1
𝑔
𝑏𝑢𝑡 𝑉𝑤1 ≈ 0
𝐻𝑒𝑎𝑑 𝑖𝑚𝑝𝑎𝑟𝑡𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑖𝑚𝑝𝑒𝑙𝑙𝑒𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 =
𝑉𝑤2𝑈2
𝑔
NB. This is not equal to the head developed by the pump H = H2 – H1
𝑀𝑎𝑛𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝜂𝑚 =
𝐻
𝑉𝑊2𝑈2 𝑔
⁄
=
𝑔𝐻
𝑉𝑤2𝑈2
𝑓𝑜𝑟 𝑖𝑑𝑒𝑎𝑙 𝑓𝑙𝑢𝑖𝑑, 𝜂𝑚 = 1
𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝜂𝑜 =
𝑇𝑜𝑡𝑎𝑙 𝑓𝑙𝑢𝑖𝑑 𝑝𝑜𝑤𝑒𝑟 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑏𝑦 𝑝𝑢𝑚𝑝
𝑆ℎ𝑎𝑓𝑡 𝑝𝑜𝑤𝑒𝑟
=
𝜌𝑔𝑄𝐻
𝑃
𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝜂𝑚𝑒𝑐ℎ =
𝑃𝑜𝑤𝑒𝑟 𝑖𝑚𝑝𝑎𝑟𝑡𝑒𝑑 𝑏𝑦 𝑖𝑚𝑝𝑒𝑙𝑙𝑒𝑟 𝑡𝑜 𝑓𝑙𝑢𝑖𝑑
𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟
=
𝜌𝑄𝑉𝑤2𝑈2
𝑃
𝑶𝒗𝒆𝒓𝒂𝒍𝒍 𝑬𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 𝜼𝒐 = 𝜼𝒎 × 𝜼𝒎𝒆𝒄𝒉
9.3 Slip in Centrifugal Pump
Slip is a phenomenon where the velocity of fluid at the outlet of the impeller is shifted in
such a way that it gives rise to a lesser actual tangential velocity than the ideal tangential
velocity. This results in lesser actual work transferred to the fluid by the impeller.
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48
Fig. 9.3.1
How does it happen?
The blade has a leading and trailing faces as it rotates. As the fluid flows past the leading
face, the flow deaccelerates and the pressure on the leading face becomes relatively high
(+ve). On the trailing face, the fluid is rather accelerated because of the curvature of the
blade resulting in lower pressure than the leading face. The difference in pressure on the
two faces of the blade results in a circulation loop around the blade, which disturbs the
radial flow through the blade passages, creating a non-uniformity in the radial flow velocity
of the fluid. This changes the velocity vector, and hence affects the tangential velocity
component at the outlet as shown in fig 9.3.1.
𝑖. 𝑒. 𝑆𝑙𝑖𝑝 ∆𝑉
𝑤 = 𝑉
𝑤 − 𝑉
𝑤
′
𝑆𝑙𝑖𝑝 𝑓𝑎𝑐𝑡𝑜𝑟 𝜎 =
𝑉
𝑤
′
𝑉
𝑤
𝐴𝑐𝑡𝑢𝑎𝑙 ℎ𝑒𝑎𝑑 𝑖𝑚𝑝𝑎𝑟𝑡𝑒𝑑 𝑡𝑜 𝑓𝑙𝑢𝑖𝑑 =
𝝈𝑽𝒘𝟐𝑼𝟐
𝒈
= 𝝈𝑯𝒕𝒉𝒆𝒐𝒓 − − − − − − − − − − − 9.2
0.85 ≤ 𝜎 ≤ 0.9
Comment
Slip can be reduced by increasing the number of blades and by reducing the passage
area. But when the number of blades is increased, frictional losses between the blade
and the fluid also increases (there will be more blade-fluid contact), which can take away
the gains made by reducing the slip. Hence there should be a compromise in the choice
of the number of blades since both slip and friction are all head losses that can reduce
the head developed by the pump.
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9.4 Characteristics/Performance of a Pump.
Refer to Fig 9.2.1, the velocity triangle at the outlet of the impeller;
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝐻𝑒𝑎𝑑 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑, (𝑚𝑎𝑥𝑖𝑚𝑢𝑚) 𝐻𝑡ℎ𝑜𝑒𝑟 =
𝑉𝑤2𝑈2
𝑔
𝑤𝑖𝑡ℎ 𝑉𝑤1 ≈ 0
𝑉𝑤2 = 𝑈2 − 𝑉𝑓2𝑐𝑜𝑡𝛽2
For a pump with rotational speed N, impeller diameter D, flow cross-sectional area normal
to flow, A,
𝑈2 = 𝜋𝐷𝑁, 𝑉𝑓2 =
𝑄
𝐴
⟹ 𝑉𝑤2 = 𝜋𝐷𝑁 −
𝑄
𝐴
𝑐𝑜𝑡𝛽2
𝐻𝑡ℎ𝑒𝑜𝑟 =
1
𝑔
𝜋𝐷𝑁 (𝜋𝐷𝑁 −
𝑄
𝐴
𝑐𝑜𝑡𝛽2) =
𝜋2
𝐷2
𝑁2
𝑔
−
𝑄(𝜋𝐷𝑁)
𝑔𝐴
𝑐𝑜𝑡𝛽2
For a given pump with fixed rotational speed,
N = constant D = constant, A = constant and 𝛽2 = constant
𝑇ℎ𝑒 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝐻𝑒𝑎𝑑 𝑯𝒕𝒉𝒆𝒐𝒓 = 𝑲𝟏 − 𝑲𝟐𝑸 − − − − − − − − − − − −9.3
𝑊ℎ𝑒𝑟𝑒 𝐾1 =
𝜋2
𝐷2
𝑁2
𝑔
𝑎𝑛𝑑 𝐾2 =
𝜋𝐷𝑁
𝑔𝐴
𝑐𝑜𝑡𝛽2
If major and minor losses are taken into account:
There are 2 types of major losses (1) shock loss at entry, hs and (2) frictional losses, hf
Shock Loss
𝒉𝒔𝒉𝒐𝒄𝒌 = 𝑲𝟑(𝑸 − 𝑸𝑫)𝟐
− − − − − − − − − − − −9.4 𝐾3 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,
QD = flowrate at designed condition, where the fluid enters the blade at an angle same as
angle at blade inlet. i.e. the fluid glides along the blade at entry without shock
Frictional Loss
ℎ𝑓 = 𝑓
𝐿
𝐷
𝑉2
2𝑔
⟹ 𝒉𝒇 = 𝑲𝟒𝑸𝟐
− − − − − − − − − − − − − 9.5 𝐾4 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
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Plotting the sum of all these 4 equations on an H-Q graph gives the actual head Hactual :
K1
H Htheor = K1 – K2Q
(1-σ)Htheor
Hactual
hshock hf
QD Q
Fig 9.4.1 Actual Head-Discharge (H-Q) characteristic graph.
9.5 Effect of Blade Outlet Angle 𝜷𝟐 on Actual H-Q Characteristic Curve
Depending on the outlet angle of the blade, the blade setting can be classified as;
i. Forward – curved blade
ii. Radial blade
iii. Backward – curved blade
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Fig 9.5.1
It can be observed that the shape of the velocity triangles are different.
𝐹𝑜𝑟 𝐹𝑜𝑟𝑤𝑎𝑟𝑑 − 𝑐𝑢𝑟𝑣𝑒𝑑 𝑏𝑙𝑎𝑑𝑒, 𝑉𝑤2 > 𝑈2, 𝛽2 > 90 𝑐𝑜𝑡𝛽2 < 0 , − 𝑣𝑒
𝐹𝑜𝑟 𝑅𝑎𝑑𝑖𝑎𝑙 𝑏𝑙𝑎𝑑𝑒, 𝑉𝑤2 = 𝑈2, 𝛽2 = 90 𝑐𝑜𝑡𝛽2 = 0 ,
𝐹𝑜𝑟 𝐵𝑎𝑐𝑘𝑤𝑎𝑟𝑑 − 𝑐𝑢𝑟𝑣𝑒𝑑 𝑏𝑙𝑎𝑑𝑒, 𝑉𝑤2 < 𝑈2, 𝛽2 < 90 𝑐𝑜𝑡𝛽2 > 0 , + 𝑣𝑒
Considering the general relation 𝐻𝑡ℎ𝑒𝑜𝑟 = 𝐾1 − 𝐾2𝑄 𝑤ℎ𝑒𝑟𝑒 𝐾2 =
𝜋𝐷𝑁
𝑔𝐴
𝑐𝑜𝑡𝛽2
Meaning
𝐹𝑜𝑟 𝐹𝑜𝑟𝑤𝑎𝑟𝑑 − 𝑐𝑢𝑟𝑣𝑒𝑑 𝑏𝑙𝑎𝑑𝑒, 𝐾2 < 0, − 𝑣𝑒 ⟹ 𝐻𝑡ℎ𝑒𝑜𝑟 = 𝐾1 + 𝐾2𝑄2
𝐹𝑜𝑟 𝑅𝑎𝑑𝑖𝑎𝑙 𝑏𝑙𝑎𝑑𝑒, 𝐾2 = 0, ⟹ 𝐻𝑡ℎ𝑒𝑜𝑟 = 𝐾1
𝐹𝑜𝑟 𝐵𝑎𝑐𝑘𝑤𝑎𝑟𝑑 − 𝑐𝑢𝑟𝑣𝑒𝑑 𝑏𝑙𝑎𝑑𝑒, 𝐾2 > 0, + 𝑣𝑒 ⟹ 𝐻𝑡ℎ𝑒𝑜𝑟 = 𝐾1 − 𝐾2𝑄2
On the H-Q graph (these exclude the major and minor losses);
H
Forward-curved
K1 Radial
Backward-curved
Fig 9.5.2 Q
If losses are taken into account,
The actual Head-Discharge (H-Q) and Power-Discharge (P-Q) graphs become;
NB that the Power-Discharge (P-Q) graph takes into account the efficiency of the pump.
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Fig 9.5.3
P-Q Graph
Forward-curved blade: - Power require decreases up to a minimum and then begins to
rise with increasing flowrate. If a pump is selected for a particular designed flowrate, and
there should be a surge in flowrate beyond the designed rating as the pump operates, the
power required for the motor to cope with the surge will increase accordingly, resulting in
the motor being overloaded which might lead to failure. So then for safe motor operation,
the flowrate must not exceed the designed rating of the pump. In choosing this type, a
safety margin in flowrate is important. This operation limitation is similar for the Radial
blade machine.
Backward-curved blade: - the curve shows an initial rise in power to a maximum value
and then begin to decrease and flowrate increases. Meaning if the designed rating
corresponds to the maximum power point, a surge in flow will require lesser power and
hence the motor can operate safely. i.e. for a backward-curved blade centrifugal pump
which is rated for a designed flowrate condition, the motor will not fail even when there is
a surge in flowrate. This phenomenon is known as self–limiting characteristic.
9.6 Pump and System Characteristics – Backward-curved blade centrifugal pump
The H-Q and η-Q characteristics curves are shown below for a given rotational speed N.
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η η-Q characteristics of pump
Designed point
H H-Q characteristics of pump
QD Q
Fig 9.6.1
QD is the pump’s designed flowrate which corresponds to its max efficiency. But in actual
operation, the operating point depends on the pumping system (which comprises the
pump, piping and fittings) resistance and the pump head. Refer to Fig 9.1a
Inlet of suction + suction pipe + pump + delivery pipe + delivery exit = Pumping System
The operating point of the pump will be decided by the pump and system characteristics.
The system characteristic take into account the all losses with the pumping system. Let
subscript 1 and 2 be for suction and delivery pipes conditions and all symbols have their
usual meanings;
𝐻𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 𝑖𝑛 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑖𝑝𝑒, ℎ1 = 𝑓1
𝐿1
𝐷1
𝑉1
2
2𝑔
+ 𝐾1
𝑉1
2
2𝑔
𝑎𝑛𝑑
𝐻𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 𝑖𝑛 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 𝑝𝑖𝑝𝑒, ℎ2 = 𝑓2
𝐿2
𝐷2
𝑉2
2
2𝑔
+ 𝐾2
𝑉2
2
2𝑔
𝑏𝑢𝑡 𝑉 =
4𝑄
𝜋𝐷2
ℎ1 = 𝑓1
𝐿1
𝐷1
5
8𝑄2
𝑔𝜋2
+ 𝐾1
8𝑄2
𝑔𝜋2𝐷1
4 = (𝑓1
𝐿1
𝐷1
5
8
𝑔𝜋2
+ 𝐾1
8
𝑔𝜋2𝐷1
4) 𝑄2
𝑎𝑛𝑑 ℎ2 = (𝑓2
𝐿2
𝐷2
5
8
𝑔𝜋2
+ 𝐾2
8
𝑔𝜋2𝐷2
4) 𝑄2
𝑇𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 = ℎ1 + ℎ2
𝐼𝑓 𝑙𝑒𝑛𝑔𝑡ℎ𝑠 𝑎𝑛𝑑 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟𝑠 𝑜𝑓 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 𝑝𝑖𝑝𝑒𝑠 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑠𝑝𝑒𝑒𝑑 𝑁
𝑇ℎ𝑒𝑛 𝑡𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑎𝑠 ℎ1 + ℎ2 = 𝐶𝑄2
𝐶 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝐻𝑒𝑎𝑑 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚, 𝑯𝒔𝒚𝒔 = 𝑯𝒔 + 𝑪𝑸𝟐
− − − −9.6 𝐻𝑠 = 𝑆𝑡𝑎𝑡𝑖𝑐 ℎ𝑒𝑎𝑑
lOMoARcPSD|21319694

54
On the H-Q graph for a given system arrangement,
η η-Q
H Designed point
System
Operating point
Hs H-Q
QD Q
Fig 9.6.2
Meaning if the pump is not attached to the system, it is most efficient at the designed
point.
When the pump is attached to a system, then the intersection of the characteristic
curves for the Head (H-Q) and the System will be the operating point as shown.
If the length and/or diameter of the pipe and system arrangement is changed, the curve
will be altered accordingly, and the operating points will also change (any change will
affect the constant C which defines the System characteristics curve) as shown below;
η
Different curves due to changes
in system setup/pipe length/
diameter, etc.
H
Hs
Fig 9.6.3 Q
lOMoARcPSD|21319694

55
NB: The operating point is decided by the System characteristic curve.
Influence of Speed, N, on Pump Characteristics – keeping system components
fixed
From similitude analysis, if pumps are in the same homologous series, then
𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑒𝑠𝑐𝑟𝑖𝑏𝑒𝑑 𝑏𝑦 𝜋1 =
𝑄
𝑁𝐷3
𝑎𝑛𝑑 𝜋2 =
𝑔𝐻
𝑁2𝐷2
𝑖. 𝑒. 𝑖𝑓 𝐷 𝑖𝑠 𝑛𝑜𝑡 𝑐ℎ𝑎𝑛𝑔𝑒𝑑,
𝑄1
𝑁1𝐷3
=
𝑄2
𝑁2𝐷3
𝑎𝑛𝑑
𝑔𝐻1
𝑁1
2
𝐷2
=
𝑔𝐻2
𝑁2
2
𝐷2
𝑄2 = (
𝑁2
𝑁1
) 𝑄1 − − − − − −(𝑖) 𝑎𝑛𝑑 𝐻2 = (
𝑁2
𝑁1
)
2
𝐻1 − − − − − − − (𝑖𝑖)
𝑓𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 (𝑖), 𝑄 ∝ 𝑁 𝑎𝑛𝑑 (𝑖𝑖) 𝐻 ∝ 𝑁2
⟹ 𝑯 ∝ 𝑸𝟐
− − − − − − − −9.7
Meaning the locus of points at different speeds N1, N2 …. form a parabola as shown
below
H
H α Q2
N1
N2
N3
Q
Fig 9.6.4
9.7 Diffuser and Cavitation
9.7.1 Diffuser of Centrifugal Pump
As fluid flows through the impeller radially outward, it gains both kinetic and pressure
energy. At the delivery of a pump high pressure energy is preferred to high kinetic energy.
lOMoARcPSD|21319694

56
The function of diffuser is to convert part of the gained kinetic energy into pressure energy,
so that a high pressure energy will be delivered at the exit of the pump. It is important to
note that not all pumps have diffusers.
As the fluid exits the diffuser, its angular momentum remains constant;
𝑉
𝑤 × 𝑟 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Assuming there is no friction, then the radial velocity is also constant, 𝑉
𝑟 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
A combination of 𝑉
𝑤𝑟 and 𝑉
𝑟 gives rise to a spiral vortex motion. That is the shape of the
volute must be designed to match the spiral streamline pattern formed in the flow process.
9.7.2 Cavitation
Cavitation occurs when absolute pressure of water falls below its saturated vapour
pressure, Pv for the water temperature. When the absolute pressure drops too low, the
water starts to boil even at the ambient operating temperature of the machine. Bubbles
of vapour form. The bubbles collapse as the pressure rises again creating very high
instantaneous pressure that can cause significant erosion of blade surfaces, thereby
reducing the performance of the machine.
In a pumping system, as the impeller rotates it sucks fluid from the sump which is at
atmospheric pressure. Because the pressure in the suction line is below atmospheric, it
draws water from the sump to the inlet of the pump. The pressure is minimum at the inlet
of the pump. In order to avoid cavitation, the minimum pressure must not be below the
vapour pressure of water at the operating temperature.
Applying Bernoulli’s equation to the suction line;
𝑃1
𝜌𝑔
+
𝑉1
2
2𝑔
+ 𝑍1 + ℎ𝑓 =
𝑃𝑎𝑡𝑚
𝜌𝑔
+ 0 + 0 𝑍0 = 0, 𝑉
𝑜 = 0, 𝑠𝑢𝑚𝑝 𝑖𝑠 𝑎 𝑙𝑎𝑟𝑔𝑒 𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟
𝐿𝑒𝑡 𝑃1 = 𝑃𝑚𝑖𝑛, 𝑡ℎ𝑒𝑛
𝑃𝑚𝑖𝑛
𝜌𝑔
=
𝑃𝑎𝑡𝑚
𝜌𝑔
− (
𝑉1
2
2𝑔
+ 𝑍1 + ℎ𝑓)
Thus if the water has to be drawn to a height Z1, overcoming frictional loss hf and generate
fluid velocity V1 from atmospheric pressure under static conditions, then
𝑃𝑚𝑖𝑛
𝜌𝑔
≥
𝑃𝑎𝑡𝑚
𝜌𝑔
− (
𝑉1
2
2𝑔
+ 𝑍1 + ℎ𝑓)
To achieve this, reduce Z1 and hf as much as possible.
A critical cavitation parameter σc is defined as
lOMoARcPSD|21319694

57
𝜎𝑐𝐻 =
𝑉1
2
2𝑔
=
𝑃𝑎𝑡𝑚
𝜌𝑔
−
𝑃𝑚𝑖𝑛
𝜌𝑔
− (+𝑍1 + ℎ𝑓)
𝜎𝑐 =
𝑉1
2
2𝑔𝐻
=
1
𝐻
(
𝑃𝑎𝑡𝑚
𝜌𝑔
−
𝑃𝑚𝑖𝑛
𝜌𝑔
− 𝑍1 − ℎ𝑓)
Defining the Thoma’s cavitation parameter, by replacing 𝑃𝑚𝑖𝑛 = 𝑃𝑣
𝜎 =
1
𝐻
(
𝑃𝑎𝑡𝑚
𝜌𝑔
−
𝑃
𝑣
𝜌𝑔
− 𝑍1 − ℎ𝑓)
To avoid cavitation, 𝑃𝑚𝑖𝑛 > 𝑃𝑣 𝑤ℎ𝑖𝑐ℎ 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡 𝜎 > 𝜎𝑐
Also, for satisfactory operation, there must be some margin between the fluid pressure
head at inlet of the impeller and the saturated vapour pressure head of the fluid. This
margin is called Net Positive Suction Head – NPSH
𝑁𝑃𝑆𝐻 = ℎ𝑖𝑛 − ℎ𝑣
ℎ𝑖𝑛 = 𝑓𝑙𝑢𝑖𝑑 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ℎ𝑒𝑎𝑑 𝑎𝑡 𝑖𝑚𝑝𝑒𝑙𝑙𝑒𝑟 𝑖𝑛𝑙𝑒𝑡, ℎ𝑣 = 𝑠𝑎𝑡 𝑣𝑎𝑝 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ℎ𝑒𝑎𝑑 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑
𝑁𝑒𝑡 𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑆𝑢𝑐𝑡𝑖𝑜𝑛 𝐻𝑒𝑎𝑑 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒, 𝑁𝑃𝑆𝐻𝐴 =
𝑃𝑎𝑡𝑚
𝜌𝑔
−
𝑃𝑣
𝜌𝑔
− 𝑍1 − ℎ𝑓
If the pump position is higher than Z1, NPSHA decreases. NPSHA can be increased by
lowering the pump (provided major and minor losses in pump do not increase)
NPSHR is the Net Positive Suction Head required by the pump in order to prevent
cavitation. This is provided by the pump manufacturer.
So to avoid vapourization and cavitation, 𝑁𝑃𝑆𝐻𝐴 > 𝑁𝑃𝑆𝐻𝑅
Example
When a lab test was carried out on a pump, it was found that for a pump total head of 36
m at a discharge of 0.05 m3/s, cavitation began when the sum of the static pressure and
velocity head at inlet was reduced to 3.5 m. the atmospheric pressure was 750 mmHg
and vapour pressure of water was 1.8 kPa. If the pump is to operate at a location where
the atmospheric pressure is 620 mmHg and the temperature is so reduced that the vapour
pressure of water is 830 Pa. what is the value of cavitation parameter is when the pump
develops the same head and discharge? Is it necessary to reduce the height of the pump?
If so, by how much?
lOMoARcPSD|21319694

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