Chapter One Handout

7/30/2019 Chapter One Handout

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Chapter One

Basic Principles of Turbomachinery

What is a Turbomachine?

A Turbomachine is a device that exchanges energy with a fluid using continuouslyflowing fluid and a blade.

If the device extracts energy from the fluid it is generally called a turbine. If the device delivers energy to the fluid it is called a compressor, fan, blower or pump

depending on the fluid used and the magnitude of the change in pressure that results.

Classification of Turbomahcines

Based on power Power generating : Turbines Power absorbing : Pumps, compressors, blowers, fans Power transmitting : fluid couplings, torque converters

Based on the direction of flowing fluid in the rotor Axial flow : Axial flow compressors , Axial turbines (propeller turbines), Axial

pumps

Radial flow : Radial turbines (Francis Turbine), Centrifugal compressors andpumps

Mixed flow Based on the type of fluid used

Liquid : Pumps, Hydraulic turbines Steam : Steam turbines Gas : Gas turbines, compressors Air : Wind turbines, compressors


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Comparison between positive displacement machines with turbomachines

The machines whose functioning depends essentially on the change of volume of acertain amount of fluid within the machine are known as positive displacement

machines.

The word positive displacement comes from the fact that there is a physicaldisplacement of the boundary of a certain fluid mass as a closed system.

This principle is utilized in practice by the reciprocating motion of a piston within acylinder while entrapping a certain amount of fluid in it.

The machine producing mechanical energy is known as reciprocating engine while themachine developing energy of the fluid from the mechanical energy is known as

reciprocating pump or reciprocating compressor.

Turbo machines differ from positive displacement machines in a way that they require arelative motion between the fluid and the moving part of the machine.

Applications


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Power generation industry: Steam turbine power plants, Gas turbine power plants,Hydro-electric power plants and wind turbine.

Transport industry: Turbo propeller and Turbo fan engines.

HVAC s: Fans, blowers Process industry and miscellaneous applications: Pumps and Compressors

Energy Equation

First law of Thermodynamics: For a system working in a cycle or

for an infinitesimal change For applications in Turbomachines the energy term will include internal energy,

potential energy and kinetic energy.

Substitution gives us the general energy equation for turbomachines

Steady flow energy equation

For steady flow processes through turbomachines, the work term in the energyequation contains shaft work and flow work. It is by applying the principle of first law of

thermodynamics for a steady flow in a control volume.


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The flow work is the work associated with the pushing the mass of fluid across thesystem boundaries.

The energy equation becomes:

But

A further simplification can be done for a compressible fluid by assuming most of theturbomachines are adiabatic and the change in potential energy is negligible ascompared to changes in enthalpy and kinetic energy

Therefore

[adiabatic energy equation]

is the stagnation enthalpy


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Stagnation enthalpy is the enthalpy a fluid would attain if it were brought to zero speedat zero elevation by a steady, adiabatic process with no external work (isentropic

process).

Suppose that our steady flow control volume is a set of streamlines describing a flow upto the nose of a blunt object

State 1: and State 2: and

The energy equation for this control volume is:

which is

(stagnation enthalpy)

In a - diagram


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Energy Transformation

Some adiabatic processes involve only energy transformation; e.g., expansion of gasesin nozzles and compression of gases in diffusers.

The adiabatic energy equation is

or

N.B.: All turbomachines are energy transfer machines In a - diagram


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Eulers equation of Energy Transfer in Turbomachinery

The energy transfer occurs only in its moving or rotating elements, the rotors.

is the absolute velocity of the fluid, is the velocity of the fluid relative to the rotorblade, and is the peripheral velocity of the blade.

The three velocities can be related in a simple vector equation

The absolute velocity has a tangential component and a radial component The linear momentum in the tangential direction is

From Newtons second law ofmotion torque is defined as the rate of change in moment

of linear momentum

Finally the work done by the rotor will be


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The specific work can be expressed as (Eulers energy equation for

turbomachines)

A simple re-arrangement can be made on the Eulers energy equation by geometricmanipulation of the velocity triangles.

Therefore:

This equation shows that the total energy transfer is composed of three components

Work and Efficiencies in Turbines

Losses are inherent in nature Efficiencies are incorporated to compensate to the losses in turbomachinery Losses in turbomachines occur due to

Mechanical losses : bearing friction, windage (frictional force between therotating shaft and air )

Fluid-rotor losses : friction between the blade and the fluid


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Ideal process in turbines is expressed by isentropic expansion (reversible) in - diagram In a - diagram

The efficiency of a turbine is defined as the ratio of the actual work to the ideal work forthe same pressure ratio ( ).

The efficiency of a turbine can be expresses in two ways: Total-to-Total efficiency: the ratio between the actual shaft work to its ideal

(isentropic) shaft work between total conditions at entry and exit.

Total-to-Static efficiency: the ratio between the actual shaft work to the ideal

shaft work between the total conditions at entry and static conditions at exit

Comparing the two efficiency equations and referring to the figure the total-to-total efficiency

is greater the total-to-static efficiency

The efficiency of a turbine is defined as the ratio of the actual work to the ideal work for

the same pressure ratio ( ).


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The efficiency of a turbine can be expresses in two ways: Total-to-Total efficiency: the ratio between the actual shaft work to its ideal

(isentropic) shaft work between total conditions at entry and exit.

Total-to-Static efficiency: the ratio between the actual shaft work to the ideal

shaft work between the total conditions at entry and static conditions at exit

Comparing the two efficiency equations and referring to the figure the total-to-total efficiency

is greater the total-to-static efficiency

Substitution gives

( )

is the ratio of specific heat capacities at constant pressure and volume respectively

And

Finite stage

A stage with a finite pressure drop is a finite stage. If the required pressure drop is more, multi staged are required, so that full expansion

will takes place and which results in more rotation and high efficiency.

and

or


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The total expansion in the turbine is divided into three stages of the same efficiency andpressure ratio

Infinitesimal stage efficiency or Polytropic efficiency

A finite stage is divided into an infinite number of small sages of the same efficiencies

Therefore

or

Finite stage efficiency

The finite stage efficiency can be expresses in terms of the polytropic efficiency

This equation is also applicable to a multi-stage turbine


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Assuming constant stage pressure ratio, the total pressure ratio is= , where is the number ofstages

Mechanical efficiency is the ratio of the turbine brake power to the indicated power

Brake power is the work output at the turbine shaft Indicated power

is the mechanical energy supplied by the rotor

N.B. Polytropic efficiency and the related efficiencies (both stage and overall) expresses by

polytropic efficiency apply for the static conditions which only shows the thermodynamic

implications of multi stages.

Example 1

Air flows through an air turbine where its stagnation pressure is decreased in the ratio 5:1. The

total to total efficiency is 0.8 and the air flow rate is 5 kg/s. If the total power output is 500 KW,

find

a) Inlet total temperature, b) the actual exit total temperature c) the actual exit static

temperature if the flow velocity is 100 m/s and d) the total to static efficiency of the device

Example 2

In a three stage turbine the pressure ratio of each stage is 2 and the stage efficiency is 75 %.

Calculate the overall efficiency and the power developed if air is initially at a temperature of

6000C flows through it at the rate of 25 kg/s. Find reheat factor

Work and Efficiencies in Compressors

The efficiency of a compressor is defined as the ratio of ideal work to its actual work forthe same pressure ratio ( )

Analogous approach to that of turbines


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In a h-s diagram

Total-to-total efficiency

Static-to-static efficiency: If magnitude of gas velocities at entry and exit of a stage arealmost equal or negligible, then their stagnation enthalpies mainly depend on static

enthalpy only.

The mechanical efficiency of a compressor is the ratio of the indicated power to thebrake power.

The steps employed to calculate efficiencies for a compressor is analogous to that of a

turbine.


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Finite stage or

Infinitesimal stage Efficiency or Polytropic efficiency

Finite stage efficiency

The above equation also applies for a multi stage compressorExample

Each stage of a four stage air compressor delivering 44 kg of air per second operates at a

pressure of 1.2, with a stage efficiency of 65 %. Calculate overall efficiency and pressure ratio.

Calculate power required to drive the compressor if air temperature at inlet is 20

0

C. Find alsoexit air temperature.

Example

An air compressor has eight stages of equal pressure ratio 1.35. The flow rate through the

compressor and its overall efficiency are 59 kg/sec and 82 % respectively. If the condition of air

at entry 1 bar and 400C, determine (a) the state of air at the compressor exit, (b) polytropic


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efficiency, (c) efficiency of each stage and (d) power required to drive the compressor and (d)

power required to drive the compressor assuming overall efficiency of the drive as 90 %, (e)

power required in each stage


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Chapter Two

Flow through cascades

While a turbomachine is rotating, it is difficult to see or study the flow conditions of the flowing

fluid. Therefore, keeping the geometric, kinematic and dynamic similarities, it is possible to

study the fluid flow conditions in a stationary cascade of blades in a model.

Cascade: is a single or an array of aerofoil blades

In a cascade view, an infinite radius of rotor is assumed and the rotor blades appear to be

arranged in a horizontal fashion.

Aerofoil blades (Nomenclature)

An aerofoil blade is a streamlined body having a thick rounded leading edge and a thin trailing

edge.

The centrifugal force along the convex surface of the blade pushes flow away from the surface

which reduces the pressure below the free stream pressure, hence termed as suction side.

The centrifugal force along the concave surface of the blade pushes flow towards the surface

which increases the pressure above the free stream pressure, hence termed as the pressure

side.


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Cascade: A single or an array of aerofoil blades

Chord line: An imaginary line joining the center of the leading edge to trailing edge

Camber line: It is a back bone lying mid-way between the upper and lower surfaces

Angle of incidence (): The angle between the air flow line and the camber lineAspect ratio: It is the ratio of the blade height to chord length of the blade

Blade pitch (S): The distance between the adjacent blades

Pitch chord ratio: The ratio of the pitch to chord length of the blade

Turbine cascades

Blades of a given shape are arranged in a different manner in the cascades of turbines and

compressors. The difference in the geometry is due to the accelerating and decelerating flows

in turbine and compressor cascades respectively.


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Stagger angle (): It is the angle between the chord line and the perpendicular axial lineCamber angles ( ): It is the angle between the chord line and the camber line. The camberangle () for a given blade is the sum of the camber angles at the entry and exit.

Blade angle ( ): It is the angle between the camber line and the axial line

At inlet At exit

Adding the above equations Air angle ( ): It is the angle between the air flow line and the axial lineAngle of incidence (): It is the angle between the air flow line and the camber angle at inlet.The figure above shows a positive angle of incidence.

Flow at large positive incidence will lead to stalling on the suction side of the blade A large negative incidence will lead to flow separation (negative stall) on the pressure

side of the blade

Deviation angle (): It is the angle between the air flow line and the camber line at exitDeflection angle (): It is the sum of the air angles at inlet and exit


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Velocity triangles for a turbine cascade

The velocity vectors and at the entry and exit are at air angles and respectively

Subscripts x and y denotes axial and tangential components respectively

The mean axial and tangential velocities are

And

assuming constant axial velocity

Blade forces

Tangential force (): This force is developed due to the change in linear momentum in thetangential direction.


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Axial force (): This force is developed from the change in linear momentum in the axialdirection and the static pressure change. In fact, the change in linear momentum in the axial

direction diminishes due to the assumption that the velocity in the axial direction is constant.

Drag force (

): The force in the direction of the air flow is called drag force

Lift force (): The force perpendicular to the air flow direction or the drag force is called the liftforce.

From the geometric relation the lift () and the drag () forces can be expressed in terms of theaxial () and tangential () force components.

The continuity equation for the control surface is

The flow is assumed to be incompressible because a stationary cascade is used and the axial

velocity is also assumed to be constant. Then,

per unit length or span of the bladeTangential Force

[ ] Substituting for ,


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But and ( )

Multiplying both sides by

To get a dimensionless tangent force coefficient

This can also be expressed in terms of exit velocity

and

is the blade to pitch ratio of a bladeAxial Force

( ) but which is per unit length or span of the blade

For incompressible flow the stagnation pressures at the entry and exit of the blade are:

and

( ) *( ) +

From the velocity triangles at inlet and exit

and


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* + * +

and * +

To get a non-dimensional force coefficient dividing both sides by

Again

is the pressure loss coefficient for the turbine cascade

Therefore the axial force coefficient

Lift force

* +

But


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From velocity triangle

Multiplying the right side with

Drag force

* +

[ ]

To find the non-dimensional coefficient divide both sides by

From velocity triangle and


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Substituting Substituting the above expression on the drag coefficient

is the pressure loss coefficient

Example: Air enters the test section of a turbine blade ( ) cascade tunnel at

(

). The pitch chord ratio of the cascade is 0.91. The average loss in

the stagnation pressure across the cascade is equivalent to 17.5 mm of W.G. Determine for this

cascade

(a) The pressure loss coefficient(b)The drag coefficient(c) Lift coefficient(d)Tangential and axial force coefficients

Chapter One Handout

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