Fundamentals of Turbomachinery

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Turbomachinery devices inject life into fluids. Principles of turbomachinery form the preliminary design tools in design of them.

Following article gives details of the video lecture.

Turbomachinery Principles

Consider following turbo machines. An axial turbine, a centrifugal machine or a pelton wheel, you can predict performance of all of these from same turbo machinery fundamentals.

Fig.1 Impeller of axial turbine, centrifugal machine and Pelton wheel

Euler Turbomachine Equation

To develop turbo machinery fundamentals consider fluid flow through channel shown below. The inlet velocity, V1 gets changed to outlet velocity V2. Velocity of fluid can be split into tangential and radial components. This is shown in following figure.

Fig.2 Velocities at inlet and outlet can be split into tangential and radial components
To make the fluid flow there should be an external torque acting on it. This torque can be derived from Newton’s 2nd law of motion, which acts as fundamental equation of turbo machinery. The torque is given by following equation, which is also called as Euler turbomachine equation.
This is the most important equation in turbo machinery.

Pump or Turbine?

If the channel is rotating at an angular velocity omega, power required to maintain the fluid flow will be torque multiplied by angular velocity.

ω times radius becomes channel velocity or blade velocity. So power required for this fluid motion can be taken as difference in product of blade velocity times tangential fluid velocity.
Vθ is positive, if it is in same direction of blade velocity. Otherwise it is negative.
Fig.3 Blade and tangential velocity of flow are shown at inlet and outlet
If we divide power by weight of fluid flowing, we will get what’s the energy head required to maintain this flow.
If power required by the fluid is positive, that means fluid is absorbing energy. Or the device is acting as a compressor. Otherwise fluid is losing energy, so the device acts as a turbine.

Vθ more precisely means, component of fluid velocity which is parallel to blade velocity.

Fig.4 Vθ is the component which is parallel to blade velocity
But determination of fluid velocity is a tricky affair in turbo machinery, since we are dealing with rotating components. For this purpose we should understand concept of velocity triangles.

Concept of Relative Velocity & Velocity Triangle

The key idea in turbo machinery is concept of relative velocity. Suppose you are standing on this rotating turbo machine. The velocity of fluid you experience while moving with it is called as relative velocity. If fluid is having an absolute velocity V, and the blade is moving with a velocity U, then relative velocity experienced by you will be as follows.

For a stationary device in order to have smooth operation, flow should be tangential to the blade. Similarly in a moving device relative velocity should be tangential to blade profile. With knowledge of direction of relative velocity and the vectorial representation of relative velocity, these 3 velocities could be drawn as shown below. This is known as a velocity triangle.

Fig.5 Velocity triangle in a turbmachine
Similar velocity triangle can be made on inlet of turbo machine. The beauty of turbo machinery is that using relatively simple analysis of inlet and outlet velocities you can predict performance of any turbo machine.

Performance of Centrifugal Pump

We will see, how to predict performance of a centrifugal pump using the concepts we developed. Here we have shown impeller of a centrifugal pump. If you know the blade geometry, you can find out blade angle at inlet and outlet. Blade angle is defined as angle opposite of blade velocity. So we can easily fix direction of relative velocity. This is shown in Fig. 6(b).

Radial component of flow velocity determines how much the volume flow rate is leaving the impeller. So you can determine Vr at outlet from this equation. Here b2 means width of the impeller.

Now construction of velocity triangle is easy. You can draw lines parallel to these velocity components. These lines are drawn in grey color in Fig. 6(b). From parallelogram constructed in Fig. 6(b)absolute velocity of flow can be easily drawn. This is shown in Fig. 6(c). From this we can find out tangential component of flow velocity, this is marked as Vθ2 in Fig. 6(c).
Fig.6 Development of inlet at outlet velocity triangles in a centrifugal pump
At inlet of centrifugal pump flow velocity will be radial. So tangential component of velocity is zero.
So energy head developed by the pump simplifies like this.
From Fig.6(c) outlet blade angle can be easily represented as follows.
You can substitute values of Vθ2 from this equation, into head equation. After substituting value of Vr2 also in that, we get most important performance equation of centrifugal pump. How energy head is varied with flow rate.
Importance of this equation in predicting performance of a centrifugal pump is elaborated in a different article. We can predict performance of an axial flow device and pelton wheel using the same concepts we developed before.

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Velocity Analysis | Mechanics

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To execute proper design of a mechanism, velocity analysis is of utmost importance. A designer’s real interest in development of mechanism is variation of outlet link velocity for a specified inlet velocity.

Check following article for detailed description of video lecture.

The interesting thing about velocity analysis is that, you can determine velocity of any link just by understanding 2 simple concepts.

  1. Concept of Rigid Body
  2. A rigid body cannot elongate or contract in any direction. That means if you take 2 points in a rigid body, it can have 2 different velocities as shown.

    Fig.1 2 points in a rigid body can have different velocities
    But since this is rigid body, velocity components parallel to the line connecting the points should be equal. If component velocity of point B is greater than A, then link will start elongating. If the case is opposite link will start contracting. Both these cases are impossible, since this is a rigid body. So velocity components of both points should be equal.
    Fig.2 Velocity components of the points along the connecting line should be equal
    This means, if we subtract velocity of A from velocity of B, the relative velocity vector will have no component, parallel to the connecting line. It will get cancelled. So relative velocity vector will be perpendicular to the connecting line. This is shown in following figure.
    Fig.3 In a rigid body relative velocity vector should be perpendicular to line connecting themr
    In short, concept 1 can be simplified as, relative velocity between any 2 points in a rigid body should be perpendicular to the line connecting them.

  3. No Penetration of Mating Surfaces
  4. Second concept is applicable to mating surfaces. Mating pair of surface will never penetrate. Red point in folloiwng figure shows the common mating point or mating line for both the links. Common normal of the mating surfaces is also shown. The same point can have different velocities on different links.

    Fig.4 Mating point is having different velocities on different links
    No penetration means velocity component of both the link velocities along common normal should be equal. If velocity component of 2 is less than velocity component of 1, then surfaces will penetrate. If opposite is the case, surfaces will detach. Both these cases are not possible for mating surfaces. So velocity components along common normal should be equal.
    Fig.5 Relative velocity vector of mating point should be perpendicular to common normal
    Since velocity components are equal, if we take a vector difference of V 1 and V 2, it should have no component along the common normal. So the relative velocity should be perpendicular to common normal.

Now we will apply these 2 concepts on different mechanisms, to do velocity analysis of them.

4 Bar Linkage Example

Let's consider 4 bar linkage shown below. We know input angular velocity and we want to find out outlet velocity.

Fig.6 Example of 4 bar linkage
Here approach is simple. We will say that bar at middle cannot expand or contract. Since we know angular velocity of first link, we can easily find out magnitude and direction of velocity of point 1.But for point 2 we know only direction of velocity. Here first concept we learned comes for help. Since this bar cannot elongate or contract, relative velocity between these points should be perpendicular to this link.
Fig.7 Procedure of 4 bar linkage
Which will lead to, magnitude of velocity at point 2. From here angular velocity of link can easily be deduced.

Cam-Cam Example

Now consider this mechanism. We know angular velocity of first cam, we want to find out angular velocity of the second cam.

Fig.8 Example of cam cam mechanism
Common normal at the time of mating is marked in figure. Consider the mating point, which lies on both the cams. We know velocity direction and magnitude of this point on first cam. But for second cam we know only direction. The direction is marked in figure.
Fig.1 Mains parts of single phase induction motor : Rotor and Stator
Here 2nd concept we learned comes to help. Since the links cannot penetrate, relative velocity should be perpendicular to common normal. So we can find out magnitude of V2, which will lead to link angular velocity.
Fig.1 Mains parts of single phase induction motor : Rotor and Stator

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  1. Velocity Analysis
  2. To execute proper design of a mechanism, velocity analysis is of utmost importance. A designer’s real interest in development of mechanism is variation of outlet link velocity for a specified inlet velocity.

  3. Degrees of Freedom - Mechanics

Degrees of freedom is the one of the most important concept in kinematics. Which means how many variables are required to determine position of a mechanism in space. In this article we will understand how to predict degrees of freedom of a mechanism.

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Working of Single Phase Induction Motors

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Single phase induction motors require just one power phase for their operation. They are commonly used in low power rating applications, in domestic as well as industrial use. This article is aimed at giving you a conceptual overview of the working of single phase motors.

The following article gives a detailed description of the video lecture.

Parts of a Single Phase Induction Motor

The main components of a single phase motor are the rotor and stator winding. The rotor is the rotating part, the stator winding helps in rotating the rotor. In Fig.1 the iron layer lamina which is fitted inside rotor is not shown, for better viewing of the rotor bars.

Fig.1 Mains parts of single phase induction motor : Rotor and Stator
The winding has got 2 parts; One main winding and an auxiliary winding. The auxiliary winding is placed perpendicular to the main winding. A capacitor is connected to the auxiliary winding.
Fig.2 Single phase motor has got 2 perpendicularly placed wingdings

Working of Single Phase Motor

To understand its working better, let’s assume there is only one coil in the main and auxiliary winding.

Fig.3 We will analyse a case with both the wingdings are having one coil
Assume no current is flowing through the auxiliary winding. AC current passing through the main winding will produce a fluctuating magnetic field.
Fig.3 AC current passing through main winding will produce a fluctuating magnetic field
The working of single phase induction motors is simple. Just put one rotor which is already rotating, inside such a magnetic field. You can note one interesting thing; the rotor will keep on rotating in the same direction.
Fig.4 A rotor with initial rotation will keep on rotating in a fluctuating magnetic field
The reason behind this phenomenon is explained in coming sections.

The Reason – Double Revolving Field Theory

The fluctuating field is equivalent to the sum of two oppositely rotating magnetic fields. This concept is known as double revolving field theory. By looking at the figure below, you can easily understand the theory. Here one fluctuating quantity is represented as a vector sum of 2 oppositely rotating quantities, for 3 time instances.

Fig.5 Fluctuating field at RHS of each figure is equivalent to sum of 2 oppositely rotating magnetic fields
The effect of the rotating magnetic field on the rotor is interesting. Since the magnetic field is varying, electricity is induced in the rotor bars due to electromagnetic induction. In Fig.6 blue arrows on the bars represent current induced. So here is a situation of current carrying bars are which are immersed in a magnetic field. This will produce a force according to Lawrence law, so the rotor will start to rotate.
Fig.6 Effect of RMF on rotor : It will induce a starting torque
But here we have got 2 such oppositely rotating magnetic fields, so the torques produced by them will be equal and opposite. The net effect will be zero torque on the rotor. So the rotor won’t start, it will simply buzz.
Fig.7 Motor will not be able to start here, since there 2 torques which are equal and opposite
But if we can somehow give this rotor an initial rotation, one torque will be greater than other. There will be a net torque in the same direction of initial rotation. As a result the loop will keep on rotating in the same direction. This is the way a single phase induction motor works.
Fig.8 An initial rotation of rotor will produce one torque greater than other

How to Provide Initial Rotation ?

But one big problem remains; how to provide such an initial rotation to the rotor ?. Nikola Tesla, a famous Yugoslav inventor suggested one ingenious solution to this problem.

If we can cancel any of the rotating fields, we will be able to start the motor. The auxiliary winding cum capacitor arrangement is used exactly for this purpose. Auxiliary winding also produces 2 oppositely revolving magnetic fields. One of them will cancel the RMF of the main winding and the other will get added up. The result will be a single magnetic field, which revolves under specific speed. This phenomenon is shown diagrammatically in the following picture.

Fig.9 Effect of auxiliary winding cum capacitor arrangement : One RMF gets cancelled other gets added up
Such a magnetic field can give starting torque to the rotor. Or the motor will self start. After the rotor has attained a specific speed, even if you cut the auxiliary winding, it will keep on rotating, as explained earlier. This cutting action is done through a centrifugal switch.

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Analysis of Beams | Shear Force & Bending Moment Diagram

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Beams are structural members, which are most commonly used in buildings. Beams have numerous other applications in case of bridges, automobiles or in mechanical systems. In this article we will see how we can do strength analysis of a beam.

Following article gives detailed description of the video lecture.

What is a Beam ?

In a beam transverse load is acted, which in fact comes from the slabs to the column or walls. It is clear from following figure that, beams are integral part of of building structure. In all of the beams load acted is transverse, as shown.

Fig.1 In beam transverse load is acted, and it is an integral part of building structure
For analysis purpose, a beam can be considered as a part of the beam column system. This way we can determine external load acting on individual beams. After determining load acting on individual beam, beam can be separated out from beam column system for further analysis.
Fig.2 Building structure consists of many beam-column systems, beam is a part of beam-column system
Length of the beam is much higher than its lateral dimensions. So axial strain developed in a beam will be very small compared to shear strain, or strain induced due to bending.This is shown in figure below.
Fig.3 Axial strain in beam is negligible compared to shear strain
So for design purpose of beams, analysis of shear force and bending moment induced are of the at most importance. The interesting thing is that you can draw shear force and bending moment distribution along any beam, by understanding what exactly is shear force and bending moment.

Both shear force and bending moment are induced in beam in order to balance external load acting on it. We will go through details of it separately.

Shear Force

Shear force is the internal resistance created in beam cross sections, in order to balance transverse external load acting on beam. Consider following beam, it does not matter from where you take a section, when you add forces acting on it, it should be in equilibrium. Shear force is induced exactly for this purpose, to bring the section to equilibrium in vertical direction. It acts parallel to cross section.

Fig.4 Shear force is induced in a section to balance the external load
So just by applying force balance in vertical direction on the free body diagram, we can determine value of shear force at a particular cross section. Usual sign convention of the shear force is as follows.
Fig.5 Usual sign convention of shear force
Now we can apply same concept in different cross section and find out how shear force varies along the length of the beam.

Bending Moment

But balance of transverse forces alone does not guarantee equilibrium of a section. There is another possibility of beam rotation, if moment acting on it is not balanced. If this is the case a bending moment will be induced in cross section of beam, to arrest this rotation. It will be induced as normal forces acting on fiber cross section as shown.

Fig.6 Bending moment is induced in section to balance external moment, section is zoomed in left figure for better viewing
Resultant of those forces will be zero, but it will produce a moment, to counter balance the external moment. So we can calculate moment induced at any cross section by balancing the external moment acting on the free body diagram.

Sign convention of bending moment is as follows.

Fig.7 Sign convention of bending moment for simply supported case
This sign convention approach is valid for simply supported beam. For cantilever case sign convention is exactly opposite to this.

With these concepts developed, we can easily calculate distribution of shear force and bending moment along the length of the beam. We will see few examples.

Example of Cantilever

Consider this case, a cantilever carrying 3 loads.

Fig.8 Analysis of cantilever carrying 3 loads
Here we can start analysis from the free end.

Section A-B

So for between A and B, if you take a section the only external force acting on it is F1. So a shear force should induce in section to balance this force. So value of shear force between A and B is F1. But force balance alone does not guarantee equilibrium of the section. There is an external moment on the section. So a bending moment will be induced in section, in order to balance the external moment. Since value of external moment is F into x, bending moment will vary linearly.

Section B-C

Between B and C effect of force F2 also comes. So shear force becomes, F1 plus F2. And in bending moment effect of F2 also gets added. Similar analysis is done between section C and D also. So SFD and BMD of this problem would look like this.

Fig.9 SFD and BMD of cantilever beam

Simply Supported Case

Now consider this problem. A simply supported beam with uniformly distributed load. First step here would be determination of reaction forces. Since the problem is symmetrical reaction forces will be equal, and will be half of total load acting on beam.

Fig.10 A simply supported beam with uniformly distributed load in it

Section A-B

Lets start analysis from point A. If you take section between point A and B, it should be in equilibrium. So shear force will have equal magnitude of Reaction force. Bending moment gives a linear variation.

Section B-C

But after point B, effect of point load and distributed load come. Effect of distributed load is something interesting. Take a section in BC. In this section, along with two point loads there is a distributed load also. This distributed load can be assumed as a point load passing through centroid of distributed load. Value of point load is U( x - L/3). And it is at a distance (x - L/3)/ 2 from section line. So shear force will have one more term, which comes from distributed load. From the equation its clear that shear force varies linearly.

You can easily predict, how bending moment varies along length, from the same force diagrams. Since this equation is quadratic it will have a parabolic shape. Same procedure is repeated in remaining section. Since this problem is symmetrical in nature, S.F.D and B.M.D would also be symmetrical. It is shown in figure below.

Fig.11 SFD and BMD of simply supported beam

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  1. How do Wind Turbines work ?

  2. Wind Turbines are the forerunners of clean energy. A huge amount of investment have been gone into this technology in the last decade. This articles illustrates working and design aspects of wind turbine in a detailed way.

  3. How does a Centrifugal pump work?

  4. Centrifugal pumps are the most preferred hydraulic pumps used in domestic and industrial world. They are used to induce flow or raise pressure of a liquid. In this article we will have a conceptual overview on working of centrifugal pumps.

  5. Working of Francis Turbine

  6. Francis turbines are the most preferred hydraulic turbines. They are the most reliable workhorse of hydroelectric power stations. It contributes about 60 percentage of the global hydropower capacity, mainly because it can work efficiently under a wide range of operating conditions. This article is aimed at giving a conceptual overview of working of Francis turbine.

  7. Wind Turbine Design

  8. Primary objective in wind turbine design is to maximize the aerodynamic efficiency, or power extracted from wind. But this objective should be met by well satisfying mechanical strength criteria and economical aspects. In this article we will see impact of number of blades, blade shape, blade length and tower height on wind turbine design.

  9. Fundamentals of Turbomachinery

  10. Turbomachinery devices inject life into fluids. Principles of turbomachinery form the preliminary design tools in design of them. In this article we will see how these principles are developed and how it is used in predicting performance of turbomachinery devices.

  11. Pelton Turbine | Working & Design Aspects

  12. Pelton turbines/wheels are suitable for power extraction, when water energy is available at high head and low flow rate. They absorb energy purely due to impulse action of water jet. In this video we will go through working principle and design aspects of Pelton turbine.

  13. Working of Kaplan Turbine

  14. Kaplan turbines derive motive force from pure reaction. They work efficiently when there is a huge water flow available. Working and design principles of Kaplan turbine are discussed elaborately in this article.

  15. Centrifugal Pumps | Design Aspects

  16. In this lecture we will learn design aspects of centrifugal pumps. More precisely we will learn how to select a centrifugal pump and motor for pumping fluid at a specified rate, for a given system.

  17. Working of Centrifugal Pumps

  18. Centrifugal pumps are most commonly used turbo machinery devices, which are used to raise pressure, or induce flow in a control volume. They are radial flow devices. Various kinds of centrifugal pumps are available in market with different construction details. But working principle behind all of them remain same. In this video we will analyze working principle of a centrifugal pump with single suction, semi open impeller.

  19. Working Principles of a Steam Turbine

  20. Steam turbines are heart of power plant, they are the devices which transform thermal energy in fluid to mechanical energy. In this video lecture working of steam turbine is explained in a logical manner.

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Pelton Turbine - Working & Design Aspects

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Pelton turbines/wheels are suitable for power extraction, when the water energy is available at high head and low flow rate. In this video we will go through working principle and design aspects of Pelton turbine.

Following article gives detailed description of the video lecture.

Pelton Turbine – The Basic Working Principle

Working principle of Pelton turbine is simple. When a high speed water jet injected through a nozzle hits buckets of Pelton wheel; it induces an impulsive force. This force makes the turbine rotate. The rotating shaft runs a generator and produces electricity.

Fig.1 Pelton wheel derives rotation from impulse force produced by the water jet
In short, Pelton turbine transforms kinetic energy of water jet to rotational energy.

Governing in Pelton Wheel

Demand of power may fluctuate over time. A governing mechanism which controls position of the spear head meets this requirement. With lowering power demand the spear head at water inlet nozzle is moved in. So that water flow rate is reduced. If power demand increases spear head is moved out this will increase the flow rate. Following figure illustrates this mechanism. The first position of the spear head produces a low flow rate, while the second position produces a high flow rate.

Fig.2 Water flow rate control in Pelton wheel by monitoring position of spear head
So in Pelton turbine synchronization between power demand and power supply is met by controlling the water flow rate. The same technique is used in other types hydroelectric turbines. If the power supply is more than the demand, then the turbine will run over speed otherwise in under speed. But such a governing mechanism in turn will balance the power supply and demand and will make sure that the turbine rotates at a constant specified RPM. This speed should also conforms to the power supply frequency. So this mechanism acts as a speed governing mechanism of Pelton wheel.

Number of Buckets in Pelton Wheel

One of the most important parameter of Pelton turbine design is number of buckets on the disk. If number of buckets is inadequate, this will result in loss in water jet. That means when one bucket departs from the water jet next bucket may not get engaged with the jet. This will result in loss in water jet for a small time duration, thus sudden drop in turbine efficiency. Following figure illustrates what happens when the number of buckets are lowered. With lowering number of buckets at some point of operation, complete water jet might be lost (3rd figure). So there should be an appropriate number of buckets, which will make sure that no water is lost (1st figure).

Fig.3 Effect of number of buckets on water-bucket interaction

Pelton Bucket - Design & Features

Most vital component of Pelton wheel is its bucket. Buckets are casted as single solid piece, in order to avoid fatigue failure. You can note that force acting on the turbine bucket is not constant with time. If you follow one particular bucket, it will have high force for a small time duration (at the time of jet impingement) after that a larger idle period where no jet interaction takes place. So the force acting on the bucket is also not constant. It varies with the time but it is having a cyclic nature. If bucket were made using pieces by welding attachment such cyclic fore will easily lead to premature fatigue failure.

Fig.4 Different views of Pelton bucket

Water jet is split into 2 equal components with help of a splitter. The special shape of bucket makes the jet turn almost 180 degree. This produces an impulsive force on bucket. Force so produced can easily be derived from Newton’s 2nd law of motion. Blade outlet angle close to 180 degree is usually used in order to maximize impulsive force.

A cut is provided on bottom portion of buckets. This makes sure that water jet will not get interfered by other incoming buckets.

Pelton – An Impulse Turbine

Since the water jet is always open to atmosphere, inlet and exit pressure of water jet will be same and will be same as atmospheric pressure. However absolute velocity of fluid will have huge drop from inlet to exit of bucket. This kinetic energy drop is the maximum energy the bucket can absorb.

So it is clear that Pelton turbine gains mechanical energy purely due to change in kinetic energy of jet, not due to pressure energy change. Which means Pelton turbine is a pure impulse machine.

Fig.5 Pressure and velocity variation across Pelton bucket
Impulse force produced by water jet is high when jet is having high velocity. Water stored at high altitude can easily produce high jet velocity. This is the reason why Pelton turbine is most suitable for operation, when water is stored at high altitude.

You can easily understand why there is a nozzle fitted at water jet injection portion. Nozzle will increase velocity of jet further, thus will aid in effective production of impulse force.

Extracting Maximum Power from Water Jet

Pelton turbine design is always aimed at extracting maximum power from water jet, or maximizing efficiency. Power extracted by the bucket, P is product of jet impulse force and bucket velocity.

So power extraction is maximum when product of impulsive force and bucket velocity is maximum. Let's consider 2 different operating conditions.

  • Buckets are Held Stationary

  • If Pelton wheel buckets are held stationary, there will be a huge impulse force produced. But power extraction will be zero since buckets are not moving.

  • Bucket Speed Same as Jest Speed

  • If buckets are moving with same speed of jet, water jet won't be able to hit the bucket. This will lead to zero impulse force. Again power extraction will be zero.

Fig.6 Relative magnitude of bucket and jet velocity is important in power extraction from fluid
In short, power extraction is zero both at zero bucket speed and when bucket speed is same as jet speed. So with respect to jet to bucket speed ratio, power extraction will vary with as shown below.
Fig.7 This graph shows how power extraction from fluid varies with respect to bucket to jet velocity ratio
It is clear from the above graph that optimum power extraction happens in between. It can be shown using Euler's turbo machinery equation that maximum power extraction happens when bucket speed is half the jet velocity. So it is always desirable to operate Pelton wheel at this condition. Pelton turbines can give efficiency as high as 90 %, at optimum working conditions.

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