NATURE OF PLASTIC DEFORMATION

NATURE OF PLASTIC DEFORMATION

The change of any dimension or shape of an object under the action of external forces is generally considered as a deformation.

When external forces are applied on an object, then the deformation along the direction of the applied force is called longitudinal deformation where as any deformation along its transverse directions are called as lateral deformations.

When external forces are applied in an object, the object will be deformed first. Due to this deformation crystal structure is also deformed and thus creating an unbalanced internal resisting force, which neutralizes the external force and a condition of equilibrium is achieved as deformation stopped.

When the deformation per unit length is small, the material shows a remarkable ability to recover its original shape and size as the external forces are removed.

Hence, as the external force is withdrawn, the deformation will be vanished.

This type of deformation is called elastic deformation and this property of the material is known as Elasticity.

During the elastic phase of deformation, no permanent change in crystal structure happens, but as the magnitude of the applied force increases, resistance due to the change or distortion of the crystal structure becomes insufficient and as a result crystal dislocation occurs.

Plastic deformation is the deformation which is permanent and beyond the elastic range of the material. Very often, metals are worked by plastic deformation because of the beneficial effect that is imparted to the mechanical properties by it.

The necessary deformation in a metal can be achieved by application of large amount of mechanical force only or by heating the metal and then applying a small force.

The deformation of metals which is caused by the displacement of the atoms is achieved by one or both of the processes called slip and twinning. These two are the prominent  mechanisms  of  plastic  deformation,  namely  slip  and  twinning.

SLIP AND TWINNING

Slip is the prominent mechanism of plastic  deformation in metals. It involves sliding of blocks of crystal over one other along definite  crystallographic planes, called slip planes.

It is analogous to a deck of cards when it is pushed from one end. Slip occurs when shear stress  applied exceeds a critical value. 

During  slip  each  atom  usually  moves  same  integral  number  of  atomic  distances  along  the  slip  plane  producing a  step  but  the  orientation  of  the  crystal remains the same.

 Generally  slip  plane  is  the  plane  of  greatest  atomic  density, and  the  slip  direction  is  the   close  packed  direction  within  the  slip  plane.

Twining : Portion  of  crystal  takes  up  an  orientation  that  is  related  to  the  orientation  of  the  rest  of  the  untwined  lattice  in  a  definite, symmetrical  way.

The  twinned  portion  of  the  crystal  is  a  mirror  image  of  the  parent  crystal.

The plane of symmetry is called twinning plane.

The  important role of twinning in plastic deformation is that it causes changes in plane orientation so that further slip can occur.

On the macroscopic scale when plastic deformation occurs, the metal appears to flow in the solid state along specific directions which are dependent on the type of processing and the direction of applied force.

The crystals or grains of the metal get elongated in the direction of metal flow. This flow of metal can be seen under microscope after polishing and suitable etching of the metal surface. These visible lines are called as “fibre flow lines".

Since the grains are elongated in the direction of flow, they would be able to offer more resistance to stresses acting across them. As a result, the mechanically worked metals called wrought products would be able to achieve better mechanical strength in specific orientation, that of the flow direction.

Since it is possible to control these flow lines in any specific direction by careful manipulation of the applied fibres. It is possible to achieve optimum mechanical properties.

The metal of course, would be weak along the flow lines. The wastage of material in metal working processes is either negligible or very small and the production rate is in general very high. These two factors give rise to the economy in production.

HOT WORKING AND COLD WORKING

The metal working processes are traditionally divided into hot working and cold working processes.

The division is on the basis of the amount of heating applied to the metal before applying the mechanical force. Those processes, working above the recrystallisation temperature, are termed as hot working processes whereas those below are termed as cold working processes.

Under the action of heat and the force, when the atoms reach a certain higher energy level, the new crystals start forming which is termed as recrystallisation.

Recrystallisation destroys the old grain structure deformed by the mechanical working, and entirely new crystals which are strain free are formed.

The grains in fact start nucleating at the points of severest deformation.

Recrystallisation temperature as defined by American Society of Metals is "the approximate minimum temperature at which complete recrystallisation of a cold worked metal occurs within a specified time".

The recrystallisation temperature is generally between one-third to half the melting point of most of the metals. The recrystallisation temperature also depends on the amount of cold work a material has already received. Higher the cold work, lower would be the recrystallisation temperature as shown in Fig. given below.

Though cold work affects the recrystallisation temperature to a great extent, there are other variables which also affect the recrystallisation temperature

In hot working, the process may be carried above the recrystallisation temperature with or without actual heating.

For example, for lead and tin the recrystallisation temperature is below the room temperature and hence working of these metals at room temperature is always hot working. Similarly for steels, the recrystallisation temperature is of the order of 1000^oC, and therefore working below that temperature is still cold working only.

In hot working, the temperature at which the working is completed is important since any extra heat left after working will aid in the grain growth, thus giving poor mechanical properties.

The effect of temperature of completion of hot working is profound. A simple heating where the grain start growing after the metal crosses the recrystallisation temperature. But, if it is cooled without any hot working, the final grain size would be larger than the grain size in the initial stage of heating.

Again, after heating, if the metal is worked before cooling the result is the reduction in size. It is due to the process of recrystallisation, that new grain will be started to form and the final grain size is reduced. This phenomena rises due to working of metal at recrystallisation, that gives rise to a large number of nucleation sites for the new crystals to form.

But if the hot working is completed much above the recrystallisation temperature the grain size start increasing and finally may end up with coarse grain size.

This increase the size of the grains occurs by a process of coalescence of adjoining grains and is a function of time and temperature.

This is not generally desirable. If the hot working is completed just above the recrystallisation temperature, then the resultant grain size would be fine. The same is schematically shown for hot rolling operation.

METAL FORMING PROCESS

METAL FORMING PROCESSES

Have you ever wonder how does a car door or a Saucepan or most of the kitchen utensils are manufactured? All of them are produced by a group of similar manufacturing processes governed by the same physical laws where large forces are applied to deform sheets of metal into a desired shapes and sizes and we classify them as Metal forming processes.

Metal forming is a general term for a large group of processes, that includes a wide variety of manufacturing processes.

Forming is the process in which the desired size and shape are obtained through the plastic deformation of a materials, i.e., when a metal is plastically or permanently deformed under the action of a strong applied force, it is in general classified as a metal forming process.

There are different types of metal forming processes, although in each case the basic principle remains same i.e. permanent deformation also known as plastic deformation of metal by the application of large force. Therefore, we can define forming , or metal forming, as the metalworking process of fashioning metal parts and objects through mechanical deformation. In this process the workpiece is reshaped without adding or removing material, and its mass remains unchanged. Forming operates on the materials science principle of plastic deformation, where the physical shape of a material is permanently deformed.

The characteristic of metal forming process is that the metal being processed is plastically deformed to shape it into a desired geometry. In order to plastically deform a metal, a force must be applied that will exceed the yield strength of the material.

When low amounts of stress are applied to a metal it will change its geometry slightly, in correspondence to the force that is exerted. Basically it will compress, stretch, and/ or bend a small amount.

The magnitude of the amount will be directly proportional to the force applied. Also the material will return to its original geometry once the force is released.

Just think of stretching a rubber band, then releasing it, and having it go back to its original shape. This is called elastic deformation. Once the stress on a metal increases past a certain point, it no longer deforms elastically, but starts to undergo plastic deformation.
In plastic deformation, the geometric change in the material is no longer directly proportional to stress and geometric changes remain after the stress is released; meaning that the material does not recover its shape.

The actual level of stress applied to a metal where elastic deformation turns to plastic deformation is called the proportional limit, and is often difficult to determine exactly.

The .002 offset convention is usually used to determine the yield point, which is taken for practical purposes as the stress level where plastic deformation, (yielding), begins to occur.

Stress and Strain Curve

It can be seen by the stress-strain graph that once the yield point of a metal is reached and it is deforming plastically, higher levels of stress are needed to continue its deformation. The metal actually gets stronger, the more it is deformed plastically. This is called strain hardening or work hardening.

As may be expected, strain hardening is a very important factor in metal forming processes. Strain hardening is often a problem that must be overcome, but many times strain hardening, when used correctly, is a vital part of the manufacturing process in the production of stronger parts.
Compared to other metalworking processes like cutting and joining metal forming tends to have more uniform characteristics across its different sub processes.

Characteristics of Metal Forming Processes

Metal forming processes are characterized by:

Very high required loads and stresses, between 50 MPa and 2500 MPa

Large, heavy, and expensive machinery in order to accommodate such high stresses and loads

Production runs with many parts, to maximize the economy of production and compensate for the expense of the machine tools.

THE CONCEPT OF FLOW STRESS

During a metal forming operation, it is important to know the force and power that will be needed to accomplish the necessary deformation. The stress- strain graph shows us that the more a work piece is deformed plastically, the more stress is needed.
The flow stress is the instantaneous value of the force necessary to continue the yielding and flow of the work material at any point during the process.
Flow stress can be considered as a function of strain. The flow stress value can be used to analyze what is going on at any particular point in the metal forming process.
The maximum flow stress may be a critical measurement in some metal forming operations, since it will specify the maximum force and power requirements for the machinery to perform the process.
The force needed at the maximum strain of the material would have to be calculated in order to determine maximum flow stress.
For different types of metal forming processes, the flow stress analysis may be different. For a process like forging, the maximum flow stress value would be very important. However, for a process like extrusion, where the metal is continuously being deformed and the different stages of the metal's deformation are occurring simultaneously, it is of interest to analyze the mean flow stress value.

THE CONCEPT OF STRAIN RATE

The strain rate for any particular manufacturing metal forming process is directly related to the speed at which deformation is occurring. A greater rate of deformation of the work piece will mean a higher strain rate. The specific process and the physical action of the equipment being used has a lot to do with strain rate. Strain rate will affect the amount of flow stress. The effect strain rate has on flow stress is dependent upon the metal and the temperature at which the metal is formed.

EFFECT OF TEMPERATURE ON METAL FORMING

Properties of a metal change with an increase in temperature. Therefore, the metal will react differently to the same manufacturing operation if it is performed under different temperatures and the manufactured part may posses different properties. For these reasons, it is very important to understand the materials that we use in our manufacturing process.

This involves knowing their behavior at various temperature ranges. In industrial metal forming manufacture, there are three basic temperature ranges at which a metal is mechanically worked or formed (which is known as metal forming process), the metal forming processes can be classified as,

COLD FORMING

HOT FORMING

WARM OR SEMI HOT FORMING

COLD FORMING / COLD WORKING OF METAL

Cold working, (or cold forming), is a metal forming process that is carried out at room temperature or a little above it.

In cold working, plastic deformation of the work causes strain hardening as discussed earlier. The yield point of a metal is also higher at the lower temperature range of cold forming. Hence, the force required to shape a part is greater in cold working than for warm working or hot working.

At cold working temperatures, the ductility of a metal is limited, and only a certain amount of shape change may be produced. Surface preparation is important in cold forming. Fracture of the material can be a problem, limiting the amount of deformation possible.

In fact, some metals will fracture from a small amount of cold forming and must be hot formed. One main disadvantage of this type of process is a decrease in the ductility of the part's material, but there are many advantages. The part will be stronger and harder due to strain hardening.

Cold forming causes directional grain orientation, which can be controlled to produce desired directional strength properties. Also, work manufactured by cold forming can be created with more accurate geometric tolerances and a better surface finish. Since low temperature metal forming processes do not require the heating of the material, a large amount of energy can be saved and faster production is possible. Despite the higher force requirements, the total amount of energy expended is much lower in cold working than in hot working.

WARM FORMING/ WORKING

Warm working, (or warm forming), is a metal forming process carried out above the temperature range of cold working, but below the recrystallization temperature of the metal. Warm working may be preferred over cold forming because it will reduce the force required to perform the operation. Also, the amount of annealing of the material that may have been necessary for the cold formed part may be less for warm working.

HOT FORMING OR HOT WORKING OF METAL

Hot working, (or hot forming), is a metal forming process that is carried out at a temperature range that is higher than the recrystallization temperature of the metal being formed.

The behavior of the metal is significantly altered, due to the fact that it is above its recrystallization temperature. Utilization of different qualities of the metal at this temperature is the characteristic of hot working.

Although many of these qualities continue to increase with increasing temperature, there are limiting factors that make overly high temperatures undesirable.

During most metal forming processes the die is often cold or slightly heated. However, the metal stock for hot working will usually be at a higher temperature relative to the die.

In the design of metal forming process, it is critical to consider the flow of metal during the forming of the work. Specific metal flow, for different forming processes, is discussed in latter sections under each specific process. For metal forming manufacturing, in general, the temperature gradient between the die and the work has a large effect on metal flow during the process.

The metal nearer to the die surfaces will be cooler than the metal closer to the inside of the part, and cooler metal does not flow as easily. High temperature gradients, within the work, will cause greater differences in flow characteristics of different sections of the metal, these could be problematic. For example, metal flowing significantly faster at the center of the work compared to cooler metal near the die surfaces that is flowing slower, can cause part defects. Higher temperatures are harder to maintain throughout the metal forming process. Work cooling during the process can also result in more metal flow variations.

Another consideration with hot forming manufacture, with regard to the temperature at which to form the part, is that the higher the temperature the more reactive the metal is likely to be. Also if a part for a hot working process is too hot then friction, caused during the process, may further increase heat to certain areas causing melting, (not good), in localized sections of the work. In an industrial hot metal working operation, the optimum temperature should be determined according to the material and the specific manufacturing process. When above its recrystallization temperature a metal has a reduced yield strength, also no strain hardening will occur as the material is plastically deformed.

Shaping a metal at the hot working temperature range requires much less force and power than in cold working. Above its recrystallization temperature, a metal also possesses far greater ductility than at its cold worked temperature. The much greater ductility allows for massive shape changes that would not be possible in cold worked parts. The ability to perform these massive shape changes is a very important characteristic of these high temperature metal forming processes.

The work metal will recrystallize, after the process, as the part cools. In general, hot metal forming will close up vacancies and porosity in the metal, break up inclusions and eliminate them by distributing their material throughout the work piece, destroy old weaker cast grain structures and produce a wrought isotropic grain structure in the part.

These high temperature forming processes do not strain harden or reduce the ductility of the formed material. Strain hardening of a part may or may not be wanted, depending upon the application. Qualities of hot forming that are considered disadvantageous are poorer surface finish, increased scale and oxides, decarburization, (steels), lower dimensional accuracy, and the need to heat parts. The heating of parts reduces tool life, results in a lower productivity, and a higher energy requirement than in cold working.

Selection Of Temperature Range For A Metal Forming Operation

Production at each of these temperature ranges has a different set of advantages and disadvantages. Sometimes, qualities that may be undesirable to one process may be desirable to another. Also, many times work will go through several processes. The goal is to design the manufacture of a part in such a way as to best utilize the different qualities to meet or enhance the specifications of the part. To produce a strong part with excellent surface finish, then a cold forming process could be a good choice. However, to produce a part with a high ductility a hot forming process may be best. Sometimes the advantages of both hot forming and cold forming are utilized when a part is manufactured by a series of processes. For example, hot working operations may first be performed on a work piece to achieve large amounts of shape change that would not be possible with cold forming due to strain hardening and limited ductility. Then the last process that completes the manufacture of the part is a cold working operation. This process does not require a significant shape change, since most of the deformation was accomplished by the hot forming process. Having a cold forming process last will finish the shape change, while strengthening the part, giving a good surface finish and highly accurate tolerances.

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Personally, I was never a believer of the Cramming Process, although I know language problems often force a student to take the help of Cramming. But, to get rid of this tandency, we wanted to establish a learning center near the place where students reside. Partially it will reduce their travelling time and get more time to study.
As per our requirement Jindal Nagar becomes the choice of our place. When we enquired about a suitable place, then we came to contact with UNIQUE CLASSES in JINDAL NAGAR. We liked the place and overtaken the UNIQUE CLASSES and merged our venture CRACKGATE EDUCATION. The new entity will be named as UNIQUE CLASSES and it will be run by us.

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I C Engines & Comptessors (5th Sem)

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Subhankar Karmakar

SAMPLE SHEET: GATE 2015; STRENGTH OF MATERIALS (MECHANICAL ENGINEERING)

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MECHANICAL ENGINEERING

TOPIC: STRENGTH OF MATERIALS

Difficulty Level: 1

SET ONE: Each question has several entries, choose the most appropriate one

01)  The intensity of stress which causes unit strain is called

            a) unit stress                                                     b) bulk modulus

            c) modulus of elasticity                                               d) principal stress

02)  Which of the following materials has poisson’s ratio more than unity

            a) steel                                                                         b) copper

            c) cast iron                                                       d) none of these

03) The change in the unit volume of a material under tension with increase in its Poisson’s ratio will

            a) increase                                                       b) decrease

            c) increase initially and then decrease              d) remain same

04) In a tensile test, near the elastic zone, the tensile strain

            a) increases more quickly                                b) decreases more quickly

            c) increases in proportion to the stress                         d) increases more slowly

05) The stress necessary to initiate yielding is

            a) considerably greater than that necessary to continue it

            b) considerably lesser than that necessary to continue it

            c) remain same to continue it

            d) can’t be predicted

06) Flow stress corresponds to

            a) fluids in motion                                           b) breaking point

            c) plastic deformation of solids                                   d) rupture stress

07) The maximum strain energy that can be stored in a body is known as

            a) impact energy                                              b) resilience

            c) proof resilience                                            d) modulus of resilience

08) Thermal stress is always

            a) tensile                                                          b) compressive

            c) tensile or compressive                                 d) none of these

09) The loss of strength in compression due to overloading is known as

            a) hysteresis                                                     b) relaxation

            c) creep                                                            d) Bouschinger effect

10) If a material expands freely due to heating, it will develop

            a) thermal stress                                               b) lateral stress

            c) creep stress                                                  d) no stress

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TOPIC: FLUID MECHANICS

Difficulty Level: 1

SET ONE: Each question has several entries, choose the most appropriate one

01) If no resistance is encountered due to displacement then such a substance is called

            a) fluid               b) real gas                 c) ideal fluid          d) visco-elastic fluid

02) The volumetric change of a fluid due to a resistance is called as

            a) volumetric strain,                             b) volumetric index

            c) compressibility                                d) cohesion

03) Mercury does not wet glass due to a property of liquids known as

            a) adhesion                                          b) cohesion

            c) viscosity                                          d) surface tension

04) The surface tension of Mercury at normal temperature compared to that of water is

            a) more                                                          b) less

            c) depends upon the glass tube                        d) equal

05) Kinematic viscosity is dependent upon

            a) pressure       b) distance       c) density         d) flow

06) Alcohol is used in manometers because

            a) it is clearly visible                            b) it provides suitable meniscus

            c) it can provide longer column due to its low density

            d) it has low surface tension

07) The buoyancy depends upon

            a) mass of liquid displaced                  b) viscosity of the liquid

            c) pressure of the displaced liquid       d) none of the above

08) The centre of gravity of the volume of the liquid displaced by an immersed body is called

            a) meta-centre                                      b) centre of gravity

            c) centre of pressure                            d) centre of buoyancy

09) Surface energy per unit area of a surface is numerically equal to

            a) atmospheric pressure                                   b) surface tension

            c) force of cohesion                                         d)  viscosity

10) Flow occurring in a pipeline when a valve is being opened is

            a) steady flow                                                 b) unsteady flow

            c) laminar flow                                                d) vortex flow

For answers contact at 09555921800

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CRACKGATE EDUCATION

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MECHANICAL ENGINEERING

TOPIC: BASIC THERMODYNAMICS

Difficulty Level: 1

SET ONE: Each question has several entries, choose the most appropriate one

01) Gas laws are applicable to

            a) Gases as well as vapours     b) Gases alone and not applicable to vapours

            c) Gases and Steam                  d) Gases and Superheated vapours

02) An ideal gas compared to a real gas at very high pressures occupies

            a) more volume           b) less volume            

c) same volume           d) can’t be predicted

03) Temperature of a gas is produced due to

            a) its heating value                   b) kinetic energy of the molecules

            c) molecular vibration              d) inter-molecular attractions

04)  According to kinetic theory of gases, the absolute zero temperature is attained when

            a) volume of the gas is zero                             b) pressure of the gas is zero

            c) kinetic energy of the molecule is zero                     d)  specific heat is zero 

05)  The quantity δQ – δW; where δQ is elemental heat transfer and δW is the elemental work    

transfer is

            a) path function                       b) point function

            c) cyclic function                     d) in-exact differential

06)  The workdone in an adiabatic process between a given pair of end states depends on

            a) the values of the endstates only,      b) the end states and specific heat ratio  γ

            c) the end states and polytropic index n,          d) none of the above.

07)  If the value of polytropic index n is high, then the compressor work between given pressure limits will be

            a) less                                      b) more

            c) no effect                              d) zero

08)  A perfect gas at 27^oC is heated at constant pressure till its volume becomes double. The final temperature will be

            a) 54^oC                                    b) 327^oC

            c) 108^oC                                  d) 600^oC

09)  Mixing of ice and water at 0^oC at atmospheric pressure is an example of

            a) reversible process                b) irreversible process

            c) quasi-static process              d) isentropic process

10)  Change in enthalpy in a closed system is equal to heat transferred if the reversible process takes place at constant

            a) pressure                               b) temperature

            c) volume                                d) entropy

GATE 2015: MECHANICAL ENGINEERING COACHING IN CHIRANJEEV VIHAR AND GOVINDPURAM GHAZIABAD

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अर्धबार्षिक परीक्षा प्रश्न पत्र  2013 

इंजीनियरिंग ड्राइंग 

प्रथम वर्ष  यांत्रिक इंजीनियरिंग 

___________________________________________________________________

[ समय : 2 घंटा ]                                                                                   पूर्णांक : 30 

               नोट : किन्ही तीन प्रश्नो के उत्तर दीजिये। प्रश्न संख्या 4 अनिवार्य है। 

                         प्रत्येक प्रश्न के अंक प्रश्न के सामने अंकित है। 

               1 ) 10 mm ऊंचाई के एकल प्रयास उर्ध्वाधर कैपिटल अक्षरों मैं निम्नलिखित वाक्य को        

                    स्वच्छता पूर्वक लिखिए।                                                                        (7. 5 )

                         "CIVIL ENGINEERING IS THE BACKBONE OF CIVILIZATION."

               2 ) एक रेखा AB , HP से 60^o का कोण बनाती है, तथा VP से 25 mm पीछे समानांतर है।  
                    यदि रेखा के सिरा 'A' H.P. से 25 mm निचे है , तथा रेखा की लम्बाई 50 mm है , तब रेखा 
                   के Front  View एबं Top View बनाइये।                                                    (7. 5 )

               3) (1 /50) निरूपक भिन्न (R. F.) वाली एक निकर्ण मापनी बनाइये जिस पर मीटर , 
                   डेसीमीटर तथा सेन्टीमीटर का मापन किया जा सके तथा 9 मीटर लम्बाई तक पढ़ने के 
                   लिए पर्याप्त हो। मापनी पर 6.47 मीटर की दूरी चिह्नित कीजिये।            (7. 5 )


               4) चित्रों मई एक समपरिमन दृश्य (Isometric View) दर्शाया गया है , दी गई दिशाओं से 
                   देखकर इनका अनुविक्षेप (Plan/ Top View) एबं  उत्सेध (Front View) तथा पार्श्व दृश्य 
                   ( Side View ) को III^rd Angle मैं बनाइये।                                                ( 15 )

ME-301: THERMODYNAMICS FOR THIRD SEMESTER; UPTU

SECTION A: Each question carry 2 marks

01) What is point function and path function?

02) Define Enthalpy.

03) What is SFEE?

04) What is internal energy of a system

05) What is vapour dome and dryness factor?

06) What is saturated liquid line and saturated vapour line?

07) What is triple point of water?

08) Define specific heats of ideal gases.

09) Write the reduced form of Vander Waals equation for real gases.

10) Define a thermodynamic system.

11) State with reasoning whether the following systems are closed, open or isolated
i) Refrigerator; ii) Pressure Cooker

12) Distinguish between isolated system and adiabatic system.

13) Explain the concept of flow work

14) What is control volume and control surface?

15) When does a real gas behave like an ideal gas?

16) What are extensive and intensive properties?

17) What is enthalpy of evaporation of steam?

18) Define degree of under-cooling and degree of superheat.

19) What is COP of a heat pump?

20) Define throttling process.

21) Define entropy.

22) Two moles of an ideal gas occupy a volume of 4.24 m³ at 400 K temperature. Find the pressure exerted by the gas.

23) Distinguish between refrigerator and heat pump.

24) What is Free Expansion?

25) Explain Zeroth law of thermodynamics.

26) Explain the concept of compressibility factor?

27) What is PMM-1.

28) What is a superheated steam?

29) Distinguish between universal gas constant and characteristic gas constant.

30) What is Exergy?

31) Distinguish between quasi-static process and reversible process.

32) What is a diathermal system boundary?

33) What is a steady flow open system?

34) What is the difference between latent heat and sensible heat?

35) What is a thermodynamic cycle?

36) Distinguish between restraint and unrestraint process.

37) What is a thermodynamic definition of work?

38) What is work of evaporation?

39) What is a pure substance?

40) What is the concept of continuum?

41) Define thermodynamic state, process and path.

42) Distinguish between thermal equilibrium and thermodynamic equilibrium.

43) What are the conditions for reversible process?

44) Distinguish between heat and work.

45) What are the differences between gas and vapour?

SECTION B: Attempt any three of the following questions. Each question contains two parts of 5 marks each. Total marks In this section is 3x10 = 30

01) a) State Zeroth law of thermodynamics and explain how it leads to the concept of temperature.

b) Explain different types of temperature scale and the relations among them.

02) a) Explain the corollaries of first law of thermodynamics.

b) 2 kg of air is confined in a rigid container of 0.42 m³ at 4 bar pressure. When heat energy of 164 kJ is added, its temperature becomes 127°C.
Find :

i) Work done by the system.
ii) Change in internal energy.
iii) Specific heat at constant volume.

03) a) Derive an expression for heat transfer and work done in a polytropic process.

b) 1.5 kg of oxygen contained in a cylinder at 4 bar pressure and 300 K expands three times its original volume in a constant pressure process. Determine
i) Initial volume, ii) Final temperature, iii) Work done by the gas, iv) Heat added and v) Change in internal energy.
; Assume C_p = 1.005 kJ/kg-K and R = 260 J/kg-K

04) a) Make steady flow energy analysis on a turbine.

b) Find the velocity and diameter at exit of a nozzle if 5 kg/s air at 9 bar and 200°C expands through the nozzle up to pressure at 1.1 bar. Approach velocity is 50 m/s.

05) a) Differentiate between absolute pressure and gauge pressure. What is a manometer?

b) An ideal gas of molecular weight 42.4 has a pressure of 10 bar and occupies a volume of 0.3 m³ at 27°C. Determine the characteristic gas constant for the ideal gas, its mass and number of moles.

06) a) Write the first law of thermodynamics for a flow process. Derive an expression for flow work.

b) Find the total work done and efficiency for a reversible Carnot cycle.

07) a) What is continuity equation in flow process?

b) 3 kg air at 2 bar pressure and 27℃ temperature has been compressed isothermally till the pressure reaches 6 bar. Next it has been heated at constant pressure and thereafter reaches the initial state by expanding adiabatically. Find the maximum Temperature reached in the cycle and total work done by the system.

08) a) Explain the Joule's experiment.

b) Prove that internal energy is a point function.

09) a) What is thermodynamic temperature scale?

b) i) A heat engine running between 300 K and 800 K generates 2000 kJ of energy. Find the total heat extracted from the source.
ii) Determine the power required to run a refrigerator that transfers 2000 kJ/min of heat from a cooled space at 0°C to the surroundings atmosphere at 27°C.

10) a) What is PMM - 2? State Kelvin-Planck statement of second law of thermodynics.

b) A heat engine running between two thermal reservoirs of 800 K and 300 K is used to power a refrigerator running between two thermal reservoirs of 325 K and 260 K. If the heat engine draws 5000 kJ heat from reservoirs at 800 K, then find the amount of heat extracted from 260 K reservoir by the refrigerator.

11) a) Explain the Vander Waal's gas equation.

b) 4 kg of steam at 16 bar occupies a volume of 0.28 m³. The steam expands at constant volume to a pressure of 8 bar. Determine the final dryness fraction, final internal energy and change in entropy.

12) a) Explain and derive Clausius Inequality.

b) 3 kg of air is heated reversibly at constant pressure of 2.5 bar from 23°C to 227°C. If the lowest available temperature is 20°C determine the increase in the available energy of air due to heating. Take C_p = 1.005 kJ/kg-K.

13) a) What is thermodynamic definitions of work? Distinguish between ∫pdV work and other types of work.

b) 3 kg of air at 1.5 bar pressure and 350 K is compressed isothermally to a pressure of 6 bar. Then heat of 350 kJ is added at constant volume. What will be the maximum temperature of air during the process? Find the total work done in the processes. Also find the change in internal energy of air.

14) a) Write the limitations of second law of thermodynamics. Prove that C_p - C_v = R

b) 10 kg of air at 300 K is stored in a totally insulated cylinder of volume 0.3 m³/kg. If 1 kg air has been taken out of the system, then what will be the value of new pressure?

15) a) Steam at 1.2 bar and a dryness fraction of 0.5 is heated at constant pressure until it becomes saturated vapour. Calculate the heat transferred per kg of steam.

b) Steam at 8 MPa and 500°C passes through a throttling process such that the pressure is suddenly dropped to 0.3 MPa. Find the expected temperature after throttling.

16) a) What are the causes of irreversibility?

b) Distinguish between a quasi-static process and reversible process.

17) a) 3 kg of air at 400 K and 4 bar pressure adiabatically mixed with 4 kg of air at 500 K and 3 bar pressure. Find the change in entropy of the universe.

b) Explain the principle of increase of entropy.

SECTION C : marks 50, 5 questions of 10 mark each. Each question contains 3 parts. Attempt any two parts out of three from each question.

01) a) Steam at 20 bar pressure and 300°C expands isentropically in a turbine to a pressure of 2 bar. Find the final condition of the steam. Also Calculate the work delivered by the turbine.

b) What is isentropic efficiency of a turbine? Calculate internal energy of steam at 6 bar pressure and 300°C.

c) Explain the steam formation process at constant pressure.

02) a) What is adiabatic mixing of two ideal gases? Derive the expressions for final temperature and pressure.

b) 5 kg of steam at 8 bar pressure and 200°C mixed with 3 kg of steam at 5 bar and dryness fraction x = 0.8 adiabatically. Find the final condition of the steam.

c) 5 kg of air at 4 bar pressure is heated at constant pressure from 300 K to 500 K. Find the change in entropy of the system.

03) a) Prove that in an adiabatic process pV^γ = Constant.

b) Polytropic compression of air from state 1 to state 2 where p₁ = 100 kPa and T₁ = 300 K, p₂ = 300 kPa and n = 1.2 where as mass of air is 3 kg. If R = 0.287 kJ/kg-K. Then find

i) heat exchange during the process
ii) change in internal energy
iii) total work done by the air
iv) change in entropy


c) A non flow reversible process occurs for which pressure and volume are correlated by the relation p = (V² + 6V), where V is the volume in m³ and pressure p is in bar. Determine work done if volume changes from 3 to 6 m³.

04) a) A gas expands according to the equation pv = 100, where " p " is the pressure in kPa and " v " is the specific volume. Initial and final pressures are 1000 kPa and 500 kPa respectively. The gas is then heated at constant volume back to it'd original pressure of 1000 kPa. Determine the net work done. Also sketch the processes in p-v coordinates.

b) What is the definition of thermodynamic work?

c) What is the efficiency of a thermodynamic cycle?

05) a) What is paddle work? Distinguish between ∫pdV work and ∫-vdp work.

b) If pV = mRT, determine whether the expression (V/T).dp + (p/T).dV is a property of a system.

c) 2 kg of air at 1 bar pressure and 300 K is compressed adiabatically to a pressure of 6 bar. Then heat of 200 kJ is added at constant pressure. What will be the maximum temperature of air during the process? Find the total work done in the processes. Also find the change in internal energy of air.

06) a) Find the expression for heat transfer in terms of work done in a polytropic process.

b) What is the specific heat C_n for a polytropic process?

c) 2 kg of air at pressure 2 bar and 300 K is compressed reversibly to 4 bar and 650 K temperature in a polytropic process. Determine the polytropic index (n) of the process.

07) a) Find an expression for mechanical work in steady flow process.

b) What is the meaning of - vdp work?

c) Air flows through a gas turbine system at a rate of 5 kg/s. It enters with a velocity of 150 m/s and an enthalpy of 1000 kJ/kg. At exit the velocity is 120 m/s and enthalpy is 600 kJ/kg. If the air passing through the turbine looses 30 kJ/kg of heat to the surroundings, determine the power developed by the system.

08) a) Write the assumptions considered in Kinetic theory of gases? Prove that C_p - C_v = R

b) Explain the law of corresponding states.

c) 10 kg of air at 300 K is stored in a cylinder of volume 0.3 m³/kg. Find the pressure exerted by air using Vander Waals gas equation. Critical properties of air are: P_c = 37.7 bar, T_c = 132.5 K, v_c = 0.093 m³/kg_mole, R = 287 J/kg-K

09) a) What are the limitations of Vander Waals gas equation? Explain reduced properties of a real gas?

b) What is a undercooled liquid and degree of undercooling? Also define enthalpy of water.

c) What are the properties of steam at critical state? Explain sublimation process and triple point line.

10) a) What are the differences between dry saturated steam and superheated steam at a same pressure? Also, explain vapourdome, saturated liquid line, saturated vapour line and critical point.

b) What are the differences between work of evaporation and enthalpy of evaporation?

c) An inventor claims to have developed a refrigeration unit which maintains −10℃ in the refrigerator which is kept at a room where the surrounding temperature is 25℃ and which has COP of 8.5. Find the claim of the inventor is possible or not.

11) a) Prove that the absolute zero temperature is impossible to achieve according to second law of thermodynamics.

b) Two reversible heat engines A and B are arranged in series. A rejects heat directly to B. Engine A receives 200 kJ at a temperature of 421℃ from the hot source while engine B is in communication with a cold sink at a temperature of 5℃. If work output of A is twice that of B, find :
(i) Intermediate temperature between A and B.
(ii) Efficiency of each engine.
(iii) Heat rejected to the sink.


c) Prove that the reversible heat engines are the most efficient.

12) a) Steam at 1 bar and a dryness fraction of 0.523 is heated in a rigid vessel until it becomes saturated vapour. Calculate the heat transferred per kg of steam.

b) Steam at 9 MPa and 600°C passes through a throttling process such that the pressure is suddenly dropped to 0.4 MPa. Find the expected temperature after throttling.

c) What will be the quality of the steam at the end of adiabatic expansion of steam at 12 bar pressure and 400°C to 1.2 bar in a turbine. Also, find the ideal work out put by the turbine.

13) a) Explain the change of entropy in a perfectly isolated system during a process in the system.

b) Explain the conditions those must be satisfied by a reversible process.

c) Two kg of water at 90℃ is mixed with three kg of water at 10℃ in a perfectly isolated system. Calculate the change in entropy of the system.

14) a) Explain the second law of thermodynamics and prove that no engine can have a 100% efficiency.

b) Explain the theoretical Carnot cycle and derive its efficiency with diagrams.

c) A reversible engine working in a cycle takes 4800 kJ of heat per minute from a source at 800 K and develops 35 kW power. The engine rejects heat to two reservoirs at 300 K and 360 K. Determine the heat rejected to each sink.

15) a) What are the causes of external irreversibility?

b) Write the first and second Tds equations and derive the expression for the change of entropy during a polytropic process.

c) Prove that reversible engines are most efficient.

16) a) Explain the second law of thermodynamics.

b) Derive the Clausius inequality.

c) Steam at 160 bar and 550℃ is supplied to a steam turbine. The expansion of steam is adiabatic with increase in entropy of 0.1 kJ/kg-K. If the exhaust pressure is 0.2 bar, calculate specific work of expansion.

17) a) 5 kg of water at 400 K is isobarically and adiabatically mixed with 3 kg of water at 500 K. Find the change in entropy of the universe.

b) Explain i) Second law efficiency, ii) Effectiveness of a system and iii) Availability of a closed system.

c) Explain the principle of increase of entropy.

18) a) Explain Helmholtz and Gibbs function.

b) Explain the concept of PMM-I and PMM-II.

c) Find an expression of exit velocity C₂ in terms of pressure ratio when air passes through a nozzle from a pressure of p₁ and temperature T₁ to a pressure p₂.

19) a) Distinguish between enthalpy and internal energy.

b) What is absolute or thermodynamic temperature? Explain briefly.

c) Two Carnot engines work in series between the source at temperature 500 K and sink at temperature 300 K. If both develop equal power, determine the intermediate temperature.

20) a) Show that two adiabatic curves on p-V diagram never intersects each other.

b) Define and classify thermodynamic systems.

c) In an isentropic flow through nozzle, air flows at the rate of 600 kg/hr. At inlet to the nozzle pressure is 2 MPa and temperature is 27℃. The exit pressure is 0.5 MPa. Initial air velocity is 300 m/s, determine
i) exit velocity of air
ii) inlet and exit area of the nozzle

THE END

AIR-FUEL MIXTURE AND STOICHIOMETRIC RATIO

AIR-FUEL MIXTURE AND STOICHIOMETRIC RATIO

CHEMICAL COMBUSTION OF FUEL

Subhankar Karmakar
Assistant Professor; SGIT
Jindal Nagar; Ghaziabad

---

Chemical Combustion is basically a rapid oxidation process of hydro-carbon fuel inside thekjm combustion chamber in the presence of air. The oxidation of fuel is basically a Exothermic or heat liberating chemical process.

In SI engines generally we use volatile hydrocarbon as fuel. The intermixing of fuel with air takes place outside the engine and the device that prepares air-fuel mixture of required mixture strength is called CARBURETION and the device is known as CARBURETTOR.

Estimation of air quantity needed for complete combustion of a given fuel

We know that any chemical reaction can be represented by the corresponding chemical equation like
CH₄ + 2O₂ = CO₂ + 2H₂O
here molecular weight of CH₄
μ_CH4 = 12 + 4x1 = 16
μ_2O2 = 2x16x2 = 64
μ_CO2 = 12 + 2x16 = 44
μ_2H2O = 2x(2+16) = 36

For complete combustion of CH₄
16 kg CH₄ needs 64 kg O₂ therefore,
1 kg of CH₄ needs (64/16) = 4 kg of O₂
For 23 kg of O₂ air needed is 100 kg
hence, for 4 kg of O₂ air needed is (100/23)x4 = 17.39 kg of air.
Air-fuel ratio will be 17.39 : 1

The ratio of air fuel mixture, needed for the complete combustion of the fuel or the chemically correct ratio of air fuel mixture required for complete combustion of the fuel is called " Stoichiometric Air fuel mixture. "

If the amount of air in the air-fuel mixture is less than the chemically correct amount of air, then the mixture is called rich mixture, where as if the quantity of air is more than the chemically corrected amount of air it is called lean mixture.

The strength of air-fuel mixture has a profound influences on the process of combustion. The required mixture strength for different operation conditions are different.

Different Operating Conditions	Required air-fuel Mixture Strength
For Max. Efficiency	17 : 1, 16.4% weak
For Max. Power	12 : 1 17.8% rich
For Starting, Idling, & Low load running	11 : 1 ~ 16 : 1 very rich mixture
For accelerated motion	13 : 1 rich mixture
For Part Load running Cruising Range	17 : 1 Lean Mixture Strength

RECIPROCATING COMPRESSORS

Q.2) Classify the compressors
(i) On the basis of operations employed
(ii) On the basis of pressure achieved
(iii) On the basis of pressure ratio
(iv) On the basis of capacity of compressors.

A.2) Depending upon different parameters, compressors can be classified on the basis of operations employed, the delivery pressure achieved, pressure ratio and capacity of compressors as follows.

On the basis of operations employed, compressors are classified into two groups:
i) :Reciprocating compressors : It uses piston cylinder arrangement and due to positive displacement of air in the cylinder, the air is compressed and delivered to a vessel called Receiver. These are capable to produce high delivery pressure with low volume flow rate.
ii) Rotary Compressors : These compressors operate at high speeds, therefore, can handle large volume flow rates compared to reciprocating compressors.

In rotary compressors, the dynamic head is imparted to the gas with the help of very high speed impeller rotating at a confined space so that the air is compressed due to centrifugal action.

On the basis of delivery pressure, compressors are classified into three categories
i) Low Pressure Compressors : Delivery pressure upto 1.1 bar
ii) Medium Pressure Compressors : Delivery pressure upto 7 bar
iii) High Pressure Compressors : Delivery pressure between 7 to 10 bar.

On the basis of pressure ratio, we can classify the devices as follows,
Fans : Pressure ratio upto 1.1
Blower : Pressure ratio upto 1.1 to 4.0
Compressors : Pressure ratio above 4.0

On the basis of capacity, compressors can be classified as follows,
Low capacity compressors : Volume flow rate upto 10 m³/min, or less
Medium capacity compressors : Volume flow rate 10 m³/min to 300 m³/min
High capacity compressors : Volume flow rate above 300 m³/min

Q.3) Find an expression for required work done to drive a compressor, when compression is,
adiabatic in nature
isothermal compression
polytropic compression

A.3) During the analysis of the operations of a reciprocating air compressor, we consider some assumptions to simplify the analysis. They are as follows
i) There is no Clearence Volume
ii) Working substance air is an ideal gas
iii) There is no frictional loss.
iv) There is no wire drawing in the valve or pipe lines.

WORK REQUIRED TO DRIVE A COMPRESSOR

Suppose, we are running a single stage air compressor, which draws air at pressure P₁ and temperature T₁ during the suction or induction process. The air thus drawn inside the cylinder then compressed to achieve a delivery pressure, P₂ by adiabatic process. In the adjacent figure, process a - b is the suction process. Process b - c is the adiabatic compression of the air from pressure P₁ to pressure P₂. Process c - d is the delivery stroke, delivering the compressed air at a pressure P₂.

W_ad = P₂V₂ + {( P₂V₂ - P₁V₁)/(γ - 1)} - P₁V₁

=> (P₂V₂ - P₁V₁){1 + 1/(γ - 1)}

=> {γ /(γ - 1)}P₁V₁ {(P₂V₂)/(P₁V₁) - 1}

=> {γ /(γ - 1)}mRT₁{(P₂/P₁)^{{(γ - 1)/γ}} - 1}

POLYTROPIC WORKDONE IN RECIPROCATING COMPRESSORS

W_poly = P₂V₂ + {( P₂V₂ - P₁V₁)/(n - 1)} - P₁V₁

=> (P₂V₂ - P₁V₁){1 + 1/(n - 1)}

=> {n/(n - 1)}P₁V₁ {(P₂V₂)/(P₁V₁) - 1}

=> {n/(n - 1)}mRT₁{(P₂/P₁)^{{(n - 1)/n}} - 1}

ISOTHERMAL WORKDONE REQUIRED TO DRIVE RECIPROCATING COMPRESSOR

W_iso = P₂V₂ + {P₁V₁ln(V₁/V₂)} - P₁V₁

but as the process is isothermal, P₁V₁ = P₂V₂

W_iso = P₁V₁ln(V₁/V₂)

W_iso = mRT₁ln r_p

V₁/V₂ = P₂/P₁ = r_p (pressure ratio)

THERMODYNAMICS: 2nd MINOR TEST AND ITS SOLUTION.

Topics: First Law of Thermodynamics, SFEE, flow work, Steam, Second Law of Thermodynamics

Total Marks: 30.

Time: 1 hr and 30 min

SECTION A: Attempt all the questions 2x3 = 6

1) What is sub-cooled or undercooled water?

2) What is degree of superheat in case of superheated steam?

3) Write the first law of thermodynamics for a open process.

SECTION B: Attempt all the questions 3x3 = 9

1) 2 kg of saturated water at 8 bar pressure has been supplied 4700 kJ of heat. Find the end condition of the steam produced. Also find the value of specific internal energy and specific entropy of the steam.

2) A stream of air with a mass rate 0.6 kg/s enters a nozzle at a pressure of 8 bar and temperature 200°C at a velocity 1.2 m/s. If the final pressure at exit is 0.3 MPa, then find the value of velocity at exit and inlet and outlet/exit diameter of the nozzle.

3) A heat engine running between two thermal reservoirs of 800 K and 300 K is used to power a refrigerator running between two thermal reservoirs of 325 K and 260 K. If the heat engine draws 5000 kJ heat from reservoirs at 800 K, then find the amount of heat extracted from 260 K reservoir by the refrigerator.

SECTION C: Attempt any three questions 5x3 = 15

1) What is flow work? Distinguish between flow work and non-flow work. Find the expression for flow work in a open system. What is SFEE?

2) What is quality of steam? Explain the terms "dryness fraction" and "wetness fraction". Calculate the specific enthalpy of steam at 9 bar pressure and 350°C temperature.

3) What will be the quality of the steam at the end of adiabatic expansion of steam at 12 bar pressure and 400°C to 1.2 bar in a turbine. Also, find the ideal work out put by the turbine.

4) Explain the second law of thermodynamics. Prove that both the statements of 2nd law of thermodynamics are equivalent to each other.

5) Explain the following terms.
i) Vapour Dome,
ii) Saturated Liquid Line,
iii) Critical Point,
iv) Saturation Temperature,
v) Reversible Heat Engine

SOLUTIONS

Topics: First Law of Thermodynamics, SFEE, flow work, Steam, Second Law of Thermodynamics

Total Marks: 30.

Time: 1 hr and 30 min

SECTION A: Attempt all the questions 2x3 = 6

1) What is sub-cooled or undercooled water?

Ans: The boiling point of water is a function of the pressure, as pressure increases, boiling point is also elevated. For a certain pressure, water has a fixed boiling temperature known as saturation temperature and denoted by t_s. If the temperature of water at a given pressure is lower than the corresponding saturation temperature i.e. t < t_s, then the water is called sub-cooled or under-cooled water.

2) What is degree of superheat in case of superheated steam?

3) Write the first law of thermodynamics for a open process.

SECTION B: Attempt all the questions 3x3 = 9

1) 2 kg of saturated water at 8 bar pressure has been supplied 4700 kJ of heat. Find the end condition of the steam produced. Also find the value of specific internal energy and specific entropy of the steam.

2) A stream of air with a mass rate 0.6 kg/s enters a nozzle at a pressure of 8 bar and temperature 200°C at a velocity 1.2 m/s. If the final pressure at exit is 0.3 MPa, then find the value of velocity at exit and inlet and outlet/exit diameter of the nozzle.

3) A heat engine running between two thermal reservoirs of 800 K and 300 K is used to power a refrigerator running between two thermal reservoirs of 325 K and 260 K. If the heat engine draws 5000 kJ heat from reservoirs at 800 K, then find the amount of heat extracted from 260 K reservoir by the refrigerator.

SECTION C: Attempt any three questions 5x3 = 15

1) What is flow work? Distinguish between flow work and non-flow work. Find the expression for flow work in a open system. What is SFEE?

2) What is quality of steam? Explain the terms "dryness fraction" and "wetness fraction". Calculate the specific enthalpy of steam at 9 bar pressure and 350°C temperature.

3) What will be the quality of the steam at the end of adiabatic expansion of steam at 12 bar pressure and 400°C to 1.2 bar in a turbine. Also, find the ideal work out put by the turbine.

4) Explain the second law of thermodynamics. Prove that both the statements of 2nd law of thermodynamics are equivalent to each other.

5) Explain the following terms.
i) Vapour Dome,
ii) Saturated Liquid Line,
iii) Critical Point,
iv) Saturation Temperature,
v) Reversible Heat Engine

COMPRESSORS AND COMPRESSED AIR

COMPRESSORS:

Q.1) What is a Compressor? What is the difference between a Compressor and a Pump? What are the practical uses of Compressed Air?

A.1) A compressor is a device which is extensively used to raise the pressure of a compressible fluid like pure air. In a compressor, the pressure is increased at the expense of work done on the fluid, which is generally provided by an electric motor, IC engines or Gas Turbines. In a compressor, fluids are compressed by reducing the specific volumes of the working fluids. Due to compression, the temperature of the fluid is also increased.

If air is used as the working fluid in a compressor and air is compressed into a high pressure by the application of work on the fluid, then it is known as Air Compressor.

PRACTICAL USES OF COMPRESSED AIR

In industry, compressed air is so widely used that it is often regarded as the fourth utility, after electricity, natural gas and water. Compressed air, commonly called Industry's Fourth Utility, is air that is condensed and contained at a pressure that is greater than the atmosphere. The process takes a given mass of air, which occupies a given volume of space, and reduces it into a smaller space. In that space, greater air mass produces greater pressure. The pressure comes from this air trying to return to its original volume. It is used in many different manufacturing operations.

Compressed air is extensively used in industrial applications like pneumatic machines, as well as in the refrigeration and air-conditioning systems or supercharging CI engines to boost the output of the engine.
01) Compressed air is extensively used to operate pneumatic tools like drills, hammers, rivetting machines etc.
02) It is used to drive Compressed Air Engine.
03) Compressed air is used to spray painting.
04) Compressor is a vital component of Air-conditioning and Refrigeration industry.
05) Very often, compressed air can be used as a means of energy storage.
06) It is used in Gas Turbine power plants.
07) It is used to Super-charging an IC engines.
08) It is used to convey or pump to flow the materials like sand or concrete slurries along a pipe line.
09) It can be used as a means to pump water through the pipe lines.
10) It is used to drive minning machineries in a fire risky zone.
11) It is also used in blast furnaces.

Q.2) Classify the compressors
(i) On the basis of operations employed
(ii) On the basis of pressure achieved
(iii) On the basis of pressure ratio
(iv) On the basis of capacity of compressors.

A.2) Depending upon different parameters, compressors can be classified on the basis of operations employed, the delivery pressure achieved, pressure ratio and capacity of compressors as follows.

On the basis of operations employed, compressors are classified into two groups
i) Reciprocating compressors : It uses piston cylinder arrangement and due to positive displacement of air in the cylinder, the air is compressed and delivered to a vessel called Receiver. These are capable to produce high delivery pressure with low volume flow rate.
ii) Rotary Compressors : These compressors operate at high speeds, therefore, can handle large volume flow rates compared to reciprocating compressors.

In rotary compressors, the dynamic head is imparted to the gas with the help of very high speed impeller rotating at a confined space so that the air is compressed due to centrifugal action.

On the basis of delivery pressure, compressors are classified into three categories
i) Low Pressure Compressors : Delivery pressure upto 1.1 bar
ii) Medium Pressure Compressors : Delivery pressure upto 7 bar
iii) High Pressure Compressors : Delivery pressure between 7 to 10 bar.

On the basis of pressure ratio, we can classify the devices as follows,
Fans : Pressure ratio upto 1.1
Blower : Pressure ratio upto 1.1 to 4.0
Compressors : Pressure ratio above 4.0

On the basis of capacity, compressors can be classified as follows,
Low capacity compressors : Volume flow rate upto 10 m³/min, or less
Medium capacity compressors : Volume flow rate 10 m³/min to 300 m³/min
High capacity compressors : Volume flow rate above 300 m³/min

QUESTION PAPER: EME-505, IC ENGINES & COMPRESSORS