ssc solved paper 2011 (conventional) · manometer is a u-shaped tube consisting of an...

1. A. The gas in a system receives heat whichcauses expansion against a constantpressure of 2 bar. An agitator in the systemis driven by an electric motor using100 W. For 4 kJ of heat supplied thevolume increase of the system in 30 secis 0.06 m3. Estimate net change in theenergy of the system.

B. Find the coefficient of performance andheat transfer rate in the condenser of arefrigerator in kJ/h which has arefrigeration capacity of 12000 kJ/h whenpower input is 0.75 kW.

2. A. What is Air standard cycle? Define, Airstandard efficiency and mention theassumptions involved with Air standardcycle.

B. Explain the purpose of Cooling systemin IC engines. Also mention differenttypes of Cooling systems in brief.

3. A. What is the principle on which Manometerworks? Also, explain its various types,with their use in brief.

B. What is Bernoulli’s Theorem? Alsomention its assumptions.

The diameter of a pipe at the section 1and 2 are 10 cm and 15 cm respectively.Find the discharge through the pipe if thevolocity of water flowing through the pipeat section 1 is 5 m/sec. Determine thevelocity at section 2.

4. A. A thick cylinder 125 mm inside diameterand 250 mm outside diameter is subjectedto an internal fluid pressure of 50 N/mm2.

Determine the maximum and minimumintensities of circumferential stress andsketch the distribution of circumferentialstress intensity and radial pressureintensity across the section.

B. Determine the maximum torque that canbe safely applied to a shaft of 200 mmdiameter if the permissible angle of twistis 1° for a length of 5 m and thepermissible shear stress is 45 N/mm2.Take Modulus of Rigidity = 0.8 × 105 45N/mm2.

5. A. Explain Tempering process and itsclassification (Austempering/Isothermalquenching and Martempering/Steppedquenching).

B. With the help of neat sketch describe arcwelding with coated electrode in detail.

6. A. The total tension on the two sides of abelt connecting two pulleys is 2 kN. Theminimum angle of embrace of the belt is150° and coefficient of friction betweenthe belt and the pulley rim is 0.25.Determine the value of the tension onboth the tight and the slack side of thebelt. Also calculate the power transmittedif the speed of the belt is 600 m/minute.

B. Draw roller follower – cam mechanismand describe the terminology

(i) base circle

(ii) pitch curve

(iii) prime circle

(iv) pressure angle and pitch point.

SSC SOLVED PAPER 2011 (CONVENTIONAL) 1 of 63

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SSC SOLVED PAPER 2011 (CONVENTIONAL)

EXPLANATORY ANSWERS

1.A. Work on the system through agitator = 100 W= 100 J/s = 100 × 30 J in 30 s = 3 kJ in 30 sExpansion work,

W1–2 = P(V2 – V1)= 2 × 105 × 0.06 = 12 × 103 N-m= 12 kJ

Net work done by the system= 12 – 3 = 9 kJ

Now, from the first law of thermodynamics,dQ = dU + dW

4 = dU + 9dU = – 5 kJ

Negative sign indicates that energy of thesystem decreases.

1.B. Refrigeration capacity, Q2 = 12000 kJ/hPower input, W = 0.75 kW (= 0.75 × 60 × 60kJ/h)Co-efficient of performance, C.O.P. :Heat transfer rate :(C.O.P.)refrigerator

=Heat absorbed at lower temperature

Work input

\ C.O.P. = 2Q 120004.44

W 0.75 60 60= =

´ ´

Fig.

Hence, C.O.P. = 4.44Hence, transfer rate in condenser = Q1

According to the first law

Q1 = Q2 + W

= 12000 + 0.75 × 60 × 60= 14700 kJ/h

Hence, heat transfer rate = 14700 kJ/h.

2.A. Air-Standard CycleAir standard cycle refers to thermodynamiccycle being studied with certain assumptions,so as to use the principles of thermodynamicsconveniently. It is the most simplified formof thermodynamic cycle under consideration.

Air Standard EfficiencyTo compare the effects of different cycles itis of paramount importance that the effect ofthe calorific value of the fuel is altogethereliminated and this can be achieved byconsidering air which is assumed to behaveas a perfect gas as the working substance inthe engine cylinder. The efficiency of engineusing air as the working medium is known asan “Air standard efficiency”. This efficiencyis oftenly called ideal efficiency.The actual efficiency of a cycle is always lessthan the air-standard efficiency of that cycleunder ideal conditions. This is taken intoaccount by introducing a new term “Relativeefficiency” which is defined as:

hrelative =Actual thermal efficiencyAir standard efficiency

The analysis of all air standard cycles isbased upon the following assumptions :Assumptions :1. The gas in the engine cylinder is a perfect

gas i.e., it obeys the gas laws and hasconstant specific heats.

2. The physical constants of the gas in thecylinder are the same as those of air atmoderate temperatures i.e., the molecularweight of cylinder gas is 29.cp = 1.005 kJ/kg K, cv = 0.718 kJ/kg K.



3. The compression and expansionprocesses are adiabatic and they takeplace without internal friction, i.e., theseprocesses are isentropic.

4. No chemical reaction takes place in thecylinder. Heat is supplied or rejected bybringing a hot body or a cold body incontact with cylinder at appropriatepoints during the process.

5. The cycle is considered closed with thesame ‘air’ always remaining in the cylinderto repeat the cycle.

2.B. We know that in case of Internal Combustionengines, combustion of air and fuel takesplace inside the engine cylinder and hot gasesare generated. The temperature of gases willbe around 2300-2500°C. This is a very hightemperature and may result into burning ofoil film between the moving parts and mayresult into seizing or welding of the same. So,this temperature must be reduced to about150-200°C at which the engine will workmost efficiently. Too much cooling is alsonot desirable since it reduces the thermalefficiency. So, the object of cooling system isto keep the engine running at its mostefficient operating temperature

It is to be noted that the engine is quiteinefficient when it is cold and hence thecooling system is designed in such a waythat it prevents cooling when the engine iswarming up and till it attains to maximumefficient operating temperature, then it startscooling.

It is also to be noted that:

(a) About 20-25% of total heat generated isused for producing brake power (usefulwork).

(b) Cooling system is designed to remove30-35% of total heat.

(c) Remaining heat is lost in friction andcarried away by exhaust gases.

There are mainly two types of coolingsystems:(a) Air cooled System and(b) Water cooled System.

Air Cooled SystemAir cooled system is generally used in smallengines say up to 15-20 kW and in aeroplane engines.In this system fins or extended surfaces areprovided on the cylinder walls, cylinder head,etc. Heat generated due to combustion in theengine cylinder will be conducted to the finsand when the air flows over the fins, heat willbe dissipated to air.The amount of heat dissipated to air dependsupon :(a) Amount of air flowing through the fins.(b) Fin surface area.(c) Thermal conductivity of metal used for

fins.

Fig. : Cylinder with Fins

Advantages of Air Cooled SystemFollowing are the advantages of air cooledsystem :(a) Radiator/pump is absent hence the

system is light.(b) In case of water cooling system there are

leakages, but in this case there are noleakages.



(c) Coolant and antifreeze solutions are notrequired.

(d) This system can be used in cold climates,where if water is used it may freeze.

Disadvantages of Air Cooled System(a) Comparatively it is less efficient.(b) It is used in aeroplanes and motorcycle

engines where the engines are exposedto air directly.

Water Cooling System

In this method, cooling water jackets arcprovided around the cylinder, cylinder head,valve seats etc. The water when circulatedthrough the jackets, it absorbs heat ofcombustion. This hot water will then becooling in the radiator partially by a fan andpartially by the flow developed by theforward motion of the vehicle. The cooledwater is again recirculatcd through the waterjackets.

Types of Water Cooling System

There are two types of water cooling system :

Thermo Siphon System

In this system the circulation of water is dueto difference in temperature (i.e. difference indensities) of water. So in this system pump isnot required but water is circulated becauseof density difference only.

Fig. Thermo Siphon System of Cooling

Pump Circulation System

In this system circulation of water is obtainedby a pump. This pump is driven by means ofengine output shaft through V-belts.

Fig.: Pump Circulation System

Water cooling system mainly consists of:(a) Radiator,(b) Thermostat valve,(c) Water pump,(d) Fan,(e) Water Jackets and(f ) Antifreeze mixtures.

Advantages and Disadvantages of WaterCooling System

Advantages(a) Uniform cooling of cylinder, cylinder

head and valves.(b) Specific fuel consumption of engine

improves by using water cooling system.(c) If we employ water cooling system, then

engine need not be provided at the frontend of moving vehicle.

(d) Engine is less noisy as compared with aircooled engines, as it has water fordamping noise.

Disadvantages(a) It depends upon the supply of water.(b) The water pump which circulates water

absorbs considerable power.(c) If the water cooling system fails then it

will result in severe damage of engine.(d) The water cooling system is costlier as

it has more number of parts. Also itrequires more maintenance and care forits parts.



3.A. Manometer :A device used to measure the pressure at anypoint in a fluid, manometers are also used tomeasure the pressure of gas and air.The term manometer is derived from theancient Greek words ‘manós’, meaning thinor rare and ‘métron’. A manometer works onthe principle of hydrostatic equilibrium andis used for measuring the pressure (staticpressure) exerted by a still liquid or gas.Hydrostatic equilibrium states that thepressure at any point in a fluid at rest is equaland its value is just the weight of theoverlying fluid. In its simplest form, amanometer is a U-shaped tube consisting ofan incompressible fluid like water or mercury.It is inexpensive and does not needcalibration.As seen in the figure, the U-shaped tube filledwith liquid measures the differential pressure,i.e., the difference in levels ‘h’ between thetwo limbs gives the pressure difference(p1 – p2) between them. When pressure isapplied at limb 1, the fluid recedes in limb 1,and its level rises in limb 2. This rise continuestill a balance is struck between the unit weightof fluid and the pressure applied. If thepressure applied at one opening; say limb 1of the U-tube, is atmospheric pressure, thedifference gives the gauge pressure at limb 2.

h = (p1 – p2)rgwhere, r = density of the liquid used in

manometerHence, rg = specific weight of the liquid

Various Types of Manometers1. U-tube Manometer2. Well type or reservoir manometer

3. Inclined manometer4. Float type manometer

U-tube Manometer

U-tube is made of glass. The tube is filledwith a fluid known as Manometer fluid.

Manometer fluid may be mercury, water etc.If the manometer is connected to samepressure source (P1 = P2) the level of themanometer will be same.

If P1 > P2 the differential pressure P2 – P1 = rhr = density of the fluid

h = height difference

While choosing the manometer fluid for aparticular application we need to rememberfollowing things.Manometer fluid should not wet the wallManometer fluid should not absorb gasManometer fluid should not react chemicallyManometer fluid should have low vaporpressureMove freely



Mercury is one of the most commonly usedmanometer fluid.

Well type ManometerIn a well type manometer, one leg is replacedby a large diameter well. Since the crosssectional area of the well is much larger thanthe other leg, when pressure is applied to thewell, the manometer liquid in the well, lowersonly slightly compared to the liquid rise inthe other leg. As a result of this, the pressuredifference can be indicated only by the heightof the liquid column in single leg.

For static balance,P2 – P1 = r(1 + A1/A2)h

whereA1 = area of smaller-diameter legA2 = area of well

If A1/A2 << 1 then P2 – P1 = rh

If the area of well is 500 or more times largerthan the area of vertical leg, the error involvedin neglecting the area is negligible.

Inclined-Tube Manometers

The inclined leg expands the scale so thatlower pressure differentials may be read easily.

Sensitivity of the manometer increases.

The scale of the manometer can be extendedgreatly by decreasing the angle of inclinedleg to a small value.

Float-Type Manometers

This is a variation of well-type manometer

Recording type manometer

Span of the measurement can be changed bychanging the diameter of the leg

A large float can be placed to generate enoughforce

3.B. Bernoulli’s theorem states that in a steadyflow of ideal incompressible fluid, the sum ofpressure head, velocity head and potentialhead is constant along a stream line providedno energy is added or taken out by externalsource.

Bernoulli’s Equation from Euler’s Equation

Bernoulli’s equation is obtained byintegrating the Euler’s equation of motion as

dpgdz vdv+ +

rò ò ò = constant

If flow is incompressible, r is constant and

\2

2p v

gz+ +r

= constant

or2

2p v

zg g

+ +r

= constant



or2

2p v

zg g

+ +r = constant

Above equation is a Bernoulli’s equation inwhich

pgr

= pressure energy per unit weight

of fluid or pressure head.v2/2g = kinetic energy per unit weight

or kinetic head.z = potential energy per unit

weight or potential head.

Assumptions

The following are the assumptions made inthe derivation of Bernoulli’s equation:(i) The fluid is ideal, i.e., viscosity is zero

(ii) The flow is steady(iii) The flow is incompressible(iv) The flow is irrotational.

Given:

At section 1,D1 = 10 cm = 0.1 m

A1 = ( )2 21D (.1)

4 4p p

=

= 0.007854 m2

V1 = 5 m/sAt section 2,

D2 = 15 cm = 0.15 m

A2 = 2(.15) 0.017674p

= m2

Fig.

Discharge through pipe is given by equation

or Q = A1 × V1

= 0.007854 × 5 = 0.03927 m3/sUsing equation, we have

A1V1 = A2V2

\ V2 = 1 1

2

A V 0.0078545.0

A 0.01767= ´

= 2.22 m/s

4.B. D = 200 mm, fs = 45 N/mm2,

q = 1° =180

p radian, L = 5 m

From TJ

=N. .

Lt q

T =N. . .J

Lt q

= 0.8 × 105 × 45 × 106

4(.2)180 32 5

p p´ ´ ´

°Tmax. = 1971.92 kN-m

5.A. TemperingQuenching of high carbon steel causesformation of martensitic structures which arehard and strong, but they are brittle.It is defined as a process in which hardenedsteel is heated to a temperature below thelower critical temperature to transform thehard and brittle martensite into ferrite andcementite. In this process ductility andtoughness is achieved by sacrificing hardnessand strength. On the basis of heatingconditions it can be classified as follows:1. Low temperature tempering (150°C -

250°C)2. Medium temperature tempering (350°C -

425°C)3. High temperature tempering (500°C -

650°C)

Objectives of Tempering

Tempering process serves the followingobjectives:1. It reduces the brittleness of hardened steel.



2. It improves toughness and ductility.3. It is used to relieve internal stress.

Classification of Tempering Process1. Austempering2. Martempering

Austempering: The process is also calledisothermal quenching. It is an isothermaltransformation process that converts austeniteto hard structure. In this process, steel is heatedto about 700°C and held at this temperaturefor some time. It is then quenched in moltensalt, brought down to about 500°C, and heldat this temperature for a prolonged period.The steel is finally quenched in water at roomtemperature (Figure). The steel thus producedis free from cracks, softer than martensite, andpossesses good impact resistance.

Fig. : Austempering

Objectives of Austempering: Steel becomesmore tough and ductile.1. Distortion and cracks developed during

quenching are the minimum.2. Martempering

Martempcring: It is also known as SteppedQuenching. In this process, steel is heated toabout 600°C and held for some time, andthen quenched in a molten salt bath down toabout 300°C. After holding it at thistemperature until the steel reaches thetemperature of the medium, the workpiece isallowed to cool gradually in air or oil (Figure).

Fig.: Martempering Process

Objectives of Martempering1. To minimise distortion and cracking.2. To relieve internal stress.

5.B. Covered Welding ElectrodesWhen molten metal is exposed to air, itabsorbs oxygen and nitrogen, and becomesbrittle or is otherwise adversely affected.A slag cover is needed to protect molten orsolidifying weld metal from the atmosphere.This cover can be obtained from the electrodecoating.The composition of the welding electrodecoating determines its usability, as well asthe composition of the deposited weld metaland the electrode specification.

The formulation of Welding electrodecoatings is based on well-establishedprinciples of metallurgy, chemistry andphysics. The coating protects the metal fromdamage, stabilizes the arc and improves theweld in other ways, which include:1. Smooth weld metal surface with even

edges.2. Minimum spatter adjacent to the weld.3. A stable welding arc.4. Penetration control.5. A strong, tough coating.6. Easier slag removal.7. Improved deposition rate.



The metal-arc electrodes may be grouped andclassified as bare or thinly coated electrodes,and shielded arc or heavy coated electrodes.The coatings of welding electrodes forwelding mild and low alloy steels may havefrom 6 to 12 ingredients, which includes:

l cellulose to provide a gaseous shield witha reducing agent in which the gas shieldsurrounding the arc is produced by thedisintegration of cellulose

l metal carbonates to adjust the basicity ofthe slag and to provide a reducingatmosphere

l titanium dioxide to help form a highlyfluid, but quick-freezing slag and toprovide ionization for the arc

l ferromanganese and ferrosilicon to helpdeoxidize the molten weld metal and tosupplement the manganese content andsilicon content of the deposited weldmetal

l clays and gums to provide elasticity forextruding the plastic coating material andto help provide strength to the coating

l calcium fluoride to provide shielding gasto protect the arc, adjust the basicity ofthe slag, and provide fluidity andsolubility of the metal oxides

l mineral silicates to provide slag and givestrength to the electrode covering

l alloying metals including nickel,molybdenum and chromium to providealloy content to the deposited weld metal

l iron or manganese oxide to adjust thefluidity and properties of the slag and tohelp stabilize the arc.

l iron powder to increase the productivityby providing extra metal to be depositedin the weld.

Light Coated Electrodes

Light coated welding electrodes have adefinite composition. A light coating has beenapplied on the surface by washing, dipping,brushing, spraying, tumbling, or wiping. The

coatings improve the characteristics of thearc stream. They are listed under the E45series in the electrode identification system.The coating generally serves the functionsdescribed below:1. It dissolves or reduces impurities such as

oxides, sulphur and phosphorus.2. It changes the surface tension of the

molten metal so that the globules of metalleaving the end of the electrode aresmaller and more frequent. This helpsmake flow of molten metal more uniform.

3. It increases the arc stability byintroducing materials readily ionized (i.e.,changed into small particles with anelectric charge) into the arc stream.

4. Some of the light coatings may producea slag. The slag is quite thin and doesnot act in the same manner as the shieldedarc electrode type slag.

Light Coated Electrode

Fig.: Arc Action Obtained with aLight Coated Electrode

Shielded Arc or Heavy Coated Electrodes

Shielded arc or heavy coated weldingelectrodes have a definite composition onwhich a coating has been applied by dippingor extrusion, The electrodes are manufacturedin three general types: those with cellulosecoatings; those with mineral coatings; andthose whose coatings are combinations ofmineral and cellulose. The cellulose coatingsare composed of soluble cotton or other formsof cellulose with small amounts of potassium,sodium, or titanium and in some cases addedminerals. The mineral coatings consist of



sodium silicate, metallic oxides clay and otherinorganic substances or combinations thereof.Cellulose coated electrodes protect themolten metal with a gaseous zone around thearc as well as the weld zone. The mineralcoated electrode forms a slag deposit. Theshielded arc or heavy coated electrodes areused for welding steels, cast iron and hardsurfacing. See figure below.

Shielded Arc Electrode

Fig.: Arc Action Obtained with aShielded Arc Electrode

Functions of Shielded Arc or Heavy CoatedElectrodes

These welding electrodes produce a reducinggas shield around the arc. This preventsatmospheric oxygen or nitrogen fromcontaminating the weld metal. The oxygenreadily combines with the molten metal,removing alloying elements and causingporosity. Nitrogen causes brittleness, lowductility and in some cases low strength andpoor resistance to corrosion.

They reduce impurities such as oxides,sulphur, and phosphorus so that theseimpurities will not impair the weld deposit.

They provide substances to the arc whichincrease its stability. This eliminates widefluctuations in the voltage so that the arc canbe maintained without excessive spattering.

By reducing the attractive force between themolten metal and the end of the electrodes,

or by reducing the surface tension of themolten metal, the vaporized and meltedcoating causes the molten metal at the end ofthe electrode to break up into fine, smallparticles.The coatings contain silicates which will forma slag over the molten weld and base metal.Since the slag solidifies at a relatively slowrate, it holds the heat and allows theunderlying metal to cool and solidify slowly.This slow solidification of the metal eliminatesthe entrapment of gases within the weld andpermits solid impurities to float to the surface.Slow cooling also has an annealing effect onthe weld deposit.The physical characteristics of the welddeposit are modified by incorporatingalloying materials in the electrode coating.The fluxing action of the slag will alsoproduce weld metal of better quality andpermit welding at higher speeds.

6.A. Initial tension, T0 = 2000 N

Angle of embrace, q = 150°

Co-efficient of friction, m = 0.25

Speed of the belt, v = 600 m/min.

Tension, T1 = ?, T2 = ?

We know that T0 = 1 2T T2+

1

2

where T tension on the tight side,and T tension on the slack side

=é ùê ú=ë û

or 2000 = 1 2T T2+

or T1 + T2 = 4000 ...(i)

Also, 1

2

TT

= eµq

=0.25 150

180epæ ö´ ´ç ÷è ø

= 1.92\ T1 = 1.92 T2 ...(ii)Substituting the above value of T1 in (i), weget

1.92 T2 + T2 = 4000



or 2.92T2 = 4000 or T2 = 1369.86 Nand T1 = 1.92 × 1369.86

= 2630.14The power transmitted,

P = ?We know that P = (T1 – T2)v

=600

(2630.14 1369.86)60

- ´

= 12602.77 W= 12.602 kW

6.B. Cam Nomenclaturel Cam profile is the actual working surface

contour of the cam. It is the surface incontact with the knife-edge, roller surface,of flat-faced follower.

l Base circle is the smallest circle drawn tothe cam profile from the radial cam center.Obviously, the cam size is dependent onthe established size of the base circle.

l Pitch curve, or pitch profile, is the pathof the trace point. In cam layout, thiscurve is often determined first and thecam profile is then established bytangents to the roller or flat-faced followersurfaces. For the elementary knife-edgefollower, the pitch curve and cam profileare the same.

l Prime circle is the smallest circle drawnto the pitch curve from the cam center. Itis similar to the base circle.

l Pressure angle is the angle (at any point)between the normal to the pitch curveand the direction of the follower motion.This angle is important in cam designbecause it represents the steepness of thecam profile, which if too large can affectthe smoothness of the action.

l Pitch point is that point on the pitchcurve having the largest pressure angle.

l Pitch circle is defined as the circle drawnthrough the pitch point with its center atthe cam center.



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PAPER-II

Previous Paper (Solved)

SSC-Junior Engineering (Mechanical) Exam 2010

1. A. With the help of figure, explain theworking of Babcock and Wilcox boiler.

B. List out the differences between:(i) Two stroke and Four stroke IC engines(ii) Petrol and Diesel engines

2. A. Explain different properties of mouldingsand.

B. With the help of figure explain differenttypes of gas flames produced in oxy-acetylene welding process.

C. With the help of neat figure explain thedifferent nomenclature of twisted drill bitused in drilling machine.

3. A. What is indexing in milling machine?Explain the procedure used in compoundindexing.

B. Mention at least five differences betweenshaper and slotting machines.

C. With the help of figure explain centerlessgrinding process.

4. A. Derive the continuity equation indifferential form.

B. Calculate the maximum allowabledischarge of water through a venturimeterthroat 5 cm, fitted in a 10 cm diameterline with its inlet at an open channel.

Assume Cd = 0.95.C. Explain the performance parameters of

centrifugal pumps.

5. A. A steel bar of rectangular section 50 mm× 30 mm and length 1.5 m is subjected toa gradually applied load of 150 kN. Findthe strain energy stored in the bar. If theelastic limit of the material of the bar is150 N/mm2, proceed to determine theproof resilience and modulus of resilience.Take E = 2 × 105 N/mm2.

B. A timber beam of rectangular section isto support a load of 20 kN uniformlydistributed over a span of 4 metres. If thedepth of the section is to be twice thebreadth and the stress in the timber is notto exceed 7 MPa, find the dimensions ofthe cross-section.

C. Derive the torsion equation for a shaftsubjected to pure torsion.

6. A. Explain:(i) Watt governor(ii) Porter governor

B. Explain about helical and bevel gears withapplications.

C. Explain the Ackermann steering gearmechanism.

EXPLANATORY ANSWERS

1.A. Babcock and Wilcox Boiler

It is a water tube boiler used in steampower plants. In this, water is circulated

68

inside the tubes and hot gases flow overthe tubes.



69

Construction of Babcock and Wilcox Boiler

Fig.: Babcock and Wilcox Boiler

Working:Now we discuss about working of Babcockand Wilcox boiler. This is a high pressure,natural circulate, water tube boiler. Theworking of this boiler is as follow.

l First water is filled in the water drumthrough feed pump. The drum is half filledwith water and the upper half is for steam.First flue is fired at the grate.

l The flue gases generate by burning offuel. These gases start flowing from oneend to another end of boiler.

l The flue gases passes by the water tubesand exchange heat with water. The bafflesare provided in the way, which deflectsthe flue gases before escaping from the

chimney. Due to this deflection, the fluegases pass more than one time throughwater tubes, which cause more heattransfer.

l The water flows from the drum to thewater tube through down take header.

l The water tube nearer to the furnaceheated more than the other, so the densityof water decrease in this tube whichcauses the flow of steam from tube todrum through uptake header. At the sametime the water from the drum enters intothe tubes through down take header.

l The circulation of water from drum totubes and again tubes to drum is natural,due to density difference.

l The steam separates in the drum at theupper half. This is saturated steam. Thissteam sends to the super heater throughsteam pipe. The steam is heated again bythe flue gases in the super heater andtaken out for process work.

l The flue gases send to the atmospherefrom the super heater.

l This process repeat until sufficientamount of steam generates. This boilercan generate 20 ton steam per hour.

Advantages:1. The steam generation rate is higher about

20 ton per hour at pressure 10 to 20 bars.2. The tubes can be replaced easily.3. The boiler can expand and contract freely.4. It is easy to repair maintenance and

cleaning.5. It drought loss is low compared to other

boiler.6. The overall efficiency of this boiler is

high.

1.B. (i) Difference between two stroke and four stroke IC engine

S. No. Four stroke engine Two stroke engine1. It has one power stroke for every two It has one power stroke for each revolution of

revolutions of the crankshaft. the crankshaft.

2. Heavy flywheel is required and engine Lighter flywheel is required and engine runsruns unbalanced because turning moment balanced because turning moment is more even



70

on the crankshaft is not even due to one due to one power stroke for each revolution ofpower stroke for every two revolutions the crankshaft.of the crankshaft.

3. Engine is heavy. Engine is light.

4. Engine design is complicated due to Engine design is simple due to absence ofvalve mechanism. valve mechanism.

5. More cost. Less cost than 4 stroke.

6. Less mechanical efficiency due to More mechanical efficiency due to less frictionmore friction on many parts. on a few parts.

7. More output due to full fresh charge Less output due to mixing of fresh charge withintake and full burnt gases exhaust. the hot burnt gases.

8. Engine runs cooler. Engine runs hotter.

9. Engine is water cooled. Engine is air cooled.

10. Less fuel consumption and complete More fuel consumption and fresh charge isburning of fuel. mixed with exhaust gases.

11. Engine requires more space. Engine requires less space.

12. Complicated lubricating system. Simple lubricating system.

13. Less noise is created by engine. More noise is created by engine.

14. Engine consists of inlet and Engine consists of inlet and exhaust ports.exhaust valve.

15. More thermal efficiency. Less thermal efficiency.

16. It consumes less lubricating oil. It consumes more lubricating oil.

17. Less wear and tear of moving parts. More wear and tear of moving parts.

18. Used in cars, buses, trucks etc. Used in mopeds, scooters, motorcyclesetc.

1.B. (ii) Difference between petrol and diesel engine

S. No. Petrol Engine Diesel Engine

1. The petrol engine works on Otto cycle The diesel engine works on diesel cycle i.e.,i.e., on constant volume. on constant pressure.

2. The air and petrol are mixed in the The fuel is fed into the cylinder by a fuelcarburetor before they enter into the injector and is mixed with air inside thecylinder. cylinder.

3. The petrol engine compresses a mixture The diesel engine compresses only a charge ofof air and petrol which is ignited by an air and ignition is done by the heat ofelectric spark. compression.

4. Compression ratio is low. Compression ratio is higher in diesel engine.

5. Less power is produced due to lower Due to higher compression ratio more power iscompression ratio. produced.

6. Petrol engine is fitted with a spark plug It is fitted with a fuel injector.

7. Burns fuel that has high volatility. Burns fuel that has low volatility.



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8. They are used in light vehicles which They are used in heavy vehicles which requirerequires less power. high power.Eg: car, jeep, motorcycle, scooters etc. Eg: bushes, trucks, locomotive etc.

9. Fuel consumption in petrol engine is high. Fuel consumption in diesel engine is less.

10. Lighter Heavier

11. Petrol engine requires frequent Overhauling of diesel engine is done after aoverhauling. long time.

12. Lesser starting problem. Greater starting problem.

13. Lower initial cost. Higher initial cost.

14. Lower maintenance cost. Higher maintenance cost.

2.A. Properties of Moulding SandsThe important properties are:1. Strength:

l The sand should have adequatestrength in its green, dry and hot states.

l Green strength is the strength of sandin the wet state and is required formaking possible to prepare and handlethe mould.

l If the metal is poured into a greenmould the sand adjacent to the metaldries and in the dry state it shouldhave strength to resist erosion and thepressure of metal.

l The strength of the sand that has beendried or basked is called dry strength.

l At the time of pouring the molten metalthe mould must be able to withstandflow and pressure of the metal at hightemperature otherwise the mould mayenlarge, crack, get washed or break.

l Strength of the moulding sand dependson:1. Grain size and shape2. Moisture content3. Density of sand after ramming

l The strength of the mould increaseswith a decrease of grain size and anincrease of clay content and densityafter ramming. The strength also goesdown if moisture content is higher thanan optimum value.

2. Permeability:

l The moulding sand must be sufficientlyporous to allow the dissolved gases,which are evolved when the metalfreezes or moisture present or generatedwithin the moulds to be removed freelywhen the moulds are poured. Thisproperty of sand is called porosity orpermeability.

3. Grain size and shape:l The size and shape of the grains in the

sand determine the application invarious types of foundry. These arethree different sizes of sand grains.1. Fine2. Medium3. Coarse

l Fine sand is used for small and intricatecastings. Medium sand is used forbenchmark and light floor works. If thesize of casting is larger coarse sand isused.

l Sand having fine, rounded grains canbe closely packed and forms a smoothsurface. Although fine-grained sandenhances mould strength.

4. Thermal stability:

l The sand adjacent to the metal issuddenly heated and undergoesexpansion. If the mould wall is notdimensionally stable under rapid



72

heating, cracks, buckling and flackingoff sand may occur.

5. Refractoriness:l Refractoriness is the property of

withstanding the high temperaturecondition moulding sand with lowrefractoriness may burn on to thecasting.

l It is the ability of the moulding materialto resist the temperature of the liquidmetal to be poured so that it does notget fused with the metal. Therefractoriness of the Silica sand ishighest.

6. Flowability:l Flowability or plasticity is the property

of the sand to respond to the mouldingprocess so that when rammed it willflow all around the pattern and takethe desired mould shape. Highflowability of sand is desirable for thesand to get compacted to a uniformdensity and to get good impression ofthe pattern in the mould.

l Flowability is also very important inmachine moulding.

l Flowability of sand increases as clayand water content are increased.

7. Sand texture:l As mentioned earlier the texture of sand

is defined by its grain size and grainsize distribution.

l The texture chosen for an applicationshould allow the required porosity,provide enough strength and producethe desired surface finish on thecasting.

8. Collapsibility:l The moulding sand should collapse

during the contraction of the solidifiedcasting it does not provide anyresistance, which may result in cracksin the castings. Besides these specificproperties the moulding material

should be cheap, reusable and shouldhave good thermal conductivity

9. Adhesiveness:l It is the important property of the

moulding sand and it is defined as thesand particles must be capable ofadhering to another body, then onlythe sand should be easily attach itselfwith the sides of the moulding boxand give easy of lifting and turningthe box when filled with the stand.

10. Reusability:

l Since large quantities of sand are usedin a foundry it is very important thatthe sand be reusable otherwise apartfrom cost it will create disposalproblems.

11. Easy of preparation and control:l Sand should lend itself to easy

preparation and control by mechanicalequipment.

12. Conductivity:

l Sand should have enough conductivityto permit removal of heat from thecastings.

2.B. Types of flames: Three basic types ofoxyacetylene flames used in oxyfuel-gaswelding and cutting operations:1. Neutral flame2. Oxidizing flame3. Carburizing, or reducing flame.

Neutral Flame: Addition of little moreoxygen give a bright whitish cone surroundedby the transparent blue envelope is calledNeutral flame (It has a balance of fuel gas andoxygen) (3200°c). Used for welding steels,aluminium, copper and cast iron.

Fig. (a): Neutral Flame



73

Oxidizing Flame : If more oxygen is added,the cone becomes darker and more pointed,while the envelope becomes shorter and morefierce is called Oxidizing flame. Has thehighest temperature about 3400°c. Used forwelding brass and brazing operation.

Fig. (b): Oxidizing Flame

Carburizing Flame: Oxygen is turned on,flame immediately changes into a long whiteinner area (Feather) surrounded by atransparent blue envelope is calledCarburizing flame (3000°c)

Fig. (c): Carburizing (reducing) Flame

2.C. Twist Drill Nomenclature

Fig.: Elements of a twist drill

The twist drill consists of mainly two partsbody and shank. Both are separated by aneck. Two long and diametrically oppositehelical grooves called flutes run throughoutthe length of the drill.

1. Body :The body is the portion of the drill whichextends from its extreme point up to the neckor shank of the drill. It consists of bodyclearance, chisel edge, face, flank, flutes, heel,land or margin, point, lip and web.(i) Body Clearance: It is the portion of the

body surface with reduced diameter whichprovides diametral clearance.

(ii) Face: It is the portion of the flute adjacentto the lip on which chip flows as it is cutfrom the workpiece.

(iii) Flank: It is the conical surface of a drillpoint which extends behind the lip tothe following flute.

(iv) Flutes: These are helical grooves cut onthe cylindrical surface of the drill andprovide the lip. They serve the followingpurposes :l Ensure easy escape and flow of chips.l Cause the chips to curl and provide

passages for their flow.l Form the lips and cutting edges on

the point.l Allow the cutting fluids to reach the

cutting edges thus reducing theirfriction.

(v) Heel: It is the edge formed by theintersection of flute surface and the bodyclearance.

R-1829 (PP)-10



74

Fig.: Twist drill nomenclature

(vi) Land: It is the cylindrically ground narrowstrip on the leading edges of drill flutes.It keeps the drill aligned. It is also knownas “margin”.

(vii) Point: It is the cone shaped sharpenedend of the drill that produces lips, faces,flanks and chisel edge of the drill.

(viii) Lips: The lips, also known as “cuttingedges”, are the edges formed by theintersection of flanks and faces. They aretwo in number with identical length andangle.

(ix) Web: It is the thickness of the drillbetween the flutes which extends frompoint towards the shank. The point endof the web forms the chisel edge.

2. Shank :The shank is the cylindrical portion of thedrill which is used to hold and drive the drill.It extends from the neck and it may be eitherstraight or papered.– Tapered shanks are used in drills of bigger

sizes.Tang: It is flattened end of the taper shankswhich fits into socket or drill holder. It ensurespositive drive of the drill from the drillspindle.

Advantages of twist drills :

The advantages of using twist drills are :1. For the same size and depth of the hole

they need less power in comparison toother forms of drills.



75

2. Cutting edges are retained in goodcondition for a fairly long time, thusavoiding the frequent regrinding of thedrill.

3. The chips and cuttings of the metal areautomatically driven out of the holethrough the flute.

4. Heavier feeds and speeds can be employedquite safely, resulting in a considerablesaving of time.

Work Holding Devices

The type of work holding device used ondrilling machines depends upon the shapeand size of the workpiece, the requiredaccuracy and the rate of production. Some ofthe work holding devices are :1. Machine vice2. V-blocks3. Strap clamps and T-bolts4. Drilling jigs5. Angle plate

Drilling Machine Operations

In addition to drilling, the followingoperations are carried out on a drillingmachine :1. Reaming 5. Spot facing2. Boring 6. Tapping3. Counter-boring 7. Trepanning.4. Counter-sinking

3.A. Different indexing methods are givenbelow:1. Index Plate2. Simple Indexing3. Compound Indexing4. Differential Indexing5. Direct Indexing6. Plain Indexing7. Indexing Operation

1. Index PlateIndex plate is round metal plate by a seriesof six or extra circles of evenly spaced holes;

the index pin on crank can put in any holein any circle. By the similar plates frequentlyfurnished through most index heads, thespacings needed for most gears, bolt heads,splines, milling cutters and so forth, can beobtain. Following sets of plates are normalequipment:(a) Brown and Sharpe type, 3 plates of 6

circles, every drilled in this way:Plate 1: 15, 16, 17, 18, 19, 20 holesPlate 2: 21, 23, 27, 29, 31, 33 holesPlate 3: 37, 39, 41, 43, 47, 49 holes

(b) Cincinnati type, one plate drill on equallysides by circles divided in this way:First side: 24, 25, 28, 30, 34, 37, 38, 39,41, 42, 43 holesSecond side: 46, 47, 49, 51, 53, 54, 57,58, 59, 62, 66 holes

2. Simple IndexingSimple indexing on milling machine is passedout through use also a plain indexing head oruniversal dividing head. This process ofindexing involves use of worm, crank, indexhead and worm wheel. Worm wheel usuallycarries 40 teeth with the worm is singlethreaded. By this arrangement, as a crankcompletes one revolution, work wheel turnthrough 1/40th of revolution. Also a wormwheel turn through 2/40 (1/20)th of revolutionand presently. Therefore, for one revolutionof workpiece, a crank wants to create 40revolutions. Holes in index plate extra helpin subdividing rotation of workpiece.

3. Compound IndexingAs the number of divisions necessary on jobis external the vary of simple indexing,method of compound indexing is use.Operation is passed out by provide two detachsimple indexing movements.(a) By turning crank in similar way as in

simple indexing.(b) Once more turning index plate and the

crank also in equal or in the conflictingdirections.



76

4. Differential IndexingAt times a number of divisions are necessarywhich cannot be achieving through simpleindexing by the index plates frequentlysupplied. To get these divisions a differentialindex head is use. Index crank is attached toworm shaft with a train of gears in its placeof through a direct coupling and with simpleindexing. Selection of these gears occupycalculations like those use in calculatingchange gear ratio for cutting threads on lathe.

5. Direct IndexingCreation of several index heads allows theworm to be detached as of the worm wheel,creation possible a faster way of indexing,call direct indexing. Index head is give by aknob which, as turned during part of arevolution, function an unconventional anddisengage the worm. Direct indexing is ableby an additional index plate fixed to indexhead spindle. Motionless plunger in indexhead fits the holes in index plate. Throughmoving the plate with hand to index directly,spindle with workpiece turn an equal distance.Direct index plates generally have 24 holesand present a rapid means of milling squares,hexagons, taps and so on. Some number ofdivisions which is factor of 24 can indexedrapidly and suitably by direct indexingmethod.

6. Plain Indexing

Following principle affect to necessaryindexing of workpieces:(a) Assume it is preferred to mill spur gear

by 8 equally spaced teeth. As, 40 turns ofindex crank will turn spindle one fullrevolve, one-eighth of 40, or else 5 turnsof crank after every cut, will space gearfor 8 teeth.

(b) Equal rule affect whether or not divisionsnecessary divide equally into 40. As, if itis preferred to index for 6 divisions, 6divided into 40 equals 6 2/3 turns also,to index for 14 spaces, 14 divided in 40

equals 2 6/7 turns. Thus, follow rule canconsequent: to agree on the number ofturns of index crank desired to take onedivision of every number of equaldivisions on workpiece, divide 40 bynumber of equivalent divisions preferredgive worm wheel has 40 teeth, which isstandard practice.

7. Indexing OperationTwo follow exemplar explain how index plateis use to get any preferred part of a entirespindle turn by plain indexing.(a) To Mill a Hexagon

By the rule over, divide 40 by 6, whichequals 6 2/3 turns, or six full turns plus2/3 of turn on any circle whose figureof holes is detachable by 3. Thus, sixfull turns of crank plus 12 spaces on an18-hole circle, or six full turns plus 26spaces on a 39-hole circle will make thepreferred rotation of workpiece.

(b) To Cut a Gear of 42 TeethBy rule over, divide 40 by 42 whichequals 40/42 or 20/21 turns, 40 spaceson 42-hole circle or 20 spaces on21-hole circle. To apply the rule given,choose a circle have number of holesdividable through the required fractionof turn concentrated to its lowly terms.The number of spaces among the holesgives the preferred fractional part of entirecircle. As counting holes, start by thefirst hole to the fore of index pin.

Fig.



77

Compound Indexing

The word compound indexing is an indicativeof compound movements of indexing crankand then plate along with crank. In this caseindexing plate is normally held stationary bya lock pin, first we rotate the indexing crankthrough a required number of holes in aselected hole circle, then crank is fixedthrough pin. It is followed by anothermovement by disengaging the rear lock pin,the indexing plate along with indexing crankis rotated in forward or backward directionthrough predetermined holes in a selectedhole circle, then lock pin is reengaged.

Following steps are to be followed forcompound indexing operation. The procedureis explained with the help of numericalexample.

Example

Let us make 69 divisions of workpiececircumference by indexing method. (Usingcompound indexing)

Solution

Follow the steps given below :(a) Factor the divisions to be make

(69 = 3 × 23) N = 69.(b) Select two hole circles at random (These

are 27 and 33 in this case, both of thehole circles should be from same plate).

(c) Subtract smaller number of holes fromlarger number and factor it as(33 – 27 = 6 = 2 × 3).

(d) Factor the number of turns of the crankrequired for one revolution of the spindle(40). Also factorize the selected holecircles.

(e) Place the factors of N and differenceabove the horizontal line and factors of40 and selected both the hole circlesbelow the horizontal line as given below.Cancel the common values.

69 = 23 × 36 = 2 × 3

40 = 2 × 2 × 2 × 527 = 3 × 3 × 333 = 3 × 11

(f ) If all the factors above the line arecancelled by those which are below theline, then the selected hole circles can beused for indexing otherwise select anothertwo hole circles. In this case there is needto select another hole circles. Let us select23 and 33 this time and repeat the step5 as indicated below.

69 = 23 × 310 = 2 × 5

40 = 2 × 2 × 2 × 5

22 = 23 × 1

33 = 11 × 3

(Difference of hole circle values)Encircled numbers below the line are theleft out numbers after cancelling thecommon factors. All the factors abovethe horizontal line are cancelled soselected hole circles with 22 and 33 holescan used for indexing.

(g) Following formula is used for indexing :

4069

= 1 2

1 2N Nn n

±

In this formula N1 = 23 and N2 = 33 (N1is always given smaller value out of two).

(h) Multiply all the remaining factors belowthe line as 2 × 2 × 11 = 44. The formulaabove will turn to

4069

=44 4423 33

-

We will neglect the +ve sign.

=21 11

1 123 33

-

The –ve sign indicates backwardmovement.



78

Difference between Shaper and Slotter

3.B. Shaper Slotter

1. Shaper is commonly used to make flat Slotter is use for cutting groove, keyways andsurfaces. slots on inside and outside surface.

2. Cutting stroke is horizontal with slower Cutting stroke is vertical with slower than thethan unused stroke. unused stroke.

3. Cutting tool is move up or down to Cutting tool is move horizontally to regulateregulate deepness of the cut. deepness of cut.

4. Cutting tool move horizontally up and Cutting tool move vertically up and down thoughdown though perform cutting process. perform cutting process.

5. Distance of tool move is familiar Distance of tool move is familiar through strokethrough stroke adjusting screw. adjusting screw.

6. Workpiece is held on a fixed bed to Workpiece is held on a fixed bed to be usuallybe usually rectangular in shape. circular in shape.

3.C. Centerless grinding is a machining processthat uses abrasive cutting to remove materialfrom a workpiece. Centerless grinding differsfrom centered grinding operations in that nospindle or fixture is used to locate and securethe workpiece; the workpiece is securedbetween two rotary grinding wheels and thespeed of their rotation relative to each otherdetermines the rate at which material isremoved from the workpiece.Centerless grinding is typically used inpreference to other grinding processes foroperations where many parts must beprocessed in a short time.

ProcessIn centerless grinding, the workpiece is heldbetween two grinding wheels, rotating in thesame direction at different speeds and aworkholding platform. One wheel, known asthe grinding wheel (stationary wheel in thediagram), is on a fixed axis and rotates suchthat the force applied to the workpiece isdirected downward, against the workholdingplatform. This wheel usually performs thegrinding action by having a higher tangentialspeed than the workpiece at the point of

contact. The other wheel, known as theregulating wheel (moving wheel in thediagram), is movable. This wheel is positionedto apply lateral pressure to the workpiece andusually has either a very rough or rubber-bonded abrasive to trap the workpiece.The speed of the two wheels relative to eachother provides the grinding action anddetermines the rate at which material isremoved from the workpiece. Duringoperation the workpiece turns with theregulating wheel, with the same linear velocityat the point of contact and (ideally) noslipping. The grinding wheel turns faster,slipping past the surface of the workpiece atthe point of contact and removing chips ofmaterial as it passes.

Fig.



79

TypesThere are three forms of centerless grinding,differentiated primarily by the methodused to feed the workpiece through themachine.

Through-feedIn through-feed centerless grinding, theworkpiece is fed through the grinding wheelscompletely, entering on one side and exitingon the opposite. The regulating the wheel inthrough-feed grinding is canted away fromthe plane of the grinding wheel in such a wayas to provide a lateral force component,feeding the workpiece through between thetwo wheels. Through-feed grinding can bevery efficient because it does not require aseparate feed mechanism; however, it can onlybe used for parts with a simple cylindricalshape.

End-feedIn end-feed centerless grinding, the workpieceis fed axially into the machine on one sideand comes to rest against an end stop; thegrinding operation is performed and then theworkpiece is fed in the opposite direction toexit the machine. End-feed grinding is bestfor tapered workpieces.

In-feedIn-feed centerless grinding is used to grindworkpieces with relatively complex shapes,such as an hourglass shape. Before the processbegins, the workpiece is loaded manually intothe grinding machine and the regulatingwheel is moved into place. The complexityof the part shapes and grinding wheel shapesrequired to grind them accurately prevent theworkpiece from being fed axially through themachine.

4.A. Continuity Equation—Differential Form

Derivation1. The point at which the continuity

equation has to be derived, is enclosedby an elementary control volume.

2. The influx, efflux and the rate ofaccumulation of mass is calculated acrosseach surface within the control volume.

Fig.: A Control Volume Appropriate to a RectangularCartesian Coordinate System

Consider a rectangular parallelopiped in theabove figure as the control volume in arectangular cartesian frame of coordinate axes.

l Net efflux of mass along x-axis must bethe excess outflow over inflow acrossfaces normal to x-axis.

l Let the fluid enter across one of suchfaces ABCD with a velocity u and adensity p. The velocity and density withwhich the fluid will leave the face

EFGH will beu

ux

¶+

¶dx and dx

x¶r

r +¶

respectively (neglecting the higher orderterms in dx).

l Therefore, the rate of mass entering thecontrol volume through face ABCD =pu dy dz.

l The rate of mass leaving the controlvolume through face EFGH will be

= udx u dx dydz

x x¶r ¶æ ö æ ör+ +ç ÷ ç ÷è ø è ø¶ ¶

= ( )u u dx dydzx¶æ ör + rç ÷è ø¶

(neglecting the higher order terms in dx)l Similarly influx and efflux take place in

all y and z directions also.



80

l Rate of accumulation for a point in aflow field

mt

¶¶

= ( )d dt t

¶ ¶rr " = "

¶ ¶

l Using, Rate of influx = Rate ofAccumulation + Rate of Efflux

rudydz + rvdxdz + rwdxdy = dt

¶r"

¶

+ udx u dx dydz

x x¶r ¶æ ö æ ör + +ç ÷ ç ÷è ø è ø¶ ¶

vdy v dy dxdz

y y¶r ¶æ ö æ ö+ r + +ç ÷ ç ÷è ø è ø¶ ¶

wdz w dz dxdy

z z¶r ¶æ ö æ ö+ r + +ç ÷ ç ÷è ø è ø¶ ¶

l Transferring everything to right side

0 =u v

u vx x y y

é ¶ ¶r ¶ ¶ræ ö æ ör + + r +ç ÷ê ç ÷è ø è ø¶ ¶ ¶ ¶ë

ww dxdydz d

z z tù¶ ¶r ¶ræ ö æ ö+ r + + "ç ÷ç ÷ ú è øè ø¶ ¶ ¶û

Þ ( ) ( ) ( ) 0u v w dt x y z

¶r ¶ ¶ ¶é ù+ r + r + r " =ê ú¶ ¶ ¶ ¶ë û

This is the Equation of Continuity for acompressible fluid in a rectangular cartesiancoordinate system.

4.B. It is well known that as the local pressure ina conflined passage drops below a certainvalue, the liquid starts evaporating ‘very fast’at the existing temperature. In order to avoidthe problems of vapour lock and dis-continuities, the pressure is not allowed todrop below a certain limit. For water, underatmospheric temperature the absolute pressureis desired to remain above 3 m of water.

Closeness to an open channel suggests that,neglecting the datum head difference, the inlethead, due to atmospheric pressure.

21 1U

2pg g

+r

= 10.33 + 0 = 10.33 m

and from the consideration of the vapour

pressure, the minimum value of2p

gr is 3 m

By the Bernoulli equation.2

1 1U2

pg g

+r

=2

2 2U2

pg g

+r

10.33 =22U

32 9.81

+´

whence the maximum velocity at the throat,

U2 = 12 m/sand the maximum actual discharge

Q = CdA2U2

Q = ( )20.95 0.05 12

4p

´ ´

= 0.0224 m3/s

The maximum mass flow rate is given by

m& = rQ = 1000 × 0.0224= 22.4 kg/s

4.C. Pump Performance ParametersCapacity, head, power, efficiency, requirednet positive suction head and specific speedare parameters that describe a pump’sperformance.

CapacityThe capacity of a pump is the amount ofwater pumped per unit time. Capacity is alsofrequently called discharge or flow rate (Q).In English units it is usually expressed ingallons per minute (gpm). In metric units it isexpressed as liters per minute (1/min) or cubicmeters per second (m3/sec).

HeadHead is the net work done on a unit weightof water by the pump impeller. It is the amountof energy added to the water between thesuction and discharge sides of the pump. Pump



81

head is measured as pressure differencebetween the discharge and suction sides ofthe pump.

Power Requirements

The power imparted to the water by the pumpis called water power. To calculate waterpower, the flow rate and the pump head mustbe known. In English units water power canbe calculated using the following equation:

WP = (Q × H)/3960 ...(1)

where:

WP = water power in horsepower units

Q = flow rate (pump capacity) in gpm

H = pump head in ft

In metric units water power is expressed inkilowatts pump capacity in liters per minute,head in meters, and the constant is 6116instead of 3960.

In any physical process there are always lossesthat must be accounted for. As a result, toprovide a certain amount of power to thewater a larger amount of power must beprovided to the pump shaft. This power iscalled brake power (brake horsepower inEnglish units). The efficiency of the pump(discussed below) determines how much morepower is required at the shaft.

BP = WP/E ...(2)

where E is the efficiency of the pumpexpressed as a fraction, BP and WP are brakepower and water power, respectively.

Efficiency

Pump efficiency is the percent of power inputto the pump shaft (the brake power) that istransferred to the water. Since there are lossesin the pump, the efficiency of the pump isless than 100% and the amount of energyrequired to run the pump is greater than theactual energy transferred to the water. Theefficiency of the pump can be calculated fromthe water power (WP) and brake power (BP):

E% = (WP/BP) × 100 ...(3)

The efficiency of a pump is determined byconducting tests. It varies with pump size,type and design. Generally, larger pumps havehigher efficiencies. Materials used for pumpconstruction also influence its efficiency. Forexample, smoother impeller finishes willincrease the efficiency of the pump.

Net Positive Suction Head

The required net positive suction head(NPSHt) is the amount of energy required toprevent the formation of vapor-filled cavitiesof fluid within the eye of impeller. Theformation and subsequent collapse of thesevapor-filled cavities is called cavitation andis destructive to the impeller.

The NPSHt to prevent cavitation is a functionof pump design and is usually determinedexperimentally for each pump. The headwithin the eye of the impeller, also called netpositive suction head available (NPSHa),should exceed the NPSHr to avoid cavitation.

Specific Speed

Specific speed is an index number correlatingpump flow, head and speed at the optimumefficiency point. It classifies pump impellerswith respect to their geometric similarity. Twoimpellers are geometrically similar when theratios of corresponding dimensions are thesame for both impellers.

This index is important when selectingimpellers for different conditions of head,capacity and speed (Figure). Usually, highhead impellers have low specific speeds andlow head impellers have high specific speeds.

There is often an advantage in using pumpswith high specific speeds. For a given set ofconditions, operating speed is higher. As aresult the selected pump can generally besmaller and less expensive. However, there isalso a trade-off since pumps operating athigher speeds will wear out faster.

R-1829 (PP)-11



82

Fig.: Theoretical pump efficiencies as functions of specific speed, impeller design and pump capacity

5.A. (a) s =3

2P 150 10100 N/mm

A 50 30´

= =´

V = 50 × 30 × 1500 = 225 × 104 mm2

\ U = ( )( )

2 24

5

(100)V 225 10

2E 2 2 10

s= ´ ´

´ ´= 56250 N-mm.

(b) Proof resilience

=2

U .V2E

ee

s=

= ( )( )

24

5

(150)225 10

2 2 10´ ´

´ ´= 144000 N-mm.

(c) Modulus of resilience

= 24

U 1440000.064 N/mm

V 225 10e = =

´

5.B. Given:Total load, W = 20 kN = 20 × 1000 N

Span, L = 4 m

Max. stress, smax = 7 N/mm2

Let b = Breadth of beam in mm

Then depth, d = 2b mm

Section modulus of rectangular beam

=2

6bd

\ L =2 3(2 ) 2

6 6b b b´

= mm3

Maximum B.M., when the simply supportedbeam carries a U.D.L. over the entire span, isat the centre of the beam and is equal to

2L WL or .

8 8w

R-1829 (PP)-11-II



83

\ M = WL 20000 48 8

´=

= 10000 Nm= 10000 × 1000 Nmm

Now using equation, we get

M = smax. Z

or 10000 × 1000 =32

76b

´

or b3 =6 10000 1000

7 2´ ´

´= 4.2857 × 106

\ b = 16.24 cm

and d = 2b = 2 × 16.24

= 32.48 cm

5.C. Torsion of ShaftsWhen equal and opposite forces are appliedtangentially to the ends of a shaft, it issubjected to a twisting moment which is equal

to the product of the force applied and theradius of the shaft. This causes the shaft eitherto remain stationary or to rotate with constantangular velocity. In either case, the stress andstrain set up in the shaft will be the same.When the shaft becomes subjected to equaland opposite torques at its two ends the shaftis said to be in torsion and as a result ofwhich the shaft will have a tendency to shearoff at every cross-section perpendicular to itslongitudinal axis. So the effect of torsion isto produce shear stress in the material of theshaft.

Torsion EquationConsider a solid shaft of radius R and lengthL. Let the end B of the shaft be subjected toa pure torque 7 as shown in Fig. (a). The lineAB on the surface of the shaft will be distortedto AB' after the application of the torque. Theangle ÐBAB¢ = f is called the shear strainand ÐBOB¢ = q is called the angle of twist.Shear strain is a measure of the distortion

Fig.

orG

Ss =

RLq

f =

\R

Ss =

GLq

...(i)

In a given shaft, under a given torque, G, qand L are constant.

RSs

=GLq

= constant

produced by the shearing force and is equalto the angle of distortion f.

\ Shear strain, f =BB RL L

¢ q=

If sS be the shear stress on the outermostsurface of the shaft and G be the modulus ofrigidity of the shaft material then

Ssf

= G



84

Thus the shear stress (ss) in the shaft isproportional to the radius of the shaft. Theshear stress is maximum at the outer surface.At the centre the shear stress is zero. Theshear stress diagram showing the variation ofshear stress from the centre to the outer surfaceof the shaft is shown in Fig. (b).

Now consider an elementary ring of the shaftat a radius r and of thickness dr as shown inFig. (c). Let the shear stress at this radiusbe sS1

.

Total force on the ring = Area of the ring ×stress on the ring

= 2pr dr × sS1

Moment of this force about the axis of theshaft

= 2pr dr × sS1 × r

= 2pr2 dr × sS1

= 22R

Sr dr rs

p ´ ´

1 GAs constant

R LSS

r

sæ ös q= = =ç ÷è ø

= 32R

Sr drs

p ´

\ sS1 =

RS r

s´

Total resisting moment of the shaft cross-section is

=R 3

02 .

RSr dr

spò

But the total resisting moment must be equalto the applied torque T

\ T =R4R 3

0 02 2

R R 4S S r

r dré ùs s

p = p ê úë ûò

=4 3R R

2 . 2 .R 4 4

SS

sp ´ = ps

=

3

3

D2 .

2 D4 16

S

S

æ öps ç ÷è ø p= s

\ T = 3D16 Sp

s

T = 4 2D

32 D Sp

´ ´ s

= 4P

2D I

32 2R RS

Ssp

´ ´ s = ´

where IP =4D

32p

= polar moment of inertia of

the section of the shaft

P

TI

=R

Ss

AgainR

Ss =

GLq

\P

TI

=G

R LSs q

=

This equation is known as torsion equation.

6.A. Gravity Controlled Centrifugal GovernorsThere are three commonly used gravitycontrolled centrifugal governors :(a) Watt governor

(b) Porter governor(c) Proell governor

Watt governor does not carry dead weight atthe sleeve. Porter governor and proellgovernor have heavy dead weight at thesleeve. In porter governor balls are placed atthe junction of upper and lower arms. In caseof proell governor the balls are placed at theextension of lower arms. The sensitiveness ofwatt governor is poor at high speed and thislimits its field of application. Porter governoris more sensitive than watt governor. Theproell governor is most sensitive out of thesethree.

Watt GovernorThis governor was used by James Watt in hissteam engine. The spindle is driven by theoutput shaft of the prime mover. The balls are



85

mounted at the junction of the two arms. Theupper arms are connected to the spindle andlower arms are connected to the sleeve asshown in Figure (a).

(a) (b)

Fig.(i) : Watt Governor

We ignore mass of the sleeve, upper and lowerarms for simplicity of analysis. We can ignorethe friction also. The ball is subjected to thethree forces which are centrifugal force (Fc),weight (mg) and tension by upper arm (T).Taking moment about point O (intersectionof arm and spindle axis), we get

FCh – mg r = 0

Since, FC = mr w2

\ mr w2h – mg r = 0

or w2 =gh

...(i)

w =2 N60p

\ h = 2 2 2

3600 894.56

4 N N

g ´=

p...(ii)

where ‘N’ is in rpm.

Fig. (ii) shows a graph between height ‘h’and speed ‘N’ in rpm. At high speed thechange in height h is very small whichindicates that the sensitiveness of thegovernor is very poor at high speeds becauseof flatness of the curve at higher speeds.

Fig. (ii): Graph between Height and Speed

Porter GovernorA schematic diagram of the porter governoris shown in Fig. (iii) (a). There are two sets ofarms. The top arms OA and OB connect ballsto the hinge O. The hinge may be on thespindle or slightly away. The lower armssupport dead weight and connect balls also.All of them rotate with the spindle. We canconsider one-half of governor for equilibrium.Let, w be the weight of the ball,

T1 and T2 be tension in upper and lowerarms, respectively,Fc be the centrifugal force,r be the radius of rotation of the ballfrom axis, andI is the instantaneous centre of the lowerarm.

Taking moment of all forces acting on theball about I and neglecting friction at thesleeve, we get

CW

F AD ID IC2

w´ - ´ - = 0

or FC = ID W ID DCAD 2 ADw +æ ö+ ç ÷è ø

or FC = ( )Wtan tan tan

2w a + a + b



86

FC = 2wr

gw

\ 2wr

gw =

W tantan 1 1

2 tanw

w

ì übæ öa + +ç ÷í ýè øaî þ

or w2 =W

tan 1 (1 K)2

gr w

ì üa + +í ýî þ

...(iii)

where K =tantan

ba

Q tan a =rh

\ w2 =W

1 (1 K)2

gh w

ì ü+ +í ýî þ

...(iv)

(a) (b)

Fig. (iii): Porter Governor

If friction at the sleeve is f, the force at thesleeve should be replaced by W + f for risingand by (W – f) for falling speed as frictionapposes the motion of sleeve. Therefore, ifthe friction at the sleeve is to be considered,W should be replaced by (W ± f). Theexpression in Eq. (iv) becomes

w2 =( )W

1 (1 K)2

fgh w

ì ü±+ +í ý

î þ...(v)

6.B. Helical Gear: Helical gears have their teethinclined to the axis of the shafts in the formof a helix, hence the name helical gears.These gears are usually thought of as highspeed gears. Helical gears can take higher

loads than similarly sized spur gears. Themotion of helical gears is smoother andquieter than the motion of spur gears.

Single helical gears impose both radial loadsand thrust loads on their bearings and sorequire the use of thrust bearings. The angleof the helix on both the gear and the must besame in magnitude but opposite in direction,i.e., a right hand pinion meshes with a lefthand gear.

Applications of Helical Gears

Helical gears are normally preferred to workunder heavy load efficiently. "When we needsilent operation such as in automobileapplications, we prefer to use helical gears assuch gears work silently and smoothly.

Areas of applications of helical gears are verylarge but let us find few applications herewhere helical gears are preferred to use.1. Helical gears are used in fertilizer

industries, Printing industries and earthmoving industries



87

2. Helical gears are also used in steel, Rollingmills, section rolling mills, power andport industries.

3. Helical gears are also used in textileindustries, plastic industries, foodindustries, conveyors, elevators, blowers,compressors, oil industries & cutters.

Bevel/Miter Gear: Intersecting but coplanarshafts connected by gears are called bevelgears. This arrangement is known as bevelgearing. Straight bevel gears can be used onshafts at any angle, but right angle is themost common. Bevel Gears have conicalblanks. The teeth of straight bevel gears aretapered in both thickness and tooth height.

Spiral Bevel Gears: In these Spiral Bevelgears, the teeth are oblique. Spiral Bevel gearsare quieter and can take up more load ascompared to straight bevel gears.

Zero Bevel Gear: Zero Bevel gears are similarto straight bevel gears, but their teeth arecurved lengthwise. These curved teeth of zerobevel gears are arranged in a manner that theeffective spiral angle is zero.

Applications of Bevel Gears

Straight bevel gears are used primarily forlow speed application with pitch linevelocities up to 300 m/min. They are widelyused in textile machines. Few of applicationsare listed below:

l drive to bobbin rail on roving machinel drive between the doffer and feed roller

on low speed carding machinesl drive from calender roller to coiler rollers,

top coiler to bottom coiler plates in card,comber and drawing machine

l drive between calender roll and lap stopmechanism-lever in lap former ofconventional blow room

Spiral bevel gears find application in sewingmachines. In hand releasing mechanisms,spiral bevel gears are used since the handmovement is jerky. Hypoid gears are almostuniversally used for automotive applications.The use of hypoid gears in automobilespermits lowering of the drive shaft and isthus advantageous in the design of cars withlow bodies. The miter gears are used in highspeed carding machine to drive the webdoffing from a motor and drive to coiler froman outer shaft. Miter gears find applicationsto coil slivers into cans.



88

PAPER-II



1. A. List out the merits and demerits of watertube boilers over fire tube boilders.

B. With the help of neat sketches explainthe working of a four-stroke diesel engine.

2. A. Explain different types of patterns usedin foundry.

B. Explain any five different operations thatcan be carried out in lathe.

3. A. Give the classification of milling machines.Also explain up and down milling.

B. Explain with figure the quick returnmechanism used in shapers.

C. Explain various parameters used inselection of grinding wheel.

4. A. Derive Bernoulli’s equation from Euler’sequation.

B. Define the following:Density, Newton’s law of viscosity,Compressibility, Surface tension, andPressure.

5. A. Explain the salient features and behaviourof stress-strain curve for a tensile materialwith the figure.

B. A rectangular beam with depth 150 mmand width 100 mm is subhjected to amaximum bending moment of 300 kNm.Determine : maximum stress in the beam,radius of curvature when the bending ismaximum and bending stress at a distanceof 40 mm from the top surface of thebeam. E for beam is 200 GPa.

C. A solid circular shaft transmits 80 kW ofpower while turning 200 revolutions perminute. Work out suitable diameter of theshaft if the shear is limited to 60 mN/m2

and the twist in the shaft is not to exceed1 degree in 2 metres of length. Assumeuniform turning moment and takemodulus of rigidity of the shaft materialC = 100 GN/m2.

6. A. Write about Grubler’s criteria for planarmechanism.

B. with neat sketch explain gear toothnomenclature.

C. Explain :(i) Turning movement diagram(ii) Flywheel of a punch press

EXPLANATORY ANSWERS

1.A. Difference between Water Tube and Fire Tube Boiler

Sr. No. Factors Water tube boiler Fire tube boiler

1. Position of water and Water flows inside tubes Flue gases are inside the tubes andflue gases. and flue gases are circulated water is circulated around the tubes.

around the tubes.

2. Floor area for the It occupies less floor area. It occupies more floor areasame power.

88



89

3. Rate of steam Higher rate of steam Lesser rate of steam generation.generation. generation.

4. Construction. It is comparatively simple It is comparatively difficult inin construction. construction.

5. Transportation. Transportation is simpler. Transportation is difficult

6. Shell diameter for the The shell diameter required The shell diameter required is large.given power. is small.

7. Treatment of water. It is not so much necessary It is more necessary.

8. Chances of explosion. More chances of explosion Less chances of explosion due tosince it handles more pressure. less pressures handled.

9. Suitability for large They are more suitable. They are less suitable.power generation.

10. Nature of firing. Generally externally fired. Generally internally fired.

11. Operating pressures. It can handle pressures as It is limited to 15-17 bar.high as 100 bar.

12. Requirements of skill. It requires more skill as well It requires less skill for efficientas careful attention. and economic working.

13. Accessibility of various It has more accessibility. The parts are not so easilyparts for cleaning, accessible.repair and inspection.

1.B. Four-Stroke Diesel Engine /CompressionIgnition Engine

It operates on theoretical diesel cycle. It isalso called constant pressure combustioncycle as the combustion of fuel takes place atconstant pressure with increase of temperature.Since ignition results due to high temperatureof compressed air, these are calledcompression Ignition (C.I.) engines. Here thecycle is completed in 2 revolutions of thecrankshaft with 4-strokes of the piston namely(1) Suction (2) Compression (3) Working orpower and (4) exhaust, which are identifiedas per the function.

In a diesel engine a fine spray of diesel isinjected into the cylinder by means of a fuelvalve called injector.

Working of Four-Stroke Diesel Engine

The working and construction of a 4-strokeC.I. engine is shown in the Fig (i). Therespective theoretical and actual P-V diagrams

are shown in the Fig (ii) (a) and (b)respectively.

(1) (2) (3) (4)

Fig. (i): Four-Strokes of a Diesel EngineI = Inlet valve

E = Exhaust valve

F = Fuel injector

OC = Crank

P = Piston

(1) Suction stroke

(2) Compression stroke

(3) Working stroke

(4) Exhaust stroke

R-1829 (PP)-12



90

Suction Stroke / Induction StrokeIt starts when the piston is at the TDC and isabout to move downwards. The crankshaft isrevolved either by the momentum of theflywheel or by the electric motor to movethe piston to BDC. The inlet valve opensand the exhaust valve is closed. Due to

suction created by the downward motion ofthe piston, only air from the atmosphere isdrawn into the cylinder. It is represented byS-l on the P-V diagrams. When the crankcompletes half a revolution and the pistonreaches BDC, suction stroke ends and theinlet valve closes.

(a) (b)Fig. (ii): Theoretical and Actual P-V Diagrams of Four-Stroke Diesel Engine

S-1 Suction At 3 Fuel supply stops1-2 Compression 3-4 Working (power)

At 2 Fuel Injection starts 4-1 Cooling

2-3 Combustion at constant pressure 1-S Exhaust

2.A. Foundry or casting is one of the metal formingprocesses in which molten metal is pouredinto a pre-shaped mould cavity and allowedto cool. On cooling the metal solidifies andacquires the configuration of the mould. Thesolidified object is called “Casting”.

The term founding refers both the solidifiedobject as well as the process. Foundry' is asection of a factory where the foundry orcasting process is carried out.

Sand Casting/Foundry Process

Foundry process consists of pattern making,making moulds, melting the metal andpouring the metal into the moulds. Thusmaking castings, founding or casting is theprocess of forming metallic products bymelting the metals pouring it into a cavity ofthe required shape, known as the “Mould”

and allowing it to solidify. In this way themolten metal takes the shape of the mould.The product of the foundry is the castingwhich may vary from a fraction of a kilogramto several tonnes. Practically all ferrous andnon-ferrous metals and alloys can be cast.A shop where moulding, casting and theirrelated processes are conducted is called“Foundry Shop”.

Types of PatternsThe type of pattern selected for a particularcasting depends upon the followingconditions.1. The shape and size of casting,2. The number of castings required, and3. The method of moulding employed.

The common types of patterns are discussedin detail in the following pages:



91

(i) Solid or Single Piece Pattern:

Fig. (i) : Solid Pattern

A solid or single piece pattern, as shown inFigure (i), has a compact form. It has no joints,partings or loose pieces in its construction. Ithas usually one broad surface that serves asa parting surface in the mould. This type ofpattern is used for a limited number ofcastings because most of the mouldingoperations like parting surface formation,cutting of gating system, providing runnersand risers, withdrawal of pattern etc. is doneby hand.

(ii) Split Pattern

The most of the patterns are not made in asingle piece because of the difficultiesencountered in moulding them. In order toeliminate this difficulty, some patterns aremade into two or more pieces. A patternconsisting of two pieces is called a two piecesplit pattern, as shown in Figure (ii). One-halfof the pattern rests in the lower pan of themoulding box known as drag and the otherhalf in the upper part of the moulding boxknown as a cope. The line of separation ofthe parts is called parting line or partingsurface.

Fig. (ii): Split Pattern

Sometimes a pattern for complex castings ismade in three parts as shown in Figure (iii).Such a pattern is called a multi-piece pattern.A three piece pattern requires a moulding

box with three parts, the middle one beingcalled as cheek.

(a) (b)Fig. (iii): Multipiece Patter

The split patterns are commonly used forcasting of spindle, cylinders, steam valvebodies, water stop-cocks and taps, bearings,small pulleys and wheels.

(iii) Match Plate PatternThe match plate patterns are used on machinemoulding for quantity production of castings.A single pattern or a number of patterns maybe mounted on a match plate. When the copeand drag portions of the split pattern aremounted on opposite sides of the wooden ormetal plate (usually aluminium plate), thepattern is called a match plate pattern. Thegates and runners are permanently fastenedto the drag side of the plate in their correctpositions in order to form a complete matchplate. When the match plate is lifted off themould, all patterns are withdrawn and thegates and risers are completed in oneoperation.Figure (iv) shows a match plate pattern uponwhich the patterns of two small dumb bellsare mounted. Though the cost of construction

Fig. (iv): Match Plate Pattern



92

of the match plate patterns is quite high, yetit is easily compensated by the high rate ofproduction, greater dimensional accuracy andminimum machining requirement of casting.

(iv) Cope and Drag Pattern

When very large castings are to be made, thecomplete mould becomes too heavy to behandled by a single operator. In order to easethis problem, the cope and drag pattern isused. It is nothing but a two-piece pattern,split on a convenient joint line. One part ismoulded in a cope and the other part in adrag of the moulding box.

(v) Loose Piece Pattern

Sometimes a pattern has to be made withprojections or overhanging parts, as shown inFigure (v). These projections make theremoval of the pattern difficult. Therefore,such projections are made in loose piecesand are fastened loosely to the main patternby means of wooden or wire dowel pins. Thesepins are taken out during the mouldingoperation. After moulding, the main patternis withdrawn first and tben the loose piece isremoved with the help of a lifter.

Fig. (v): Loose Piece Pattern

(vi) Gated Pattern

The gated pattern, as shown in Figure (vi) isused for mass production of small castings.When a number of small patterns are placedin a single mould, then each pattern may beprovided with a gate with it. It consists ofpieces of wood or metal fixed to the patternsto form the runner and rising channels in themould. In this way, full supply of the moltenmetal flows into every part of the mould.

The gated pattern eliminates hand cutting ofgates and thus makes the moulding easy. If agroup of patterns is to be placed in one mould,the gate pattern has a further function ofholding the patterns in the proper positionwith respect to each other.

Fig. (vi): Gated Pattern

(vii) Shell Pattern

The shell pattern, as shown in Figure (vii), isused largely for drainage fittings and pipework. This type of pattern is usually made ofmetal mounted on a plate and parted alongthe centre line, the two sections beingaccurately doweled together. The short bendsare usually moulded and cast in pairs. Theshell pattern is a hollow construction likeshell. The outside shape is used as a patternto make the mould while the inside is usedas a core box for making cores.

Fig. (vii): Shell Pattern

2.B. A lathe is the forerunner of all machine tools.It is the most important machine used in anyworkshop. A lathe removes the material byrotating the workpiece against a single pointcutting tool. The parts to be machined can beheld between two rigid supports calledcentres, or by some other device such as a



93

chuck or face plate, that is screwed or securedto the end of the spindle.

Operations Performed by a LatheTurning: Turning involves various processesof removal of material from the outer surfaceof a workpiece to obtain finished surfaces,when the job rotates against a single pointcutting tool. The surfaces may be of uniformdiameter, stepped, tapered or contoured.Facing: Facing is the process of making flatsurfaces on a lathe. The job is held on a faceplate or chuck and the tool is fed at rightangles to the bed to obtain flat surfaces.Drilling: This is the process of making holesin the workpiece with the help of drills. Thedrill is held in the tailstock and the drillingoperation is carried out by advancing thedrill in the workpiece by rotating the handleof the tailstock. On a lathe, drilling isgenerally done in the centre of the workpiece.Reaming: It is the process of enlarging holesto accurate sizes. Reaming is always carriedout after drilling. It is similar to the drillingprocess—the reamer is held in the tailstockto carry out the reaming operation.Milling: On a lathe, the milling cutter is heldin the Deadstock and the workpiece isclamped in movable vice. The millingoperation is carried out by a cutter revolvingagainst the workpiece. This process is usedfor milling small workpieces only, where amilling machine cannot be used.Grinding: On a lathe, the workpiece is heldbetween the centres and the grindingoperation is carried out by mounting the toolpost grinder on the compound slide. Thegrinding operation is carried out after roughturning, to provide an accurate finish to theworkpiece by removing a small amount ofmaterial.Threading: The process of cutting threadson a workpiece is known as threading.External threading is the process of cuttingthreads on the outside surface of theworkpiece. Internal threading is the process

of cutting threads on the inside surface orpart of a hole.Spring Winding: The process of making acoil spring is known as spring winding. It issimilar to threading and is done by guidingthe wire around a mandrel so that the turnsarc a fixed distance apart.Spinning: It is the process of producing athin circular symmetrical article by pressingthe rotating part with a blunt tool. Pans maybe formed either from flat discs or frompreviously drawn parts.Roll Forming: It is another form of thespinning process, in which a hardened tool isused under high pressure. In this process thematerial is considerably deformed and itsthickness reduced by upto 80 per cent.

3.A. Milling MachinesMilling machine is a machine tool in whichmetal is removed by means of a revolvingcutter with many teeth, each tooth having acutting edge which removes metal from aworkpiece.

l In a milling machine the work issupported by various methods on theworkable and may be fed to the cutterlongitudinally, transversely or vertically.

l In the milling process, the workpiece isnormally fed into a rotating cutting toolknown as milling cutter. Equally spacedperipheral teeth on the cutter come incontact with the workpiece intermittentlyand machine the workpiece. This is calledintermittent cutting. In some specialmilling machines, the workpiece remainsstationary and the cutter is fed into theworkpiece.

l Milling machines are used to produceparts having flat as well as curved shapes.Intricate shapes, which cannot beproduced on the other machine tools, canbe made on the milling machines.

l This machine in perhaps next to the lathein importance.



94

Generally there are two types of millingprocesses, namely :1. Up milling (or conventional milling)

process2. Down milling (or climb milling) process.

1. Upmilling (or conventional milling)process. Refer to Fig. (i)

In “upmilling process”, the workpiece is fedopposite to the cutter’s tangential velocity.Each tooth of the cutter starts the cut withzero depth of cut, which gradually increasesand reaches the maximum value as the toothleaves the cut. The chip thickness at the startis zero increases to maximum at the end ofthe cut.

l The action of the cutter forces theworkpiece and the table against thedirection of table feed, thus each ‘tooth’enters a clean metal gradually thus theshock load on each tooth is minimised.

Fig. (i): Up Milling Process

Disadvantages :1. When making deep cuts, such as in heavy

slotting operations, the cutter tends topull the workpiece out of the vice or thefixture since the cutting force is directedupward at an angle ; This requires securedclamping of workpiece.

2. Owing to typical nature of the cut,difficulty is experienced in pouringcoolant on the cutting edge and as aresult, chips accumulate at the cuttingzone and may be carried over with thecutter, thus spoiling the surface finish.The surface becomes slightly wavy, asthe cut does not begin as soon as thecutter touches the workpiece.

2. Down milling (or climb milling) process:Refer to Fig. (ii)

In “down milling process” the workpiece isfed in the same direction as that of the cutter'stangential velocity. The cutter enters the topof the workpiece and removes the chip thatgets progressively thinner as the cutter toothrotates.The advantages of this milling process are :(i) The cutting force tends to hold the work

against the machine table, permittinglower clamping forces.

Fig. (ii): Down Milling Process

(ii) This process produces better finish anddimensional accuracy.

(iii) The coolant can be fed easily. The chipsare also disposed off conveniently andthey do not interfere with the cutting.Thus the machined surface of theworkpiece is not spoiled.

l Down or climb milling is used only onmaterials that are free of scale and othersurface imperfections that would damagethe cutters.

Classification of Milling MachinesThe milling machines are broadly classifiedas follows :

1. Column and knee type(i) Horizontal milling machine.

(ii) Vertical milling machine,(iii) Universal milling machine.(iv) Omniversal milling machine.

2. Manufacturing or fixed bed type(i) Simplex milling machine.

(ii) Duplex milling machine.



95

(iii) Triplex milling machine.3. Planer type4. Special type(i) Rotary table milling machine.

(ii) Drum milling machine.(iii) Planetary milling machine.(iv) Pantograph, profiling and tracer

controlled milling machine.

3.B. Quick Return Mechanism of a Shaper(Crank and Slotted Lever Type)The crank and slotted lever quick returnmotion mechanism for a shapcr is shown inFig. (i). An electric motor drives the pinion ata uniform angular speed which, in turn, drivesa bull gear The crank CB is contained in thebull gear and may be adjusted in length byscrew mechanism. The crank rotates at auniform angular speed with the bull gear aboutthe fixed centre C. A sliding block is attachedto the crankpin at B which is fitted within theslotted lever. This sliding block slides in theguideways of the slotted lever, the bottomend of which is pivoted at A to the base ofthe shaper while the upper end P is forkedand connected to the ram by a short link PR.

Fig. (i): Crank and Slotted Lever QuickReturn Motion Mechanism

When the bull gear rotates, the crank and thesliding block attached to the crank pin alsorotates and at the same time slides up anddown in the slotted lever causing it tooscillate about the pivoted point A. The shortlink PR transmits motion from AP to ramwhich carries the tool and reciprocate alongthe line of stroke R1R2 as shown in Fig. (ii).Thus in a shaper. the rotary motion of thecrank or bull gear is converted intoreciprocating motion of the ram. The line ofstroke (i.e., R1R2) is perpendicular to the lineAQ produced.

Fig. (ii): Principle of Quick Return Motion

The principle of quick return motion can beeasily understood from the configurationdiagram of the mechanism as shown inFig. (ii). In the extreme, positions, the slottedlever AP occupies the positions AP1 and AP2which are tangential to the crank-pin circleand the cutting tool is at the end of thestroke. The forward or cutting stroke occurswhen the crank rotates from the positionsCB1 to CB2 (or through an angle b) in theclockwise direction. The backward or returnstroke occurs when the crank rotates from theposition CB2 to CB1 (or through an angle a)in the clockwise direction. Since the crankrotates at a uniform speed, therefore,



96

Time of cutting strokeTime of return stroke

=360

b b=

a ° - bLength of stroke = R1R2 = P1P2 = 2P1Q

= CB2AP

AC´

3.C. Grinding is a metal cutting operationperformed by means of a rotating abrasivetool, called “grinding wheel”. Such wheelsare made of fine grains of abrasive materialsheld together by a bonding material, called a“bond”. Each individual and irregularlyshaped grain acts as a cutting element (asingle point cutting tool).

Selection of Grinding Wheels

The proper selection of a grinding wheel isimportant to ensure rapid work, good surfacefinish and increased wheel life. To getoptimum results, the various elements thatinfluence the process need consideration. Thefactors that influence the selection of agrinding wheel can be classified as (a)constant factors and (b) variable factors.Constant factors depend upon the material ofthe workpiece. the amount of material to beremoved, the area of contact and the finishand accuracy required. Variable factorsdepend upon the speed of the grinding wheel,the speed of the workpiece and the skill ofthe operator.

The various factors that need considerationfor selection of a grinding wheel are abrasives,grain size and shape, type of bond, bondstrength and hardness. A brief description ofthese elements follows.

Abrasive: The selection of an abrasivedepends upon the material to be ground.Silicon carbide (SiC) and aluminium oxide(Al2O3) are abrasives commonly used forgrinding wheels. Silicon carbide is used forhard materials while aluminium oxide is usedfor soft materials.

Grain size: The conventional practicefollowed in grinding is to use coarse-grained

wheels for soft materials and fine-grainedwheels for hard materials. Medium sizes areused for operations requiring stock removaland finish. For fine finish, soft wheels arepreferred.

Grain size is determined by the mesh numberand can be broadly divided into very coarse,coarse, medium, fine and very fine. The meshnumber denotes the number of meshes perlinear inch (25.4 mm) of the screen, throughwhich the grains pass when graded aftercrushing. Table (i) shows the mesh number ofvarious grains used in grinding.

Table (i)

Grain Mesh number

Very coarse 6-14Coarse 16-30Medium 36-60Fine 80-120Very fine 150-240

Grade: The grade refers to the hardness of awheel. A hard material resists wear and tearand increases wheel life. Grains are heldtogether by binding materials. The bindingmaterials must hold the abrasive until it iscompletely used. On the basis of hardness,grinding wheels can be classified as very hard,hard, medium, soft and very soft. The hardnessof grinding wheels is denoted by letters Ato Z.

Table (ii): Classification of grindingwheels on the basis of hardness

Very soft A to GSoft H to KMedium L to QHard P to SVery hard T to Z

Structure: The structure of a grinding wheelrepresents the voids between the abrasives.The material of the grinding wheel has amarked effect on the structure. The chips ofa harder material are smaller in size and the



97

rate of metal removal is low. For suchpurposes, wheels of lesser porosity are used.Wheels of greater porosity are used forremoval at a higher rate. The structure isdenoted by the numbers 1-15.

Area of contact: The area of contact betweenthe grinding wheel and the workpiece largelyaffects the grain size and grade. The area ofcontact is large in internal grinding andsurface grinding. When the area of contact islarge, the total effect of the forces is distributedover a large area, resulting in lesser pressure.Thus, softer grinding wheels are used forinternal grinding, while harder wheels are usedfor external grinding.

Wheel speed: The speed of a grinding wheelis influenced by the grade and the bond. Thehigher the speed of a grinding wheel, thesofter it is. However, the speed of grindingwheels cannot be increased beyondpermissible limits. Recommended wheelspeeds for different types of grinding wheelsare shown in Table (iii).

Table (iii)

Type of wheel Wheel speed (m/min)

Vitrified wheels 1200-2000Resinoid bonded wheels 2000-3000Hand grinding of tools 1200-1500Hand grinding of carbide tools 1000-1500Automatic grinding of HSS tools 1080-1500

Work speed: The speed at which theworkpiece traverses across the wheel face isknown as the work speed. The higher thespeed of work, the greater is the wear and tearof the wheel. If the work speed is low, thewheel wear is also low. However, low speedresults in local overheating, producesdeformation and lowers the hardness ofworkpieces by producing tempering treatment.Most grinding machines are provided withvariable speed mechanisms. As the diameterof the wheel decreases, the speed needs to beincreased accordingly to provide optimumworking conditions.

4.A. Euler’s Equation of MotionThis is equation of motion in which the forcesdue to gravity and pressure are taken intoconsideration. This is derived by consideringthe motion of a fluid element along a stream-line as :Consider a stream-line in which flow is takingplace in s-direction as shown in Fig. Considera cylindrical element of cross-section dA andlength ds. The forces acting on the cylindricalelement are:1. Pressure force pdA in the direction of

flow.

2. Pressure force pp ds dA

s¶æ ö+ç ÷è ø¶

opposite

to the direction of flow.

3. Weight of element rgdAds.

(a) (b)Fig. (i): Forces on a fluid element

Let q is the angle between the direction offlow and the line of action of the weight ofelement. The resultant force on the fluidelement in the direction of s must be equal tothe mass of fluid element × acceleration inthe direction s.

\ A A cosp

pd p ds dA gd dss

¶æ ö- + - r qç ÷è ø¶= rdAds × as ...(i)

R-1829 (PP)-13



98

where as is the acceleration in the directionof s.

Now as = ,dvdt

where v is a function of s and t.

=v ds v v v vs dt t s t

¶ ¶ ¶ ¶+ = +

¶ ¶ ¶ ¶ds

vdt

ì ü=í ýî þQ

If the flow is stedy,

vt

¶¶

= 0

\ as =v v

s¶¶

Substituting the value of as in equation (i)and simplifying the equation, we get

A A cos Ap v

dsd g d ds d dss s

¶ ¶- - r q = r ´

¶ ¶

Dividing by A, cosp v v

dsd gs s

¶ ¶r - - q =

r¶ ¶

or cos 0p v

g vs s

¶ ¶+ q + =

r¶ ¶

But from Fig. (i), we have cosdzds

q =

\1

0dp dz vdv

gds ds ds

+ + =r

or 0dp

gdz vdv+ + =r

or 0dp

gdz vdv+ + =r

...(ii)

Equation (ii) is known as Euler’s equation ofmotion.

Bernoulli’s Equation From Euler’sEquationBernoulli’s equation is obtained byintegrating the Euler’s equation of motion(ii) as

dpgdz vdv+ +

rò ò ò = constant

If flow is incompressible, r is constant and

\2

2p v

gz+ +r

= constant

or2

2p v

zg g

+ +r = constant

or2

2p v

zg g

+ +r = constant ...(iii)

Equation (iii) is a Bernoulli’s equation inwhich

pgr

= pressure energy per unit weight

of fluid or pressure head.v2/2g = kinetic energy per unit weight

or kinetic head.z = potential energy per unit weight

or potential head.

4.B. Density rDensity, also called as mass density, is definedas the mass of a substance per unit volume.For a fluid continuum, density must be acontinuous function of space. If a fluidelement enclosing a point P has a volume dVand mass dm, then the density is given by

r =V 0Lt

V Vm dm

dd ®

d=

dObviously, density is a local fluid propertywhich may vary from point to point as wellas at a point with the passage of time.

It is customary in the study of thermo-fluiddynamics to use the terms ‘specific volume’.The specific volume is the volume per unitmass of the fluid. It follows, therefore, thatthe specific volume v is the inverse of themass density r ;

v =1r

Pressure pThe pressure at a point in a fluid continuumis defined as the normal compressive forceper unit area at that point. In other words, thecompressive normal stress is referred to as thepressure ;

R-1829 (PP)-13-II



99

p = – s

It may be noted that a fluid can withstandhigh compressive stresses or pressure but it isweak in tension. It is for this reason that thepressure in a fluid is always referred abovethe absolute value since negative absolutepressure would imply tension.

Atmospheric pressure may also serve as asuitable reference for pressure measurement.Pressure above the atmospheric value is calledgauge pressure and that below it is calledvacuum or sub-atmospheric pressure.Relationship between the pressure stated fordifferent references is shown in Fig. (i).

Fig. (i): Pressure References

Newton’s Law of Viscosity:According to the Newton’s law of viscosity,the shear stress on a fluid element layer isdirectly proportional to the rate of strain, theconstant of proportionality being called thecoefficient of viscosity.

Consider, for example, a fluid containedbetween two parallel plates, distance h apart,as shown in Fig. (ii). If the lower plate is heldstationary and the upper plate is moved at asmall but constant velocity u by theapplication of a force F, the fluid will besubjected to shearing forces and the velocitywill vary across the gap. A typical velocityprofile is shown. Consider the velocity profile,as shown by an enlarged view in the vicinityof a point P. The rate of strain at the point isgiven by

e =du

dy

Velocity Profiledudy

t = m

(Enlarged View at P)Fig. (ii): Relationship between Stress

and Rate of Strain

and the shear stresses across the layer isindicated as t. According to the Newton's lawof viscosity,

t µ e or t µdu

dywhich may be written as

t =du

dym

where the constant of proportionality m iscalled the coefficient of viscosity.

CompressibilityThe measure of compressibility of a fluid isthe bulk modulus of elasticity K defined asthe compressive stress per unit volumetricstrain. Alternatively, the coefficient ofcompressi-bility b is defined as volumetricstrain per unit compressive stress, i.e., theinverse of the bulk modulus K.

b =1K

Consider the piston and cylinder arrangementshown in Fig. (iii), let the volume and pressure



100

at any instant of compression be v and prespectively. The compressibility curve drawnbetween the compressive stress p and thevolumetric strain dv/v is also shown. Bydefinition,

K =1

/ /dp dp dp

dv v d d= = r =

r r r b

because,

dvv

=dr

-r

since v µ 1/r

Fig. (iii)

Units of the bulk modulus of compressibilityK can be derived from its definition. Sincethe units of compressive stress p are N/m2

and the volumetric strain dv/v is dimen-sionless, units of K must be N/m2. Values ofK for different fluids are expressed in terms ofkN/m2 or MN/m2 for the sake of convenience.

Surface Tension s

It is the property of the apparent tension effectwhich occurs at the interface of a liquid anda gas or at the interface of two immiscibleliquids. The origin of surface tension isexplained by considering the mechanism ofcohesive forces within a liquid as in Fig. (iv).

Fig. (iv): Origin of Surface Tension

The surface tension decreases as thetemperature of the liquid increases becausethe intermolecular cohesive forces decrease.At the critical temperature of a liquid, thesurface tension becomes zero and theboundary between the liquid and its vapourvanishes!

The force of surface tension is a line force ;it is expressed as force per unit length drawnon the free surface of a liquid. Thephenomenon of surface tension isalternatively explained by the existence offree-surface energy at the interface of a liquidand another fluid. The surface tension islikewise expressed as free surface energy perunit length of the surface.

Water in contact with air has a surface tensionof 0.073 N/m at 20°C. Mercury in contactwith air has a surface tension of 0.5 N/m. Thesurface tension of water is considerablyreduced by adding a small quantity of anorganic solute such as soap or detergent. Thesurface tension of water increases if a saltsuch as sodium chloride is dissolved in it.Organic liquids such as alcohol have surfacetension varying between 0.02 and 0.03 N/m.

5.A. The behaviour of a material subjected to anincreased tensile load is studied by testing aspecimen in a tensile testing machine andplotting the stress strain diagram. Stress-straindiagrams of different materials vary widely.However, it is possible to distinguish somecommon characteristics among various stress-strain diagrams of various groups of materials.It is observed that broadly the materials canbe divided into two categories on the basis ofthese characteristics: ductile materials andbrittle materials.

Ductile materials such as steel and manyalloys of other metals have the ability toyield at normal temperatures. The plotbetween strain and the corresponding stressof a ductile material is represented graphicallyby a tensile test diagram. Figure shows astress vs. strain diagram for steel in which the



101

stress is calculated on the basis of originalarea of a steel bar. Most of other engineeringmaterials show a similar pattern to a varyingdegree. The following are the salient featuresof the diagram:

Fig.

l When the load is increased gradually,the strain is proportional to load or stressupto a certain value. Line OP indicatesthis range and is known as the line ofproportionality. Hooke’s law is applicablein this range. The stress at the end pointP is known as the proportional limit.

l If the load is increased beyond the limitof proportionality, the elongation isfound to be more rapid, though thematerial may still be in the elastic state,i.e., on removing the load, the strainvanishes. This elongation with arelatively small increase in load is causedby slippage of the material along obliquesurfaces and is mainly due to shearstresses. The point E depicts the clasticlimit. Hooke’s law cannot be applied inthis range as the strain is not proportionalto stress. Usually, this point is very nearto P and many times the differencebetween P and E is ignored and thereforeelastic limit is taken as the limit ofproportionality.

l When the load is further increased, plasticdeformation occurs, i.e., on removing theload, the strain is not fully recoverable.At point Y, metal shows an appreciable

strain even without further increasing theload. Actually, the curve drops slightlyat this point to Y¢ and the yielding goesup to the point Y¢¢. The points Y¢ and Y¢¢are known as the upper and lower yieldpoints respectively. The stress-strain curvebetween Y and Y¢ is not steady.

l After the yield point further straining ispossible only by increasing the load. Thestress-strain curve rises up to the point U,the strain in the region Y to U is about100 times that from O to Y¢. The stressvalue at U is known as the ultimate stressand is mostly plastic which is notrecoverable.

l If the bar is stressed further, it begins toform a neck, or a local reduction in cross-section occurs. After this, somewhat lowerloads are sufficient to keep the specimenelongating further. Ultimately, thespecimen fractures at point R. It is notedthat fracture occurs along a cone-shapedsurface at about 45° with the originalsurface of the specimen indicating thatshear is primarily responsible for thefailure of ductile materials.

l If the load is divided by the original areaof the cross-section, the stress is knownas the nominal stress. This is lesser at therupture load than at the maximum load.However, the stress obtained by dividingwith the reduced area of cross-section isknown as the actual or true stress and isgreater at the maximum load. It is shownin the figure by the dotted line.

In brittle materials such as cast iron, glassand stone, etc., rupture occurs without anyappreciable change in the rate of elongationand there is no difference between theultimate strength and the rupture strength.The strain at the time of rupture is muchsmaller for brittle materials as compared toductile materials. There is no neck formationof the specimen of a brittle material and therupture occurs along a perpendicular surface



102

to the load indicating that normal stresses areprimarily responsible for the failure of brittlematerials.

5.B. (i) Given data :b = 100 mm, h = 150 mmE = 200 GPam = 300 kNm

Moment of inertia

I =3

12bh

I =30.1 (0.15)

12´

Distance of N.A. from the top surface of thebeam :

y = h/2 =0.15

0.075m2

=

= 0.000028125 m4

Using the relation a = My/I

s =3300 10 0.075

0.000028125´ ´

= 8 × 108 N/m2

(ii) Radius of Curvature

R =9

8

E.y 200 10 0.075

8 10

´ ´=

s ´= 18.75 m

100 mm

150 mmNA

35 mm

40 mm

75 mm

(iii) Using the relation, M/I = s/y = s1/y1

s1 = My1/I =3 3300 10 35 10

0.000028125

-´ ´ ´

= 373.33 MN/m2

5.C. P =2 NT

60 1000p´

Þ 80 =2 200 T

60 1000p ´ ´

´

Þ T =80 60 1000

3821.66 N m2 200´ ´

= -p ´

First case: Considering allowable shear stress(60 MPa),

T =3D

16t ´ p ´

Þ 3821.66 = 60 × 106 × 36 D16p

´

D = 0.0687 m = 68.7 mmSecond case : Considering angle of twist (1°),

P

TT

=GLq

Þ4

3821.66

D32p

´ =

9100 10 1 /1802

´ ´ ´ p

D = 0.081745 m= 81.745 mm

From the two cases we find that suitablediameter for the shaft is 81.745 mm (i.e.,greater of the two value)

6.A. Grubler’s Criterion

Grubler's criterion for planar mechanisms oflower pairs incorporated the number of linksn and the number of joints j into an equationvalid for constrained motion i.e., F = 1 for thechain. Grubler’s criterion applies tomechanisms with only single degree offreedom joints where the overall mobility ofthe mechanism is unity i.e., F = 1 and j2 = 0.We find Grubler's criterion for planarmechanisms with constrained motion asfollows:

1 = 3 (n – 1) –2j1 – 0

i.e., 1 = 3n – 3 – 2j1

i.e., 3n – 2j1 – 4 = 0 ...(i)

This equation can also be obtained as follows :



103

Fig.: Rigid Link 2 in Planar Motion withRespect to Reference link 1.

Consider two links in planar motion. Link 1,with co-ordinate system Oxy is taken as thereference plane. The moving plane or link 2carries a line AB containing a point A. Theposition of the link 2 with respect to link 1is determined from the specification of threevariables, the coordinates xA and yA of thepoint A and the inclination q of the line AB.These three variables are further identified asthe degrees of freedom of a body in a planarmotion; for such a body maximum degrees offreedom are 3. If xA and yA are made invariantby a revolute connection at A, two restraints‘r’ to motion are imposed on link 2 and itsdegree of freedom drops to F = 1; its positionis now given by only a single variable q. Arevolute pair connection is thus characterisedby either F = 1 or r = 2.

Suppose that we have a closed chain of nlinks, connected by j1 revolute pairs anddegrees of freedom F. Each link initiallypossesses three degrees of freedom beforeconnection to any other link, hence the totaldegrees-offreedom number 3n. On choosingone link as a reference for all others, i.e.,fixing one link to create a mechanism, n – 1moving links remain and the degrees offreedom 3(n – l). Each revolute connectionmeans the loss of two degrees of freedom;with j1 connections, there is a loss of 2j1degrees of freedom. Since we are concernedwith mechanisms with only single degrees offreedom joints, therefore j2 = 0. The resulting

degrees of freedom (F) of a planar n-linkmechanism is given by

F = 3(n – 1) – 2j1

A mechanism with constrained motion hasF = 1, therefore

1 = 3(n – l) – 2j,i.e., 3n – 2j1 – 4 = 0This equation is same as equation (i).Rearranging the terms, produces

2j1 – 3n + 4 = 0 ...(ii)Equation (ii) is called as the classical form ofthe Grubler’s criterion. Here, it is importantto note that Grubler's criterion involves onlythe number of links n and the revoluteconnections j1 and pays no attention to thelink dimensions or other geometric features.As the j1 and a are to be whole numbers, theGrubler’s criterion can be satisfied only if nis even.Applications of the equation (ii) to variousclosed kinematic chains will allow certaindeductions as mentioned below :For possible linkages made of binary linksonlyn = 4, j1 = 4, No excess turning pairn = 6, j1 = 7, One excess turning pairn = 8, j1 = 10, Two excess turning pairs

and so on.Thus, with the increase in the number of links,the number of excess turning pairs goes onincreasing. To get the required number ofturning pairs from the required number ofbinary links is not possible. Therefore, theexcess or the additional pairs or joints can beobtained from the links having more thantwo ends i.e., ternary or quaternary links, etc.

6.B. Gear Tooth Nomenclature

Addendum: The addendum is the height bywhich a tooth of a gear projects beyond(outside for external, or inside for internal)the standard pitch circle or pitch line; also,the radial distance between the pitch diameterand the outside diameter.



104

Addendum Circle: The addendum circlecoincides with the tops of the teeth of a gearand is concentric with the standard (reference)pitch circleand radially distant from it by theamount of the addendum. For external gears,the addendum circle lies on the outsidecylinder while on internal gears the addendumcircle lies on the internal cylinder.Base Circle: The base circle of an involutegear is the circle from which involute toothprofiles are derived.Base Pitch (Pb) – is the circular pitch in theplane of rotation at the base circle.Pb = cos f.p/Pd (f = Pressure Angle, Pd =Diametral Pitch)Bottom Land – is the surface of a gearbetween the flanks of two adjacent teetch.For external gears it is measured in the rootand for internal gears it is measured at theminor tip of the tooth.Circular Pitch (p) is the distance along thepitch circle or pitch line between corres-ponding profiles of adjacent teeth.

Clearance: Clearance is the radial distancefrom top of the tooth to the bottom of thetooth space in the mating gear.Dedendum (b) is the depth of a tooth spacebelow the pitch line. It is normally greaterthan the addendum of the mating gear toprovide clearance.Dedendum Circle: Dedendum circle is theinner most profile circle. Dedendum is theradial distance between the pitch circle andthe dedendum circle.Face Width: Face width is length of toothparallel to axes.Pitch Circle is the circle derived from anumber of teeth and a specified diametral orcircular pitch. Circle on which spacing ortooth profiles is established and from whichthe tooth proportions are constructed.Top Land is the surface of the top the tooth.Tooth Thickness: The thickness of the toothmeasured on the pitch circle. It is the lengthof an arc and not the length of a straight line.

Fig.: Gear, Footh, Nomenclature

6.C(i). The turning moment diagram also knownas crank-effort diagram is the graphicalrepresentation of the turning moment orcrank-effort for various positions of the crank.It is plotted on Cartesian coordinates, inwhich the turning moment is taken as theordinate and crank angle as abscissa.

From, x = ( ){ }0.52 2(1 cos ) sinr n né ù- q + - - qê úë ûLength of stroke, L = 2r,

\Lx

= ( ){ }0.52 20.5 (1 cos ) sinn né ù- q + - - qê úë ûx/L is called the displacement constant.



105

Procedure for Determination of TurningMoment Diagram

The following steps may be followed todetermine the turning moment diagram:

1. Determine the pressure of the workingfluid on both sides of the piston eitherfrom the indicator diagram or fromtheoretical calculations. Calculate the netload on the piston. Ps.

2. Calculate the acceleration of the piston.ap.

3. Calculate the force due to reciprocating

parts, .R .pa

g

4. Calculate the piston effort. PE = Ps ± R

— .R .pa

g Remember that second term is

zero for a horizontal engine. For a verticalengine, take the positive sign before Kfor outstroke and the negative sign forinstroke.

5. Calculate the crank effort,

CE =sin 2

PE sin .2

rn

qé ù´ q +ê úë ûTurning Moment Diagram for a VerticalSteam Engine

The turning moment diagram for a verticalsteam engine is shown in Fig. (i). The hatchedarea below the mean torque shows deficientenergy and the area above the mean torque isthe surplus energy. One cycle is completedduring 360º of the crank rotation.

Fig. (i): Turning moment diagram for avertical steam engine

Turning Moment Diagram for a FourStroke I.C. Engine

The turning moment diagram for a four strokei.e., engine is shown in Fig.(ii). Here onecycle is completed during 720° of the crankrotation. Energy is supplied mainly duringthe expansion stroke. This diagram is morenon-uniform as compared to Fig.(ii).

Fig. (ii): Turning Moment diagramfor a four stroke I.C. Engine

Uses of Turning Moment Diagram

The uses of turning moment diagram are:1. The area under the turning moment

diagram represents work done per cycle.This area multiplied by number of cyclesper second gives the power developedby the engine in Watts.

2. By dividing the area of the turningmoment diagram with the length of thebase, we get the mean turning moment.This enables us to find the fluctuation ofenergy.

3. The maximum ordinate of the turningmoment diagram gives the maximumtorque to which the crankshaft issubjected to. This enables us to find thediameter of the crank shaft.

6.C.(ii): Flywheel of a Punch Press

Flywheels are used to reduce the speedfluctuations during the working cycle of a

R-1829 (PP)-14



106

machine. They increase their rotational speedwhen there is an excess of energy and decreasetheir rotational speed to release energy whenthere is not enough available. Flywheels servea function similar to accumulators used inpneumatic or hydraulic circuits, whichmaintain nearly constant fluid pressure whilethe demand varies. In piston engines, theflywheel compensates for the strokes whenenergy is consumed rather than created duringthe engine cycle, thus allowing the crankshafttorque to be delivered at close to constantspeed. In case of punch presses, like the onein Figure, the actual punching occurs for onlya small fraction of the machine cycle, causinga strongly fluctuating load torque. To limitthe size of the motor and also to alleviate itsspeed fluctuation (electric motors are knownto work best at certain rpm), the energydelivered during the actual punch issupplemented by the energy released by theflywheel as it slows down from a maximumangular velocity wmax right before the punch,to a minimum angular velocity wmin rightafter the punch ends.

Here, we will consider the problem of selectingthe electric motor and that of sizing theflywheel required by a punch press. Theflywheel is assumed mounted on thecrankshaft, which is driven by the motor viaa speed reducer as shown in Figure. Themechanism of the press is a centric crank-slider with the crank length OA = 0.15 m,coupler length AB = 0.5 m, and punch-headlength BP = 0.15 m. The press punchesd = 65 mm diameter holes into h = 20 mm

thick aluminum stock of shear strengthSsy = 140 MPa, at a rate of np = 80 holes perminute. The punch begins when thedisplacement s of the punch head equals ss

O

y

Aq w,

x

Motor Flywheels

ss

sf

F

Stock

B

P

iMT

Fig.: Schematic of a crank-slider punch press.

and ends when s equals sf (see Figure). Weassume that sf = 0.75 m and correspondinglyss = sf – h = 0.73 m. Using the principle ofvirtual work, we will find the resisting torqueat the crankshaft as a function of the crankangle q and then we will evaluate the requiredaverage motor torque and its correspondingpower. Then we will calculate the moment ofinertia I of a flywheel for which the motorspeed fluctuation ranges between 1740 and1580 rpm.



107

PAPER-II



1. A. What do you understand by characteristiccurves of centrifugal pump? Draw thefigure.

B. What do you mean by specific speed ofa turbine? Explain.

C. What will be the force exerted by(i) direct impact of a jet on a stationary

flat plate?(ii) oblique impact of a jet on a stationary

flat plate?Explain with neat sketches.

2. A. Show that the air standard efficiency ofOtto cycle depends on compression ratioonly.

B. Describe working of a simple plain tubecarburettor with the help of a neat sketch.

3. A. A load of 270 kN is applied on a shortconcrete column 250 mm × 250 mm. Thecolumn is reinforced with 8 bars of 16mm diameter. If the modulus of elasticityfor steel is 18 times that of concrete, findthe stresses in concrete and steel.If the stress in concrete shall not exceed4 N/mm2, find the area of steel requiredso that the column may support a load of400 kN.

B. Draw the B.M. and S.F. diagrams for theoverhanging beam carrying loads asshown in figure given below.Mark the values of principal ordinatesand locate the point of contraflexure.

4. A. Define inversion. Write its properties andimportance.

B. A capstan and a rope are used in a railwaygoods yard for moving trucks. The capstanruns at 50 r.p.m. The rope from the line oftrucks makes 2.75 turns around the capstanat a radius of 20 cm and the free end ofthe rope is pulled with a force of 147.15 N.Determine the pull on the trucks, thepower taken by the trucks, and the powersupplied by the capstan Take µ = 0.25.

5. A. Define and explain with proper sketches,the following lathe operations:(i) Grooving

(ii) ChamferingB. Explain Taper turning on lathe in datail.C. Determine the angle at which the

compound rest would be swivelled forcutting a taper on a workpiece having alength of 150 mm and outside diameter80 mm. The smallest diameter on thetapered end of the rod should be 50 mmand the required length of the taperedportion is 80 mm.

6. A. Explain centreless grinders. Give sketchesfor external centreless grinding and writeabout (i) through feed (ii) infeed and(iii) end feed.

B. Calculate the time required to drill a25 mm diameter hole in a workpiecehaving thickness of 60 mm to thecomplete depth. The cutting speed is14 m/min and feed is 0.3 mm/rev. Assumelength of approach and overtravel as5 mm.

107



108

EXPLANATORY ANSWERS

1.A. Characteristic Curves of Centrifugal Pump

Characteristic or performance curve aredesigned to predict the behaviour of a pumpwhen it is working under different operatingconditions. The curves are drawn from theresults of a number of tests on the centrifugalpump. The following are the importantcharacteristic curves for pumps :1. Main characteristic curves2. Operating characteristic curves3. Constant efficiency4. Constant head and constant discharge

curves

1. Main Characteristic CurvesIn order to draw main characteristic curves,tests are conducted at different speed andfor each speed, the discharge is varied.Then, for each discharge, head and shaftpower are measured and the overallefficiency is calculated. Figure (i) showsthe main characteristics curves of a pump.

Fig. (i): Main characteristic curves of a pump

2. Operating Characteristic Curves

The operating characteristic curves maybe obtained when the pump is operated atthe design speed. The speed is keptconstant and the variation of dischargewith respect to manometric head, powerand efficiency are plotted. Figure (ii)

shows operating characteristic curves of apump.

Fig. (ii): Operating characteristiccurves of a pump

3. Constant Efficiency CurvesThese curves are used to determine therange of operation of a pump at givenefficiency. These curves are called Muschelcurves for the pump or iso-efficiency curve.These curves are obtained from the maincharacteristic curves of discharge versusefficiency and head versus dischargecurves. Horizontal lines indicatingconstant efficiency are plotted on thedischarge versus efficiency curves, asshown in Figure (iii). The points at whichthese horizontal lines cut the efficiencycurves at various speeds are transferred to

Fig. (iii): Constant efficiency curves of a pump



109

the corresponding head versus dischargecurves. Then, the points having the sameefficiency are joined to obtain constantsmooth efficiency curves.

4. Constant Head and Constant DischargeCurves

The constant head curves may be obtainedwhen the pump is operated at a constanthead. Discharge versus speed curve isobtained by measuring the discharge atvarious speeds while keeping the headconstant. The constant discharge curvesmay be obtained when the pump isoperated at constant discharge. Headversus speed curve is obtained bymeasuring the head at various speeds,while keeping the discharge constant.Figure (iv) shows the constant head andconstant discharge curves of a pump.

Fig. (iv): Constant head and constantdischarge curves of a pump

1.B. Specific Speed of a Turbine

The specific speed of a turbine is defined asthe speed of a turbine which is identical inshape, geometrical dimensions, blade angles,gate opening etc., with the actual turbine butof such a size that it will develop unit horsepower when working under unit head.

Specific speed Ns = 5/4

N P

H

(where P is in kW and H in metres)

[Ns (S.I. Units) = 0.86 Ns (metric)]

Specific speed plays an important role forselecting the type of the turbine. Also theperformance of a turbine can be predicted byknowing the specific speed of the turbine.

To compare the characteristics of machinesof different types, it is necessary to know acharacteristic of an imaginary machineidentical in shape. The imaginary turbine iscalled a specific turbine. The specific speedprovides a means of comparing the speed ofall types of hydraulic turbines on the basis ofhead and horse power capacity.

If a runner of high specific speed is used fora given head horse power output, the overallcost of installation is lower. The selection oftoo high specific speed reaction runner wouldreduce the size of the runner to such an extentthat the discharge velocity of water into thethroat of draft tube would be excessive. Thisis objectionable because a vacuum may becreated in the extreme case.

The runner of too high specific speed withavailable head increases the cost of turbineon account of high mechanical strengthrequired. The runner of too low specific speedwith low available head increases the cost ofgenerator due to the low turbine speed.

An increase in specific speed of turbine isaccompanied by lower maximum efficiencyand greater depth of excavation of the drafttube. In choosing a high specific speedturbine, an increase in cost of excavation offoundation and draft tube should beconsidered in addition to the efficiency. Theweighted efficiency over the operating rangeof turbine is more important in the selectionof a turbine instead of maximum efficiency.

Table gives the specific speeds for variousturbines.



110

Table: Specific Speeds

Specific Speed (Ns)

Type of Turbine M.K.S. Units S.I. Units

Slow 4–10 3.5–8.5Impulse (Palton) Normal 10–25 8.5–21.5

Fast 25–60 21.5–51.5

Radial and mixed flow Slow 60–150 51.5–130(Francis and Deriaz) Normal 150–250 130–215

Fast 250–400 215–345

Slow 300–450 255–385Axial flow (Kaplan) Normal 450–700 385–600

Fast 700–1000 600–860

ìïíïî

ìïíïî

ìïíïî

1.C (i). Impact of Jet on a Stationary Vertical Plate

Figure shows a jet which strikes perpendi-cularly on a vertical stationary plate. The jetafter striking the plate leaves it tangentiallyas shown in the figure. In other words, the jetgets deflected through 90°. Considering theplate to be smooth, the loss due to friction isneglected. Further, if there is no change ofelevation, i.e., the jet strikes and leaves theplate at the same elevation, the striking aswell as the leaving velocity of the jet remainsthe same. This is also evident from theapplication of Bernoulli's equation.

Figure: Impact of jet on stationary vertical plate

Let,a = the area of jet

d = diameter of the jetV = striking as well as leaving velocity

of the jet on the vertical stationaryplate

w = specific weight of liquidg = acceleration due to gravity

r =wg

æ öç ÷è ø

be the mass density of water

Now, the quantity of water striking the plateper second is

Q = 2V4

a d Vpæ ö= ç ÷è ø

The mass of water striking the plate is givenby

rQ =w

aVg

æ öç ÷è ø

After striking the plate, the jet gets deflectedthrough an angle of 90°, and hence thecomponent of velocity of jet leaving the jetis zero. The dynamic force P exerted by thejet on the plate is the mass flow ratemultiplied by the change in velocity, i.e. thechange of momentum.Mathematically, it is expressed as:

P =Mass

Time (s) × Change in velocity



111

or P = Q V(V 0)w

ag

r = -

P =2

2VV

waa

g= r

Sincew

g = Mass density r

1.C (ii). Impact of Jet on a Stationary InclinedPlate

The plate is now placed at an angle of q tothe direction of the jet as shown in Figure.The water coming from the jet after theimpact leaves the plate tangentially in alldirections as shown. It is also assumed thatno frictional loss occurs on the impact, andthe small difference of elevation of incomingjet and the outgoing tangential flow due toinclination is neglected. The velocity of thejet perpendicular to the plate before impact isV sin q, and after the impact it is zero as theplate is stationary. Now, applying the impulse-momentum equation, the dynamic forceexerted by the jet on the plate is the productof the mass of jet water impacted and thechange of velocity perpendicular to the plate.Mathematically,

P = ( )Q V Vsin 0w

ag

r = q -

Therefore, P =2

2Vsin V sin

waa

gq = r q

Figure: Impact of jet on stationary inclined plate

There is no resultant force acting in thedirection of the plate. The dynamic force P

can be resolved into two components: oneparallel to the direction of jet and the otherone perpendicular to the direction of the jet,i.e. x and y directions, respectively. Thesetwo components are :

Px =2

2VPsin sin

wag

q = q

Py =2V

Pcos sin coswa

gq = q q

= raV2 sin q cos q

2.A. Otto Cycle (Constant Volume Cycle)

This cycle is composed of four internallyreversible processes, two adiabatic and twoconstant volume processes. The p-v and T-sdiagrams are shown in Fig. (i). The variousprocesses are:Process 1-2 : Isentropic compression.Process 2-3 : Constant volume heat addition.Process 3-4 : Isentropic expansion.Process 4-1 : Constant volume heat rejection.This cycle is used for spark ignition (petrol)engines.

(a) P-v diagram (b) T-s diagram

Fig.: Otto Cycle

Consider 1 kg of air flowing through thecycle. Since the air in the cylinder acts as aclosed system, from first law of thermo-dynamics for isentrophic compression andexpansion, we have

q – w = Du

For constant volume heat supplied andrejection process, since w = 0.

q = Du = cvDT



112

Heat supplied = qs = q2–3 = cv(T3 – T2)Heat rejected qr = q4–1 = cv(T4 – T1)Work done per cycle,

wnet = qs – qr

= cv(T3 – T2) – cv(T4 – T1)Air standard (or thermal) efficiency

=Work done

Heat supplied

ha or hth ; ( ) ( )( )

3 2 4 1

3 2

T T T T

T Tv v

v

c c

c

- - --

= 4 1

3 2

T T1

T T-

--

Now, T2 = 111 1

2

T Tv

rv

g -=

Where, r = 1 1

2 c

v vv v

=

= compression ratio

T3 = 2 32

2

TTp

pp

= a

= apT1rg –1

Fig. (ii): Otto cycle thermal efficiencyv’s compression ratio

Where, ap = 3 3

2 2

TT

pp

= = pressure ratio

T4 =1

3 11

T Tpr

g -æ ö = aç ÷è ø

\ ha =1 1

th 1 11 1

T T1

T Tp

p r rg - g -

a -h = -

a -

= 1

11

( 1)p

p r g -

a --

a -

= 1

11

r g --

The air standard efficiency of Otto cycledepends on compression ratio only andincreases as compression ratio increases(Fig. ii). In actual engine working on Ottocycle, the compression ratio varies from 5 to8. This engine is used for spark ignitionengines working on petrol.

2.B. Simple Carburettor

The simple carburettor is also called the simpleplain-tube carburettor. The main parts of thesimple carburettor are the float chamber andthe venturi tube. The details are shown inFigure (i).

Fig. (i): Simple Carburettor

The float chamber consists of a containerinto which petrol can flow from the fuel tankthrough the inlet which is provided with aneedle valve which can open or shut theinlet. The float as the name indicates, floatson the surface of the liquid fuel in the floatchamber. A lever connected rigidly to thefloat turns about a hinge on the other side ofwhich is the needle valve. When the float



113

rises with the level of fuel in the chamber, thelever rotates about the hinge pin. The otherside of the lever pushes the needle in thevalve against its seat and shuts off furtherinflow of fuel into the float chamber. Theflow of fuel into the float chamber isautomatically controlled by the combinedaction of the float, the hinged lever and theneedle valve.

The venturi tube is a circular pipe with anarrow section throat or constriction at thecentre. Protruding into the central cross-section of the venturi tube is a very narrowpipe with a nozzle at its end. The other endof the narrow pipe is connected to the bottomof the float chamber. The level of the tip ofthe nozzle in the venturi is slightly above thelevel of fuel in the float chamber.

One end of the venturi tube is connected tothe air intake from the atmosphere throughthe air cleaner. The other end of the tube isconnected to the intake manifold of theengine. During the suction stroke atmosphericair is drawn into the venture tube. At theconstriction the velocity of flow increasesand the pressure decreases. Due to thedepression, the fuel from the nozzle rises andflows into the gushing air stream. The finefuel droplets from the spray start vaporizing.Downstream the mixture of air, fuel dropletsand some fuel vapour enter the mixingchamber behind the throttle valve. Thethrottle valve is a circular flap hinged on itsdiameter, which controls the flow of air intothe engine by operation of the accelerationpedal. As the throttle valve opens more andmore, the vacuum pressure at the throat of theventuri tube increases and fuel is sprayedinto the air stream. The mixture chamber helpsin mixing the fuel droplets in the air andformation of some fuel vapour. The degree ofvaporization of fuel depends on many factorslike the properties of the fuel, the temperaturein the manifold, the depression in the venturitube, the temperature of the incoming air and

primarily on the degree of atomization or thesize of the fuel droplets.

3.A. Given:Area of concrete column

= 250 × 250 = 62500 mm2

Area of steel bars,As = 1607.68 mm2

\ Area of concrete,

Ac = 62500 – 1607.68

= 60892.32 mm2

Total load on column,

P = 270 kN = 270000 N

E for steel = 18 × E for concrete

or, Es = 18 Ec

Let, ss = Stress in steel in N/mm2

sc = Stress in concrete in N/mm2

Now, strain in steel = Strain in concrete

or,E

s

s

s =E

c

c

s StressStrain

Eæ ö=ç ÷è øQ

or, ss =E

18E

sc c

c

s = s ...(i)

Also, we know that

Total load = Load on steel + Load onconcrete

or, P = ss . As + sc . Ac

(Q Load = Stress × Area)

(Q ss = 18 sc)

or, 270000 = 18sc × 1607.68 + sc

× 60892.32

= 89830.56 sc

\ sc =270000

89830.56 = 3 N/mm2

Substituting this value in equation (i), we get

ss = 18 × 3 = 54 N/mm2

Area of steel required so that column maysupport a load of 400 kN and maximum stressin concrete is 4 N/mm2.

R-1829 (PP)-15



114

Let, As* = Area of steel required

Area of concrete = 62500 – As*

ss* = Stress in steel in N/mm2

sc* = Stress in concrete = 4 N/mm2

P* = Total load = 400 kN = 400000 N

We know that

ss* = 18 sc*

= 18 × 4 (Q sc* = 4)

= 72 N/mm2

Also, we know that

Total load = load on steel + Load on concrete

or, P* = ss* × As* + ss* × Ac*

or, 400000 = 72 × As* + 4 × (62500 – As*)

= 72 × As* + 4 × 62500 – 4As*

or, 400000 – 4 × 62500 = 68As*

or, 150000 = 68As*

\ As* =1,50000

68 = 2206 mm2.

3.B. First calculate the reactions RA and RB

Upward forces = Downward forces

RA + RB = 20 × 7 + 10

= 150 kN

Taking moments of all forces about A, we get

RB × 5 = 20 × 7 × 3.5 + 10 × 7

= 560 kNm

RB = 112 kN : RA = 38 kN

Shear force diagram

Shear force Shear force towards

At point Left of the section Right of the section

A — 20 × 7 + 10 – 112= 38 kN

B 38 – 20 × 5 = –62 kN 20 × 2 + 10 = 50 kN

C 38 + 112 – 20 × 7 —= 10 kN

Bending Moment Diagram:MA = 0

5 m 2 m

40

621.9m

38A

DB C

+50

RA

A

Re

B C

10 kN2 kN/m

( )a

( )b

S.F. diagram

–

+

4.A. Inversion is a type of transformation thatmoves points from the inside of a circle tothe outside and from the outside of a circleto the inside using a specific rule. It can beused to create images such as the one on thispage, but more importantly it can be used tosimplify certain proofs. Inversion is also amethod for understanding and solving theProblem of Apollonius.

Inversion of a Point

The inversion transformation is defined bythe rule that the distance from the center ofthe circle, O, to the original point, P¢, timesthe distance from O to the inverse point, P,equals the length of the radius of the circle,k, squared. In mathematical notation:| OP || OP¢ | = k2.

This applet can be seen on its own page at:Inversion of A Point to the left you can seeexactly how inversion works by moving thered point which you can think of as P¢ andwatching the blue inverse point, P, movedepending on the rule: | OP || OP¢ | = k2.

4.B. Given N = 50 r.p.m.; n = 2.75 turns;r = 20 cm;F = 147.15 N; m = 0.25

Here, total angle of lapq = 2.75 × 2p = (5.5)p

Therefore, limiting ratio of tensions,

R-1829 (PP)-15-II



115

T1/T2 = e0.25 × 5.5p= e4.3197 = 75.165

(Note: The grooves in capstan permit contactwith ropes over the entire groove surface andas such, groove angle 2a loses its significance)

Sol.

Angle of lap, q = 2.75 truns= 2.75 × 2p radians= 5.5p radians

Pull on the free end = T2 = 147.15 N

Speed of capstan = 50 rpm

Coefficient of friction = 0.25

Radius of capstan r = 20 cm

Þ Diameter = 40 cm

Let T1 be pull in rope on the side of trucks.

Tension ratio T1/T2 = emq = e5.5p × 0.25

= 74.967

\ T1 = 74.967 T2

= 74.967 × 147.15

= 11031.4 N

Velocity of rope =N 0.4 50

60 60dp p ´ ´

=

= 0.33p m/s

\ Power supplied, P = (T1 – T2)v

= (11031.4 – 147.15)

× 0.33p watts

= 11380.8 watts

= 11.38 kW

5.A. (i) Grooving: It is the process of reducing thediameter of work piece over a very narrowsurface is called as grooving. Usually, thetool used for this application is called asgrooving tool and it has an edge with a smallwidth. The spindle speed and feed are kept

slightly lower than the plain turning. Thegrooving tool fixed on the tool post is fedagainst the work to get a required shape andsize as shown in Fig.

Fig.: Grooving Operation

(ii) Chamfering: The process of removingsharp edges from the edge of the work piecesurface is called as chamfering. This operationprotects the labour from the sharp edges andburrs. It is usually performed after all theoperations are completed. For chamferingoperation, the tool used must have cuttingedge approximately set at an angle of 45°.This operation can be performed on bolt andnut ends, ends of cylindrical surfaces andends of shafts etc.

Fig. Chamfering

5.B. Taper turning is an operation of generatingconical surfaces externally as well asinternally. The angle formed by the path ofthe tool with the axis of the work piece shouldcorrespond to the half taper angle.

It is very much essential that during taperturning, the tool cutting edge should beaccurately set on the centre line of the workpiece.



116

A taper may be turned by any one of thefollowing methods.(1) Taper turning by Set Over the tail stock(2) Taper turning by swiveling the Compound

Rest(3) Taper turning by a Taper Turning

Attachment(1) Taper turning by Set Over the tail

stock : The principle of taper turning bythis method is shifting the axis of rotationof the work piece at an angle to the latheaxis and feeding the tool parallel to thelathe axis. The angle through which theaxis of rotation is shifted is half the angleof taper. For this, the body of the tailstock is made to slide on its base towardsor away from the operator by asset overscrew.

Fig.: Taper Turning by Set-over Method

Due to limitation of set over, this method issuitable for turning small taper on long jobs.

Set over =( )LX D2X1

d-

Where,L = Length of WorkD = Larger Diameterd = Smaller Diameter1 = Length of taper

The main disadvantage of this method isthat the live and dead centres are notequally stressed and the wear is notuniform.

(2) Taper turning by swiveling the CompoundRest: This method employs the principlethat the work piece is rotated along thelathe axis and the tool is fed at an angleto the axis of rotation of the work piece.The tool is held in a tool post which ismounted on the compound rest.

Fig.: Taper Turning by SwivellingCompound Rest

The compound rest is mounted on thecross slide, which is graduated in degreesand can be swiveled or clamped at anydesired position.

The compound rest is set at an angle equalto half the taper angle. After this compoundslide screw is rotated which causes thetool to be fed at desired angle generatingcorresponding taper.

Limitations :(i) The use of this method is limited to turn

a short taper due to limited movement ofthe compound rest.

(ii) This method gives low productioncapacity and poor surface finish sincethe tool is moved purely by hand.

(3) Taper turning by a taper turningattachment: The principle of taperturning by this method is that a taperattachment is used to guide a tool in astraight path which is set an angle to theaxis of rotation of the work piece. The



117

work piece is rotated between the centresor in a chuck.

Fig.: Taper Turning by TaperTurning Attachment

A taper turning attachment is illustrated inFig.

It consists of a frame which is attached to therear end of the lathe bed. It supports a guidebar (graduated in degrees) pivoted at thecentre. The bar may be held at the desiredangle with the lathe axis.

When the taper turning attachment is used,the cross slide is made free from the leadscrew. The rear end of the cross slide istightened.

During engagement of longitudinal feed, thetool on the cross slide follows angular path.The required depth of cut is obtained by acompound slide which is placed perpendicularto the lathe axis. The angle of the guide barshould be half the taper to be worked on thework piece.

The maximum angle of the rotation of guidebar is 10° to 12° on either sides of centre line.

Advantages

The advantages of taper turning by taperturning attachment are :(1) The taper turning can be performed in

one setting without much loss of time.

(2) Various length of work pieces can beworked once the taper is set.

(3) Very steep taper can be turned that can'tbe done by other methods.

(4) Accurate taper on large number of workpieces can be worked.

(5) The internal tapers can be done with ease.

5.C. Given,

D1 = 80 mm, D2 = 50 mm, L = 80 mm (withusual notations).

Substituting these values in Eq., we get

tan a =80 502 80

-´

or, a = 10.62°

The compound rest should be swivelled at10.62°

Note: The data on length of job is a redundantdata.

6.A. Centreless grinders are basically cylindricalgrinders only but they differ from centre-typecylindrical grinders in that the work, insteadof being held between centres, is supportedby a combination of a grinding wheel, aregulating wheel and a work rest blade. Workis revolved and traversed across the face togrinding wheel while being supported on thework rest blade. The cutting pressure of thegrinding wheel and regulating wheel keepsthe workpiece in contact with the work restblade. As mentioned, the workpiece is notonly revolved but also simultaneously givenan axial movement by the regulating wheeland guides so as to pass between the wheels.For this, the axis of regulating wheel isinclined at an angle a (2 to 10°) vertically.Amount of metal to be removed determinesas to how many times a workpiece has to passbetween the wheels. Block diagram of acentreless grinder is given in Fig.



118

Fig.: Block Diagram of a Centreless Grinder

Through-feed GrindingIn the case of through-feed grinding (Fig. a)the regulating wheel is kept slightly inclinedto the grinding wheel axis and the workpieceis steered past the wheel.

Fig.(a): Centreless Grinding – through-feed

The rate of longitudinal feed Vs can becalculated from the following equation:

Vs = pD.nr sina, mm/min

where, D = diameter of the regulatingwheel, mm

nr = rpm of the regulating wheela = angle of inclination of the

regulating wheel which variesfrom 1° to 8°.

The above equation does not include anycorrection for slip which is around 2% andhas the effect of reducing the output. Thegenerally given grinding allowance is about0.04 – 0.4 mm on the diameter. Low speed ofthe regulating wheel combined with a largerinclination is used for roughing to reduce thegrinding time. Higher speeds with lessinclination is used for finish grinding to givethe wheel more time to cut the material and

longer sparking time to produce a betterfinish. Through-feed grinding is adopted forstraight cylindrical surfaces. Bars longer thanthe grinding wheel can be centreless groundby the through-feed method. Short cylindricalparts can be continuously ground, severalparts being in simultaneous engagement withthe wheels.

Infeed GrindingInfeed centreless grinding (Fig. b) is similarto cylindrical plunge grinding and there isno relative axial movement between the workand the grinding wheel. However, theregulating wheel is given a tilt of 20¢-30¢ toassure that the resultant thrust holds the workagainst a shoulder. This is used for grindingparts which have a shoulder, a head or someportion larger than the ground diameter. Theprocess is also used for work with multiplediameters, taper, or various profiles, grooves,etc. The length of the ground portion of theworkpiece is less than the width of the wheel.If the grinding length is more than the widthof the wheel, one end of the job is supportedby an outboard roller support while the otherportion is being ground.

Fig. (b): Infeed Centreless Grinding

End-Feed GrindingEnd-feed grinding (Fig. c) is used mainly fortapered work. The work is traversed to a fixedend stop. Either the grinding wheel, or theregulating wheel, or both should be trued toobtain the required taper. The work advance,after contact with the wheels, can be controlled



119

by the movement of the loader pusher rodacting alone or in combination with thetraversing action of the inclined regulatingwheel.

Fig. (c): End-feed Centreless Grinding

6.B. Given dataD = 20 mm, Lj = 60 mm, v = 14 m/min,f = 0.3 mm/rev. with usual notations.Substituting the data in the equation, wecalculate the drill rotational speed, that is

N = 178 rpm.Length of tool travel = Lj + Length ofapproach and over travel.

= 60 + 5= 65 mm

Substituting the values of f, L and N inEquation, we get

t = 65/(0.3 × 178)= 1.21 minute.



ssc solved paper 2011 (conventional) · manometer is a u-shaped tube consisting of an...

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