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Question and answers

Examination:
Que.No Marks
Q )

4

Question:

State one application of each : v-belt drive, flat belt drive, chain drive and gear drive.


Answer:

 Application of Belts: 1. V- Belt drive – In I.C. Engine power transmission from crankshaft pulley to water pump pulley. 2. Flat Belt drive – Floor mill 3. Chain drive – motor cycle 4. Gear drive – In automotive gear boxes

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Q 3 a )

4

Question:

State the norms of Bharat stage III and IV.


Answer:

In year 2010 – Bharat Stage III Emission Norms for 2-wheelers, 3-wheelers and 4-wheelers for entire country whereas Bharat Stage – IV (Equivalent to Euro IV) for 13 major cities for only 4- wheelers. Bharat Stage IV also has norms were implemented for 4-wheelers for 13 major cities for only 4-wheelers. Currently, BS-IV petrol and diesel are being supplied in whole of Northern India covering Jammu and Kashmir, Punjab, Haryana, Himachal Pradesh, Uttarakhand, Delhi and parts of Rajasthan and western UP. The rest of the country has BS-III grade fuel.

 

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Q 4a)(iii)

4

Question:

State the effect of keyway on the strength of the shaft.


Answer:

The keyway is a slot machined either on the shaft or in hub to accommodate the key. It is cut by vertical or horizontal milling cutter. A little consideration will show that the keyway cut into the shaft reduces the load carrying capacity of the shaft. This is due to the stress concentration near the corners of the keyway and reduction in the crosssectional area of the shaft. It other words, the torsional strength of the shaft is reduced. The following relation for the weakening effect of the keyway is based on the experimental results by H.F. Moore.

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Q 5 c )

8

Question:

Draw psychrometric chart with all the property lines and represent following psychrometric processes : i) Sensible heating ii) Sensible cooling with dehumidification iii) Humidification iv) Dehumidification.


Answer:

Psychometric chart representing various psychometric processes: i) Sensible Heating

 

 

ii) Sensible Cooling with dehumidification

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Q 6 b )

4

Question:

Define : i) Free air delivered ii) Compressor capacity iii) Swept volume iv) Pressure ratio, w.r.to compressor.


Answer:

Define i) Free air delivered (FAD) – It is volume of air delivered under the condition of temperature and pressure existing at compressor intake, i.e. volume of air delivered at surrounding air temperature & pressure. In absence of any given free air conditions these are generally taken as 1.101325 bar and 150 c. ii) Compressor capacity – It is quantity of free air actually delivered by compressor in m3 /min. iii) Swept volume – It is the volume of air taken during sanction stroke. It is expressed in m3 . iv) Pressure ratio – It is defined as delivery pressure to suction pressure.

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Examination: 2017 SUMMER
Que.No Marks
Q 1 a )

2

Question:

Define inversion with example.


Answer:

When one of the links is fixed in a kinematic chain, it is called a mechanism. So we can obtain as many mechanisms as the number of links in a kinematic chain by fixing, in turn, different links in a kinematic chain. This method of obtaining different mechanisms by fixing different links in a kinematic chain is known as inversion of the mechanism.

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Q 1a)(a)

4

Question:

What is function of (i) oil reservoir (ii) pressure relief valve, (iii) direction control valve, (iv) filters ?


Answer:

What is function of (i) Oil Reservoir – To store the Hydraulic oil for the circuit (ii) Pressure Relief Valve- To release the extra pressure whenever not required by system (iii) Direction Control Valve- To give the direction to the actuator (iv) Filters- To filter the foreign particle from the oil and to separates sub-micron level contamination

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Q 1a)(a)

4

Question:

     Enlist uses of compressed air (any four).              


Answer:

Following are the applications of compressed air 1) To drive air motors in coal mines. 2) To inject fuel in air injection diesel engines. 3) To operate pneumatic drills, hammers, hoists, sand blasters. 4) For cleaning purposes. 5) To cool large buildings. 6) In the processing of food and farm maintenance. 7) In vehicle to operate air brake. 8) For spray painting in paint industry.

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Q 1a)(b)

4

Question:

Draw and explain working of pressure reducing valve.


Answer:

Draw and explain working of pressure reducing valve. The main function of pressure reducing valve is to reduce the pressure in particular branch of the circuit to different level as demanded by consumer in that branch. Construction: It consists of spool and spring housed in the bore of valve body. Spring compression can be adjusted by pressure setting screw. Port P is pressure port connected to pump. Port A is consumer port requiring reduced pressure.

 

Working: As shown in normal position, port P is supplying oil to consumer port A. If the main supply is below the set pressure, there will be continuous flow from P to A. Hence normally this valve is open. When outlet pressure rises to valve setting then oil will flow through ‘passage x’ and will act on

spool and spool will shift to right thereby partly closing the port A. Now only enough flow will pass through port ‘A’ so that consumer connected to A will receive reduced pressure.

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Q 1a)(b)

4

Question:

What are the advantages of multistaging ?


Answer:

1. The air can be cooled in between two cylinders 2. The power required is less 3. Mechanical balance is good 4. Reduced leakage losses 5. More volumetric efficiency 6. High pressure range 7. Comparatively lighter in construction

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Q 1a)(c)

4

Question:

Explain piston pump with neat sketch.


Answer:

It’s an inclined plate in axial piston pump on which all pistons are connected through piston rod. This swash plate is usually inclined. Use – It is helps to reciprocate the piston of axial piston pump while the cylinder block is rotating Working: Motor drives the shaft, which in turn rotates the entire cylinder block. The pistons are connected to inclined swash plate through piston rod. Now since swash plate is inclined and block is rotating, the piston reciprocates inside the barrel. The reciprocating motion of piston causes suction and delivery of fluid through inlet and outlet ports which come in front of outlet of piston. If we change the angle of swash plate i.e. θ if a) θ = 0 then no flow of oil, because pistons are at same level. When θ = 0 swash plate is vertical. No reciprocation of piston, hence no flow. b) θ = max or + ve, then x will be stroke length which is maximum and there will be maximum forward flow. c) θ = - ve, then ‘x’ i.e. stroke length will be maximum in reverse direction and hence there will be reverse flow. By changing the swash plate angle we can vary the stroke length of the piston. and also output flow can be changed.

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Q 1a)(c)

4

Question:

Classify gas turbines (any four)


Answer:

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Q 1a)(d)

4

Question:

What is an accumulator ? Why accumulator is necessary for huge hydraulic pressers ?


Answer:

makes available to the system whenever required. Necessity of accumulator for every huge hydraulic press: Oil requirement of the system is not continuous but intermittent. A hydraulic press needs the oil only during the lifting operation. While, during the lowering of load, holding the load or during idle period, it doesn’t not require any supply of oil. During such period, the hydraulic pump has to stop or delivered oil must be drained back to the sump. But frequent starting and stopping of pump is not desirable as it reduces the pump life. Also, draining the pressurized oil is wastage of power which reduces the system efficiency. The remedy to above problem is to use accumulator in the system.

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Q 1a)(d)

4

Question:

Define : i) Tonnage of refrigeration ii) Coefficient of performance.


Answer:

i) Tonnage of Refrigeration – is defined as the amount of refrigeration effect produced by uniform melting of one ton (1000Kg) of ice from and at 00 in 24 hours. .

ii) Coefficient of performance- is the ratio of heat extracted in refrigerator to work done on the refrigerant

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Q 1a)(ii)

8

Question:

What is a cotter joint? State any four applications of a cotter joint? Why taper is provided on cotter joint?


Answer:

COTTER JOINT

Cotter Joint: " A cotter joint is temporary joint and used to connect two coaxial rods or bars which are subjected to axial tensile and or compressive forces."

It consist of

1) spigot : It is the male part of the joint , it has a rectangular slot for passing the cotter through it. Spigot  has a collar which rests against the socket end.

2) socket :It is the female part of the joint, it also has a rectangular slot for passing the cotter through it. It has a circular hole in which spigot fits.

3) cotter : is a wedge shaped piece of metal which actually connects two parts which are non rotating.

cotter joint

Cotter Joint Applications:

1) Lewis foundation bolt

2) connection of the piston rod to cross head of a reciprocating steam engine.

3) valve rod & its stem 4) piston rod to the trail end in an air pump.

5) Cycle pedal sprocket wheel.

Cotter joint taper why and how much?

Cotter is a flat wedge shaped metal piece which is used to connect two rods which transmit the force but without rotation. The force may be axial and of tensile or compressive nature. Cotter is fitted in the tapered slot and remains in its position because of  wedge action. This happens because of taper.

Because of taper,

i) It is simple to remove the cotter and  dismantle the joint parts.

ii) Taper ensures tightness of the joint in operation and it  prevents slackening of the parts.

Generally the value  of taper on cotter is 1 in 48 to 1 in 24.

1 in 48 means that there will be reduction of 1 mm in size after the length of 48 mm, and 1 in 24 means there will be reduction in size of cotter by 1 mm after 24 mm.

 

Link to other chapters in machine design

http://mechdiploma.com/elements-machine-design-syllabus22564

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Q 1a)(iii)

8

Question:

A hollow shaft for a rotary compressor is to be designed to transmit maximum torque of 4750 N-m. The shear stress in the shaft is limited to 50 MPa. Determine the inside outside diameter of the shaft if the ratio of inside to outside diameter of the shaft is 0.4.


Answer:

Design of Hollow shaft: Given Data: T =4750 N-m = 4750 X 103 N-mm , Ʈ = 50 N/mm2 , K=Di/Do =0.4 The hollow shaft is designed on the basis of strength from the derived torsion equation. 4750 X 103 N-mm ---- Thus Do= 79.18 mm 80 mm ( Say ) Di = 0.4 x Do = 0.4 x 80 = 32 mm

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Q 1 b )

2

Question:

List the inversions for double slider crank mechanism.​


Answer:

Inversions of Double Slider Crank Chain :

1. Elliptical trammels.

2. Scotch yoke mechanism.

3. Oldham’s coupling.

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Q 1b)(a)

6

Question:

Write the causes and remedies for the following : (i) Excess heat in oil (ii) Noisy pump (iii) Low pressure in system.


Answer:

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Q 1b)(a)

6

Question:

Give classification of IC engines (any six).


Answer:

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Q 1b)(b)

6

Question:

What is pressure compensated flow control valve ? Explain with sketch.


Answer:

In any hydraulic circuit there are slight variations in presence of oil. When pressure changes the rate of flow changes but many circuits requires constant flow regardless of input or output pressure variations in the circuit then the pressure compensated FCV is used. It consists of hollow cylinder shaped poppet at the bottom of which there is a fixed orifice. There is a spring inside a poppet as shown in fig. Pressurized oil entering through the inlet port will apply full force on the bottom of the poppet and will try to compress the spring by shifting the poppet to right the poppet will move to right and will close the outlet port. Then movement of the poppet toward right will stop. Now flow of oil through the orifice will start. Oil will occupy the bore of cylinder this flow of oil will equalize the pressure on both ends of the poppet. The poppet will then balance. During the process of poppet balancing, spring will expand and poppet will move toward left thereby uncovering the outlet port. A balance will automatically be established between quantity of oil through orifice and quantity of oil going out through the outlet port even if the pressure of incoming oil changes, the rebalancing will established automatically and constant flow of oil will come out

 

 

 

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Q 1b)(b)

6

Question:

Explain Morse test.


Answer:

Morse Test

Morse test is a method to measure the frictional power of a multicylinder SI engine.

Morse Test – This test carried out on multi cylinder I.C. engine. In this test, first engine is allowed to run at constant speed and brake power of engine is measured when all cylinders are working and developing indicated power. (Considering Four cylinders)

I1+I2+I3+I4 = (BP)engine +(F1+F2+F3+F4)

Where I1, I2, I3 and I4 – Indicated power of four cylinders

(BP)engine – Brake power of engine when all cylinders are working

F1, F2, F3, F4Frictional power of all four cylinders

Then the first cylinder is cut off by short circuiting spark plug in case S.I. engine (or cutting fuel supply in case C.I. engine). This causes the speed to drop due to non firing of first cylinder. It should be noted that although first cylinder is not producting power still it is moving up and down so its frictional power must be considered. This speed is once again maintained to its original value by reducing load on the engine

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Q 1b)(i)

6

Question:

 i) What are the factors to be considered for selection of materials for design of machine elements?


Answer:

Factors to be considered for selection of material for design of machine elements a) Availability: Material should be available easily in the market. b) Cost: the material should be available at cheaper rate. c) Manufacturing Consideration: the manufacturing play a vital role in selection of material and the material should suitable for required manufacturing process. d) Physical properties: like colour, density etc. f) Mechanical properties: such as strength, ductility, Malleability etc. g) Corrosion resistance: it should be corrosion resistant.

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Q 1b)(ii)

6

Question:

Design a bushed pin type flexible coupling for connecting a motor shaft to a pump shaft for the following service conditions. Power to be transmitted = 40 KW. Speed of the motor shaft = 1000 RPM. Diameter of the motor shaft = 50 mm Diameter of the pump shaft = 45 mm The bearing pressure in the rubber bush and allowable stress in the pins are to be limited to 0.45 N/mm2 and 25 MPa respectively


Answer:

 

 

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Q 1 c )

2

Question:

Define sliding pair with example.


Answer:

Sliding pair :

When the two elements of a pair are connected in such a way that one can only slide relative to the other, the pair is known as a sliding pair. The piston and cylinder, cross-head and guides of a reciprocating steam engine, ram and its guides in shaper, tail stock on the lathe bed etc. are the examples of a sliding pair. A little consideration will show that a sliding pair has a completely constrained motion.

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Q 1 d )

2

Question:

Define centripetal and tangential acceleration.


Answer:

Centripetal acceleration:

The centripetal acceleration is the rate of change of tangential velocity. When an object is moving with uniform acceleration in circular direction, it is said to be experiencing the centripetal acceleration.

Tangential acceleration:

Tangential acceleration is a measure of how the tangential velocity of a point at a certain radius changes with time. Tangential acceleration is just like linear acceleration, but it’s particular to the tangential direction, which is relevant to circular motion.

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Q 1 e )

2

Question:

Find the velocity of point B and midpoint C of link AB shown in Figure (1).                                                                                   


Answer:

Velocity of point B & C :

Vb = AB x wAB = 0.35 x 50 = 17.5 m/s

Vc = AC x wAB = 0.175 x 50 = 8.75 m/s

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Q 1 f )

2

Question:

Classify the cam.


Answer:

Classification of cam:

1. Radial or disc cam: In radial cams, the follower reciprocates or oscillates in a direction perpendicular to the cam axis. The cams as shown in above Fig. are all radial cams.

2. Cylindrical cam: In cylindrical cams, the follower reciprocates or oscillates in a direction parallel to the cam axis. The follower rides in a groove at its cylindrical surface. A cylindrical grooved cam with a reciprocating and an oscillating follower is shown in Fig. below (a) and (b) respectively.

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Q 1 g )

2

Question:

Define following terms with respect to cam and follower : (i) Prime circle (ii) Pitch circle (iii) Pressure angle (iv) Trace point  


Answer:

i. Prime circle: It is the smallest circle that can be drawn from the centre of the cam and tangent to the pitch curve. For a knife edge and a flat face follower, the prime circle and the base circle are identical. For a roller follower, the prime circle is larger than the base circle by the radius of the roller.

ii. Pitch circle: It is a circle drawn from the centre of the cam through the pitch points.

iii. Pressure angle: It is the angle between the direction of the follower motion and a normal to the pitch curve. This angle is very important in designing a cam profile. If the pressure angle is too large, a reciprocating follower will jam in its bearings.

iv. Trace point: It is a reference point on the follower and is used to generate the pitch curve. In case of knife edge follower, the knife edge represents the trace point and the pitch curve corresponds to the cam profile. In a roller follower, the centre of the roller represents the trace point.

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Q 1 h )

2

Question:

What are the limitations of knife edge follower ?


Answer:

Limitations of knife edge follower are:

1. Excessive wear due to small area of contact between cam & follower surfaces.

2. In this follower a considerable thrust exists between the follower and guide.

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Q 1 i )

2

Question:

Define self-energizing and self-locking brake. 


Answer:

Self energizing & Self Locking brake

Rn x X = PL + μaRn

Rn = Normal reaction, P = Applied force, L = lever length

X = Distance of block from hinge, μ= coefficient of friction, a = distance of drum from hinge

In the above equation when frictional force adds to the breaking torque. In other words, the frictional torque and braking torque are in the same direction its a self locking brake.

In the above equation when X < μa, P becomes negative

Hence, P is not required for braking and brake gets applied on its own. It is called as self energizing brake.

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Q 1 i )

2

Question:

Write down the formula of length of belt for open belt drive and cross belt drive.


Answer:

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Q 1 i )

2

Question:

List the methods to reduce the slip in belt and pulley.


Answer:

Methods to reduce the slip in belt and pulley:

1. Vertical belt drive should be avoided.

2. In horizontal belt drive the upper side should be kept as loose side.

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Q 1 k )

2

Question:

Define law of gearing.


Answer:

Law of Gearing: The law of gearing states that the angular velocity ratio of all gears of a meshed gear system must remain constant also the common normal at the point of contact must pass through the pitch point.

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Q 1 l )

8

Question:

What is factor of safety? State its importance in design of machine elements.


Answer:

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Q 1 m )

2

Question:

What are the limitations of shoe brake ?


Answer:

Limitations of a shoe brake :

1. Heavy side thrust causes bending of the shaft.

2. More wear & tear as the contact surface is large.

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Q 1 n )

2

Question:

Define uniform wear theory and uniform pressure theory.


Answer:

Uniform Wear theory:

When the product of pressure and area of the contacting surface transmitting load is taken as constant to determine the axial force & torque, it is termed as uniform wear theory as it is assumed that wear along the surface is uniform.

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Q 1 o )

2

Question:

State effects of imbalance in machine.


Answer:

Effects of imbalance in machine

1. Imbalance imparts vibratory motion to the frame of the machine.

2. Produces noise which leads to human discomfort.

3. Detrimental effects on the machine performance & structural integrity of the machine foundation.

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Q 2 a )

8

Question:

In an Ideal ottocycle the air at the beginning of isentropic compression is at 1.01325 bar and 20°C. The compression ratio is 8. If the heat added during constant volume process is 250 kJ/kg. Determine : a) Maximum temperature in the cycle b) Air standard efficiency c) Work done per cycle d) Heat rejected


Answer:

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Q 2 a )

4

Question:

Draw a neat sketch and explain working of beam engine.


Answer:

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Q 2 a )

8

Question:

Design a knuckle joint to transmit 150 KN. The design stresses may be taken as 75 MPa in tension, 60 MPa in shear and 150 MPa in compression.


Answer:

Design of knuckle joint: Step 1) Diameter of Rod: d : =? Consider tensile failure of Rod 1. P =σt x A , 150 x 103 = 75 x π/4 xd2 , d = 50.4 mm 52 mm ( say)

Using Imperial relations Diameter of Knuckle pin Outside

 

 

 

 

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Q 2 a )

8

Question:

Compare meter-in-circuit with meter-out-circuit, draw neat sketch of meter-in-circuit.


Answer:

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Q 2 b )

8

Question:

Explain vapour compression refrigeration cycle on T-S and p-h charts (for superheated vapourat the end of compression)  


Answer:

Vapour Compression Refrigeration Cycle

The P-H and T-S diagram for the simple vapor compression refrigeration cycle is shown in the figure for vapour entering the compressor is in dry saturation condition The dry and saturated vapour entering the compressor at point 1 that vapour compresses isentropic ally from point 1 to 2 which increases the pressure from evaporator pressure to condenser pressure At point 2 the saturated vapour enters the condenser where heat is rejected at constant pressure, due to rejection of heat decreases the temperature and change of phase takes place i.e. latent heat is removed and reaches to liquid saturation temperature at point 3 then this liquid refrigerant passed through expansion valve where liquid refrigerant is throttle keeping the enthalpy constant and reducing the pressure.

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Q 2 b )

8

Question:

The following data refers to a trial conducted on 4-stroke petrol engine Air-fuel ratio (by mass) = 15.5 : 1 Heat value of fuel = 48000 kJ/kg Mechanical efficiency = 82% Air standard efficiency = 54% Relative efficiency = 70% Volumetric efficiency = 80% Speed = 240 rpm Brake power = 75 kW Calculate : i) Compression ratio ii) Indicated thermal efficiency iii) Brake specific fuel consumption.


Answer:

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Q 2 b )

2

Question:

Explain with neat sketch how to find the velocity of a slider in slider crank mechanism by Klein’s construction.


Answer:

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Q 2b)(ii)

8

Question:

Draw neat sketch of a protected type flanged coupling showing all details.


Answer:

distance to some degree. c) For effective conjugate action i.e for maintaining a constant velocity ration, in case of involute gearing system, the center distance can be changed without affecting angular velocity ratio. d)In involute gearing as the path of contact is a straight line and the pressure angle is constant .

Sketch of Protected type flanged coupling with details :

 

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Q 2b)(l)

8

Question:

(i) State and explain two most important reasons for adopting involute curves for a gear tooth profile


Answer:

 

 

For power transmission gears, the tooth form most commonly used the involute profile as a)Involute gears can be manufactured easily: Since the rack in an involute system has straight sides and since the generating cutters usually have rack profile, these cutters can be easily manufactured. Involute gears can be produced more accurately and at a lesser cost. b) The gearing has a feature that enables smooth meshing despite the misalignment of center

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Q 2 c )

8

Question:

What is function of filters ? Classify the filters and draw any two types of filters


Answer:

What is function of filters? Classify the filters and draw any two types of filters. Function of filter: To remove foreign particles from the oil and remove submicron particles dissolved in the oil. Classification of filters: 1- Full flow filter 2- Proportional flow filter Full flow filter: Incurs a large pressure drop. A relief valve is needed which cracks when the filter becomes blocked. Proportional flow filter: Localised low pressure area is formed at the venturi. The fluid is drawn from the filter due to the pressure difference. low pressure drop

 

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Q 2 c )

8

Question:

Differentiate between reciprocating air compressor and rotary air compressor.........


Answer:

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Q 2 c )

4

Question:

Draw and explain in short, types of followers used in cam and follower.


Answer:

Types of followers :

The followers may be classified as discussed below:

1. According to the surface in contact.

(a)Knife edge follower. When the contacting end of the follower has a sharp knife edge, it is called a knife edge follower.

(b) Roller follower. When the contacting end of the follower is a roller, it is called a roller follower.

(c) Flat faced or mushroom follower. When the contacting end of the follower is a perfectly flat face, it is called a flat faced follower and when the flat faced follower is circular, it is then called a mushroom follower.

2.According to the motion of the follower.

(a) Reciprocating or translating follower. When the follower reciprocates in guides as the cam rotates uniformly, it is known as reciprocating or translating follower.

(b) Oscillating or rotating follower. When the uniform rotary motion of the cam is converted into predetermined oscillatory motion of the follower, it is called oscillating or rotating follower.

3. According to the path of motion of the follower.

(a) Radial follower. When the motion of the follower is along an axis passing through the centre of the cam, it is known as radial follower

(b) Off-set follower. When the motion of the follower is along an axis away from the axis of the cam centre, it is called off-set follower.

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Q 2c)(i)

8

Question:

Why are bushes of softer material inserted in the eyes of levers?


Answer:

the forces acting on the boss of lever & the pin are equal & opposite .There is a relative motion between the pin & the lever and bearing pressure becomes design criteria. The projected area of the pin is d1 x l1therefore Reaction R= P (d1 x l1 ). A softer material like phosphorous bronze bush with 3 mm thick is fitted in eyes to reduce the friction. & bear a bearing pressure upto5 to 10 N/mm2. Bushes are cheaper and can be easily replaceable.

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Q 2c)(ii)

8

Question:

Explain the following types of stresses  a) Transverse shear stress b) Compressive stress c) Torsional shear stress


Answer:

 

Explanation of stresses :

a) Transverse shear stress: When a section is subjected to two equal & opposite forces acting tangentially across the section such that it tends to shear off across the section. The stress is produced is called as transverse stress For Single shearing, Shear stress Ʈ = W/A For Double shearing, Shear stress Ʈ = W/2A

b)Compressive stress: When a body is subjected to equal & opposite axial push forces, the stress produced is called as compressive stress. It is denoted by “ σc”

 

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Q 2 d )

4

Question:

Explain condition for maximum power transmission.


Answer:

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Q 2 e )

4

Question:

Explain the compound gear train with neat sketch and write down the velocity ratio’s equation.


Answer:

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Q 2 f )

4

Question:

A multiplate clutch has three pairs of contact surfaces. The outer and inner radii of the contact surfaces are 100 mm and 50 mm respectively. The maximum axial spring force is limited to 1.25 kN. If the co-efficient of friction is 0.35 and assuming uniform wear, find the power transmitted by the clutch at 1600 rpm.


Answer: -----------------------------------------------------------------------------------------------------
Q 3 )

4

Question:

Represent Brayton cycle on PV and TS diagram. Name the processes completing the cycle


Answer:

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Q 3 a )

4

Question:

Explain 4-way-3 position direction control valve used in hydraulic system.


Answer:

4-way-3 position direction control valve used in hydraulic system is known as 4X3 DC Valve. The valve has four ports and three positions. Following figure shows the Normal and working positions of DCV. Spool of this valve is having three positions. The spool is so selected because we have to obtain third position also called as ‘Closed Centre Position’ This position is shown in figure. We have shifted the spool in such a manner that all ports are closed to each other. Mo flow from port P to port A or B and no flow from port A and B to R. When DC valve attains this position, pressured oil returns to reservoir via pressure relief valve. The closed center position of DC valve is suitable for immediate closing of movement of actuator.

Position- I

Position- II

Closed center position

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Q 3 a )

4

Question:

Differentiate between mechanism and machine.


Answer:

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Q 3 b )

4

Question:

Design a foot brake lever from the following data: Length of lever from C.G. of the spindle to the point of application of the load = 1 meter. Max. load on the foot plate = 800 N Overhang from the nearest bearing = 100 mm Permissible tensile and shear stress = 70 MPa.


Answer:

Methods of reducing stress concentration in cylindrical members with holes . Stress concentration can be reduced in cylindrical members with holes by providing additional holes in vicinity of holes as shown in fig. (ii). Fig (i) Showing cylindrical member with hole at center having stress line in disturb manner at vicinity of hole and component will fail at hole so for fig (i) ,stress concentration is more . fig. (ii) members shoulder having additional hole in vicinity of hole and therefore stress line maintain spacing between them so here stress concentration is less. Design of foot lever : Given data: L=1 m =1000 mm , P=800 N , σt =70 N/mm2 , Ʈ =70 N/mm2 , Assume B=3t Step 1) Considering shaft is under pure torsion , therefore

 

 

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Q 3 b )

4

Question:

Explain gear pump with neat sketch.


Answer:

It consists of one external and one internal meshing gear pair. External gear is connected to electric motor and hence is driving gear. Internal gear or ring gear is driven gear which rotates in same direction as that of external gear. Between two gear a spacer called ‘crescent’ is located which is a stationary pieces connected to housing. Inlet and outlet ports are located in end plates. External gear (driving gear) drives the internal gear (Ring Gear). Portion where teeth start meshing, a tight seal is created near port the vacuum is created due to quick un meshing and oil enters from oil tank through inlet port. Oil is trapped between the internal and external gear teeth on both sides of crescent (spacer) and is then carried from inlet to outlet port. Meshing of gear near outlet port reduces the volume or gap and oil gets pressurized. These pumps make very less noise.

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Q 3 b )

4

Question:

A petrol engine has a cylinder of diameter 60 mm and stroke 100 mm. If the mass of charge admitted per cycle is 2×10 – 4 kg. Find volumetric efficiency of the engine.


Answer:

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Q 3 b )

4

Question:

Explain the working of Whitworth quick return mechanism.


Answer:

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Q 3 c )

4

Question:

Explain any four criteria for selection of hydraulic pump in hydraulic system.


Answer:

1) Pressure: It is the basic selection criteria. Pump pressurizes the hydraulic oil to the level required by actuator. When pressures up to 150 bars are required then gear pumps can be selected. For pressure of 150 to 250 vane pump is suitable and for above 500 bar pressure piston pumps are useful. 2) Flow of pressurized oil: Volumetric output of pump is measures in LPM. The flow of oil decides the speed of actuator. The displacement can also be changed for variable displacement pumps. 3) Speed of pump: The speed of pump is decided by rated capacity of the manufacturer. If wrong speed is selected for pump then efficiency and working of hydraulic system may get hamper.

4) Efficiency of the pump:The selected pump must have good efficiency. We can consider following efficiencies: 1) Volumetric 2) Mechanical 3) Overall 5) Oil compatibility: The oils used in pump should be compatible with the material of the pump. If wrong oil gets selected then pump will not work to its rated performance

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Q 3 c )

4

Question:

Explain with neat sketch two way catalytic converter.


Answer:

C) Catalytic Convertor: -. A catalytic converter is cylindrical unit about the size of small silencer and is installed into exhaust system of vehicle. It converts the harmful gases from the engine into harmless gases and escapes them into atmosphere. Inside converter there is honeycomb structure of ceramic or metal which is coated with alumina base material and therefore a second coating of precious metal platinum, palladium or rhodium or combination of same. As a result of catalytic reaction, the exhaust gases pass over the converter substance, the toxic gases such CO, HC and NOx are converted into harmless CO2, H2 and N2. Two way catalytic converter: Through catalytic action a chemical changes converts carbon monoxide (CO) and hydrocarbon (HC) into carbon dioxide (CO2) and water (oxidation)..

 

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Q 3 c )

4

Question:

In slider crank mechanism, the length of crank OB and connecting rod AB are 130 mm and 500 mm respectively. The centre of gravity G of the connecting rod is 275 mm from slider A. The crank speed is 750 rpm in clockwise. When crank has turned 45 from inner dead centre position determine (i) velocity of slider ‘A’ (ii) velocity of centre of gravity of connecting rod ‘G’.


Answer:

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Q 3 d )

4

Question:

Compare welded joints with screwed joints. (Any six points)


Answer:

Consideration in design of key: 1) Power to be transmitted. 2) Tightness of fit 3) Stability of connection 4) Cost 5) Crushing failure of key: 6) shearing failure of key 7) Material of key ,shaft should be same but key should be weaker than shaft .

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Q 3 d )

4

Question:

Name any four components of pneumatic system. What are the factors considered while selecting them ?


Answer:

1) Compressor: a)Pressure Requirement b) Volume of Air c) Compressor configuration 2) Actuators: a) According to maximum pressure b) According to application – Linear/Rotary c) shape and size of actuator 3) Air Receiver: a) Storage capacity b) Material of the tank 4) FRL unit: a) According to working environment b) According to pressure required at hand tools 5) DCV: a) According to maximum pressure of system. b) According to actuator configuration c) According to application- One hand /Two hand d) According to actuation method suitable for application

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Q 3 d )

4

Question:

Differentiate between closed cycle and open cycle gas turbine


Answer:

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Q 3 d )

4

Question:

Find the width of the belt, necessary to transmit 7.5 kW to a pulley 300 mm diameter, if the pulley makes 1600 rpm and the co-efficient of friction between the belt and pulley is 0.3. Assume the angle of contact as 180o and the maximum tension in the belt is not to exceed 8 N/mm width.


Answer:

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Q 3 e )

4

Question:

A shaft 30 mm. diameter is transmitting power at a maximum shear stress of 80 MPa. If a pulley is connected to the shaft by means of a key, find the dimension of the key so that stress in the key is not to exceed 50 MPa and length of the key is 4 times the width.


Answer:

Comparison of welded joints with screwed joint. 1) Welded Joint is rigid & permanent. Screwed joint is temporary. 2) Cost of welded assembly is lower than that of screwed joints. 3) Strength of welded structure is more than screwed joints. 4) For welding joints, highly skilled worker are required 5) Welded joints are tight & leak proof as compared to Screwed joints. 6) Welded joint is very difficult to inspect compared to other joints.

 

 

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Q 3 e )

4

Question:

Draw labelled sketch of air lubricator.


Answer:

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Q 3 e )

4

Question:

Explain the working of Watt governor with neat diagram.


Answer:

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Q 3 e )

4

Question:

Explain the effect of superheating and subcooling on the performance of vapour compression cycle


Answer:

Effect of superheating: As shown in the figure a & b the effect of superheating is to increase the refrigerating effect, but this increase in the refrigerating effect is at the cost of increase in amount of work spent to attain upper pressure limit. Since the increase in work is more as compared to increase in refrigerating effect, therefore overall effect of superheating is to give a low value of C.O.P.

ii) Effect of sub-cooling: sub-cooling is the process of cooling the liquid refrigerant below the condensing temperature for a given pressure. In figure the process of sub-cooling is shown by 2’-3’. As is evident from the figure the effect of sub-cooling is to increase the refrigerating effect. Thus sub-cooling results in increase of C.O.P provided that no further energy has to be spent to obtain the extra cold coolant required.

 

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Q 3 f )

4

Question:

Explain the working of centrifugal clutch with neat sketch.


Answer:

Centrifugal Clutch

Centrifugal clutch is a clutch that uses centrifugal force to connect two concentric shafts, with the driving shaft nested inside the driven shaft.

Centrifugal clutch

• A centrifugal clutch is a clutch that uses centrifugal force to connect two concentric shafts, with the driving shaft nested inside the driven shaft.

• It consists of number of shoe on the inside of a rim of pulley. The outer surface of pulley is covered with friction material.

• These shoes move radially in guides.

• As the speed of the shaft increase, the centrifugal force on the shoes increases.

• When the centrifugal force is less than the spring force, the shoes remain in the same position as when the driving shaft was stationary, but when the centrifugal force is equal  to the spring force, the shoes are just floating.

• When the centrifugal force exceeds the spring force, the shoes move outward and come into contact with the driven member presses against it.

• The force with which the shoe presses against the driven member is the difference of the centrifugal force and the spring force.

• The increase of speed causes the shoe to press harder and enable more torque to be transmitted.


Alternate diagram and description

It consists of a number of shoes on the outside of a rim of the pulley. The outer surface of the shoes are covered with friction material. These shoes can move radially in guides and are held together against the boss or spider on the driving shaft, by means of a spring. The springs exert a radially inward force on shoe. Under rotation, the shoe experiences a centrifugal force acting outward. This centrifugal force is directly related (Proportional to) speed of rotation. At a particular speed, the centrifugal force overcomes the force of the spring and the shoe moves outwards and comes in contact with the driven member and presses against it. The net force due to the effect of centrifugal force and spring force causes power to be transmitted.

Centrifugal clutch


Videos explaining Centrifugal clutch working

 

If you watch carefully animation. As the speed increases the shoes move out. Causing the contact with the external wheel and thus transmitting motion.

Centrifugal clutch


Advantages and Limitations of Centrifugal-clutch

Advantages:

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Q 4 a )

4

Question:

Explain the working of freewheel mechanism of bicycle with sketch.


Answer:

A freewheel mechanism on a bicycle allows the rear wheel to turn faster than the pedals. If there is no freewheel on a bicycle, a simple ride could be exhausting, because one could never stop pumping the pedals. And going downhill would be downright dangerous, because the pedals would turn on their own, faster than one could keep up with them.

Power Train of a bicycle: The power train of a simple bicycle consists of a pair of pedals, two sprockets and a chain. The pedals are affixed to one sprocket — the front sprocket, which is mounted to the bike below the seat. The second sprocket is connected to the hub of the rear wheel. The chain connects the two sprockets. When you turn the pedals, the front sprocket turns. The chain transfers that rotation to the rear sprocket, which turns the rear wheel, and the bicycle moves forward. The faster you turn the pedals, the faster the rear wheel goes, and the faster the bike goes.

Coasting: At some point — when going downhill, for instance — speed is high enough so that the rear wheel is turning faster than the pedals. That's when coasting: we stop working the pedals and let the bike's momentum keep moving forward. It's the freewheel that makes this possible. On a bicycle, instead of being affixed to the wheel, the rear sprocket is mounted on a freewheel mechanism, which is either built into the hub of the wheel — a "freehub" — or attached to the hub, making it a true freewheel.

Now when you have to move forward, the pawl acts like a hook and gets locked with the teeth - called ratchet and transmits the torque. The complete mechanism is called ratchet and pawl mechanism.

But when you reverse pedal, it falls back and becomes "free". A spring prevents it from falling permanently. This is the reason why you hear the distinct "click-click" sound when you reverse pedal. Also, there are multiple "pawls" placed along the circumference too.

 

 

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Q 4a)(a)

4

Question:

What are actuators ? Draw a double acting cylinder.


Answer:

Actuator - Actuators are those components of hydraulic / pneumatic system, which produces mechanical work output. They develop force and displacement, which is required to perform any specific task. An actuator is used to convert the energy of the fluid back into mechanical power.

 

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Q 4a)(a)

6

Question:

a) Define : i) Stroke ii) Bore iii) Piston speed iv) MEP (Mean Effective Pressure).


Answer:

a) i) Stroke – Distance travelled by piston from one dead Centre to other dead Centre (Say TDC to BDC). ii) Bore:- The nominal Inner diameter of engine cylinder is called cylinder bore. iii) Piston Speed- Distance traveled by piston in one minute.(= 2LN m/min.) iv) The Mean Effective Pressure (MEP) :-It is a fictitious pressure that, if it operated on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle. OR The average pressure acting on the piston which will produce the same output as is done by the varying pressure during the cycle

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Q 4a)(b)

4

Question:

Explain pressure relief valve in pneumatic system.


Answer:

The pressure relief valves are used to protect the system components from excessive pressure. Its primary function is to limit the system pressure within a specified range. It is normally a closed type and it opens when the pressure exceeds a specified maximum value by diverting pump flow back to the tank. The simplest type valve contains a poppet held in a seat against the spring force as shown in Figure. This type of valves has two ports; one of which is connected to the pump and another is connected to the tank. The fluid enters from the opposite side of the poppet. When the system pressure exceeds the preset value, the poppet lifts and the fluid is escaped through the orifice to the storage tank directly. It reduces the system pressure and as the pressure reduces to the set limit again the valve closes.

( Pressure switch in Pneumatic or Pressure relief valve in hydraulic system can be considered)

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Q 4a)(b)

4

Question:

Explain with sketch working of screw compressor.


Answer:

A rotary-screw compressor is a type of gas compressor that uses a rotary-type positivedisplacement mechanism. They are commonly used to replace piston compressors where large volumes of high-pressure air are needed, either for large industrial applications or to operate highpower air tools. Rotary-screw compressors use two meshing helical screws, known as rotors, to compress the gas. In a dry-running rotary-screw compressor, timing gears ensure that the male and female rotors maintain precise alignment. In an oil-flooded rotary-screw compressor, lubricating oil bridges the space between the rotors, both providing a hydraulic seal and transferring mechanical energy between the driving and driven rotor. Gas enters at the suction side and moves through the threadsas the screws rotate. The meshing rotors force the gas through the compressor, and the gas exits at the end of the screws.

 

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Q 4a)(c)

4

Question:

Explain with sketch, a pneumatic circuit for speed control of bidirectional motor.


Answer:

Varying the rate of flow of oil will vary the speed of the actuator. Speed control is possible using meter in circuit, meter out circuit, bleed off circuit or by placing flow control before the DCV. Speed control of bi-directional air motor: Bi-directional air motor rotates in clockwise as well as anti-clockwise direction. The speed of bi-directional motor is controlled as shown in fig. The speed control of motor by using variable two flow control valves having built-in check valve and 4x3 DC valve having zero position or central hold position with Pilot S1 and S2. ( Lever / Push button / Solenoid may be used ) When Pilot S2 is operated, port P will be connected to port A of air motor and motor will start rotating in clockwise direction. Its speed can be controlled by using variable flow control valve F1. Port B of motor will be connected to exhaust R and air in motor will be exhausted through port R via DC valve. When Pilot S1 is operated, pressure port P will be connected to port B of motor and naturally motor will start rotating in anticlockwise direction. Port A will be connected to port R and air in the motor will be exhausted through port R via DC valve.

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Q 4a)(c)

4

Question:

Classify gas turbines on the following basis :  i) Working cycle ii) Application iii) Cycle of operation iv) Fuels


Answer:

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Q 4a)(d)

4

Question:

State any two applications of 3 × 2 DC valve. Draw symbol for the same.


Answer:

To start, stop and change the direction of motion of a Single acting cylinder. (Clamping of Job)

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Q 4a)(d)

4

Question:

Name the refrigerants used for : i) Water cooler ii) Domestic refrigerator iii) Ice plant iv) Cold storage.


Answer:

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Q 4a)(i)

4

Question:

a) Attempt any THREE of the following: 12 (i) Give the composition of : 1) 35Mn 2 Mo28 2) 30Ni 4 Crl and 3) 25 Cr 3 Mo 55


Answer:

Composition in percentage 1) Carbon-0.3-0.4 %, manganese 0.5 % and molybdenum 2.8 % 2) Carbon-0.26-0.34, ,Nickel 1 % and Chromium 0.25 % 3) Carbon-0.2-0.3,chromium 0.75 % and molybdenum 5.5 %

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Q 4a)(ii)

4

Question:

Define following terms with respect to springs : 1) Free length 2) Solid height 3) Spring rate 4) Spring index


Answer:

Definition of 1) Free length-it is a length of spring in unloaded condition 2) Solid height-it is a length of spring in fully loaded condition 3) Spring rate-load per unit deflection 4) Spring index- ratio of mean diameter of coil to diameter of wire

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Q 4a)(iii)

4

Question:

Explain effect of keyways on strength of shaft. Name one type of key which does not affect strength of shaft


Answer:

Effect of keyways – when the keyways are cut on the shafts, material is removed at the skin, there by weakening the cross section of the shaft. Stress concentration effect is also serious at the corner of the keyways. Thus the shaft become weak. Type of key- Hollow saddle key or Tangent key

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Q 4a)(iv)

4

Question:

Define following terms w.r.t. bolts: 1) Major diameter 2) Minor diameter 3) Pitch 4) Lead


Answer:

Definition w.r.t. bolts 1) Major dia.- dia. Of imaginary cylinder parallel with the crest of the thread ,it is the distance from crest to crest largest dia. of an external or internal thread 2) Minor dia.-dia. Of imaginary cylinder which just touches the roots of an external thread or smallest dia.of an external or internal screw thread 3) Pitch-distance from a point on one thread to the corresponding point on the next thread. 4) lead- distance between two corresponding points on the same helix

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Q 4 b )

4

Question:

In a four bar mechanism ABCD link AD is fixed and the crank AB rotates at 10 radians per second in clockwise, lengths of the links are AB = 60 mm, BC = CD = 70 mm, DA = 120 mm, when angle DAB = 60 and both B and C lie on the same side of AB, find angular velocities of BC and CD link.


Answer:

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Q 4b)(a)

6

Question:

Explain with neat sketch the working of variable displacement vane pump.


Answer:

) Explain with neat sketch working of variable displacement vane pump. In a hydraulic system the flow rate of the pump needs to be variable this can be easily achieved by varying the rpm of the electric motor. Other method is displacement of a vane inside the pump and therefore its delivery is proportional to the eccentricity between the rotor axis and cam ring. Changing the geometric position of the ring relative to the rotor center will change the delivery volume as per system need. Main components of the vane pumps are: 1. Hardened cam ring 2. Rotor 3. Vanes 4. Screw for position adjustment 5. Thrust bearing 6. Stop

 

Working : The rotor containing the vanes is positioned eccentric or off-center with regard to cam ring by means of the adjusting screw hence when the rotor is rotated, in increasing and decreasing volume can be created inside the cylinder bore. If the screw is adjusted slightly so that the eccentricity of the rotor to the cam ring is not sufficient the flow will be less where as with higher eccentricity the delivery volume will be increased with the screw adjustment back completely out the cam ring naturally centers with a rotor and no pumping will be the eccentricity will be zero.

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Q 4b)(b)

6

Question:

Compare pressure relief valve and pressure reducing valve.


Answer: -----------------------------------------------------------------------------------------------------
Q 4b)(i)

6

Question:

(i) Explain different causes of gear tooth failure and suggest possible remedies to avoid such failures


Answer:

Causes-1) Bending failure-every gear tooth acts as a cantilever. If the total repetitive dynamic load acting on the gear tooth is greater than the beam strength of the gear tooth then the gear tooth will fail in bending remedies-module and face width of the gear is adjusted so that the beam strength is greater than the dynamic load 2)Pitting-surface fatigue failure which occurs due to many repetition of Hertz contact stresses , failure occurs when the surface contact stresses are higher than the endurance limit of the material. It starts with the formation of pits which continue to grow resulting in the rupture of the tooth surface. Remedies- dynamic gear tooth load the of gear tooth between the gear tooth should be less than the wear strength 3)Scoring-the excessive heat is generated when there is a excessive surface pressure,

high speed or supply of lubricant fails. it is stick-slip phenomenon in which alternate shearing and welding takes place rapidly at high spots. Remedies- by proper designing of the parameters such as speed, pressure and proper flow of lubricant, so that the temperature at the rubbing faces is within the permissible limits. 4)Abrasive wear- the foreign particles of lubricants such as dirt, dust or burr enter between the tooth and damage the form of tooth Remedies- by providing filters for the lubricating oil or using high viscosity lubricant oil which unable the formation of thicker oil film and hence permits easy passage of such particles with ought damage of gear tooth surface. 5)Corrosive wear- due to presence of corrosive elements such as additives present in the lubricating oils. Remedies- proper anti corrosive additives should be used.

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Q 4b)(ii)

6

Question:

Explain the importance of Aesthetic considerations in design by giving any two examples.


Answer:

Each product is to be design to perform a specific function or a set of functions to the satisfaction of customers. In a present days of buyer’s market, with a number of products available in the market are having most of the parameters identical,the appearance of the product is often a major factor in attracting the customer. For any product, there exists a relationship between the functional requirement and the appearance of a product. The aesthetic quality contributes to the performance of the product, through the extent of contribution varies from product to product. The job of industrial designer is to create new shapes and forms for the product which are aesthetically appealing. For ex.(1) The chromium plating of automobile components improves the corrosion resistance along with the appearance.(2) the aerodynamic shape of the car improves the performance as well as gives the pleasing appearance

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Q 4b)(ii)

6

Question:

Explain the working of two stage reciprocating compressor. Show work saved on PV diagram.


Answer:

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Q 4b)(l)

6

Question:

 Explain how the heat balance sheet for an IC engine is prepared ?


Answer:

i) Heat Balance Sheet :-The complete record of heat supplied and heat rejected during a certain time(Say one minute)by an IC engine is entered in a tabulated form called as heat balance sheet. i) Heat supplied by the fuel= Mf x C  where Mf= mass of fuel supplied in Kg/min C = Lower calorific value of fuel kj/kg  

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Q 4 c )

4

Question:

What are the advantages of ‘V’ belt drive over flat belt drive ?


Answer:

Advantages of V-belt drive over flat belt drive :

1. The V-belt drive gives compactness due to the small distance between the centres of pulleys.

2. The drive is positive, because the slip between the belt and the pulley groove is negligible.

3. Since the V-belts are made endless and there is no joint trouble, therefore the drive is smooth.

4. It provides longer life, 3 to 5 years.

5. It can be easily installed and removed.

6. The operation of the belt and pulley is quiet.

7. The belts have the ability to cushion the shock when machines are started.

8. The high velocity ratio (maximum 10) may be obtained.

9. The wedging action of the belt in the groove gives high value of limiting ratio of tensions. Therefore the power transmitted by V-belts is more than flat belts for the same coefficient of friction, arc of contact and allowable tension in the belts.

10. The V-belt may be operated in either direction with tight side of the belt at the top or bottom. The centre line may be horizontal, vertical or inclined.

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Q 4 d )

4

Question:

Explain the working of flywheel with the help of turning moment diagram.


Answer:

Working of Flywheel with the help of Turning moment diagram:

A flywheel used in machines serves as a reservoir, which stores energy during the period when the supply of energy is more than the requirement, and releases it during the period when the requirement of energy is more than the supply.

The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of operation. Consider the turning moment diagram for a single cylinder double acting steam engine as shown in Fig. We see that the mean resisting torque line AF cuts the turning moment diagram at points B, C, D and E. When the crank moves from a to p, the work done by the engine is equal to the area aBp, whereas the energy required is represented by the area aABp. In other words, the engine has done less work (equal to the area a AB) than the requirement. This amount of energy is taken from the flywheel and hence the speed of the flywheel decreases. Now the crank moves from p to q, the work done by the engine is equal to the area pBbCq, whereas the requirement of energy is represented by the area pBCq. Therefore, the engine has done more work than the requirement. This excess work (equal to the area BbC) is stored in the flywheel and hence the speed of the flywheel increases while the crank moves from p to q.

Similarly, when the crank moves from q to r, more work is taken from the engine than is developed. This loss of work is represented by the area C c D. To supply this loss, the flywheel gives up some of its energy and thus the speed decreases while the crank moves from q to r. As the crank moves from r to s, excess energy is again developed given by the area D d E and the speed again increases. As the piston moves from s to e, again there is a loss of work and the speed decreases. The variations of energy above and below the mean resisting torque line are called fluctuations of energy. The areas BbC, CcD, DdE, etc. represent fluctuations of energy.

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Q 4 e )

4

Question:

Explain the working of internal expanding brake with neat sketch.


Answer:

Internal Expanding shoe brake:

An internal expanding brake consists of two shoes S1 and S2. The outer surface of the shoes are lined with some friction material (usually with Ferodo) to increase the coefficient of friction and to prevent wearing away of the metal. Each shoe is pivoted at one end about a fixed fulcrum O1and O2 and made to contact a cam at the other end. When the cam rotates, the shoes are pushed outwards against the rim of the drum. The friction between the shoes and the drum produces the braking torque and hence reduces the speed of the drum. The shoes are normally held in off position by a spring . The drum encloses the entire mechanism to keep out dust and moisture. This type of brake is commonly used in motor cars and light trucks.

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Q 4 f )

4

Question:

A shaft has number of collars integral with it. The external diameter of the collars is 400 mm and the shaft diameter is 250 mm. If the uniform intensity of pressure is 0.35 N/mm2 and its co-efficient of friction is 0.05; find (i) power absorbed in overcoming friction when shaft rotates at 105 rpm and carries a load of 150 kN, and (ii) number of collars required.


Answer:

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Q 5 a )

8

Question:

List the factors to be considered for selecting the pipe while designing the pneumatic system. Give specification of pipes for the pneumatic system.


Answer:

List the Factors to be considered for selecting the pipe while designing pneumatic system. Give specification of pipes for the pneumatic system. Factors to be considered while selecting the pipe for pneumatic system 1. Pressure of compressed air in the line. 2. Total flow rate per unit time through the line. 3. Permissible pressure drop in the line. 4. Type of tube material and type of line fittings. 5. Length and diameter of tube or other pipelines. 6. Working environment. Pipe Size Specifications: Generally pipe size is specified in three ways 1. Nominal Pipe Size (NPS) : This number indicates the base diameter of pipe in inches. e.g. : ½ inch, ¾ inch, 1 inch, 1 ½ inch etc. 2. Schedule Number (SCH): This number is based on wall thickness, greater the SCH, greater will be the wall thickness of pipe. A schedule number indicates the approximate value of SCH = 1000 * P / S where P – service pressure & S – Allowable stress SCH Number 40, 80 and 160 are widely used. 3. Pipes are also classified as Standard (STD), Extra strong (XS) size, Double Extra strong (XXS) size based on strength

 

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Q 5 a )

8

Question:

A cam with 40 mm minimum diameter rotates in clockwise at uniform speed and has to give the following motion to a roller follower 15 mm diameter : (i) Follower to complete outward stroke of 40 mm during 120o of cam rotation with uniform velocity. (ii) Follower to dwell for 60o of cam rotation. (iii) Follower will return to its initial position during 120 of cam rotation with uniform acceleration and retardation. (iv) Follower will dwell for remaining 60o of cam rotation. Draw the profile of cam, if the axis of follower passes through the axis of cam. 


Answer:

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Q 5 a )

8

Question:

Explain the working of 4-stroke petrol engine with neat sketch.


Answer:

4 Stroke petrol engine

4 STROKE PETROL ENGINE  refers to its use in petrol engines, gas engines, light, oil engine and heavy oil engines in which the mixture of air fuel are drawn in the engine cylinder. Since ignition in these engines is due to a spark, therefore they are also called spark ignition engines. In four stroke cycle engine, cycle is completed in two revolutions of crank shaft or four strokes of the piston. Each stroke consists of 1800 of crankshaft rotation. Therefore, the cycle consists of 7200 of crankshaft rotation.

Cycle consists of following four strokes

1) Suction Stroke

2) Compression Stroke

3) Expansion or Power Stroke

4) Exhaust Stroke

SUCTION STROKE: In this Stroke the inlet valve opens and proportionate fuel-air mixture is sucked in the engine cylinder. Thus the piston moves from top dead centre (T.D.C.) to bottom dead centre (B.D.C.). The exhaust valve remains closed through out the stroke.

COMPRESSION STROKE: In this stroke both the inlet and exhaust valves remain closed during the stroke. The piston moves towards (T.D.C.) and compresses then closed fuel-air mixture drawn. Just before the end of this stroke the operating. plug initiates a spark which ignites the mixture and combustion takes place at constant pressure.

4 stroke petrol engine

 

4 stroke petrol engine animation

 

 

POWER STROKE OR EXPANSION STROKE: In this stroke both the valves remain closed during the start of this stroke but when the piston just reaches the B.D.C .the exhaust valve opens. When the mixture is ignited by the spark plug the hot gases are produced which drive or throw the piston from T.D.C. to B.D.C. and thus the work is obtained in this stroke.

EXHAUST STROKE: This is the last stroke of the cycle. Here the gases from which the work has been collected become useless after the completion of the expansion stroke and are made to escape through exhaust valve to the atmosphere. This removal of gas is accomplished during this stroke. The piston moves from B.D.C. to T.D.C. and the exhaust gases are driven out of the engine cylinder; this is also called scavenging

Links to other pages of power engineering

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Q 5a)(i)

8

Question:

(i) Show that the efficiency of a self locking screw is less than 50%


Answer:

(i) efficiency of screw η = tan α / tan(α +φ) And for self locking screws, φ ≥α or α≤ φ Efficiency ≤ tan(φ) /tan(φ +φ) ≤ tan φ/tan 2 φ ≤tan φ/ (2 tan φ/(1-tan2 φ)) ≤tan φ X (1-tan2 φ)/ (2 tan φ) ≤ ½-tan2 φ/2 From this expression efficiency of self locking screw is less than 50%

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Q 5a)(ii)

8

Question:

What is self locking property of threads and where it is necessary?


Answer:

self locking property of the threads-if φ > α the torque required to lower the load will  be positive, indicating that an effort is applied to lower the load. if friction angle is greater than the helix angle or coefficient of friction is greater than the tangent of helix angle applications- for very large use of screw in threaded fastener, screws in screw top container lids, vices, C-clamps and screw jacks

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Q 5 b )

8

Question:

State the methods used to improve thermal efficiency of gas turbine and explain any one.


Answer:

Methods to improve thermal efficiency of gas turbine Regeneration – This is done by preheating the compressed air before entering to the combustion chamber with the turbine exhaust in a heat exchanger, thus saving fuel consumption..

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Q 5 b )

8

Question:

Draw and explain pneumatic meter in circuit to control of speed extension.


Answer:

.Draw and Explain pneumatic meter in circuit to control of speed extension. In meter in pneumatic circuit flow control valve with check valve is fitted between DCV and actuator. For speed control of actuator during extension stroke, FCV with check valve is fitted on piston side of the actuator as shown in figure. With a meter-in circuit, fluid enters into the actuator at a controlled rate. Pneumatic circuit diagram for meter-in flow-control circuit is as shown in figure. In this circuits, the rate of flow of compressed air into the cylinder is controlled by flow control

 

valve. FCV is placed at inlet of the cylinder. Cap end port “C” is inlet for extension and rod end port “R” is inlet for retraction. Working: In first position of 4/2 DCV, compressed air flows from P to A and B to T. this flow is through flow control valve, the flow is controlled and hence piston extends slowly. In second position 4/2 DCV, compressed air flows P to B and A to T. this flow is through check valve. This is free flow. Hence the piston retracts at higher speed, Which is not controlled.

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Q 5 b )

8

Question:

In the toggle mechanism as shown in Fig. (2), D is constrained to move on a horizontal path. The dimensions of various links are AB = 200 mm, BC = 300 mm, OC = 150 mm and BD = 450 mm. The crank OC is rotating in a counter clockwise direction at a speed of 180 rpm. Find, for given configuration (1) velocity and (2) acceleration of ‘D’.


Answer:

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Q 5b)(i)

8

Question:

(i) The extension springs are in considerably less use than compression springs. Why?


Answer:

(i) it is easier to overextend the extension spring. Compression springs will bottom out before the overextend. Also it seems like the tensile strength will be weaker at the attachment point for the extension spring, making it generally larger and more cumbersome to correct the deficiency

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Q 5b)(ii)

8

Question:

Explain the terms self locking and overhauling of screw.


Answer:

self locking property - torque required to lower the load, T= Wtan(φ - α)xd/2 self locking property of the threads-if φ > α the torque required to lower the the load will be positive, indicating that an effort is applied to lower the load. if friction angle is greater than the helix angle or coefficient of friction is greater than the tangent of helix angle(2marks) Over hauling of screwsin the above expression, if φ< α,then the torque required to lower the load will be negative. The load will start moving downward without the application of any torque, such a condition is known as over hauling of screws.

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Q 5 c )

8

Question:

Explain with neat sketch (position based) working of sequencing circuit for two double acting Air cylinders.


Answer:

Explain with neat sketch (position based ) working of sequencing circuit for two double acting Air cylinders. Pneumatic double acting cylinders can be operated sequentially using a sequence valve or by using position based method. In pneumatics, use of sequence valve is not popular. Position based sequencing is possible using roller operated DCV or solenoid operated DCV. Various components required for Position based sequencing using roller operated DCV are as follows. I. Double acting cylinder - 02 Nos. II. 3/2 roller operated DCV – 02 Nos. III. 4/2 or 5/2 DCV – 01 No. IV. FRL Unit, Compressed air supply, hose pipes etc. Components are connected as shown in figure.

 

Working: In the first position of lever of 4/2 DCV (5/2 DCV can be used), the DAC extends. By the end of extension of first DAC, the cam presses roller of valve LV1 hence compressed air flows to second DAC, and second DAC extends. When the lever of 4/2 DCV is shifted to second position, DAC retracts. By the end of retraction of first DAC, the cam presses roller valve LV2, hence compresses air flows to second DAC and second DAC retracts.

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Q 5 c )

8

Question:

A leather belt is required to transmit 7.5 kW from a pulley 1.2 m in diameter running at 250 rpm. The angle of contact is 165o and the co-efficient of friction between the belt and the pulley is 0.35. If the safe working stress for the leather belt is 2 MPa, density of leather is 1050 kg/m3 and the thickness of belt is 10 mm, determine the width of belt, taking centrifugal tension into account.


Answer:

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Q 5c)(i)

8

Question:

Define following terms as applied to rolling contact bearings: 1) Basic static load rating 2) Basic dynamic load rating 3) Limiting speed


Answer:

(i) definition of (1) Basic static load rating-static radial load or axial load which corresponds to a total permanent deformation of the ball and race,at the most heavily stressed contact,equal to 0.001times the ball diameter. (2) basic dynamic load rating- the constant stationary radial load or a constant axial load which a group of of apparently bearings with stationary outer ring can endure for a rating life of one million revolutions with only 10% failure. (3)Limiting speed- it is the empirically obtained value for the maximum speed at bearings can be continuously operated without failing from seizure or generation of excessive heat.

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Q 5c)(ii)

8

Question:

List important physical characteristics of good bearing material.


Answer:

Physical characteristics of good bearing material- compressive strength, fatigue strength, embeddability, bondability, corrosion resistant, thermal conductivity, thermal expansion, conformability

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Q 6 a )

4

Question:

Draw profiles to square and Acme threads with full details. Which one is stronger?


Answer:

hread is stronger-

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Q 6 a )

4

Question:

The following results were obtained during Morse test on 4 stroke petrol engine Brake power developed when all cylinders are working = 16.2 kw Brake power developed with cylinder no. 1 cut off = 11.5 kw Brake power developed with cylinder no. 02 cutoff = 11.6 kw Brake power developed with cylinder no. 03 cutoff = 11.68 kw Brake power developed with cylinder no. 04 cutoff = 11.5 kw Calculate mechanical efficiency of the engine.


Answer:

Brake Power Engine (BP)engine = 16.2 kW Brake Power developed when 1st Cylinder cut-off ( BP )2,3,4 = 11.5 kW Brake Power developed when 2nd Cylinder cut-off ( BP )1,3,4 = 11.6 kW Brake Power developed when 3 rdCylinder cut-off ( BP )1,2,4 = 11.68 kW Brake Power developed when 4 thCylinder cut-off ( BP )1,2,3 = 11.5 kW Indicated Power of 1st cylinder IP1 = (BP)engine - ( BP )2,3,4 = 16.2 – 11.5 = 4.7 kW IP2 = (BP)engine - ( BP )1,3,4 = 16.2 – 11.6 = 4.6 kW IP3 = (BP)engine - ( BP )1,3,4 = 16.2 – 11.68 = 4.52 kW IP4 = (BP)engine - ( BP )1,2,3 = 16.2 – 11.5 = 4.7 kW Indicated Power of Engine IP = IP1 + IP2 + IP3 + IP4 = 4.7 + 4.6 + 4.52 + 4.7 = 18.52 kW Mechanical Efficiency of the Engine ηm = ( BP / IP ) x 100 = ( 16.2 / 18.52 ) x 100 = 87.47 %

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Q 6 a )

4

Question:

Draw speed control of single acting cylinder pneumatic circuit using 3 × 2 DC valve


Answer:

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Q 6 a )

4

Question:

Draw a neat sketch of Oldham’s coupling and explain the working of it.


Answer:

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Q 6 b )

4

Question:

A helical valve spring is to be designed for an operating load range of approximately 135 N. The deflection of the spring for the load range is 7.5 mm. Assume spring index of 10. Permissible shear stress for the material of the spring = 480 MPa and its modulus of rigidity = 80 KN/mm2. Design the spring. Take Wahle’s factor 4 4 4 1 . , C C C 0 615 = - - + ‘C’ being the spring index


Answer:

given load W= 135N Deflection ᵟ =7.5mm Spring index c=10 Permissible shear stress Ʈ=480 MPa Modulus of rigidity G =80 KN/mm2 Wahl’s factor K =4C-1/4C-4 +0.615/C=4X10-1/4X10-4 +0.615/10=1.14 (1)Mean dia. Of the spring coil (1 mark) Maximum shear stress, Ʈ = Kx 8WC/π d 2 480 = 1.14x 8x135x10/3.142xd2 d = 2.857mm from table we shall take a standard wire of size SWG 3 having diameters (d) =2.946mm mean dia. Of the spring coil D= CXd =10x2.946=29.46 mm outer dia. Of the spring coil Do =D+d=29.46+2.946=32.406mm (2) number of turns of the spring coil (n) (1 mark) Deflection ᵟ= 8WC3 n/Gd 7.5 =8x135X103x n/ 80000xd n =1.64 say 2 For square and ground end n’ =n+2=2+2=4 (3) free length of spring (1 mark) =Lf =n’d+ ᵟ + 0.15 x ᵟ=4x2.496+7.5+0.15xx7.5=18.609mm (4) pitch of the coil (1 mark) p= free length/n’-1=18.609/4-1=6.203mm

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Q 6 b )

4

Question:

What is the necessity of purification of air ? How to remove oil, moisture and dust from air?


Answer:

The air sucked by the compressor is not clean. It contains various types of solid, liquid and gaseous contaminants such as dust, dirt, moisture etc. The presence of contaminants may have high damaging effects such as corrosion, wear and tear on the finely finished mating surfaces of pneumatic components. Air lines may get chocked or damaged. Therefore, purification of air by removing oil, moisture and dust is done to protect the pneumatic system from failure, so that the system should work efficiently. 1) Particulate Filters ( Dry Air Filters ) Particulate filters are used to remove dust and particles out of the air. This will allow air to travel faster in the piping system and prevent clogs. The main element in this filtration is the membrane. The membrane acts like a gate which only lets air pass through while anything bigger gets blocked by the membrane material.

2) Coalescing Filters Coalescing filters are used to capture oil and tiny moisture droplets and prevent condensate from developing in the system. This will prolong the life of the piping system and other components by avoiding rust. The main component used is the flow of the air. The filter may contain a membrane element in it as well but altering the flow of the air in a tight space causes condensate or oil to gather at the bottom of the filter.

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Q 6 b )

4

Question:

State any four reasons of failure of pneumatic seals.


Answer:

1. Incompatibility of seal material with operating system 2. Excessive heat 3. Excessive load 4. Excessive clearance 5. Excessive pressure 6. Improper fitting 7. Improper groove geometry 8. Abrasion 9. Wear

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Q 6 b )

4

Question:

State any four reasons of failure of pneumatic seals.


Answer:

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Q 6 b )

4

Question:

Define following terms : Fluctuation of energy, co-efficient of fluctuation of energy, co-efficient of fluctuation speed, maximum fluctuation of energy.


Answer:

Fluctuations of energy: The variations of energy above and below the mean resisting torque line are called fluctuations of energy.

Coefficient of fluctuation of energy: It may be defined as the ratio of the maximum fluctuation of energy to the work done per cycle. Mathematically, Coefficient of fluctuation of energy, E = Maximum fluctuation of energy/Work done per cycle

Coefficient of fluctuation of speed: The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is called the coefficient of fluctuation of speed.

Maximum fluctuation of energy: Δ E = Maximum energy – Minimum energy = (E + a1) – (E + a1 – a2 + a3 – a4) = a2 – a3 + a4

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Q 6 c )

4

Question:

A bracket as shown in Figure No. 1 is fixed to a vertical steel column by means of five standard bolts.  


Answer:

Horizontal component of 45 KN, WH= 45Sin 600 =45x 0.866=38971N and vertical component of 45 KN, Wv,= 45xcos600 =45x0.5=22500N Direct tensile load in each bolt,Wt1= WH /5=38971/5=7794.20N Turning moment due to WH about G TH = WHx 25=38971x25=974275N (anticlockwise) direct shear load on each bolt =Ws=Wv/5 =22500/5=4500N Turning moment due to Wv about edge of the bracket, Tv= Wvx175=22500x175=3937500N-mm( clockwise( clockwise) Net turning moment =3937500-974275=2963225N---------(I) total moment of the load on the bolts @ th tilting edge = 2w x(L1) 2 + 2w x(L2) 2 =2xwx(50)2 + 2xwx(150)2 = 50000 w N-mm-----(II) from equations (I) and(II) 2963225N=50000 w N- w= 592.645 N max. tensile load on each of the upper bolt, Wt2= wL2 =592.645x150=88896.75 N tensile load on each of the upper bolt, Wt = Wt1+ Wt2 =7794.20+ 88896.75=96690.95N equivalent tensile load =Wte=1/2(Wt+ √‾(Wt)2 + 4(Ws)2 =1/2 ( 96690.95+97108.91)=96899.93 N Tensile load on each bolt = ∏/4(dc)2 x 6t =0.7854x(dc)2 x 70 dc = 41.98 mm from coarse series the standard core dia. Is 49.0177 mm and corresponding size of the bolt is M56 thickness of the arm of the bracket cross sectional area of the arm A = bXt =100x t

section modulus of the arm, Z = 1/6 t (b)2 = 1/6 xtx(100)2 =1666.67 xt direct tensile stress 6t1 = WH/A = 38971/100t =389.71/t bending stress 6t2 = MH/Z = 208/t bending stress 6t3 = Mv /Z = 2632.49/t net tensile stress, 6t1 +6t2 + 6t3 = 3230.20/t max. tensile stress , 6t max. 6t/2+ ½ √‾(6t)2 + 4(Ʈ)2 =70 t = 46.36 mm

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Q 6 c )

4

Question:

Draw the schematic diagram of turbojet engine.


Answer:

Turbo Jet Engine

 

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Q 6 c )

4

Question:

(c) What are the advantages of pneumatic system over hydraulic systems ?


Answer:

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Q 6 c )

4

Question:

Explain the working of rope brake dynamometer with neat sketch


Answer:

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Q 6 d )

4

Question:

What are rolling contact bearings? State their advantages over sliding contact bearings.


Answer:

Rolling contact bearing- contact between the surfaces is rolling ,it is antifriction bearing Advantages (any six) (1)low starting and running friction except at very high speed (2) ability to withstand momentary shock loads (3) accuracy of shaft alignment (4) low cost of maintenance (5) reliability of service (6) easy to mount and erect (7) cleanliness (8) small overall dimension

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Q 6 d )

4

Question:

Define : i) WBT ii) DPT iii) DBT iv) Degree of saturation.


Answer:

i) WBT: Wet bulb Temperature twb : It is the temperature recorded by a thermometer when its bulb is covered by a wet cloth exposed to the air. ii) DPT: Dew point temperature tdp :It is the temperature of air recorded by thermometer, when the moisture (water vapour) present in its, begins to condensed. iii) DBT: Dry Bulb Temperature tdb : It is the temperature of air recorded by ordinary thermometer with a clean, dry sensing element . iv) Degree of Saturation (μ):Degree of saturation is defined as ‘the ratio of mass of water vapour associated with unit mass of dry air to mass of water vapour associated with saturated unit mass of dry air at same temperature. 

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Q 6 d )

4

Question:

Compare positive displacement pump with Rotodynamic pump.


Answer:

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Q 6 d )

4

Question:

Explain the working of single plate clutch with neat diagram.


Answer:

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Q 6 e )

4

Question:

State the strength equation of double parallel fillet weld and single transverse fillet weld with neat sketches


Answer:

Strength equation of double parallel fillet weld= throat area x allowable shear stress P= 2x 0.707x Sw x lwx Ʈ =1.414 x Sw x lwx Ʈ (1mark) Strength equation of single transverse fillet weld

P =throat area x allowable tensile stress P= 0.707x Sw x lw x σ

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Q 6 e )

4

Question:

Explain the working of simple vapour absorption refrigeration system.


Answer:

 

Working of Simple Vapor absorption system: A Simple Vapor absorption system consists of evaporator, absorber, generator, condenser, expansion valve, pump & reducing valve. In this system ammonia is used as refrigerant and solution is used is aqua ammonia. Strong solution of aqua ammonia contains as much as ammonia as it can and weak solution contains less ammonia. The compressor of vapor compressor system is replaced by an absorber, generator, reducing valve and pump. The heat flow in the system at generator, and work is supplied to pump. Ammonia vapors coming out of evaporator are drawn in absorber. The weak solution containing very little ammonia is spread in absorber. The weak solution absorbs ammonia and gets converted into strong solution. This strong solution from absorber is pumped into generator. The addition of heat liberates ammonia vapor and solution gets converted into weak solution. The released vapor is passed to condenser and weak solution to absorber through a reducing valve. Thus, the function of a compressor is done by absorber, a generator, pump and reducing valve. The simple vapor compressor system is used where there is scarcity of Electricity and it is very useful at partial and full load.

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Q 6 e )

4

Question:

What are the various types of Hoses used in pneumatic system


Answer:

Hoses are flexible connecting tubes or pipes to connect actuators, control valves. Different layers of hose 1) Inner tube 2) Reinforcement 3) Outer protective cover Hoses: are flexible vessels that are constructed of multiple layers of different materials. Fittings for hoses are often not permanent, since the hose itself is often replaced in time due to wear

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Q 6 e )

4

Question:

State reasons for balancing of rotating elements of machine. Explain balancing concept.


Answer:

Reasons for balancing of rotating elements of machine: The balancing of the moving parts both rotating and reciprocating of such machine is having greater importance. Because, if these parts are not balanced properly then the unbalanced dynamic forces can cause serious consequences, which are harmful to the life of the machinery itself, the human beings and all the property around them. These unbalanced forces not only increase the load on the bearings and stresses in various members, but also produces unpleasant and dangerous vibrations in them.

Concept of balancing: When a mass moves in circular pitch, it experience a centripetal acceleration which generates a force acting towards the center of rotation. An equal and opposite force which is acting radially outwards which is called centrifugal force. This force is the disturbing force for the system. The magnitude of this force remains constant but the direction goes on changing with the rotation of mass. The centrifugal force , on a rotating machine can be expressed mathematically as follows:

Fc = m. ω².r Newton

Where, m = Mass of rotating part in kg,

Ω = angular speed of this part in rad/sec, and r = Distance of the center of gravity of mass from the axis of rotation of part in m.

For the balance of rotating masses, it is the centrifugal force which is to be balanced. This type of problem is very common in steam turbine rotors, engine crank shafts, rotory compressors and centrifugal pumps.

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Q 6 f )

4

Question:

Four masses A, B, C and D are attached to a shaft and revolve in the same plane. The masses are 12 kg, 10 kg, 18 kg and 15 kg respectively and their radii of rotations are 40 mm, 50 mm, 60 mm and 30 mm. The angular position of the masses B, C and D are 60O, 135O ,and 270O from the mass ‘A’. Find the magnitude and position of the balancing mass at a radius of 100 mm. Use graphical method only.


Answer:

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Examination: 2017 WINTER
Que.No Marks
Q 1 a )

2

Question:

Define machine design.


Answer:

Machine design is the process of selection of the materials, shapes, sizes and arrangements of
mechanical elements so that the resultant machine will perform the prescribed task. OR
Machine Design is the creation of new and better machines and improving the existing ones

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Q 1a)(a)

4

Question:

Draw p-v and T-S diagram for Diesel cycle. Name the processes involved in it.


Answer:

Diesel Cycle on P-V and T-S diagram :

Processes : 1-2 : Isentropic compression 2-3 : Heat addition at constant pressure 3-3 Isentropic expansion 4-1 Heat rejection at constant volume

 

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Q 1a)(a)

2

Question:

Pappu Define kinematic link and kinematic chain. 


Answer:

a) Kinematic link: Each part of a machine, which moves relative to some other part, is known as a kinematic link (or simply link) or element. Kinematic Chain: When the kinematic pairs are coupled in such a way that the last link is joined to the first link to transmit definite motion (i.e. completely or successfully constrained motion), it is called a kinematic chain.

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Q 1a)(a)

2

Question:

Define kinematic link and kinematic chain.


Answer:

a) Kinematic link: Each part of a machine, which moves relative to some other part, is known as a kinematic link (or simply link) or element. Kinematic Chain: When the kinematic pairs are coupled in such a way that the last link is joined to the first link to transmit definite motion (i.e. completely or successfully constrained motion), it is called a kinematic chain.

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Q 1a)(a)

4

Question:

State the four advantages and disadvantages of screw pump.


Answer:

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Q 1a)(b)

4

Question:

Draw actual valve timing diagram for 4-stroke petrol engine.


Answer:

Valve timing diagram of four stroke diesel engine

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Q 1a)(b)

2

Question:

State types of cams.


Answer:

b) Types of cam: 1. Radial or disc cam 2. Cylindrical cam

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Q 1a)(b)

4

Question:

Explain the construction of 4/2 poppet valve with neat sketch & symbol.


Answer:

4/2 puppet valve Figure shows a cross sectional schematic view of a poppet type 4/2 direction control valve. Inside the valve housing, a number of bores are engraved and interconnected through number of valve elements. The ports ‘P’, ‘R’, ‘A’, and ‘B’ shown in the diagram are designated as ‘Ppressure port, ‘A’ and ‘B’ – cylinder port and ‘R’ – exhaust port. In the position shown in the sketch, it is found that ‘P’ connects to ‘A’ and ‘B’ to ‘R’, When the elements are actuated by means of the push button, they are unseat and ‘P’ connects to ‘B’ and ‘A’ to ‘R’. The rated size of the valve depends on the cross-section of the valve port. Through proper shaping of the fluid ports or canals, the loss of pressure may be minimized. The actuating elements of the spool in zero position are spring controlled and for accurate controlling may be designed as pressure compensated

 

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Q 1a)(c)

4

Question:

Classify air compressors


Answer:

Classification of Air compressors:

1. According to principle: a) Reciprocating air compressors b) Rotary air compressors 2. According to the capacity a. Low capacity air compressors b. Medium capacity air compressors c. High capacity air compressors 3. According to pressure limits a. Low pressure air compressors b. Medium pressure air compressors c. High pressure air compressors 4. According to method of connection a. Direct drive air compressors b. Belt drive air compressors c. Chain drive air compressors

 

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Q 1a)(c)

2

Question:

State law of gearing.


Answer:

Gearing law (Law of gearing) :

Gearing law states that, "The law of gearing states that the angular velocity ratio of all Gears of a meshed gear system must remain constant also the common normal at the point of contact must pass through the pitch point."

gearing law

Gearing law illustration

As illustrated in above animation the common normal at the point of contact passes through the pitch point. Gearing law must be followed in order to two gears transmit motion form one to another.

In order to have a constant angular velocity ratio for all positions of the wheels, it is must that the point P must be the fixed point (called pitch point) for the two wheels. In other words it can be said that , the common normal at the point of contact between a pair of teeth should always pass through the pitch point for  proper working.

This is the fundamental condition which must be satisfied while designing the profiles for the teeth of gear wheels, it is also known as the law of gearing.

Gearing law explination with diagram

 

 

 

 

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Q 1a)(c)

4

Question:

State the essential properties of hydraulic fluids.


Answer:

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Q 1a)(d)

4

Question:

Explain any two mounting methods of cylinder.


Answer:

1) Centreline mounting Centreline mounts are used to take care of thrust that can occur linearly or along a centreline with the cylinder. Proper alignment is essential to prevent compound stresses that may cause excessive friction and bending, as piston extends. Additional holding strength may be essential with long stroke cylinders. 2) Foot mounting It consists of mounting the cylinder with the help of side end lungs or side covers. These mountings are used where cylinders are to be mounted on to surface parallel to the axis of cylinder

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Q 1a)(d)

4

Question:

Classify gas turbine on the basis of a) Cycle of operation b) Thermodynamic cycle c) Application d) Combustion process


Answer:

Classification of gas turbine on the basis of

a. Cycle of operation 1. Open cycle 2. Closed cycle

b. Thermodynamic cycle 1. Brayton or Joules cycle 2. Atkinson cycle 3. Erricsson cycle

c. Application 1. For supercharging of IC engine 2. For locomotive propulsion 3. For ship propulsion 4. Industrial application 5. Air craft engines 6. Electric power generation

d. Combustion process 1. Continuous combustion 2. Explosion combustion

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Q 1a)(d)

2

Question:

State the types of chains & sprockets.


Answer:

Types of Chains & Sprockets: The chains, on the basis of their use, are classified into the following three groups : 1. Hoisting and hauling (or crane) chains, 2. Conveyor (or tractive) chains, and 3. Power transmitting (or driving) chains. Sprockets: 1. Taper lock sprockets 2.Pilot bore sprocket 3.Platewheel sprocket

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Q 1a)(e)

2

Question:

State the function of flywheel in I.C. Engine.


Answer:

A flywheel used in machines serves as a reservoir, which stores energy during the period when the supply of energy is more than the requirement, and releases it during the period when the requirement of energy is more than the supply. In other words, a flywheel controls the speed variations caused by the fluctuation of the engine turning moment during each cycle of operation.

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Q 1a)(f)

2

Question:

State the function of governor.


Answer:

The function of a governor is to regulate the mean speed of an engine, when there are variations in the load e.g. when the load on an engine increases, its speed decreases, therefore it becomes necessary to increase the supply of working fluid. On the other hand, when the load on the engine decreases, its speed increases and thus less working fluid is required. The governor automatically controls the supply of working fluid to the engine with the varying load conditions and keeps the mean speed within certain limits

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Q 1a)(g)

2

Question:

Compare brakes and dynamometers. (any two points)


Answer:

Compare Brakes & Dynamometers: A dynamometer is a mechanical device used to indirectly measure the power output of a prime mover like an engine or a motor. Examples: hydraulic brake dynamometer, eddy current dynamometer, prony brake dynamometer. A brake is a mechanical device usually found in automobiles that helps in decelerating a vehicle and brings it to a complete stop. Examples: internal expanding shoe brake, single and double shoe brake, simple and differential band brake.

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Q 1a)(h)

2

Question:

Why is balancing of rotating parts necessary for high speed engines ?


Answer:

Reasons for balancing of rotating elements of machine: The balancing of the moving parts both rotating and reciprocating of such machine is having greater importance. Because, if these parts are not balanced properly then the unbalanced dynamic forces can cause serious consequences, which are harmful to the life of the machinery itself, the human beings and all the property around them. These unbalanced forces not only increase the load on the bearings and stresses in various members, but also produces unpleasant and dangerous vibrations in them.

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Q 1a)(i)

2

Question:

(a) Define : (i) Spherical pair (ii) Higher pair


Answer:

a) Single plate clutch b) Multi plate clutch c) Cone clutch d) Centrifugal clutch

ii) Classification of follower:

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Q 1 b )

4

Question:

State any four types of friction clutch, along with its application each.


Answer:

(Types of clutches: Two marks, applications Two marks) Types of clutches: a) Single plate clutch b) Multi plate clutch c) Cone clutch d) Centrifugal clutch Applications: a) Single plate clutch: Heavy vehicles, four-wheeler such as car, truck, bus b) Multi plate clutch: Two wheelers, mopeds, scooters, bikes c) Cone clutch: Machine tools, automobiles, press work d) Centrifugal clutch: mopeds, Luna

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Q 1 b )

4

Question:

Define slip and creep with reference to belt drive. Also state their effect on velocity ratio.


Answer:

V- Belt drive – air compressor, machine tools (drilling machine)  Flat belt drive - lathe headstock, floor mill, stone crusher unit  Gear drive – gear box of vehicles, cement mixing unit, machine tools, I.C. Engine, differential of automobile, dial indicator  Chain drive – Bicycle, cranes, Hoists, bikes

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Q 1 b )

2

Question:

  Give the composition of :- (i) FeE220: (ii) 20C8


Answer:

(i) FeE220: Steel having yield strength of 220 N/mm2 . (ii) 20C8 : Carbon steel containing 0.15 to 0.25 percent (0.2 percent on average) carbon and 0.60 to 0.90 percent (0.80 percent on average) manganese.

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Q 1 b )

2

Question:

(b) Define : (i) Radial follower (ii) Off-set follower


Answer:

1. According to the surface in contact:  Knife-edge follower  Roller follower  Flat faced or mushroom follower  Spherical follower 2. According to the motion of the follower:  Reciprocating or translating follower  Oscillating or rotating follower 3. According to the path of motion of follower:

Radial follower  Off-set followe

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Q 1b)(a)

6

Question:

Draw general layout of hydraulic system and explain its working.


Answer:

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Q 1b)(a)

4

Question:

Define completely constrained motion and successfully constrained motion with neat sketch. State one example of each.


Answer:

a) 1. Completely constrained motion: When the motion between a pair is limited to a definite direction irrespective of the direction of force applied, then the motion is said to be a completely constrained motion. For example, the piston and cylinder (in a steam engine) form a pair and the motion of the piston is limited to a definite direction (i.e.it will only reciprocate) relative to the cylinder irrespective of the direction of motion of the crank. Examples: 1. The motion of a square bar in a square hole 2. the motion of a shaft with collars at each end in a circular hole,

2. Successfully constrained motion: When the motion between the elements, forming a pair, is such that the constrained motion is not completed by itself, but by some other means, then the motion is said to be successfully constrained motion. Consider a shaft in a foot-step bearing as shown in Fig. The shaft may rotate in a bearing or it may move upwards. This is a case of incompletely con-strained motion. But if the load is placed on the shaft to prevent axial upward movement of the shaft, then the motion of the pair is said to be successfully constrained motion. Examples:1. The motion of an I.C. engine valve (these are kept on their seat by a spring) 2. The piston reciprocating inside an engine cylinder 3. Shaft in a foot step bearing

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Q 1b)(a)

6

Question:

Draw a neat labelled sketch of fuel injection pump. Give its function.


Answer:

Fuel injection pump : Fuel injection pump is used widely for the supply of fuel under high pressure in diesel engines

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Q 1b)(b)

6

Question:

With a neat sketch explain pressure compensated flow control valve. Draw symbol of it.


Answer:

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Q 1b)(b)

4

Question:

State function of clutch. Explain working principle of clutch.


Answer:

b) Function of the Clutch 1. Function of transmitting the torque from the engine to the drive train. 2. Smoothly deliver the power from the engine to enable smooth vehicle movement. 3. Perform quietly and to reduce drive-related vibration. WORKING PRINCIPLE OF CLUTCH It operates on the principle of friction. When two surfaces are brought in contact and are held against each other due to friction between them, they can be used to transmit power. If one is rotated, then other also rotates. One surface is connected to engine and other to the transmission system of automobile. Thus, clutch is nothing but a combination of two friction surfaces

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Q 1b)(b)

6

Question:

Explain regeneration method to improve thermal efficiency of gas turbine with the help of flow diagram and T-S diagram.


Answer:

Regenerative method to improve thermal efficiency in gas turbines : The exhaust gases a lot of heat as their temperature is far above the ambient temperature . The heat of exhaust gases can be used to heat the air coming from the compressor thus reducing the mass of the fuel supplied in the combustion chamber as shown in the figure. This method is called regenerative method.

 

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Q 1b)(ii)

8

Question:

2) Define : a) Ductility b) Toughness c) Creep


Answer:

a) Ductility: the property of material which enables it to be drawn into thin wire under the action of tensile load is called as ductility. b) Toughness: The property which resists the fracture under the action of impact loading is called as toughness. Toughness is energy for failure by fracture. c) Creep: when a component is subjected to constant stress at a high temperature over a long period of time ,it will undergo a slow& permanent deformation called creep Or it is defined as “slow and progressive deformation of material with time under constant stress at elevated temperature. E.g : Bolts & pipes in thermal power plants

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Q 1 c )

2

Question:

What do you mean by crowning of pulleys in flat belt drive ? State its use.


Answer:

1. As no slip takes place, hence, perfect velocity ratio is obtained (Positive drive). 2. Chain drive gives high transmission efficiency (up to 98 %). 3. Chain drive may be used when the distance between the shafts is less. 4. Chain is made up of metal which would occupy less space as compared with belt or rope drive. 5. Ability to transmit power to several shafts by one chain. 6. Load on the shaft is less and long life.

 

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Q 1 c )

2

Question:

State four types of loads acting on machine elements


Answer:

(i) Dead or steady load (ii) Live or variable load (iii) Suddenly applied or shock load (iv) Impact load

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Q 1 d )

2

Question:

What do you mean by creep?


Answer:

When a machine part is subjected to a constant stress at high temperature for a long period of time, it will undergo a slow and permanent deformation called ‘creep’. This property is considered in designing internal combustion engines, boilers and turbines

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Q 1 d )

2

Question:

Define initial tension in belt drive & state its effect.


Answer:

1. Manufacturing cost of chains is relatively high 2. The chain drive needs accurate mounting and careful maintenance 3. High velocity fluctuations especially when unduly stretched 4. Chain operations are noisy as compared to belts.

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Q 1 e )

2

Question:

Define Ergonomics.


Answer:

Ergonomics is defined as the scientific study of the man – machine working environment relationship and the application of anatomical, physiological, psychological principles to solve the problems arising from this relationship.

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Q 1 e )

2

Question:

Define fluctuation of speed and fluctuation of energy in case of flywheel.


Answer:

Fluctuation of speed: It is the difference between the maximum and minimum speed of Flywheel. Fluctuation of speed = (N1 – N2) rpm N1 – maximum speed, N2 -- minimum speed Fluctuation of energy: It is the difference between the maximum and minimum energy of Flywheel. Maximum energy of Flywheel I ω1 2 Minimum energy of Flywheel = I ω2 2 Fluctuation of energy = I (ω1 2 - ω2 2 ) in N-m or J I – moment of inertia of flywheel = mk2 where, m – mass of the flywheel, kg and k - radius of gyration of flywheel, m2 ω1 – Maximum Angular velocity, rad/sec ω2 – Minimum Angular velocity, rad/sec

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Q 1 f )

2

Question:

Give two applications of knuckle joint.


Answer:

(i) A knuckle joint is used to connect two rods which are under the action of tensile loads. However, if the joint is guided, the rods may support a compressive load. (ii) Its use may be found in the link of a cycle chain, tie rod joint of roof truss, valve rod joint with eccentric rod, pump rod joint, tension link in bridge structure and lever and rod connections of various types.

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Q 1 f )

2

Question:

Define the sensitivity in relation to governer. State its significance.


Answer:

The function of governor is to regulate the mean speed of the engine, when there are variations in the load. Governor automatically adjusts and controls the supply of fuel / working fluid to the engine with the varying load conditions and keeps the mean speed within the certain desired limits. e.g. When the load on an engine increases, its speed decreases, therefore it becomes necessary to increase the supply of fuel or working fluid. The configuration of the governor changes and valve is moved to increase the supply of working fluid. Conversely, when the load on the engine

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Q 1 g )

2

Question:

Define following terms of spring:


Answer:

(i) Spring rate: The spring rate is defined as the load required per unit deflection of the spring. It is also
known as spring stiffness or spring constant. Mathematically,
Spring rate, k = W / δ
Where,
W = Load
δ = Deflection of the spring
(ii) Spring index: The spring index is defined as the ratio of the mean diameter of the coil to the
diameter of the wire. Mathematically,
Spring index, C = D / d
Where,
D = Mean diameter of the coil
d = Diameter of the wire

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Q 1 h )

2

Question:

How do you express the life of bearings?


Answer:

The life of an individual bearing is defined as the total number of revolutions (or the number of hours at a given constant speed) which the bearing can complete before the evidence of fatigue failure develops on the balls or races.  The bearing life can be defined by rating life.  The rating life of a group of apparently identical bearing is defined as the number of revolutions (or the number of hours at a given constant speed) that 90 percent of a group of bearings will complete or exceed before the first evidence of fatigue failure develops. It is also known as L10 life.

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Q 1 h )

2

Question:

State the adverse effect of imbalance of rotating elements of machine.


Answer:

The process of providing the second mass in order to counter act the effect of the centrifugal force of the disturbing mass is called balancing. In order to prevent the bad effect of centrifugal force of disturbing mass, another mass (balancing) is attached to the opposite side of the shaft at such a position, so as to balance the effect of centrifugal force of disturbing mass. This is done in such a way that the centrifugal forces of both the masses are made equal and opposite. Methods of balancing:  Balancing of rotating masses 1) Balancing of a single rotating mass by a single rotating mass in the same plane 2) Balancing of a single rotating mass by two masses rotating in the different planes * Disturbing mass lies in a plane between the planes of balancing masses * Disturbing mass lies in a plane on one end of the planes of balancing masses 3) Balancing of different masses rotating in the same plane 4) Balancing of different masses rotating in the different planes  Balancing of reciprocating masses

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Q 1 i )

2

Question:

Draw the different thread profiles used for power screws.


Answer:

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Q 1 j )

2

Question:

State four types of keys.


Answer:

(i) Sunk key: Rectangular sunk key, Square sunk key and Parallel sunk key (ii) Gib-head key (iii) Feather key Any four (iv) Woodruff key types of keys (v) Saddle key: Flat saddle key, Hollow saddle key (vi) Tangent key (vii) Round key

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Q 1 k )

2

Question:

Give two examples, where screwed joints are preferred over welded joints.


Answer:

i) Cylinder head of the engine. (ii) Machine foundation. (iii) Assembly of fan, couplings. Any two examples (iv) Connect two bogies of the train with the turn buckle. (v) Structural bridges, pressure vessels, fly press (vi) Assembly of crank shaft and connecting rod.

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Q 1 l )

2

Question:

State any four applications of rolling contact bearings.


Answer:

(i) Industrial and automotive gear boxes. (ii) Electric motors and machine tool spindles. (iii) Small size centrifugal pumps. (iv) Automobile front and rear axles

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Q 1 m )

2

Question:

What are the requirement of a good coupling?


Answer:

A good coupling should have the following requirements: (i) It should be easy to connect and disconnect. (ii) It should transmit the full power from one shaft to another shaft without losses. (iii) It should hold the shafts in perfect alignment. (iv) It should reduce the transmission of shock loads from one shaft to another shaft. (v) It should have no projecting parts

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Q 1 n )

2

Question:

Draw stress – strain diagram for brittle material.


Answer:

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Q 2 a )

8

Question:

Reciprocating air compressor draws 6 kg of air per minute at 25°C. It compresses the air polytropically and delivers it at 105°C. Find the work done by the compressor and air power. Also find mechanical efficiency if shaft power is 14 kW. Assume R = 0.287 kJ/kg°K and n = 1.3.


Answer:

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Q 2 a )

8

Question:

Explain with neat sketch the working of hydraulic circuit for milling machine.


Answer:

Working of hydraulic circuit for milling machine. Hydraulic circuit for milling machine is comparatively different from other circuits. Table movement of milling machine is required to be adjustable for different feeds for different type of work. Therefore for both strokes of the cylinder, on both ends of cylinder flow control valves are used. Another feature of this circuit is that there are two pumps 1. Main pump – low pressure high discharge 2. Booster pump - high pressure low discharge The function of booster pump is to boost the hydraulic pressure to a higher level than given by main pump. Reason behind using this type is to save power as well as use of high pressure high discharge pump is avoided. 4/3 DCV used manually operated stroke length of cylinder is adjustable through limit switch. In centre position of 4/3 DCV all the ports are close therefore, total hydraulic system is lock.. position (I) pump flow is given to cylinder blank end and extension starts and oil from rod end is discharge to tank. In (II) position, pump flow diverted to rod end for retraction and blank end side flow pass to tank

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Q 2 a )

4

Question:

Differentiate between machine and structure.


Answer:

Sl. No
Machine
Structure

1
All parts / links have relative motion
No relative motion between the links

2
It transforms the available energy into some useful work
No energy transformations

3
The kinematic link of a machine may transmit both power and
The member of the structure transmit forces only

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Q 2 a )

4

Question:

What is a machine ? Differentiate between a machine and a structure.


Answer:

Sl. No
Machine
Structure

1
All parts / links have relative motion
No relative motion between the links

2
It transforms the available energy into some useful work
No energy transformations

3
The kinematic link of a machine may transmit both power and motion
The member of the structure transmit forces only

4
Examples: I.C. Engine, Machine tools, steam engine, type writer, etc.
Example: Truss of roof, frame of machine, truss of bridge

5
Studied under 'Dynamics'
Studied under 'Statics'

 

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Q 2 a )

8

Question:

Explain various failures to be considered in designing a cotter joint along with the necessary sketches and strength equations.


Answer:

 

It consist of 3 elements: i. Socket ii. Spigot iii. Cotter Where, d= End diameter of rod d1= Diameter of spigot/Inside diameter of socket d2= Diameter of spigot collar D1= Outer diameter of socket D2= Diameter of socket collar C=Thickness of socket collar t1= Thickness of spigot collar t= thickness of cotter b= Mean width of cotter a= Distance of end of slot to the end of spigot P= Axial tensile/compressive force σt , σc , τ= Permissible tensile,

compressive, shear stress for the component materia

 

 

 

 

 

 

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Q 2 b )

8

Question:

Name any eight pipe or tube fitting with their application.


Answer:

1) Adaptor – To connect two pipes of different diameters. 2) Coupling- To connect two pipes of same diameters. 3) Tee- To connect two pipes with one pipe. 4) Cross- To connect two pipes in crosswise. 5) Elbow- To divert the flow between two pipes at right angle. 6) Hex nipple- To connect two pipes internally with the help of hexagonal nut. 7) 450 Elbow- To divert the flow between two pipes at 450 angle. 8) Reducer- To connect two pipes of different diameters and it will increase or reduce the pressure of flow

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Q 2 b )

8

Question:

List various types of air motors. Explain vane type air motor with neat sketch.


Answer:

Various Types of Air Motors 1. Vane Motor 2. Gerotor Motor 3. Turbine Motor 4. Piston Motor Construction: It consists of simple Vane rotor which is having slots in which vanes (flat piece of steel) slides freely. The rotor is eccentrically located inside the stator housing. Working: When pressurized air comes in through inlet port, the pressure of air distributes equal in all directions. Since vane is sliding freely in slots of rotator, the vane comes in to way of pressurized air and air pushes the vanes so that rotor starts rotating with speed. The used low pressure air is exhausted through exhaust port. This is unidirectional motor. Since vanes are freely sliding in slots, there is possibility of leakage of air. With the help of these motors we can achieve the speeds up to 25000 r.p.m.

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Q 2 b )

4

Question:

Explain with the neat sketch working of crank and slotted lever quick return mechanism.


Answer:

Crank and slotted lever quick return motion mechanism: This mechanism is mostly used in shaping machines, slotting machines and in rotary internal combustion engines. In this mechanism, the link AC (i.e. link 3) forming the turning pair is fixed, as shown in fig. The link 3 corresponds to the connecting rod of a reciprocating steam engine. The driving crank CB revolves with uniform angular speed about the fixed centre C. A sliding block attached to the crank pin at B slides along the slotted bar AP and thus causes AP to oscillate about the pivoted point A. A short link PR transmits the motion from AP to the ram which carries the tool and reciprocates along the line of stroke R1R2. The line of stroke of the ram (i.e. R1R2) is perpendicular to AC produced

In the extreme positions, AP1 and AP2 are tangential to the circle and the cutting tool is at the end of the stroke. The forward or cutting stroke occurs when the crank rotates from the position CB1 to CB2 (or through an angle β) in the clockwise direction. The return stroke occurs when the crank rotates from the position CB2 to CB1 (or through angle α) in the clockwise direction. Since the crank has uniform angular speed,

 

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Q 2 b )

4

Question:

Describe with neat sketch the working of scotch yoke mechanism.


Answer:

Crank and slotted lever quick return motion mechanism. This mechanism is mostly used in shaping machines, slotting machines and in rotary internal combustion engines. In this mechanism, the link AC (i.e. link 3) forming the turning pair is fixed, as shown in fig. The link 3 corresponds to the connecting rod of a reciprocating steam engine. The driving crank CB revolves with uniform angular speed about the fixed centre C. A sliding block attached to the crank pin at B slides along the slotted bar AP and thus causes AP to oscillate about the pivoted point A. A short link PR transmits the motion from AP to the ram which carries the tool and reciprocates along the line of stroke R1R2. The line of stroke of the ram (i.e. R1R2) is perpendicular to AC produced

 

In the extreme positions, AP1 and AP2 are tangential to the circle and the cutting tool is at the end of the stroke. The forward or cutting stroke occurs when the crank rotates from the position CB1 to CB2 (or through an angle β) in the clockwise direction. The return stroke occurs when the

 

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Q 2 b )

8

Question:

State the theories of elastic failure. Explain maximum normal stress theory and maximum shear stress theory with equations.


Answer:

The principal theories of failure for a member are as follows:(Any four) (i) Maximum principal or normal stress theory (ii) Maximum shear stress theory (iii) Maximum principal or normal strain theory (iv) Maximum strain energy theory (v) Maximum distortion energy theory Maximum normal stress theory  According to this theory, the elastic failure occurs when the greatest principal stress reaches the elastic limit value in a simple tension test irrespective of the value of other two principal stresses.  Taking factor of safety (F. S.) into consideration, the maximum principal or normal stress (σt) is given by, σt = σyt / F. S. (for ductile materials) σt = σu / F. S. (for brittle materials) where, σyt = Yield point stress in tension as determined from simple tension test σu = Ultimate stress  This theory ignores the possibility of failure due to shear stress, therefore it is not used for ductile

 However, for brittle materials which are relatively strong in shear but weak in tension and compression, this theory is generally used.  This theory is also known as maximum principal stress theory or Rankine’s theory. Maximum Shear Stress Theory  According to this theory, the failure or yielding occurs at a point in a member when the maximum shear stress reaches a value equal to the shear stress at yield point in a simple tension test. Mathematically, τmax = τyt / F. S. where, τmax = Maximum shear stress τyt = Shear stress at yield point as determined from simple tension test F. S = Factor of safety  Since the shear stress at yield point in a simple tension test is equal to one half the yield stress in tension, therefore τmax = σyt / (2 x F. S.)  This theory is mostly used for designing members of ductile materials.  This theory is also known as Guest’s theory or Tresca’s theory.

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Q 2 c )

8

Question:

The following results were obtained during Morse test on 4-stroke petrol engine. Brake power developed when all cylinders working = 16.2 kW Brake power developed when 1 st cylinder cutoff = 11.5 kW Brake power developed when 2 nd cylinder cutoff = 11.6 kW Brake power developed when 3 rd cylinder cutoff = 11.68 kW Brake power developed when 4 th cylinder cutoff = 11.57 kW Calculate mechanical efficiency and friction power.


Answer:

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Q 2 c )

8

Question:

What is seal ? Classify seals according to shape. State the factors for seal selection.


Answer:

Seal: The seal is an agent or element which prevents leakage of oil from hydraulic elements and protects the system from dust and dirt. Classification of seals based on shape:- a) ‘O’ Ring seal b) ‘V’ Ring seal c) U-packing seal d) T- ring seal e) Cup seal

Factors for seal selection: 1) Type of fluid used in system 2) Maximum temperature of system in working condition 3) Functional reliability expected 4) Cost of seal 5) Working pressure of system 6) Environmental condition

 

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Q 2 c )

4

Question:

Define linear velocity, angular velocity, absolute velocity and state the relation between linear velocity and angular velocity


Answer:

Relation between linear and angular velocity: V = ω.r

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Q 2 c )

4

Question:

Explain the inter-relation between linear and angular velocity, linear and angular acceleration with suitable example.


Answer:

 

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Q 2c)(i)

8

Question:

(i) State and describe in brief about four ergonomic considerations in the designing of machine elements.


Answer:

The different areas covered under the ergonomics are: 1. Communication between the man (user) and the machine. 2. Working environment. 3. Human anatomy and posture while using the machine. 4. Energy expenditure in hand and foot operations. Communication between man and machine  The machine has a display unit and a control unit.  A man (user) receives the information from the machine display through the sense organs.  He (or she) then takes the corrective action on the machine controls using the hands or feet.  This man-machine closed loop system in influenced by the working environmental factors such

as: lighting, noise, temperature, humidity, air circulation, etc. Working Environment  The working environment affects significantly the man-machine relationship.  It affects the efficiency and possibly the health of the operator.  The major working environmental factors are: Lighting, Noise, Temperature, Humidity and air circulation. Ergonomics Considerations in Design of Controls  The control devices should be logically positioned and easily accessible.  The control operation should involve minimum and smooth moments.  The control operation should consume minimum energy.  The controls should be painted in proper colour to attract the attention. Ergonomics Considerations in the Design of Displays  The scale should be clear and legible.  The size of the numbers or letters on the scale should be taken appropriate.  The pointer should have a knife-edge with a mirror in a dial to minimize the parallax error while taking the readings.  The scale should be divided in a linear progression such as 0 – 10 – 20 – 30… and not as 0 – 5 – 25 – 45…..  The number of subdivisions between the numbered divisions should be as less as possible.  The numbering should be in clockwise direction on a circular scale, from left to right on a horizontal scale and from bottom to top on a vertical scale.

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Q 2c)(ii)

8

Question:

(ii) How will select bearing from manufacturer catalogue?


Answer:

The following steps must be adopted in selecting the bearing from the manufacturer’s catalogue: 1. Calculate the radial and axial load reaction (Fa and Fr) acting on the bearing. 2. Decide the diameter of the shaft on which the bearing is to be mounted. 3. Select the proper size of bearing suitable for given application, specified with speed and available space. 4. Find the basic static rating Co of the selected bearing from the catalogue. 5. Calculate the ratio (Fa / VFr) and (Fa / Co). 6. Find the value of x and y i. e. radial and thrust factor from the catalogue. These values depend upon (Fa / VFr) and (Fa / Co).

7. Find the value of load factor or application factor ‘Ka‘from the catalogue. 8. Calculate the equivalent dynamic load by using relation, Pe = (XVFa + YFa ) Ka 9. Calculate the approximate bearing life in hours from the type of bearing, operation and type of machinery that depends upon application. 10. Calculate the required basic dynamic capacity for the bearing by using relation, L10 = (C / Pe ) a  or

 

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Q 2 d )

4

Question:

Explain the Klein’s construction to determine velocity and acceleration of single slider crank mechanism


Answer:

 

If ωAO is the angular velocity of the crank, then Linear velocity’s of the links is given byVAO = ωAO x AO, VAP = ωAO x AM, VPO = ωAO x MO Acceleration of the links is given bya r AO = ω 2 AO x AO, a r AP = ω 2 AO x AC, a t AP = ω 2 AO x CN, aPO = ω 2 AO x NO

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Q 2 d )

4

Question:

Explain the Klein’s construction to determine velocity and acceleration of a link in an I.C. engine mechanism.


Answer:

Let OC be the crank and PC the connecting rod of a reciprocating steam engine, as shown in Fig. Let the crank makes an angle θ with the line of stroke PO and rotates with uniform angular velocity ω rad/s in a clockwise direction. The Klien’s velocity and acceleration diagrams are drawn as discussed below:

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Q 2 e )

4

Question:

Draw the labelled displacement, velocity and acceleration diagrams for a follower when it moves with uniform velocity.


Answer:

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Q 2 e )

4

Question:

Draw the labelled displacement, velocity and acceleration diagrams for a follower when it moves with simple harmonic motion.


Answer:

Roller follower is preferred over knife edge follower  Knife-edge of the follower will cause the wear of the cam.  Higher load on the small contact area the follower likely to cause wear at the tip of Knifeedge due to more stresses.  Knife-edge follower practically not feasible for higher torque / load applications.  More friction due to sliding motion of the knife-edge follower and hence, more maintenance.  Roller follower on the other hand produces smooth operation with less wear and tear of both cam and follower.  Pure rotational motion of roller follower causes less friction and less loss of power.  Considerable side thrust exists between knife-edge follower and the guide.

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