4

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

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.

2

Define inversion with example.

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.

4

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

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

4

What are the advantages of multistaging ?

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

4

Classify gas turbines (any four)

4

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

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

8

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.

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

6

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

6

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

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

6

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

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.

2

Define sliding pair with example.

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.

2

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

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

2

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

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.

2

Define self-energizing and self-locking brake. 

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.

2

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

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.

8

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

2

Define uniform wear theory and uniform pressure theory.

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.

8

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.

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

8

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

8

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

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.

8

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.

8

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

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

4

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

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.

8

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

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.

4

Explain condition for maximum power transmission.

4

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.

4

Differentiate between mechanism and machine.

4

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.

4

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.

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

4

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’.

4

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

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

4

Differentiate between closed cycle and open cycle gas turbine

4

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

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

4

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.

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.

4

Explain the working of Watt governor with neat diagram.

4

Explain the working of freewheel mechanism of bicycle with sketch.

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.

4

What are actuators ? Draw a double acting cylinder.

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.

4

Explain pressure relief valve in pneumatic system.

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)

4

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

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.

4

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

4

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

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

4

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

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

6

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

) 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.

6

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

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.

6

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

4

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

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.

4

Explain the working of internal expanding brake with neat sketch.

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.

8

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. 

8

Explain the working of four stroke petrol engine with neat sketch. OR Explain with diagram how the Four stroke petrol engine works.

Four Stroke petrol engine

FOUR 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 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

A multi cylinder Four stroke petrol engine looks like this.

Links to other pages of power engineering

8

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

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

8

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’.

8

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

(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

8

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.

8

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

(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.

4

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.

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 %

4

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

hread is stronger-

4

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

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.

4

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

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

4

State any four reasons of failure of pneumatic seals.

4

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

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

4

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

4

Explain the working of single plate clutch with neat diagram.

4

Compare positive displacement pump with Rotodynamic pump.

4

Explain the working of simple vapour absorption refrigeration system. Or Explain with neat sketch vapour absorption refrigeration system.

Vapour absorption refrigeration system

vapour absorption refrigeration system is an energy efficient system of achieving refrigeration effect.

Vapor absorption refrigeration system is schematically demonstrated in following diagram.

Vapour absorption refrigeration system working:

Vapor absorption refrigeration 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.

============================Answer Ends Here=========================

For further understanding of the Vapor absorption refrigeration system use following material.

4

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

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

2

Define machine design.

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

4

State the four advantages and disadvantages of screw pump.

2

Pappu Define kinematic link and kinematic chain. 

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.

4

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

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

2

State law of gearing.

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 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

4

Classify air compressors

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

4

Explain any two mounting methods of cylinder.

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

2

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

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.

2

Compare brakes and dynamometers. (any two points)

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.

2

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

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

ii) Classification of follower:

4

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

(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

2

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

(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.

4

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

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

6

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

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.

4

State function of clutch. Explain working principle of clutch.

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

2

State four types of loads acting on machine elements

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

2

What do you mean by creep?

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

2

Define Ergonomics.

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.

2

Give two applications of knuckle joint.

(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.

2

Define following terms of spring:

(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

2

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

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

2

State four types of keys.

(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

2

State any four applications of rolling contact bearings.

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

2

Draw stress – strain diagram for brittle material.

4

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

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'

8

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

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

8

Name any eight pipe or tube fitting with their application.

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

8

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

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.

4

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

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,

4

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

4

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

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

8

(ii) How will select bearing from manufacturer catalogue?

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

4

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

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

4

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