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.

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.

Enlist uses of compressed air (any four).

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.

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

Classify gas turbines (any four)

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

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

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

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

 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.

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.

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

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.

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.

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.

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

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.

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

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

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.

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.

(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

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

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

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.

Explain condition for maximum power transmission.

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.

Differentiate between mechanism and machine.

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.

Explain gear pump with neat sketch.

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.

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’.

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

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

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 .

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

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.

Draw labelled sketch of air lubricator.

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.

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.

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)

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

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

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

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

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.

(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.

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

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.

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.

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. 

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

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

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’.

(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

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.

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.

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 %

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

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

Define following terms :

Fluctuation of energy, co-efficient of fluctuation of energy, co-efficient of fluctuation speed, maximum fluctuation of energy.

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

State any four reasons of failure of pneumatic seals.

Explain the working of rope brake dynamometer with neat sketch

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

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

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. 

Compare positive displacement pump with Rotodynamic pump.

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.

What are the various types of Hoses used in pneumatic system

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

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

State the four advantages and disadvantages of screw pump.

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.

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

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

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

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

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.

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.

(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:

(b) Define :

(i) Radial follower (ii) Off-set follower

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

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

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

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

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

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

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

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

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

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.

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.

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

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

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

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

Draw stress – strain diagram for brittle material.

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.

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

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

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

Describe with neat sketch the working of scotch yoke 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

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

Or

Explain Maximum shear stress theory. {Maximum shear stress theory is also called as?}

Theories of Failures

Maximum shear stress theory is explained below

The principal theories of failure for a member are as follows:

(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 (Maximum principal stress theory or Rankines 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 (Guest’s theory or Tresca’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.

===================Answer to question ends here=======================

For further understanding and details follow following material.

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.

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

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

(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

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

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:

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

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.

A pulley rotating at 50 m/s transmits 40 kW. The safe pull in belt is 400 N/cm width of belt. The angle of lap is 170º. If coefficient of friction is 0.24, find required width of belt.

Data: Initial tension, To = 2000 N, coefficient of friction, µ = 0.3, Angle of lap, θ = 1500 = 1500 x П / 180 = 2.618 rad, Smaller pulley radius, R = 200 mm, hence, D = 400 mm, Speed of smaller pulley, N = 500 r.p.m. We know that the velocity of the belt, v = П = П = 10.47 m/sec (01 mark) Let T1 = Tension in the belt on the tight side, N Let T2 = Tension in the belt on the slack side, N We know that, T0 = Hence, 2000 = (T1 + T2) / 2 Thus, (T1 + T2) = 4000 N ....................... (1) We also know that, = therefore, = or = 2.2 ............. (2) From equations 1 and 2, T1 = 2750 N and T2 = 1250 N (02 marks) Power transmitted by belt, P = [T1 - T2] v = [2750 - 1250] 10.47 = 15700 watts = 15.7 kW

In a four bar chain ABCD, AD is fixed and is 150 mm long. The crank AB is 40 mm long and rotates at 120 r.p.m. clockwise, while the link CD = 80 mm oscillates about D. BC and AB are of equal length. Find the angular velocity of link CD when angle BAD = 60.

Explain MPFI with neat sketch.

Attempt any FOUR MPFI : MPFI means Multipoint Injection System in which each cylinder has number of injector to supply / spray the fuel in cylinders.

In a rigid flaninged complended to  transmit 20K.W at 700.r.p.m

 

Differentiate supercharging and turbocharging in I.C. engine.

What are the considerations in design of dimensions of formed and parallel key having rectangular cross section?

 

Explain slip and creep phenomenon in belts.

Define slip and creep in the belt drive Slip --- Slip is defined as insufficient frictional grip between pulley (driver/driven) and belt. Slip is the difference between the linear velocities of pulley (driver/driven) and belt. Creep ----- Uneven extensions and contractions of the belt when it passes from tight side to slack side. There is relative motion between belt and pulley surface, this phenomenon is called creep of belt.

Explain with neat sketch the working of meter out hydraulic circuit.

Meter OUT Circuit A typical meter out circuit is shown in figure. Here the flow control valve is installed in the return line metering the fluid being discharged. In that way, this circuit also gives the control over the actuating speed. But this way of control offers altogether different characteristics to the circuit. Now, the circuit pressure has to overcome the load resistance and the pressure drop across the flow control valve. However, as the flow control valve is on the right side of the piston, the differential area will cause rise in the pressure. This increased pressure helps to overcome the pressure drop across the flow control valve. As the system pressure required will be relatively low, it makes this circuit marginally more efficient on the extend stroke. Initially, the compensatory spool is fully open, and full pump flow is passed into the cylinder until piston moves forward building up pressure at the flow control valve. The compensatory spool will now come into operation and restricts the flow to its correct value. Thus, there is an initial flow surge before the compensatory spool adjusts as in the case of ‘meter-in’ When using meter-out system, the pressure in the rod-end of the cylinder must be carefully considered. With meter-out speed control, the quantity of oil leaving the cylinder is controlled. When the cylinder is extending, the oil from the rod-end is metered which a smaller quantity than that is flowing into the full bore end. Consequently, under extend conditions; meter-out flow control is not as sensitive as meter-in control. When the cylinder is retracting, the reverse is true. Meter-out circuits are best where negative loads may occur, because back pressure is maintained on the exhaust side of the actuator preventing erratic motion. Meter-out circuits provide accurate speed control even with reversing loads. However, as with the meter-in system, considerable heat will be generated when used with a fixed delivery pump and a wide range of piston speeds. Applications: Drilling, Boring, Reaming and tapping operations.

Draw neat labelled sketch of (i) Internal gear pump (ii) Gerotor pump

Write the procedure for balancing of a single rotating mass by single masses rotating in the same plane.

Procedure :Balancing of a Single Rotating Mass By a Single Mass Rotating in the Same Plane Consider a disturbing mass m1 attached to a shaft rotating at ω rad/s as shown in Fig. Let r1 be the radius of rotation of the mass m1 (i.e. distance between the axis of rotation of the shaft and the centre of gravity of the mass m1). We know that the centrifugal force exerted by the mass m1 on the shaft, FCl= m1.ω2 . r1 . . . (i) This centrifugal force acts radially outwards and thus produces bending moment on the shaft. In order to counteract the effect of this force, a balancing mass (m2) may be attached in the same plane of rotation as that of disturbing mass (m1) such that the centrifugal forces due to the two masses are equal and opposite.

What is FRL ? State the function of each component of FRL.

FRL Unit: It is service unit used in Pneumatic system which is combination of three devices named as Filter, Regulator and Lubricator. Function of FRL unit Filter (F) – 1) To remove the micron and sub-micron particles present in the entering air of compressor. It is Used to separate out contaminants like dust, dirt p[articles from the compressed air Regulator (R) – In pneumatic system the pressure of compressed air may not stable due to possibility of line fluctuation. Hence there is a need to maintain and regulate the air pressure. This function is performing by regulator. Lubricator (L) – Sliding components like spool, a pneumatic cylinder has sliding motion between parts. It may cause friction and wear and tear at mating parts. To reduce friction, lubricating oil particles are added in the compressed air with the help of lubricator.

State advantages and disadvantages of chain drive over belt drive

Advantages of chain drive over belt drive (Any four)

a) No slip takes place in chain drive as in belt drive there is slip.

b) Occupy less space as compare to belt drive.

c) High transmission efficiency.

d) More power transmission than belts drive.

e) Operated at adverse temperature and atmospheric conditions.

f) Higher velocity ratio.

g) Used for both long as well as short distances.

Disadvantages of chain drive: 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

Define the following w.r.to. I.C. engine.
i) Indicated power
ii) Brake power
iii) Volumetric efficiency
iv) BSFC

(i) Indicated Power: The total power developed by combustion of fuel in the combustion chamber is called indicated power.

(ii) Brake Power: The power developed by an engine at the output shaft is called brake power.

(iii) Volumetric efficiency: It is defined as the ratio of the actual volume of the charge admitted into the cylinder to the swept volume of the piston is known as volumetric efficiency.

( iv) Brake specific fuel consumption: It is the mass of fuel consumed per kw developed per hour, and is a criterion of economical

List any four applications of pneumatic rotary actuator. Draw the symbol for variable speed bidirectional air motor.

In all pneumatic power tools like screw drivers, angle grinders, straight grinders. To rotate conveyor belts in food industry. Power device in printing press machine Agitators and mixers Vibrators. symbol for variable speed bidirectional air motor

What is the Function of Time Delay valve?

Time delay valve function

Time delay valve function need:

Time delay valve function : In certain  applications like machining or press operation it is necessary that certain operation be delayed by some fraction even after pressing valve. This objective is achieved by Time delay valve.

Time delay valve function working:

Time delay valve is a combination valve used to set the operation time as per the requirement. The time delay can be increased or decreased by adjusting the flow through the non-return flow control valve. The change invariably increases or decreases the time taken to fill and pilot actuates the direction control valve. Time delay valve is a combination of a pneumatically actuated 3/2 direction control valve, an air reservoir and a throttle relief valve. The time delay function is obtained by controlling the air flow rate to or from the reservoir by using the throttle valve. Adjustment of throttle valve permits fine control of time delay between minimum and maximum times. In pneumatic time delay valves, typical time delays in the range 5-30 seconds are possible.

The time delay can be extended with the addition of external reservoir. Time delay valve, NC type. The constructions of an on-delay timer (NC) type in the normal and actuated are shown in Figure It can be seen that 3/2 DCV operates in the on delay mode permanently. But, in some designs, the valve can be operated in the off-delay mode by connecting the check valve in reverse direction. Time delay valve  is shown in diagram below.

Time delay valve function diagram:

Pneumatic time delay valve is used when time based sequencing is required. Construction and symbol of valve is shown in figure below. It is simply a 3/2 direction control valve, which is impulse operated from one side. Time delaying is achieved by delaying the impulse actuation. The valve has an in-built air reservoir and in-built non-return flow control valve. When impulse is applied to port “Z” for valve actuation, the air passes through needle control valve, which controls the rate of flow, further the valve spool is not actuated until the air reservoir is filled completely and pressure is built-up. (This time difference between impulse application at “Z” and actual spool actuation is the “delay”, which is adjustable through needle valve). This valve is found applicable in time based sequencing, which is common industrial application, e.g. clamping, indexing, feed motions etc.

ALTERNATE DIAGRAM

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Time delay valve function additional information and Diagrams are given below.

Draw speed control pneumatic circuit for bi-directional air motor.

Fig. Speed control of bi-directional air motor Bi-directional air motor rotates in clockwise as well as anti-clockwise direction. The speed of bidirectional 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 lever L1 and L2. When lever L1 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 lever L2 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.

Justify that slider crank mechanism is a modification of the basic four bar mechanism with neat sketch.

Justification for a single slider crank mechanism is a modification of four bar chain mechanism is as given below. 1) Single slider mechanism has four kinematic links – crank, connecting rod, frame and slider and four bar mechanism has crank, coupler, frame and a follower. 2) A follower in four bar mechanism is replaced by a slider. 3) A four bar mechanism has 4 turning pairs and single slider crank mechanism has also four pairs, but one of the turning pairs is replaced by a sliding pairs. 4) A four bar mechanism rotary motion of the crank into oscillating motion of the follower whereas in single slider motion is converted in sliding motion of the piston

Explain variable displacement axial piston pump with neat sketch.

Fig. Variable displacement axial piston pump (Sketch 2 Marks and Explanation 2 Marks) Construction and Working: 1. It consists of swash plate which has angular surface with reference to the cylinder block axis. It is used to obtain reciprocating movement of pistons in the cylinder bores. 2. The two or more cylinders are mounted parallel to the axis of driving shaft, the piston rod ends are attached to the angular surface of swash plate with the help of shoe and shoe plate. 3. When driving shaft is rotated it will cause reciprocating movements of pistons in cylinders depending upon the angular surface movement with respect cylinder barrel. 4. It will cause suction of the oil in one cylinder while discharge of high pressure oil in another cylinder. This cycle is repeated for cylinders to give high pressure oil through discharge ports

Explain working of counterbalance hydraulic circuit with neat diagram.

Counterbalance valves are commonly used to counterbalance a weight or external force or counteract a weight such as a platen or a press and keep it from freefalling.Figure1.16 illustrates  the use of a counterbalance or back-pressure valve to keep a vertically mounted cylinder in the upward position while the pump idles, that is, when the DCV is in its center position. During the downward movement of the cylinder, the counterbalance valve is set to open at slightly above the pressure required to hold the piston up (a check valve does not permit flow in this direction). The control signal for the counterbalance valve can be obtained from the blank end or rod end of the cylinder. If derived from the rod end, the pressure setting of the counterbalance valve equals the ratio of the load to the annulus area of the piston. If derived from the blank end, the pressure setting equals the ratio of load to the area of piston. This pressure is less and hence usually it has to be derived from the blank end. This permits the cylinder to be forced downward when pressure is applied on the top. The check valve is used to lift the cylinder up as the counterbalance valve is closed in this direction. The directional control valve unloads the pump.

Compare flywheel and governor.

Explain with neat sketch construction and working of eddy current dynamometer.

Eddy Current Dynamometer : It consists of a stator on which are fitted a number of electromagnets and a rotor disc made of copper or steel and coupled to the output shaft of the engine. When the rotor rotates, eddy currents are produced in the stator due to magnetic flux set up by the passage of field current in the electromagnets. These eddy currents oppose the motion of the rotor thus loading the engine. The eddy currents are dissipated in producing heat so that this type of dynamometer also requires some cooling arrangements. The torque is measured similar to absorption dynamometers i.e. with the help of moment arm. The load is controlled by regulating the current in the electromagnets.

Four masses attached to a shaft and their respective radii of rotation are given as :
m 1 = 180 kg m 2 = 300 kg m 3 = 230 kg m 4 = 260 kg
r 1 = 0.2 m r 2 = 0.15 m r 3 = 0.25 m r 4 = 0.3 m
The angles between successive masses are 45, 75 and 135. Find the
position and magnitude of the balance mass required, it its radius of rotation is
0.2 m. The masses revolve in same plane.

Given : m1 = 180 kg, m2 = 300 kg, m3 = 230 kg, m4 = 260 kg r1 = 0.2 m, r2 = 0.15 m, r3 = 0.25 m, r4 = 0.3 m ϴ1 = 45, ϴ2 = 75, ϴ = 135 The centrifugal forces are given by - m1r1 = 36, m2r2 = 45, m3r3 = 57.5, m4r4 = 78

From vector diagram the resultant force is at 60 to the mass m1 and is represented by ar ar = 12 kg m Therefore mb * rb = 12 kgm Balancing mass mb = 12/0.2 = 60 kg at an angle of 2400 with the direction of m1 mass

List different types of pressure regulator valves ? Explain any one with neat sketch.

Pressure-relief valve.Pressure-reducing valve. Unloading valve Counterbalance valve Pressure-sequence valve Brake valve.

Working: The compressed air pressure from FRL unit acts against the poppet (valve element) through inlet of pressure relief valve. When the force of air is greater than the spring force then poppet gets lifted from the valve seat and valve opens. Thus the excessive pressurized air will get release to the atmosphere through port R. (03 marks) OR Pressure regulation in pneumatics is vital for the correct operation of circuits and for damage prevention to circuit components. As you would imagine all pneumatic components have a maximum operating pressure. Types: A) According to no. of stages Single stage and two stage regulator B) According to pressure relief Non- relieving type and relieving type pressure regulator Non-relieving pressure regulator Non-relieving pressure regulators work by restricting flow rather then venting it should over pressure occur. The regulator restricts flow when the pressure gets too high because the pressure acts on the diaphragm forcing i t up against the spring pressure, the diaphragm has what is called a ‘poppet’ attached on the end of it which is drawn up with the diaphragm and restricts the passing air flow.

The crank of a slider crank mechanism rotates clockwise at a constant speed
of 300 rpm. The crank is 150 mm and the connecting rod is 600 mm long.
Determine :
(i)
linear velocity and acceleration of the mid-point of the connecting rod,
and
(ii)
angular velocity and angular acceleration of the connecting rod, at a
crank angle of 45 from inner dead centre position.

Note: Since D is the midpoint of AB, therefore d' is also midpoint of vector b' a'. Join o' d'. The vector o' d' represents the acceleration of midpoint D of the connecting rod i.e. aD.

By measurement, we find that aD = vector o' d' = 117 m/s2

Given: Lift = 40 mm Rise = ¼ x 360 = 900 Fall =1/6 x 360 = 600 Dwell = 1/10 x 360 =360

Develop a pneumatic circuit for operation of two DA cylinders such that one operates after other using travel dependant sequencing.

Answer: a pneumatic circuit for operation of two DA cylinders such that one operates after other using travel dependant sequencing as shown in fig.

Draw the profile of a cam to raise a valve with S.H.M. through 40 mm in
of revolution, keep it fully raised through 1/10
th
1 th
4
revolution and to lower it
with uniform acceleration and retardation in 1/6 th revolution. The valve
remains closed during the rest of the revolution. The diameter of roller is
20 mm and minimum radius of cam to be 30 mm. The axis of the valve rod
passes through the axis of cam shaft.

Fig shows a displacement diagram and a cam profile for the roller follower.

Explain pneumatic impulse circuit with neat sketch.

Figure 1 shows a symbol circuit of an impulse-valve controlled double acting pneumatic cylinder (A). The position of the impulse-valve (3), which is controlled by the start/stop-valve (1) and the end position valve (2), determines if the cylinder piston shall make a positive stroke and negative stroke. Positive piston stroke is initiated by manual activation of the start valve (1). Negative piston stroking takes place when valve (2) is activated by the cylinder rod at the position a1.

Two pulley, one 450 mm diameter and the other 200 mm diameter are on
parallel shafts 1.95 m apart and are connected by a crossed belt. Find the
length of the belt required and the angle of contact between the belt and each
pulley.
What power can be transmitted by the belt when the larger pulley rotates at
200 rpm, if the maximum permissible tension in the belt is 1 kN and the
co-efficient of friction between the belt and pulley is 0.25 ?

State at least four advantages and disadvantages of pneumatic systems.

Pneumatic system Advantages Disadvantages

Pneumatic system Advantages Disadvantages are given in details below.

Pneumatic system Advantages

1) Infinite availability of the source

Air being freely and easily available everywhere there is no scarcity of the source. Just need some filtration for its use in the pneumatic system. As compared to the Hydraulic oil it is very cheap.

2) Safe and clean

As compared to hydraulic and other systems air is very safe and clean in operation. Food industry, pharmaceuticals industry ect. prefer pneumatic system over other because leakage of air doesn't causes any issues. The system is relatively clean , no dirt accumulates due to oil leakage like in hydraulic system.

3) Less Operating and Maintenance cost

Pneumatic system operates with very less resources and is easy to maintain.

4) The transfer of power and the speed is very easy to set up .

The speed and power is controlled by pressure and flow of air, which can be easily controlled by knobbed controls. Just varying the pressure changes the force and varying air flow changes the speed of the actuator.

5) Can be stored and Easy utilized .

Compressed air can be easily stored in a reservoir tank. It retains its pressure over long time. Also it is harmless even it leakages.

Pneumatic System Disadvantages

1) Requires installation of air-producing equipment

Every pneumatic system needs constent supply of compressed air, which is produced by a compressor.

2) Can easily leak

Leakage of air is one of the main issue in pneumatic system. Air can easily leak from the joints and threadings.

3) Potential noise

Exhausting air produces large noise as compared to other power transmitting systems.

4) Low operating pressure

As compared to Hydraulic system where the pressure above 500 bar is possible to produce, pneumatic system max pressure is limited in most cases to 10-20 bar. Hence very heavy work can not be done with pneumatic system.

State types of gear train and explain any one.

i) Types of gear trains 1) Simple gear train 2) Compound gear train 2) Epicyclic gear train 4) Inverted gear train Simple gear train. When there is only one gear on each shaft, it is known as simple gear train. The gears are represented by their pitch circles. When the distance between the two shafts is small, the two gears are made to mesh with each other to transmit motion from one shaft to the other Epicyclic gear train: A simple epicyclic gear train is shown in Fig. where a gear A and the arm C have a common axis at O1about which they can rotate. The gear B meshes with gear A and has its axis on the arm at O2, about which the gear B can rotate. If the arm is fixed, the gear train is simple and gear A can drive gear B or vice- versa, but if gear A is fixed and the arm is rotated about the

axis of gear A (i.e. O1), then the gear B is forced to rotate upon and around gear A. Such a motion is called epicyclic and the gear trains arranged in such a manner that one or more of their members move upon and around another member are known as epicyclic gear trains.

Compound Gear Train When there are more than one gear on a shaft, it is called a compound train of gear. Whenever the distance between the driver and the driven or follower has to be bridged over by intermediate gears and at the same time a great (or much less) speed ratio is required, then the advantage of intermediate gears is intensified by providing compound gears on intermediate shafts

Reverted Gear Train When the axes of the first gear (i.e. first driver) and the last gear (i.e. last driven or follower) are co-axial, then the gear train is known as reverted gear train. We see that gear 1 (i.e. first driver) drives the gear 2 (i.e. first driven or follower) in the opposite direction.

Draw symbol of unloading valve and sequence valve.

A simple band brake is operated by lever 40 cm long. The brake drum
diameter is 40 cm and brake band embrance 5/8 of its circumference. One end
of band is attached to a fulcrum of lever while other end attached to pin 8 cm
from fulcrum. The co-efficient of friction is 0.25. The effort applied at the end
of lever is 500 N. Find braking torque applied if drum rotates anti-clockwise
and acts downwards.

Simple band brake:

Given: Length of lever l = 40 cm = 0.4m, diameter d = 40 cm = 0.4, µ = 0.25, b =0 .08 m ϴ = Angle of wrap = 5/8 x 360 = 225 x π /180 = 3.93 rad Braking torque = (T1 –T2) x r T1/T2 = e µϴ = e 0.25 x 3.93 = 2.67 Taking moments about fulcrum P x l = b x T1 500 x 0.40 = 0.08 x T1 T1 = 2500 N T2 = 2500 / 2.67 = 936.3 N Braking Torque = (2500 – 936.3) x 0.2 = 312.74 N-m

Design consideration while designing the spur Gear

1) The power to be transmitted 2) The velocity ration or speed of gear drive. 3) The central distance between the two shafts 4) Input speed of the driving gear. 5) Wear characteristics of the gear tooth for a long satisfactory life. 6) The use of space & material should be economical. 7) Efficiency & speed ratio 8) Cost

Draw only a neat labelled sketch of window air-conditioner.

Sketch of window air conditioner

State any four area of Application of spring:

1) To cushion, absorb or control energy to external load : Car springs, Railway buffers 2) To store Energy : Watches Toys 3) To Measure forces : Spring Balances, Gauges ,Engines 4) To provide clamping force in Jigs & fixtures. 5) To apply forces as in brakes, clutches & spring loaded valve.

Enlist applications of hydraulic system.

1 Industrial: Plastic processing machineries, steel making and primary metal extraction applications, automated production lines, machine tool industries, paper industries, loaders, crushes, textile machineries, R & D equipment and robotic systems etc. 2 Mobile hydraulics: Tractors, irrigation system, earthmoving equipment, material handling equipment, commercial vehicles, tunnel boring equipment, rail equipment, building and

construction machineries and drilling rigs etc. 3 Automobiles: It is used in the systems like breaks, shock absorbers, steering system, wind shield, lift and cleaning etc. 4 Marine applications: It mostly covers ocean going vessels, fishing boats and navel equipment. 5 Aerospace equipment: There are equipment and systems used for rudder control, landing gear, breaks, flight control and transmission etc. which are used in airplanes, rockets and spaceships. Any four applications

Explain pneumatic circuit for speed control of single acting cylinder with neat sketch.

Pneumatic cylinders can be directly controlled by actuation of final directional control valve as shown in fig. These valves can be controlled manually or electrically. This circuit can be used for small cylinders as well as cylinders which operates at low speeds where the flow rate requirements are less. When the directional control valve is actuated by push button, the valve switches over to the open position, communicating working source to the cylinder volume. This results in the forward motion of the piston. When the push button is released, the reset spring of the valve restores the valve to the initial position [closed]. The cylinder space is connected to exhaust port there by piston retracts either due to spring or supply pressure applied from the other port.

Explain single cylinder 4-stroke I.C. engine using turning moment diagram.

A turning moment diagram for a four stroke cycle internal combustion engine, we know that in a four stroke cycle internal combustion engine, there is one working stroke after a crank has turned through two revolution i.e.7200 . Since the pressure inside the engine cylinder is less than the atmospheric pressure during suction stroke therefore a negative loop is formed. During the compression stroke, the work is done on gases, therefore a higher negative loop is obtained.

During the expansion or working stroke, the fuel burns and the gases expand, therefore a positive loop is obtained. In this stroke the work done is by the gases. During exhaust stroke, the work is done on the gases, therefore negative loop is formed. It may be noted that effect of inertia forces on the piston is taken is account.

Define following terms w.r.t. air compressor. i) FAD ii) Compression ratio.

i) FAD – It is the volume of air delivered under the intake conditions of temperature and pressure.  ii) Compression ratio – It is defined as the ratio of absolute discharge pressure to the absolute inlet pressure.

Draw ‘Valve timing diagram for 4-stroke cycle diesel engine.

Define (i) Pressure angle (ii) Pitch point related to cam.

(i) 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 angel is too large, a reciprocating follower will jam in its bearing.
(ii) Pitch point: It is point on pitch curve having the maximum pressure angle.

Define:
(i) Coefficient of fluctuation of speed.
(ii) Coefficient of fluctuation of energy.

(i) Coefficient of fluctuation of speed: Coefficient of fluctuation of speed is defined as the  ratio of the maximum fluctuation of speed to the mean speed. It is denoted by C s.
Mathematically,
C s = (N 1 – N 2 ) /N
Where, N 1 = maximum speed in rpm;

N 2 =minimum speed in rpm; N = mean speed in rpm

(ii) Coefficient of fluctuation of energy: Coefficient of fluctuation of energy may be defined as the ratio of the maximum fluctuation of energy to the work done per cycle. It is denoted by Ce.
Mathematically,
Ce = Maximum fluctuation of energy/ Work done per cycle.

Draw line diagram of porter governor

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

The high speed of engines and other machines is a common phenomenon now-a-days. It
is, therefore, very essential that all the rotating and reciprocating parts should be completely
balanced as far as possible. If these parts are not properly balanced, the dynamic forces are set
up. These forces not only increase the loads on bearings and stresses in the various members, but
also produce unpleasant and even dangerous vibrations. The balancing of unbalanced forces is
caused by rotating masses, in order to minimize pressure on the main bearings when an engine is
running.

Explain with neat sketch the constructional features of ‘Three Way Catalytic Converter’.

- Three way convertor uses thin coating of platinum, palladium and rhodium over a support metal (generally alumina) & acts on all three major constituents of exhaust gas pollution i. e. hydrocarbons, carbon monoxide & oxides of nitrogen, oxidizing these to water, carbon dioxide & free hydrogen & nitrogen respectively. - It operates in two stages, the first convertor stage uses rhodium to reduce the NO2 in the exhaust into nitrogen & oxygen. In second stage convertor platinum or palladium acts as oxidation catalyst to change HC & CO into harmless water & CO2. - For supplying the oxygen required in the second stage air is fed into the exhaust after the first stage. - Reactions within catalyst produce additional heat that reaches temperature of 900oC, which is required for the catalytic converter to operate at complete efficiency. To safeguard from this high temperature, the catalytic converter is made of stainless steel &special heat shields are also used.

Explain: (i) Uniform pressure theory. (ii) Uniform wear theory in clutches and bearing.

(i) Uniform pressure theory:
 When the mating component in clutch, bearing are new, then the contact between
surfaces may be good over the whole surface.
It means that the pressure over the rubbing surfaces is uniform distributed.
 This condition is not valid for old clutches, bearings because mating surfaces may
have uneven friction.
The condition assumes that intensity of pressure is same.
P = W/A =Constant; where, W= load, A= area
(ii) Uniform wear theory in clutches and bearings:

When clutch, bearing become old after being used for a given period, then all
parts of the rubbing surfaces will not move with the same velocity.
The velocity of rubbing surface increases with the distance from the axis of the
rotating element.
It means that wear may be different at different radii and rate of wear depends
upon the intensity of pressure (P) and the velocity of rubbing surfaces (V).
 It is assumed that the rate of wear is proportional to the product of intensity of
pressure and velocity of rubbing surfaces.
This condition assumes that rate of wear is uniform;
P*r = Constant; where, P = intensity of pressure, r = radius of rotation.

Compare cross belt drive and open belt drive on the basis of:

(i) Velocity ratio.

(ii) Direction of driven pulley.

(iii) Length of belt drives

(iv) Application.

Draw stress-strain diagram for ductile material stating salient points

Stress-Strain diagram for ductile material stating salient points (1 Mark for diagram, 3 Marks for description)

.

Proportional limit (A): The stress is proportional to strain. Beyond point A, the curve slightly deviates from the straight line. It is thus obvious, that Hooke's law holds good up to point A and it is known as Proportional limit.

Elastic limit (B): If the load is increase between point A and B, the body will regain its original shape when load is removed; it means body possesses elasticity up to point B, known as Elastic Limit. Upper yield point (C): If the material is stressed beyond point B, the plastic stage will reach and the material will start yielding known as Upper Yield Point. Lower yield point (D): Further addition of small load drops the stress-strain diagram to point D, as soon as the yielding start, this point ‘D’ is known as Lower yield point. Ultimate stress point (E): After the end of yielding, if the load is increase beyond point ‘D’, there is increase in stresses up to point E and thus maximum value of stresses at point ‘E’ is called as Ultimate Stress point. Breaking Stress point (F): After the specimen has reached the ultimate stress, a neck is formed, which decreases the cross-sectional area of the specimen. The stress corresponding to point F is known as Breaking stress.

State any four factors to be considered while selecting the coupling.

Following factors should be consider while selecting coupling 1) Cyclic operation 2) Duration or life 3) Misalignment of shafts 4) Required torque and desired speed 5) Direction of rotation 6) Protection against overload 7) Operating conditions

 It is desired to compress 15 m3 of air per minute from 1.0132 bar to 10 bar. Calculate minimum power required to drive the compressor having two stages and compared it the power required for single stage compression. Assume value of index for compression process to be 1.3 for both cases also assume the condition for maximum efficiency

 State any four factors that govern ‘factor of safety’.

a. Reliability of applied load. b. The extent of simplifying assumptions. c. The certainty as to exact mode of failure. d. Reliability of properties of material and change in these properties during service. e. Extent of stress concentration. f. The reliability of test results to actual machine parts. g. The extent of initial stresses set up during manufacturing. h. The extent of loss of life, if failure occurs.

Represent following processes on Psychrometric chart. i) Heating with humidification ii) Sensible heating. iii) Sensible cooling iv) Evaporative cooling.

Draw neat sketch showing the details of cotter joint. State strength equations for each component with suitable failure cross-sectional area.

It consist of 3 elements i. Socket ii. Spigot iii. Cotter Where, d= End diameter of rodd1= Diameter of spigot/ID 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 material

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

Linear Velocity: It may be defined as the rate of change of linear displacement of a body with respect to the time. Since velocity is always expressed in a particular direction, therefore it is a vector quantity. Mathematically, linear velocity, v = ds/dt

Angular Velocity: It may be defined as the rate of change of angular displacement with respect to time. It is usually expressed by a Greek letter ɷ (omega). Mathematically, angular velocity, ɷ = dƟ /dt

Absolute Velocity: It is defined as the velocity of any point on a kinematic link with respect to fixed point. Relation between v and ɷ: V = r. ɷ Where V = Linear velocity. ɷ = angular velocity. r = radius of rotation.

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

We have already discussed that the velocity diagram for given configuration is a triangle OCP as shown in Fig. If this triangle is rotated through 90°, it will be a triangle oc1 p1, in which oc1 represents VCO (i.e. velocity of C with respect to O or velocity of crank pin C) and is parallel to OC, op1 represents VPO (i.e. velocity of P with respect to O or velocity of cross-head or piston P) and is perpendicular to OP, and c1p1 represents VPC (i.e. velocity of P with respect to C) and is parallel to CP. A little consideration will show that the triangles oc1p1 and OCM are similar. Therefore,

Thus, we see that by drawing the Klein’s velocity diagram, the velocities of various points may be obtained without drawing a separate velocity diagram. Klien’s acceleration diagram: The Klien’s acceleration diagram is drawn as discussed below: 1. First of all, draw a circle with C as centre and CM as radius. 2. Draw another circle with PC as diameter. Let this circle intersect the previous circle at K and L. 3. Join KL and produce it to intersect PO at N. Let KL intersect PC at Q. This forms the quadrilateral CQNO, which is known as Klien’s acceleration diagram. We have already discussed that the acceleration diagram for the given configuration is as shown in Fig. We know that (i) o'c' represents CO ar (i.e. radial component of the acceleration of crank pin C with respect to O ) and is parallel to CO; (ii) c'x represents PC ar (i.e. radial component of the acceleration of crosshead or piston P with respect to crank pin C) and is parallel to CP or CQ; (iii) xp' represents PC at (i.e. tangential component of the acceleration of P with respect to C ) and is parallel to QN (because QN is perpendicular to CQ); and (iv) o'p' represents aPO (i.e. acceleration of P with respect to O or the acceleration of piston P) and is parallel to PO or NO. A little consideration will show that the quadrilateral o'c'x p' is similar to quadrilateral CQNO . Therefore,

A shaft runs at 80 rpm & drives another shaft at 150 rpm through belt drive. The diameter of the driving pulley is 600 mm. Determine the diameter of the driven pulley in the following cases: (i) Taking belt thickness as 5 mm. (ii) Assuming for belt thickness 5 mm and total slip of 4%.

Ans.: Given data; N1 = 80 rpm. N2 =150 rpm. D1= 600 mm. S = 4 % To find; D2 =?;

(i) Case I: Taking t = 5 mm. Velocity ratio, (V.R.) N2/N1 = (D1 + t)/ (D2 + t)

150/80 = (600 + 5)/ (D2 + 5)

Therefore, diameter of driven pulley D2 = 317.66 mm ~ 318mm

(ii) Case II: Assuming for belt thickness 5 mm and total slip of 4%. Velocity ratio, (V.R.) N2/N1 = {(D1 + t)/ (D2 + t)} × {1- (S/100)}

 150/80 = {(600 + 5) / (D2 + 5)} × {1- (4/100)

Therefore, diameter of driven pulley D2 = 304.76 mm ~ 305 mm

State any four advantages of standardization.

i. It saves effort of design of engineers to design and manufacture new machines, as standard components are readily designed by experts. ii. It ensures certain minimum specified quality. iii. It help in manufacturing the components on mass production. iv. Easy and quick replacement of the components is possible. v. Interchangeability of components is possible. vi. It helps in manufacturing the components quickly and economically. vii. Effective utilization of resources. viii. It also contributes to ensure the safety

Classify gas turbine on the basis of i) working cycle ii) application iii) cycle of operation iv) fuel used

1. On the basis of combustion process a) Constant pressure type b) Constant volume or explosion type 2. On the basis of path of working substance a) Open cycle gas turbine b) Closed cycle gas turbine 3. On the basis of action of expanding gases a) Impulse gas turbine b) Impulse reaction gas turbine 4. On the basis of direction of flow a) Axial flow b) Radial flow

Prove that, for a square key, the permissible crushing stress is twice the permissible shear stress.

Explain epicyclic gear train with neat sketch.

In case of Epicyclic Gear train, the axis of shafts on which gears are mounted may have a relative motion between them, unlike other gear trains. This gives advantage that, very high or low velocity ratio can be obtained compared to simple and compound gear trains; in the small space. In above sketch, if gears A and B are rotating and arm RS is fixed, then it behaves like simple gear train. However, when Arm C rotates and gear A is fixed, then train becomes epicyclic. It is also known as planetary gear train. Applications- Differential gears of the automobiles, back gear of lathe, hoists, pulley blocks

Draw a labelled sketch of multiplate clutch and state its applications.

Applications- Automobiles like scooters, motorcycles, textile and paper industries, machine tools

Describe ‘bolt of uniform strength’ with neat sketch

When an ordinary bolt of uniform diameter is subjected to shock load stress concentration across at the weakest part of the bolt i.e. threaded portion (as shown in figure a), it means that greater portion of energy will be absorbed at the region of threaded part and it may cause the failure of threaded portion

There are two methods to achieve bolts of uniform strength i. Turn down shank diameter of bolt equal or lesser than the core diameter of thread (dc) as shown in figure (b) and it gives bolt of uniform strength. ii. In this method an axial hole is drilled to the head as far as threaded portion such that area of shank become equal to the root area of thread as shown in figure (c). Where,d1= Diameter of hole to be drill do= Nominal diameter dc= Core diameter

Write the procedure of balancing single rotating mass when it balance mass is rotating in the same plane as that of disturbing mass.

Explain in brief the constructional features of MPFI engine

Absence of Venturi – No Restriction in Air Flow/Higher Vol. Eff./Torque/Power • Hot Spots for Preheating cold air eliminated / Denser air enters • Manifold Branch Pipes not concerned with Mixture Preparation • Better Acceleration Response • Fuel Atomization Generally Improved. • Use of Greater Valve Overlap • Use of Sensors to Monitor Operating Parameters/Gives Accurate Matching of Air/fuel Requirements: Improves Power, Reduces fuel consumption and Emissions • Precise in Metering Fuel in Ports • Precise Fuel Distribution Between Cylinders • Fuel Transportation in Manifold not required so no Wall Wetting • Fuel Surge During Fast Cornering or Heavy Braking Eliminated • Adaptable and Suitable For Supercharging • Increased power and torque.

State any four effect of detonation

Effects of detonation  (1) Noise – As intensity of detonation increases, the sound intensity increases & it is harmful. (2) Mechanical damage – shock waves are so violent that it may cause mechanical damage like breaking of piston. It increases the rate of wear erosion of piston. (3) Pre-ignition – Due to local overheating of spark plug & this pre-ignition increases detonation. (4) Power output & efficiency decreases - Power output & thermal efficiency decreases due to abnormal combustion. (5) Increase in heat transfer – Temperature of cylinder in detonating engine is higher than in non – detonating engine, hence increases the heat transfer. (6) Carbon deposits- Detonation results in increased carbon deposits.

Define Endurance limit and draw typical S-N curve for steel.

Endurance Limit: It is defined as maximum value of the completely reversed bending stress which a polished standard specimen can withstand without failure, for infinite number of cycles (usually 107 cycles).It is known as endurance or fatigue limit (ϭe).

State any four applications of spring.

1) To cushion, absorb or control energy to external load : Car springs, Railway buffers 2) To store Energy : Watches Toys 3) To Measure forces : Spring Balances, Gauges ,Engines 4) To provide clamping force in Jigs & fixtures. 5) To apply forces as in brakes, clutches & spring loaded valve.

Justify with neat sketch elliptical trammel as an inversion of double slider crank chain.

Elliptical trammel :

Since Elliptical trammel consist of two turning pairs and two sliding pairs, it is inversion of double slider crank chain. This instrument is used for drawing ellipses. This inversion is obtained by fixing a slotted plate (link 4) as shown in fig. It has got two right angled grooves cut into it. 1-2 is turning pair                                     2-3 is turning pair                                                                           1-4 is sliding pair                                      3-4 is sliding pair.................................. [2 M]

A petrol engine working on constant volume cycle has compression ratio of 8 and consume 1 kg of air per minute, if minimum and maximum temp. during cycle is 300 °K and 2000 °K respectively. Find power developed by engine. Assume γ = 1.4 and Cv = 0.71 kJ/kg °K.

State any six design considerations while designing the spur gear

1) The power to be transmitted 2) The velocity ration or speed of gear drive. 3) The central distance between the two shafts 4) Input speed of the driving gear. 5) Wear characteristics of the gear tooth for a long satisfactory life. 6) The use of space & material should be economical. 7) Efficiency & speed ratio 8) Cost

Explain construction and working of eddy current dynamometer.

Construction and Working of Eddy current dynamometer :

A rotor having the following properties :

m1 = 4 kg                                           r1 = 75 mm                             θ1 = 45o               

m2 = 3 kg                                           r2 = 85 mm                             θ2 = 135o

m3 = 2.5 kg                                       r3 = 50 mm                             θ3 = 240o

Determine the amount of the countermass at a radial distance of 75 mm required for the static balance.

Data :

Calculations:

i) Velocity of crank AO:

V_AO = (r X ω ) X (480 X 20)

a_ab  = l(ab) X Scale = 4.3 X 40 X 10³

 a_ab = 172 X 10 mm/sec²

State and explain Law of Gearing.

Law of Gearing

Law of gearing states that the common normal at the point of contact between a pair of teeth must always pass through the pitch point for all positions of mating gear. This law forms the basis for the gear profile design. This is a must condition for the two gears to perform properly.

Law of gearing Proof

Consider the portions of two gear teeth in mesh.o1 and o2 are centre points,

Let K= point of contact

TT = COmmon tangent at the point of contact K

N'N' = common tangent at the pont of contact K

O1m and O2N are perpendicular to common normal N'N'.

Law of gearing animation

As shown in the diagram below the common normat at the point of contact between a pair of teeth must always pass through the pitch point for all positions of mating gears.

This is the fundamental condition which must be satisfied while designing the profiles of teeth for gears.

This law is must for a gearing pair to perform properly. The animation clearly demonstrates the blue line which traces the path of the point of contact.

Link to other topics of Theory of machines is given below.

• Unit-1-Fundamentals and type of mechanisms
• Unit-2-Velocity and Acceleration in Mechanisms
• Unit-3-Cams and Followers
• Unit-4-Belt,Chain and Gear Drives
• Unit-5-Brakes and Clutches
• Unit-6-Flywheel,Governor and Balancing
• Question Paper And Solution