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.

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

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

Draw and explain working of pressure reducing valve.

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

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

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

Explain piston pump with neat sketch.

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

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

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

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

Give classification of IC engines (any six).

Explain Morse test conducted to find the frictional power of a multi-cylinder petrol engine

Morse Test

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

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

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

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

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

F1, F2, F3, F4 – Frictional power of all four cylinders

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

 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.

Draw a neat sketch and explain working of beam engine.

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

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

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

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.

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

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

Position- I

Position- II

Closed center position

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.

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

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

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

Explain with neat sketch two way catalytic converter.

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

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

Differentiate between closed cycle and open cycle gas turbine

Draw labelled sketch of air lubricator.

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.

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)

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

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

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

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

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.

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

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

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

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

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

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

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.

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

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

hread is stronger-

State any four reasons of failure of pneumatic seals.

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

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

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

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

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

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

Draw the schematic diagram of turbojet engine.

Turbo Jet Engine

Explain the working of single plate clutch with neat diagram.

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. 

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

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

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

Fc = m. ω².r Newton

Where, m = Mass of rotating part in kg,

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

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

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.

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

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.

State the four advantages and disadvantages of screw pump.

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

Valve timing diagram of four stroke diesel engine

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

State the essential properties of hydraulic fluids.

State the types of chains & sprockets.

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

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:

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

(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

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

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

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

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

Define initial tension in belt drive & state its effect.

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 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.

Differentiate between machine and structure.

Sl. No
Machine
Structure

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

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

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

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

Sl. No
Machine
Structure

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

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

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

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

5
Studied under 'Dynamics'
Studied under 'Statics'

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

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

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

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.

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

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

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

(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

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 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.

Differentiate supercharging and turbocharging in I.C. engine.

In a slider crank mechanism, the length of crank OB and connecting rod AB are 125 mm and 500 mm respectively. The centre of gravity G of the connecting rod is 275 mm from the slider. The crank speed is 600 rpm clockwise. When the crank has turned 45 from the inner dead centre position, determine :

(i) Velocity of slider ‘A’,

(ii) Velocity of the point ‘G’ graphically.

State the advantages of lubricant additives (any four).

Additives  (1) Detergents – To keep engine parts, such as piston and piston rings, clean & free from deposits. (2) Dispersants – To suspend & disperse material that could form varnishes, sludge etc that clog the engine. (3) Anti – wear – To give added strength & prevent wear of heavily loaded surfaces such as crank shaft rods & main bearings. (4) Corrosion inhibitors – To fight the rust wear caused by acids moisture. Protect vital steel & iron parts from rust & corrosion. (5) Foam inhibitors – control bubble growth, break them up quickly to prevent frothing & allow the oil pump to circulate oil evenly. (6)Viscosity index improver – added to adjust the viscosity of oil. (7) Pour point depressant - improves an oil ability to flow at very low temperature

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.

Define stress concentration. What are the causes of stress concentration? State any four methods of reducing stress concentration with neat sketches.

Stress concentration: Whenever a machine component changes the shape of its cross section, the simple stress distribution no longer holds good and the neighbourhood of the discontinuity is different. This irregularity in the stress distribution caused by abrupt changes of form is called ‘stress concentration’.

Causes of stress concentration The various causes of stress concentration are as follows: (i) Abrupt change of cross section (ii) Poor surface finish (iii) Localized loading (iv) Variation in the material properties Methods of reducing stress concentration The presence of stresses concentration cannot be totally eliminated but it can be reduced, so following are the remedial measures to control the effects of stress concentration. 1. Provide additional notches and holes in tension members. a) Use of multiple notches. b) Drilling additional holes. 2. Fillet radius, undercutting and notch for member in bending. 3. Reduction of stress concentration in threaded member. 4. Provide taper cross-section to the sharp corner of member

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

State the advantages of closed cycle gas turbine over open cycle gas turbine (any four).

Advantages of closed cycle gas turbine over open cycle gas turbine: (i) It has higher thermal efficiency for the same minimum and maximum temperature limits and for the same pressure ratio. (ii) Since the heating is external, any kind of fuel even solid fuel having low calorific value may be used. (iii) There is no corrosion due to circulation of combustion product. (iv) As the system is a closed one there is no loss of the working fluid. (v) The size of the turbine will be smaller compared to an open cycle gas turbine of the same output. (vi) The regulation is more simple. (vii) The heat transmission coefficient in the exchanger is better due to the increase in suction pressure. (viii) Loss due to fluid friction is less due to higher Reynolds number.

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.

Fig 2 show a C.I. bracket to carryof a shaft 

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

Explain the term swept volume (V s ) w.r.to.
i) I.C. engine
ii) Reciprocating air compressor

(i)  Swept Volume (VS) w.r.t I.C.Engine: The volume swept through by the piston in moving between top dead centre and bottom dead centre is called swept volume or piston displacement. It is denoted by VS. It is equal to the area of the piston multiplied by its stroke length. Therefore, Swept Volume = π/4xD2 xL

Where D = bore of the cylinder in m, and

L = stroke length in m.

(ii) Swept Volume (VS) w.r.t. Reciprocating air compressor: It is the actual volume of air taken in during suction stroke. It is expressed in m3 . Swept volume when expressed in m3 /min, it is known as piston displacement.

Draw a neat block diagram of “vapour compression cycle”. Show the direction of flow ofrefrigerant.

Block diagram of Vapour Compression cycle :-

Explain w.r.to. dual cycle i) cutoff ratio ii) pressure ratio.

Duel cycle:

A helical compeersion imperssion speraing 

 

What is meant by catalytic converter ? Briefly explain with the help of neat sketch.

Catalytic converter:

harmless gases. Catalytic converter is used in exhaust emission in control system to convert CO, NOx, HC and other harmful gases to harmless gases. A Catalytic converter consists of a cylindrical unit of small size like a small silencer and is installed into the exhaust system of a vehicle. It is placed between the exhaust manifold and the silencer. Inside the cylindrical tube i.e. converter there is a honey comb structure of a ‘ceramic or metal’ which is coated with ‘alumina base’ material and there after a second coating of precious metals ‘platinum, palladium or rhodium’ or combination of the same. This second coating serves as a catalyst. A catalyst is a substance which causes a chemical reaction intro the gases. When the exhaust gases pass over the converter substance, the toxic gases as CO, HC & NOx are converted into harmless gases as CO2, H2 & N2. 

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.

Design of screw jack

 

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

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

State the methods to improve efficiency of air compressor. Explain two stage air compressor with perfect intercooling with neat sketch.

Following are the methods to improve efficiency of air compressor 1. Cooling cylinder by spraying water during compression stroke. 2. Circulation of water surrounding to cylinder by providing jackets 3. Installing inter cooler between two cylinders 4. Providing greater fins on cylinder 5. By selecting suitable material for cylinder 6. By providing suitable choice of cylinder proportions i.e. short stroke and large bore in construction with sleeve valve Two stage reciprocating air compressor :

Two stage reciprocating compressor

Multistage compression refers to the compression process completed in more than one stage i.e. a part of compression occurs in one cylinder ( L.P. cylinder) and subsequently compressed air is sent to subsequent cylinders ( H.P. cylinder) for further compression.  Figure shows the schematic of two stage compressor with intercooler between stages. The total work requirement for running this shall be algebraic summation of work required for low pressure (LP) and high pressure (HP) stages. The size of HP cylinder is smaller than LP cylinder as HP cylinder handles high pressure air having smaller specific volume. Intake temp of air at LP =intake temp of air at HP for perfect intercooling.

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.

State the applications of reciprocating compressor and rotary compressor (4 each).

Applications of Reciprocating Compressor  1. In spray painting shop. 2. In workshop for cleaning machines. 3. For operation of pneumatic tool like rock drill, vibrator etc. 4. In automobile service station to clean vehicle. 5. To drive air motors in coal mines. 6. Food and beverage industry

Applications of Centrifugal Compressor  1. In gas turbines and auxiliary power units. 2. In automotive engine and diesel engine turbochargers and superchargers. 3. In pipeline compressors of natural gas to move the gas from the production site to the consumer. 4. In oil refineries, natural gas processing, petrochemical and chemical plants. 5. Air-conditioning and refrigeration and HVAC: Centrifugal compressors quite often supply the compression in water chillers cycles. 6. In air separation plants to manufacture purified end product gases. 7. In oil field re-injection of high pressure natural gas to improve oil recovery.

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 ?

List out different pollutants in exhaust gases of petrol and diesel engine ? Briefly explain theireffects on human beings and environments (atleast four).

The major air pollutants emitted by petrol & diesel engines are CO2, CO, HC, NOx, SO2, smoke & lead vapour.

Effect of CO:  Carbon monoxide combines with hemoglobin forming carboy 

hemoglobin ,which reduces oxygen carrying capacity of blood.  This leads to laziness, exhaustion of body & headache.  Prolong exposure can even leads to death.  It also affects cardiovascular system, thereby causing heart problem Effect of CO2: Causes respiratory disorder & suffocation.

Effect of NOx: It causes respiration irritation, headache, bronchitis, pulmonary emphysema, impairment of lungs, and loss of appetite & corrosion of teeth to human body.

Effect of HC: • It has effect like reduced visibility, eye irritation, peculiar odour & damage to vegetation & acceleration the cracking of rubber products. • It induce cancer, affect DNA & cell growth are know a carcinogens.

Effect of SO2: It is toxic & corrosive gas, human respiratory track of animals, plants & crops.

Effect of Keyway on strength of shaft

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.

Sketch a psychrometric chart and show the following properties of air on it. i) DBT lines ii) WBT lines iii) Specific volume lines iv) Relative humidity lines

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

Enlist the hydraulic oil manufacturer’s in India.

Castrol-Shell-Indian oil

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 different uses of compressed air.

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) For spray painting in paint industry.

8) In automobile & railway braking systems.

9) To operate air tools like air guns.

10) To hold & index cutting tools on machines like milling

State the applications of gas turbine (any four).

Following are the applications of gas turbine

1. It is used for electric power generation.

2. It is used for locomotive propulsion.

3. It is used for ship propulsion.

4. Gas turbine is used in aircrafts.

5. It is used for supercharging for heavy duty Diesel engines.

6. Used in turbo jet and turbo-propeller engine.

7. It is used for various industrial purpose such as in steel industry, oil and other chemical industry.

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.

The shaft running at 125 r.p.m. transmits 440 kW. Find the diameter of shaft (d) if allowable shear stress in shaft material is 55 N/mm2 and the angle of twist must not be more than 1 on a length of 16(d). The modulus of rigidity G = 0.80  105 N/mm

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

Draw a labeled sketch of quick return mechanism of shaper and explain its working?

Quick return mechanism

Quick return mechanism : Crank and slotted lever quick return motion mechanism. This quick return mechanism is mostly used in shaping machines, slotting machines and in rotary internal combustion engines. In this quick return 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.

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

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Quick return mechanism Alternate answer :

Links -

1.Slider

2. Crank

3.Frame

4.Slotted Lever

 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.

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

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

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

In a slider-crank mechanism, the crank is 480 mm long and rotates at 20 rad/sec in the counter-clockwise direction. The length of the connecting rod is 1600 mm. when the crank turns 60 from the inner-dead centre. Determine the velocity of the slider by relative velocity method.

Draw a neat sketch of bell crank lever. Enlist steps in designing the bell crank lever

Enlist the four effects of subcooling on the performance of V.C.C. refrigeration cycle.

The process of cooling refrigerant below condensing temperature for a given pressure is known as sub cooling. -Due to sub cooling the refrigerating effect increases or for same refrigerating effect the circulation rate refrigerant decreases and therefore COP of system increases. Thus sub cooling is desirable & is done to increase refrigerating effect & COP of system

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

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

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

Explain in brief the importance of ‘Super Charging’.

Supercharging: The power output of an engine depends upon the amount of air inducted in cylinder per unit time, the degree of utilization of this air and the thermal efficiency of the engine. The amount of air inducted per unit time can be increased by increasing the engine speed or by increasing the density of air at intake. As engine speed increases, the inertia load, engine friction, bearing load increases. The method of increasing the inlet air density is called as supercharging. It is usually employed to increase the power output of the engine. The increase of the amount of air inducted per unit time by supercharging is obtained mainly to burn greater amount of fuel in a given engine. Supercharging increases the power output for a given weight and bulk of engine, compensate for loss of power due to altitude and obtain more power from the existing engine

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

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 :

A screw jack is used to lift a load of 50 kN through a maximum lift of 200 mm. The material used for a screw is steel of allowable stresses in tension and compression as 100 N/mm2 and 50 N/mm2 respectively. The pitch of screw is 8 mm. The nut is made of phosphor bronze with allowable stresses as 50 N/mm2 and 45 N/mm2 in tension and crushing. The allowable shear stress for nut material is 40 N/mm2 . The allowable bearing pressure between nut and screw is not to exceed 20 N/mm2 . If the coefficient of friction between screw and nut is 0.14, design the screw and nut

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