8

Fig.1Show of hacksaw The belt is assembled with tensoion   

4

Explain the working of counter balance valve in hydraulic circuit.

Counter Balance Valve It is basically a relief valve but it is used to set up a back pressure in a circuit to prevent load from falling. They are frequently employed in vertical presses, loaders, lift trucks and other machines that must maintain a particular position or hold a suspended load. In such applications, the counterbalance valve creates a back pressure to prevent the movement of piston rod of cylinder

Figure shows a typical counterbalance valve. At the present pressure (due to load) acting at port A, the valve remains closed under the spring force. The fluid in the port A is trapped thereby prevents the movement of the load. When the pressure in the port A increases beyond certain value, it acts on the spool from downward direction. The spool moves against the spring force and provides the passage for the fluid to tank. This allows descending of the load. As this valve gets actuated line pressure, it is known as direct operated counterbalance valve.

4

Explain with neat sketches only (i) Methods of reducing stress concentration in cylindrical members with shoulders. (ii) Methods of reducing stress concentration in cylindrical members with holes.  

4

Discuss pilot operated check valve with neat sketch.

When pilot signal of pressurized oil is used to control movement of poppet in the check valve, it is called as pilot operated check valve. It is used when no flow characteristics of the valve is desired only for a portion of the system cycle. Figure shows the pilot operated check valve. A pilot piston is introduced below moving poppet. This pilot piston can move up by introducing pilot signal. Working: In normal position there is no flow from (A) to (B) because the movable valve poppet has blocked the flow. Now pilot signal is given through port (C). This oil will push up the pilot piston upwards, thereby compressing springs (S1). The piston rod of pilot piston will push the movable poppet in upward direction thereby compressing the spring (S2). Now the flow from (A) to (B) will start. As and when we cut-off the pilot signal the flow from (A) and (B) will continue. When pilot signal will be cut-off, spring S1 and S2 will expand and moving poppet will again block the flow from (A) to (B).

8

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

4

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.

4

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

4

Draw the neat sketch of diaphragm clutch and explain its working.

Diaphragm Spring Type Single Plate Clutch A diaphragm spring type clutch is shown in fig. where shows the clutch in the engaged position and in the disengaged position. It is seen from the above figures that the diaphragm spring is supported on a fulcrum retaining ring so that any section through the spring can be regarded as a simple lever. The pressure plate E is movable axially, but it is fixed radically with respect to the cover. This is done by providing a series of equally spaced lugs cast upon the back surface of the pressure plate. The drive from the engine flywheel is transmitted through the cover, pressure plate and the friction plate to the gear box input shaft. The clutch is disengaged by pressing the clutch pedal which actuates the release fingers by means of a release ring. This pivots the spring about its fulcrum, relieving the spring load on the outside diameter, thereby disconnecting the drive.

4

Explain the working principle of turbojet with neat sketch.

Turbojet engine working principle

Turbojet engine working principle : shows the schematic of turbojet engine. It has a diffuser section at inlet for realizing some compression of air passing through this section. Due to this air reaching compressor section has pressure more than ambient pressure. This action of partly compressing air by passing it through diffuser section is called “ramming action” or “ram effect”. Subsequently compressor section compresses air which is fed to combustion chamber and fuel is added to it for causing combustion. Combustion products available at high pressure and temperature are then passed through turbine and expanded there. Thus, turbine yields positive work which is used for driving compressor. Expanding gases leaving turbine are passed through exit nozzle where it is further expanded and results in high velocity jet at exit. This high velocity jet leaving nozzle is responsible for getting desired thrust for propulsion.

Turbojet engine working principle is demonstrated below.

The turbojet is an air breathing engine which takes air and then adds heat to it before expanding it.

4

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

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

4

Give detailed classification of followers.

Types of followers The followers may be classified as discussed below:

1. According to the surface in contact. (a)Knife edge follower. When the contacting end of the follower has a sharp knife edge, it is called a knife edge follower. (b) Roller follower. When the contacting end of the follower is a roller, it is called a roller follower. (c) Flat faced or mushroom follower. When the contacting end of the follower is a perfectly flat face, it is called a flat faced follower and when the flat faced follower is circular, it is then called a mushroom follower. (d) Spherical faced follower. When the contacting end of the follower is of spherical shape, it is called a spherical faced follower.

2.According to the motion of the follower. (a) Reciprocating or translating follower. When the follower reciprocates in guides as the cam rotates uniformly, it is known as reciprocating or translating follower. (b) Oscillating or rotating follower. When the uniform rotary motion of the cam is converted into predetermined oscillatory motion of the follower, it is called oscillating or rotating follower.

3. According to the path of motion of the follower. (a) Radial follower. When the motion of the follower is along an axis passing through the centre of the cam, it is known as radial follower (b) Off-set follower. When the motion of the follower is along an axis away from the axis of the cam centre, it is called off-set follower.

8

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

4

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

4

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

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

4

What is the Function of Time Delay valve?

Time delay valve function

Time delay valve function need:

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

Time delay valve function working:

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

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

Time delay valve function diagram:

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

ALTERNATE DIAGRAM

====================================================

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

4

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

Fig. Speed control of bi-directional air motor Bi-directional air motor rotates in clockwise as well as anti-clockwise direction. The speed of bidirectional motor is controlled as shown in fig. The speed control of motor by using variable two flow control valves having built-in check valve and 4x3 DC valve having zero position or central hold position with lever L1 and L2. When lever L1 is operated, port P will be connected to port A of air motor and motor will start rotating in clockwise direction. Its speed can be controlled by using variable flow control valve F1. Port B of motor will be connected to exhaust R and air in motor will be exhausted through port R via DC valve. When lever L2 is operated, pressure port P will be connected to port B of motor and naturally motor will start rotating in anticlockwise direction. Port A will be connected to port R and air in the motor will be exhausted through port R via DC valve.

4

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

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

6

Explain variable displacement axial piston pump with neat sketch.

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

6

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.

4

Compare flywheel and governor.

4

A flat foot step bearing 225 mm in diameter supports a load of 7500 N. If the co-efficient of friction is 0.09 and the shaft rotates at 600 rpm, calculate the power lost in friction.

Problem on Foot step bearing D = 225 mm = 0.225 m W = 7500 N µ = 0.09 N = 600 rpm ω = 2 π N / 60 =62.83 rad/sec Uniform pressure condition Frictional torque T = 2/3 µ W R = 50.625 Nm Power lost in friction = T x ω = 50.625 x 62.83 = 3180.8 W --------Ans Uniform wear condition Frictional torque T = 1/2 µ W R = 37.98 Nm Power lost in friction = T x ω = 37.98 x 62.83 = 2385.57 W -------- Ans

8

Parallel and transverse Weld  

8

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

8

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

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

8

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.

8

Explain pneumatic impulse circuit with neat sketch.

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

8

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.

4

State at least four advantages and disadvantages of pneumatic systems.

Pneumatic system Advantages Disadvantages

Pneumatic system Advantages Disadvantages are given in details below.

Pneumatic system Advantages

1) Infinite availability of the source

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

2) Safe and clean

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

3) Less Operating and Maintenance cost

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

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

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

5) Can be stored and Easy utilized .

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

Pneumatic System Disadvantages

1) Requires installation of air-producing equipment

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

2) Can easily leak

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

3) Potential noise

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

4) Low operating pressure

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

4

Explain the following terms :- i) Daltons law of partial pressures  ii) Relative humidity

8

Draw turning moment diagram for single cylinder four stroke I.C. Engine.

ii) Turning Moment Diagram:

8

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

Simple band brake:

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

8

Identyfiy the metrial and its compotion A) X10Cr 18 Ni9 Mo 4 Si 2 B) XT72W18Cr4V1:  

A) X10Cr 18 Ni9 Mo 4 Si 2 : High Alloy steel having carbon 0.10% , chromium 18%, nickel 9 % ,Molybdenum 4% & silicon 2% B) XT72W18Cr4V1: high speed tool steel having carbon 0.72% ,chromium 4% , tungsten 18% , vanadium 1% 

4

Enlist the hydraulic oil manufacturer’s in India.

Castrol-Shell-Indian oil

4

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

Sketch of window air conditioner

8

State of the Classification of shaft coupling :

4

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

4

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.

4

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.

4

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

2

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.

2

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.

2

Draw line diagram of porter governor

2

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.

6

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.

6

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

4

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.

4

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

4

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,

-----------------------------------------------------------------------------------------------------------------------------------------------

Quick return mechanism Alternate answer :

Links -

1.Slider

2. Crank

3.Frame

4.Slotted Lever

8

 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.

4

What are the types of kinematic pair ? Give its examples.

8

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

8

A belt pulley is fastened to a 90 mm diameter shaft running at 300 r.p.m. by means of a key 20 mm wide and 140 mm long. Allowable stress for the shaft and key material are 40 N/mm2 in shear and 100 N/mm2 in crushing. Find the power transmitted and the depth of the key required.

4

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,

4

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

4

State any four advantages of standardization.

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

4

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

4

Explain epicyclic gear train with neat sketch.

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

4

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

4

Why a coupling should be placed as close to a bearing as possible

Coupling should be placed as close to a bearing because of following reasons i. It gives minimum vibrations ii. Bending load on the shaft can be minimized iii. It increases power transmission stability. iv. To avoid deflections of shaft.

4

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

4

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

4

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.

4

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.

4

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).

4

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.

4

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]

6

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.

6

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

4

Explain construction and working of eddy current dynamometer.

Construction and Working of Eddy current dynamometer :

4

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 :

8

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

8

Calculations:

i) Velocity of crank AO:

V_AO = (r X ω ) X (480 X 20)

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

 a_ab = 172 X 10 mm/sec²

8

A railway wagon having 1500 kg mass and moving at 1 m/s velocity dashes against a bumper consisting of two helical springs of spring index 6. The springs, which get compressed by 150 mm while resisting a dash made of spring steel having allowable shear stress of 360 N/mm2 and modulus of rigidity 8.4  104 N/mm2 . Design the helical coil spring with circular crosssection of spring wire.

8

Explain the working of ‘Turbo-Prop’ engine with neat sketch.

The main components of turbo-prop engine are a propeller, gear reduction unit, a compressor, a combustor, gas turbine and the nozzles. In this engine 80 to 90% of the total propulsive thrust is generated by the gas turbine and the remainder is developed by the expansion of the gases in nozzles. Due to this the power generated in the gas turbine is used for driving the compressor and the propeller, while in case of turbojet engines the turbine power is only used to drive the compressor and the auxiliaries. The gas turbine drives the propeller through the reduction gear unit and it draws a large amount of air. A large part of this air drawn by the propeller is passed through the ducts around the engine and the remainder is compressed in the diffuser by ram compression and further in the compressor. Fuel is burnt in the combustor and the resultant high temperature gases are expanded in the turbine and finally in the nozzles. The total thrust developed is the sum of thrust developed by the propeller and the nozzle. Unlike the turbojet engines the turboprop engines are widely used for commercial and military air crafts, due to their low specific fuel consumption and high flexibility of operation at reasonably high speed.

8

Show that the efficiency of self locking screw is less than 50%

4

Draw a neat sketch of leaf spring of semi-elliptical type and name its parts.

Sketch of Leaf Spring of semi elliptical Type ….Diagram+ Names :

4

Explain steep and creep phenomenon in belts.

Slip of Belt:- Ans:- When driver pulley rotates firm grip between its surface and the belt. This firm grip between pulley and belt is because of friction and known as frictional grip. If this frictional grip becomes insufficient to transmit the motion of pulley to belt. Then there will be. 1) Forward motion of driver pulley without carrying belt called as slip on driving side.

2) Some forward motion of belt without carrying driven pulley this is called as slip on driver side. The difference between linear speed of rim of pulley and belt on the pulley is known as slip of belt. The velocity ratio considering slip is given by:-

Creep of Belt:- The belt moves from driving pulley is known as Tight side and belt moves from driving pulley to driver pulley as slack side.

Tension on both i.e. on tight sides and slack side is not equal ( T1> T2 ) . The belt material is elastic material which elongates more on Tight side than the slack side resulting in unequal stretching on both sides of drive. A certain portion of belt when passes from slack side to tight side extends and certain portion of belt when contracts, passes from tight side to slack side because of relative motion. The relative motion between belt and pulley surface due to unequal stretching of two sides of drives is known as creep.

8

A simple band brake shown in figure 2 is applied to a shaft carrying a flywheel of mass 250 kg and of radius of gyration 300 mm. The shaft speed is 200 rpm. The drum diameter is 200 mm and the coefficient of friction is 0.25. The dimensions a and l are 100 mm and 280 mm respectively and the angle β  = 135o. Determine (i) the brake torque when a force of 120 N is applied at the lever end. (ii) the number of turns of the flywheel before it comes to rest. (iii) the time taken by flywheel to come to rest.

A simple band brake drum:

4

Explain different forms of threads with their relative advantages and applications.

Square threads

Square threads are the most commonly used thread form for the power screws. Following table gives you various thread forms and comparisons.

Sq. threads Advantages and disadvantages

The advantages of sq. threads are as follows:

1) Efficiency of sq. threads is more than trapezoidal threads

2) There is no side thrust or radial pressure.

The disadvantages of sq. threads are,

1) Sq. threads are difficult to manufacture than trapezoidal threads.

2) The wear of sq. threads can not be compensated as it can be done in trapezoidal.

3) The thread thickness at core is less than trapezoidal, hence sq. threads have less load carrying capacity.

Square threads and other forms diagrams

4

Represent Carnot cycle on P-V and T-S diagram

4

Compare between window air conditioner and split air conditioner (any four)

4

Explain any one method to improve thermal efficiency of gas turbine with the help of block diagram.

+

4

Explain with neat sketch working of lobe type air compressor.

Lobe type air compressor: it is a rotary type of compressor consisting of two rotors which are driven externally. One rotor is connected to drive and second is connected to gear. These two rotors have two or three lobes having epicycloids, hypocycloid or involutes profiles. In the figure two lobes compressor is shown with a inlet arrangement and receiver. A very small clearance is maintained between surfaces so that wear is prevented. Air leakage through this clearance decreases efficiency of this compressor. During rotation a volume of air V at atmospheric pressure is trapped between left hand rotor and casing . this air is positively displaced with change in volume until space is opened to high pressure region. At this instant some high pressure air rushes back from the receiver and mixed with the blower air until both pressure are equalized .

4

A two stage air compressor with perfect intercooling takes in air at 1 bar pressure and 27 °C. The law of compression in both the stages is Pv1.3 = constant. The compressed air is delivered at 9 bar from the H.P. Cylinder to an air receiver. Calculate per kg. of air i) The minimum work done. ii) The heat rejected to the intercooler.

4

Write any four applications oil hydraulic systems

Write any four applications of oil hydraulic systems 1. Earth Moving equipments 2. Broaching machine 3. CNC/VMC/HMC Machines. 4. Hydraulic thread rolling machine 5. Hydraulic press brake. 6. Material handling equipments 7. Hydraulic thread rolling machine 8. Hydraulic cranes

2

Define Kinematic link with one example.

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. Example – any one Example of machine element, (e.g. shaft, spindle, gear, crank, belt, pulley, key etc. )

4

What are the effects of contaminants in the oil?

Following are the effects of contaminants in the oil 1) Contaminants in oil make fluid improper or even hazardous for reuse. 2) Excessive heat gets generated during operation of the hydraulic circuit. 3) Electromagnetic radiation contaminated hydraulic system often generates noise thereby polluting the environment. 4) The system operates at lesser efficiency than the desired.

4

Draw a general layout of pneumatic system and state the function of components

1) Air inlet filter – Free air from the atmosphere enters the compressor through an air-inlet filter which will essentially keep out the dust and dirt from entering the system. 2) Compressor: It is used to compress the air from atmosphere pressure to the desired higher pressure level. It can be single stage or multistage in operation. 3) Cooler: Removes the heat generated during the process of operation. 4) Pressure switch and control unit: Maintains the pressure in the receiver in the predetermined range by starting and stopping the prime mover. 5) Moisture separator: Cooling air in the cooler results in condensation of vapor in the air. The condensate in the form of water droplets are separated from air. 6) Service unit: Filter – Separates sub-micron level contamination. Regulator – Bring the pressure of air from receiver pressure to the device pressure. Lubricator – Adds lubricants to air. 7) Pipe Line: They carry the compressed air from one location to another. 8) Control Valves: They are required to control the air direction, pressure and flow rate. They are responsible for the smooth and precise control of the pneumatic actuator, and also for the safe operation of the system. 9) Actuator: They will convert the high pressure energy of the compressed air into mechanical force or do useful work. Actuators can either be pneumatic cylinders to provide linear motion or pneumatic motors to provide rotary motion.

2

State the advantages of roller follower over knife edge follower.

Advantages of roller follower over knife edge follower a) Roller follower has less wear and tear than knife edge follower. b) Power required for driving the cam is less due to less frictional force between cam and follower.

4

Draw neat labeled sketches of Acme and square thread profile and state its relative characteristics.

Characteristics of Acme thread : (i)thread angle is 290 (ii) permit the use of split nut (iii)easy to manufacture (iv) max. bursting pressure on the thread Characteristics of Square thread : (i) zero profile thread angle (ii) minimum bursting pressure on the nut

2

Give four advantages of chain drive over belt drive.

Advantages of chain drive over belt drive 

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

b) Occupy less space as compare to belt drive.

c) High transmission efficiency.

d) More power transmission than belts drive.

e) Operated at adverse temperature and atmospheric conditions.

f) Higher velocity ratio. g) Used for both long as well as short distances

2

Define fluctuation of energy and coefficient of fluctuation of energy.

a) Fluctuation of energy -- The difference of maximum and minimum kinetic energy of flywheel is known as Fluctuation of energy

b) Coefficient of fluctuation of energy -- - It is defined as the ratio of the maximum fluctuation of energy to the work done per cycle. It is denoted by ke = (E1 - E2)/work done per cycle

6

A four cylinder engine running at 1200 rpm delivers 20 kW. The average torque when one cylinder was cut is 110 N.m. Find the indicated thermal efficiency if the calorific value of the fuel is 43 MJ/Kg and the engine uses 360 gm. of gasoline (fuel) per kW. hr.

6

State and explain main considerations in machine design.

Main considerations in machine design Type of loads and stresses caused by the load: the load on a machine component, may act in several ways, due to which, the internal stresses are set up. Mechanism: the successful operation of any machine depends largely upon the simplest arrangement of the parts, which will give desired motion Selection of material: designer should know the deep knowledge of properties of materials and behavior under working conditions

Convenient and economical features: the designed machine must be convenient to operate and cost wise economical for the customer Use of standard part: reduced the overall cost Safety of operation: to avoid accidental hazards, care should be taken by designer Workshop facilities: a designer should be familiar with the limitations of his employer’s workshop, in order to avoid necessity of vendors Number of machines to be manufactured Cost of construction and assembly: designed machine should be cheap and easy to assem6ble Frictional resistance and lubrication: designer should provide necessary lubrication to the parts, where there is a sliding, rolling and rotating motion

4

State any four inversions of single slider crane chain. Describe any one with neat sketch.

1.Reciprocating engine, Reciprocating compressor;

2. Whitworth quick return mechanism, Rotary engine,

3.Slotted crank mechanism, Oscillatory engine.

4.Hand pump, pendulum pump.

1.Reciprocating engine, Reciprocating compressor link 1 is fixed

2.Whitworth quick return mechanism, Rotary engine​ link 2 is fixed,

3.- Oscillatory engine. Slotted crank mechanism. link 3 is fixed

4. Hand pump, pendulum pump. link 4 is fixed

6

Explain with neat sketch the working of rotary spool type DC valve.

A rotary spool valve consists of a rotating spool which aligns with ports in stationary valve casing, so that fluid is directed to required port. A/B/P/R are the ports in casing. The port ‘P’ is a pressure port though which pressurized oil is coming in the valve. ‘R’ port is the port through which used oil is returning to oil tank. From fig port p is connected to port B and port A is connected to port R

4

The central distance two shaft is 4m having two pulleys with diameter having 500mm and 700mm respectively find the length of belt required - (1) for open belt drive (2) for cross belt drive

8

Explain with the help of neat sketches three basic types of lever. State one application of each type.

In the first type of levers, the fulcrum is in between the load and effort. In this case, the effort arm is greater than load arm, therefore M.A. obtained is more than 1 Application: Bell crank levers used in railway signaling arrangement, rocker arm in I.C. Engines , handle of a hand pump, hand wheel of a punching press, beam of a balance, foot lever (any 1) In the second type of levers, the load is in between the fulcrum and effort. In this case, the effort arm is more than the load arm, therefore M.A. is more than 1. Application: levers of loaded safety valves, wheel barrow, nut cracker (any1) In the third type of levers, the effort is in between the fulcrum and load. Since the effort arm, in this case, is less than the load arm, therefore M.A. is less than 1 Application: a pair of tongs, the treadle of sewing machine