4

A flat belt drive is required to transmit 35 kW from a pulley of 1.5 m effective diameter running at speed of 300 rpm. The angle of contact is spread over 11/24 of the circumference co-efficient of friction for the surface is 0.3. Determine the maximum tension in the belt. 

4

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.

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

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.

4

Differentiate supercharging and turbocharging in I.C. engine.

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

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

4

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

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

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

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.

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.

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

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.

4

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

Block diagram of Vapour Compression cycle :-

4

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

Duel cycle:

8

A helical compeersion imperssion speraing   

6

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. 

6

Draw superimposed p-v diagram of Otto cycle, Diesel cycle and Dual cycle to compare their efficiencies for same compression ratio (R c ) and heat rejection (Q r ).

Superimposed P-V Diagram of Otto, Diesel & Duel Cycle: A comparison of the cycles (Otto, Diesel and Dual) on the p-v and T-s diagrams for the same compression ratio and heat supplied is shown in the Fig. Since all the cycles reject their heat at the same specific volume, process line from state 4 to 1, the quantity of heat rejected from each cycle is represented by the appropriate area under the line 4 to 1 on the T-s diagram. As is evident from the cycle which has the least heat rejected will have the highest efficiency. Thus, Otto cycle is the most efficient and Diesel cycle is the least efficient of the three cycles

8

Design of screw jack  

4

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.

4

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

8

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

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

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

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

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

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.

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

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 ?

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.

8

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.

4

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

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

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

4

Enlist the hydraulic oil manufacturer’s in India.

Castrol-Shell-Indian oil

4

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.

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

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.

4

Represent ‘Brayton Cycle’ on P-V and T-S diagram.

4

Enlist different uses 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.

2

Enlist the types of constrained motion. Draw a label sketch of any one

Types of constrained motion:
(i)Completely constrained motion.
(ii)Incompletely constrained motion.
(iii)Successfully constrained motion.

2

How are drives classified?

Classification of drives:
(i) Belt drives.
(ii) Chain drives.
(iii) Rope.
(iv) Gear drives.

2

Write any two disadvantages of chain drive.

Disadvantages of chain drives:
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.

2

State the application of (i) Disc brake (ii) Internal expanding brake

(i) Disc brake: Used in two wheelers as well as in four wheelers.
(ii) Internal expanding brake: Used in motor cars, light trucks, two wheelers etc.

6

Explain in brief, how ‘Morse test’ is carried out ?

First engine is allowed to run at constant speed and brake power of engine is measured when all four cylinder working. (IP1 + IP2 + IP3 + IP4) = (BP) 1234 + (FP) 1234  Where, IP- indicated power. BP – brake power develop. FP – frictional power.

4

State inversions of double slider crank chain. Explain Oldham's coupling with neat sketch

Inversions of double slider crank chain:
i.Scotch Yoke mechanism.
ii.Oldham’s coupling.
iii. Elliptical trammel.

Oldham’s coupling:
An Oldham’s coupling is used for connecting two parallel shafts whose axes are at a
small distance apart. The shafts are coupled in such a way that if one shaft rotates, the other shaft
also rotates at the same speed. This inversion is obtained by fixing the link 2, as shown in Fig.
The shafts to be connected have two flanges (link 1 and link 3) rigidly fastened at their ends by
forging. The link 1 and link 3 form turning pairs with link 2. These flanges have diametrical slots
cut in their inner faces, as shown in Fig. The intermediate piece (link 4) which is a circular disc,
have two tongues (i.e. diametrical projections) T1 and T2 on each face at right angles to each
other. The tongues on the link 4 closely fit into the slots in the two flanges (link 1 and link 3).
The link 4 can slide or reciprocate in the slots in the flanges.

When the driving shaft A is rotated, the flange C (link 1) causes the intermediate piece
(link 4) to rotate at the same angle through\ which the flange has rotated, and it further rotates the
flange D (link 3) at the same angle and thus the shaft B rotates. Hence links 1, 3 and 4 have the
same angular velocity at every instant. A little consideration will show that there is a sliding
motion between the link 4 and each of the other links 1 and 3.

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

6

What is stress concentration ? State the remedial measures to control the effect of stress concentration with neat sketches

i. Stress Concentration: Whenever a machine component changes the shape of its cross-section, the simple stress distribution no longer holds good and the neighborhood of the discontinuity is different. This irregularity in the stress distribution caused by abrupt changes of form is called stress concentration. It occurs for all kinds of stresses in the presence of fillets, notches, holes, keyways, splines, surface roughness or scratches etc.

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 as shown in fig (a) a)Use of multiple notches. b)Drilling additional holes as shown in fig(b) 2. Fillet radius, undercutting and notch for member in bending. 3. Reduction of stress concentration in threaded members as shown in fig(c) 4. Provide taper cross-section to the sharp corner of member as shown in fig(d)

4

Write the design procedure for turn buckle. (Any four steps)

Write any four equation s in the design of turn buckle with relevant sketches

4

Why square threads are preferred over V-thread for power transmission ?

. Square threads are preferred over V-thread for power transmission because of following points.  1) Square thread has the greatest efficiency as its profile angle is zero. 2) It produces minimum bursting pressure on the nut. 3) It has more transmission efficiency due to less friction. 4) It transmits power without any side thrust in either direction. 5) It is more smooth and noiseless operation.

8

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

8

Why taper is provided on cotter ? State recommended values of taper.

a. When cotter is driven through the slots, it fit, fight due to wedge action. This ensures tightness of joint in operation and present loosening of the parts. b. Due to taper, it is easy to remove the cotter and dismantle the joint. The normal value of taper varies from 1 in 48 to 01 in 24 and it may increase to 1 in 8

8

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

4

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.

8

Following observations were made during a trial on 4-stroke, single cylinder engine running at 240 rpm having brake wheel diameter 1.5 meter. Duration of trial 30 min. Fuel consumption 6 liter C.V. of fuel 42000 kJ/kg Sp. gravity 0.8 IMEP 550 kPa Brake load 150 kg Spring balance reading 15 kg Cylinder diameter 30 cm Stroke length 45 cm Jacket cooling water 11 kg/min Temp. rise in jacket water 36°C Determine : i) I.P. and B.P. ii) Heat balance sheet on percentage basis.

4

 Draw neat sketch of radial cam with follower and show on it  (i) Base circle. (ii) Pitch point. (iii) Prime Circle. (iv) Cam profile

4

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.

4

In a slider crank mechanism, crank AB = 20 mm & connecting rod BC = 80 mm. Crank AB rotates with uniform speed of 1000 rpm in anticlockwise direction. Find (i) Angular velocity of connecting rod BC (ii) Velocity of slider C. When crank AB makes an angle of 60 degrees with the horizontal. Draw the configuration diagram also. Use analytical method.

4

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

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

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

What is ‘Scavenging’ ? List any two types of ‘scavenging’.

Scavenging: At the end of expansion stroke the combustion chambers of a two stroke engine is left full of products of combustion. This is because unlike four stroke engines, there is no exhaust stroke available to clear the cylinder of burnt gases in two stroke engine the process of clearing the cylinder after the expansion stroke is called scavenging process. The scavenging systems are as follows : 1. Uniform scavenging system 2. Cross scavenging system 3. Loop or reverse scavenging 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

What are the different types of follower motion ? Also draw displacement diagram for uniform velocity.

Different types of follower motions –

The follower during its travel may have one of the following motions:-

Uniform velocity, Simple harmonic motion, Uniform acceleration and retardation, Cycloidal motion.

Displacement Diagram of Uniform Velocity:

4

An engine has piston diameter 15 cm, length of stroke 40 cm and mean effective pressure 5 bar. Engine makes 120 power strokes per minute. Find mechanical efficiency if brake power is 5 kW.

4

Explain the term w.r.t. I.C. engine. i) Mean Effective Pressure (MEP) ii) Cut off ratio.

( i ) Mean effective pressure – Defined as the average pressure acting on the piston which will produce the same output as is done by the varying pressure during the cycle.

( ii ) Cut off ratio – Fuel is injected into combustion chamber where only air compressed and is at high temperature. Fuel is injected for a duration of time, say T. The piston would not have reached the bottom dead centre in time T. The fuel is cut off when the volume is, say V2. The clearance volume is V1. The ratio V2/V1 is cut off ratio.

4

State the effect of key-way on the strength of shaft with suitable diagram

+

4

State any four advantages and disadvantages of welded joints over riveted joints.

1. The welded structures are usually lighter than riveted structures. This is due to the reason, that in welding, gussets or other connecting components are not used. 2. The welded joints provide maximum efficiency (may be 100%) which is not possible in case of riveted joints. 3. Alterations and additions can be easily made in the existing structures. 4. As the welded structure is smooth in appearance, therefore it looks pleasing. 5. In welded connections, the tension members are not weakened as in the case of riveted joints. 6. A welded joint has a great strength. Often a welded joint has the strength of the parent metal itself. 7. Sometimes, the members are of such a shape (i.e. circular steel pipes) that they afford difficulty for riveting. But they can be easily welded. 8. The welding provides very rigid joints. This is in line with the modern trend of providing rigid frames. 9. It is possible to weld any part of a structure at any point. But riveting requires enough clearance. 10. The process of welding takes less time than the riveting. Disadvantages

1. Since there is an uneven heating and cooling during fabrication, therefore the members may get distorted or additional stresses may develop. 2. It requires a highly skilled labour and supervision. 3. Since no provision is kept for expansion and contraction in the frame, therefore there is a possibility of cracks developing in it. 4. The inspection of welding work is more difficult than riveting work

6

Name any four additives used in lubricants ? State their advantages

(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.

6

Describe the importance of aesthetic considerations in design related to shape, colour and surface finish.

1) The shape should not be like blocks but various forms like sculpture, streamlined, aerodynamic, taper should be used. 2) The component should be symmetrical at lean about one axis. 3)proper shape of a product help to make the product more attractive. 4) The shape of the product should be regular, even & proportionate Regarding Colour: ……….Any 2 pt : 1 M Each 1) The colour and shape of component should be such that in should attract appeal and impress customer. 2) The colour should match with conventions, moods e.g. red for danger, gray for dull, yellow for cautions, green for safe etc. 3) Too bright colour should be avoided. 4) The colour should be compatible with conventional ideas of the operator. Regarding Surface finish ……….Any 2 pt : 1 M Each 1) Products with better surface finish are always aesthetically pleasing’ 2) The surface coating processes like spray painting, anodizing, electroplating etc greatly the aesthetic appeal of product

4

Differentiate between flywheel and governor.

Difference between Flywheel and Governor :

4

A multiplate disc clutch transmits 55 kW of power at 1800 rpm. Coefficient of friction for the friction surfaces is 0.1. Axial intensity of pressure is not to exceed 160 kN/m2 . The internal radius is 80 mm and is 0.7 times the external radius. Find the number of plates needed to transmit the required torque.

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

Explain with neat block diagram the working of ‘Vapour Absorbtion Cycle’.

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

8

Draw the profile of cam operating a roller reciprocating follower with the following data : Minimum radius of cam = 25 mm lift = 30 mm Roller diameter = 15 mm The cam lifts the follower for 120owith SHM followed by a dwell period of 30o. Then the follower lowers down during 150o of the cam rotation with uniform acceleration and deceleration followed by a dwell period.

ii) cam profile:

8

What do you mean by ‘Perfect Intercolling’ ? Explain with the help of P.V. diagram.

Intercooling : In perfect intercooling the temperature of air after passing out of intercooler is same as that of the temperature of air before compression of LP cylinder. 

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

State any four advantages of ball bearings over plain journal bearings.

1) The ball bearings have a far smaller contact area and thus have a lower frictional drag coefficient. 2) Due to less frictional drag means better response and less power consumption. 3) The turbo can spool up much faster, which reduces turbo-lag and offers a major performance advantage over journal bearing turbochargers at lower to mid turbocharger speeds. 4) The reduced contact area of the ball bearings means that it requires far less lubrication, allowing for lower oil pressure feeds. 5) The ball bearing more reliable. 6) Less expensive. 7) Ball bearings generate less heat and require simple and inexpensive lubrication - by oil ring, oil mist, or oil bath method.

4

State the following term : i) Tonnes of refrigeration ii) COP

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

Why majority of air compressors available in the market are multi staged ? Explain

Multi-stage air compressors feature many benefits and so, they are mostly used in the market. Some of those features are given below 1. Higher air pressures are achievable by multi-staging (about 175 PSI against 120 PSI in single stage) 2. It requires less power for running 3. Light weight cylinders can are used 4. Leakages are less 5. Overall discharge temperatures are lower 6. Intercooler increases the efficiency of unit 7. It has a greater durability 8. Many multi-stage air compressors have the crankcase cast separate from the pump cylinders, which makes it easier to repair. 9. Multi-stage compressors Air compressors can perform (are suitable) many different functions in industrial applications

8

A conical pivot with angle of cone as 100o, supports a load of 18 kN. The external radius is 2.5 times the internal radius. The shaft rotates at 150 rpm. If the intensity of pressure is to be 300 kN/m2 and coefficient of friction as 0.05, what is the power lost in working against the friction ?

4

Determine the size of bolt in the cylinder head of a steam engine. The engine cylinder has a bore of 400 mm and the maximum steam pressure to which the cylinder is subjected is 1.5 N/mm2 . Cylinder head is held on the cylinder by 16 number of bolts. The permissible tensile stress for the bolt material is 25 N/mm2

Bolt size will be M 30 or M32

4

State any four disadvantages of rolling bearings as compared to journal bearings

Disadvantages of rolling bearing as compared to Journal Bearing: 1) Initial cost is very high 2) Noisy in normal operation. 3) Shock capacity is less. 4) Finite life due to failure by fatigue. 5) Dirt & metal chips can enter the bearing & may lead it to failure. 6) Occupies greater diametral space compared to journal bearing.

4

State four assumptions made for air standard cycle.

Assumption made in air standard cycle Following assumption made in actual cycle to analysis as air standard cycle. 1. The working fluid is perfect gas. 2. There is no change in mass of the working medium. 3. All the process that constitutes the cycle is reversible. 4. Heat is assumed to be supplied from a constant high temperature source and not from chemical reaction during the cycle. 5. There are no heat losses. 6. The working medium has constant specific heats throughout the cycle.

4

Compare SI and CI engine on the basis of i) fuel used, ii) Compression ratio, iii) Weight, iv) Noise and vibration

Difference between SI and CI engines

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.

2

Name different mechanisms generated from a single slider crank chain.

Different mechanism generated by single slider crank chain mechanism. 

a) Reciprocating engine, Reciprocating compressor

b) Whitworth quick return mechanism, Rotary engine,

c) Slotted crank mechanism, Oscillatory engine

d) Hand pump, pendulum pump or Bull engine

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 the hydraulic circuit showing control of DA cylinder. Using 4 × 2 DC valve. Explain the working in brief.

Draw the hydraulic circuit showing control of DA cylinder using 4 X 2 DC Valve. Explain the working in brief. Fig shows the circuit used to control DA cylinder using 4 X 2 DC Valve. The operation is described as follows: 1) When the 4/2 way DC valve is in its open center position pump oil flows from port P to port A to the blank end of the cylinder, extending the piston rod against a load. The oil in the rodend of the cylinder is free to flow back to the tank via port B and T. 2) When DC valve is activated then it engaged in cross connection of the ports. The cylinder retracts as the oil flows from ports P and B to the rod end. Oil in the blank end is returned to the tank via ports A and T.

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

State the effect of centrifugal tension on power transmission.

Effect of centrifugal tension on power transmission:

As the belt passes over the pulley with high velocity, centrifugal force is produced on the belt, which tends to act on the belt. This force tries to move the belt away from the pulley.

This force is given by,

T_C = m x V²

There is no effect of centrifugal tension on power transmitted.

2

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

Adverse effect of imbalance of rotating elements:

a) Vibration, noise and discomfort,

b) Machine accuracy get disturbed,

c) Power losses,

d) More maintenance

6

Explain with neat sketch working of non dispersive infra red (NDIR) gas analyser.

Non dispersive infra red gas analyzer ( NDIR) : The working principle of infra red gas exhaust gas analyzer is as shown in figure . It works on the principle of hetero atomic gases absorbs infra red energy at distinct and separated wavelength. The absorbed energy raises the temperature and pressure of confined gas. This enables to measure contents of hydro carbon and carbon monoxide. This is a faster method of gas analysis. The standard sample is filled in reference cell R . the sample of gas under testing is filled in cell S . The detector cell D is filled with specific gas to be measured, say CO2 . the detector cell is divided into two compartments by diaphragm. It is very sensitive. Initially infra red energy in both compartment is same and indicator reading is zero. The sample is connected to exhaust gas. This lowers pressure on sample side. It will absorb energy in proportion to concentration of CO2 in sample and detector gives percentage of CO2 present in the samp0le.

6

Draw actual hydraulic system and explain its working.

Draw actual hydraulic system and explain its working. Oil hydraulics system uses pressurized oil which is circulated through various components of the hydraulic system to perform the given task. The various components of hydraulic system have to perform its intended function and they are arranged to form a layout of the system as per sequence of operation of hydraulic system. This arrangement of various hydraulic system components as per the nature of equipment/machine is known as actual layout of the system. The oil from reservoir is cleaned by the filter and sucked by the pump when driven by the motor. The pump increases the pressure of oil and high pressure oil is then passed through relief valve to drain excess pressure. Now oil is circulated to the direction to control to the actuator. The oil moves the piston and piston rod to give output force/motion. This motion/force is then utilized for performing the work/task. The oil from the relief valve and outlet of the actuator transferred to the reservoir through drain line and recirculated in the hydraulic system

4

Compare multiplate clutch with cone clutch on the following basis. (1) Power Transmission (2) Size

Comparison of multiplate clutch and Cone clutch:

6

A hollow shaft is required to transmit 50 kW power at 600 rpm. Calculate its inside and outside diameters if its ratio is 0.8. Consider yield strength of material as 380N/mm2 and factor of safety as 4.

Given : P= 50 KW = 50000W Speed = 600rpm k=Di/do = 0.8 σyt= 380 N/mm2 Factor of safety= 4 Design stress σt=σyt/fos =380/4 =95 Shear stress = τ =σt/2 = 95/2 =47.5N/mm2 Torque transmitted by hollow shaft T = P x 60/2πN T = 50000 x 60/2π x600 T = 795.67 N-m T= 795670 Nmm T= π/16 Xτ X do3 (1-k 4 ) 795670 =π/16 X 47.5 X do3 ( 1-0.84 ) Do3=144529.313 Do = 53 mm say 55 mm Di = 0.8X 55 = 44mm

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

8

Differentiate vapour compression and vapour absorption refrigeration system. (min. eight points of difference)

Differences between Vapour Absorption and Vapour Compression refrigeration system