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ESE (ME) Paper II Mock Test - 2 - Mechanical Engineering MCQ


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30 Questions MCQ Test - ESE (ME) Paper II Mock Test - 2

ESE (ME) Paper II Mock Test - 2 for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The ESE (ME) Paper II Mock Test - 2 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The ESE (ME) Paper II Mock Test - 2 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for ESE (ME) Paper II Mock Test - 2 below.
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ESE (ME) Paper II Mock Test - 2 - Question 1

Statement l: Required degree of subcooling and superheating may not be possible if one were to rely only on heat transfer between the refrigerant and external heat source and sink.
Statement ll: The degree of subcooling will always be less than the degree of superheating in liquid-suction heat exchange.

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 1
  • If the temperature of the refrigerant at the exit of the evaporator in VCRs is not sufficiently superheated, then it may get superheated by exchanging heat with the surroundings as it flows through the connecting pipelines (useless superheating), which is detrimental to system performance. One way of achieving the required amount of subcooling and superheating is by the use of a liquid-suction heat exchanger (LSHX).
  • An LSHX is a counterflow heat exchanger in which the warm refrigerant liquid from the condenser exchanges heats with the cool refrigerant vapor from the evaporator. The figures given below show the schematic of single-stage VCRs with a liquid-suction heat exchanger and the modified VCRS cycle on P-h diagrams.

If we assume that there is no heat exchange between the surroundings and the LSHX and negligible kinetic and potential energy changes across the LSHX, then, the heat transferred between the refrigerant liquid and vapor in the LSHX, QLSHX is given as,
⇒ QLSHX = mr (h3 – h4) = mr (h1 – h6) … (1)
Where,
mr = Mass flow rate of the refrigerant
If we take average values of specific heats for the vapor and liquid, then we can write the above equation as.
⇒ cp,l(T3 – T4) = cp,v (T1 – T6)
We know the specific heat of liquid is larger than that of vapor.
⇒ (T3 – T4) < (T1 – T6)
This means that the degree of subcooling (T3-T4) will always be less than the
degree of superheating, (T1-T6).

Additional Information
The effectiveness of the LSHX, εLSHX can be defined as the ratio of actual heat transfer rate in the LSHX to maximum possible heat transfer rate, then:

ESE (ME) Paper II Mock Test - 2 - Question 2

Which of the following statements does NOT apply to the volumetric efficiency of a reciprocating air compressor?

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 2

Concept:

  • In actual compressors, a small clearance is left between the cylinder head and piston to accommodate the valves and to take care of thermal expansion and machining tolerances.
  • A large clearance volume results in low volumetric efficiency and hence large cylinder dimensions increase the contact area between the piston and cylinder and so, increases friction and work.

Volumetric efficiency is given by:

Where c = Clearance ratio

From above equation,

Volumetric efficiency decreases when

  1. Outlet pressure Increases
  2. Clearance ratio Increases

Volumetric efficiency increases when

  1. 'n' increases
  2. Inlet temperature increases

As the clearance ratio increases, the volumetric efficiency of the reciprocating compressor decreases as the pressure ratio increases.

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ESE (ME) Paper II Mock Test - 2 - Question 3

A simply supported beam is subjected to a uniformly distributed load. Which of the following statements are true?
i. Maximum or minimum shear force occurs where the radius of curvature is zero.
ii. Maximum or minimum bending moment occurs where the shear force is zero
iii. Maximum or minimum bending moment occurs where the radius of curvature is zero
iv. Maximum bending moment and maximum shear force occur at the same section

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 3

For section AB

For equilibrium, ∑M0 = 0

For M to be maximum or minimum
⇒ dM/dx = 0
Hence dM/dx = V = 0
If the radius of curvature is zero (i.e. R = 0)
M/EI = 1/R ⇒ Bending Moment is Maximum/Infinity
We know that, F = dM/dx  ⇒ The slope of bending moment diagram is the shear force
If Bending Moment is considered Maximum, Slope of BMD is zero ⇒ Shear force is minimum
If Bending Moment is considered Infinity, Slope of BMD is Maximum ⇒ Shear force is maximum

ESE (ME) Paper II Mock Test - 2 - Question 4

Consider the following statements regarding nano-materials.

1. They have a large fraction of surface atoms.

2. They have high surface energy.

3. They have large imperfections compare to the bulk material.

4. They are used in fuel cells.

Out of these statements:

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 4

The most important property of nano material is that they have large surface to volume ratio due to this most of the atoms lies on the surface of nano material. Since they have high surface to volume ratio, they are highly reactive.

→ Nano material have reduced imperfection compare to bulk material due to small size.

Application of Nano-material:

1) Fuel cell

2) Catalysis

3) Medicine

4) Biomedical engineering

5) Sensors

6) Rechargeable batteries

7) Next generation computer chips

ESE (ME) Paper II Mock Test - 2 - Question 5

A link OB is rotating with a constant angular velocity of 2 rad/s in counter clockwise direction and a block is sliding radially outward on it with an uniform velocity of 0.75 m/s with respect to the rod, as shown in the figure below.If OA = 1 m, the magnitude of the absolute acceleration of the block at location A in m/s2 is

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 5

Concept:
Corriolis acceleration c) is acted on block is
αc = 2vω
Radial acceleration (αr)
towards center is
αr = rω2
Calculation:
Given:
ω = 2 rad/sec, v = 0.75 m/s, r = 1 m
Corriolis acceleration c) is
αc = 2 × v × ω
αc = 2 × 0.75 × 2 = 3 m/s2
Radial acceleration (αr) is:
αr = r × ω2

αr = 1 × 22 = 4 m/s2
Resultant acceleration (a) is:


Important Point
Tangential acceleration = α × r

ESE (ME) Paper II Mock Test - 2 - Question 6

Statement I: An increase in current in EDM results in increased MRR
Statement II: An increase in spark frequency at constant current results in an improved surface finish.

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 6

MRR in practice increases with an increase in working voltage, current, pulse on time and decreases with an increase in pulse off time.

The figure drawn below shows the effect of the frequency of sparking at a constant current on the surface finish. Referring to this diagram, it can be noticed that an increase in spark frequency at constant current results in improved surface finish since the energy available for the material removal during a given period of time is shared by a large number of sparks, so the size of the crater reduced.

Both the statements are true and statement 2 is not a proper explanation of statement 1.

ESE (ME) Paper II Mock Test - 2 - Question 7

Consider the following statements:
In the critical path method of constriction planning. ‘Free Float’ can be
1. greater than total float
2. greater than independent float
3. equal to total float
4. less than independent float
Of these statements

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 7

Free float is the portion of positive total float by which an activity can be delayed without affecting succeeding activity


It is greater than or equal to independent float and can be equal to total float.

ESE (ME) Paper II Mock Test - 2 - Question 8

FRL stands for

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 8

Concept:
FRL Unit

  • Filter, regulator, lubricator (FRL) assemblies are pre-packaged or modular assemblies of air filters, pressure regulators, and gauges.
  • Air leaving a compressor is hot, dirty, and wet and can cause damage to equipment and tools if it is not filtered.

Fitter

  • The filter cleans compressed air by trapping solid particles and separating liquids, such as oil and water, that are trapped in the compressed air.
  • They remove contaminants from pneumatic systems, preventing damage to equipment and reducing production losses due to contaminant-related downtime.

Regulator

  • Pressure regulators control fluid pressure in compressed air systems.
  • Regulators are also known as pressure-reducing valves (PRVS).

Lubricator

  • Lubricators add controlled quantities of oil into the compressed air system to reduce the friction between moving components within air tools and other equipment that are powered by the system.
  • Adding lubrication oil to the system also clears compressor oils that travel through the system in vapor form.

ESE (ME) Paper II Mock Test - 2 - Question 9

Consider the following statements regarding cams:
1. Base circle is the smallest circle tangent to the cam profile (contour) drawn from the centre of rotation of
a radial cam.
2. Pitch curve is the curve drawn by the trace point assuming that the cam is fixed, and the trace point of
the follower rotates around the cam.
3. Pitch circle is the circle passing through the pitch point and concentric with the base circle.
4. The smallest circle drawn tangent to the pitch curve is known as the prime circle.
Which of the above statements are correct?

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 9

Cam profile:

  • It is the surface area of the cam where the follower touches. This is the actual working curve of the cam.

Base Circle:

  • It is the smallest circle tangent to the cam profile (contour) drawn from the centre of rotation of a radial cam.

Trace point:

  • It is a reference point on the follower to trace the cam profile such as the knife-edge and centre of the roller of a roller follower.

Pitch Curve:

  • It is the curve drawn by the trace point assuming that the cam is fixed, and the trace point of the follower rotates around the cam.

Pressure Angle:

  • The pressure angle, representing the steepness of the cam profile, is the angle between the normal to the pitch curve at a point and the direction of the follower motion.
  • It varies in magnitude at all instants of the follower motion.
  • A high value of the maximum pressure angle is not desired as it might jam the follower in the bearings.
  • When pressure angle becomes zero, All transmitted force goes into the motion of the follower and the direction of the axis of the transmission.

  • When the pressure angle becomes 90 degrees there will be no motion of the follower.

  • Therefore, we would like to maintain the pressure angle as low as possible.

  • The pressure angle is inversely proportional to the base circle diameter.

  • As we increase the size of the cam the pressure angle decreases, therefore to maintain the pressure angle as low as possible the base circle should be as big as possible.

Pitch Point:

  • It is the point on the pitch curve at which the pressure angle is maximum.

Pitch Circle:

  • It is the circle passing through the pitch point and concentric with the base circle.

Prime Circle:

  • The smallest circle drawn tangent to the pitch curve is known as the prime circle.

​Working circle:

  • The working circle is considered to be the distance equal to the distance from the centre of base circle to the highest point on the cam.
  • It can also be concluded that it is the smallest circle from the cam center through the cam profile curve.


Additional Information
Cam and cam follower system:

  • A cam is a mechanical member used to impart desired motion to a follower by direct contact.
  • The cam may be rotating or reciprocating whereas the follower may be rotating, reciprocating or oscillating.
  • It is used in automatic machines, IC engines, machine tools, printing control mechanisms.

Elements of cam follower system:

  • A driver member known as cam.
  • A driven member called the follower.
  • A frame that supports the cam and guides the follower
ESE (ME) Paper II Mock Test - 2 - Question 10

Wear rating of a long shoe drum brake is

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 10

Shoe or Block Brake:

  • In a shoe brake, the rotating drum is brought in contact with the shoe by suitable force.
  • The contact surface of the shoe is coated with friction material.
  • The wear rating of a long shoe drum brake is directly proportional to the work done by friction force as power.

​⇒ Wear ∝ friction power ∝ Frictional torque ∝ Frictional force

  • The force needed to secure contacts is supplied by a lever.
  • When a force F is applied to the shoe, the frictional force is proportional to the applied force, Fr = μ × F, where μ depends on the frictional material and the geometry of the shoe.

Brake factor = Fr/P
Frictional torque = Tf = Fr × r
Taking moments about the hinge:
Mhinge = 0
⇒ P × L = N + Fr × a
And, Fr × a + P × L = N
⇒ P × L = N - Fr × a
⇒ P = N/L x [b -  μa]
∴ For counterclockwise rotation of the drum
P = N/L x [b - μa]

ESE (ME) Paper II Mock Test - 2 - Question 11

Total Quality Management (TQM) and Quality assurance are the responsibility of everyone involved in designing and manufacturing of the product. Who among the following pioneers has NOT been quality control heightened?

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 11

Burrhus Frederic Skinner (B. F. Skinner) was not been related to quality management but he was an American psychologist, behaviourist, author, inventor and social philosopher.
Deming (Dr. W. Edwards Deming)-

  • Deming’s theory of management details the steps that must be taken to transform a company’s quality culture. It is a theory that means it is insufficient to simply solve problems that arise.
  • A culture of continuous improvement must be established and maintained with the overall goal of achieving customer satisfaction. Along with the fourteen basic points of his theory of management, Deming also defines what he calls the deadly sins and diseases that virtually every company in the West is being crippled by.

Taguchi (Dr. Genichi Taguchi) -

  • Also known as Robust Design, Taguchi Methods include both design of experiments concepts, and a particular philosophy for design in a more general sense (e.g. manufacturing design). Taguchi sought to improve the quality of manufactured goods and advocated the notion that ‘quality’ should correspond to low variance.

Juran (Dr. Joseph Juran)-

  • Dr. Juran creates the “Pareto Principle,” also known as the 80-20 principle, to help separate the “vital few” from the “useful many” in their activities.
  • Juran’s Trilogy shows how an organization can improve every aspect and business results by better understanding of the relationship between processes that plan, control and improve quality.
ESE (ME) Paper II Mock Test - 2 - Question 12

An oil of viscosity 10 poise flows between two parallel fixed plates which are kept at a distance of 50 mm apart. What is the rate of flow of oil between the plates if the drop of pressure in a length of 1.2 m be 0.3 N/cm2 and the width of the plate is 200 mm?

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 12

Concept:
Volume Flow Rate in case of Laminar Flow through Parallel Plate is given by

where ΔP = Pressure Difference, h = Distance between parallel plates, b = Width of Plate, L = Length of Plate,  μ = Viscosity
Calculation:
Given:
ΔP = 0.3 N/cm2 = 0.3 × 104 N/m2, h = 50 mm = 0.05 m, b = 200 mm = 0.2 m, L = 1.2 m,  μ = 10 Poise = 1 Pa.s

Q = 5.208 × 10-3 m3/sec = 5.208 L/s

ESE (ME) Paper II Mock Test - 2 - Question 13

Consider the following statements regarding the Euler turbine equation:
1. The axial component produces bending of the shaft which is taken by the journal bearings.
2. The axial force only can cause the rotation of the runner and produce work.
3. The tangential force only can cause the rotation of the runner and produce work.
4. The radial component produces bending of the shaft which is taken by the journal bearings.
Which of the above statements are correct?

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 13

Euler Turbine Equation:

  • The Euler turbine equation relates the power added to or removed from the flow, to the characteristics of a rotating blade row. The equation is based on the concepts of conservation of angular momentum and conservation of energy.
  • The fluid velocity at the turbine entry and exit can have three components in the tangential, axial and radial directions of the rotor.
  • This means that the fluid momentum can have three components at the entry and exit.
  • This also means that the force exerted on the runner can have three components.
  • Out of these, the tangential force only can cause the rotation of the runner and produce work.
  • The axial component produces a thrust in the axial direction, which is taken by suitable thrust bearings.
  • The radial component produces bending of the shaft which is taken by the journal bearings.
  • Thus it is necessary to consider the tangential component for the determination of work done and power produced.
  • The work done or power produced by the tangential force equals the product of the mass flow, tangential force and tangential velocity.
  • As the tangential velocity varies with the radius, the work done also will vary with the radius.
  • The help of the moment of momentum theorem is used for this purpose. It states that the torque on the rotor equals the rate of change of moment of momentum of the fluid as it passes through the runner

Assumptions:

  • The flow is steady.
  • The heat and work interaction between the control volume and its surroundings take place at a constant rate.
  • Velocity is uniform over any area normal to the flow.
  • There is no friction or other losses in the system.
  • The fluid is assumed to have the perfect guide through the flow system. This implies that there is infinite number of thin blades on the runner.
  • Fluid enters and leaves the vane in a direction tangential to the vane tip at inlet and outlet.

Angular Velocity of the wheel (rotational Speed) (rad/sec), ω = 2 × π × N/60

The power produced can be expressed as due to three effects. These are the dynamic, centrifugal and acceleration effects. Consider the general velocity triangles at inlet and exit of turbine runner as shown above:

Let

V1, V2 Absolute velocities at inlet and outlet.

 Relative velocities at inlet and outlet.

U1, U2 Tangential velocities at inlet and outlet.

The tangential component of absolute velocities at inlet and outlet.

Tangential Momentum of the fluid at entry =

Moment of momentum or angular momentum at entry =

Similarly Angular Momentum at the outlet =

T = Torque on the wheel=change of angular momentum

Workdone=rate of energy transferred = (torque х angular velocity)
But, ω × r1 = U1 and ω × r2 = U2

∴ Work done

Euler's head 

ESE (ME) Paper II Mock Test - 2 - Question 14

Which one of the following statements is NOT correct?

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 14

Vacuum efficiency -

  • It is defined as the ratio of the actual vacuum in the condenser as recorded by the vacuum gauge to the ideal vacuum.

where Actual vacuum = Barometric pressure - Actual condenser pressure
Ideal vacuum = Barometric pressure - Ideal pressure (pi)
Ideal pressure (pi) is that which corresponds with condensate temperature or with the temperature of
steam entering the condenser. It can be read from steam tables.
Condensers :

  • The Condenser removes the heat from the refrigerant carried from the evaporator and added by the compressor and converts the vapor refrigerant into liquid refrigerant.
  • It is a Heat Exchanger in which heat transfer takes place from high-temperature vapor to a low-temperature liquid with help of cooling like air or water.
  • The function of the condenser is to condense exhaust steam from the steam turbine by rejecting the heat of vaporization to the cooling water passing through the condenser.

Surface condensers -

  • Surface condensers are basically a shell and tube heat exchangers consisting of water boxes for directing the flow of cooling water to and from horizontal tubes.
  • In surface condensers, the cooling water and exhaust steam do not come in direct contact with each other.
ESE (ME) Paper II Mock Test - 2 - Question 15

Match the following:


Select the correct matching using the code given below:

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 15

Rankine cycle -
The Rankine cycle is the ideal cycle for vapor power plants; it includes the following four reversible processes:

ESE (ME) Paper II Mock Test - 2 - Question 16

The concentration ratio of as high a value as 3000 can be obtained by

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 16

'Concentration ratio' (CR) is the ratio of the effective area of the aperture to the surface area of the absorber. The value of CR may change from unity (for flat-plate collectors) to a thousand (for parabolic dish collectors). The CR is used to classify collectors by their operating temperature range.

A solar thermal energy collector is equipment in which solar energy is collected by absorbing radiation in an absorber and then transferring it to a fluid.

In general, there are two main types of collectors:


Flat-plate solar collector -

  • It has no optical concentrator. Here, the collector area and the absorber area are numerically the same, the efficiency is low, and temperatures of the working fluid can be raised only up to 100oC.

Concentrating-type solar collector:

  • Here the area receiving the solar radiation is several times greater than the absorber area and the efficiency is high. Mirrors and lenses are used to concentrate the sun's rays on the absorber, and the fluid temperature can be raised up to 500oC.
  • Compound Parabolic Collector with Plane Receiver Reflectors is curved segments that are parts of two parabolas. The CR varies from 3 to 10. For a CR of 10, the acceptance angle is 11.5o, and tracking adjustment is required after a few days to ensure collection of 8 hours a day.

Cylindrical parabolic collector -

  • The reflector is in the form of a trough with a parabolic cross-section in which the image is formed on the focus of the parabola along a line. The aperture area ranges from 1 m2 to 6 m2, where the length is more than the aperture width. The CR range is from 10 to 30.

Central tower receiver -

  • The Central Receiver is a field of heliostats (an array of minors) that reflect solar radiation to a centrally located receiver on a tower. Heliostats follow the sun to harness maximum solar heat. The CR range is from 100 to 3000. Water flowing through the receiver absorbs heat to produce steam which operates a Rankine cycle turbo generator to generate electrical energy
ESE (ME) Paper II Mock Test - 2 - Question 17
A rod of length 2 m and diameter 50 mm is elongated by 5 mm when an axial force of 400 kN is applied. The modulus of elasticity of the material of the rod will be nearly
Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 17

Concept:

The stress and strain under uniaxial load are given by

Young’s modulus is given by

Calculation:

Given l = 2m, d = 50 mm, Δl = 5 mm, P = 400 kN;

Nearest option is 82 GPa

ESE (ME) Paper II Mock Test - 2 - Question 18

A cylindrical storage tank has an inner diameter of 600 mm and a wall thickness of 18 mm. The transverse and longitudinal strains induced are 255 × 10-6 mm/mm and 60 × 10-6 mm/mm, and if G is 77 GPa, the gauge pressure inside the tank will be

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 18

Concept:

Longitudinal strain in a thin cylinder is given by:

ϵL =

Transverse or hoop strain in a thin cylinder is given by:

ϵH =


 

Modulus of elasticity is given as:

E = 2G (1 + μ)

Calculation:

Given:

The inner diameter of the pipe, D = 600 mm

The thickness of the pipe, t = 18 mm

Shear modulus (G) = 77 GPa

Longitudinal strain (ϵL ) = 60 × 10-6 mm/mm

Transverse strain (ϵH) = 255 × 10-6 mm/mm

Now,

∴ μ = 0.3

Modulus of elasticity is given as:

E = 2G (1 + μ)

E = 2 × 77 × 103 (1 + 0.3)

E = 200.2 × 103 MPa

Also,

p = 3.6 MPa
Thus, the gauge pressure inside the tank will be 3.6 MPa.

ESE (ME) Paper II Mock Test - 2 - Question 19

In the study of phase diagrams, the rule which helps to calculate the relative proportions of liquid and solid material present in the mixture at any given temperature is known as

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 19

Concept:
The lever rule is a rule used to determine the mole fraction or the mass fraction of each phase of a binary equilibrium phase diagram.
It can be used to determine the fraction of liquid and solid phases for a given binary composition and temperature that is between the liquidus and solidus line.Lever rule is given as

Percentage of one phase = xz/xy × 100

Similarly for another phase = zy/xy × 100

ESE (ME) Paper II Mock Test - 2 - Question 20

What is the standard value of solar constant adopted by World Radiation Centre?

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 20

Solar constant:

  • Solar constant, the total radiation energy received from the sun per unit of time per unit of area on a theoretical surface perpendicular to the sun’s rays and at earth’s mean distance from the sun. It is most accurately measured from satellites where atmospheric effects are absent.
  • The value of the constant is approximately 1.366 kW/m2 or 1367 W/m2.
  • The “constant” is fairly constant, increasing by only 0.2% at the peak of each 11-year solar cycle.
ESE (ME) Paper II Mock Test - 2 - Question 21

In single basin, double effect scheme, power is generated

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 21

Tidal Powerplant Working principle:

The gravitational attraction of the moon and the sun upon the rotating earth produces tides in the sea. The rise and fall of tides occur twice a day.

The highest level of tidal water is called high tide or flood tide, whereas the lowest level of tidal water is called low tide or ebb tide. The level difference between high and low tides is called the tidal range.

In a tidal energy power plant, we collect the water during high tides in an artificial basin and release it during the period of low tides.

Single Basin Tidal Power Plant Working:

In a single basin tidal power plant, the powerhouse is located at the mouth of the basin. The hydraulic turbine in the powerhouse only operates during the discharge of water from the basin during low tide. The basin is filled again during the high tide. Intermittent operation is the main disadvantage of this system.

However, in a double cycle system, power generation is possible during the low as well as high tide. The direction of flow during the low and high tide alternates and production of power takes place during both the emptying and filling cycle of the basin.

Double Basin Tidal Power Plant Working:

In a double basin tidal power plant, two basins at different levels are made, and a dam is provided in between them. Inlet and outlet sluice gates are made in the dam. The water level in the upper basin is maintained above the level of water in the lower basin.

The upper basin is filled with water during high tide, and the lower basin is evacuated during the low tide. Therefore, a permanent head is produced between the upper and lower basins.

ESE (ME) Paper II Mock Test - 2 - Question 22

In a solar passive space heating system, the south-facing thick wall is called

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 22

Passive solar heating:
Passive solar heating allows homeowners to collect, absorb, and distribute solar energy right through their windows. By using materials that hold onto heat during the day and disperse it at night, homeowners can keep their houses comfortable 24/7 even in spaces that would normally get cool.
This method is called “passive” solar heating because it doesn’t take any solar panels or technology to make it happen–just sunshine itself.

Trombe wall:
A Trombe wall is a massive south-facing wall that is painted dark color in order to absorb thermal energy from incident sunlight and covered with a glass on the outside with an insulating air gap between the wall and the glaze.

Additional Information
Passive Solar Design Basics:

A complete passive solar design has five elements:

  • Aperture/Collector
  • Absorber
  • Thermal Mass
  • Distribution
  • Control
ESE (ME) Paper II Mock Test - 2 - Question 23

In order to burn a fuel completely, which of the following basic conditions must be fulfilled?

1. Supply enough air for complete combustion of fuel.

2. Secure low turbulence for thorough mixing of fuel and air.

3. Maintain a furnace temperature high enough to ignite the incoming fuel air mixture.

4. Provide a furnace volume large enough to allow time for combustion to be completed.

Select the correct answer using the code given below:

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 23

Combustion:

It is an exothermic process that releases light and heat. It is supported by oxygen. If oxygen is available in a sufficient amount then complete combustion is achieved otherwise partial combustion occurs.

Following things are required to achieve combustion:

  • Supply enough air for complete combustion of fuel.
  • Maintain a furnace temperature high enough to ignite the incoming fuel-air mixture.
  • Provide a furnace volume large enough to allow time for combustion to be completed.
  • High turbulence is needed for the proper mixing of fuel and air so that it can be burn completely.
ESE (ME) Paper II Mock Test - 2 - Question 24

The constant pressure gas turbine works on

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 24

Brayton cycle:
Gas turbines operate on the Brayton cycle/Joule cycle. The Joule cycle consists of four internally reversible processes:

  • Isentropic compression (in a compressor)
  • Constant-pressure heat addition
  • Isentropic expansion (in a turbine)
  • Constant-pressure heat rejection

Rankine Cycle:
Process 1 – 2: Isentropic compression
Process 2 – 3: Isobaric heat addition
Process 3 – 4: Isentropic expansion
Process 4 – 1: Isobaric heat rejection

Though the Rankine cycle is a constant pressure cycle, it does not come under gas-turbine as the working fluid is water here.
Atkinson Cycle:
In the case of the Atkinson Cycle heat is supplied at constant volume and rejected at constant pressure.
Stirling cycle:
A Stirling cycle consists of two reversible isothermal and two reversible constant volume (isochoric) processes.

ESE (ME) Paper II Mock Test - 2 - Question 25

In a thermoelectric thermometer for t°C temperature, the emf is given as: E = 0.003t - 5 × 10-7 t2 + 0.5 × 10-3 volts. The thermometer is having a reference junction at the ice point and is calibrated at the ice point and steam point. What is the temperature shown by the thermometer for a substance at 30°C?

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 25

Given equation:
E = 0.003t - 5 × 10-7t2 + 0.5 × 10-3 volts
At ice point i.e. 0°C
E = 0.5 × 10-3 Volts
At steam point 100°C
E = (0.003 × 100) - (5 × 10-7 × 1002) + 0.5 × 10-3 volts
E = 0.2955 Volts
At 30°C
E = (0.003 × 30) - (5 × 10-7 × 302) + 0.5 × 10-3 volts
E = 0.09005 Volts



⇒ t = 30.393°C

ESE (ME) Paper II Mock Test - 2 - Question 26

A body is moving with a velocity of 2 m/s. After 4 seconds, the velocity of the body becomes 5 m/s. The acceleration of the body is

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 26

Concept:
Equations of motion:
v = u + at

v2 = u2 + 2as
where, u = initial velocity, v = final velocity, a = acceleration/retardation, s = displacement.
Calculation:
Given:
Initial velocity = 2 m/s, Final velocity = 5 m/s, t = 4 sec
v = u + at
5 = 2 + a(4)
∴ a = 0.75 m/s2

ESE (ME) Paper II Mock Test - 2 - Question 27

Consider the following statements for the symmetric beam under pure bending :
1. In the elastic range, the normal stress varies linearly with the distance from the neutral surface.
2. As long as the stresses remain in the elastic range, the neutral axis passes through the centroid of the section.
3. If stresses are in the plastic range, the neutral axis passes through the centroid of the section.
Which of the above statements is/are correct?

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 27

Properties of Pure Bending:

When loaded by a bending force, the beam bends so that the inner surface is in compression and the outer surface is in tension. The neutral plane is the surface within the beam between these zones, where the material of the beam is not under stress (either compression or tension).

As there is no lengthwise stress force on the neutral plane, there is no strain so when the beam bends, the length of the neutral plane remains constant.

The neutral axis is an axis in the cross-section of a beam (a member resisting bending) or shaft along which there are no longitudinal stresses or strains.

If the section is symmetric, isotropic, and is not curved before a bend occurs, then the neutral axis is at the geometric centroid.

All fibers on one side of the neutral axis are in a state of tension, while those on the opposite side are in compression.

As long as the stresses are under elastic range, the extension or compression does not effect the location of neutral axis, whereas if the stresses goes beyond the elastic range the shape of the beam is distorted, such that the neutral axis is also displaced.

From Pure Bending equation:


we can conclude that normal stress varies linearly with distance from the neutral axis.

ESE (ME) Paper II Mock Test - 2 - Question 28

Consider the following statements for ductile fracture:

1. The material undergoes substantial plastic deformation with high energy absorption before fracture.

2. Presence of cracks on the surface of material initiates this type of failure.

3. Fracture occurs due to necking.

Which of the above statements is/are correct?

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 28

Ductile Fracture:

  • Most ductile metals fracture preceded by a moderate amount of necking, followed by the formation of voids, cracks and finally shear. Hence we can say that ductile fracture takes place after necking with little sound.
  • This gives a characteristic cup-and-cone fracture. In this central interior region has an irregular and fibrous appearance, which signifies plastic deformation.
  • Different progressive stages of ductile fracture are shown in figure.
  • As the cross-sectional areas is getting reduced at subsequent stages high energy is absorbed before fracture happens.

Additional Information
Brittle Fracture:

  • The brittle fracture that takes place with little or no preceding plastic deformation. It occurs, often at unpredictable levels of stress, by rapid crack propagation.
  • The direction of crack propagation is very nearly perpendicular to the direction of applied tensile stress. This fracture is also said to be trans granular because crack propagates through grains.
  • Thus, it has a grainy or faceted texture.
  • Most brittle fractures occur in a transgranular manner. However, the brittle fracture can occur in an intergranular manner i.e. crack propagates along grain boundaries.
ESE (ME) Paper II Mock Test - 2 - Question 29

Which one of the following theories gives satisfactory results for brittle materials ?

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 29

The relation between the theories of failure, their suitable materials, and graphical representation is given below.

ESE (ME) Paper II Mock Test - 2 - Question 30

Consider the following statements regarding defects in forgings

1. Flakes are internal breaks or ruptures occurring in some grades of alloy steel.

2. Die shift is caused by misalignment between the top and bottom forging dies.

3. Fins and rags are small projections or loose metal driven into the surface of the forging.

Which of the above statements are correct ?

Detailed Solution for ESE (ME) Paper II Mock Test - 2 - Question 30

Defects in forging:

  • A mismatch occurs in drop forging when the dies are incorrectly aligned, and results in a lateral displacement between portions of the forging.
  • Scale pits are shallow depressions caused by hot removing scale from the dies. The scale is subsequently worked into the surface of the forging.
  • An unfilled section is similar to misrun in casting and occurs when metal does not completely fill the die cavity. It is usually caused by using insufficient metal or insufficient heating of metal.
  • Defects resulting from improper heating and cooling of the forging such as burnt metal, decarburized steel, and flakes.
  • These defects are found in all metals which are heated to the plastic stage and then shaped.
  • Defects resulting from the melting practice, such as dirt, slag and blowholes.
  • Defects resulting from improper forging such as seams, cracks, laps etc.
  • Ingot defects such as pipes, cracks, scabs and segregation.
  • Fins and rags are small projections or loose metal driven into the surface of the forging happens due to poor die-design.
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