ESE (ME) Paper II Mock Test - 4 - Mechanical Engineering MCQ

The Triple point is a line on the P-V diagram where all the three phases solid, liquid and gases exist in equilibrium. At a pressure below the triple point line, the substance cannot exist in the liquid phase and the substance when heated, transforms from solid to vapour by absorbing the latent heat of sublimation from the surroundings.

The triple point is merely the point of intersection of the sublimation and vaporization curves. It has been found that on a ‘p-T’ diagram the triple point is represented by a point and on a ‘p-v’ diagram it is a line, and on a ‘u-v’ diagram it is a triangle. In the case of ordinary water, the triple point is at a pressure of 4.58 mm Hg and a temperature of 0.01°C.

Concept:
Shape Factor is defined as the fraction of radiation energy leaving a surface that reaches another surface,
for example, F₁₂ means the fraction of energy leaving the surface 1 reaches surface 2.
Summation Rule: F₁₁ + F₁₂ = 1 and F₂₁ + F₂₂ = 1
Reciprocity theorem: A₁ F₁₂ = A₂ F₂₁ (A is the surface area)
Calculation:
Given:
Let us consider the circular part of the hemisphere as one and the spherical part as surface 2

From summation rule F₁₁ + F₁₂ = 1
Radiation energy leaving the surface 1 will not strike the surface again, therefore, F₁₁ = 0
∴ F₁₂ = 1
Now, from reciprocity theorem
A₁F₁₂ = A₂F₂₁
πR² = F₂₁ × 2πR²
F₂₁ = 0.5
From F₂₁ + F₂₂ = 1 ⇒ F22 = 0.5

Concept:

Hoop stress (σh) for a thin spherical shell is given by:

Calculation:

Given:

d = 200 mm, P = 2 MPa = 2 N/mm², σh = 40 MPa = 40 N/mm2

Concept:
Current - Duty cycle relation is given by
I²D = Constant
From the above relation,
I_r²D_r = I_d²D_d
where I_r and D_r are rated current and duty cycle, I_d and D_d are desired current and duty cycle respectively.
Calculation:
Given:
I_r = 250 A, D_r = 100%, D_d = 50%, I_d = ?
Now, we know that
Ir²Dr = Id²Dd
250² × 1 = I_d² × 0.5
∴ I_d = 353.55 A

Using Torsion equation,

Where,

T = Applied Torque

I_P = polar moment of inertia

τ = shear stress

G = Shear modulus

θ = Angle of rotation of the shaft

L = Length of the shaft

The atomic packing factor is defined as the ratio of the volume occupied by the average number of atoms in a unit cell to the volume of the unit cell.
Mathematically, Atomic Packing Factor (APF):

Characteristics of various types of structures are shown in the table below:

For Cubic Unit Cell
N_av = N_c/8 + N_f/2 + N_i/1
Nav = Average no of atoms in unit cell, Nc = No of corner atoms, Ni = No of interior atoms, Nf = No of face centre atoms
Calculation:
No of atoms in f.c.c unit cell = 4

for FCC a = 2√2 r where a is side of the cube and r is atomic radius.

APF = 0.74
For BCC:
Nav = 8/8 + 0 + 1/1 = 2
√3a = 4r
Put all values in equation 1
(APF)BCC = 0.6

Concept:

For Hartnell governor:

s = stiffness of the spring, h = movement of the sleeve

a = length of the vertical or ball arm of the lever

b = length of the horizontal or sleeve arm of the lever

Calculation:

Given: m = 2.5 kg, r₂ = 20 cm, r₁ = 14 cm, b = 10 cm, a = 20 cm

The maximum material Limit (MML) in case of the hole is the lower limit of the hole.

Or the 'GO' condition,

Given dimension of hole =

base size of hole = 40 mm

upper limit of hole = 40 + 0.06 = 40.06 mm

lower limit of hole = 40 + 0.02 = 40.02 mm

So MML for hole = lower limit of hole

and 'NO GO' condition or minimum material limit or the upper limt of hole = 40.06 mm

Concept:

Calculation:
Given:
TFC = Rs.40000
P = Rs.10
VC = Rs.5
Calculation:
Break even point = 40000/10−5
∴ Break-even point = Rs.8000

Concept:

• Resolver is a robotic sensory device which is analogical in nature.
• It is an internal state sensor, whose output is proportional to the angle of a rotating element with respect to a fixed element point.
• It is a form of 'synchro' in which the stator and rotor windings of the resolver are displaced mechanically at 90° to each other.
• It consists of a single rotor and two stator windings.

Resolvers
A resolver is another type of analog device whose output is proportional to the angle of a rotating element with respect to a fixed element. In its simplest form a resolver has a single winding on its rotor and a pair of windings on its stator. If the rotor is excited with a signal of the type A sin(ωt) the voltage across the two pairs of stator terminals will be

• V_s1(t) = A sin(ωt) sin θ
• V_s2(t) = A sin(ωt) cos θ

where θ is the angle of the rotor with respect to the stator. This signal may be used directly, or it may be converted into a digital representation using a device known as a "resolver-to-digital" converter. Since a resolver is essentially a rotating transformer it is important to remember that an ac signal must be used for excitation. If a de signal were used there would be no Output signal.
Characteristics:

• It consists of devices used to measure position, velocity and acceleration of robot joints or end effector.
• The resolver converts available A.C signal into digital signal.
• It can be used a s servo system.
• It eliminates the separate input and output transformers.
• It can be used a san incremental encoder.
• A resolver produces torque output
• It performs all the synchronous functions.

The endurance limit of the specimen is given by
S_e = K_a K_b K_c K_d S_e'
where K_a = Surface finish factor, K_b = Size factor, K_c = Reliability factor, K_d = Modifying factor to account
for stress concentration, S_e = Endurance limit stress of a particular mechanical component subjected to reversed bending stress (N/mm²), S_e' = Endurance limit the stress of a rotating beam specimen subjected
to reversed bending stress (N/mm²)

• When the surface finish is poor, there are scratches and geometric irregularities on the surface. These surface scratches serve as stress raisers and result in stress concentration. The endurance limit is reduced due to the introduction of stress concentration on these scratches.
• When the machine part is larger greater is the probability that a flaw exists somewhere in the component, the chances of fatigue failure originating at these flaws are more. The endurance limit, therefore, reduces with the increasing size of the component.
• The greater the likelihood that a part will survive, the more is the reliability factor. The reliability factor K_c depends upon the reliability that is used in the design of the component.

∴ The endurance limit is reduced due to stress concentration.

Pearlite:

• It is formed by decomposition of austenite at 723° C.
• The decomposition of austenite leads to the formation of a eutectoid mixture of 88% α - ferrite and 12% cementite called pearlite.
• The pearlite constituent consists of alternate lamellae of ferrite and cementite and contain 0.8% carbon.
• For microstructure of a 0.2% carbon, steel would consists of a quarter of pearlite and three quarters of ferrite.
• Ferrite is soft and ductile and pearlite is of medium hardness and it imparts mechanical strength to steel.
• The higher the carbon content, the higher would be the pearlite content and hence higher mechanical strength.
• Conversely, when the pearlite content increases, the ferrite content decreases and hence the ductility is reduced.
• Pearlite is formed when homogeneous austenite steel is slowly cooled in either air (called Normalizing) or furnace atmospheres (called annealing) below eutectoid point (temperature 723°C and 0.8%C).
• Structure will be coarser in case of furnace atmosphere and finer in case of cooling in air.
• When coarseness of pearlite increases, the layers resist slipping very less, relative to each other. Hence steel become more ductile.
• Both coarse pearlite and fine pearlite α-Ferrite + Fe3C phases are present.

​Eutectoid reaction:
At 727° C, 0.8 % Carbon:

Spheroidite:

• It consists of sphere like cementite particles in an α - ferrite matrix.
• Spheroidite is formed when carbon steel is heated for more than 30 hours at a temperature of more than 690°C.
• Spheroidite is relatively soft and allows more formability.
• It finds wide application in railroad tracks, tyre cords, bridge cables etc.

Concept:
Dynamic load carrying Capacity:

• It is defined as the radial load in radial bearing (Thrust load in thrust bearing) that can be carried for a minimum life of one million revolutions.
• It is based on the assumption that the inner race is rotating and the outer race is stationary.

Equivalent / Actual bearing Load:

• The equivalent dynamic load is defined as the contact radial load in radial bearings (Thrust load in thrust bearing), which if applied to the bearing would give the same life as that which the bearing would attain in actual condition.

​Bearing life under variable load:
​Life, (L / L₉₀ / L₁₀ ) = (C/P_e)ⁿ million revolution
n = 3 for ball bearing, n = 10/3 for roller bearing, C = basic dynamic load, P_e = Dynamic load
Calculation:
Given:
life = 1 × 10⁶ rev = 1 million revolution, Radial load (P_e1) = 8000 Kg, P_e2 = 4000 kg, n = 3 (ball bearing)

⇒ L₂ = 8 × L₁
⇒ L₂ = 8 ×10⁶ rev.

Concept:

The operating characteristics curve for an attribute sampling plan is a graph of fraction defective in a lot against the probability of acceptance.

For different sampling plans, the operating characteristics curve will be different. To construct we should know the mathematical probability of accepting lots with varying percent defectives.

Acceptance Quality Level (AQL):

• AQL represents the maximum proportion of defectives that the consumer finds definitely acceptable.
• Ideally, at AQL the probability of acceptance should be one i.e.100%. The probability of rejecting the lot at AQL represents the producer’s risk.

Rejection Quality Level (RQL):

• RQL represents the proportion of defects that the consumer finds definitely unacceptable. The probability of accepting a lot at the RQL level represents the consumer’s risk.
• It is also called Lot Tolerance Percentage Defective (LTPD).

Concept:
A gear pump is a type of positive displacement (PD) pump.
It moves a fluid by repeatedly enclosing a fixed volume using interlocking cogs or two gears, transferring it mechanically using a cyclic pumping action. It delivers a smooth pulse-free flow proportional to the rotational speed of its gears.

These pumps have a disadvantage of small leakage due to a gap between teeth and the pump housing.

Vane pumps generate a pumping action by tracking of vanes along the casing wall. In vane pump vanes are located on the slotted rotor. The rotor is eccentrically placed inside the casing. When the prime mover rotates the rotor, the vanes are thrown outward due to centrifugal force. In vane pumps leakage is less and its volumetric efficiency is around 95%.

Concept:

A special version of the jointed arm robot is the SCARA stands for selective compliance assembly robot arm, and this configuration provides substantial rigidity of the robot in the vertical direction, but compliance in the horizontal plane. This makes it ideal for many assembly tasks

Concept:

It is asked least work required, the work will be minimum only when refrigerator operates on reversed Carnot cycle.

For reversed Carnot cycle,

COP = T_L/T_H−T_L = Q_L/W

Calculation:

Given T_L = -15°C = 258 K, T_H = 30°C = 303 K, Q_L = 1.75 kJ/s;

⇒ W = 0.3 kJ/s = 0.3 kW

A cycloidal is the locus of a point on a circle rolling on a straight line.
Mathematically,
Acceleration in the cycloidal profile is expressed by:

where:
h = Maximum follower displacement
ω = Angular velocity of cam
ϕ = Angle for the maximum follower displacement for cam rotation
This acceleration at θ = ϕ/4
Maximum acceleration is

Concept:
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 and here the motion of the first gear and the last gear is like.
The reverted gear trains are used in automotive transmissions, lathe back gears, industrial speed reducers, and clocks.

Additional Information
Simple gear train: When every gear axes are fixed relative to the fixed frame and, there is only one gear on each shaft. The number of shafts may be any.

Compound gear train: When any shaft has more than one gear (first shaft or second shaft or intermediate shaft) i.e. two gear have the same angular velocity, the gear drive is called compound gear drive.

Epicyclic gear train: When the gear train is having a relative motion of axes it is called the epicyclic gear train. The axis of at least one of the gears also moves relative to the frame.

Concept:
The factor of safety is based on yield point. So, we have to use Soderberg equation i.e.

where, σ_m = mean stress, σ_v = amplitude stress, S_yt = yield strength, S_e = endurance limit stress, N = factor of safety

Calculation:
Given:
Width of the plate w = 120 mm, S_yt = 300 MPa, S_e = 225 MPa, N = 1.5
Maximum value of the applied force P₁ = 250 kN, Minimum value of the applied force P₂ = 100 kN
Let t be the thickness of the plate
Now Maximum stress σ_max = = MPa
Minimum stress σ_min = = MPa
Mean stress, σ_m = σ_m 
Amplitude stress σ_a = σ_v
Now by applying Sodeburg equation,

⇒ t = 11.45 mm ≈ 12 mm
Important Point

For Brittle materials, we will use Goodman's Theory i.e.

Concept:

In the combustion chamber is where the combustion takes place. Here, heat is added to the jet engine in the Brayton Cycle. Compressed air flows from the compressor into the chamber and ignites after being mixed with the fuel.

The principal requirements of a combustion chamber are

• Low weight and small frontal area
• Low pressure loss.
• Stable and efficient combustion
• Thorough mixing of hot and cold fluid streams to give a uniform temperature distribution (sufficiently high) for continuous combustion
• A high velocity and a high fuel-air ratio will give a rich blowout. Which means that oxygen is displaced by fuel, which lowers the temperature of the flame and in some cases distinguishes it. So it should not have high rate of combustion.
• A lean blowout is where not enough fuel is given to the flame and could also cause it to be distinguished, this is also used for lowering the engine RPM.
• Because of turbine material limitations, only a limited amount of fuel can be burnt in the combustion chamber. The exhaust products downstream of the turbine still contain a considerable amount of excess oxygen.

Calculation:

Given

Output = 35 MW; η = 80%

⇒ Heat supplied = 35/0.80 = 43.75 MW;

Heating value of coal = 26 MJ/kg

⇒ Mass flow rate of fuel supplied = 43.75/26 = 1.6827 kg/s;

Given coal has sulphur content of 3.6 %

⇒ Mass flow rate of sulphur = 1.6827 × 0.036 = 0.0605 kg/s;

Moles of sulphur = 0.0605/32 = 1.89 × 10^-3 kmol/s;

A particular limestone to be fed to it at a calcium-sulphur molar ratio of 3.0

Moles of calcium (CaCO₃) required = 3 × 1.89 × 10^-3 kmol/s = 5.679 × 10^-3 kmol/s;

Mass of CaCO₃ required = 0.5679 kg/s;

Mass of limestone required = 0.5679/0.85 = 0.668 kg/s = 2405 kg/hr;

Concept:
The energy balance equation for heat exchanger gives:
Heat gain by cool body = Heat loo by hot body
⇒ Increase in the enthalpy of cold water = Decrease in enthalpy of hot air
Calculation:
Given:
= 50 kg/min, C_pw= 4.186 kJ/kg.K, Inlet water temperature, T_ci = 50˚C, Exit water temperature, T_co = 110˚C
∴ Net enthalpy change in water = Net enthalpy change in water = C_pw (T_co- T_ci)
= 50 × 4.186 × (110-50)
= 12558 kJ/min ⇒ 12.6 MJ/min

Concept:

A nacelle is a cover housing that houses all of the generating components in a wind turbine, including the generator, gearbox, drive train, and brake assembly.

Rotor Solidity:

• Solidity is one of the most important parameters that can affect the performance of straight-bladed wind turbines. The solidity σ, states a relation between the blade area and the turbine swept area.
• Solidity is the ratio of total rotor planform area to total swept area.

Low solidity (0.10) = high speed, low torque.
High solidity (>0.80) = low speed, high torque.
The solidity of American Multiblade: 0.7
The solidity of savonious rotor: 1

Concept:
Unit Quantities:
If the speed, discharge, and power developed by a turbine under a head are known, then by using unit quantities the speed, discharge, and power developed by the same turbine under a different head can be obtained easily. They are as follows:

Calculation:
Given:
P₁ = 8000 kW, N₁ = 1000 rpm, H₁ = 30 m, H₂ = 18 m


N₂ = 774.6 rpm

Concept:
If σ_x and σ_y are normal stress on vertical and horizontal plane respectively and this plane is accompanied by shear stress τ_xy then normal stress and shear stress on plane a-a, which is inclined at an angle θ from plane of σy.

where θ taken from Y-axis or σy

Calculation:

Given:

σ_x = -160 MPa, σ_y = -100 MPa, & τ_xy = 80 MPa.

Here θ = 60°


σ₆₀ = −130 + 15 + 69.282 = −45.71

Bolt Failure Analysis: In this bolts are tested in different conditions and loadings and results are recorded for better understanding of failure.
According to bolt failure analysis:

• 15% failures occur at the fillet under the head
• 20% failures occur at the end of threads on the shank.
• 65% failures occur in the threads that are in contact with the nut

There are different milling operations:
Form/Profile milling:
This milling process is used for machining those surfaces which are of irregular shapes. The form milling cutter used has the shape of its cutting teeth conforming to the profile of the surface to be produced.

Plain or slab milling:
Slab milling is used to machine flat and horizontal surfaces. Here plain milling cutter is used, which is held in the arbor and rotated. The table is moved upwards to give the required depth of cut.

Face milling:
It is an operation that can be used for machining a flat surface, perpendicular to the axis of the cutter. It may use periphery cutters or face cutters.

According to the first law of thermodynamics:

When a system undergoes a thermodynamic cycle then the net heat supplied to the system from the surrounding is equal to net work done by the system on its surroundings.

Given:

Q₁₂ = 180 kJ, Q₂₃ = -85 kJ, W₂₃ = -35 kJ, U₁ = 105 kJ, W₃₁ = ?

Process 1 – 2 constant volume W₁₂ = 0

Q₁₂ = W₁₂ + ΔU₁₂

Q₁₂ = U₂ – U₁

Q₁₂ + U₁ = U₂

∴ U₂ = 285 kJ.

Process 2 – 3 constant pressure

Q₂₃ = W₂₃ + ΔU₂₃

∴ Q₂₃ – W₂₃ = U₃ – U₂

∴ U₃ = -85 + 35 + 285 = 235 kJ

Process 3 – 1 adiabatic process

Q₃₁ = 0

Q₃₁ = W₃₁ + (U₁ – U₃)

∴ W₃₁ = -(U₁ – U₃) = -(105 – 235) = 130 kJ.

Check ΣW₁₂₃ = ΣQ₂₃

ΣW₁₂₃ = 130 – 35 = 95 kJ

ΣQ₂₃ = 180 – 85 = 95 kJ