ESE (ME) Paper II Mock Test - 5 - Mechanical Engineering MCQ

Concept:
Combined equations of first and second law of thermodynamics
Tds = du + Pdv
Tds = dh – vdP
These equations are applicable for both reversible and irreversible process and for the closed and open system as
well.
du = C_vdT
dh = C_pdT
C_v = specific heat at constant volume, C_p = specific heat at constant pressure

Calculation:

From first equation
Tds = du + Pdv
For constant volume, dv = 0
And, du = C_vdT
∴ 1^st equation becomes: Tds = C_vdT

From second equation
Tds = dh – vdP
For constant pressure, dP = 0 & dh = C_pdT
Tds = C_pdT

Therefore, ratio of slope of constant pressure and constant volume lines at point of intersection:

Concept:

Rankine cycle with superheating is shown in the figure.

• Direct effects that can be inferred
• Increased heat addition and rejection
• Due to increase heat addition, mean temperature of heat addition increases
• The increase in mean temperature of heat addition increases the efficiency
• Heat rate in inversely proportional to efficiency. Hence, it decreases

Therefore, all the given statements are correct.

Concept:
(COP)_{Heat Pump} = 1 + (COP)_Refrigerator
Refrigerator: A refrigerator is a device which works on reverse Carnot cycle and extracts heat from a lower temperature body to keep the temperature of the body lower than the surrounding temperature. And by taking work input it transfers heat to the higher temperature body or surrounding.

Refrigerating Effect R.E.= QL
Work input = Q_H - Q_L
For a reversible engine, Q_H/Q_L = T_H/T_L
Coefficient of Performance (COP) 
(COP)_Refrigerator = Q_L/Q_H - Q_L = T_L/T_H - T_L

Heat Pump: A heat pump is a device which works on reversed Carnot cycle and transfers heat from a lower temperature body to a higher temperature body. A heat pump maintains a body at a temperature higher than the surrounding temperature or a lower temperature body.

(COP)_{Heat Pump} = 1 + (COP)_Refrigerator
The difference between the COP of heat pump and refrigerator
COP_{Heat Pump} - COP_Refrigerator = 1

Properties
All measurable characteristics of a system are known as properties.
Eg. Pressure, volume, temperature, etc.
There are two types of properties:
Extensive property

• Those properties which depend on mass are known as extensive properties.
• Examples are volume, energy, enthalpy, etc.

Intensive property

• Those properties which don't depend on mass are known as intensive properties.
• Examples are pressure, temperature, density, viscosity.

Important Points

• The ratio of two extensive properties is an intensive property.
• Specific properties are intensive properties. For example specific volume, specific energy etc.
• If property divides with space then it is extensive property otherwise the property will be intensive.
• Property is a point function.
• Property is independent of past history.
• In a cycle change in property is equal to zero.
• Properties are exact differential.

Note: Heat, work, and entropy generation are path functions, they are not properties.

Concept:
Flow-through Pipes: In flow-through pipe different type of losses takes place.
Major losses are due to viscous (friction) resistance through the wall, and head loss due to friction is given by Darcy's Weichback equation:

​
where, f = friction factor, L = length of pipe, D = diameter of the pipe, Q = pipe flow
Minor losses these losses are due to change in momentum of the fluid due to the sudden expansion or contraction,
change in the shape of the pipe.
Compound Pipes: Types of pipe connection are:
In Series Connection of pipe discharge is same and losses are different in each pipe.

• Q₁ = Q₂ = Q₃ = ................ = Q
• Equivalent head h_f = h_f1 + h_f2 + h_f2 ....

In Parallel Connection of pipe discharge is different and losses are same in each pipe.

• Q = Q₁ + Q2 + Q₃ ..... + Q
• Equivalent head h_f = h_f1 = h_f2 = h_f2 ......

Calculation:
Given:
D₁ = 10 cm, D₂ = 40 cm , same length L₁ = L₂ = L, same friction factor f₁ = f₂ = f
In parallel Connection Head loss in pipe are equal:
h_f1 = h_f2
⇒

Concept:
The entropy generation for a plane wall shown below is given by

ΔS = Q(1/T₂ - 1/T₁)
Calculation:
Given Q = 1500 W, T₁ = 27°C = 300 K, ΔS = 0.25 W/K;
Now the entropy generated will be
ΔS = Q(1/T₂ - 1/T₁) ​​​​​​​⇒ 0.25 = 1500(1/T₂ - 1/300)
⇒ T₂ = 285.71 K;
⇒ T₂ = 285.71 - 273 = 12.71°C;​​​​​​​

Concept:
Statement - I: Temperature is continuous always

• In a single slab, temperature shows a linear profile, which is shown as-

−kA∫dT = Q∫dx

• Hence it is evident that temperature profile in any single slab is linear and continuous.
• Also if we talk about two slabs the interface temperature is the same.
• Hence, if we talk about composite slab containing two slabs, the temperature is always continuous in the second slab and since the interface temperature is the same, therefore we can say that temperature is continuous along with the two slabs or nay ‘n’ number of slabs.

Statement - II: Temperature gradient is not Continuous

• In a single slab, the temperature gradient is given as-

dT/ dx = − QkA
Since heat flow and area of cross-section is same in both slabs so,

• Therefore temperature gradient is constant in single slab having constant thermal conductivity which is the usual case.
• If 2 slabs are having different thermal conductivity they will have different gradient and hence at the interface gradient will be non-continuous.

Statement - III: Heat flow Is not Continuous

• Heat flow is constant across any slab in a composite slab and therefore is continuous. So, statement III is not correct.

Hence, the only statement I and II are correct. So, the correct option is an option (b)

Pool Boiling:

• Pool boiling refers to the situation where the liquid above the hot surface is essentially stationary and the of liquid motion near the surface is due to free convection and the mixing induced by bubble growth and detachment.

Fim condensation:

• In the Fim condensation, the condensate wets the surface and forms a liquid film on the surface that slides down under the influence of gravity.
• Film condensation results in low heat transfer rates as the film of condensate impedes the heat transfer.
• The thickness of the film formed depends on many parameters including orientation of the surface, viscosity, rate of condensation, etc.
• The film increases the thermal resistance to heat flow between the surface and the vapor. The rate of heat transfer is reduced because of this resistance.
• In general, film condensation occurs when a vapor free from impurities condenses on the clean and uncontaminated surface.

Boiling:

• By the definition, the evaporation at solid liquid surface is termed as boiling.

Saturated Boiling:

• Here the temperature of the liquid slightly exceeds the saturation temperature. The bubbles form at the surface area are then propelled through the liquid by buoyancy forces, eventually escaping from the free surfaces.

Concept:
Vapor absorption system:

Basic Vapour Absorption Refrigeration System

• The basic absorption cycle employs two fluids, the absorbate or refrigerant, and the absorbent
• The most common fluids are Water/Ammonia as the refrigerant and lithium bromide/ water as the absorbent
• These fluids are separated and recombined in the absorption cycle
• In the absorption cycle, the low-pressure refrigerant vapour is absorbed into the absorbent releasing a large amount of heat (Absorber pressure is equal to evaporator pressure)
• The liquid refrigerant/absorbent solution is pumped to a high-operating pressure generator using significantly less electricity than that for compressing the refrigerant for an electric chiller
• Heat is added at the high-pressure generator from a gas burner, steam, hot water or hot gases
• The added heat causes the refrigerant to desorb from the absorbent and vaporize
• The vapours flow to a condenser, where heat is rejected and condense to a high-pressure liquid
• The liquid is then throttled through an expansion valve to the lower pressure in the evaporator where it evaporates by absorbing heat and provides useful cooling
• The remaining liquid absorbent, in the generator, passes through a valve, where its pressure is reduced, and then is recombined with the low-pressure refrigerant vapours returning from the evaporator so the cycle can be repeated.
• From the given diagram, the total heat is,
• Q_generator + Q_evaporator = Q_absorber + Q_condensor

​Important Point
In the vapour absorption cycle the compressor is replaced by

• Absorber
• Pump
• Generator or desorber

Concept:

• An elliptical termal is an inversion of a double slider four Bar link Mechanism in which the slotted plate is fixed.
• Here Link is extended as shown in the figure. So here we can imagine that the position of the imaginary pen to draw the ellipse is at point C so the locus of point C will be an ellipse traced by an imaginary pen at point C(x, y).

Here,

The semi-major axis = AC (Distance between imaginary pen position and slider pin which is at a larger distance that is Slider A)

The semi-minor axis = BC (Distance between imaginary pen position and slider pin which is at a smaller distance that is Slider B)

Calculation:

Given:

AB = 100 mm; BC = 150 mm,

Let the Point C (Imaginary Pen position) ≡ (x, y)

Here,

AC = AB + BC

⇒ AC = 100 + 150

⇒ AC = 250 mm

By using the above diagram,

x = BC × cosθ = 150 × cosθ …(1)

y = AC × sinθ

y = (100 + 150) × sinθ

y = 250 × sinθ …(2)

By eliminating θ from equation (1) and (2),

we get locus of point C as,

Here, the length of the major axis of the ellipse = 2 × AC = 500 mm

Also, the length of the minor axis of the ellipse = 2 × BC = 300 mm

Statement P:
The ratio of the actual damping coefficient (c) to the critical damping coefficient (cc) is known as damping factor or damping ratio.

• Overdamped System: ζ > 1
• Underdamped: ζ < 1
• Critical Damping: ζ = 1: The displacement will be approaching to zero in the shortest possible time. The system does not undergo a vibratory motion.

Statement Q:

Logarithmic decrement:

If the system executes n cycles:

From the above formula, we can say that the damping factor can be calculated with the help of logarithmic decrement.

Statement R:

In the case of damping due to dry friction between moving surfaces resisting the force of constant magnitude acts opposite to the relative motion.

Statement S:

Viscous drag force is

F_D = Cẋ

Where c = damping coefficient; ẋ = relative velocity

Hence we can say that Drag Force is directly proportional to relative velocity and not it’s square.

So option S is incorrect

• Since the area of cross-section of the threaded portion of the bolt is less than that of the area of cross-section of the shaft. So, the stresses (σ) induced in the threaded portion will be higher.

Now for a bolt under induced stress (σ), strain energy (U) stored in the bolt will be proportional to the square of induced stress (σ),
⇒ U ∝ σ²
Hence, the threaded portion of the bolt will have a higher stress-absorbing capacity than the shank portion.

• Tensile, shear, and compressive stresses are induced on the bolt without preheating.
• In an eccentric loaded bolted joint in shear, the bolt closer to the applied load will experience the highest induced stress so, it will be worst loaded.

Isochronism:

• A governor with a range of speed zero is known as an isochronous governor.
• During this situation, for all positions of the sleeve or the balls, the governor has the same equilibrium speed.

Effort of a governor:
The effort of a governor is the mean force acting on the sleeve to raise or lower it for a given change of
speed.
∴ at constant speed i.e. for an isochronous governor, the resulting force acting on the governor is zero.

The quantity δQ/T  does not depend upon the path. It depends upon the initial and final state of the process. So, the quantity δQ/T must be the property. this property is known as entropy which is represented by 'ds'
From the second law of thermodynamics for any process,
⇒ ds ≥ δQ/T    …(1)
For an irreversible process,
⇒ ds > δQ/T    …(2)
For any cyclic process,

Using equation (2),
So, Clausius inequality for the irreversible heat engine is given by,

For an irreversible heat engine,

The heat content of the irreversible heat engine will be positive as heat is given to the system for heat
engine
 ⇒ dQ > 0
Important Point

Concept:
Total work done = (Potential work) + (Kinetic work) = (m × g × h) + 1/2 x m x v²
Power =
Calculation:
Given:
Amount of water raised = 6000 litres / minute, Height, h = 2 m, Velocity, v = 10 m/s, g = 9.81 m/s²
Volume = 6000 litres, Density, ρ = m/v
⇒ 1000 = m/ 6000 x 10^-3 (∵ 1 Litre = 10^-3 m³)
⇒ m = 6000 Kg
∴ Total work done in 1 minute = (m × g × h) + 1/2 x m x v² 
= 6000 x 9.81 x 2 + 1/2 x 6000 x 10² 
⇒ Total work done in 1 minute = 417720 J
Power = 417720/60 = 6962 watt = 6962/746 = 9.33 hp

Concept:
Given:
Specific Gravity of Oil (S_o) = 0.9
Specific Gravity of Mercury (S_Hg) = 13⋅6
Difference in Mercury Level (h) = 15 cm
acceleration due to gravity as 9⋅81 m/s²
Density of Water (ρ_ω) = 1000 kg/m³
Density of Oil (ρ​​​​​​​_o) = ρ​​​​​​​_ω x S_o = 900 kg/m³
Density of Mercury (ρ​​​​​​​_Hg) = ρ​​​​​​​_ω x S_Hg = 13600 kg/m³

Assume points A and B are at the same level.
So, by equating the pressure on both limbs over section xx, we have
P_A + ρ_og(x + h) = P_B + ρ_ogx + ρH_ggh
P_A + (9.81*900)(x + 0.15) = P_B + (900*9.81*x) + (9.81*13600*0.15)
P_A - P_B = 20012.4 - 1324.35
P_A - P_B = 18688.05 N/m²

Flammable Gas:

• Flammable Gas is a gas that burns in the presence of an oxidant when provided with a source of ignition.
• Flammable gasses can include methane, butane, acetylene, ammonia, hydrogen, propane, and propylene.
• Flammable Gases are explosive in nature when they are mixed with air or oxygen in the right proportion.

Concept:
Kaplan Turbine:

• Kaplan turbine is an Axial Flow Reaction Turbine.
• It offers High Discharge and Low Head.
• In this turbine water flow parallel to the axis of rotation of the shaft.

Discharge (Q) for a Kaplan Turbine is given by

where D_o = Tip Diameter, D_h = Hub Diameter,V_f1 = Flow velocity

Flow Ratio: The ratio of Velocity of Flow at the inlet (V_f1) to the velocity given √2gh, is known as flow ratio.

Flow Ratio,

where h = Head on Turbine

Calculation:

Given:

Power Output, P = 2000 kW, Head, h = 8 m, Speed, N = 100 rpm, Discharge, Q = 250 m³/s, Tip Diameter, D_o = 6.5 m, Hub to Tip Ratio = 0.43

Discharge,

V_f1 = 9.243 m/s

Flow Ratio,

Flow Ratio, Ψ = 0.737

Concept:

• The stepper motor is a device that produces rotation through equal angles, the so-called steps, for each digital pulse supplied to its input.
• Stepper motors can be used to give controlled rotational steps but also can give continuous rotation with their rotational speed controlled by controlling the rate at which pulses are applied to it to cause stepping. This gives a very useful controlled variable speed motor which finds many applications.

Additional InformationStepper Motors:

• The name stepper is used because this motor rotates through a fixed angular step in response to each input current
• pulse received by its controller.
• As industrial motors are used to convert electric energy into mechanical energy but they cannot be used for precision positioning of an object or precision control of speed without using closed-loop feedback.
• Stepping motors are ideally suited for situations where either precise positioning or precise speed control or both are required in automation systems.
• Apart from stepping motors, other devices used for the above purposes are synchros and resolvers as well as dc/ac servomotors.
• The unique feature of a stepper motor is that its output shaft rotates in a series of discrete angular intervals or steps, one step being taken each time a command pulse is received. When a definite number of pulses are supplied, the shaft turns through a definite known angle.
• Stepper motors rotate in fixed steps. They use open-loop control and are operated by having a controller generate pulses that are input to a driver, which in turn supplies the drive current to the motor.
• Such motors develop torques ranging from 1 µN-m (in a tiny wristwatch motor of 3 mm diameter) up to 40 N-m in a motor of 15 cm diameter suitable for machine tool applications.
• Their power output ranges from about 1 W to a maximum of 2500 W.
• The only moving part in a stepping motor is its rotor which has no windings, commutator, or brushes.
• This feature makes the motor quite robust and reliable.
• Applications: Used for operation control in computer peripherals, the textile industry, IC fabrications, robotics, etc.

Concept:
The power in the jet with respect to a datum at the jet will be

Calculation:
Given Diameter (d) = 120mm = 0.12 m, Velocity (V) = 40 m/s, ρ = 1000 kg/m³;
A = π/4 × d² = 0.01131m²
Power in the jet with respect to datum is given by
P = 1/2 × ρ × A × V³
P = 1/2 × 1000 × 0.01131 × (40³) = 362.0571 kW
The nearest option is 360 kW

Concept:

Streamline is a line everywhere tangent to the velocity vector at a given instant

Path line is the actual path traversed by a given fluid particle.

The streak line is the locus of particles that have earlier passed through a prescribed point.

For steady flow, streamlines, pathlines, and streak lines are identical because:

• For a steady flow, the velocity vector at any point is invariant with time.
• The path line of the particles with different identities passing through a point will not differ.
• The path line could coincide with one another in a single curve which will indicate the streak line too.

So, both the statements are correct.

Concept:

Starting Jet Valve:

Starting jet valves are essentially part of a carburettor in some automotive engines. They control the fuel flow in the starting throttle position providing a richer mixture, which is important for cold starts. However, these are not manually operated. They work automatically, depending on the throttle position and engine requirements, so this is not the correct answer.

Compensating Jet Valve:

Compensating jet valves are also part of the carburettor setup, used more for carburettor calibration and maintaining the proper air-fuel mixture across different driving conditions. They aid in delivering the fuel at varying amounts based on the throttle position and engine load but are not specifically provided for starting in cold weather.

Choke Valve:

The choke valve is a device that restricts the airflow in the carburettor of an engine to enrich the fuel-air mixture at startup, especially in cold conditions. When the choke is activated, less air flows into the carburettor. This increases the amount of fuel in the mixture that gets into the engine. The enriched mixture improves combustion on cold engine start due to the higher fuel volume. As the engine warms up to optimal operating temperature, the choke is gradually opened (or automatically regulated in some systems) allowing more air into the carburettor and hence returning to a normal fuel/air mixture for efficient operation.

Auxiliary Air Valve:

An auxiliary air valve, also known as an idle air control valve in modern fuel-injected engines, is designed to allow additional airflow into the engine during cold start conditions to regulate the engine's idle speed. Their operation is reliant on engine temperature sensors; the colder the engine is, the more air the valve allows to bypass the throttle body to raise the idle speed. As the engine warms, the valve progressively closes until it fully closes when the engine is at operating temperature. Despite this valve being involved in cold start scenarios, it doesn't get manually operated to start the engine in cold conditions, rather it is an automatic process.

In summary, it's the choke valve that is manually used during cold weather conditions to aid in starting an engine. It helps increase the concentration of fuel in the air-fuel mixture making it easier to ignite and starting the engine more reliably in cold conditions.

Concept:
Cooling tower: It is a heat rejection device that used to cool the water stream by evaporating some of the water flow into an air stream.
Adiabatic evaporative cooling: Water cools due to evaporation and air gets heated and humidified. It indicates Adiabatic Humidification in the figure below.

Sensible cooling: It is the process of extracting sensible heat to the air with an evaporator. The wet-bulb temperature and total heat content of the air decreases and the relative humidity increases. There is no change in dew point or moisture content.

Humidification/Dehumidification: ω (moisture content) increases in humidification and decreases in dehumidification but DBT remains constant.

Chemical dehumidification: In this process, Dry Bulb Temperature (DBT) increases and specific humidity (ω) decreases.

Important Point

• In a cooling tower, the temperature of the air at the inlet is Wet Bulb Temperature.

Concept:

A heliostat is a device that includes a mirror, usually a plane mirror, which turns so as to keep reflecting sunlight toward a predetermined target, compensating for the sun's apparent motions in the sky.

Heliostats are also defined as the reflected mirrors which are employed for exploiting solar energy

Uses of Heliostat:

• Heliostats are used for day lighting or for the production of concentrated solar power, usually to generate electricity.
• They are also sometimes used in solar cooking.
• A few are used experimentally, or to reflect motionless beams of sunlight into solar telescopes.
• Before the availability of lasers and other electric lights, heliostats were widely used to produce intense, stationary beams of light for scientific and other purposes.

CONCEPT:

• Stress: Stress is the ratio of the load or force to the cross-sectional area of the material to which the load is applied.
  • The standard unit of stress is N/m².
• Strain: Strain is a measure of the deformation of the material as a result of the force applied.
  • The strain is a unitless quantity.
• Hooke's law states that within the proportional limit the stress applied on a body is directly proportional to strain produced.

⇒ Strain ∝ Stress
⇒ Stress = E × Strain (Where E = modulus of elasticity)
⇒ σ = E ϵ

Calculation:
Given:

Percentage by volume fibre = 65%

Percentage by volume epoxy resin = 35%

Relative density of the fibre = 1.48

Relative density of resin =1.2

% volume of fibre = V_F/V_F + V_R

% volume of resin = V_R/V_R + V_F

Also, Volume = mass/density

∴ m_F/m_R = 2.2904
m_F + m_R = 100

∴ m_F = 69.609 ≃ 70
Hence, the percentage weight of fibre will be nearly 70%.

Steam Power Plant:

A steam power plant continuously converts the energy stored in fossil fuels (coal, oil, natural gas) or fissile fuels (uranium, thorium) into shaft work and ultimately into electricity.

The working fluid is water which is sometimes in the liquid phase and sometimes in the vapor phase during its cycle of operations.

Rankine Cycle:

Rankine cycle:

Rankine cycle is an ideal cycle of heat engine which uses water and steam as a working fluid to generate power with the help of a steam turbine.

Supercritical Boiler:

• A steam generator or boiler which operates at pressures above then the critical pressure of working fluid (220 bar for water) is called a supercritical boiler.
• Critical pressure is the pressure at which liquid water immediately converts into steam without any latent heat transfer.
• They are also called a once-through boiler because the entire process of heating, steam formation and superheating is done in a single continuous tube. Feeds pumps feed water to economizer from where it goes to the radiant heating section at the pressure above then the critical pressure. So it does not contain headers and steam separators
• It is a drumless boiler.
• Benson boiler is an example of a supercritical boiler.

Deaerator

• Deaerator removes dissolved gases like O2, CO2 etc. from feed water which is used for steam generation in the boiler.
• It is done by heating water to its boiling temperature.

Concept:
Slip:
Slip of a pump is defined as the difference between the theoretical discharge and actual discharge of the pump.

Discharge by single-acting reciprocating pump:
Q_theroretical = ALN/60

Calculation:
Given:
Qactual = 0.00736 m³/s, N = 50 rpm, d = 200 mm, L = 300 mm.

Concept:

The Archimedes principle states that the buoyant force on a submerged body is equal to the weight of liquid displaced by the body and acts vertically upward through the centroid of the displaced volume.

If the body get sinked then,

Buoyancy force = Weight of the bodies

Calculation:

Given:

γ = 25 kN/m³, Volume of beam = 0.1 m³, SG = 0.6, g = 9.81 m/s²

As per the problem, the concrete, as well as beam, get sinked, therefore,

Buoyancy force = Weight of the bodies

Let the volume of concrete be 'V' m³

ρ_waterV_totalg = γV_concrete + ρ_beamV_beamg

10³ × (V + 0.1) × 9.81 = (25 × 10³ × V) + (0.6 × 10³ × 0.1 × 9.81)

9810V + 981 = 25000V + 588.6

(25000 - 9810)V = 981 - 588.6 ⇒ 392.4

15190V = 392.4

V = 0.02583278472 m³.

Weight of concrete = γV_concrete

Weight of concrete = 25 × 0.02583278472 = 0.645 kN

Concept:
Interplanar spacing:

where a = lattice constant.
Calculation:
Given:
d_hkl = d₂₂₀, h = 2, k = 2, l = 0, and a = 0.361 nm.
Interplanar spacing: