Previous Year Paper

UPPSC AE Mechanical 2013 Official Paper II (Previous Year Paper)

Held in 2013

100 questions120 minutes
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Mechanical Engineering (100 questions)

Question 1

Water at 42°C is sprayed into a stream of air at atmospheric pressure, dry bulb temperature of 40°C and a wet bulb temperature of 20°C. The air leaving the spray humidifier is not saturated. Which of the following statements is true?

  1. AOption (a)

    Air gets cooled and humidified

  2. BOption (b)

    Air gets heated and humidified

  3. COption (c)

    Air gets heated dehumidified

  4. DOption (d)

    Air gets cooled and dehumidified

View answer & explanation

Correct answer

Answer: (b)B.

Air gets heated and humidified

Why this answer is correct

Concept:

Tsp = Temperature of water sprayed = 42°C, DBT = Dry bulb temperature of air = 40°C

WBT = Dry bulb temperature of air = 20°C

Now,

Here we see that Tsp > DBT, Hence air gets heated. 

Also, water is added to it, so air gets humidified.

Hence in this process air gets heated and humidified.

Question 2

In an ideal vapour compression refrigeration cycle, the specific enthalpy of refrigerant (kJ/kg) at the following stages is given as

Inlet of condenser = 283

Outlet of condenser = 116

Exit of evaporator = 232

The COP is

  1. AOption (a)

    2.27

  2. BOption (b)

    2.75

  3. COption (c)

    3.27

  4. DOption (d)

    3.75

View answer & explanation

Correct answer

Answer: (a)A.

2.27

Why this answer is correct

The given specific enthalpies are

Exit of evaporator (h1) = 232 kJ/kg, Inlet of condenser (h2) = 283 kJ/kg,

Exit of condenser (h3) = 116 kJ/kg

h4 = 116 kJ/kg      --- From p-h curve we can see that h3 = h4.

Now,   

\(COP =\frac{{Refrigerationeffect}}{{Workdone}}\)

COP=h1−h4h2−h1{\rm{COP}} = \frac{{{\rm{h}}1 - {\rm{h}}4}}{{{\rm{h}}2 - {\rm{h}}1}}

Substitute the values, we get

COP=232−116283−232{\rm{COP}} = \frac{{232 - 116}}{{283 - 232}}

COP=11651{\rm{COP}} = \frac{{116}}{{51}}

∴ COP = 2.27

Question 3

During the chemical dehumidification process of air

  1. AOption (a)

    dry bulb temperature and specific humidity decreases

  2. BOption (b)

    dry bulb temperature increases and specific humidity decreases

  3. COption (c)

    dry bulb temperature decreases and specific humidity increases

  4. DOption (d)

    dry bulb temperature and specific humidity increases

View answer & explanation

Correct answer

Answer: (b)B.

dry bulb temperature increases and specific humidity decreases

Why this answer is correct

Concept:

If a mass of moist air in an airtight vessel heated to a higher temperature, then relative humidity of the air decreases.

Important Points

Heating and Humidifying: A Simultaneous increase in both the dry bulb temperature and humidity ratio of the air.

Cooling and Dehumidifying: It involves the removal of water from the air as the air temperature falls below the dewpoint temperature.

Sensible heating: During this process, the moisture content of air remains constant and its temperature increases as it flows over a heating coil.

Sensible cooling: During this process, the moisture content of air remains constant, but its temperature decreases as it flows over a cooling coil.

In sensible heating/cooling, specific humidity is constant while relative humidity varies. In heating/cooling with chemical humidification/ dehumidification, both specific humidity and relative humidity varies.

Question 4

Dew point temperature is the temperature at which condensation begins when the air is cooled at constant

  1. AOption (a)

    Volume

  2. BOption (b)

    Entropy

  3. COption (c)

    Pressure

  4. DOption (d)

    Enthalpy

View answer & explanation

Correct answer

Answer: (c)C.

Pressure

Why this answer is correct

Explanation:

Dew point temperature (DPT): 

Water vapour in air exists in the superheated state (state 1) and the air is highly unsaturated. When it gets cool at constant pressure, it meets the saturation vapour curve where the moisture of condensation begins.

The temperature at this point is known as the dew point temperature.

Dry bulb temperature (DBT): 

It is the temperature of moist air measured by ordinary thermometer

Wet bulb temperature (WBT): 

It is the temperature of air shown by a thermometer whose bulb is covered with a wet cloth.

Additional Information

Normally for unsaturated air: DBT > WBT > DPT

But for saturated air: DBT = WBT = DPT

Question 5

For air with a relative humidity of 80% _____.

  1. AOption (a)

    Dry bulb temperature is less than the wet bulb temperature

  2. BOption (b)

    Dew point temperature is less than wet bulb temperature

  3. COption (c)

    Dew point and wet bulb temperatures are equal

  4. DOption (d)

    Dry bulb and dew point temperatures are equal

View answer & explanation

Correct answer

Answer: (b)B.

Dew point temperature is less than wet bulb temperature

Why this answer is correct

Concept:

The psychometric chart is

For air with Relative humidity 80%, the dew point will be less than wet bulb temperature.

General trend of temperatures is:

DPT ≤ WBT ≤ DBT

At saturation curve i.e. RH = 100%

DPT = WBT = DBT

Question 6

In the window air conditions the expansion device used is

  1. AOption (a)

    Capillary tube

  2. BOption (b)

    Thermostatic expansion valve

  3. COption (c)

    Automatic expansion valve

  4. DOption (d)

    Float valve

View answer & explanation

Correct answer

Answer: (a)A.

Capillary tube

Why this answer is correct

Concept:

Function of Expansion Device: It reduces the pressure from the condenser to the evaporator and it supplies the flow of refrigerant to the evaporator as per the desired capacity.

Capillary Tube: It is a narrow tube of constant cross-section area. The pressure drop in the capillary tube is achieved due to frictional resistance and due to the acceleration of the fluid in the tube. The pressure drop is directly proportional to the length of the capillary tube and inversely proportional to the diameter of the capillary tube. It is used in the domestic refrigerators, water coolers, and window Air conditioners.

Question 7

One ton of refrigeration is equivalent to SI unit of

  1. AOption (a)

    1 kW

  2. BOption (b)

    2.5 kW

  3. COption (c)

    3.5 kW

  4. DOption (d)

    5 kW

View answer & explanation

Correct answer

Answer: (c)C.

3.5 kW

Why this answer is correct

Concept:

  • A ton of refrigeration (TR), also called a refrigeration ton (RT), is a unit of power used, especially to describe the heat-extraction capacity of refrigeration and air conditioning equipment.
  • It is defined as the rate of heat transfer that results in the melting of 1 ton of pure ice at 0°C in 24 hours.
  • The unit conversions is:
  • 1 TR = 210 kJ/min = 3.5 kW
  • 1 kCal = 4.18 kJ
  • ∴ 1 kJ = 0.239 kCal
  • ∴ 210 kJ = 50.19 kCal
  • ∴ 1 TR = 50.19 kCal/min = 3011.52 kCal/hr

Question 8

Efficiency of a Carnot engine is 75%. If the cycle direction is reverse, COP of the heat pump working on reversed Carnot cycle will be

  1. AOption (a)

    1.33

  2. BOption (b)

    0.75

  3. COption (c)

    0.33

  4. DOption (d)

    1.75

View answer & explanation

Correct answer

Answer: (a)A.

1.33

Why this answer is correct

Concept:

Efficiency of Carnot cycle η=WQH=TH−TcTH=0.75\eta=\frac{W}{Q_H}=\frac{T_H-T_c}{T_H}=0.75

By reversing the Carnot cycle we can have either heat pump or refrigerator.

Calculation:

COP of heat pump = COP=QHW=THTH−TC=1η=10.75=1.33COP=\frac{Q_H}{W}=\frac{T_H}{T_H-T_C}=\frac{1}{\eta }=\frac{1}{0.75}=1.33

Question 9

As an index of comfort, the temperature of saturated air at which a person would experience the same feeling of comfort as experienced in the actual unsaturated environment is called the

  1. AOption (a)

    Comfort temperature

  2. BOption (b)

    Effective temperature

  3. COption (c)

    Wet bulb temperature

  4. DOption (d)

    Soothing temperature

View answer & explanation

Correct answer

Answer: (b)B.

Effective temperature

Why this answer is correct

Concept:

  • The temperature at which motionless saturated air would induce, in a sedentary worker wearing ordinary indoor clothing, the same sensation of comfort as that induced by the actual conditions of temperature, humidity, and air movement.
  • The effective temperature is used as a guide in air-conditioning practice, and, on the comfort chart (American Society of Heating and Air Conditioning Engineers), it appears as a family of curves that serves as one coordinate in defining comfort zones.

Question 10

If the specific humidity of most air remains the same but its DBT increases, its DPT

  1. AOption (a)

    remains the same

  2. BOption (b)

    increases

  3. COption (c)

    decreases

  4. DOption (d)

    may increase or decrease depending on its relative humidity

View answer & explanation

Correct answer

Answer: (a)A.

remains the same

Why this answer is correct

Every psychometric chart includes vertical lines that represent the dry bulb temperatures, diagonals line represents wet bulb temperature and curved lines represent relative humidity.

Dry Bulb Temperature: Actual temperature of gas or mixture of gases

Wet Bulb temperature: Temperature obtained by an accurate thermometer having a wick moistened with distilled water

Dew point temperature: Temperature at which the liquid droplets just appear when the moist air is cooled continuously.

Relative humidity along the saturation line is 100%.

From the Psychrometric Chart, we can conclude that at constant specific humidity, DPT remains constant and is independent of DBT.

Question 11

In a vapour compression cycle, a good refrigerator should have a

  1. AOption (a)

    large latent heat of vaporization at condenser pressure

  2. BOption (b)

    large latent heat at evaporator pressure

  3. COption (c)

    condenser pressure close to critical pressure

  4. DOption (d)

    low critical pressure

View answer & explanation

Correct answer

Answer: (b)B.

large latent heat at evaporator pressure

Why this answer is correct

Concept:

In a vapour compression cycle, a good refrigerator should have:

  • The refrigerant should have a low boiling point and low freezing point.
  • It must have low specific heat and high latent heat because high specific heat decreases the refrigerating effect per kg of refrigerant and high latent heat at low temperature increases the refrigerating effect per kg of refrigerant.
  • It must have high critical pressure and temperature to avoid large power requirements.
  • It should have a low specific volume to reduce the size of the compressor.
  • It must have high thermal conductivity to reduce the area of heat transfer in the evaporator and condenser.
  • It should be non-flammable, non-explosive, non-toxic and non-corrosive.
  • It should not have any bad effects on the stored material or food when any leak develops in the system.

Question 12

R-12 is preferred over R-22 in deep freezer because

  1. AOption (a)

    It has lower operating pressure

  2. BOption (b)

    It gives higher COP

  3. COption (c)

    It is miscible with oil over a large range of 

  4. DOption (d)

    All of the above

View answer & explanation

Correct answer

Answer: (c)C.

It is miscible with oil over a large range of 

Why this answer is correct

Concept:

R-12 or Freon 12 is used in domestic refrigerators and freezers, liquid chillers, dehumidifiers, ice makers, water coolers, water fountains and transport refrigeration. The wide range of applications of the refrigerant are due to its safe properties.

Properties of R12 and its Advantages

  1. Safe properties: Refrigerant R12 is nontoxic, nonflammable, and non-explosive.

  2. Suitable for a wide range of operating conditions: R12 has the boiling point of -29.8°C due to which it condenses at the moderate pressures at the atmospheric temperature. This means the discharge pressure of the compressor should be only moderate. This helps in using the compressor of a low compression ratio that has higher efficiency.

  3. Miscibility with oil: Refrigerant R12 is miscible with the compressor oil under all the operating conditions. So, there is no problem of the oil return back to the compressor.

R12 has a low refrigerating effect.

Question 13

Low grade fuels have

  1. AOption (a)

    low moisture content

  2. BOption (b)

    low ash content

  3. COption (c)

    low calorific value

  4. DOption (d)

    high carbon content

View answer & explanation

Correct answer

Answer: (c)C.

low calorific value

Why this answer is correct

Concept:

The thermo technical characteristics of low grade fuels:

  • high moisture and ash content
  • low calorific value
  • sticking and congealing during transportation
  • fragility
  • small fractional composition

Question 14

Which of the following does not use ambient air for propulsion?

  1. AOption (a)

    Turbo jet

  2. BOption (b)

    Pulse jet

  3. COption (c)

    Turbo-prop

  4. DOption (d)

    Rocket

View answer & explanation

Correct answer

Answer: (d)D.

Rocket

Why this answer is correct

1. Turbo jet:

  • The turbojet is an air breathing jet engine, typically used in aircraft. It consists of a gas turbine with a propelling nozzle.
  • The gas turbine has an air inlet, a compressor, a combustion chamber, and a turbine (that drives the compressor).
  • The compressed air from the compressor is heated by the fuel in the combustion chamber and then allowed to expand through the turbine.
  • The turbine exhaust is then expanded in the propelling nozzle where it is accelerated to high speed to provide thrust.

2. Pulse jet:

  • The pulse jet is a type of jet engine in which combustion occurs in pulses.
  • A pulsejet engine can be made with few or no moving parts and is capable of running statically (i.e. it does not need to have air forced into its inlet typically by forward motion).
  • Pulsejets are used today in target drone aircraft, flying control line model aircraft.

3. Turbo-prop:**

  • In its simplest form a turboprop consists of an intake, compressor, combustor, turbine and a propelling nozzle. Air is drawn into the intake and compressed by the compressor. Fuel is then added to the compressed air in the combustor, where the fuel-air mixture then combusts.
  • The hot combustion gases expand through the turbine. Some of the power generated by the turbine is used to drive the compressor.

4. Rocket:

  • Rocket engines produce thrust by the expulsion of an exhaust fluid that has been accelerated to high speed through a propelling nozzle.
  • The fluid is usually a gas created by high pressure (150-to-4,350-pound-per-square-inch (10 to 300 bar)) combustion of solid or liquid propellants, consisting of fuel and oxidiser components, within a combustion chamber.
  • As the gases expand through the nozzle, they are accelerated to very high (supersonic) speed, and the reaction to this pushes the engine in the opposite direction.

Question 15

Humidity ratio can be given in terms of partial pressures of dry air (Pa) and water vapour (Pv) as

  1. AOption (a)

    0.622;(PvPa)0.622;\left( {\frac{{{P_v}}}{{{P_a} {}}}} \right)

  2. BOption (b)

    0.622;(PaPv)0.622;\left( {\frac{{{P_a}}}{{{P_v} {}}}} \right)

  3. COption (c)

    0.622;(PvP−Pa)0.622;\left( {\frac{{{P_v}}}{{{P} - {P_a}}}} \right)

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

0.622;(PvPa)0.622;\left( {\frac{{{P_v}}}{{{P_a} {}}}} \right)

Why this answer is correct

Concept:

Humidity ratio/Specific humidity (ω ) (kg/kg) of a given moist air sample is defined as the ratio of the mass of water vapour (mw) to the mass of dry air (ma) contained in the sample.

w=mwma=0.622PvP−Pvw = \frac{{{m_w}}}{{{m_a}}} = 0.622\frac{{{P_v}}}{{P - {P_v }}}

Total pressure is given by:

P = Pv + Pa

where p = atmospheric pressure, pa = partial pressure of dry-air and Pv = partial pressure of water-vapour.

Relative humidity (ϕ):

It is defined as the ratio of partial pressure of water vapour (pv)in a mixture to the saturation pressure (ps) of pure water at the same temperature of the mixture.

ϕ=pvpvs\phi = \frac{{{p_v}}}{{{p_{vs}}}}

Question 16

If the air is passed over the cooling coils then this process is termed as

  1. AOption (a)

    sensible heating

  2. BOption (b)

    cooling with humidification

  3. COption (c)

    cooling with dehumidification

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (c)C.

cooling with dehumidification

Why this answer is correct

Basic Processes in Conditioning of Air:

Sensible heating: Moisture content of air remains constant so specific humidity is constant, temperature increases as it flows over a heating coil

Sensible cooling: Moisture content of air remains constant so specific humidity is constant, but its temperature decreases as it flows over a cooling coil

Cooling and dehumidification: This process involves lowering of both the air temperature and the specific humidity. This process is commonly used in summer air conditioning in which air passes over a cooling coil. When moist air is cooled below its dew point, vapour is condensed from the air resulting in simultaneous cooling and dehumidification.

Heating and Humidification: During winter it is essential to heat and humidify the room air for comfort. It is done by first sensibly heating the air and then adding water vapour to the air stream through steam nozzles, as a result, both the temperature and humidity ratio of air increases.

Cooling & humidification: Air temperature drops and its humidity increases. 

Heating and de-humidification: This process can be achieved by using a hygroscopic material, which absorbs the water vapour from the moisture. If this process is thermally isolated, then the enthalpy of air remains constant, as a result, the temperature of air increases.

Question 17

COP of air refrigerator is related with COP of vapour compression refrigerator as

  1. AOption (a)

    (COP)air > (COP)vap.c.

  2. BOption (b)

    (COP)air < (COP)vap.c.

  3. COption (c)

    (COP)air = (COP)vap.c.

  4. DOption (d)

    None of these

View answer & explanation

Correct answer

Answer: (b)B.

(COP)air < (COP)vap.c.

Why this answer is correct

Concept:

  • The Vapour compression cycle consists of phase change from gas to liquid in condenser and the compressor pumps liquid phased refrigerant.
  • Making it more efficient as compression work per unit of mass flow is low. It is generally used in refrigeration purposes and air conditioning in buildings.
  • The Air cycle uses air as a working fluid and doesn't have phase change in any part of the cycle. Meaning air always remains in the gas state.
  • Hence the fluid circulation per KW of cooling is very high. Also, the compression work is more.
  • One of the very few applications that satisfy its suitability is air conditioning in aircraft.

So (COP)air < (COP)vap.c

Question 18

In an air craft refrigeration system the pressure at the cooling turbine outlet is equal to

  1. AOption (a)

    ambient pressure

  2. BOption (b)

    cabin pressure

  3. COption (c)

    pressure at inlet to compressor

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

cabin pressure

Why this answer is correct

Concept:

In an aircraft, cooling systems are required to keep the cabin temperatures at a comfortable level. Even though the outside temperatures are very low at high altitudes, still cooling of cabin is required due to:

  • Large internal heat generation due to occupants, equipment etc.
  • Heat generation due to skin friction caused by the fast moving aircraft.
  • At high altitudes, the outside pressure will be sub-atmospheric. When air at this low pressure is compressed and supplied to the cabin at pressures close to atmospheric, the temperature increases significantly.
  • For example, when outside air at a pressure of 0.2 bar and temperature of 223 K (at 10000 m altitude) is compressed to 1 bar, its temperature increases to about 353 K. If the cabin is maintained at 0.8 bar, the
  • temperature will be about 332 K. This effect is called as ram effect. This effect adds heat to the cabin, which needs to be taken out by the cooling system.

Question 19

During a simple sensible heating process, the relative humidity ______.

  1. AOption (a)

    Increases

  2. BOption (b)

    Decreases

  3. COption (c)

    Is zero

  4. DOption (d)

    Remains constant

View answer & explanation

Correct answer

Answer: (b)B.

Decreases

Why this answer is correct

Sensible heating: It is the process of increasing the Dry bulb temperature at constant specific humidity.

Effects of sensible heating:

i) Dry bulb temperature increases

ii) Sp. Humidity or humidity ratio remains constant

iii) Dew point temperature is constant

iv) Relative humidity decreases

v) Enthalpy increases

vi) Specific Volume Increases

vii) Wet bulb temperature increases

Question 20

The Kelvin-Planck statement of the second law of thermodynamics deals with the conversion of:

  1. AOption (a)

    heat into work

  2. BOption (b)

    work into heat

  3. COption (c)

    mass

  4. DOption (d)

    volume

View answer & explanation

Correct answer

Answer: (a)A.

heat into work

Why this answer is correct

Concept:

Kelvin-Plank gives the concept of heat engine hence, it is a conversion of heat into work. According to Kelvin-Planck statement of second law “It is impossible to construct a device (engine) operating in a cycle that will produce no effect other than extraction of heat from a single reservoir and convert all of it into work.”

Clausius statement: It is impossible to construct a system which will operate in a cycle, transfers heat from the low-temperature reservoir (or object) to the high-temperature reservoir (or object) without any external effect or work interaction with the surrounding.

Question 21

Thermodynamic work is the product of

  1. AOption (a)

    Two intensive properties

  2. BOption (b)

    Two extensive properties

  3. COption (c)

    An intensive property and change in an extensive property

  4. DOption (d)

    An extensive property and change in an intensive property

View answer & explanation

Correct answer

Answer: (c)C.

An intensive property and change in an extensive property

Why this answer is correct

Explanation:

Intensive Property: These are the properties of system which are independent of mass under consideration. For e.g. Pressure, Temperature, density

Extensive Properties: The properties which depend on the mass of system under consideration.

For e.g Internal Energy, Enthalpy, Volume, Entropy

Note: All specific properties are intensive properties. For e.g. specific volume, specific entropy etc.

Question 22

Air is compressed adiabatically in a steady flow process with negligible change in potential and kinetic energy. The work done in the process in given by

  1. AOption (a)

    −∫pdv- \smallint pdv

  2. BOption (b)

    +∫pdv+ \smallint pdv

  3. COption (c)

    −∫vdp- \smallint vdp

  4. DOption (d)

    +∫vdp+ \smallint vdp

View answer & explanation

Correct answer

Answer: (c)C.

−∫vdp- \smallint vdp

Why this answer is correct

Steady flow energy equation for a compressor from the above figure is given by:

h1 + dQ = h2 + dW ----(i)

Neglecting the changes of potential and kinetic energy. From the property relation

Tds = dh - νdp

For a reversible process,

Tds = dQ

So,                                          

dQ = dh - νdp    ----(ii)

If consider the process is reversible adiabatic then

dQ = 0

From equation (i) and (ii),      

h1 - h2 = dW  

dh = h2 - h1 = - dW    ----(iii)

And,                                        

dh = νdp    ----(iv)

From equation (iii) and (iv),   

-dW = νdp

W = -∫ νdp

Negative sign shows the work is done on the system (compression work) for initial and Final stage.

Question 23

A heat engine is supplied with 250 kJ/s of heat at constant fluid temperature of 227°C. The heat is rejected at 27°C. The cycle is reversible, if the amount of heat rejected is  

  1. AOption (a)

    273 kJ/s

  2. BOption (b)

    200 kJ/s

  3. COption (c)

    180 kJ/s

  4. DOption (d)

    150 kJ/s

View answer & explanation

Correct answer

Answer: (d)D.

150 kJ/s

Why this answer is correct

Concept:

For reversible heat engine efficiency

η=1−TRTS\begin{array}{l} \eta = 1 - \frac{{{T_R}}}{{{T_S}}}\end{array}

η=WQS=1−QRQS \begin{array}{l} \eta = \frac{W}{{{Q_S}}} = 1 - \frac{{{Q_R}}}{{{Q_S}}}\ \end{array}

Calculation:

η=1−TRTS=1−300500=0.4 η=WQS=1−QRQS QR=150;kW\begin{array}{l} \eta = 1 - \frac{{{T_R}}}{{{T_S}}} = 1 - \frac{{300}}{{500}} = 0.4\ \eta = \frac{W}{{{Q_S}}} = 1 - \frac{{{Q_R}}}{{{Q_S}}}\ {Q_R} = 150;kW \end{array}

∴ Q = 150 kW or 150 kJ/s

Question 24

The sequence of processes that eventually returns the working substance to its original state is known as

  1. AOption (a)

    Event

  2. BOption (b)

    Process

  3. COption (c)

    Thermodynamic property

  4. DOption (d)

    Thermodynamic cycle

View answer & explanation

Correct answer

Answer: (d)D.

Thermodynamic cycle

Why this answer is correct

Concept:

Thermodynamic cycle:

It can be defined as a series of state changes such that the final and initial state is identical.

Thermodynamic process:

It is a passage of a thermodynamic system from an initial state to the final state of the thermodynamic equilibrium. The initial and final states of the process are the defining elements. 

Thermodynamic property:

A thermodynamic property is any property that is measurable, and whose value describes a state of the system. Some e.g of the thermodynamic property are pressure, temperature, viscosity, density. Properties are point function i.e. these do not depend on the path followed.

Question 25

If the dryness fraction of a sample by throttling calorimeter is 0.8 and that by separating calorimeter is also 0.8, then the actual dryness fraction of sample will be taken as

  1. AOption (a)

    0.8

  2. BOption (b)

    0.8\sqrt {0.8}

  3. COption (c)

    0.64

  4. DOption (d)

    0.5

View answer & explanation

Correct answer

Answer: (c)C.

0.64

Why this answer is correct

Concept:

Throttling Calorimeter: A throttling calorimeter is based on the principle of throttling the wet steam so that it becomes superheated.

Separating Calorimeter: It is a vessel used initially to separate some of the moisture from the steam, to ensure superheat conditions after throttling.

Combined Separating and Throttling Calorimeter

A combined separating and throttling calorimeter is therefore found most suitable for accurate measurement of dryness of steam reducing the drawback of Separating Calorimeter (i.e. water particles from wet steam are not fully separated) and the drawback of Throttling Calorimeter (i.e. not suitable for very wet steam).

x1 = dryness fraction of steam considering separating calorimeter

x2 = dryness fraction of steam considering the throttling calorimeter

x = actual dryness fraction of steam in the sample

Calculation:

Actual dryness fraction of steam in the sample is:

X = x1 × x2

Given:             

x 1 = 0.8, x2 = 0.8   

Calculation:       

x = 0.8 × 0.8

∴ X = 0.64

Question 26

Thermodynamic equilibrium is completely defined by the specifications of

  1. AOption (a)

    Internal energy

  2. BOption (b)

    Enthalpy

  3. COption (c)

    Generalized displacements

  4. DOption (d)

    All of the above

View answer & explanation

Correct answer

Answer: (d)D.

All of the above

Why this answer is correct

Explanation:

The state of a system in which properties are definite as long as external conditions are unchanged is called an equilibrium state.

Thermodynamic equilibrium:

The system is said to be in thermodynamic equilibrium if the conditions for the following three equilibrium is satisfied:

  1. Mechanical equilibrium: 

When there are no unbalanced forces within the system and between the system and the surrounding, the system is said to be under mechanical equilibrium.

This means that the pressure of the system is con**stant everywhere.

  1. Chemical equilibrium: 

The system is said to be in chemical equilibrium when there are no chemical reactions going on within the system or there is no transfer of matter from one part of the system to other due to diffusion.

  1. Thermal equilibrium: 

When the temperature of the system is uniform and not changing throughout the system and also in the surroundings, the system is said to be in thermal equilibrium.

This means that the temperature of the system is constant everywhere.

A system is said to be in thermodynamic equilibrium if it is simultaneously in mechanical, thermal, and chemical equilibrium.

Therefore, thermodynamic equilibrium is completely defined by the specification of internal energy (since it is a function of temperature only), enthalpy (enthalpy (H) = Internal energy (U) + PV), and generalized displacements (i.e. due to unbalanced forces).

Question 27

Gas expands for a definite volume in a closed vessel. The maximum work will be done when the process is at constant

  1. AOption (a)

    Volume

  2. BOption (b)

    Temperature

  3. COption (c)

    Pressure

  4. DOption (d)

    Enthalpy

View answer & explanation

Correct answer

Answer: (c)C.

Pressure

Why this answer is correct

Explanation:

If a process starts from the same initial state and goes to the same final volume then work done will be maximum for an isobaric process, i.e. constant pressure process.

The different processes given in the P-V diagram are:

1: Isochoric process

2: Adiabatic process

3: Isothermal process

4: Isobaric process

∴ It is evident from the P-V diagram that the area under the curve is maximum during the constant pressure process.

Question 28

Which conversion is incorrect?

  1. AOption (a)

    1 kWh = 3.6 × 106 Nm

  2. BOption (b)

    1 Nm = 0.238 × 10-3 kcal

  3. COption (c)

    1 HP hr = 0.75 kWh

  4. DOption (d)

    1 kcal = 4.1868 Nm

View answer & explanation

Correct answer

Answer: (d)D.

1 kcal = 4.1868 Nm

Why this answer is correct

Concept:

• 1 kilowatt hour (kWh) = 3,600,000.00 newton meters (N-m)

• 1 newton meter (N-m) = 0.000238 kilocalories (kcal)

• 1 horsepower hour (Hp hr) = 0.75 kilowatt hours (kWh)

• 1 kilocalorie (kcal) = 4,186.80 newton meters (N-m)

Question 29

In an air standard Diesel cycle at fixed compression ratio and fixed value of adiabatic index (γ )

  1. AOption (a)

    thermal efficiency increases with increase in heat addition cut-off ratio

  2. BOption (b)

    thermal efficiency decreases with increase in heat addition cut-off ratio

  3. COption (c)

    thermal efficiency remains same with increase in heat addition cut-off ratio

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

thermal efficiency decreases with increase in heat addition cut-off ratio

Why this answer is correct

Concept:

\({\eta _{th}}; = ;1 - \frac{1}{{r^{ \gamma -1}}}\left{ {\frac{{r_c^\gamma - 1}}{{\gamma \left( {{r_c} - 1} \right)}}} \right}\)

From the above equation, it is observed that the thermal efficiency of the diesel engine can be increased by increasing the compression ratio r, by decreasing the cut - off ratio rc, or by using gas with a large value of γ.

Hence we can say that from the given options, the second option is correct because the thermal efficiency decreases with an increase in heat addition cut-off ratio.

Question 30

In Rankine cycle, the work output from the turbine is given by

  1. AOption (a)

    change of internal energy between inlet and outlet

  2. BOption (b)

    change of enthalpy between inlet and outlet

  3. COption (c)

    change of entropy between inlet and outlet

  4. DOption (d)

    change of temperature between inlet and outlet

View answer & explanation

Correct answer

Answer: (b)B.

change of enthalpy between inlet and outlet

Why this answer is correct

Concept:

Rankine cycle is a reversible cycle which have two constant pressure and two isentropic processes. These are four processes in the Rankine cycle:

Process 1 – 2: Isentropic compression

Process 2 – 3: Isobaric heat addition

Process 3 – 4: Isentropic expansion

Process 4 – 1: Isobaric heat rejection

Apply Steady Flow energy equation:

For boiler: Qin = h3 – h2

For turbine: Wturb,out = h3 – h4 (change of enthalpy between inlet and outlet)

For Condenser: Qout = h4 – h1

For Pump: Wpump,in = h2 – h1

Question 31

For a closed system, undergoing an expansion process according to the law PVn = constant, the work output.

  1. AOption (a)

    increases with increase in ‘n'

  2. BOption (b)

    increases with decrease in ‘n’

  3. COption (c)

    is maximum when n = 1

  4. DOption (d)

    is independent of ‘n’

View answer & explanation

Correct answer

Answer: (b)B.

increases with decrease in ‘n’

Why this answer is correct

Concept:

Polytropic Process is represented by

PVn = C

  • n = 0 ⇒ P = C ⇒ Constant Pressure Process (Isobaric Process)
  • n = 1 ⇒ PV = C ⇒ Constant Temperature Process (Isothermal process)
  • n = γ ⇒ PVγ = C ⇒ Adiabatic Process
  • n = ∞ ⇒ V = C ⇒ Constant Volume Process (Isochoric process)

From the above graph, we can see that in the expansion process, the work done increases as the value of n decreases.

Question 32

Law of degradation of energy says that unavailable energy is gradually decreasing due to

  1. AOption (a)

    increase in reversible processes

  2. BOption (b)

    increase in irreversible processes

  3. COption (c)

    increase in unavailable energy

  4. DOption (d)

    None of these

View answer & explanation

Correct answer

Answer: (a)A.

increase in reversible processes

Why this answer is correct

Concept:

  • The available energy of a system decreases as its temperature or pressure decreases and approaches to that of surroundings.
  • When heat is transferred from a system, its temperature decreases, and hence the quality of the energy also decreases.
  • The degradation is more for energy loss at a higher temperature.
  • Quantity wise the losses may be the same but the quality-wise the losses are different.
  • If the first law is the law of conservation of energy, then the second law is called the law of degradation of energy as it is the second law which tells us that energy always degrades quality-wise
  • The Law of degradation of energy says that unavailable energy is gradually decreasing due to an increase in reversible processes.

Available Energy: Maximum amount of work extracted from a heat input in a cycle.

Wmax (Available energy)= Qs-Tods

Unavailable energy: Minimum amount of energy that is rejected to a sink.

Unavailable energy=Tods

<sub> </sub> <sub> </sub>

As we can see in the above figure that when there is an increase in reversible processes (lesser will be irreversibility in the form of friction, heat dissipation) resulting in greater work output and decrease in heat rejection i.e. unavailable energy.

Question 33

For the same compression ratio, the efficiency of Brayton cycle is

  1. AOption (a)

    equal to that of Diesel cycle

  2. BOption (b)

    equal to that of Otto cycle

  3. COption (c)

    equal to that of Dual cycle

  4. DOption (d)

    greater than that of Diesel cycle

View answer & explanation

Correct answer

Answer: (b)B.

equal to that of Otto cycle

Why this answer is correct

Concept:

Brayton Cycle:

The Brayton cycle (or Joule cycle) represents the operation of a gas turbine engine.

The cycle consists of four processes

  • Process a - b: Reversible adiabatic compression in the inlet and compressor
  • Process b - c: Isobaric heat addition (fuel combustion)
  • Process c – d: Reversible adiabatic expansion in the turbine and exhaust nozzle
  • Process d - a: Cool the air at constant pressure back to its initial condition.
<br>

The efficiency of the Brayton cycle:

ηb=1−1(rp)γ−1γ{\eta _{\rm{b}}} = 1 - \frac{1}{(r_p)^{\frac{\gamma -1}{\gamma}}}

Where, rp = (pressure ratio), compression ratio = (Pb/Pa) = (Pc/Pd )

Otto cycle:

Otto cycle consists of four processes. They are as follows:

Process 1-2: Reversible adiabatic or Isentropic compression

Process 2-3: Constant Volume Heat Addition

Process3-4: Isentropic (reversible adiabatic) expansion.

Process 4-1: Constant Volume Heat Rejection

The efficiency of the Otto cycle:

ηotto=1−1rγ−1{\eta _{otto}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}

Where r = compression ratio =,,V2V1= ,,\frac{{V2}}{{V1}}

When compression ratio is the same in both the Otto cycle and Brayton cycle i.e. r = rp

Then, ηOtto = ηb.

∴ Option 2 is correct.

Question 34

If the temperature at the turbine inlet is kept constant, the net output of a simple gas turbine plant would

  1. AOption (a)

    increase with increasing pressure ratio

  2. BOption (b)

    decrease with increasing pressure ratio

  3. COption (c)

    first increase and then decrease with increasing pressure ratio

  4. DOption (d)

    remains unaffected with changes in pressure ratio

View answer & explanation

Correct answer

Answer: (c)C.

first increase and then decrease with increasing pressure ratio

Why this answer is correct

Concept:

The Brayton cycle is a theoretical cycle for simple gas turbine. Ideal Brayton cycle consists of two isentropic and two constant pressure processes.

As the plot depicts for a fixed turbine inlet temperature T3, the net work output per cycle increases with the pressure ratio initially, it reaches a maximum value and then it starts to decrease.                                                        

For a fixed values of Tmax and Tmin pressure ratio for maximum work done by gas turbine:

(rp )max work = (Tmax/Tmin)k/2(k-1)

where, k = specific heat ratio**, Tmax **= Maximum temperature in gas turbine cycle

 Tmin = Minimum temperature in the gas turbine cycle

Question 35

When the relationship between Reynolds number and the friction factor is represented by a straight line, the flow is said to be

  1. AOption (a)

    isentropic

  2. BOption (b)

    laminar

  3. COption (c)

    turbulent

  4. DOption (d)

    vortex

View answer & explanation

Correct answer

Answer: (b)B.

laminar

Why this answer is correct

Concept:

For a laminar fully developed flow through pipes, the friction factor, f is found from the exact solution of the Navier-Stokes equation is given by:

f=64Ref = \frac{{64}}{{Re}}

Where ,  f = Friction factor, Re = Reynolds Number

Thus, the friction factor is inversely proportional to Reynolds Number. So, after plotting in the log plain it is like a straight line coming downwards. So, that is what you see for the laminar flow in the moody chart.

Question 36

At the point of boundary layer separation

  1. AOption (a)

    Shear stress is maximum

  2. BOption (b)

    Shear stress is zero

  3. COption (c)

    Velocity is negative

  4. DOption (d)

    Density variation is maximum

View answer & explanation

Correct answer

Answer: (b)B.

Shear stress is zero

Why this answer is correct

  • When the pressure goes on increasing in the direction of flow, the pressure force acts against the direction of flow in the boundary layer and hence thickenings the boundary layer more rapidly.
  • This and the boundary shear bring the fluid in the boundary layer to rest and cause backflow.
  • Due to this the boundary layer no more sticks to the boundary but is shifted away from the boundary.

This phenomenon is called as “Boundary layer separation”.

A favorable pressure gradient is one in which the pressure decreases in the flow direction (i.e., dp/dx < 0)

On the other hand, an adverse pressure gradient is one in which pressure increases in the flow direction (i.e., dp/dx > 0)

At the verge of separation (∂u∂y)y=0{\left( {\frac{{\partial u}}{{\partial y}}} \right)_{y = 0}} is zero

∴ Shear stress τ=μ(∂u∂y)y=0\tau = \mu {\left( {\frac{{\partial u}}{{\partial y}}} \right)_{y = 0}} is also zero

Question 37

A potential function exists for

  1. AOption (a)

    steady flow only

  2. BOption (b)

    two dimensional irrotational flow only

  3. COption (c)

    irrotational flow of fluid whether compressible or incompressible

  4. DOption (d)

    irrotational flow of incompressible fluids only

View answer & explanation

Correct answer

Answer: (c)C.

irrotational flow of fluid whether compressible or incompressible

Why this answer is correct

Concept:

Velocity Potential function: This function is defined as a function of space and time in flow such that the negative derivation of this function with respect to any direction gives the velocity of the fluid in that direction.

If velocity potential (ϕ) exist, there will be a flow.

u=−∂ϕ∂xu = - \frac{{\partial ϕ }}{{\partial x}}

v=−∂ϕ∂yv = - \frac{{\partial ϕ }}{{\partial y}}

w=−∂ϕ∂zw = - \frac{{\partial ϕ }}{{\partial z}}

Now, Angular velocity is given by:

ωz=12(∂v∂x−∂u∂y){\omega _z} = \frac{1}{2}\left( {\frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}}} \right)

ωz=12[∂∂x(−∂ϕ∂y)−∂∂y(−∂ϕ∂x)]{\omega _z} = \frac{1}{2}\left[ {\frac{\partial }{{\partial x}}\left( { - \frac{{\partial ϕ }}{{\partial y}}} \right) - \frac{\partial }{{\partial y}}\left( { - \frac{{\partial ϕ }}{{\partial x}}} \right)} \right]

ωz=12[−∂2ϕ∂x∂y;+∂2ϕ∂y∂x;]{\omega _z} = \frac{1}{2}\left[ { - \frac{{{\partial ^2}ϕ }}{{\partial x\partial y}}; + \frac{{{\partial ^2}ϕ }}{{\partial y\partial x}};} \right]

Since ϕ is a continuous function

∂2ϕ∂x∂y=;∂2ϕ∂y∂x;\frac{{{\partial ^2}ϕ }}{{\partial x\partial y}} = ;\frac{{{\partial ^2}ϕ }}{{\partial y\partial x}};

Therefore, ωz  = 0

It is important to note that, for all irrotational flow, a velocity potential function ϕ exists whether the flow is compressible or incompressible.

Important points

  • Stream function exists only when compressibility is negligible, irrespectively of whether the motion is rotational or irrotational.
  • This is because the existence of stream function is merely a consequence of the assertion of the continuity of incompressible flow.

Question 38

Which property of mercury is the main reason for its use in barometers?

  1. AOption (a)

    High density

  2. BOption (b)

    Negligible capillarity effect

  3. COption (c)

    Very low vapour pressure

  4. DOption (d)

    Low compressibility

View answer & explanation

Correct answer

Answer: (c)C.

Very low vapour pressure

Why this answer is correct

Concept:

  • Mercury has very low vapor pressure and hence it is an excellent fluid to be used in a barometer.
  • Another reason for using mercury is it doesn't wet the glass, good conductor of heat and it expands evenly with temperature.

Question 39

In case of fluid flow through pipes, cavitation is caused by

  1. AOption (a)

    high pressure

  2. BOption (b)

    high velocity

  3. COption (c)

    low pressure below a limit

  4. DOption (d)

    weak material of pipe

View answer & explanation

Correct answer

Answer: (c)C.

low pressure below a limit

Why this answer is correct

Concept:

Cavitation is the phenomenon of formation of vapour bubbles of a flowing liquid in a region where the pressure of the liquid falls below the vapour pressure of the fluid and sudden collapsing of these bubbles in the region of higher pressure.

Cavitation:

  • It is the formation of gas bubbles of a flowing liquid in a region where the pressure of the liquid falls below its vapour pressure i.e. low pressure.
  • It is the phenomenon of formation of vapour bubbles of a flowing liquid in a region where the pressure of the liquid falls below the vapour pressure of the fluid and sudden collapsing of these bubbles in the region of higher pressure.
  • In centrifugal pumps, the cavitation may occur at the inlet of the impeller of the pump or at the suction side of the pumps, where the pressure is considerably reduced.
  • Hence if the pressure at the suction side of the pump drops below the vapour pressure of the liquid then the cavitation may occur.

Question 40

A stream function

  1. AOption (a)

    is a mathematical function which has no physical equivalence

  2. BOption (b)

    is defined only for steady and incompressible flow

  3. COption (c)

    satisfies laplace equation for rotational motion

  4. DOption (d)

    may not remain constant for a streamline

View answer & explanation

Correct answer

Answer: (b)B.

is defined only for steady and incompressible flow

Why this answer is correct

Concept:

The stream function in a two-dimensional flow automatically satisfies the continuity equation.

Stream Function: It is the scalar function of space and time.

The partial derivative of stream function with respect to any direction gives the velocity component perpendicular to that direction. Hence it remains constant for a streamline

u=−∂ψ∂yu = - \frac{{\partial \psi }}{{\partial y}} & v=∂ψ∂xv = \frac{{\partial \psi }}{{\partial x}}

Stream function defines only for the two-dimensional flow which is steady and incompressible..

Properties of stream function:

  1. If ψ exists, it follows continuity equation and the flow may be rotational or irrotational.
  2. If ψ satisfies the Laplace equation, then the flow is irrotational.
<br>

Continuity equation:

∂;u∂;x+∂;v∂;y=0\frac{{\partial {\rm{;u}}}}{{\partial {\rm{;x}}}} + \frac{{\partial {\rm{;v}}}}{{\partial {\rm{;y}}}} = 0

i.e.;∂;∂;x(−∂ψ∂y)+∂;∂;y(∂ψ∂x)=0i.e.;\frac{{\partial {\rm{;}}}}{{\partial {\rm{;x}}}}\left( { - \frac{{\partial \psi }}{{\partial y}}} \right) + \frac{{\partial {\rm{;}}}}{{\partial {\rm{;y}}}}\left( {\frac{{\partial \psi }}{{\partial x}}} \right) = 0

⇒  −∂2ψ∂x∂y+∂2ψ∂x∂y=0 - \frac{{{\partial ^2}\psi }}{{\partial x\partial y}} + \frac{{{\partial ^2}\psi }}{{\partial x\partial y}} = 0

Question 41

For the flow to occur between two points in a pipeline, the differential pressure between these points should be more than

  1. AOption (a)

    surface friction

  2. BOption (b)

    viscosity force

  3. COption (c)

    frictional force

  4. DOption (d)

    All of the above

View answer & explanation

Correct answer

Answer: (d)D.

All of the above

Why this answer is correct

Concept:

For the flow to occur between two points in a pipeline the differential pressure between these points should be more than

  • Surface friction
  • Viscosity force
  • Frictional force

So that the pressure force of fluid can overcome the surface friction, viscous effects and frictional resistances allowing the flow to take place in a pipeline between any two points in a pipeline.

Question 42

Fluid is flowing in a curved path without any external impressed contact force. This flow is known as

  1. AOption (a)

    free vortex flow

  2. BOption (b)

    forced vortex flow

  3. COption (c)

    radial flow

  4. DOption (d)

    spiral flow

View answer & explanation

Correct answer

Answer: (a)A.

free vortex flow

Why this answer is correct

Concept:

Free Vortex Flow: In free vortex flow the liquid particles describe circular paths about a fixed vertical axis without any external torque acting on them.

Example: The flow of water through the hole in the bottom of a washbasin.

Forced vortex flow: In this type of flow, the vessel containing the liquid is forced to rotate about the fixed vertical axis with the help of some external torque.

Question 43

In flow through a pipe, the transition from laminar to turbulent flow does not depend on

  1. AOption (a)

    Density of fluid

  2. BOption (b)

    Length of pipe

  3. COption (c)

    Diameter of pipe

  4. DOption (d)

    Velocity of the fluid

View answer & explanation

Correct answer

Answer: (b)B.

Length of pipe

Why this answer is correct

Explanation:

Reynold's number: 

  • It is a dimensionless number that determines the nature of the flow of liquid through a pipe or flat plate.
  • It is defined as the ratio of the inertial force to the viscous force for a flowing fluid.

Reynold's number is written as Re.

Re=Inertial;forceViscous;force⇒ρVxμ=Vxν{Re} = \frac{{{\rm{Inertial;force}}}}{{{\rm{Viscous;force}}}}\Rightarrow\frac{ρ Vx}{μ}=\frac{Vx}{ν}

where ρ = density of fluid, V = velocity of flowing fluid, μ = dynamic viscosity of fluid, ν = kinematic viscosity of fluid and x = characteristic diameter.

where x=4APx=\frac{4A}{P}

and A = area of cross-section and P = perimeter.

Question 44

In the region of boundary layer on a flat plate surface where velocity is not zero, the viscous force is

  1. AOption (a)

    less than inertial force

  2. BOption (b)

    more than inertial force

  3. COption (c)

    equal in magnitude

  4. DOption (d)

    not predictable

View answer & explanation

Correct answer

Answer: (b)B.

more than inertial force

Why this answer is correct

Explanation:

  • In the boundary layer, viscous forces dominate over inertial forces.
  • The viscous forces are those associated with transverse velocity gradients such as in the boundary layer while inertial forces dominate outside the boundary layer, as the flow outside boundary layer is the inviscid flow region.

Question 45

The magnitude of water hammer in the flow of a liquid through a pipe does not depend upon

  1. AOption (a)

    length of pipe

  2. BOption (b)

    elastic properties of pipe material

  3. COption (c)

    temperature of liquid

  4. DOption (d)

    time of valve closure

View answer & explanation

Correct answer

Answer: (c)C.

temperature of liquid

Why this answer is correct

Concept:

When a liquid flowing through a long pipe is suddenly brought to rest by closing the valve at the end of a pipe, then a pressure wave of high intensity is produced behind the valve. This pressure wave of high intensity has the effect of hammering action on the walls of the pipe. This phenomenon is known as water hammer or hammer blow.

The magnitude of water hammer depends upon

(a) The length of the pipeline,

(b) The elastic properties of the pipe material,

(c) The elastic properties of the liquid flowing through the pipe, and

(d) The time at which the valve is closed.

Question 46

Compressibility effect can be treated as negligible when Mach number

  1. AOption (a)

    upto 0.2

  2. BOption (b)

    upto 0.5

  3. COption (c)

    less than 1

  4. DOption (d)

    1

View answer & explanation

Correct answer

Answer: (a)A.

upto 0.2

Why this answer is correct

Concept:

Mach;Number=Inertia;forceelastic;force=Inertia;forceCompressible;force{\rm{Mach;Number}} = \frac{{Inertia;force}}{{elastic;force}} = \frac{{Inertia;force}}{{Compressible;force}}

  • Incompressible flow
  • M < 0.3
  • From the above point, it is clear that the compressibility effect can be treated as negligible when the Mach number is up to 0.2
  • Subsonic flow
  • 0.3 < M < 1
  • Sonic flow
  • M = 1
  • Supersonic flow
  • 1 < M < 5
  • Hypersonic flow
  • M > 5

Question 47

A body is called streamline body when

  1. AOption (a)

    it is symmetrical about the axis along the free stream

  2. BOption (b)

    surface of the body coincides with the streamlines

  3. COption (c)

    flow is laminar around it

  4. DOption (d)

    it produces no drag for flow around it

View answer & explanation

Correct answer

Answer: (b)B.

surface of the body coincides with the streamlines

Why this answer is correct

Explanation:

streamlined body:

It is defined as the body whose surface coincides with the streamline when the body is placed in the flow. Flow separation takes place only at the trailing edge in such cases.

  • The boundary layer in such bodies must be attached to the surface of the body for a long time as possible and hence behind a streamlined body wake formation zone will be very small and consequently, pressure drag will be very small.
  • Thus, the total drag on the streamlined body will be due to friction (shear) only.
  • A streamlined body is a shape that lowers the friction drag between a fluid, like air and water, and an object moving through that fluid. Drag is a force that slows down motion. Friction drag is a special kind of drag.

Question 48

Mach number is the ratio of

  1. AOption (a)

    elastic force to gravity force

  2. BOption (b)

    viscous force to elastic force

  3. COption (c)

    inertial force to surface tension

  4. DOption (d)

    inertial force to compressibility force

View answer & explanation

Correct answer

Answer: (d)D.

inertial force to compressibility force

Why this answer is correct

Concept:

Mach number (M): It is defined as the ratio of inertia force to compressibility force.

M=Vc=VdPdρM = \frac{V}{c} = \frac{V}{{\sqrt {\frac{{dP}}{{d\rho }}} }}

  • Mach number is defined as the ratio of the speed of the fluid flow to the speed of sound in the same medium under similar conditions of temperature and pressure.
  • If Mach Number is greater than 1, the body is called as Supersonic whereas it is said to be moving with Subsonic speed, if it is less than 1.
  • If the Mach Number is greater than 5, the body is known as Hypersonic.

Reynolds number (Re): It is defined as the ratio of inertia force to viscous force

Re=ρVlμRe = \frac{{\rho Vl}}{\mu }

Euler number (Eu): It is defined as the ratio of inertia force.to pressure force.

Eu=VPρEu = \frac{V}{\sqrt{\frac{P}{\rho}}}

Froude number (Fr): It is defined as the ratio of inertia force to gravity force.         

Fr=VgLFr = \frac{V}{{\sqrt {gL} }}

Weber number (We): It is defined as the ratio of the inertia force to surface tension force.

We=ρV2lσWe = \frac{{\rho {V^2}l}}{\sigma }

Question 49

For a linear distribution of velocity in the boundary layer on a flat plate, the ratio of displacement thickness to nominal thickness is

  1. AOption (a)

    14\frac{1}{4}

  2. BOption (b)

    12\frac{1}{2}

  3. COption (c)

    18\frac{1}{8}

  4. DOption (d)

    16\frac{1}{6}

View answer & explanation

Correct answer

Answer: (b)B.

12\frac{1}{2}

Why this answer is correct

Concept:

Nominal boundary thickness as δ 

Displacement thickness is given as

\({δ ^*} = \mathop \smallint \limits_o^δ \left( {1 - \frac{u}{U}} \right)dy\)

Calculation:

\({δ ^*} = \mathop \smallint \limits_o^δ \left( {1 - \frac{y }{δ}} \right)dy\)

δ∗=δ−δ2{δ ^*} =δ - \frac{δ }{2}= δ2\frac{δ }{2}

∴ The ratio of displacement thickness to nominal thickness will be ​ = δ2δ\frac{{\frac{δ }{2}}}{δ }  

∴ After simplification, the answer will be 1/2.

Question 50

In case of laminar flow through pipe, the ratio of total kinetic energy of fluid passing per second to the energy value obtained on the basis of average velocity is

  1. AOption (a)

    1.2

  2. BOption (b)

    1.54

  3. COption (c)

    2

  4. DOption (d)

    2.37

View answer & explanation

Correct answer

Answer: (c)C.

2

Why this answer is correct

Concept:

The governing equation is

uu0=1−(rR)2\frac{u}{{{u_0}}} = 1 - {\left( {\frac{r}{R}} \right)^2}

Calculation:

The relation between the maximum velocity (um) and the average velocity (u0) is

um=u02=0.5u0{u_m} = \frac{{{u_0}}}{2} = 0.5{u_0}

Now,

Let α be the ratio of the total kinetic energy of fluid passing per second to the energy value obtained on the basis of average velocity.

∴ α=∫u3dAAum3α = \frac{{\smallint {u^3}dA}}{{Au_m^3}}

∴ α=∫u0[1−(rR)2]3(2πr)drπR2(0.5)3u03α = \frac{{\smallint {u_0}{{\left[ {1 - {{\left( {\frac{r}{R}} \right)}^2}} \right]}^3}\left( {2\pi r} \right)dr}}{{\pi {R^2}{{\left( {0.5} \right)}^3}u_0^3}}

Integrating the above equation

∴ α = 2

Note: The derivation of the above equation is a standard result and can be remembered directly as a relation.

Question 51

Sonic velocity will have a low value in the medium having

  1. AOption (a)

    low value of coefficient of compressibility

  2. BOption (b)

    high value of coefficient of compressibility

  3. COption (c)

    high bulk modulus of elasticity

  4. DOption (d)

    homogenous composition

View answer & explanation

Correct answer

Answer: (b)B.

high value of coefficient of compressibility

Why this answer is correct

Concept:

The acoustic velocity can alternatively be expressed with Hook's Law as

C = (K / ρ)1/2

Where, C = speed of sound, K = Bulk modulus of elasticity, ρ = density of medium

C ∝ K1/2

Now, Coefficient of compressibility is defined as inverse of Bulk modulus of elasticity,

β=1K{\rm{\beta }} = \frac{1}{K}

Hence,                                    

C∝β−12{\rm{C}} \propto {{\rm{\beta }}^{ - \frac{1}{2}}}

Therefore**,**

Sonic velocity in fluid is

  • Directly proportional to square root of Bulk modulus of medium.
  • Inversely proportional to square root of coefficient of compressibility of medium.

Question 52

An isentropic flow is one which is

  1. AOption (a)

    adiabatic and reversible

  2. BOption (b)

    isothermal only

  3. COption (c)

    adiabatic only

  4. DOption (d)

    adiabatic and irreversible

View answer & explanation

Correct answer

Answer: (a)A.

adiabatic and reversible

Why this answer is correct

Concept:

  • An adiabatic process is a process in which there is no heat interaction or heat transfer through the system boundary. The adiabatic process can be reversible or irreversible.
  • However, an isentropic process is a process during which entropy of the system must remain constant, which means entropy does not change.
  • In fluid dynamics, an isentropic flow is a fluid flow that is adiabatic as well as reversible i.e. no heat is added to the flow and no energy transformations take place due to friction or any dissipative effects.

Question 53

The size of a venturimeter is specified by

  1. AOption (a)

    fluid pressure

  2. BOption (b)

    discharge

  3. COption (c)

    pipe diameter and throat diameter 

  4. DOption (d)

    length of venturimeter 

View answer & explanation

Correct answer

Answer: (c)C.

pipe diameter and throat diameter 

Why this answer is correct

Concept:

  • A venturimeter is a device used for measuring the rate of flow of a fluid of a liquid flowing through a pipe

  • The venturimeter always have a smaller convergent portion and larger divergent portion

  • The size of the venturimeter is specified by its pipe diameter as well as throat diameter.

  • This is done to ensures a rapid converging passage and a gradual diverging passage in the direction of flow to avoid the loss of energy due to separation

  • In the course of flow through the converging part, the velocity increases in the direction of flow according to the principle of continuity, while the pressure decreases according to Bernoulli’s theorem

  • The velocity reaches its maximum value and pressure reaches its minimum value at the throat

  • Subsequently, a decrease in the velocity and an increase in the pressure take place in course of flow through the divergent part

  • The angle of convergence ≈ 20°, Angle of divergence = 6° - 7° → It should be not greater than 7° to avoid flow separation

Question 54

In a flow field at the stagnation point

  1. AOption (a)

    pressure is zero

  2. BOption (b)

    total energy is zero

  3. COption (c)

    pressure head is equal to velocity head

  4. DOption (d)

    All the velocity head is converted into pressure head

View answer & explanation

Correct answer

Answer: (d)D.

All the velocity head is converted into pressure head

Why this answer is correct

Concept:

  • A stagnation point is a point in a flow field where the local velocity of the fluid is zero.
  • The Bernoulli equation shows that the static pressure is highest when the velocity is zero and hence static pressure is at its maximum value at stagnation points. Because all the velocity head is converted into the pessure head at stagnation point.
  • This static pressure is called the stagnation pressure.

Question 55

Which two forces are most important in laminar flow between parallel plates?

  1. AOption (a)

    Inertial and viscous

  2. BOption (b)

    Viscous and pressure

  3. COption (c)

    Gravity and pressure

  4. DOption (d)

    Pressure and inertial

View answer & explanation

Correct answer

Answer: (b)B.

Viscous and pressure

Why this answer is correct

Explanation:

In laminar flow between parallel plates, the two most important forces are viscous forces and pressure forces.

Viscous forces arise due to the friction between the fluid layers and act in the direction of decreasing velocity. They are responsible for the parabolic velocity profile observed in laminar flow, where the velocity is highest at the center of the channel and decreases towards the walls.

Pressure forces are the forces acting due to the difference in pressure between the two plates. They are responsible for driving the flow from the high-pressure region to the low-pressure region.

The other options are incorrect because:

Option 1: Inertial and viscous is not correct because inertial forces are only significant in turbulent flow, where the fluid motion is highly chaotic and irregular. In laminar flow, inertial forces are negligible.

Option 3: Gravity and pressure is not correct because gravity is only significant in flow situations where the fluid density is not uniform, such as in natural convection. In laminar flow between parallel plates, the fluid density is assumed to be uniform, and gravity does not play a significant role.

Option 4: Pressure and inertial are not correct because inertial forces are not significant in laminar flow, as explained earlier. The most important forces in laminar flow between parallel plates are viscous forces and pressure forces.

Question 56

A high value of thermal diffusivity represents

  1. AOption (a)

    high storage, less conduction of heat

  2. BOption (b)

    less storage, more conduction of heat

  3. COption (c)

    There is always equal amount of conduction and storage since it is a property

  4. DOption (d)

    It has no relavance

View answer & explanation

Correct answer

Answer: (b)B.

less storage, more conduction of heat

Why this answer is correct

Concept:

Thermal diffusivity of material is given as, α=kρc\alpha = \frac{k}{{\rho c}}. It is the property of a material.

Larger the value of α, faster heat will diffuse through the material. It means less heat will be stored and more conduction will occur.

A high value of α could result either from a high value of thermal conductivity or low value of thermal heat capacity ρc. Thermal diffusivity α has units of square meters per second (m2/s).

Question 57

What happens when the thickness of insulation on a pipe exceeds the critical value?

  1. AOption (a)

    Heat transfer rate increases

  2. BOption (b)

    Heat transfer rate decreases

  3. COption (c)

    Heat transfer rate remains constant

  4. DOption (d)

    None of these

View answer & explanation

Correct answer

Answer: (b)B.

Heat transfer rate decreases

Why this answer is correct

Concept:

Pipe with insulation can be assumed as cylinder. 

  • The value of Router for which the heat transfer rate is maximum is called the critical radius of insulation.
  • The thickness up to which heat flow increases and after which heat flow decreases is termed as Critical thickness.
  • The insulation radius at which resistance to heat flow is minimum and consequently heat flow rate is maximum is called “critical radius”. And it starts decreasing after the critical thickness of insulation.

Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h.

The critical radius of insulation for a cylindrical body:

rcr,cylinder=kh{r_{cr,cylinder}} = \frac{k}{h}

Critical radius of insulation for a spherical shell:

rcr,sphere=2kh{r_{cr,sphere}} = \frac{{2k}}{h}

The rate of heat transfer from the cylinder increases with the addition of insulation for r2 < rcr, reaches a maximum when r2 = rcr, and starts to decrease for r2 > rcr. Thus, insulating the pipe may increase the rate of heat transfer from the pipe instead of decreasing it when r2 < rcr.

Question 58

For flow of fluid over a heated plate, the following fluid properties are known:

Viscosity = 0.001 Pa.s, Sp. Heat at constant pressure = 1 kJ/kg-K, thermal conductivity = 1W/mK.

The hydrodynamic boundary layer thickness at a specified location on the plate if 1 mm, the thermal boundary layer thickness at the same location is

  1. AOption (a)

    0.001 mm

  2. BOption (b)

    0.01 mm

  3. COption (c)

    1 mm

  4. DOption (d)

    10 mm

View answer & explanation

Correct answer

Answer: (c)C.

1 mm

Why this answer is correct

Concept:

Prandtl Number (Pr) = kinematic;viscositythermal;diffusivity\frac{{kinematic;viscosity}}{{thermal;diffusivity}}

Calculation:

∴Pr=να ∴Pr=;μCpk ∴Pr=0.001×10001\begin{array}{l} \therefore Pr = \frac{\nu }{\alpha }\ \therefore Pr = ;\frac{{\mu C{\rm{p}}}}{k}\ \therefore {\rm{Pr}} = \frac{{0.001 \times 1000}}{1} \end{array}

∴ Pr = 1

Where**, µ** = dynamic viscosity, Cp = Heat at constant pressure, k = Thermal Conductivity

Also, 

δδt\frac{\delta }{{\delta t}} = Pr1/3      

∴ 𝞭 = 𝞭t   (∵ Pr = 1) 

∴ Thermal boundary layer thickness: 𝞭t = 1 mm

Question 59

Which one of the following configurations has the highest fin effectiveness?

  1. AOption (a)

    Thin, closely spaced fins

  2. BOption (b)

    Thin, widely spaced fins

  3. COption (c)

    Thick widely spaced fins

  4. DOption (d)

    Thick, closely spaced fins

View answer & explanation

Correct answer

Answer: (a)A.

Thin, closely spaced fins

Why this answer is correct

Explanation:

Effectiveness (ϵ)

Effectiveness is defined as the ratio of heat transfer rate with a fin to heat transfer rate without fin.

\(\epsilon = \frac{{{{\dot Q}{with;fin}};}}{{{{\dot Q}{without;fin}}}}\)

For a long fin

\(\epsilon = \frac{{{{\dot Q}{with;fin}};}}{{{{\dot Q}{without;fin}}}} = \sqrt {\frac{{KP}}{{h{A_c}}}} \)

where K = thermal conductivity of fin, P = perimeter of fin

h = heat transfer coefficient, Ac = cross-section area of fin

From the above formula, we can conclude:

Case 1: If PAc\frac{P}{{{A_c}}} increases, the effectiveness of fin increases.

  • Ac must be small (thin fin) and fins must be closed (not too close).
  • If fins are too closed it will abstract flow of air which will decrease the effectiveness of fin.

Case 2: if K increases, the effectiveness of fin increases.

  • If K is large, the temperature drop along the length will be less.
  • And the temperature drop between fin and surrounding will be more and heat transfer will be more.

Case 3: if h decreases, the effectiveness of fin increases.

  • Fins are more effective under free convection ( h is less).

Question 60

In a condenser of a power plant, the steam condenses at a temperature of 60o C. The cooling water enters at 30o C and leaves at 45o C. The logarithmic mean temperature difference (LMTD) of the condenser is

  1. AOption (a)

    16.2°C

  2. BOption (b)

    21.6°C

  3. COption (c)

    30°C

  4. DOption (d)

    37.5°C

View answer & explanation

Correct answer

Answer: (b)B.

21.6°C

Why this answer is correct

Concept:

Flow configuration in the condenser as shown below.

ΔT1 = Th,i - Tc,i,  ΔT2 = Th,o - Tc,o

LMTD=ΔT1 − ΔT2lnΔT1ΔT2LMTD=\frac{{\Delta}T_1~-~{\Delta}T_2}{ln{\frac{\Delta{T_1}}{\Delta{T_2}}{}}}

 

Calculation:

Given:

Th,i = Th,o = 60oC, Tc,i = 30oC, Tc,o = 45oC

 Therefore, ΔT1=30∘C,ΔT2=15∘C,{\rm{Δ }}{T_1} = 30^\circ C,{\rm{Δ }}{T_2} = 15^\circ C,

Now, we know that LMTD=ΔT1−ΔT2In(ΔT1ΔT2)=30−15In(3015)=21.6∘CLMTD = \frac{{{\rm{Δ }}{T_1} - {\rm{Δ }}{T_2}}}{{In\left( {\frac{{{\rm{Δ }}{T_1}}}{{{\rm{Δ }}{T_2}}}} \right)}} = \frac{{30 - 15}}{{In\left( {\frac{{30}}{{15}}} \right)}} = 21.6^\circ C

Question 61

In a heat exchanger, the temperature of the hot fluid decreases while the temperature of the cold fluid increases. The increase and decrease following:

  1. AOption (a)

    A quadratic law

  2. BOption (b)

    A linear law

  3. COption (c)

    A cubic law

  4. DOption (d)

    An exponential law

View answer & explanation

Correct answer

Answer: (d)D.

An exponential law

Why this answer is correct

Concept:

In a heat exchanger, the temperature of the hot fluid decreases while the temperature of the cold fluid increases. The increase and decrease follow an exponential law.

It can be parallel as well as counter flow.

Now,

Effectiveness of a parallel flow heat exchanger:

ϵparallel=1−exp⁡[−NTU(1+R)]1+R{\epsilon_{parallel}} = \frac{{1 - \exp \left[ { - NTU\left( {1 + R} \right)} \right]}}{{1 + R}}

Effectiveness of a counter flow heat exchanger:

ϵcounter=1−exp⁡[−NTU(1−R)]1−Rexp⁡[−NTU(1−R)]{\epsilon_{counter}} = \frac{{1 - \exp \left[ { - NTU\left( {1 - R} \right)} \right]}}{{1 - R\exp \left[ { - NTU\left( {1 - R} \right)} \right]}}

From the above explanation, we can clearly see that it follows the exponential law.

Question 62

This substance has the minimum value of thermal conductivity

  1. AOption (a)

    Air

  2. BOption (b)

    Water

  3. COption (c)

    Plastic

  4. DOption (d)

    Rubber

View answer & explanation

Correct answer

Answer: (a)A.

Air

Why this answer is correct

Concept:

Among the following options given value of thermal conductivity of air is minimum.

KAir≃0.02{K_{Air}} \simeq 0.02

The thermal conductivity of a material can be defined as the rate of heat transfer through a unit thickness of the material per unit area per unit temperature difference.

The thermal conductivity of a material is a measure of the ability of the material to conduct heat. A high value for thermal conductivity indicates that the material is a good heat conductor, and a low value indicates that the material is a poor heat conductor or insulator. T

The thermal conductivities of some common materials at room temperature are given as:

MaterialThermal conductivity
Diamond2300
Silver410
Copper385
Gold317
Aluminium202
Nickel93
Iron73
Lead35
Mercury8.54
Glass0.78
Brick0.72
Water0.613
Human skin0.37
Wood0.17
Soft rubber0.13
Glass Fiber0.043
Air0.026

Question 63

Lumped parameter analysis for transient heat conduction is essentially valid of

  1. AOption (a)

    Bi < 0.1

  2. BOption (b)

    0.1 < Bi < 0.5

  3. COption (c)

    1 < Bi < 10

  4. DOption (d)

    Bi → ∞

View answer & explanation

Correct answer

Answer: (a)A.

Bi < 0.1

Why this answer is correct

Concept:

The lumped heat capacity analysis is such an analysis in which temperature is the only function of time and not a function of space.

T ≠ f (space)

T = f (Time) only

For lumped parameter analysis

Biot Number < 0.1

Biot;No=Internal;Conductive;Resistance;(ICR)External;Convective;Resistance;(ECR){\rm{Biot;No}} = \frac{{{\rm{Internal;Conductive;Resistance;}}\left( {{\rm{ICR}}} \right)}}{{{\rm{External;Convective;Resistance;}}\left( {{\rm{ECR}}} \right)}}

∴ for Biot No → 0, we should have Internal Conductive Resistance (ICR) as negligible.

Question 64

Cork is a good thermal insulator because

  1. AOption (a)

    Its density is low

  2. BOption (b)

    It is porous

  3. COption (c)

    It can be powdered

  4. DOption (d)

    It is flexible

View answer & explanation

Correct answer

Answer: (b)B.

It is porous

Why this answer is correct

Concept:

Q=kAdTdx⇒dTdx=qk;Q = kA\frac{{dT}}{{dx}} \Rightarrow \frac{{dT}}{{dx}} = \frac{q}{k};

For same q, is k is low (i.e. for an insulator), dT/dx will be large i.e. there will be a large temperature difference across the wall.

If k is high (i.e. for a conductor), dT/dx will be small i.e. there will be a small temperature difference across the wall.

Material with high thermal conductivities are good conductors of heat, whereas materials with low thermal conductivities are good thermal insulators.

Conduction of heat occurs most readily in pure metals, less so in alloys and much less readily in non-metals.

The very low thermal conductivities of certain thermal insulators e.g. cork is due to their porosity, the air trapped within the material acting as an insulator.

Question 65

Unsteady state of heat flow occurs in

  1. AOption (a)

     Flow of heat through furnace walls

  2. BOption (b)

    Flow of heat through insulated pipe with constant surface temperature

  3. COption (c)

     Annealing of castings

  4. DOption (d)

    Flow of heat through refrigerator walls

View answer & explanation

Correct answer

Answer: (c)C.

 Annealing of castings

Why this answer is correct

Concept:

In unsteady state heat transfer, the temperature varies with time as well as location within the system.

T = f(x, y, z, t) and ∂T∂t≠0\frac{{\partial T}}{{\partial t}} \ne 0

In steady state heat transfer, the temperature at any location within the system does not vary with time. The temperature in steady state is the function of space coordinates while it is independent of time.

T = f(x, y, z) and ∂T∂t=0\frac{{\partial T}}{{\partial t}} = 0

Except in case of annealing in all other case temperature does not change with time at any location.

Hence annealing of castings is the correct answer

Question 66

The temperature inside a furnace is usually measured by:

  1. AOption (a)

    Mercury thermometer

  2. BOption (b)

    Optical pyrometer

  3. COption (c)

    Alcohol thermometer

  4. DOption (d)

    Any of the above

View answer & explanation

Correct answer

Answer: (b)B.

Optical pyrometer

Why this answer is correct

Concept:

To measure the high temperature inside a furnace, the options are:

The upper limit for thermistor devices is around 100°C (212°F) and RTDs are limited to around 750°C (1382°F). That leaves thermocouples and infrared pyrometers or imagers as the most suitable devices for measuring temperatures above 1000°C.

Hence Optical pyrometer is used to measure the temperature inside a furnace.

Question 67

Heat is mainly transferred by conduction, convection and radiation in

  1. AOption (a)

    Insulated pipes carrying hot water

  2. BOption (b)

    Refrigerator freezer coil

  3. COption (c)

    Boiler furnaces

  4. DOption (d)

    Condensation of steam in a condenser

View answer & explanation

Correct answer

Answer: (c)C.

Boiler furnaces

Why this answer is correct

Concept:

A steam boiler is designed to absorb the maximum amount of heat released from the process of combustion.

Heat transfer within a steam boiler is accomplished by three methods: radiation, convection, and conduction.

The heating surface in the furnace area receives heat primarily by radiation. The remaining heating surface in the steam boiler receives heat by convection from the hot flue gases. Heat received by the heating surface travels through the metal by conduction, heat is then transferred from the metal to the water by convection.

Question 68

The density of water is maximum at

  1. AOption (a)

    20°C

  2. BOption (b)

    4°C

  3. COption (c)

    0°C

  4. DOption (d)

    -4°C 

View answer & explanation

Correct answer

Answer: (b)B.

4°C

Why this answer is correct

Concept:

The maximum density of water is 103 kg m-3 at 4 oC temperature.

  • The maximum water density occurs at 4°C, since two opposing effects are in equilibrium at this temperature.
  • The water molecules in ice are in a crystal lattice that has plenty of empty space. When the ice melts into liquid water, the structure breaks and the liquid density increases.
  • The molecules move more quickly and get further apart at temperatures well above zero. Density decreases as the temperature increases. The water also contains several ice-like clusters at temperatures close to 0°C.

Question 69

Which non-metallic body is expected to have highest value of emissivity?

  1. AOption (a)

    Iron oxide

  2. BOption (b)

    Carbon

  3. COption (c)

    Ice

  4. DOption (d)

    Paper

View answer & explanation

Correct answer

Answer: (c)C.

Ice

Why this answer is correct

Concept:

Values of some important bodies is given below

MaterialEmissivity
Ice0.97
Paper0.86
Aluminium foil0.03
Brick0.90
Glass0.95
Silver0.04

Question 70

The rate of heat transfer by conduction in pipes at critical radius is

  1. AOption (a)

    equal to the rate of heat transfer by convection and is maximum

  2. BOption (b)

    equal to the rate of heat transfer by convection and is minimum

  3. COption (c)

    greater than the rate of heat transfer by convection

  4. DOption (d)

    less than the rate of heat transfer by convection

View answer & explanation

Correct answer

Answer: (a)A.

equal to the rate of heat transfer by convection and is maximum

Why this answer is correct

Explanation:

Concept of the critical thickness of insulation:

  • In a plane wall thicker the insulation, lower the heat transfer rate. This is due to the fact the outer surface have always the same area.
  • But in cylindrical and spherical coordinates, the addition of insulation also increases the outer surface area which decreases the convection resistance at the outer surface.
  • At the same time conduction resistance increases due to increase in insulation thickness. So, to optimize the balance of conductive and convective resistance, a critical radius is decided. At this critical radius, the equivalent conduction and convection resistance will be minimum and heat transfer will be maximum.
  • Moreover, in some cases, a decrease in the convection resistance due to the increase in surface area can be more significant than an increase in conduction resistance due to thicker insulation.

As a result, the total resistance may actually decrease resulting in increased heat flow.

Thickness up to which heat flow increases and beyond which heat flow decreases is termed as critical thickness. For cylinders and spheres, it is the critical radius (rc ).

That can be derived from the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h.

For cylinder

rc=;kh{{\bf{r}}_{\bf{c}}} = ;\frac{k}{h}

For sphere

rc=;2kh{{\bf{r}}_{\bf{c}}} = ;\frac{{2k}}{h}

Question 71

The heat transfer coefficient over the surface of a pin fin decreases, then

  1. AOption (a)

    its effectiveness will decrease

  2. BOption (b)

    its effectiveness will increase

  3. COption (c)

    its effectiveness will remain unchanged

  4. DOption (d)

    its effectiveness will first increase and then decrease

View answer & explanation

Correct answer

Answer: (b)B.

its effectiveness will increase

Why this answer is correct

Concept:

It is defined as the ratio of the actual heat transfer that takes place from the fin to the heat that would be dissipated from the same surface area without fin.

εfin = heat;transfer;with;finheat;transfer;without;fin\frac{{heat;transfer;with;fin}}{{heat;transfer;without;fin}}

ε=hpkAcΔThAcΔT{\rm{\varepsilon }} = \frac{{\sqrt {hpk{A_c}} {\rm{\Delta }}T}}{{h{A_c}{\rm{\Delta T}}}}

ε=kPhA{\rm{\varepsilon }} = \sqrt {\frac{{{\rm{kP}}}}{{{\rm{hA}}}}}

εfin∝1h{{\rm{\varepsilon }}_{{\rm{fin}}}} \propto \frac{1}{{\sqrt h }}

Where, h = Heat transfer coefficient, k = Thermal conductivity of fin, P = Perimeter of a fin, Ac = Cross-sectional area of a fin

Effectiveness is inversely proportional to the heat transfer coefficient. Hence if the heat transfer coefficient over the surface of a pin fin decreases, then its effectiveness will increase.

∴ Fins are never used with water or condensation heat transfer where the ‘h’ is relatively high and thus low effectiveness.

Significance:

The effectiveness of fin tells about the percentage increase in heat transfer rate while using a fin as compared to without fin. If the effectiveness of fin is low, then fins are not worth to be used because they will not be helpful in increasing the heat transfer rate.

Question 72

Critical thickness of insulation for spheres is given by _____.

  1. AOption (a)

    k/h

  2. BOption (b)

    k/4h

  3. COption (c)

    h/2k

  4. DOption (d)

    2k/h

View answer & explanation

Correct answer

Answer: (d)D.

2k/h

Why this answer is correct

Concept:

The thickness up to which heat flow increases and after which heat flow decreases is termed as Critical thickness.

The insulation radius at which resistance to heat flow is minimum and consequently heat flow rate is maximum is called “critical radius”.

Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h.

The critical radius of insulation for a cylindrical body:

rcr,cylinder=kh{r_{cr,cylinder}} = \frac{k}{h}

The critical radius of insulation for a spherical shell:

rcr,sphere=2kh{r_{cr,sphere}} = \frac{{2k}}{h}

The rate of heat transfer from the cylinder increases with the addition of insulation for r2 < rcr, reaches a maximum when r2 = rcr, and starts to decrease for r2 > rcr. Thus, insulating the pipe may increase the rate of heat transfer from the pipe instead of decreasing it when r2 < rcr.

Question 73

In a cylinder under steady state conduction with uniform heat generation, the temperature gradient at half the radius location will be

  1. AOption (a)

    one half of that at surface

  2. BOption (b)

    one fourth of that at surface

  3. COption (c)

    twice that at surface

  4. DOption (d)

    four times that at surface

View answer & explanation

Correct answer

Answer: (a)A.

one half of that at surface

Why this answer is correct

Concept:

For a cylinder under steady state conduction with uniform heat generation:

Heat generated in the rod = Heat conducted radially at the surface

q˙×πr2l=−k;(2πrl)(dTdr)\dot q \times \pi {r^2}{\rm{l}} = - {\rm{k;}}\left( {2{\rm{\pi rl}}} \right)\left( {\frac{{{\rm{dT}}}}{{{\rm{dr}}}}} \right)

dTdr=−q˙r2k\frac{{dT}}{{dr}} = - \frac{{\dot qr}}{{2k}}

Temperature gradient is proportional to radius.

∴ The temperature gradient at half the radius location (i.e. at r = R/2) will be one half of that at surface (i.e. at r = R)

Question 74

For the quick response of a thermocouple

  1. AOption (a)

    its wire diameter should be large

  2. BOption (b)

    the convective heat transfer coefficient should be high

  3. COption (c)

    the specific heat should be high

  4. DOption (d)

    the density should not be very small

View answer & explanation

Correct answer

Answer: (b)B.

the convective heat transfer coefficient should be high

Why this answer is correct

Concept:

The response of a thermocouple is defined as the time required for the thermocouple to reach the source temperature when it is exposed to it.

Referring to the lumped-parameter solution for transient heat conduction

ln⁡(T0−T∞Ti−T∞)=−hAsρVCpt\ln \left(\frac{{T_0 - T_\infty }}{{{T_i} - T_\infty }}\right) = - \frac{{h{A_s}}}{{\rho V{C_p}}}t

where,

Ti = Initial temperature of body

T =  temperature of source temperature

h = convective heat transfer coefficient

As = surface area of body

ρ = density of body

V = volume of body

Cp = specific heat of body

It is evident that larger the parameter hA/ρVCp, the faster the exponential term will reach zero or more rapid will be the response of the thermocouple. A larger value of hA/ρVCp can be obtained either by increasing the value of convective coefficient, or by decreasing the wire diameter, density and specific heat.

Question 75

If Nusselt number is 390, Reynolds number is 39 and Prandtl number is 20, then Stanton number will be

  1. AOption (a)

    780

  2. BOption (b)

    200

  3. COption (c)

    2

  4. DOption (d)

    0.5

View answer & explanation

Correct answer

Answer: (d)D.

0.5

Why this answer is correct

Concept:

The Stanton number, St, is a dimensionless number that measures the ratio of heat transferred into a fluid to the thermal capacity of fluid.

Calculation:

St=hρucp=NuRe.Pr=39039×20=0.5St = \frac{h}{{\rho u{c_p}}} = \frac{{Nu}}{{Re.Pr}} = \frac{{390}}{{39 \times 20}} =0. 5

Question 76

The temperature of a solid surface is raised from 227°C to 727°C. The emissive power of the body will change from E1 to E2 such that E2/E1

  1. AOption (a)

    400

  2. BOption (b)

    16

  3. COption (c)

    4000

  4. DOption (d)

    1600

View answer & explanation

Correct answer

Answer: (b)B.

16

Why this answer is correct

Concept:

According to Stefan-Boltzmann’s law:

E = µ T⁴

E = Energy of radiation emitted by surface of body (W/m2)

T = Temperature in Kelvin.

T₁ = 227°C = 227 + 273 = 500 K

T₂ = 727°C = 727 + 273 = 1000 K

Now,

E2E1=T24T14\frac{{{E_2}}}{{{E_1}}} = \frac{{T_2^4}}{{T_1^4}}

∴ E2/E1 = (1000/500)⁴

∴ E2/E1 = (2)⁴

∴ E2/E1 = 16

Question 77

For an opaque body sum of absorptivity and reflectivity is

  1. AOption (a)

    0

  2. BOption (b)

    1.0

  3. COption (c)

    less than 1.0

  4. DOption (d)

    greater than 1.0

View answer & explanation

Correct answer

Answer: (b)B.

1.0

Why this answer is correct

Concept:

When thermal radiation falls onto an object,

  • The radiation will be absorbed by the surface of the object, causing its temperature to change
  • The radiation will be reflected from the surface of the body, causing no temperature change
  • The radiation will pass completely through the object, causing no temperature change

Absorptivity (α) is a measure of how much of the radiation is absorbed by the body

Reflectivity (ρ) is a measure of how much is radiation is reflected

Transmissivity (τ) is a measure of how much radiation passes through the object.

Each of these parameters is a number that ranges from 0 to 1.

For any given wavelength (λ), αλ + ρλ + τλ = 1. Objects may have different values of each of these parameters at different wavelengths.

Now,

For an opaque body, transmissivity is 0 since no wavelength is transmitted through it.

∴ For an opaque body sum of absorptivity and reflectivity is 1

Question 78

Efficiency of a Diesel cycle will approach to Otto cycle when

  1. AOption (a)

    Diesel engine will operate at high speed

  2. BOption (b)

    cut-off period of diesel cycle is reduced to zero

  3. COption (c)

    diesel fuel is balanced with petrol

  4. DOption (d)

    None of these

View answer & explanation

Correct answer

Answer: (b)B.

cut-off period of diesel cycle is reduced to zero

Why this answer is correct

Concept:

ηotto=1−1rγ−1{\eta _{otto}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}

ηdiesel=1−1rγ−1αγ−1γ(α−1){\eta _{diesel}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}\frac{{{\alpha ^\gamma } - 1}}{{\gamma \left( {\alpha - 1} \right)}}

When cut off ratio (α) is equal to zero then the efficiency of Diesel cycle approaches to Otto cycle efficiency.

Question 79

A gas turbine cycle with heat exchanger and reheating improves

  1. AOption (a)

    only the thermal efficiency

  2. BOption (b)

    only the specific power output

  3. COption (c)

    both thermal efficiency and specific power output

  4. DOption (d)

    neither thermal efficiency not specific power output

View answer & explanation

Correct answer

Answer: (c)C.

both thermal efficiency and specific power output

Why this answer is correct

Concept:

As we know Due to Regeneration efficiency is increased and due to Reheating work output will increase. Hence in case of a gas turbine with heat exchanger and Reheating both Thermal efficiency and specific power output will improve.

Various Graphs of Gas turbine cycle with heat exchanger and Reheating:

t = ratio of turbine inlet temperature to turbine outlet temperature

r = pressure ratio of cycle, W = specific work output, η = efficiency of cycle

Consider a gas turbine cycle with heat exchanger and reheating.

Temperature-entropy graph for this is shown below

Fig : gas turbine cycle with heat exchanger and reheating

Fig : Work output Vs pressure ratio of gas turbine cycle with heat exchanger and reheating

Fig : Efficiency Vs pressure ratio of gas turbine cycle with heat exchanger and reheating

Question 80

The ideal efficiency of simple gas turbine cycle depends upon

  1. AOption (a)

    pressure ratio

  2. BOption (b)

    cut-off ratio

  3. COption (c)

    both (a) and (b)

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

pressure ratio

Why this answer is correct

Concept:

Processes: -

1-2: isentropic compression

2-3: constant pressure energy addition

3-4: isentropic expansion

4-1: constant pressure energy rejection

Energy added, Q1 = m Cp (T3 - T2)

Energy rejected, Q2 = m Cp (T4 - T1)

Thermal efficiency, 

η=Q1−Q2Q1=1−T4−T1T3−T2\eta = \frac{{{Q_1} - {Q_2}}}{{{Q_1}}} = 1 - \frac{{{T_4} - {T_1}}}{{{T_3} - {T_2}}}

η=1−T1[(T4T1)−1]T2[(T3T2)−1]\eta = 1 - \frac{{{T_1}\left[ {\left( {\frac{{{T_4}}}{{{T_1}}}} \right) - 1} \right]}}{{{T_2}\left[ {\left( {\frac{{{T_3}}}{{{T_2}}}} \right) - 1} \right]}}

The pressure ratio of the Brayton cycle, rp is defined as, 

rP=P2P1{r_P} = \frac{{{P_2}}}{{{P_1}}}

The processes 1-2 and 3-4 are isentropic. Hence,

T2T1=(P2P1)(γ−1)γ\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{\left( {\gamma - 1} \right)}}{\gamma }}}

T3T4=(P3P4)(γ−1)γ\frac{{{T_3}}}{{{T_4}}} = {\left( {\frac{{{P_3}}}{{{P_4}}}} \right)^{\frac{{\left( {\gamma - 1} \right)}}{\gamma }}}

We get,

T2T1=T3T4\frac{{{T_2}}}{{{T_1}}} = \frac{{{T_3}}}{{{T_4}}}

η=1−T1T2=1−(P1P2)γ−1γ\eta = 1 - \frac{{{T_1}}}{{{T_2}}} = 1 - {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}

η=1−(1rp)γ−1γ\eta = 1 - {\left( {\frac{1}{{{r_p}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}

Hence, Efficiency of simple gas turbine cycle depends on

Pressure ratio

Specific heat capacity ratio

Question 81

The area of a p-v diagram for a Carnot cycle represents

  1. AOption (a)

    heat supplied

  2. BOption (b)

    heat rejected

  3. COption (c)

    work done

  4. DOption (d)

    temperature drop

View answer & explanation

Correct answer

Answer: (c)C.

work done

Why this answer is correct

Carnot cycle consists of two reversible isothermal and two isentropic process.

Carnot cycle is one of the best-known reversible cycles. The Carnot cycle is composed of four reversible processes.

  • Reversible Isothermal Expansion (process 1-2)
  • Reversible adiabatic expansion (process 2-3)
  • Reversible isothermal compression (process 3-4)
  • Reversible adiabatic compression (process 4-1)

Fig. P-V and T-S diagrams of Carnot Cycle

∵ We know that 

Work done (W) = PΔV

∴ The area under the PV diagram represents the work done.

Question 82

For a given set of operating pressure limits of a Rankine cycle the highest efficiency occurs for

  1. AOption (a)

    Saturated cycle

  2. BOption (b)

    Superheated cycle

  3. COption (c)

    Reheat cycle

  4. DOption (d)

    Regenerative cycle

View answer & explanation

Correct answer

Answer: (d)D.

Regenerative cycle

Why this answer is correct

Concept:

1.  Superheated Rankine cycle :    

  • The average temperature at which heat is added to the steam can be increased without increasing the boiler pressure by superheating the steam to high temperatures. Thus, efficiency increases.
  • Superheating the steam to higher temperatures has another very desirable effect: It decreases the moisture content of the steam at the turbine exit.
<br>

2. Reheat Rankine cycle :

The advantages of using a reheat cycle are

  • higher thermal efficiency
  • reduced feed water pump power
  • smaller condenser
  • smaller boiler
  • long life of the turbine
  • less handling of the fuel and firing requirement.
<br>

3. Regenerative Rankine cycle

The advantages of Regeneration cycle:

  1. Heat supplied to boiler becomes reduced.
  2. Thermal efficiency is increased since the average temperature of heat addition to the cycle is increased.
  3. Due to bleeding in the turbine, erosion of turbine due to moisture is reduced.

Question 83

Which process is responsible for the production of energy in the sun?

  1. AOption (a)

    Nuclear fission reaction

  2. BOption (b)

    Nuclear fusion reaction

  3. COption (c)

    exothermal chemical reaction

  4. DOption (d)

    All of the above

View answer & explanation

Correct answer

Answer: (b)B.

Nuclear fusion reaction

Why this answer is correct

Concept:

 

  • The sun generates its energy through a process called nuclear fusion.
  • This process happens in the sun's core, where the extreme temperature (around 15 million degrees Celsius) and pressure cause hydrogen atoms to come together to form helium in a series of fusion reactions.
  • Specifically, the primary fusion process in the sun is the proton-proton chain, which converts four protons (hydrogen nuclei) into a helium nucleus, two positrons, and two neutrinos along with energy in the form of gamma rays.
  • Nuclear fusion reactions release a tremendous amount of energy because a small amount of the mass of the hydrogen atoms is converted directly into energy, according to Einstein's equation E=mc^2. This is the energy that powers the sun and other stars.
  • While nuclear fission reactions and exothermic chemical reactions can also release energy, these are not the processes that supply energy in the sun. Nuclear fission, which involves the splitting of heavy atoms, is not responsible for the Sun's energy production. Exothermic chemical reactions, where a reaction releases energy usually in the form of heat, are not nearly powerful enough to explain the sheer amount of energy the Sun outputs.

Question 84

Terrestrial radiation has a wavelength in the range of

  1. AOption (a)

    0.2 μm to 4 μm

  2. BOption (b)

    0.2 μm to 0.5 μm

  3. COption (c)

    0.380 μm to 0.760 μm

  4. DOption (d)

    0.29 μm to 2.3 μm

View answer & explanation

Correct answer

Answer: (d)D.

0.29 μm to 2.3 μm

Why this answer is correct

Concept:

Terrestrial Radiation:

  • The solar radiation received at the earth’s surface after passing through the atmosphere are terrestrial radiation .
  • The solar radiation received at the surface of the earth is entirely different from the radiation received outside the atmosphere due to absorption and scattering in atmosphere.
  • A fraction of radiation reaching the earth’s surface is reflected back into the atmosphere and subjected to atmospheric phenomenon and the remaining is received by the earth surface.
  • Most of the terrestrial solar energy received by the earth’s surface lies in the range of 0.29 µm to 2.5 µm.
<br>

Points to remember:

Absorption: As solar radiation passes through atmosphere short -wave ultraviolet rays are absorbed by the ozone in the atmosphere and long wavelength infrared radiation are absorbed by carbon di-oxide and the water vapour present in the atmosphere.

Scattering: The air molecules, water vapour, dust particle scatter the solar radiation as it passes through the atmosphere. A part of this scattered radiation also reaches atmosphere.

Question 85

A solar thermal collector

  1. AOption (a)

    collects solar energy and reflects it back

  2. BOption (b)

    absorbs the solar radiation and dissipates it to the ambient

  3. COption (c)

    collects and converts solar energy into electrical energy

  4. DOption (d)

    collects and converts the solar energy into thermal energy and delivers it to the next stage of the system

View answer & explanation

Correct answer

Answer: (d)D.

collects and converts the solar energy into thermal energy and delivers it to the next stage of the system

Why this answer is correct

Solar thermal collector:

It is a device which is used to collect the heat by absorbing the sunlight. It is mainly used to collect the solar energy and then convert into thermal energy.

The most commonly used solar thermal collector is flat plate collector.

Flat plate solar collector: The flat-plate solar collectors are probably the most fundamental and most studied technology for solar-powered domestic hot water systems. The Sun heats a dark flat surface, which collects as much energy as possible, and then the energy is transferred to water, air, or other fluid for further use.

The solar radiation is absorbed by the black plate and transfers heat to the fluid in the tubes. The thermal insulation prevents heat loss during fluid transfer; the screens reduce the heat loss due to convection and radiation to the atmosphere

The main components of a typical flat-plate solar collector:

  • Black surface - absorbent of the incident solar energy
  • Glazing cover - a transparent layer that transmits radiation to the absorber, but prevents radiative and convective heat loss from the surface
  • Tubes containing heating fluid to transfer the heat from the collector
  • Support structure to protect the components and hold them in place
  • Insulation covering sides and bottom of the collector to reduce heat losses
<br>

Advantages:

  • Easy to manufacture
  • Low cost
  • Collect both beam and diffuse radiation
  • Permanently fixed (no sophisticated positioning or tracking equipment is required)
  • Little maintenance
<br>

The efficiency of flat plate collectors is usually 61%.

Question 86

Solar cell is basically

  1. AOption (a)

    A voltage source, controlled by a flux radiation

  2. BOption (b)

    A current source, controlled by a flux radiation

  3. COption (c)

    An uncontrolled current source

  4. DOption (d)

    An uncontrolled voltage source

View answer & explanation

Correct answer

Answer: (b)B.

A current source, controlled by a flux radiation

Why this answer is correct

Concept:

Solar cell:

  • Solar photovoltaic (PV) systems convert solar energy directly into electrical energy.
  • The basic conversion device used is known as a solar photovoltaic cell or solar cell.
  • A solar cell is basically an electrical current source, driven by a flux of radiation as the current output energy is dependent on solar radiation.

 

Equivalent circuit representation:

We can represent the solar cell as a current source in parallel with a diode as shown below.

When you shine a light on the solar cell, the light generates a current and we represent this current by a current source.

The value of this current source is directly proportional to the light shining on the diode i.e. the current is dependent on the radiation of the light.

Question 87

The working fluid used in an MHD system coupled to a fast breeder reactor is

  1. AOption (a)

    hot flue gases

  2. BOption (b)

    seeded inert gas

  3. COption (c)

    liquid metal inert has

  4. DOption (d)

    liquid metal only

View answer & explanation

Correct answer

Answer: (d)D.

liquid metal only

Why this answer is correct

MHD generator is a device used for converting heat energy of fuel directly into electrical energy without conventional electric generator.

MHD system works on principle of Faradays laws of electromagnetic induction.

Closed system MHD systems are classified in 2 types:

Seeded Inert Gas Closed-Cycle MHD System:

  • This MHD system works on Brayton cycle with inert carrier gas (Argon/helium).In seeded inert gas system, electrical conductivity is maintained in the working fluid by ionization of a seed material.
  • The inert gas is compressed and heated in primary heat exchanger.
<br>

Small quantity of seed material (e.g., cesium) is them added to make up for the loss of seed through leakage etc. At lower temperature, low cost seed material can also be a used but at low temperature, efficiency of the system reduces.

Close-Cycle Liquid Metal MHD System:

  • In this system, a liquid metal (Potassium) is used as a working fluid.
  • The liquid potassium after being heated in the ‘breeder reactor’ is passed through the ‘nozzle’ where its velocity is increased.
  • The vapours formed due to nozzle action are separated in the ‘separated and condensed and them pumped back to the reactor.
  • Then the liquid metal with high velocity is passed through ‘MHD generator’ to produce D.C. power.
<br>

The liquid potassium coming out of MHD generator is passed through the heat exchanger’ (boiler) to use its remaining heat to run a turbine-generator set and then pumped back to the reactor.

Question 88

For the same heat added and the same compression ratio, ________.

  1. AOption (a)

    Otto cycle is more efficient than diesel cycle.

  2. BOption (b)

    Diesel cycle is more efficient than Otto cycle.

  3. COption (c)

    Both Diesel and Otto cycle are equally efficient.

  4. DOption (d)

    Cannot be determined.

View answer & explanation

Correct answer

Answer: (a)A.

Otto cycle is more efficient than diesel cycle.

Why this answer is correct

Concept:

η=1−Heat;RejectedHeat;Supplied\eta = 1 - \frac{{Heat;Rejected}}{{Heat;Supplied}}

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 equation

The cycle which has the least heat rejected will have the highest efficiency.

Thus, the Otto cycle is the most efficient and the Diesel cycle is the least efficient of the three cycles.

ηotto>ηdual>ηdiesel{\eta _{otto}} > {\eta _{dual}} > {\eta _{diesel}}

For the Same Compression Ratio and the Same Heat Inputηotto>ηdual>ηdiesel{\eta _{otto}} > {\eta _{dual}} > {\eta _{diesel}}
For Constant Maximum Pressure and Heat Suppliedηdiesel>ηdual>ηotto{\eta _{diesel}} > {\eta _{dual}} > {\eta _{otto}}

Question 89

Consider the following emissions of an I.C. engine.

  1. CO2
  2. HC
  3. NOx
  4. Particulate

Which of these emissions causes photochemical smog?

  1. AOption (a)

    1 and 4

  2. BOption (b)

    1 and 2

  3. COption (c)

    2 and 3

  4. DOption (d)

    3 and 4

View answer & explanation

Correct answer

Answer: (c)C.

2 and 3

Why this answer is correct

Concept:

  • In photochemical smog, nitrogen oxides, and hydrocarbons are primary pollutants whereas ozone and acrolein are secondary pollutants.
  • A chemical reaction between solar ultraviolet radiation and an atmosphere polluted with hydrocarbons and oxides of nitrogen causes photochemical smog. This is especially common from automobile exhaust. Smog can happen both during the day and at night, but photochemical smog only happens in the presence of sunlight.
  • Photochemical smog is composed of primary and secondary pollutants.
  • Primary pollutants, which include nitrogen oxides and volatile organic compounds, are introduced into the atmosphere via vehicular emissions and industrial processes.
  • Photochemical smog is a type of smog produced when ultraviolet light from the sun reacts with nitrogen oxides in the atmosphere. It is visible as a brown haze, and is most prominent during the morning and afternoon, especially in densely populated, warm cities.
  • Nitrogen oxides are compounds of nitrogen and oxygen produced during combustion. Ozone is produced when nitrogen oxides, pollutants in cars' industrial exhausts, react with products of hydrocarbons.

Question 90

Consider the following statements:

Knock in the S.I. the engine can be reduced by

  1. Supercharging
  2. Retarding the spark
  3. Using a fuel of long straight-chain structure
  4. Increasing the engine speed.

Of these correct statements are

  1. AOption (a)

    1 and 2

  2. BOption (b)

    2 and 3

  3. COption (c)

    1, 3 and 4

  4. DOption (d)

    2 and 4

View answer & explanation

Correct answer

Answer: (d)D.

2 and 4

Why this answer is correct

Concept:

When the spark is advanced, burning gas is compressed by the rising piston and therefore temperature and pressure are increased, thus, the tendency to knock increases.

In spark ignition (SI) engines, the possibility of knocking can be reduced by decreasing compression ratio as a reduction in compression ratio will result in the generation of lower peak temperature. In SI engine, knocking occurs before TDC and because of early combustion of charge/more than one flame propagation.

Supercharging in SI and CI engine:

Supercharging increases the pressure and temperature of the charge at the end of compression. This reduces ignition delay in CI engine, thereby the combustion becomes smooth, and the tendency for knocking is avoided. In SI engine the short ignition delay promotes detonation.

Hence supercharging is preferred in diesel engine than in petrol engine.

Factors affecting knocking in SI engine:

i) Compression ratio:- Higher compression ratio will result in the higher temperature of the charge, the tendency to knock will increase.

ii) Inlet air temperature:- The effect of inlet air temp is the same as compression ratio. High inlet temp of the air-fuel mixture will increase knocking.

iii) Advancing the spark:- Advancing the spark will result in more compression during sparking. The increase in temperature during the sparking will be higher. Hence tendency to knock will increase and so by retarding the spark, the tendency to knock will decrease.

iv) Coolant water temperature:- By increasing coolant water temperature, less heat will be carried from the engine which will increase the engine temperature. So the knocking will increase.

v) Engine Speed: TO avoid the knocking in SI engine the engine speed should be high.

vi) Octane rating: It should be as high as possible

Question 91

Which of the following is considered to be superior quality coal for power plants?

  1. AOption (a)

    Bituminous coal

  2. BOption (b)

    Peat

  3. COption (c)

    Coke

  4. DOption (d)

    Lignite

View answer & explanation

Correct answer

Answer: (a)A.

Bituminous coal

Why this answer is correct

Concept:

Quality of coal is mainly depend on its calorific value.

Peat

  • Calorific value : about 17000 kJ/kg
  • Contains less than 40 to 55 per cent carbon.
  • Contains sufficient volatile matter and lot of moisture [more smoke and more pollution].
  • Left to itself, it burns like wood, gives less heat, emits more smoke and leaves a lot of ash.
<br>

Lignite

  • calorific value: 8000kJ/kg to 15000 kJ/kg
  • Lower grade coal.
  • 40 to 55 per cent carbon.
  • Moisture content is high (over 35 per cent).
<br>

Bituminous coal

  • calorific value : about 30000 kJ/kg
  • 40 to 80 per cent carbon.
  • Moisture and volatile content (15 to 40 per cent)
  • Calorific value is very high due to high proportion of carbon and low moisture.
<br>

Coke

  • Calorific value : about 25000 kJ/kg

Question 92

A curve showing the variation of load on a power station with respect to time is known as

  1. AOption (a)

    Load curve

  2. BOption (b)

    ​Load duration curve

  3. COption (c)

    Diversity factor

  4. DOption (d)

    Load factor

View answer & explanation

Correct answer

Answer: (a)A.

Load curve

Why this answer is correct

Explanation:

The load on a power station varies from time to time due to uncertain demands of the consumers and is known as variable load on the station.

Load Curves: The curve showing the variation of load on the power station with respect to (w.r.t) time is known as a load curve.

Average load: The average of loads occurring on the power station in a given period (day or month or year) is known as average load or average demand.

Load Duration Curve: When the load elements of a load curve are arranged in the order of descending magnitudes, the curve thus obtained is called a load duration curve.

Diversity factor: The ratio of the sum of individual maximum demands to the maximum demand on power station is known as diversity factor.

Load Factor: The ratio of average load to the maximum demand during a given period is known as load factor

Question 93

The capacity of generators being installed in super thermal power plant is

  1. AOption (a)

    100 MW

  2. BOption (b)

    200 MW

  3. COption (c)

    400 MW

  4. DOption (d)

    500 MW

View answer & explanation

Correct answer

Answer: (d)D.

500 MW

Why this answer is correct

Concept:

In India, the highest size of generators in power plant unit is 500 megawatts, which is called ‘super thermal unit ’while outside India the highest size unit is 1300 megawatts.

Advantages:

  • Larger of the size of the unit, the better it is as the cost per KW decreases, if you move from 100 MW unit to 500 MW units. They also consume less fuel, the automatic advantage is less pollution.
<br>

Disadvantages:

  • If one such 500 MW unit gets means faulted and stops working, it is a big shock to the system and many industries have to suffer very much loss. The second reason is money, as approximately 4 crores are needed for 1 MW, so it requires a huge amount of capital investment.

Question 94

Fuel injection pressure in solid injection system is approximately in the range of

  1. AOption (a)

    < 10.5 bar

  2. BOption (b)

    10.5 – 21 bar

  3. COption (c)

    30 – 50 bar

  4. DOption (d)

    200 – 246 bar

View answer & explanation

Correct answer

Answer: (d)D.

200 – 246 bar

Why this answer is correct

Classification of Injection System

1.      Air Injection System 2.      Solid Injection System

 1. Air Injection System

  • In this Injection system, the fuel will be injected into the cylinder by means of compressed air. So it needs an additional compression system for the Injection System.
  • Due to this additional system, the weight of the engine will increase which will result in the reduced brake power output.
  • This type of fuel injection system is used very less nowadays because of the bulky design of the engine. Since it needs an additional mechanical compressor system.
<br>

The following are the advantages of the Air Injection system

  • High viscosity fuel can be used which are comparatively less expensive than the fuel used by the Solid Injection System.
  • Can achieve higher Mean Effective Pressures (MEP) due to the good mixing of fuel with the air.
<br>

2. Solid Injection System

  • In this Injection system, the fuel will be directly injected into the cylinder without the aid of the compressed air as like in the Air Injection System.
  • This can be also called as the Airless Mechanical Injection System.
  • In this Solid Injection System, There are different types of Injection Systems are there.
  1. Individual pump and Nozzle system
  2. Unit Injector system
  3. Common rail system
  4. Distributor system

High pressure pumps are used in all the systems with pressure above 200 bar.

Question 95

The thermal efficiency of a gas turbine cycle with ideal regenerative heat exchanger is

  1. AOption (a)

    equal to work ratio

  2. BOption (b)

    less than work ratio

  3. COption (c)

    more than work ratio

  4. DOption (d)

    unpredictable

View answer & explanation

Correct answer

Answer: (a)A.

equal to work ratio

Why this answer is correct

Concept:

Work ratio is defined as net work output and work done by turbine

Work ratio = Net;work;outputpositive;workdone;(Turbine;work)\frac{{Net;work;output}}{{positive;workdone;\left( {Turbine;work} \right)}}

WR=Cp[T3−T4]−Cp[T2−T1]Cp[T3−T4]WR = \frac{{{C_p}\left[ {{T_3} - {T_4}} \right] - {C_p}\left[ {{T_2} - {T_1}} \right]}}{{{C_p}\left[ {{T_3} - {T_4}} \right]}}

WR=1−[T2−T1][T3−T4]WR = 1-\frac{{{}{{} {}} {}\left[ {{T_2} - {T_1}} \right]}}{{{}\left[ {{T_3} - {T_4}} \right]}}

Efficiency is defined as the ratio of net work output to the energy supplied to the cycle.

Heat added in system =Cp[T3−T2′] = {C_p}\left[ {{T_3} - {T_2'}} \right]

Head rejected = Cp [T4' – T1]

For ideal Regenerator

T2 = T4' and T2' = T4 

η=1−QRejectedQadded\eta = 1 - \frac{{{Q_{Rejected}}}}{{{Q_{added}}}}

η=1−Cp[T4′−T1]Cp[T3−T2′]\eta = 1 - \frac{{{C_p}\left[ {{T_4'} - {T_1}} \right]}}{{{C_p}\left[ {{T_3} - {T_2'}} \right]}}

η=1−Cp[T2−T1]Cp[T3−T4]\eta = 1 - \frac{{{C_p}\left[ {{T_2} - {T_1}} \right]}}{{{C_p}\left[ {{T_3} - {T_4}} \right]}}

η=1−[T2−T1][T3−T4]\eta = 1 - \frac{{\left[ {{T_2} - {T_1}} \right]}}{{\left[ {{T_3} - {T_4}} \right]}}

Hence, Efficiency is equal to work ratio

Question 96

The ratio of work done to the energy supplied to rotor in a turbine stage is called

  1. AOption (a)

    blade efficiency

  2. BOption (b)

    stage efficiency

  3. COption (c)

    nozzle efficiency

  4. DOption (d)

    None of these

View answer & explanation

Correct answer

Answer: (a)A.

blade efficiency

Why this answer is correct

Concept:

Blade or diagram efficiency is defined as the ratio of work done on the blades to the energy supplied to the blades.

\({\eta b} = \frac{{Work;done;on;blades}}{{Energy;supplied;to;blades}} = \frac{{2\left( {{V{w1}} + {V_{w2}}} \right)u}}{{V_1^2}}\)

Blade speed ratio is the ratio of blade speed to a steam speed

ρ=uV1\rho = \frac{u}{{{V_1}}}

ηb will be maximum when:

dηbdρ=0⇒ρ=cos⁡α12=cos⁡30∘2=0.433\frac{{d{\eta _b}}}{{d\rho }} = 0 \Rightarrow \rho = \frac{{\cos {\alpha _1}}}{2} = \frac{{\cos 30^\circ }}{2} = 0.433

α1 is of the order 18° to 22°

ηb,max=cos⁡2α1;{\eta _{b,max}} = {\cos ^2}{\alpha _1};

Question 97

The diagram efficiency is highest for simple impulse turbine stage having smooth and symmetrical blade when blade steam speed ratio can be given as

  1. AOption (a)

    cos α1

  2. BOption (b)

    cos⁡α14\frac{{\cos {\alpha _1}}}{4}

  3. COption (c)

    cos⁡α12\frac{{\cos {\alpha _1}}}{2}

  4. DOption (d)

    None of these

View answer & explanation

Correct answer

Answer: (c)C.

cos⁡α12\frac{{\cos {\alpha _1}}}{2}

Why this answer is correct

Concept:

Blade or diagram efficiency is defined as the ratio of work done on the blades to the energy supplied to the blades.

\({\eta b} = \frac{{Work;done;on;blades}}{{Energy;supplied;to;blades}} = \frac{{2\left( {{V{w1}} + {V_{w2}}} \right)u}}{{V_1^2}}\)

Blade speed ratio is the ratio of blade speed to a steam speed

ρ=uV1\rho = \frac{u}{{{V_1}}}

ηb will be maximum when:

dηbdρ=0⇒ρ=cos⁡α12\frac{{d{\eta _b}}}{{d\rho }} = 0 \Rightarrow \rho = \frac{{\cos {\alpha _1}}}{2}

Question 98

What will happen to the volumetric efficiency with increasing pressure ratio in case of single stage compression in compression?

  1. AOption (a)

    Decreases

  2. BOption (b)

    Increases

  3. COption (c)

    Remains unaffected

  4. DOption (d)

    None of these

View answer & explanation

Correct answer

Answer: (a)A.

Decreases

Why this answer is correct

Concept:

The volumetric efficiency of a compressor is the ratio of the actual free air delivered to the displacement of the compressor. The expression for a single stage reciprocating compressor can be derived as follows:

Let,

ηv = Volumetric efficiency

Vs =Swept volume

Va=Actual volume

Vc=Clearance volume

The volumetric efficiency is given by the expression,

=1+C−C(P2P1)1n = 1 + C - C{\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{1}{n}}}

The following factors affect volumetric efficiency:

  • Increase in clearance volume causes a decrease in volumetric efficiency.
  • Increase in compression ratio also decreases the volumetric efficiency.
  • Piston ring leakage allows gas from compression chamber to escape the piston and enter the other end of cylinder. This reduces the volumetric efficiency.
  • Valve slip leads to reversal of gas flow through the valves. This leads to volume loss at the inlet and outlet valves. Thus, decreasing the volumetric efficiency.

Question 99

The compression work requirement is minimum in case of compression process being

  1. AOption (a)

    Adiabatic

  2. BOption (b)

    Isochoric

  3. COption (c)

    Isothermal

  4. DOption (d)

    Hyperbolic

View answer & explanation

Correct answer

Answer: (c)C.

Isothermal

Why this answer is correct

Concept:

The work done during compression of air is given by 

dw = ∫ vdP

That is the area under curve projected on P–axis (on PV plot)

Now, between two same pressure limits P1 and P2 (P2 > P1), the final volume of the air under various processes is

Say P2P1=r\frac{P_2}{P_1}=r,

Process-1 (Isothermal):

P1V1 = P2V2

V2=V1r{V_2} = \frac{{{V_1}}}{r}

Process-2 (Adiabatic):

P1V1γ=P2V2γP_1V_1^{\gamma}=P_2V_2^{\gamma}

V2=V1r1γ{V_2} =\frac{V_1}{r^{\frac{1}{\gamma}}}

Comparing denominators, (V2) Isothermal < (V2) Adiabatic

Now plotting the processes on the PV plot, we get

From the figure, it is clearly evident that the area under the curve (work done) is minimum for isothermal compression and maximum for isentropic compression.

∴ work input is minimum for isothermal compression

Note:

Though work done is zero for an isochoric process, there is no change in volume, hence no compression. At constant volume, either heat rejection or heat addition takes place.

Question 100

If a certain mass of moist air in an air tight vessel is heated to a higher temperature, then

  1. AOption (a)

    Specific humidity of the air increases

  2. BOption (b)

    Specific humidity of the air decreases

  3. COption (c)

    Relative humidity of the air increases

  4. DOption (d)

    Relative humidity of the air decreases

View answer & explanation

Correct answer

Answer: (d)D.

Relative humidity of the air decreases

Why this answer is correct

Explanation:

If the dry bulb temperature is increased, then it's compulsory that the relative humidity will decrease, since the saturation pressure increase.

When the temperature is increased, the specific humidity (w) is constant.

100 questions in the full paper