1. In airstandard Otto cycle the terminal pressures at the end of compression, heat release and expansion are respectively p_{2}, p_{3}, p_{4}. If the corresponding values are p_{2}’, p_{3}’, p_{4}’, taking into account the effect of variable specific heat and dissociation of the working fluid, then
(a) p_{2} < p_{2}’ and p_{3} > p_{3}’
(b) p_{3} < p_{3}’ and p_{4} > p_{4}’
(c) p_{2} > p_{2}’, p_{3} > p_{3}’ and p_{4} < p_{4}’
(d) p_{2} > p_{2}’ and p_{3} > p_{3}’
(2 Mark, 1989)
2. An air standard Diesel cycle consist of:
(a) Two adiabatic and two constant volume processes.
(b) Two constant volume and two isothermal processes.
(c) One constant pressure, one constant volume and two adiabatic processes
(d) One constant pressure, one constant volume and two isothermal processes
(2 Mark, 1990)

Ans: c
3. The power output from a spark ignition engine is varied by:
(a) Changing the ignition timing
(b) Regulating the amount of airfuel inducted.
(c) Regulating the amount of airfuel mixture.
(d) Regulating the amount of fuel.
(2 Mark, 1990)

Ans: c
5. For determining the ignition quality of compression ignition engine fuels, the reference fuels used are
(a) Isooctane and nheptane
(b) Cetane and αmethylnapthalene
(c) Hexadecane and nheptane
(d) Cetane and iso octane
(2 Mark, 1991)

Ans: b
6. If air fuel ratio of the mixture in petrol engine is more than 15:1
(a) NO_{x }is reduced
(b) CO_{2 }is reduced
(c) HC is reduced
(d) CO is reduced
(2 Mark, 1991)

Ans: a
7. BHP of a diesel engine can be increased by
(a) Increasing the pressure of intake air
(b) Increasing the temperature of intake air
(c) Increasing the density of intake air
(d) Decreasing the density of intake air
(2 Mark, 1991)

Ans: c
8. Alcohols are unsuitable as diesel engine fuels because
(a) the cetane number of alcohol fuels is very low which prevents their ignition by compression
(b) the cetane number of alcohol fuels is very high which prevents their ignition by compression
(c) the cetane number of alcohol fuels is constant which prevents their ignition by compression
(d) None of the above
(2 Mark, 1992)

Ans: a
9. Brake thermal efficiency of the three types of reciprocating engines commonly used in road vehicles are given in the increasing order as
(a) 2 stroke SI engine, 4 stroke SI engine, 4 stroke CI engine
(b) 2 stroke SI engine, 4 stroke CI engine, 4 stroke SI engine
(c) 4 stroke SI engine, 2 stroke SI engine, 4 stroke CI engine
(d) 4 stroke CI engine, 4 stroke SI engine, 2 stroke CI engine
(2 Mark, 1992)

Ans: a
10. Knocking tendency in a SI engine reduces with increasing
(a) Compression ratio
(b) Wall temperature
(c) Super charging
(d) Engine speed
(2 Mark, 1993)

Ans: d

Ans: A – 4, B – 5, C – 2, D – 1

Ans: A – 3, B – 1, C – 4, D – 2
12. In order to burn 1 kilogram of CH_{4} completely, the minimum number of kilograms of oxygen needed is (take atomic weights of H, C and O as 1, 12 and 16 respectively).
(a) 3
(b) 4
(c) 5
(d) 6
(2 Mark, 1995)

Ans: b
Explanation:
CH_{4} + 2O_{2} = CO_{2} + 2H_{2}O
1 mole (16 gram) CH_{4 }requires 2 mole (64 gram) O_{2 }to completely burn.
Amount of oxygen requires to burn 1 kg CH_{4 }= (64/16) * 1 = 4 kg
13. A cycle consisting of two reversible isothermal processes and two reversible isobaric processes is known as
(a) Atkinson cycle
(b) Stirling cycle
(c) Brayton cycle
(d) Ericsson cycle
(1 Mark, 1996)

Ans: d
15. Air (C_{P} = 1kJ/kgK, γ = 1.4) enters a compressor at a temperature of 27°C. the compressor pressure ratio is 4. Assuming an efficiency of 80%, the compressor work required in kJ/kg is
(a) 160
(b) 172
(c) 182
(d) 225
(2 Mark, 1998)

Ans: c
Explanation:
Air temperature at the exit of compressor:
{ T }_{ 2 }={ T }_{ 1 }\times { r }_{ p }^{ \frac { \gamma 1 }{ \gamma } }=> { T }_{ 2 }=300\times { 4 }^{ \frac { 1.41 }{ 1.4 } } = 445.8 K
Ideal Compressor work (W)= { c }_{ p }\left( { T }_{ 2 }{ T }_{ 1 } \right) = 1(445.8 – 300) = 145.8 K
Actual compressor work = \frac { W }{ \eta } =\frac { 145.8 }{ 0.8 } = 182 kJ/kg
16. With increasing temperature of intake air, IC engine efficiency
(a) decreases
(b) increases
(c) remains same
(d) depends on other factors
(1 Mark, 1998)

Ans: a
Explanation:
If the temperature of intake air increases, then the density of intake area decreases. It decreases the work output of the engine.
So, the IC engine efficiency decreases.
17. An IC engine has a bore and stroke of 2 units each. The area to calculate heat loss can be taken as
(a) 4π
(b) 5π
(c) 6π
(d) 4
(2 Mark, 1998)

Ans: c
Explanation:
Total area to calculate heat loss from the engine:
A = \pi dL+2\times \frac { \pi }{ 4 } { d }^{ 2 }
=> A = \pi \times 2\times 2+2\times \frac { \pi }{ 4 } \times 2^{ 2 }=6\pi
18. An air breathing aircraft is flying at an altitude where the air density is half the value at ground level. With reference to the ground level, the airfuel ratio at this altitude will be
(a) \sqrt [ 3 ]{ 2 }
(b) \sqrt { \frac { 1 }{ 2 } }
(c) 2
(d) 4
(2 Mark, 1998)

Ans: b
Explanation:
Air fuel ratio \alpha \sqrt { density.of.air }
=> \frac { { \left( AFR \right) }_{ a } }{ { \left( AFR \right) }_{ g } } =\frac { \sqrt { \frac { 1 }{ 2 } { \rho }_{ g } } }{ \sqrt { { \rho }_{ g } } } =\sqrt { \frac { 1 }{ 2 } }
19. The silencer of an internal combustion engine
(a) reduces noise
(b) decreases brake specific fuel consumption (BSFC)
(c) increases BSFC
(d) has no effect on its efficiency
(2 Mark, 1998)

Ans: a
21. In a spark ignition engine working on the ideal Otto cycle, the compression ratio is 5.5. The work output per cycle (i.e., area of the PV diagram) is equal to 23.625 × 10^{5 }× V_{c} J, where V_{c} is the clearance volume in m^{3}. The indicated mean effective pressure is
(a) 4.295 bar
(b) 5.250 bar
(c) 86.870 bar
(d) 106.300 bar
(2 Mark, 2001)

Ans: c
Explanation:
Given, compression ratio = 5.5
=> \frac { { v }_{ s }+{ v }_{ c } }{ { v }_{ c } } = 5.5
=> { v }_{ s }=4.5{ v }_{ c }
The work output per cycle = { p }_{ m }\times { v }_{ s }
=> { 23.625\times { 10 }^{ 5 }\times { v }_{ c }=P }_{ m }\times { v }_{ s }
=> { 23.625\times { 10 }^{ 5 }\times { v }_{ c }=P }_{ m }\times 4.5{ v }_{ c }
=> P_{ m } = 5.25 bar
22. For a spark ignition engine, the equivalence ratio (ϕ) of mixture entering the combustion chamber has values.
(a) ϕ < 1 for idling and ϕ > 1 for peak power conditions.
(b) ϕ > 1 for both idling and peak power conditions.
(c) ϕ > 1 for idling and ϕ < 1 for peak power conditions.
(d) ϕ < 1 for both idling and peak power conditions.
(1 Mark, 2003)

Ans: b
23. A diesel engine is usually more efficient than a spark ignition engine because
(a) diesel being a heavier hydrocarbon, releases more heat per kg than gasoline
(b) the air standard efficiency of diesel cycle is higher than the Otto cycle, at a fixed compression ratio
(c) the compression ratio of a diesel engine is higher than that of an SI engine
(d) self ignition temperature of diesel is higher than that of gasoline
(1 Mark, 2003)

Ans: c
24. An automobile engine operates at a fuel air ratio of 0.05, volumetric efficiency of 90% and indicated thermal efficiency of 30%. Given that the calorific value of the fuel is 45 MJ/kg and the density of air at intake is 1 kg/m^{3}, the indicated mean effective pressure for the engine is
(a) 6.075 bar
(b) 6.75 bar
(c) 67.5 bar
(d) 243 bar
(2 Mark, 2003)

Ans: a
Explanation:
Volumetric efficiency \left( { \eta }_{ v } \right) =\frac { Actual.volume }{ Swept.volume }
=> 0.9=\frac { { V }_{ ac } }{ { V }_{ s } } => { V }_{ ac }=0.9\times { V }_{ s }
Again, { m }_{ a }={ \rho }_{ air }\times { V }_{ ac }=1\times 0.9\times { V }_{ s }=0.9{ V }_{ s }
Given, \frac { { m }_{ f } }{ { m }_{ a } } =0.05
=> { m }_{ f }=0.05\times 0.9{ V }_{ s }=0.045{ V }_{ s }
Given, { \eta }_{ thermal } = 30%
=> \frac { W }{ Q } =\frac { { p }_{ mep }\times { V }_{ s } }{ { m }_{ f }\times C.V } =0.3
=> \frac { { p }_{ mep }\times { V }_{ s } }{ 0.045{ V }_{ s }\times \left( 45\times { 10 }^{ 6 } \right) } =0.3
=> { p }_{ mep }=0.3\times 0.045\times 45\times { 10 }^{ 6 } = 6.075 bar
25. At the time of starting, idling and low speed operation, the carburettor supplies a mixture which can be termed as
(a) lean
(b) slightly leaner than stoichiometric
(c) stoichiometric
(d) rich
(1 Mark, 2004)

Ans: d
26. During a Morse test on a 4 cylinder engine, the following measurements of brake power were taken at constant speed.
All cylinders firing 3037 kW
Number 1 cylinder not firing 2102 kW
Number 2 cylinder not firing 2102 kW
Number 3 cylinder not firing 2100 kW
Number 4 cylinder not firing 2098 kW
The mechanical efficiency of the engine is
(a) 91.53%
(b) 85.07%
(c) 81.07%
(d) 61.22%
(2 Mark, 2004)

Ans: c
Explanation:
Given, Break power = 3037 kW
IP_{1 }= 3037 – 2102 = 935 kW
IP_{2 }= 3037 – 2102 = 935 kW
IP_{3 }= 3037 – 2100 = 937 kW
IP_{4 }= 3037 – 2098 = 939 kW
Total indicated power of the cylinder = IP_{1 }+IP_{2 }+ IP_{3 }+ IP_{4}
IP = 935 + 935 + 937 + 939 = 3746 kW
Mechanical efficiency of the engine = \frac { BP }{ IP } =\frac { 3037 }{ 3746 } = 81.07%
28. An ideal air standard Otto cycle has a compression ratio of 8.5. If the ratio of the specific heats of air (γ) is 1.4, what is the thermal efficiency (in percentage) of the Otto cycle?
(a) 57.5
(b) 45.7
(c) 52.5
(d) 95
(2 Mark, 2002)

Ans: a
Explanation:
Efficiency \left( \eta \right) =1\frac { 1 }{ { r }^{ \gamma 1 } }
=> \eta =1\frac { 1 }{ { 8.5 }^{ 1.41 } } = 57.5%
29. Which one of the following is NOT a necessary assumption for the airstandard Otto cycle?
(a) All processes are both internally as well as externally reversible.
(b) Intake and exhaust processes are constant volume heat rejection processes.
(c) The combustion process is a constant volume heat addition process.
(d) The working fluid is an ideal gas with constant specific heats.
(1 Mark, 2008)

Ans: b
Explanation:
Intake process is not heat rejection process.
30. For an engine operating on air standard Otto cycle, the clearance volume is 10% of the swept volume. The specific heat ratio of air is 1.4. the air standard cycle efficiency is
(a) 38.3%
(b) 39.8%
(c) 60.2%
(d) 61.7%
(2 Mark, 2003)

Ans: d
Explanation:
Given, v_{c} = 0.1 * v_{s}
Compression ratio (r) = \frac { { v }_{ c }+{ v }_{ s } }{ { v }_{ c } }
=> r = \frac { { 0.1v }_{ s }+{ v }_{ s } }{ { 0.1v }_{ s } } =\frac { 1.1{ v }_{ s } }{ 0.1{ v }_{ s } } = 11
Air standard efficiency \left( \eta \right) =1\frac { 1 }{ { r }^{ \gamma 1 } }
=> \eta =1\frac { 1 }{ { 11 }^{ 1.41 } } = 61.7%
31. An engine working on air standard Otto cycle has a cylinder diameter of 10 cm and stroke length of 15 cm. The ratio of specific heats for air is 1.4. If the clearance volume is 196.3 cc and the heat supplied per kg of air per cycle is 1800 kJ/kg, the work output per cycle per kg of air is
(a) 879.1 kJ
(b) 890.2 kJ
(c) 895.3 kJ
(d) 973.5 kJ
(2 Mark, 2004)

Ans: d
Explanation:
Swept volume \left( { v }_{ s } \right) =\frac { \pi }{ 4 } { D }^{ 2 }L =
=> { v }_{ s }=\frac { \pi }{ 4 } { \times 10 }^{ 2 }\times 15 = 1178.09 cc
Compression ratio (r) = \frac { { v }_{ c }+{ v }_{ s } }{ { v }_{ c } } =\frac { 196.3+1178.09 }{ 196.3 } = 7
Efficiency of the cycle \left( \eta \right) =1\frac { 1 }{ { r }^{ \gamma 1 } }
=> \eta =1\frac { 1 }{ { 7 }^{ 1.41 } } = 0.5408
Again, \eta =\frac { W }{ Q }
=> 0.5408=\frac { W }{ 1800 }
=> W = 973.44 kJ/kg
32. A reversible thermodynamic cycle containing only three processes and producing work is to be constructed. The constraints are:
(i) there must be one isothermal process,
(ii) there must be one isentropic process,
(iii) the maximum and minimum cycle pressures and the clearance volume are fixed, and
(iv) polytropic processes are not allowed.
Then the numbers of possible cycles are
(a) 1
(b) 2
(c) 3
(d) 4
(2 Mark, 2005)
Common Data for Questions 33, 34:
In two air standard cycles – one operating on the Otto and the other on the Brayton cycle – air is isentropically compressed from 300 to 450 K. Heat is added to raise the temperature to 600 K in the Otto cycle and to 550 K in the Brayton cycle.
33. If η_{o} and η_{B} are the efficiencies of the Otto and Brayton cycles, then
(a) η_{o} = 0.25, η_{B} = 0.18
(b) η_{o} = η_{B} = 0.33
(c) η_{o} = 0.5, η_{B} = 0.45
(d) it is not possible to calculate the efficiencies unless the temperature after the expansion is given.
(2 Mark, 2005)
34. If W_{o} and W_{B} are work outputs per unit mass, then
(a) W_{o} > W_{B}
(b) W_{o} < W_{B}
(c) W_{o} = W_{B}
(d) it is not possible to calculate the work outputs unless the temperature after the expansion is given.
(2 Mark, 2005)

Ans: a
Explanation:
Heat addition for Otto cycle = c_{v}(T_{3} – T_{2}) = 0.718(600 – 450) = 107.7 kJ/kg
Heat addition for Bryton cycle = c_{p}(T_{3} – T_{2}) = 1.0(600 – 500) = 100 kJ/kg
From the previous question, { \eta }_{ otto }={ \eta }_{ Bryton }
=> \frac { { W }_{ o } }{ { \left( { Q }_{ H } \right) }_{ o } } =\frac { { W }_{ B } }{ { \left( { Q }_{ H } \right) }_{ B } }
We know, { \left( { Q }_{ H } \right) }_{ o } > { \left( { Q }_{ H } \right) }_{ B }
So, { W }_{ o } > { W }_{ B }
35. Group I shows different heat addition processes in power cycles. Likewise, Group II shows different heat removal processes. Group III lists power cycles. Match items from Groups I, II and III.
(a) P – S – 5, R – U – 3, P – S – 1, Q – T – 2
(b) P – S – 1, R – U – 3, P – S – 4, P – T – 2
(c) R – T – 3, P – S – 1, P – T – 4, Q – S – 5
(d) P – T – 4, R – S – 3, P – S – 1, P – S – 5
(2 Mark, 2006)

Ans: a
Explanation:
36. The stroke and bore of a four stroke spark ignition engine are 250 mm and 200 mm respectively. The clearance volume is 0.001 m^{3}. If the specific heat ratio γ = 1.4, the airstandard cycle efficiency of the engine is
(a) 46.40%
(b) 56.10%
(c) 58.20%
(d) 62.80%
(2 Mark, 2007)

Ans: c
Explanation:
Swept volume = \left( { v }_{ s } \right) =\frac { \pi }{ 4 } { D }^{ 2 }L
=> { v }_{ s }=\frac { \pi }{ 4 } \times { 0.2 }^{ 2 }\times 0.25 = 0.00785 m^{3}
Compression ratio (r) = \frac { { v }_{ c }+{ v }_{ s } }{ { v }_{ c } }
=> r =\frac { 0.001+0.00785 }{ 0.001 } = 8.85
Air standard efficiency \left( \eta \right) =1\frac { 1 }{ { r }^{ \gamma 1 } }
=> \eta =1\frac { 1 }{ { 8.85 }^{ 1.41 } } = 0.582 = 58.2%
Common Data for Questions 37 & 38:
A thermodynamic cycle with an ideal gas as working fluid is shown below.
37. The above cycle is represented on Ts plane by
(2 Mark, 2007)

Ans: c
Explanation:
The processes such as constant volume, Constant pressure, and adiabatic are correctly plotted in option ‘c’.
38. If the specific heats of the working fluid are constant and the value of specific heat ratio γ is 1.4, the thermal efficiency (%) of the cycle is
(a) 21
(b) 40.9
(c) 42.6
(d) 59.7
(2 Mark, 2007)

Ans: a
Explanation:
The cycle given in the question is Lenoir cycle. Efficiency of Lenoir cycle is:
\eta =1\gamma \left( \frac { { r }_{ p }^{ 1/\gamma }1 }{ { r }_{ p }1 } \right)Here, { r }_{ p }=\frac { 400 }{ 100 } =4
Now, \eta =11.4\left( \frac { { 4 }^{ \frac { 1 }{ 1.4 } }1 }{ 41 } \right) = 0.21 = 21%
39. In an airstandard Otto cycle, the compression ratio is 10. The condition at the beginning of the compression process is 100 kPa and 27°C. Heat added at constant volume is 1500 kJ/kg, while 700kJ/kg of heat is rejected during the other constant volume process in the cycle. Specific gas constant for air = 0.287 kJ/kgK. The mean effective pressure (in kPa) of the cycle is
(a) 103
(b) 310
(c) 515
(d) 1032
(2 Mark, 2009)

Ans: d
Explanation:
Using the ideal gas equation:
{ v }_{ 1 }=\frac { R{ T }_{ 1 } }{ { p }_{ 1 } }=> { v }_{ 1 }=\frac { 0.287\times 300 }{ 100 } = 0.861 m^{3}/kg
{ v }_{ 2 }=\frac { { v }_{ 1 } }{ r } =\frac { 0.861 }{ 10 } = 0.0861 m^{3}/kg
Specific workone (W) = HA – HR = 1500 – 700 = 800 kJ/kg
Mean effective pressure (P_{mean}) = \frac { W }{ { v }_{ 1 }{ v }_{ 2 } } =\frac { 800 }{ 0.8610.0861 } = 1032.39 kPa
40. A turbocharged fourstroke direct injection diesel engine has a displacement volume of 0.0259m^{3} (25.9 litres). The engine has an output of 950 kW at 2200 rpm. The mean effective pressure in MPa is closest to
(a) 2
(b) 1
(c) 0.2
(d) 0.1
(2 Mark, 2010)

Ans: a
Explanation:
Displacement volume per second (V_{s})= \frac { ALn }{ 60 }
As the engine is four stroke:
So, n = \frac { N }{ 2 } =\frac { 2200 }{ 2 } = 1100
Now, V_{s }= \frac { 0.0259\times 1100 }{ 60 } = 0.4748 m^{3}/s
Power = { P }_{ mean }{ \times { V }_{ s } }
=> 950 * 1000 = P_{mean }* 0.4748
=> P_{mean }= 2 MPa
41. The crank radius of a single–cylinder I. C. engine is 60mm and the diameter of the cylinder is 80mm. The swept volume of the cylinder in cm^{3} is
(a) 48
(b) 96
(c) 302
(d) 603
(2 Mark, 2012)

Ans: d
Explanation:
The stroke length (L) = 2r = 120 mm = 12 cm
diameter of the cylinder (D)= 80 mm = 8 cm
Swept volume (V_{s}) = \frac { \pi }{ 4 } { D }^{ 2 }L
=> V_{s }= \frac { \pi }{ 4 } \times { 8 }^{ 2 }\times 12=603 cm^{3}
42. In an airstandard Otto cycle, air is supplied at 0.1 MPa and 308 K. The ratio of the specific heats (γ) and the specific gas constant (R) of air are 1.4 and 288.8 J/kg.K, respectively. If the compression ratio is 8 and the maximum temperature in the cycle is 2660 K, the heat (in kJ/kg) supplied to the engine is ______.
(2 Mark, 20014[1])

Ans: 1409.6
Explanation:
In process 1 2:
\frac { { T }_{ 2 } }{ { T }_{ 1 } } ={ \left( \frac { { v }_{ 1 } }{ { v }_{ 2 } } \right) }^{ \gamma 1 }={ 8 }^{ 0.4 }=> { T }_{ 2 }=308\times { 8 }^{ 0.4 }=707.6K
Here, { c }_{ v }=\frac { R }{ \gamma 1 } =\frac { 288.8 }{ 0.4 } = 722 j/kgK
Heat supplied is given by (Q) = c_{v}(T_{3} – T_{2}) kJ/kg
=> Q = 722 * (2660 – 707.6) = 1409.6 kJ/kg
43. A diesel engine has compression ratio of 17 and cutoff takes place at 10% of the stroke. Assuming ratio of specific heats (γ) as 1.4, the airstandard efficiency (in percent) is ___.
(2 Mark, 20014[3])

Ans: 59.6
Explanation:
v_{3} – v_{2} = 0.1(v_{1} – v_{2})
=> \frac { { v }_{ 3 } }{ { v }_{ 2 } } 1=0.1\left( \frac { { v }_{ 1 } }{ { v }_{ 2 } } 1 \right)
=> \frac { { v }_{ 3 } }{ { v }_{ 2 } } =0.1\left( 171 \right) +1
=> Cutoff ratio (r_{c}) = 2.6
Efficiency \left( \eta \right) =1\frac { 1 }{ { r }^{ \gamma 1 } } .\frac { { r }_{ c }^{ \gamma }1 }{ \gamma \left( { r }_{ c }1 \right) }
=> \eta =1\frac { 1 }{ { 17 }^{ 1.41 } } .\frac { { 2.6 }^{ 1.4 }1 }{ 1.4\left( 2.61 \right) } = 0.596 = 59.6%
44. In a compression ignition engine, the inlet air pressure is 1 bar and the pressure at the end of isentropic compression is 32.42 bar. The expansion ratio is 8. Assuming ratio of specific heats (γ) as 1.4, the air standard efficiency (in percent) is ____.
(2 Mark, 20014[4])

Ans: 59.6
Explanation:
Expansion ratio (r_{e}) = 8
Compression ratio (r) = \frac { { V }_{ 1 } }{ { V }_{ 2 } } ={ \left( \frac { { p }_{ 2 } }{ { p }_{ 1 } } \right) }^{ \frac { 1 }{ \gamma } }={ \left( \frac { 32.42 }{ 1 } \right) }^{ \frac { 1 }{ 1.4 } } = 12
Cutoff ratio (r_{c}) = \frac { r }{ { r }_{ e } } =\frac { 12 }{ 8 } =1.5
This is diesel cycle.
Efficiency \left( \eta \right) =1\frac { 1 }{ { r }^{ \gamma 1 } } .\frac { { r }_{ c }^{ \gamma }1 }{ \gamma \left( { r }_{ c }1 \right) }
=> \eta =1\frac { 1 }{ { 12 }^{ 1.41 } } .\frac { { 1.5 }^{ 1.4 }1 }{ 1.4\left( 1.51 \right) } = 0.596 = 59.6%
45. Air enters a diesel engine with a density of 1.0 kg/m^{3}. The compression ratio is 21. At steady state, the air intake is 30 × 10^{3} kg/s and the network output is 15 kW. The mean effective pressure (in kPa) is _____.
(1 Mark, 20015[1])

Ans: 525
Explanation:
Volume at the inlet (V_{1}) = \frac { m }{ \rho } =\frac { 30\times { 10 }^{ 3 } }{ 1 } =30\times { 10 }^{ 3 } m^{3}/s
Compression ratio (r) = \frac { { V }_{ 1 } }{ { V }_{ 2 } } = 21
=> { V }_{ 2 }=\frac { { V }_{ 1 } }{ r } =\frac { 30\times { 10 }^{ 3 } }{ 21 } =1.428\times { 10 }^{ 3 } m^{3}/s
Mean effective pressure = \frac { W }{ { V }_{ 1 }{ V }_{ 2 } }
=> P_{mean}= \frac { 15 }{ 30\times { 10 }^{ 3 }1.428\times { 10 }^{ 3 } } = 525 kPa
46. For the same values of peak pressure, peak temperature and heat rejection, the correct order of efficiencies for Otto, Dual and Diesel cycles is
(a) { \eta }_{ otto } > { \eta }_{ dual } > { \eta }_{ diesel }
(b) { \eta }_{ diesel } > { \eta }_{ dual } > { \eta }_{ otto }
(c) { \eta }_{ dual } > { \eta }_{ diesel } > { \eta }_{ otto }
(d) { \eta }_{ diesel } > { \eta }_{ otto } > { \eta }_{ dual }
(1 Mark, 20015[2])
47. An airstandard Diesel cycle consists of the following processes:
12: Air is compressed isentropically.
23: Heat is added at constant pressure.
34: Air expands isentropically to the original volume.
41: Heat is rejected at constant volume.
If γ and T denote the specific heat ratio and temperature, respectively, the efficiency of the cycle is
(a) 1 \frac { { T }_{ 4 }{ T }_{ 1 } }{ { T }_{ 3 }{ T }_{ 2 } }
(b) 1 \frac { { T }_{ 4 }{ T }_{ 1 } }{ \gamma \left( { T }_{ 3 }{ T }_{ 2 } \right) }
(c) 1 \frac { \gamma \left( { T }_{ 4 }{ T }_{ 1 } \right) }{ { T }_{ 3 }{ T }_{ 2 } }
(d) 1 \frac { { T }_{ 4 }{ T }_{ 1 } }{ \left( \gamma 1 \right) \left( { T }_{ 4 }{ T }_{ 1 } \right) }
(1 Mark, 20015[3])

Ans: b
Explanation:
\eta =\frac { W }{ { Q }_{ H } } =\frac { { Q }_{ H }{ Q }_{ L } }{ { Q }_{ H } } =1\frac { { Q }_{ L } }{ { Q }_{ H } }=> \eta =1\frac { { c }_{ v }\left( { T }_{ 4 }{ T }_{ 1 } \right) }{ { c }_{ p }\left( { T }_{ 3 }{ T }_{ 2 } \right) }
=> \eta =1\frac { { T }_{ 4 }{ T }_{ 1 } }{ \gamma \left( { T }_{ 3 }{ T }_{ 2 } \right) }
48. Propane (C_{3}H_{8}) is burned in an oxygen atmosphere with 10% deficit oxygen with respect to the stoichiometric requirement. Assuming no hydrocarbons in the products, the volume percentage of CO in the products is ___.
(1 Mark, 20016[1])

Ans: 14.29
Explanation:
The stoichiometric reaction:
{ C }_{ 3 }{ H }_{ 8 }+{ 5O }_{ 2 }\rightarrow { 3CO }_{ 2 }+4{ H }_{ 2 }OAt 10% deficit oxygen, the reaction becomes:
{ C }_{ 3 }{ H }_{ 8 }+{ 4.5O }_{ 2 }\rightarrow { 2CO }_{ 2 }+CO+4{ H }_{ 2 }OVolume percentage of CO in product side =\frac { Number.of.molecules.of.CO }{ Total.number.of.molecules.in.product.side } \times 100 = \frac { 1 }{ 2+1+4 } \times 100 = 14.28%
49. Air contains 79% N_{2} and 21% O_{2} on a molar basis. Methane (CH_{4}) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N_{2} in the products is _____.
(2 Mark, 20017[1])

Ans: 73.83
Explanation:
Air contains 21% of oxygen an 79% of nitrogen.
So, 1 mole of 0_{2 }= \frac { 79 }{ 21 } mole of N_{2}
Stoichiometric equation:
{ CH }_{ 4 }+2\left( { O }_{ 2 }+\frac { 79 }{ 21 } { N }_{ 2 } \right) \rightarrow { CO }_{ 2 }+2{ H }_{ 2 }O+2\times \frac { 79 }{ 21 } { N }_{ 2 }With 50% excess air the equation is:
{ CH }_{ 4 }+1.5\times 2\left( { O }_{ 2 }+\frac { 79 }{ 21 } { N }_{ 2 } \right) \rightarrow { CO }_{ 2 }+2{ H }_{ 2 }O+3\times \frac { 79 }{ 21 } { N }_{ 2 }+{ O }_{ 2 }% of nitrogen in product is: \frac { Number\quad of\quad { N }_{ 2 }\quad molecules }{ Total\quad number\quad of\quad molecules\quad at\quad output }
=> % of N_{2 }= \frac { 3\times 3.76 }{ 3\times 3.76+1+2+1 } = 73.83
50. An engine working on air standard Otto cycle is supplied with air at 0.1 MPa and 35^{0}C. The compression ratio is 8. The heat supplied is 500 kJ/kg. Property data for air: c_{p} = 1.005 kJ/kgK, c_{v} = 0.718 kJ/kgK, R = 0.287 kJ/kgK. The maximum temperature (in K) of the cycle is _________ (correct to one decimal place).
(2 Mark, 20018[1])

Ans: 1403.98
Explanation:
Given, compression ratio (r) = \frac { { V }_{ 2 } }{ { V }_{ 1 } } = 8
For the process 12: \frac { { T }_{ 2 } }{ { T }_{ 1 } } ={ \left( \frac { { V }_{ 1 } }{ { V }_{ 2 } } \right) }^{ \gamma 1 }
=> \frac { { T }_{ 2 } }{ 35+273 } ={ \left( 8 \right) }^{ 1.41 }
=> { T }_{ 2 } = 707.6 K
In process 23:
c_{v}(T_{3} – T_{2}) = 500
=> 0.718(T_{3} –707.6) = 500
=> T_{3 }= 1403.98 K
51. A vehicle powered by a spark ignition engine follows air standard Otto cycle (γ =1.4). The engine generates 70 kW while consuming 10.3 kg/hr of fuel. The calorific value of fuel is 44,000 kJ/kg. The compression ratio is _______ (correct to two decimal places).
(2 Mark, 20018[2])

Ans: 7.61
Explanation:
Efficiency of otto cycle (\eta) =\frac { { W }_{ net } }{ { Q }_{ in } }
=> \eta =\frac { { W }_{ net } }{ \dot { m } \times CV } =\frac { 70 }{ \frac { 10.3 }{ 3600 } \times 44000 } = 0.556
Again, \eta =1\frac { 1 }{ { r }^{ \gamma 1 } }
=> 0.556=1\frac { 1 }{ { r }^{ 1.41 } }
=> compression ratio (r) = 7.61
52. An air standard Otto cycle has thermal efficiency of 0.5 and the mean effective pressure of the cycle is 1000 kPa. For air, assume specific heat ratio γ = 1.4 and specific gas constant R = 0.287 kJ/kg.K. If the pressure and temperature at the beginning of the compression stroke are 100 kPa and 300 K, respectively, then the specific network output of the cycle is ___ kJ/kg (round off to two decimal places).
(2 Mark, 20019[2])

Ans: 709
Explanation:
Ideal gas equation, Pv = mRT
Total volume (V_{T}) = \frac { mRT }{ P }
=> V_{T }= \frac { 0.287\times 300 }{ 100 } = 0.861 m^{3}/kg
Thermal efficiency of otto cycle \left( \eta \right) =1\frac { 1 }{ { r }^{ \gamma 1 } }
=> 0.5=1\frac { 1 }{ { r }^{ 1.41 } }
=> compression ratio (r) = 5.656
=> \frac { { V }_{ T } }{ { V }_{ C } } =5.656
=> { V }_{ C }=\frac { 0.861 }{ 5.656 } = 0.152 m^{3}/kg
Swept volume (V_{S}) = V_{T} – V_{C}
V_{S} = 0.861 – 0.152 = 0.709 m^{3}/kg
Specific network output of the cycle:
W_{net} = P_{mep }* V_{S }= 1000 * 0.709 = 709 kJ/kg