The document contains multiple choice and short answer questions related to thermodynamics and heat engines. Some sample questions include identifying the process represented by lines on a T-s diagram, defining key thermodynamic cycles like Otto and Diesel, and calculating efficiency and heat supplied for ideal Brayton cycles. Short answer questions provide definitions for terms like refrigerating capacity and differentiate concepts such as normal and shear stress. Numerical problems involve truss analysis and calculating properties of gas turbine cycles.

1. Yasser Siddiqi

2. MCQs
 An isothermal expansion process for a vapor on a Ts
diagram is shown by
 A horizontal line
 A vertical line
 Downward sloping curve
 Upward sloping curve

3.  The heat supply process in a diesel cycle is
 isochoric
 isobaric
 isothermal
 isentropic

4.  If the maximum temperature of Source is 200oC
and of the sink is 100oC, the Carnot efficiency of
the cycle will be
 0.5
 Less than 0.5
 Greater than 0.5
 None of the above

5.  In a Otto cycle, if the compression ratio decreases,
its efficiency
 Decreases
 Increases
 Remains same
 None of the above

6.  In an Electrolux refrigeration system, a weak
solution of ammonia in water flows from
 Separator to Absorber
 Absorber to Separator
 Generator to Separator
 Evaporator to Absorber

7.  Refrigerating effect is the heat removed in which of
the following pieces of equipment;
 Evaporator
 Condenser
 Compressor
 None of the above

8.  Which of the following parts is found in a Petrol
engine but not in a Diesel engine;
 Piston
 Engine bearing
 Fuel injector
 Spark plug

9.  During one revolution of the crankshaft, the four
stroke engine completes;
 Half stroke of piston
 One stroke of piston
 Two strokes of piston
 Four strokes of piston

10.  When a car is pulled with a rope, the rope is
subjected to;
 Normal tensile stress
 Normal compressive stress
 Shear stress
 None of the above

11.  The normal stress acting in the direction of the
axis of a slender member is called;
 Bearing stress
 Shear stress
 Axial stress
 None of the above

12.  COP for a reverse heat engine is always
 Greater than one
 Less than one
 zero
 None of the above

13.  The most efficient possible heat engine cycle is
 Otto cycle
 Brayton Cycle
 Carnot Cycle
 None of the above

14.  At compressor inlet in a vapor compression cycle,
the desired condition of the vapor is;
 Dry saturated vapor
 Wet vapor with high wetness fraction
 Wet vapor with low dryness fraction
 None of the above

15.  On a Ts diagram for a single phase fluid which line
slopes less steeply (tends towards horizontal)
 Constant Volume line
 Constant Pressure line
 Constant Temperature line
 None of the above

16.  During the heating process in a gas turbine, the
pressure of the working fluid;
 Decreases
 Increases
 Remains same
 None of the above

17.  A good refrigerant has a
 Boiling point below target temperature
 A low density in vapor form
 A low specific enthalpy of vaporization
 None of the above

18.  Air fuel mixture in a Petrol engine is prepared
inside the;
 Governor
 Cylinder
 Carburetor
 Fuel Pump

19. SAQs
 Define second law of Thermodynamics in context
of reversed heat engine.
 It is impossible to construct a device that operating in
a cycle will produce no effect other than transfer of
heat from a cooler to a hotter body.

20.  Define unit of refrigerating capacity.
 The standard unit of refrigeration in vogue is ton
refrigeration or simply ton denoted by the symbol TR.
It is equivalent to the production of cold at the rate at
which heat is to be removed from one US tonne of
water at 32oF to freeze it to ice at 32oF in one day or 24
hours. Thus
 1 TR = ( 1 x 2,000 lb x 144 Btu/lb ) / 24 hr
 = 12,000 Btu/hr = 200
Btu/min

21.  Why carburetor is not used in a Diesel engine?
 Carburetor is used in a petrol engine to prepare air fuel
mixture to enter into the cylinder during a suction
stroke. Whereas in a diesel engine only air is sucked in
which is compressed to a relatively high pressure and
temperature and then diesel is sprayed into it through
a fuel injector.

22.  Differentiate between a normal and a shear stress.
 The stress on a surface is an internally distributed force
system that can be resolved into two components:
normal (perpendicular) to the imaginary cut surface,
called normal stress, and tangent (parallel) to the
imaginary cut surface, called shear stress.

23.  What is the primary difference between a
refrigerator and a heat pump?
 Both the heat pump and the refrigerator are actually
reversed heat engines. A reversed heat engine lifts heat
energy from a low temperature energy reservoir and
throws it into high temperature energy reservoir. The
difference between the two is the difference of
purpose. A refrigerator is used to produce cold i.e. lift
heat energy from low temperature energy reservoir
whereas a heat pump is used to produce heat i.e. throw
heat into high temperature energy reservoir.

24.  What is the purpose of using Hydrogen gas in
Electrolux refrigeration system?
 Hydrogen gas is filled in the evaporator-absorber
circuit of Electrolux refrigeration system at a partial
pressure slightly less that total pressure of the system
to allow liquid nitrogen flowing from condenser to
evaporator to expand and attain thermodynamics
properties corresponding to that low partial pressure.

25.  What is ‘deflector piston’ and in which type of
engine it is used?
 The deflector piston is used in a two stroke engine.
When air fuel mixture enters the cavity between piston
and cylinder head, it finds exhaust port open. To
minimize leakage of fresh charge through the exhaust
port, the head of the piston is designed in a way to
deflect fresh charge towards top of the cylinder. In this
way, the fresh charge helps in pushing out the burnt
gases, a process known as scavenging.

26.  What is the difference between the pressure and
the stress.
 Pressure is generally applied by some fluid on its solid
boundary, whereas stress is induced inside a solid body
when it is subjected to some external loading.

27.  Define the Second Law of Thermodynamics in the
context of a Forward Heat Engine.
 It is impossible for a heat engine to produce a net work
output in a complete cycle if it exchanges heat with a
single reservoir.

28.  How do you define ‘work ratio’ for the Brayton
cycle?
 For Brayton cycle, work ratio is defined as Net work
output divided by Gross Work output.
 Net Work output = Turbine work – Compressor
work
 Gross Work output = Turbine work

29.  What is ‘under-cooling’ in a vapor compression
cycle? Show diagrammatically.
 When a refrigerant is condensed in a condenser it
reaches saturated liquid line. To make use of the
temperature difference between condensed refrigerant
and the cooling water, the condensation process in
further extended in the liquid region. This extension
of cooling process in the liquid region is called under-
cooling.

31.  Define bearing stress?
 The compressive normal stress that is produced when
one real surface presses against another is called the
bearing stress.

32. Numericals 1
 In a gas turbine unit, air is drawn at 1.02 bar and 15oC,
and is compressed to 6.12 bar. Calculate the thermal
efficiency and the heat supplied of the ideal Brayton
cycle, when the maximum cycle temperature is limited
to 800oC. (Take Cp = 1.005 kJ/kg K and γ = 1.4 for air)

34.  Pressure ratio. rp = P2/ P1 = 6.12/1.02 = 6
 For an isentropic process
 T1/T2 = (P1/P2)ɣ-1/ɣ
 T2 = T1 (P2/P1) ɣ-1/ɣ = 288 (6) 0.4/1.4 = 480.5 K

 Thermal efficiency, η = 1 – 1 / rpɣ-1/ɣ = 40.1%

 Heat supplied, Q23 = Cp(T3 – T2) = 1.005(1073 – 480.5) =
595.46 kJ/kg

35. Q2
 All members of the truss shown in the figure have a
cross-sectional area of 500 mm2. Determine the axial
stress in CD and EF.

36.  Constructing equilibrium equations for forces in
vertical direction at point D, we get
 NDC sin 45o – 21 kN = 0 or NDC = 29.7 kN
 Axial stress, σDC = NDC / Am = 29.7 x 103 / 500 x 10-6 =
59.4 x 106 N/m2 = 59.4 MPa (T)


37.  Taking moments about point C, we get
 -21 kN (2m) – NEF (2m) = 0
 Or NEF = - 21 kN
 Axial stress, σEF = NEF / Am = -21 x 103 / 500 x 10-6 = -42
x 106 N/m2 = 42 MPa (C)

38. Q3
 Air enters the compressor of a gas turbine at 85 kPa
and 0 oC. If the pressure ratio is 6 and the maximum
temperature is 1000 oC, find (a) the thermal efficiency
and (b) the heat supplied for the associated Brayton
cycle. (Take Cp = 1.005 kJ/kg K and γ = 1.4 for air)

40.  For an isentropic process
 T1/T2 = (P1/P2)ɣ-1/ɣ
 T2 = T1 (P2/P1) ɣ-1/ɣ = 273 (6) 0.4/1.4 = 455.5 K

 Thermal efficiency, η = 1 – 1 / rpɣ-1/ɣ = 40%

 Heat supplied, Q23 = Cp(T3 – T2) = 1.005(1273 – 455.5) =
821.58 kJ/kg

41. Q4
 Truss analysis showed the forces at joint A given in
figure. Determine the sequence in which the three
members at joint A should be assembled to minimize
the shear stress in the pin.

42.  Three different sequences are possible for the given
joint i.e. BCD, BDC, and CBD.
 We have to compute shear stress at two different
sections for every possible configuration and compare
it with the other.
Possible configuration Shear stress at sec 1 Shear stress at sec 2
BCD 32.77/Ap 22.94/Ap
BDC 32.77/Ap 40/Ap
CBD 40/Ap 22.94/Ap

43.  Where Ap is cross-sectional area of pin that is a
constant. Comparing the above-mentioned three
configurations it is evident that configuration BCD is
the preferred one since it results in lesser value of
maximum shear stress.

44. Q5
 A maximum temperature of 1600 oC is possible in an
Otto Cycle in which air enters the compression process
at 85 kPa and 30 oC. Find the heat addition and the
thermal efficiency of the cycle, if the compression
ratio is 6. (Use Cv = 0.718 kJ/kg K and γ = 1.4 for air)

46.  For an isentropic process from 1 to 2
 T1 / T2 = (v2/v1)ɣ-1
 T2 = T1 (v1/v2) ɣ-1 = 303 (6)0.4 = 620.4 K

 Thermal Efficiency, η = 1 – 1 / rv ɣ-1 = 1 – 1 / 60.4 =
51.1%

 Heat Supplied, Q = Cv (T3 – T2) = 0.718 (1873 –
620.4) = 899.3 kJ/kg

47. Q6
 Two possible joint configurations are to be evaluated.
The forces on joint in a truss were calculated and a
magnified view is shown in the figure. The pin
diameter is 20 mm. Determine which joint assembly is
better by calculating the maximum shear stress in the
pin for each case.

48.  Area of the pin = Ap = (π/4) * d2 = 3.14 x 10-4 m2
 We have to compute shear stress at 3 sections of both
configurations i.e. between every two consecutive
forces.
Section Shear stress in Shear stress in
Configuration 1 Configuration 2
1 32.68/Ap 32.68/Ap
2 58.31/Ap 43.97/Ap
3 30/Ap 30 / Ap

49.  Shear stress is maximum at section 2 in both of the
configurations. Comparing the two values it is
concluded that Configuration 2 is preferred since it
results in lesser value of maximum shear stress.