This document provides a summary of key concepts in engineering thermodynamics for a third semester mechanical engineering course. It covers topics like the second law of thermodynamics, Carnot cycle efficiency, heat pumps, refrigerators, coefficient of performance, availability analysis, and vapor compression refrigeration cycles. The document is compiled by an assistant professor and contains 33 questions and answers on these thermodynamics topics.

1. ME 6301
ENGINEERING
THERMODYNAMICS
unit - ii
[FOR THIRD SEMESTER B.E MECHANICAL
ENGINEERING STUDENTS]
COMPILED BY
BIBIN.C
ASSISTANT PROFESSOR
DEPARTMENT OF MECHANICAL ENGINEERING
RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
GUMMIDIPOONDI TALUK
TIRUVALLUR DIST

2. ME 6301 - ENGINEERING THERMODYNAMICS
UNIT II - SECOND LAW AND AVAILABILITY ANALYSIS
1. State the Kelvin – Plank statement of second law of thermodynamics
Kelvin – Plank states that it is impossible to construct a heat engine working on
cyclic process, whose only purpose is to convert all the heat energy given to it into an equal
amount of work.
OR
It is impossible to construct a heat engine to produce network in a complete cycle if
it exchanges heat from a single reservoir at single fixed temperature.
2. State Clausius statement of second law of thermodynamics.
It states that heat can flow from hot body to cold without any external aid but heat
cannot flow from cold body to hot body without any external aid.
OR
It is impossible for a self-acting machine working in a cyclic process, to transfer heat
from a body at lower temperature to a body at a higher temperature without the aid of an
external agency.
3. State Carnot theorem.
It states that no heat engine operating in a cycle between two constant temperature
heat reservoirs can be more efficient than a reversible engine operating between the same
reservoirs.
4. What is absolute entropy(Third law of Thermodynamics)?
The entropy measured for all perfect crystalline solids at absolute zero temperature
is known as absolute entropy.
5. What are the Corollaries of Carnot theorem?
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3. ME 6301 - ENGINEERING THERMODYNAMICS
i. In all reversible engine operating between the two given thermal reservoirs
with fixed temperature, have the same efficiency.
ii. The efficiency of any reversible heat engine operating between two
reservoirs is independent of the nature of the working fluid and depends
only on the temperature of the reservoirs.
6. Define Heat pump.
A heat pump is a device, which is working in a cycle and transfers heat from lower
temperature to higher temperature.
7. Define Heat engine.
Heat engine is a machine, which is used to convert the heat energy into mechanical
work in a cyclic process.
OR
A heat engine is a device which is used to convert the thermal energy into
mechanical energy.
8. What are the assumptions made on heat engine?
i. The source and sink are maintained at constant temperature.
ii. The source and sink has infinite heat capacity.
9. What is the difference between a heat pump and a refrigerator?
Heat pump is a device which operating in cyclic process, maintains the temperature
of a hot body at a temperature higher than the temperature of surroundings.
A refrigerator is a device which operating in a cyclic process, maintains the
temperature of a cold body at a temperature lower than the temperature of the
surroundings.
10.Define the term COP?
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4. ME 6301 - ENGINEERING THERMODYNAMICS
Co-efficient of performance is defined as the ratio of heat extracted or rejected to
work input.
inputWork
RejectedorextractedHeat
=COP
11.Write the expression for COP of a heat pump and a refrigerator?
COP of heat pump
12
2
T
T
)(
T
=COP HP
−
COP of Refrigerator
12
1
T
T
)(
T
=COP REF
−
12.What is the relation between COPHP and COP ref?
1)()( +REFHP COP=COP
13.Why Carnot cycle cannot be realized in practical?
i. In a Carnot cycle all the four processes are reversible but in actual practice there
is no process is reversible.
ii. There are two processes to be carried out during compression and expansion.
For isothermal process the piston moves very slowly and for adiabatic process
the piston moves as fast as possible. This speed variation during the same
stroke of the piston is not possible.
iii. It is not possible to avoid friction moving parts completely.
14.Name two alternative methods by which the efficiency of a Carnot cycle can be
increased.
i. Efficiency can be increased as the higher temperature T2 increases.
ii. Efficiency can be increased as the lower temperature T1 decreases.
15.Why a heat engine cannot have 100% efficiency?
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5. ME 6301 - ENGINEERING THERMODYNAMICS
For all the heat engines there will be a heat loss between system and surroundings.
Therefore we can’t convert all the heat input into useful work.
16.When will be the Carnot cycle efficiency is maximum?
Carnot cycle efficiency is maximum when the initial temperature is 0°K.
17.What are the processes involved in Carnot cycle.
Carnot cycle consist of
i. Reversible isothermal compression
ii. Isentropic compression
iii. Reversible isothermal expansion
iv. Isentropic expansion
18.Write the expression for efficiency of the Carnot cycle.
2
12
T
T T
=Carnot
−
η
19.What are the limitations of Carnot cycle?
i. No friction is considered for moving parts of the engine.
ii. There should not be any heat loss.
20.Define availability.
The maximum useful work obtained during a process in which the final condition of
the system is the same as that of the surrounding is called availability of the system.
21.Define available energy and unavailable energy.
Available energy is the maximum thermal useful work under ideal condition. The
remaining part, which cannot be converted into work, is known as unavailable energy.
22.Explain the term source and sink.
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6. ME 6301 - ENGINEERING THERMODYNAMICS
Source is a thermal reservoir, which supplies heat to the system and sink is a
thermal reservoir, which takes the heat from the system.
23.What do you understand by the entropy principle?
The entropy of an isolated system can never decrease. It always increases and
remains constant only when the process is reversible. This is known as principle of
increase in entropy or entropy principle.
24.Power requirement of a refrigerator is _________
Ans: Inversely proportional to cop
25.In SI Units, one ton of refrigeration is equal to __________
Ans: 210KJ/min
26.The capacity of a domestic refrigerator is in the range of __________
Ans: 0.1 to 0.3 tonnes.
27.The vapour compression refrigerator employs the __________cycle
Ans: Reversed Carnot
28.In vapour compression cycle the condition of refrigerant is dry saturated vapour
________
Ans: Before entering the compressor
29.Define the unit for refrigeration
Unit of refrigeration is expressed in terms of tonne of refrigeration (TR). A tonne of
refrigeration is defined as the quantity of heat required to be removed form one tonne of
water at 0oC to convert into ice at 0oC in 24 hours.
30.What is the unit of refrigeration?
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7. ME 6301 - ENGINEERING THERMODYNAMICS
The capacity of refrigeration is expressed in tonnes of refrigeration (TOR).
1 tonnes of refrigeration = 210 kJ/min (or) = 3.5 kJ/sec (kW)
A tonne of refrigeration is defined as the quantity of heat to be removed in order to
form one tonne of ice at 0oC in 24 hours.
31.Define refrigeration effect.
The amount of heat extracted in a given time is known as refrigeration effect.
32.What is the refrigeration effect of the refrigerant?
Refrigeration effect is the total heat removed from the evaporator by the refrigerant.
It is called as Tonne of Refrigeration of kW.
33.Define COP of refrigeration.
The COP of a refrigeration system is the ratio of net refrigeration effect to the
work required to produce the effect.
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