Me 6301 Engineering Thermodynamics Short Questions and Answers - Unit 2

 ME

6301

ENGINEERING  THERMODYNAMI  THERMO DYNAMICS CS uNIT - II [FOR THIRD SEMESTER B.E MECHANICAL  ENGINEERING  ENGINE ERING STUDENTS S TUDENTS] ] COMPILED BY

BIBIN.C ASSISTANT PROFESSOR DEPARTMENT OF MECHANICAL ENGINEERING RMK COLLEGE OF ENGINEERING AND TECHNOLOGY GuMMIDIPOONDI TALuK  TIRuVALLuR DIST DI ST

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 e nergy 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?

COMPILED BY BIBIN, AP/MECH, RMKCET

Page 2

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. ii.

The source and sink are maintained at constant temperature. 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 10. Define the term COP? COMPILED BY BIBIN, AP/MECH, RMKCET

Page 3

ME 6301 - ENGINEERING THERMODYNAMICS Co-efficient of performance is defined as the ratio of heat extracted or rejected to work input. COP =

Heat extracted or Rejected  Work input

11.Write 11. Write the expression for COP of a heat pump and a refrigerator? COP of heat pump (COP ) HP =

COP of Refrigerator (COP ) REF  =

T2 T2 − T 1 T1 T2 − T 1

12.What 12. What is the relation between COP HP and COP ref ?

(COP ) HP

= (COP ) REF  +1

13.Why 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 moves as fast as possible. This speed variation variation during the same stroke of the piston is not possible.

iii.

It is not possible to avoid friction moving parts completely.

14.Name 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 15. Why a heat engine cannot have 100% efficiency? COMPILED BY BIBIN, AP/MECH, RMKCET

Page 4

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 16. When will be the Carnot cycle efficiency is maximum? Carnot cycle efficiency is maximum when the initial temperature is 0° K.

17.What 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 18. Write the expression for efficiency of the Carnot cycle. η Carnot  =

T2 − T 1 T2

19.What 19. What are the limitations of Carnot cycle? i. ii.

No friction is considered for moving parts of the engine. There should not be any heat loss.

20.Define 20. Define availability. The maximum useful work obtained during a process in which the final condition of the system is the same as a s that of the surrounding is called availability of the system.

21.Define 21. Define available energy and unavailable energy. Available energy is the maximum thermal useful useful work under ideal ideal condition. The remaining part, which cannot be converted into work, is known as unavailable energy. 22.Explain 22. Explain the term source and sink.

COMPILED BY BIBIN, AP/MECH, RMKCET

Page 5

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 23. What do you understand by the entropy principle? The entropy of an isolated isolated system can never decrease. It always increases increases and remains constant only only when the process is reversible. reversible.

This is known known as principle principle of

increase in entropy or entropy principle.

24.Power 24. Power requirement requirement of a refrigerator is _________ Ans: Inversely proportional to cop

25.In 25. In SI Units, one ton of refrigeration is equal to __________ __________ Ans: 210KJ/min

26.The 26. The capacity of a domestic refrigerator is in the range of __________ Ans: 0.1 to 0.3 tonnes.

27.The 27. The vapour compression refrigerator employs the __________cycle Ans: Reversed Carnot

28.In 28. In vapour compression cycle the condition of refrigerant is dry saturated vapour ________ Ans: Before entering the compressor 29.Define 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 30. What is the unit of refrigeration? COMPILED BY BIBIN, AP/MECH, RMKCET

Page 6

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 31. Define refrigeration effect. The amount of heat extracted in a given time is known as refrigeration effect.

32.What 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 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.

COMPILED BY BIBIN, AP/MECH, RMKCET

Page 7