1. Limitation of the 1st law of thermodynamics
First law of thermodynamic states that Energy can neither be created nor be destroyed but it can be converted from one form to another. It is the law of conservation of energy.
According to this law every conservation of energy is possible.
like, Work  Heat & Heat  Work.
But in practically
Work  Heat , But Heat  Work(natural way)

2. Example-1 A cup of hot tea left alone in a cooler room will cool off.
The process satisfied since heat lost by the hot tea is equal to heat gained by the surroundings air. It satisfied the 1st law of thermodynamics.
Reverse of the process is not possible.

3. Example-2
Body A is at higher temperature T1 & Body B at lower temperature T2 .
If these body are brought in contact , heat Q will be transferred from body A to B till the temperature of both becomes equal.
Amount of heat lost by body A will be equal to the amount of heat lost by body B.
Heat never flow from lower temperature body to higher temperature body spontaneously .

4. Example-3

5. Other Examples
Water flows from higher level to a lower level but the flow can not be reversed.
Two gases will be mixed spontaneously when placed in an isolated system but the mixed gases can not be separate without external work.
Flywheel rotating at constant speed , it has kinetic energy. If the flywheel is stopped by applying brakes at that time generate the friction and also heat produce cause of friction. If we supplying same amount of energy to the flywheel , it would not possible to attain the same kinetic energy.

6. Limitation of the 1st law .
The process can not be proceed in a particular direction only.
e.g. heat transfer take place from hot body to cold body only. But heat can not be flow naturally from cold body to hot body.
All processes involving conversion of heat into work and vice-versa are not equivalent. The work can not be fully converted in to heat, but conversion of heat completely into work is not possible.
First law provides a necessary but not a sufficient condition for a process to ocure.

7. Thermal Reservoir.
A thermal reservoir is the part of environment which can exchange heat energy with the system.
It has sufficiently large capacity and its temperature is not affected by the quantity of heat transferred to or from it.
The temperature of a heat reservoir remain constant. The changes that do take place in the thermal reservoir as heat enters or leaves are so slow and so small that processes within it are quasistatic.
The reservoir at high temperature which supplies heat to the system is called HEAT SOURCE.
For example: Boiler Furnace, Combustion chamber, Nuclear Reactor.
The reservoir at low temperature which receives heat from the system is called HEAT SINK.
For example: Atmospheric Air, Ocean, river

8. Define the following terms:
1. Thermal reservoir,
2. Heat engine,
3. Heat pump

9. HEAT ENGINE.
A heat engine is such a thermodynamics system that operates in a cycle in which heat is transferred from heat source to heat sink.
For continuous production of work. Both heat and work interaction take place across the boundary of the engine.
It receive heat Q1 from a higher temperature reservoir at T1. It converts part of heat Q1 into mechanical work W1.
It reject remaining heat Q2 into sink at T2.
There is a working substance which continuously flow through the engine to ensure continuous/cyclic operation.

10. Performance of HE
Measured by thermal efficiency which is the degree of useful conversion of heat received into work.
hth = Net work output/ Total Heat supplied
= W/Q1 = (Q1 – Q2) / Q1
hth = 1 – Q2/Q1
= 1 – T2/T1; Since Q1/Q2 = T1/T2
Or, Thermal efficiency is defined as the ratio of net work gained (output) from the system to the heat supplied (input) to the system.

11. Heat Pump:
Heat pump is the reversed heat engine which removes heat from a body at low temperature and transfer heat to a body at higher temperature.
It receive heat Q2 from atmosphere at temperature T2 equal to atmospheric temperature.
It receive power in the form of work ‘W’ to transfer heat from low temperature to higher temperature.
It supplies heat Q1 to the space to be heated at temperature T1.

12. Performance of HP It is measured by coefficient of performance (COP).
Which is the ratio of amount of heat rejected by the system to the mechanical
work received by the system.
(COP)HP = Q1/W
= Q1/(Q1 – Q2)
= T1/ (T1 – T2).

13. Refrigerator
The primary function of a heat pump is to transfer heat from a low temperature system to a high temperature system, this transfer of heat can be utilized for two different purpose, either heating a high temperature system or cooling a low temperature system.
Depending upon the nature of use.
The heat pump is said to be acting either as a heat pump or as a refrigerator.
If its purpose is to cause heating effect it is called operating as a H.P. And if it is used to create cold effect, the HP is known to be operating as a refrigerator.

14. Performance of Refrigerator
(COP)ref = Heat received/ Work Input
= Q2/W = Q2/(Q1 – Q2)
(COP)ref = Q2/(Q1 – Q2)
= T2/(T1 – T2)
(COP)HP = (COP)ref + 1
COP is greater when heating a room than when cooling it.

15. second law of thermodynamics
1. Kelvin Planck Statement
2. Clausius statement
3. Concept of perpetual motion m/c of second kind
4. Principle of degradation of energy
5. Principle of increase of entropy
Kelvin Plank is applicable to HE while the clausius statement is applicable to HP.

16. Kelvin Plank Statement
It is impossible to construct such a H.E. that operates on cyclic process and converts all the heat supplied to it into an equivalent amount of work.
The following conclusions can be made from the statement
1. cyclic engine can not converts whole of heat into equivalent work.
2. There is degradation of energy in a cyclic heat engine as some heat has to be degraded or rejected.
Thus second law of thermodynamics is called the law of degradation of energy.
For satisfactory operation of a heat engine there should be a least two heat reservoirs source and sink.

17. Consider an engine shown in Fig
Consider an engine shown in Fig. which receives heat Q1 from high temperature heat reservoir at T1 , rejects heat Q2 to low temperature heat reservoir and dose the work. W= Q1 - Q2 The efficiency of the engine, hth = 1 – Q2/Q1 Since Q2 < Q1 the efficiency of the engine is always less then the unity.

18. Clausius statement
It is impossible to construct such a a device that operates on cyclic process and allows transfer of heat from a colder body to a hotter body without the aid of an external agency.
Heat energy can not flow body at lower temperature to a body at higher temperature , however , such heat transferred can be achieve in practice by expanding some of work energy on the system .e.g in case of refrigerator shown in fig.

19. Perpetual motion Machine(P.M.M)
Definition :- “Any device that violate either the first law and the second law of thermodynamics is called the perpetual motion machine.” Two types:- (1) perpetual motion machine of first kind(PMM1) (2) perpetual motion machine of second kind (PMM2)

20. (1) perpetual motion machine of first kind(PMM1)
A machine or a device which violates the first law of thermodynamics is called the PMM-1.
Such a machine will give continuous work without receiving energy from other system . In other word such a machine will create energy thus violating the first law of thermodynamics

21. (2) perpetual motion machine of second kind(PMM2)
A machine which violate the second law of thermodynamics is called the perpetual motion machine of second kind(PMM2).
Such a machine will exchange heat from a single reservoir and produce equal amount of work .
Refer the fig. , if Q2 = 0 , then ,W = Q1, its efficiency is 100 % and it violates Kelvin – Planck statement of second law of thermodynamics .
Second Law:- “ It is impossible construct a perpetual motion machine of second kind”

22. Equivalent of Kalvin Plank and Clausius statement
The Kalvin -plank and clausius statements of the second law and are equivalent in all respect.
The equivalence of the statement will be proved by the logic and violation of one statement leads to violation of second statement and vice versa.

23. Violation of Clausius statement
Consider a heat pump which violates the clausius statement shown in fig. i.e. The heat pump receives the heat Q2 from law temperature reservoir and heat reject the same heat to high temperature reservoir without any external work. (W=0)

24. Introduction an engine which receives heat Q1(Q1 > Q2) from high temperature reservoir, rejects heat Q2 and dose the work W=Q1 – Q2.
Combine the heat pump and engine into one system. The net energy transfers of the combined system are shown in fig.

25. We find that the combined system operates as an engine in which it receives heat from single high temperature heat reservoir as (Q1 – Q2) and converts into equal amount of work without any heat rejection.
Thus violating the Kelvin-Planck statement of the 2nd law of thermodynamics also.

26. Violation of K-P Statement
Consider an engine (PMM-2) which violates the Kelvin-Planck statements by receiving heat from a single high temperature heat reservoir at T1 and converts completely into work.
(W1 = Q1).
Now introducing a heat pump which is driven by the engine.
On the expense of work W1 = Q1 supplied by the engine, it extract the heat Q2 from low temperature heat reservoir and rejects heat (Q2+Q1) to high temperature heat reservoir.

27. Combine the engine and heat pump into one system working between two heat reservoir as shown in fig.
We find that the sole effect of combined system is to transfer heat Q2 from low temperature heat reservoir, T2 to higher temperature heat reservoir without any external work thus violating the Clausius statement also.

28. Carnot Cycle
Sadi carnot; based on second law of thermodynamics introduced the concepts of reversibility and cycle in He show that the temperature of heat source and heat sink are the basis for determining the thermodynamics efficiency of a reversible cycle.
He showed that all such cycles must reject heat to the sink and efficiency is never 100%.
To show a non existing reversible cycle, Carnot invented his famous but a hypothetical cycle known as Carnot cycle.
Carnot cycle consist of two isothermal and two reversible adiabatic or isentropic operation.

30. Carnot Cycle Process

31. Efficiency Of Carnot Engine

33. Negative sign indicate that the heat transferred from the system to surroundings.

35. Carnot Theorem
No heat engine operating in a cycle between two given thermal reservoir, with fixed temperature can be more efficient than a reversible engine operating between the same thermal reservoir.
Thermal efficiency hth = Work out/Heat supplied
Thermal efficiency of a reversible engine (hrev)
hrev = (T1 – T2)/T1
No engine can be more efficient than a reversible Carnot engine i.e hrev > hth

37. Absolute Thermodynamic T Scale

38. Absolute Thermodynamics T Scale

41. Meaning of Absolute zero on Thermodynamic scale

44. Carnot Efficiency Condition: 1. If T1 = T2; No work, = 0
= (Heat added – Heat rejected) / Heat added
= [mRT2lnV2/V1 – mRT4lnV3/V4 ]/ mRT2lnV2/V1
= 1- T1/T2
Condition:
1. If T1 = T2; No work, = 0
2. Higher the temperature diff, higher the efficiency
3. For same degree increase of source temperature or decrease in sink temperature carnot efficiency is more sensitive to change in sink temperature.

45. Clausius inequality
When ever a closed system undergoes a cyclic process, the cyclic integral is less than zero (i.e., negative) for an irreversible cyclic process and equal to zero for a reversible cyclic process.

46. The efficiency of a reversible H. E
The efficiency of a reversible H.E. operating within the temperature T1 & T2 is given by:
(i)

47. Now the efficiency of an irreversible H. E
Now the efficiency of an irreversible H.E. operating within the same temperature limit T1 & T2 is given by
(ii)

48. Combine equation (i) and (ii); we get
The equation for irreversible cyclic process may be written as:
I = Amount of irreversibility of a cyclic process.

49. Numerical:-1
Heat pump is used for heating the premises in winter and cooling the same during summer such that temperature inside remains 25ºC. Heat transfer across the walls and roof is found 2MJ per hour per degree temperature difference between interior and exterior. Determine the minimum power required for operating the pump in winter when outside temperature is 1ºC and also give the maximum temperature in summer for which the device shall be capable of maintaining the premises at desired temperature for same power input.

51. Numerical:-2
A reversible heat engine operates between temperature 8000C and 5000C of thermal reservoir. Engine drives a generator and a reversed carnot engine using the work output from the heat engine for each unit equality. Reversed Carnot engine abstracts heat from 5000C reservoir and rejected that to a thermal reservoir at 7150C. Determine the heat rejected to the reservoir by the reversed engine as a fraction of heat supplied from 8000C reservoir to the heat engine. Also determine the heat rejected per hour for the generator output of 300KW.

53. Numerical:-3
A reversible heat engine operates between two reservoirs at temperature of 6000C and 400C.The Engine drives a reversible refrigerator which operates between reservoirs at temperature of 400C and – 200C. The heat transfer to the heat engine is 2000KJ and net work output of combined engine refrigerator plant is 360KJ. Evaluate the heat transfer to the refrigerator and the net heat transfer to the reservoir at 400C.