1. Applied Thermodynamics

2. 1. Air Standard Power Cycles
Introduction
Two important applications of thermodynamics are power generation and refrigeration.
Both are usually accomplished by systems that operate on thermodynamic cycles.
Hence the thermodynamic cycles are usually divided into two general categories, viz., “power cycles” and “ refrigeration cycles”;
Power or refrigeration cycles are further classified as “ gas cycles” and “ vapour cycles” ;

3. In case of gas cycles, the working substance will be in gaseous phase throughout the cycle, where as in vapour cycles, the working substance will be in liquid phase in one part of the cyclic process and will be in vapour phase in some other part of the cycle;
Thermodynamic cycles are also classified as “ closed cycles” and “ open cycles”.
In closed cycles, the working fluid is returned to its original state at the end of each cycle of operation and is recirculated.

4. In an open cycle, the working substance is renewed at the end of each cycle instead of being re-circulated.
In automobile engines, the combustion gases are exhausted and replaced by fresh air-fuel mixture at the end of each cycle.
Though the engine operates in a mechanical cycle, the working substance does not go through a complete thermodynamic cycle.

5. Basic Considerations in the Analysis of Power Cycles
The cycles encountered in actual devices are difficult to analyze because of the presence of friction, and the absence of sufficient time for establishment of equilibrium conditions during the cycle.
In order to make an analytical study of a cycle feasible, we have to make some idealizations by neglecting internal Irreversibilities and complexities.
Such cycles resemble the actual cycles closely but are made up of internal reversible processes.
These cycles are called ideal cycles.

6. Air Standard Cycles
In gas power cycles, the working fluid will be in gaseous phase throughout the cycle.
Petrol engines (gasoline engines), diesel engines and gas turbines are familiar examples of devices that operate on gas cycles.
All these devices are called “ Internal combustion engines” as the fuel is burnt within the boundaries of the system.
Because of the combustion of the fuel, the composition of the working fluid changes from a mixture of air and fuel to products of combustion during the course of the cycle.

7. The actual gas power cycles are complex.
However, considering that air is predominantly nitrogen which hardly undergoes any chemical reaction during combustion, the working fluid closely resembles air at all times.
The actual gas power cycles are complex.
In order that the analysis is made as simple as possible, certain assumptions have to be made.
These assumptions result in an analysis that is far from correct for most actual combustion engine processes, but the analysis is of considerable value for indicating the upper limit of performance.

8. Air standard assumptions
The working medium is a perfect gas with constant specific heats and molecular weight corresponding to values at room temperature.
No chemical reactions occur during the cycle The heat addition and heat rejection processes are merely heat transfer processes.
The processes are reversible.
Losses by heat transfer from the apparatus to the atmosphere are assumed to be zero in this analysis.
The working medium at the end of the process (cycle) is unchanged and is at the same condition as at the beginning of the process (cycle).
i.e Changes in kinetic and potential energies of the working substance are very small and hence negligible.

9. Air standard Carnot Cycle
The Carnot cycle is represented on P-v and T-s diagrams as in Fig.
The Carnot cycle is composed of four totally reversible processes: isothermal heat addition, isentropic expansion, isothermal heat rejection, and isentropic compression.
The Carnot cycle is the most efficient cycle that can be executed between a heat source at temperature and a sink at temperature , and its thermal efficiency is expressed as

10. Process 1 – 2: Reversible Adiabatic Compression
Process 1-2: In this, air is compressed isentropically from volume During this process heat rejected is zero. i.e.,
P: Increases from p1 to p2
V: Decreases from V1 to V2
T: Increases from T1 to T2
S: Remains same.
1W2 = =
1Q2 = 0 or

11. Process 2 -3: Isothermal Heat Addition
In this air is heated isothermally
so that volume increases and
Temperature remains constant.
Amount of heat supplied is equal
to the work done by the gas.
P: Decreases from p2 to p3
V: Increases from V2 to V3
T: Remains same.
S: Increases from S2 to S3
2W3= p2V2 ln = mRT2 ln
2Q3 = p2V2 ln

12. Reversible Adiabatic Expansion
Process 3 – 4:
Reversible Adiabatic Expansion
This is isentropic(Adiabatic) expansion
process.
Heat supplied during the process is zero. i.e., P: Decreases from p3 to p4
V: Increases from V3 to V4
T: Decreases from T3 to T4
S: Remains same.
3W4 = =
3Q4 = 0

13. Process 4 – 1:
Isothermal Heat Rejection
P: Increases from p4 to p1
V: Decreases from V4 to V1
T: Remains same.
S: Decreases from S4 to S1
4W1= p4V4 ln = mRT4 ln
4Q1 = p4V4 ln

16. And also,

18. Mean Effective Pressure Mean effective pressure may
be defined as the theoretical
pressure which, if it is maintained
constant throughout the volume
change of the cycle, would give the
same work output as that obtained from the cycle.
Or it is the constant pressure which produces the same work output while causing the piston to move through the same swept volume as in the actual cycle.

19. Mean effective Pressure:
When the piston moves from TDC to BDC, the air inside expands
resulting in work output. If Pm1 is the average pressure on the piston during
this stroke, the average force on the piston is 
Where d = diameter of piston or cylinder bore 
Work output = average force on piston X stroke length 
During the return stroke, as the piston moves from BDC to TDC, air is
compressed requiring work input of the average pressure on the piston
during this stroke is Pm2, the work input is given by;   
Where Pm is known as mean effective pressure and is the swept volume.
Usually the net work output is in kJ, volume in m3 and mean effective
pressure in bar.

20. Stirling cycle When a confined body of gas (air, helium, whatever) is
heated, its pressure rises.
This increased pressure can push on a piston and do work.
The body of gas is then cooled, pressure drops, and the
piston can return.
The same cycle repeats over and over, using the same
body of gas.
That is all there is to it. No ignition, no carburetion, no valve
train, no explosions.
Many people have a hard time understanding the Stirling
because it is so much simpler than conventional internal
combustion engines.

21. The Stirling cycle is represented on P-v and T-s diagrams as in Fig.
It consists of two isothermal processes and two isochors. 
Process 1-2: In this air is heated isothermally so that volume increases from Temperature remains constant.
Amount of heat supplied is equal to the work done by the gas.

22. Process 3-4: In this air is compressed isothermally from volume
Stirling Cycle:
Process 2-3: This is constant volume heat rejection process. Temperature decreases from pressure decreases from the heat rejected during the process is given by,
Process 3-4: In this air is compressed isothermally from volume
During this process heat rejected is equal to the work done by the gas.

23. This is constant volume heat addition process.
Stirling Cycle:
Process 4-1:
This is constant volume heat addition process.
Temperature increase from The heat added during the process is given by,

25. The Efficiency of the cycle:
Due to heat transfers at constant volume processes, the efficiency of the Stirling cycle is less than that of the Carnot cycle.
However if a regenerative arrangement is used such that,
i.e., the area under 2-3 is equal to the area under 4 -1 on T-s diagram, then the efficiency,

26. Otto cycle OR Constant volume cycle:
The Otto cycle is the ideal cycle for spark-ignition reciprocating engines.
It is named after Nikolaus A. Otto, who built a successful four-stroke engine in 1876.
This cycle is also known as constant volume cycle as the heat is received and rejected at constant volume.
The cycle consists of two adiabatic processes and two constant volume processes as shown in P-v and T-s diagrams.

27. Otto cycle OR Constant volume cycle: Process 1-2:
In this air is compressed isentropically from V1 to V2 Temperature increases from T1 to T2.
Since this is an adiabatic process heat rejected is zero. i.e.
  Process 2-3:
In this air is heated at constant volume and temperature increases from T2 to T3.
Heat supplied during this process is given by,

28. Otto cycle OR Constant volume cycle: Process 3-4:
In this air is expanded isentropically from V3 to V4 and temperature decreases from T3 to T4. Since this is an adiabatic process, the heat supplied is zero. i.e.,
Process 4-1:
In this air is cooled at constant volume and temperature decreases from T4 to T1. Heat rejected during this process is equal to change in internal energy and is given by,

29. The Efficiency of the cycle: Efficiency of the cycle is given by,.
Considering isentropic expansion process 3-4, Or
Considering isentropic compression process 1-2, Or
Substituting for in eqn (1) Or 1
Where, r = compression OR expansion ratio and

30. Mean effective pressure: We know that for Otto cycle,
the pressure ratio

32. Diesel cycle OR Constant pressure cycle:
The Diesel cycle is the ideal cycle for Compression Ignition reciprocating engines.
The CI engine was first proposed by Rudolph Diesel.
The Diesel cycle consists of one constant pressure heating process, one constant volume cooling process and two adiabatic processes as shown in P-v and T-s diagrams.
This cycle is also known as constant pressure cycle because heat is added at constant pressure.

33. Diesel cycle OR Constant pressure cycle:
Process 1-2:
During this process air is compressed adiabatically and volume decreases from V1 to V2 Heat rejected during this process is zero. i.e.,
Process 2-3:
During this process air is heated at constant pressure and temperature rises from T2 to T3 Heat supplied during this process is given by,

34. Diesel cycle OR Constant pressure cycle:
Process 3-4:
During this process air is expanded adiabatically and volume increases from V3 to V4.
Heat supplied during the process is zero. i.e.,
Process 4-1:
In this air is cooled at constant volume and temperature decreases from T4 to T1 .
Heat rejected during this process is given by,

35. The Efficiency of the cycle: The efficiency of the cycle is given by,
Considering process 1-2,
Cut-off ratio,
Let, compression ratio,
Expansion ratio,

36. Considering process 2-3,
Considering process 3-4,
in eqn (1), we get
Substituting for

37. Mean effective pressure:
we know that work done per kg in Diesel cycle is given by,
And the mean effective pressure is given by:

38. Expression for cut-off ratio:
Let ‘k’ be the cut-off in percentage of stroke (from
We know that,

39. Dual combustion or Limited pressure or Mixed cycle:
This cycle is a combination of Otto and Diesel cycles.
It is also called semi-diesel cycle because semi-diesel engines work on this cycle.
In this cycle heat is absorbed partly at constant volume and partly at constant pressure.
It consists of two reversible adiabatic or isentropic, two constant volume and a constant pressure processes as shown in P-v and T-s diagrams. 3 4 5

40. Dual combustion or Limited pressure or Mixed cycle:4 3
Process 1-2:
The air is compressed reversibly and adiabatically from temperature T1 to T2 .
No heat is rejected or absorbed by the air.
Process 2-3:
The air is heated at constant volume from T2 to T3.
Heat absorbed by the air is given by, 5

41. Dual combustion or Limited pressure or Mixed cycle:
Process 3-4:
The air heated at constant pressure from temperature T3 to T4.
The heat supplied by the fuel or heat absorbed by the air is given by,
Process 4-5:
The air is expanded reversibly and adiabatically from temperature T4 to T5 . No heat is absorbed or rejected during the process.
Process 5-1: The air is now cooled at constant volume from temperature T5 to T1 . Heat rejected by the air is given by, 4 3 5

42. The Efficiency of the cycle: The efficiency of the cycle is given by,5
Let, compression ratio,
Cut-off ratio,
Pressure ratio,
Expansion ratio,

43. Considering process 1-2,
Considering process 2-3,
Considering process 3-4,
Considering process 4-5,
Substituting for
in (1)

44. Mean effective pressure:
We know that work done per kg in dual cycle is given by,
And the mean effective pressure is given by:
Note:
1) For Otto cycle
2) For Diesel cycle

45. Comparison between Otto, Diesel and Dual combustion cycles
The important variables which are used as the basis for comparison of the cycles are compression ratio, peak pressure, heat supplied, heat rejected and the net work output.
In order to compare the performance of the Otto, Diesel and Dual combustion cycles some of these variables have to be fixed.

46. Comparison with same compression ratio and heat supply:

47. The comparison of these cycles for the same compression ratio and same heat supply are shown in on both p – V and T – S diagrams.
In these diagrams, cycle represents Otto Cycle, cycle 1-2-3’-4’-1 represents diesel cycle and cycle 1-2”-3”-4”-1 represents the dual combustion cycle for the same compression ratio and heat supply.

48. From the T-S diagram, it can be seen that area 5236 = area 522”3”6” = area 523’6’ as this area represents the heat supply which is same for all the cycles.
All the cycles start from the same initial point 1 and the air is compressed from state 1 to state 2 as the compression ratio is same.

49. It is seen from the T-s diagram, that for the same heat supply, the heat rejection in Otto cycle (area 5146) is minimum and heat rejection in Diesel cycle (area 514’6’) is maximum. Consequently Otto cycle has the highest work output and efficiency. Diesel cycle has the least efficiency and dual cycle has the efficiency between the two.

50. Therefore for the same compression ratio and same heat rejection, Otto cycle is the most efficient while the Diesel cycle is the least efficient.
It can also be seen from the same diagram that q3>q2>q1
We know that thermal efficiency is given by
1 – heat rejected/heat supplied
Thermal efficiency of these engines under given circumstances is of the following order
Diesel>Dual>Otto
Hence in this case it is the diesel cycle which shows greater thermal efficiency.

51. Problem 1 
In an Otto cycle, the upper and lower limits for the absolute temperature respectively are T1 and T2.
Show that for the maximum work, the ratio of compression should have the value

52. Solution: Process 1-2 is reversible adiabatic

53. Work done = Heat added - Heat rejected
In the above equation T3, T1 and Cv are constants.
Therefore for maximum work

55. Problem 2
An engine working on Otto cycle in which salient points are 1,2,3 and 4 has upper and lower temperature limits T3 and T1.
If the maximum work per kg of air is to be done, show that the intermediate temperatures are given by

56. Solution: For maximum work/kg in an Otto cycle

58. Problem 3
An engine working on the otto cycle has a suction pressure of 1 bar and a pressure of 14 bar at the end of compression.
Find Compression ratio, Clearance volume as a percentage of cylinder volume
The ideal efficiency and MEP if the pressure at the end of combustion is 21 bar.
Solution:
Given: P1 = 1 bar, P2 = 14 bar, P3 = 21 bars

61. Problem 4 In a constant volume cycle the pressure at the end of compression is 15 times that at the start, the temperature of air at the beginning of compression is 37° C and the maximum temperature attained in the cycle is 1950°C. Find, (i) the compression ratio (ii) thermal efficiency of the cycle (iii) heat supplied per kg of air (iv) the work done per kg of air Solution: Given: P2/P1 = 15 , T1 = 37ºC = 310 K T3 = 1950ºC = 2223 K

63. Heat supplied = Cv(T3-T2) = 0.72(2223 - 671.66)
= KJ/kg of air
Work done = 0.54 x
= KJ/kg of air

64. Problem 5 An air standard Diesel cycle has a compression ratio of 18 and the heat transferred to the working fluid per cycle is 2000 kJ/kg. At the beginning of the compression stroke, the pressure is 1 bar and the temperature is 300 K. Calculate the thermal efficiency. Given: rc = 18 P1 = 1 bar T1 = 300 K

65.  Heat transferred = Cp(T3 – T2)
T3 = K

66. (i) Pressure and temperatures at all salient points of the cycle.
Problem 6 
An engine with 200 mm cylinder diameter and 300 mm stroke length, works on the theoretical Diesel cycle. The initial pressure and temperature of air are 1 bar and 27° C. The cut off is at 8% of the stroke and compression ratio is 15. Determine
(i) Pressure and temperatures at all salient points of the cycle.
(ii) theoretical air standard efficiency.
(iii) mean effective pressure.
(iv) power developed if there are 400 working strokes per minute.

67. Solution: 
Given:
rc = 15,
P1 = 1 bar,
T1 = 27º C
d = 200 mm, L = 300 mm

72. Problem 7
  In a dual combustion cycle the compression ratio is 14, maximum pressure is limited to 55 bar.
The cut-off ratio is Air is admitted at a pressure of 1 bar. Find the thermal efficiency and M.E.P of the cycle.
 Solution: (Given):
rc = 14
P1 = 1 bar
P3 = 55 bar
Cut off ratio =  =1.07

77. Problem 8
  From the PV diagram of an engine working on the Otto cycle, it is found that the pressure in the cylinder after 1/8th of the compression stroke is executed is 1.4 bar. After 5/8th of the compression stroke, the pressure is 3.5bar. Compute the compression ratio and the air standard efficiency. Also if the maximum cycle temperature is limited to 1000.C, find the net work out put

82. Problem 9
An air standard diesel cycle has a compression ratio of 16. The temperature before compression is 27°C and the temperature after expansion is 627°C. Determine:
i) The net work output per unit mass of air
ii) Thermal efficiency
iii) Specific air consumption in kg/kWh.

86. Problem 10
  The compression ratio of a compression ignition engine working on the ideal Diesel cycle is 16. The temperature of air at the beginning of compression is 300K and the temperature of air at the end of expansion is 900K. Determine
i) cut off ratio
ii) expansion ratio and
iii) the cycle efficiency

90. Problem 11 An air standard limited pressure cycle has a compression ratio of 15 and compression begins at 0.1 MPa, 40°C. The maximum pressure is limited to 6 MPa and heat added is MJ/kg. Compute (i) the heat supplied at constant volume per kg of air (ii) the heat supplied at constant pressure per kg of air (iii) the work done per kg of air (iv) the cycle efficiency (v) cut off ratio and (vi) the m.e.p of the cycle