1. A pplied Thermodynamics 1

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” ; 2

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. 3

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. 4

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. 5

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. 6

7. 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. 7

8. Air standard assumptions 1.The working medium is a perfect gas with constant specific heats and molecular weight corresponding to values at room temperature. 2.No chemical reactions occur during the cycle. The heat addition and heat rejection processes are merely heat transfer processes. 3.The processes are reversible. 4.Losses by heat transfer from the apparatus to the atmosphere are assumed to be zero in this analysis. 5.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. 8

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 9

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 V 1 to V 2 T: Increases from T 1 to T 2 S: Remains same. 1 W 2 = = 1 Q 2 = 0 or 10

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 p 2 to p 3 V: Increases from V 2 to V 3 T: Remains same. S: Increases from S 2 to S 3 2 W 3 = p 2 V 2 ln = mRT 2 ln 2 Q 3 = p 2 V 2 ln 11

12. Process 3 – 4: Reversible Adiabatic Expansion This is isentropic(Adiabatic) expansion process. Heat supplied during the process is zero. i.e., P: Decreases from p 3 to p 4 V: Increases from V 3 to V 4 T: Decreases from T 3 to T 4 S: Remains same. 3 W 4 = = 3 Q 4 = 0 12

13. Process 4 – 1: Isothermal Heat Rejection P: Increases from p 4 to p 1 V: Decreases from V 4 to V 1 T: Remains same. S: Decreases from S 4 to S 1 4 W 1 = p 4 V 4 ln = mRT 4 ln 4 Q 1 = p 4 V 4 ln 13

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15. 15

16. And also, 16

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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. 18

19. Mean effective Pressure: When the piston moves from TDC to BDC, the air inside expands resulting in work output. If P m1 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 P m2, the work input is given by; Where P m is known as mean effective pressure and is the swept volume. Usually the net work output is in kJ, volume in m 3 and mean effective pressure in bar. 19

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. 20

21. Stirling Cycle: 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. 21

22. 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. 22

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

24. 24

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, 25

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. 26

27. Otto cycle OR Constant volume cycle: Process 1-2: In this air is compressed isentropically from V 1 to V 2 Temperature increases from T 1 to T 2. 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 T 2 to T 3. Heat supplied during this process is given by, 27

28. Otto cycle OR Constant volume cycle: Process 3-4: In this air is expanded isentropically from V 3 to V 4 and temperature decreases from T 3 to T 4. 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, 28

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 Where, r = compression OR expansion ratio and. 1 29

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

31. 31

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. 32

33. Diesel cycle OR Constant pressure cycle: Process 1-2: During this process air is compressed adiabatically and volume decreases from V 1 to V 2 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 T 2 to T 3 Heat supplied during this process is given by, 33

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

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

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

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: 37

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

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. 4 5 3 39

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

41. Dual combustion or Limited pressure or Mixed cycle: Process 3-4: The air heated at constant pressure from temperature T 3 to T 4. 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 T 4 to T 5. No heat is absorbed or rejected during the process. Process 5-1: The air is now cooled at constant volume from temperature T 5 to T 1. Heat rejected by the air is given by, 3 4 5 41

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

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

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 44

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. 45

46. Comparison with same compression ratio and heat supply: 46

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 1- 2-3-4-1 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. 47

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. 48

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. 49

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. 50

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 51

52. Solution: Process 1-2 is reversible adiabatic Process 3-4 is reversible adiabatic 52

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

54. 54

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 55

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

57. 57

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 58

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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 61

62. 62

63. Heat supplied = C v (T 3 -T 2 ) = 0.72(2223 - 671.66) =1116.96 KJ/kg of air Work done = 0.54 x 1116.96 = 603.16 KJ/kg of air 63

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: r c = 18 P 1 = 1 bar T 1 = 300 K 64

65. Heat transferred = C p (T 3 – T 2 ) 2000 = 1.005(T3 -953.3)] T3 = 2943.34 K 65

66. 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. 66

67. Solution: Given: r c = 15, P 1 = 1 bar, T 1 = 27º C d = 200 mm, L = 300 mm 67

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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 1.07. Air is admitted at a pressure of 1 bar. Find the thermal efficiency and M.E.P of the cycle. Solution: (Given): r c = 14 P 1 = 1 bar P 3 = 55 bar Cut off ratio =  =1.07 72

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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 77

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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. 82

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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 86

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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 1.675 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 90

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95. 3. GAS TURBINES AND JET PROPULSION Introduction: Gas turbines are prime movers producing mechanical power from the heat generated by the combustion of fuels. They are used in aircraft, some automobile units, industrial installations and small – sized electrical power generating units. A schematic diagram of a simple gas turbine power plant is shown below. This is the open cycle gas turbine plant. 95

96. Working: Air from atmosphere is compressed adiabatically (idealized) in a compressor (usually rotary) i.e., Process 1–2. This compressed air enters the combustion chamber, where fuel is injected and undergoes combustion at constant pressure in process 2–3. The hot products of combustion expand in the turbine to the ambient pressure in process 3–4 and the used up exhaust gases are let out into the surroundings. 96

97. The compressor is usually coupled to the turbine, so that the work input required by the compressor comes from the turbine. The turbine produces more work than what is required by the compressor, so that there is net work output available from the turbine. Since the products of combustion cannot be re–used, real gas turbines work essentially in open cycles. The p–v and T–s diagrams of such a plant are shown above. 97

98. The compressor is usually coupled to the turbine, so that the work input required by the compressor comes from the turbine. The turbine produces more work than what is required by the compressor, so that there is net work output available from the turbine. Since the products of combustion cannot be re–used, real gas turbines work essentially in open cycles. The p–v and T–s diagrams of such a plant are shown above. 98

99. Brayton Cycle: This is the air– standard cycle for the gas turbine plant. It consists of two reversible adiabatic processes and two reversible isobars (constant pressure processes). The p–v and T–s diagrams of a Brayton Cycle are as shown. 99

100. Process 1 - 2: Reversible adiabatic compression. 2 – 3: Reversible constant pressure heat addition. 3 – 4: Reversible adiabatic expansion. 4 – 1: Reversible constant pressure heat rejection. A schematic flow diagram of this somewhat hypothetical gas turbine plant is shown below. 100

101. Though this plant works on a closed cycle, each of the four devices in the plant is a steady–flow device, in the sense that there is a continuous flow of the working fluid (air) through each device. Hence, the steady–flow energy equation is the basis for analysis, and can be applied to each of the four processes. Neglecting changes in kinetic and potential energies, the steady flow energy equation takes the from Q – W = ∆h = C p.∆T (Since air is assumed to be an ideal gas) Process 1 – 2 is reversible adiabatic, hence Q 1-2 = 0 W 1-2 = - C p.∆T = - C p (T 2 – T 1 ): - ve, work input Work of compression W c = |W 1 – 2 | = C p (T 2 -T 1 ) 101

102. Process 2–3 is a constant pressure process Heat added, Process 3-4 is again reversible adiabatic, +ve work output. 102

103. Process 4-1 is also a constant pressure process : -ve, i.e., heat is rejected Heat rejected, Q 2 = |Q 4-1 | = C p (T 4 -T 1 ) Therefore the cycle efficiency, 103

104. Therefore the cycle efficiency, For isentropic process 1-2, & for process 3-4, 104

105. Since p 3 = p 2 and p 4 = p 1 Compression ratio, 105

106. Therefore, Pressure ratio, Thus it can be seen that, for the same compression ratio, A closed cycle turbine plant is used in a gas – cooled nuclear reactor plant, where the source is a high temperature gas cooled reactor supplying heat from nuclear fission directly to the working fluid (gas/air). 106

107. Comparison between Brayton Cycle and Otto cycle:- For the same compression ratio, and nearly same net work output (represented by the area inside the p–v diagram), the Brayton cycle handles larger range of volume and smaller range of pressure than does the Otto cycle. 107

108. A Brayton cycle is not suitable as the basis for the working of reciprocating type of devices (Piston– Cylinder arrangements). A reciprocating engine cannot efficiently handle a large volume flow of low pressure gas. The engine (Cylinder) size becomes very large and friction losses become excessive. Otto cycle therefore is more suitable in reciprocating engines. 108

109. However, a Brayton cycle is more suitable than an Otto cycle, as a basis for a turbine plant. An I.C. engine is exposed to the highest temperature only intermittently (for short way during each cycle), so that there is time enough for it to cool. On the other hand, a gas turbine, being a steady flow device, is continuously exposed to the highest temperature. 109

110. Metallurgical considerations, therefore limit the maximum temperature that can be used. Moreover, in steady flow machines, it is easier to transfer heat at constant pressure than at constant volume. Besides, turbines can be efficiently handle large volume of gas flow. In view of all these, the Brayton cycle more suitable as the basis for the working of gas turbine plants. 110

111. Effect of irreversibility’s in turbine/compressor: In the ideal Brayton cycle, compression and expansion of air are assumed to be reversible and adiabatic. In reality, however, irreversibility’s do exist in the machine operations, even though they may be adiabatic. Hence the compression and expansion processes are not really constant entropy processes. Entropy tends to be increase (as per the principle of increase of entropy). 111

112. Effect of irreversibility’s in turbine/compressor: The T–s diagram of a Brayton cycle subject to irreversibility’s will be as shown. Irreversibility’s result in a reduction in turbine output by (h 4 -h 4S ) and in an increase in the compressor input by (h 2 – h 2S ). Hence the output reduces by the amount (h 4 –h 4S )+ (h 2 –h 2s ). Though heat input is also reduced by (h 2 -h 2s ), the cycle efficiency is less than that of an ideal cycle. The extent of losses due to irreversibility’s can be expressed in terms of the turbine and compressor efficiencies. 112

113. Turbine efficiency, Compressor efficiency, 113

114. Methods of improving the efficiency of Brayton cycle: Use of regeneration: The efficiency of the Brayton cycle can be increased by utilizing part of the energy of exhaust air from the turbine to preheat the air leaving the compressor, in a heat exchanger called regenerator. This reduces the amount of heat supplied Q 1 from an external source, and also the amount of heat rejected Q 2 to an external sink, by an equal amount. Since W net = Q 1 - Q 2 and both Q 1 and Q 2 reduce by equal amounts, there will be no change in the work output of the cycle. 114

115. 115

116. Heat added Q 1 = h 3 –h 2’ = C p (T 3 – T 2’ ) Heat rejected Q 2 = h 4’ – h 1 = C p (T 4’ – T 1 ) Turbine output W T = h 3 – h 4 = C p (T 3 – T 4 ) Compressor input W C = h 2 – h 1 = C p (T 2 – T 1 ) Regeneration can be used only if the temperature of air leaving the turbine at 4 is greater than that of air leaving the compressor at 2. In the regenerator, heat is transferred from air leaving the turbine to air leaving the compressor, thereby raising the temperature of the latter. The maximum temperature to which compressed air at 2 can be heated is equal to the temperature of turbine exhaust at 4. 116

117. This, however, is possible only in an ideal regenerator. In reality, T 2’ <T 4. The ratio of the actual temperature rise of compressed air to the maximum possible rise is called effectiveness of the regenerator. 117

118. With a regenerator, since W net remains unchanged, but Q 1 reduces, efficiency η = W net /Q 1 increases. This is also evident from the fact that the mean temperature of heat addition increases and the mean temperature of heat rejection reduces with the use of the regenerator, and efficiency is also given by 118

119. With regenerator, In the regenerator, Heat lost by hot air = Heat gained by cold air i.e., With an ideal regenerator, 119

120. therefore, 120

121. For a fixed ratio, the cycle efficiency decreases with increasing pressure ratio. In practice, a regenerator is expensive, heavy and bulky and causes pressure losses, which may even decrease the cycle efficiency, instead of increasing it. 121

122. 2.Multistage compression with inter cooling: 122

123. In this arrangement, compression of air is carried out in two or more stages with cooling of the air in between the stages. The cooling takes place in a heat exchanger using some external cooling medium (water, air etc). Shown above is a schematic flow diagram of a gas turbine plant with two-stage compression with inter cooling. 123

124. 1-2: first stage compression (isentropic) 2-3: inter cooling (heat rejection at constant pressure) 3-4: second stage compression (isentropic) 4-3: constant pressure heat addition 5-6: isentropic expansion 6-1: constant pressure heat rejection. 124

125. Air, after the first stage compression is cooled before it enters the second stage compressor. If air is cooled to a temperature equal to the initial temperature (i.e., if T 3 =T 1 ), inter cooling is said to be perfect. In practice, usually T 3 is greater than T 1. Multistage compressor with inter cooling actually decreases the cycle efficiency. This is because the average temperature of heat addition T add is less for this cycle 1-2-3-4-5-6 as compared to the simple Brayton cycle 1-4’-5-6 with the initial state 1. (refer fig). Average temperature of heat rejection T rej also reduces, but only marginally. 125

126. Hence efficiency is less for the modified cycle. However, if a regenerator is also used the heat added at lower temperature range (4 to 4’) comes from exhaust gases from the turbine. So there may be an increase in efficiency (compared to a simple Brayton cycle) when multi– stage compression with inter cooling is used in conjunction with a regenerator. For a gas turbine plant using 2–stage compression without a generator, Q 1 = h 5 - h 4 = C p (T 5 - T 4 ) W T = h 5 - h 6 = C p (T 5 -T 6 ) W C = (h 2 - h 1 ) + (h 4 - h 3 ) = C p [(T 2 - T 1 ) + (T 4 - T 3 )] 126

127. W C = (h 2 - h 1 ) + (h 4 - h 3 ) = C p [(T 2 - T 1 ) + (T 4 - T 3 )] W net = W T – W C = C p [(T 5 - T 6 ) – {(T 2 - T 1 ) + (T 4 - T 3 )}] 127

128. 3) Multi-Stage expansion with reheating: 128

129. Here expansion of working fluid (air) is carried out in 2 or more stages with heating (called reheating) in between stages. The reheating is done in heat exchangers called Reheaters. In an idealized cycle, the air is reheated, after each stage of expansion, to the temperature at the beginning of expansion. The schematic flow diagram as well as T-s diagram for a gas turbine plant where in expansion takes place in two turbine stages, with reheating in between, are shown. Multi-Stage expansion with reheating, by itself, does not lead to any improvement in cycle efficiency. In fact, it only reduces. 129

130. However, this modification together with regeneration may result in an increase in cycle efficiency. It can be seen from the T-s diagram that the turbine exhaust temperature is much higher when multi stage expansion with reheating is used, as compared to a simple Brayton cycle. This makes the use of a regenerator more effective and may lead to a higher efficiency. Heat added Q 1 = (h 3 - h 2 ) + (h 5 - h 4 ) = C p (T 3 - T 2 ) + C p (T 5 - T 4 ) Turbine output W T = (h 3 - h 4 ) + (h 5 - h 6 ) = C p (T 3 - T 4 ) + C p (T 5 - T 6 ) Compressor input W C = h 2 - h 1 = C p (T 2 - T 1 ) 130

131. Ideal Regenerative cycle with inter cooling and reheat: Considerable improvement in efficiency is possible by incorporating all the three modifications simultaneously. Let us consider a regenerative gas turbine cycle with two stage compression and a single reheat. The flow diagram and T-S diagram of such an arrangement is shown. Idealized Regenerative Brayton cycle with two stage compression with inter cooling and also two stage expansion with reheating – ideal regenerator, equal pressure ratios for stages, no irreversibilities, perfect inter cooling and reheating. 131

132. Heat added Q 1 = C p (T 5 - T 4’ ) + C p (T 7 - T 6 ) Turbine output W T = C p (T 5 - T 6 ) + C p (T 7 - T 8 ) Compressor input W C = C p (T 2 - T 1 ) + C p (T 4 - T 3 ) If perfect inter cooling, no irreversibilities, equal pressure ratios for stages and ideal regenerator are assumed, T 1 =T 3, T 2 =T 4 =T 8 ’, T 5 =T 7 and T 6 =T 8 =T 4 ’ 132

133. Then, Q 1 = C p (T 5 - T 4’ ) + C p (T 7 – T 6 ) = C p (T 5 - T 6 ) + C p (T 5 - T 6 ) = (T 5 - T 6 ) Q 2 = C p (T 8’ - T 1 ) + C p (T 2 - T 3 ) = C p (T 2 - T 1 ) + C p (T 2 - T 1 ) =2 C p (T 2 - T 1 ) 133

134. . 134

135. It can be seen from this expression that the efficiency decreases with increasing pressure ratio r p. 135

136. Effect of pressure Ratio r p on simple Brayton Cycle:- That means, the more the pressure ratio, the more will be the efficiency. Temperature T 1 (=T min ) is dependent on the temperature of surroundings. Temperature T 3 (=T max ) is limited by metallurgical considerations and heat resistant characteristics of the turbine blade material. For fixed values of T min and T max, the variation in net work output, heat added and efficiency with increasing pressure ratio r p can be explained with the help of a T-s diagram as shown. 136

137. For low pressure ratio, the net work output is small and the efficiency is also small (Cycle 1 – 2 – 3 - 4). In the limit, as r p tends 1, efficiency tends to zero (net work output is zero, but heat added is not zero). As the pressure ratio increases, the work output increases and so does the efficiency. However, there is an upper limit for r p when the compression ends at T max. As r p approaches this upper limit (r p ) max, both net work output and heat added approach zero values. However, it can be seen that the mean temperature heat addition T add approaches T max, while the mean temperature of heat rejection approaches T min, as r p comes close to (r p ) max. 137

138. Hence cycle efficiency, given by approaches the Carnot efficiency i.e., r p - (r p ) max When the compression ends at T max i.e., when state point 2 is at T max. When r p =r pmax, 138

139. The variation of net work output W net with pressure ratio r p is shown below. As r p increases from 1 to (r p ) max, W net increases from zero, reaches a maximum at an optimum value of r p i.e., (r p ) opt and with further increase in r p, it reduces and becomes zero when r p = r pmax 139

140. Pressure Ratio for maximum net work output:- W net = C p [(T 3 - T 4 ) - (T 2 - T 1 )] T 3 = T max & T 1 = T min 140

141. Condition for maximum W net is i.e., It can be seen that, 141

142. Maximum net work output Corresponding to r p = (r p ) opt i.e., when W net is maximum, cycle efficiency is 142

143. Open Cycle Gas Turbine Plants: In practice, a gas turbine plant works on an open cycle. Air from atmosphere is first compressed to a higher pressure in a rotary compressor, which is usually run by the turbine itself, before it enters the combustion chamber. Fuel is injected into the combustion chamber where it undergoes combustion. The heat released is absorbed by the products of combustion and the resulting high temperature; high pressure products expand in the turbine producing work output. 143

144. The used up combustion products (exhaust gases) are let out into the atmosphere. In the ideal case, compression and expansion are assumed to be isentropic and combustion is assumed to take place at constant pressure. The schematic flow diagram and p-v and T-s diagrams of an open cycle gas turbine plant are as shown. 144

145. Advantages and disadvantages of closed cycle over open cycle Advantages of closed cycle: 1.Higher thermal efficiency 2.Reduced size 3.No contamination 4.Improved heat transmission 5.Improved part load  6.Lesser fluid friction 7.No loss of working medium 8.Greater output and 9.Inexpensive fuel. 145

146. Disadvantages of closed cycle: 1.Complexity 2.Large amount of cooling water is required. This limits its use of stationary installation or marine use 3.Dependent system 4.The wt of the system pre kW developed is high comparatively,  not economical for moving vehicles 5.Requires the use of a very large air heater. 146

147. Problems: 1. In a Gas turbine installation, the air is taken in at 1 bar and 15 0 C and compressed to 4 bar. The isentropic  of turbine and the compressor are 82% and 85% respectively. Determine (i) compression work, (ii) Turbine work, (iii) work ratio, (iv) Th. . What would be the improvement in the th.  if a regenerator with 75% effectiveness is incorporated in the cycle. Assume the maximum cycle temperature to be 825 0 K. Solution: P 1 = 1 bar T 1 = 288 0 KP 2 = 4 bar T 3 = 825 0 K  C = 0.85  t = 0.82 147

148. Case1: Without Regeneration: Process 1-2s is isentropic i.e., But Process 3-4s is isentropic But 148

149. (i) Compressor work, W C = C P (T 2 – T 1 ) = 1.005 (452.87 – 288) = 165.69 kJ/kg (ii) Turbine work, W t = C P (T 3 – T 4 ) = 1.005 (825 – 603.57) = 222.54 kJ/kg (iii) Work ratio = = 0.255 (iv) Thermal Efficiency , = 15.2% 149

150. Case2: With Regeneration: We have effectiveness,   T 5 = 565.89 0 K  Heat supplied, Q­ H 1 = Q 5-3 = C P (T 3 – T 5 ) = 1.005 (825 – 565.89) = 260.4 kJ/kg = 0.218  Improvement in  th due to regenerator = 0.436 i.e., 43.6% 150

151. 2.The maximum and minimum pressure and temperatures of a gas turbine are 5 bar, 1.2 bar and 1000K and 300K respectively. Assuming compression and expansion processes as isentropic, determine the  th (a)when an ideal regenerator is incorporated in the plant and (b) when the effectiveness of the above regenerator is 75%. Solution: P 2 = P 3 = 5 barP 1 = P 4 = 1.2 bar T 3 = 1000KT 1 = 300K 151

152. Process 1-2s is isentropic i.e., Process 3-4s is isentropic i.e., 152

153. Ideal regenerator: i.e., T 5 = T 4  Heat supplied = C P (T 3 – T 5 ) = 1.005 [1000 – 664.88] = 336.79 kJ/kg W net = W T – W C = C P (T 3 – T 4 ) – C P (T 2 – T 1 ) = 1.005 [1000 – 664.88 – 451.21 + 300] = 183.91 = 0.546 or 54.6% 153

154. Regenerator with  = 0.75 i.e., i.e.,  Heat supplied, Q H = C P (T 3 – T 5 ) = 1.005 (1000 – 611.46) = 390.48kJ/kg = 0.471 or 47.1% 154

155. 3.Solve the above problem when the adiabatic efficiencies of the turbine and compressor are 90% and 85% respectively. 4. A gas turbine plant uses 500kg of air/min, which enters the compressor at 1 bar, 17 0 C. The compressor delivery pressure is 4.4 bar. The products of combustion leaves the combustion chamber at 650 0 C and is then expanded in the turbine to 1 bar. Assuming isentropic efficiency of compressor to be 75% and that of the turbine to be 85%, calculate (i) mass of the fuel required /min, of the CV of fuel is 39000KJ/Kg. (ii)net power output (iii)Overall thermal efficiency of the plant. Assume C P =1.13KJ/Kg-K,  =1.33 for both heating and expansion. 155

156. Solution: P 1 = 1 barT 1 = 290 0 KP 2 = 4.4 barT 3 = 923 0 K  C = 0.75  t = 0.85, W N = ?,  th ? Calorific Value = 39000 kJ/kg Process 1-2s is isentropic compression i.e., But i.e., 156

157. Process 3-4s is isentropic expansion i.e., But i.e., (i) We have  = 6.21kg/min 157

158. (ii) W N = ? Compressor work, W C = C P (T 2 – T 1 ) = 1.005 (494.03 – 290) = 205.05 kJ/kg Turbine work, W T = C P (T 3 – T 4 ) = 1.13 (923 – 681.76) = 272.6 kJ/kg  W N = W T – W C = 67.55 kJ/kg  Net work output per minute = = (500+6.21) (67.55) = 34194.49 kJ/min  Power output = 569.91 kW 158

159. (iii)  th = ? Heat supplied, Q H = C P (T 3 – T 2 ) = 1.33 (923 – 494.03) = 570.53 kJ/kg = 0.118 or 11.8% 159

160. 5. A gas turbine cycle having 2 stage compression with intercooling in between stages and 2 stages of expansion with reheating in between the stages has an overall pressure ratio of 8. The maximum cycle temperature is 1400 0 K and the compressor inlet conditions are 1 bar and 27 0 C. The compressors have  s of 80% and turbines have  s of 85%. Assuming that the air is cooled back to its original temperature after the first stage compression and gas is reheated back to its original temperature after 1 st stage of expansion, determine (i) the net work output (ii) the cycle  th. 160

161. Solution: T 5 = 1400 0 KT 1 = 300 0 K, P 1 = 1 bar  C1 = 0.8 =  C2,  t1 =  t2 = 0.85,T 3 = T 1,T 7 = T 5 For maximum work output, For process 1-2, = 300 (2.83) 0.286 = 403.95 0 K But 161

162. Since T 3 = T 1 and We have T 4s = T 2s = 403.95 0 K Also since  C1 =  C2,T 4 = T 2 = 429.9 0 K  Compressor work, W C = C P (T 2 – T 1 ) + C P (T 4 – T 3 ) = 2 C P (T 2 – T 1 ) = 2 (1.005) (429.9 – 300) = 261.19 kJ/kg For process 5 – 6, 162

163. But Since T 7 = T 5 and, then T 8 = T 6 Since  t1 =  t2,T 6 = T 8 = 1093.76 0 K  Turbine work, W t = C P (T 5 – T 6 ) + C P (T 7 – T 8 ) = 2 C P (T 5 – T 6 ) = 2 (1.005) (1400 – 1093.76) = 615.54 kJ/kg  W N = W T – W C = 354.35 kJ/kg 163

164.  th = ? Heat Supplied, Q H = C P (T 5 – T 4 ) + C P (T 7 – T 6 ) = 1.005 (1400 – 429.9 + 1400 – 1093.76) = 1282.72 kJ/kg = 0.276 or 27.6% 164

165. 6. Determine the  of a gas turbine having two stages of compression with intercooling and two stages of expansion with reheat. Given that the pressure ratio is 4, minimum temperature of the cycle 27 0 C and maximum temperature of the cycle is 600 0 C, when  t,  C and regenerator  are equal to 80%. ( Home work) 7. A two stage gas turbine cycle receives air at 100 kPa and 15 0 C. The lower stage has a pressure ratio of 3, while that for the upper stage is 4 for the compressor as well as the turbine. The temperature rise of the air compressed in the lower stage is reduced by 80% by intercooling. Also, a regenerator of 78% effectiveness is used. The upper temperature limit of the cycle is 1100 0 C. The turbine and the compressor  s are 86%. Calculate the mass flow rate required to produce 6000kW. 165

166. Solution: P 1 = 1 barT 1 = 288 0 K  IC = 0.8 ε =  reg = 0.78, T 5 = 1373 0 K,  C1 =  C2 =  t1 =  t2 = 0.86, if P = 6000 kW Process 1-2s is isentropic compression  T 2s = 288 (3) 0.286 = 410.75 0 K But Also, 166

167. Process 3-4s is 2 nd stage isentropic compression  T 4s = 316.54 (4) 0.286 = 470.57 0 K But Process 5-6s is 1 st stage isentropic expansion 167

168. But Process 6-7 is reheating, assume T 7 = T 5 = 1373 0 K Process 7-8s is 2nd stage isentropic expansion i.e., But Regenerator is used to utilizes the temperature of exhaust gases i.e.,  T x = 931.65 0 K 168

169. We have, Compressor work: W C = C P (T 2 – T 1 ) + C P (T 4 – T 3 ) = 1.005 (430.73 – 288 + 495.64 – 316.54) = 323.44 kJ/kg Also, Turbine work : W T = C P (T 5 – T 6 ) + C P (T 7 – T 8 ) = 1.005 (1373 – 986.51 + 1373 – 1054.63) = 708.38 kJ/kg  Net work output, W N = W T - W C = 384.95 kJ/kg But, power produced, i.e., 6000 x 1000 = 384.95 x 1000  = 15.59 kg/sec We have, heat supplied, Q H = C P (T 5 – T x ) + C P (T 7 – T 6 ) = 1.005 (1373 – 931.65 + 1373 – 986.51) = 831.98 kJ/kg 169

170. 8. In a gas turbine plant working on Brayton cycle, the inlet conditions are 1 bar and 27 0 C. The compression of air is carried out in two stages with a pressure ratio of 2.5 for each stage with intercooling to 27 0 C. The expansion is carried out in one stage with a pressure ratio of 6.25. The maximum temperature in the cycle is 800 0 C. The  of turbine and both compression stages are 80%. Determine (i) compressor work, (ii) Turbine work, (iii) Heat supplied, (iv) cycle , (v) cycle air rate. Hint: P 1 = 1 barP 4 = P 5 = 6.25 bar, P 3 = P 2 = 2.5 bar 170

171. 9. The pressure ratio of an open cycle constant pressure gas turbine is 6. The temperature range of the plant is 15 0 C and 800 0 C. Calculate (i)  th of the plant, (ii) Power developed by the plant for an air circulation of 5 kg/s, (iii) Air fuel ratio, (iv) specific fuel consumption. Neglect losses in the system. Use the following data: for both air and gases: C P 1.005 kJ/kg 0 K and  = 1.4. Calorific value of the fuel is 42000 kJ/kg,  C = 0.85,  t = 0.9 and combustion  of 0.95. 10. In a G.T. unit with two stage compression and two stage expansion the gas temperature at entry to both the turbines are same. The compressors have an intercooler with an effectiveness of 83%. The working temperature limits are 25 0 C and 1000 0 C, while the pressure limits are 1.02 bar and 7 bar respectively. Assuming that the compression and expansion processes in the compressors and turbine are adiabatic with  C of 84% and  t of 89% for both the stages. Calculate (i) the air-fuel ratio at the combustion chambers if the calorific value of the fuel is 38500 kJ/kg, (ii) Power output in kW for an air flow rate of 1kg/s and (iii) overall cycle . 171

172. 11. In a reheat gas turbine cycle, comprising one compressor and two turbine, air is compressed from 1 bar, 27 0 C to 6 bar. The highest temperature in the cycle is 900 0 C. The expansion in the 1 st stage turbine is such that the work from it just equals the work required by the compressor. Air is reheated between the two stages of expansion to 850 0 C. Assume that the isentropic  s of the compressor, the 1 st stage and the 2 nd stage turbines are 85% each and that the working fluid is air and calculate the cycle . Solution: P 1 = 1 barT 1 = 300KP 2 = 6 bar T 3 = 1173KW T1 = W C T 5 = 1123K  C = 0.85  t1 =  t2 = 0.85 172

173. We have process 1-2 is isentropic i.e.,  Compressor work, W C = C P (T 2 – T 1 ) = 1.005 (536 – 300) = 237 kJ/kg From data, W T1 = W C = 237 kJ/kg = C P (T 3 – T 4 )  T 4 = 937 kJ/kg 173

174. Process 3-4 is isentropic i.e., From T-S diagram, intermediate pressure, P 4 = P 5 = 2.328 bar Process 5-6s is isentropic in the 2 nd stage turbine 174

175.  W T2 = C P (T 5 – T 6 ) = 1.005 (1123 – 918) = 206 kJ/kg  Net work output = W T – W C = (W T1 + W T2 ) – W C = 206 kJ/kg Net heat transfer or heat supplied, Q = Q H + Q R = C P (T 3 – T 2 ) + C P (T 5 – T 4 ) = 640 + 187 = 827 kJ/kg  Cycle efficiency, 175

176. 12. In a simple gas turbine unit, the isentropic discharge temperature of air flowing out of compressor is 195 0 C, while the actual discharge temperature is 240 0 C. Conditions of air at the beginning of compression are 1 bar and 17 0 C. If the air-fuel ratio is 75 and net power output from the unit is 650kW. Compute (i) isentropic  of the compressor and the turbine and (ii) overall . Calorific value of the fuel used is 46110 kJ/kg and the unit consumes 312 kg/hr of fuel. Assume for gases C P = 1.09 kJ/kg-K and  = 1.32 and for air C P = 1.005 kJ/kg-K and  = 1.4. 176

177. Solution: T 2S = 195+273 = 468 KT 2 = 240+273 = 513K T 1 = 290KP 1 =1barA/F = 75, Power output = W net = W T­ – W C = 650kW  C = ?  T = ?  cycle = ? CV = 46110 kJ/kg, C Pg = 1.09 kJ/kg-k,  g = 1.30, C Pa = 1.005 kJ/kg-K,  a = 1.4 We have, Compressor Efficiency, Also, = 75 (0.0867) = 6.503 kg/s 177

178. Applying SFEE to the constant pressure heating process 2-3, 0.0867 (46110) = (6.503 + 0.0867) 1.09 (T 3 – 513)  T 3 = 1069.6K Also,  T 4S = 712.6K. Further, i.e., 650 = (6.503 + 0.0867) 1.09 (1069.6 – T 4 ) – 6.503 (1.005) (513 – 290)  T 4 = 776K 178

179. Now, Turbine Efficiency, And, Or 179