Successfully reported this slideshow.
Upcoming SlideShare
×

# Unit11

11,727 views

Published on

• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

### Unit11

1. 1. THE STEAM POWER CYCLE J2006/11/1 UNIT 11THE STEAM POWER CYCLE OBJECTIVESGeneral Objective : To understand and apply the concept of steam power cycle in thermodynamicsSpecific Objectives : At the end of the unit you will be able to:  define, derive, calculate and differentiate the following heat engine cycle: • Carnot cycle • Rankine cycle
2. 2. THE STEAM POWER CYCLE J2006/11/2 INPUT11.0 INTRODUCTION What is the most common working fluid used in heat engine cycle? S team is the most common working fluid used in heat engine cycles because of its many desirable characteristic, such as low cost, availability, and high enthalpy of vaporization. Other working fluids used include sodium, potassium, and mercury for high-temperature applications and some organic fluids such as benzene and the freons for low-temperature applications. The majority of this chapter is devoted to the discussion of steam power plants, which produce most of the electric power in the world today. Steam power plants are commonly referred to as coal plants, nuclear plants or natural gas plants, depending on the type of fuel used to supply heat to the steam. But the steam goes through the same basic cycle in all of them. Therefore, all can be analysed in the same manner. In this chapter it can be shown that there is an ideal theoretical cycle which is the most efficient conceivable; this cycle is called the Carnot cycle. The highest thermal efficiency possible for a heat engine in practice is only about half that of the ideal theoretical Carnot cycle, between the same temperature limits. This is due to the irreversibilities in the actual cycle, and to the deviations from the ideal cycle, which are made for various practical reasons. The choice of a power plant in practice is a
3. 3. THE STEAM POWER CYCLE J2006/11/3 compromise between thermal efficiency and various factors such as the size of the plant for a given power requirement, mechanical complexity, operating cost and capital cost. Fig 11.0 Model of a steam plant
4. 4. THE STEAM POWER CYCLE J2006/11/411.1The Carnot cycle From the Second Law of Thermodynamics it can be derived that no heat engine can be more efficient than a reversible heat engine working between the same temperature limits. Carnot, a French engineer, has shown in a paper written in 18241 that the most efficient possible cycle is one in which all the heat supplied is supplied at one fixed temperature, and all the heat rejected is rejected at a lower fixed temperature. The cycle therefore consists of two isothermal processes joined by two adiabatic processes. Since all processes are reversible, then the adiabatic processes in the cycle are also isentropic. The cycle is most conveniently represented on a T-s diagram as shown in Fig. 11.1. Q41 4 1 Boiler Turbine W12 Compressor Condenser 3 2 W34 Q23 T 4 1 T1 T2 3 2 B s A Fig 11.1 The Carnot cycle1 This paper, called ‘Reflection on the Motive Power of Heat’ was written by Carnot before the enunciation ofthe First and Second Laws of Thermodynamics. It is a remarkable piece of original thinking, and it laid thefoundations for the work of Kelvin, Clausius and others on the second law and its corollaries.
5. 5. THE STEAM POWER CYCLE J2006/11/5 A brief summary of the essential features is as follows: 4 to 1: The heat energy is supplied to the boiler resulting in evaporation of the water, therefore the temperature remains constant. 1 to 2: Isentropic expansion takes place in the turbine or engine. 2 to 3: In the condenser, condensation takes place, therefore the temperature remains constant. 3 to 4: Isentropic compression of the wet steam in a compressor returns the steam to its initial state. The plant required and the numbers referring to the state points for the Carnot cycle is shown in Fig. 11.1. The steam at the inlet to the turbine is dry saturated. The steam flows round the cycle and each process may be analysed using the steady flow energy equation where changes in kinetic energy and potential energy may be neglected. i.e. h1 + Q = h2 + W In this statement of the equation the subscripts 1 and 2 refer to the initial and final state points of the process; each process in the cycle can be considered in turn as follows: Boiler: h4 + Q41 = h1 + W41 Therefore, since W = 0, Q451 = h1 – h4 (11.1) Turbine: The expansion is adiabatic (i.e. Q = 0), and isentropic (i.e. s1 = s2), and h2 can be calculated using this latter fact. Then h1 + Q12 = h2 + W12 ∴ W12 = ( h1 – h2) (11.2)
6. 6. THE STEAM POWER CYCLE J2006/11/6 Condenser: h2 + Q23 = h3 + W23 Therefore, since W = 0 Q23 = h3 – h2 i.e. Q23 = - ( h2 – h3 ) ∴ Heat rejected in condenser = h2 – h3 (11.3) Compressor: H3 + Q34 = h4 + W34 The compression is isentropic ( i.e. s3 = s4 ), and adiabatic ( i.e. Q = 0 ). ∴ W34 = ( h3 – h4 ) = -( h4 – h3 ) i.e. Work input to pump = ( h4 – h3 ) (11.4) 11.1.1 Thermal efficiency of Carnot cycle The thermal efficiency of a heat engine, defined in chapter 9, was shown to be given by the equation, Q2 η = 1− Q1 In the Carnot cycle, with reference to Fig. 11.1, it can be seen that the heat supplied is given by the area 41BA4, i.e. Q1 = area 41BA4 = T1(sB- sA) Similarly the heat rejected, Q2, is given by the area 23AB2, i.e. Q2 = area 23AB2 = T(sB – sA) Hence we have T2 ( s B − s A ) Thermal efficiency of Carnot cycle, η carnot = 1 − T1 ( s B − s A )
7. 7. THE STEAM POWER CYCLE J2006/11/7 T2 i.e. η carnot = 1 − (11.5) T1 Net work output or η carnot = Heat supplied in the boiler ( h1 − h2 ) − (h4 − h3 ) i.e. η Carnot = (11.6) h1 − h4 11.1.2 The work ratio for Carnot cycle The ratio of the net work output to the gross work output of the system is called the work ratio. The Carnot cycle, despite its high thermal efficiency, has a low work ratio. net work Work ratio = (11.7) gross work (h1 − h2 ) − ( h4 − h3 ) i.e. Work ratio = (11.8) (h1 − h2 ) The work output of the Carnot cycle also can be found very simply from the T-s diagram. From the first law, ΣQ = ΣW therefore, the work output of the cycle is given by W = Q1 – Q2 Hence for the Carnot cycle, referring to Fig. 11.1, Wcarnot = area 12341 = (T1 - T2)(sB – sa) (11.9)
8. 8. THE STEAM POWER CYCLE J2006/11/8 This cycle is never used in practice owing to: 1. The difficulty in stopping the condensation at 3, so that subsequent compression would bring the state point to 4. 2. A very large compressor would be required. 3. Compression of wet steam in a rotary compressor is difficult as the water tends to separate out. 4. Friction associated with the expansion and compression processes would cause the net work done to be very small as compared to the work done in the turbine itself. The Carnot cycle is modified to overcome the above difficulties and this modified cycle, known as the Rankine cycle, is widely used in practice. T Thermal efficiency of Carnot cycle,
9. 9. THE STEAM POWER CYCLE J2006/11/9 Example 11.1 What is the highest possible theoretical efficiency of a Carnot cycle with a hot reservoir of steam at 200oC when the cooling water available from condenser is at 10oC? Solution to Example 11.1 From equation 11.5, T2 η carnot = 1 − T1 10 + 273 = 1− 200 + 273 283 = 1− 473 i.e. Highest possible efficiency = 1 – 0.598 = 0.402 or 40.2 % Example 11.2 A steam power plant operates between a boiler pressure of 42 bar and a condenser pressure of 0.035 bar. Calculate for these limits the cycle efficiency and the work ratio for a Carnot cycle using wet steam. Solution to Example 11.2 A Carnot cycle is shown in the figure given in the next page. T1 saturation temperature at 42 bar = 253.2 + 273 = 526.2 K T2 saturation temperature at 0.035 bar = 26.7 + 273 = 299.7 K
10. 10. THE STEAM POWER CYCLE J2006/11/10 T 4 526.2 1 299.7 2 3 s Then from equation 11.5 T1 − T2 η carnot = T1 526.2 − 299.7 = 526.2 = 0.432 or 43.2 % Also, Heat supplied = h1-h4 = hfg at 42 bar = 1698 kJ/kg Then, W η carnot = = 0.432 Q ∴ W = 0.432 x 1698 i.e. W = 734 kJ/kg To find the gross work of the expansion process it is necessary to calculate h2, using the fact that s1 = s2. From the Steam Tables, h1 = 2800 kJ/kg and s1 = s2 = 6.049 kJ/kg K From the equation s2 = 6.049 = sf2 + x2sfg2 = 0.391 + x2 8.13 6.049 − 0.391 ∴ x2 = = 0.696 8.13
11. 11. THE STEAM POWER CYCLE J2006/11/11 Then, h2 = hf2 + x2hfg2 = 112 + 0.696 x 2438 = 1808 kJ/kg Hence, from equation 11.2, W12 = (h1 – h2) = (2800 – 1808) = 992 kJ/kg Therefore, using equation 11.7, net work Work ratio = gross work 734 = 992 = 0.739
12. 12. THE STEAM POWER CYCLE J2006/11/12 Activity 11A TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 11.1 Describe the suitable components of a simple close cycle steam plant as illustrated in the figure below. Q41 4 1 (a) (b) W12 (d) (c) 3 2 W34 Q23 11.2 A steam power plant operates between a boiler pressure of 40 bar and a condenser pressure of 0.045 bar. Calculate for these limits the cycle efficiency and the work ratio for a Carnot cycle using wet steam.
13. 13. THE STEAM POWER CYCLE J2006/11/13 Feedback To Activity 11A 11.1 The components of a simple close cycle steam plant as illustrated in the figure are: a) The boiler, where water is converted into steam at a constant pressure and temperature by the heat energy is received from the combustion of the fuel. b) The engine or turbine, in which the steam expands to a low pressure causing work energy to be available. c) The condenser, in which heat energy flows from the low pressure steam into the condenser cooling water, resulting in the steam being condensed. d) The feed pump or the compressor, which returns the water into the boiler. 11.2 A Carnot cycle is shown in the figure below. T1 saturation temperature at 40 bar = 250.3 + 273 = 523.3 K T2 saturation temperature at 0.045 bar = 31.0 + 273 = 304.0 K T 4 523.3 1 304.0 2 3 s
14. 14. THE STEAM POWER CYCLE J2006/11/14 Then from equation 11.5 T1 − T2 η carnot = T1 523.3 − 304 = 523.3 = 0.419 or 41.9 % Also, Heat supplied = h1-h4 = hfg at 42 bar = 1714 kJ/kg Then, W η carnot = = 0.419 Q ∴ W = 0.419 x 1714 i.e. W = 718.2 kJ/kg To find the gross work of the expansion process it is necessary to calculate h2, using the fact that s1 = s2. From the tables, h1 = 2801 kJ/kg and s1 = s2 = 6.070 kJ/kg K From the equation s2 = 6.070 = sf2 + x2sfg2 = 0.451 + x2 7.98 6.070 − 0.451 ∴ x2 = = 0.704 7.98 Then, h2 = hf2 + x2hfg2 = 130 + 0.704 x 2428 = 1839.3 kJ/kg Hence, from equation 11.2, W12 = (h1 – h2) = (2801 – 1839.3) = 961.7 kJ/kg
15. 15. THE STEAM POWER CYCLE J2006/11/15 Therefore, using equation 11.7, net work Work ratio = gross work 718.2 = 961.7 = 0.747
16. 16. THE STEAM POWER CYCLE J2006/11/16 INPUT11.2 Rankine Cycle In the Rankine cycle, the exhaust steam is completely condensed into water in the condenser. Many of the impracticalities associated with the Carnot cycle can be eliminated by condensing it completely in the condenser, as shown schematically on a T-s diagram in Fig. 11.3. The cycle that results is the Rankine cycle, is the ideal cycle for vapour power plants. The ideal Rankine cycle does not involve any internal irreversibility and consists of the following processes: 4,5 to 1: Constant pressure heat addition in a boiler 1 to 2: Isentropic expansion taking place in the turbine or engine 2 to 3: Constant pressure heat rejection in the condenser 3 to 4: Isentropic compression of water in the feed pump
17. 17. THE STEAM POWER CYCLE J2006/11/17 From a comparison made between Fig. 11.1 and Fig. 11.2, the similarities between the Carnot and the Rankine cycles can be clearly seen. In the Rankine cycle, the exhaust steam is completely condensed into water in the condenser. It actually follows the isentropic expansion in the turbine. This water is then pumped into the boiler by a boiler feed pump. After the feed pump, since the water is not at the saturation temperature corresponding to the pressure, some of the heat energy supplied in the boiler is taken up by the water as sensible heat before evaporation can begin. This results in the boiler process being no longer completely isothermal; the process is, therefore, irreversible, causing the Rankine cycle to be an irreversible cycle and to have a lower efficiency than the Carnot cycle. Q451 5 4 1 Boiler Pump Turbine W12 Condenser W34 2 3 Q23 T p1 5 1 T1 p2 4 T2 3 2 A s B Fig 11.2 The Rankine cycle
18. 18. THE STEAM POWER CYCLE J2006/11/18 The plant required and the numbers referring to the state points for the Rankine cycle is shown in Fig. 11.2. The steam at the inlet to the turbine may be wet, dry saturated, or superheated, but only the dry saturated condition is shown in Fig. 11.2. The steam flows round the cycle and each process may be analysed using the steady flow energy equation where changes in kinetic energy and potential energy may be neglected. i.e. h1 + Q = h2 + W In this statement of the equation the subscripts 1 and 2 refer to the initial and final state points of the process; each process in the cycle can be considered in turn as follows: Boiler: h4 + Q451 = h1 + W Therefore, since W = 0, Q451 = h1 – h4 (11.10) Turbine: The expansion is adiabatic (i.e. Q = 0), and isentropic (i.e. s1 = s2 ), and h2 can be calculated using this latter fact. Then h1 + Q12 = h2 + W12 ∴ W12 = ( h1 – h2) (11.11) Condenser: h2 + Q23 = h3 + W23 Therefore, since W = 0 Q23 = h3 – h2 i.e. Q23 = - ( h2 – h3 ) ∴ Heat rejected in condenser = h2 – h3 (11.12)
19. 19. THE STEAM POWER CYCLE J2006/11/19 Pump: h3 + Q34 = h4 + W34 The compression is isentropic ( i.e. s3 = s4 ), and adiabatic ( i.e. Q = 0 ). ∴ W34 = ( h3 – h4 ) = -( h4 – h3 ) i.e. Work input to pump = ( h4 – h3 ) (11.13) This is the feed pump term, and as the quantity is small as compared to the turbine work, W12. The feed pump is usually neglected, especially when the boiler pressures are low. The net work done in the cycle, W = W12 + W34 i.e. W = ( h1 – h2) – ( h4 – h3 ) (11.14) Or, if the feed pump work is neglected, W = ( h1 – h2 ) (11.15) 11.2.1 Thermal efficiency of Rankine cycle Rankine efficiency, Net work output ηR = (11.16) Heat supplied in the boiler (h1 − h2 ) − (h4 − h3 ) i.e. η R= h1 − h4 (h1 − h2 ) − (h4 − h3 ) or η R= (11.17) (h1 − h3 ) − (h4 − h3 ) If the feed pump term, (h4 – h3) is neglected, equation (11.17) becomes (h1 − h2 ) η R= (11.18) (h1 − h3 )
20. 20. THE STEAM POWER CYCLE J2006/11/20 When the feed pump term is included, it is necessary to evaluate the quantity, W34. From equation (11.13) Pump work = -W34 = νf (p4 – p3) (11.19) 11.2.2 The work ratio for Rankine cycle It has been stated that the efficiency of the Carnot cycle is the maximum possible, but that the cycle has a low work ratio. Both efficiency and work ratio are criteria of performance. The work ratio is defined by net work Work ratio = gross work (h1 − h2 ) − ( h4 − h3 ) i.e. Work ratio = (h1 − h2 ) (h1 − h2 ) − v f ( p 4 − p3 ) or Work ratio = (11.20) (h1 − h2 ) 11.3 Specific steam consumption Another criterion of performance in steam plant is the specific steam consumption. It relates the power output to the steam flow necessary to produce steam. The steam flow indicates the size of plant with its component part, and the specific steam consumption is a means whereby the relative sizes of different plants can be compared. The specific steam consumption is the steam flow in kg/h required to develop 1 kW, i.e. W x (specific steam consumption, s.s.c.) = 1 x 3600 kJ/h (where W is in kJ/kg),
21. 21. THE STEAM POWER CYCLE J2006/11/21 3600 i.e. s.s.c. = kg/kwh W 3600 or. s.s.c. = kg/kwh (11.21) (h1 − h2 ) − ( h4 −h 3 ) Example 11.3 A steam power plant operates between a boiler pressure of 42 bar and a condenser pressure of 0.035 bar. Calculate for these limits the cycle efficiency, the work ratio, and the specific steam consumption for a Rankine cycle with dry saturated steam at entry to the turbine. Solution to Example 11.3 The Rankine cycle is shown in the figure below. T 42bar 5 1 0.035 bar 526.2 4 299.7 3 2 s A B As in example 11.2 h1= 2800 kJ/kg and h2 = 1808 kJ/kg Also, h3 = hf at 0.035 bar = 112 kJ/kg Using equation 11.18, with v = vf at 0.035 bar
22. 22. THE STEAM POWER CYCLE J2006/11/22 Pump work = -W34 = νf (p4 – p3) = 0.001 x ( 42 – 0.035) x 102 = 4.2 kJ/kg Using equation 11.11 W12 = h1 – h2 = 2800 – 1808 = 992 kJ/kg Then using equation 11.17 (h1 − h2 ) − (h4 − h3 ) η R= (h1 − h3 ) − (h4 − h3 ) (992) − (4.2) = (2800 − 112) − (4.2) = 0.368 or 36.8 % Using equation 11.7 net work Work ratio = gross work 992 - 4.2 = 992 = 0.996 Using equation 11.21 3600 s.s.c. = (h1 − h2 ) − (h4 −h 3 ) 3600 = (992) − ( 4.2) = 3.64 kg/kW h
23. 23. THE STEAM POWER CYCLE J2006/11/23 Activity 11B TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 11.3 Based on the diagram below, describe the four-stage processes that represent a steam plant operating on an ideal Rankine cycle. Q451 5 4 1 Boiler W12 Pump Turbine W34 3 2 Condenser Q23 11.4 A steam power plant operates between a boiler pressure of 40 bar and a condenser pressure of 0.045 bar. Calculate for these limits the cycle efficiency, the work ratio, and the specific steam consumption for a Rankine cycle with dry saturated steam at entry to the turbine.
24. 24. THE STEAM POWER CYCLE J2006/11/24 Feedback To Activity 11B 11.3 The ideal Rankine cycle does not involve any internal irreversibility and consists of the following four-stage processes: Stage 1. 4 to 1: Constant pressure heat addition in a boiler. Stage 2. 1 to 2: Isentropic expansion that takes place in the turbine or engine. Stage 3. 2 to 3: Constant pressure heat rejection in the condenser. Stage 4 3 to 4: Isentropic compression of water in the feed pump 11.4 The Rankine cycle is shown in the figure below. T 40bar 5 1 0.045 bar 523.3 4 3040. 3 2 A s B As in activity 11.2 h1= 2801 kJ/kg and h2 = 1839.3 kJ/kg
25. 25. THE STEAM POWER CYCLE J2006/11/25 Also, h3 = hf at 0.045 bar = 130 kJ/kg Using equation 11.18, with v = vf at 0.045 bar Pump work = -W34 = νf (p4 – p3) = 0.001 x ( 40 – 0.045) x 102 = 4.0 kJ/kg Using equation 11.11 W12 = h1 – h2 = 2801 – 1839.3 = 961.7 kJ/kg Then using equation 11.17 (h1 − h2 ) − (h4 − h3 ) η R= (h1 − h3 ) − (h4 − h3 ) (961.7) − (4.0) = (2801 − 130) − (4.0) = 0.359 or 35.9 % Using equation 11.7 net work Work ratio = gross work 961.7 - 4.0 = 961.7 = 0.996 Using equation 11.21 3600 s.s.c. = (h1 − h2 ) − (h4 −h 3 ) 3600 = (961.7) − (4.0) = 3.76 kg/kW h
26. 26. THE STEAM POWER CYCLE J2006/11/26 SELF-ASSESSMENT You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback to Self-Assessment on the next page. If you face any problem, discuss it with your lecturer. Good luck. 1. Explain why the Rankine cycle and not the Carnot cycle is taken as the ideal cycle for steam plant. Sketch the T-s diagram for these cycles when using steam as the working fluid. 2. What is the highest thermal efficiency possible for a Carnot cycle operating between 210o C and 15o C. 3. A steam power plant operates between a boiler pressure of 30 bar and a condenser pressure of 0.04 bar. Calculate for these limits the cycle efficiency, the work ratio and the specific steam consumption for a Carnot cycle using wet steam. 4. A steam power plant operates between a boiler pressure of 30 bar and a condenser pressure of 0.04 bar. Calculate for these limits the cycle efficiency, the work ratio, and the specific steam consumption for a Rankine cycle with dry saturated steam at entry to the turbine. 5. In a steam power plant, dry saturated steam enters the turbine at 47 bar and is expanded isentropically to the condenser pressure of 0.13 bar. Determine the Rankine cycle efficiency when a) the feed pump work is neglected b) the feed pump work is taken into account
27. 27. THE STEAM POWER CYCLE J2006/11/27 Feedback to Self-Assessment Have you tried the questions????? If “YES”, check your answers now.1. Carnot cycle is never used in practice owing to: 1. The difficulty in stopping the condensation at 3, so that subsequent compression would bring the state point to 4. 2. A very large compressor would be required. 3. Compression of wet steam in a rotary compressor is difficult as the water tends to separate out. 4. Friction associated with the expansion and compression processes would cause the net work done to be very small as compared to the work done in the turbine itself. The Carnot cycle is modified to overcome the above difficulties and this modified cycle, known as the Rankine cycle, is widely used in practice. T T p1 T1 4 1 1 T1 5 p2 4 T2 T2 3 2 3 2 A B s s Carnot cycle Rankine cycle
28. 28. THE STEAM POWER CYCLE J2006/11/28 2. 40.37 % 3. 40.4 %, 0.771, 4.97 kg/kW h 4. 34.6 %, 0.997, 3.88 kg/kW h 5. 33.67 %, 33.55 % CONGRATULATION S!!!!…May success be with you always…