Lecture note steam power cycle

726 views

Published on

Nota subjek Applied Thermo

Published in: Education, Technology, Business
0 Comments
2 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
726
On SlideShare
0
From Embeds
0
Number of Embeds
4
Actions
Shares
0
Downloads
64
Comments
0
Likes
2
Embeds 0
No embeds

No notes for slide

Lecture note steam power cycle

  1. 1. 12/9/2013 1 Vapor Power Cycles Vapor Power Plant 1 Introduction Chapter Objective The objective of this chapter is to carry out first law and second law analysis on a vapor power plant in which the working fluid is alternatively vaporized and condensed as it completes a thermodynamics cycle. What is a Vapor Power Plant? A vapor power plant is a thermodynamics heat engine used to produce mechanical power output from energy sources such as fossil fuel, nuclear (uranium) or solar energy. Vapor Power Plant 2
  2. 2. 12/9/2013 2 Introduction Cont’d The power output is typically used to drive electrical generator, to produce electricity for our everyday use. The plant uses water as a working fluid, which will be alternately vaporized and condensed as it undergoes a complete thermodynamics cycle. Note: The processes taking place in actual power generating system are complicated. To carry out thermodynamics study on the system, we will develop a simplified model of the system. Vapor Power Plant 3 Vapor Power Plant 4 Simplified Model for Analysis Figure 1 A simplified model for a fossil-fuel vapor power plant
  3. 3. 12/9/2013 3 Simplified Model for Analysis • Subsystem A: This is where the energy conversion process occurs. Heat energy, obtained from thermal energy sources such as fossil-fuel and nuclear, is converted into mechanical work, in a form of a rotating shaft. • Subsystem B: This subsystem supplies the energy required to vaporize the liquid water. In fossil-fuel plants, this is accomplished by heat transfer from the hot gases produced from the combustion of the fossil-fuel, to the liquid water passing through the tubes and drums in the boiler. • Subsystem C: It comprises of a cooling water circuit. Cooling water is used to cool off the wet vapor exiting the turbine, thus condensing it back into the liquid water. The hot cooling water is sent to a cooling tower, where the heat energy taken up in the condenser is rejected to the atmosphere. • Subsystem D: It comprises of an electric generator, which is connected to the turbine via a shaft. The shaft rotates as the steam expands to lower pressure through the turbine. It drives the generator, which produces electrical power output. Vapor Power Plant 5 Focus of the Analysis Vapor Power Plant 6 We will focus our analysis on Subsystem A, shown in Figure 2. Figure 2 Subsystem A of the plant
  4. 4. 12/9/2013 4 Basic Components Vapor Power Plant 7 The functions of the basic components of Subsystem A are: a) Boiler: to transform liquid water into vapor (steam) of high pressure and temperature. b) Turbine-Generator: to transform the kinetic energy of the vapor into mechanical power (rotating shaft). The mechanical power is used to drive an electric generator, to produce electricity. c) Condenser: to cool off the wet vapor exiting the turbine and transform it back into the liquid water. d) Feed-water Pump: to deliver the water exiting the condenser back into the boiler, thus completing one thermodynamics cycle. Revision • The 1st Law of Thermodynamics Net heat received by any cyclic device is the same with the net work produced. • The 2nd Law of Thermodynamics It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work. Certain amount of heat must be rejected to the surrounding. Vapor Power Plant 8
  5. 5. 12/9/2013 5 The 1st Law of Thermodynamics Vapor Power Plant 9 ∑ ∑= WQ netnet WQ = netWQQ =− ⋯21 The 2nd Law of Thermodynamics Vapor Power Plant 10 Cyclic device Q1 Q2 Wt
  6. 6. 12/9/2013 6 Revision Isentropic Process A process during which the entropy remains constant. It also can be recognized as internally reversible, adiabatic process. Vapor Power Plant 11 )./(0 12 KkgkJssors ==∆ Energy Balance for Steady-flow Systems The first law or energy balance relation for a general steady-flow system is, For single-stream (one-inlet-one exit) systems, the inlet and exit states are denoted by subscripts 1 and 2 for simplicity. The mass flow rate through the entire control volume remains constant ( ) and is denoted by . Then the energy balance for single- stream steady-flow systems becomes 21 •• = mm ∑∑       + ∇ +−      + ∇ +=− •••• inleteachfor i i ii exiteachfor e e ee gzhmgzhmWQ 22 22 Vapor Power Plant 12 • m ( )      −+ ∇−∇ +−=− ••• 12 2 1 2 2 12 2 zzghhmWQ
  7. 7. 12/9/2013 7 Energy Balance for Steady-flow Systems If the fluid experiences a negligible change in its kinetic and potential energies as it flows through the control volume (that is, ∆KE = 0, ∆PE = 0), then the energy equation for a single-stream steady-flow system reduces further to (1) Vapor Power Plant 13 ( )12 hhmWQ −=− ••• Performance of Steam Plant 1. Specific steam consumption (s.s.c.) Defined as the steam flow rate in kg/hr required to develop 1 kWatt of power output. (2) The lower the s.c.c the more compact the steam plant. 2. Work ratio (wr) Defined as the ratio of the net work produced by the plant to the work produced by the turbine, i.e., (3) where Wt = turbine work Wp = pump or compressor work ( ) 3600 3600 . . . / . net net t p m s s c kg kW hr W W W W = = = − Vapor Power Plant 14 t pt t net W WW W W wr − ==
  8. 8. 12/9/2013 8 Performance of Steam Plant 3. Thermal efficiency (ηth) Defined as the ratio of net work produced by the plant to the amount of heat added to the working fluid in the boiler i.e., (4) Vapor Power Plant 15 %100%100 x Q WW x Q W in pt in net th − ==η Performance of Steam Plant 4. Isentropic efficiency (ηis) The actual expansion and pumping processes are adiabatic but not reversible. Thus, they are not isentropic. For the expansion process in the turbine, (5) For the pumping process in the feed-pump, (6) ( ) ( )3 ' 4 34 , hh hh workActual workIsentropic pis − − ==η Vapor Power Plant 16 ( ) ( )21 ' 21 , hh hh workIsentropic workActual tis − − ==η T s 1 2’23 4 4’ Figure 3
  9. 9. 12/9/2013 9 Performance of Steam Plant 5. Back work ratio Defined as the ratio of the work supplied to the feed-water pump to the work produced by the turbine, i.e., (7) 6. Efficiency ratio Defined as, (8) efficiencycycleRankineIdeal cycleactualofefficiencythermal =ratioη Vapor Power Plant 17 t p W W bwr = • The most efficient cycle is the Carnot cycle for given temperatures of source and sink (T1 and T2). • It also can be called as the ideal heat engine. • The cycle for a wet vapor is shown in Figure 4 and a brief summary of the essential features is as follows, 4 to 1: heat is supplied at constant temperature and pressure. 1 to 2: the vapor expands isentropically from the high pressure and temperature to the low pressure. 2 to 3: the vapor, which is wet at 2, has to be cooled to state point 3. 3 to 4: isentropic compression from 3 to 4. From 4 the cycle is repeated. The Carnot Cycle Vapor Power Plant 18 ss3 = s4 s1 = s2 Figure 4
  10. 10. 12/9/2013 10 • Although the Carnot cycle is the most efficient cycle, there are several problems such as, 1. the work ratio is low 2. it is difficult to stop the condensation at point 3 and then compressed it just to state 4 3. the steam at outlet from the turbine is wet, so it can damage the turbine blades 4. the 3 to 4 process is bad for the compressor because two phases compression is not practical. • The efficiency of the cycle can be calculated by, (9) The Carnot Cycle 1 2 1 2 11 T T or Q Q thth −=−= ηη Vapor Power Plant 19 Figure 4 Repeated ss3 = s4 s1 = s2 Vapor Power Plant 20 The Rankine Cycle Basic (Ideal/actual) With Superheat (Ideal/actual) Reheat Cycle (Ideal/actual) Regenerative Cycle With Open- type Feedwater Heater (Ideal/actual) Regenerative Cycle With Closed- type Feedwater Heater (Ideal/actual)
  11. 11. 12/9/2013 11 Basic Rankine Cycle The working fluid undergoes a thermodynamics cycle as it flows through each component of the plant. The cycle is called an ideal Rankine cycle, and is shown on a temperature-entropy (T-s) diagram, in Figure 5. Vapor Power Plant 21 Basic Rankine Cycle Vapor Power Plant 22 Figure 2 Repeated Figure (2-3) Repeated Figure 5 An ideal Rankine Cycle
  12. 12. 12/9/2013 12 Basic Rankine Cycle By comparing with the Carnot Cycle, this basic Rankine Cycle is more practical since, it is more convenient to allow condensation process to proceed to completion. In addition, the working fluid is water, and this can be conveniently pumped to boiler pressure as shown at point 4. Vapor Power Plant 23 Basic Rankine Cycle The Cycle Analysis The steam flows round the cycle and each process is analyzed using the steady flow energy equation (sfee). The changes in kinetic energy and potential energy are neglected. Vapor Power Plant 24
  13. 13. 12/9/2013 13 Basic Rankine Cycle The Cycle Analysis a) Boiler Since there is no work interaction between the working fluid and the surrounding, W = 0. Hence, the amount of heat added to the working fluid in the boiler is (10) b) Turbine Since the expansion process is assumed to be isentropic (reversible adiabatic), then Q = 0. Thus the amount of mechanical work produced by the turbine is (11) Vapor Power Plant 25 ( ) kgkJhhqq in /4114 −==− Figure (2-3) Repeated ( ) kgkJhhww t /2121 −==− Basic Rankine Cycle The Cycle Analysis c) Condenser There is no work interaction, so that W = 0. Hence, the amount of heat rejected from the working fluid to the cooling water is (12) d) Feed-water pump Since the pumping process is assumed to be isentropic (reversible adiabatic), then Q = 0. Thus the amount of work supplied to the feed-water pump is (13) where v is the specific volume (m3/kg) of water at pressure p3. Vapor Power Plant 26 ( ) kgkJhhqq out /3232 −==− Figure (2-3) Repeated ( ) ( ) kgkJppvhhww p /3433443 −=−==−
  14. 14. 12/9/2013 14 Example 1 A steam power plant operates between a boiler pressure of 50 bar and a condenser pressure of 0.035 bar. Calculate for these limits the thermal efficiency, the work ratio, and the specific steam consumption: a) for a Rankine cycle with dry saturated steam at entry to the turbine b) for a Rankine cycle with the turbine isentropic efficiency of 85 %. Sketch the cycle on a T-s diagram. Vapor Power Plant 27 The Solutions ( ) ( ) ( ) ( ) ( ) ( ) 441.0 11552794 2.87011559.17852794 41 3421 = − −−− = − −−− = − == hh hhhh q ww q w in ct in net thη Vapor Power Plant 28 a) h1 = hg@50 bar = 2794 kJ/kg s1 = sg@50 bar = s2 = 5.973 kJ/kgK = sf + x2.sfg@0.035 bar = 0.391 + x2 (8.130) x2 = 0.6866 h2 = hf + x2.hfg@0.035 bar = 112 + 0.6866 (2438) = 1785.9 kJ/kg s4 = sf@50 bar = s3 = 2.921 kJ/kg = sf + x3.sfg@0.035 bar = 0.391 + x3 (8.130) x3 = 0.311 h3 = hf + x3.hfg@0.035 bar = 112 + 0.311 (2438) = 870.2 kJ/kg h4 = hf@50 bar = 1155 kJ/kg T s 1 23 4 50 bar 0.035 bar
  15. 15. 12/9/2013 15 The Solutions ( ) ( ) ( ) ( ) ( ) ( ) 717.0 9.17852794 2.87011559.17852794 21 3421 = − −−− = − −−− = − = hh hhhh w ww wr t ct ( ) ( ) ( ) ( ) hrkWkg hhhhww css ct ./98.4 2.87011559.17852794 3600 36003600 ... 3421 = −−− = −−− = − = Vapor Power Plant 29 T s 1 23 4 50 bar 0.035 bar The Solutions ( ) ( ) ( ) ( ) ( ) kgkJxhhxhhw hh hh w w tisactual Isentropic actual tis /9.8569.1785279485.021, ' 21 21 ' 21 , =−=−=−=→ − − == ηη ( ) ( ) ( ) ( ) ( ) kgkJ hh hhw hh hh w w cis actual actual Isentropic cis /335 85.0 2.8701155 , 34 3 ' 4 3 ' 4 34 , = − = − =−=→ − − == η η kgkJhhq kgkJhkgkJhh in /8.15882.12052794 /2.12052.870335/335 ' 41 ' 3 ' 44 =−=−= =+=→=− 328.0 8.1588 3359.856 14 = − = − == −q ww q w ct in net thη Vapor Power Plant 30 b) hrkWkg w css w w wr net t net ./9.6 3359.856 36003600 ... 609.0 9.856 3359.856 = − == = − == T s 1 23 4 50 bar 0.035 bar 2’ 4’
  16. 16. 12/9/2013 16 The Solutions ( ) ( ) ( ) 375.0 41 3421 14 = − −−− = − == − hh hhhh q ww q w pt in net thη ( ) ( ) kgkJxppvhhwp /9.410035.050001.0 2 34334 =−=−≈−= kgkJWhh p /9.1169.411234 =+=+= Vapor Power Plant 31 c) ( ) ( ) ( ) ( ) ( ) hrkWkg hhhhw css hh hhhh w w wr net t net ./59.3 36003600 ... 995.0 3421 21 3421 = −−− == = − −−− == T s 1 23 4 50 bar 0.035 bar The Solutions 85.0 21 ' 21 , = − − == hh hh w w isentropic actual tisη ( ) kgkJhhhh /9.85685.0 21 ' 21 =−=−→ ( ) ( ) ( ) 32.0 41 34 ' 21 14 = − −−− = − == − hh hhhh q ww q w pt in net thη ( ) ( ) ( ) ( ) ( ) hrkWkg hhhhw css hh hhhh w w wr net t net ./23.4 36003600 ... 845.0 34 ' 21 ' 21 34 ' 21 = −−− == = − −−− == Vapor Power Plant 32 d) T s 1 23 4 50 bar 0.035 bar 2’
  17. 17. 12/9/2013 17 Rankine Cycle with Superheating • Improvement in the basic Rankine cycle. • Steam temperature at inlet to the turbine is increased at boiler pressure, thus increasing the average temperature of heat addition. • Increase the cycle efficiency. • Steam exits the turbine is more dry, i.e., the dryness fraction, x2 of the steam increases. • Specific steam consumption drops. Vapor Power Plant 33 Rankine Cycle with Superheating Technique of Superheating The saturated steam exiting the boiler is passed through a second bank of smaller tubes located within the boiler, which is heated by the hot gases from the furnace. Vapor Power Plant 34
  18. 18. 12/9/2013 18 Rankine Cycle with Superheating Vapor Power Plant 35 Hot well Receiver Collecting steam from others boiler Provide storage for the condensate Degree of superheat, ∆T Superheated temperature, Tsh Saturation temperature Ts at boiler pressure 1 Figure 6 Ideal Rankine cycle with superheating Example 2 Reconsider the vapor power cycle of Example 1. Calculate it’s thermal efficiency and s.s.c. if the steam exiting the boiler is heated to 500oC before entering the turbine. Assume the pump work is small and can be neglected. Sketch the cycle on a T-s diagram. Vapor Power Plant 36
  19. 19. 12/9/2013 19 The Solutions Vapor Power Plant 37 s 0.035 bar h1 = 3433 kJ/kg s2 = s1 = 6.975 kJ/kg = sf + x2.sfg@0.035 bar = 0.391 + x2 (8.310) x2 = 0.81 h2 = hf + x2.hfg@0.035 bar = 112 + 0.81 (2438) = 2086.8 kJ/kg h3 = hf@0.035 bar = 112 kJ/kg h4 = h3 pump work neglected ( ) ( ) ( ) ( ) 405.0 1123433 8.20863433 41 21 = − − = − − = − == hh hh q ww q w in pt in net thη ( ) hrkWkg hhww css pt ./67.2 2.1346 3600 36003600 ... 21 = = − = − = Rankine Cycle with Reheating • Improvement in the Superheat Rankine cycle. • The average heat addition is increased in another way. • As the steam is expanded in the turbine it is withdrawn at the point where it just about becomes wet and then reheated to a high temperature. • The dryness fraction of the steam exiting the turbine stages is further increased, which is the desired effect. • The steam is reheated at constant pressure in process 2 to 3. • The specific steam consumption is improved (decrease). • Usually, the steam is reheated to the inlet temperature of the high-pressure turbine. Vapor Power Plant 38
  20. 20. 12/9/2013 20 Rankine Cycle with Reheating Vapor Power Plant 39 3 4 5 6 Figure 7 A steam power plant with reheating process Low pressure turbine 1 2 3 45 6 T Figure 8 Ideal Rankine cycle with superheat and reheat Rankine Cycle with Reheating The Cycle Analysis a) Heat supplied The heat added to the working fluid is given by (14) b) Work output The work developed by the turbine is given by (15) Vapor Power Plant 40 ( ) ( ) kgkJhhhhq qqq totalin totalin /2361, 3216, −+−= += −− ( ) ( ) kgkJhhhhw www totalt totalt /3421, 4321, −+−= += −− Low-pressure turbine 1 2 3 45 6 T Figure 8 Repeat High-pressure turbine Reheating
  21. 21. 12/9/2013 21 Rankine Cycle with Reheating The Cycle Analysis c) Work input The work supplied to the feed-water pump is given by (16) Vapor Power Plant 41 ( ) ( ) kgkJppvhhwwp /5655665 −=−== − Low-pressure turbine 1 2 3 45 6 T Figure 8 Repeated High-pressure turbine Reheating Example 3 In a reheat Rankine cycle, steam enters the high-pressure turbine at 15 MPa and 600°C and is condensed in a condenser at a pressure of 10 kPa. If the moisture content of the steam at the exit of the low-pressure turbine is not to exceed 10.4 percent, determine: a) the pressure at which the steam should be reheated b) thermal efficiency of the cycle Assume that the steam is reheated to the inlet temperature of the high-pressure turbine. Vapor Power Plant 42
  22. 22. 12/9/2013 22 Example 3 Vapor Power Plant 43 Example 4 (Test 1 – 2004/05-1) A steam power plant operates on the actual Rankine cycle between boiler pressure of 60 bar and condenser pressure of 0.1 bar. Steam enters the high-pressure turbine at 500oC and is expanded to a pressure of 16 bar. Steam is then reheated to 400oC before it expands in the low-pressure turbine to the condenser pressure. The isentropic efficiencies for each turbine are 90% and the pump is 100%. Show a complete schematic diagram of the plant and the cycle on a T-s diagram. Considering the pump work determine, (i) the net work output of the plant (kJ/kg); (ii) the heat supplied to the plant (kJ/kg); (iii) the thermal efficiency of the plant (%); (iv) the specific steam consumption of the plant (kg/kWhr); (v) the thermal efficiency of the plant, if it operates on the ideal Rankine cycle (%). Vapor Power Plant 44
  23. 23. 12/9/2013 23 Example 4 Vapor Power Plant 45 160 bar P2 = P3 = 16 bar 2 3 4 5 6 60 bar 5 4 6 1 2 3 60 bar 0.1 bar 2’ 4’ 15 bar ToC s 1 2 3 6 0.1 bar 60 bar 16 bar 2’ 4 4’ 5 500 400 Example 5 (Test 1 – 2005/06-1) A steam power plant operates on the actual Rankine cycle. Steam enters the high-pressure turbine at 40 bar and 350oC and leaves at 10 bar. Steam is then reheated at constant pressure to a temperature of 350oC before it expands to 0.03 bar in the low-pressure turbine. If the isentropic efficiencies of both turbines are 84% and 80%, respectively, calculate (i) the work output per kg of steam; (ii) the heat supplied per kg of steam; (iii)the thermal efficiency; (iv)the specific steam consumption. Show the cycle on a T-s diagram. For this case, neglect the feed pump term. Vapor Power Plant 46
  24. 24. 12/9/2013 24 Example 5 Vapor Power Plant 47 140 bar P2 = P3 = 10 bar 2 3 4 5 6 40 bar 5 4 6 1 2 3 40 bar 0.03 bar 2’ 4’ 10 bar 0.03 bar 0.03 bar 350 The Regenerative Cycle Vapor Power Plant 48 What is Regeneration Process? • In a regenerative cycle, the feed-water is preheated in a feed-water heater (FWH), using some amount of steam bled off the turbine, before it is delivered back into the boiler. This is shown in Figure 9. • The preheating process occurs in the FWH at a constant pressure. The steam required for heating the feed- water is bled off the turbine at certain bleeding pressure, Pbleed.
  25. 25. 12/9/2013 25 The Regenerative Cycle Vapor Power Plant 49 Figure 9 Ideal regenerative Rankine cycle power plant The Regenerative Cycle Vapor Power Plant 50 Purpose of Regeneration Process • The main purpose of regeneration process is to increase the thermal efficiency of the cycle. • If the feed-water is preheated before entering the boiler, then less heat will be required to transform the feed-water into steam, in the boiler. • As a result, thermal efficiency of the plant increases.
  26. 26. 12/9/2013 26 The Regenerative Cycle Vapor Power Plant 51 Types of Feed-water Heater (FWH) There are two types of feed-water heater or heat exchanger: an open-type and a closed-type. a) Open-type Feed-water Heater • An open-type FWH is basically a “mixing chamber”. • The feed-water is preheated by direct mixing with the steam extracted from the turbine. • The plant can use more than one open feed-water heater. • Each open-type FWH requires one extra pump. Open-type Feedwater Heater Vapor Power Plant 52 Figure 10 Ideal regenerative cycle using open-type FWH3 1 4 2 Condenser Pump 1 Pump 2 steam exits open FWH = saturated liquid
  27. 27. 12/9/2013 27 Analysis of open-type FWH The mass of the steam extracted from the boiler, y, is determined by doing an energy balance on the feed-water heater. i.e. Solve to give, (17) The choice of Bled-Off Pressure (18) The Regenerative Cycle ( ) ( )outin hmhm ∑∑ = .. ( ) ( ) ( ) 326 .1.1. hhyhy =−+ ( ) ( )26 23 hh hh y − − = bleedbleed condsboilers bleed Tatpressuresaturatedp TT T =→ + = 2 ,, Vapor Power Plant 53 Open FWH (1 – y) 2 3 (1) 6 (y) The Regenerative Cycle Vapor Power Plant 54 Open FWH (1 – y) 2 3 (1) 6 (y) Note: y is chosen so that the condition of point 3 is saturated liquid. 3 2 1 4 s Figure 10 Repeated
  28. 28. 12/9/2013 28 Analysis of the cycle a) Heat supplied The amount of heat added to the working fluid, qin = q4-5 = (h5 – h4) (19) b) Turbine work The total amount of work produced by the turbine, wt = (h5 – h6)+(1-y)(h6 – h7) (20) The Regenerative Cycle Vapor Power Plant 55 The Regenerative Cycle Analysis of the cycle c) Heat rejected The heat rejected to the cooling water in the condenser, qout = q7-1 = (1-y)(h7 – h1) (21) d) Feed-water pump work The total amount of work supplied to the feed-water pumps, wp = (1-y)(h2 – h1)+(h4 - h3) (22) Note: The heat added to the working fluid in the FWH = (h3 – h2) Vapor Power Plant 56
  29. 29. 12/9/2013 29 Example 5 A steam power plant operates on the ideal regenerative cycle with one feed-water heater. Steam enters the turbine at 15 MPa and 600°C and is condensed to a condenser pressure of 10 kPa. Some steam is extracted from the turbine at a pressure of 1.2 MPa and enters the open-type feed-water heater. Determine: (a) the amount of steam extracted from the turbine, and, (b) thermal efficiency of the cycle. Vapor Power Plant 57 Example 5 Vapor Power Plant 58 Pump 11 Pump 1
  30. 30. 12/9/2013 30 The Regenerative Cycle Vapor Power Plant 59 b) Closed-type Feed-water Heater • Each open-type FWH requires one extra pump, thus, the plant cost increases. • This can be improved by using closed-type FWH. • The feed-water does not mix freely with the bled off steam. • There is only a heat transfer from the steam to the feed-water in the closed-type FWH. • Since there is no mixing, the bled off steam can be at different pressure than the feed-water. • The condensate exiting the closed-type FWH (state 7) is throttled back into the condenser and mix with the feed-water in the condenser. • The mixture of condensate and feed-water is then pumped back into the boiler to repeat the cycle. Closed-type Feedwater Heater Vapor Power Plant 60 Throttle valve Figure 12 Ideal Rankine cycle With closed-type FWH Figure 11 s Ideally, T9 = T7 Pbleed = p6
  31. 31. 12/9/2013 31 Analysis of closed-type FWH Mass of Bled-Off Steam By performing energy balance on the closed-type FWH, i.e., Simplify gives the fraction of the steam bled off, (23) The Regenerative Cycle ( ) ( )∑∑ = outin hmhm .. ( ) ( ) ( ) ( ) 9746 .1..1. hhyhhy +=+ ( ) ( )76 49 hh hh y − − = Vapor Power Plant 61 s Figure 12 Repeated T9 = T7 Pbleed = P6 The Regenerative Cycle Analysis of the Cycle For 1 kg of steam flowing through the cycle, a) Heat supplied The heat supplied to the working q9-1 = (1)(h1–h9) (24) b) Turbine work The work produced by the turbine w1-2 = (1)(h1–h6)+(1-y)(h6-h2) (25) c) Heat rejected The heat rejected to the cooling water, q2-3 = (1-y)(h2-h8)+(1)(h8-h3) (26) Vapor Power Plant 62 s Figure 12 Repeated T9 = T7 Pbleed = P6
  32. 32. 12/9/2013 32 The Regenerative Cycle Analysis of the Cycle d) Feed-water pump work The feed-water pump work input, w3-4 = v3(p4-p3) (27) Vapor Power Plant 63 s Figure 12 Repeated T9 = T7 Pbleed = P6 The Regenerative Cycle Alternate Scheme Sometimes, two closed-type FWH are used as shown in Figure 13. Vapor Power Plant 64 y1 kg (y1 + y2) kg Figure 13 Regenerative plant using two closed-type FWH (1 – y1 – y2) kg Ideally, T6 = T11 T5 = T9
  33. 33. 12/9/2013 33 The Regenerative Cycle Mass of Extracted Steam The mass of steam extracted for feed-water heater 1, y1.h7 + h5 = y1.h11 + h6 (28) h6 = h11 = h12 and that for feed-water heater 2, y2.h8 + y1.h12 + h4 = (y1 + y2)h9 + h5 (29) h5 = h9 = h10 Note: in liquid region, h = hf@T Vapor Power Plant 65 Figure 13 Repeated (1 – y1 – y2) kg Ideally, T6 = T11 T5 = T9 The Regenerative Cycle Cycle Performance a) Heat supplied The heat supplied to the working fluid, qin = q6-1 = (1).(h1+h6) (30) b) Work output The work output of the turbine wt = (h1-h7)+ (1-y1)(h7-h8)+(1-y1-y2) (h8-h2) (31) Vapor Power Plant 66 Figure 13 Repeated (1 – y1 – y2) kg Ideally, T6 = T11 T5 = T9
  34. 34. 12/9/2013 34 The Regenerative Cycle Comparison Comparisons between open-type and closed-type feed-water heaters are summarized as follows: Aspects Open-type Closed-type Design Simpler More complex Cost Cheaper More expensive Heat Transfer Good Less effective Feed Pump Many Single/Less Vapor Power Plant 67 Example 6 An ideal regenerative steam power plant operates between a pressure limit of 40 bar and 0.2 bar. The steam enters the turbine at 450°C and exits the turbine with a dryness fraction of 0.86. Some amount of steam is extracted from the turbine at a pressure of 4 bar and enters a closed-type feed-water heater. Neglecting feed pump work, determine: (a) the mass of steam extracted from the turbine, (b) thermal efficiency of the cycle, (c) specific steam consumption, and (d) condenser heat load. Vapor Power Plant 68
  35. 35. 12/9/2013 35 Example 6 Vapor Power Plant 69 Throttle valve s Ideally, T9 = T7 4 bar 450oC 40 bar 0.2 bar Example 7 Consider a steam power plant that operates on an ideal reheat- regenerative Rankine cycle with one open feed-water heater, one closed feed-water heater, and one reheater. Steam enters the turbine at 140 bar and 520oC and is condensed in the condenser at a pressure of 0.1 bar. Some steam is extracted from the high pressure (H.P) turbine at 40 bar for the closed feed-water heater, and the remaining steam is reheated at the same pressure to 520oC. The condensate exiting the closed feed-water heater and is throttled to the open feed-water heater at 7 bar. Some steam of the low-pressure (L.P) turbine is extracted from the turbine at 7 bar for the open feed-water heater. Saturated liquid from the open feed-water heater is pumped to 140 bar before entering the closed feed-water heater and passes to the boiler. Determine (a) the fraction of steam extracted from each turbine as well as (b) the thermal efficiency of the cycle. Vapor Power Plant 70
  36. 36. 12/9/2013 36 Example 8 Stim meninggalkan dandang pada 140 bar dan 520°C. Stim kemudiannya memasuki turbin peringkat pertama dan keluar pada tekanan 40 bar. Sebahagian stim yang keluar dari turbin ini dibekalkan kepada sebuah pemanas air suapan tertutup, dan selebihnya dipanaskan semula kepada suhu 520°C. Stim yang dipanaskan semula memasuki turbin peringkat kedua dan dikembangkan kepada tekanan pemeluwap 0.1 bar. Sebahagian stim dijujuh dari turbin peringkat kedua pada tekanan 7 bar untuk digunakan oleh pemanas air suapan terbuka. Stim jujuhan daripada turbin peringkat pertama memeluwap apabila ia melalui pemanas tertutup dan didikitkan ke pemanas terbuka pada 7 bar. Cecair tepu air yang meninggalkan pemanas terbuka melalui sebuah pam yang meningkatkan tekanan kepada 140 bar sebelum memasuki pemanas tertutup dan dialirkan kepada dandang. Dengan anggapan kitar adalah unggul dan mengabaikan kerja pam, tentukan (a) peratus stim memasuki turbin peringkat pertama yang memasuki pemanas tertutup, (b) peratus stim memasuki turbin peringkat kedua yang dialirkan kepada pemanas terbuka, dan (c) kecekapan kitar. Vapor Power Plant 71 Example 8 Vapor Power Plant 72 520oC 12 (y1) 10 11
  37. 37. 12/9/2013 37 Vapor Power Plant 73 (1) 7 T3 = 420oC 4 (y1) 11 12 P2 = P3 = 4 Mpa 1 High- Pressur e turbine 10 8 6 Heater P 2 Heater P 1 3 5 9 Low- Pressure turbine P4 = 800 kPa (y2) P5 = 10 kPa (1 – y1 – y2) 2Boiler Condense r P1 = 13 Mpa T1 = 600oC

×