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- 1. James Li # 26 3/27/2015 Raj, Rishi ME 43000 Spring 2015 Thermal Project # 1: Optimal Thermal Cycle for Steam Turbine Power Plant 1
- 2. Abstract: The purpose of this project is to make a design to optimize the performance a steam turbine using the ideal Rankine cycle. For this project, five different Rankine cycles will be used to analyze the thermal efficiency of the turbine with a minimum moisture content of 13 %. The given inlet power and inlet pressure of 150,000 KW and 2000 PSI respectively and the reheat temperature maximizes at 1000 degrees Fahrenheit. The turbine will analyze first with the standard Ideal Rankine cycle, followed by adding another turbine using the same cycle with a reheat, and finally by adding three water heaters in a process known as regeneration that conserves fuel and improve work output. The efficiency and cost effectiveness will then be calculated to determine the best design for the steam turbine. Table of Contents Nomenclature ……………………………………………………………………………………………. Pages 2-3 Introduction to the Steam Turbine …………………………………………………………………… Pages 3-4 Rankine Cycle Background……………………………………………………………………………… Pages 4-5 Sample Calculations …………………………………………………………………………………… Pages 5-22 • Step 1: Ideal Rankine Cycle [page 6] • Step 2: Ideal Rankine Cycle with Reheat [page 7] • Step 3: Ideal Rankine Cycle with Regeneration (cycle 3: one open feed water heater) [page 11] • Step 3: Ideal Rankine Cycle with Regeneration (cycle 4: two open feed water heater) [page 14] • Step 3: Ideal Rankine Cycle with Regeneration (cycle 5: four open feed water heater) [page 17] • Cost Effectiveness [Page 22] Excel Graphs and Spreadsheets…………………………………………………………………………… Pages 22 - 28 Discussions and conclusions………………………………………………………………………………Pages 28 - 30 Appendix ……………………………………………………………………………………………………… References ………………………………………………………………………………………………… Nomenclature P = Pressure (Psi) 2
- 3. T = Temperature (degrees Fahrenheit) h = enthalpy (btu/lbm) s = entropy (btu/lbm.R) v = specific volume (ft^3/lbm) x = Steam Quality m = mass flow rate (lbm/s) q (in) = entering heat (btu/lbm) q (out) = exiting heat (btu/lbm) Wp = Work done by the pump (btu/lbm) η = efficiency f = saturated liquid state g = saturated vapor state fg = difference between saturated liquid state and saturated vapor state Introduction to the Steam Turbine The purpose of this design project is to find the most cost effective and efficient steam turbine cycle for operating a steam turbine power plant. A steam turbine is a machine that provides mechanical energy by converting pressurized steam into said energy when the steam hits the turbine’s rotating blades. When the pressurized steam flows past and hits the rotating blade, it cools and transfers its thermal energy into the turbine and rotates the shaft up to 3600 rpm to power the electricity generator. The thermal energy generated from the cooling steam can extract the mechanical work needed to operate the electricity generators via power cycles such as the Rankine, Rankine Reheat, and regenerative cycles. A steam turbine is also compact due to having the steam flow spin the turbine blade continuously and expanding the steam in order to drive a machine. This negates the need for a push-pull action or a piston operation, making the turbine useful for operating machines where space is limited, such as on a ship. The cycle used for this turbine would be the Ideal Rankine cycle and the cycle will be divided into multiple stages in order to find the most efficient stage to operate the power plant. For this design, the turbine 3
- 4. will operate in three stages. The first stage of this turbine is based on the ideal Rankine cycle with a steam quality minimum at around 87% or x = 0.87. The second stage of this turbine involves implementing a reheat to the Rankine cycle of turbine. The cycle reheat involves separating the cycle into a high pressure stage and a low pressure stage where the steam will be reheated during the high pressure stage prior to entering the low pressure stage. The third stage consists of three different cycles. The first cycle of the third stage involves adding an open feed water heater to the high pressure turbine. The second cycle involves adding another open feed water heater to the intermediate pressure turbine and the third and final cycle involves adding two open feed water heater to the low pressure turbine. For each stage, the efficiency will be calculated, analyzed and compared with the other stages, after which the income of the stages will be calculated in order to determine whether or not the improvements in efficiency would be worth designing a particular stage and cycle of the turbine. Rankine Cycle Background As mentioned earlier, the cycle used to operate the steam turbine power plant is the Rankine cycle. The Rankine cycle is a variant of the Carnot cycle that is less efficient but more practical. The Carnot cycle is a highly efficient vapor power cycle consisting of two isentropic (1 2 and 3 4) and two isothermal (2 3 and 4 1) processes as shown below: Figure 1: Carnot Cycle 4
- 5. Despite this cycle’s efficiency, its process is highly idealized due to its pumping process being a mixture of liquid and vapor. Modern technology cannot yet handle a liquid-vapor mixture process yet. The moisture content generated from the steam quality is too high and corrodes the turbine’s blade rapidly. These issues make the cycle not achievable in real life thus, this is where the Rankine cycle comes into play. The Rankine cycle eliminates the liquid-vapor mixture and moisture content problems by vaporizing the steam at a constant pressure instead of at a constant temperature like the Carnot cycle does. This means that the steam is superheated in the boiler and condenses in the condenser so the liquid-vapor do not mix and the moisture content is reduced. While this makes the Rankine Cycle more practical in real life, it needs some adjustments to achieve its optimal efficiency and that is where its stages come in. The Parameters for this Particular design is given below. • Power Plant output: 150,000 KW • Turbine inlet Pressure: 2000 Psi • Reheat inlet temperature: 1000 degrees Fahrenheit • Maximum moisture level (1-x): 13% or 0.13 The calculations will explain the steps necessary to achieve the highest efficiency for each stage with the given parameters Sample Calculations: 5
- 6. Step 1: Ideal Rankine Cycle Figure 2: The Ideal Rankine Cycle Cycle Steps • 1 2: Isentropic Pump Compression (s1=s2) • 2 3: Constant pressure heat addition inside of the boiler (P2=P3) • 3 4: Isentropic turbine steam expansion (s3=s4) • 4 1: Constant pressure heat addition inside of the condenser (P4=P1) State 1 Test Pressure: P1 = 1 Psi (first value) v1 = 0.01614 ft3/lbm (Appendix 2 Table A-5E) h1 = 69.72 Btu/lbm (Appendix 2 Table A-5E) State 2 P2 = 2000 Psi wp (work) = v1*(P2-P1)*(144/778) = 5.971717 Btu/lbm h2 = h1 + wp = 75.69172 Btu/lbm State 3 P3 = P2 = 2000 Psi T3 = 1000 degrees Fahrenheit h3 = 1474.9 Btu/lbm (Appendix 2 Table A-6E) s3 = 1.5606 Btu/(lbm*R) State 4 P4 = P1 = 1 Psi s4 = s3 = 1.5606 Btu/(lbm*R) (Appendix 2 Table A-6E) 6
- 7. sg4 = 1.9776 (Appendix 2 Table A-5E) sfg4 = 1.84496 (Appendix 2 Table A-5E) hg4 = 1105.4 Btu/lbm (Appendix 2 Table A-5E) hfg4 = 1035.7 (Appendix 2 Table A-5E) 1-x4 (moisture content) = (sg4-s4)/(sfg4) = 0.2260 h4 = hg4 – (1-x4)*hfg4 = 871.3086 Btu/lbm x4 (Steam quality) = 1 – (1-x4) = 0.7740 efficiency η = 1 – (h4-h1)/(h3-h2) = 0.4271 = 42.71 % The process shown in the chart above is then repeated for test pressures 5, 10, 15, 20, 25, and 30 Psi and the efficiency would be calculated accordingly (see table 1). Based on the trends shown (see Figures 6 and 7), steam quality increases and efficiency decreases as pressure goes up The exit pressure used to proceed to the next step and the efficiency of this cycle would be interpolated at steam quality: x4 = 0.87 (87%). The exit pressure and efficiency interpolated at x4 = 0.87 was approximately 17 Psi and 34.20 % percent respectively. With step 1 calculated, it is time to proceed to step 2 to improve upon this cycle. Step 2: Ideal Rankine Cycle with Reheat Figure 3: Ideal Rankine Cycle with Reheat The Ideal Rankine Cycle with reheating involves adding another turbine to the previous cycle. For this stage, the steam reenters the boiler and passes through the second turbine after passing through the first turbine (state 7
- 8. 4) with excess moisture eliminated. The second turbine has a pressure at around 10 – 40 percent ([0.1-0.4] * Turbine inlet pressure) of the first turbine. The reheating also improves the steam quality due to the decrease in excess moisture, therefore improving operational life and efficiency. 1 2: Isentropic pump compression (s1=s2) 2 3: Constant pressure heat addition inside of the boiler (P2=P3) 3 4: Isentropic expansion inside of the first high pressure turbine (s3=s4) 4 5: Constant pressure heat addition inside of the boiler (P4=P5) 5 6: Isentropic expansion inside of the second low pressure turbine (s5=s6) 6 1: Constant pressure heat rejection inside of the condenser (P6=P1) Case 1: Constant turbine exit pressure at x = 0.87 17 Psi and variable reheat pressure ([0.1-0.4] * Turbine inlet pressure) State 1 P1 = 17 Psi h1 = 187.21 Btm/lbm (Interpolated from table A-5E) v1 = 0.016764 ft3/lbm (Interpolated from table A-5E) sf = 0.322548 Btu/(lbm*R) (Interpolated from table A-5E) sfg = 1.4154 Btu/(lbm*R) (Interpolated from table A-5E) hfg = 965.654 Btu/lbm (Interpolated from table A-5E) State 2 P2 = 2000 Psi h2 = h1 + wp = h1 + v1*(P2-P1)*(144/778) = 193.363 Btu/lbm State 3 P3 = P2 = 2000 Psi T3 = 1000 degrees Fahrenheit h3 = 1474.9 Btu/lbm (extracted from table A-6E) s3 = 1.5606 Btu/(lbm*R) (extracted from table A-6E) State 4 P4 = 200 Psi (first value) s4 = s3 = 1.5606 Btu/(lbm*R) h4 = 1211.253 Btu/lbm (interpolated from table A-6E) State 5 P5 = P4 – 200 Psi T5 = T3 = 1000 degrees Fahrenheit 8
- 9. h5 = 1529.6 Btu/lbm (extracted from table A-6E) s5 = 1.843 Btu/(lbm*R) (extracted from table A-6E) State 6 P6 = P1 = 17 Psi s6 = s5 = 1.843 Btu/lbm x6 = (s6-sf)/sfg = 1.07422 h6 = hf + x6*hfg = 1224.53553 Btu/lbm Efficiency q(in) = (h3-h2) + (h5-h4) = 1599.884 Btu/lbm q(out) = h6-h1 = 1037.32553 Btu/lbm η = 1 – [q(out)/q(in)] = 0.351625 The process shown in the chart above is then repeated for P4 = 275, 350, 400, 450, 500, 600, 700, and 800 Psi and the efficiency is calculated accordingly (see table 2). The optimal Efficiency for this process was 35.733 % at exit pressure 500 Psi (see Figure 8). The pressure at optimal efficiency will then be used for step 2 case 2. Case 2: Retaining exit pressure from Case 1 and changing the exit pressure to obtain efficiency at x = 0.87 State 1 P1 = 1 Psi (First value) h1 = 69.72 Btu/lbm (extracted from table A-5E) hfg = 1035.7 Btu/lbm (extracted from table A-5E) v1 = 0.01614 ft3/lbm (extracted from table A-5E) State 2 P2 = 2000 Psi wp (work) = v1*(P2-P1)*(144/778) = 5.971717 Btu/lbm h2 = h1 + wp = 75.69172 Btu/lbm State 3 P3 = P2 = 2000 Psi T3 = 1000 degrees Fahrenheit h3 = 1474.9 Btu/lbm (extracted from table A-6E) s3 = 1.5606 Btu/(lbm*R) State 4 P4 = 500 psi s4 = s3 = 1.5606 Btu/(lbm*R) T4 = 602.9091 degrees Fahrenheit (interpolated from table A-6E) h4 = 1300.334 Btu/lbm (interpolated from table A6-E) 9
- 10. State 5 P5 = P4 = 500 Psi T5 = T3 = 1000 degrees Fahrenheit s5 = 1.7376 Btu/(lbm*R) (extracted from Table A-6E) h5 = 1521 Btu/lbm (extracted from Table A-6E) sf = 0.13262 Btu/(lbm*R) (extracted from Table A-6E) sfg = 1.84495 Btu/(lbm*R) (extracted from Table A-6E) State 6 P6 = P1 = 1 Psi x = (s6-sf)/sfg = 0.8669931 h6 = h1 + (x*hfg) = 970.708 Btu/lbm Efficiency q(in) = (h3-h2)/(h5-h4) = 1619.874 Btu/lbm q(out) = h6 – h1 = 900.988 Btu/lbm η = 1 – [q(out)/q(in)] = 0.443791 The process from the case 2 chart was then repeated for P1 = 2 6 Psi and the efficiency is calculated accordingly (see table 3). The optimal Efficiency and Exhaust pressure for this process was 44.37 % and 1 Psi at x = 0.87 respectively (see Figures 9 and 10). After the efficiency and exhaust pressures are calculated, Step 3 will be initiated using the exhaust pressure as P1 in order to improve efficiency. Step 3: Ideal Rankine Cycle with Regeneration (cycle 3: one open feed water heater) 10
- 11. Figure 4: Ideal Rankine Cycle with Regeneration (one water heater) The first Regenerative Ideal Rankine Cycle involves adding an open feed water heater to the high pressure turbine of the reheated Rankine cycle in order to further boost the cycle’s efficiency. By adding an open feed water heater, the average temperature also increases. The temperature increase reduces the moisture content, thus limiting potential moisture damage to the turbine. The process of the Regenerative Ideal Rankine Cycle with one open feed water heater is shown on the next page. 1 2: Isentropic compression in the first pump 2 3: Constant pressure heat addition inside the open feed water heater 3 4: Isentropic compression in the second pump 11
- 12. 4 5: Constant pressure heat addition inside the boiler 5 6: Isentropic steam expansion in high pressure turbine 6 3: Constant pressure heat regeneration at the open feedwater heater 6 7: Constant pressure heat addition inside the boiler 7 8: Isentropic steam expansion inside of the low pressure turbine 8 1: Constant pressure heat rejection inside of the condenser State 1 P1 = 1 Psi (from step 2 case 2) h1 = 69.72 Btu/lbm (extracted from table A-5E) hfg = 1035.7 Btu/lbm (extracted from table A-5E) v1 = 0.01614 ft3/lbm (extracted from table A-5E) State 2 P2 = 500 Psi (from step 2 case 1) wp = v1*(P2-P1)*(144/778) = 1.490689 Btu/lbm h2 = h1 + wp = 71.2107 Btu/lbm State 3 P3 = P2 = 500 Psi h3 = 449.51 Btu/lbm (extracted from table A-5E) v3 = 0.01975 ft3/lbm (extracted from table A-5E) s3 = 0.649 Btu/(lbm*R) (extracted from table A-5E) State 4 P4 = 2000 Psi wp2 = v3*(P4-P3)*(144/778) = 5.48329 Btu/lbm h4 = h3 + wp2 = 454.9932905 Btu/lbm State 5 P5 = 2000 Psi T5 = 1000 degrees Fahrenheit s5 = 1.5606 (table A-6E) State 6 P6 = 500 Psi (from cycle 2 case 2) s6 = s5 = 1.5606 Btu/(lbm*R) h6 = 1300.333818 Btu/lbm (interpolated from Table A-6E) State 7 P7 = P6 = 500 Psi T7 = 1000 degrees Fahrenheit h7 = 1521 Btu/lbm (Table A-6E) 12
- 13. s7 = 1.7376 Btu/(lbm*R) (Table A-6E) State 8 P8 = P1 = 1 Psi s8 = s7 = 1.7376 Btu/lbm sf8 = 0.13263 Btu/lbm (table A-5E) sfg8 = 1.85595 (table A-5E) x8 = (s8-sf8)/(sfg8) = 0.8699931 h8 = h1 + x8*hfg = 970.708 Btu/lbm efficiency y = (h3-h2)/(h6-h2) = 0.30778 (fraction of steam extracted) q (in) = (h3-h4) + (1-y) * (h7-h6) = 1172.656 Btu/lbm q (out) = (1-y) * (h8-h1) = 623.6821 Btu/lbm η = 1 – [q(out)/q(in)] = 0.468146 = 46.8146 % Mass Flow Rate Wnet = q (in) – q(out) = 548.974 Btu/lbm m = Power*3412/Wnet = 932284.1883 lbm/hr The efficiency showed an improvement over the previous cycle at x = 0.87, giving an efficiency of 46.8146 %. The Mass Flow Rate was also calculated to be 932284.1883 lbm/hr (see table 4). Step 3: Ideal Rankine Cycle with Regeneration (cycle 4: two open feed water heater) 13
- 14. Figure 5: Ideal Rankine Cycle with Regeneration (two water heaters) The second Regenerative Ideal Rankine Cycle involves adding another open feed water heater to the intermediate pressure turbine of the reheated Rankine cycle to give another enhancement to the previous cycle’s efficiency. The process of this cycle is shown below: 1 2: Isentropic compression inside the first pump 2 3: Constant pressure heat addition inside of the second open feed water heater 3 4: Isentropic compression inside of the second pump 4 5: Constant pressure heat addition inside of the first open feed water heater 5 6: Isentropic compression inside of the third pump 6 7: Constant pressure heat addition inside of the boiler 7 8: Isentropic steam expansion inside of the low pressure turbine 8 9: Constant pressure heat addition inside of the boiler 8 5: Constant pressure heat regeneration inside of the first open feed water heater 9 11: Isentropic steam expansion inside of the intermediate pressure turbine 10 3: Constant pressure heat regeneration inside of the second open feed water heater 14
- 15. 11 1: Constant pressure heat rejection inside of the condenser State 1 P1 = 1 Psi (from cycle 2 case 2) h1 = 69.72 Btu/lbm(Table A-5E) v1 = 0.01614 ft3/lbm (Table A-5E) s1 = 0.13262 Btu/(lbm*R) (Table A-5E) State 2 P2 = 250 Psi (assumed between 1 and 500 Psi) s2 = s1= 0.13262 Btu/(lbm*R) wp2 = v1*(P2-P1)*(144/778) = 0.74385 Btu/lbm h2 = h1 + wp2 = 70.46385 Btu/lbm State 3 P3 = P2 = 250 Psi h3 = 376.09 Btu/lbm (Table A-5E) v3 = 0.01865 ft3/lbm (Table A-5E) s3 = 0.56784 Btu/(lbm*R) (Table A-5E) State 4 P4 = 500 Psi (from step 2 case 1) s4 = s3 = 0.56784 Btu/(lbm*R) wp4 = v3*(P4-P3) * (144/778) = 0.86298 Btu/lbm h4 = h3 + wp4 = 376.952982 Btu/lbm State 5 P5 = P4 = 500 Psi h5 = 449.51 Btu/lbm (Table A-5E) v5 = 0.01975 ft3/lbm (Table A-5E) s5 = 0.649 Btu/(lbm*R) (Table A-5E) State 6 P6 = 2000 psi s6 = s5 = 0.649 Btu/(lbm*R) wp6 = v5*(P6-P5)*(144/778) = 5.48329 Btu/lbm h6 = h5 + wp6 = 454.9933 Btu/lbm State 7 P7 = P6 = 2000 Psi T7 = 1000 degrees Fahrenheit h7 = 1474.9 Btu/lbm (Table A-6E) s7 = 1.5606 Btu/(lbm*R) (Table A-6E) State 8 P8 = P4 = 500 Psi s8 = s7 = 1.5606 Btu/(lbm*R) 15
- 16. h8 = 1300.333818 (Interpolated from table A-6E) State 9 P9 = P8 = 500 Psi T9 = 1000 degrees Fahrenheit h9 = 1521 Btu/lbm (table A-6E) s9 = 1.7376 Btu/(lbm*R) (table A-6E) State 10 P10 = P3 = 250 Psi s10 = s9 = 1.7376 Btu/(lbm*R) h10 = 1419.881944 (Interpolated from table A-6E) State 11 P11 = P1 = 1 Psi s11 = s10 = 1.7376 Btu/(lbm*R) sf = 0.13262 Btu/(lbm*R) (Table A-5E) sfg = 1.84495 Btu/(lbm*R) (Table A-5E) x = (s11-sf)/sfg = 0.869931434 hfg = 1035.7 (Table A-5E) h11 = h1 + x*hfg = 970.7079867 Btu/lbm Efficiency y = (h5-h4)/(h8-h4) = 0.078577565 z = (1-y)*(h3-h2)/(h10-h2) = 0.20869054 q(in) = (1-y)*(h9-h8) + (h7 – h8) = 1223.23348 Btu/lbm q(out) = (1-y-z)*(h11-h1) = 642.1628748 Btu/lbm η = 1 – [q(out)/q(in)] = 0.475028 = 47.5028 % Mass Flow Rate Wnet = q(in) – q(out) = 581.0706053 Btu/lbm m = Power*3412/Wnet = 880787.9719 lbm/hr The efficiency and mass flow rate for this cycle are 47.5028 % and 880787.9719 lbm/hr at x = 0.87 respectively (see table 5). The efficiency has improved and the mass flow rate has decreased from the previous cycle Step 3: Ideal Rankine Cycle with Regeneration (cycle 5: four open feed water heater) 16
- 17. Figure 6: Ideal Rankine Cycle with Regeneration (four water heaters) The third and final Regenerative Ideal Rankine Cycle involves adding two open feed water heater to the lower pressure turbine of the reheated Rankine cycle to give a third enhancement to the previous cycle’s efficiency. The process of this cycle is shown on the below: 1 2: Isentropic compression at the first pump 2 3: Constant pressure heat addition at the fourth open feed water heater 3 4: Isentropic compression at the second pump 17
- 18. 4 5: Constant pressure heat addition at the third open feed water heater 5 6: Isentropic compression at the third pump 6 7: Constant pressure heat addition at the second open feed water heater 7 8: Isentropic compression at the fourth pump 8 9: Constant pressure heat addition at the first open feed water heater 9 10: Isentropic compression at the fifth pump 10 11: Constant pressure heat addition inside of the boiler 11 12: Isentropic steam expansion inside of the high pressure turbine 12 9: Constant pressure heat regeneration at the first open feed water heater 12 13: Constant pressure heat addition at the boiler 14 7: Constant pressure heat regeneration at the second open feed water heater 13 15: Isentropic steam expansion inside of the intermediate pressure turbine 16 5: Constant pressure heat regeneration at the third open feed water heater 17 3: Constant pressure heat regeneration at the fourth open feed water heater 15 18: Isentropic steam expansion inside of the low pressure turbine State 1 P1 = 1 Psi (from cycle 2 case 2) h1 = 69.72 Btu/lbm(Table A-5E) v1 = 0.01614 ft3/lbm (Table A-5E) s1 = 0.13262 Btu/(lbm*R) (Table A-5E) State 2 P2 = 50 Psi (Assumed for low pressure turbine) s2 = s1 = 0.13262 Btu/(lbm*R) Wp1 = v1*(P2-P1)*(144/778) = 0.14638 Btu/lbm h2 = h1 + Wp1 = 69.8663 Btu/lbm State 3 P3 = P2 = 50 Psi h3 = 250.21 Btu/lbm(Table A-5E) v3 = 0.01727 ft3/lbm (Table A-5E) 18
- 19. s3 = 0.41125 Btu/(lbm*R) (Table A-5E) State 4 P4 = 100 Psi (Assumed for low pressure turbine) s4 = s3 = 0.41125 Btu/(lbm*R) Wp2 = v3*(P4-P3)*(144/778) = 0.159825 Btu/lbm h4 = h3 + Wp2 = 69.8663 Btu/lbm State 5 P5 = P4 = 100 Psi h5 = 298.51 Btu/lbm (Table A-5E) v5 = 0.01774 ft3/lbm (Table A-5E) s5 = 0.47427 Btu/(lbm*R) (Table A-5E) State 6 P6 = 250 Psi (Assumed for intermediate pressure turbine) s6 = s5 = 0.47427 Btu/(lbm*R) Wp3 = v6*(P6-P5)*(144/778) = 0.49252 Btu/lbm h6 = h5 + Wp3 = 299.0025244 Btu/lbm State 7 P7 = P6 = 250 Psi h7 = 376.09 Btu/lbm (Table A-5E) v7 = 0.01865 ft3/lbm (Table A-5E) s7 = 0.56784 Btu/(lbm*R) (Table A-5E) State 8 P8 = 500 Psi (from step 2 cycle 1) s8 = s7 = 0.56784 Btu/(lbm*R) Wp4 = v8*(P8-P7)*(144/778) = 0.862982 Btu/lbm h8 = h7 + Wp4 = 376.952982 Btu/lbm State 9 P9 = P8 = 500 Psi h9 = 449.51 Btu/lbm (Table A-5E) v9 = 0.01975 ft3/lbm (Table A-5E) s9 = 0.649 Btu/(lbm*R) (Table A-5E) State 10 P10 = 2000 Psi s10 = s9 = 0.649 Btu/(lbm*R) Wp5 = v10*(P10-P9)*(144/778) = 5.483290488 Btu/lbm 19
- 20. h10 = h9 + Wp5 = 454.9932905 Btu/lbm State 11 P11 = P10 = 2000 Psi T11 = 1000 degrees Fahrenheit v11 = 0.39479 ft3/lbm (Table A-6E) h11 = 1474.9 Btu/lbm (Table A-6E) s11 = 1.5606 Btu/(lbm*R) (Table A-6E) State 12 P12 = P9 = 500 Psi s12 = s11 = 1.5606 Btu/(lbm*R) h12 = 1300.333818 Btu/lbm (interpolated from table A-6E) m12 = (h9-h8)/(h12-h8) = 0.078577 State 13 P13 = P12 = 500 Psi T13 = 1000 degrees Fahrenheit v13 = 1.70094 ft3/lbm (Table A-6E) h13 = 1521 Btu/lbm (Table A-6E) s13 = 1.7376 Btu/(lbm*R) (Table A-6E) m13 = 1- m12 = 0.9214 State 14 P14 = P7 = 250 Psi s14 = s13 = 1.7376 Btu/(lbm*R) h14 = 1419.994431 Btu/lbm (interpolated from table A-6E) m14 = (1 - m13)*(h7-h6)(/(h14-h6) = 0.063363642 State 15 (Irrelevant for this particular calculation) State 16 P16 = P5 = 100 Psi s16 = s14 = 1.7376 Btu/(lbm*R) h16 = 1308.273494 Btu/lbm (interpolated from table A-6E) m16 = (1-m13-m14)*(h5-h4)/(h16-h4) = 0.039046183 State 17 P17 = P3 = 50 Psi s17 = s16 = 1.7376 Btu/(lbm*R) h17 = 1245.05 Btu/lbm (interpolated from table A-6E) m17 = (1-m13-m14-m16)*(h4-h3)/(h17-h3) = 0.125685634 State 18 P18 = P1 = 1 Psi s18 = s17 = 1.7376 Btu/(lbm*R) x18 = (s18-s1)/sfg = (s18-s1)/ 1.84495 = 0.869931434 20
- 21. h18 = h1 + x18*hfg = h1 + x18*1035.7 = 970.7079867 Btu/lbm Efficienc y q(in) = (h11-h10) + (1-m12)*(h13-h12) = 1223.23348 Btu/lbm q(out) = (1-m12-m14-m16-m17)*(h18- h1) = 624.6792759 Btu/lbm η = 1 – [q(out)/q(in)] = 0.489321306 = 48.93213061 % Mass Flow Rate Wnet = q(in) – q(out) = 598.5542041 Btu/lbm m = power*3412/Wnet = 855060.4047 (lbm/hr) The efficiency and mass flow rate for this cycle are 48.9321% and 855060.4047 lbm/hr at x = 0.87 respectively (see table 6). The efficiency has improved and the mass flow rate has decreased again from the previous cycle. [Note: Mass flow rates for step 4 has already been calculated in step 3] Cost Effectiveness Formula: annual income = Power (Kw) * 8765.81 (hours/year) * (Efficiency/100) * $0.16 (cost of electricity in kwh) Cycle Type Efficiency (%) Annual income ($ in Millions) Profits gained ($ in Millions) Ideal Rankine Cycle 34.2011 71.95 N/A Ideal Rankine Cycle with reheating 44.3745 93.35 21.4 21
- 22. Ideal Rankine Cycle with 1 Feed water Heater 46.8146 98.49 26.54 Ideal Rankine Cycle with 2 Feed water Heater 47.5208 99.97 28.02 Ideal Rankine Cycle with 4 Feed water Heater 48.3213 101.66 29.71 Based on the cost effectiveness calculation in the chart above, it is safe to conclude that reheating and adding feed water heaters to the turbine will improve its efficiency Excel Graphs and Spreadsheets Test Pressure P1 (Psi) P2 = P3 T3 v1 1 2000 1000 0.01614 5 2000 1000 0.01641 10 2000 1000 0.01659 15 2000 1000 0.01672 20 2000 1000 0.01683 25 2000 1000 0.01692 30 2000 1000 0.017 h1 wp h2 h3 s3 = s4 hg4 hfg4 sg4 sfg4 69.72 5.971717 75.69172 1474.9 1.5606 1105.4 1035.7 1.9776 1.84495 130.18 6.059466 136.2395 1474.9 1.5606 1130.7 1000.5 1.8438 1.60894 161.25 6.110579 167.3606 1474.9 1.5606 1143.1 981.82 1.7875 1.50391 181.21 6.142988 187.353 1474.9 1.5606 1150.7 969.47 1.7549 1.44441 196.27 6.167827 202.4378 1474.9 1.5606 1156.2 959.93 1.7319 1.39606 208.52 6.185152 214.7052 1474.9 1.5606 1160.6 952.03 1.7141 1.3606 218.93 6.198663 225.1287 1474.9 1.5606 1164.1 945.21 1.6995 1.33132 1- x4 (moisture content) h4 steam quality (x4) efficiency 0.226022385 871.308615 0.773977615 0.4271127 0.176016508 954.595484 0.823983492 0.384149 0.15087339 994.969488 0.84912661 0.3623753 0.134518592 1020.28826 0.865481408 0.3483125 0.122702463 1038.41423 0.877297537 0.3381774 0.112817874 1053.194 0.887182126 0.3297275 0.104332542 1065.48384 0.895667458 0.322633 Steam quality At 87% (0.87) Exit pressure = 16.91 --> 17 Psi Efficiency = 0.3420114291= 34.2011 % Table 1: Step 1 22
- 23. Steam quality versus exit pressure y = 0.0357Ln(x) + 0.7705 R 2 = 0.9944 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9 0.92 0 5 10 15 20 25 30 35 Exit Pressure (Psi) SteamQuality Figure 6: Steam Quality vs. Exit Pressure Efficiency versus exit pressurey = -0.0307Ln(x) + 0.4301 R 2 = 0.9946 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0 5 10 15 20 25 30 35 Exit Pressure (Psi) Efficiency Figure 7: Efficiency vs. Exit Pressure 23
- 24. Exit pressure at x = 0.87 (Pe) = 17 Psi Pa = 15 Psi Pb = 20 Psi hf1 = 181.21 btu/lbm hf2 = 196.21 btu/lbm hf = h1 = (Pe -Pa)*(hf2-hf1)/(Pb-Pa) + hf1 = 187.21 btu/lbm vf2 = 0.01683 ft^3/lbm vf1 = 0.01672 ft^3/lbm vf = (Pe -Pa)*(vf2-vf1)/(Pb-Pa) + vf1 = 0.016764 P2 = P3 = 2000 Psi h2 = h1 + wp = h1 + [vf*(P2-Pe)]*(144/778) = 193.363 btu/lbm T3 = 1000 degrees Fahrenheit h3 = 1474.9 Btu/lbm s3 = 1.5606 sf1 = 0.31370 Btu/lbm-R sf2 = 0.33582 Btu/lbm-R sfg1 =1.44441 Btu/lbm-R sfg2 = 1.39606 Btu/lbm-R hfg1 = 969.47 Btu/lbm hfg2 = 959.93 Btu/lbm sf = (Pe-Pa)*(sf2-sf1)/(Pb-Pa) + sf1 = 0.322548 Btu/lbm-R sfg = (Pe-Pa)*(sfg2-sfg1)/(Pb-Pa) + sfg1 = 1.4154 Btu/lbm-R hfg = (Pe-Pa)*(hfg2-hfg1)/(Pb-Pa) + hfg1 = 965.654 Btu/lbm- P4 = P5 (0.1-0.4)*P(given) h4 h5 s5 = s6 x6 200 1211.253 1529.6 1.843 1.074220715 275 1240.311 1527.4 1.8068 1.048644906 350 1263.607 1525.3 1.7791 1.029074467 400 1276.906 1523.9 1.7636 1.018123499 450 1289.302 1522.4 1.7499 1.008444256 500 1300.334 1521 1.7376 0.999754133 600 1320.431 1518.1 1.716 0.984493429 700 1337.987 1515.2 1.6974 0.971352268 800 1353.809 1512.2 1.6812 0.95990674 h6 = hf + (x6)*hfg q(in) q(out) efficiency 1224.53553 1599.884 1037.32553 0.351625 1199.838148 1568.626 1012.628148 0.354449 1180.939875 1543.23 993.729875 0.356071 1170.365029 1528.531 983.155029 0.356797 1161.01823 1514.635 973.8082296 0.357067 1152.626578 1502.203 965.4165777 0.357333 1137.890018 1479.206 950.6800181 0.357304 1125.200203 1458.75 937.9902029 0.35699 1114.147783 1439.928 926.9377832 0.356261 Optimal Efficiency 35.7333% at 500 Psi (P4) Table 2: Step 2 Case 1 24
- 25. Pressure versus Efficiency 0.351 0.352 0.353 0.354 0.355 0.356 0.357 0.358 0 100 200 300 400 500 600 700 800 900 Pressure (Psi) Efficiency Series1 Figure 8: Pressure versus Efficiency P1 = P6 P2 = P3 P4 = P5 h1 = hf hfg 1 2000 500 69.72 1035.7 2 2000 500 94.02 1021.7 3 2000 500 109.4 1012.8 4 2000 500 120.9 1006 5 2000 500 130.18 1000.5 6 2000 500 138.02 995.88 v1 = vf wp (work) h2 T3 = T5 h3 s3 = s4 0.01614 5.971717 75.69172 1000 1474.9 1.5606 0.01623 6.002013 100.022 1000 1474.9 1.5606 0.0163 6.024882 115.4249 1000 1474.9 1.5606 0.01636 6.044032 126.944 1000 1474.9 1.5606 0.01641 6.059466 136.2395 1000 1474.9 1.5606 0.01645 6.071192 144.0912 1000 1474.9 1.5606 T4 h4 s5 = s6 h5 sf sfg x 602.9090909 1300.334 1.7376 1521 0.13262 1.84495 0.869931 602.9090909 1300.334 1.7376 1521 0.17499 1.74444 0.895766 602.9090909 1300.334 1.7376 1521 0.2009 1.68489 0.912048 602.9090909 1300.334 1.7376 1521 0.21985 1.64225 0.924189 602.9090909 1300.334 1.7376 1521 0.23488 1.60894 0.933981 602.9090909 1300.334 1.7376 1521 0.24739 1.58155 0.942247 h6 q(in) q(out) efficiency 970.708 1619.874 900.988 0.443791 1009.224 1595.544 915.2041 0.4264 1033.122 1580.141 923.7219 0.415418 1050.635 1568.622 929.7345 0.407292 1064.628 1559.327 934.4484 0.400736 1076.384 1551.475 938.3645 0.395179 Efficiency at x = 0.87 Exhaust pressure at x = 0.87 0.443745318 1 Psi 44.37% Table 3: Step 2 Case 2 25
- 26. Pressure versus steam quality 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0 1 2 3 4 5 6 7 Pressure (Psi) SteamQuality Figure 9: Pressure vs. Steam Quality Pressure vs efficiency 0.39 0.4 0.41 0.42 0.43 0.44 0.45 0 1 2 3 4 5 6 7 Pressure (Psi) Efficiency Figure 10: Pressure vs. Efficiency 26
- 27. P1 = P8 (Psi) P2 = P3 = P6 = P7 h1 = hf hfg v1 = vf wp = work h2 = h1 + wp 1 500 69.72 1035.7 0.01614 1.490689 71.21068874 h3 = hf3 v3 = vf3 s3 = s4 = sf P4 = P5 (Psi) wp2 h4 = h3 + wp2 449.51 0.01975 0.649 2000 5.48329 454.9932905 T5 h5 s5 = s6 h6 T7 h7 s7 = s8 1000 1474.9 1.5606 1300.333818 1000 1521 1.7376 sf8 sfg8 x8 h8 y 0.13262 1.84495 0.869931 970.708 0.30778 q(in) q (out) efficiency 1172.656 623.6821 0.468146 efficiency at .87 (%) Wnet (btu/lbm) Mass Flow Rate (lbm/hr) 46.81459012 548.974236 932284.1883 Table 4: Step 3 Cycle 3 P1 h1 = hf v1 = vf s1 = sf 1 69.72 0.01614 0.13262 P2 s2 wp2 (work) h2 250 0.13262 0.743850694 70.46385069 P3 h3 v3 s3 250 376.09 0.01865 0.56784 P4 s4 wp4 h4 500 0.56784 0.862982005 376.952982 P5 h5 v5 s5 500 449.51 0.01975 0.649 P6 s6 wp6 h6 2000 0.649 5.483290488 454.9932905 P7 T7 h7 s7 2000 1000 1474.9 1.5606 P8 s8 h8 500 1.5606 1300.333818 P9 T9 h9 s9 500 1000 1521 1.7376 P10 s10 h10 250 1.7376 1419.881944 P11 s11 sf sfg 1 1.7376 0.13262 1.84495 x hfg h11 0.869931434 1035.7 970.7079867 y z q (out) q (in) 0.078577565 0.20869054 642.1628748 1223.23348 27
- 28. efficiency at 0.87 Wnet (btu/lbm) Mass flow rate (lbm/hr) 0.475028369 581.0706053 880787.9719 47.50283693 Table 5: Step 3 Cycle 4 P1 h1 v1 s1 1 69.72 0.01614 0.13262 P2 s2 Wp1 h2 50 0.13262 0.146380257 69.86638026 P3 h3 v3 s3 50 250.21 0.01727 0.41125 P4 s4 Wp2 h4 100 0.41125 0.159825193 250.3698252 P5 v5 h5 s5 100 0.01774 298.51 0.47427 P6 s6 Wp3 h6 250 0.47427 0.492524422 299.0025244 P7 v7 h7 s7 250 0.01865 376.09 0.56784 P8 s8 Wp4 h8 500 0.56784 0.862982005 376.952982 P9 v9 h9 s9 500 0.01975 449.51 0.649 P10 s10 Wp5 h10 2000 0.649 5.483290488 454.9932905 P11 T11 v11 h11 2000 1000 0.39479 1474.9 s11 1.5606 P12 s12 h12 m12 500 1.5606 1300.333818 0.078577565 P13 T13 v13 h13 500 1000 1.70094 1521 s13 m13 1.7376 0.921422435 P14 s14 h14 m14 250 1.7376 1419.994432 0.063363642 28
- 29. P16 s16 h16 m16 100 1.7376 1308.273494 0.039046183 P17 s17 h17 m17 50 1.7376 1245.05 0.125685634 P18 s18 x18 h18 1 1.7376 0.869931434 970.7079867 efficiency (%) q(in) q(out) Wnet 0.489321306 1223.23348 624.6792759 598.5542041 48.93213061 Mass flow rate (lbm/hr) 855060.4047 Table 6: step 3 cycle 5 Discussions and Conclusion It is shown through the project that one can improve the efficiency of standard Rankine cycle via reheating (34.2 to 44.37 percent) and adding feed water heaters (34.2 to 46.8146, 47.5208, and 48.3213 with one, two and four water heaters respectively) to the turbine. The design proved that adding more open feed water heaters to the turbine cycle will improve the efficiency of the steam turbine and lower the turbine’s mass flow rate. The improved efficiency resulted in increased annual incomes and the diminished mass flow rate resulted in less steam required to obtain the same amount of work. However, it should be noted that efficiency gains starts to decrease as more feed water heaters are added until it tapers off at a certain point. This will eventually leaves potential gains not worth the cost effectiveness at a certain amount of feed water heaters being added. Based on the calculations in step 3, it is shown in the cost effectiveness section that the 5th cycle, the Ideal Rankine Cycle with 4 Feed water Heater, produced that most gains out of all the other cycles. However in terms actual cost effectiveness, the 4th cycle, the Ideal Rankine Cycle with 2 Feed water Heater, is the most effective because water heater cost extra money to maintain and the gains from adding four more water heaters basic Rankine cycle turbine is barely anymore than the gains from merely adding two water heaters to said Rankine cycle turbine (29.71 million dollars versus 28.02 million dollars respectively), meaning the cost needed to maintain essentially double the amount of water heaters, outweighs the minimal potential gains produced. 29
- 30. In conclusion, this project was very helpful in demonstrating how a steam turbine power plant works. The process in calculating the Rankine cycle’s efficiency proved that the Power Plant’s Performance can be improved via reheating and regeneration, showing that the theory behind optimizing the power plant’s performance has indeed been confirmed. Appendix 30
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- 37. References [1]: Thermodynamics, an Engineering Approach. Yunus A. Cengel & Michael A. Boles. Seventh edition. McGraw Hill Publications [2]: Lecture notes: ME: 43000 Thermo-Fluid Systems Analysis and Design, Professor Rishi Raj. [3]: www.google.com , Google images of the Rankine cycle [4]: Woodford, Chris (19th July 2014), Steam Turbine, Explain that Stuff!, retrieved from http://www.explainthatstuff.com/steam-turbines.html 37