 Ch 10. Rankine Cycle Multimedia Engineering Thermodynamics RankineCycle Reheat Regeneration Cogeneration
 Chapter 1. Basics 2. Pure Substances 3. First Law 4. Energy Analysis 5. Second Law 6. Entropy 7. Exergy Analysis 8. Gas Power Cyc 9. Brayton Cycle 10. Rankine Cycle Appendix Basic Math Units Thermo Tables Search eBooks Dynamics Statics Mechanics Fluids Thermodynamics Math Author(s): Meirong Huang Kurt Gramoll ©Kurt Gramoll THERMODYNAMICS - CASE STUDY SOLUTION A new system has just started operating in Neon's power plant. An analysis needs to be done to make sure that the cycle works well. The thermal efficiency and the net work output need to be determined. Assumptions: All the components in the cycle operate at steady state. Kinetic and potential energy changes are negligible. The Actual Cycle T-s Diagram of the Actual Rankine Cycle Model the cycle in the new section of the power plant as a Rankine cycle. The schematic and the T-s Diagram of this cycle is shown on the left. (1) Determine the work output of the new system The net work output of the cycle equals the difference between the turbine work output and the pump work input.       wnet,out = wturb,out - wpump,in The enthalpy at each state needs to be determined first, which can be obtained from the water table. State 1: Compressed liquid water        P1 = 9 kPa      T1 = 38oC (given)        v1 = 0.001008 m3/kg State 3: Compressed liquid water       P3 = 15.9 MPa  T1 = 45oC (given)       h3 = 187.8 kJ/kg State 4: Superheated vapor       P4 = 15.2 MPa  T4 = 625oC (given)       h4 = 3641.8 kJ/kg       s4 = 6.7961 kJ/(kg-K) State 5: Superheated vapor       P5 = 15 MPa  T5 = 600oC (given)       h5 = 3582.3 kJ/kg       s5 = 6.6776 kJ/(kg-K) State 6: Saturated mixture       P6 = 10 kPa (given)       s6s = s5 = 6.6776 kJ/(kg-K)       h6s= 2114.9 kJ/kg where s6s and h6s are the entropy and enthalpy at state 6 if process 5-6 is isentropic. Since the isentropic efficiencies of the pump and turbine are given, the turbine work and the pump work can be determined as following:  Therefore, the net work output per mass is       wnet,out = wturb,out - wpump,in             = 1,173.9 - 19.0 = 1,154.9 kJ/kg The power produced by the new system is determined from (2) Determine the thermal efficiency Heat input to the cycle from the boiler is determined from       qin = h4 - h3 = 3,641.8 - 187.8 = 3,454 kJ/kg The thermal efficiency of the cycle is       ηth = wnet,out/qin = 1,154.9/3,454 = 33% The Ideal Rankine Cycle T-s Diagram of the Ideal Rankine Cycle (3) Determine the thermal efficiency of the ideal Rankine cycle The properties at each state of the ideal Rankine cycle can be obtained from the water table. State 1: Saturated water       P1 = 10 kPa (given)       h1 = 191.83 kJ/kg       s1 = 0.6493 kJ/(kg-K)       v1 = 0.001010 m3/kg State 2: Compressed liquid water       P2 = 16 MPa (given)       s2 = s1 = 0.6493 kJ/(kg-K) State 3: Superheated vapor       T3 = 600oC (given)       P3 = P2 = 16 MPa       h3 = 3,641.8 kJ/kg       s3 = 6.6988 kJ/(kg-K) State 4: Saturated mixture       P4 = P1 = 10 kPa         s4 = s3 = 6.6988 kJ/(kg-K)       h4 = 2,121.7 kJ/kg Pump work input:       wpump,in = v1(P2-P2)                    = 0.001010(16,000 - 10) = 16.2 kJ/kg Turbine work output:       wturb,out = h3 - h4                   = 3,641.8 - 2,121.7 = 1,520.1 kJ/kg Hence, the net work output of the ideal cycle is       wnet,out = wturb,out - wpump,in                   = 1,520.1 - 16.2 = 1,503.9 kJ/kg Heat input to the ideal cycle can be determined from       qin = h3 - h2 = 3,641.8 - 174.4 = 3,467.4 kJ/kg The thermal efficiency of the cycle is then determined as       ηth = wnet,out/qin = 1,503.9/3,467.4 = 43.4% Compared with the actual Rankine cycle, the ideal cycle has a higher thermal efficiency.