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9. Daemons>Vapor Power Cycles> Examples


 
                                               EXAMPLE-1
In a steam power plant operating on a reheat Rankine cycle, steam enters the HP turbine at 15 MPa and 620oC and is condensed in the condenser at a pressure of 15 kPa. If the moisture content in the turbine is not to exceed 10%, determine (a) the reheat pressure and (b) the thermal efficiency of the cycle.

What-if scenario: (c) How would the answers change if the moisture content in the turbine were not to exceed 15%? 




Time Saver: To reproduce the visual  solution, copy and paste  these TEST-codes on the I/O Panel  of the appropriate daemon, then click the Load and Super-Calculate buttons.

This is one of the cases, where you have to use the Super-Iterate button after Super-Calculate to continue iterations between the state and device panels.
 

 

# TEST>Daemons>Systems>Open>SteadyState>Specific>
#   PowerCycle>PC Model

   States { 
      State-1:  H2O;
      Given:       { p1= 0.015 MPa;   x1= 0.0 %;   Vel1= 0.0 m/s; 
           z1= 0.0 m;   mdot1= 1.0 kg/s;   }

      State-2:  H2O;
      Given:       { p2= 15.0 MPa;   s2= "s1" kJ/kg.K; 
            Vel2= 0.0 m/s;   z2= 0.0 m;   }

      State-3:  H2O;
      Given:       { p3= "p2" MPa;   T3= 620.0 deg-C; 
            Vel3= 0.0 m/s;   z3= 0.0 m;   }

      State-4:  H2O;
      Given:       { p4= "p5" MPa;   s4= "s3" kJ/kg.K; 
            Vel4= 0.0 m/s;   z4= 0.0 m;   }

      State-5:  H2O;
      Given:       { T5= "T3" deg-C;   s5= "s6" kJ/kg.K; 
            Vel5= 0.0 m/s;   z5= 0.0 m;   }

      State-6:  H2O;
      Given:       { p6= "p1" MPa;   x6= 90.0 %;   Vel6= 0.0 m/s; 
            z6= 0.0 m;   }
  }

 Analysis {
      Device-A:  i-State =  State-1;  e-State =  State-2; Mixing: true
      Given: { Qdot= 0.0 kW;   T_B= 25.0 deg-C;   }

      Device-B:  i-State =  State-2, State-4; 
            e-State =  State-3, State-5; Mixing: false
      Given: { Wdot_ext= 0.0 kW;   T_B= 25.0 deg-C;   }

      Device-C:  i-State =  State-3;  e-State =  State-4; Mixing: true
      Given: { Qdot= 0.0 kW;   T_B= 25.0 deg-C;   }

      Device-D:  i-State =  State-5;  e-State =  State-6; Mixing: true
      Given: { Qdot= 0.0 kW;   T_B= 25.0 deg-C;   }

      Device-E:  i-State =  State-6;  e-State =  State-1; Mixing: true
      Given: { Wdot_ext= 0.0 kW;   T_B= 25.0 deg-C;   }
 }
 



Step 1: Launch
the appropriate open cycle
daemon.
 
 
 
 

Step 2: Set up the cycle.
 
 
 

Step 3: Calculate the states.
 
 
 
 
 

 

Solution

Launch the open cycle daemon located at the page  TEST. Daemons. Systems. Open. Steady. Specific. PowerCycles.  PhaseChange.

Let us set up the cycle as follows: Device-A: isentropic pumping from State-1 to State-2 ; Device-B : constant pressure boiler with State-2 and State-4 as inlets and  State-3 and State-5 as exits; Device-C : high pressure isentropic turbine from State-3 to State-4 ; Device-D : low pressure isentropic turbine from State-4 to State-5 ; Device-E : constant pressure heat rejection from State-4 to State-1 .

State-1: Enter mdot1 (assume 1 kg/s),  p1 (15 kPa) and x1 (0%). Calculate

State-2: Enter p2  (15 MPa), s2 ('=s1'), and Calculate.

State-3: Enter p3 ('=p2'), T3 (620C) and Calculate.

State-4: Enter p4 ('=p5'), s4 ('=s3'), and Calculate. Note that p5 is not yet known.

State-5: Enter T5 ('=T3'), s5 ('=s6') and Calculate . Note that s6 is not yet known.

State-6: Choose 'More...' from the state selector to add more states to the menu. Choose State-6. Enter p6 ('=p1'), x6 (90%) and Calculate

Super-Calculate to propagates s6 back to State-5 and then p5 back to State-4, thus completing evaluation of all states. It is always a good practice to draw a T-s or some other thermodynamic plot to visualize the calculated states before proceeding to other panels.
 



 
Fig. 1 Image of State-4.  Only after State-6 is Calculated, is State-4 updated by Super-Calculate.

 

Step 4: Analyze the open steady devices.
On Device Panel work on the five devices.

Device-A: Select State-1 and State-2 as the i1- and e1-States , enter Qdot=0, and   Calculate.   The pumping power  is calculated as -15.17 kW. 

Device-B:  Select State-2 and State-4 as the i1- and e1-States ,  and  State-3 and State-5 as the e1- and e2-States , Calculate.   Click on the 'non-mixing' radio button. Enter Wdot_ext=0. The heat transfer is 3831 kW (note that we could have broken the boiler into two devices, steam generator and superheater) . 

Device-C: Select State-3 and State-4 as the i1- and e1-States , enter Qdot=0, and   Calculate.   The work is calculated as Wdot_ext=360 kW.

Device-D: Select State-5 and State-6 as the i1- and e1-States ,  enter Qdot=0 kW and   Calculate. The work is calculated as Wdot_ext=1349 kW.

On the Cycle panel, no further work is necessary. The thermal efficiency is calculated as eta_th=44.2%

Use  Super-Calculate to produce detailed output and the TEST-Code. Use  Super-Iterate   for further iteration if the solution is not complete. 

This is one of the rare instances where the Super-Iterate button has been used. One could avoid that by working in modules.  Working on State-1,3,4 and Device-d  (a Super-Calculate among them will completely evaluate State-1) first will reduce the need for iterations.



Step 5: For the what-if study, change a variable, Calculate and Super-Calculate.
For the parametric study, change x6 to 85%,  Calculate and Super-Calculate to obtain the new efficiency as 43.24% (the efficiency increases with moisture content for cycles without superheat).


Fig. 2 Image of Device Panel (Device-B, the boiler).  Notice that the Non-Mixing option must be turned on for the
non-mixing flow of a heat exchanger.


 
                                                    EXAMPLE-2





A combined gas turbine-steam power plant.has a net power output of 50 MW. Air enters the compressor of the gas turbine at 100 kPa, 300 K, and has a compression ratio of 12 and an isentropic efficiency of 85%. The turbine has an isentropic efficiency of 90% and has the inlet conditions of 1200 kPa and 1400 K, and an exit pressure of 100 kPa. The air from the turbine exhaust passes through a heat exchanger and exits at 400 K. On the steam turbine side, steam at 8 MPa,  400oC enters the turbine, which has an isentropic efficiency of 85%, and expands to the condenser pressure of 8 kPa. Saturated water enters the pump, which has an isentropic efficiency of 80% at 8 kPa. Determine (a) the ratio of mass flow rates in the two cycles, (b) the mass flow rate of air if the net power is 50 MW, (c) the thermal efficiency.

What-if scenario: (d) How would the thermal efficiency change if the compression ratio is increased to 15?

Solution See Ex-3 in Gas Power Cycle chapter.


 

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Copyright 1998-: Subrata Bhattacharjee