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8. Daemons>Gas Power Cycles> Examples

   
                                           EXAMPLE-1

A gas turbine power plant operates on a simple Brayton cycle with air as the working fluid. The air enters the turbine at 1 MPa and 1000 K and leaves at 125 kPa and 610 K. Heat is rejected to the surroundings at a rate of 8000 kW and the air flow rate is 25 kg/s. Assuming a compressor efficiency of 80%, determine (a) the net power output and (b) the thermal efficiency. Assume constant c_p.

What-if scenario: (c) How would the answers change if the compressor efficiency were increased to 90%? 




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.
 

 

# TEST>Daemons>Systems>open>Process>
#          Specific>PowerCycle>PG Model

 States    { 
     State-1:  Air;
     Given:       { p1= 1.0 MPa;   T1= 1000.0 K;   Vel1= 0.0 m/s;   z1= 0.0 m;   mdot1= 25.0 kg/s;   }

     State-2:  Air;
     Given:       { p2= 125.0 kPa;   T2= 610.0 K;   Vel2= 0.0 m/s;   z2= 0.0 m;   mdot2= "mdot1" kg/s;   }

     State-3:  Air;
     Given:       { p3= "p2" kPa;   Vel3= 0.0 m/s;   z3= 0.0 m;   mdot3= "mdot1" kg/s;   }

     State-4:  Air;
     Given:       { p4= "p1" kPa;   s4= "s3" kJ/kg.K;   Vel4= 0.0 m/s;   z4= 0.0 m;   mdot4= "mdot1" kg/s;   }

     State-5:  Air;
     Given:       { p5= "p4" kPa;   Vel5= 0.0 m/s;   z5= 0.0 m;   j5= "j3+(j4-j3)/.8" kJ/kg;   mdot5= "mdot1" kg/s;   }
    }

  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;  e-State =  State-3;  Mixing: true;
     Given: { Qdot= -8000.0 kW;   Wdot_ext= 0.0 kW;   T_B= 25.0 deg-C;   }

     Device-C:  i-State =  State-3;  e-State =  State-5;  Mixing: true;
     Given: { Qdot= 0.0 kW;   T_B= 25.0 deg-C;   }

     Device-D:  i-State =  State-5;  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

Start the open cycle daemon with the perfect gas model at TEST. Daemons. Systems. Open. Steady. Specific.PowerCycles. PerfGas.

Let us set up the cycle as follows: Device-A: expansion from State-1 to State-2; Device-B: constant pressure heat rejection from State-2 to State-3 ; Device-C : compression from State-3 to State-5 with  State-4 being the isentropic state ; Device-D : constant pressure heat addition from State-5 to State-1 .

State-1-5: Evaluate the states as described on the TEST codes. Note how the turbine efficiency is used to evaluate j3.



 
Fig. 1  Image of State-5 with the calculated states plotted on a T-s diagram. 

 

Step 4: Analyze the four open and steady devices.


















Step 5: For the what-if study, change a variable, Calculate and Super-Calculate.
On Process Panel, work on the four devices.

Device-A:   Select State-1 and State-2 as the i1- and e1-States , enter Qdot=0, and   Calculate.   

Device-B:   Select State-2 and State-3 as the i1- and e1-States,  and   Calculate.   Enter  Qdot=-8000 kW, Wdot_ext=0 .  

Device-C:   Select State-3 and State-5 as the i1- and e1-States, enter Qdot=0, and   Calculate.   The work is calculated as Wdot_ext=9799 kW.

Device-D:   Select State-5 and State-1 as the i1- and e1-States,  enter  Wdot_ext=0 and   Calculate.

Use  Super-Calculate to update all the states and devices. The thermal efficiency is calculated as eta_th=22.9%. On the Cycle panel, no further work is necessary as cycle variables are automatically calculated.

For the parametric study, go to State Panel and change j3  to '=j1+(j2-j1)/.9 '. Calculate the State, Super-Calculate and Super-Iterate to update all calculations. The new answer is: eta_th=28.6%.


Fig.  1.2 Image of the Device Panel. Only a single inlet and exit are used in this device.

 
 EXAMPLE-2 Air enters steadily the first compressor of the gas turbine shown in Fig. 7.28 at 100 kPa and 300 K with a mass flow rate of 50 kg/s. The pressure ratio across the two-stage compressor and turbine is 15. The intercooler and reheater each operates at an intermediate pressure given by the square root of the product of the first compressor and turbine inlet pressures. The inlet temperature of each turbine is 1500 K and that of the second compressor is 300 K. The isentropic efficiency of each compressor and turbine is 80% and the regenerator effectiveness is also 80%. Determine (a) the thermal efficiency. (b) What-if-Scenario How would the thermal efficiency of the cycle change if the turbine and compressor efficiency increased to 90%? Use the ideal gas model for air.

What-if scenario: (c) How would the thermal efficiency of the cycle change if the turbine and compressor efficiency increased to 90%? Use the ideal gas (IG) model for air.


 






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.

Step 1: Launch the open cycle
daemon with IG model. 

Step 2: Calculate the principal states as described in the codes. Note how p2 is entered in terms of p1 and p5 so that only one property has to be changed during the parametric study. If change in ke cannot be neglected, j should be used instead of h in defining the isentropic efficiencies. Notice how h3, h6, h9, and h12 use the isentropic efficiency and h13 uses the regenerator effectiveness h14, on the other hand, is obtained from an energy balance on the regenerator.

Step 3:  Set up the devices A-I (Device-I is the heat exchanger necessary to close the loop) by picking appropriate inlet and exit states and interning the known heat or work interaction. (Qdot =0 or Wdot_ext=0). For the regenerator (Device-D) select the 'Non-Mixing' option (radio buttons) since the two streams do not mix. Also Qdot, left as an unknown, must be calculated as zero since h14 was entered based on the assumption that the regenerator is adiabatic. If in a parametric study, Qdot becomes non-zero that would mean that the regenerator cannot work under the chosen configuration, possibly because the compressor output may become hotter than the turbine exhaust.

Step 4: The net power and thermal efficiency are calculated on the Cycle Panel as
19.33 MW and 44.2%  respectively.

Step 5: For the parametric study change p5 to a new value, press the Enter key, Super-Calculate, and obtain the new answer in Cycle Panel. Repeat as many time as needed.

p5/p1      eta_th        Wdot_net
 

 5          0.426         13.78
10         0.444         17.64
15         0.442         19.33
20         0.436         20.30
25         0.430         20.80
30         0.423         21.23



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

 
States    { 
     State-1:  Air;
     Given:       { p1= 100.0 kPa;   T1= 300.0 K;   Vel1= 0.0 m/s;   z1= 0.0 m;   mdot1= 50.0 kg/s;   }

     State-2:  Air;
     Given:       { p2= "sqrt(p1*p5)" kPa;   s2= "s1" kJ/kg.K;   Vel2= 0.0 m/s;   z2= 0.0 m;   mdot2= "mdot1" kg/s;   }

     State-3:  Air;
     Given:       { p3= "p2" kPa;   h3= "h1+(h2-h1)/0.8" kJ/kg;   Vel3= 0.0 m/s;   z3= 0.0 m;   mdot3= "mdot1" kg/s;   }

     State-4:  Air;
     Given:       { p4= "p2" kPa;   T4= "T1" K;   Vel4= 0.0 m/s;   z4= 0.0 m;   mdot4= "mdot1" kg/s;   }

     State-5:  Air;
     Given:       { p5= "15*p1" kPa;   s5= "s4" kJ/kg.K;   Vel5= 0.0 m/s;   z5= 0.0 m;   mdot5= "mdot1" kg/s;   }

     State-6:  Air;
     Given:       { p6= "p5" kPa;   h6= "h4+(h5-h4)/0.8" kJ/kg;   Vel6= 0.0 m/s;   z6= 0.0 m;   mdot6= "mdot1" kg/s;   }

     State-7:  Air;
     Given:       { p7= "p6" kPa;   T7= 1400.0 K;   Vel7= 0.0 m/s;   z7= 0.0 m;   mdot7= "mdot1" kg/s;   }

     State-8:  Air;
     Given:       { p8= "p2" kPa;   s8= "s7" kJ/kg.K;   Vel8= 0.0 m/s;   z8= 0.0 m;   mdot8= "mdot1" kg/s;   }

     State-9:  Air;
     Given:       { p9= "p8" kPa;   h9= "h7-(h7-h8)*0.8" kJ/kg;   Vel9= 0.0 m/s;   z9= 0.0 m;   }

     State-10:  Air;
     Given:       { p10= "p9" kPa;   T10= "T7" K;   Vel10= 0.0 m/s;   z10= 0.0 m;   mdot10= "mdot1" kg/s;   }

     State-11:  Air;
     Given:       { p11= "p1" kPa;   s11= "s10" kJ/kg.K;   Vel11= 0.0 m/s;   z11= 0.0 m;   }

     State-12:  Air;
     Given:       { p12= "p11" kPa;   h12= "h10-(h10-h11)*0.8" kJ/kg;   Vel12= 0.0 m/s;   z12= 0.0 m;   mdot12= "mdot1" kg/s;   }

     State-13:  Air;
     Given:       { p13= "p6" kPa;   h13= "h6+(h12-h6)*.8" kJ/kg;   Vel13= 0.0 m/s;   z13= 0.0 m;   mdot13= "mdot1" kg/s;   }

     State-14:  Air;
     Given:       { p14= "p1" kPa;   h14= "h12-(h13-h6)" kJ/kg;   Vel14= 0.0 m/s;   z14= 0.0 m;   mdot14= "mdot1" kg/s;   }
    }

  Analysis    {
     Device-A:  i-State =  State-1;  e-State =  State-3;  Mixing: true;
     Given: { Qdot= 0.0 kW;   T_B= 25.0 deg-C;   }

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

     Device-C:  i-State =  State-4;  e-State =  State-6;  Mixing: true;
     Given: { Qdot= 0.0 kW;   T_B= 25.0 deg-C;   }

     Device-D:  i-State =  State-6, State-12;  e-State =  State-13, State-14;  Mixing: false;
     Given: { Wdot_ext= 0.0 kW;   T_B= 25.0 deg-C;   }

     Device-E:  i-State =  State-13;  e-State =  State-7;  Mixing: true;
     Given: { Wdot_ext= 0.0 kW;   T_B= 25.0 deg-C;   }

     Device-F:  i-State =  State-7;  e-State =  State-9;  Mixing: true;
     Given: { Qdot= 0.0 kW;   T_B= 25.0 deg-C;   }

     Device-G:  i-State =  State-9;  e-State =  State-10;  Mixing: true;
     Given: { Wdot_ext= 0.0 kW;   T_B= 25.0 deg-C;   }

     Device-H:  i-State =  State-10;  e-State =  State-12;  Mixing: true;
     Given: { Qdot= 0.0 kW;   T_B= 25.0 deg-C;   }

     Device-I:  i-State =  State-14;  e-State =  State-1;  Mixing: true;
     Given: { Wdot_ext= 0.0 kW;   T_B= 25.0 deg-C;   }
    }



 

 
                                                    EXAMPLE-3





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 efficiently 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?




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.

Step 1: Launch the open cycle
daemon with the PC/IG model. 

Step 2: Let us base the analysis on an air flow rate of 1 kg/s. Once the net power is calculated, the actual flow rate can be found from the proportionality between power and mass flow rate.

Step 3: Calculate the gas turbine states as described on the TEST-Codes. Notice how j's are used in the definition of isentropic efficiencies, although, with negligible ke and pe, h could be used instead of j. The advantage of using j is that if the duct areas were available, ke would be automatically included in the analysis. Note that the variable Model is automatically set to 2 as air is chosen as the working fluid.

Step 4:  We start the steam cycle calculation from State-8 leaving mdot8 as an unknown, which is to be determined from a device analysis of the heat exchanger.  

Step 5: All the states in the steam cycle are calculated except the mass flow rate is an unknown in each state.

Step 6:  Set up Device-A for the heat exchanger.  Enter Qdot=Wdot_ext=0 and make sure to select the non-mixing button. A Calculate produces mdot8 which is posted back to State-8. A Super-Calculate updates all the states.

Step 7: Set up one device each for every work producing/consuming device as described on the TEST-Codes.

Step 8: The net power can be calculated simply as
 -364.6 +672  +146.3 -1.5 = 452.8 kW.

Step 8: The heat added can be found from Device-C as 851.3 kW. The thermal efficiency, therefore, is  452.8/851.3 =
53.2%.

Step 9: To produce 50 MW of power, the air mass flow rate must be 10000/452.8 =
22.1 kg/s.

Step 10: In State-2 change p2 to 15*p1, press the Enter key and Super-Calculate. In a few
seconds all the calculations are updated. The new Wdot_net= 433.3 kW and Qdot_in=804.6 kW. The new thermal efficiency is 433.3/804.6= 53.9%. The net power is lower, but the efficiency has slightly improved.

Step 3.11: Can you figure out why Wdot_net is not equal to Qdot_net in this problem?

# TEST>Daemons>Systems>Open>SteadyState>Specific>
#   PowerCycle>PC/IG

 
States    { 
     State-1:  H2O, Air;
     Given:       { p1= 100.0 kPa;   T1= 300.0 K;   Vel1= 0.0 m/s;   z1= 0.0 m;   mdot1= 1.0 kg/s;   Model1= 2.0 UnitLess;   }

     State-2:  H2O, Air;
     Given:       { p2= "12*p1" kPa;   s2= "s1" kJ/kg.K;   Vel2= 0.0 m/s;   z2= 0.0 m;   mdot2= "mdot1" kg/s;   Model2= 2.0 UnitLess;   }

     State-3:  H2O, Air;
     Given:       { p3= "p2" kPa;   Vel3= 0.0 m/s;   z3= 0.0 m;   j3= "j1+(j2-j1)/.85" kJ/kg;   mdot3= "mdot1" kg/s;   Model3= 2.0 UnitLess;   }

     State-4:  H2O, Air;
     Given:       { p4= "p3" kPa;   T4= 1400.0 K;   Vel4= 0.0 m/s;   z4= 0.0 m;   mdot4= "mdot1" kg/s;   Model4= 2.0 UnitLess;   }

     State-5:  H2O, Air;
     Given:       { p5= 100.0 kPa;   s5= "s4" kJ/kg.K;   Vel5= 0.0 m/s;   z5= 0.0 m;   mdot5= "mdot1" kg/s;   Model5= 2.0 UnitLess;   }

     State-6:  H2O, Air;
     Given:       { p6= "p5" kPa;   Vel6= 0.0 m/s;   z6= 0.0 m;   j6= "j4-(j4-j5)*.90" kJ/kg;   mdot6= "mdot1" kg/s;   Model6= 2.0 UnitLess;   }

     State-7:  H2O, Air;
     Given:       { p7= "p6" kPa;   T7= 400.0 K;   Vel7= 0.0 m/s;   z7= 0.0 m;   mdot7= "mdot1" kg/s;   Model7= 2.0 UnitLess;   }

     State-8:  H2O, Air;
     Given:       { p8= 8000.0 kPa;   T8= 400.0 deg-C;   Vel8= 0.0 m/s;   z8= 0.0 m;   Model8= 1.0 UnitLess;   }

     State-9:  H2O, Air;
     Given:       { p9= 8.0 kPa;   s9= "s8" kJ/kg.K;   Vel9= 0.0 m/s;   z9= 0.0 m;   mdot9= "mdot8" kg/s;   Model9= 1.0 UnitLess;   }

     State-10:  H2O, Air;
     Given:       { p10= "p9" kPa;   Vel10= 0.0 m/s;   z10= 0.0 m;   j10= "j8-(j8-j9)*.85" kJ/kg;   mdot10= "mdot8" kg/s;   Model10= 1.0 UnitLess;   }

     State-11:  H2O, Air;
     Given:       { p11= "p10" kPa;   x11= 0.0 %;   Vel11= 0.0 m/s;   z11= 0.0 m;   mdot11= "mdot8" kg/s;   Model11= 1.0 UnitLess;   }

     State-12:  H2O, Air;
     Given:       { p12= "p8" kPa;   s12= "s11" kJ/kg.K;   Vel12= 0.0 m/s;   z12= 0.0 m;   mdot12= "mdot7" kg/s;   Model12= 1.0 UnitLess;   }

     State-13:  H2O, Air;
     Given:       { p13= "p12" kPa;   Vel13= 0.0 m/s;   z13= 0.0 m;   j13= "j11+(j12-j11)/.8" kJ/kg;   mdot13= "mdot8" kg/s;   Model13= 1.0 UnitLess;   }
    }

 Analysis    {

     Device-A:  i-State =  State-6, State-13;  e-State =  State-7, State-8;  Mixing: false;
     Given: { Qdot= 0.0 kW;   Wdot_ext= 0.0 kW;   T_B= 25.0 deg-C;   }

     Device-B:  i-State =  State-1;  e-State =  State-3;  Mixing: true;
     Given: { Qdot= 0.0 kW;   T_B= 25.0 deg-C;   }

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

     Device-D:  i-State =  State-4;  e-State =  State-6;  Mixing: true;
     Given: { Qdot= 0.0 kW;   T_B= 25.0 deg-C;   }

     Device-E:  i-State =  State-8;  e-State =  State-10;  Mixing: true;
     Given: { Qdot= 0.0 kW;   T_B= 25.0 deg-C;   }

     Device-F:  i-State =  State-11;  e-State =  State-13;  Mixing: true;
     Given: { Qdot= 0.0 kW;   T_B= 25.0 deg-C;   }

     Device-G:  i-State =  State-11;  e-State =  State-13;  Mixing: true;
     Given: { Qdot= 0.0 kW;   T_B= 25.0 deg-C;   }
   }


 

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