Gas Turbines, Steam Power and Refrigeration

(The Open Cycle daemons build upon the open steady daemons. Visit Open Steady pages first.

States	Closed Process	Closed Steady	Open Steady
Open Process	Closed Cycle		States-II
HVAC	Combustion	Equilibrium	Gas Dynamics

(Each section above is divided into two sub-sections - Manual and Applications.)
Manual

(Gas Power, Vapor Power and Refrigeration cycles )

Daemons>Open Cycle> Applications

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 the net power output and the thermal efficiency. Assume constant c_p.

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

Time Saver: To reproduce the visual solution, copy and paste
this TEST-code on the I/O panel
of the appropriate daemon, click the Load button, and then the Super-Calculate button and Super-Iterate button.

# HOME>Daemons>Systems>open>Process>
# Specific>PowerCycle>PerfGas

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 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 HOME. 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. This state is completely evaluated only after Super-Calculate.

Step 4: Analyze the four open and steady devices.

On the Analysis panel, work on the four devices.

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

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

Device-C: Load 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: Load 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.

Step 5: For the What-If study, change a variable, Calculate and Super-Calculate.

For the parametric study, go to the States 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. 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 model for air.

Time Saver: To reproduce the visual solution, copy and paste
this TEST-code on the I/O panel
of the appropriate daemon, click the Load button, and then the Super-Calculate button.

Step 1: Launch the Open Cycle
Daemon with IG model.

Step 2: Calculate the principal states as described in the code. Note how p2 is entered in terms of p1 and p5 so that only 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 uses the isentropic efficiency and h13 uses the regenerator effectivess. 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 necessarry to close the loop) by picking approriate inlet and exite states and enterning the known heat or work interaction. (Qdot =0 or Wdot_ext=0). For the regenerator (Device-D) select the 'Non-Mixing' button since the two streams do not mix. Also Qdot, left as an unknown, must be calculated as zero since h14 was enterered 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 maybe hotter than the turbine exhaust.

Step 4: The net power and thermal efficiency are calculated on the Cycle Panel as calculated simply 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 from the 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

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

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

In a steam power plant operating on a reheat Rankine cycle, steam enters the HP turbine at 15 MPa and 620^oC 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: 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
this TEST-code on the I/O panel
of the appropriate daemon, click the Load button, and then the Super-Calculate button.

This is one of the cases, where you have to use the Super-Iterate button after a Super-Calculate.

# HOME>Daemons>Systems>Open>SteadyState>Specific>
# PowerCycle>PhaseChange

   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 Cycle
Daemon.

Step 2: Set up the cycle.

Step 3: Calculate the States.

Solution

Answering the six questions described in the Approach section leads you to the appropriate daemon page for a Rankine cycle: HOME. Daemons. Systems. Open. Steady. Specific.PowerCycles. PhaseChange .

Let us set Device-A to ; : constant pressure boiler with State-2 State-4 State-3 State-5 Device-C turbine from to ; : low pressure isentropic turbine State-4 State-5 Device-E rejection from to . up the cycle as follows: color="#000099">: isentropic pumping from State-1 State-2 Device-B and as inlets and and as exits; : high pressure isentropic State-3 State-4 Device-D from to ; : constant pressure heat State-4 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.

A Super-Calculate at this point propagates s6 back to State-5 and then p5 back to State-4, thus completely evaluating all the states. It is always a good Applications to draw a T-s or some other thermodynamic plot to visualize the calculated states before proceeding to the Analysis window.

Fig. 3 Image of State-4. Only after State-6 is Calculated, a Super-Calculate propagates p5 back
to State-4 and State-4 is completely evaluated.

Step 4: Analyze the open steady devices.

On the Analysis panel, work on the five devices.

Device-a: Select Device-a . Load 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 Device-b . Load 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 Device-c . Load 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 Device-d . Load 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. 4: Image of the Device Panel (Device-B, the boiler). Notice that the Non-Mixing button
must be turned on for the mass balance equation to be simplified.

Fig. 2.3: Image of the I/O Panel. The top part contains the TEST-Code

EXAMPLE-4

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 efficienty 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, 400^oC 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: How would the thermal efficiency change if the compression ratio is increased to 15?

Time Saver: To reproduce the visual solution, copy and paste
this TEST-code on the I/O panel
of the appropriate daemon, click the Load button, and then the Super-Calculate button.

Step 1: Launch
the Open Cycle
Daemon with 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 porportionality 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 ke neglecgted, j=h. However, if the duct areas were available, ke could be included in the analysis without any further changes. 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 +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 sedonds all the calculations are updated. The new Wdot_net= 433.3 kW and Q_in=804.6. 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?

# HOME>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;   }
    }

EXAMPLE-5

Consider a two-stage R-12 refrigeration system operating between 0.15 MPa and 1 MPa. The refrigerant leaves the condenser as saturated liquid and is throttled to a flash chamber operating at 0.4 MPa. The vapor from the flash chamber is mixed with the refrigerant leaving the low-pressure compressor and the mixture is compressed by the high-pressure compressor to the condenser pressure. The liquid in the flash chamber is throttled to the evaporator pressure where the cooling load is handled through evaporation. Assume the refrigerant leaves the evaporator and mixer as saturated vapor.

Determine the COP.

What-if scenario: How would the COP change if the intermediate pressure were increased to 0.6 MPa?

Time Saver: To reproduce the visual solution, copy and paste
this TEST-code on the I/O panel
of the appropriate daemon, click the Load button, and then the Super-Calculate button.

Because of the complexity of the problem you may have to use the Super-Iterate button if Super-Calculate does not perform enough iteration.

# HOME>Daemons>Systems>Open>SteadyState>Specific>
# RefriCycle>PhaseChange

States {
State-1: R-12;
Given: { p1= 0.15 MPa; x1= 100.0 %; Vel1= 0.0 m/s; z1= 0.0 m; }

State-2: R-12;
Given: { p2= 0.4 MPa; s2= "s1" kJ/kg.K; Vel2= 0.0 m/s; z2= 0.0 m; }

State-3: R-12;
Given: { p3= "p2" MPa; x3= 100.0 %; Vel3= 0.0 m/s; z3= 0.0 m; mdot3= "x6*mdot9" kg/s; }

State-4: R-12;
Given: { p4= 1.0 MPa; s4= "s9" kJ/kg.K; Vel4= 0.0 m/s; z4= 0.0 m; }

State-5: R-12;
Given: { p5= "p4" MPa; x5= 0.0 %; Vel5= 0.0 m/s; z5= 0.0 m; }

State-6: R-12;
Given: { p6= "p2" MPa; h6= "h5" kJ/kg; Vel6= 0.0 m/s; z6= 0.0 m; }

State-7: R-12;
Given: { p7= "p6" MPa; x7= 0.0 %; Vel7= 0.0 m/s; z7= 0.0 m; mdot7= "(1-x6)*mdot9" kg/s; }

State-8: R-12;
Given: { p8= "p1" MPa; h8= "h7" kJ/kg; Vel8= 0.0 m/s; z8= 0.0 m; }

State-9: R-12;
Given: { p9= "p2" MPa; Vel9= 0.0 m/s; z9= 0.0 m; mdot9= 1.0 kg/s; }
}

Analysis {
Device-A: i-State = State-9; e-State = State-4; Mixing: true
Given: { Qdot= 0.0 kW; T_B= 25.0 deg-C; }

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

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

Device-D: i-State = State-6; e-State = State-3, State-7; Mixing: true
Given: { Qdot= 0.0 kW; Wdot_ext= 0.0 kW; T_B= 25.0 deg-C; }

Device-E: i-State = State-7; e-State = State-8; Mixing: true
Given: { Wdot_ext= 0.0 kW; T_B= 25.0 deg-C; }

Device-F: i-State = State-8; e-State = State-1; Mixing: true
Given: { Wdot_ext= 0.0 kW; T_B= 25.0 deg-C; }

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

Device-H: i-State = State-2, State-3; e-State = State-9; Mixing: true
Given: { Qdot= 0.0 kW; Wdot_ext= 0.0 kW; T_B= 25.0 deg-C; }
}

Step 4.1: Launch
the appropriate Cycle
Daemon.

Step 4.2: Set up the cycle.

Step 4.3: Calculate the States.

Solution

Answering the six questions described in the Approach section leads you to the appropriate daemon page for the reverse-Rankine refrigeration cycle: HOME. Daemons. Systems. Open. Steady. Specific.RefriCycles. PhaseChange .

Let us set up the cycle as follows : Device-A: isentropic compression from State-9 to State-4 ; Device-B : constant pressure heat rejection from State-4 to State-5 ; Device-C : isenthalpic expansion from State-5 to State-6 ; Device-D : constant pressure phase separation from inlet at State-6 to exits at State-3 ; Device-E : isenthalpic expansion from State-7 to State-8 ; Device-F : constant pressure heat addition from State-8 to State-1 ; Device-G : isentropic compression from State-1 State-2 ; Device-H State-3 into State-9 .

The mass flow rate being an unknown, we will take 1 kg/s through the top loop as the basis for this analysis, i.e., mdot9=1 kg/s. In that case mdot3=x6*mdot9 and mdot7=(1-x6)*mdot9.

In evaluating the states, we can follow two approaches. Utilize known information on devices or put them off until the Device Analysis part and let the daemon export the device info onto the affected States. For instance, in an isentropic device (say, Device-a) operating between State-1 and -2, we can enter s2 as '=s1'. Or alternatively, we could enter Qdot=0 and Sdot_gen=0 (i.e. adiabatic and reversible) in the analysis of Device-a. The Super-Calculate operation, in that case, would solve the entropy-balance equation for Device-a, conclude that s2=s1, substitute s1 for s2 in State-2, and recalculate State-2. For complex cycles such as this one, we will follow the first approach.

State-1-8: Enter the known values or relations as described in the TEST-Codes and Calculate the states fully or partially.

Fig. 5 Image of Device-D . Solution of the energy equation yields mdot3 which is posted back
on State-3. Super-Calculate calculates State-3 and 7 completely and later solves the entropy
equation to evaluate Sdot_gen.

Step 4: Analyze all six open, steady devices.

Device-A through H: For each device select a letter, load the anchor states (see the TEST-Codes above), enter the known device variable (Qdot or Wdot_ext) and Calculate.

Use Super-Calculate followed by a Super-Iterate to update all the States and Devices. The COP is calculated as COP=349%. The mass flow rate for the bottom cycle is calculated as mdot1=0.77 kg/s.

Super-Iterate is sometimes necessary to continue the iterations between the State and Analysis Panels just in case Super-Calculate, with a fixed number of iterations, is not sufficient.

Step 5: For the what-if study, change a variable, Calculate and Super-Calculate.

For the parametric study, go to the States Panel and change p2 to 0.6 MPa. Calculate the State, Super-Calculate and Super-Iterate to update all calculations. The new COP is calculated as 332%.

Fig. 4.2 Image of the Cycle Panel. Cycle variables are automaticaly calculated
once the cycle is complete.

EXAMPLE-6

In a gas refrigeration system air enters the compressor at 10^o C and 50 kPa and the turbine at 50^oC and 250 kPa. The mass flow rate is 0.08 kg/s. Assuming variable specific heat, determine (a) the rate of cooling, (b) the net power input and (c) the COP.

What-if scenario: How would the answers change if the working fluid were helium instead?

Time Saver: To reproduce the visual solution, copy and paste
this TEST-code on the I/O panel
of the appropriate daemon, click the Load button, and then the Super-Calculate button.

# HOME>Daemons>Systems>Closed>Process>
# Specific>RefrigCycle>IdealGas;

States {
               State-1: Air;
               Given:       { p1= 50.0 kPa;   T1= 10.0 deg-C;
                          Vel1= 0.0 m/s;   z1= 0.0 m;   mdot1= 0.08 kg/s;   }

               State-2: Air;
               Given:       { p2= 250.0 kPa;   s2= "s1" kJ/kg.K;
                          Vel2= 0.0 m/s;   z2= 0.0 m;   }

               State-3: Air;
               Given:       { p3= "p2" kPa;   T3= 50.0 deg-C;
                          Vel3= 0.0 m/s;   z3= 0.0 m;   }

               State-4: Air;
               Given:       { p4= "p1" kPa;   s4= "s3" kJ/kg.K;
                          Vel4= 0.0 m/s;   z4= 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; e-State = State-3; Mixing: true
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-4; e-State = State-1; Mixing: true
Given: { Wdot_ext= 0.0 kW; T_B= 25.0 deg-C; }
}

Step 1: Launch
the appropriate Cycle
Daemon.

Step 2: Set up the cycle.

Step 3: Calculate the States.

Solution

Answering the six questions described in the Approach section leads you to the appropriate daemon page for a Brayton cycle: HOME. Daemons. Systems. Open. Steady. Specific.RefriCycles. IdealGas .

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

State-1-4: Enter known values or relations and Calculate
each state fully or partially.

Fig. 4.1 Image of the Cycle Panel for a refrigeration/heat pump cycle.

Step 4: Analyze the four open and steady devices.

On the Analysis panel, work on the four devices.

Device-A through D: As described in the TEST-Codes, identify each device by a unique letter, load the anchor states, enter the known device variables (Qdot and Wdot_ext) and Calculate.

Use Super-Calculate to produce the TEST-code and the detailed output. On the Cycle panel, no further work is necessary. The COP is calculated as 1.72. Now change the gas to helium in the State panel and Super-Calculate .
The new value for COP is 1.11.

Your input!

You will find more examples on closed cycles on the Slide Show and the Home.TEST.Problems.Chapter08 pages. If you detect an error or any inconsistent instructions on this page, or would like to see more examples on a particular topic, please write to me using the Home.Comments page. Your input will be appreciated.


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