Charging/Discharging Analysis Software - First and Second Law Analysis of Charging and Evacuation Process

Ex A 0.2 m³ tank initially contains saturated vapor of R-12 at 1 MPa. The tank is charged to 1.2 MPa, x=0 % from a supply line that carries R-12 at 1.5 MPa, 30^oC. Determine (a) the final temperature, (b) the heat transfer, and (c) the entropy generation if the surroundings temperature is same as the supply line temperature. (d) What-if scenario: How would the answers change if the supply line had a pressure of 2 MPa instead?

Solution The system is open and goes through a process from a beginning-state to a final-state during the charging process. Launch the open-process TESTapp located in the following page: TEST. TESTapps. Systems. Open. Process.Phase-Change and select R-12 as the working fluid.

Step-by-step solution (show/hide)

#
# TEST-code: The code describes the solution procedure in a nut shell and can recreate the visual solution.
# TESTapp Path: TEST>TESTapps>Systems>Open>Process>PC Model ;

States {
         State-1: R-12;
         Given:       { p1= 1.0 MPa;   x1= 100.0 %;   Vel1= 0.0 m/s;
              z1= 0.0 m;   Vol1= 0.2 m^3;   }

         State-2: R-12;
         Given:       { p2= 1.2 MPa;   x2= 0.0 %;   Vel2= 0.0 m/s;
              z2= 0.0 m;   Vol2= "Vol1" m^3;   }

         State-3: R-12;
         Given:       { p3= 1.5 MPa;   T3= 30.0 deg-C;   Vel3= 0.0 m/s;
              z3= 0.0 m;   }
}

Analysis {
         Process-A:    ie-State = State-3, State-Null;
                                 bf-State = State-1, State-2;
         Given: { W_ext= 0.0 kJ;   T_B= 30.0 deg-C;   }
}

In this charging problem the uniform open system, which can be represented by a single state at any given instant, executes an open process as it goes from a clearly defined b-State to a unique f-State. During the process the inlet at i-State remains steady (notice the strategic location of the i-State in the figure). We first evaluate the three states ( State-1, 2 and 3) as best as possible from the given data, do a process analysis, and then update all calculations with the Super-Calculate button.

State-1: Enter p1, x1(=100%), Vol1, and Calculate. The mass is calculated as 11.46 kg.

State-2: Enter p2, x2, Vol2('=Vol1'), and Calculate. The temperature is calculated as T2=49.3^oC kg.

State-3: Enter p3, T3, and Calculate.

On the Analysis panel, load State-1 as the b-State, State-2 as the f-State and State-3 as the i-State. Enter the known process variable W_ext (=0) and T_B (= 30^o C). A Calculate and Super-Calculate produce Q=2804 kJ and S_gen=0.626 kJ/K .

For the what-if study, go to States panel; choose State-3, change p3 to 2 MPa, Calculate and Super-Calculate. All the answers are updated. The new values are Q=2714 kJ and S_gen=0.922 kJ/K .

Fig. 5b.1.1 Process panel. The system sketch adjusts to the selected anchor states.

Ex A 0.5 m3 tank initially contains saturated liquid water at 200^oC. A valve in the bottom of the tank is opened and half the liquid is drained. Heat is transferred from a source at 300^oC to maintain constant temperature inside the tank. Determine (a) the mass of the water discharged, (b) the heat transfer and (c) the entropy generation. (d) What-if scenario: How would the answers change if all the saturated liquid was drained out?

Solution The system is open and goes through a process from a beginning-state to a final-state during the discharge process. Launch the open-process TESTapp located in the following page: TEST. TESTapps. Systems. Open. Process. PhaseChange .

Step-by-step solution (show/hide)

#
# TEST-code: The code describes the solution procedure in a nut shell and can recreate the visual solution.
# TESTapp Path: TEST>TESTapps>Systems>Open>Process>PhaseChange;

States {
          State-1: H2O;
          Given:       { T1= 200.0 deg-C;   x1= 0.0 %;
                Vel1= 0.0 m/s;   z1= 0.0 m;   Vol1= 0.5 m^3;   }

          State-2: H2O;
          Given:       { T2= "T1" deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;
                m2= "m1/2" kg;   Vol2= "Vol1" m^3;   }

          State-3: H2O;
          Given:       { T3= "T2" deg-C;   x3= 0.0 %;   Vel3= 0.0 m/s;
                z3= 0.0 m;   }
}

Analysis {
          Process-A:    ie-State = State-Null, State-3;
                                  bf-State = State-1, State-2;
          Given: { W_ext= 0.0 kJ;   T_B= 300 deg-C;   }
}

In this discharge problem the uniform open system, which can be represented by a single state at any given instant, executes an open process as it goes from a clearly defined b-State to an unique f-State. During the process the discharge at e-State remains steady. We first evaluate the three states ( State-1, 2 and 3) as much as possible from the given data, do a process analysis, and then update all calculations with the help of the Super-Calculate button.

State-1: Enter T1, x1, Vol1, and Calculate. The mass is calculated as 432 kg.

State-2: Enter T2, m2 ('=m1/2'), Vol2 ('=Vol1'), and Calculate. The quality, although half the liquid has been drained, is calculated as 0.92 % (note that the volume fraction is about 50%).

State-3: Enter T3 ('=T2', the exit state is drawn before the valve), x3 (=0% as saturated liquid is being drained), and Calculate.

On the Analysis panel, load State-1 as the b-State, State-2 as the f-State and State-3 as the e-State. Enter the known process variable W_ext (=0) and T_B (= 300^o C). A Calculate Super-Calculate produce m_e=216 kg, Q=3843 kJ and S_gen=1.42 kJ/K .

For the what-if study, go to States panel and choose State-2. When all the liquid is drained out the water becomes saturated vapor, i.e., x2=100%. Uncheck m2 (make it an unknown) and enter this new value of x2. A Calculate and Super-Calculate produce m_e=428 kg, Q=7616 kJ and S_gen=2.807 kJ/K .

Fig. 5b.2.1 I/O panel after a Super-Calculate.

Preassure cooker with vapor and liquid water

Ex A 0.2 ft³ pressure cooker has an operating pressure of 40 psia. Initially 50% of the volume is filled with vapor and the rest with liquid water. (a) Determine the heat transfer necessary to vaporize all the water in the cooker. (b) What-if scenario: How would the answer change if the operating pressure was raised to 60 psia instead?

Step-by-step solution (show/hide)

#
# TEST-code: The code describes the solution procedure in a nut shell and can recreate the visual solution.
# TESTapp Path: TEST>TESTapps>Systems>Open>Process>PhaseChange Model;

States {
            State-1: H2O;
            Given:       { p1= 40.0 psia;   y1= 50.0 %;   Vel1= 0.0 ft/s;
                    z1= 0.0 ft;   Vol1= 0.2 ft^3;   }

State-2: H2O;
Given: { p2= "p1" psia; x2= 100.0 %; Vel2= 0.0 ft/s;
z2= 0.0 ft; Vol2= "Vol1" ft^3; }

State-3: H2O;
Given: { p3= "p1" psia; x3= 100.0 %; Vel3= 0.0 ft/s;
z3= 0.0 ft; }
}

Analysis {
            Process-A: ie-State = State-Null, State-3;
                                  bf-State = State-1, State-2;
             Given: { W_ext= 0.0 ft.lbf;   T_B= 77.0 deg-F;   }
}

In this discharge problem the uniform open system, which can be represented by a single state at any given instant, executes an open process as it goes from a clearly defined b-State to a unique f-State. During the process the discharge at e-State remains steady. We first evaluate the three states ( State-1, 2 and 3) as much as possible from the given data, do a process analysis, and then update all calculations with the help of the Super-Calculate button.

Because the problem is in English system, click on the English radio button.

State-1: Enter p1, y1 (volume fraction is 50%), Vol1, and Calculate. The mass is calculated as 5.841 lbm.

State-2: Enter p2 ('=p1' maintained by the valve), x2 (100%), Vol2 ('=Vol1'), and Calculate. The mass is calculated as 0.019 lbm.

State-3: Enter p3 ('=p1'), x3 (=100% as saturated vapor is being released), and Calculate.

On the Analysis panel, load State-1 as the b-State, State-2 as the f-State and State-3 as the e-State. Enter the known process variable W_ext (=0) and T_B (= 300^o C). A Calculate and Super-Calculate produce m_3=5.82 lbm .

For the what-if study, go to States panel, choose State-1 and change p1 to 60 psia. A Calculate and Super-Calculate produce m_e=m_3=5.74 lbm and Q=5270 Btu.

Note the negative value of S_gen which results because of the choice of the default value of T_B. If T_B is changed to a realistic value, say the flame temperature responsible for the heating, S_gen will become positive. A value of T_B=2000F produces (in the second part of the problem) a S_gen=1 Btu/R.

Ex Air at 100 kPa, 30 deg-C enters a completely evacuated insulated cylinder of volume 1 m³ until the pressure inside becomes 100 kPa. Determine (a) the final temperature, (b) the entropy generated during the filling process. (c) What-if scenario: How would the answers change if the volume of the tank was twice as large?

Step-by-step solution (show/hide)

#
# TEST-code: The code describes the solution procedure in a nut shell and can recreate the visual solution.
# TESTapp Path: TEST>TESTapps>Systems>Open>Process>IdealGas Model;

States {
            State-1: Air;
            Given:       { p1= 0.0 kPa; Vel1= 0.0 m/s;   z1= 0.0 m;
                    m1= 0.0 kg;   Vol1= 1.0 m^3;   }

            State-2: Air;
            Given:       {p2= 100.0 kPa;   Vel2= 0.0 m/s;   z2= 0.0 m;
                    Vol2= "Vol1" m^3;   }

            State-3: Air;
            Given:       { p3= 100.0 kPa;   T3= 30.0 deg-C;
                    Vel3= 0.0 m/s;   z3= 0.0 m;   }
}

Analysis {
            Process-A: ie-State = State-3, State-Null;
                                  bf-State = State-1, State-2;
            Given: { Q= 0.0 kJ;   W_ext= 0.0 kJ; T_B= 25.0 deg-C;   }
}

Fig. 5b.4.1 Process panel after Super-Calculate operation.

Ex As the valve is opened, oxygen at 20 MPa, 15 deg-C enters a 5 L insulated rigid tank, initially containing oxygen at 200 kPa and 20 deg-C. When the pressure inside reaches 20 MPa, the valve is closed. Determine (a) the final temperature, (b) the final mass, and (b) the exergy destroyed during the process if the outside conditions are 100 kPa and 25 deg-C. Treat oxygen as a perfect gas.

Note that while State-1 is fully known, only pressure and volume are known about the final state (State-2). Without either the final temperature (T2) or the mass that enters (m2-m1), it is not possible to calculate the final state.

One way to solve the problem is to take advantage of iteration. On the process panel, set Q as unknown and W_ext=0. Now guess T2 and use Super-Calculate to find Q. Use a few trial and error and we can establish that a T2 close to 403 K produces Q close to zero. The entropy generated is calculated as 0.2734 kJ/K. The exergy destroyed, therefore, must be 0.2734*298 = 81.47 kJ.

Step-by-step solution (show/hide)

#
# TEST-code: The code describes the solution procedure in a nut shell and can recreate the visual solution.
# TESTapp Path: TEST>TESTapps>Systems>Open>Process>PerfGas;

States {
State-1: O2;
Given: { p1= 200.0 kPa; T1= 20.0 deg-C; Vel1= 0.0 m/s; z1= 0.0 m; Vol1= 5.0 L; }

State-2: O2;
Given: { p2= 20000.0 kPa; T2= 404.0 K; Vel2= 0.0 m/s; z2= 0.0 m; Vol2= "Vol1" L; }

State-3: O2;
Given: { p3= 20000.0 kPa; T3= 15.0 deg-C; Vel3= 0.0 m/s; z3= 0.0 m; }
}

Analysis {
Process-A: ie-State = State-3, State-Null; bf-State = State-1, State-2;
Given: { W_ext= 0.0 kJ; T_B= 25.0 deg-C; }
}

The process panel showing calculated entropy generation

Fig. 5b.5.1 The process panel showing Q very close to zero and the calculated entropy generation.