Open Processes - TEST Tutorial

(The Open Process daemons build upon the state and closed-process daemons. Visit States pages first.)

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

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

(Open system going through a process. This page builds upon the States and Closed Process pages.)

Daemons>Open Process> Applications

EXAMPLE-1

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, (c) (b) the entropy generation if the surroundings temperature is same as the supply line temperature.

What-if scenario: How would the answers change if the supply line had a pressure of 2 MPa 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 followed by the Super-Calculate button.

# HOME>Daemons>Systems>Open>Process>Phase-Change;

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;   }
}

Step 1: Launch
the appropriate Open Process
Daemon for R-12

Step 2: Calculate the b-, f- and i- States

Step 3: Load the States on the Process panel, enter process variables, Calculate and Super-Calculate.

Step 4: Change a parameter, Calculate and Super-Calculate.

Solution

Answering the questions described in the Approach section leads you to the appropriate daemon page: HOME. Daemons. Systems. Open. Process.PhaseChange . and select R-12 as the working fluid.

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 an 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 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 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. 1.1: Image of the Process Panel. The system sketch adjusts to the selected anchor states.

EXAMPLE-2

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.

What-if scenario: How would the answers change if all the saturated liquid were drained out?

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 followed by the Super-Calculate button.

# HOME>Daemons>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;   }
}

Step 1: Launch the appropriate Open Process daemon for H2O

Step 2: Calculate the b, f and e States

Step 3: Calculate the Process and Super-Calculate
to obtain the desired answers

Step 4: For the what-if study, change a variable, Calculate and Super-Calculate

Solution

Answering the questions described in the Approach section leads you to the appropriate daemon page: HOME. Daemons. Systems. Open. Process. PhaseChange .

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. 2.1 Image of I/O panel after a Super-Calculate.

EXAMPLE-3

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. Determine the heat transfer necessary to vaporize all the water in the cooker.

What-if scenario: (a) How would the answer change if the operating pressure were raised to 60 psia 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 followed by the Super-Calculate button.

# HOME>Daemons>Systems>Open>Process>PhaseChange;

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;   }
}

Step 1: Launch the appropriate Open Process Daemon for H2O

Step 2: Choose English system

Step 3: Calculate b, f and e States

Step 4: Enter known process variables. A Calculate and Super-Calculate produce the desired answers

Step 5: Change a parameter, Calculate and Super-Calculate the new answer

Step 6: If S_gen is calculated as part of the solution, try to understand what the 2nd-Law is telling us about this process.

Solution

Answering the questions described in the Approach section leads you to the appropriate daemon page: HOME. Daemons. Systems. Open. Process. PhaseChange .

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.

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.

EXAMPLE-4

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.

What-if scenario: How would the answers change if the volume of the tank were twice as large?

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 followed by the Super-Calculate button.

The detailed output generated by this code is shown below. The final temperature is T2=150.5 deg-C. Note that if the code is pasted on the corresponding perfect gas daemon, the final temperature calculated is 141.3 deg-C. The difference between the ideal gas and perfect gas assumptions is insignificant if the temperature change in a problem is relatively small (under 100 deg-C).

# HOME>Daemons>Systems>Open>Process>IdealGas;

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;   }
}

#
#
# States
#
# State-1: Air > IdealGas;
#   Given:       Vel1= 0.0 m/s;   z1= 0.0 m;   m1= 0.0 kg;
#   Vol1= 1.0 m^3;
#   Calculated:   rho1= 0.0 kg/m^3;   v1= Infinity m^3/kg;   s1= 0.0 kJ/kg.K;
#   e1= 0.0 kJ/kg;   j1= 0.0 kJ/kg;   MM1= 28.97 kg/kmol;
#   R1= 0.28698653 kJ/kg.K;
#
# State-2: Air > IdealGas;
#   Given:       p2= 100.0 kPa;   Vel2= 0.0 m/s;   z2= 0.0 m;
#   Vol2= "Vol1" m^3;
#   Calculated:   T2= 149.72232 deg-C;   rho2= 0.8240038 kg/m^3;   v2= 1.2135867 m^3/kg;
#   u2= 5.1699 kJ/kg;   h2= 126.528564 kJ/kg;   s2= 7.221223 kJ/kg.K;
#   e2= 5.1699 kJ/kg;   j2= 126.528564 kJ/kg;   m2= 0.8240038 kg;
#   MM2= 28.97 kg/kmol;   R2= 0.28698653 kJ/kg.K;
#
# State-3: Air > IdealGas;
#   Given:       p3= 100.0 kPa;   T3= 30.0 deg-C;   Vel3= 0.0 m/s;
#   z3= 0.0 m;
#   Calculated:   rho3= 1.1494256 kg/m^3;   v3= 0.8699997 m^3/kg;   u3= -81.83012 kJ/kg;
#   h3= 5.169855 kJ/kg;   s3= 6.884009 kJ/kg.K;   e3= -81.83012 kJ/kg;
#   j3= 5.169855 kJ/kg;   m3= 0.8240037937606582 kg;   MM3= 28.97 kg/kmol;
#   R3= 0.28698653 kJ/kg.K;
#
# 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;
#   Calculated: S_gen= 0.27786583 kJ/K;

Fig. 4.1: Image of the detailed report produced on the I/O panel.

Your input!

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.


Manual