Process Analysis Software - First and Second Law Analysis of Thermodynamics Processes

Tutorial (TOC) > Daemons > Closed Processes > Examples

Ex-1 An insulated piston cylinder device contains 1 kg of N₂ at 100 kPa and 20^oC. The gas is now electrically heated to four times the original volume at constant pressure. Modeling nitrogen as an ideal gas, determine (a) the final temperature, (b) electrical work, (c) entropy generation, (d) reversible work, and (e) the irreversibility during the process. Assume the ambient temperature and pressure to be 25^oC and 100 kPa. (f) What-if scenario: How would the answers change if the initial temperature was 0^oC?

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# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Closed> Process>Generic>Uniform>IG-Model;

States    {
    State-0: N2;
    Given:       { p0= 100.0 kPa;   T0= 25.0 deg-C;   Vel0= 0.0 m/s;   z0=0.0 m;   }

    State-1: ;
    Given:       { p1= 100.0 kPa;   T1= 20.0 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   m1= 1.0 kg;   }

    State-2: ;
    Given:       { p2= "p1" kPa;   Vel2= 0.0 m/s;   z2= 0.0 m;   Vol2= "4*Vol1" m^3;   }
    }

Analysis    {
    Process-A: b-State = State-1; f-State = State-2;
    Given: { Q= 0.0 kJ;   T_B= 25.0 deg-C;   }
    }

Solution The unsteady closed system goes through a process from a begin-state to a finish-state. Launch the appropriate closed process daemon by navigating to TEST. Daemons. Systems. Closed. Process. Generic. Uniform. IdealGas page.

In this problem the uniform closed system, which can be represented by a single state at any given instant, executes a closed process as it goes from a clearly defined b-State to a unique f-State. Designating these anchor states as State-1 and State-2, we evaluate them from the given data as best as possible before doing a process and exergy analysis.

State-1: Enter m1, T1, p1, and Calculate.

State-2: Enter p2 ('=p1'), Vol2('=Vol1*4'), and Calculate.

On Process Panel panel, load State-1 as the b-State and State-2 as the f-State. Enter the only known process variable Q (=0) and T_B as 25^o C. Calculate and Super-Calculate to produce W_O (electrical work is the only 'other' work) as -977.5 kJ and S_gen=1.5135 kJ/K as well as all other process variables. Go to State Panel to find the final temperature T2 evaluated as 899^oC.

For the exergy (availability) analysis, go back to State Panel and evaluate the dead state, State-0, from the known atmospheric pressure (100 kPa) and temperature (25^o C) . Go to Exergy Panel. All exergy variables are automatically calculated for the selected process in Process Panel. The reversible work W_rev is found as -526 kJ and irreversibility I as 451 kJ .

For the what-if study, go to State Panel, choose State-1, change T1, press the Enter key, and Super-Calculate . All the answers are updated. Another way to do the same is to modify TEST-codes on I/O Panel, Load and Super-Calculate.

Fig. 1.1 Image of Process Panel in Ex. 1

Ex-2 A rigid container of volume 1 m³ contains ammonia at 100 kPa and 90% quality. If 200 kJ of heat is transferred to the tank, determine (a) the final pressure. (b) What-if scenario: How would the answer change if the amount of heat transfer was 400 kJ?

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# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Closed> Process>Generic>Uniform>PC-Model;

States {
            State-1: Ammonia;
            Given:       { p1= 100.0 kPa;   x1= 90.0 %;
                    Vel1= 0.0 m/s;   z1= 0.0 m;   Vol1= 1.0 m^3;   }

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

           Analysis {
            Process-A: b-State = State-1; f-State = State-2;
            Given: { Q= 200.0 kJ;   W_O= 0.0 kJ;   T_B= 25.0 deg-C;   }
}

Solution The unsteady closed system goes through a process from a begin-state to a finish-state. Launch the appropriate closed-process daemon by navigating to TEST. Daemons. Systems. Closed. Process. Generic. Uniform. Phase-Change page.

In this problem the uniform closed system, which can be represented by a single state at any given instant, executes a closed process going from a clearly defined b-State (State-1) to an unique f-Staten (State-2).

State-1: Enter Vol1, p1, x1 and Calculate.

State-2: Enter Vol2 ('=Vol1'), and Calculate.

On Process Panel, load State-1 as the b-State and State-2 as the f-State. Enter the known process variables W_O(=0) and Q (=200 kJ). Calculate and then Super-Calculate to produce the final pressure p2 as 135 kPa.

To evaluate the effect of heat transfer on the final pressure in the tank, change Q to the new value, Calculate and Super-Calculate to produce the final pressure p2 as 193 kPa .

Fig. 2.1: Image of State-2 with a superposed T-v diagram. Notice the gray background color of variables e2 and m2,
indicating postings by Process Panel.

Ex-3 A mixture of 50% N2 and 50% H2 (by volume) undergoes a polytropic process with n=1.2 in a piston-cylinder device from p1=1atm, T1=30^oC to p2=5 atm. Employing the ideal gas model, determine (a) the work and (b) heat transfer per unit mass. (c) What-if scenario: How would the answer change if the gas was pure N2?

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# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Closed> Process>Generic>Uniform>IG/IG

States    {
    State-1: N2, H2;
    Given:       { p1= 1.0 atm;   T1= 30 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   m1= 1.0 kg;   y_A1= 0.5 fraction;   }

    State-2: N2, H2;
    Given:       { p2= 5.0 atm;   Vel2= 0.0 m/s;   z2= 0.0 m;   m2= "m1" kg;   Vol2= "Vol1*(p1/p2)^(1/1.2)" m^3;   y_A2= "y_A1" fraction;   }
    }

Analysis    {
    Process-A: b-State = State-1; f-State = State-2;
    Given: { W_O= 0.0 kJ;   T_B= 25.0 deg-C;   }
    }

We have a uniform mixture, identifiable by a single State at any instant, undergoing a closed process. Let us represent the b-State and the f-State with State-1 and State-2 respectively.

State-1: Choose N2 and H2 as Gas-A and Gas-B. Enter y_A1=0.5 (volume fraction is same as mole fraction), m1 (=1 kg), p1 and T1. Calculate.

State-2: Calculate (with a calculator) Vol2 from the polytropic equation p1*Vol1^1.2=p2*Vol2^1.2 as 0.439 m³. Enter Vol2, or type in the expression '=Vol1*(p1/p2)^(1/1.2)' . Also enter m2 ('=m1'), p2 and y_A2 ('=y_A1'). Calculate.

On Process Panel, load State-1 and State-2 as the b- and f-States. Notice how W_B is automatically calculated to produce Q = -129 kJ. Also notice that the polytropic coefficient is calculated on Process Panel.

To obtain a solution for a different mixture, change y_A1 to the new value, 100% in this case, Calculate and Super-Calculate to produce Q =-69 kJ and and W_B= -138 kJ. Notice how severely the energy transfers change with the mixture composition.

Ex-4 An insulated rigid tank has two compartments, one 10 times larger than the other, divided by a partition. At the beginning the smaller side contains 4 kg of H2O at 200 kPa and 90^oC, and the other side is evacuated. Determine (a) the final temperature and (b) the exergy destroyed (irreversibility) if the partition is broken. Assume the surrounding temperature to be 25^oC. (c) What-if scenario: If the larger chamber was 100 times larger, how would that affect the result?

Solution The unsteady closed system goes through a process from a composite begin-state (vacuum+water) to a uniform finish-state. Launch the mixing, closed-process daemon by navigating to TEST. Daemons. Systems. Closed. Process. Generic. NonUniformMixing. PhaseChange page. H2O is the default fluid.

In this problem the non-uniform closed system, which can be represented by two states ( State-1 and State-2) at the beginning and a single state (State-3 ) at the end of the process. We calculate the three states from the given data, do a process analysis, and Super-Calculate the final temperature.

#
# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Closed>Process>Generic> NonUniformMixing>PC-Model;

States {
             State-1: H2O;
              Given:       { p1= 200.0 kPa;   T1= 90.0 deg-C;
                        Vel1= 0.0 m/s;   z1= 0.0 m;   m1= 4.0 kg;   }

              State-2: H2O;
              Given:       { p2= 0.0 kPa;   Vel2= 0.0 m/s;
                     z2= 0.0 m;   m2= 0.0 kg;   Vol2= "10*Vol1" m^3;   }

              State-3: H2O;
              Given:       { Vel3= 0.0 m/s;   z3= 0.0 m;
                        Vol3= "Vol1+Vol2" m^3;   }
}

Analysis {
Process-A: b-State = State-1, State-2; f-State = State-3;
Given: { Q= 0.0 kJ; W= 0.0 kJ; T_B= 25.0 deg-C; }
}

State-1: Enter m1=4 kg, p1=200 kPa, T1=90^oC and Calculate.

State-2: To define vacuum in TEST, any two of the properties mass, pressure and temperature must be set to zero. Enter m2 (=0), p1=0 and Vol2 ('=10*Vol1'). Calculate.

State-3: Enter Vol3 as '=Vol1+Vol2', and Calculate. At this point the state cannot be fully evaluated.

On Process Panel, load State-1 as the bA-State , State-2 as the bB-State and State-3 as the f-State. Enter the known process variables W (=0) and Q (=0). Calculate to produce T3=87.94^oC , p3=64.77 kPa , and S_gen=0.0072 kJ/K (Look for the exact value on the Message Panel as you move the pointer over S_gen widget) . Therefore, I=298*0.0072 =2.15 kJ.

In State Panel, change Vol2 to '=100*Vol1'. Calculate and Super-Calculate to yield the entropy generation as 0.05897 kJ/K. Therefore, the new value for I=298*0.05897 =17.57 kJ.

Ex-5 A 10 m³ insulated rigid tank containing hydrogen at 20^oC and 200 kPa is connected to another 0.2 m³insulated rigid tank containing carbon-dioxide at 100^oC and 600 kPa. The valve is opened and the combined system is allowed to reach thermal equilibrium. Determine (a) the final pressure, (b) heat transfer and (c) the entropy generation. Assume constant specific heats. (d) Also determine the heat transfer necessary to bring the mixture back to the ambient temperature of 20^oC. (e) What-if scenario: How would the answers change if both the gases was CO₂?

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# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Closed>Process>Generic>NonUniformMixing>PG/PG Model;

States    {
    State-1: H2, CO2;
    Given:       { p1= 200.0 kPa;   T1= 19.999994 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   Vol1= 10.0 m^3;   x_A1= 1.0 fraction;   }

    State-2: H2, CO2;
    Given:       { p2= 600.0 kPa;   T2= 100 deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;   Vol2= 0.2 m^3;   x_A2= 0.0 fraction;   }

    State-3: H2, CO2;
    Given:       { Vel3= 0.0 m/s;   z3= 0.0 m;   Vol3= "Vol1+Vol2" m^3;   x_A3= "m1/(m1+m2)" fraction;   }

    State-4: H2, CO2;
    Given:       { T4= 20 deg-C;   Vel4= 0.0 m/s;   z4= 0.0 m;   Vol4= "Vol3" m^3;   x_A4= "X_A3" fraction;   }
    }

Analysis    {
    Process-A: b-State = State-1, State-2; f-State = State-3;
    Given: { Q= 0.0 kJ;   W= 0.0 kJ;   T_B= 25.0 deg-C;   }

    Process-B: b-State = State-3; f-State = State-4;
    Given: { W= 0.0 kJ;   T_B= 25.0 deg-C;   }
    }

Solution The unsteady closed system goes through a mixing process from a composite begin-state to a uniform finish-state. Launch the non-uniform, mixing, closed-process daemon located at TEST. Daemons. Systems. Process. Generic. NonUniformMixing. PG/PG Model .

The PG/PG model is selected because the specific heats of the two gases are constant. In this problem the non-uniform closed system, which can be represented by two states ( State-1 and State-2 ) at the beginning and a single state ( State-3 ) at the end of the process. We calculate the three states from the given data, do a process analysis, and Super-Calculate the answers.

State-1: Choose H2 and CO2 as Gas-A and Gas-B, enter x_A1=1, Vol1=10 m3, p1=200 kPa, T1=20 deg^oC and Calculate. The mass is calculated as 1.64 kg.

State-2: Enter x_A2=0, Vol2=0.2 m3, p2=600 kPa, T2=100 deg^o C and Calculate. The mass is calculated as 1.7 kg.

State-3: Enter x_A3 as '=m1/(m1+m2)' and Vol3 as '=Vol1+Vol2', and Calculate. At this point the State cannot be fully evaluated. Notice that the mixture is heavily biased towards hydrogen (look at molar mass, for instance).

On Process Panel, load State-1 as the bA-State and State-2 as the bB-State and State-3 as the f-State. Enter the known process variable W (=0). Calculate and Super-Calculate to produce T3=25^oC, p3=209 kPa , and S_gen=1.423 kJ/K . Entropy is generated in a mixing process.

State-4: Enter x_A4 as '=x_A3', Vol4 as '=Vol3', T4= 20^oC and Calculate.

On Process Panel, choose Process-B . Load State-3 as the bA-State , State-4 as the bA-State and State-3 as the f-State. Enter the known process variable W (=0) and Q (=0). Calculate and Super-Calculate to produce Q=-89.5 kJ. Notice that S_gen is not determined for Process-B . It is because, this is a rare instance where a Super-Calculate does not finish up the job. More iterations must be performed through Super-Iterate to complete the solution.

Suppose Process-A did not take place at all, and the tanks allowed to cool without ever having the valve opened. Would Q be the same in that situation. In this new process, Process-C , State-2 will cool down to State-5 , a new State with Vol5=Vol2, x_A5=x_A2, m5=m2. The first tank, being at the same temperature as the ambient, will not participate in any heat transfer with the surroundings. A process between State-2 and State-5 with W=0 produces Q=-89.5 kJ , identical to the answer in Process-B.

Fig. 5.1 Image of Process Panel in Ex. 5. Complete mixing between sub-systems A and B results in a single f-state in Ex. 4
and 5.

Ex-6 Two rigid tanks, each holding steam, are connected by a valve. Tank A is insulated and contains 0.2 m³ of steam at 500 kPa and a quality of 0.8. Tank B is not insulated and contains 3 kg of steam at 200 kPa and 250^oC. The valve is now opened and steam flows from tank A to B until the pressure in tank A drops to 300 kPa. During this process 500 kJ of heat is rejected from tank B to the surroundings at 10^oC. Assuming steam in tank to have undergone a reversible adiabatic process (i.e., isentropic process), determine (a) the final temperature in each tank, and (b) the entropy generated during the process.

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# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Closed>Process>Generic>NonUniformSemiMixed>PC-Model;

   States    {
    State-1: H2O;
    Given:       { p1= 500.0 kPa;   x1= 0.8 fraction;   Vel1= 0.0 m/s;   z1= 0.0 m;   Vol1= 0.2 m^3;   }

    State-2: H2O;
    Given:       { p2= 200.0 kPa;   T2= 250.0 deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;   m2= 3.0 kg;   }

    State-3: H2O;
    Given:       { p3= 300.0 kPa;   s3= "s1" kJ/kg.K;   Vel3= 0.0 m/s;   z3= 0.0 m;   Vol3= "Vol1" m^3;   }

    State-4: H2O;
    Given:       { Vel4= 0.0 m/s;   z4= 0.0 m;   m4= "m1+m2-m3" kg;   Vol4= "Vol2" m^3;   }
    }

Analysis    {
    Process-A: b-State = State-1, State-2; f-State = State-3, State-4;
    Given: { Q= -500.0 kJ;   W= 0.0 kJ;   T_B= 10.0 deg-C;   }
    }

Solution The unsteady closed system goes through a process from a composite begin-state to a composite finish-state with partial mixing between the sub-systems. Since mixing is not complete, launch the semi-mixing, non-uniform, closed-process daemon by navigating to the TEST. Daemons. Systems. Process. Generic. Semi-Mixing. Phase-Change Model page.

Two states (State-1 and State-2) are necessary to describe the bA and bB states, and two states (State-3 and State-4) for the fA and fB states. Evaluate these four states as described in TEST-codes above. In Process Panel load the states, enter the process variables (Q=-500 kJ, W=0 and T_B=10 deg-C and Super-Calculate to completely evaluate the four states. The desired answers are calculated as
T3=133.5^o C, T4=163.8^o C , and S_gen=0.78 kJ/K . Entropy is generated in this process.

Fig. 6.1 Image of Process Panel in Ex. 6. Partial mixing is allowed between sub-systems A and B; therefore, the final state
is also a composite state (fA and fB states) unlike the fully mixed case shown in Fig. 5.1.

Ex-7 A block of copper of mass 1 lbm is heated to a temperature of 300^o F and is dropped into an insulated container containing 5 lbm of liquid water at 70^oF. Determine (a) the temperature of water and copper after equilibrium is reached. (b) Also obtain the entropy generated during the process. (c) What-if scenario: How would the answers change if the block had a mass of 2 lbm, instead?

Solution

The unsteady closed system goes through a process from a composite begin-state to a composite finish-state without any possibility of mixing between the sub-systems. Launch the non-mixing, non-uniform, closed-process daemon located at Daemons. Systems. Closed. Process. Generic. NonUniformUnmixed. Solid/Liquid.

#
# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Closed>Process>Generic> NonUniformUnMixed>SL/SL Model;

States    {
    State-1: Copper(Cu), Water(L);
    Given:       { T1= 300.0 deg-F;   Vel1= 0.0 ft/s;   z1= 0.0 ft;   m1= 1.0 lbm;   Model1= 1.0 UnitLess;   }

    State-2: Copper(Cu), Water(L);
    Given:       { T2= 70.0 deg-F;   Vel2= 0.0 ft/s;   z2= 0.0 ft;   m2= 5.0 lbm;   Model2= 2.0 UnitLess;   }

    State-3: Copper(Cu), Water(L);
    Given:       { T3= 74.0 deg-F;   Vel3= 0.0 ft/s;   z3= 0.0 ft;   m3= "m1" lbm;   Model3= 1.0 UnitLess;   }

    State-4: Copper(Cu), Water(L);
    Given:       { T4= "T3" deg-F;   Vel4= 0.0 ft/s;   z4= 0.0 ft;   m4= "m2" lbm;   Model4= 2.0 UnitLess;   }
    }

Analysis    {
    Process-A: b-State = State-1, State-2; f-State = State-3, State-4;
    Given: { W= 0.0 ft.lbf;   T_B= 77.0 deg-F;   }
    }

In this problem the non-uniform closed system, the copper block (part-A) in water (part-B), involves four states with the composite begin-state described by State-1 or bA-State for copper and State-3 or bB-State for water. Similarly State-2 or fA-State for copper and State-4 or fB-State for water describe the composite finish-state. We calculate the two begin states from the given data. Note that although T3=T4, neither State-3 not State-4 are completely known. In such situations, we guess a T3 and calculate Q and iterate until Q is close to the given value, 0 in this case.

Select English system.

State-1: Choose copper as the working substance from the left menu (Model-1). Enter m1 (1 lbm) and T1 (300^oF), and Calculate.

State-2: Choose water as the working substance from the right menu (Model-2). Enter m2 (5 lbm) and T2 (70^oF), and Calculate.

State-3: Choose copper as the working substance from the left menu (Model-1). Enter m3 (=m1) and a guessed value for T3, say, 100 deg-F. Calculate.

State-4: Choose water as the working substance from the right menu (Model-2). Enter T4 (=T3), and Calculate.

On Process Panel, load State-1 as the bA-State, State-2 as the bB-State, State-3 as the fA-State, and State-4 as the fB-State. Enter the known process variable W (=0). A Calculate produces Q=131 Btu , a positive value indicating that this heat must be added to the tank to obtain the final temperature of 100^oF. Go back to State-2 and change T2 to a lower value, press the Enter key and Super-Calculate. Repeat until Q is close to zero. A few iterations leads to T3 calculated as74.16 ^oF .

For the what-if study, change m1 to 2 lbm at State-1. A Calculate registers the changes made in State-1. A Super-Calculate yields Q=-20.9 Btu . The trial-and-error procedure outlined above must be repeated to obtain the finial temperature of 78^oF.

Note that for solids or liquids temperature alone determines properties u and s, and, therefore, information on pressure is not necessary to solve this problem.

Fig. 7.1 Image of Process Panel in Ex. 7. No mixing is allowed between sub-systems A and B.