Open Steady Systems - TEST Tutorial

Tutorial (TOC) > Daemons > Exergy Analysis > Examples

Ex-1 A granite rock (density 2700 kg/m³, specific heat 1.017 kJ/kg.K) with m = 5000 kg, heats up to T = 45^oC during daytime due to solar heating. Assuming the surroundings to be at 20^oC. Determine the maximum amount of useful work that could be extracted from the rock.

#
# Daemon Path: TEST>Daemons>States>System>SL Model
#

States {
State-0: Granite;
Given: { p0= 100.0 kPa; T0= 20.0 deg-C; Vel0= 0.0 m/s; z0= 0.0 m; }

State-1: Granite;
Given: { p1= "p0" kPa; T1= 45.0 deg-C; Vel1= 0.0 m/s; z1= 0.0 m; m1= 5000.0 kg; }
}

Solution Although the rock goes through a closed process, a process analysis is not required to evaluate the maximum useful work that can be extracted out of the system, which is the stored exergy, a property of the system. To evaluate the stored exergy, launch the system state daemon located at TEST. Daemons. States.System.SL Model. The SL model is appropriate for a solid or a liquid that has no possiblity of a phase change.

Select Granite from the working substance selector. Evaluate the dead state (state-0) and then the heated state (state-1).

State-0: Enter p0 (100 kPa), T0 (25^oC), and Calculate. The dead state is calculated.

State-1: Enter p1 as '=p0', T1 (45^oC), m1 (5000 kg), and Calculate.

The specific stored exergy, phi1, is calculated as part of state-1 as shown in the image below. On the I/O panel, you can calculate the total stored energy as '=m1*phi1'= 5130 kJ.

Ex-2 Steam enters steadily a turbine through a duct of diameter 0.2 m. The steam velocity is 90 m/s, the pressure is 14 MPa, and the temperature is 600^oC. Steam exits the turbine through a duct of diameter 0.8 m with a pressure of 500 kPa and a temperature of 180^oC. Heat loss from the turbine to the atmosphere at 100 kPa, 25^oC occurs at a rate of 1000 kW. Determine (a) the exergetic (2nd law) efficiency and (b) the rate of exergy destruction in the turbine's universe.

# Daemon Path: TEST>Daemons>Systems>Open>SteadyState>Generic>
# SingleFlow>PC Model
#

States {
State-0: H2O;
Given: { p0= 100.0 kPa; T0= 25.0 deg-C; Vel0= 0.0 m/s; z0= 0.0 m; }

State-1: H2O;
Given: { p1= 14.0 MPa; T1= 600.0 deg-C; Vel1= 90.0 m/s; z1= 0.0 m; A1= "PI*.2^2/4" m^2; }

State-2: H2O;
Given: { p2= 0.5 MPa; T2= 180.0 deg-C; z2= 0.0 m; mdot2= "mdot1" kg/s; A2= "PI*.8*.8/4" m^2; }
}

Analysis {
Device-A: i-State = State-1; e-State = State-2;
Given: { Qdot= -1000.0 kW; T_B= 25.0 deg-C; }
}

Solution The turbine, an open device with a single flow, is working at steady state. Therefore, launch the open steady daemon located at TEST. Daemons. Systems. Open. SteadyState. Generic. Single-Flow. PC Model.

H2O is the default fluid. Let State-1 and State-2 represent the i-state and e-state respectively. Also, let State-0 represent the dead state.

State-0: Enter p0 (100 kPa), T0 (25^oC), and Calculate. The dead state is calculated.

State-1: Enter A1 as '=PI*.4*.4/4', Vel1 (90 m/s), p1 (14 MPa), T1 (600^oC), and Calculate. The mass flow rate is calculated as 105 kg/s.

State-2: Enter A2 as '=PI*.8*.8/4', p2 (500 kPa), T2 (180^o C), and Calculate. The exit velocity is calculated as 84.8 m/s.

On the device panel, load State-1 as the i-state and State-2 as the e-state. Enter the known device variables Qdot (=-1000 kW) and T_B (25^o C). Calculate and Super-Calculate to produce Wdot_ext as 81 MW. Switch to the exergy panel where each term of the exergy balance equation is evaluated and displayed. The useful output is Wdot_u=Wdot_ext=81 MW and the net exergy input into the turbine is calculated as Psidot_net=89.9 MW. Therefore, the exergetic efficiency is 81/89.9=90.1%. The rate of exergy destruction is also displayed as Idot=8.67 MW (see image of the exergy panel below).

Ex-3 Refrigerant-12 enters steadily an adiabatic compressor as a mixture of saturated vapor and saturated liquid with a quality of 95% and pressure 80 kPa at a rate of 1 m³/min, and exits at 2 MPa. The compressor has an adiabatic efficiency of 75%. Assuming the surrounding conditions to be 100 kPa, 25^oC, determine (a) the actual power, (b) the rate of entropy generation and (c) the second-law efficiency. Neglect ke and pe.

What-if scenario: (a) How would the answers change if the ambient temperature was 15^oC instead? (b) How would the answers change if R-12 was replaced with R-134a?

# TEST>Daemons>Systems>Open>SteadyState>Generic>
# SingleFlow>PC-Model;
#

     States    {
    State-0: R-12;
    Given:       { p0= 100.0 kPa;   T0= 25.0 deg-C;   Vel0= 0.0 m/s;   z0= 0.0 m;   }

    State-1: R-12;
    Given:       { p1= 0.08 MPa;   x1= 95.0 %;   Vel1= 0.0 m/s;   z1= 0.0 m;   Voldot1= 1.0 m^3/min;   }

    State-2: R-12;
    Given:       { p2= 2.0 MPa;   s2= "s1" kJ/kg.K;   Vel2= 0.0 m/s;   z2= 0.0 m;   }

    State-3: R-12;
    Given:       { p3= "p2" MPa;   Vel3= 0.0 m/s;   z3= 0.0 m;   }
    }

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

    Device-B: i-State = State-1; e-State = State-3;
    Given: { Qdot= 0.0 kW;   Wdot_ext= -6.440787 kW;   T_B= 25.0 deg-C;   }
    }

Solution This is also a problem involving an open, steady system. Navigate to TEST. Daemons. Systems. Open. SteadyState. Generic. SingleFlow page and launch the daemon by choosing the PC model. Select R-12 as the working fluid .

Let State-1 represent the i-State, State-2 the isentropic exit state, State-3 the actual exit state, and State-0 the dead state.

State-0: Enter p0 (100 kPa) and T0 (25^oC), and Calculate.

State-1: Enter p1 (80 kPa), x1 (95%), Voldot1 (1m3/min), and Calculate. Leave Vel1 and z1 at their default value of zero. The mass flow rate is calculated as 0.089 kg/s . Note that the inlet area calculated is ridiculously high. To obtain a realistic A1, Vel1 has to have a realistic value.

State-2: Enter p2 (2 MPa), s2 as '=s1', leave Vel1 and z1 at their default value and Calculate. The temperature T2 is calculated as 79.4^oC .

State-3: Enter p3'=p2' and leave Vel2 and z2 at their default value of zero. eta_adb=Wdot_ext,s/Wdot_ext = (j1-j3)/(j1-j2). Therefore, j3=j1-(j1-j2)/eta_ad . Enter j3 as '=j1-(j1-j2)/0.75 and Calculate. T3 is calculated as 100^oC .

In the device panel, load State-1 as the i-State and State-2 as the e-State for Device-A, the isentropic compressor. Enter Qdot=0 and Calculate the isentropic power as Wdot_ext=-4.83 kW. Select Device-B for the actual compressor, load State-1 and State-3 as the anchor states, Qdot (=0), Wdot_ext as '=-4.83/0.75', Calculate and Super-Calculate. State-3 is updated to produce T3 as 100^oC . Also for Device-B, the actual compressor, Sdot_gen=0.0044 kW/K.

An alternative way to solve this problem could be to evaluate State-3 completely using the definition of adiabatic efficiency as j3=j1-(j1-j2)/eta_ad .

In the exergy panel, all exergy variables are calculated automatically as shown in the figure below. The second-law efficiency can be calculated from eta_II= (exergy captured)/(exergy supplied)= Psidot_net/Wdot_u =79.8%.

For the what-if study, go to States panel, choose State-0, change T0 to its new value (15^o C) and Calculate . Get back to the Availability panel and do a Super-Calculate . All the answers are updated. The new value for eta_II=80.4%. For the second part, select the working fluid on the States panel to be R-134a. A Super-Calculate yields: Wdot_ext=-6.19 kW and Sdot_gen=0.0044 kW/K.

Ex-4 An insulated piston cylinder device contains 20 L of O₂ at 300 kPa and 100^oC. It is now heated for 1 min by a 200W resistance heater placed inside the cylinder. The pressure of O₂ is maintained constant during the process. Determine (a) the change in stored exergy, and (b) irreversibility during the process. Assume the surroundings to be at 100 kPa, 25^oC, and variable specific heats for oxygen.

# TEST>Daemons>Systems>Open>SteadyState>Generic>
# SingleFlow>PC-Model;
#

States {
State-0: O2;
Given: { p0= 100.0 kPa; T0= 25.0 deg-C; Vel0= 0.0 m/s; z0= 0.0 m; }

State-1: O2;
Given: { p1= 300.0 kPa; T1= 100.0 deg-C; Vel1= 0.0 m/s; z1= 0.0 m; Vol1= 20.0 L; }

State-2: O2;
Given: { p2= "p1" kPa; Vel2= 0.0 m/s; z2= 0.0 m; m2= "m1" kg; }
}

Analysis {
Process-A: b-State = State-1; f-State = State-2;
Given: { Q= 0.0 kJ; W_O= -12.0 kJ; T_B= 25.0 deg-C; }
}

Solution The gas trapped in the piston-cylinder assembly undergoes a closed process. Therefore, we launch the closed process daemon located at TEST. Daemons. Systems. Closed. Process. Generic. Uniform page and select the IG model to allow for variation of specific heats with temperature. Select oxygen (O2) as the working fluid.

Let State-1 represent the b-State, State-2 the isobaric f-state, and State-0 the dead state.

State-0: Enter p0 (100 kPa) and T0 (25^oC), and Calculate.

State-1: Enter p1 (300 kPa), T1 (100^oC), Vol1 (20 L), and Calculate. Leave Vel1 and z1 at their default value of zero.

State-2: Enter p2 ('=p1'), and m2 ('=m1').

In the process panel, load State-1 as the b-State and State-2 as the f-State. Also enter Q=0, and W_O=-200*60/1000=-12 kJ (negative because work is going in). W_B is already posted as 3.19 kJ. A Calculate followed by Super-Calculate produces the detailed solution. In the exergy panel, some of the answers are posted as part of the exergy balance - Delta_Phi is the change in stored exergy, 2.2 kJ, and I is the irreversibility, 7.67 kJ (see figure below).

For the what-if scenario, simply change Vol1 to 50 L, press the Enter key, and Super-Calculate. On the exergy panel the irreversibility is updated as 8.67 kJ.