Turbines, Pumps, Compressors, Nozzles, Diffusers Analysis Software - First and Second Law Analysis of Open and Steady Systems

Tutorial (TOC) > Daemons > Open Steady > Examples

Ex-1 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. Determine (a) the velocity of the steam at the exit, (b) the mass flow rate in kg/s, (c) the power produced by the turbine if the heat lost is 1 MW, and (d) the rate of entropy generation if the ambient temperature is 25^oC. (e) What-if scenario: How would the answers change if the exit duct diameter was doubled?

Solution The system is open and steady, involves a single flow, and does not belong to specific topics. Navigate to TEST. Daemons. Systems. Open. SteadyState. Generic. Single-Flow page and click on Phase-Change (PC) model to launch the daemon.

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

    States {
             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= "3.14*.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; }
}

H2O is the default fluid. Let State-1 and State-2 represent the i-state and e-state respectively.

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 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 and Sdot_gen=29 kW/K as well as all other device variables.

For the what-if study, go to State Panel, choose State-2, change the exit area to '=PI*1.6^2/4', Calculate and Super-Calculate . All the answers are updated. Note the exit velocity decreases to 21.2 m/s , while the power output is hardly affected with the new value as 81.5 MW.

Fig. 1.1 Image of the Device Panel of the Systems...SingleFlow.H2O daemon.

Ex-2 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. (d) What-if scenario: How would the answers change if the ambient temperature was 15^oC instead? (e) How would the answers change if R-12 was replaced with R-134a?

#
# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: 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 The system is open and steady, involves a single flow, and does not belong to specific topics. Navigate to TEST. Daemons. Systems. Open. SteadyState. Generic. SingleFlow. PhaseChange page and launch the daemon by choosing the PC model. Select the R-12 as the working fluid .

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

On 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 .

On Exergy Panel, all exergy variables are calculated automatically. The second-law efficiency can be calculated from eta_II= (exergy captured)/(exergy supplied)= Psidot_net/Wdot_u =79.8%. Note that Device-B must be selected in Device Panel for the exergy analysis of the actual device.

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.

Fig. 2.1 Image of Exergy Panel for the actual compressor (Device-B) in Ex. 2.

Ex-3 Air enters an insulated diffuser operating at steady state with a pressure of 80 kPa, a temperature of 0^oC and a velocity of 250 m/s. (a) Determine the percent increase in the area necessary to raise the exit pressure to 100 kPa. Assume the diffuser to be reversible and air to have constant specific heats. (b) What-if scenario: How would the answer change if the air temperature was 40^oC instead?

Solution The system is open and steady, involves a single flow, and does not belong to specific topics. Launch the single-flow daemon at the following address: TEST. Daemons. Systems. Open. SteadyState. Generic. SingleFlow. PerfGas .

#
# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState> Generic>SingleFlow>PG Model

   States {
               State-1: Air;
               Given:       { p1= 80.0 kPa;   T1= 0.0 deg-C;
                            Vel1= 250.0 m/s;   z1= 0.0 m;   mdot1= 1.0 kg/s;   }

State-2: Air;
Given: { p2= 100.0 kPa; z2= 0.0 m; }
}

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

Let State-1 and State-2 represent the i-State and the e-State respectively. Note that no mass flow rate is given in the problem. We will assume a mdot of 1 kg/s and check that the answers produced by this analysis do not depend on our choice of flow rate.

State-1: Enter p1 (80 kPa), T1 (0^oC), Vel1 (250 m/s), mdot1 (1 kg/s) and Calculate. The inlet area A1 is calculated as 39.2 cm2 .

State-2: Enter p2 (100 kPa), initialize Vel2 (unknown), and Calculate. The state is, obviously, not fully evaluated.

On Device Panel, load State-1 as the i-State and State-2 as the e-State. Enter the known device variable Qdot (0 kW), Wdot_ext (0 kW), and Sdot_gen (=0 because of no irreversibility). The value of T_B does not affect the entropy balance equation because the diffuser is adiabatic. A Calculate and Super-Calculate produce the exit state.

Get back to State-2 to find a completely evaluated state (note the gray background of the properties exported from the device panel). The exit area is calculated as
51.5 cm2 , an increase of about 31.4 % .

Select State-2 , change mdot1 to 2 kg/s. Update all calculations using a Calculate followed by a Super-Calculate . The inlet and exit areas, both double without affecting the answer.

For the what-if study, change T1 to 40^oC. A Calculate and Super-Calculate yield A1 and A2 as 44.9 cm2 and 66.0 cm2 respectively, an increase of 47.0 %.

Ex-4 A pump operating at steady state raises oil from 25 ft underground where the pressure is 15 psia and temperature is 20^oC, and delivers it 100 ft above ground at a pressure of 60 psia. (a) Determine the minimum pumping power for a mass flow rate of 10 lbm/s. Neglect any change in kinetic energy and heat transfer. (b) What-if scenario: How would the answer change if the supply pressure was reduced to 45 psia?

Solution The system is open and steady, involves a single flow, and does not belong to specific topics. Launch the single-flow daemon found at TEST. Daemons. Systems. Open. SteadyState. Generic. SingleFlow. SolidLiquid. Let us assume the pump to be reversible and adiabatic.

#
# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState> Generic>SingleFlow>SL Model

    States {
          State-1: Oil(L);
          Given:       { p1= 15.0 psia;   T1= 20.0 deg-C;
                Vel1= 0.0 ft/s;   z1= -25.0 ft;   mdot1= 10.0 lbm/s;   }

          State-2: Oil(L);
          Given:       { p2= 60.0 psia;   Vel2= 0.0 ft/s;
                z2= 100.0 ft;   }
}

Analysis {
          Device-A: i-State = State-1; e-State = State-2;
          Given: { Qdot= 0.0 Btu/min;   T_B= 77.0 deg-F;
                Sdot_gen= 0.0 Btu/R.min;   }
}

Let State-1 represent the i-State (at the surface of the liquid in the supply reservoir where the pressure and height are known) and State-2 (at the surface of the water where the pressure and height are known) the e-State . Let us fix the datum at the pump location so that z1=-25 ft . Switch to English system and choose Oil as the working fluid.

State-1: Enter mdot1 (10 lbm/s), p1 (15 psia), T1(20^oC), z1 (-25 ft), and Calculate. Leave Vel1 at its default value of zero (which produces a ridiculously large A1. This has no effect on the solution. For a realistic A1 change Vel1 to a reasonable value).

State-2: Enter p2 (60 psia), z2 (100 ft), and Calculate.

On Device Panel, load State-1 as the i-State and State-2 as the e-State. Enter the known device variable Qdot (=0), Sdot_gen (=0 for a pump with no irreversibility or wasted power), and T_B as 25^oC (its value does not affect the calculations as the heat transfer is assumed to be zero). A Calculate and Super-Calculate produce Wdot_ext=-4.35 hp. Notice that T2 is evaluated as part of the solution. The problem could be solved more directly by entering s2 as '=s1'. Although we have assumed the pump to be adiabatic, a more general analysis shows that the adiabatic assumption is unnecessary.

For the what-if study, go to State Panel, choose State-2, change p2 to its new value (45 psia) and Calculate . Get back to the device panel and Super-Calculate . All the answers are updated. The new value for Wdot_ext=-3.66 hp.

If the problem involves H2O, the appropriate daemon could be chosen as pure liquid (SL model) or phase-change (PC) mixture. In the PC model, liquid can be represented by saturated liquid (x=0).

Ex-5 100 m³/min of N2 at 100 kPa, 30^o C is mixed with 50 m³/min of CO₂ at 200^oC and 100 kPa in an adiabatic mixing chamber. Determine (a) the final temperature and (b) the rate of generation of entropy. Assume the gases to behave as ideal gases and no pressure drop in the mixing chamber. Neglect any pressure drop or changes in KE or PE in the mixing chamber. (c) What-if scenario: How would the entropy generation change if the pressure inside the chamber was 200 kPa instead?

Solution The system is open and steady, involves two flows that mix, and does not belong to specific topics. Launch the multi-flow, mixing daemon located at TEST. Daemons. Systems. Open. SteadyState. Generic. FlowMixed. IG/IG. Since two different ideal gases are involved the IG/IG binary mixture model is suitable in this problem. For manual solution, the PG/PG model is the recommended model.

#
# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState>Generic> MultiFlowMixed>IG/IG;

States   {
   State-1: N2, CO2;
   Given:       { p1= 100.0 kPa;   T1= 30 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   Voldot1= 100.0 m^3/min;   x_A1= 1.0 fraction;   }

   State-2: N2, CO2;
   Given:       { p2= "p1" kPa;   T2= 200.0 deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;   Voldot2= 50.0 m^3/min;   x_A2= 0.0 fraction;   }

   State-3: N2, CO2;
   Given:       { p3= "p1" kPa;   Vel3= 0.0 m/s;   z3= 0.0 m;   x_A3= "mdot1/(mdot1+mdot2)" fraction;   }
   }

Analysis   {
   Device-A: i-State = State-1, State-2; e-State = State-3; Mixing: true;
   Given: { Qdot= 0.0 kW;   Wdot_ext= 0.0 kW;   T_B= 25.0 deg-C;   }
}

Solution The system is open and steady, involves a single flow, and does not belong to specific topics. Launch the multi-flow, mixing daemon located at TEST. Daemons. Systems. Open. SteadyState. Generic. FlowMixed. IG/IG. Since two different ideal gases are involved the IG/IG binary mixture model is suitable in this problem. For manual solution, the PG/PG model is the recommended model.

Let State-1 and State-2 represent the two inlets, i1-State and i2-State respectively, and State-3 represent the e-State. Choose N2 from the first gas-selector (gas-A) and CO2 from the second gas-selector (gas-B).

State-1: Enter x_A1 (1), T1 (30^oC), p1(100 kPa), Voldot1 (100 m3/min), and Calculate. A pure nitrogen flow is created with a mass flow rate calculated as 1.85 kg/s.

State-2: Enter x_A2 (0), T2 (200^oC), p2('=p1'), Voldot2 (50 m3/min), and Calculate. The mass flow rate of pure CO2 is calculated as 0.93 kg/s.

State-3: Enter x_A3 ('=mdot1/(mdot1+mdot2)), p3 (=p1), and Calculate. The state cannot be fully evaluated at this stage.

On Device Panel, load State-1 as the i1-State, State-2 as the i2-State, and State-3 as the e1-State (note that the daemon can handle up to two exit states). Enter the known device variable Qdot (0 kW) and Wdot_ext (0 kW). The value of T_B does not affect the entropy balance equation because the chamber is adiabatic. Calculate and Super-Calculate to produce the exit state and the unknown device variable, Sdot_gen=0.4657 kW/K. Go to State Panel to find T3=83.5^oC .

For the what-if study, change p1 to 200 kPa. A Calculate and Super-Calculate yield Sdot_gen=0.9307 kW/K.

The daemon allows two exits, either of which could be used as the exit of the device. The two exits are useful for separators where a single flow enters and separates into two different flows.

Fig. 5.1 Image of Device Panel of the mixing daemon. Note that State-e2 is left alone as there is only one exit in this mixing chamber.

Ex-6 Air enters a heat exchanger, operating at steady state, at 1 MPa and 600 K at a mass flow rate of 1000 kg/min and leaves without any significant pressure loss. Heat is transferred to the air at a rate of 4000 kW. Nitrogen, flowing separately, enters at 150 kPa and 800 K and leaves at 120 kPa and 600 K. Determine (a) the exit temperature of air, (b) the mass flow rate of nitrogen and (c) the rate of entropy generation. Neglect any change in kinetic energy and heat transfer. (d) What-if scenario: How would the answers change if the heat transfer had to be augmented to 5000 kJ/min?

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

States   {
   State-1: Air, N2;
   Given:       { p1= 1000.0 kPa;   T1= 600.0 K;   Vel1= 0.0 m/s;   z1= 0.0 m;   mdot1= 1000.0 kg/min;   x_A1= 1.0 fraction;   }

   State-2: Air, N2;
   Given:       { p2= "p1" kPa;   Vel2= 0.0 m/s;   z2= 0.0 m;   x_A2= "x_A1" fraction;   }

   State-3: Air, N2;
   Given:       { p3= 150.0 kPa;   T3= 800.0 K;   Vel3= 0.0 m/s;   z3= 0.0 m;   x_A3= 0.0 fraction;   }

   State-4: Air, N2;
   Given:       { p4= 120.0 kPa;   T4= 600.0 K;   Vel4= 0.0 m/s;   z4= 0.0 m;   mdot4= "mdot3" kg/min;   x_A4= "x_A3" fraction;   }
   }

Analysis   {
   Device-A: i-State = State-1, State-3; e-State = State-2, State-4; 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-2; Mixing: false;
   Given: { Qdot= 4000.0 kW;   Wdot_ext= 0.0 kW;   T_B= 25.0 deg-C;   }

   Device-C: i-State = State-3; e-State = State-4; Mixing: false;
   Given: { T_B= 25.0 deg-C;   }
   }

Solution The system is open and steady, involves two flows that do not mix, and does not belong to specific topics. Launch the multi-flow, non-mixing daemon located at TEST. Daemons. Systems. Open. SteadyState. Generic. FlowUnmixed. IG/IG Model . Select Air as gas-A and Nitrogen as gas-B.

In this problem there are essentially three devices depending on the analyst's point of view.

Device-A : The entire heat exchanger with State-1 representing the i1-State and State-2 the e1-State , State-3 the i2-State and State-4 representing the e2-State .

Device-B : The air stream with a single inlet and exit, State-1 representing the i1-State and State-2 representing the e1-State.

Device-C : The nitrogen stream with a single inlet and exit, State-3 representing the i2-State and State-4 representing the e2-State.

State-1 (Air inlet): Select Air as gas-A with x_A=1. Enter p1 (1 MPa), T1(600 K), and mdot1 (1000 kg/min) (leave Vel1 and z1 to their default value of 0) and Calculate voldot1 as 2.87 m3/s.

State-2 (Air exit): Enter x_A as '=x_A1' , p2 as '=p1', and mdot2 as '=mdot1'. Calculate. Nothing more is known about this state at this point.

State-3 (N2 inlet): To create a pure flow of Nitrogen, enter x_A=0. Also enter p3 (150 kPa) and T3 (800 K), and Calculate.

State-4 (N2 exit): Enter x_A4=x_A3. Enter p4 (120 kPa), T4 (600 K), and Calculate.

On Device Panel set up the following three devices.

Device-A : Load State-1 as the i1-State, State-2 as the e1-State, State-3 as the i2-State and State-4 as the e2-State. Enter Qdot (=0, heat transfer in the heat exchanger is interanal), and Wdot_ext (=0). Calculate.

Device-B : Load State-1 as the i1-State and State-2 as the e1-State. Enter Qdot (=4000 kW) and Wdot_ext (=0). A Calculate and Super-Calculate produce T2= 822 K, mdot3= 1083 kg/min and Sdot_gen= 1.11 kW/K

Note that we do not need Device-C in this problem. However, if you create Device-C with and enter Wdot_ext=0, you will get Qdot=-4000 kW.

For the what-if study, go to Device-B , change Qdot to the new value, Calculate and Super-Calculate. All the answers are updated. The new value for T2=874 K . Note that p2 remains an unknown as it depends on the friction inside the tube. If p2 was given, the device variable, Sdot_gen would have been determined.

Suppose T4 is given in terms of T2, say, T4=T2+20, and T2 is also an unknown in the problem. Obviously State-2 and 4 could not be calculated directly. In such situation, the solution must be iterative - you guess T2 and Super-Calculate to produce a desired Qdot (or any other known device variable). Although it sounds complicated, convergence is achieved very quickly. You also gain physical insight since the iterative solution turns out to be a what-if study for the problem.

Fig. 6.1 Image of Device Panel for a non-mixing multi-flow daemon. Note that the Non-Mixing option is selected.

Ex-7 Air enters a heat exchanger, operating at steady state, at 250 kPa and 127 deg-C at a volume flow rate of 25 m^3/s and exits at 75 deg-C without any loss of pressure. Water, flowing separately, enters at 100 kPa and 25 deg-C and leaves at 100 kPa and 40 deg-C. Determine (a) the mass flow rate of water and (b) the rate of entropy generation. Neglect heat transfer and any change in ke/pe. (c) What-if scenario: How would the answers change if the heat exchanger lost heat at the rate of 10 kW? Assume the atmospheric temperature to be 25 deg-C.

#
# TEST-codes: The codes describe the solution procedure in a nut shell and can recreate the visual solution.
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState> Generic>FlowUnmixed>IG/SL

States    {
    State-1: Air, Water(L);
    Given:       { p1= 250.0 kPa;   T1= 127.0 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   Voldot1= 25.0 m^3/s;   Model1= 1.0 UnitLess;   }

    State-2: Air, Water(L);
    Given:       { p2= "p1" kPa;   T2= 75.0 deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;   mdot2= "mdot1" kg/s;   Model2= 1.0 UnitLess;   }

    State-3: Air, Water(L);
    Given:       { p3= 100.0 kPa;   T3= 25.0 deg-C;   Vel3= 0.0 m/s;   z3= 0.0 m;   Model3= 2.0 UnitLess;   }

    State-4: Air, Water(L);
    Given:       { p4= "p3" kPa;   T4= 40.0 deg-C;   Vel4= 0.0 m/s;   z4= 0.0 m;   Model4= 2.0 UnitLess;   }
    }

Analysis    {
    Device-A: i-State = State-1, State-3; e-State = State-2, State-4; Mixing: false;
    Given: { Qdot= 0.0 kW;   Wdot_ext= 0.0 kW;   T_B= 25.0 deg-C;   }
    }

Solution The system is open and steady, involves two flows that do not mix, and does not belong to specific topics. Modeling the water with the solid/liquid (SL) model and air with the ideal gas (IG) model, launch the ... Systems. Open. SteadyState. Generic. FlowUnmixed. IG/SL daemon.

Setup the four states as described in the TEST-codes above. For each state you must select the working fluid carefully. Load the states on Device Panel, enter Qdot=Wdot_ext=0, Calculate, and Super-Calculate. The mass flow rate of water is found to be 45.78 kg/s and the entropy generation rate 1.71 kW/K .

For the what-if study, go to Device-A , change Qdot to -10 kW, Calculate, and Super-Calculate. The new value for mdot3= 45.6 kg/s and Sdot_gen= 1.71 kW/K. Could you explain why the entropy generation rate in the system's universe does not change?