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Ex-1 An ideal Otto cycle has a compression ratio of 9. At the beginning of compression, air is at 100 kPa, 27oC. The pressure is doubled during the constant-volume heat addition process. Determine (a) the thermal efficiency, (b) the net work output, and (c) the MEP.  Assume variable c_p. (d) What-if scenario: How would the answers change if the compression ratio was increased to 10? 


Solution Detailed analysis of cyclic processes executed by a closed system is handled in the specific branch of the closed process daemons. Launch the closed-cycle daemon located at  TEST. Daemons. Systems. Closed. Process. Specific.Cycles. IdealGas.





 

 

#
# 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> Specific>ReciprocatingCycles>IG Model

   States { 
              State-1:  Air;
              Given:   { p1= 100.0 kPa;   T1= 27.0 deg-C; 
                          Vel1= 0.0 m/s;  z1= 0.0   m;   m1= 1.0 kg; }

              State-2:  Air;
              Given: { s2= "s1" kJ/kg.K;   Vel2= 0.0 m/s;   z2= 0.0 m;  m2= "m1" kg;   v2= "v1/9" m^3;   }

              State-3:  Air;
              Given: { p3= "p2*2" kPa;   Vel3= 0.0 m/s;   z3= 0.0 m;  m3= "m2" kg;   v= "v2" m^3;   }

              State-4:  Air;
              Given: { s4= "s3" kJ/kg.K;   Vel4= 0.0 m/s;   z4= 0.0 m;  m4= "m3" kg;   v4= "v1" m^3;   }
             }

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

              Process-B:  b-State =  State-2;  f-State =  State-3; 
              Given: { T_B= 25.0 deg-C;   }

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

              Process-D:  b-State =  State-4;  f-State =  State-1; 
              Given: { T_B= 25.0 deg-C;   }
 }
 



 

 

Let us set up the cycle as follows: Process-A: isentropic compression from State-1 to State-2 ; Process-B : constant volume heat addition from State-2 to State-3 ; Process-C : isentropic expansion from State-3 to State-4 ; Process-D : constant volume heat rejection from State-4 to State-1 .

State-1: Enter m1 (assume 1 kg), T1, p1, and Calculate.

State-2: Enter s2 ('=s1'), v2 ('=v1/9'), and Calculate.

State-3: Enter p3 ('=p2*2'), v3 ('=v2'), and Calculate.

State-4: Enter s4 ('=s3'), v4 ('=v1'), and Calculate.
 



 
Fig. 1.1 Image of State-2.  The State and Process panels are identical to those found in closed-process daemons.

 

On Process Panel, work on the four processes.

Process-A: Select Process-A . Select State-1 and State-2 as the b- and f-States , enter Q=0, and Calculate.   The work is calculated as W_B=-299.5 kJ.

Process-B: Select Process-B . Select State-2 and State-3 as the b- and f-States ,  and Calculate.   The heat transfer is calculated as Q=599.7 kJ.

Process-C: Select Process-C . Select State-3 and State-4 as the b- and f-States, enter Q=0, and   Calculate. The work is calculated as W_B=627.3 kJ.

Process-D: Select Process-D . Select State-4 and State-1 as the b- and f-States ,  and   Calculate.   The heat transfer is calculated as Q=-270.8 kJ.


Fig. 1.2 Image of the Process-A. You do not have to enter W_B as it is automatically calculated. For a known polytropic
process, the polytropic coefficient n can be entered in this panel.

On the Cycle Panel, cycle variables are automatically calculated if the cycle has been completed.  The thermal efficiency is calculated as eta_th=54.8% . Super-Calculate to obtain MEP=428.2 kPa. 

For the parametric study, go to State Panel and change v2 to '=v1/10'. Calculate the state and  Super-Calculate to update all calculations. The new answers are: eta_th=56.3% and MEP=455.5 kPa . 



Fig. 1.3 Image of the Cycle Panel. 

 


Ex-2 An air standard Diesel cycle has a compression ratio of 15 and cut-off ratio of 3. At the beginning of the compression process, air is at 97 kPa and 30°C. Using the IG model for air, determine (a) the temperature after the heat addition process, (b) the thermal efficiency, and (c) the work output in kJ/kg. (c) What-if scenario: How would the answers change if the cut-off ratio was increased to 4? 


Solution Detailed analysis of cyclic processes executed by a closed system as in a Diesel cycle is handled in the specific branch of the closed process daemons. Launch the closed-cycle daemon located at  TEST. Daemons. Systems. Closed. Process. Specific.Cycles. IdealGas.





 

 

#
# 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> Specific>ReciprocatingCycles>IG Model

States    { 
      State-1:  Air;
      Given:       { p1= 97.0 kPa;   T1= 30.0 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   m1= 1.0 kg;   }

      State-2:  Air;
      Given:       { v2= "v1/15" m^3/kg;   s2= "s1" kJ/kg.K;   Vel2= 0.0 m/s;   z2= 0.0 m;   m2= "m1" kg;   }

      State-3:  Air;
      Given:       { p3= "p2" kPa;   v3= "v2*3" m^3/kg;   Vel3= 0.0 m/s;   z3= 0.0 m;   }

      State-4:  Air;
      Given:       { v4= "v1" m^3/kg;   s4= "s3" kJ/kg.K;   Vel4= 0.0  m/s;   z4= 0.0 m;   }
    }

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

      Process-B:  b-State =  State-2;  f-State =  State-3;
      Given: { T_B= 25.0 deg-C;   }

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

      Process-D:  b-State =  State-4;  f-State =  State-1;
      Given: { T_B= 25.0 deg-C;   }
    }

 



 

 

Let us set up the cycle as follows: Process-A: isentropic compression from State-1 to State-2 ; Process-B : constant pressure heat addition from State-2 to State-3 ; Process-C : isentropic expansion from State-3 to State-4 ; Process-D : constant volume heat rejection from State-4 to State-1 .

State-1: Enter m1 (assume 1 kg), T1, p1, and Calculate.

State-2: Enter s2 ('=s1'), v2 ('=v1/15'), and Calculate.

State-3: Enter p3 ('=p2'), v3 ('=v2*3', cut off ratio is 3), and Calculate. The temperature is calculated as T3=2266°C.

State-4: Enter s4 ('=s3'), v4 ('=v1'), and Calculate.
 


On Process Panel, work on the four processes.

Process-A: Select Process-A . Select State-1 and State-2 as the b- and f-States , enter Q=0, and Calculate.   The work is calculated as W_B=-415.8 kJ.

Process-B: Select Process-B . Select State-2 and State-3 as the b- and f-States ,  and Calculate. The heat transfer is calculated as Q=2042 kJ.

Process-C: Select Process-C . Select State-3 and State-4 as the b- and f-States, enter Q=0, and   Calculate. The work is calculated as W_B=928 kJ.

Process-D: Select Process-D . Select State-4 and State-1 as the b- and f-States ,  and   Calculate.   The heat transfer is calculated as Q=-1044 kJ.


On the Cycle Panel, cycle variables are automatically calculated if the cycle has been completed.  The thermal efficiency is calculated as eta_th=48.86% and the net work output as W_net=998 kJ (see image of the cycle panel below).

For the parametric study, go to State Panel and change v3 to '=v2*4'. Calculate the state and  Super-Calculate to update all calculations. The new answers are: T3=3112°C, eta_th=44.42% and W_net=1371 kJ/kg, which can be found on the cycle panel or the I/O panel as shown below.





Ex-3 An ideal Stirling engine using helium as the working fluid operates between the temperature limits of 38°C and 850°C and pressure limits of 102 and 1020 kPa. Assuming the mass used in the cycle is 1 kg, determine (a) the thermal efficiency of the cycle, and (b) the net work.


Solution Detailed analysis of cyclic processes executed by a closed system as in a Diesel, Otto, or Stirling cycle is handled in the specific branch of the closed process daemons. Helium being a perfect gas (constant c_p), launch the closed-cycle daemon located at  TEST. Daemons. Systems. Closed. Process. Specific.Cycles. PG Model.





 

 

#
# 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> Specific>ReciprocatingCycles>PG Model

States {
State-1: He;
Given: { p1 = 102.0 kPa; T1 = 38 deg-C; Vel1 = 0.0 m/s; z1 = 0.0 m; m1 = 1 kg; }

State-2: He;
Given: {p2 = 1020 kPa; T2 = "T1" deg-C; Vel2 = 0.0 m/s; z2 = 0.0 m; m2 = "m1" kg; }

State-3: He;
Given: { T3 = 850.0 deg-C; Vel3 = 0.0 m/s; z3 = 0.0 m; m3 = "m1" kg; Vol3 = "Vol2" m^3; }

State-4: He;
Given: { T4 = "T3" deg-C; Vel4 = 0.0 m/s; z4 = 0.0 m; m4 = "m1" kg; Vol4 = "Vol1" m^3; }
}

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

Process-B: b-State = State-2; f-State = State-3;
Given: { T_B = 25.0 deg-C; }

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

Process-D: b-State = State-4; f-State = State-1;
Given: { T_B = 25.0 deg-C; }
}
 



 

 

Let us set up the cycle as follows: Process-A: isothermal compression from State-1 to State-2 ; Process-B : constant volume heat addition from State-2 to State-3 ; Process-C : isothermal heat addition from State-3 to State-4 ; Process-D : constant volume heat rejection from State-4 to State-1 .

Evaluate the four states as described in the TEST-codes.
 


On Process Panel, work on the four processes.

Select Process-A . Select State-1 and State-2 as the b- and f-States , and Calculate. The work is calculated as W_B=-1489 kJ.

Select Process-B . Select State-2 and State-3 as the b- and f-States ,  and Calculate. The heat transfer is calculated as Q=2531 kJ.

Select Process-C . Select State-3 and State-4 as the b- and f-States, and   Calculate. The work is calculated as W_B=5375 kJ.

Select Process-D . Select State-4 and State-1 as the b- and f-States ,  and   Calculate.   The heat transfer is calculated as Q=-2531 kJ.


On the Cycle Panel, cycle variables are automatically calculated without taking into account regeneration. To correctly specify the transfer of heat from Process-B (2-3)to Process-D (4-1), select the regeneration Donor and Receiver as shown in the image below. The thermal efficiency is calculated as eta_th=72.3% and the net work output as W_net=3886 kJ (see image of the cycle panel below).

 



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