Internal Combustion Engines

(The Closed Cycle daemons build upon the closed-process daemons. Visit Closed Process pages first.)

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

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

(IC Engines operates on closed cycles such as Otto or Diesel cycles. This page builds upon the Closed Process pages.)

Daemons>Closed Cycle> Applications

EXAMPLE-1

An ideal Otto cycle has a compression ratio of 9. At the beginning of compression, air is at 100 kPa, 27^oC. 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.

What-if scenario: How would the answers change if the compression ratio were increased to 10?

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, and then the Super-Calculate button.

# HOME>Daemons>Systems>Closed>Process>
# Specific>ReciprocatingCycles>IdealGas

   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; }
}

Step 1: Launch
the appropriate Cycle
Daemon.

Step 2: Set up the cycle.

Step 3: Calculate the States.

Solution

Answering the six questions described in the Approach section leads you to the appropriate daemon page: HOME. Daemons. Systems. Closed. Process. Specific.Cycles. IdealGas .

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 ('=v1/9'), 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. Use of algebraic relations (s2=s1, m2=m1, etc.) facilitates parametric
studies.

Step 4: Calculate the four processes.

On the Analysis panel, work on the four processes.

Process-a: Select Process-a . Load 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 . Load 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 . Load 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 . Load 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.

Step 5: Calculate the cycle variables.

Step 6: For the What-If study, change a variable, Calculate and Super-Calculate.

On the Cycle Panel, Calculate the cycle variables. The thermal efficiency is calculated as eta_th=54.8% . However to obtain MEP a Super-Calculate is ncecssary which produces MEP=428.2 kPa .

For the parametric study, go to the States 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.

Fig. 1.4: The output generated by the Super-Calculate on the I/O Panel.

Your input!

You will find more examples on closed cycles on the Slide Show and the Home.TEST.Problems.Chapter08 pages. 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