Matlab Simulation Internal Combustion Engine

1.5

CO

2

CO 1.4

O

1.3

) K g k / J1.2 k ( c

NO O

p

2

1.1

1 0.9 0

N2 1000

2000

3000 T (K)

4000

5000

3.5 H O 2

3

) 2.5 K g k / J (k

OH H/10

cp 2

H /10 2

1.5

1 0

1000

2000

3000 T (K)

4000

5000

5 4.5

Heywood a =a =0 4

5

4 )3.5 K g k / J 3 k ( c

p

C H

8 18

CH

4

2.5 CH OH

2

3

1.5 1 300

400

500

600

700 T (K)

800

900

1000

10

0

N

2

CO −1

10 n o ti c a rf e l o m −2 10

O

2

H2O

NO

CO

2

OH

H

O

10

−3

0

H

2

0.5

1

1.5

2 phi

2.5

3

0

10

N

2

CO −1

10 n o ti c a fr e l o m −2 10

O

2

H O 2

NO

CO

2

OH

H

O

−3

10

0

H

2

0.5

1

1.5 phi

2

2.5

3

7 6 5 ) a P 4 (M re u 3 s e r p 2 1 0 −180

−90 0 90 crank angle (degrees ATC)

180

3000

burned gas unburned gas

2500

)2000 (K e r a rtu1500 e p m e t1000 500 0 −180

−90 0 90 crank angle (degrees ATC)

180

0 90 −90 crank angle (degrees ATC)

180

600 450 300 ) J ( rk 150 o w 0

−150 −300 −180

300 250 200 ) J ( r e f150

n s ra t100 t a e h 50 0 −50 −180

−90 0 90 crank angle (degrees ATC)

180

−90 0 90 crank angle (degrees ATC)

180

0 −1 −2 ) J (−3 e g a k−4 a e l t a e−5 h −6 −7 −8 −180

5

12

x 10

10

2

8 ) m / 6 (W x u tfl 4 a e h 2 0 −2 −180

−90 0 90 crank angle (degrees ATC)

180