AME 436

Ideal 4-stroke Otto cycle process"
Compression ratio r = V 2 /V 1 = V 2 /V 3 = V 5 /V 4 = V 6 /V 7
Stroke Process Name Constant Mass in
cylinder
Other info
A 1 → 2 Intake P Increases P 2 = P 1 ; T 2 = T 1
At 1, exhaust valve closes,
intake valve opens
B 2 → 3 Compression s Constant P 3 /P 2 = r γ ; T 3 /T 2 = r (γ-1)
At 2, intake valve closes
--- 3→ 4 Combustion V Constant T 4 = T 3 + fQ R /C v ;
P 4 /P 3 = T 4 /T 3
At 3, spark fires
C 4 → 5 Expansion s Constant P 4 /P 5 = r γ ; T 4 /T 5 = r (γ-1)
--- 5 → 6 Blowdown V Decreases P 6 = P ambient ;
T 6 /T 5 = (P 6 /P 5 ) (γ-1)/γ
At 5, exhaust valve opens,
exhaust gas blows
down; gas remaining in
cylinder experiences ≈
isentropic expansion
D 6 → 7 Exhaust P Decreases P 7 = P 6 ; T 7 = T 6
AME 436 - Lecture 8 - Spring 2013 - Ideal cycle analysis
5
P-V & T-s diagrams for ideal Otto cycle"
Model shown is open cycle, where mixture is inhaled, compressed,
burned, expanded then thrown away (not recycled)
In a closed cycle with a fixed (trapped) mass of gas to which heat is
transferred to/from, 6 → 7, 7 → 1, 1 → 2 would not exist, process
would go directly 5 → 2 (Why dont we do this Remember heat
transfer is too slow!)
Pressure (atm)
7.0
6.0
5.0
4.0
3.0
2.0
1.0
Compression Combustion Expansion
Blowdown Intake Exhaust
Intake start 1 2
3 4 5
6 7
P-V diagram
Temperature (K)
Compression Combustion Expansion
Blowdown Intake Exhaust
Close T-s cycle 1 2
3 4 5
6 7
1200
T-s diagram
1000
800
600
400
200
0.0
0.E+00 1.E-04 2.E-04 3.E-04 4.E-04 5.E-04 6.E-04
Cylinder volume (m^3)
0
-100 0 100 200 300 400 500 600 700
Entropy (J/kg-K)
AME 436 - Lecture 8 - Spring 2013 - Ideal cycle analysis
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