Thermodynamics II course taught at air university Islamabad by Sijal Ahmed Memon. Otto and diesel cycle analysis

1. THERMODYNAMICS II
LECTURE 3
OTTO AND DIESEL CYCLES
1

2. Introduction
• Internal Combustion Engine (IC engine)
• Spark Ignition (SI) and Compression Ignition (CI) engine
• Corresponding cycles are otto cycle and diesel cycle
• Combination of otto and diesel cycle is know as dual combustion cycle
– Gasoline engines or Spark ignition  spark plug
– Diesel engines or compression ignition  fuel injection
2

3. Two Stroke and Four Stroke Engine
• 4-Stroke Engine
– Four strokes (intake, compression, power stroke, exhaust) are executed for
every two revolutions of the crankshaft, and one thermodynamic cycle
• 2-Stroke Engine
– Two strokes (intake, compression, power stroke, and exhaust) are executed
for every one revolution of the crankshaft, and one thermodynamic cycle
• Advantages/Disadvantages of Two stroke
– Less efficient than four stroke due to mixing of exhaust and intake fresh
charge
– Simple and inexpensive, high power-to-weight ratio and power-volume-ratio
– Suitable for motocycles, chains saws and lawn movers
3

4. Terminology and Real Cycle
4

5. Overview of Reciprocating Engine
5
Compression ratio or volume
ratio (not the pressure ratio!)
Mean Effective Pressure (MEP)
To compare performance of
same size engines

6. Mean Effective Pressure
• To compare engines of same size
• More MEF means more work done
6

7. Otto Cycle
• Developed by otto in 1876.
7
• 1-2 Isentropic compression from BDC to TDC
• 2-3 Isochoric heat input (combustion)
• 3-4 Isentropic expansion (power stroke)
• 4-1 Isochoric heat rejection (exhaust)
2 3
BDC
TDC
 12 2 1W m u u 
 23 3 2Q m u u 
 34 3 4W m u u 
 41 4 1Q m u u 
T
s
1
2
3
4
P
1
2
3
4
BDCTDC
v
1 4

8. Efficiency of During Otto Cycle
8
34 12 4 1
,ASC
23 3 2
1net
th
in
W W W u u
Q Q u u

 
   

Thermal Efficiency
Compression Ratio
1 4
2 3
v v
CR
v v
 
11
,cold ASC
2
1 1 k
th
T
CR
T
 
   
P
1
2
3
4
BDCTDC
v
T
s
1
2
3
4
𝐶𝑣 = 0.718
𝑘𝐽
𝑘𝑔 − 𝑘
𝐶 𝑝 = 1.005
𝑘𝐽
𝑘𝑔 − 𝑘
𝑘 =
𝐶 𝑝
𝐶𝑣
= 1.4