1. EGR 334 Thermodynamics Chapter 9: Sections 1-2
Lecture 33:
Gas Power Systems:
The Otto Cycle
Quiz Today?

2. Today’s main concepts:
Understand common terminology of gas power cycles.
Be able to explain the processes of the Otto Cycle
Be able to perform a 1st Law analysis of the Otto Cycle and determine its thermal efficiency.
Be able to discuss limitations of the Otto cycle compared to real spark ignition power systems.
Be able to state the assumptions of standard air analysis.
Reading Assignment:
Read Chapter 9, Sections 3-4
Homework Assignment:
Problems from Chap 9: 1, 4, 11, 14

3. Two types of internal combustion engine
Sec 9.1 : Introducing Engine Terminology
Two types of internal combustion engine
Spark Ignition (lower power & lighter)
Compression Ignition (spontaneous combustion)
Terminology
Stroke : The distance the piston moves in one direction
Top Dead Center : The piston has minimum volume at the top of the stroke.
Bottom Dead Center : The piston has maximum volume at the bottom of the stroke.
Clearance Volume : Min vol
Displacement Volume : Max vol. – Min vol.
Compression Ratio: Max vol. / min vol.

4. Intake stroke : fill cylinder
Sec 9.1 : Introducing Engine Terminology
Four Stroke Cycle : Two revolutions : Combusts mix of hydrocarbons + O2
Intake stroke : fill cylinder
Spark cycle : fill with fuel and air mixture
Compression cycle : fill with air
Compression stroke : p , T , V , Win
Spark cycle : spark near end of stroke
Compression cycle : inject fuel
Power stroke  : gas expands
Exhaust stroke  : spent gas is exhausted
U--Tube video of 4 stroke:

5. Two Stroke Cycle : Two revolutions Power/Exhaust:
Sec 9.1 : Introducing Engine Terminology
Two Stroke Cycle : Two revolutions
Power/Exhaust:
The piston is forced down
@ exhaust port, spent gas leaves
Piston continues down and compresses air/fuel in crank case
Compressed charge enters cylinder
Intake/Compression
Piston moves up compressing charge
Draws vacuum in crank case.
Two Stroke Animation:

6. Mean Effective Pressure (mep)
Sec 9.1 : Introducing Engine Terminology
Mean Effective Pressure (mep)
Air –standard Analysis
(A simplification used to allow
for thermodynamic analysis)
Assumptions:
-- Fixed amount of air modeled as
closed system -- Air is treated as Ideal Gas
-- Constant cp (cold air-standard)
-- Combustion is modeled as a heat
transfer to system, Exhaust as
heat flow out of system
-- All processes internally reversible

7. Ideal Gas Model Review State Equation: Energy Relationships:
Chap 3: Quality Polytropic Process
Ideal Gas Model Relations:
Ideal Gas Model Review
State Equation:
Energy Relationships:
where
or look up values for k, cv, and cp on Table A-20
Entropy Relationships:
where so values are found on Table A-22
Special case: isentropic process where s1 = s2 then
( assuming constant specific heats)
( vr and pr for use with Table A-22)

8. Assumption: At top dead center, heat addition occurs instantaneously
Sec 9.2 : Air-Standard Otto Cycle
Assumption: At top dead center, heat addition occurs instantaneously
Otto Cycle: comprised of 4 internally reversible processes
Process 1 – 2 : Isentropic compression of air (compression stroke).
Process 2 – 3 : Constant volume heat transfer to the air from an external source while piston is at top dead center (ignition)
Process 3 – 4 : Isentropic expansion (power stroke)
Process 4 – 1 : Completes cycle by a constant volume process in which heat is rejected from the air while piston is at bottom dead center
ignition
power
ignition
compression
power
compression
exhaust
exhaust

9. Otto Cycle analysis Processes 1–2: ∆s = 0 and Q = 0
Sec 9.2 : Air-Standard Otto Cycle
Otto Cycle analysis
Closed system energy balance :
Processes 1–2: ∆s = 0 and Q = 0
Process 3–4: ∆s = 0 and Q = 0
Processes 2–3 : ∆V = 0 and W = 0
Processes 4-1 : ∆V = 0 and W = 0

10. Otto Cycle Thermal Efficiency:
Sec 9.2 : Air-Standard Otto Cycle
Otto Cycle Thermal Efficiency:
Thermal Efficiency can also be related
to the compression ratio:
clearance
displacement
As the compression ration, r, , the efficiency, η, 

11. Example (9.11): An air-standard Otto cycle has a compression ratio of At the beginning of compression, p1 = 85 kPa and T1 = 32°C. The mass of air is 2 g, and the maximum temperature in the cycle is 960 K. Determine
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure, in kPa.
State 1 2 3 4
T (K)
305
960
p (kPa) 85
u (kJ/kg) v
T3= 960 K
p1=85 kPa
T1=305 K

12. State 1: Using Ideal Gas Law: and Table A-22:
Example (9.11):
State 1 2 3 4
T (K)
305
960
p (kPa) 85
u (kJ/kg) v
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure, in kPa.
State 1: Using Ideal Gas Law:
and Table A-22:
State 2: Using compression ratio, ideal gas law, and Table A-22: 

13. State 3: using v3 = v2, ideal gas law, and Table A22:
Example (9.11):
State 1 2 3 4
T (K)
305
960
p (kPa) 85
u (kJ/kg)
217.67 v
1.0298
0.1373
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure, in kPa.
State 3: using v3 = v2, ideal gas law, and Table A22:
State 4: Using v4=v1:

14. then for isentropic process 1-2 using k = 1.361 and Table A-22:
Example (9.11):
State 1 2 3 4
T (K)
305
960
p (kPa) 85
2006.7
u (kJ/kg)
217.67
725.02 v
1.0298
0.1373
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure
, in kPa.
also knowing for isentropic processes
then for isentropic process 1-2 using k = and Table A-22:
and for isentropic process 3-4:

15. Heat added during process 2-3:
Example (9.11):
State 1 2 3 4
T (K)
305
631
960
464
p (kPa) 85
1319
2006.7
129.3
u (kJ/kg)
217.67
458.55
725.02
328.5 v
1.0298
0.1373
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure, in kPa.
Heat added during process 2-3:
Heat rejected during process 4-1:

16. Net Work over the cycle:
Example (9.11):
State 1 2 3 4
T (K)
305
631
960
464
p (kPa) 85
1319
2006.7
129.3
u (kJ/kg)
217.67
458.55
725.02
328.5 v
1.0298
0.1373
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure, in kPa.
Net Work over the cycle:
Cycle Efficiency:
Compare to

17. Mean Effective Pressure:
Example (9.11):
State 1 2 3 4
T (K)
305
631
960
464
p (kPa) 85
1319
2006.7
129.3
u (kJ/kg)
217.67
458.55
725.02
328.5 v
1.0298
0.1373
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure, in kPa.
Mean Effective Pressure:
where

18. End of Slides for Lecture 33