A gasoline engine receives air at 10C 100 kPa, having a compression ratio of 9:1 by volume. The heat addition by combustion gives the highest temperature as 2500 K. Use cold air properties to find the highest cycle pressure, the specific energy added by combustion, and the mean effective pressure.
A gasoline engine receives air at 10C 100 kPa, having a compression ratio of 9:1 by volume. The heat addition by combustion gives the highest temperature as 2500 K. Use cold air properties to find the highest cycle pressure, the specific energy added by combustion, and the mean effective pressure.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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A gasoline engine receives air at 10C 100 kPa, having a compression ratio of 9:1 by volume. The heat addition by combustion gives the highest temperature as 2500 K. Use cold air properties to find the highest cycle pressure, the specific energy added by combustion, and the mean effective pressure.
![Otto Cycle
Ideal Gas Model used with specific heats as f(T)
STATE POINTS
1 - Compression In
2 - Combustion Begin
3 - Expansion Begins
4- Expansion Ends
5- Compression In
PROCESSES
Compression
Combustion
Power Stroke
Heat Rejection
Eta
P [kPa]
100.0
2167.4
7946.3
Uin [kJ/kg]
0.0
0.0
0.0
0.0
CR=
T[°C]
10.0
408.7
2226.8
T[K]
9
283.2
681.9
2500.0
1083.1
Uout [kJ/kg] Q [kJ/kg]
0.0
0.0
0.0
0.0
Sums =
v [m³/kg] s [kJ/kg-K] u [kJ/kg]
0.8126
0.0813
W [kJ/kg]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7aa760e1-e2a9-4749-a30f-edcca023ea58%2F94554a95-dca0-4e40-be45-0ad482a36912%2Flxfpoom_processed.png&w=3840&q=75)
Transcribed Image Text:Otto Cycle
Ideal Gas Model used with specific heats as f(T)
STATE POINTS
1 - Compression In
2 - Combustion Begin
3 - Expansion Begins
4- Expansion Ends
5- Compression In
PROCESSES
Compression
Combustion
Power Stroke
Heat Rejection
Eta
P [kPa]
100.0
2167.4
7946.3
Uin [kJ/kg]
0.0
0.0
0.0
0.0
CR=
T[°C]
10.0
408.7
2226.8
T[K]
9
283.2
681.9
2500.0
1083.1
Uout [kJ/kg] Q [kJ/kg]
0.0
0.0
0.0
0.0
Sums =
v [m³/kg] s [kJ/kg-K] u [kJ/kg]
0.8126
0.0813
W [kJ/kg]
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