Calculate Thermal Indicated Thermal Efficiency Of An Otto Engine

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SUMMARY

The indicated thermal efficiency of an Otto engine was calculated to be 61.1% based on a fuel consumption of 13.45 cc/min at 1730 rpm, producing approximately 4.335 kW. The fuel's heat content was determined to be 34.8 kJ/L, leading to a total heat input of 468 kJ. The maximum ideal efficiency for a compression ratio of 10.5:1 is approximately 62.81%, indicating that the calculated efficiency is close to the theoretical limit. The analysis suggests that the engine operates efficiently, utilizing about 50% less fuel than expected.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically related to engine efficiency
  • Familiarity with the Otto engine cycle and its parameters
  • Knowledge of fuel consumption metrics and calorific values
  • Basic proficiency in performing efficiency calculations using work and heat input
NEXT STEPS
  • Research the impact of compression ratios on engine efficiency in Otto engines
  • Learn about advanced thermodynamic cycles and their efficiencies
  • Explore methods for optimizing fuel consumption in internal combustion engines
  • Investigate the use of dynamometers for accurate engine performance testing
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Engineers, automotive technicians, and students studying mechanical engineering or thermodynamics, particularly those focused on internal combustion engine performance and efficiency optimization.

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Homework Statement



What is the indicated thermal efficiency of an engine?

Fuel used in total 42.61 cc/min @ 1730 rpm.

Fuel used to get to 1730 rpm = 29.16 cc/min no load.

Fuel left = 13.45 cc/min to do approx 4.335 kw at 1730 rpm approx 40.95 kph

Fuel heat content 34.8 kj/l

Approx 62kj / kw

According to the graph on http://web.mit.edu/16.unified/www/SPRING/thermodynamics/notes/node25.html"
The max ideal engine efficiency is about 62.81% efficient for a compression ratio of 10.5:1 when you look at the bit map with paint.

I worked it out on total fuel and got 36% but the engine uses approx 50% less fuel.
So I used the fuel used to create the power at 1730 rpm.

Homework Equations



Work over heat input.

The Attempt at a Solution



4.335 kw/min = 286 kj

13.45 cc/min = 0.01345 L * 34.8 kj = .46806 Mj = 468 kj

So 286kj / 468 kj = .611 * 100 = 61.1 % indicated thermal efficiency

My test maybe a few % low on consumption when compared to the ADR mileage test which I used the dyno figure for the mass of my car and a 30% efficient drive train.

Note: when I have finished checking and tidying up the raw data it will be in your favor slightly If I have not missed something or made a big error...LOL I wish I had lots of video camera's.

Thanks for all your help so far and in advance People from the Physics Forum.
 
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