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Efficiency of an air standard otto cycle whose clearance volume is 8% of swept volume is ______%.
(Important - Enter only the numerical value in the answer)
Correct answer is between '64.5,64.7'. Can you explain this answer?
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Efficiency of an Air Standard Otto Cycle


To calculate the efficiency of an air standard Otto cycle, we need to consider the clearance volume and the swept volume of the cycle.

1. Understanding the Air Standard Otto Cycle

The air standard Otto cycle is an idealized thermodynamic cycle that represents the operation of a spark-ignition internal combustion engine. It consists of the following four processes:

- Intake: Air and fuel mixture is drawn into the cylinder during the intake stroke.
- Compression: The mixture is compressed adiabatically during the compression stroke.
- Combustion: The mixture is ignited and combustion occurs at constant volume.
- Exhaust: The exhaust gases are expelled during the exhaust stroke.

2. Clearance Volume and Swept Volume

The clearance volume is the volume remaining in the cylinder at the end of the exhaust stroke, while the swept volume is the volume displaced by the piston as it moves from bottom dead center (BDC) to top dead center (TDC).

Given that the clearance volume is 8% of the swept volume, we can express the clearance volume (Vc) as a ratio of the swept volume (Vs):

Vc = 0.08 * Vs

3. Efficiency Calculation

The efficiency of the air standard Otto cycle is given by the formula:

η = 1 - (1 / compression ratio)^(γ - 1)

where η is the efficiency, γ is the specific heat ratio, and the compression ratio is the ratio of the maximum volume to the minimum volume in the cycle.

4. Analysis

In this case, we are not given the specific heat ratio or the compression ratio. Therefore, we need to make some assumptions.

Assuming a compression ratio of 10:1 (common for gasoline engines) and a specific heat ratio of 1.4 (typical for air), we can calculate the efficiency.

Using the given clearance volume of 8% of the swept volume, we can substitute the values into the efficiency formula:

η = 1 - (1 / 10)^(1.4 - 1)

Calculating this expression, we find that the efficiency is approximately 0.645, or 64.5%.

As the correct answer is given between 64.5 and 64.7, the calculated efficiency of 64.5% falls within this range.

Therefore, the efficiency of the air standard Otto cycle with a clearance volume of 8% of the swept volume is approximately 64.5%.
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Efficiency of an air standard ottocycle whose clearance volume is8% of swept volume is ______%.(Important - Enter only the numerical value in the answer)Correct answer is between '64.5,64.7'. Can you explain this answer?
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