Problem 11 - College Physics

Question

\(\cdot\) What compression ratio \(r\) must an Otto cycle have to achieve an ideal efficiency of 65.0\(\%\) if the gas used in the chamber has \(\gamma=1.40 ?\)

Step-by-Step Solution

Answer

The compression ratio must be approximately 12.08.

Step 1: Understanding the Compression Ratio in Otto Cycle

The Otto cycle is an idealized cycle for internal combustion engines. Its efficiency \( \eta \) is related to the compression ratio \( r \) and the specific heat ratio \( \gamma \) of the gas by the formula \( \eta = 1 - \frac{1}{r^{\gamma-1}} \).

Step 2: Efficiency Equation Setup

We know that we want an efficiency \( \eta \) of 65%, which means \( \eta = 0.65 \). We'll substitute this into our Otto cycle efficiency equation: \( 0.65 = 1 - \frac{1}{r^{1.40-1}} \).

Step 3: Rearranging the Equation

Rearrange the equation to solve for \( r \): \[0.65 = 1 - \frac{1}{r^{0.40}}\] This becomes \[\frac{1}{r^{0.40}} = 1 - 0.65\] which simplifies to \[\frac{1}{r^{0.40}} = 0.35.\]

Step 4: Solving for Compression Ratio r

To solve for \( r \), take the reciprocal and then raise both sides to the power of \( \frac{1}{0.40} \): \[ r^{0.40} = \frac{1}{0.35} \] Calculate \( \frac{1}{0.35} \), then raise that number to the power of \( \frac{1}{0.40} \). \[ r = \left( \frac{1}{0.35} \right)^{\frac{1}{0.40}}. \]

Step 5: Final Calculation and Result

Compute \( \frac{1}{0.35} \approx 2.8571 \) and then raise it to the power of 2.5: \[ r \approx 2.8571^{2.5} \approx 12.08. \] So, the required compression ratio \( r \) is approximately 12.08.

Key Concepts

Compression Ratio

The compression ratio in an Otto cycle is a critical parameter in determining the efficiency of internal combustion engines. The compression ratio, denoted as \( r \), is the ratio of the volume of the combustion chamber from its largest capacity to its smallest capacity. In simpler terms, it measures how much the engine compresses the air-fuel mixture before ignition. A higher compression ratio usually leads to better engine efficiency as it allows more of the energy from the fuel to be converted into mechanical energy. However, there are limitations since very high compression ratios can cause knocking, which is a harmful engine condition. Understanding and optimizing the compression ratio is essential for achieving the desired efficiency level in engines using the Otto cycle.

Internal Combustion Engines

Internal combustion engines are devices that convert the chemical energy in fuel into mechanical energy. This process occurs within the engine's combustion chamber, where the fuel is burned and expands rapidly, moving the pistons. The Otto cycle specifically refers to the theoretical model for spark-ignition engines, like the ones in most cars. These engines typically have four strokes: intake, compression, power, and exhaust.

• The intake stroke draws in the air-fuel mixture.
• The compression stroke compresses the mixture.
• The power stroke ignites the mixture, creating an explosion that drives the piston.
• Finally, the exhaust stroke expels the combustion gases.

Each stroke is carefully timed to maximize efficiency and power output. By analyzing these processes, engineers can tweak various parameters, including the compression ratio, to enhance performance and efficiency.

Specific Heat Ratio

The specific heat ratio, often represented by the Greek letter \( \gamma \), is an important property of gases used in thermodynamics. It is the ratio of the specific heat at constant pressure \( (C_p) \) to the specific heat at constant volume \( (C_v) \). For air and many other diatomic gases, \( \gamma \) is typically about 1.40. The value of \( \gamma \) plays a significant role in the behavior of gases when they are compressed or expanded, particularly in processes where heat exchange is not allowed, also known as adiabatic processes. In the context of an Otto cycle, the specific heat ratio affects how efficiently the engine converts fuel into work. A higher specific heat ratio means the working gas can achieve a greater increase in pressure during the compression of the same volume, thus enhancing engine efficiency.

Efficiency Equation

The efficiency of an Otto cycle is largely determined by two factors: the compression ratio \( r \) and the specific heat ratio \( \gamma \). The efficiency equation for an Otto cycle is given by:\[\eta = 1 - \frac{1}{r^{\gamma-1}}\]where \( \eta \) represents the efficiency. This equation shows how the relationship between the compression ratio and the specific heat ratio influences the theoretical maximum efficiency of the engine. As the compression ratio increases, \( \eta \) approaches nearer to 1, indicating better efficiency. However, as noted earlier, practical limits exist, as overly high compression ratios can lead to engine knocking. The task in the original exercise was to find the compression ratio that achieves an efficiency of 65%, given \( \gamma = 1.40 \). Through calculations, it was determined that a compression ratio of approximately 12.08 is required to meet this efficiency. Understanding the efficiency equation helps in designing and optimizing engines to balance performance and reliability.