(a) Calculate the theoretical efficiency for an Otto-cycle engine with g = 1.40 and r = 9.50.

Verified step by step guidance

1. Identify the given values: the specific heat ratio (\(\gamma\)) is 1.40, and the compression ratio (\(r\)) is 9.50.
2. Recall the formula for the efficiency (\(\eta\)) of an Otto cycle, which is given by \(\eta = 1 - \frac{1}{r^{\gamma - 1}}\).
3. Substitute the given values into the efficiency formula. Replace \(\gamma\) with 1.40 and \(r\) with 9.50.
4. Calculate the exponent \(\gamma - 1\) which is 1.40 - 1 = 0.40.
5. Evaluate the expression \(r^{\gamma - 1}\) which is \(9.50^{0.40}\), and then compute \(1 - \frac{1}{9.50^{0.40}}\) to find the efficiency.

Key Concepts

Otto Cycle

The Otto cycle is a thermodynamic cycle that describes the functioning of a gasoline engine. It consists of two adiabatic processes and two isochoric processes, where the fuel-air mixture is compressed and then ignited, producing work. Understanding the Otto cycle is essential for analyzing engine efficiency and performance.

Efficiency of an Engine

The efficiency of an engine is a measure of how well it converts the energy from fuel into useful work. For an Otto cycle engine, the theoretical efficiency can be calculated using the formula: η = 1 - (1/g^(r-1)), where g is the specific heat ratio and r is the compression ratio. This efficiency is crucial for evaluating engine performance.

Specific Heat Ratio (g)

The specific heat ratio, denoted as g (gamma), is the ratio of the specific heat at constant pressure (Cp) to the specific heat at constant volume (Cv). It is a critical parameter in thermodynamics that influences the behavior of gases during compression and expansion processes in engines. For air, g is typically around 1.40, which is relevant for calculating the efficiency of an Otto cycle engine.