Calculate the theoretical efficiency for an Otto-cycle engine with $g equals 1.40$ and $r equals 9.50$ .

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Understand the Otto cycle: The Otto cycle is a thermodynamic cycle that describes the functioning of a typical spark ignition piston engine. It consists of two isochoric processes (constant volume) and two adiabatic processes (no heat exchange).

Identify the formula for the efficiency of an Otto cycle engine: The efficiency $ \eta $ of an Otto cycle engine is given by the formula $ \eta = 1 - \frac{1}{r^{\gamma - 1}} $, where $ r $ is the compression ratio and $ \gamma $ is the specific heat ratio.

Substitute the given values into the formula: You are given $ \gamma = 1.40 $ and $ r = 9.50 $. Substitute these values into the efficiency formula: $ \eta = 1 - \frac{1}{9.50^{1.40 - 1}} $.

Simplify the expression: Calculate $ 1.40 - 1 $ to get $ 0.40 $. Then, compute $ 9.50^{0.40} $ to find the value of the denominator.

Calculate the efficiency: Subtract the result from 1 to find the theoretical efficiency of the Otto-cycle engine.

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Otto Cycle

The Otto cycle is a thermodynamic cycle that describes the functioning of a typical spark-ignition piston engine. It consists of two adiabatic and two isochoric processes. Understanding the Otto cycle is crucial for calculating the efficiency of engines that operate on this cycle, such as gasoline engines.

Specific Heat Ratio (γ)

The specific heat ratio, denoted as γ (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, affecting the efficiency of cycles like the Otto cycle. For air, γ is typically around 1.40, which influences the theoretical efficiency calculation.

Compression Ratio (r)

The compression ratio, denoted as r, is the ratio of the maximum to minimum volume in the cylinder of an internal combustion engine. It is a key factor in determining the efficiency of the Otto cycle, as higher compression ratios generally lead to higher efficiencies. In this question, r is given as 9.50, which is used in the efficiency formula.