Short Answer:

The efficiency equation for the Otto cycle shows how much of the input heat energy is converted into useful work in an ideal spark-ignition engine (like in petrol engines). The thermal efficiency (η) of the Otto cycle depends only on the compression ratio (r) and the specific heat ratio (γ) of the working fluid.

The formula is:
η = 1 − (1 / r^(γ−1))
This equation means that a higher compression ratio gives better efficiency, and gases with higher γ (like air) also help increase efficiency. This formula is used for ideal conditions without losses.

Detailed Explanation:

Efficiency equation for the Otto cycle

The Otto cycle is the ideal thermodynamic cycle for spark-ignition internal combustion engines, which are commonly used in petrol vehicles. It consists of four processes: two isentropic (adiabatic) and two constant volume processes.

The thermal efficiency of the Otto cycle is the percentage of heat energy input that is converted into useful work output during the cycle. In real engines, some energy is always lost due to friction, heat transfer, and other factors, but this equation gives the theoretical maximum efficiency under ideal conditions.

Derivation of the Efficiency Equation

In the Otto cycle, heat is added and rejected at constant volume. The efficiency (η) of the cycle is calculated as:

η = 1 − (Q_out / Q_in)

Where:

Q_in is the heat added during the constant-volume process (combustion)
Q_out is the heat rejected during the exhaust stroke

From thermodynamics, for an ideal gas undergoing constant-volume processes:

Q_in = Cv × (T3 − T2)
Q_out = Cv × (T4 − T1)

So,

η = 1 − (T4 − T1) / (T3 − T2)

Now, using isentropic relations and assuming ideal gas behavior:

T2 / T1 = r^(γ−1)
T3 / T4 = r^(γ−1)

Where:

r = Compression ratio = V1 / V2
γ = Specific heat ratio = Cp / Cv (usually 1.4 for air)

After substituting and simplifying, we get the Otto cycle efficiency equation:

η = 1 − (1 / r^(γ−1))

Meaning of the Equation

The equation shows that efficiency depends only on r and γ.
Higher r (compression ratio) = more efficient cycle.
Higher γ = better efficiency (air has γ ≈ 1.4).

This is why petrol engines with higher compression ratios are more efficient, but very high compression can cause knocking, so practical limits are used.

Key Points

Compression ratio (r) is the most important factor in efficiency.
The cycle assumes no heat loss, no friction, and ideal gas behavior.
It is useful for comparing engines under theoretical conditions.
Real engine efficiencies are lower due to mechanical and thermal losses.

Example Calculation

Let’s say a petrol engine has a compression ratio of 8:1 and γ = 1.4.
Then efficiency = 1 − (1 / 8^0.4) ≈ 1 − (1 / 2.297) ≈ 1 − 0.435 ≈ 0.565 or 56.5%

In practice, real engines have efficiencies of around 25% to 35%, due to practical losses.

Conclusion

The efficiency equation for the Otto cycle is given by
η = 1 − (1 / r^(γ−1)),
which shows the ideal thermal efficiency of a petrol engine based on its compression ratio (r) and specific heat ratio (γ). It helps engineers design better engines by understanding how higher compression ratios lead to better fuel efficiency, though real engines cannot reach this ideal due to various mechanical and thermal losses.