Question

In an ideal Otto cycle, 800 kJ/kg is transferred to air during the constant volume heat addition process and 381 kJ/kg is removed during the constant volume heat rejection process. The thermal efficiency in % of the cycle is _________.

Hint:

To calculate thermal efficiency, subtract the heat rejected from the heat added and then divide by the heat added.

Answer 1

Correct Answer: 52.2

The thermal efficiency \( \eta \) of an ideal Otto cycle is given by the formula: \[ \eta = \frac{Q_{\text{in}} - Q_{\text{out}}}{Q_{\text{in}}} \] Where:
- \( Q_{\text{in}} = 800 \, \text{kJ/kg} \) is the heat added to the system,
- \( Q_{\text{out}} = 381 \, \text{kJ/kg} \) is the heat removed.
Substituting the values: \[ \eta = \frac{800 - 381}{800} = \frac{419}{800} = 0.52375 \] Thus, the thermal efficiency is \( \boxed{52.2} % \).