Question

An air standard Otto cycle has a compression ratio of 6 and a mean effective pressure of 1000 kPa. Assume that the specific heat ratio \(\gamma\) is 1.4 and specific gas constant \(R\) as 0.287 kJ/kgK for the air. If the pressure and temperature at the beginning of the compression stroke are 100 kPa and 300 K, respectively, then the specific work output of the cycle is ________________ kJ/kg (rounded off to one decimal place).

Hint:

In Otto cycle problems, mean effective pressure simplifies work evaluation without requiring full state analysis.

Answer 1

Correct Answer: 716.5

Mean effective pressure (MEP) relates work output and displacement volume: \[ W = \text{MEP} \times (v_1 - v_2) \] For an Otto cycle, \[ \frac{v_1}{v_2} = r = 6 \] Using ideal gas law at compression start: \[ v_1 = \frac{RT_1}{P_1} = \frac{0.287 \times 300}{100} = 0.861\ \text{m}^3/\text{kg} \] \[ v_2 = \frac{v_1}{6} = 0.1435 \] Displacement volume: \[ v_1 - v_2 = 0.7175\ \text{m}^3/\text{kg} \] Thus work output: \[ W = 1000\ \text{kPa} \times 0.7175 = 717.5\ \text{kJ/kg} \] This lies within the expected range: \[ \boxed{716.5\ \text{to}\ 718.5\ \text{kJ/kg}} \]
Final Answer: 716.5–718.5 kJ/kg