## Question Restatement **Given:** - Net work done by Otto cycle, \( W_{net} = 418.17 \) kJ/kg - Compression ratio, \( r = 8 \) - Initial air conditions: \( T_1 = 17^\circ C = 290 \) K, \( P_1 = 100 \) kPa **Asked:** Calculate the **mean effective pressure (MEP)**. --- ## Step 1: Mean Effective Pressure Formula \[ \text{MEP} = \frac{W_{net}}{v_{d}} \] Where: - \( W_{net} \) = Net work per cycle per kg - \( v_{d} \) = Specific displacement volume per kg --- ## Step 2: Find Displacement Volume per kg For an Otto cycle: \[ r = \frac{V_{max}}{V_{min}} \] \[ v_d = V_{max} - V_{min} = V_{min}(r - 1) \] From the ideal gas law at the start: \[ P_1 V_{max} = R T_1 \implies V_{max} = \frac{R T_1}{P_1} \] where \( R = .287 \) kJ/kg·K for air. So, \[ V_{min} = \frac{V_{max}}{r} \] \[ v_d = V_{max} - V_{min} = V_{max}\left(1 - \frac{1}{r}\right) \] --- ## Step 3: Plug in the Values ### Calculate \( V_{max} \): \[ V_{max} = \frac{R T_1}{P_1} = \frac{.287 \times 290}{100} = \frac{83.23}{100} = .8323 \text{ m}^3/\text{kg} \] ### Calculate \( v_d \): \[ v_d = V_{max}\left(1 - \frac{1}{r}\right) = .8323 \left(1 - \frac{1}{8}\right) = .8323 \times \frac{7}{8} = .8323 \times .875 = .7288 \text{ m}^3/\text{kg} \] --- ## Step 4: Calculate MEP \[ \text{MEP} = \frac{418.17}{.7288} = 573.74 \text{ kPa} \] --- ## Step 5: Choose the Correct Option - **574 kPa** --- ## Final Answer **The correct option is:** \[ \boxed{574\ \text{kPa}} \]