# Otto Cycle Problem Solution Let's break down your Otto cycle problem step by step. ## Given Data - **Displacement volume, \( V_{disp} \):** 125 cc = 0.000125 m³ - **Compression ratio, \( r \):** 9.8 - **Initial pressure, \( P_1 \):** 0.81666667 bar = 81,666.67 Pa - **Initial temperature, \( T_1 \):** 7°C = 280.15 K - **Maximum temperature, \( T_3 \):** 1050°C = 1323.15 K - **Engine speed, \( N \):** 2100 RPM - **4-stroke engine** - **Cold-air-standard:** Assume constant specific heats at room temperature (for air): - \( c_v = 0.718 \) kJ/kg·K - \( c_p = 1.005 \) kJ/kg·K - \( R = 0.287 \) kJ/kg·K --- ## 1a. Net Work for the Cycle (\( W_{cycle} \)) ### Step 1: Calculate Volumes - **Clearance volume \( V_{c} \):** \[ r = \frac{V_1}{V_2} = \frac{V_{c} + V_{disp}}{V_{c}} \implies V_{c} = \frac{V_{disp}}{r-1} \] \[ V_{c} = \frac{0.000125}{9.8-1} = \frac{0.000125}{8.8} = 0.0000142~\text{m}^3 \] - **Total volume at start of compression (\( V_1 \)):** \[ V_1 = V_c + V_{disp} = 0.0000142 + 0.000125 = 0.0001392~\text{m}^3 \] - **Volume after compression (\( V_2 \)):** \[ V_2 = V_c = 0.0000142~\text{m}^3 \] ### Step 2: Calculate Mass of Air From the ideal gas law: \[ m = \frac{P_1 V_1}{R T_1} \] Plug in values (convert \( R \) to J/kg·K: \( R = 287~\text{J/kg·K} \)): \[ m = \frac{81666.67 \times 0.0001392}{287 \times 280.15} \] \[ m \approx \frac{11.373}{80323} \approx 0.0001416~\text{kg} \] ### Step 3: Otto Cycle Temperatures #### Compression (1 → 2, isentropic): \[ T_2 = T_1 \cdot r^{k-1} \] Assume \( k = \gamma = \frac{c_p}{c_v} = 1.4 \): \[ T_2 = 280.15 \times (9.8)^{0.4} = 280.15 \times 2.518 \approx 705.8~\text{K} \] #### Heat addition (2 → 3, constant volume): Given \( T_3 = 1323.15~\text{K} \). #### Expansion (3 → 4, isentropic): \[ T_4 = T_3 \cdot \left(\frac{1}{r}\right)^{k-1} = T_3 \cdot r^{-(k-1)} \] \[ T_4 = 1323.15 \times (9.8)^{-0.4} = 1323.15 / 2.518 \approx 525.5~\text{K} \] --- ### Step 4: Net Work (per cycle) \[ W_{cycle} = Q_{in} - Q_{out} \] - **Heat added (2 → 3):** \[ Q_{in} = m c_v (T_3 - T_2) \] \[ Q_{in} = 0.0001416 \times 718 \times (1323.15 - 705.8) \] \[ Q_{in} = 0.0001416 \times 718 \times 617.35 \approx 62.74~\text{J} \] - **Heat rejected (4 → 1):** \[ Q_{out} = m c_v (T_4 - T_1) \] \[ Q_{out} = 0.0001416 \times 718 \times (525.5 - 280.15) \] \[ Q_{out} = 0.0001416 \times 718 \times 245.35 \approx 24.92~\text{J} \] - **Net work:** \[ W_{cycle} = Q_{in} - Q_{out} = 62.74 - 24.92 = \boxed{37.8~\text{J}} \] --- ## 1b. Thermal Efficiency (\( \eta \)) \[ \eta = 1 - \frac{Q_{out}}{Q_{in}} = 1 - \frac{24.92}{62.74} = 1 - 0.397 = \boxed{0.603} \text{ or } 60.3\% \] Alternatively, for an ideal Otto cycle: \[ \eta_{Otto} = 1 - \frac{1}{r^{k-1}} = 1 - \frac{1}{(9.8)^{0.4}} = 1 - \frac{1}{2.518} = 0.603 \] --- ## 1c. Power Developed (\( \dot{W} \)), 4-Stroke Engine - **Number of cycles per revolution (4-stroke):** 0.5 - **Number of cycles per minute:** \( 2100 \times 0.5 = 1050 \) - **Number of cycles per second:** \( 1050 / 60 = 17.5 \) \[ \dot{W} = W_{cycle} \times \text{cycles/sec} = 37.8 \times 17.5 = 661.5~\text{W} \] \[ \dot{W} = \boxed{0.66~\text{kW}} \] --- ## 2. Work of Compression Stroke (\( W_{12} \)) For isentropic compression: \[ W_{12} = m c_v (T_2 - T_1) \] \[ W_{12} = 0.0001416 \times 718 \times (705.8 - 280.15) \] \[ W_{12} = 0.0001416 \times 718 \times 425.65 \approx 43.4~\text{J} \] --- # Summary Table | Item | Value | |-----------------------|---------------| | Net work per cycle | 37.8 J | | Thermal efficiency | 60.3% | | Power (\(\dot{W}\)) | 0.66 kW | | Compression work \(W_{12}\) | 43.4 J | --- ## Key Otto Cycle Equations - \( r = \frac{V_1}{V_2} \) - \( \eta = 1 - \frac{1}{r^{k-1}} \) - \( Q_{in} = m c_v (T_3 - T_2) \) - \( Q_{out} = m c_v (T_4 - T_1) \) - \( W_{cycle} = Q_{in} - Q_{out} \) --- ## Diagram  *Otto cycle P-V diagram. [Source: Wikipedia](https://en.wikipedia.org/wiki/Otto_cycle)* --- **Let me know if you need any steps explained in more detail!**