# Otto Cycle Problem: Step-by-Step Solution Given: - Engine type: Four-stroke gasoline (Otto cycle) - Engine displacement: \(4.2\, \text{L}\) (total swept volume) - RPM: \(200\) - Inlet (state 1): \(P_1 = 85\, \text{kPa}\), \(T_1 = 280\, \text{K}\) - After combustion (state 3): \(T_3 = 200\, \text{K}\) - Maximum pressure: \(P_3 = 5\, \text{MPa}\) - Assume air-standard analysis; use constant specific heats at room temp: \(c_v = .718\, \text{kJ/kg.K}\), \(c_p = 1.005\, \text{kJ/kg.K}\), \(R = .287\, \text{kJ/kg.K}\) - Find: 1. Compression ratio (\(r\)) 2. Cycle efficiency (\(\eta\)) 3. Exhaust temperature (\(T_4\)) 4. Otto cycle sketch, \(P-v\), and \(T-s\) diagrams --- ## 1. **Otto Cycle Overview** The air-standard Otto cycle consists of 4 processes (see diagram): 1. **1→2:** Isentropic compression 2. **2→3:** Constant-volume heat addition 3. **3→4:** Isentropic expansion 4. **4→1:** Constant-volume heat rejection --- ## 2. **Step 1: Find Compression Ratio \(r\)** ### **Relevant Equations** - For isentropic process (ideal gas): \[ \frac{T_2}{T_1} = r^{k-1} \] where \(k = \frac{c_p}{c_v} \approx 1.4\) - At state 3 (after combustion): \[ \frac{P_3}{P_2} = \frac{T_3}{T_2} \] (since 2→3 is at constant volume: \(\frac{P}{T}\) is constant). ### **Find \(T_2\)** Let \(T_1 = 280\, \text{K}\), \(T_3 = 200\, \text{K}\), \(P_3 = 5\, \text{MPa}\). First, find \(r\). #### **Step 1a. Let \(r\) be unknown** \[ T_2 = T_1 \cdot r^{k-1} \] #### **Step 1b. Find \(P_2\)** Isentropic compression: \[ \frac{P_2}{P_1} = r^k \implies P_2 = P_1 \cdot r^k \] #### **Step 1c. Find \(P_3\)** Constant volume heat addition: \[ \frac{P_3}{P_2} = \frac{T_3}{T_2} \implies P_3 = P_2 \cdot \frac{T_3}{T_2} \] But \(P_3\) is given (\(5\, \text{MPa}\)), so substitute everything in: \[ P_3 = P_2 \cdot \frac{T_3}{T_2} = [P_1 r^k] \cdot \frac{T_3}{T_1 r^{k-1}} = P_1 \frac{T_3}{T_1} r^{k - (k-1)} = P_1 \frac{T_3}{T_1} r \] Solve for \(r\): \[ P_3 = P_1 \frac{T_3}{T_1} r \implies r = \frac{P_3}{P_1} \cdot \frac{T_1}{T_3} \] Plug in the numbers: - \(P_3 = 5\, \text{MPa} = 500\, \text{kPa}\) - \(P_1 = 85\, \text{kPa}\) - \(T_1 = 280\, \text{K}\) - \(T_3 = 200\, \text{K}\) \[ r = \frac{500}{85} \cdot \frac{280}{200} \] Calculate each term: - \(\frac{500}{85} \approx 58.82\) - \(\frac{280}{200} = .\) So, \[ r = 58.82 \times .14 \approx 8.23 \] **Final compression ratio:** \[ \boxed{r = 8.2} \] (Rounded to one decimal) --- ## 3. **Step 2: Find Cycle Efficiency (\(\eta\))** For the ideal Otto cycle: \[ \eta = 1 - \frac{1}{r^{k-1}} \] Plug in \(r = 8.2\), \(k = 1.4\): \[ r^{k-1} = 8.2^{.4} \] Calculate \(8.2^{.4}\): - \(\ln 8.2 = 2.104\) - \(.4 \times 2.104 = .8416\) - \(e^{.8416} \approx 2.32\) So, \[ \eta = 1 - \frac{1}{2.32} = 1 - .431 = .569 \] or \[ \boxed{56.9\%} \] --- ## 4. **Step 3: Find Exhaust Temperature (\(T_4\))** \(T_4\) is after isentropic expansion (3→4): \[ \frac{T_4}{T_3} = \left(\frac{V_3}{V_4}\right)^{k-1} = \frac{1}{r^{k-1}} \] (because \(V_4/V_3 = r\)) \[ T_4 = T_3 \cdot \frac{1}{r^{k-1}} \] We already calculated \(r^{k-1} = 2.32\): \[ T_4 = 200\, \text{K} \cdot \frac{1}{2.32} \approx 200 \cdot .431 = 862\, \text{K} \] **Final exhaust temperature:** \[ \boxed{T_4 = 862\, \text{K}} \] --- ## 5. **Cycle Sketch: \(P\)-\(v\) and \(T\)-\(s\) Diagrams** ### **Otto Cycle Diagrams** #### P-v Diagram (Pressure vs. Specific Volume) ```plaintext 2 3 *---------* / \ / \ *---------------* 1 4 ``` - 1→2: Isentropic compression (steep up) - 2→3: Constant volume (vertical up) - 3→4: Isentropic expansion (steep down) - 4→1: Constant volume (vertical down) #### T-s Diagram (Temperature vs. Entropy) ```plaintext 3 * / \ / * 2 4 *-----* 1 ``` - 1→2: Isentropic (vertical up) - 2→3: Heat addition (horizontal right) - 3→4: Isentropic (vertical down) - 4→1: Heat rejection (horizontal left) --- ## 6. **Summary Table** | State | \(P\) (kPa) | \(T\) (K) | Process | |-------|-------------|-----------|-----------------------------| | 1 | 85 | 280 | Intake | | 2 | \(P_1 r^k\) | \(T_1 r^{k-1}\) | Isentropic compression | | 3 | 500 | 200 | Constant-volume combustion | | 4 | -- | 862 | Isentropic expansion | --- ## 7. **Final Answers** 1. **Compression Ratio:** \(\boxed{8.2}\) 2. **Thermal Efficiency:** \(\boxed{56.9\%}\) 3. **Exhaust Temperature:** \(\boxed{862\, \text{K}}\) --- ## 8. **Key Equations Used** - \( r = \dfrac{P_3}{P_1} \cdot \dfrac{T_1}{T_3} \) - \( \eta = 1 - \dfrac{1}{r^{k-1}} \) - \( T_4 = T_3 \cdot \dfrac{1}{r^{k-1}} \) --- ## 9. **References** - Moran, M.J., Shapiro, H.N., Fundamentals of Engineering Thermodynamics. - Cengel, Y.A., Boles, M.A., Thermodynamics: An Engineering Approach. --- **If you need the calculation for mass, volumes, or actual \(P_2\), \(P_4\), let me know!**