Let's break down the problem step by step: ### **Given:** - **Engine:** Petrol (Otto cycle) - **Compression ratio, \( r \) = 18:1** - **Inlet conditions:** \( T_1 = 15^\circ C = 288\,K \), \( P_1 = 1\,bar \) - **Maximum temperature = ?** - **\( T_4 = 650^\circ C = 923\,K \)** - **Similar dimensions as Example 10-5(B):** (Assume swept volume, clearance volume, or masses are the same as a previous example; will use symbols for now.) Assume **air-standard cycle** (\( c_v = 0.718\,kJ/kg\,K, \gamma = 1.4 \)). --- ## **Step 1: Temperatures at Key Points** ### **(a) Compression—Find \( T_2 \):** For isentropic compression (1-2): \[ T_2 = T_1 \cdot r^{\gamma-1} \] \[ T_2 = 288 \cdot 18^{0.4} \] Calculate \( 18^{0.4} \): \[ \ln(18) = 2.8904 \implies 0.4 \cdot 2.8904 = 1.1562 \implies e^{1.1562} \approx 3.178 \] \[ T_2 = 288 \times 3.178 \approx 915\,K \] --- ### **(b) Heat Addition—Find \( T_3 \) (Maximum Cycle Temperature):** For air-standard Otto cycle, heat is added at constant volume: \[ \frac{T_3}{T_2} = \frac{T_4}{T_1} \] But **this is not generally true** for the Otto cycle. Let’s recall the process: - **1-2:** Isentropic compression - **2-3:** Heat addition at constant volume \(\rightarrow T_3\) is the **maximum temperature** - **3-4:** Isentropic expansion - **4-1:** Heat rejection at constant volume But: From **isentropic expansion (3-4)**, \[ \frac{T_4}{T_3} = \left(\frac{V_3}{V_4}\right)^{\gamma-1} \] But \( V_3 = V_2 \), \( V_4 = V_1 \), and \( r = \frac{V_1}{V_2} \implies \frac{V_3}{V_4} = \frac{V_2}{V_1} = \frac{1}{r} \) So, \[ \frac{T_4}{T_3} = \left(\frac{1}{r}\right)^{\gamma-1} \implies T_3 = T_4 \cdot r^{\gamma-1} \] \[ T_3 = 923 \times 3.178 \approx 2932\,K \] ## **Answer 1: Maximum Cycle Temperature:** \[ \boxed{2932\,K} \] --- ## **Step 2: Otto Cycle Efficiency** \[ \eta_{Otto} = 1 - \frac{1}{r^{\gamma-1}} \] \[ \eta_{Otto} = 1 - \frac{1}{3.178} = 1 - 0.3146 = 0.685 \text{ or } 68.5\% \] But the problem says "assuming the temperature at point 4 is 650°C", so let's check the **actual efficiency** based on heat supplied and rejected: ### **Heat supplied:** \[ q_{in} = c_v (T_3 - T_2) \] ### **Heat rejected:** \[ q_{out} = c_v (T_4 - T_1) \] ### **Thermal efficiency:** \[ \eta = 1 - \frac{q_{out}}{q_{in}} \] Plug in the values: \[ q_{in} = 0.718 \times (2932 - 915) = 0.718 \times 2017 = 1449\,kJ/kg \] \[ q_{out} = 0.718 \times (923 - 288) = 0.718 \times 635 = 456\,kJ/kg \] \[ \eta = 1 - \frac{456}{1449} = 1 - 0.315 = 0.685 \text{ or } 68.5\% \] So the answer is the same as the air-standard efficiency (as expected). ## **Answer 2: Thermal Efficiency** \[ \boxed{68.5\%} \] --- ## **Step 3: Net Work Output and Work Ratio** Assume **unit mass of air** (1 kg). ### **Net Work Output:** \[ w_{net} = q_{in} - q_{out} = 1449 - 456 = 993\,kJ/kg \] --- ### **Work Ratio** For an Otto cycle, the **work ratio** is usually defined as: \[ \text{Work Ratio} = \frac{\text{Net Work Output}}{\text{Total Heat Supplied}} \] \[ = \frac{993}{1449} = 0.685 \] --- ## **Summary Table** | Quantity | Value | |-------------------------------|----------------| | Maximum temperature (\(T_3\)) | \(2932\,K\) | | Thermal efficiency | \(68.5\%\) | | Net work output | \(993\,kJ/kg\) | | Work ratio | \(0.685\) | --- ## **Final Answers** ### **Maximum temperature in the cycle:** \[ \boxed{2932\,K} \] ### **Thermal efficiency:** \[ \boxed{68.5\%} \] ### **Net work output:** \[ \boxed{993\,kJ/kg} \] ### **Work ratio:** \[ \boxed{0.685} \] --- **If you need the net work output for a specific mass or actual engine dimensions, please provide the displacement or air mass per cycle.**