# Atkinson Cycle: Step-by-Step Solution ## Given Data - **State 1 Pressure \(P_1\):** 150 kPa - **State 1 Temperature \(T_1\):** 300 K - **Compression Ratio \(r_c\):** 9 - **Expansion Ratio \(r_e\):** 14 - **Air-Standard (Low-Temperature) Assumption** - **Required:** Heat added during combustion (\(Q_{in}\)) --- ## 1. Atkinson Cycle Overview The Atkinson cycle consists of 4 processes (using air-standard assumptions): 1. **1→2:** Isentropic Compression (\(V_1 \to V_2\), \(T_1 \to T_2\)) 2. **2→3:** Constant-Volume Heat Addition (\(V_2 = V_3\), \(T_2 \to T_3\)) 3. **3→4:** Isentropic Expansion (\(V_3 \to V_4\), \(T_3 \to T_4\)) 4. **4→1:** Constant-Pressure Heat Rejection (\(P_4 = P_1\), \(T_4 \to T_1\)) --- ## 2. State Points and Volume Relations Let: - \(V_1\): Initial volume - \(V_2\): Volume after compression - \(V_3\): Volume after heat addition - \(V_4\): Volume after expansion ### Compression Ratio \[ r_c = \frac{V_1}{V_2} = 9 \] ### Expansion Ratio \[ r_e = \frac{V_4}{V_3} = 14 \] - In Atkinson cycle, **expansion > compression** (\(V_4 > V_1\)), so - \(V_4 = r_e \cdot V_3\) ### Constant-Volume Heat Addition \[ V_2 = V_3 \] --- ## 3. Isentropic Relations (for Air, \(\gamma \approx 1.4\)) ### 3.1. From 1→2 (Isentropic Compression) \[ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1} = r_c^{\gamma-1} \] \[ T_2 = T_1 \cdot r_c^{\gamma-1} \] ### 3.2. From 3→4 (Isentropic Expansion) \[ \frac{T_4}{T_3} = \left(\frac{V_3}{V_4}\right)^{\gamma-1} = \left(\frac{1}{r_e}\right)^{\gamma-1} \] \[ T_4 = T_3 \cdot r_e^{-(\gamma-1)} \] --- ## 4. Finding Temperatures at Key Points ### 4.1. \(T_2\) (After Compression) \[ T_2 = T_1 \cdot r_c^{\gamma-1} \] \[ T_2 = 300 \cdot 9^{.4} \] Calculate \(9^{.4}\): \[ \ln(9) = 2.1972 \] \[ .4 \times 2.1972 = .87888 \] \[ e^{.87888} \approx 2.407 \] \[ T_2 = 300 \times 2.407 \approx 722.1\, \text{K} \] ### 4.2. \(T_3\) (After Heat Addition) Since heat is added at constant volume, let’s denote \(Q_{in}\) as the heat added. \[ Q_{in} = m c_v (T_3 - T_2) \] But we need to relate \(T_3\) to known quantities. To do that, we need to find a relation using the expansion ratio. ### 4.3. \(T_4\) (After Expansion) \[ T_4 = T_3 \cdot r_e^{-(\gamma-1)} \] \[ T_4 = T_3 \cdot 14^{-.4} \] Calculate \(14^{-.4}\): \[ \ln(14) = 2.6391 \] \[ -.4 \times 2.6391 = -1.0556 \] \[ e^{-1.0556} \approx .3483 \] \[ T_4 = T_3 \times .3483 \] --- ## 5. Closing the Loop (Using Constant Pressure Process 4→1) From state 4 to 1, **constant pressure**: \[ \frac{T_1}{V_1} = \frac{T_4}{V_4} \] \[ T_1 V_4 = T_4 V_1 \] \[ \frac{V_4}{V_1} = \frac{T_4}{T_1} \] But, \(V_4 = r_e V_3\), \(V_1 = r_c V_2 = r_c V_3\) (since \(V_2 = V_3\)): \[ \frac{V_4}{V_1} = \frac{r_e V_3}{r_c V_3} = \frac{r_e}{r_c} \] Thus, \[ \frac{T_4}{T_1} = \frac{r_e}{r_c} \] \[ T_4 = T_1 \cdot \frac{r_e}{r_c} \] \[ T_4 = 300 \cdot \frac{14}{9} \approx 466.7 \, \text{K} \] Recall from above: \[ T_4 = T_3 \times .3483 \] Set equal: \[ T_3 \times .3483 = 466.7 \] \[ T_3 = \frac{466.7}{.3483} \approx 1339.9\, \text{K} \] --- ## 6. Calculate Heat Added For **unit mass** of air (\(m = 1\)), with \(c_v = .718\, \text{kJ/kg K}\): \[ Q_{in} = c_v (T_3 - T_2) \] \[ Q_{in} = .718 \cdot (1339.9 - 722.1) \] \[ Q_{in} = .718 \cdot 617.8 \] \[ Q_{in} \approx 443.6\, \text{kJ/kg} \] --- ## 7. Final Answer ### **The amount of heat addition required for combustion is:** \[ \boxed{443.6\ \text{ kJ/kg}} \] --- ## Summary Table | State | Temp (K) | Volume Relation | |-------|----------|---------------------------| | 1 | 300 | \(V_1\) | | 2 | 722.1 | \(V_2 = V_1/9\) | | 3 | 1339.9 | \(V_3 = V_2\) | | 4 | 466.7 | \(V_4 = 14 V_3\) | --- ## Key Equations - \(T_2 = T_1 \cdot r_c^{\gamma - 1}\) - \(T_4 = T_3 \cdot r_e^{-(\gamma - 1)}\) - \(T_4 = T_1 \cdot \frac{r_e}{r_c}\) - \(Q_{in} = c_v (T_3 - T_2)\) --- > **If you need a detailed \(P-V\) or \(T-S\) diagram, let me know!**