Let's solve the Rankine cycle problem step by step: --- ### **Given:** - Rankine cycle for an **ideal gas** (not steam!) - Four **reversible processes**: 1. **1→2:** Isobaric expansion (constant pressure) 2. **2→3:** Adiabatic expansion (isentropic) 3. **3→4:** Isobaric compression (constant pressure) 4. **4→1:** Adiabatic compression (isentropic) - Need to analyze the cycle: --- ## **a) Draw the (p, V) diagram for the Rankine cycle for an ideal gas** ### **Steps:** 1. **Isobaric expansion (1→2):** - Pressure constant, volume increases. 2. **Adiabatic expansion (2→3):** - Pressure and volume both decrease, but entropy stays constant. 3. **Isobaric compression (3→4):** - Pressure constant (lower than 1→2), volume decreases. 4. **Adiabatic compression (4→1):** - Pressure and volume both increase, entropy stays constant. #### **Diagram** - The (p, V) diagram forms a closed loop: - **1→2:** horizontal line (constant p, V↑) - **2→3:** curve down left (adiabatic, V↑, p↓) - **3→4:** horizontal line (constant p, V↓) - **4→1:** curve up right (adiabatic, V↓, p↑) --- ## **b) Work performed in each process and per cycle** ### **Work for each process:** #### **1→2 (Isobaric expansion):** \[ W_{12} = P_1 (V_2 - V_1) \] #### **2→3 (Adiabatic expansion):** \[ W_{23} = \int_{V_2}^{V_3} P \, dV \] For ideal gas and adiabatic process: \[ PV^\gamma = \text{constant} \] \[ W_{23} = \frac{P_2 V_2 - P_3 V_3}{\gamma - 1} \] #### **3→4 (Isobaric compression):** \[ W_{34} = P_3 (V_4 - V_3) \] #### **4→1 (Adiabatic compression):** \[ W_{41} = \int_{V_4}^{V_1} P \, dV = \frac{P_4 V_4 - P_1 V_1}{\gamma - 1} \] ### **Net work per cycle:** \[ W_{cycle} = W_{12} + W_{23} + W_{34} + W_{41} \] #### **In terms of enthalpies:** - For an ideal gas, work (and heat) can be related to enthalpy changes for isobaric and adiabatic processes: - Isobaric: \( Q = H_2 - H_1 \) - Adiabatic: \( Q = 0 \), \( \Delta H = C_p(T_{end} - T_{start}) \) --- ## **c) Heat provided by the hot reservoir (Q⁺)** Heat input is during **isobaric expansion (1→2):** \[ Q_{in} = Q_{12} = H_2 - H_1 \] Heat rejected (3→4): \[ Q_{out} = Q_{34} = H_4 - H_3 \] So, \[ Q^+ = Q_{12} = H_2 - H_1 \] --- ## **d) Efficiency of the Rankine cycle** \[ \eta_{R} = \frac{W_{cycle}}{Q^+} \] But, \[ W_{cycle} = (H_2 - H_1) + (H_3 - H_2) + (H_4 - H_3) + (H_1 - H_4) \] \[ W_{cycle} = (H_2 - H_1) + (H_4 - H_3) + (H_3 - H_2) + (H_1 - H_4) \] But since adiabatic steps have \( Q = 0 \), only net enthalpy difference over the isobaric legs matter. So, \[ \eta_{R} = \frac{(H_2 - H_1) + (H_4 - H_3)}{H_2 - H_1} \] But since \( H_4 < H_3 \), \( H_4 - H_3 \) is negative (work done out), so: \[ \eta_{R} = 1 - \frac{Q_{out}}{Q_{in}} = 1 - \frac{H_4 - H_3}{H_2 - H_1} \] --- ### **Final Answers Summarized** #### **a) (p, V) diagram:** Draw a rectangle with two isobaric and two adiabatic curves as sides. #### **b) Work per process:** - \( W_{12} = P_1(V_2 - V_1) \) - \( W_{23} = \frac{P_2V_2 - P_3V_3}{\gamma-1} \) - \( W_{34} = P_3(V_4 - V_3) \) - \( W_{41} = \frac{P_4V_4 - P_1V_1}{\gamma-1} \) #### **c) Heat input:** - \( Q^+ = Q_{12} = H_2 - H_1 \) #### **d) Efficiency:** \[ \boxed{ \eta_{R} = 1 - \frac{H_4 - H_3}{H_2 - H_1} } \] Where \( H_1, H_2, H_3, H_4 \) are the enthalpies at states 1, 2, 3, 4 respectively. --- **If you need further numerical examples or step-wise calculation with values, let me know!**