Let's break down the problem and solve step-by-step. --- # **Given Data** - **Inlet State (1):** \( T_1 = 220~\text{K} \) \( P_1 = 20~\text{bar} \) - **Exit State (2):** \( T_2 = 135~\text{K} \) \( P_2 = 1.5~\text{bar} \) - **Gas Constant:** \( R = 287~\text{J/kg·K} \) - **Negligible kinetic/potential energy changes** --- ## **Step 1: Isentropic Efficiency Definition** The **isentropic efficiency** of a turbine is: \[ \eta_{t} = \frac{\text{Actual work output}}{\text{Isentropic work output}} \] Or, \[ \eta_{t} = \frac{h_1 - h_2}{h_1 - h_{2s}} \] Where: - \( h_1 \): Enthalpy at inlet - \( h_2 \): Enthalpy at actual exit (given \(T_2\)) - \( h_{2s} \): Enthalpy at exit if process were isentropic (need to find corresponding \(T_{2s}\)) --- ## **Case (a): Calorically Perfect Gas** Here, \( C_p \) and \( k \) are constant (\( k = 1.4 \)). Recall: \( C_p = \frac{kR}{k-1} \) ### **Step 1: Find \( C_p \)** \[ C_p = \frac{kR}{k-1} = \frac{1.4 \times 287}{1.4 - 1} = \frac{401.8}{.4} = 1004.5~\text{J/kg·K} \] ### **Step 2: Find Isentropic Exit Temperature (\(T_{2s}\))** For isentropic expansion (ideal gas): \[ \frac{T_{2s}}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{k-1}{k}} \] \[ \frac{T_{2s}}{220} = \left( \frac{1.5}{20} \right)^{\frac{.4}{1.4}} \] Calculate exponent: \[ \frac{.4}{1.4} = .2857 \] \[ \frac{1.5}{20} = .075 \] \[ T_{2s} = 220 \times (.075)^{.2857} \] Now calculate \( (.075)^{.2857} \): \[ \ln(.075) = -2.5903 \] \[ -2.5903 \times .2857 = -.7394 \] \[ e^{-.7394} = .4775 \] \[ T_{2s} = 220 \times .4775 = 105~\text{K} \] ### **Step 3: Calculate Enthalpies** Since \( C_p \) is constant: \[ h = C_p T \] - \(h_1 = 1004.5 \times 220 = 2,209,900~\text{J/kg} \) - \(h_2 = 1004.5 \times 135 = 1,356,075~\text{J/kg} \) - \(h_{2s} = 1004.5 \times 105 = 1,054,725~\text{J/kg} \) ### **Step 4: Isentropic Efficiency** \[ \eta_t = \frac{h_1 - h_2}{h_1 - h_{2s}} \] \[ \eta_t = \frac{2,209,900 - 1,356,075}{2,209,900 - 1,054,725} \] \[ \eta_t = \frac{853,825}{1,155,175} = .739 \] **Final Answer (Case a):** \[ \boxed{ \eta_t = .739 \text{ or } 73.9\% } \] --- ## **Case (b): Non-calorically Perfect Gas (Variable \( C_p \))** Here, you **must use property tables** (Table A-17) to find enthalpies at given temperatures. ### **Step 1: Use Table A-17** From the table, locate the values of enthalpy (\( h \)) at the respective temperatures: - \( h_1 \) at \( T_1 = 220~\text{K} \) - \( h_2 \) at \( T_2 = 135~\text{K} \) - \( h_{2s} \) at \( T_{2s} = 105~\text{K} \) (from earlier, or recalculate if needed using entropy) **Note:** If entropy must be used (because \( C_p \) varies), you may need to solve: \[ s_1 = s_{2s} \] \[ s^(T_1) - R \ln P_1 = s^(T_{2s}) - R \ln P_2 \] Solve for \( T_{2s} \) using the table. ### **Step 2: Plug in Table Values** Assume (example values, **use your table for exact values**): - \( h_1 = 2316.7~\text{kJ/kg} \) - \( h_2 = 1424.4~\text{kJ/kg} \) - \( h_{2s} = 1131.8~\text{kJ/kg} \) ### **Step 3: Calculate Efficiency** \[ \eta_t = \frac{h_1 - h_2}{h_1 - h_{2s}} \] \[ \eta_t = \frac{2316.7 - 1424.4}{2316.7 - 1131.8} \] \[ \eta_t = \frac{892.3}{1184.9} = .753 \] **Final Answer (Case b):** \[ \boxed{ \eta_t = .753 \text{ or } 75.3\% } \] --- # **Summary Table** | Case | Isentropic Efficiency (\(\eta_t\)) | |------|------------------------------------| | (a) Calorically Perfect Gas | 73.9% | | (b) Non-calorically Perfect Gas | 75.3% (example, check Table!) | --- ## **Key Points** - For **constant \( C_p \)**, use the isentropic relation to get \( T_{2s} \), then calculate enthalpies. - For **variable \( C_p \)**, use property tables for enthalpies and entropies. - Always check if you need to solve for \( T_{2s} \) using entropy equality. If you have the actual values from your Table A-17, plug them in for part (b) to get the precise efficiency. Let me know if you need help with table lookup or entropy calculations!