# Turbine Problem—Step-by-Step Solution Given Data: - **Inlet Pressure, \( P_1 \):** 130 kPa - **Inlet Temperature, \( T_1 \):** 830 K - **Inlet Velocity, \( V_1 \):** 65 m/s - **Inlet Area, \( A_1 \):** 3.5 m² - **Exit Pressure, \( P_2 \):** 85 kPa - **Exit Velocity, \( V_2 \):** 10 m/s - **Actual Turbine Power, \( \dot{W}_{\text{actual}} \):** 38,000 kW - **Heat Transfer Rate, \( \dot{Q} \):** -1500 kJ/s (to surroundings, so negative) - **Neglect Potential Energy Changes** - **Gas Constant for Air, \( R \):** 0.287 kJ/kg-K - **Assume Variable Specific Heats (use \( h(T) \) and \( s(T) \) as needed)** Let's solve each part step by step. --- ## (a) Mass Flow Rate at Inlet and Exit \([kg/s]\) **Thought Process:** Use the ideal gas law to find the density at the inlet and then apply the continuity equation (\( \dot{m} = \rho_1 V_1 A_1 \)). At steady state with one inlet and one exit and negligible accumulation, \( \dot{m}_1 = \dot{m}_2 \). ### Step 1: Find Inlet Density \[ \rho_1 = \frac{P_1}{R T_1} \] - \( P_1 = 130\,\text{kPa} = 130,000\,\text{Pa} \) - \( R = 0.287\,\text{kJ/kg-K} = 287\,\text{J/kg-K} \) - \( T_1 = 830\,\text{K} \) \[ \rho_1 = \frac{130,000}{287 \times 830} = \frac{130,000}{238,210} \approx 0.545\,\text{kg/m}^3 \] ### Step 2: Calculate Mass Flow Rate \[ \dot{m}_1 = \rho_1 V_1 A_1 = 0.545 \times 65 \times 3.5 = 124.2\,\text{kg/s} \] ### Step 3: Mass Flow Rate at Exit At steady state, and assuming negligible leakage: \[ \dot{m}_2 = \dot{m}_1 = 124.2\,\text{kg/s} \] **Final Answers:** - **Inlet Mass Flow Rate:** \(\boxed{124.2\ \text{kg/s}}\) - **Exit Mass Flow Rate:** \(\boxed{124.2\ \text{kg/s}}\) --- ## (b) Actual Exit Enthalpy \([kJ/kg]\) and Temperature \([K]\) **Thought Process:** Apply the steady-flow energy equation including kinetic energy, enthalpy, work, and heat. Solve for \( h_2 \), then use property tables or \( h(T) \) for air to get \( T_2 \). ### Step 1: Steady Flow Energy Equation \[ \dot{W}_\text{shaft} = \dot{m}(h_1 - h_2) + \dot{m}\left(\frac{V_1^2 - V_2^2}{2}\right) + \dot{Q} \] Solve for \( h_2 \): \[ h_2 = h_1 + \frac{V_1^2 - V_2^2}{2} + \frac{\dot{Q} - \dot{W}_\text{shaft}}{\dot{m}} \] ### Step 2: Find \( h_1 \) for \( T_1 = 830\,K \) For air at \( T_1 = 830\,K \), \( h_1 \approx 870.8\,\text{kJ/kg} \) (from air tables). ### Step 3: Plug in Values - \( V_1 = 65\,\text{m/s} \), \( V_2 = 10\,\text{m/s} \) - Kinetic energy change per kg: \[ \frac{V_1^2 - V_2^2}{2} = \frac{65^2 - 10^2}{2} = \frac{4225 - 100}{2} = \frac{4125}{2} = 2062.5\,\text{J/kg} = 2.0625\,\text{kJ/kg} \] - \( \dot{Q} = -1500\,\text{kJ/s} \), \( \dot{W}_\text{shaft} = 38,000\,\text{kW} \), \( \dot{m} = 124.2\,\text{kg/s} \) - \( \frac{\dot{Q} - \dot{W}_\text{shaft}}{\dot{m}} = \frac{-1500 - 38,000}{124.2} = \frac{-39,500}{124.2} = -318.1\,\text{kJ/kg} \) Now, \[ h_2 = 870.8 + 2.0625 - 318.1 = 554.8\,\text{kJ/kg} \] ### Step 4: Find Exit Temperature Look up \( T_2 \) in air tables for \( h_2 \approx 554.8\,\text{kJ/kg} \): For air: - \( h \approx 555\,\text{kJ/kg} \rightarrow T_2 \approx 580\,K \) **Final Answers:** - **Actual Exit Enthalpy:** \(\boxed{554.8\,\text{kJ/kg}}\) - **Actual Exit Temperature:** \(\boxed{580\,\text{K}}\) --- ## (c) Entropy Production \([kJ/K\cdot s]\) & Physical Feasibility **Thought Process:** Apply the entropy balance for a control volume. ### Step 1: Entropy Balance \[ \dot{S}_{\text{gen}} = \dot{m}(s_2 - s_1) - \frac{\dot{Q}}{T_{\text{b}}} \] - \( T_b \): boundary temperature for heat transfer (given as inlet T, 830 K) ### Step 2: Find \( s_1 \) and \( s_2 \) - For \( T_1 = 830\,K \), \( P_1 = 130\,kPa \) - For \( T_2 = 580\,K \), \( P_2 = 85\,kPa \) From air tables: - At 830 K: \( s^0_1 = 2.203\,\text{kJ/kg-K} \) - At 580 K: \( s^0_2 = 1.664\,\text{kJ/kg-K} \) Calculate: \[ s_1 = s^0_1 - R \ln\left(\frac{P_1}{P_0}\right) \] \[ s_2 = s^0_2 - R \ln\left(\frac{P_2}{P_0}\right) \] But since both are on the same reference, the difference is: \[ s_2 - s_1 = (s^0_2 - s^0_1) + R \ln\left(\frac{P_1}{P_2}\right) \] \[ s_2 - s_1 = (1.664 - 2.203) + 0.287 \ln\left(\frac{130}{85}\right) \] \[ = -0.539 + 0.287 \times \ln(1.529) \] \[ \ln(1.529) = 0.424 \] \[ s_2 - s_1 = -0.539 + 0.287 \times 0.424 = -0.539 + 0.122 = -0.417\,\text{kJ/kg-K} \] ### Step 3: Plug Values into Entropy Generation \[ \dot{S}_{\text{gen}} = 124.2 \times (-0.417) - \frac{-1500}{830} \] \[ = -51.78 + 1.807 = -49.97\,\text{kJ/K-s} \] **Interpretation:** Negative entropy production is not physically possible. This means the process as described is not physical; there is an error in the data or assumptions. In reality, entropy generation must be positive due to the second law. **Final Answers:** - **Entropy Production:** \(\boxed{-50.0\,\text{kJ/K-s}}\) (not possible) - **Physical?** No. The turbine is **not physical** because the calculated entropy generation is negative, which violates the second law of thermodynamics. --- ## (d) Ideal Exit Enthalpy \([kJ/kg]\) **Thought Process:** For the ideal (isentropic) process, entropy is constant (\( s_2 = s_1 \)) and the exit pressure is the same as the actual. Find \( h_{2s} \). ### Step 1: Isentropic Expansion Given \( s_1 \), find \( T_{2s} \) such that \[ s^0(T_{2s}) - R \ln\left(\frac{P_2}{P_0}\right) = s_1 \] Or, \[ s^0(T_{2s}) = s_1 + R \ln\left(\frac{P_2}{P_1}\right) \] \[ R \ln\left(\frac{85}{130}\right) = 0.287 \times \ln(0.654) = 0.287 \times (-0.425) = -0.122 \] \[ s^0(T_{2s}) = 2.203 - 0.122 = 2.081\,\text{kJ/kg-K} \] ### Step 2: Find \( T_{2s} \) and \( h_{2s} \) From air tables, \( s^0 = 2.081 \) corresponds to \( T_{2s} \approx 670\,K \). At 670 K, \( h_{2s} \approx 673.8\,\text{kJ/kg} \). **Final Answer:** - **Ideal Exit Enthalpy:** \(\boxed{673.8\,\text{kJ/kg}}\) --- ## (e) Ideal Exit Power \([kW]\) **Thought Process:** Use the energy balance with no irreversibilities, i.e., use isentropic enthalpy drop. \[ \dot{W}_{\text{ideal}} = \dot{m} (h_1 - h_{2s}) + \dot{m} \left(\frac{V_1^2 - V_2^2}{2}\right) + \dot{Q} \] \[ \dot{W}_{\text{ideal}} = 124.2 \times (870.8 - 673.8) + 124.2 \times 2.0625 - 1500 \] \[ = 124.2 \times 197 + 256.2 - 1500 \] \[ = 24,468 + 256.2 - 1500 = 23,224\,\text{kW} \] **Final Answer:** - **Ideal Exit Power:** \(\boxed{23,200\,\text{kW}}\) --- ## (f) Isentropic Efficiency **Thought Process:** Isentropic efficiency for a turbine: \[ \eta_{\text{turbine}} = \frac{\text{Actual Work Output}}{\text{Isentropic Work Output}} \] Note: The actual power output is much higher than the ideal one (which is not possible). \[ \eta_{\text{turbine}} = \frac{38,000}{23,224} = 1.637 \] **Final Answer:** - **Isentropic Efficiency:** \(\boxed{1.64}\) (not possible) --- ## (g) Ideal Exit Cross-sectional Area \([m^2]\) **Thought Process:** Find density at isentropic exit, then use continuity equation. ### Step 1: \( \rho_{2s} = \frac{P_2}{R T_{2s}} \) - \( P_2 = 85,000\,\text{Pa} \), \( R = 287\,\text{J/kg-K} \), \( T_{2s} = 670\,\text{K} \) \[ \rho_{2s} = \frac{85,000}{287 \times 670} = \frac{85,000}{192,290} = 0.442\,\text{kg/m}^3 \] ### Step 2: Area \[ A_{2s} = \frac{\dot{m}}{\rho_{2s} V_2} = \frac{124.2}{0.442 \times 10} = \frac{124.2}{4.42} = 28.1\,\text{m}^2 \] **Final Answer:** - **Ideal Exit Area:** \(\boxed{28.1\,\text{m}^2}\) --- ## (h) Summary