Let's break down and **solve the problem step by step**, referencing the steam tables as needed. --- ## **Given Data** - **Steam enters turbine:** - Pressure, \( P_1 = 20 \text{ MPa} \) - Temperature, \( T_1 = 560^\circ \text{C} \) - **Turbine exit:** - Pressure, \( P_2 = .5 \text{ bar} \) - **Condenser exit:** - Pressure, \( P_3 = .4 \text{ bar} \) - Temperature, \( T_3 = 75^\circ \text{C} \) - **Pump exit:** - Pressure, \( P_4 = 20.1 \text{ MPa} \) - **Isentropic efficiencies:** - Turbine: \( \eta_{\text{turbine}} = 81\% = .81 \) - Pump: \( \eta_{\text{pump}} = 85\% = .85 \) - **Cooling water:** - Enters condenser at \( 20^\circ \text{C} \) - Exits condenser at \( 38^\circ \text{C} \) - Mass flow rate: \( \dot{m}_{cw} = 70.7 \text{ kg/s} \) --- ## **Steps** ### **1. Draw the T-s Diagram** - Sketch the Rankine cycle: 1 (turbine inlet) → 2 (turbine exit) → 3 (condenser exit) → 4 (pump exit) → 1. --- ### **2. Find Properties at Each State** #### **State 1: Turbine Inlet (\( P_1 = 20 \) MPa, \( T_1 = 560^\circ \)C)** From **superheated steam tables**: - At \( P = 20 \) MPa, \( T = 560^\circ \)C: - **Enthalpy, \( h_1 \) ≈ 353.9 kJ/kg** - **Entropy, \( s_1 \) ≈ 7.0215 kJ/kg·K** --- #### **State 2: Turbine Exit (\( P_2 = .5 \) bar = .05 MPa, \( s_2 \))** **Isentropic expansion:** - \( s_2' = s_1 \) From **steam tables at .5 bar**: - \( s_f = .5926 \) - \( s_g = 8.3951 \) Since \( s_1 \) is between \( s_f \) and \( s_g \), the steam is **mixture** at exit. **Quality (\( x \)) at .5 bar:** \[ x_2' = \frac{s_1 - s_f}{s_g - s_f} = \frac{7.0215 - .5926}{8.3951 - .5926} = \frac{6.4289}{7.8025} \approx .824 \] **Enthalpy at .5 bar:** - \( h_f = 340.54 \) kJ/kg - \( h_g = 2643. \) kJ/kg So, \[ h_2' = h_f + x_2'(h_g - h_f) = 340.54 + .824 (2643. - 340.54) \] \[ h_2' = 340.54 + .824 \times 2302.46 = 340.54 + 1898.63 = 2239.17~\text{kJ/kg} \] **Actual turbine exit enthalpy:** \[ \eta_{\text{turbine}} = \frac{h_1 - h_2}{h_1 - h_2'} \implies h_2 = h_1 - \eta_{\text{turbine}} (h_1 - h_2') \] \[ h_2 = 353.9 - .81 \times (353.9 - 2239.17) \] \[ h_2 = 353.9 - .81 \times 1291.73 = 353.9 - 1046.30 = 2484.6~\text{kJ/kg} \] --- #### **State 3: Condenser Exit (\( P_3 = .4 \) bar, saturated liquid)** From **steam tables at .4 bar**: - \( h_3 = h_f = 268.41~\text{kJ/kg} \) - \( s_3 = s_f = .3758~\text{kJ/kg·K} \) --- #### **State 4: Pump Exit (\( P_4 = 20.1 \) MPa)** **Isentropic Pump Work:** \[ v_f \text{ at .4 bar} = .00103~\text{m}^3/\text{kg} \] \[ w_{\text{pump, s}} = v (P_4 - P_3) \] \[ = .00103 \times (20100 - .4)~\text{kPa} = .00103 \times 20100 = 20.703~\text{kJ/kg} \] \[ h_4' = h_3 + w_{\text{pump, s}} = 268.41 + 20.703 = 289.11~\text{kJ/kg} \] **Actual pump work:** \[ \eta_{\text{pump}} = \frac{h_4' - h_3}{h_4 - h_3} \implies h_4 = h_3 + \frac{h_4' - h_3}{\eta_{\text{pump}}} \] \[ h_4 = 268.41 + \frac{20.703}{.85} = 268.41 + 24.36 = 292.77~\text{kJ/kg} \] --- ## **3. Mass Flow Rate of Steam (\( \dot{m}_{steam} \))** \[ \dot{Q}_{out} = \dot{m}_{cw} c_p (T_{out} - T_{in}) \] \[ = 70.7 \times 4.18 \times (38 - 20) = 70.7 \times 4.18 \times 18 \] \[ = 70.7 \times 75.24 = 5322~\text{kW} \] \[ \dot{Q}_{out} = \dot{m}_{steam} (h_2 - h_3) \] \[ \implies \dot{m}_{steam} = \frac{5322}{2484.6 - 268.41} = \frac{5322}{2216.19} \approx 2.40~\text{kg/s} \] --- ## **4. Thermal Efficiency** \[ \eta = \frac{W_{net}}{Q_{in}} \] \[ Q_{in} = \dot{m}_{steam} (h_1 - h_4) \] \[ Q_{in} = 2.40 \times (353.9 - 292.77) = 2.40 \times 3238.13 = 7771.5~\text{kW} \] \[ W_{net} = \dot{m}_{steam} \left[ (h_1 - h_2) - (h_4 - h_3) \right] \] \[ = 2.40 \left[ (353.9 - 2484.6) - (292.77 - 268.41) \right] = 2.40 \left[ 1046.3 - 24.36 \right] = 2.40 \times 1021.94 = 2452.7~\text{kW} \] \[ \eta = \frac{2452.7}{7771.5} = .3155 \approx 31.6\% \] --- ## **Summary Table** | State | P (MPa) | T (°C) | h (kJ/kg) | s (kJ/kg·K) | x | |-------|---------|--------|-----------|-------------|-------| | 1 | 20 | 560 | 353.9 | 7.0215 | - | | 2 | .05 | ? | 2484.6 | - | .824 | | 3 | .04 | 75 | 268.41 | .3758 | | | 4 | 20.1 | ? | 292.77 | - | | --- ## **Final Answers** ### **a. Mass flow rate of steam:** \[ \boxed{2.40~\text{kg/s}} \] ### **b. Thermal efficiency:** \[ \boxed{31.6\%} \] --- ### **T-s Diagram** - Draw a standard Rankine cycle with points labeled 1–4. --- **Note:** All enthalpy and entropy values are from typical steam tables and may vary slightly based on the version used. For precise values, refer to your textbook's property tables.