2. A BWR system with a one-stage moisture separation is shown in Fig. 0.2. The conditions in Table 0.1 may be used. WT2 High pressure turbine Reactor 2 9 Feedwater pump WP2 WTI 3 Low pressure turbine 4 Moisture separator Condenser OFWH 7 6 WP1 Main condensate pump Figure 0.2: Schematic of BWR Plant Table 0.1: State Conditions for Problem 2 Points p (kPa) Condition 1 6,890 Saturated vapor 2 1,380 3 1,380 Saturated vapor 4 1,380 Saturated liquid 5 6.89 6 6.89 Saturated liquid 7 1,380 8 1,380 9 6,890 Turbine isentropic efficiency = 90% Pump isentropic efficiency = 85% Environment temperature = 30C 2 1. Calculate the cycle thermal efficiency. 2. Recalculate the cycle thermal efficiency, assuming that the pumps and turbines have isentropic efficiency of 100%. Thermodynamic properties of water and steam: p = 6,890 kPa: Tsat = 2,774.05 kJ/kg; 284.76 C; h = 1,261.97 kJ/kg; hg Sf 3.11247 kJ/(kg-K); sg = 5.82273 kJ/(kg-K) p = 1,380 kPa: Tsat = 194.37C; h = 826.96 kJ/kg; hg = 2,788.39 kJ/kg; sf = 2.27714 kJ/(kg-K); sg = 6.47256 kJ/(kg-K) p = 6.89 kPa: Tsat 38.71 C; hf = 162.12 kJ/kg; hg = 2,571.19 kJ/kg; sf = 0.555081 kJ/(kg-K); sg = 8.28006 kJ/(kg-K); v = 1.00739 10 m/kg p = 1,380 kPa and h = 256.67 kJ/kg (or similar value) (Subcooled liquid water): vf = 1.0170710-3 m/kg p = 1,380 kPa and h = 266.27 kJ/kg (or similar value) (Subcooled liquid water): vf = 1.0183210-3 m/kg Note: This problem is modified from Prob. 6.3 in the text.