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Elliott Shortcut Selection Method for Steam Turbines  325

              The extraction-to-exhaust portion of this turbine therefore operates
            on steam conditions of 250 psig/600°F/4-in Hga. The TSR (extraction-
            to-exhaust) is 9.35 lb/kWh. The energy available to the extraction-to-
            exhaust section is therefore:

                     ΔH i = (3413 Btu/kWh) ÷ (9.35 lb/kWh) = 365 Btu/lb


            The blade velocity for 35-in nominal diameter staging with a 1-in blade
            height will be:


                             V b =π(35 + 1)(4500)/720 = 706 ft/s

              If all staging is of the Rateau type in this portion of the turbine:


                                  V j = 706/.46 = 1535 ft/s

            ΔH i per stage is therefore:

                                  2         2
                               Vj       1535         2
                       ΔH i =       =         = (6.85) = 47.0 Btu/lb
                                       224
                              224
            The number of Rateau stages in this section would therefore be:
                         Total energy available   365
                                                =      = 7.7 (say 8)
                       Energy removed per stage   47.0
              Our turbine will, therefore, contain one 25-in diameter Curtis stage
            followed by eight 35-in diameter Rateau stages with the extraction
            opening after the Curtis stage.
              Using metric data, we would observe Fig. 15.10 and proceed as fol-
            lows:
              Assume: Shaft power and speed: 18,500 kW at 4500 r/min
                      Steam conditions: 40 bar (gage)/400°C/150 mbar (abs.)
                      Extraction requirements:
                      70,000 kg/h at 17.25 bar (gauge)
              First tabulate the TSR for an absolute inlet pressure of 41.0 bar at 400°C:
              TSR to extraction (p g = 17.25 bar, p a = 18.26 bar):
              h 1 = 3211.8 kJ/kg at s = 6.756 kJ/(kg K)
              From steam table, at the same entropy,
              at p a = 18.0 bar, h = 2991.5 kJ/kg
              at p a = 18.5 bar, h = 2998.2 kJ/kg
            Therefore, at 18.26 bar, h 2 = 2995.0 kJ/kg
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