A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one open feedwater heater, one closed feedwater heater, and one reheater. The fractions of stream extracted from turbines and the thermal efficiency of the cycle are to be determined.

Assumptions:

• All the components in the cycle operate at steady state.
• Kinetic and potential energy changes are negligible.

(1) Determine the fraction of steam extracted from the turbines

The enthalpies at various states and the pump work per unit mass of fluid flowing through them can be determined by using the water tables.

State1: saturated water
      P₁ = 10 kPa ( given)
      h₁ = 191.83 kJ/kg
      v₁ = 0.00101 m³/kg

State 2: compressed water
      P₂ = 1 MPa (given)
      w_pump,in = v₁ (P₂ - P₁)
                   = 0.00101(1,000 - 10) = 1.0 kJ/kg
      h₂ = h₁ + w_pump,in = 192.83 kJ/kg

State 3: saturated water
      P₃ = 1 MPa (given)
      h₃ = 762.81 kJ/kg
      v₃ = 0.00113 m³/kg

State 4: compressed water
      P₄ = 16 MPa (given)
      w_pump,in = v₃ (P₄ - P₃)
                   = 0.00113(16,000 - 1,000) = 16.9 kJ/kg
      h₄ = h₃ + w_pump,in = 762.81+ 16.9 = 779.71 kJ/kg

State 5: saturated water
      P₅ = 16 MPa (given)
      h₅ = 1,650.1 kJ/kg

State 6: superheated vapor
      P₆ = 16 MPa (given)
      T₆ = 600^oC (given)
      h₆ = 3,569.8 kJ/kg
      s₆= 6.6988 kJ/(kg-K)

State7: superheated vapor
      P₇ = 5 MPa (given)
      s₇ =s₆= 6.6988 kJ/(kg-K)
      h₇ = 3,222.4 kJ/kg

State 8: superheated vapor
      P₈ = 5 MPa (given)
      T₈ = 600^oC (given)
      h₈ = 3,665.6 kJ/kg
      s₈= 7.2731 kJ/(kg-K)

State 9: superheated vapor
      P₉ = 1 MPa (given)
      s₉ =s₈= 7.2731 kJ/(kg-K)
      h₉ = 3,138.1 kJ/kg

State 10: saturated mixture
      P₁₀ = 10 kPa (given)
      s₁₀ =s₈= 7.2731 kJ/(kg-K)
      x₈ = (s₈ - s_{f@10 kPa})/s_{fg@10 kPa} = 88.3%
      h₁₀ = h_{f@10 kPa}+ x₈h_{fg@10 kPa} = 2,304.76 kJ/kg

State 11: saturated water
      P₁₁ = 5 MPa (given)
      h₁₁ = 1,154.2 kJ/kg

State 12: The trap is an throttling device
      h₁₂ = h₁₁ = 1,154.2 kJ/kg

Open Feedwater Heater

Closed Feedwater Heater

Assume y kg steam is extracted from the high-pressure turbine per kg mass of water flowing through the boiler. Also assume z kg steam is extracted from the low-pressure turbine per kg mass of water flowing through the boiler. Then y and z can be determined from the energy balance of the closed and open feedwater heaters.

Closed feedwater heater:
      y(h₇ - h₁₁) = 1(h₅ - h₄)
      y = (1650.1 - 779.71)/(3222.4 -1154.2 )
         = 0.42 kg/kg

Open feedwater heater:
      (1)h₃ = (z)h₉ + (1 - y - z)h₂ + (y)h₁₂
      z =0.056 kg/kg

(2) Determine the thermal efficiency of the cycle

The power output from the high-pressure and low-pressure turbine for 1 kg of mass flowing through the boiler is:

      w_turb,out = (1)(h₆ - h₇) + (1 - y)(h₈ - h₉)
                      + (1 -y - z )(h₉ - h₁₀)
                  = (1)( 3,569.8 - 3,222.4) +
                         (1 -0.42)(3,665.6 - 3,138.1)+
                         (1 - 0.42 - 0.056)(3,138.1 - 2,304.7)
                  = 1,090.0 kJ/kg

The pump input works have been determined in (1) as

      w_pump,in = (1 - y -z)1.0 + 16.9
                   = 17.4 kJ/kg

So the total net work output fro 1 kg mass of water flowing through the boiler is
      w_net,out = w_turb,out - w_pump,in
                 = 1,090.0 - 17.4 = 1,072.6 kJ/kg

The total heat input to the cycle equals the sum of the heat input in the boiler and the heat input from the reheater.

      q_in = (1)(h₆ - h₅) + (1 - y)(h₈ - h₇)
           = (1)(3,569.8 - 1,650.1) +
             (1 - 0.42)(3,665.6 - 3,222.44)
           = 2,176.7 kJ/kg

The thermal efficiency is
η_th = w_net,out/q_in = 1,072.6/ 2,176.7 = 49.3%