A new system has just started operating in Neon's power plant. An analysis needs to be done to make sure that the cycle works well. The thermal efficiency and the net work output need to be determined.

Assumptions:

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

The Actual Cycle

T-s Diagram of the Actual Rankine Cycle

Saturated Water Temperature Table
Saturated Water Pressure Table
Superheated Steam Table

Model the cycle in the new section of the power plant as a Rankine cycle. The schematic and the T-s Diagram of this cycle is shown on the left.

(1) Determine the work output of the new system

The net work output of the cycle equals the difference between the turbine work output and the pump work input.
      w_net,out = w_turb,out - w_pump,in

The enthalpy at each state needs to be determined first, which can be obtained from the water table.

State 1: Compressed liquid water
       P₁ = 9 kPa      T₁ = 38^oC (given)
       v₁ = 0.001008 m³/kg

State 3: Compressed liquid water
      P₃ = 15.9 MPa  T₁ = 45^oC (given)
      h₃ = 187.8 kJ/kg

State 4: Superheated vapor
      P₄ = 15.2 MPa  T₄ = 625^oC (given)
      h₄ = 3641.8 kJ/kg
      s₄ = 6.7961 kJ/(kg-K)

State 5: Superheated vapor
      P₅ = 15 MPa  T₅ = 600^oC (given)
      h₅ = 3582.3 kJ/kg
      s₅ = 6.6776 kJ/(kg-K)

State 6: Saturated mixture
      P₆ = 10 kPa (given)
      s_6s = s₅ = 6.6776 kJ/(kg-K)
      h_6s= 2114.9 kJ/kg

where s_6s and h_6s are the entropy and enthalpy at state 6 if process 5-6 is isentropic.

Since the isentropic efficiencies of the pump and turbine are given, the turbine work and the pump work can be determined as following:
      

Therefore, the net work output per mass is
      w_net,out = w_turb,out - w_pump,in
            = 1,173.9 - 19.0 = 1,154.9 kJ/kg

The power produced by the new system is determined from
      

(2) Determine the thermal efficiency

Heat input to the cycle from the boiler is determined from
      q_in = h₄ - h₃ = 3,641.8 - 187.8 = 3,454 kJ/kg

The thermal efficiency of the cycle is
      η_th = w_net,out/q_in = 1,154.9/3,454 = 33%

The Ideal Rankine Cycle



T-s Diagram of the
Ideal Rankine Cycle

(3) Determine the thermal efficiency of the ideal Rankine cycle

The properties at each state of the ideal Rankine cycle can be obtained from the water table.

State 1: Saturated water
      P₁ = 10 kPa (given)
      h₁ = 191.83 kJ/kg
      s₁ = 0.6493 kJ/(kg-K)
      v₁ = 0.001010 m³/kg

State 2: Compressed liquid water
      P₂ = 16 MPa (given)
      s₂ = s₁ = 0.6493 kJ/(kg-K)

State 3: Superheated vapor
      T₃ = 600^oC (given)
      P₃ = P₂ = 16 MPa
      h₃ = 3,641.8 kJ/kg
      s₃ = 6.6988 kJ/(kg-K)

State 4: Saturated mixture
      P₄ = P₁ = 10 kPa  
      s₄ = s₃ = 6.6988 kJ/(kg-K)
      h₄ = 2,121.7 kJ/kg

Pump work input:
      w_pump,in = v₁(P₂-P₂)
                   = 0.001010(16,000 - 10) = 16.2 kJ/kg

Turbine work output:
      w_turb,out = h₃ - h₄
                  = 3,641.8 - 2,121.7 = 1,520.1 kJ/kg

Hence, the net work output of the ideal cycle is
      w_net,out = w_turb,out - w_pump,in
                  = 1,520.1 - 16.2 = 1,503.9 kJ/kg

Heat input to the ideal cycle can be determined from
      q_in = h₃ - h₂ = 3,641.8 - 174.4 = 3,467.4 kJ/kg

The thermal efficiency of the cycle is then determined as
      η_th = w_net,out/q_in = 1,503.9/3,467.4 = 43.4%

Compared with the actual Rankine cycle, the ideal cycle has a higher thermal efficiency.