A cogeneration power plant is used to generate power as well as supply heat for a district heating system. The mass flow in the cycle and the utilization factor are needed to be determined.

Assumptions:

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

The Schematic of the Cogeneration Power Plant an d the T-s Diagram of the Ideal Cogeneration Rankine Cycle

Model the cycle as an ideal cogeneration Rankine cycle. The schematic of the cogeneration power plant and its T-s diagram are shown on the left.     

(1) Determine the mass flow rate in the cycle

Assume the mass flow rate in the cycle is , then the mass flow rate of steam extracted from the turbine to the heat exchanger is and the mass flow rate of steam through the condenser is , where y is the fraction of steam extracted to the heat exchanger per kg of mass flowing through the boiler.

The community needs 1,100 kW heat from the power plant. Since the heat exchanger has a 85% efficiency, the energy balance of the heat exchanger is,

Also, the net work output from the cycle is 5,500 kW.

All the enthalpies at state 1 to state 6 can be obtained from the water tables. Then solving the above equations simultaneously can get the mass flow rate .

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

State 2: compressed water
      P₂ = 16 MPa (given)
      w_pump,in = v₁ (P₂ - P₁) = 16.15 kJ/kg
      h₂ = h₁ + w_pump,in = 207.98 kJ/kg

State 4: superheated vapor
      P₄ = 16 MPa (given)
      T₄ = 600^oC (given)
      h₄ = 3569.8 kJ/kg
      s₄= 6.6988 kJ/(kg-K)

State 5: superheated vapor
      P₅ = 5 MPa (given)
      s₅ =s₄= 6.6988 kJ/(kg-K)
      h₅ = 3222.44 kJ/kg

State 6: saturated mixture
      P₆ = 10 kPa (given)
      s₆ =s₄= 6.6988 kJ/(kg-K)
      x = (s - s_{f@10 kPa})/s_{fg@10 kPa} = 80.7%
      h₆ = h_{f@10 kPa}+ xh_{fg@10 kPa} = 2121.67 kJ/kg

State 7: saturated water
      P₇ = 5 MPa (given)
      h₇ = 1153.36 kJ/kg
      v₇ = 0.0012849 m³/kg

State 8: compressed water
      P₈ = 16 MPa (given)
      w_pump,in = v₇(P₈ - P₇) = 14.13 kJ/kg
      h₈ = h₇ + w_pump,in = 1167.49 kJ/kg

Substituting the enthalpies to the above equations and solving for , yields,

       = 4.24 kg/s
      = 0.635 kg/s
      y = 0.15 kg/kg

(2) the utilization factor of the cogeneration plant

The definition of the utilization factor of the cogeneration plant is

Net work output from the power plant is given as 5,500 kW. The process heat is:

Before determine the total heat input, the enthalpy at state 3 needs to be obtained first. Water at state 2 and state 8 mixes in the mixing chamber at constant pressure. The energy balance of the mixing chamber is

The total heat input to the cycle is

Hence, the utilization factor of the cogeneration plant is

      ε_u = (5,500+1,294.1)/(13,644) = 49.8%