The ideal cycle for vapor power plants is the Rankine cycle. The simple ideal Rankine cycle does not involve any internal irreversibilities and consists of the following four processes:

1-2 Isentropic compression in a pump: Water is pumped from low pressure to high pressure, which is the boiler’s operating pressure. At this stage the fluid is in liquid state so the pump requires little input energy.

2-3 Constant pressure heat addition in a boiler: the high pressure liquid enters a boiler where it is heated at constant pressure by an external heat source to become dry saturated vapor. Steam tables can be used to calculate the input energy.

3-4 Isentropic expansion in a turbine: the dry saturated vapor enters a turbine where it expands isentropically, generating power by rotating the shaft connected to an electric generator. Steam leaves the turbine as a liquid-vapor mixture with high energy levels.

4-1 Constant pressure heat rejection in a condenser: Condensers are heat exchangers that typically use lakes, rivers or the atmosphere where heat is rejected to. At constant pressure, the steam is condensed in a condenser, to become saturated liquid. Then, steam leaves the condenser as saturated liquid and enters the pump, completing the cycle.

## Energy analysis of the Simple Ideal Rankine Cycle

Pump, boiler, turbine and condenser are the four devices of an ideal Rankine cycle and operate in a steady-flow regime. The kinetic and potential energy variations of the steam are so small, when compared to the work and heat transfer, and therefore they are neglected.

The pump and the turbine are isentropic which means that no entropy is generated, maximizing the work output. The boiler and the condenser do not involve any work. The conservation of energy for each device can be expressed as:

Pump (q = 0): w_{pump,in} = h_{2} – h_{1 }or w_{pump,in} = v(P2 – P1) ; h1 = h_{f @ P1} and v_{1} = v_{f @ P1}

Boiler (w = 0): q_{in} = h_{3} – h_{2}

Turbine (q = 0): w_{turb,out} = h_{3 }– h_{4}

Condenser ( w = 0): q_{out} = h_{4} – h_{1}

The thermal efficiency of the Rankine cycle is determined from:

## Example

Consider a steam power plant operating on the simple ideal Rankine cycle. Steam enters the turbine at 3 MPa and 350°C and is condensed in the condenser at a pressure of 75 kPa. Determine the thermal efficiency of this cycle.

This example was solved using Engineering Equation Solver, or just EES, but steam charts can be used also.

"State 1" P1=75 [kPa] x1=0 h1=Enthalpy(Steam;x=x1;P=P1) s1=Entropy(Steam;x=x1;P=P1) v1=Volume(Steam;x=x1;P=P1) "State 2" P2=3000[kPa] s2=s1 h2=Enthalpy(Steam;s=s2;P=P2) "State 3" P3=3000 [kPa] T3=350 [C] h3=Enthalpy(Steam;T=T3;P=P3) s3=Entropy(Steam;T=T3;P=P3) "State 4" P4=75 [kPa] s4=s3 h4=Enthalpy(Steam;s=s4;P=P4) x4=Quality(Steam;P=P4;s=s4) "Pump work" w_pump=v1*(P2-P1) "Turbine work" w_turb=h3-h4 "Boiler work" q_in=h3-h2 "Condenser work" q_out=h4-h1 "Thermal efficiency" eta_th=(w_turb-w_pump)/q_in

The thermal efficiency of the simple Rankine Cycle in the example is 0,26 or 26%.