A steam power plant operates on the simple ideal Rankine cyc

A steam power plant operates on the simple ideal Rankine cycle between the pressure limits of 10 kPa and 10 MPa, with a turbine inlet temperature of 500 degree C. The rate of heat transfer in the boiler is 800 kJ/s. Disregarding the pump work, find the power output of this plant.

Solution

To compute the quality at the turbine exit, we recognize that this exit state is defined by the condenser pressure of 10 kPa and an isentropic process such that s_out = s_in = s(10 MPa, 500^oC) = 6.5995 kJ/kgK. The outlet quality is thus found from the following equation. xout = 0.793

In order to compute the efficiency, we need the enthalpy values at all the state points. Following a conventional Rankine cycle calculation, we find the properties at state one as those of a saturated liquid at the condenser pressure: h₁ = h_f(10 kPa) = 191.81 kJ/kg and v₁ = 0.001010 m³/kg. The isentropic pump work, |w_p| = v₁(P₂ – P₁) where P₂ is the same as the inlet pressure to the turbine, 10 MPa = 10,000 kPa. Thus,

wP1 = 10.09 kJ/kg

We then find h₂ = h₁ + |w_P1| = 191.81 kJ/kg + 10.09 kJ/kg = 201.90 kJ/kg.

h₃ = h(10 MPa, 500^oC) = 3375.1 kJ/kg.             s₃ = s(10 MPa, 500^oC) = 6.5995 kJ/kgK.

As noted above, state 4 is in the mixed region with P₄ = 10 kPa and x₄ = 0.793. We thus find the enthalpy from the quality as h₄ = h_f(P₄ = 10 kPa) + x₄ h_fg(P₄ = 10 kPa) = 191.81 + (0.793)(2392.1 kJ/kg) or h₄ = 2089.7 kJ/kg.

The heat input to the steam generator, q_h = h₃ – h₂ = 3375.1 kJ/kg – 201.90 kJ/kg = 3173.2 kJ/kg

Mass flow : (800 kJ/s)/(3173.2 kg/s) = 0.252 kg/s

The condenser heat rejection, q_l = |h₁ – h₄| = |191.81 kJ/kg – 2089.7 kJ/kg| = 1897.9 kJ/kg.

The net work, w = q_h - |q_L| = 3173.2 kJ/kg| - 1897.9 kJ/kg = 1275.3 kJ/kg.
Power Output = 1275.3*.252 = 321.27 kJ/s

The efficiency = w / q_H = (1275.3 kJ/k ) / (3173.2 kJ/kg) or h = 40.2%.