Q2. Consider an ideal Rankine cycle where the boiler operates at 12.5 MPa and thecondenser operates at 40 kPa. The steam is superheated after the boiler to 600 oC. Theturbine produces 100 MW of shaft work. Determine:a) The mass flow rate of water through the turbine (in kg/s).b) The work required for the pump.Q3. For the process in Q2. Determine:a) The heat input into the boiler (in MW).b) The heat out for the condenser (in MW)c) The thermal efficiency of the cycle (in %).

A2.

a) To determine the mass flow rate of water through the turbine, we first need to find the specific enthalpy at the inlet and outlet of the turbine.

From the steam tables, we find that the specific enthalpy at the inlet of the turbine (state 3) is h3 = 3625.3 kJ/kg (at 12.5 MPa and 600°C). The specific enthalpy at the outlet of the turbine (state 4) can be found by interpolating the steam tables at 40 kPa, which gives h4 = 2724.7 kJ/kg.

The work done by the turbine is the difference in specific enthalpy times the mass flow rate:

100 MW = m_dot*(h3 - h4)

Solving for m_dot gives:

m_dot = 100 MW / (h3 - h4) = 100,000 kW / (3625.3 kJ/kg - 2724.7 kJ/kg) = 47.6 kg/s

b) The work required for the pump can be found using the specific volume at the inlet of the pump (state 1) and the pressure difference across the pump.

From the steam tables, we find that the specific volume at state 1 is v1 = 0.001043 m^3/kg (at 40 kPa). The work done by the pump is:

W_pump = v1*(P2 - P1) = 0.001043 m^3/kg * (12.5 MPa - 40 kPa) = 12.4 kJ/kg

Q3.

a) The heat input into the boiler is the difference in specific enthalpy between the outlet and inlet of the boiler times the mass flow rate:

Q_in = m_dot*(h3 - h2) = 47.6 kg/s * (3625.3 kJ/kg - h2)

We need to find h2, the specific enthalpy at the outlet of the pump. This can be approximated as h2 = h1 + W_pump = 2724.7 kJ/kg + 12.4 kJ/kg = 2737.1 kJ/kg.

So, Q_in = 47.6 kg/s * (3625.3 kJ/kg - 2737.1 kJ/kg) = 423.4 MW

b) The heat out for the condenser is the difference in specific enthalpy between the inlet and outlet of the condenser times the mass flow rate:

Q_out = m_dot*(h4 - h1) = 47.6 kg/s * (2724.7 kJ/kg - h1)

We need to find h1, the specific enthalpy at the inlet of the condenser. This can be approximated as h1 = h4 = 2724.7 kJ/kg.

So, Q_out = 47.6 kg/s * (2724.7 kJ/kg - 2724.7 kJ/kg) = 0 MW

c) The thermal efficiency of the cycle is the net work done divided by the heat input:

η = (W_turbine - W_pump) / Q_in = (100 MW - 12.4 kJ/kg * 47.6 kg/s) / 423.4 MW = 0.235 or 23.5%