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
Question
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
Solution
To solve this problem, we need to make use of the steam tables and the principles of the Rankine cycle.
a) The mass flow rate of water through the turbine (in kg/s).
First, we need to find the enthalpy at the exit of the turbine. From the steam tables, we can find that the enthalpy at the entrance of the turbine (state 3) is 3625.3 kJ/kg (at 12.5 MPa and 600 oC). The enthalpy at the exit of the turbine (state 4) can be approximated as the enthalpy of saturated steam at 40 kPa, which is 2574.7 kJ/kg.
The work done by the turbine is the difference in enthalpy between the entrance and exit of the turbine, which is W_turbine = h3 - h4 = 3625.3 kJ/kg - 2574.7 kJ/kg = 1050.6 kJ/kg.
The power output of the turbine is given as 100 MW, or 100,000 kJ/s. Therefore, the mass flow rate of the water through the turbine can be found by dividing the power output by the work done per kg of steam:
m_dot = W_turbine / (h3 - h4) = 100,000 kJ/s / 1050.6 kJ/kg = 95.1 kg/s.
b) The work required for the pump.
The work required for the pump is the difference in enthalpy between the exit and entrance of the pump. The enthalpy at the entrance of the pump (state 1) can be approximated as the enthalpy of saturated liquid at 40 kPa, which is 167.57 kJ/kg. The enthalpy at the exit of the pump (state 2) can be found using the formula h2 = h1 + v1*(P2-P1), where v1 is the specific volume of the liquid at state 1 (0.001010 m^3/kg), P2 is the pressure at the exit of the pump (12.5 MPa), and P1 is the pressure at the entrance of the pump (40 kPa).
Substituting these values, we get h2 = 167.57 kJ/kg + 0.001010 m^3/kg * (12.5 MPa - 40 kPa) = 167.57 kJ/kg + 12.46 kJ/kg = 180.03 kJ/kg.
Therefore, the work required for the pump is W_pump = h2 - h1 = 180.03 kJ/kg - 167.57 kJ/kg = 12.46 kJ/kg.
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