The steam enter the turbine (3) of the steam plant at 8 MPa and 500°C. The steam extends in an isentropic process and leaves the turbine (4) at 200 kPa, then enters and leaves the condenser at constant pressure (You may treat the steam as a saturated liquid when it leaves the condenser). If the work of the pump is 7 kj/kg and both the kinetic and the potential energies are negligible. Compute:
1- The states of the steam when it enters and leaves the turbine.
2- The work done by the turbine in kj/kg.
3- The net power of the plant if the steam flows with 30 kg/s.
4- The amount of heat transfer of both the boiler and the condenser.
5- The efficiency of the cycle.
6- The two processes (3-4, 4-1) plotted on T-s diagram.
efficiency=Wnet/Qin * 100
∑Qnet = ∑ Wnet
The Attempt at a Solution
1-[/B] P3=8 MPa, T3=500°C
Using Tables: T3>TS ----> The inlet state is superheated vapor.
P4=200 kPa, s3=s4=6.7240 kg/kgK (Isentropic Process)
Using Tables: sf < s4 < sg ---> The exit state is mixture of saturated liquid and vapor.
2- Since q3-4=0 in the turbine, and both KE and PE are negligible, we get:
W3-4=h3-h4 (The first law of thermodynamics, h is the enthalpy)
h3= 3398.3 kj/kg, h4=2545.86 kj/kg
3- Wnet=W3-4+Wpump=652.44+7=659.44 kj/kg
W`net=Wnet * 30 kg/s = 19783.2 kj/s
4- Since W4-1=0 in the condenser, then q4-1=h1-h4
h1=504.7 kj/kg, h4=2545.86 kj/kg
Since qnet=Wnet --> The law of closed cycle
5- eff. = Wnet/qboiler * 100 = 24.41%