Question 7.11: Problem: Steam enters an insulated turbine at 7 MPa and 500 ...
Problem: Steam enters an insulated turbine at 7 MPa and 500 °C and leaves at 75 kPa with a mass flow rate of 10 kg / s. Assuming that expansion in the turbine is reversible, what is the work output?
Find: Work output of turbine W.
Known: Inlet pressure P_1 = 7 MPa, inlet temperature T_1 = 500 °C, exit pressure P_2 = 75 kPa, mass flow rate \dot{m} = 10 kg /s.
Diagram:
Assumptions: Expansion is reversible, process is adiabatic due to insulated turbine, together make an isentropic process.
Governing Equation: For a turbine \dot{W}=\dot{m}(h_2-h_1)
At the inlet pressure P_1 = 7 MPa the saturation temperature of steam T_{sat} = 285.88 °C (Appendix 8b), and since T_1> T_{sat} the steam is superheated. From the superheated steam tables (Appendix 8c) specific entropy s_1 = 6.7975 kJ / kgK and specific enthalpy h_1 = 3410.3 kJ / kg. We assume that the expansion is adiabatic and reversible, which means that it is isentropic so that s_1 = s_2. It is convenient to show an isentropic process on a T‐s diagram where it appears as a vertical line. The initial state (1) lies in the superheated vapour region. It is not clear as yet whether state (2), which lies on the 75 kPa isobar, lies above or below the vapour dome – whether the exhaust from the turbine is superheated steam or a saturated mixture. From the saturated water tables (Appendix 8b) at P_2 = 75 kPa, specific entropy of fluid s_f = 1.2130 kJ / kgK and gas s_g = 7.4564 kJ / kgK. Since s_f< s_2< s_g, the final state is a saturated vapour–liquid mixture. We can therefore draw the isentropic expansion line on a T‐s diagram:
We can find the quality at state 2:
x_2=\frac{s_2-s_f}{s_g-s_f}=\frac{6.7975 \ kJ/kgK-1.2130 \ kJ/kgK}{7.4564 \ kJ/kgK-1.2130 \ kJ/kgK}=0.894 \ 46.
Then we can solve for the final specific enthalpy,
h_2=h_f+x_2h_{fg} ,
At 75 kPa specific enthalpy of fluid h_f = 384.39 kJ.kg and latent heat of vaporisation h_{fg} = 2278.6 kJ / kg from Appendix 8b, so
h_2= 384.39 \ kJ/kg+0.89446 \times (2278.6 \ kJ/kg)=2422.5 \ kJ/kg, \\ \dot{W}=\dot{m}(h_2-h_1)=10 \ kg/s(2422.5 \ kJ/kg-3410.3 \ kJ/kg), \\ \dot{W}= -9878.0 \ kW=-9.88 \ MW.
The negative sign means that the work is done by the system on the surroundings.
Answer: The work output from the turbine is 9.88 MW.