Question 9.2: Problem: Superheated steam at a pressure of 6 MPa and temper...
Problem: Superheated steam at a pressure of 6 MPa and temperature of 400 °C enters the turbine of a Rankine cycle and leaves at a condenser pressure of 30 kPa. Find the thermal efficiency of the cycle if (a) the turbine is isentropic and (b) the turbine isentropic efficiency is 92%.
Find: Thermal efficiency of the Rankine cycle for (a) an isentropic turbine η_{R,s} (b) a non‐ isentropic turbine η_{R,a}.
Known: Turbine inlet pressure P_2 = P_3 = 6 MPa, turbine inlet temperature T_3 = 400 °C, condenser pressure P_1 = P_4 = 30 kPa, turbine isentropic efficiency η_t = 92%.
Process Diagram: The Rankine cycle on a T‐s diagram (Figure E9.2).
Assumptions: Expansion through the turbine is isentropic.
Governing Equations:
Thermal efficiency of Rankine cycle η_{th,Rankine}=1-\frac{(h_4-h_1)}{(h_3-h_2)}
Isentropic efficiency η_t=\frac{h_3-h_4}{h_3-h_{4,s}}
We need to find the enthalpy at each of the four states marked in the process diagram.
For superheated steam at P_3 = 6 MPa and T_3 = 400 °C (Appendix 8c), specific enthalpy h_3 = 3177.2 kJ / kg and specific entropy s_3 = 6.5408 kJ / kgK.
For saturated water at P_4 = 30 kPa (Appendix 8b), specific enthalpy of liquid h_{4,f} = 289.23 kJ / kg and vapour h_{4,g} = 2625.3 kJ / kg, and specific entropy of liquid s_{4,f} = 0.9439 kJ / kgK and vapour s_{4,g} = 7.7686 kJ / kgK. If expansion through the turbine is isentropic, s_{4s} = s_3 = 6.5408 kJ / kgK, and the quality of the mixture in the turbine would be
x_{4s}=\frac{s_{4s}-s_{4,f}}{s_{4,g}-s_{4,f}}=\frac{6.5408 \ kJ/kgK -0.9439 \ kJ/kgK}{7.7686 \ kJ/kgK-0.9439 \ kJ/kgK}=0.820 \ 09.
The enthalpy at the turbine exit would be
h_{4s}=h_{4,f}+x_{4s}(h_{4,f}-h_{4,f}), \\ h_{4s}=289.23 \ kJ/kg+0.8201 \times (2625.3 \ kJ/kg-289.23 \ kJ/kg)=2205.0 \ kJ/kg.
The enthalpy at the condenser exit is h_1 = h_{4,f} = 289.23 kJ / kg.
The enthalpy at the pump exit (from the solution of) is h_2 = 295.33 kJ / kg.
The isentropic efficiency of a turbine can be used to find the actual turbine enthalpy:
h_4=h_3-η_t(h_3-h_{4s}), \\ h_4= 3177.2 \ kJ/kg -0.92 \times (3177.2 \ kJ/kg-2205.0 \ kJ/kg)=2282.8 \ kJ/kg.
The thermal efficiency of the isentropic Rankine cycle is
η_{th, Rankine}=1-\frac{(h_{4s}-h_1)}{(h_3-h_2)}=1-\frac{2205.0 \ kJ/kg-289.23 \ kJ/kg}{3177.2 \ kJ/kg -295.33 \ kJ/kg}=0.335 \ 23.
The thermal efficiency of the Rankine cycle with a turbine with 92% isentropic efficiency is
η_{th, Rankine}=1-\frac{(h_4-h_1)}{(h_3-h_2)}=1-\frac{2282.8 \ kJ/kg-289.23 \ kJ/kg}{3177.2 \ kJ/kg -295.33 \ kJ/kg}=0.308 \ 24.
Answer: The thermal efficiency of the isentropic Rankine cycle is 33.5%, while the thermal efficiency with a turbine with 92% isentropic efficiency is 30.8 %.