Question 9.1: Problem: Saturated steam at a pressure of 6 MPa enters the t...
Problem: Saturated steam at a pressure of 6 MPa enters the turbine of a Rankine cycle and leaves at a condenser pressure of 30 kPa. What is the back work ratio? Find the thermal efficiency of the cycle and compare it to that of a Carnot cycle operating between the same temperatures.
Find: Back work ratio bwr of the Rankine cycle, thermal efficiency of the Rankine cycle η_{th,Rankine}, thermal efficiency of a Carnot cycle η_{th,Carnot} operating between the same temperatures.
Known: Turbine inlet pressure P_2 = P_3 = 6 MPa, condenser outlet pressure P_1 = P_4 = 30 kPa.
Process Diagram: The Rankine cycle on a T‐s diagram (Figure E9.1).
Assumptions: Expansion through the turbine is isentropic so ∆S_{34} = 0.
Governing Equations:
Back work ratio of Rankine cycle bwr=\frac{(h_2-h_1)}{(h_3-h_4)}
Thermal efficiency of Rankine cycle η_{th , Rankine}=1-\frac{(h_4-h_1)}{(h_3-h_2)}
We need to find the enthalpy at each of the four states marked in the process diagram.
For saturated steam at P_3 = 6 MPa (Appendix 8b), specific enthalpy of vapour h_3 = h_g = 2784.3 kJ / kg and specific entropy of vapour s_3 = s_g = 5.8892 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. Expansion through the turbine is isentropic, so that s_4 = s_3 = 5.8892 kJ / kgK; so the quality of the mixture in state 4 can be found:
x_4= \frac{s_4-s_{4,f}}{s_{4,g}-s_{4,f}}= \frac{5.8892 \ kJ/kgK-0.9439 \ kJ/kgK}{7.7686 \ kJ/kgK-0.9439 \ kJ/kgK}=0.724 \ 62.
The enthalpy at the turbine exit is then
h_4=h_{4,f}+x_4(h_{4,g}-h_{4,f}) , \\ h_4= 289.23 \ kJ/kg+0.7246 \times (2625.3 \ kJ/kg-289.23 \ kJ/kg)=1982.0 \ kJ/kg.
The enthalpy at the condenser exit is h_1 = h_{4,f} = 289.23 kJ / kg.
The enthalpy at the pump exit is
h_2=h_1+w_p=h_1+ν_1(P_2-P_1).
For saturated water at P_1 = 30 kPa (Appendix 8b), specific volume ν1= 0.001022 m³ / kg, then
h_2= 289.23 \ kJ/kgK+0.001 \ 022 \ m^3/kg \times (6000 \ kPa-30 \ kPa)=295.33 \ kJ/kg.
The back work ratio is
bwr=\frac{(h_2-h_1)}{(h_3-h_4)}=\frac{295.33 \ kJ/kg – 289.23 \ kJ/kg}{2784.3 \ kJ/kg-1982.0 \ kJ/kg}=0.007 \ 603 \ 1.
The thermal efficiency of a Rankine cycle is
η_{th,Rankine}=1- \frac{(h_4-h_1)}{(h_3-h_2)}=1- \frac{1982.0 \ kJ/kg-289.23 \ kJ/kg}{2784.3 \ kJ/kg-295.33 \ kJ/kg}=0.319 \ 89.
For a Carnot engine using the same working fluid pressures, at 6 MPa T_H = T_{sat} = 275.64 °C = 548.79 K, and 30 kPa T_C = T_{sat} = 69.10 °C = 342.25 K, so the Carnot efficiency would be
η_{th,Carnot}=1-\frac{T_C}{T_H}=1-\frac{342.25 \ K}{548.79 \ K}=0.376 \ 36.
Answer: The back work ratio is 0.760%, and the thermal efficiency of the Rankine cycle is 32.0%, while the related thermal efficiency of the Carnot cycle operating between the same temperatures is 37.6%.