Question 9.4: Problem: A Rankine cycle with regeneration is supplied super...
Problem: A Rankine cycle with regeneration is supplied superheated steam at a pressure of 6 MPa and temperature of 400 °C. The steam expands to a pressure of 0.8 MPa in the first stage of a turbine, after some of it is diverted into an open feedwater heater. The remaining steam expands in the second stage of the turbine, passes through a condenser at a pressure of 30 kPa, and leaves as saturated liquid that is pumped into the feedwater heater where it mixes with the steam that was removed. Saturated liquid leaves the feedwater heater at 0.8 MPa. Find (a) the fraction of steam extracted and (b) the thermal efficiency of the cycle.
Find: (a) Fraction of stream f extracted from the feedwater heater, (b) thermal efficiency
η_{th, \text{Rankine}} of the cycle.
Known: Turbine inlet pressure P_3 = P_7 = 6 \ MPa, turbine inlet temperature T_3 = 400 °C, feedwater heater pressure P_5 = P_6 = 0.8 \ MPa, condenser pressure P_1 = P_4 = 30 \ kPa.
Process Diagram: The Rankine cycle with regeneration on a T‐s diagram (Figure E9.4).
Assumptions: Expansion through the turbines is isentropic so ∆S_{35} = 0 \text{ and } ∆S_{34} = 0.
Governing Equations:
Feedwater mass flow fraction f = \frac{h_6 – h_2}{h_5 – h_2}
Work of turbines with regeneration w_t = (h_3 – h_5) + (1 – f)(h_5 – h_4)
Work to pumps with regeneration w_p = (h_7 – h_6) + (1 – f)(h_2 – h_1)
Heat to boiler with regeneration q _H = h_3 – h_7
Thermal efficiency of Rankine cycle η_{th, \text{Rankine}} = \frac{w_t – w_p}{q_H}
For superheated steam at P_3 = 6 \ MPa \text{ 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_5 = 0.8 \ MPa (Appendix 8b), specific enthalpy of liquid h_{5,f} = 721.11 \ kJ \ / \ kg \text{ and vapour } h_{5,g} = 2769.1 \ kJ \ / \ kg, and specific entropy of liquid s_{5,f} = 2.0462 \ kJ \ / \ kgK \text{ and vapour } s_{5,g} = 6.6628 \ kJ \ / \ kgK. If expansion through the first stage of the turbine is isentropic, s_5 = s_3 = 6.5408 \ kJ \ / \ kgK, then the quality of the mixture in the first stage is
x_5 = \frac{S_5 – S_{5,f}}{S_{5,g} – S_{5,f}} = \frac{6.5408 \ kJ \ / \ kgK – 2.0462 \ kJ \ / \ kgK}{6.6628 \ kJ \ / \ kgK – 2.0462 \ kJ \ / \ kgK} = 0.973 \ 57 .
The enthalpy at the turbine exit is then
h_5 = h_{5,f} + x_5 \left( h_{5,g} − h_{5, f} \right) ,
h_5 = 721.11 \ kJ \ / \ kg + 0.97357 \times ( 2769.1 \ kJ \ / \ kg – 721.11 \ kJ \ / \ kg) = 2715.0 \ kJ \ / \ kg.
For saturated water at P_4 = 30 \ kPa (Appendix 8b), specific enthalpy of liquid h_{4,f} = 289.23 \ kJ \ / \ kg \text{ and vapour } h_{4,g} = 2625.3 \ kJ \ / \ kg, and specific entropy of liquid s_{4,f} = 0.9439 \ kJ \ / \ kgK \text{ and vapour } s_{4,g} = 7.7686 \ kJ \ / \ kgK. If expansion through the second stage of the turbine is isentropic, s_4 = s_3 = 6.5408 \ kJ \ / \ kgK, and quality in the second turbine is
x_4 = \frac{S_4 – 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 is then
h_4 = h_{4,f} + x_4 \left( h_{4,g} − h_{4, f} \right) ,
h_4 = 289.23 \ kJ \ / \ kg + 0.82009 \times ( 2625.3 \ kJ \ / \ kg – 289.23 \ kJ \ / \ kg) = 2205.0 \ kJ \ / \ kg.
The enthalpy at the condenser exit at 30 kPa is h_1 = h_f = 289.23 kJ / kg.
The enthalpy at the exit of pump 1 is
h_2 = h _1 + w_{p,1} = h_1 + v_1 (P_2 – P_1).
Using saturated water at P_1 = 30 kPa (Appendix 8b), specific volume v_1 = v_f = 0.001 \ 022 \ m^3 / kg, then
h_2 = 289.23 \text{ kJ / kgK} + 0.001 \ 022 \ m^3 / kg \times (6000 \ kPa – 30 \ kPa) = 295.33 \text{ kJ / kg}
The enthalpy at the exit of the feedwater heater is h_6 = h_{5,f} = 721.11 kJ / kg.
The enthalpy at the exit of pump 2 is
h_7 = h_6 + w_{p,1} = h_6 + v_6 (P_7 – P_6)
For saturated water at P_6 = 0.8 MPa (Appendix 8b), specific volume v_6 = v_f = 0.00115 m3 \ / \ kg, then
h_7 = 721.11 \text{ kJ / kgK} + 0.001115 \ m^3 / kg \times (6000 \ kPa – 800 \ kPa) = 726.91 \text{ kJ / kg} .
(a) The mass fraction of steam extracted for the feedwater heater can now be solved for:
f= \frac{h_6 – h_2}{h_5 – h_2} = \frac{721.11 \text{ kJ / kg } – 295.33 \text{ kJ / kg }}{2715.0 \text{ kJ / kg } – 295.33 \text{ kJ / kg }} = 0.175 \ 97 .
(b) The work done by the turbine is
w_t = (h_3 – h_5) + (1 – f) (h_5 – h_4)
w_t = (3177.2 \text{ kJ / kg } – 2715.0 \text{ kJ / kg }) + (1 – 0.17597) (2715.0 \text{ kJ / kg } – 2205.0 \text{ kJ / kg }) =882.46 \text{ kJ / kg }
The total work done by both pumps is
w_p = (h_7 – h_6) + (1 – f) (h_2 – h_1)
w_p = (726.91 \text{ kJ / kg } – 721.11 \text{ kJ / kg }) + (1 – 0.175 \ 97) (295.33 \text{ kJ / kg } – 289.23 \text{ kJ / kg }) = 10.827 \text{ kJ / kg }
The heat input to the boiler is
q_H = (h_3 – h_7) = 3177.2 \text{ kJ / kg } – 726.91 \text{ kJ / kg } = 2450.3 \text{ kJ / kg },
The thermal efficiency of the Rankine cycle is then
η_{th, \text{Rankine}} = \frac{w_t – w_p}{q_H} = \frac{882.46 \text{ kJ / kg} – 10.827 \text{ kJ / kg}}{2450.3 \text{ kJ / kg} } = 0.355 \ 73 .
Answer: (a) The mass fraction of steam extracted is 17.6% and (b) the thermal efficiency of the Rankine cycle is 35.6%.