Question 9.4: Problem: A Rankine cycle with regeneration is supplied super...

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%.