Explanation Draw the T-s diagram for the ideal regenerative Rankine cycle as shown in Figure (1). For the steam power plant which functions on a reheat Rankine cycle, pressure, specific enthalpy, specific entropy, specific volume, and temperature at i th state be P i , h i , s i , v i , and T i . Express the specific work input of pump I to the system. w p I ,in = v 1 ( P 2 − P 1 ) (I) Express the enthalpy of steam at state 2. h 2 = h 1 + w p I , in (II) Express the specific work input of pump II to the system. w p I I ,in = v 3 ( P 4 − P 3 ) (III) Express the enthalpy of steam at state 2. h 4 = h 3 + w p I I , in (IV) Express the quality of steam at state 6. x 6 = s 6 − s f s f g (V) Here, specific entropy of wet steam at 0.4 MPa is s f , and specific entropy of vaporization at 0.4 MPa is s f g . Express the specific enthalpy of steam at state 6. h 6 = h f + x 6 h f g (VI) Here, specific enthalpy of wet steam at 0.4 MPa is h f , and specific enthalpy of vaporization at 0.4 MPa is h f g . Express the quality of steam at state 7. x 7 = s 7 − s f s f g (VII) Here, specific entropy of wet steam at 20 kPa is s f , and specific entropy of vaporization at 20 kPa is s f g . Express the specific enthalpy of steam at state 7. h 7 = h f + x 7 h f g (VIIII) Here, specific enthalpy of wet steam at 20 kPa is h f , and specific enthalpy of vaporization at 20 kPa is h f g . Express the energy balance equation for the system. E ˙ in − E ˙ out = 0 (IX) Here, energy flow at the inlet is E ˙ in , and energy flow at the outlet is E ˙ out . Express the specific heat input in the regenerative Rankine cycle. q in = h 5 − h 4 (XI) Here, specific heat input of the cycle is q in . Express the specific heat output in the regenerative Rankine cycle. q out = ( 1 − y ) ( h 7 − h 1 ) (XII) Here, specific heat input of the cycle is q out . Express the exergy destruction for the regenerative cycle. x destroyed,cycle = T 0 ( q out T L − q in T H ) (XIII) Here, the source temperature is T L , sink temperature is T H , and ambient temperature is T 0 . Conclusion: Refer to Table A-5, “Saturated water-Pressure table”, obtain the following properties at the pressure of 20 kPa . h 1 = 251 .42 kJ / kg v 1 = 0.001017 m 3 / kg Substitute 0.001017 m 3 / kg for v 1 , 20 kPa for P 1 , and 400 kPa for P 2 in Equation (I). w p I ,in = 0.001017 m 3 / kg ( 400 − 20 ) kPa ( 1 kJ 1 kPa ⋅ m 3 ) = 0.39 kJ / kg Substitute 251 .42 kJ / kg for h 1 , and 0.39 kJ / kg for w p I ,in in Equation (II). h 2 = 251 .42 kJ / kg + 0.39 kJ / kg = 251.81 kJ / kg Refer to Table A-5, “Saturated water-Pressure table”, obtain the following properties at the pressure of 0.4 MPa . h 3 = 604 .66 kJ / kg v 3 = 0.001084 m 3 / kg Substitute 0.001084 m 3 / kg for v 3 , 400 kPa for P 3 , and 6 , 000 kPa for P 4 in Equation (III). w p I I ,in = 0.001084 m 3 / kg ( 6 , 000 − 400 ) kPa ( 1 kJ 1 kPa ⋅ m 3 ) = 6.07 kJ / kg Substitute 604 .66 kJ / kg for h 3 , and 6.07 kJ / kg for w p I I ,in in Equation (IV). h 4 = 604.66 kJ / kg + 6.07 kJ / kg = 610.73 kJ / kg Refer to Table A-6, “superheated water-table”, obtain the following properties for the pressure and temperature of of 6 MPa and 450 ° C respectively. h 5 = 3,302 .9 kJ / kg s 5 = 6 .7219 kJ / kg ⋅ K Since, the entropy at state 5 is equal to the entropy at state 6 s 5 = s 6 =6 .7219 kJ / kg ⋅ K

8th Edition

ISBN: 9780100257054

Author: CENGEL

Publisher: YUZU

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Chapter 10.9, Problem 63P

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The exergy destruction for the ideal regenerative Rankine cycle

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Chapter 10.9, Problem 62P

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EBK THERMODYNAMICS: AN ENGINEERING APPR

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