Explanation Draw the T-s diagram of the ideal Rankine cycle as shown in figure (1). Write the expression for the work input: w p , i n = v 1 ( P 2 − P 1 ) (I) Here, specific volume at inlet is v 1 , pressure at state point 2 is P 2 , and pressure at state point 1 is P 1 . Write the expression for the enthalpy at state 2 using work input. h 2 = h 1 + w p , i n (II) Here, enthalpy at inlet is h 1 . Write the expression for the entropy relation at state 6. s 6 = s f − x 6 s f g (III) Here, entropy of saturated liquid is s f , vapor quality at state 6 is x 6 , and entropy of vapor mixture is s f g . Write the expression for the enthalpy relation at state 6. h 6 = h f − x 6 h f g (IV) Here, enthalpy of saturated liquid is s f and enthalpy of vapor mixture is s f g . Write the formula for the heat input for the process 2-3. q 23 , i n = h 3 − h 2 (V) Here, enthalpy at state 3 is h 3 and enthalpy at state 2 is h 2 . Write the formula for the heat input for the process 4-5. q 45 , i n = h 5 − h 4 (VI) Here, enthalpy at state 5 is h 5 and enthalpy at state 4 is h 4 . Write the formula for the heat output of the turbine. q o u t = h 6 − h 1 (VII) Here, enthalpy at state 6 is h 6 and enthalpy at state 1 is h 1 . Write the formula for the exergy destruction for the process 2-3. x d e s t r o y e d 2 , 3 = T 0 ( s 3 − s 2 + q 23 T R ) (VIII) Write the formula for the exergy destruction for the process 4-5. x d e s t r o y e d 45 = T 0 ( s 5 − s 4 + q R 45 T R ) (IX) Write the formula for the exergy destruction for the process 6-1. x d e s t r o y e d 6 , 1 = T 0 ( s 1 − s 6 + q R 61 T R ) (X) Conclusion: Refer to Table A-5, “Saturated water—Pressure table”, obtain the entropy of saturated liquid at the pressure of 20 kPa . h 1 = h f @ 20 kPa = 251 .42 kJ / kg v 1 = v f @ 20 kPa = 0.00107 m 3 / kg s 1 = s f @ 20 kPa = 0.8320 kJ / kg ⋅ K Substitute 0.001017 m 3 / kg for v 1 , 6 MPa for P 2 , 2 MPa for P 1 in equation (I). w p , i n = ( 0.001017 m 3 / kg ) ( 6 MPa − 2 MPa ) = ( 0.001017 m 3 / kg ) ( 6 000 kPa − 2000 kPa ) ( 1 kJ 1 kPa ⋅ m 3 ) = 6.08 kJ / kg Substitute 251.41 kJ / kg for h 1 and 6.08 kJ / kg for w p , i n in equation (II). h 2 = 251.41 kJ / kg + 6.08 kJ / kg = 257.50 kJ / kg Refer to Table A-6, “Superheated water”, obtain the following properties at the pressure of 6 MPa and temperature of 400 ° C . h 3 = 3178 .3 kJ / kg s 3 = 6 .5432 kJ / kg ⋅ K Refer to Table A-6, “Superheated water”, obtain the following properties at the pressure of 2 MPa and temperature of 400 ° C . h 5 = 3248 .4 kJ / kg s 5 = 7 .1292 kJ / kg ⋅ K Substitute 7.1292 kJ / kg ⋅ K for s 6 , 0.8320 kJ / kg ⋅ K for s f , and 7.0752 kJ / kg ⋅ K for s f g in equation (III) 7.1292 kJ / kg ⋅ K = 0.8320 kJ / kg ⋅ K − x 6 ( 7

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ISBN: 9781307434316

Author: CENGEL

Publisher: INTER MCG

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Chapter 10.9, Problem 60P

To determine

The exergy during the each process of the reheat Rankine cycle

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Chapter 10.9, Problem 59P

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Chapter 10 Solutions

THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I

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