Explanation Draw the T-s diagram of the ideal Rankine cycle as shown in Figure (1). Write the expression for the specific work input to the system. w p,in = v 1 ( P 2 − P 1 ) (I) Here, pressure at state 1 is P 1 , pressure at state 2 is P 2 , and specific volume of steam at state 1 is v 1 . Write the expression the enthalpy of steam at state 2. h 2 = h 1 + w p , in (II) Here, specific enthalpy of steam at state 1 is h 1 , and specific enthalpy of steam at state 2 is h 2 . Write the expression the quality of steam at the end of heat rejection process. x 4 = s 4 − s f s f g (III) Here, specific entropy of steam at 50 kPa is s f , specific entropy of vaporization at 50 kPa is s f g , and specific entropy of the saturated liquid at 3 MPa is s 4 . Write the expression the enthalpy of steam at state 4. h 4 = h f + x 4 h f g (IV) Here, specific enthalpy of vaporization of steam at 3 MPa is h f g . Write the expression the specific heat input to the Rankine cycle. q in = h 3 − h 2 (V) Here, specific heat input to the cycle is q in , specific enthalpy of steam at state 3 is h 3 . Write the expression the specific heat output of the Rankine cycle. q out = h 4 − h 1 (VI) Here, specific heat output of the cycle is q out . Write the expression the exergy destruction for the process. x d e s t r o y e d ( i j ) = T 0 ( s i − s j + q R , i j T R ) (VII) Here, entropy at state i is s i , entropy at state j is s j , heat transferred during the process is q R , i j , and temperature of the heat source is T R , and ambient temperature is T 0 . Conclusion: From Table A-5, “Saturated water-Pressure table”, select the enthalpy ( h 1 = h f ) and specific volume ( v 1 ) at the pressure of 50 kPa and as 340 .54 kJ / kg and 0.001030 m 3 / kg respectively. Substitute 0.001030 m 3 / kg for v 1 , 50 kPa for P 1 , and 3 , 000 kPa for P 2 in Equation (I). w p,in = 0.001030 m 3 / kg ( 3 , 000 − 50 ) kPa ( 1 kJ 1 kPa ⋅ m 3 ) = 3.04 kJ / kg Substitute 340 .54 kJ / kg for h 1 , and 3.04 kJ / kg for w p,in in Equation (II). h 2 = 340 .54 kJ / kg + 3.04 kJ / kg = 343.58 kJ / kg Refer to Table A-6, “Superheated water”, obtain the following properties at the the pressure of 3 MPa and temperature of 300 ° C as, h 3 = 2994 .3 kJ / kg s 3 = 6.5412 kJ / kg ⋅ K From Table A-5, “Saturated water-Pressure table”, select the entropy of saturated liquid ( s f ) and entropy of vapor mixture ( s f g ) at the pressure of 50 kPa and as 1.0912 kJ / kg ⋅ K and 6.5019 kJ / kg ⋅ K respectively. Substitute 6.5412 kJ / k g ⋅ K for s 4 , 1.0912 kJ / k g ⋅ K for s f , and 6

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Chapter 10.9, Problem 58P

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