Explanation Draw the T-s diagram with steam as working fluid for the ideal Rankine cycle as shown in Figure (1). The simple Rankine cycle involves following processes and following conditions: Processes: 1. Process 1 to 2 – Isentropic compression in a Pump 2. Process 2 to 3 – Heat addition at constant pressure in a Boiler 3. Process 3 to 4 – Isentropic expansion in a Turbine 4. Process 4 to 1 – Heat rejection at constant pressure in a Condenser Conditions: The pressure is constant for the process 2 to 3 and the process 4 to 1. P 2 = P 3 and P 4 = P 1 The entropy is constant for the process 1 to 2 and the process 3 to 4. s 1 = s 2 and s 3 = s 4 Here, water is the working fluid. 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 for 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 for 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 10 kPa is s f , specific entropy of vaporization at 10 kPa is s f g , and specific entropy of the saturated liquid at 10 MPa is s 4 . Write the expression for the specific enthalpy of steam at state 4. h 4 = h f + x 4 h f g (IV) Here, specific enthalpy of vaporization of steam at 10 MPa is h f g . Write the expression for 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 for 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 for 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: Refer Table A-5, “Saturated water-Pressure table”, the enthalpy ( h 1 ) and specific volume ( v 1 ) at state 1 corresponding to the pressure of 10 kPa is 191 .81 kJ / kg and 0.00101 m 3 / kg respectively. Substitute 0.00101 m 3 / kg for v 1 , 10 kPa for P 1 , and 10 MPa for P 2 in Equation (I). w p,in = 0.00101 m 3 / kg ( 10 , 000 − 10 ) kPa ( 1 kJ 1 kPa ⋅ m 3 ) = 10.09 kJ / kg Substitute 191 .81 kJ / kg for h 1 , and 10.09 kJ / kg for w p,in in Equation (II). h 2 = 191 .81 kJ / kg + 10.09 kJ / kg = 201.90 kJ / kg Refer Table A-6, “Superheated water”, the entropy ( s 3 ) at state 3 corresponding to the pressure of 10 MPa and the temperature of 500 ° C is 6.5995 kJ / kg ⋅ K respectively. Refer Table A-5, “Saturated water-Pressure table”, the entropy of saturated liquid ( s f ) and the entropy of vapor mixture ( s f g ) at state 4 corresponding to the pressure of 10 kPa is 0

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Chapter 10.9, Problem 59P

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Chapter 10.9, Problem 58P

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