Explanation Draw the T − s diagram for the given 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 formula for work done by the pump during process 1-2. w p,in = v 1 ( P 2 − P 1 ) (I) Here, the specific volume is v , the pressure is P . Write the formula for enthalpy ( h ) at state 2. h 2 = h 1 + w p,in (II) The quality of water at the exit of the turbine is expressed as follows. x 4 = s 4 − s f , 4 s f g , 4 (III) The enthalpy and entropy at state 4 is expressed as follows. h 4 = h f , 4 + x 4 h f g , 4 (IV) Here, the enthalpy is h , the entropy is s , the quality of the water is x , the suffix f indicates the fluid condition, the suffix f g indicates the change of vaporization phase; the suffix 4 indicates the state 4. Write the formula for heat in ( q in ) and heat out ( q out ) of the cycle. q in = h 3 − h 2 (V) q out = h 4 − h 1 (VI) Write the formula for the exergy destruction for the process. x destroyed ( ij ) = T 0 ( s i − s j + q R , ij T R ) (VII) Here, the entropy at state i is s i , the entropy at state j is s j , the heat transferred during the process is q R , i j , the temperature of the heat source is T R , and ambient temperature is T 0 . At state 1: The water exits the condenser as a saturated liquid at the pressure of 10 kPa . Hence, the enthalpy, specific volume, and entropy at state 1 is as follows. h 1 = h f @ 10 kPa v 1 = v f @ 10 kPa s 1 = s f @ 10 kPa Refer Table A-5, “Saturated water-Pressure table”. The enthalpy ( h 1 ) , specific volume ( v 1 ) and entropy ( s 1 ) at state 1 corresponding to the pressure of 10 kPa is 191.81 kJ/kg , 0.001010 m 3 /kg , and 0.6492 kJ/kg ⋅ K respectively. At state 3: (Turbine inlet) The steam enters the as superheated vapor. Refer Table A-6, “Superheated water”. The enthalpy ( h 3 ) and entropy ( s 3 ) at state 3 corresponding to the pressure of 10 MPa and the temperature of 500 ° C is as follows. h 3 = 3375.1 kJ/kg s 3 = 6.5995 kJ/kg ⋅ K At state 4: (Condenser inlet) Refer Table A-5, “Saturated water-Pressure table”. Obtain the following properties corresponding to the pressure of 10 kPa . h f , 4 = 191.81 kJ/kg h f g , 4 = 2392.1 kJ/kg s f , 4 = 0.6492 kJ/kg ⋅ K s f g , 4 = 7.4996 kJ/kg ⋅ K The ambient temperature is 17 ° C ( 290 K ) . Conclusion: Substitute 0.001010 m 3 /kg for v 1 , 10 MPa for P 2 and 10 kPa for P 1 in Equation (I). w p,in = ( 0.001010 m 3 /kg ) ( 10 MPa − 10 kPa ) = ( 0.001010 m 3 /kg ) [ ( 10 MPa × 10 3 kPa 1 MPa ) − 10 kPa ] = 0.001010 m 3 /kg ( 9990 kPa ) = 10.0899 kPa ⋅ m 3 /kg × 1 kJ 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)

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Author: CENGEL

Publisher: MCG

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Chapter 10.9, Problem 64P

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