Explanation Draw the T − s diagram of the given regenerative Rankine cycle as shown in Figure 1. Here, water (steam) is the working fluid of the regenerative Rankine cycle. The cycle involves two pumps. Write the formula for work done by the pump during process 1-2. w pI,in = v 1 ( P 2 − P 1 ) (I) Here, the specific volume is v , the pressure is P , and the subscripts 1 and 2 indicates the process states. Write the formula for enthalpy ( h ) at state 2. h 2 = h 1 + w pI,in (II) Write the formula for work done by the pump during process 3-4. w pII,in = v 3 ( P 4 − P 3 ) (III) Here, the specific volume is v , the pressure is P , and the subscripts 3 and 4 indicates the process states. Write the formula for enthalpy ( h ) at state 4. h 4 = h 3 + w pII,in (IV) At state 6: The steam expanded to the pressure of 0.4 MPa and the steam is at the state of saturated mixture. The quality of water at the exit of the turbine (state 6) is expressed as follows. x 6 = s 6 − s f , 6 s f g , 6 (V) The enthalpy at state 6 is expressed as follows. h 6 = h f , 6 + x 6 h f g , 6 (VI) 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 subscript 6 indicates the state 6. At state 7: The steam enters the condenser at the pressure of 20 kPa and at the state of saturated mixture. The quality of water at state 7 is expressed as follows. x 7 = s 7 − s f , 7 s f g , 7 (VII) The enthalpy at state 7 is expressed as follows. h 7 = h f , 7 + x 7 h f g , 7 (VIII) Here, the subscript 7 indicates the process state 7. Write the formula for heat in ( q in ) and heat out ( q out ) of the cycle. q in = h 5 − h 4 (IX) q out = ( 1 − y ) ( h 7 − h 1 ) (X) Here, the mass fraction steam extracted from the turbine to the inlet mass of the boiler ( m ˙ 6 / m ˙ 3 ) is y . Write the general equation of energy balance equation. E ˙ in − E ˙ out = Δ E ˙ system (XI) Here, the rate of net energy inlet is E ˙ in , the rate of net energy outlet is E ˙ out and the rate of change of net energy of the system is Δ E ˙ system . At steady state the rate of change of net energy of the system ( Δ E ˙ system ) is zero. Δ E ˙ system = 0 Refer Equation (XI). Write the energy balance equation for open feed water heater. E ˙ in − E ˙ out = 0 E ˙ in = E ˙ out ∑ m ˙ in h in = ∑ m ˙ out h out m ˙ 6 h 6 + m ˙ 2 h 2 = m ˙ 3 h 3 (XII) Rewrite the Equation (XII) in terms of mass fraction y . y h 6 + ( 1 − y ) h 2 = 1 h 3 y h 6 + h 2 − y h 2 = h 3 y ( h 6 − h 2 ) = h 3 − h 2 y = h 3 − h 2 h 6 − h 2 (XIII) Write the formula for exergy destruction of the cycle. x destroyed, cycle = T 0 ( q out T L − q in T H ) (XIV) Here, the surrounding temperature is T 0 , the sink temperature is T L , and the source temperature is T H . The surrounding temperature is equal to the sink temperature. T 0 = T L At state 1: (Pump I inlet) The water exits the condenser as a saturated liquid at the pressure of 20 kPa . Hence, the enthalpy and specific volume at state 1 is as follows. h 1 = h f @ 20 kPa v 1 = v f @ 20 kPa 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 20 kPa is 251.42 kJ/kg and 0.001017 m 3 /kg respectively. At state 3: (Pump II inlet) The water exits the open feed water heater as a saturated liquid at the pressure of 0.4 MPa ( 400 kPa ) . Hence, the enthalpy and specific volume at state 3 is as follows. h 3 = h f @ 400 kPa v 3 = v f @ 400 kPa Refer Table A-5, “Saturated water-Pressure table”. The enthalpy ( h 3 ) and specific volume ( v 3 ) at state 3 corresponding to the pressure of 0.4 MPa ( 400 kPa ) is 604.66 kJ/kg and 0.001084 m 3 /kg respectively. At state 5: The steam enters the turbine as superheated vapor. Refer Table A-6, “Superheated water”. The enthalpy ( h 5 ) and entropy ( s 5 ) at state 5 corresponding to the pressure of 6 MPa ( 6000 kPa ) and the temperature of 450 ° C is as follows. h 5 = 3302.9 kJ/kg s 5 = 6.7219 kJ/kg ⋅ K Refer Figure 1. s 5 = s 6 = s 7 = 6.7219 kJ/kg ⋅ K At state 6: The steam expanded to the pressure of 0.4 MPa ( 400 kPa ) and in the state of saturated mixture. Refer Table A-5, “Saturated water-Pressure table”. Obtain the following properties corresponding to the pressure of 0.4 MPa ( 400 kPa ) . h f , 6 = 604.667 kJ/kg h f g , 6 = 21334 kJ/kg s f , 6 = 1.7765 kJ/kg ⋅ K s f g , 6 = 5.1191 kJ/kg ⋅ K At state 7: The steam enters the condenser at the pressure of 20 kPa and at the state of saturated mixture. Refer Table A-5, “Saturated water-Pressure table”. Obtain the following properties corresponding to the pressure of 20 kPa . h f , 7 = 251.42 kJ/kg h f g , 7 = 2357.5 kJ/kg s f , 7 = 0.8320 kJ/kg ⋅ K s f g , 7 = 7.0752 kJ/kg ⋅ K Conclusion: Substitute 0.001017 m 3 / kg for v 1 , 20 kPa for P 1 , and 400 kPa for P 2 in Equation (I). w pI,in = ( 0.001017 m 3 / kg ) ( 400 kPa − 20 kPa ) = 0

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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Author: CENGEL

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Solution Summary: The author illustrates the regenerative Rankine cycle by drawing the T-s diagram of water (steam) as the working fluid of the cycle.

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