(a) 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 ) η P (I) Here, the specific volume is v , the pressure is P , the isentropic efficiency of the pump is η 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 ) η P (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 9 s : The steam enters the condenser at the pressure of 15 kPa and at the state of saturated mixture. The quality of water at state 9 s is expressed as follows. x 9 s = s 9 s − s f , 9 s s f g , 9 s (V) The enthalpy at state 9 s is expressed as follows. h 9 s = h f , 9 s + x 9 s h f g , 9 s (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 9 s indicates the ideal process state 9 s . Write the general equation of energy balance equation. E ˙ in − E ˙ out = Δ E ˙ system (VII) 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 (VII). 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 ˙ 8 h 8 + m ˙ 2 h 2 = m ˙ 3 h 3 (VIII) Rewrite the Equation (VIII) in terms of mass fraction y . y h 8 + ( 1 − y ) h 2 = 1 h 3 y h 8 + h 2 − y h 2 = h 3 y ( h 8 − h 2 ) = h 3 − h 2 y = h 3 − h 2 h 8 − h 2 (IX) The mass fraction y is the ratio of mass fraction steam extracted from the turbine to the inlet mass of the boiler ( m ˙ 8 / m ˙ 3 ) . At state 1: (Pump I inlet) The water exits the condenser as a saturated liquid at the pressure of 15 kPa . Hence, the enthalpy, and specific volume at state 1 is as follows. h 1 = h f @ 15 kPa v 1 = v f @ 15 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 15 kPa is 225.94 kJ/kg , and 0.001014 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 0.6 MPa ( 600 kPa ) . Hence, the enthalpy, and specific volume at state 3 is as follows. h 3 = h f @ 600 kPa v 3 = v f @ 600 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 0.6 MPa ( 600 kPa ) is 670.38 kJ/kg , and 0.001101 m 3 /kg respectively. At state 5: (Before reheating- H.P. Turbine inlet) 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 10 MPa ( 10000 kPa ) and the temperature of 500 ° C is as follows. h 5 = 3375.1 kJ/kg s 5 = 6.5995 kJ/kg ⋅ K Refer Figure 1. s 5 = s 6 s = 6.5995 kJ/kg ⋅ K At state 6 s : The steam expanded to the pressure of 1 MPa ( 1000 kPa ) and in the state of superheated vapor. Refer Table A-6, “Superheated water”. The enthalpy ( h 6 s ) at state 6 s corresponding to the pressure of 1 MPa ( 1000 kPa ) and the entropy of 6.5995 kJ/kg ⋅ K is as follows. h 6 s = 2783.8 kJ/kg The isentropic efficiency for the process 5-6 is expressed as follows. η T = h 5 − h 6 h 5 − h 6 s h 6 = h 5 − η T ( h 5 − h 6 s ) (X) At state 7: (After reheating- L.P. Turbine inlet) Reheating occurs at the pressure of 1 MPa ( 1000 kPa ) and the steam enters the turbine as superheated vapor at 500 ° C . Refer Table A-6, “Superheated water”. The enthalpy ( h 7 ) and entropy ( s 7 ) at state 7 corresponding to the pressure of 1 MPa ( 1000 kPa ) and the temperature of 500 ° C is as follows. h 7 = 3479.1 kJ/kg s 7 = 7.7642 kJ/kg ⋅ K Refer Figure 1. s 7 = s 8 s = s 9 s = 7.7642 kJ/kg ⋅ K At state 8 s : The steam is extracted at the pressure of 0.6 MPa ( 600 kPa ) for open feed water heater. The extracted steam is at the state of superheated vapor. Refer Table A-6, “Superheated water”. The enthalpy ( h 8 s ) at state 8 s corresponding to the pressure of 0.6 MPa ( 600 kPa ) and the entropy of 7.7642 kJ/kg ⋅ K is as follows. h 8 s = 3310 (b) To determine The thermal efficiency of the cycle.

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Author: CENGEL

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Chapter 10.9, Problem 101RP

(a)

To determine

The fraction of steam extracted for regeneration.

(b)

To determine

The thermal efficiency of the cycle.

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Chapter 10.9, Problem 100RP

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EBK THERMODYNAMICS: AN ENGINEERING APPR

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The fraction of steam extracted for regeneration.

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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The fraction of steam extracted for regeneration.

The thermal efficiency of the cycle.

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