(a) Explanation The T − s diagram with given state is shown below. Figure-(1) Write the expression for the isentropic efficiency for the 1 st stage turbine. η T I = h 1 − h 2 h 1 − h 2 s (I) Here, the enthalpy at point 1 is h 1 and the enthalpy at point 2 is h 2 . Write the expression for the isentropic efficiency for the 2 nd stage turbine. η T I I = h 2 − h 3 h 2 − h 3 s (II) Here, the enthalpy at point 1 is h 1 and the enthalpy at point 2 is h 2 . Write the expression for the enthalpy at point 5. h 5 = h 4 + v 4 ( p 5 − p 4 ) η p I (III) Here, the isentropic efficiency of the 1 st pump is η p I the pressure at the point 5 is p 5 , the pressure at the point 4 is p 4 , the specific volume at point 4 is v 4 , the specific enthalpy at state 5 is h 5 and the specific enthalpy at state 4 is h 4 . Write the expression for the enthalpy at point 7. h 7 = h 6 + v 6 ( p 7 − p 6 ) η p I I (IV) Here, the isentropic efficiency of the 2 nd pump is η p I I ,the pressure at the point 7 is p 7 , the pressure at the point 6 is p 6 , the specific volume at point 6 is v 6 , the specific enthalpy at state 7 is h 7 and the specific enthalpy at state 6 is h 6 . Write the expression for the fraction of the mass of steam entering in to feed water heater before entering into second stage turbine. y = ( h 6 − h 5 ) ( h 2 − h 5 ) (V) Write the expression for te turbine work. W T m ˙ = ( h 1 − h 2 ) + ( 1 − y ) ( h 2 − h 3 ) (VI) Write the expression for the pump work. W P m ˙ = ( h 7 − h 6 ) + ( 1 − y ) ( h 5 − h 4 ) (VII) Write the expression for the heat addition for 1 st stage turbine. Q ˙ 1 st m ˙ = h 1 − h 7 (VIII) Here, the enthalpy at point 1 is h 1 . Write the expression for thermal efficiency. η t h = W T m ˙ − W P m ˙ Q ˙ 1 st m ˙ (IX) Conclusion: Refer to Superheated steam table to obtain specific enthalpy as 3506.2 kJ / kg and the specific entropy 6.6840 kJ / kg ⋅ K at temperature 560 ° C and pressure 12 MPa and specific enthalpy as 2823.3 kJ / kg , at pressure 1 MPa and the specific entropy as 6.9174 kJ / kg ⋅ K at pressure 1 MPa , specific enthalpy of the liquid as 152.74 kJ / kg at 1 MPa and the specific entropy of the liquid as 0.5249 kJ / kg ⋅ K at pressure 1 MPa . Refer to saturated steam table to obtain specific enthalpy of liquid as 151.53 kJ / kg , specific enthalpy of vaporization 2415.9 kJ / kg , and the specific entropy of liquid 0.521 kJ / kg ⋅ K and specific entropy of vaporization as 7.8094 kJ / kg ⋅ K and specific volume of liquid 1.0064 × 10 − 3 m 3 / kg and dryness fractions as 0.8191 and 0.8760 and specific enthalpies as 2130.4 kJ / kg and 2267.9 kJ / kg , and specific entropy as 7.3620 kJ / kg ⋅ K at pressure 6 kPa , the specific enthalpy of the liquid as 762.81 kJ / kg , the specific entropy as 2.1387 kJ / kg ⋅ K and the specific volume of the saturate liquid as 1.1273 × 10 − 3 m 3 / kg . Substitute 3506.2 kJ / kg for h 1 , 0.83 for η T 1 , 2823.3 kJ / kg for h 2 s in Equation (I). 0.83 = 3506.2 kJ / kg − h 2 3506.2 kJ / kg − 2823.3 kJ / kg h 2 = 3506.2 kJ / kg − 583.40 kJ / kg h 2 = 2923 kJ / kg Substitute 2923 kJ / kg for h 2 , 0.83 for η T I I , 2130.4 kJ / kg for h 3 s in Equation (II). 0.83 = 2923 kJ / kg − h 3 2923 kJ / kg − 2130.4 kJ / kg h 3 = 2923 kJ / kg − 657 (b) To determine The mass flow rate into the first turbine stage. (c) To determine The rate of entropy production in the open feedwater heater.

8th Edition

ISBN: 9781119391630

Author: MORAN

Publisher: WILEY

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Chapter 8.6, Problem 43P

(a)

To determine

The cycle efficiency.

(b)

To determine

The mass flow rate into the first turbine stage.

(c)

To determine

The rate of entropy production in the open feedwater heater.

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Chapter 8.6, Problem 42P

Chapter 8.6, Problem 44P

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