(a) Explanation Write the expression for the turbine work per unit mass. W ˙ t m ˙ = ( h 1 − h 2 ) (I) Here, the turbine work per unit mass is W ˙ t , the mass of the vapor is m ˙ , the enthalpy at state 1 is h 1 and the enthalpy at state 2 is h 2 . Write the expression for the enthalpy at state 2. h 2 = h f 2 + x 2 ( h g 2 − h f 2 ) (II) Here, the quality at state 2 is x 2 , the saturated vapor enthalpy is h f 2 and the saturated liquid enthalpy is h g 2 . Write the expression for the pump work input per unit mass. W ˙ p m ˙ = ( h 4 − h 3 ) (III) Here, the pump work per unit mass is W ˙ p , the enthalpy at state 3 is h 3 and the enthalpy at state 4 is h 4 . Write the expression for the isentropic pump efficiency. η p = ( h 4 s − h 3 ) ( h 4 − h 3 ) (IV) Here, the isentropic pump efficiency is η p . Write the expression for the enthalpy at state 4. h 4 = h 3 + ( h 4 s − h 3 ) η p (V) Write the expression for the back work ratio of the cycle. b w r = ( W ˙ p m ˙ ) ( W ˙ t m ˙ ) (VI) Here, the back work ratio of the cycle is b w r . Write the expression for interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (VII) Here, the variables denote by x and y are temperature and specific heat of constant pressure. Conclusion: Refer to table A-4 “properties of saturated water vapor” to obtain the enthalpy at temperature T 1 = 440 ° C and pressure p 1 = 7.5 Mpa as 3253.9 kJ / kg Refer to table A-2 “properties of superheated water (liquid-vapor) temperature table” to obtain the saturated liquid enthalpy and saturated vapor enthalpy at 440 ° C and 6 kpa as 151.53 kJ / kg and 2567.4 kJ / kg respectively. Substitute 2567.4 kJ / kg for h g 2 , 151.53 kJ / kg for h f 2 and 0.92 for x 2 in Equation (II). h 2 = 151.53 kJ / kg + 0.92 ( 2567.4 kJ / kg − 151.53 kJ / kg ) = 151.53 kJ / kg + 2222.60 kJ / kg = 2374.1 kJ / kg Substitute 2374.1 kJ / kg for h 2 and 3253.9 kJ / kg for h 1 in Equation (I). W ˙ t m ˙ = ( 3253.9 kJ / kg − 2374.1 kJ / kg ) = 879.8 kJ / kg Here, the enthalpy at state 3 is equal to the saturated liquid enthalpy at state 2, that is, ( h 3 = h f 2 ) . Refer to table A-5 “properties of the compressed liquid water” to obtain the entropies at pressure 7.5 Mpa as s 3 = 0.5210 kJ / kg ⋅ K , s 1 = 0.2950 kJ / kg ⋅ K and s 2 = 0.5696 kJ / kg ⋅ K (b) To determine The back work ratio for the cycle. The specific enthalpy at state 4 using approximation method.

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 6.13, Problem 176P

(a)

To determine

The back work ratio for the cycle.

The specific enthalpy at state 4 using interpolation method.

(b)

To determine

The back work ratio for the cycle.

The specific enthalpy at state 4 using approximation method.

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Chapter 6.13, Problem 175P

Chapter 6.13, Problem 177P

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The back work ratio for the cycle. The specific enthalpy at state 4 using interpolation method.