Explanation The T-s diagram with all given states is shown below. Figure-(1) Write the expression for the specific entropy at point 2. s 2 = s f + x 2 s f g (I) Here, the specific entropy of the saturated water is s f , the specific entropy of the water vapor mixture is s f g and the dryness fraction corresponding to point 2 is x 2 . Write the expression for the specific enthalpy at point 2. h 2 = h f 2 + x 2 h f g 2 (II) Here, the specific enthalpy of the saturated water is h f 2 , the specific enthalpy of the water vapor mixture is h f g and the dryness fraction is x 2 corresponding to point 2. Write the expression for the specific entropy at point 3. s 3 = s f 3 + x 3 s f g (III) Here, the specific entropy of the saturated water is s f 3 , the specific entropy of the water vapor mixture is s f g and the dryness fraction is x 3 corresponding to point 3. Write the expression for the specific enthalpy at point 3. h 3 = h f 3 + x 3 h f g 3 (IV) Here, the specific enthalpy of the saturated water is h f 3 and the specific enthalpy of the water vapor mixture is h f g 3 and the dryness fraction is x 3 corresponding to point 3. Write the expression for the enthalpy at point 5. h 5 = h 4 + v 4 ( p 5 − p 4 ) (V) Here, 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 ) (VI) Here, 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 ) (VII) Write the expression for the turbine work. W T m ˙ = ( h 1 − h 2 ) + ( 1 − y ) ( h 2 − h 3 ) (VIII) Here, the mass flow rate of the steam is m ˙ . Write the expression for the pump work. W P m ˙ = ( h 7 − h 6 ) + ( 1 − y ) ( h 5 − h 4 ) (IX) Write the expression for the net work. W n e t m ˙ = W T m ˙ − W P m ˙ (X) Write the expression for energy balance in condenser. m ˙ ( h 3 − h 4 ) ( 1 − y ) = m ˙ w c p w ( Δ T c w ) (XI) Here, the mass flow rate of water in condenser is m ˙ w , specific heat capacity of water is c p w , and the temperature increase in cooling water is Δ T c w . Write the expression for energy loss in cooling tower. ( e o u t − e i n ) l o s s = m ˙ w m ˙ [ c p w Δ T c w − c p w T i n ln T o u t T i n ] (XII) Write the exergy destruction in feed water heater. ( Φ D e s t r u c t i o n i n h e a t e r ) = T o [ s 6 − y s 2 − ( 1 − y ) s 5 ] (XIII) Here, specific entropy at state 6 is s 6 , specific entropy at state 2 is s 2 , the surrounding temperature is T o , and specific entropy at state 5 is s 5 . Write the expression for exergy destruction in condenser. ( Φ D e s t r u c t i o n i n c o n d e n s e r ) = T o [ ( 1 − y ) ( s 4 − s 3 ) + m ˙ w m ˙ c p w ln T o u t T i n ] (XIV) Here, specific entropy at state 4 is s 4 , and specific entropy at state 3 is s 3 . Conclusion: Refer to Table A-4 “Properties of Superheated Water Vapor table” to obtain specific enthalpy ( h 1 ) as 3465.4 kJ / kg and the specific entropy ( s 1 ) as 6.5132 kJ / kg ⋅ K at temperature 560 ° C and pressure 160 bar . Refer to Table A-3 “Properties of Saturated Water-Pressure table” to obtain the specific enthalpy ( h f ) of liquid as 173.9 kJ / kg , specific enthalpy of vaporization ( h f g ) as 2403.2 kJ / kg , the specific entropy of liquid ( s f ) 0.593 kJ / kg ⋅ K , specific entropy of vaporization ( s f g ) as 7.657 kJ / kg ⋅ K , and specific volume as 1.008 × 10 − 3 m 3 / kg at pressure 0.08 bar . Refer to Table A-3 “Properties of Saturated Water-Pressure table” to obtain specific enthalpy ( h f ) of liquid as 762.6 kJ / kg , specific enthalpy of vaporization ( h f g ) as 2013.6 kJ / kg , the specific entropy ( s f ) of liquid 2.138 kJ / kg ⋅ K , specific entropy ( s f g ) of vaporization as 4.445 kJ / kg ⋅ K and specific volume ( v f ) of liquid as 1.127 × 10 − 3 m 3 / kg at pressure 10 bar . Substitute 6.5132 kJ / kg ⋅ K for s 2 , 2.138 kJ / kg ⋅ K for s f and 4.445 kJ / kg ⋅ K for s f g in Equation (I). 6.5132 kJ / kg ⋅ K = ( 2.138 kJ / kg ⋅ K ) + x 2 × ( 4.445 kJ / kg ⋅ K ) x 2 = ( 6.5132 kJ / kg ⋅ K ) − ( 2.138 kJ / kg ⋅ K ) ( 4.445 kJ / kg ⋅ K ) x 2 = 0.984 Substitute 762.6 kJ / kg for h f 2 , 0.984 for x 2 and 2013.6 kJ / kg for h f g 2 in Equation (II). h 2 = 762.6 kJ / kg + 0.984 × 2013.6 kJ / kg = 762.6 kJ / kg + 1981.3824 kJ / kg = 2744 kJ / kg Substitute 6.5132 kJ / kg ⋅ K for s 3 , 0.593 kJ / kg ⋅ K for s f 3 and 7.657 kJ / kg ⋅ K for s f g in Equation (III). 6.5132 kJ / kg ⋅ K = ( 0.593 kJ / kg ⋅ K ) + x 3 × ( 7.657 kJ / kg ⋅ K ) x 3 = ( 6.5132 kJ / kg ⋅ K ) − ( 0.593 kJ / kg ⋅ K ) ( 7.657 kJ / kg ⋅ K ) x 3 = 0.7731 Substitute 173.9 kJ / kg for h f 3 , 0.7731 for x 3 and 2403.2 kJ / kg for h f g 3 in Equation (IV). h 3 = 173.9 kJ / kg + 0.7731 × 2403.2 kJ / kg = 173.9 kJ / kg + 1857.9 kJ / kg = 2031.8 kJ / kg Substitute 173.9 kJ / kg for h 4 , 1.008 × 10 − 3 m 3 / kg for v 4 , 1000 kPa for p 5 and 8 kPa for p 4 in Equation (V). h 5 = 173.9 kJ / kg + ( 1.008 × 10 − 3 m 3 / kg ) × ( 1000 kPa − 8 kPa ) = 173

Fundamentals of Engineering Thermodynamics, Binder Ready Version

8th Edition

ISBN: 9781118820445

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

Publisher: WILEY

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Chapter 8.6, Problem 91P

To determine

The rate of exergy transfer.

The energy loss in cooling water.

The exergy destruction in feed water heater.

The exergy destruction in condenser.

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Chapter 8.6, Problem 90P

Chapter 8.6, Problem 92P

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Fundamentals of Engineering Thermodynamics, Binder Ready Version

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Chapter 8 Solutions

The rate of exergy transfer.

The energy loss in cooling water.

The exergy destruction in feed water heater.

The exergy destruction in condenser.