Explanation Figure-(1) shows the T - S diagram for the Rankine cycle. Figure-(1) In Figure-(1), x -axis shows the entropy change and y -axis shows the temperature. Write the expression for dryness fraction at state 2. x 2 = s 1 − s f ( s f g ) (I) Here, entropy at state 1 is s 1 , entropy at saturated liquid state is s f , dryness fraction is x 2 , and difference in entropy at saturated vapor state is s f g . Write the expression for enthalpy at state 2. h 2 = h f + x 2 h f g (II) Here, enthalpy at state 2 is h 2 , enthalpy at saturated liquid state is h f , enthalpy change at saturated vapor state is h f g and dryness fraction is x 2 . Write the expression for dryness fraction at state 3. x 3 = s 2 − s f ( s f g ) (III) Here, entropy at state 3 is s 3 , entropy at saturated liquid state is s f , dryness fraction is x 3 , difference in entropy at saturated vapor state is s f g . Write the expression for enthalpy at state 3. h 3 = h f + x 3 h f g (IV) Here, enthalpy at state 3 is h 3 , enthalpy at saturated liquid state is h f , enthalpy change at saturated vapor state is h f g and dryness fraction is x 3 . Write the expression for dryness fraction at state 4. x 4 = s 3 − s f ( s f g ) (V) Here, entropy at state 4 is s 4 , entropy at saturated liquid state is s f , dryness fraction is x 4 , difference in entropy at saturated vapor state is s f g . Write the expression for enthalpy at state 4. h 4 = h f + x 4 h f g (VI) Here, enthalpy at state 4 is h 4 , enthalpy at saturated liquid state is h f , enthalpy change at saturated vapor state is h f g and dryness fraction is x 4 . Write the expression for the pump work for the process (5-6). W p 1 = ∫ − v d p = − v f ∫ d p = − v f ( p 6 − p 5 ) (VII) Here, pump work is W p 1 , volume at saturated vapor phase is v f , pressure at state 5 is p 5 and pressure at state 6 is p 6 . Write the energy equation for the pump for the process 5-6. Q + h 5 = w p 1 + h 6 0 + h 5 = w p 1 + h 6 (VIII) Here, heat transfer is Q , enthalpy at state 5 is h 5 and enthalpy at state 6 is h 6 Since the pump is isolated, so the heat transfer to and from the pump is zero, that is Q = 0 . Write the expression for the pump work for the process 7-8. W p 2 = ∫ − v d p = − v f ∫ d p = − v f ( p 8 − p 7 ) (IX) Here, pump work is W p 2 , volume at saturated vapor phase is v 3 , pressure at state 7 is p 7 and pressure at state 8 is p 8 . Write the energy equation for the pump work for the process 7-8. Q + h 7 = w p 2 + h 8 0 + h 7 = w p 2 + h 8 (X) Here, enthalpy at state 7 is h 7 and enthalpy at state 8 is h 8 Write the expression for extraction of steam at state 2. 0 = y ( h 2 − h 10 ) − ( h 8 − h 9 ) y = ( h 9 − h 8 ) ( h 2 − h 10 ) (XI) Here, enthalpy at state 9 is h 9 , extraction of steam at state 2 is y and enthalpy at state 10 is h 10 . Write expression for mass flow rate at state 2 m ˙ 2 = y m ˙ 1 (XII) Here, mass flow rate at state 1 is m ˙ 1 . Write expression for mass flow rate at state 3 m ˙ 3 = y ′ m ˙ 1 (XIII) Write expression for mass flow rate at state 6 m ˙ 6 = ( 1 − y − y ′ ) m ˙ 1 (XIV) Write expression for exergy destruction in close feed water heater ( Δ Φ ˙ ) c l o s e h e a t e r = T o [ m ˙ 2 ( s 10 − s 2 ) + m ˙ 1 ( s 9 − s 8 ) ] (XV) Here, entropy at state 10 is s 10 , entropy at state 9 is s 9 , entropy at state 2 is s 2 , entropy at state 8 is s 8 ,surrounding temperature is T o . Write expression for exergy destruction in open feed water heater ( Δ Φ ˙ ) o p e n h e a t e r = T o [ m ˙ 1 s 7 − m ˙ 3 s 3 − m ˙ 6 s 6 − m ˙ 2 s 11 ] (XVI) Here, entropy at state 7 is s 7 , entropy at state 3 is s 3 , entropy at state 6 is s 6 , entropy at state 11 is s 11 ,surrounding temperature is T o . Write the expression for rate of exergy flow in steam generator. ( E ˙ 1 − E ˙ 2 ) s t e a m g e n e a r t o r = m ˙ 1 [ ( h 1 − h 9 ) − T o ( s 1 − s 9 ) ] (XVII) Write expression for fraction of flow exergy in open heater α = ( Δ Φ ˙ ) c l o s e h e a t e r ( E ˙ 1 − E ˙ 2 ) s t e a m g e n e a r t o r (XVIII) Write expression for fraction of flow exergy in close heater β = ( Δ Φ ˙ ) c l o s e h e a t e r ( E ˙ 1 − E ˙ 2 ) s t e a m g e n e a r t o r (XIX) Conclusion: From Table A-4, “Properties of superheated water vapor”, obtain 3465.4 kJ / kg for h 1 , and 6.5132 kJ / kg ⋅ K for s 1 at pressure 16 Mpa and inlet temperature 560 ° C . From Table A-2, “Properties of superheated water vapor”, obtain 6.5132 kJ / kg ⋅ K for s 2 , 3015.4 kJ / kg for h f , 101.8 kJ / kg for h f g , 6.4553 kJ / kg ⋅ K for s f and 0.16 kJ / kg ⋅ K for s f g at pressure 4 Mpa for p 2 . Substitute 6.5132 kJ / kg ⋅ K for s 1 , 6.4553 kJ / kg ⋅ K for s f and 0.16 kJ / kg ⋅ K for s f g in Equation (I). 6.5132 kJ / kg ⋅ K = 6.4553 kJ / kg ⋅ K + x 2 0.16 kJ / kg ⋅ K x 2 = 0.0579 kJ / kg ⋅ K 0.16 kJ / kg ⋅ K x 2 = 0.36 Substitute 0.36 for x 2 , 3015.4 kJ / kg for h f and 101.8 kJ / kg for h f g in Equation (II). h 2 = 3015.4 kJ / kg + 0.36 ( 101.8 kJ / kg ) = 3015.4 kJ / kg + 36.648 kJ / kg = 3050.864 kJ / kg From Table A-3, “Properties of superheated water vapor” obtain 6.5132 kJ / kg ⋅ K for s 2 , 561.47 kJ / kg for h f , 2163.8 kJ / kg for h f g , 1.6718 kJ / kg ⋅ K for s f , 1.0732 × 10 − 3 m 3 / kg for V 3 and 5.3201 kJ / kg ⋅ K for s f g at pressure 0.3 Mpa . Substitute 6.5132 kJ / kg ⋅ K for s 2 , 1.6718 kJ / kg ⋅ K for s f and 5.3201 kJ / kg ⋅ K for s f g in Equation (III). 6.5132 kJ / kg ⋅ K = 1.6718 kJ / kg ⋅ K + x 3 5.3201 kJ / kg ⋅ K x 3 = 4.8312 kJ / kg ⋅ K 5.3201 kJ / kg ⋅ K x 3 = 0.91 Substitute 0.91 for x 3 , 561.47 kJ / kg for h f and 2163.8 kJ / kg for h f g in Equation (IV). h 3 = 561.47 kJ / kg + 0.91 ( 2163.8 kJ / kg ) = 561.47 kJ / kg + 1969.058 kJ / kg = 2530.528 kJ / kg From Table A-3, “Properties of saturated water vapor”, obtain 6.5132 kJ / kg ⋅ K for s 3 , 173.88 kJ / kg for h f , 2403.1 kJ / kg for h f g , 0.5926 kJ / kg ⋅ K for s f , 1.0084 × 10 − 3 m 3 / kg for V 4 and 7.6361 kJ / kg ⋅ K for s f g at pressure 8 kpa . Substitute 6.5132 kJ / kg ⋅ K for s 3 , 0.5926 kJ / kg ⋅ K for s f and 7.6361 kJ / kg ⋅ K for s f g in Equation (V). 6.5132 kJ / kg ⋅ K = ( 0.5926 kJ / kg ⋅ K ) + x 4 ( 7.6361 kJ / kg ⋅ K ) x 4 = 5.9386 kJ / kg ⋅ K 7.6361 kJ / kg ⋅ K x 4 = 0.7753 Substitute 0.7753 for x 4 , 173.88 kJ / kg for h f and 2403.1 kJ / kg for h f g in Equation (VI). h 4 = 173.88 kJ / kg + 0.7753 ( 2403.1 kJ / kg ) = 173.88 kJ / kg + 1863.12 kJ / kg = 2037.003 kJ / kg From Table A-3, “Properties of saturated water vapor”, obtain 173.88 kJ / kg for h 5 at pressure 0

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 87P

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The fraction of flow energy increase in steam generator for open feed water pump.

The fraction of flow energy increase in steam generator for closed feed water pump.

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Chapter 8.6, Problem 86P

Chapter 8.6, Problem 88P

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Chapter 8 Solutions

Fundamentals of Engineering Thermodynamics, Binder Ready Version

Ch. 8.6 - Prob. 1E Ch. 8.6 - Prob. 2E Ch. 8.6 - Prob. 3E Ch. 8.6 - Prob. 4E Ch. 8.6 - Prob. 5E Ch. 8.6 - Prob. 6E Ch. 8.6 - Prob. 7E Ch. 8.6 - 8. What is the relationship between global climate...Ch. 8.6 - Prob. 9E Ch. 8.6 - Prob. 10E

Ch. 8.6 - Prob. 11E Ch. 8.6 - Prob. 12E Ch. 8.6 - Prob. 13E Ch. 8.6 - Prob. 1CU Ch. 8.6 - Prob. 2CU Ch. 8.6 - 3. The component of the Rankine cycle in which the...Ch. 8.6 - 4. A cycle that couples two vapor cycles so the...Ch. 8.6 - 5. The ratio of the pump work input to the work...Ch. 8.6 - 6. A shell-and-tube-type recuperator in which the...Ch. 8.6 - Prob. 7CU Ch. 8.6 - Prob. 8CU Ch. 8.6 - Prob. 9CU Ch. 8.6 - Prob. 10CU Ch. 8.6 - 11. An example of an external irreversibility...Ch. 8.6 - Prob. 12CU Ch. 8.6 - Prob. 13CU Ch. 8.6 - Prob. 14CU Ch. 8.6 - 15. A direct-contact–type heat exchanger found in...Ch. 8.6 - 16. The component of a regenerative vapor power...Ch. 8.6 - Prob. 17CU Ch. 8.6 - 18. A Rankine cycle that employs an organic...Ch. 8.6 - Prob. 19CU Ch. 8.6 - Prob. 20CU Ch. 8.6 - Prob. 21CU Ch. 8.6 - Prob. 22CU Ch. 8.6 - Prob. 23CU Ch. 8.6 - 24. The purpose of deaeration is ______________. Ch. 8.6 - Prob. 25CU Ch. 8.6 - Prob. 26CU Ch. 8.6 - Prob. 27CU Ch. 8.6 - Prob. 28CU Ch. 8.6 - 29. The total cost associated with a power plant...Ch. 8.6 - Prob. 30CU Ch. 8.6 - Prob. 31CU Ch. 8.6 - Prob. 32CU Ch. 8.6 - Prob. 33CU Ch. 8.6 - Prob. 34CU Ch. 8.6 - Prob. 35CU Ch. 8.6 - Prob. 36CU Ch. 8.6 - Prob. 37CU Ch. 8.6 - Prob. 38CU Ch. 8.6 - Prob. 39CU Ch. 8.6 - 40. For a vapor power cycle with and , the...Ch. 8.6 - Prob. 41CU Ch. 8.6 - Prob. 42CU Ch. 8.6 - Prob. 43CU Ch. 8.6 - Prob. 44CU Ch. 8.6 - Prob. 45CU Ch. 8.6 - Prob. 46CU Ch. 8.6 - Prob. 47CU Ch. 8.6 - Prob. 48CU Ch. 8.6 - Prob. 49CU Ch. 8.6 - 50. In a binary cycle, energy discharged by heat...Ch. 8.6 - Prob. 1P Ch. 8.6 - Prob. 2P Ch. 8.6 - Prob. 3P Ch. 8.6 - Prob. 6P Ch. 8.6 - 8.7 Water is the working fluid in an ideal Rankine...Ch. 8.6 - Prob. 8P Ch. 8.6 - 8.10 Water is the working fluid in an ideal...Ch. 8.6 - Prob. 12P Ch. 8.6 - Prob. 13P Ch. 8.6 - 8.14 On the south coast of the island of Hawaii,...Ch. 8.6 - Prob. 15P Ch. 8.6 - 8.17. Water is the working fluid in a Rankine...Ch. 8.6 - 8.19 Water is the working fluid in a Rankine...Ch. 8.6 - Prob. 20P Ch. 8.6 - Prob. 21P Ch. 8.6 - 8.22 Superheated steam at 8 MPa and 480°C leaves...Ch. 8.6 - Prob. 23P Ch. 8.6 - Prob. 25P Ch. 8.6 - Prob. 26P Ch. 8.6 - 8.27 Steam is the working fluid in the ideal...Ch. 8.6 - Prob. 28P Ch. 8.6 - Prob. 29P Ch. 8.6 - Prob. 30P Ch. 8.6 - Prob. 31P Ch. 8.6 - 8.32 An ideal Rankine cycle with reheat uses water...Ch. 8.6 - Prob. 33P Ch. 8.6 - 8.34 Steam at 4800 lbf/in.2, 1000℉ enters the...Ch. 8.6 - Prob. 35P Ch. 8.6 - Prob. 37P Ch. 8.6 - 8.38 For the cycle of Problem 8.37, reconsider the...Ch. 8.6 - Prob. 39P Ch. 8.6 - Prob. 40P Ch. 8.6 - Prob. 41P Ch. 8.6 - Prob. 42P Ch. 8.6 - Prob. 43P Ch. 8.6 - Prob. 44P Ch. 8.6 - Prob. 45P Ch. 8.6 - Prob. 46P Ch. 8.6 - Prob. 47P Ch. 8.6 - 8.48 For the cycle of Problem 8.47, investigate...Ch. 8.6 - Prob. 49P Ch. 8.6 - Prob. 50P Ch. 8.6 - Prob. 51P Ch. 8.6 - 8.52 As indicated in Fig. P8.52, a power plant...Ch. 8.6 - Prob. 53P Ch. 8.6 - Prob. 54P Ch. 8.6 - Prob. 55P Ch. 8.6 - Prob. 56P Ch. 8.6 - Prob. 57P Ch. 8.6 - Prob. 58P Ch. 8.6 - Prob. 59P Ch. 8.6 - Prob. 60P Ch. 8.6 - Prob. 61P Ch. 8.6 - Prob. 63P Ch. 8.6 - Prob. 64P Ch. 8.6 - Prob. 65P Ch. 8.6 - Prob. 66P Ch. 8.6 - 8.67 Water is the working fluid in a Rankine cycle...Ch. 8.6 - Prob. 68P Ch. 8.6 - Prob. 69P Ch. 8.6 - Prob. 70P Ch. 8.6 - 8.72 Water is the working fluid in a...Ch. 8.6 - Prob. 73P Ch. 8.6 - Prob. 74P Ch. 8.6 - Prob. 75P Ch. 8.6 - 8.76 A binary vapor power cycle consists of two...Ch. 8.6 - A binary vapor cycle consists of two Rankine...Ch. 8.6 - Prob. 78P Ch. 8.6 - Prob. 79P Ch. 8.6 - Prob. 80P Ch. 8.6 - 8.81 Figure P8.81 shows a combined heat and power...Ch. 8.6 - 8.82 Figure P8.82 shows a cogeneration cycle that...Ch. 8.6 - Prob. 83P Ch. 8.6 - 8.84 The steam generator of a vapor power plant...Ch. 8.6 - 8.85 Determine the exergy input, in kJ per kg of...Ch. 8.6 - 8.86 In the steam generator of the cycle of...Ch. 8.6 - Prob. 87P Ch. 8.6 - 8.88 Determine the rate of exergy input, in Btu/h,...Ch. 8.6 - Prob. 89P Ch. 8.6 - Prob. 90P Ch. 8.6 - Prob. 91P Ch. 8.6 - 8.92 Figure P8.92 provides steady-state operating...Ch. 8.6 - 8.93 Steam enters the turbine of a simple vapor...

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Chapter 8 Solutions

The fraction of flow energy increase in steam generator for open feed water pump.

The fraction of flow energy increase in steam generator for closed feed water pump.