(a) To determine The net power developed. Explanation Figure-(1) shows the T-s diagram for the Rankine cycle. Figure-(1) The 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, specific entropy at state 1 is s 1 , specific entropy at saturated liquid state is s f , dryness fraction is x 2 , specific entropy of the vaporization is s f g . Write the expression for specific enthalpy at state 2. h 2 = h f + x 2 h f g (II) Here, specific enthalpy at state 2 is h 2 , specific enthalpy at saturated liquid state is h f , specific enthalpy of vaporization 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, specific entropy at state 3 is s 3 , specific entropy at saturated liquid state is s f , dryness fraction is x 3 , specific entropy of vaporization is s f g . Write the expression for enthalpy at state 3. h 3 = h f + x 3 h f g (IV) Here, specific enthalpy at state 3 is h 3 , specific enthalpy at saturated liquid state is h f , specific enthalpy of vaporization 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, specific entropy at state 4 is s 4 , specific entropy at saturated liquid state is s f , dryness fraction is x 4 , specific entropy of vaporization is s f g . Write the expression for enthalpy at state 4. h 4 = h f + x 4 h f g (VI) Here, specific enthalpy at state 4 is h 4 , specific enthalpy at saturated liquid state is h f , specific enthalpy of vaporization 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 f ( p 6 − p 5 ) (VII) Here, pump work is W p 1 , volume at saturated liquid 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). h 5 = w p 1 + h 6 (VIII) Here, specific enthalpy at state 5 is h 5 and specific enthalpy at state 6 is h 6 . Write the expression for the pump work for the process (7-8). W p 2 = − v f ( p 8 − p 7 ) (IX) Here, pump work is W p 2 , volume at saturated liquid phase is v f , 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). h 7 = w p 2 + h 8 (X) Here, specific enthalpy at state 7 is h 7 and specific enthalpy at state 8 is h 8 . Write the expression for extraction of steam at state 2. y = ( h 9 − h 8 ) ( h 2 − h 10 ) (XI) Here, specific enthalpy at state 9 is h 9 , extraction of steam at state 2 is y and specific enthalpy at state 10 is h 10 . Write the expression for mass and energy balance to open feed water heater. ( 1 − y − y ′ ) h 6 + y h 11 + y ′ h 3 − h 7 = 0 (XII) Write the expression for turbine work output. W T = ( h 1 − h 2 ) + ( 1 − y ) ( h 2 − h 3 ) + ( 1 − y − y ′ ) ( h 3 − h 4 ) (XIII) Here, extraction of steam at state 3 is y ′ . Write the expression for pump work input. W P = ( 1 − y − y ′ ) W p 1 + W p 2 (XIV) Write the expression for net power developed. W ˙ n e t = m ˙ ( W T + W P ) (XV) Here, mass flow rate is m ˙ . Conclusion: From Table A-4, “Properties of superheated water vapor” to obtain 3465.4 kJ / kg for h 1 , and 6.5132 kJ / kg ⋅ K for s 1 at pressure 16 MPa and temperature 560 ° C . From Table A-2, “Properties of superheated water vapor” 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 and saturation temperature. 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” to 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 f 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 f 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 (b) To determine The heat transfer rate to the working fluid passing through the steam generator. (c) To determine The thermal efficiency. (d) To determine The mass flow rate of the cooling water passing through the condenser.

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832301

Author: SHAPIRO

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 8.6, Problem 67P

(a)

To determine

The net power developed.

(b)

To determine

The heat transfer rate to the working fluid passing through the steam generator.

(c)

To determine

The thermal efficiency.

(d)

To determine

The mass flow rate of the cooling water passing through the condenser.

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Chapter 8.6, Problem 66P

Chapter 8.6, Problem 68P

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A tensile specimen made of hot-rolled AISI 1020 steel is loaded to point corresponding to a strain of 43%. 60 Su = 66 ksi Stress σ (ksi) 20 Sy = 39 ksi Se = 36 ksi Hot-rolled 1020 steel F 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 Strain € (%) 0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 Area ratio R 0.1 0.2 0.3 0.4 0.5 Area reduction A, What value of area reduction is applicable to this location? 0.6

Chapter 8 Solutions

Fundamentals of Engineering Thermodynamics

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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