To determine The outputs, losses and destructions of exergy in the cycle. Explanation The following figure shows the T - s diagram of the cycle. Figurer-(1) Write the expression for dryness fraction. x 6 s = s 5 − s f 6 s f g 6 (I) Here, specific entropy of liquid at point 6 is s f 6 , specific entropy of liquid vapor mixture at state 6 is s f g 6 , specific entropy at point 5 is s 5 and dryness factor at state 6 s is x 6 s . Write the expression for enthalpy at state 6. h 6 = h f 6 + x 6 ( h f g 6 ) (II) Here, specific enthalpy of liquid at point 6 is h f 6 , specific enthalpy of liquid vapor mixture is h f g 6 . Write the expression for enthalpy at state 8. h 8 = h 7 + v 7 ( p 8 − p 7 ) (III) Here, specific enthalpy at state 8 is h 8 , pressure at state 7 is p 7 and pressure at state 8 is p 8 . Write the expression for specific enthalpy at state 10. h 10 = h 9 + v 9 ( p 10 − p 9 ) η p (IV) Here, specific enthalpy at state 9 is h 9 , pressure at state 10 is p 10 the isentropic efficiency of pump is η p and pressure at state 9 is p 9 . Write the expression for mass fraction extracted at first stage. y ′ = h 11 − h 10 h 2 − h 12 (V) Here, specific enthalpy at state 11 is h 11 , specific enthalpy at state 12 is h 12 , specific enthalpy at state 10 is h 10 and specific enthalpy at state 2 is h 2 . Write the expression for mass fraction extracted at second stage. y ′ ′ = ( h 9 − h 8 ) + y ′ ( h 8 − h 13 ) ( h 5 − h 8 ) (VI) Here, specific enthalpy at state 9 is h 9 , specific enthalpy at state 8 is h 8 , specific enthalpy at state 13 is h 13 and specific enthalpy at state 5 is h 5 . Write expression for mass flow rate entering first stage of turbine. m ˙ 1 = W n e t [ ( h 1 − h 2 ) + ( 1 − y ′ ) ( h 4 − h 5 ) + ( 1 − y ′ − y ′ ′ ) ( h 5 − h 6 ) − ( 1 − y ′ − y ′ ′ ) ( h 8 − h 7 ) − ( h 10 − h 9 ) ] (VII) Here, mass flow rate of the steam is m ˙ 1 , power output of the cycle is W n e t , enthalpy at state 1 is h 1 the specific enthalpy at state 4 is h 4 the specific enthalpy at state 5 is h 5 the specific enthalpy at state 6 is h 6 and the specific enthalpy at state 7 is h 7 . Write expression for rate of exergy input in steam generator and reheater. Φ ˙ = m ˙ 1 [ ( h 1 − h 11 ) − T o ( s 1 − s 11 ) + ( 1 − y ′ ) { ( h 4 − h 3 ) − T o ( s 4 − s 3 ) } ] (VIII) Write expression for rate of exergy loss from cycle in condenser. Φ ˙ l o s s = m ˙ 1 [ ( 1 − y ′ − y ′ ′ ) { ( h 6 − h 7 ) − T o ( s 6 − s 7 ) } ] (IX) Here, specific entropy at 6 is s 6 , specific entropy at 7 is s 7 , specific enthalpy at 6 is h 6 and temperature of surrounding is T o . Write expression for rate of exergy destruction in the turbine. Ψ T u r b i n e = T o m ˙ 1 [ y ′ s 2 + ( 1 − y ′ ) s 3 + y ′ ′ s 5 + ( 1 − y ′ − y ′ ′ ) s 6 − s 1 − ( 1 − y ′ ) s 4 ] (X) Here, specific entropy at 5 is s 5 , specific entropy at 3 is s 3 , specific enthalpy at 4 is h 4 , and specific enthalpy at 4 is h 4 . Write expression for rate of exergy destruction in pump 1. Ψ p u m p ( 1 ) = T o m ˙ 1 [ ( 1 − y ′ − y ′ ′ ) ( s 8 − s 7 ) ] (XI) Here, specific entropy at 8 is s 8 . Write expression for rate of exergy destruction in pump 2. Ψ p u m p ( 2 ) = T o m ˙ 1 ( s 10 − s 9 ) (XII) Here, specific entropy at 9 is s 9 and specific entropy at 10 is s 10 . Write expression for rate of exergy destruction in close feed water heater. Ψ C l o s e h e a t e r = T o m ˙ 1 [ ( s 11 − s 10 ) + y ′ ( s 12 − s 2 ) ] (XIII) Here, specific entropy at 12 is s 12 . Write expression for rate of exergy destruction in open feed water heater. Ψ O p e n h e a t e r = T o m ˙ 1 [ s 9 − y ′ ′ s 5 − ( 1 − y ′ − y ' ' ) s 8 − y ′ s 13 ] (XIV) Here, specific entropy at 13 is s 13 Write expression for rate of exergy destruction in trap. Ψ T r a p = T o m ˙ 1 y ′ ( s 13 − s 12 ) (XV) Conclusion: Refer to Table A-4E “Properties of superheated water vapor” to obtain specific enthalpy ( h 1 ) at state 1 as 1493.5 Btu / lb and specific entropy ( s 1 ) at state 1 as 1.6094 Btu / lb ⋅ ° R corresponding to pressure 1400 lb / in 2 and temperature 1000 ° F . Since the process 1-2s is isentropic process so s 1 = s 2 s . Since the points 2 s and 2 are on the same constant pressure line on the T - s diagram, so p 2 s = p 2 . Refer table A-4E “Properties of superheated water vapor”, to obtain the specific enthalpy at the pressure p 2 = 500 lb / in 2 and specific entropy s 1 = 1.6094 Btu / lb ⋅ ° R by using interpolation method of two variables. Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (XVI) Here, the variables denote by x and y are specific entropy and specific enthalpy. Show the values of specific enthalpies corresponding to the specific entropy of 1 .5585 Btu / lb ⋅ ° R and 1 .6112 Btu / lb ⋅ ° R as in below table. Specific entropy Specific enthalpy x 1 = 1 .5585 Btu / lb ⋅ ° R y 1 = 1298.3 Btu / lb x 2 = 1.6094 Btu / lb ⋅ ° R y 2 = ? x 3 = 1 .6112 Btu / lb ⋅ ° R y 3 = 1356.7 Btu / lb Substitute 1 .5585 Btu / lb ⋅ ° R for x 1 , 1.6094 Btu / lb ⋅ ° R for x 2 , 1 .6112 Btu / lb ⋅ ° R for x 3 , 1298.3 Btu / lb for y 1 , and 1356.7 Btu / lb for y 3 in Equation (XVI). y 2 = ( 1.6094 Btu / lb ⋅ ° R − 1 .5585 Btu / lb ⋅ ° R ) ( 1356.7 Btu / lb − 1298.3 Btu / lb ) ( 1 .6112 Btu / lb ⋅ ° R − 1 .5585 Btu / lb ⋅ ° R ) + ( 1298.3 Btu / lb ) = ( 0.0509 Btu / lb ⋅ ° R ) ( 58.4 Btu / lb ) ( 0.0527 Btu / lb ⋅ ° R ) + ( 1298.3 Btu / lb ) = ( 56.4053 Btu / lb ) + ( 1298.3 Btu / lb ) = 1354.70 Btu / lb So, the specific enthalpy at state 2 s is h 2 s = 1354.70 Btu / lb . Write the expression for enthalpy at point 2. h 2 = h 1 − η t ( h 1 − h 2 s ) (XVII) Here, the specific enthalpy at point 1 is h 1 and the isentropic efficiency of turbine is η t . Substitute 1493.5 Btu / lb for h 1 , 0.85 for η t and 1354.71 Btu / lb for h 2 s in Equation (XVII). h 2 = ( 1493.5 Btu / lb ) − 0.85 ( 1493.5 Btu / lb − 1354.71 Btu / lb ) = ( 1493.5 Btu / lb ) − ( 117.97 Btu / lb ) = 1375.55 Btu / lb Show the values of specific entropies corresponding to the specific enthalpy of 1356.7 Btu / lb and 1412.1 Btu / lb as in below table. Specific enthalpy Specific entropy x 1 = 1356.7 Btu / lb y 1 = 1 .6112 Btu / lb ⋅ R x 2 = 1375.55 Btu / lb y 2 = ? x 3 = 1412.1 Btu / lb y 3 = 1 .6571 Btu / lb ⋅ R Substitute 1356.7 Btu / lb for x 1 , 1375.55 Btu / lb for x 2 , 1412.1 Btu / lb for x 3 , 1 .6112 Btu / lb ⋅ R for y 1 , and 1 .6571 Btu / lb ⋅ R for y 3 in Equation (XVI). y 2 = ( 1375.55 Btu / lb − 1356.7 Btu / lb ) ( 1 .6571 Btu / lb ⋅ R − 1 .6112 Btu / lb ⋅ R ) ( 1412.1 Btu / lb − 1356.7 Btu / lb ) + ( 1 .6112 Btu / lb ⋅ R ) = ( 18.85 Btu / lb ) ( 0 .0459 Btu / lb ⋅ R ) ( 55.4 Btu / lb ) + ( 1 .6112 Btu / lb ⋅ R ) = ( 0.0156 Btu / lb ⋅ R ) + ( 1 .6112 Btu / lb ⋅ R ) = 1.6268 Btu / lb ⋅ R So, the specific entropy at state 2 is 1.6268 Btu / lb ⋅ R . Refer to Table A-4E “Properties of superheated water vapor” to obtain enthalpy ( h 1 ) at state 1 as 1466.5 Btu / lb and entropy ( s 1 ) at state 1 as 1.6987 Btu / lb ⋅ ° R corresponding to pressure 500 lb / in 2 and temperature 900 ° F . Since the process 4-5 is isentropic process so s 4 = s 5 . Since 5 and 4 point on same constant pressure line on the T − s diagram so p 2 s = p 2 . Refer table A-4E “Properties of superheated water vapor” to obtain the specific enthalpy at the pressure p 2 = 120 lb / in 2 and specific entropy s 1 = 1.6987 Btu / lb ⋅ ° R by using interpolation method of two variables. Show the values of specific enthalpies corresponding to the specific entropy of 1 .6868 Btu / lb ⋅ ° R and 1 .7371 Btu / lb ⋅ ° R as in below table. Specific entropy Specific enthalpy x 1 = 1 .6868 Btu / lb ⋅ ° R y 1 = 1277.1 Btu / lb x 2 = 1.6987 Btu / lb ⋅ ° R y 2 = ? x 3 = 1 .7371 Btu / lb ⋅ ° R y 3 = 1327.8 Btu / lb Substitute 1 .6868 Btu / lb ⋅ ° R for x 1 , 1.6987 Btu / lb ⋅ ° R for x 2 , 1 .7371 Btu / lb ⋅ ° R for x 3 , 1277.1 Btu / lb for y 1 and 1327.8 Btu / lb for y 3 in Equation (XVI). y 2 = ( 1.6987 Btu / lb ⋅ ° R − 1 .6868 Btu / lb ⋅ ° R ) ( 1327.8 Btu / lb − 1277.1 Btu / lb ) ( 1 .7371 Btu / lb ⋅ ° R − 1 .6868 Btu / lb ⋅ ° R ) + ( 1277.1 Btu / lb ) = ( 0.0119 Btu / lb ⋅ ° R ) ( 50.7 Btu / lb ) ( 0.0503 Btu / lb ⋅ ° R ) + ( 1277.1 Btu / lb ) = ( 11.9946 Btu / lb ) + ( 1277.1 Btu / lb ) = 1289.094 Btu / lb So, the specific enthalpy at state 5 s is 1289.094 Btu / lb . Write the expression for enthalpy at point 5. h 5 = h 4 − η t ( h 4 − h 5 s ) (XVIII) Here, the specific enthalpy at point 4 is h 4 , the specific enthalpy at point 5 s is h 5 s and the isentropic efficiency of turbine is η t . Substitute 1466.5 Btu / lb for h 4 , 0.85 for η t and 1289.09 Btu / lb for h 5 s in Equation (XVIII). h 5 = ( 1466.5 Btu / lb ) − 0.85 ( 1466.5 Btu / lb − 1289.09 Btu / lb ) = ( 1466.5 Btu / lb ) − ( 150 Btu / lb ) = 1315.70 Btu / lb Show the specific entropy at pressure p 2 = 120 lb / in 2 and specific enthalpy h 2 = 1315.70 Btu / lb as in Table. Show the values of specific entropies corresponding to the specific enthalpy of 1277.1 Btu / lb and 1327.8 Btu / lb . Specific enthalpy Specific entropy x 1 = 1277.1 Btu / lb y 1 = 1 .6868 Btu / lb ⋅ R x 2 = 1315.70 Btu / lb y 2 = ? x 3 = 1327.8 Btu / lb y 3 = 1 .7371 Btu / lb ⋅ R Substitute 1277.1 Btu / lb for x 1 , 1315.70 Btu / lb for x 2 , 1327.8 Btu / lb for x 3 , 1 .6868 Btu / lb ⋅ R for y 1 , and 1 .7371 Btu / lb ⋅ R for y 3 in Equation (XVI). y 2 = ( 1315.70 Btu / lb − 1277.1 Btu / lb ) ( 1 .7371 Btu / lb ⋅ R − 1 .6868 Btu / lb ⋅ R ) ( 1327.8 Btu / lb − 1277.1 Btu / lb ) + ( 1 .6868 Btu / lb ⋅ R ) = ( 38.6 Btu / lb ) ( 0 .0503 Btu / lb ⋅ R ) ( 50.7 Btu / lb ) + ( 1 .6868 Btu / lb ⋅ R ) = ( 0.03829 Btu / lb ⋅ R ) + ( 1 .6112 Btu / lb ⋅ R ) = 1.7151 Btu / lb ⋅ R So, the specific entropy at state 5 is 1.7151 Btu / lb ⋅ R . Refer to Table A-3E “Properties of saturated water-pressure table” to obtain isentropic entropy of saturated liquid ( s f 6 ) as 0.1750 Btu / lb ⋅ ° R , isentropic entropy of liquid-vapor ( s f g 6 ) as 1.7448 Btu / lb ⋅ ° R , the specific volume v f 7 of liquid as 0.01623 ft 3 / lb , isentropic enthalpy of saturated liquid ( h f 6 ) as 94.02 Btu / lb and isentropic enthalpy of liquid-vapor ( h f g 6 ) as 1022.1 Btu / lb corresponding to state 6 at pressure 2 lbf / in 2 . Since the process 5-6 s is isentropic process, so the entropies at states 5 and 6 s are equal, that is s 6 = s 5 s = 1.7251 Btu / lb ⋅ ° R Substitute 1.7251 Btu / lb ⋅ ° R for s 5 s , 0.1750 Btu / lb ⋅ ° R for s f 6 s and 1.7448 Btu / lb ⋅ ° R for s f g 6 s in Equation (II). x 6 s = ( 1.7251 Btu / lb ⋅ ° R − 0.1750 Btu / lb ⋅ ° R ) ( 1.7448 Btu / lb ⋅ ° R ) = ( 1.5501 Btu / lb ⋅ ° R ) ( 1.7448 Btu / lb ⋅ ° R ) = 0.8884 Substitute 94.02 Btu / lb for h f 6 , 0.8884 for x 6 s and 1022.1 Btu / lb for h f g 6 in Equation (III). h 6 = ( 94.02 Btu / lb ) + ( 0.8884 ) ( 1022.1 Btu / lb ) = ( 94.02 Btu / lb ) + ( 908.033 Btu / lb ) = 1002.053 Btu / lb Write the expression for enthalpy at point 6. h 6 = h 5 − η t ( h 5 − h 6 s ) (XIX) Here, the specific enthalpy at point 5 is h 5 , the specific enthalpy at point 6s is h 6 s and the isentropic efficiency of turbine is η t . Substitute 1315.70 Btu / lb for h 5 , 0.85 for η t and 1000.02 Btu / lb for h 6 s in Equation (XIX). h 6 = ( 1315.70 Btu / lb ) − 0.85 ( 1315.70 Btu / lb − 1000.02 Btu / lb ) = ( 1315.70 Btu / lb ) − ( 268.328 Btu / lb ) = 1047.37 Btu / lb Substitute 94.02 Btu / lb for h 7 , 0.01623 ft 3 / lb for v 7 , 2 lbf / in 2 for p 7 , 0.85 for η p , and 120 lbf / in 2 for p 8 in Equation (III). h 8 = ( 94.02 Btu / lb ) + ( 0.01623 ft 3 / lb ) ( 120 lbf / in 2 − 2 lbf / in 2 ) 0.85 = ( 94.02 Btu / lb ) + ( 0

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832318

Author: MORAN

Publisher: WILEY

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Chapter 8.6, Problem 88P

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Chapter 8.6, Problem 87P

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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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