(a) To determine Net exergy increase of the air passing through the gas turbine combustor. Explanation The figure below shows the T − s diagram of the given cycle. Figure-(1) The process 1-2 is an isentropic compression process, process 2-3 is an isobaric heat addition process, 3-4 is an isentropic expansion process and 4-1 is an isobaric heat rejection process. Write the expression for the specific exergy destroyed in turbine. e ˙ d t = T 0 [ ( s 4 0 − s 3 0 ) − R ln ( p 4 p 3 ) ] (I) Here, the surrounding temperature is T 0 , the specific entropy at state 4 is s 4 0 , the specific entropy at state 3 is s 3 0 , the gas constant is R , the pressure at state 4 is p 4 and the pressure at state 3 is p 3 . Write the expression for the specific exergy destroyed in compressor. e ˙ d c = T 0 [ ( s 2 0 − s 1 0 ) − R ln ( p 2 p 1 ) ] (II) Here, the specific entropy at state 2 is s 2 0 , the specific entropy at state 1 is s 1 0 , the pressure at state 2 is p 2 and the pressure at state 1 is p 1 . Write the expression for the net exergy developed in the cycle. W ˙ n e t = m ˙ [ ( h 3 − h 4 ) − ( h 2 − h 1 ) ] W ˙ n e t m ˙ = ( h 3 − h 4 ) − ( h 2 − h 1 ) (III) Here, the specific enthalpy at state 1 is h 1 , the specific enthalpy at state 4 is h 4 , the specific enthalpy at state 2 is h 2 and the specific enthalpy at state 3 is h 3 . Write the expression for the net exergy loss in process 4-1. m ˙ ( e f 4 − e f 1 ) = m ˙ [ ( h 4 − h 1 ) − T 0 ( s 4 0 − s 1 0 − R ln ( p 4 p 1 ) ) ] ( e f 4 − e f 1 ) = ( h 4 − h 1 ) − T 0 ( s 4 0 − s 1 0 − R ln ( p 4 p 1 ) ) (IV) Here, specific flow exergy at state 1 is e f 1 , specific exergy flow at state 4 is e f 4 , the specific enthalpy at state 1 is h 1 , the specific enthalpy at state 4 is h 4 , the surrounding temperature is T 0 , the specific entropy at state 4 is s 4 0 , the specific entropy at state 1 is s 1 0 , the gas constant is R , the pressure at state 4 is p 4 and the pressure at state 1 is p 1 . Write the expression for the net exergy input to the heat exchanger. m ˙ ( e f 3 − e f 2 ) = m ˙ [ ( h 3 − h 2 ) − T 0 ( s 3 0 − s 2 0 − R ln ( p 3 p 2 ) ) ] e f 3 − e f 2 = ( h 3 − h 2 ) − T 0 ( s 3 0 − s 2 0 − R ln ( p 3 p 2 ) ) (V) Here, specific flow exergy at state 3 is e f 3 , and specific exergy flow at state 2 is e f 2 . Conclusion: Convert the unit of temperature at state 1 from ° F to ° R . T 1 = 80 ° F = ( 80 ° F + 459.67 ) ° R = 539.67 ° R ≈ 540 ° R Convert the unit of temperature at state 2 from ° F to ° R . T 2 = 514 ° F = ( 514 ° F + 459.67 ) ° R = 973.67 ° R ≈ 974 ° R Convert the unit of temperature at state 3 from ° F to ° R . T 3 = 1540 ° F = ( 1540 ° F + 459.67 ) ° R = 1999.67 ° R ≈ 2000 ° R Convert the unit of temperature at state 4 from ° F to ° R . T 4 = 917 ° F = ( 917 ° F + 459.67 ) ° R = 1376.67 ° R ≈ 1377 ° R Refer to table A-22E “Ideal gas properties of air” to obtain the specific enthalpy as 129.06 Btu / lb and specific entropy as 0.60078 Btu / lb ⋅ ° R with respect to 540 ° R . The specific enthalpy at state 2 is evaluated by using interpolation. Refer to table A-22E “Ideal gas properties of air” to obtain the specific enthalpy ( h 21 ) as 231.06 Btu / lb with respect to temperature T 21 as 960 ° R and the specific enthalpy ( h 22 ) as 236.02 Btu / lb with respect to temperature T 22 as 980 ° R . Write the expression for interpolation from two variables. T 2 − T 21 T 22 − T 21 = h 2 − h 21 h 22 − h 21 (VI) Here, the temperature is T and the specific enthalpy is h . Substitute 231.06 Btu / lb for h 21 , 960 ° R for T 21 , 236.02 Btu / lb for h 22 , 980 ° R for T 22 and 974 ° R for T 2 in Equation (VI). 974 ° R − 960 ° R 980 ° R − 960 ° R = h 2 − 231.06 Btu / lb 236.02 Btu / lb − 231.06 Btu / lb 14 20 = h 2 − 231.06 Btu / lb 4.96 Btu / lb h 2 − 231.06 Btu / lb = 3.472 Btu / lb h 2 ≈ 234.45 Btu / lb The specific entropy at state 2 is evaluated by using interpolation. Refer to table A-22E “Ideal gas properties of air” to obtain the specific entropy s 21 as 231.06 Btu / lb with respect to temperature T 21 as 960 ° R and the specific entropy s 22 as 236.02 Btu / lb with respect to temperature T 22 as 980 ° R . Write the expression for interpolation from two variables. T 2 − T 21 T 22 − T 21 = s 2 0 − s 21 s 22 − s 21 (VII) Here, the temperature is T and the specific entropy is s . Substitute 0.7403 Btu / lb ⋅ ° R for s 21 , 960 ° R for T 21 , 0.7454 Btu / lb for s 22 , 980 ° R for T 22 and 974 ° R for T 2 in Equation (VII). 974 ° R − 960 ° R 980 ° R − 960 ° R = s 2 0 − 0.7403 Btu / lb ⋅ ° R 0.7454 Btu / lb − 0.7403 Btu / lb ⋅ ° R 14 20 = s 2 0 − 0.7403 Btu / lb ⋅ ° R 0.0051 Btu / lb ⋅ ° R s 2 0 − 0.7403 Btu / lb ⋅ ° R = 0.00357 Btu / lb ⋅ ° R s 2 0 ≈ 0.7437 Btu / lb ⋅ ° R Refer to table A-22E “Ideal gas properties of air” to obtain the specific enthalpy as 504.71 Btu / lb and specific entropy as 0.93205 Btu / lb ⋅ ° R with respect to 2000 ° R . The specific enthalpy at state 4 is evaluated by using interpolation. Refer to table A-22E “Ideal gas properties of air” to obtain the specific enthalpy h 41 as 332.48 Btu / lb with respect to temperature T 41 as 1360 ° R and the specific enthalpy h 42 as 342.9 Btu / lb with respect to temperature T 42 as 1400 ° R . Write the expression for interpolation from two variables. T 4 − T 41 T 42 − T 41 = h 4 − h 41 h 42 − h 41 (VIII) Here, the temperature is T and the specific enthalpy is h . Conclusion: Substitute 332.48 Btu / lb for h 41 , 1360 ° R for T 41 , 342.9 Btu / lb for h 42 , 1400 ° R for T 42 and 1377 ° R for T 4 in Equation (VIII). 1377 ° R − 1360 ° R 1400 ° R − 1360 ° R = h 4 − 332.48 Btu / lb 342.9 Btu / lb − 332.48 Btu / lb 17 40 = h 4 − 332.48 Btu / lb 10.42 Btu / lb h 4 − 332.48 Btu / lb = 4.4285 Btu / lb h 4 ≈ 336 (b) To determine The exergetic efficiency of gas turbine.

Appendices to accompany Fundamentals of Engineering Thermodynamics, 8e

8th Edition

ISBN: 9781118957219

Author: Michael J. Moran, Howard N. Shapiro

Publisher: WILEY

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Chapter 9.14, Problem 58P

(a)

To determine

Net exergy increase of the air passing through the gas turbine combustor.

(b)

To determine

The exergetic efficiency of gas turbine.

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Chapter 9.14, Problem 57P

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Appendices to accompany Fundamentals of Engineering Thermodynamics, 8e

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