Explanation Write the expression for entropy change by using Redlich-Kwong equation for state 1 and state 2. s 2 − s 1 = R ¯ ln ( v 2 − b v 1 − b ) + a 2 b T 3 / 2 ln ( v 2 ( v 1 + b ) v 1 ( v 2 + b ) ) (I) Here, the specific entropy at state 1 is s 1 , the specific entropy at state 2 is s 2 , universal gas constant is R ¯ the specific volumes of ethylene at initial state is v 1 , the specific volumes of ethylene at final state is v 2 , the constants are a and b and temperature of ethylene is T c . Write the expression for the constant value a . a = a ′ R ¯ 2 T c 5 / 2 p c (II) Here, the critical pressure is p c . Write the expression for the constant value b . b = b ′ R ¯ T c p c (III) Here, p c and the critical temperature is T c and the constants are a ′ and b ′ . Write the expression for the initial specific volume by using ideal gas equation. v 1 = R ¯ T 1 p 1 (IV) Here the initial pressure is p 1 and the initial temperature is T 1 . Write the expression for the final specific volume by using Ideal gas equation. v 2 = R ¯ T 2 p 2 (V) Here the final pressure is p 2 and the final temperature is T 2 . Write the expression for the change in specific internal energy. u 2 − u 1 = 3 a 2 b T 1 / 2 ln ( v 2 ( v 1 + b ) v 1 ( v 2 + b ) ) (VI) Here, specific internal energy at initial state is u 1 and specific internal energy at final state is u 2 . Write the expression for the change in specific enthalpy. h 2 − h 1 = u 2 − u 1 + ( p 2 v 2 − p 1 v 1 ) (VII)) Here, specific enthalpy at final state is h 2 , specific enthalpy at initial state is h 1 pressure at final state is p 2 and pressure at initial state is p 1 . Conclusion: Refer to the Table A-24 “Constants for the Vander-Waals, Redlich-Kwong and Benedich-Webb-Rubin Equation of state” to obtain the value of Redlich-Kwong constants for ethylene as a ′ = 0.42748 and b ′ = 0.08664 . a ′ = 0.42748 b ′ = 0.08664 Refer to the Table A-21 “Critical properties of selected elements” to obtain the values of critical pressure ( p c ) as 51.2 bar and critical temperature ( T c ) as 283 K of ethylene. Substitute 8314 J / kmol ⋅ K for R ¯ , 283 K for T c , 51.2 bar for p c and 0.42748 for a ′ in Equation (II). a = 0.42748 × ( 831 4 J / kmol ⋅ K ) 2 × ( 283 K ) 5 / 2 ( 51.2 bar ) = 29548527.34 ( N ⋅ m / kmol ⋅ K ) 2 × ( 283 K ) 5 / 2 ( 51.2 bar ) ( 1 × 10 5 N / m 2 1 bar ) = ( 3.98108 × 10 13 N 2 ⋅ m 2 / kmol 2 ) ( 51.2 × 10 5 N / m 2 ) K 1 2 = 7775565.227 N m 2 m 6 kmol 2 K 1 2 a = 77.7556 × 10 5 N m 2 m 6 kmol 2 K 1 2 ( 1 bar 10 5 N / m 2 ) = 77.7556 bar m 6 kmol 2 K 1 2 = 77.7556 bar ( m 3 kmol ) 2 K 1 2 Substitute 8314 J / kmol ⋅ K for R ¯ , 283 K for T c , 51.2 bar for p c and 0.42748 for a ′ in Equation (III). b = 0

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

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

Publisher: WILEY

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Chapter 11.9, Problem 75P

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The change in specific enthalpy in kJ/kmol for ethylene between 400 K,1 bar and 400 K,100 bar.

The change in specific entropy in kJ/kmol for ethylene between 400 K,1 bar and 400 K,100 bar.

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Chapter 11.9, Problem 73P

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Ch. 11.9 - Prob. 1E Ch. 11.9 - Prob. 2E Ch. 11.9 - 3. What is an advantage of using the Redlich–Kwong...Ch. 11.9 - To determine the specific volume of superheated...Ch. 11.9 - Prob. 5E Ch. 11.9 - Prob. 6E Ch. 11.9 - Prob. 7E Ch. 11.9 - Prob. 8E Ch. 11.9 - Prob. 9E Ch. 11.9 - Prob. 10E

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The change in specific enthalpy in kJ / kmol for ethylene between 400 K,1 bar and 400 K , 100 bar . The change in specific entropy in kJ / kmol for ethylene between 400 K,1 bar and 400 K , 100 bar .

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The change in specific enthalpy in kJ/kmol for ethylene between 400 K,1 bar and 400 K,100 bar. The change in specific entropy in kJ/kmol for ethylene between 400 K,1 bar and 400 K,100 bar.

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The change in specific enthalpy in kJ/kmol for ethylene between 400 K,1 bar and 400 K,100 bar.

The change in specific entropy in kJ/kmol for ethylene between 400 K,1 bar and 400 K,100 bar.