Explanation The figure shown below shows the Redlich-Kwong process on T − S diagram. Figure (1) Write the expression for change in enthalpy. h x − h 1 = ( u x − u 1 ) + p 2 v 2 − p x v x (I) Here, enthalpy at state 1 is h 1 , enthalpy at state 2 is h 2 , internal energy at state x is u x , internal energy at state 1 is u 1 , pressure at state 2 is p 2 , pressure at state x is p x , volume at state 2 is v 2 and volume at state x is v x . Write the expression for change in internal energy. ( u x − u 1 ) = ∫ v 1 v 2 [ T ( ∂ p ∂ T ) v − p ] d v (II) Here, change in pressure with respect to T with volume v as a constant is ( ∂ p ∂ T ) , pressure is p and temperature is T . Write the expression for pressure by the Benedict–Webb-Rubin Equation. p = [ R T v + ( B R T − A − C T 2 ) 1 v − 2 + ( b R T − a ) v − 3 + a α v − 6 + c v − 3 T 2 ( 1 + γ v − 2 ) exp ( − γ v − 2 ) ] (III) Here, Benedict–Webb-Rubin constants are A , B , C , γ , α , a , b and c , and gas constant is R . Differentiate the Equation (III) with respect to T with volume v as constant. ∂ p ∂ T = ∂ ∂ T [ R T v + ( B R T − A − C T 2 ) 1 v − 2 + ( b R T ) v − 3 + a α v − 6 + c v − 3 T 2 ( 1 + γ v − 2 ) exp ( − γ v − 2 ) ] = [ R v + ( B R − C T 3 ) 1 v − 2 + ( b R − a ) v − 3 − 2 c v − 3 T 3 ( 1 + γ v − 2 ) exp ( − γ v − 2 ) ] T ( ∂ p ∂ T ) v = [ R T v + ( B R T − C T 2 ) 1 v − 2 + ( b R T ) v − 3 − 2 c v − 3 T 2 ( 1 + γ v − 2 ) exp ( − γ v − 2 ) ] (IV) Subtract Equation (III) from Equation (IV). T ( ∂ p ∂ T ) v − p = [ [ R T v + ( B R T − C T 2 ) 1 v − 2 + ( b R T ) v − 3 − 2 c v − 3 T 2 ( 1 + γ v − 2 ) exp ( − γ v − 2 ) ] − [ R T v + ( B R T − A − C T 2 ) 1 v − 2 + ( b R T − a ) v − 3 + a α v − 6 + c v − 3 T 2 ( 1 + γ v − 2 ) exp ( − γ v − 2 ) ] ] = [ ( A + 3 C T 2 ) 1 v − 2 + a v − 3 − a α v − 6 − 3 c v − 3 T 2 ( 1 + γ v − 2 ) exp ( − γ v − 2 ) ] (V) Substitute [ ( A + 3 C T 2 ) 1 v − 2 + a v − 3 − a α v − 6 − 3 c v − 3 T 2 ( 1 + γ v − 2 ) exp ( − γ v − 2 ) ] for T ( ∂ p ∂ T ) v − p in Equation (II). ( u x − u 1 ) = ∫ v 1 v 2 [ ( A + 3 C T 2 ) 1 v − 2 + a v − 3 − a α v − 6 − 3 c v − 3 T 2 ( 1 + γ v − 2 ) exp ( − γ v − 2 ) ] d v = [ ( A + 3 C T 2 2 ) ( 1 v x − 1 v 2 ) + a 2 ( 1 v x 2 − 1 v 2 2 ) − a α 5 ( 1 v x 5 − 1 v 2 5 ) − 3 c T 2 ∫ v x v 2 1 v − 3 exp ( − γ v − 2 ) − 3 c γ T 2 ∫ v x v 2 1 v − 3 exp ( − γ v − 2 ) ] = { ( A + 3 C T 2 2 ) ( 1 v x − 1 v 2 ) + a 2 ( 1 v x 2 − 1 v 2 2 ) − a α 5 ( 1 v x 5 − 1 v 2 5 ) − 1 2 γ [ exp − γ v 2 2 − exp − γ v x 2 ] − 1 2 γ 2 [ ( ( γ v 2 2 ) + 1 ) exp ( − γ

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 11.9, Problem 76P

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The specific enthalpy of methane in kJ/kmol between 300 K1 atm and 400 K 200 atm.

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Chapter 11 Solutions

Fundamentals of Engineering Thermodynamics, Binder Ready Version

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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