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Unformatted text preview: T Ar 403.15K = n total n N2 n Ar + = x 1 n N2 n total = x 2 n Ar n total = x 1 0.615 = x 2 0.385 = Cv Ar 3 2 R ⋅ = Cv N2 5 2 R ⋅ = Cp Ar Cv Ar R + = Cp N2 Cv N2 R + = Find T after mixing by energy balance: T T N2 T Ar + 2 = (guess) Given n N2 Cv N2 ⋅ T T N2 ( 29 ⋅ n Ar Cv Ar ⋅ T Ar T ( 29 ⋅ = T Find T ( ) = T 273.15 K ⋅ 90degC = Find P after mixing: P P N2 P Ar + 2 = (guess) Chapter 11  Section A  Mathcad Solutions 11.1 For an ideal gas mole fraction = volume fraction CO2 (1): x 1 0.7 = i 1 2 .. = N2 (2): x 2 0.3 = ∆ S R i x i ln x i ( 29 ⋅ ∑ ⋅ = ∆ S 5.079 J mol K ⋅ = Ans. 11.2 For a closed, adiabatic, fixedvolume system, D U =0. Also, for an ideal gas, D U = Cv D T. First calculate the equilibrium T and P. n N2 4 mol ⋅ = T N2 75 273.15 + ( ) K ⋅ [ ] = P N2 30 bar ⋅ = n Ar 2.5 mol ⋅ = T Ar 130 273.15 + ( ) K ⋅ = P Ar 20 bar ⋅ = T N2 348.15K = 290 Ans. ∆ S 1411 J s K ⋅ = ∆ S R molarflow total ⋅ i y i ln y i ( 29 ⋅ ∑ ⋅ = y 2 0.776 = y 2 molarflow H2 molarflow total = y 1 0.224 = y 1 molarflow N2 molarflow total = molarflow total 319.409 mol sec = molarflow total molarflow N2 molarflow H2 + = molarflow H2 mdot H2 molwt H2 = molarflow N2 mdot N2 molwt N2 = molwt H2 2.016 gm mol ⋅ = molwt N2 28.014 gm mol ⋅ = mdot H2 0.5 kg s ⋅ = mdot N2 2 kg s ⋅ = 11.3 Ans. ∆ S 38.27 J K = ∆ S ∆ S N2 ∆ S Ar + ∆ S mix + = ∆ S mix 36.006 J K = ∆ S mix n total R i x i ln x i ( 29 ⋅ ∑ ⋅ ⋅ = ∆ S Ar 9.547 J K = ∆ S Ar n Ar Cp Ar ln T T Ar ⋅ R ln P P Ar ⋅ ⋅ = ∆ S N2 11.806 J K = ∆ S N2 n N2 Cp N2 ln T T N2 ⋅ R ln P P N2 ⋅ ⋅ = Calculate entropy change by twostep path: 1) Bring individual stream to mixture T and P. 2) Then mix streams at mixture T and P. P 24.38bar = P Find P ( ) = n N2 n Ar + ( 29 R ⋅ T ⋅ P n N2 R ⋅ T N2 ⋅ P N2 n Ar R ⋅ T Ar ⋅ P Ar + = Given 291 W ideal ∆ H T σ ∆ S ⋅ = W ideal 2483.4 J mol = Ans. 11.5 Basis: 1 mole entering air. y 1 0.21 = y 2 0.79 = η t 0.05 = T σ 300 K ⋅ = Assume ideal gases; then ∆ H = The entropy change of mixing for ideal gases is given by the equation following Eq. (11.25). For UNmixing of a binary mixture it becomes: ∆ S R y 1 ln y 1 ( 29 ⋅ y 2 ln y 2 ( 29 ⋅ + ( 29 ⋅ = ∆ S 4.273 J mol K ⋅ = By Eq. (5.27): W ideal T σ ∆ S ⋅ = W ideal 1.282 10 3 × J mol = By Eq. (5.28): Work W ideal η t = Work 25638 J mol = Ans. 11.4 T 1 448.15 K ⋅ = T 2 308.15 K ⋅ = P 1 3 bar ⋅ = P 2 1 bar ⋅ = For methane: MCPH 448.15 308.15 , 1.702 , 9.081 10 3 ⋅ , 2.164 10 6 ⋅ , 0.0 , ( 29 4.82200 = MCPS 448.15 308.15 , 1.702 , 9.081 10 3 ⋅ , 2.164 10 6 ⋅ , 0.0 , ( 29 4.79051 = For ethane: MCPH 448.15 308.15 , 1.131 , 19.225 10 3 ⋅ , 5.561 10 6 ⋅ , 0.0 , ( 29 7.59664 = MCPS 448.15 308.15 , 1.131 , 19.225 10 3 ⋅...
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This note was uploaded on 01/29/2011 for the course CHEM 101 taught by Professor Brown during the Spring '10 term at The University of Akron.
 Spring '10
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