Chapter11_A

# Chapter11_A - Chapter 11 Section A Mathcad Solutions 11.1...

This preview shows pages 1–5. Sign up to view the full content.

n Ar 2.5 mol := T Ar 130 273.15 + () K := P Ar 20 bar := T N2 348.15K = T Ar 403.15K = i1 2 .. := 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 TT N2 n Ar Cv Ar T Ar T = T Find T := Chapter 11 - Section A - Mathcad Solutions 11.1 For an ideal gas mole fraction = volume fraction CO2 (1): x 1 0.7 := V 1 0.7m 3 := N2 (2): x 2 0.3 := V 2 0.3m 3 := 2 .. := P 1bar := T 25 273.15 + K := n P i V i RT := n 40.342mol = Sn R i x i ln x i := S 204.885 J K = Ans. 11.2 For a closed, adiabatic, fixed-volume system, U =0. Also, for an ideal gas, U = Cv T. First calculate the equilibrium T and P. n N2 4 mol := T N2 75 273.15 + K [] := P N2 30 bar := 341

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
molarflow total 319.409 mol sec = molarflow total molarflow N2 molarflow H2 + := molarflow H2 mdot H2 molwt H2 := molarflow N2 mdot N2 molwt N2 := i1 2 .. := molwt H2 2.016 gm mol := molwt N2 28.014 gm mol := mdot H2 0.5 kg sec := mdot N2 2 kg sec := 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 () := S Ar 9.547 J K = S Ar n Ar Cp Ar ln T T Ar Rln P P Ar := S N2 11.806 J K = S N2 n N2 Cp N2 ln T T N2 P P N2 := Calculate entropy change by two-step 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 + R T P n N2 R T N2 P N2 n Ar R T Ar P Ar + = Given (guess) P P N2 P Ar + 2 := Find P after mixing: T 273.15 K 90degC = 342
MCPS mix 6.161 = H R MCPH mix T 2 T 1 () := H 7228 J mol = S R MCPS mix ln T 2 T 1 Rln P 2 P 1 R2 0.5 ln 0.5 + := The last term is the entropy change of UNmixing S 15.813 J mol K = T σ 300 K := W ideal HT σ S := W ideal 2484 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 H0 = The entropy change of mixing for ideal gases is given by the equation following Eq. (11.26). For UNmixing of a binary mixture it becomes: y 1 molarflow N2 molarflow total := y 1 0.224 = y 2 molarflow H2 molarflow total := y 2 0.776 = SR molarflow total i y i ln y i := S 1411 J sec K = Ans. 11.4 T 1 448.15 K := T 2 308.15 K := P 1 3 bar := P 2 1 bar := For methane: MCPH m MCPH T 1 T 2 , 1.702 , 9.081 10 3 , 2.164 10 6 , 0.0 , := MCPS m MCPS T 1 T 2 , 1.702 , 9.081 10 3 , 2.164 10 6 , 0.0 , := For ethane: MCPH e MCPH T 1 T 2 , 1.131 , 19.225 10 3 , 5.561 10 6 , 0.0 , := MCPS e MCPS T 1 T 2 , 1.131 , 19.225 10 3 , 5.561 10 6 , 0.0 , := MCPH mix 0.5 MCPH m 0.5 MCPH e + := MCPH mix 6.21 = MCPS mix 0.5 MCPS m 0.5 MCPS e + := 343

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
F i Z i 1 P i := Fi is a well behaved function; use the trapezoidal rule to integrate Eq.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 33

Chapter11_A - Chapter 11 Section A Mathcad Solutions 11.1...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online