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δ12 2B 12 ⋅B11 −B22 −:= B12 52 cm3 mol ⋅:=B22 1523−cm3 mol ⋅:=B11 963−cm3 mol ⋅:= BUBL P calculations with virial coefficients:(b) y1x1 () 0.808=Pbubl x1 () 85.701 kPa=x10.75:= y1x1 () 0.731=Pbubl x1 () 80.357 kPa=x10.50:= y1x1 () 0.562=Pbubl […]

K20.5798=K2exp ∆G− RT⋅ ⎛ ⎜ ⎝ ⎞ ⎠ :=∆G 5.892 103 ×J mol = ∆G∆H298 T T0 ∆H298 ∆G298 − () ⋅− R IDCPH T0T,∆A,∆B,∆C,∆D, () ⋅+ … R−T⋅IDCPS T0T,∆A,∆B,∆C,∆D, () ⋅+ … := ∆D 1.164−105 ⋅:=∆C 0.0:=∆B 0.540−10 3− […]

For CH3OH(3): Tc3 512.6kelvin:= Pc3 80.97bar:= ω30.564:= By Eq. (11.67) and data from Tables E.15 & E.16. φ30.6206 0.9763ω3 ⋅:= φ30.612= For H2(2), the reduced temperature is so large that it may be assumed ideal: φ = 1. Therefore: i13..:= […]

0.2 0.3 0.4 0.5 0.6 2.086 2.084 Gε () 105 2.082 ε ε0.3 0.31,0.6..:= εe0.45308=εeFind εe () := εe Gεe () d d 0J mol ⋅=Given εe0.5:= Guess: Gε () 1ε− 2 ⎛ ⎜ ⎝ ⎞ ⎠395790−()⋅J mol ⋅ε 2192420−200240−()⋅J […]

∆H298 411153−285830−425609−92307−()−[]J⋅:= ∆H298 1.791−105 ×J= NaOH(s) + HCl(g) —> NaCl(s) + H2O(l) (1) NaOH(inf H2O) —> NaOH(s) + inf H2O (2) HCl(9 H2O) —> HCl(g) + 9 H2O(l) (3) NaCl(s) + inf H2O —> NaCl(inf H2O) (4) —————————————————————————————- NaOH(inf H2O) […]

∆H11.509 kJ mol =∆H1104.8 kJ kg ⋅21.01 kJ kg ⋅− ⎛ ⎜ ⎝ ⎞ ⎠18.015⋅gm mol ⋅:= H2O @ 5 C —–> H2O @ 25 C (1) LiCl(3 H2O) —–> LiCl + 3 H2O (2) LiCl + 4 H2O —–> […]

Λ21 T() V1 V2 exp a21− RT⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅:=Λ12 T() V2 V1 exp a12− RT⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅:= a21 1351.90 cal mol ⋅:=a12 775.48 cal mol ⋅:= V2 18.07 cm3 mol ⋅:=V1 75.14 […]

P-x,y Diagram from Margules Equation fit to GE/RT data. 0 0.2 0.4 0.6 0.8 36 38 P-x data P-y data P-x calculated P-y calculated Pi kPa x1iy1i ,X1j ,y1calcj , (d) Consistency Test: δGERTiGeRT x1ix2i , () GERTi −:= δlnγ1γ2iln […]

γ1x1x2,() exp x2 Λ12 x1 x2 Λ12 ⋅+ Λ21 x2 x1 Λ21 ⋅+ − ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ x1 x2 Λ12 ⋅+ () := γ2x1x2,() exp x1−Λ12 x1 x2 Λ12 ⋅+ x2 […]

Calculate EXPERIMENTAL values of activity coefficients and excess Gibbs energy. Psat219.953 kPa⋅:=Psat184.562 kPa⋅:= Vapor Pressures from equilibrium data: y21y 1 − () → ⎯ ⎯ ⎯ :=x21x 1 − () → ⎯ ⎯ ⎯ := Calculate x2 and y2: i1n..:=n10=n […]

To find the maximum, set dVE/dx1 = 0 and solve for x1. Then use x1 to find VEmax. (b) Vbar2 () Ex1 () 2ab−2b c−()⋅x1 ⋅+ 3c⋅x1 () 2 ⋅+ ⎡ ⎣ ⎤ ⎦ ⋅= Vbar1 () Ex2 () 2a2b⋅x1 […]

nAr 2.5 mol⋅:= TAr 130 273.15+()K⋅:= PAr 20 bar⋅:= TN2 348.15 K=TAr 403.15 K=i12..:= ntotal nN2 nAr +:= x1 nN2 ntotal := x2 nAr ntotal := x10.615=x20.385= Find T after mixing by energy balance: TTN2 TAr + 2 := (guess) […]

b)For water as solvent: Ms18.015 gm mol := 10.36 Acetone: Psat1T() e 14.3145 2756.22 T degC 228.060+ − kPa⋅:= Acetonitrile Psat2T() e 14.8950 3413.10 T degC 250.523+ − kPa⋅:= a) Find BUBL P and DEW P values T 50degC:= x10.5:= […]

y10.33:= T 100 degC⋅:= Guess: x10.33:= P 100 kPa⋅:= Given x1Psat1T()⋅1x 1 − () Psat2T()⋅+ P= (c) Given: x10.33:= P 120 kPa⋅:= Guess: y10.5:= T 100 degC⋅:= Given x1Psat1T()⋅1x 1 − () Psat2T()⋅+ P= x1Psat1T()⋅y1P⋅= y1 T ⎛ ⎜ ⎝ […]

APPENDIX— EVOLUTION OF A TEXTBOOK Introduction to Chemical Engineering Thermodynamics HENDRICK C. VAN NESS Rensselaer Polytechnic Institute •Troy, NY Rarely does a textbook remain in print for anything ap- proaching 55 years. Introduction to Chemical Engineering Thermodynamics is the only […]

(b) Steps 3–2 and 1–4 (Fi g . 8.2) are isentropic, for which S3=S2 and S1=S4 . Thus by Eq. 6.82): x3 S2 Sliq− Svap Sliq− := x3 0.971=x4 S1 Sliq− Svap Sliq− := x4 0.302= (c) Heat addition, Step […]

t8t95+:=t9 190 t7 − 2t7 +:=t7tsat ∆T67 K +:= CP 272.0 230.2− 10 kJ kg K⋅ ⋅:=β 1 Vliq 1.023 1.012− 20 ⎛ ⎜ ⎝ ⎞ ⎠ ⋅cm3 gm K⋅ ⋅:= We apply Eq. (7.25) for the calculation of the […]

For isentropic expansion, S’3S2 := x’3 S’3Sliq − ∆Slv := x’30.855= Chapter 8 – Section A – Mathcad Solutions 8.1 With reference to Fig. 8.1, SI units, At point 2: Table F.2, H23531.5:= S26.9636:= At point 4: Table F.1, H4209.3:= […]

(guess) τ1.1:= The actual final temperature is now found from Eq. (6.91) combined with E q (4.7), written: The actual enthalpy change from Eq. (7.17): ∆H’ 1158.8 J mol = ∆H’ ∆Hig RT c ⋅HRB TrPr ,ω, () HRB Tr0 […]

These are sufficiently close, and we conclude that: xS0.925=xH0.924= The trial values given produce: xS 6.7093 Sl− Sv Sl− :=xH Hv 801.7−.75 Hl⋅− .75 Hv Hl−()⋅ := The two equations for x are: Sv 7.1293:=Sl 1.5276:= Hv 2706.0:=Hl 503.7:= If […]

Interpolation in Table F.2 at P = 525 kPa and S = 7.1595 kJ/(kg*K) yields: H22855.2 kJ kg ⋅:= V2531.21 cm3 gm ⋅:= mdot 0.75 kg sec ⋅:= With the heat, work, and potential-energy terms set equal to zero and […]

The process is the same as that of Example 6.8, except that the stream flows out rather than in. The energy balance is the same, except for a sign: m1257.832 kg=V17.757 10 3− ×m3 kg = m1 Vtank V1 :=V1Vliq […]

A 1.637:= B22.706 10 3− ⋅ K := C6.915−10 6− ⋅ K2 := Solve energy balance for final T. See Eq. (4.7). τ1:= (guess) Given HRRAT⋅τ1− () ⋅B 2T2 ⋅τ 21− () ⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ C […]

(c) Tc647.1 K⋅:= Pc220.55 bar⋅:= ω 0.345:= Tr1 T1 Tc := Pr1 P1 Pc := Tr2 T2 Tc := Pr2 P2 Pc := Tr1 1.11752=Pr1 0.13602=Tr2 0.63846=Pr2 0.01066= The generalized virial-coefficient correlation is suitable here ∆H∆Hig RT c ⋅HRB Tr2 […]

6.8 Isobutane: Tc408.1 K⋅:= Zc0.282:= CP2.78 J gm K⋅ ⋅:= P14000 kPa⋅:= P22000 kPa⋅:= molwt 58.123 gm mol ⋅:= Vc262.7 cm3 mol ⋅:= Eq. (3.63) for volume of a saturated liquid may be used for the volume of a compressed […]

5.34 E 110 volt⋅:= i 9.7 amp⋅:= Tσ300 K⋅:= Wdotmech 1.25−hp⋅:= Wdotelect iE⋅:= Wdotelect 1.067 103 ×W= At steady state: Qdot Wdotelect +Wdotmech + t Ut d d =0= Qdot TσSdotG + t St d d =0= Qdot Wdotelect −Wdotmech […]

5.4 (a) TC303.15 K⋅:= TH623.15 K⋅:= ηCarnot 1TC TH −:= η 0.55 ηCarnot ⋅:= η 0.282=Ans. Chapter 5 – Section A – Mathcad Solutions 5.2 Let the symbols Q and Work represent rates in kJ/s. Then by Eq. (5.8) ηWork […]

nCO2∆H1148a ⋅nO2∆H1148b ⋅+ 0= The combined heats of reaction must be zero: ∆H1148b 2.249−105 ×J mol = ∆H1148b ∆H298b R MCPH 298.15K 1148.15K,∆A,∆B,∆C,∆D, () ⋅1148.15K 298.15K−()⋅+ …:= ∆D 1.899 105 ×=∆C0:=∆B 9.34−10 4− ×=∆A 0.429−= ∆D i niDi ⋅ () […]

n39.781=n32.6 79 21 ⋅:= n22.6=n22 1.3⋅:= n11:= Moles methane Moles oxygen Moles nitrogen Entering: FURNACE: Basis is 1 mole of methane burned with 30% excess air. CH4 + 2O2 = CO2 + 2H2O(g) 4.29 This value is for the constant-V […]

PROPRIETARY MATERIAL only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. . © 2005 The McGraw-Hill Companies, Inc. All rights reserved. Limited distribution permitted Ans. 4.2 (a) […]

Pr1 P1 Pc := Tr2 T2 Tc := Pr2 P2 Pc := Tr1 0.62=Pr1 0.03=Tr2 0.88=Pr2 3.561= From Fig. (3.16): ρr1 2.69:= ρr2 2.27:= 3.54 For ethanol: Tc513.9 K⋅:= T 453.15 K⋅:= Tr T Tc := Tr0.882= Pc61.48 bar⋅:= P […]

Table 3.1 αTrω, () 1 0.480 1.574ω+ 0.176ω2 − () 1T r 2 − ⎛ ⎜ ⎝ ⎞ ⎠ ⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ := qT r () Ψα Trω, () ⋅ ΩTr ⋅ := Eq. (3.54) βTrPr […]

a bit of algebra leads to 3.5 κabP⋅+=a 3.9 10 6− ⋅atm 1− ⋅:= b 0.1−10 9− ⋅atm 2− ⋅:= P11 atm⋅:= P23000 atm⋅:= V1ft 3 ⋅:= (assume const.) Combine Eqs. (1.3) and (3.3) for const. T: Work V P1 […]

(d) For the restoration process, the change in internal energy is equal but of opposite sign to that of the initial process. Thus Q∆Utotal −:= Q 1.715−kJ=Ans.Ans. (e) In all cases the total internal energy change of the universe is […]

MCPH T0 T,A,B,C,D,()AH 2T0 T,B,()+H3T0 T,C,()+H4T0 T,D,()+ () ≡ Enthalpy H4T0 T,D,() D τT0 T,() T0 K ⎛ ⎜ ⎝ ⎞ ⎠ 2 ⋅ ≡ H3T0 T,C,() C 3 T0 K ⎛ ⎜ ⎝ ⎞ ⎠ 2 ⋅τT0 T,() 2τT0 […]

1.7 Pabs ρg⋅h⋅Patm += ρ13.535 gm cm3 ⋅:= g 9.832 m s2 ⋅:= h 56.38cm:= Patm 101.78kPa:= Pabs ρg⋅h⋅Patm +:= Pabs 176.808kPa=Ans. 1.8 ρ13.535 gm cm3 ⋅:= g 32.243 ft s2 ⋅:= h 25.62in:= Patm 29.86in_Hg:= Pabs ρg⋅h⋅Patm +:= Pabs […]

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