ALGORITMO PARA EL CALCULO DEL FLASH ISOTERMICO CUANDO LOS VALORES DE K DEPENDE DE LA COMPOSICION

Estimacion Xi & Yi

X1= 0.02 Y1= 0.35 Z1= 0.333X2= 0.25 Y2= 0.33 Z2= 0.333X3= 0.73 Y3= 0.32 Z3= 0.333

Calcular Ki=

Pri= P/Pci Psistema= 7.35Tri = T/Tci Tsistema= 594.6

Pci (PSI Tci (°R) Pri Tri w(fact. De comp) AiACETONA 693.7 917 0.01059536 0.648418757 0.3055 0.043ETANOL 925.3 929.3 0.00794337 0.639836436 0.6341 0.037AGUA 3206.7 1165.1 0.00229208 0.51034246 0.3477 0.027

Zv=LOG V0iL= LOG V(0)iL + wi*LOG V (1)iLLOG V(0)iL= A0+(A1/Tri)+LOG V (1)iL=

ACETONA ETANOLA0 2.05135A1 -2.10899 LOG V(0)iL= 0.699511959 LOG V(0)iL= 0.78245272A2 0 LOG V (1)iL= -1.356660865 LOG V (1)iL= -1.42241399A3 -0.19396 LOG V0iL= 0.285052065 LOG V0iL= -0.11949999A4 0.02282 V0iL= 1.927756005 V0iL= 0.75945144A5 0.08852A6 0 ViL= ec. 5-19A7 -0.00872 =A8 -0.00353A9 0.00203 (fi)1= 0.987560605 ψ f(ψ)A10 -4.23893 (fi)2= 0.990876589 0 -0.001A11 8.6508 (fi)3= 0.996476291 0.1 -0.00031506A12 -1.2206 0.2 0.0089773A13 -3.15224 K1= 1.9824 0.3 0.02658864A14 -0.025 K2= 0.7751 0.4 0.05282275

K3= 0.3589 0.5 0.08860250.6 0.135616380.7 0.196635230.8 0.276126250.9 0.38145976

1 0.52543524

Bi0.000193 A 0.03590.000146 B 0.0001320.000053

0.99

AGUA

LOG V(0)iL= 0.51133475LOG V (1)iL= -2.61983476LOG V0iL= -0.39958179V0iL= 0.39849071

1.- CALCULAR Pri & Tri

°R °F °KPri= P/Pci Psistema= 7.35Tri = T/Tci Tsistema= 594 134.6 330

Pci (PSI) Tci (°R) Pri Tri w(factor de compresibilidad)ACETONA 681.8944 914.58 0.010778795 0.64947845 0.3055ETANOL 925.848 929.16 0.007938668 0.6392871 0.6341AGUA 3197.8496 1165.14 0.00229842 0.50980998 0.3477

2.- CALCULAR V(0)iL & V(1)Il Ec. 5-12,13

ACETONA ETANOLA0 2.05135A1 -2.10899 LOG V(0)1L= 0.69692373 LOG V(0)2L= 0.77999806A2 0 LOG V (1)1L= -1.34864717 LOG V (1)2L= -1.42668041A3 -0.19396 LOG V01L= 0.28491202 LOG V02L= -0.12465999A4 0.02282 V01L= 1.92713448 V02L= 0.75048153A5 0.08852A6 0A7 -0.00872A8 -0.00353A9 0.00203A10 -4.23893A11 8.6508A12 -1.2206A13 -3.15224A14 -0.025

3.- CALCULAR V0iL Ec. 5-11

ACETONA ETANOL AGUA0.284912022 LOG ν 02L= -0.12465999 LOG ν 03L= -0.4070159

ν 01L= 1.92713448 v 02L= 0.75048153 ν 03L= 0.39172754

4.- CALCULAR ViL 5-19 pag 723, intro a la termo. En ing. Quimica /simth van nessR[(ft^3)(psia)/(lbmol)(°R)]= 10.73

ACETONA ETANOL AGUAV1L= 1.660509606 V2L= 1.094775 V3L= 0.3558833

5.- SUPONER Xi, Yi & V/F

LOG ν 01L=

X1= 0.02 Y1= 0.35 Z1= 0.333X2= 0.25 Y2= 0.33 Z2= 0.333X3= 0.73 Y3= 0.32 Z3= 0.333

6.- CALCULAR Φi Ec. 5-5 FRACCION DE VOLUMEN MOLARACETONA ETANOL AGUAΦ1= 0.058602904 Φ2= 0.48296163 Φ3= 0.45843547

7.- CALCULAR γiL Ec. 5-7R= 1.987elemento υ (cm^3/mol)acetona 9.566 73.5etanol 12.915 58.4agua 18 18.1

ln(γ1L)= 3.370947044 ln(γ2L)= 0.40592827 ln(γ3L)= 0.24023992γ1L= 29.10607875 γ2L= 1.5006949 γ3L= 1.27155419

8.- CALCULAR Ai & Bi Ec. 4-44,45

A1= 0.026435726 B1= 0.00019577A2= 0.022732047 B2= 0.00014648A3= 0.012582453 B3= 5.31805E-05

9.- CALCULAR A & B Ec. 4-42,43

A= 0.020780465B= 0.000133875

10.- CALCULAR Z (RAÍZ SUPERIOR DE LA Ec. 4-41)

Z = 0.99

ϕ1= 0.992979351ϕ2= 0.996604312ϕ3= 1.004474749

K1= 56.48790975 k1= 56.0913279K2= 1.130081201K3= 0.495883836

δ (parametro de solubilidad) (cal/cm^3)^1/2

11.- CALCULAR ϕiv Ec. 4-72 (COEFICIENTE DE FUGACIDAD)

12.- CALCULAR Ki Ec. 4-29

ψ0= 0.86-0.105808130.480338509

ψ1= 1.08027825

0.006835031

13.- CALCULAR ψ

f(ψ) = f'(ψ) =

AGUA

LOG V(0)3L= 0.50591539LOG V (1)3L= -2.62562925LOG V03L= -0.4070159V03L= 0.39172754


°R °FPri= P/Pci Psistema= 3.17E+04Tri = T/Tci Tsistema= 995.4 535.7

Pci (PSI) Tci (°R) Pri Tri w(factor de compresibilidad)OXIGENO 736.9 278.6 43.07233003 3.57286432 0.025ETILENO 742.2 509.5 42.76475344 1.95368008 0.0872OXIDO DE ETILENO 1043.4 842.7 30.41978148 1.18120328 0.2121


ACETONAA0 2.05135A1 -2.10899 LOG V(0)1L= 4.31675626A2 0 LOG V (1)1L= -118.504425A3 -0.19396 LOG V01L= 1.35414563A4 0.02282 V01L= 22.6019356A5 0.08852A6 0A7 -0.00872A8 -0.00353A9 0.00203A10 -4.23893A11 8.6508A12 -1.2206A13 -3.15224A14 -0.025


ACETONA ETANOL1.35414563364003 LOG ν 02L= 0.83811999

ν 01L= 22.60193562563 v 02L= 6.88842593


ACETONA ETANOLV1L= 9.92131285341733 V2L= 2.99351542


X1= 0.0568 Y1= 0.3741X2= 0.2453 Y2= 0.345X3= 0.6978 Y3= 0.2798

LOG ν 01L=

6.- CALCULAR Φi Ec. 5-5 FRACCION DE VOLUMEN MOLARACETONA ETANOLΦ1= 0.422580466892785 Φ2= 0.55064409

7.- CALCULAR γiL Ec. 5-7R(cal*g/gmol*°K)= 1.987elemento υ (cm^3/mol)OXIGENO 4 28.4ETILENO 5.801 61OXIDO DE ETILENO 10.271 49

ln(γ1L)= 0.0347556830177674 ln(γ2L)= 0.02283701γ1L= 1.03536672022359 γ2L= 1.02309977


A1= 0.0205489221263455 B1= 3.29302E-05A2= 0.0220802117564484 B2= 5.97922E-05A3= 0.0198314465428594 B3= 7.03467E-05


A= 0.0208538635661326B= 5.26304965625651E-05


Z = 0.99

ϕ1= Err:502ϕ2= Err:502ϕ3= Err:502

K1= Err:502 k1= 23.401292K2= Err:502K3= Err:502

ψ0= 0.86Err:502Err:502




13.- CALCULAR ψ

f(ψ) = f'(ψ) =

ψ1= Err:502

Err:502

°K

553

w(factor de compresibilidad)

ETANOL AGUA

LOG V(0)2L= 1.93012306 LOG V(0)3L= -0.1752772LOG V (1)2L= -12.5229709 LOG V (1)3L= -0.99451581LOG V02L= 0.83811999 LOG V03L= -0.386214V02L= 6.88842593 V03L= 0.41094718

AGUALOG ν 03L= -0.386214ν 03L= 0.41094718

pag 723, intro a la termo. En ing. Quimica /simth van ness

AGUAV3L= 0.0511698

Z1= 0.333Z2= 0.333Z3= 0.333

AGUAΦ3= 0.02677544

υ (cm^3/mol)

ln(γ3L)= 1.16506358γ3L= 3.20612672


Psistema= 7.35Pri= P/Pci °R °F °KTri = T/Tci Tsistema= 540 80.33 300



ACETONA ETANOLA0 2.05135A1 -2.10899 LOG V(0)1L= 0.38485488 LOG V(0)2L=A2 0 LOG V (1)1L= -1.83259108 LOG V (1)2L=A3 -0.19396 LOG V01L= -0.17500169 LOG V02L=A4 0.02282 V01L= 0.66834131 V02L=A5 0.08852A6 0A7 -0.00872A8 -0.00353A9 0.00203A10 -4.23893A11 8.6508A12 -1.2206A13 -3.15224A14 -0.025


ACETONA ETANOL AGUA-0.17500169 LOG ν 02L= -0.75227459 LOG ν 03L=

ν 01L= 0.668341315 v 02L= 0.17689901 ν 03L=


ACETONA ETANOL AGUAV1L= 1.521993447 V2L= 1.00413543 V3L=


X1= 0.0077 Y1= 0.7227 Z1=X2= 0.447 Y2= 0.1958 Z2=X3= 0.5448 Y3= 0.0813 Z3=

LOG ν 01L=

6.- CALCULAR Φi Ec. 5-5 FRACCION DE VOLUMEN MOLARACETONA ETANOL AGUAΦ1= 0.018376946 Φ2= 0.70383303 Φ3=

7.- CALCULAR γiL Ec. 5-7R(cal*g/gmol*°K)= 1.987elemento υ (cm^3/mol)acetona 9.566 73.5etanol 12.915 58.4agua 18 18.1

ln(γ1L)= 2.723749985 ln(γ2L)= 0.17881993 ln(γ3L)=γ1L= 15.2373551 γ2L= 1.1958054 γ3L=


A1= 0.026752558 B1= 0.00021534A2= 0.023004491 B2= 0.00016113A3= 0.012733254 B3= 5.84986E-05


A= 0.024873567B= 0.000191933


Z = 1

ϕ1= 0.996740515ϕ2= 0.996826186ϕ3= 0.998145911

K1= 10.21705629 K1= 10.1837539K2= 0.212210313 K2= 0.2115368K3= 0.153260846 K3= 0.15297669

SUPUESTA ψ0= 0.336516237.55906E-060.854758357




13.- CALCULAR ψ

f(ψ) = f'(ψ) =

ψ1= 0.336507386 ERROR= -8.84351E-06

14.- ITERAR PARA OBTENER V/F

SUPUESTA ψ0= 0.336516237.55906E-060.854758357

ψ1= 0.336507386 ERROR= -8.84351E-06

15.- COMPROBAR Xi & Yi

X1= 0.081186067 Y1= 0.82948261X2= 0.45312535 Y2= 0.09615787X3= 0.465696143 Y3= 0.07137298

CRITERIO DE ACEPTACION .0001EPSILON= -0.07348607EPSILON= -0.00612535EPSILON= 0.079103857

NORMALIZAR

X1 NORM= 0.081185453 Y1 NORM= 0.83196731X2 NORM= 0.453121925 Y2 NORM= 0.09644591X3 NORM= 0.465692622 Y3 NORM= 0.07158678

f(ψ) = f'(ψ) =


AGUA

0.46238055 LOG V(0)3L= 0.10023185-1.91555769 LOG V (1)3L= -3.16211392-0.75227459 LOG V03L= -0.999235160.17689901 V03L= 0.10017627

-0.999235160.10017627


0.32516944

0.3330.3330.333

0.27779003

0.423354241.52707516


°R °F °KPri= P/Pci Psistema=Tri = T/Tci Tsistema= 995.4 553

Pci (PSI) Tci (°R) Pri Tri w(factor de compresibilidad)ACETONA 681.8944 914.58 1.459756819 #REF! 0.3055ETANOL 925.848 929.16 1.075122482 #REF! 0.6341AGUA 3197.8496 1165.14 0.311271675 #REF! 0.3477


ACETONA ETANOLA0 2.05135A1 -2.10899 LOG V(0)1L= #REF! LOG V(0)2L=A2 0 LOG V (1)1L= #REF! LOG V (1)2L=A3 -0.19396 LOG V01L= #REF! LOG V02L=A4 0.02282 V01L= #REF! V02L=A5 0.08852A6 0A7 -0.00872A8 -0.00353A9 0.00203A10 -4.23893A11 8.6508A12 -1.2206A13 -3.15224A14 -0.025


ACETONA ETANOL AGUA#REF! LOG ν 02L= #REF! LOG ν 03L=

ν 01L= #REF! v 02L= #REF! ν 03L=


ACETONA ETANOL AGUAV1L= #REF! V2L= #REF! V3L=


X1= 0.08118545 Y1= 0.83196731 Z1=X2= 0.45312192 Y2= 0.09644591 Z2=X3= 0.46569262 Y3= 0.07158678 Z3=

LOG ν 01L=

6.- CALCULAR Φi Ec. 5-5 FRACCION DE VOLUMEN MOLARACETONA ETANOL AGUAΦ1= #REF! Φ2= #REF! Φ3=


ln(γ1L)= #REF! ln(γ2L)= #REF! ln(γ3L)=γ1L= #REF! γ2L= #REF! γ3L=


A1= #REF! B1= #REF!A2= #REF! B2= #REF!A3= #REF! B3= #REF!


A= #REF!B= #REF!


Z = 1

ϕ1= #REF!ϕ2= #REF!ϕ3= #REF!

K1= #REF! K1= #REF!K2= #REF! K2= #REF!K3= #REF! K3= #REF!

SUPUESTA ψ0= 0.33651623#REF!#REF!




13.- CALCULAR ψ

f(ψ) = f'(ψ) =

ψ1= #REF! ERROR= #REF!

14.- ITERAR PARA OBTENER V/F

SUPUESTA ψ0= 0.03045353#REF!#REF!

ψ1= #REF! ERROR= #REF!

15.- COMPROBAR Xi & Yi

X1= #REF! Y1= #REF!X2= #REF! Y2= #REF!X3= #REF! Y3= #REF!

CRITERIO DE ACEPTACION .0001EPSILON= #REF!EPSILON= #REF!EPSILON= #REF!

NORMALIZAR

X1 NORM= #REF! Y1 NORM= #REF!X2 NORM= #REF! Y2 NORM= #REF!X3 NORM= #REF! Y3 NORM= #REF!

f(ψ) = f'(ψ) =


AGUA

#REF! LOG V(0)3L= #REF!#REF! LOG V (1)3L= #REF!#REF! LOG V03L= #REF!#REF! V03L= #REF!

#REF!#REF!


#REF!

0.3330.3330.333

#REF!

#REF!#REF!


°R °F °KPri= P/Pci Psistema= 7.35Tri = T/Tci Tsistema= 566.1 106.4 314.3



ACETONA ETANOLA0 2.05135A1 -2.10899 LOG V(0)1L= 0.5435553 LOG V(0)2L=A2 0 LOG V (1)1L= -1.58910686 LOG V (1)2L=A3 -0.19396 LOG V01L= 0.05808315 LOG V02L=A4 0.02282 V01L= 1.14309718 V02L=A5 0.08852A6 0A7 -0.00872A8 -0.00353A9 0.00203A10 -4.23893A11 8.6508A12 -1.2206A13 -3.15224A14 -0.025


ACETONA ETANOL AGUA0.058083152 LOG ν 02L= -0.43497189 LOG ν 03L=

ν 01L= 1.143097176 v 02L= 0.36730608 ν 03L=


ACETONA ETANOL AGUAV1L= 1.589409262 V2L= 1.04827755 V3L=


X1= 0.333 Y1= 0.65176355 Z1=X2= 0.333 Y2= 0.13986961 Z2=X3= 0.333 Y3= 0.20836683 Z3=

LOG ν 01L=

6.- CALCULAR Φi Ec. 5-5 FRACCION DE VOLUMEN MOLARACETONA ETANOL AGUAΦ1= 0.53375955 Φ2= 0.35203529 Φ3=


ln(γ1L)= 0.540074874 ln(γ2L)= 0.13619495 ln(γ3L)=γ1L= 1.716135351 γ2L= 1.14590526 γ3L=


A1= 0.026595178 B1= 0.00020541A2= 0.02286916 B2= 0.0001537A3= 0.012658347 B3= 5.58015E-05


A= 0.023170048B= 0.000167007


Z = 1

ϕ1= 0.997030246ϕ2= 0.997077335ϕ3= 0.998238631

K1= 1.96755262 K1= 1.96170947 Y1=K2= 0.422131715 K2= 0.42089797 Y2=K3= 0.629262724 K3= 0.62815436 Y3=

f(T)= 0.005309371si la funcion es negativa aumentar la tempratura si la funfion es positiva disminuir la temperatura





AGUA

0.62388255 LOG V(0)3L= 0.30625268-1.66985403 LOG V (1)3L= -2.89464254-0.43497189 LOG V03L= -0.700214530.36730608 V03L= 0.19942769

-0.700214530.19942769


0.34007586

0.3330.3330.333

0.11420516

1.14733423.14978502

0.65519502 Y1normalizada= 0.65173472 0.651763550.14056986 Y1normalizada= 0.13982747 0.139869610.20954449 Y1normalizada= 0.20843781 0.208366831.00530937 1

ALGORITMO PARA EL CALCULO DEL FLASH ISOTERMICO CUANDO LOS VALORES DE K DEPENDE DE LA COMPOSICION

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Transcript of ALGORITMO PARA EL CALCULO DEL FLASH ISOTERMICO CUANDO LOS VALORES DE K DEPENDE DE LA COMPOSICION