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U.M.S.N.H. FACULTAD DE INGENIERÍA QUÍMICA FISICOQUÍMICA I RESOLUCIÓN DE PROBLEMAS DE MEZCLAS DE GASES REALES

MEZCLAS DE GASES REALES PROFESOR: M.C. LUIS NIETO LEMUS

ALUMNO: DAVID MOLINA PÉREZ

MATRÍCULA: 1645561B

PRIMER SEMESTRE, GRUPO: 02

MORELIA, MICHOACÁN A 30 DE OCTUBRE DEL 2019

1.Trescientas libras-masa por hora de una mezcla que contienen 10% en mol de propano, 20% de n-butano y 70% de n-pentano se vaporiza completamente en una tubería. La temperatura y la presión a la salida son de 515° F y 600 lbf/𝑝𝑢𝑙𝑔2 abs. ¿Cuál es la velocidad del vapor en pie/s a la salida de la tubería en esas condiciones el diámetro interno es de dos pulgadas? Emplear: a) Ecuación de Peng-Robinson b) Ecuación de Soave Redlich Kwong a) Ecuación de Peng-Robinson: 𝑅𝑇

𝑎𝑚

P=𝑉−𝑏𝑚 − 𝑉 2 +2𝑏𝑚𝑉−𝑏𝑚2 α1=0.45723553

2 𝑅 2 𝑇𝑐𝑖

𝑃𝑐𝑖 𝑅𝑇𝑐

b1=0.07779607= 𝑃𝑐𝑖

𝑇

α1=[1+(0.37464+1.54226ω-0.26992𝜔2 )(1-𝑇𝑅10.5 )]2 Ṽ3 - (

𝑅𝑇

𝑎𝛼

-b)Ṽ2 + ( 𝑃 – 3𝑏 2 𝑃

Ṽ=Ṽ3 - (

𝑅𝑇 𝑃

2𝑏𝑅𝑇 𝑃

𝑎𝛼

-b)Ṽ2 + ( 𝑃 – 3𝑏 2 -

) v+

2𝑏𝑅𝑇 𝑃

𝑅𝑇𝑏 𝑃

-

𝑎𝑏𝛼

𝑅𝑇𝑏

) v+

𝑃

𝑃

-

TR=𝑇𝑐

=0

𝑎𝑏𝛼 𝑃

αm=∑∑YiYj√(α, αi)(α,αj) Cuando

i=1, 2,3

j=1,2,3

αm=𝑌𝑖 2 α.α+𝑌22 α2. α2 + 𝑌33 α3α3+2YiY2√(α1 α1)(α2,α2) +2YiY3√(α1 α1)(α3,α3) +2YiY2√(α2 α2)(α3,α3) αm=(0.10)2 (10.0431)(1.02660)+ (0.20)2 (14.85168)(1.01111)+ (0.70)2 (20.4253)(1.00405 )+2(20) √(10.04316)(1.02660) (0.08206 𝑙.

b1=0.07779607

=0.056310

41.9245

(0.08206 𝑙.

b2=0.07779607 b3=0.07779607

𝑎𝑡𝑚 )(369.8𝑘) 𝑚𝑜𝑙𝑘

𝑎𝑡𝑚 )(425.1𝑘) 𝑚𝑜𝑙𝑘

=0.072438

37.4636

(0.08206 𝑙.

𝑎𝑡𝑚 )(469.7𝑘) 𝑚𝑜𝑙𝑘

33.2593

=0.090156

bm =0.2189049 α1=0.45723553

(0.08206 𝑙.

𝑎𝑡𝑚 )(369.7𝑘) 𝑚𝑜𝑙𝑘

41.9245

=10.043126

α1=0.45723553 α1=0.45723553

(0.08206 𝑙.

𝑎𝑡𝑚 )(425.1𝑘) 𝑚𝑜𝑙𝑘

37.46636

(0.08206 𝑙.

𝑎𝑡𝑚 )(469.7𝑘) 𝑚𝑜𝑙𝑘

33.2593

=14.85168

=20.042353 𝑇

α1=[1+(0.37464+1.54226ω-0.26992𝜔2 )(1-𝑇𝑅10.5 )]2 α1=1.02660 Ṽ=Ṽ3 - (

𝑅𝑇 𝑃

α2=1.0111 𝑎𝛼

-b)Ṽ2 + ( 𝑃 – 3𝑏 2 -

TR=𝑇𝑐

α3=1.00405 2𝑏𝑅𝑇 𝑃

𝑅𝑇𝑏

) v+

𝑃

-

𝑎𝑏𝛼 𝑃

Ṽ=Ṽ3 -(0.809429) Ṽ2 (0.16828) (-0.04668) Ṽ=3Ṽ2 -1.7388558-0.16828 Vmsup(L/mol) 1.08833403 1.06866491 1.06800816 1.06800744 1.06800744 1.06800744 1.06800744 1.06800744

F(V) 0.0293596 0.00091918 1.0075E-06 1.2151E-12 2.2204E-16 -3.0531E-16 2.2204E-16 -3.0531E-16

F'(V) 1.49267455 1.39959752 1.39652977 1.3965264 1.3965264 1.3965264 1.3965264 1.3965264

Vmcal 1.06866491 1.06800816 1.06800744 1.06800744 1.06800744 1.06800744 1.06800744 1.06800744

b) Ecuación de Soave Redlich Kwong 𝑅𝑇

𝑎𝑚

P=𝑉−𝑏=𝑣(𝑣+𝑏𝑚)

a=

1 3

2 𝑅 2 𝑇𝑐𝑖

=

3

( √2−1) 𝑅𝑖𝑐𝑖

bi=

𝑃𝑐𝑖

𝑎( √2−1)

𝑇

αi= [1+(0.48508+1.55171wi-0.15613wi)(1-TR𝑖 0.5)]2 ;TRi=𝑇𝑐𝑖 am=∑∑YiYi√(𝑎𝑖𝛼𝑖)(𝑎𝑗𝛼𝑗)

bm=∑Yibi 𝑅1

α=2.3392=.4274 b=0.08664 4𝑄

‫𝑑𝜋=ע‬2

Q=FṼ

α1=

α2= α3=

2 𝑅 2 𝑇𝑐𝑖

𝑃𝑐𝑖

2 𝑅 2 𝑇𝑐𝑖

𝑃𝑐𝑖 𝑚

M=∑YiMi

F=𝑀

(.4274)(0.08206)2 (369.8)2 41.9245

(.4274)(0.08206)2 (425.1)2 3704636

(.4274)(0.08206)2 (469.7)2 33.2593

=9.4921

=13.8760

=19.090862

(0.08664)(0.08206)2 (369.8)2

b1=

41.9245

(0.08664)(0.08206)2 (425.1)2

b2=

37.4636

(0.08664)(0.08206)2 (469.8)2

b3=

33.2593

=0.06271

=0.08067

=0.100426

α1=[1+(0.48508+1.55171ω-0.15613ω)(1-TR^0.5)] α=[1+(0.48508+1.55171)=1.52084 α2=1.6339 α3=1.7688

3

= 𝑃𝑐𝑖

bm=(.10)(0.06271)+(.20)(0.08067)+(.70)(0.100426) bm =0.9270 am =∑∑YiYi√(𝑎𝑖𝑎𝑖)(𝑎𝑗𝑎𝑗) am=(0.14436+0.723649+3.09103+11.10928+6.18206+16.54628 = 37.696660 am=27.5925065 F (Ṽ)=Ṽ3 -

Vm sup 1.08833403 7.99584835 5.49693496 3.85823268 2.81163534 2.19065114 1.89198255 1.81049225 1.80444451

𝑅𝑇 𝑃

𝑎𝑎

Ṽ2 +(- 𝑃 -𝑏 2

F(V) -1.65486077 433.440443 127.391047 36.9604694 10.3399994 2.59353047 0.46251278 0.02992367 0.00015863

𝑅𝑇𝑏𝑚 𝑃

)Ṽ-

𝑎𝑎𝑏 𝑃

F'(V) 0.23957399 173.451568 77.7389822 35.3148895 16.6509861 8.68363988 5.67567894 4.94790892 4.89548624

Vm cal 7.99584835 5.49693496 3.85823268 2.81163534 2.19065114 1.89198255 1.81049225 1.80444451 1.8044121

2. Una mezcla de equimolar de nitrógeno y metano se comprime y isotérmicamente de 1 atm y 0 °F a 50 atm ¿Qué volumen ocupan 10 libras masa de esta mezcla en las dos condiciones? a) Empleando la ecuación de estado Soave Redlich-Kwong b) Mediante la ecuación de estado de Peng Robinson c) Aplicando la ley de estados correspondientes: el método pseudocrítico Datos: P=50atm y 1 atm T=0° F Y1= 0.5

255.37K Y2= 0.5

M= 10 lbm N2: pc 33.50 atm Tc 126.05 K W= 0.038

CH4 Pc 46.58758 atm Tc 190.6 K W= 0.012

a) Ecuación de Soave Redlich Kwong a=

1 3

9( √2−1)

bi= Tr = 1.339 𝛼 = [1 + (0.48508 + 1.55771 𝑊1 − 0.15613𝑊1 2 )(1 − 𝑇1 − 1.3390.5 )]2 = 1.2963 𝛼 = [1 + (0.48508 + 1.55771 𝑊1 − 0.15613𝑊1 2 )(1 − 𝑇1 − 1.3390.5 )]2 = 1.2537 Am= 1.21321

𝑅𝑇𝐶1

b=8𝑃𝐶1 b1=0.02908

b2=0.027915

bm=∑Y1b1 a1=5.63265

a2=9.4196

am=(Y1a10.5+Y2𝑎20.5 )2=1.21321 𝑅𝑇

𝑧𝑚

P=Ṽ−𝑏𝑚 - Ṽ2 𝑚

F(Ṽ)=Ṽ3 − (𝑏𝑚 +

𝑅𝑇 𝑃

)Ṽ2 +

𝑎𝑚 𝑃

𝑎𝑚𝑏𝑚

Ṽ-

𝑝

𝛼 = [1 + (0.37464 + 1.54226 𝑊1 − 0.26992𝑊1 2 )(1 − 𝑇1 − 1.3390.5 )]2 = 1.2963 𝛼 = [1 + (0.48508 + 1.55771 𝑊1 − 0.15613𝑊1 2 )(1 − 𝑇1 − 1.3390.5 )]2 = 1.2537

F (V) =Ṽ3 -0.419113224Ṽ2 +0.01178540657V-0.000677335143 F (V) =3Ṽ2 -0.838226448+0.011178540657´

Vsup

Fsup

F'sup

Vcal

20.9556622

13.1148356

1299.86553

20.9455728

20.9455728

-3.39180615

1298.65364

20.9481846

20.9481846

-2.24821456

1298.97971

20.9499154

20.9499154

-1.4890451

1299.1958

20.9510615

20.9510615

-0.9861749

1299.33892

20.9518205

20.9518205

-0.65310672

1299.43369

20.9523231

20.9523231

-0.43251765

1299.49645

20.9526559

20.9526559

-0.28642871

1299.53802

20.9528763

20.9528763

-0.18968135

1299.56554

20.9530223

20.9530223

-0.12561158

1299.58377

20.9531189

20.9531189

-0.08318264

1299.59584

20.9531829

20.9531829

-0.05508513

1299.60383

20.9532253

20.9532253

-0.03647835

1299.60912

20.9532534

20.9532534

-0.02415658

1299.61263

20.953272

20.953272

-0.01599688

1299.61495

20.9532843

𝑚𝑇

NT= 𝑀 = 205.9172 mol V=4314.639101 L

F (V) =Ṽ3 -20.9556622Ṽ2 +0.627453344257 Ṽ -0.03386675715 F´ (V) =3Ṽ2 -41.9113244+0.011178540657

Vsup 1 atm 0.41911322 0.39637499 0.31223916 0.27111683 0.25227714 0.24767831 0.24740658 0.24740565 0.24740565

Fsup

F'sup

0.00426208 0.01587511 0.00330246 0.00080617 0.00013333 7.0504E-06 2.3915E-08 2.7845E-13 1.9516E-18

0.1874413 0.18868433 0.08030827 0.04279098 0.02899188 0.02594685 0.0257709 0.0257703 0.0257703

𝑚𝑇

NT= 𝑀 = 205.9172 mol V= 50.94713788 L

Vcal 0.39637499 0.31223916 0.27111683 0.25227714 0.24767831 0.24740658 0.24740565 0.24740565 0.24740565

b) Peng Robinson Datos: P=50atm y 1 atm T=0° F Y1= 0.5

255.37K Y2= 0.5

M= 10 lbm N2: pc 33.50 atm Tc 126.05 K W= 0.038

CH4: Pc 46.58758 atm Tc 190.6 K W= 0.012 F (V) =Ṽ3 -0.419113224Ṽ2 +0.01178540657V-0.000677335143 F (V) =3Ṽ2 -0.838226448+0.011178540657

3. Un compresor maneja 1000 lbm/hora de una mezcla de tano y propano que contienen 67% de etano en peso. El gas sale del compresor a 50 atm y 100 °C. Calcule el volumen de gas que sale del compresor por hora. a) Empleando la ecuación de estado de Van der Waals b) Usando la ecuación usando la ley de estados correspondientes: el método por pseudocrítico c) Ecuación de estado de Redlich-Kwong d) Ecuación de Peng Robinson Datos: 1000lbm/h P=50atm T=100ºC

373.15ºK

m=125.7999g/s y1=.7485

y2=.2514

Conversiones: 1𝑎𝑡𝑚

1𝑎𝑡𝑚

48.42 bar1.03𝑏𝑎𝑟=47.0097087atm

42.48bar1.03𝑏𝑎𝑟=4102427atm

a) Ecuación de Van der Waals 𝑅𝑇𝐶1

b=8𝑃𝐶1 b1=0.06661

b2=0.09197

bm=∑Y1b1 bm=(bm1+bm2)=(0.0498+0.02312) 27

ax=64 a1=5.63265

bm=0.072921

𝑅𝑇𝐶𝑖 𝑃𝐶1

a2=9.4196

am=(Y1a10.5+Y2𝑎20.5 )2=6.4922 𝑅𝑇

𝑧𝑚

P=Ṽ−𝑏𝑚 - Ṽ2 𝑚

F(Ṽ)=Ṽ3 − (𝑏𝑚 +

𝑅𝑇 𝑃

)Ṽ2 +

𝑎𝑚 𝑃

𝑎𝑚𝑏𝑚

Ṽ-

F (V) =Ṽ3 -0.68533478Ṽ2 +0.129844V-9.4683593 F (V) =3Ṽ2 -1.37066956V+0.129844

𝑝

b) Ley de estados correspondientes: método pseudocrítico. Tpc=∑yiTci

PV=2mRT Tpc=∑yiTci Ppc=∑yiPcI

ppc=∑yiPci

Ppc=

𝑅(∑𝑌𝑖𝑍𝑐𝑖)𝑇𝑝𝑐 ∑𝑦𝑖𝑉𝑐𝑖

(.7485)(305.3)(.2514)(369.8)=228.517+92.9677=321.4847 (.7485)(47.0097087)(.2514)(41.2427)=35.186+10.3684=46.1044 𝑇

TR=𝑇𝑃𝑐=

373.15°𝐾 321

𝑃

=1.16 ⸗ 1.2

50𝑎𝑡𝑚

PR=𝑃𝑝𝑐=46.1044 =-8.1

A partir de la gráfica: Zn=.93 𝑎𝑡𝑚

V= V=

4𝑄 𝜋𝑑2

𝑍𝑛𝑅𝑇 (.93)(82.06𝑐𝑚𝑚𝑜𝑙𝑘)(375.15𝐾) 𝑃

=

𝑉

Q=

50𝑎𝑡𝑚

F=

𝑡

𝑛

=569.5448𝑐𝑚3 /mol F=

Q=FV

𝑡

𝑚

M=∑YiMi

𝑀

M=YiMi+Y2M2= (.7485)(30.05)+(.2514)(44.0602) Ṁ=22.4924+11.07673=33.5688 g/mol 𝑚

125.9979𝑔/𝑠

F=𝑀 =33.5688𝑔/𝑚𝑜𝑙=3.7534 mol/s Q= (3.7534 mol/s)(569.5448 𝑐𝑚3 /mol) =2137.7424 𝑉

V= ∴ V=Vn 𝑛

n=

𝑐𝑚3 𝑠

𝑚 𝑀

1453593 𝑔

n=23.5688 𝑔/𝑚𝑜𝑙=13512.3388 mol 𝐶𝑀3

𝐿

V=(569.5448𝑀𝑂𝐿 )(13012.338 mol)=7695882.321𝑐𝑚3 =7695.88823ℎ

C) Ecuación de estado de Redlich-Kwong 𝑅 2 𝑇𝑐10.5 𝑎1 = 0.4274 = 99.72603 9 𝑃𝑐1 𝑎2 = 0.4274

𝑅 2 𝑇𝑐20.5 = 183.54782 9 𝑃𝑐2

𝑏1 = 0.08664

𝑅𝑇𝑐1 = 0.046173 𝑃𝑐1

𝑏2 = 0.08664

𝑅𝑇𝑐2 = 0.063749 𝑃𝑐2

Am=11841142 𝑏𝑚 = ∑

𝑦𝑖𝑏𝑖 = 𝑦1𝑏1 + 𝑦2𝑏2

𝑏𝑚 = 0.0505914 F(V)=0 F(V)=𝑉 3 -

𝑅𝑇 𝑃

𝑉 2 +(

𝑅𝑏𝑇 𝑃

𝑎

𝑎𝑏

- 𝑏 2 + 𝑃𝑇)Ṽ -𝑃√𝑇

D) Ecuación de Peng Robinson 𝑅 2 𝑇𝑐12 𝑎1 = 0.457223553 = 6.107436 𝑃𝑐1

𝑎2 = 0.457223553

𝑅 2 𝑇𝑐22 = 10.213619 𝑃𝑐2

𝑏1 = 0.07779607

𝑅𝑇𝑐1 = 0.046173 𝑃𝑐1

𝑏2 = 0.07779607

𝑅𝑇𝑐2 = 0.063749 𝑃𝑐2

𝑏𝑚 = ∑ = 𝑦𝑖𝑏𝑖 = 𝑦1𝑏1 + 𝑦2𝑏2 𝑏𝑚 = 0.050594 𝛼 = [1 + (0.37464 + 1.54226 𝑊1 − 0.26992𝑊1 2 )(1 − 𝑇𝑅1 0.5 )]2 𝜶𝟏 = 𝟏. 𝟓𝟐𝟖𝟖𝟔𝟔 𝜶𝟐 = 𝟏. 𝟑𝟖𝟔𝟑𝟔𝟖 𝑎𝑚 = 𝑌12 𝑎1 𝛼1 + 2𝑌1 𝑌2 √(𝑎1 𝛼1 )(𝑎2 𝛼2 + 𝑌22 𝑎2 𝛼2 𝑎𝑚 = 10.456125 F (V)=𝑉 3 - (

𝑅𝑇 𝑃

− 𝑏𝑚)Ṽ+(

𝑎𝑚 𝑃

- 3𝑏𝑚2 -

2𝑏𝑚𝑅𝑇 𝑃

)Ṽ + -

𝑅𝑇𝑏𝑚 𝑃

-

𝑎𝑚𝑏𝑚 𝑃

F(V)= 𝑉 3 -0.56182026𝑉 2 +0.029045647V+1.5676X10−3 -9.3448 X10−3 F(V)=3𝑉 3 -1.1236405V+0.02904567

4. Un gas natural tiene la siguiente composición en peso Metano ((CH4)

85%

Etano (C2H6)

5%

Nitrógeno (N2)

10%

El gas se comprime a 600 lb/𝑝𝑢𝑙𝑔2 abs para enviarlo por una tubería a 60 °F. Un millón de pies cúbicos (medidos a 1 atm y 60 °F por hora) es comprimido y enviado por la tubería que tiene un diámetro interno de un pie. Calcule la velocidad del gas en la tubería en pie/s. a) b) c) d)

Empleando la ecuación de estado de Van der Waals Mediante la ecuación de estado de Redlich Kwong Ley de estados correspondientes: el método pseudocrítico Usando la ecuación de estado de Soave Redlich Kwong (SRK)

CH4 C2H6 N2

Tc(K) 190.6 305.25 126.05

Pc(atm) 46.58787 49.36554 33.50

Vc cm/mol 48.6 145.5 89.2

Y3 0.85 0.05 0.1

Datos 𝐶𝐻4 = 0.85 𝑌𝑖 𝑃 = 600 𝑃𝑆𝐼𝑎 = 40.8441𝑎𝑡𝑚 𝐶2𝐻6 = 0.5 𝑌𝑖 𝑇 = 60𝑜 𝐹 = 288.7055𝐾 𝑁2 = 0.10 𝑌𝑖 𝐷 = 1 𝑝𝑖𝑒 = 0.3048𝑚 𝑉̇ =

106 𝑓𝑡 3 7865.49376𝐿 → 1𝑎𝑡𝑚 → 60𝑜 𝐹 = 𝑉̇ = → 1𝑎𝑡𝑚 → 288.7055𝐾 ℎ 𝑠

Zc 0.286 0.287 0.289

a) Ecuación de Van Der Waals 𝑃= 27 𝑅 2 𝑇𝑐𝑖 𝑎𝑖 = 64 𝑃𝑐𝑖 𝑏𝑖 =

𝑅𝑇𝑐𝑖 8𝑃𝑐𝑖

𝑅𝑇 𝑎𝑚 − 2 𝑉̂ − 𝑏𝑚 𝑉̂ 2

2

𝑎𝑚 = [∑ 𝑌𝑖 𝑎𝑖 ] 𝑖

𝑏𝑚 = ∑ 𝑌𝑖 𝑏𝑖 𝑖

2

𝑎𝐶𝐻4

27 𝑅 2 𝑇𝑐𝐴 𝐿2 𝑎𝑡𝑚 = = 2.25 64 𝑃𝑐𝐴 𝑚𝑜𝑙 2 2

𝑎C2H6

27 𝑅 2 𝑇𝑐𝐵 𝐿2 𝑎𝑡𝑚 = = 5.49 64 𝑃𝑐𝐵 𝑚𝑜𝑙 2 2

𝑎𝑁2

27 𝑅 2 𝑇𝑐𝐶 𝐿2 𝑎𝑡𝑚 = = 1.39 64 𝑃𝑐𝐶 𝑚𝑜𝑙 2

𝑅𝑇𝑐𝐴 𝐿 = 0.0428 8𝑃𝑐𝐴 𝑚𝑜𝑙 2 𝑅𝑇𝑐𝐵 𝐿 𝑏C2H6 = = 0.0638 8𝑃𝑐𝐵 𝑚𝑜𝑙 2 𝑅𝑇𝑐𝐶 𝐿 𝑏𝑁2 = = 0.0391 8𝑃𝑐𝐶 𝑚𝑜𝑙 2 𝑏𝐶𝐻4 =

𝑋𝑤𝑖 𝑋𝑤𝑖 𝑀𝑖 𝑀𝑖 𝑌𝑖 = = 𝑋𝑤𝑖 𝑋𝑤𝑖 ∑𝑖 ∑𝐶𝑖=𝐴 𝑀𝑖 𝑀𝑖

2

𝑎𝑚 = [∑ 𝑌𝑖 𝑎𝑖 ] 𝑖

𝑌𝐶𝐻4

𝑋𝑤𝐴 𝑀𝐴 = = 0.9101 𝑋𝑤𝑖 ∑𝐶𝑖=𝐴 𝑀𝑖

𝑌C2H6

𝑋𝑤𝐵 𝑀𝐵 = = 0.0286 𝑋𝑤𝑖 ∑𝐶𝑖=𝐴 𝑀𝑖

𝑋𝑤𝐶 𝑀𝐶 = = 0.06132 𝑋𝑤𝑖 ∑𝐶𝑖=𝐴 𝑀𝑖

𝑌𝑁2

2 𝑋𝑤𝑖 𝐿2 𝑎𝑡𝑚 𝑀𝑖 𝑎𝑚 = [∑ 𝑌𝑖 𝑎𝑖 ] = [∑ 𝑎𝑖 ] 2.2633 𝑚𝑜𝑙 2 𝐶 𝑋𝑤𝑖 𝑖 𝑖 ∑𝑖=𝐴 𝑀𝑖 2

𝑏𝑚 = ∑ 𝑌𝑖 𝑏𝑖 = 0.0437 𝑖

𝑓(𝑉̂ ) = 𝑉̂ 3 − 𝑉̂ 2 (𝑏𝑚 +

𝐿 𝑚𝑜𝑙

𝑅𝑇 𝑎𝑚 𝑏𝑚 𝑎𝑚 ) + 𝑉̂ ( ) − 𝑃 𝑃 𝑃

𝑓(𝑉̂ ) = 𝑉̂ 3 − 0.62321𝑉̂ 2 + 0.0555𝑉̂ − 2.3922𝑥10−3 𝑓´(𝑉̂ ) = 3𝑉̂ 2 − 1.24642𝑉̂ + 0.0555 𝑉̂𝑠𝑢𝑝

𝑅𝑇 𝐿 = 0.580039 𝑃 𝑚𝑜𝑙

𝑓(𝑉̂ ) 𝑉̂𝑐𝑎𝑙 = 𝑉̂𝑠𝑢𝑝 − ( ) 𝑓´(𝑉̂ )

Vsup 0.580039

F(Vm)

F’(V)

Vmcal

0.01522308 0.34177351 0.53549756

0.535497563 0.00212751 0.24822805 0.52692678 0.526926781 7.1601E-05 0.23159342 0.52661762 0.526617615 9.1498E-08 0.23100161 0.52661722 0.526617219 1.5006E-13 0.23100085 0.52661722 𝑉̂𝑐𝑎𝑙 = 0.52661722 𝑛̇ = 𝑉̇ =

𝐿 𝑚𝑜𝑙

𝑃𝑉 𝑉̇ = (𝑉̂ ∙ 𝑛̇ ) 𝑅𝑇

10000000𝑓𝑡 3 7865.49376𝐿 → 1𝑎𝑡𝑚 → 60𝑜 𝐹 = 𝑉̇ = → 1𝑎𝑡𝑚 → 288.7055𝐾 ℎ 𝑠

𝑛̇ =

𝑃𝑉 𝑚𝑜𝑙 = 322.00102 𝑅𝑇 𝑠

𝑣=

4𝑉̇ 𝜋𝐷2

𝑣=

4𝑉̇ 𝑚 = 0.23961 2 𝜋𝐷 𝑠

𝑉̇ = (𝑉̂ ∙ 𝑛̇ ) = 174.8374

𝑣 = 0.7861

𝑓𝑡 𝑠

𝐿 𝑠

0.01748374

𝑚3 𝑠

b) Ecuación de Redlich Kwong 𝑅𝑇 𝑎𝑚 − 𝑉̂ − 𝑏𝑚 √𝑇𝑉̂ (𝑉̂ + 𝑏𝑚)

𝑃=

𝑅 2 𝑇𝑐𝑖 𝑎𝑖 = 3 9( √2 − 1) 𝑃𝑐𝑖 1

2.5

𝑎𝑚 = ∑ ∑ 𝑌𝑖 𝑌𝑗 √𝑎𝑖 𝑎𝑗 𝑖

𝑗

3

𝑏𝑖 =

√2 𝑅𝑇𝑐𝑖 3 𝑃𝑐𝑖

𝑏𝑚 = ∑ 𝑌𝑖 𝑏𝑖 𝑖

𝐶𝐻4 = 190.6𝐾 C2H6 = 305.3𝐾 𝑁2 = 126.2𝐾 𝐶𝐻4 = 45.99𝑏𝑎𝑟 C2H6 = 48.72𝑏𝑎𝑟 𝑁2 = 34𝑏𝑎𝑟 𝑎𝐶𝐻4

C2H6 = 48.0829𝑎𝑡𝑚 𝑁2 = 33.5554𝑎𝑡𝑚

𝑅 2 𝑇𝑐𝐴 = 3 9( √2 − 1) 𝑃𝑐𝑖

𝑎C2H6 𝑎𝑁2

𝐶𝐻4 = 45.3886𝑎𝑡𝑚

1

2.5

𝑅 2 𝑇𝑐𝐵 = 3 9( √2 − 1) 𝑃𝑐𝐵 1

𝑅 2 𝑇𝑐𝐶 = 3 9( √2 − 1) 𝑃𝑐𝐶 1

= 31.80823 2.5

2.5

= 97.50018

𝐿 √2 𝑅𝑇𝑐𝐵 = 0.02985 3 𝑃𝑐𝐵 𝑚𝑜𝑙 3

𝑏C2H6 = 3

𝑏𝑁2 =

𝐿2 𝑎𝑡𝑚𝐾 0.5 𝑚𝑜𝑙 2

𝐿2 𝑎𝑡𝑚𝐾 0.5 = 15.3485 𝑚𝑜𝑙 2

3

𝑏𝐶𝐻4 =

𝐿2 𝑎𝑡𝑚𝐾 0.5 𝑚𝑜𝑙 2

𝐿 √2 𝑅𝑇𝑐𝐵 = 0.04514 3 𝑃𝑐𝐵 𝑚𝑜𝑙

𝐿 √2 𝑅𝑇𝑐𝐶 = 0.02674 3 𝑃𝑐𝐶 𝑚𝑜𝑙

𝑎𝑚 = ∑ ∑ 𝑌𝑖 𝑌𝑗 √𝑎𝑖 𝑎𝑗 𝑖

𝒊 = 𝟏, 𝟐, 𝟑

𝒋 = 𝟏, 𝟐, 𝟑

𝑗

𝑋𝑤𝑖 𝑋𝑤𝑖 𝑀𝑖 𝑀𝑖 𝑌𝑖 = = 𝑋𝑤𝑖 𝑋𝑤𝑖 ∑𝑖 ∑𝐶𝑖=𝐴 𝑀𝑖 𝑀𝑖 𝑌𝐶𝐻4

𝑋𝑤𝐴 𝑀𝐴 = = 0.9101 𝑋𝑤𝑖 ∑𝐶𝑖=𝐴 𝑀𝑖

𝑌C2H6

𝑋𝑤𝐵 𝑀𝐵 = = 0.0286 𝑋𝑤𝑖 ∑𝐶𝑖=𝐴 𝑀𝑖 𝑌𝑁2

𝑋𝑤𝐶 𝑀𝐶 = = 0.06132 𝑋𝑤𝑖 ∑𝐶𝑖=𝐴 𝑀𝑖

𝑋𝑤𝑗 𝑋𝑤𝑖 𝑀𝑗 𝑀𝑖 𝑎𝑚 = 𝑎𝑚 = ∑ ∑ 𝑌𝑖 𝑌𝑗 √𝑎𝑖 𝑎𝑗 = ∑ ∑ (𝑌𝑖 = ) 𝑌𝑗 = 𝑎𝑎 𝑋𝑤𝑖 𝑋𝑤𝑗 √ 𝑖 𝑗 𝐶 𝐶 ∑𝑖=𝐴 ∑𝑗=𝐴 𝑖 𝑗 𝑖 𝑗 𝑀𝑖 ( 𝑀𝑗 ) = 35.734

𝐿2 𝑎𝑡𝑚𝐾 0.5 𝑚𝑜𝑙 2

bm = ∑ Yi bi = .030097 i

L mol

𝑅𝑇 −𝑅𝑏𝑚 𝑇 𝑎𝑚 𝑏𝑚 𝑎𝑚 𝑓(𝑉̂ ) = 𝑉̂ 3 − 𝑉̂ 2 ( ) + 𝑉̂ ( − 𝑏𝑚 2 + )− 𝑃 𝑃 𝑝√𝑇 𝑃√𝑇 𝑓(𝑉̂ ) = 𝑉̂ 3 − 0.58004𝑉̂ 2 + 0.033127𝑉̂ − 1.549701𝑥10−3 𝑓´(𝑉̂ ) = 3𝑉̂ 2 − 1.16008𝑉̂ + 0.033127 𝑉̂𝑠𝑢𝑝

𝑅𝑇 𝐿 = 0.580039 𝑃 𝑚𝑜𝑙

𝑉̂𝑐𝑎𝑙 = 𝑉̂𝑠𝑢𝑝 − (

𝑓(𝑉̂ ) ) 𝑓´(𝑉̂ )

Vsup

F(V)

F'(V)

Vmcal

0.580039

0.01766491

0.36957108

0.53224058

0.532240578

0.00254121

0.26552545

0.52267008

0.52267008

9.2246E-05

0.24633993

0.52229561

0.522295615

1.3848E-07

0.24560043

0.52229505

0.522295051

3.1376E-13

0.24559932

0.52229505 𝑉̂𝑐𝑎𝑙 = .52229505

𝑛̇ =

𝑃𝑉 ; 𝑉̇ = (𝑉̂ ∙ 𝑛̇ ) 𝑅𝑇

𝑉̇ =

10000000𝑓𝑡 3 7865.49376𝐿 → 1𝑎𝑡𝑚 → 60𝑜 𝐹 = 𝑉̇ = → 1𝑎𝑡𝑚 → 288.7055𝐾 ℎ 𝑠

𝑃𝑉 𝑚𝑜𝑙 𝑛̇ = = 322.00102 𝑅𝑇 𝑠 𝑣=

4𝑉̇ 𝜋𝐷2

𝑣=

4𝑉̇ 𝑚 = .2305 2 𝜋𝐷 𝑠

𝑉̇ = (𝑉̂ ∙ 𝑛̇ ) = 168.17954

→ 𝑣 = 0.75616

𝑓𝑡 𝑠

𝐿 𝑚3 → 0.0168179 𝑠 𝑠

𝐿 𝑚𝑜𝑙

c) Ley de Estados Correspondientes: Método Pseudocritico Datos: 𝐶𝐻4 = 0.85 𝑌𝑖 𝑃 = 600 𝑃𝑆𝐼𝑎 = 40.8441𝑎𝑡𝑚 𝐶2𝐻6 = 0.5 𝑌𝑖 𝑇 = 60𝑜 𝐹 = 288.7055𝐾 𝑁2 = 0.10 𝑌𝑖 𝐷 = 1 𝑝𝑖𝑒 = 0.3048𝑚

𝑃𝑉̂ = 𝑍𝑅𝑇 → 𝑉 = 𝑇𝑅 =

𝑍𝑚𝑅𝑇 𝑃

𝑇 → 𝑇𝑃𝑐 = ∑ 𝑌𝑖𝑇𝑐 𝑖 𝑇𝑃𝐶

𝑃𝑝𝑐 = ∑ 𝑌 𝑖𝑃𝑐𝑖

𝑃𝑝𝑐 =

𝑃𝑅 =

𝑃 𝑃𝑝𝑐

𝑅(∑ 𝑌𝑖𝑍𝑐𝑖)𝑇𝑝𝑐 ∑ 𝑌𝑖𝑉𝑐𝑖

𝑇𝑝𝑐 = 𝑌1 𝑇𝑐1 + +𝑌2 𝑇𝑐2+ 𝑌3 𝑇𝑐3 𝑇𝑝𝑐 = (0.85)(190.6) + (0.05)(305.25) + (0.1)(126.05) 𝑇𝑝𝑐 = 189.8775 𝐾 𝑇𝑟 =

𝑇 288.7055 𝐾 = = 1.520 𝑇𝑝𝑐 189.8775 𝐾

𝑃𝑝𝑐 =

𝑅(∑ 𝑖𝑌𝑖𝑍𝑐𝑖)𝑇𝑝𝑐 = 77.4799 ∑ 𝑌𝑖𝑉𝑐𝑖

𝑃𝑝𝑐 = (82.06)(0.28595)(189.8775 𝐾)/57.505 ∑ 𝑌𝑖𝑉𝑐𝑖 = (0.85)(48.6) + (0.05)(145.5) + (0.1)(89.2) = 57.505 ∑ 𝑌𝑖𝑍 = (0.85)(0.286) + (0.05)(0.279) + (0.1)(0.289) = 0.28595 𝑃

40.8274

PR= 𝑃𝑝𝑐 = 77.4799 = 0.526 𝑍 = 0.95 𝑉̂ =

𝑍𝑚𝑅𝑇 (0.95)(82.06)(288.705) = 𝑃 40.8274

𝑉̂ = 551.2615

𝑐𝑚3 𝑚𝑜𝑙

→

𝑉̂ = 0.5512615

𝑚3 𝑚𝑜𝑙

𝑛̇ =

𝑃𝑉 ; 𝑉̇ = (𝑉̂ ∙ 𝑛̇ ) 𝑅𝑇

𝑉̇ =

106 𝑓𝑡 3 7865.49376𝐿 → 1𝑎𝑡𝑚 → 60𝑜 𝐹 = 𝑉̇ = → 1𝑎𝑡𝑚 → 288.7055𝐾 ℎ 𝑠

𝑛̇ =

𝑃𝑉 𝑚𝑜𝑙 = 322.00102 𝑅𝑇 𝑠

𝑣=

4𝑉̇ 𝜋𝐷2

𝑣=

4𝑉̇ 𝑚 = 0.243397 2 𝜋𝐷 𝑠

𝑣 = 0.798537

𝑓𝑡 𝑠

𝑉̇ = (𝑉̂ ∙ 𝑛̇ ) = 177.50676

𝐿 𝑠

0.017759676

𝑚3 𝑠

d) Ecuación de Soave Redlich Kwong 𝑅𝑇 𝑎𝑚 − 𝑉̂ − 𝑏𝑚 𝑉̂ (𝑉̂ + 𝑏𝑚)

𝑃=

𝑅 2 𝑇𝑐𝑖 𝑎𝑖 = 3 9( √2 − 1) 𝑃𝑐𝑖 1

2

𝑎𝑚 = ∑ ∑ 𝑌𝑖 𝑌𝑗 √𝑎𝑖 𝑎𝑗 𝑖

𝑗

3

𝑏𝑖 =

√2 𝑅𝑇𝑐𝑖 3 𝑃𝑐𝑖

𝑏𝑚 = ∑ 𝑌𝑖 𝑏𝑖 𝑖 2

𝑇 0.5 2) ∝𝑖 = [1(. 48508 + 1.5517𝜔𝑖 + .15613𝜔𝑖 (1 − ( ) )] 𝑇𝑐𝑖 𝐶𝐻4 = 190.6𝐾 C2H6 = 305.3𝐾 𝑁2 = 126.2𝐾

𝐶𝐻4 = 45.99𝑏𝑎𝑟 C2H6 = 48.72𝑏𝑎𝑟 𝑁2 = 34𝑏𝑎𝑟

𝐶𝐻4 = 45.3886𝑎𝑡𝑚 C2H6 = 48.0829𝑎𝑡𝑚 𝑁2 = 33.5554𝑎𝑡𝑚

𝐶𝐻4 = 0.012 C2H6 = 0.100 𝑁2 = 0.038 𝑎𝐶𝐻4

𝑎C2H6 𝑎𝑁2

2

𝑅 2 𝑇𝑐𝐴 𝐿2 𝑎𝑡𝑚 = 3 = 2.3039 𝑚𝑜𝑙 2 9( √2 − 1) 𝑃𝑐𝑖 1

2

𝑅 2 𝑇𝑐𝐵 𝐿2 𝑎𝑡𝑚 = 3 = 5.5801 𝑚𝑜𝑙 2 9( √2 − 1) 𝑃𝑐𝐵 1

2

𝑅 2 𝑇𝑐𝐶 𝐿2 𝑎𝑡𝑚 = 3 = 1.3663 𝑚𝑜𝑙 2 9( √2 − 1) 𝑃𝑐𝐶 1

3

𝐿 √2 𝑅𝑇𝑐𝐵 = 0.02985 3 𝑃𝑐𝐵 𝑚𝑜𝑙

𝑏𝐶𝐻4 =

3

𝑏C2H6 =

𝐿 √2 𝑅𝑇𝑐𝐵 = 0.04514 3 𝑃𝑐𝐵 𝑚𝑜𝑙

3

𝑏𝑁2 =

𝐿 √2 𝑅𝑇𝑐𝐶 = 0.02674 3 𝑃𝑐𝐶 𝑚𝑜𝑙 2

∝𝐶𝐻4 = [1(. 48508 + 1.5517𝜔𝐴 + .15613𝜔𝐴

2)

𝑇 0.5 (1 − ( ) )] = 1.02682 𝑇𝑐𝐴 2

𝑇 0.5 2) ∝C2H6 = [1(. 48508 + 1.5517𝜔𝐵 + .15613𝜔𝐵 (1 − ( ) )] = 1.0005 𝑇𝑐𝐵 2

∝𝑁2 = [1(. 48508 + 1.5517𝜔𝐵 + .15613𝜔𝐵

𝑎𝑚 = ∑ ∑ 𝑌𝑖 𝑌𝑗 √(𝑎𝑖 ∝𝑖 )(𝑎𝑗 ∝𝑗 ) 𝑖

2)

𝑇 0.5 (1 − ( ) )] = 1.14296 𝑇𝑐𝐵

𝒊 = 𝟏, 𝟐, 𝟑

𝒚

𝒋 = 𝟏, 𝟐, 𝟑

𝑗

𝑋𝑤𝑖 𝑋𝑤𝑖 𝑀𝑖 𝑀𝑖 𝑌𝑖 = = 𝑋𝑤𝑖 𝑋𝑤𝑖 ∑𝑖 ∑𝐶𝑖=𝐴 𝑀𝑖 𝑀𝑖 𝑌𝐶𝐻4

𝑋𝑤𝐴 𝑀𝐴 = = 0.9101 𝑋𝑤𝑖 ∑𝐶𝑖=𝐴 𝑀𝑖

𝑌C2H6

𝑋𝑤𝐵 𝑀𝐵 = = 0.0286 𝑋𝑤𝑖 ∑𝐶𝑖=𝐴 𝑀𝑖

𝑋𝑤𝐶 𝑀𝐶 𝑌𝐶 = = 0.06132 𝑋𝑤𝑖 𝐶 ∑𝑖=𝐴 𝑀𝑖

𝑎𝑚 = 𝑎𝑚 = ∑ ∑ 𝑌𝑖 𝑌𝑗 √(𝑎𝑖 ∝𝑖 )(𝑎𝑗 ∝𝑗 ) 𝑖

𝑗

𝑋𝑤𝑗 𝑋𝑤𝑖 𝑀𝑗 𝑀𝑖 = ∑ ∑ (𝑌𝑖 = ) 𝑌𝑗 = (𝑎 ∝ )(𝑎 ∝ ) 𝑋𝑤𝑖 𝑋𝑤𝑗 √ 𝑖 𝑖 𝑗 𝑗 𝐶 𝐶 ∑𝑖=𝐴 ∑𝑗=𝐴 𝑖 𝑗 𝑀𝑖 ( 𝑀𝑗 ) = 2.5976

𝐿2 𝑎𝑡𝑚 𝑚𝑜𝑙 2

𝑏𝑚 = ∑ 𝑌𝑖 𝑏𝑖 = 0.030097 𝑖

𝑓(𝑉̂ ) = 𝑉̂ 3 − 𝑉̂ 2 (𝑏𝑚 ) + 𝑉̂ (𝑏𝑚 2 ) −

𝐿 𝑚𝑜𝑙

𝑅𝑇 2 𝑎𝑚 𝑎𝑚 𝑏𝑚 (𝑉̂ − 𝑉̂ 𝑏𝑚 ) + 𝑉̂ ( ) − 𝑃 𝑃 𝑃

𝑓(𝑉̂ ) = 𝑉̂ 3 − 0.610137𝑉̂ 2 + 0.08196123𝑉̂ − 1.914106𝑥10−3 𝑓´(𝑉̂ ) = 3𝑉̂ 2 − 1.220274 + 0.08196123 𝑉̂𝑠𝑢𝑝

𝑅𝑇 𝐿 = 0.580039 𝑃 𝑚𝑜𝑙

𝑓(𝑉̂ ) 𝑉̂𝑐𝑎𝑙 = 𝑉̂𝑠𝑢𝑝 − ( ) 𝑓´(𝑉̂ )

Vsup

F(V)

F'(V)

Vmcal

0.58004

0.04

0.38

0.48747

0.48747

0.01

0.20

0.44302

0.44302

0.00

0.13

0.43075

0.43075

0.00

0.11

0.42981

0.42981

0.00

0.11

0.42980

0.42980

0.00

0.11

0.42980

𝑉̂𝑐𝑎𝑙 = 0.4298 𝑛̇ =

𝐿 𝑚𝑜𝑙

𝑃𝑉 ; 𝑉̇ = (𝑉̂ ∙ 𝑛̇ ) 𝑅𝑇

𝑉̇ =

10000000𝑓𝑡 3 7865.49376𝐿 → 1𝑎𝑡𝑚 → 60𝑜 𝐹 = 𝑉̇ = = 1𝑎𝑡𝑚 = 288.7055𝐾 ℎ 𝑠

𝑛̇ =

𝑃𝑉 𝑚𝑜𝑙 = 322.00102 𝑅𝑇 𝑠

𝑉̇ = (𝑉̂ ∙ 𝑛̇ ) = 138.3960 𝑣=

𝑣=

𝐿 𝑠

0.0138396

4𝑉̇ 𝜋𝐷2

4𝑉̇ 𝑚 = 0.1897 𝜋𝐷2 𝑠

𝑣 = 0.62228

𝑓𝑡 𝑠

𝑚3 𝑠