Tarea XI Termo
Tarea XI Termo
Tarea XI Termo
35
a) Aplicando la ecuacion (11.63)
b) Suponiendo que la mezcla es una solucion ideal, estipule que todos los Kij = 0
Solucion
IJ Tcij (°K) Zcij Vcij (cc/mol) Pcij (bar) Wij Trij
11 305.30 0.279 145 48.8397 0.1 1.0585
22 369.80 0.276 200 42.4283 0.152 0.8739
12 336.01 0.278 171.028891 45.3263 0.126 0.9617
lnφ1 φ1 lnφ2 φ2 f1 f2
-0.15965373 0.85243891 -0.37217815 0.6892314422 8.95060861 13.4400131
Solucion ideal
φ1 φ2 f1 f2
1.0000 1.0000 10.5 19.5
, P=30 bar y Y1=0.35
Propieades Criticas
Vc/10-6 cc/mol Zc W
145 0.279 0.1
200 0.276 0.152
T N2 Ar2 diferencia
363.15 12471 12471 0
Calculando entropia
1 (bar cc/mol*K) = 10^-1 J
(bar cc/K) (J/K)
AS --- N2 118.06119126 11.8061191
AS --- Ar2 -95.47146357 -9.54714636
AS mix 360.06353307 36.0063533
AS 382.65326075 38.2653261
Cv Cp Xi*LnXi
207.85 290.99 -0.29877404
124.71 207.85 -0.3675044
-0.66627844
R (bar cc/mol*K) 83.14
Componente Flujo (Kg/s) Flujo (g/s) Pm (g/mol) flujo molar Yi
(mol/s)
Nitrogeno 2 2000 28.014 71.392875 0.224
Hidrogeno 0.5 500 2.016 248.015873 0.776
319.408748 1
1 (bar cc/mol*K) = 10^-1 J
∆𝑆=𝑅∗⟨𝐶_𝑃^𝑖𝑔 ⟩_𝑠/
datos
T(°K) 308.15 R(J/mol°k) 8.3145
To(°K) 448.15
P1 (bar) 1
P2 (bar) 1
comp compcicion A B C D τ
metano 0.5 1.702 0.009081 -2.164E-06 0 0.6876046
etano 0.5 1.131 0.019225 -5.561E-06 0
𝑤_𝑖𝑑𝑒𝑎𝑙=∆𝐻−𝑇_𝜎 ∆𝑆
Cp S CpH ΔS ΔH w ideal
3.75617414 4.82299919 -15.9987405 -7228.41754 -2428.79538
6.07361702 7.59664212
4.91 6.21
𝑂 (𝜏+1)+𝐶/3 𝑇_0^2
))
𝐵^𝑜=0.083−0.422⁄( 〖𝑇𝑟〗 ^1.6
0.8
φi , φG(Pi)
φi
φG(Pi)
0.6
φi , φG(
φi
φG(Pi)
0.6
0.4
0 100 200 300 400 500 600
Pi (bar)
ln〖∅ =∫1▒ 〖 (𝑧−1) 𝑑𝑃/𝑃=𝑃𝑟/𝑇𝑟 (𝐵^𝑜+𝜔𝐵^𝑙 ) 〗〗
Bo Bl
-0.16587211 0.09599481
400
300
fi , fG(Pi)
200
fi
fG(Pi)
100
200
fi , fG(
fi
fG(Pi)
100
0
0 100 200 300 400 500 600
Pi (bar)
ln〖∅ =∫1▒ 〖 (𝑧−1) 𝑑𝑃/𝑃=
DATOS
T (°K) 600
P (bar) 300
Propieades Criticas
Componente Tc (°K) Pc (bar) W Tr Pr
SO2 430.8 78.84 0.245 1.39275766 3.80517504
Pr φ1 1.35424724
Tr 3 3.80517504 5
1.3 1.2853 1.3868
1.39275766 φ1
1.4 1.2942 1.4488
〖∅ =∫1▒ 〖 (𝑧−1) 𝑑𝑃/𝑃=𝑃𝑟/𝑇𝑟 (𝐵^𝑜+𝜔𝐵^𝑙 ) 〗〗
∅=∅^0 〖 (∅^1) 〗 ^𝜔
φ fi GR/RT
0.72399817 217.199451 -0.32296642
ln〖∅ =∫1▒ 〖
A) DATOS
T (°C)P 280 T (°K) 553.15
P (bar) 20 P (bar) 20 𝐵^𝑜=0.083−0.422⁄( 〖𝑇𝑟〗
Propieades Criticas
Componente Tc (°K) Pc (bar) W Tr Pr
isubutileno 417.9 40 0.194 1.32364202 0.5
B) DATOS
T (°C) 280 T (°K) 553.15
P (bar) 100 P (bar) 100
Propieades Criticas
Componente Tc (°K) Pc (bar) W Tr Pr
isubutileno 417.9 40 0.194 1.32364202 2.5
Pr φ1 = 1.46193447
Tr 2 2.5 3
1.3 1.776 1.2853
1.3236420196 M
1.4 1.1858 1.2942
ln〖∅ =∫1▒ 〖 (𝑧−1) 𝑑𝑃/𝑃=𝑃𝑟/𝑇𝑟 (𝐵^𝑜+𝜔𝐵^𝑙 ) 〗〗
Bo Bl lnφ φ fi
-0.18645086 0.08602242 -0.06412705 0.93788583 18.7577167
Según el profe
Bo Bl lnφ φ fi
-0.18645086 0.08602242 -0.32063525 0.7256879 72.56879
∅=∅^0 〖 (∅^1) 〗 ^𝜔
φ fi
0.75595561 75.5955608
𝐵^𝑜=0.083−0.422⁄( 〖𝑇𝑟〗 ^1.6 ) ; 𝐵^𝑙=0.139−0.172⁄( 〖𝑇𝑟〗
Pr,sat = Pi,sat/Pc
A) DATOS PV (bar) = presion de vap
T (°C) 110 T (°K) 383.15 PV (bar) R
P (bar) 275 P (bar) 275 5.267 83.14
Propieades Criticas
Componente Tc (°K) Pc (bar) Vc Zc W Tr
Ciclopentano 511.8 45.02 258 0.273 0.196 0.74863228
Pi,sat = 5.267
lnφi,sat -0.10534662
φi,sat 0.90001251
fi 11.7841834
Propieades Criticas
Componente Tc (°K) Pc (bar) Vc Zc W Tr
isubutileno 420 40.43 239.3 0.277 0.191 0.93607143
Pi,sat = 25.83
lnφi,sat -0.27495583
φi,sat 0.75960567
fi 20.28535
𝑉_𝑖^𝑠𝑎𝑡=𝑉_𝑐
〖𝑍𝑐〗 ^((1−𝑇𝑟)^(2
Pr,sat Vi,sat Bo Bl
0.11699245 107.545777 -0.58762969 -0.44123049
Pr,sat Vi,sat Bo Bl
0.63888202 133.299189 -0.38604884 -0.08800253
𝑉_𝑖^𝑠𝑎𝑡=𝑉_𝑐
〖𝑍𝑐〗 ^((1−𝑇𝑟)^(2⁄7) )
A) DATOS
T (°C)P 100 T (°K) 373.15
P (bar) 35 P (bar) 35
K 0 Y1 0.21
Y2 0.43
R(cc*bar/(mol*K)) 83.14 Y3 0.36
Solucion
IJ Tcij (°K) Zcij Vcij (cc/mol)
11 190.60 0.286 98.6
12 241.23 0.2825 120.5334969683
13 265.49 0.281 143.3784693222
21 241.23 0.283 120.5334969683
22 305.30 0.279 145.5
23 336.01 0.278 171.3078099766
31 265.49 0.281 143.3784693222
32 336.01 0.278 171.3078099766
33 369.80 0.276 200
B) DATOS
T (°C)P 100 T (°K) 373.15
P (bar) 35 P (bar) 35
K 0 Y1 0.21
Y2 0.43
R(cc*bar/(mol*K)) 83.14 Y3 0.36
Solucion
IJ Tcij (°K) Zcij Vcij (cc/mol)
11 190.60 0.286 98.6
12 241.23 0.2825 120.5334969683
13 265.49 0.281 143.3784693222
21 241.23 0.283 120.5334969683
22 305.30 0.279 145.5
23 336.01 0.278 171.3078099766
31 265.49 0.281 143.3784693222
32 336.01 0.278 171.3078099766
33 369.80 0.276 200
Propieades Criticas
Componente Tc (°K) Pc (bar) Vc/10-6 cc/mol Zc W
Metano 190.6 45.99 98.6 0.286 0.012
Etano 305.3 48.72 145.5 0.279 0.1
Propano 369.8 42.48 200 0.276 0.152
φ1 φ2 φ3
1.01919679 0.88065833 0.774943262
f1 f2 f3
7.4910964 13.2539079 9.7642851008
Propieades Criticas
Componente Tc (°K) Pc (bar) Vc/10-6 cc/mol Zc W
Metano 190.6 45.99 98.6 0.286 0.012
Etano 305.3 48.72 145.5 0.279 0.1
Propano 369.8 42.48 200 0.276 0.152
φ1 φ2 φ3
0.97712404 0.88034828 0.75919127
f1 f2 f3
7.18186166 13.2492416 9.56580997
A) 𝐺^𝐸∕𝑅𝑇=( 〖− 2.6𝑥 〗 _1− 〖 1.8𝑥 〗 _2 ) 𝑥_1 𝑥_2
𝑥_2=1−𝑥_1
𝐺^𝐸/𝑅𝑇=−1.8𝑥_1+2 〖𝑥 _1 〗 ^2+
ln〖𝛾 _1=𝐺^𝐸/𝑅𝑇+ 〗 (1−𝑥_1 ) ln〖𝛾
(𝑑 𝐺^𝐸/𝑅𝑇)/(𝑑𝑥_1 )
_2=𝐺^𝐸/𝑅𝑇−𝑥_1 (𝑑 𝐺^𝐸/𝑅𝑇)/(𝑑𝑥_1 ) 〗
(𝑑 𝐺^𝐸/𝑅𝑇)/(𝑑𝑥_1 )=−1.8+2𝑥_1+2.4 〖𝑥 _1 〗 ^2
𝐺^𝐸/𝑅𝑇=−1.8𝑥_1+ 〖𝑥 _1 〗 ^2+0.8 〖𝑥 _1 〗 ^3
𝑥 _1 〗 ^3+1.6 〖𝑥 _1 〗 ^4
DATOS
T (°C) 75 T (°K) 348.15 Componente Yi
P (bar) 2 P (bar) 2 propano 0.5
R 83.14 n-pentano 0.5
〖𝑑𝐵〗 _𝑖𝑗/𝑑𝑇=[■8(−276&−466@−466&−809)]
HR (bar SR (bar
HR/RT SRR cc/(mol ) cc/(mol k)) HR (J/mol) SR (J/mol K)
-0.12023984 -0.08539812 -3480.365 -7.1 -348.0365 -0.71
^ = 𝑃(𝐵_11+𝑦_2^2
ln〖∅ ̂ _1=𝑃/𝑅𝑇
ln ∅ 1 ( 𝐵 + 𝑦 2 𝛿 )𝛿_12 𝑑𝑜𝑛𝑑𝑒
11 2 12
)𝑑𝑜𝑛𝑑𝑒
〗 𝛿 =2 𝐵 − 𝐵
𝛿_12=2𝐵_12−𝐵_
12 12
𝜔𝑖 +𝜔 𝑗
𝜔_𝑖𝑗=(𝜔_𝑖+𝜔_𝑗)/2 𝑅𝑇
𝜔𝑖𝑗 = 〖𝑇𝑐〗 _𝑖𝑗=( 〖𝑇𝑐〗 _𝑖 〖) 𝑇𝑐〗 _𝑗 )^(1∕2) (1−𝐾_𝑖𝑗) ∅ ̂_1=𝑓 ̂_1/(𝑌_1𝑓^𝑃) , ∅ ̂_2=𝑓 ̂
2 𝑓^ _
ln〖∅ ̂ ln ^ 2= 𝑃(𝐵_22+𝑦_1^2
_2=𝑃/𝑅𝑇
∅ ( 𝐵22+ 𝑦 1 𝛿 12)𝛿_12 ) 〗 ∅
2 ^ 1= 1 , ∅ ^ 2= 2
𝑅𝑇 𝑌1𝑃 𝑌2𝑃
〖𝑍𝑐〗 _𝑖𝑗=(
𝑍𝑐〖 𝑍𝑐𝑍𝑐 + 𝑍𝑐〖𝑗 𝑍𝑐〖𝑇𝑐〗
〗𝐵_𝑖𝑗=(𝑅
𝑖_𝑖+ 〗 _𝑗)/2 𝑅 𝑇𝑐
_𝑖𝑗)/ 〖𝑖𝑗𝑃𝑐〗𝑜 _𝑖𝑗 (𝐵^𝑜+𝜔_𝑖𝑗
𝑙 𝐵^𝑙 )
𝑖𝑗 =
2 𝐵 𝑖𝑗 = ( 𝐵 + 𝜔𝑖𝑗 𝐵 )
𝑃𝑐 𝑖𝑗
_𝑖𝑗)/𝑉𝑐 𝑉𝑐〗 _𝑗 〗 ^(1∕3))/2)^3
3
〖𝑉𝑐〗
〖𝑃𝑐〗 _𝑖𝑗=( 〖𝑍𝑐〗 _𝑖𝑗𝑍𝑐 𝑅 𝑇𝑐 𝑖𝑗〖〖
_𝑖𝑗=((
𝑅 𝑖𝑗〖𝑇𝑐〗 〗 _𝑖 〗
〖𝑉𝑐〗 𝑉𝑐
1/3
_𝑖𝑗^(1∕3)+ 〖〖
+𝑉𝑐
1 /3
𝑃𝑐𝑖𝑗 =
𝑉𝑐 𝑖𝑗
𝑉𝑐 𝑖𝑗 =( 𝑖
2
𝑗
)
𝐵^𝑜=0.083−0.422⁄( 〖𝑇𝑟〗 _𝑖𝑗^1.6 ) ; 𝐵^𝑙=0.139−0.172⁄( 〖𝑇
𝑇=𝑇𝑐∗𝑇𝑟
𝑃=𝑃𝑐∗𝑃𝑟
𝑑𝑜𝑛𝑑𝑒
𝑑𝑜𝑛𝑑𝑒 𝛿 12=2 𝐵 12 − 𝐵11 − 𝐵 12
𝛿_12=2𝐵_12−𝐵_11−𝐵_12
∅ ̂_1=𝑓 ̂_1/(𝑌_1𝑓^𝑃) , ∅ ̂_2=𝑓 ̂
𝑓^ _2/(𝑌_2 𝑃)
∅^ 1= 1 , ∅ ^ 2= 2
𝑌1𝑃 𝑌2𝑃
6 ) ; 𝐵^𝑙=0.139−0.172⁄( 〖𝑇𝑟〗 _𝑖𝑗^4.2 ) ; 〖𝑇𝑟〗 _𝑖𝑗=𝑇⁄ 〖𝑇𝑐〗 _𝑖𝑗 ; 〖𝑃𝑟〗 _𝑖𝑗=𝑃⁄ 〖𝑃𝑐〗 _𝑖𝑗
=𝑇𝑐∗𝑇𝑟
=𝑃𝑐∗𝑃𝑟