Chemistry">
Sistema Alcanfor-Undecano
Sistema Alcanfor-Undecano
Sistema Alcanfor-Undecano
Procesos de separación II
Profesor: ALEJANDRE
SANCHEZ, ELIEZER
Trabajo final
Sistema alcanfor-undecano
25/06/2021
ALUMNOS:
Ivan Alberto Quintero Silva
Erick Guadalupe Mendoza Arriola
Elvia Monserrath Rodriguez Alejo
Alejandra Gonzalez Jimenez
Jose de Jesus Martinez Gomez
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Para analizar
Realice el diseño y los cálculos para una columna de destilación de platos perforados
que incluye todos sus equipos de apoyo (re boiler, condensador e intercambiador de
calor) y los equipos de almacenamiento (tanque de alimentación (48 horas de flujo),
tanque de fondos, tanque de destilado) y equipos de servicios (bombas), el diseño y
los cálculos debe de contener los siguientes puntos:
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ALCANFOR
𝑪𝟏𝟎 𝑯𝟏𝟔 𝑶
Masa molar: 𝟏𝟓𝟐. 𝟐𝟑𝒈𝒓⁄𝒎𝒐𝒍
Punto de ebullición: 209 °𝑐
Densidad: 990 𝑘𝑔⁄𝑚3
𝑪𝒑 𝒈𝒂𝒔
𝐶𝑝𝑔𝑎𝑠 = 𝐴 + 𝐵𝑇 + 𝐶𝑇 2 + 𝐷𝑇 3 + 𝐸𝑇 4
𝐶𝑝𝑔𝑎𝑠 = −145.746 + (1.3060𝑥100 )(482.15) + (−9.5043𝑥10−4 )(482.15)2 +
(3.282𝑥10−7 )(482.15)3 + (−3.9497𝑥10−11 )(482.15)4
𝐶𝑝𝑔𝑎𝑠 = 288.52 𝐽⁄𝑚𝑜𝑙 ∙ 𝑘
𝑪𝒑 𝑳í𝒖𝒊𝒅𝒐
𝐶𝑝𝑙𝑖𝑞 = −9.18198 + (7.25532𝑥100 )(482.15) +
(−1.447𝑥10−2 )(482.15)2 +)1.0320𝑥10−5 )(482.15)3
𝐽
𝐶𝑝𝑙𝑖𝑞 = 1281.399 ⁄𝑚𝑜𝑙 ∙ 𝑘
UNDECANO
𝐶11 𝐻24
𝑔𝑟
Masa molar: 156.31 ⁄𝑚𝑜𝑙
Punto de ebullición: 196 °𝑐
𝑔
Densidad: 74 ⁄𝑐𝑚3
𝑪𝒑 𝒈𝒂𝒔
𝐶𝑝𝑔𝑎𝑠 = 125.212 + (3.1401𝑥10−1 )(469.15)+)7.913𝑥10−4 )(469.15)2 +
(−9.1410𝑥10−7 )(469.15)3 + (2.7568𝑥10−10 )(469.15)4
𝐽
𝐶𝑝𝑔𝑎𝑠 = 365.66 ⁄𝑚𝑜𝑙 ⋅ 𝑘
𝐶𝑝 𝐿í𝑞𝑢𝑖𝑑𝑜
𝐶𝑝 𝐿𝑖𝑞 = 94.169 + (1.7806𝑥100 )(469.15) + (−4.6303𝑥10−3 )(469.15)2 +
(4.4675𝑥10−6 )(469.15)3
𝐽
𝐶𝑝𝑙𝑖𝑞 = 423.348 ⁄𝑚𝑜𝑙 ⋅ 𝑘
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𝑝 𝐵
𝑙𝑜𝑔10 = 𝐴 + + 𝐶𝑙𝑜𝑔10 𝑇 + 𝐷𝑇 + 𝐸𝑇 2 𝑃𝑟𝑒𝑠𝑖ó𝑛 𝑣𝑎𝑝𝑜𝑟
𝑇
kg⁄
F = 10,000 hr
XF = 0.38
XD = 0.95
XW = 0.09
L
= 0.89
V
TF = 140°c
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BALANCE DE MATERIA
𝑃𝑚𝑚 = (156.31)(0.5) + (152.23)(0.5) = 154.27
𝐹 = (10,000)(1⁄154.27) = 64.82 𝑘𝑚𝑜𝑙 ⁄ℎ𝑟
𝐹 =𝐷+𝑊
64.32 = 𝐷 + 𝑊(1)
ENTALPÍA LÍQUIDA
𝐇𝐋 = 𝐗 𝐀 𝐂𝐏𝐀 𝚫𝐓 + (𝟏 − 𝐗 𝐀 )𝐂𝐏𝐁 𝚫𝐓 + 𝐇𝐦
HL = (0.45)(288.52)(93.86 − 50°c) + (1 − 0.45)(365.66)(93.86 − 50)
J
HL = 14515.33
Kmol
ENTALPÍA DE VAPOR
𝐻𝑉 = (𝑌𝐴 (𝐶𝑃𝐴 Δ𝑇 − 𝜆𝐴 )) + (1 − 𝑌𝐴 )(𝐶𝑃𝐵 Δ𝑇 + 𝜆𝐵 ) + Δ𝐻𝑚
𝐻𝑉 = (0.66)((288.52)(93.86 − 50) + (30820))(+(1 − 0.66)((138)(93.86 − 50) +
33330
𝐾𝐽
𝐻𝑉 = 94059.247
𝐾𝑚𝑜𝑙
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Cp LÍQUIDO DE LA MEZCLA
𝐶𝑝𝑙 = (1281.399)(0.5) + (423.348)(0.5)
𝐾𝐽
𝐶𝑝𝑙 = 852.3735 ⁄𝐾𝑚𝑜𝑙
(𝐻𝑉 − 𝐻𝐿 ) + (𝐶𝑝𝑙𝑚 (𝑇𝐵 − 𝑇𝐹 ))
𝑞=
𝐻𝑉 − 𝐻𝐿
(94059.42 − 14515.33) + (852.3735)(93.86 − 54)
𝑞=
(94059.42 − 14515.33)
27519.5582
𝑞=
25632.91
𝑞 = 1.074
9 1.074
𝑚𝑞 = = = 14.51
𝑞 − 1 1.074 − 1
𝑎𝑟𝑐𝑡𝑎𝑔(𝑚𝑞 )
𝑎𝑟𝑐𝑡𝑎𝑔(14.51) = 86.0575°
PUNTO DE ENTIQUECIMIENTO
𝑽𝒏+𝟏 = 𝑯𝑽𝒏+𝟏 = 𝑳𝒏 𝒉𝒍𝒏 + 𝑫𝒉𝒍𝑫
𝑄𝐷 = 𝑉1 𝐻𝑉 − 𝐿ℎ𝐼𝐷 − 𝐷ℎ𝑙𝐷 (𝑐𝑎𝑙𝑜𝑟 𝑐𝑒𝑑𝑖𝑑𝑜)
𝐷ℎ𝑙𝐷 + 𝑄𝐷
𝑄1 = (𝑝𝑢𝑛𝑡𝑜 𝑑𝑒 𝑒𝑛𝑟𝑖𝑞𝑢𝑒𝑐𝑖𝑚𝑖𝑒𝑛𝑡𝑜)
𝐷
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MOLES DE VAPOR
𝑉 =𝐿+𝐷
𝑉 = 𝐿 + 22.08
𝑉 = 178.64 + 22.08
𝑉 = 200.72 𝐾𝑚𝑜𝑙⁄ℎ
DISEÑO DE LA COLUMNA
𝟒𝑽 (𝟐𝟐. 𝟎𝟒)𝑻/𝟕𝟔𝟎
𝑫=√
𝟑𝟔𝟎𝟎𝝅𝑼𝑷(𝟐𝟕𝟑)
DIÁMETRO DE LA COLUMNA
𝑷𝑳 − 𝑷𝑽
𝑼 = 𝑲√
𝑷𝑳
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931.7632 − 5.1303
𝑈 = 0.35√ = 0.34903
931.7632
CALCULO DE DIÁMETRO
𝒀𝑨 𝑷 (𝟎. 𝟕𝟎𝟐𝟗)(𝟕𝟔𝟎)
𝑷°𝑨 = = = 𝟏𝟎𝟐𝟕. 𝟑𝟏𝟓𝟑 𝒎𝒎𝑯𝒈
𝑿𝑨 𝟎. 𝟓
𝑽𝟏 = 𝟐𝟒𝟖. 𝟒𝟏𝟔𝟔 𝑲𝒎𝒐𝒍⁄𝒉
𝟒𝑽 (𝟐𝟐. 𝟎𝟒)𝑻/𝟕𝟔𝟎
𝑫=√
𝟑𝟔𝟎𝟎𝝅𝑼𝑷(𝟐𝟕𝟑)
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𝟒(𝟐𝟒𝟖.𝟒𝟏𝟔𝟔)(𝟐𝟐.𝟒))𝟏𝟖𝟔.𝟖)(𝟕𝟔𝟎)
𝑫=√ = 𝟎. 𝟗𝟗𝟖𝟏
𝟑𝟔𝟎𝟎𝝅(𝟎.𝟓𝟒𝟖𝟒𝟖)(𝟏𝟎𝟐𝟕.𝟑𝟏𝟓𝟑)(𝟐𝟕𝟑)
431.7632 − 5.1303
𝑈 = (0.55)√ = 0.544848
431.7632
**las condiciones para una separación de 60 cm el diámetro debe ser > 0.5 pero ≤ 1𝑀
por lo que SÍ se cumple.
*Consultando el manual:
Una lámina calibre 12 tiene como espesor 0.1046
0.1046
0.4 < < 0.7
3/16
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𝑀1 𝐺1
𝑅𝐸 = ℎ𝑉𝑊 𝑉𝑎 𝜌𝑔 𝑆𝑐𝑙 = 𝐷𝐺 =
𝜌1 𝐷𝐵 𝑀1 𝑉𝐶
𝐵
𝐶𝑇 + 𝐷𝑇 2 )
𝑀1 = 10(𝐴 +
𝑇
Donde: 𝐴 = 13.9237 𝐵 = 2.6296𝑥10 𝐶 = 2.5331𝑥10−2 𝐷 = 1.897𝑥10−5
3
Sustituyendo:
2.6296𝑥103
𝑀1 = 10(−13.4237 + (2.535𝑥10−2 )(339.15) + (−1.8972𝑥10−6 )(339.15)
66+273.15
𝑘𝑔⁄
𝑀1 = 0.0017322 𝑚⋅𝑠
𝑻 𝒏
𝝈 = 𝑨(𝟏 −
)
𝑻𝒄
𝑨 = 𝟓𝟑. 𝟕𝟕 𝒏 = 𝟏. 𝟎𝟗𝟔𝟎 𝑻𝒄 = 𝟔𝟑𝟏
𝟑𝟑𝟗. 𝟏𝟓 𝟏.𝟎𝟗𝟔𝟎
𝝈𝟏 = (𝟓𝟑. 𝟕𝟕)((𝟏 − ) ) = 𝟐𝟑. 𝟏𝟕𝟕𝟏
𝟔𝟑𝟑𝟏. 𝟗
≈ 𝟎. 𝟎𝟐𝟑𝟏𝑵
𝟑𝟑𝟗. 𝟏𝟓 𝟏.𝟐𝟐𝟐𝟐
𝝈𝒎 = (𝟕𝟕. 𝟗𝟏𝟐)((𝟏 − ) ) = 𝟑𝟒. 𝟒𝟗𝟑𝟔
𝟔𝟗𝟕
= (𝟎. 𝟎𝟐𝟑𝟏)(𝟎. 𝟓𝟐) + (𝟎. 𝟎𝟑𝟒𝟒)(𝟏 − 𝟎. 𝟓𝟐)
𝒌𝒈
= 𝟎. 𝟎𝟐𝟓𝟖 ⁄ 𝟐
𝒔
𝟏 𝑻
𝑫𝑨𝑩 = 𝟏. 𝟏𝟕𝟑𝒙𝟏𝟎−𝟏𝟔 (𝝓𝑴𝑩 )𝟐
𝑴𝑩 𝑽𝟎.𝟔
𝑨
𝟏 𝟗𝒎𝟐
−𝟏𝟔 𝟒𝟓𝟗.𝟏𝟓 −
𝑫𝑨𝑩 = 𝟏. 𝟏𝟕𝟑𝒙𝟏𝟎 ((𝟏)(𝟏𝟐𝟒. 𝟏𝟒)) 𝟐
(𝟎.𝟎𝟓𝟏𝟔𝟕𝟐𝟎)(𝟎.𝟏𝟕𝟎𝟐)𝟎.𝟔
= 𝟏. 𝟎𝟑𝟗𝟔𝒙𝟏𝟎 𝒔
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CALOR SUMINISTRADO
PUNTO DE EMPOBRECIMIENTO
𝑊𝐻𝐿𝑊 − 𝑄𝑅
𝑄 ′′ =
𝑊
(29.1773)(63479)−19828980.23𝐾𝐽⁄ℎ 𝐾𝐽⁄
𝑄 ′′ = = −467615.7714 𝑚𝑜𝑙
29.1773 𝐾𝑚𝑜𝑙⁄ℎ
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CALOR CEDIDO
𝑄𝐷 = 𝑉1 𝐻𝑉 − 𝐿ℎ𝐿𝐷 − 𝐷ℎ𝐿𝐷
𝑄𝐷 = (200.72)(94059.247) − (178.64)(548.3) − (22.08)(548.64)
𝐾𝐽
𝑄𝐷 = 701791.72 ⁄ℎ
PUNTO DE ENRIQUECIMIENTO
𝐷ℎ𝑙𝐷 + 𝑄𝐷
𝑄´ =
𝐷
(22.08)(548.3){701791.72 𝐾𝐽
𝑄´ = = 32332.345 ⁄𝐾𝑚𝑜𝑙
(22.08)
PUNTO DE EMPOBRECIMIENTO
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𝜋𝑟 2 𝑛 𝑐(𝑟 − ℎ)
𝐴𝑆𝑒𝑐𝑐𝑖𝑟 = −
360 2
ÁREA REACTIVA
𝐴𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑎 = 𝐴𝑐𝑖𝑟𝑐𝑢𝑙𝑜 − 2𝐴𝑆𝑒𝑐𝑐𝑖𝑟 − 2𝐴𝑠𝑒𝑟𝑑𝑖𝑠
𝐴𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑎 = 𝜋(0.49905)(0.49905) − (2(0.068592) − (2(0.06986)(0.000714248))
𝐴𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑎 = 0.6352𝑚2
𝑉 = 248.4166 𝐾𝑚𝑜𝑙⁄ℎ
248.4166𝐾𝑚𝑜𝑙 120.3648 1 1 1 𝑚
𝑉= ( )( )( )( ) = 2.5487
ℎ 1 3600 5.1303 0.6352) 𝑠
248.4166 120.3648 1 1 1
𝑉𝑐𝑜𝑙 = ( )( )( )( ) = 2.0692 𝑚⁄𝑠
ℎ 1 3600 5.1303 0.7824
ℎ𝑤 𝑉𝐴 𝜌𝑔 (0.095)(2.5487)(5.1303)
𝑅𝐸𝑀 = = = 690.6999
𝜇1 (0.001703784)
𝜇1 0.001703784
𝑆𝐶𝐿 = = = 1758.4063
𝜌1 𝐷𝐴𝐵 𝐿 (931.7632)(1.0346𝑥10−9
𝐺1 0.0285
𝐷𝐺 = = = 8.0840
𝑀1 𝑉𝑐 (0.001703784)(2.0692)
𝑀1 0.001703784
𝑆𝐶𝐿 = = = 1758.9063
𝜌1 𝐷𝐴𝐵𝐿 (931.7632)(1.0396𝑥10−9 )
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PLATOS REALES
𝑁𝑒𝑡𝑎𝑝𝑎𝑠𝑟𝑒𝑎𝑙𝑒𝑠 = 5 𝑒𝑡𝑎𝑝𝑎𝑠 𝑡𝑒ó𝑟𝑖𝑐𝑎𝑠 (1 + 0.8289)
INTERCAMBIADOR DE CALOR
𝟒𝒎𝑯 𝟒
𝑵𝑻 = = 𝒌𝒈 = 𝟐𝟖. 𝟗𝟑 ≈ 𝟐𝟗 𝒕𝒖𝒃𝒐𝒔
𝑷𝒏 𝝅(𝒅𝒊)𝟐 𝑼𝒎 ⁄ 𝟑 (𝝅)(𝒎)𝟐 (𝟏.𝟓)
𝒎
𝝅(𝒅𝒊)𝟐
𝑨𝑻 = [ ] 𝑵𝑻
𝟒
𝝅(𝟎. 𝟎𝟏𝟓𝟕𝟒𝟖)𝟐
𝑨𝑻 = [ ] (𝟐𝟗) = 𝟎. 𝟎𝟎𝟓𝟔𝟒𝟖𝒎𝟐
𝟒
DIÁMETRO DE CORAZA
8𝑖𝑛 = 0.2031𝑚
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𝐷𝑆𝐶𝑅 0.2032
𝑁𝑇𝐶 =
= = 8.5342
𝑃𝑇 0.02381
𝐴𝑆 = [𝐷𝑆𝐶𝑅 − (𝑁𝑇𝐶 (𝑑𝑜))]𝐵 = [0.2032 − (8.5342(0.01905))]0.1524 = 0.006191 𝑚2
𝑑𝑜(𝑚𝑐) (0.01905)(13.8888)
NRS = = = 42651.1925
𝜇𝑐 (𝐴𝑠 ) (1.002𝑥10−3 )(0.006191)
FACTOR DE LA PELÍCULA
2
𝑚𝑛 𝐾𝑛 3
ℎ𝑗 = 𝐽𝑖(𝐶𝑝𝑛 ) [ ] [ ]
𝐴𝑠 𝐶𝑝𝑛 (𝜇𝑛
2
8.3333649.2𝑥10−3 3 𝑊
ℎ𝑗 = (0.005487)(4182) [ ] [(4182)(0.5042𝑥10−13)] = 14083.2523
0.006191 𝑚2 𝑘
DIÁMETRO EQUIVALENTE
𝜋(𝐷𝑜)2 ) 𝜋(0.01905)2
4((𝑃𝑇)2 −(( (0.02381)2 −(
4 4
𝐷𝐸 = = 4[ ] = 0.01884𝑚
𝜋(𝐷𝑜) 𝜋(0.01905)
REYNOLDS EQUIVALENTE
𝐷𝑒(𝑚𝑐 ) 0.01884(13.8888)
𝑁𝑅𝐸 = = = 42181.0219
𝜇𝑐 (𝐴𝑠) (1.002𝑥10−3 )(0.001691)
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0.37(𝑘𝑐)
ℎ𝑜 = ( ) (𝑁𝑅𝐸 )0.55 (𝑃𝑟𝑐 )1/3
𝐷𝑒
0.36(598.4𝑥10−3 𝑤
ℎ𝑜 = ( ) (42181.0219)0.55 (7.006)1/3 = 7653.3393
0.01884 𝑚2 𝑘
𝑟
1 1 𝑑0 𝑟𝑜 ln( 𝑜 )
𝑟𝑖
=( )+( )+( )
𝑈𝑐 ℎ𝑜 𝑛𝑖 𝑑𝑖 𝑘𝑚
0.009525
1 1 0.01905 0.009525 ln( )
0.007874
=( )+( )+( ) = 0.0002465
𝑈𝑐 7653.3393 (14083.2523)(0.015748 (60.5)
𝑤
𝑉𝑐 = 4056.3741
𝑚2 𝐾
𝐹 𝑚𝐾
(𝑁)000175 ℎ𝑓𝑡 2 = 3.082𝑥10−2
𝐵𝑇𝑈 𝑤
1 𝐵𝑇𝑈 𝑤
= 5.678
(𝐻𝑓𝑡 2 𝐹 ) 𝑚2 𝐾
1 1
= + 𝑅𝑤
𝑈𝑓 𝑈𝑐
1 1
= + 3.082𝑥10−5 = 0.0002773
𝑈𝑓 4063.3941
𝑤
𝑈𝑓 = 3605.625 2
𝑚 𝐾
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(8.3333)(4182)(55 − 40)
𝑇𝐶2 = + 20 = 28.99°𝑐
(13.8888)(4183)
(40)−(55)
𝑃 = (20)−(55) = 0.4285 𝐿𝑀𝑇𝐷𝑐𝑜𝑟𝑟 = 𝐿𝑀𝑇𝐷(𝐹)
(20)−(28.99)
𝑅= (40)−(50)
= 0.5993 𝐿𝑀𝑇𝐷𝑐𝑜𝑟𝑟 = 22.87(0.96) = 21.95°
CALOR CEDIDO
𝑄 = 𝑚𝑛 𝐶𝑝𝑛 (𝑇𝑛1 − 𝑇𝑛2 )
𝑄 = (8.3333)(4182)(55 − 40) = 522747.909 𝐽/𝑠
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𝑄 𝑄 𝐴𝑓
𝐴𝑓 = 𝐴𝑐 = %𝐸𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑖𝑎 = (2 − ( )) 100
𝑈𝑓 𝐿𝑀𝑇𝐷𝐶𝑂𝑅𝑅 𝑈𝑐 𝐿𝑀𝑇𝐷𝐶𝑂𝑅𝑅 𝐴𝑐
522747.904
𝐴𝑓 = = 6.6035𝑚2
(3605.6257)(21.9552)
522747.904
𝐴𝑓 = = 5.8696𝑚2
(4056.3941)(21.9552)
6.6035
%𝐸𝑓𝑖𝑒𝑐𝑖𝑒𝑐𝑛𝑐𝑖𝑎 = (2 − ) (100) = 98.61%
5.8696
LONGITUD DE LA TUBERÍA
𝐴𝑓 6.6035
𝐿= = = 3.8047𝑚 = 2.95𝑚
𝑁𝑇 𝜋𝑑𝑜 (29)(𝜋)(0.01905)
1/2
𝐶𝐿 𝐴𝑓 (𝑃𝑅)2 𝑑𝑜
𝐷𝑆 = 0.637 ( √ )[ ]
𝐶𝑇𝑃 𝐿
1/2
0.87 6.6035(125)2 (0.01905)
𝐷𝑆 = 0.637 (√ )[ ] = 0.1616𝑚 ≈ 6.36 𝑖𝑛
0.9 2.95
𝐷𝑆𝐶𝑅 = 8 𝑖𝑛 ≈ 0.2032 𝑚
FACTOR DE FANING
𝑚𝑐 (𝐺𝑠)2
𝐺𝑠 = Δ𝑃𝑤 = 4𝐹𝑓 ( ) 𝑁𝑐
𝐴𝑠 2𝜌𝑐
13.8888 𝑘𝑔
𝐺𝑠 = = 2243.3855 ⁄𝑚2 𝑠
0.006191
2243.3855 𝑘𝑔
Δ𝑃𝑤 = 4(0.1005) ( ) 9 = 9120.7140
2(498.2 𝑚𝑠 2
NÚMERO DE DEFLECTORES
𝐿
𝑁𝑏 = 𝐿𝐵 = (0.2)𝐷𝑆𝐶𝑅 𝐿𝐵 = (1)(0.2032) = 0.2032𝑚
𝐿𝐵
2.95
𝑁𝑏 = = 14.51 ≈ 15 𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑜𝑟𝑒𝑠
0.2032
ESPESOR DE CORAZA
𝐷𝑖á𝑚𝑒𝑡𝑟𝑜 = 8′′
𝑡𝑏𝑎𝑓𝑓𝑙𝑒 = 1/2"
CAÍDA DE PRESIÓN
𝜋(𝐷𝑆𝐶𝑅 )2 𝜋(0.2032)2
𝐴𝑤 = [ ] %𝑐𝑜𝑟𝑡𝑒 = [ ] (0.25) = 0.008107𝑚2
4 4
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𝑁𝑊
Δ𝑃𝑠 = [((𝑁𝑏 − 1)Δ𝑃𝑊 𝑅𝑏 ) + 𝑁𝑏 Δ𝑃𝑊 ]𝑅𝑖 + 2Δ𝑃𝑊 (1 + ) 𝑅𝑏
𝑁𝑐
2
Δ𝑃𝑠 = [((15 − 1)(9120.714)(0.7)) + (6160.428)](0.4) + 2(9120.7140) (1 + ) (0.7)
9
Δ𝑃𝑠 = 88322.3219 𝑃𝑎 ≈ 0.8716 𝑎𝑡𝑚
REBOILERS
((𝑀)(𝐶𝑝)(Δ𝑇)
𝑊𝑠 =
(ℎ𝑓𝑔)(ℎ)
𝑘𝑔⁄
𝑀 = (373.73𝐾𝑚𝑜𝑙 ) (22.5 𝐾𝑚𝑜𝑙 ) = 840915 𝑘𝑔
𝐶𝑝 = [(0.15)(115.13) + (0.85)(75.15)](22.5)
𝑇 = (351.4 − 298) = 53.4𝐾
1 𝑘𝑔
ℎ𝑓𝑔 = [(0.15)(115.13) + (0.85)(75.15)] ( ) = 3.6
22.5 𝑘𝑔 𝑘 ℎ
(8409.15)(3.6)(53.4)
𝑊𝑠 = = 36.86 𝑘𝑔/ℎ
(1827.34)(24)
𝑣𝑎𝑝
𝑊𝑠 = 884.64 𝑘𝑔 + 𝑑í𝑎
𝑑
CONDENSADOR
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𝑄𝐷 11796843.78 𝑘𝑔
𝑀𝑐𝑤 = = = 80596.0496
𝐶𝑝(𝑇 − 𝑇) (4.182)(55 − 20) 𝑑í𝑎
DISEÑO DE TANQUE
21𝐾𝑔 1422.33𝑝𝑠𝑖
𝑆=( )( ) = 29,868.93 𝑝𝑠𝑖 ≈ 30 𝐾𝑝𝑠𝑖
𝑚𝑚2 𝐾𝑔⁄
1 𝑚𝑚2
PRESIÓN DE DISEÑO
FACTOR DE CORROSIÓN
E = 0.85
𝑃 346.5 𝑝𝑠𝑖
𝐹= = = 0.041
𝐶𝐸𝑆 (0.3325)(0.85)(29,868.93 𝑝𝑠𝑖 )
40,000 Lt = 1,412.58 𝑓𝑡 3
4𝑉 4(1,412.58𝑓𝑡 3 )
𝐿= = = 28.10𝑓𝑡 = 337.22 𝑖𝑛
𝜋𝐷2 𝜋(8 𝑓𝑡)2
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𝑃𝑅
𝑡= +𝐶
𝑆𝐸 − 0.6𝑃
(346.5𝑝𝑠𝑖 )(48𝑖𝑛)
𝑡= + 0.3325
(29868.93𝑝𝑠𝑖 )(0.85) − 0.6(346.5𝑝𝑠𝑖 )
15
= 0.9930 𝑖𝑛 = 𝑖𝑛
16
(346.5𝑝𝑠𝑖 )(48𝑖𝑛)
𝑡= + 0.3325
(29868.93𝑝𝑠𝑖 )(0.85) − 0.6(346.5𝑝𝑠𝑖 )
11
= 0.6604 𝑖𝑛 = 𝑖𝑛
16
ACCESORIOS
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Ventilació
nnn.
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