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    I-Dus: Empenxe

    Uploaded by

    Daniel Nuñez
    This document provides information to solve a mass balance problem for a distillation column. It gives flow rates and compositions for feeds to the column including benzene, toluene, and xylenes. It then outlines a three-equation, three-unknown system to determine the minimum reflux rate and distillate and bottoms flows. Key values given include feed rates of 394.6824 kmol/h of benzene and 0.0434 kmol/h of xylenes.

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd

    I-Dus: Empenxe

    Uploaded by

    Daniel Nuñez
    8 views10 pages
    This document provides information to solve a mass balance problem for a distillation column. It gives flow rates and compositions for feeds to the column including benzene, toluene, and xylenes. It then outlines a three-equation, three-unknown system to determine the minimum reflux rate and distillate and bottoms flows. Key values given include feed rates of 394.6824 kmol/h of benzene and 0.0434 kmol/h of xylenes.

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    Original Title

    7d15

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    This document provides information to solve a mass balance problem for a distillation column. It gives flow rates and compositions for feeds to the column including benzene, toluene, and xylenes. It then outlines a three-equation, three-unknown system to determine the minimum reflux rate and distillate and bottoms flows. Key values given include feed rates of 394.6824 kmol/h of benzene and 0.0434 kmol/h of xylenes.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    8 views10 pages

    I-Dus: Empenxe

    Uploaded by

    Daniel Nuñez
    This document provides information to solve a mass balance problem for a distillation column. It gives flow rates and compositions for feeds to the column including benzene, toluene, and xylenes. It then outlines a three-equation, three-unknown system to determine the minimum reflux rate and distillate and bottoms flows. Key values given include feed rates of 394.6824 kmol/h of benzene and 0.0434 kmol/h of xylenes.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 10

    I-Dus

    Wan -
    -

    come 97
    empenxe
    aredado
    W D
    S

    99,92
    %. Reing:
    1000 Asmol/h L
    B enzen

    Liavido sat,
    Lo/p = 1.2

    Benieno
    Zen 0.397
    =

    B 101.3
    =

    Pan

    Toluene 270 0.167


    =

    coment [c0 0.432


    =
    =< vi

    B %, Rezug:99,99
    Cureng

    101=2.25
    &
    Bence -

    & Tolving -

    TC I
    =

    0.21
    =

    & aumeno -40)


    Destilado

    DXben=Froen ·
    F: zbeu
    Dess

    =
    0.9992-1000.0.397

    394,6824 kmolh
    =

    Dxcrme: 21-FraumesFords) ·

    F.2 cume

    = 21-0.9999): 1000.
    0,439

    =0.0434 Emol/n

    BXxo = ?
    Wonders
    B
    XBen,B (1
    =
    -

    0.9992). 1000.0.397

    =0.3173 kmolh

    B
    Xunen, 0.9999
    =
    x1000 x
    0,434
    435.959
    =

    kmoyh

    B X
    01B
    +

    =
    9 1
    =

    Salim AC1-9) =

    ligudesaturado y 1
    =

    Valin 100 (1-1) =


    0
    =

    0
    a
    =

    -0.0.397+0.167, 0.4
    6
    =

    1 4 0.21 4
    -

    ~clave ligero
    101=2.25
    &

    ①, 1.2161 Bence

    I
    = -

    & Tolving -

    TC I
    =

    82 0.3373
    =

    & aumeno -40)


    =

    0.21
    I ↓
    Nota:ambos o esten entra
    clave pesado
    c de los caves
    101=2.25
    &
    Bence -

    & Tolving -

    TC I
    =

    0.21
    =

    & aumeno -40)

    Usere los 2 gara armar

    es sistema de eca. con

    incongnitt DX101: Umin

    Umin=xDXBensD+ADXyosDID C+vi -40 -

    4
    Umin=sgeson isthe +04)(
    0.21 -1.2141

    osi
    Umin=4.anD +
    +

    1 0.33730.27 0.3373
    -
    -

    ⑳
    + D
    Umin=Lmin

    Umin=Lmin+2DXen
    +DXaumans (DoD

    Umin=(min
    + (394,4824 0.0434 Bx+0B)
    +
    +

    ③
    Resolviendo el Sistema 3 x3

    Umt=564,11 kmolh
    DX+ 01, D 44,63
    =

    Lmin 102.74=

    Amolh
    D =
    DXDen, DXcumansD+DXoD
    +

    D (394,4824 0.0434
    =
    +
    +

    64.65)
    D
    =

    461.374 kmol/n

    Determinand a FR+ol

    DXtasD=FRN,D ·
    F-2501
    64,45= FRT,D: 1000. 0.867

    FRIGD =

    0.387

    C1-0,387) 0.613
    FR+01, B
    =

    los Fondo, qura


    Retorno mus

    se el
    verificer que cungle
    Balance Global F D+B
    =
    Wonders
    B
    XBen,B (1
    =
    -

    0.9992). 1000.0.397

    =0.3173 kmolh

    B
    Xunen, 0.9999
    =
    x1000 x
    0,434
    435.959
    =

    kmoyh

    B x
    01, FRiolsB x0.197
    +

    B = X 1080

    =0.613x1000 0,147
    x

    =102, 371 kmol/n

    B
    BXBeng +
    =

    BxcomejB+ BXx01, B
    => 0:3174 +

    435.954 102.371
    +

    538.645
    =
    F D B
    =
    +

    F 461.374
    =
    338.645
    +

    F 1000.02
    =

    kmd/n8F
    deliz Ser
    1000

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