01 parte 01 diseño de reactores ad 2014

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Page 1: 01 Parte 01 Diseño de Reactores AD 2014

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b% 6+ 0U'(+5%/$ ') '$ ').+*0 ').+5%0$+(%0f

Z$.0$5') '6 M+6+$5' '$ '6 )-M &06-1'$ ')f

DD

dN  j 

dt = 0

F  j ( y)!F 

 j ( y+" y)+ r

 j "V   = 0

!V   = A! y

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701M%$+$*0 6+) *0) j6_1+) '5-+5%0$')f

b%'$.0$5')

DC

F  j ( y)!F  j ( y+" y)" y

#$% &

'(= ! Ar j 

! y" 0

dF  j 

dy= Ar j  dF  j 

dV = r j 

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!K 

>+U%*'Y *' ('+55%/$

DK 

Z5-+5%/$ ;'$'(+6 *'6 M+6+$5' *' 106')!K  >'+5.0( %$.'(1%.'$.'

DK  >'+5.0( *' 1'Y56+ 501U6'.+ !"#$

CK 

>'+5.0( *' [-J0 U%)./$ &'$FK  >'+5.0( '1U+5+*0 &($

DF

G! \+6+$5') *' 106')

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30) W4> V 7bX> )0$ -_6%Y+*0) U+(+

('U(')'$.+( )%).'1+) 50$ ('+55%/$ '$ -$+

S+)'f )%).'1+) T010;@$'0)K b%$ '1M+(;0^'H%).'$ ('+55%0$') Q-' )-5'*'$ '$

5+.+6%Y+*0(')K

Z$ ').0) 5+)0)^ )%).'1+) )/6%*0 k [-%*0^ 6+

(+U%*'Y *' ('+55%/$ ').e ('S'(%*+ + 6+ 1+)+

*'6 5+.+6%Y+*0(^ l^ V $0 +6 &06-1'$ *'6

('+5.0(K

D]

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Z6 *')+((0660 *'6 M+6+$5' U+(+ -$ W\> ')

)%1%6+( +6 Q-' )' *')+((066/ U+(+ -$ W4>f

8-'*+$*0f

DI

dF  A

dW = r A

'

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b%).'1+) + +$+6%Y+(f

Dd

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XT%) ;0+6 m%66 M' +5T%'&'* MV -)%$; + $-1M'( 0S

U(0M6'1) .T+. +(' 0U'$c'$*'* .0 &+(%0-)*';('')K

DB

K

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DE

K

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CG

K

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!K 

<'O$%5%/$ *' 50$&'()%/$

DK 

Z5-+5%0$') *' *%)'=0!K  >'+5.0( %$.'(1%.'$.'

DK  >'+5.0( *' 1'Y56+ 501U6'.+ !"#$

CK 

>'+5.0( *' [-J0 U%)./$ &'$FK  >'+5.0(') '$ )'(%' 

C!

GD 70$&'()%/$ V X+1+=0 *'6 >'+5.0(

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!K 

<'O$%5%/$ *' 50$&'()%/$

DK 

Z5-+5%0$') *' *%)'=0!K  >'+5.0( %$.'(1%.'$.'

DK  >'+5.0( *' 1'Y56+ 501U6'.+ !"#$

CK 

>'+5.0( *' [-J0 U%)./$ &'$FK  >'+5.0(') '$ )'(%' 

CD

GD 70$&'()%/$ V X+1+=0 *'6 >'+5.0(

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3+ '5-+5%/$ *' 50$&'()%/$ V 0.(+)

'5-+5%0$') ').+(e$ ('S'(%*+) +

<'O$%5%/$ *' 50$&'()%/$ 

CC

 A+b

a B!

c

aC +

a D

 X  A  =

Moles de A que reaccionan

Moles de A alimentadas

 X   !  X  A

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!K 

>+U%*'Y *' ('+55%/$

DK 

Z5-+5%/$ ;'$'(+6 *'6 M+6+$5' *' 106')!K  >'+5.0( %$.'(1%.'$.'

DK  >'+5.0( *' 1'Y56+ 501U6'.+ !"#$

CK 

>'+5.0( *' [-J0 U%)./$ &'$FK  >'+5.0(') '$ )'(%' 

CF

GD 70$&'()%/$ V X+1+=0 *'6 >'+5.0(

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W0( 60 .+$.0

b' 50$&%'(.' '$f

C]

 N  A  =  N 

 A0 !  N 

 A0 X   =  N 

 A0  1!  X ( )

dN  A

dt = r

 AV 

 N  A0

dX 

dt =   !r A( )V 

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!K 

>+U%*'Y *' ('+55%/$

DK 

Z5-+5%/$ ;'$'(+6 *'6 M+6+$5' *' 106')!K  >'+5.0( %$.'(1%.'$.'

DK  >'+5.0( *' 1'Y56+ 501U6'.+ !"#$

CK 

>'+5.0( *' [-J0 U%)./$ &'$FK  >'+5.0(') '$ )'(%' 

CI

GD 70$&'()%/$ V X+1+=0 *'6 >'+5.0(

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Z6 M+6+$5' '$ -$ 7bX>f

:$&06-5(+$*0 6+ *'O$%5%/$ *' 50$&'()%/$

b' 0M_'$' 6+ '5-+5%/$ *' *%)'=0

Cd

V   =F  A0

 ! F  A

!r A

F  A   = F  A0 !F  A0 X 

V   =F  A0

 !   F  A0

 ! F  A0 X ( )

!r A

V   =F  A0 X 

!r A( )

salida

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!K 

>+U%*'Y *' ('+55%/$

DK 

Z5-+5%/$ ;'$'(+6 *'6 M+6+$5' *' 106')!K  >'+5.0( %$.'(1%.'$.'

DK  >'+5.0( *' 1'Y56+ 501U6'.+ !"#$

CK 

>'+5.0( *' [-J0 U%)./$ &'$FK  >'+5.0(') '$ )'(%' 

CB

GD 70$&'()%/$ V X+1+=0 *'6 >'+5.0(

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Z6 M+6+$5' '$ -$ W4>f

:$&06-5(+$*0 6+ *'O$%5%/$ *' 50$&'()%/$

b' 0M_'$' 6+ '5-+5%/$ *' *%)'=0

CE

F  A   = F  A0 !F  A0 X 

!dF  A

dV =

!r A

F  A0

dX 

dV 

= !r A

V   = F  A0

dX 

!r A0

 X 

 " 

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!K  >+U%*'Y *' ('+55%/$

DK 

Z5-+5%/$ ;'$'(+6 *'6 M+6+$5' *' 106')!K  >'+5.0( %$.'(1%.'$.'

DK  >'+5.0( *' 1'Y56+ 501U6'.+ !"#$

CK 

>'+5.0( *' [-J0 U%)./$ &'$FK  >'+5.0(') '$ )'(%'

FG

GD 70$&'()%/$ V X+1+=0 *'6 >'+5.0(

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Z$ 05+)%0$') ') $'5')+(%0 +$+6%Y+( )%).'1+)

'$ 60) Q-' 6+ 50((%'$.' *' )+6%*+ *' -$

('+5.0( ') 6+ +6%1'$.+5%/$ *'6 )%;-%'$.'K

3+ 50$&'()%/$^ *')*' 6+ +6%1'$.+5%/$

U(%$5%U+6 T+).+ 6+ )+6%*+ *' 5+*+ ('+5.0(f

F!

F  A1

  = F  A0

 ! F  A0 X 

1

F  A2

  = F  A0

 ! F  A0

 X 2

F  A3

  = F  A0

 ! F  A0 X 

3

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FD

F  A1

  = F  A0

 !F  A0 X 

1

F  A2   = F  A0 !F  A0 X 2

F  A3

  = F  A0

 !F  A0 X 

3

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W+(+ '6 U(%1'( ('+5.0(

W+(+ '6 )';-$*0

FC

V   = F  A0dX 

!r A0

 X 1

 " 

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n U+(+ '6 .'(5'(0f

FF

V   =F  A0

  X 2 !  X 

1( )!r

 A2

V   = F  A0

dX 

!r A X 2

 X 3

 " 

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F]

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FI

T 6 h * 6

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Fd

XT' ;0+6) h X0 *'&'60U 5('+_&'

.T%$o%$; )o%66)

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FB

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FE

Th¡

rcaction

is

essentially irreversible.

Estimate

the

time required to saponify 95l.

ofthe ester at

¿lO"C.

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¡ll

h

has

been

mentioned

that

gaseous

reactions

are

more

suitably

carried out

on

a

commercial

scale

h

llow

equipment

than

in batch reactors. Consider the

following example.

Watsont

has

studied the

lhrmrl

(noncatalytic)

cracking

of

butenes

at

I

atm

pressure

in

a flow

reaclo¡. The rate quation

dücnnined

from

his experimental

data is

losk.:-60'ffi

-15.27

4.5757

'

t K,

M.

Watson,

Chem.

Eng. Progr.,

i11,229

(1948).

206 cnrurc¡l

ENGTNEERTNG

KrNETrcs

where

*

butenes

cracked,

g

mol/(h)(liter)(atm),

and

Z is in

degrees Kelvin.

Alrhough

the

feed

consists

of

a

number

ofdillerent

butenes

and the products vary from coke to butadiene,

the

irreversible

reaction may be

considered

first order,

CaH"

-

CnH"

.r

1¡,

It

is desired

to

crack butenes

in a batch+ype

re¿ctor which

will operate

at 1200"F

and

will

be

equipped

for efficient

agitation,

The initial

charge to

the

reactor will

consist of

1

lb

mole ofbutenes

and

l0 lb

moles

of steam.

Under these

conditions the

change

in

total

moles during

the

course of the

reaction can

be neglected.

(a)

Determine

the time required for a

conversion

of

30/o

of the

butenes,

(b)

Determine

the reactor

volume

required. (c)

Suppose

that the initial

charge

consists

of l0 moles

of

steam

per

mole of hydrocarbon,

as

before,

but

this time

the hydrocarbon

fraction

contains

ffi mol

\

butenes and

tlo'/.

butadiene.

The butadiene

may

undergo

two reactions:

cracking

and

polymerization

to the dimer.

Assuming

that the rates

of these reactions

are known, outline

a method

of determining

the conversion

of butenes

and

of

butadiene

for

a

given

reaction

time.