TB Pu HDau

8/14/2019 TB Pu HDau

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THIT B PHN NG

trong cng nghip ho du.

Ngi son : PGS,TS Trn Cng Khanh.

Trng i hc Bch khoa H Ni.


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I-GII THIU V TBP:

I.1-Gii thiu: -TBP- h thng thit b thc hin cc phn ng ho hc to ra snphm ca mt qu trnh sn xut,do quyt nh nng sut (do vntc phn ng r ) v hiu qu ( chuyn ho X v chn lc S)ca sn xut.

-Vn tc phn ng chuyn ho cht i:

Ri = dNi / Vdt . ( 1 . 1 )Trong : Ni-S mol ca cht i, du cng l to thnh (sn phm phnng), du tr l tiu hao (cht phn ng).

V-Th tch ca h thng

Khi th tch khng i ta c Ci = Ni / V, do pt (1.1) thnh:

Ri = dCi / dt . ( 1 . 2 )


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- chuyn ho ca cht i:

Xi

= (C0

- C1

) / C0

= 1 - C1

/C0

. ( 1 . 3 )

Trong : C0- nng cht phn ng i i vo ( hay nng banu)

C1- nng cht phn ng i i ra ( hay nng cui )

- chn lc i vi sn phm i:

Si = Ci / Cj j = 1, n ( 1 . 4 ) Trong : Ci -nng ca sn phm i trong hn hp phn ng .

Cj -tng nng cc sn phm trong hn hp phnng


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V TR H THNG TBP TRONG S CNG NGH

Trong s cng ngh TBP nm v tr nh sau :

H.1.1- Lu khi ca cng ngh sn xut .

Trong h thng thit b chun b hn hp phn ng,tch v tinh ch sn phm c th gm mt s lng lncc thit b thc hin cc qu trnh chuyn khi v truynnhit nh chng luyn, hp th,hp ph, trch ly, un nng,lm lnh, ngng t ...m sinh vin lm quen trong mnhc "Qu trnh v thit b ho hc ".

H thng tchv tinh chsn phm

H thngchun bhn hpphn ng

THIT BPHNNG

Nguyn liu cha chuyn ho

Ng.liu 2

Ng.liu 1

Sn phm


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I.2-c im :

- a dng Do iu kin phn ng rt khc nhau:

*Nhit phn ng c th t nhit phng n 800-9000C, c bit c th n 1300-15000C. ng thi phi c

nhng gii php hp l cp hay gii nhit phn ng. *p sutc th t p sut kh quyn ( 0,1 MPa ) n 70MPa.

Trong nhiu phn ng pha kh thng dng p sut khong 2-3 MPa gim th tch TBP, tng cng vn tc phn ng

v h s trao i nhit vi thnh thit b.Vi mi p sut cn c dng hnh hc ca thit b ph hp :hnh ng, hnh cu chu p sut tt hn hnh hp, mt phng.


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Phn ng trong thit b c th tin hnh cc

trng thi pha khc nhau:*ng th: kh, lng

*Cc h d th kh-rn, kh-lng, lng-rn,

lng-lng*Cc h ba pha kh-lng-rn, lng-lng-rn,

kh-lng-lng ...

Tnh a dng ca TBP cn do tng hng,trn th gii c nhng hng c cng ngh, xctc v h thng TBP ring ca mnh


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C IM TBP

-Phc tp Do trong TBP cc qu trnh hohc (phn

ng ) v vt l( chuyn khi: dng chy ,

khuch tn, v cc qu trnh nhit: truyn nhit,to v thu nhit ) xy ra an xen v nhhngln nhau

Trong , cc qu trnh vt l thng tuyn

tnh vi nhit , cn cc phn ng ho hcph thuc vo nhit dng hm m theophng trnh Arrhnius ( phi tuyn ).


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I.3- Phn loi TBP :

1/Theo ch lm vic :a/Thit b lm vic gin on *Ch dng cho pha lng .

*Cc bc ca qu trnh: np liu, unnng, tin hnh phn ng, lm ngui v tho

sn phm, c thc hin trong mt thit b.

Do cc thng s nh nng , nhit , psut ...thay i theo thi gian.

V d: tin hnh phn ng trong thit b loithng c khuy nng cht phn ng thayi theo thi gian nh hnh 1.2.


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H.1.2-M hnh TBP lm vic gin on v thay inng theo thi gian

Cht phn ng

vo gin on

Sn phmtho gin on

0

Nng cht phn ng

Thi gian t

CA0

CAt


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b/Thit b lm vic na gin on:

*Cht phn ng: mt cht cho gin on, mt chtcho lin tc.

Cht cho gin on thng l cht lng, v d cht A.

Cht cho lin tc thng l cht kh hay c th lcht lng, v d cht B. Phn ng: A + B C

Vi mc ch lun ngho cht B trong hn hp phnng trnh phn ng ph: B + C D

Hay vn tc to nhit ca phn ng ph hp vikh nng gii nhit ca thit b .


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*Nng A v B thay i theo thi gian phn ng nh hnh1.3

S mol NB vo

CAt

CBt

Cht A voGin on

Cht B voLin tc CA0

CB0Thi gian t

H.1.3-M hnh TBP lm vic na gin on v thay inng cht phn ng trong thit b.

Nng Cht phn ng


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c/Thit b lm vic lin tc

*y l loi thit b thng gp trong cngnghip vi qui m sn xut ln.

*Trng thi dng(steady state ): l trngthi t c ca TBP sau khi m my mt

thi gian, trng thi ny cc thng s ca qutrnh khng thay i theo thi gian t ,lc snphm thu c c cht lng n nh. T khim my n trng thi dng ta c giai on qu, thi gian qu ph thuc vo ch dng

chy trong thit b v phc tp ca h thngTBP.


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Thi gian lu trung bnh:

Thi gian lu thc ca cht phn ng trong thit b khc nhau, ph thuc vo ch dng chy. Ta c thi gian lu trungbnh theo nh ngha sau:

tTB = VR / FV

Trong : tTB -Thi gian lu trung bnh. [ h ].VR -Th tch TBP. [ m3 ].

FV -Lu lng ca dng . [ m3 / h ].


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2/Theo ch dng chy :

*M hnh y ltng :

-L m hnh dngchy trong thit bchuyn ng tnh tin

theo th t trc saunh chuyn ng capit-tng trong xi lanh

V do nng chtphn ng thay i t

t, bt u u vol CA0 n u ra l CALnh hnh 1.4

L

CA0 CAL

L

Nng cht phn ng

CA0

CAL

Chiu ding phn ng

L

CA0

CAL

0

H.1.4-M hnh LT v thay i nng cht phn ng trong thit b.


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*M hnh khuyl tng : -L m hnhdng chy trong

thit b ckhuy trnmnh, cht

phn ng i voc trn ln

ng u ngaytc khc trongthit b, do nng cht

phn ng thayi t ngt

ti u vo cathit b nh

hnh 1.5

CA1CA1

CA0

Nng cht phn ng

To phn ng

Cht A vo

CA0 Cht A raC

A1

H.1.5-M hnh KLT v thay i nng trong thit b


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M hnh khuy l tng

*Cng do khuy trn nng cht phn ngtrong khp thit b ng u v bng u ra lC1.

*Do nng cht phn ng trong thit b thp

(nht l khi chuyn ho X yu cu cao ) nnvn tc phn ng thp v do nng sutTBP theo m hnh khuy ltng thp hny l tng.

Ni mt cch khc, m bo chuyn hoX nh nhau thit b theo m hnh khuy l tngcn c th tch VR ln hn nhiu so vi m hnhy l tng, c bit khi X yu cu cao.


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* m bo nng sut thit b cao vi mhnh khuy l tng h thng nhiu thitb khuy ni tip c s dng nh hnh1.6.

To phn ng

CAn

CA(n-1)

CA3

CA2

CA1CA0

CA0 CA1 CA2 CA3 CA(n-1) CAn

1 2 3 n

0

H.1.6-M hnh h thng n thit b KLT ni tip v thay inng cht phn ng theo tng thit b


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* hnh 1.6 nng cht phn ng thay i tngbc t u vo n u ra ca h thng, khi n

ln (gii hn khi n ) s thay i nng cht phn ng ging nh trng hp yltng (hnh 1.4 ).

Trong thc tin cng nghip s thit b n trong h

thng thng t 4 n 10 ngoi vic m bonng sut ca h thng thit b cn phn bthi gian lu ngu hn, nhit cthiu chnh khc nhau tng thit b v cht phn

ng th hai c th cho vo t t theo tng thit btheo yu cu


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3/Theo trng th pha:

*H ng th: Cn c khuy trn ng u v nng cc cu t v nhit trong thit b phn ng.

*H d th: i vi h ny cn ch to b mt tip xc phaln tng cng vn tc phn ng.

-H d th lng - lng: Cn c khuytrn tt , to nh tng c b mt tip xc ln.

-H d th kh -lng: m bo b mt tip xc pha ca h ny cn khuytrn hoc si bt hoc dng m rn c b mt ring ln.

-H d th kh - rn v lng - rn: B mt tip xc pha l b mtca cht rn , do vy cht rn ( l xc tc ) thng l vt liuxp c b mt ring ln hoc c phn tn cao.

Trong cng nghip cng hay gp h nhiu pha: kh - lng - rn,lng - lng - rn


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4/Theo ch nhit: *on nhit:

-Khng c b phn trao i nhit .-Hay s dng v n gin, cho cc phn ng chiu ng nhit thp hay t nhy vi s thay inhit .

*ng nhit: -Thng gp cc thit b c khuy trn tt.

- Trong ti liu i khi gi thit b phn ng xc

tc kh - rn dng ng chm c b mt trao inhit ln l thit b ta ng nhit, tuy vy thit b loi ny vn tn ti gradien nhit theong knh v hng trc ng xc tc .


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*T nhit:

-Hay dng khi c th v n gin v kinh t .

-Phn ng to nhit ln v c kh nng traoi nhit phn ng vi nguyn liu vo tnhit m phn ng c th tin hnh .

*Ch nhit theo qui hoch:

-Thng dng t ch nhit ti u choqu trnh phn ng .

-Gp thit b loi ng , thit b c nhiu ngnon nhit v h thng nhiu thit b khuy nitip .


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II/ THI GIAN LU -Thi gian lu ca cht phn ng trong

thit b l thi gian tin hnh phn ng, vvy n nh hng quan trng n chuynho X v chn lc S.

V d: ta c phn ng ni tip ABC, trong B l sn phm chnh v C l sn phm ph,nng ca A, B v C ph thuc vo thi gian

phn ng nh hnh 2.1


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Thi gian phn ng

CC1

CB1

CA1

CA0

0 t1

H.2.1-S thay i nng cc cht ca phn ng ni tipA B C theo thi gian phn ng

Nng

A

B

C

CB2

CA2

CC2

t2


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Gi s ti thi im t1 nng cc chtA, B, C l CA1, CB1 v CC1 ta c:

chuyn ho: X1=1- CA1/CA0.

chn lc: S1=CB1/ ( CB1 + CC1 ).

-Ph thuc vo ch dng chy, thigian lu ca cc phn t ca dng voc th rt khc nhau. Do vy cn nghincu phn b thi gian lu ca chngtrong thit b.


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II.1-THI GIAN LU TRONG CC M HNH L TNG

1/Hm phn b TGL:

-Hm E(t): nh ngha: Phn ca dng vo c thi

gian lu trong thit b t t n t+dt l

E(t).dt. T nh ngha trn ta c:

= 1 ( 2.1 )

= tTB ( 2.2)

0

.).( dtttE

0

E(t).dt


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-Hm F(t):

nh ngha: F(t) l phn ca dng c thi gian lu trong

thit b nh hn t. *T nh ngha ny F(t) l phn ca dng i vo thit b thi

im t = o v n thi im t ra khi thit b.

*Phn ca dng i vo thit b thi im t = 0 n thi im

t vn cn trong thit b l [1-F(t)].T nh ngha v F(t) ta c:

Khi t = 0 th F(t) = 0

Khi t = th F(t) = 1 ( 2.3 )

F(t) = hay E(t) =dF(t)/dt (2.4 )''

).( dttE

t

0


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II.2-Gi tr ca hm phn b TGL trong ccm hnh l tng:

-M hnh LT - T nhngha ca m hnh ny vnh ngha ca hm phn bF(t) ta c gi tr ca hm

phn b TGL nh sau: Khi 0 < t < tTB th F(t) = 0

Khi t > tTB th F(t) = 1(

2.5 ) Ta c ng biu din F(t)

theo t nh hnh 2.2

ttTB

F(t)

1

0

H.2.2-S ph thuc ca F(t) m hnh LTvo thi gian t


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-M hnh KLT -T nh ngha ca m hnhny v ca F(t) ta thy rng do khuy trn

mnh trong thit b cc phn t ca dngva vo c trn u khp trong th tchca thit b.

Do vy kh nng i ra ca cc phn thin c trong thit b l nh nhau, khngphn bit phn t vo trc hay vo sau.Ni cch khc, xc sut xut hin ca ra

ca nhng phn t hin c trong thit b lnh nhaukhng phn bit lch s cachng.


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Nh vy: C th c phn t va mi vo c mt ca ra v ra khi thit b nn TGL bng 0

V c th c phn t vo thit b t lu

mi ra khi thit b nn TGL ca nhng phnt ny bng .

Ngha l TGL ca cht phn ng trong mhnh KLT khng ng u v phn b t 0n .


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Theo l thuyt xc sut ta c mnh sau: xc sut ca nhng phn t c TGLtrong thit b l t+dt gm xc sut canhng phn t c TGL trong thit b l tv xc sut ca nhng phn t c TGLl dt, nh vy c th vit:

[ 1 - F(t+dt) ] = [ 1- F(t)].[ 1 - F(dt)] (2.6)


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[ 1 - F(t+dt) ] = [ 1- F(t)] . [ 1 - F(dt)] ( 2.6 ) F(t+dt) = F(t) + dF(t)V kh nng ra khi thit b ca nhng phn t l nh nhau, ta

c:

F(dt) = FV .dt / VR = dt / tTB

Thay vo phng trnh ( 2.6 ): 1 - F(t) - dF(t) = [ 1 - F(t) ] . [ 1 - dt/ tTB]

- dF(t) = - dt / tTB + F(t) . dt/ tTB= - dt/ tTB . [ 1 - F(t) ]

dF(t) / [ 1 - F(t) ] = dt/ tTB ( 2.7 )


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Gii phng trnh vi phn ( 2.7 ):

Gii phng trnh vi phn (2.7 ), ta c:

ln [ 1 - F(t) ] = - t/ tTB + C.

T iu kin ban u t=0 th F(t)=0 c C = 0

1 - F(t) = e-t/tTBdo F(t) = 1 - e-t/tTB . ( 2.8 )

ng biu din hm F(t) ca m hnh KLT theophng trnh ( 2.8 ) c trnh by hnh 2.3


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1 - F(t)

F(t)

ttTB

0,632

F(t)

1,0

0,5

0,0

H.2.3-S ph thuc ca F(t) m hnh KLT vo thi gian t


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Nh vy, do khuy TGL ca cc phn t

dngvo rt khc nhau, phn b t 0n .

Lm gim chuyn ho X (v do gim nng sut thit b ) v chn lc

S (i vi cc qu trnh phn ng nitip).

t chuyn ho nh nhau so vi mhnh LT thit b loi ny cn c th tch

ln hn, s chnh lch ph thuc vo chuyn ho X yu cu.


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H thng nhiu thit b khuy ni tip lin tuc:

M hnh h thng nhiu thit b khuy ni tip khcphc c nhc im trn vi s thit b n ln( v l thuyt khi n v trong thc t n c th t 5n 10 ).

Gi tr hm F(t) trong h thng n thit b khuy nitip lin tc c th c xc nh bng php tnhcn bng vt cht, c:

F(t) = 1 - e- nt/tTB

[1+(nt/tTB

)+(nt/tTB

)2/2+...+(nt/tTB

)n-1/(n-1)] .

( 2.19)


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Hnh 2.5 biu din gi tr hm phn b F(t) ca h thng thit b KLT nitip lm vic lin tc ph thuc vo t (trc honh c n v l =t/tTB).

1,0

F

(t)

0,01 2

n=1

3

5

1020

n=

H.2.5-Hm F(t) ph thuc vo s thit b ntrong h thng KLT lin tc, ni tip.

=t/tTB


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T hnh 2.5 thy r h thng thit b KLT

ni tip khi s thit b n tng khong phnb TGL ca cc phn t ca dng vohp li( quanh vng t/tTB=1 ).

V khi n ta c ng biu din F(t)ca h thng tng t nh m hnhLT: TGL ca cc phn t ca dng ngu v bng tTB.


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II.3-THI GIAN LU TRONG THIT B THC -Thit b thc loi thng c khuy:

*Nu c khuy trn mnh v nhtmi trng phn ng khng qu ln cth coi nh m hnh LT . Trong thc tin

sn xut cng nhip d thc hin iu ny,nh thit b nitr ho hydrocacbon thm,nng cc cu t c trn ln hon tonsau 8 giy trong khi TGL trung bnh thngtrn 10 pht.


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*Khi mi trng c nht qu cao tac trng thi trung gian gia hai mhnh LT v KLT.

* h thng nhiu thit b khuy nitip, nu trong mi thit b s khuy trnkhng c hon ton th nh hngca n c th b qua c, coi nh

LT.


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-Thit b thc loi ng:

*ng biu din F(t) c dng ch S dotrn ln theo hng trc ng bi ccnguyn nhn sau:

1/ Khuy trn i lu do cc dngchuyn ng xoy gy nn.

2/ Do gradien vn tc dng theo tit dinngang ca ng.

3/ Khuy trn do khuch tn phn t,

thng nh hng ny khng ln.


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* nghin cu hin tng trn ln ca dng chytrong ng gy nh hng n TGL trong thit b v chuyn ho, chia vct trn ln thnh hai thnh phn:

Hng trc ng vi h s trn dc Dl

Hng ng knh ng vi h s trn ngang Dr

Dl v Dr c xc nh bng chun s Peclet (k hiu

Pe ) nh sau:Pel = v . L / Dl v Per = v . D / Dr ( 2.20 )

Trong : v - Vn tc di ca dng chy.

L - Chiu di ca ng phn ng.D - ng knh ng hay ng knh ca

ht trong ng.


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a/ng rng, chy dng: -Gradien vn tc dng theo tit din ngang

l nguyn nhn ch yu gy nn s sai khcTGL trong ng rng chy dng, khuch tni lu v khuch tn phn t b, c th b

qua. -Tnh F(t): Vn tc dng phn b trn

tit din ng theo mt parabol, ph thucvo khong cch r n tm ng l vr (hnh

2.6)


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r

vrr0

H.2.6-M hnh gradien vn tc dng theohng ng knh ca ng rng, chy dng


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-Tnh F(t) ca ng rng, chy dng

vr

= ( 2FV

/.r0

2 )[ 1 - ( r / r0

)2 ] (2.21)Trong r0 l bn knh ca ng.

Do , TGL ca phn ca dng cch trc ngmt khong r l

tr = L/vr = .r02 (L / 2FV).[ 1 - (r/r0)2 ] = VR/2FV[ 1 - (r/r0)2]tr = ( tTB / 2 ) [ 1 - (r/r0)2 ] ( 2.22 )

Thay r = 0 c TGL ca dng ti trc ng l

t tm(r=0) = tTB / 2

Thay r = r0 c TGL ca dng ti thnh ng lt thnh(r=r0) =


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-Tnh F(t) ca ng rng , chy dng C th xem phn ca dng chy nm cch tm

l r+dr (hnh vnh khn 2r.dr ) c vn tcnh nhau l vr, t nh ngha ca hm F(t) ta c:

dF(r) = vr . 2r.dr / FV ( 2.23) Thay vr t phng trnh (2.21) v rt gn,c :

dF(r) = 4 [ 1 - (r/r0)2 ] r.dr / r0

chuyn dF(r) thnh dF(t) phi dng phngtrnh (2.22) .


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T (2.22) ta c [ 1 - (r/r0)2 ] =tTB / 2t , ly vi phnpt (2.22) ta rt ra:

rdr = r02 . tTB.dt / 4 t2

Thay vo pt (2.23) ,c:

dF(t) = tTB2.dt / 2 . t3 ( 2.24)

T xc nh F(t) cho trng hp ngrng chy dng nh sau:

F(t) = tTB2/2. dt/t3 t = tTB/2, .

F(t)= 1 - (tTB/t)2/4 ( 2.25 )


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F(t)

0,5

1,00,75

0,0tTB/2

tTB t

H.2.7-ng biu din ca F(t) trongng rng, chy dng


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TBP thcb/ ng rng,chy ri :

* ch ny khuch tn i lu do dngchy xoyl chnh .

*Vn tc dng phn b theo tit din ngc dng hnh thang ( hnh 2.8 ), ngha l tr

lp bin st thnh ng, vn tc dng coi nhng u .

* Khi gi tr Re cng ln ( ln hn 10 4 )chiu dy lp bin cng mng v th tch lp

bin so vi FV rt nh,c th b qua,ta c ch dng chy ta m hnh LT .


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Lp bin

FV

H.2.8-M hnh phn b vn tc dng chy

theo tit din ng ch chy ri


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c/ng phn ng c lp ht tnh ( ht xc tcrn, thng gp trong thc t ):

*Lp ht tng cng hin tng

khuch tn trong thit b theo chai hng trc ng v ng knhng

c bit theo hng ng knhtheo m hnh nh hnh 2.9.


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dht

H.2.9-M hnh lch ngang khi chyqua lp ht tnh


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Theo m hnh ny khi dng chy qua mtlp ht b lch ngang dht/2 v qua n lpht s lch ngang ndht/2. V nh vy,

lch ngang lm cho vn tc dng v TGLng u hn.

*Nh vy, khi Dng/dht 10 v L lp ht /dht

10 ch dng chy c coi nh LT.


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II.4-NH HNG CA KHUCH TN DCN CHUYN HO X CA PHN NG

*Trn dc trc ng phn ng lm cho TGL khngng u v do nh hng n chuyn ho X .

*Dng thc trong thit b c th chia thnh hai phn:

-Dng LT vi vn tc di l v .

-Dng khuch tn dc vi h s khuch tntheo hng trc l Dl v chun s Peclet vi k hiu

Pel:

Pel = v . L / Dl

L-l chiu di ca ng phn ng .


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*Gi s:-ng phn ng c tit din l S (hnh2.10).

-Tin hnh phn ng chuyn ho cht A

s.phm *Thc hin php tnh cn bng vt chtcho VR = S . l: A vo = A ra + A phn ng + A tch t.

(2.26) H.2.10-M hnh ng phn ng c khuch tn

dc .

A ra do dng LTv . S . CA(l+l)

A ra do khuch tn

A vo do dng LTV S.CAl

A vo do khuch tn-Dl.S.(CA/l)l

A ra do dng LTS.v.CA(l+l)

A ra do dng LTS.v.CA(l+l)

A ra do khuch tn-Dl.S.(CA/l)l+l

l

S

lL

l+ll


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A tch t = VR . dC/dt, trng thi dng A tch t bng 0.

Ta c:(A ra - A vo)dng LT+ (A ra - A vo)khuch tn + A phn ng = 0.

Thay cc gi tr ca A vo,ta c:

v.S(CA(l+l)-CAl)+(-Dl.S)[(CA/l)l+l -(CA/l)l]+S.l(-CA/t)=0

Chia cho S.l , c:

v.(C A(l+l) - CAl)/l - Dl[(CA/l) l+l - (CA/l)l ]/l +(-CA/t) = 0

Thay (-CA/t) = RA = k.CAn v cho l0 ,c:

v (CA/ l) - Dl (2CA/l2 ) + k .CAn = 0 ( 2.27 ) y l m hnh ton mt th nguyn miu t qu trnh

trong ng phn ng ch nh hng bi khuch tn theohng dc trc.


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v (CA/l) - Dl (2CA/l2 ) + k .CAn = 0 ( 2.27 )

*Chuyn ( 2.27) thnh dng khng c th nguyn:

- t i lng chiu di l z = l/L, ta c: z = l/L , do z/l = 1/L

CA/l = (CA/z) . z/l = (1/L) . CA/z

2CA/l2 = [1/L. (CA/z) / z ]. z/l = (1/L2).2CA/z2

T pt (2.27) thnh:

(v/L).CA/z - (Dl/L2 ).2CA/z2 + kCAn = 0. ( 2 . 28 )

Chia hai v pt (2.28) cho -L/v = -tTB :

(Dl/L.v).2CA/z2 - CA/z - k.tTB.CAn = 0.(2.29)

(Dl/L.v) . 2CA/z2 - CA/z - k . tTB . CAn = 0. ( 2 . 29 )


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( l/ ) CA/ CA/ tTB CA 0 ( 9 )* Gii pt (2.29) vi iu kin b :- ti u vo z = 0 , CA = CA0

- ti u ra z = 1 , CA = CAL

c CAL v t tnh chuyn ho XA, . . XA ph thuc vo 3 thng sk , tTB v Pel

= L.v/Dl .

Vi phn ng bc 1 (n = 1) gii c : CAL/CA0 = 1 - XA

= 4a.exp(Pel/2)/[(1+a)2exp(a.Pel/2) -

(1-a)2exp(a.Pel/2)]

( 2 . 30 ) Vi a = lTB/Pe4kt1+


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H.2.11-Biu din VRthc/VRLT ph thuc vo Pel,

k.tTB v X caphn ng bc 1.


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59

H.2.12-Biu din VRthc/VRLT vo Pel,k.t

TB.C

0v X ca phn ng bc 2


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Thc nghim cho thy rng chun skhuch tn dc Pel ch yu ph thucvo chun s Re = v.dng./.

Hnh sau l kt qu thc nghim camt s tc giv s ph thuc caDl/v.dng = (Dl/v.L).L/dng vo Re

T c th rt ra tng quan sau:Dl/v.dng = 3.107/(Re)2,1+1,35/(Re)0,125.

H 2 13-S liu thc nghim s ph thuc D /v d


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H.2.13 S liu thc nghim s ph thuc Dl/v.dngvo Re ca mt s tc gi.


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II.4THC NGHIM XC NH GI TR HM PHNB TGL THIT B THC.1/ Thc nghim xc nh hm phn b TGL E(t) v

F(t):

*Nguyn tc:

-Tit = 0 bt u cho cht ch th vo. -Xc nh nng cht ch th u ra

theo thi gian t.

-X l tn hiu nng ra,xc lp F(t). Thng cho ch th vo theo 3 dng tn

hiu: dng bc cp, dng xung v dnghnh sin.


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a/Tn hiu vo dng bc cp, xc nh hm F(t):

-Tn hiu cho ch th vo dng bc cp:

= 0 khi t 0 .

Cch th(z=0)

{ ( 2 . 31 )

= C0 khi t > 0 .

-Miu t: Ti t = 0 bt u cho ch th i vo

thit b v duy tr trong sut thi gian sau (hnh 2.11 ). Nh vy khi t > 0, lng ch th vothit b khng i, bng FV.C0.

Xc lp hm F(t): Gi s nng ch th ra


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Xc lp hm F(t): Gi s nng ch th ra thi im t l Ct, phn ca ch th ra tithi im ny l FV.Ct. T nh ngha ca

hm F(t) ta c:F(t) = FV.Ct / FV.C0 = Ct / C0. ( 2 . 32 )

Tn hiu vo

Nng

chth

ra

Nng Ch th

C0

00

t

Ct

H.2.11-Tn hiu vo dng bc cp v nng ch th i ra theo thi gian t


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b/ Tn hiu vo dng xung vung, xc nh hm E(t):

-Tn hiu cho ch th vodng xung:

= 0 khi t 0 .

Cch th (z=0)

=C0 khi 0t0 .

Miu t: Ti t = 0 bt ucho ch th i vo thit btrong thi gian ngn t0

( yu cu t0


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66

Xc lp hm E(t): Gi s ton b lng ch th cho vo thit

b dng xung vung l q0, ti thi imt nng ch th dng ra l Ct, trongkhonh khc dt th tch dng ra l FV.dtv lng ch th i ra l C

t

.FV

.dt, chnhl lng ch th c trong thit b t t nt+dt. T nh ngha v E(t) lng ch th bng q0.E(t).dt, ta c:

q0 . E(t) . dt = FV . Ct . dt. ( 2.34 )

T : E(t) = FV . Ct / q0 ( 2.35 )


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ta xc nh hm E(t) cn bit q0, lmvic ny ly tch phn phng trnh (2.34):

,theo (2.1):

Ta c: q0= Ct.t

T :

E(t) = Ct / Ct.t (2 . 36)

=0 0

tV0 .dt.CF.E(t).dtq

=0 0

tV0.dt. CF. E(t).dtq

=0 0

tV0 .dt.CF.E(t).dtq

=0

1E(t).dt

=0 0

VtV .F.dtC.F

=0 0

tt/C.dtC


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Bi tp 2.4 - Cho tn hiu vo dng xung, nng chth ra o c nh bng sau , xc nh hm E(t) .

t , ph : 0 5 10 15 20 25 30 35

C , g/l : 0 3 5 5 4 2 1 0

Li gii:

-Hm E(t) c tnh theo pt (2.36) :

-Trc tin tnh Ct . t = 5(3+5+5+4+2+1) = 100.

-Vy E(t) = Ct / 100, kt qu nh bng sau:

t , ph : 0 5 10 15 20 25 30 35

E(t).102: 0 3 5 5 4 2 1 0


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-ng biu din nng ch th ra theo t nh hnh 2.13.

-Din tch gii hn bi ng biu din v trc honh lgi tr ca tch phn, cn din tch cc ch nht l gi trca tng.

5

4

3

2

1

0

Nng ch th

g/l

t,ph5 10 15 20 25 30 35

H.2.13-ng biu din nng ch th ra theo thi gian.


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2/ Xc nh chuyn ho X quagi tr hm phn b TGL: Nh nghin cu trn, dng i ra khi thit

b lm vic lin tc khng phi l dng ngnht v TGL v nh vy, nng ca cht

phn ng dng ra ( t tnh chuyn hoX = 1 - Cra/C0 ) l nng trung bnh ca ccphn t ca dng vi TGL trong thit b khcnhau . Do c th vit:

Cra = Cng.t.E(t).dt = Cng.t.E(t).tt = 0, (2.37)


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02/10/0871

Trong :C nguyn t c th tnh c t phng

trnh ng hc ca phn ng. E(t).dt - Phn ca dng c TGL t t n

t+dt [ t nh ngha ca E(t) ].

T tnh c chuyn ho X:

X =1-Cra/C0=1-( /C0)

( 2 . 38 ) .

ttECnguyento

).(.0


72/256

02/10/0872

Bi tp 2.5- Phn ng bc 1 chuyn ho chtA, hng s vn tc k = 0,307 ph-1, c tin

hnh trong TBP c hm phn b TGL E(t)nh bi tp 2.4. Xc nh chuyn ho X. Li gii: - Phng trnh vn tc ca phn ng bc 1:

-dCA / dt = kC, gii vi iu kin khi t = 0 th CA= CA0,, c C A nguyn t = CA0 . e-kt , thay vopt(2.37):

CA ra = CA0 . t .( 2 . 39 )

.E(t).e0

kt


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Lp bng tnh phng trnh (2.39):

t , ph E(t) k.t e-kt e-kt . E(t) . t5 0,03 1,53 0,2154 0,0323

10 0,05 3,07 0,0464 0,011615 0,05 4,60 0,0100

0,0025 20 0,04 6,14 0,00210,0004 25 0,02 7,68

0,0005 0,00005 30 0,019,21 0,0001 0

CAra - C A0= e-kt . E(t) . t = 0,047

XA = 1 - CAra/CA0 = 1 - 0,047 = 0,953.


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Bi tp 2.6- Tin hnh phn ng bc 1 nh bi tp 2.5Bi tp 2.6- Tin hnh phn ng bc 1 nh bi tp 2.5trong thit b phn ng c TGL trung bnh nh bi tptrong thit b phn ng c TGL trung bnh nh bi tp

2.4 nhng dng chy theo ch LT.2.4 nhng dng chy theo ch LT.Li gii: -Thi gian lu trung bnh ca thit b bi tp 2.4

c xc nh theo cng thc (2.2).

tTB

=

=5(0,03.5+0,05.10+0,05.15+0,04.20+0,02.25+0,01.30)=15 ph.

-Thit b theo m hnh LT c TGL ng u v bngtTB. Vi phn ng bc 1 t pt dCA/dt=kCA, c:

CA / CA0 = e-ktTB = e-0,3.15 = 0,01. XA = 1 - CA/CA0 = 1 - 0,01 = 0,99 .

=00

.).(.).( tttEdtttE


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III/THIT B PHN NG LOITHNG C KHUY

- Thng dng cho pha lng, c bit cc qutrnh d th lng - lng, lng - rn, lng - kh vlng - rn - kh tng cng tip xc cc pha.

- Do khuy nn nng , nhit ng u trongkhp thit b, nhng nh ni trn, TGL thit b lm vic lin tc phn b t 0 n .

- Khuy tng cng h s trao i nhit gia mitrng phn ng vi thnh thit b.

- M hnh l KLT.


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III.1- Thit b lm vic gin on: - Ch dng cho pha lng, qui m sn xut nh.

- Tnh th tch thit b VR:

VR = Vngy m.(1 + z).tm / 24 . ( 3 . 1 ) -Nu th tch thit b VR chn, ta tnh s thit b n:

n = Vngy m.(1 + z).tm / 24.VR. . ( 3 . 2 ) Trong :

Vngy m - Th tch sn phm i ra trong 24 gi

z - H s d tr phng cc s c bt thng.Thng z c gi tr t 0,10 n 0,15, khi thit b phc tphay lm vic nhit v p sut caoc th ly t 0,15n 0,20.

Thit b l i i


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Thit b lm vic gin on - H s y . Thng vi thit b phn ng c khuy ly

gi tr t 0,75 n 0,80 ; trng hp mi trng phn ng tobt c th ly n 0,40.

tm- Thi gian tin hnh mt m, gm thi gian chun bv thi gian tin hnh phn ng.

Trong :

-t chun b = t np liu + t un nng nguyn liu t nhit u n nhit phn ng + t lm ngui sn phm + t thosn phm + t lm sch thit b cho m sau ...

- t phn ng c xc nh t cc thng tin sau:

*Tnh theo phng trnh vn tc vi X xc nh.

*Theo s liu thc nghim c, v d th t-X thc nghim.

III 2 Thi b l i li


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III.2-Thit b lm vic lin tc:

*Tnh ton theo m hnh KLT.

*Dng phng php i s v phng phpdng hnh.

a/Phng php i s:

-Lp phng trnh cn bng cht cho thit bth i ( hnh 3.1 ):

Ci-1

.FV(i-1)

= Ci

.FVi

+ Ri

.VRi

+ dCi

.VRi

/dt.( 3 . 3 )


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Thit b lm vic lin tc

CA(i-1)

FV(i-1) FVi

CAi

CAi

VRi

TiTi-1

Ri=ki.f(CAi) ki=k0.e

-E/RTi

TBP th i

H.3.1-Cc thng s ca thit b th iloi thng c khuy


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Ci-1.F V(i-1) = Ci.FVi + Ri.VRi + dCi.VRi/dt. ( 3 . 3 )

Trong :Ci-1 : Nng cht phn ng i vo thit b th i.

Ci : Nng cht phn ng trong thit b i v l nng ira.

F V(i-1) v FVi : Lu lng ca dng vo v ra.

VRn : Th tch thit b th i.

dCi.VRi/dt : Lng cht phn ng tch t trong thit b i.

Ri = ki . f(Ci) : Vn tc phn ng thit b th i.


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*Vi phn ng chuyn ho cht A sn phm:

-Bc 0 : Ri = ki ; pt (3.4) thnh:

C i-1 = Ci + ki . tTBi . ( 3 . 4 )'

-Bc1 : Ri = ki.Ci ; pt (3.4) thnh:

C i-1 = Ci + ki . tTBi . Ci hay

C i-1/Ci = 1+ ki . tTBi . ( 3 . 4 )''

-Bc2 : Ri = ki.Ci2; pt (3.4) thnh:

C i-1 = Ci + ki . tTBi.Ci2 ( 3 . 4 )'''


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02/10/0883

Bi tp 3.1: Tnh chuyn ho X ca phn ng bc1 nh bi tp 2.5, tin hnh trong thit b thng ckhuy c TGL trung bnh l 15 pht nh bi tp 2.4

Li gii:

Vi i = 1, t phng trnh (3.4)" ta c:

C0/C1 = 1 + 0,307 . 15

X = 1 - C1/C0 = 1 - 1/( 1 + 0,307.15 ) = 0,822

So snh kt qu tnh ton vi bi tp 2.5v 2.6 thy rng cng mt iu kin ,thit b theo ch KLT c chuyn hothp nht.


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02/10/0884

-Lp phng trnh cn bng cht cho hthng n thit b khuy ni tip:

Dng phng trnh (3.4) v bt u t thit b1 n n

*Khi th tch cc thit b trong h thng khng

bng nhau:

VRi VRj tTBi tTBj

V nhit cc thit b khng bng nhau

Ti Tj ki kj .


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02/10/0885

Phn ng bc 0:

i = 1 C0 - C1 = k1 . tTB1

i = 2 C1 - C2 = k2 . tTB2

i = 3 C2

- C3

= k3

. tTB3

... ......

i = n C n-1 - Cn = kn . tTBn .

Cng cc phng trnh theo tng v c:C0 - Cn = ki.tTBi i = 1, n

Xn0= ki.tTBi / C0 ( 3 . 5 )

Phn ng bc 1:Phn ng bc 1:


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i = 1 C0 / C1 = 1 + k1 . tTB1

i = 2 C1 / C2 = 1 + k2 . tTB2 i = 3 C2 / C3 = 1 + k3 . tTB3 ... ... i = n C n-1 / Cn = 1 + kn . tTBn . Nhn cc phng trnh theo tng v, c:

C0 / Cn = . T chuyn ho thit b n i vi phn ng bc 1

l:

Xn1= 1 - Cn / C0 = 1 - 1 / .( 3 . 6 )

=

+

n

1i

TBii ).tk(1

=

+n

1i

TBii ).tk(1

Phn ng bc 2:


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02/10/08 87

Gii pt (3.4)''' c:

C i = (- 1 ) / 2ki.tTBi .

i = 1 C1= (- 1 ) / 2k1.tTB1i = 2 C2= (- 1 ) / 2k2.tTB2Thay C1 vo ta c:

C2 = [-1 ]/2k2.tTB2 .( 3 . 7 )

141 + iii Ctk

01141 Ctk+

12241 Ctk+

111122 2/)411(41 tkCtktk o++


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02/10/0888

*Khi th tch cc thit b trong h thng bng nhau:

VRi = VRj tTBi = tTBj = tTB .

v nhit cc thit b bng nhau:

Ti = Tj ki = kj = k .

chuyn ho ca phn ng bc 0 trong h

thng n thit b ni tip theo cng thc (3.5) thnh: Xn0 = ( C0 - Cn ) / C0 = n.k.tTB / C0 . ( 3 . 5 )'

chuyn ho ca phn ng bc 1 trong h

thng n thit b ni tip theo cng thc (3.6) thnh: Xn1 = 1 - Cn/C0 = 1 - ( 1 + k . tTB )-n . ( 3 . 6 )'

b/Ph h d h h


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02/10/0889

b/Phng php dng hnh: Phng trnh (3.4) c th vit li dng:

Ri = -Ci / tTBi + C i-1 / tTBi . ( 3 . 8 )

V tri ca pt (3.8) l hm s xc nh ca

Ci: Ri = ki .f( Ci ), ta v ng biu dinca R ph thuc vo C ( hnh 3.2 ).

V phi l hm s tuyn tnh ca Ci, giaoim ca hai ng biu din lnghim ca phng trnh (3.8).


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02/10/0890

Phng php dng hnh

ng thng biu din v phi c h sgc l -1/tTBi.

ng thng u tin ( i = 1 ) bt u t

trc honh ti C0 vi h s gc l -1/tTB1giao im ca ng thng ny ving cong R c to ng vi C1 v R1,

l nng cht phn ng v vn tc phnng thit b 1.

Phng php dng hnh


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02/10/0891

Phng php dng hnh

T C1 trn trc honh v ng thng vi h

s gc l -1/tTB2, ta c C2 v R2 ... v c thtip tc n khi t Cn C yu cu.

Bi ton c gii, ta tm c s thit b n

ca h thng cng nh cc thng s nng cht phn ng C v vn tc phn ng R trongtng thit b .

Nu nhit cc thit b khc nhau, ki kj,ng vi mi nhit phi v ng biu din

Ri = ki . f (C) tng ng .

Phng php dng hnh


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02/10/08 92

C3Cn C0C2 C1

Rn

R3

R2

R1

Vn tc

p , R

Nng

cht p, C

Tg 1 = -1/tTB1Tg 2 = -1/tTB2

Tg 3 = -1/tTB3...

H.3.2-Phng php dng hnh gii bi ton hthit b khuy ni tip, lin tc.

123


93/256

02/10/0893

Bi tp 3.2-

Phn ng thun nghch 2A B + C, hng svn tc phn ng thun kth = 10 m3/kmol.h,hng s cn bng Kcb = 16, nng ban uCA0 = 1,5 kmol/m3, CB0 = CC0 = 0, tin hnh

trong thit b phn ng loi thng c khuy vilu lng dng FV = 100 m3/h. chuyn hoX yu cu bng 80% trng thi cn bng. Tnh:

a/Th tch TBP khi tin hnh trong mt

thit b VR1. b/S thit b n ca h thng nu ly VRn =

VR1/10.

Li gii:


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02/10/0894

g

a/t nng ca B v C trng thi cnbng l CBcb = CCcb = Ccb.

Ta c: CAcb = 1,5 - 2Ccb.

Kcb = C2cb / (1,5 - 2Ccb )2 =16.

Gii c Ccb = 0,667 kmol/m3.

X yu cu bng 80% trng thi cn bngnn nng yu cu ca B v C nh sau:

CByu cu = C Cyu cu = 0,80.Ccb=0,80.0,667 = 0,533

kmol/m3.


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02/10/0895

Vy CAyu cu = 1,5 - 2 . 0,533 = 0,434 kmol/m3.

Phng trnh vn tc ca phn ng thun nghchchuyn ho cht A c dng:

RA = kth.CA2 - kng . CB .CC

= kth ( CA2 - CB . CC/ Kcb ) v kng = kth/Kcb.

Ta c:

RA = 10 ( CA2

- CB .CC /16 ). ( 3.9 )


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02/10/0896

Vn tc phn ng trong h thng phn ng mt thit b l:

RA(1) = 10 [ (0,434)2 - (0,533)2/16 ]

= 1,7 kmol/m3.h.

T pt c trng (3.4) vi i = 1 ta c:

tTB1 = VR1 / FV = ( C0 - C1 ) / R1 . Do : VR1 = FV . ( C0 - C1 ) / R1 .

Thay s vo , ta c:

VR1 = 100 ( 1,5 - 0,434 ) / 1,7 = 62,7 m3

.


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02/10/0897

b/Nu ly VRn = 1/10 VR1 = 6,27 m3, tm sthit b n ca h thng khuy ni tip ta dng

phng php dng hnh. V ng biu dinRA ph thuc vo CA theo phng trnh vn tc(3.9): cho CA cc gi tr t 0 n 1,5 kmol/m3

Xc nh CB , CC tng ng v tnh RA nh bng sau:

CA, 1,5 1,3 1,1 0,9 0,7 0,5 0,3

C

B,C

C0 0,1 0,2 0,3 0,4 0,5 0,6

RA 22,5 16,9 12,1 8,1 4,9 2,5 0,68kmol/m3.h


98/256

02/10/0898

T CA = CA0 = 1,5 kmol/m3 v ng thng

to vi trc honh gc :tg = - 1/tTB= - 100/6,27 = -15,9

ng thng ny ct ng cong RA

im c to CA1 = 0,94 kmol/m3

v RA1 = 8,9 kmol/m3.h


99/256

02/10/0899

T CA1 tip tc v ng thng song song vi

ng thng trc( v cng h s gc ) ta cgiao im th 2 vi to tng ng l CA2 v RA2( hnh 3.3 ), c th tip tc c kt qu nh sau:

CA1 = 0,94 kmol/m3 RA1 = 0,89 kmol/m3.h

CA2 = 0,68 RA2 =4,52

CA3 = 0,52 RA3 = 2,65

CA4 = 0,42 RA4 = 1,70


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02/10/08100

R

20

10

[kmol/m3.h]

0,5 1,0 1,5

CAKmol/m3

12

34

1

2

3

4

R2=4,52;C2=0,68

R3=2,65;C3=0,52

R4=1,70;C4=0,42

R1=8,90;C1=0,94

Tg=-1/tTB=-15,9

H.3.3-Phng php dng hnh gii bi tp3.2


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02/10/08101

Nh vy CA4 < CA yu cu , ta c s thit btrong h thng n = 4 .

Cng cn lu rng th tch cc thit b

trong h thng l 4 . 6,27 = 25 m3

, nhhn nhiu so vi th tch phn ng khidng mt thit b.

(iu ny cn c gii thch?)

IV/C TRNG NHIT TRONG THIT B


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02/10/08102

IV/C TRNG NHIT TRONG THIT BPHN NG

Phn ng ho hc lun c hiu ngnhit, thng kh ln c th thay

i nhit ca qu trnh. Nhit l thng s cng tnh, nh

hng mnh n vn tc phn ng.

Do khng ch nhit phn nglin quan n lm vic an ton cathit b v qu trnh sn xut.


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02/10/08103

Cc gii php duy tr ch nhit cho phn ng

nh hng n kt cu, hnh dng ca thit bphn ng.

Trng thi dng ca h TBP ch t c khi

tho mn pt sau: QR = QS ( 4 . 1 )

QR - Nhit phn ng trong n v thi gian.

QS - Nhit trao i trong n v thi gian.

IV 1-Ch nhit ti u:


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02/10/08104

IV.1 Ch nhit ti u: y l nhn t lun c ch n trong

sn xut cng nghip, c bit khi cngsut ln, m bo nng sut cngnh chuyn ho X ca qu trnh phnng.

Ch nhit ti u ph thuc vo ctrng nhit ng, ng hc ca phnng cng nh tnh nng ca xc tc

Cho nn ch nhit ti u a dng, tutng trng hp c th.


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02/10/08105

Vi d: C phn ng thun nghch, to nhit: A B

Hng s vn tc phn ng thun: k1

= k10

. e-E1/RT

Hng s vn tc phn ng nghch: k2 = k20 . e-E2/RT

Ta c: RA = k1 . CA - k2 . CB

= k10..exp(-E1/RT).CA0(1 - XA) - k20.exp(-E2/RT). CA0 . XA

Tm ch nhit ti u m bo vn tc phnng ln nht bng cch cho dRA / dT = 0:

dRA/dT =

k10.CA0.(1-XA).e-E1/RT.E1/RT2 - k20.CA0.XA.e-E2/RT.E2/RT2 = 0 .


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02/10/08106

k10.(1-XA).e -E1/RT(t).E1 = k20. XA.e -E2/RT(t).E2 .

Ly loga c hai v: ln [ k10.(1-XA).E1] - E1/RT(t) =

ln ( k20.XA.E2 ) - E2/RT(t) .

T :

(E2 - E1)/RT(t) = ln [ k20.E2.XA/k10.E1.(1-XA).

T(t) = (E2 - E1)/Rln[k20.E2.XA/k10.E1.(1 - XA)].

( 4 . 2 )


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02/10/08107

Nh vy,ch nhit ti u trng hp phnng thun nghch, to nhit ph thuc vo

chuyn ho XA nh cng thc (4.2)

Lc u khi XA cn thp nn tin hnh phnng nhit cao m bo vn tc ln,

sau khi XA tng ln cn gim dn nht . Trong thc tin sn xut, phn ng c c

trng ny ( nh SO2 + 1/2O2 SO3 + q ) c

thc hin trong thit b trong TBP c nhiungn xc tc on nhit c trao i nhittrung gian gia cc ngn nh hnh 4.1 .


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02/10/08108

T0

T1

T2

T3

T4

T5

T6

T7

T8

T9

Nguyn liu Sn phm

T9T7

T5

X2

X3

X4

X1

X5

H.4.1- Cc lp xc tc on nhit ( p to nhit ) vi lm lnh trung gian .

Bng cch nh vy qu trnh phn ng trong thit b c thc hintheo ng dch-dc quanh ch nhit ti u ( hnh 4.2 ).


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02/10/08 109

theo ng dch dc quanh ch nhit ti u ( hnh 4.2 ).

TT0

T1T2

T3T4

T5T6

T7T8

T9

Xcb

Tt=f(X)

X1

X2

X3

X4

X5

X

H.4.2- ng biu din s thay i nhit v chuyn hoX thit b nh hnh 4.1 i vi phn ng thun nghch to nhit.

IV.2-Cc gii php duy tr ch


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02/10/08110

IV.2 Cc gii php duy tr ch nhit ti u trong thit b

phn ng:a/Trao i nhit qua thnh:

-Thit b loi thng c khuy vi v bc ngoiv ng xon trong thit b.

-Thit b ng chm vi hng nghn ng vi bmt trao i nhit rt ln, thng dng cho

phn ng pha kh to nhiu nhit.

b/Dng tc nhn mang nhit l kh, lng hay rn:


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02/10/08111

g g , g y

-Kh:Kh tr, kh chy nhit cao ... v

thng dng hi nc ( c sn trong nhmy v d tch khi sn phmhydrocacbon bng cch lm lnh ngng tv phn ly, ngoi ra hi nc cn gimto cc bm vo thnh thit b ).

-Lng: Dung mi hay cc cht tr c tronghn hp phn ng.


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02/10/08112

-Rn:Vt liu rn chu nhit nh gm, s,cc vt liu silict ... v c xc tc rn,in hnh l qu trnh crcking xc tc lphn ng thu nhit, xc tc mau mt hot

tnh do b ph cc b mt nn phi dnglp xc tc chuyn ng v tun hongia thit b phn ng v thit b ti sinhbng phn ng t cc to nhit nh hnh 4.4


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02/10/08 113

TBTS+Q TBP-Q

Xc tc nng

Xc tc ngui

H.4.4-Qu trnh s dng nhit ca crcking xt

c/Tng phn on nhit c trao i nhit


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02/10/08 114

/ g p trung gian:

-Ging m hnh 4.1. -in hnh l thit b reforming xc tc,

hydro ho xc tc ... -Vi TBP thng c khuy nh hnh 4.5.

T0

T1

H.4.5-H thng thit b thng c khuyvi lm lnh trung gian

d/iu chnh nhit


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02/10/08115

d/iu chnh nhit ban u T0 i vi

cc qu trnh onnhit.

e/Bay hi cu t cnhit si xp x

nhit phn ngvi h thng lmlnh ngng t v chocht lng quay v

TBP ( hnh 4.6 ).

Lng

Hi

H.4.6-Gii nhit phn ng bng cchBay hi cu t nh

ChtPhn ng Sn phm

/B


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02/10/08116

g/Bmtun honhn hpphn ngqua thit

b trao inhit( h.4.7).

Sn phm

Nguyn liu

TBP

H.4.7-Gii nhit bng cch tun hon hn hpphn ng qua thit b trao i nhit

IV 3 t h t hit


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02/10/08117

IV.3-c tnh t nhit:

-L cc qu trnh t xy ra sau khi c "mi ". -iu kin t nhit:

Phn ng to nhit ln c th a nhit

hn hp phn ng t nhit ban u T0n nhit lm vic ca thit b.

C kh nng trao i nhit gia sn phm hay

khi phn ng vi nguyn liu i vo. iu kinny ph thuc vo c trng v kt cu cathit b phn ng.


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02/10/08118

-Cc loi thit b c kh nng t nhit:

a/Thit b loi thng c khuy:Nh khuy nguyn liu i vo

c trn ln vi khi phnng nng c trong thit b,nu phn ng to nhit ln th

qu trnh tin hnh trong loi thitb ny c kh nng t nhit saukhi c "mi" (hnh 4.8a ).


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02/10/08119

T0

T0

T1

T1

T0

Mt cangn la

Tf

a b

H.4.8-Thit b phn ng t nhita- Thng c khuy. b- Ngn la

b/Thit b phn ng dng ngn la: Thit b


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02/10/08120

b/Thit b phn ng dng ngn la: Thit bdng ny thng c s dng tin hnhcc phn ng to nhit cao nh qu trnh cloho ( t hydro trong mi trng clo snxut HCl ) hay qu trnh oxy ho mt phnhydrocacbon sn xut axtylen.

Trong khng gian cht hp ca ngn la nhit

thay i t T0 rt thp n nhit cao c th t bc chy, ngha l ta c gradien vnhit , v nng , v th ho hc rt ln(c th t hng vn n v ), nh qu trnhtruyn nht, chuyn cht v hot ho xy ra

mnh m trn mt ca ngn la n c tht nhit (hnh 4.8b ).


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02/10/08121

c/Thit b phn ng loi ng:

- c th t nhit cn c s ( hnh 4.9b ).trao i nhit gia nguyn liu vo visn phm phn ng (hnh 4.9a) hay vikhi phn ng

- trng hp b/ c th iu chnh cnhit lm vic ca lp xc tc cng nhnhit T1 theo thi gian lm vic ca xc

tc bng cch iu chnh thng s .


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02/10/08122

T0 T1

T2

T1T0

T3

T3

FV (1-)FV

FV

T0

T2

T2T0 T2

T1

H.4.9- Thit b phn ng loi ng c kh nng t nhit.a-Trao i nhit vi sn phm.b-Trao i nhit vi xc tc

c/Qu trnh nm phm vi khuch tn ngoi:


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02/10/08123

c/Qu t p uc t go

- Thng gp khi tin hnh phn ng xc tc

kh- rn, to nhit nhit cao, lc vn tcphn ng ln v chm nht l cc qu trnh khuchtn qua lp bin gia pha kh v b mt xc tc rn( hnh 4.10a).

- Do tr lc khuch tn ch yu lp bin,

hnh nh cng tng t vi qu trnh truyn nhitgia b mt xc tc v pha kh, nn nhit bmt xc tc Ts cao hn nhiu so vi nhit trong pha kh Tg .

- Trn b mt xc tc nng qu trnh phn

ng tip tc t tin hnh.

Qu trnh cng nghip:


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02/10/08124

- Qu trnh cng nghip:

+Oxy ho NH3 thnh oxit nit trnxc tc dng li Pt-Rh nhit 800 -8500C.

+Amonoxi ho CH4 thnh HCN

trn xc tc dng li Pt-Rh 1000 -12000C.

+Oxihydro ho mtanol thnh

fomaldhit trn xc tc Ag 650 - 7200C( hnh 4.10b).


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02/10/08125

- qu trnh c th tin hnh t nhit, khi

m my cn phi un nng hn hpmtanol - khng kh n 300 - 4000C. Sau ,khi lp xc tc l cc tinh th bc t trn ling nng n nhit phn ng th dng unnng, kt thc qu trnh "mi " phn ng.

-TBP cn c mng phng n v thit blm lnh ng chm t st ngay sau lp xctc lm lnh nhanh hn hp sn phm tnhit phn ng xung 200 - 3000C, gim

phn hu fomalehit to thnh nhit caotheo phn ng: CH2O CO + H2

M h


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02/10/08126

CAg

CAs

Ts

Tg

Dng kh Lpbin

CAs > Tg

B mt xc tc

Mng phng n

un nng inMi phn ng

Xc tc Ag

Lm lnhng chm

Hn hp mtanol-khng kh

a b

H.4.10- c trng t nhit ca qu trnh phn ng xc tc phm vi khuchtn ngoi. a-M hnh khuch tn ngoi. b-TBP oxyhidro ho mtanol.

IV.4-c trng nhit ca TBP loi thng c khuy:


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02/10/08127

*Nhit ng u khp thit b.

*C kh nng t nhit.

a/Cc trng thi dng:

iu kin dng: QR = QS ( 4 . 3 )

Trong QR - nhit phn ng trong n v thi gian.QS - nhit trao i trong n v thi gian.

gii pt (4.3) ta dng phng php dng hnh : kho stng biu din ca QR v QS theo T, nghim ca pt l

giao im ca hai ng biu din.

+Kho st QR :Gi h b 1 h h ht A t hit


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02/10/08128

Gi s c phn ng bc 1 chuyn ho cht A, to nhit,khng thun nghch, tin hnh trong TBP loi thng ckhuy th tch V

R.

S mol cht A chuyn ho trong n v thi gian l NA, do :

QR = NA . ( - HR )

NA = k . CA . VR

CA l nng trong thit b, rt ra t pt (3.4)''

CA = CA0 / ( 1 + k . tTB ) . T :

NA = k.CA0.VR/(1+k.tTB ) = k0.e-E/RT.CA0.VR/(1+k0.e-E/RT.tTB)

NA = k0.CA0.VR / ( e E/RT + k0.tTB ) . Do :

QR = k0 . CA0 . VR . (-HR) / ( e E/RT + k0 . tTB ) ( 4 . 4 )

ng biu din ca QR ph thuc vo T theot (4 4) d h S i khi


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02/10/08 129

pt (4.4) c dng ch S, ngoi ra khi ccthng s khc nh VR, FV thay i ta c

mt h ng cong S trong mt phng vitrc to QR - T nh hnh 4.11.

T

QR

FV tng

H.4.11- ng biu din QR theo T khi FV thay i

+Kho st Q :


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02/10/08130

+Kho st QS: QS c hai thnh phn:

- Trao i nhit vi nguyn liu vo, nhit nguyn liu t T0 T.

-Trao i nhit qua thnh.

QS = FV. .CP( T - T0 ) + K.F. ( T - TC ). ( 4 . 5 ) hay QS = ( FV. .CP+ K.F ) T - ( FV. .CP.T0 + K.F.TC ) Trong : - Khi lng ring ca hn hp phn ng.

CP- Nhit dung ring ca hn hp phn ng.T - Nhit phn ng.

T0- Nhit vo. K - H s trao i nhit qua thnh.

F - B mt trao i nhit ca thit b. TC- Nhit thnh thit b.

ng biu din ca QS theo T l


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02/10/08 131

g QSng thng, khi F thay i ta c cc

ng thng vi h s gc thay i( hnh 4.12a ), khi T0 thay i c ccng thng song song (hnh 4.12b).

QS

K.F=

0

K.F tng

T T

QS

T0 tng

ba

H.4.12- ng biu din QS theo T.a- Khi KF thay i. b- Khi T0 thay i.

QQ


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02/10/08132

1

2

3

QSQR

T

T3T2

QR1

QR3

H.4.13- Cc trng thi dng ca TBP thng c khuy

Nh vy, c th c 3 giao im gia


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02/10/08133

y, g gng QR dng ch S v ng thng

QS, ngha l c 3 trng thi dng cah thng ( hnh 4.13 ).Trng thi dng 1c T1 thp v QR1,QS1 u thp, phn ng

hu nh khng xy ra. Trng thi dng 3 nhit cao T3 do c chuynho cao , l iu ta mong i. Trng thi

dng 2 khng n nh, s c kho st mc 3/.

b/Trng thi dng ph thuc vo


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02/10/08134

g g p chiu thay i ca cc thng s v

bin php mi phn ng: a/Gi s FV thay i, cc thng s

khc c nh:

QR l cc ng cong A, B, C, D vE, QS coi nh khng i,c cc trng

thi dng t 1 n 9, ng vi nhit dng t T1n T9 nh hnh 4.14.


135/256

02/10/08135

T

AB

C

D

E

QR QS

4

32

1

5

6

7 89

H.4.14-Cc trng thi dng khi FV thay i

FV tng


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02/10/08136

-Khi FV tng t thp ln cao ta c cc

ng QR ln lt l A, B, C, D v E, cctrng thi dng l 9, 8, 7, 6 v 1.

-Ngc li, khi FV gim t cao xung,cc ng QR ln lt l E, D, C, B v A,cc trng thi dng l 1, 2, 3, 4 v 9.

Nh vy, trng thi dng ph thucvo chiu thay i ca FV nh hnh

4.15

T


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02/10/08137

Tdng

FVEDCBA

9

8 7

6

3

4

2 1

H.4.15- Cc nhit dng ph thuc vo chiuthay i ca FV


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02/10/08138

T rt ra cch mi phn ng:

Khi m my ban u dng lu lngdng FV thp phn ng tin hnh

trng thi dng c chuyn ho vnhit dng cao ( T9 ), sau tngdn FV n nng sut thit k ( nhngkhng th tng qu ng D ).

b/Gi s T0 thay i, cc thng s khc c nh:


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02/10/08139

0 y g

QR khng i, QS l cc ng thng song

song , ta c cc trng thi dng t 1 n 9nh hnh 4.16a.

- Khi T0 tng t thp ln, QS ln lt l ccng thng A, B, C, D v E, cc trng thi dng

l 1, 2, 3, 4 v 9. -Ngc li, khi T0 t cao gim xung, QS ln lt

l cc ng thng E, D, C, B v A, cc trngthi dng l 9, 8, 7, 6 v 1.

Nh vy, trng thi dng ph thuc vochiu thay i ca T0 nh hnh 4.16b.


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02/10/08140

A B C D E987

6

5

4

321

T0 tng

T

QR QS

1 2 3

4

6

7 89

Tdng

T0ba

H.4.16- Cc trng thi dng khi T0 thay i.a- Cc trng thi dng.b-Cc nhit dng theo chiu thay i ca T0

T rt ra bin php mi phn ng:


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02/10/08141

khi m my un nng nguyn liu vo QS

n qu ng D(v d ng E) t trngthi dng 9

Sau c th gim (hay ngng) un nng i vo trng thi dng 8, 7, 6 (ng vi ccng QS l D, C v B) l cc trng thi dngc nhit v chuyn ho cao, m khng bri vo trng thi dng c nhit v chuyn ho thp( tuy rng vi cc ng QS l

B, C v D ta c th gp cc trng thi dng 2, 3v 4).

c/ n nh ca TBP:


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02/10/08142

-Trng thi dng 5 ( hnh 4.14 v 4.16a )

khng n nh Nu c mt nhiu lon no , nh khi T > T5,

lc QR > QS, h thng c un nng bngnhit phn ng v n T7

Ngc li khi T < T5 th QR < QS, h b ngui iv v T3.

T c th rt ra iu kin n nh ca cctrng thi dng l:

dQS / dT > dQR / dT ( 4 . 6 )

IV.5-c trng nhit ca thit bl i


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02/10/08143

loi ng:

-Do khng c khuy trn nn nhit ca thit b loi ny thay i t imny n im khc trong thit b.

-ng thng c tit din trn phn bnhit trong ng c trc i xng ltrc tm ng, do vy c th din t

hnh nh phn b nhit trong khnggian thit b theo hai thng s l v r.

a/Phn b nhit trong ng phn ng:


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02/10/08144

a/Phn b nhit trong ng phn ng:

Phn ng to nhit khng c trao i nhit quathnh, chy ri hay c lp ht xc tc tnh ( mhnh LT ):

-Nhit ng u theo hng ng knhng ( hng r ).

-Nhit thay i theo hng trc ng( hng l ).

-Ph thuc vo vn tc phn ng ( hot tnhxc tc ) ta c hnh nh nhit nh hnh 4.17

T


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02/10/08145

T

X L

L

21 3

1

0

T0

Tn1 2 3 1-Xc tc hot tnhcao

2-Xc tc hot tnhtrung bnh

3-Xc tc hot tnhthp

H.4.17-S thay i nhit v chuyn ho

trong ng phn ng on nhit vi xc tc chot tnh khc nhau

*Phn ng to nhit c trao i nhit qua thnh:


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02/10/08146

-Nhit thay i theo c hai hng l v r.

-C ch truyn nhit trong lp ht c 3 thnhphn:

+Dn nhit ca lp ht: b v ht xp v din

tip xc nh.+Dn nhit ca lu th: pha kh thng nh.

+Truyn nhit do dng lu th chuyn ngmang nhit: thng ng vai tr ch yu.

-Tm ng c nhit cao nht ( phn ng )


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02/10/08147

thu nhit c hnh nh ngc li ).

-Trn tm ng thng tn ti Tmax , niphn ng xy ra mnh m do nhit cao, c gi l "im nng'' hay "vngphn ng'' v c ch theo di trong

qu trnh lm vic ca thit b cng nh khithit k TBP.

-Mt hnh nh v phn b nhit vi cc

ng ng mc T c th din t hnh4.18


148/256

02/10/08148

T5FV,T0

T0 T1 T2 T3 T4

H.4.18-Phn b nhit trong ng, phn ng tonhit c trao i nhit qua thnh

T5 > T4 > T3 > T2 > T1 > T0

T5 : Tmax

Cn tnh thm thnh phn khuch tn theo hng ngknh vi h s khuch tn ngang l D v a vo pt mt th


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02/10/08149

knh vi h s khuch tn ngang l Drv a vo pt mt thnguyn (2.27).

+Tnh thnh phn kt ngang ( k hiu l A ) cho nguyn tth tch 2r.dr.dl theo hnh 4.19

r dr

-Dr.2rdl.C/r -Dr.2(r+dr)dl.(C+C/r.dr)/rdl

H.4.19-M hnh tnh khuch tn ngang cho nguyn tth tch thit b 2r.dr.dl

C C+C/r.dr

A = Lng ra - Lng vo


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02/10/08150

= -Dr.2(r+dr)dl.(C+C/r.dr)/r

- (-Dr).2rdl. C/r.

A/2dl= -Dr [(r+dr).(C+C/r.dr)/r - r.C/r].

A/2dl= -Dr ( r.2C/r2.dr + C/r.dr + 2C/r2.dr2 ).

S hng 2C/r2.dr2 qu b, b qua v chia tipcho rdr:

A/2r.dr.dl = -Dr ( 2C/r2 + 1/r.C/r ) (4 . 7)

a thnh phn khuch tn ngang A / 2r.dr.dl t (4.7) vo pt(2 27):


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02/10/08151

(2.27):

v.C/ l - Dl.2C/ l2 - D

r(2C/r2 + 1/r.C/r) + R = 0. ( 4 . 8 )

Qu trnh truyn nhit tng t chuyn cht, nn c th vit:

v.CP.T/ l - l.2T/ l2 - r (2T/r2 + 1/r.T/r)+ ( -HR)i . Ri = 0. i= 1, n. ( 4 .

9 )

Trong :+ CP-Nhit dung ring ca hn hp phn ng.

+ l -H s truyn nhit theo hng trc ng.

+ r -H s truyn nhit theo hng ng knh

ng.+(-HR)i-Nhit ca phn ng th i.

Nh vy:


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v.C/l -Chuyn cht vi dng chy c vn tcv.

v.CP.T/l -Truyn nhit do dng cht mangnhit.

- l.2T/l2 -Truyn nhit theo hng trc ng.- r(2T/r2 + 1/r.T/r) -Truyn nhit theo

hng ng knh ng.

(-HR)i.Ri -Tng nhit phn ng chnh v ph.

b/im nng Tmax:


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02/10/08153

-Vi phn ng to nhit, c trao inhit qua thnh, tn ti Tmax trn trc

tm ng nh hnh (4.18).

-V tr v gi tr Tmax -i vi mt qutrnh phn ng, kch thc ng v iukin trao i nhit xc nh-ph thucvo hot tnh xc tc (ngha l phthuc vo vn tc phn ng ) nh hnh4.20

T


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02/10/08154

T0

1 23

Tmax

TmaxTmax

T

L

1-Xc tc mi, HT cao.2-Xc tc gim HT.

3-Xc tc HT thp.

H4.20-ng biu din nhit tm ng phnng theo chiu di l vi xc tc c HT khc nhau.

-Xc nh cc thng s nh hng n Tmax: Dng F i qua nguyn t th tch S dl nhit


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Dng FV i qua nguyn t th tch S.dl, nhit thay i l dT v nng cht phn ng thay

i mt i lng l dC (hnh 4.21)

FV

dl

SP

H.4.21-M hnh cn bng nhit ca dng FVi qua nguyn t th tch S.dl

Ta c pt cn bng nng lng ca dng

F dh + dQ = 0 ( 4 10 )


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- FV. .dh + dQ = 0 ( 4 . 10 ) Trong dh- Thay i entalpi ca dng

dh = CP.dT + HR.dC CP- Nhit dung ring ca dng. HR- Entalpi ca phn ng.

- Khi lng ring ca hn hp phn ng. dQ- Nhit trao i qua thnh ( a vo ly du +, raly du - ).

Gi s phn ng to nhit, lm lnh ngoi thnh ng phn ng:

dQ = K ( T T ) P dl


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dQ = -K.( T - TC ).P.dl

K- H s trao i nhit qua thnh.TC- Nhit ca thnh.

P- Chu vi ca ng phn ng.

Ta c bn knh thu lc ca ng phn ng:

RTL = S/P P = S/RTL.

dQ = -K.(T - TC).S.dl / RTL.

Thay vo pt (4.10)

-FV..(CP.dT + HR.dC) - K.(T - TC).S.dl / RTL = 0. ( 4 . 11 )

Phng trnh cn bng vt cht trong S.dl vi vntc phn ng l R:


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tc phn ng l R:

FV..dC = S.dl.R S.dl = FV..dC / R.

Thay S.dl vo pt (4.11) v chia cho FV.

-CP.dT - HR.dC - K.(T - TC).dC / RTL.R = 0.

( 4 . 12 )

Ti T = Tmax th dT = 0, ta c:

-HR.dC - K.(Tmax- TC).dC /RTL.R = 0. T

Tmax = TC + (-HR).R.RTL / K ( 4 . 13 )

T (4.13): Tmax t l thun vi nhit phn ng (-

HR), vn tc phn ng (R) v ng knh ng


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HR), vn tc phn ng (R) v ng knh ngphn ng (RTL).Vi phn ng c hiu ng nhitv khong nhit thch hp khc nhau cnthit k kch thc ng phn ng ph hp:

-Khi nhit phn ng trong khong 50 n

200 Kj/mol, ng knh ng phn ng c th t20 n 70 mm ph thuc vo nhy ca sthay i chn lc hay ca hot tnh xc tcvo nhit .

-Cn lu rng khi gim ng knh ngca thit b ng chm th s ng s tng mnhv do gi thnh thit b tng vt ln.

T (4.13): T t l th i t h R


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Tmax t l thun vi vn tc phn ng R,

ngha l vi hot tnh xc tc. Trong trnghp cn thit c th s dng xc tc c hottnh thp (hay xc tc c) u ng phn ng gim gi tr ca Tmax .

Tmax t l nghch vi h s trao i nhit Knn cn lu vic s dng tc nhn trao inhit cng nh lm sch b mt trao i nhit m bo K c gi tr ln.

T (4.13): Cui cng Tmax t l thun vi nhit


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Cui cng Tmax t l thun vi nhit

ca cht ti nhit TC. Tuy vy, TBP nhit cht ti nhit TC

qu thp s lm ngui lp xc tc v do

tc phn ng cng nh nng sutthit b thp.

iu lin quan n c im nhynhit ca TBP loi ng nh phn sau.

Chnh t c im ny phi tin hnh gii


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Chnh t c im ny phi tin hnh gii

nhit cho cc phn ng to nhit nhit xp x nhit phn ng.

V d, khi nhit phn ng l 3000C nh

qu trnh oxy ho mtanol thnhfomaldhit trn xc tc oxit Fe-Mo philm lnh bng du nng 260-2700C( hnh 4.22).

V Hi nc


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FV

3000C

Nc mm

Hi nc

Du nng

260-2700C

H.4.22-S gii nhit phn ng

c/ nhy nhit caTBP loi ng:


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y g

- nhy thng s l hin tng xy ra khimt thng s no c thay i tng i nhnhng h thng TBP c s thay i ln, vithng s l nhit ta c nhy nhit.

-Mt v d i vi phn ng bc1, khngthun nghch, to nhit tm thy nhy canhit im nng Tmax i vi cc thng sTC, K/RTL v T on nhit.

Hnh 4.23 din t s thay i nhit Tmaxkhi nhit ban u ca hn hp T0=340 K,cn TC thay i t 300 n 342,5 K

T,K


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10 20 30 tTB,S

T,K

420

380

340

3001

2

3

456

Tmax=440

420430 1-TC=300,0 K2-TC=320,0 K3-TC=335,0 K4-TC=337,5 K

5-TC=340,0 K6-TC=342,5 K

H.4.23- nhy ca Tmax vo TC

Nhn hnh v ta thy mi n khi TC=335 K,nhit trong ng phn ng t thay i theo


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nhit trong ng phn ng t thay i theo

chiu di ng, c th ni lp xc tc b ngui vvn tc phn ng thp, do chuyn ho Xkhng t yu cu.

Khi tng TC thm 2,5 K (TC=337,5 K) ng

biu din nhit ti tm ng c cc i caohn T0 n 80 K (Tmax=420 K).

Khi tng TC ln mt t na (TC=342,5 K) nhit

im nng vt nhit ban u n 100K.

. d/ n nh ca TBP loi ng:


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-Khc vi TBP loi thng c khuy, phn

ng tin hnh trong thit b loi ng lun lunn nh. Do c im ca thit b loi ny mts ri lon no ti mt th tch nh ca ngphn ng khng lan rng ra qui m khp thitb m s b dng chy ko ra khi zn phn

ng, trng thi dng ca h thng quay vtrng thi c.

- TBP loi ng c trao i nhit giasn phm i ra v nguyn liu vo ( hay c traoi nhit gia lp xc tc vi nguyn liu vo )c th c trng thi dng khng n nh

V-TBP XC TC KH-RN ( XC TCRN ). -H phn ng vi xc tc rn rt hay gp trong


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-H phn ng vi xc tc rn rt hay gp trongcng nghip ho hc ni chung v ring i vicng nghip ho du.

-Cc vn c t ra i vi h phnng d th ny l:

B mt tip xc pha.Cc bin php duy tr ch nhit ti u

cho qu trnh: rt quan trng v nh hng ncu to v hnh dng ca thit b.

Trng thi lp xc tc: lp tnh, lp ri,lp cun theo dng kh.

V.1-B mt tip xc pha v hiu sut s dng b mtbn trong ca xc tc:


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Gii php m bo b mt tip xc pha ca

h kh rn l s dng xc tc xp c b mtbn trong rt pht trin, t 1-2 m2/g n gn1000 m2/g.

a/Cc c trng ca xc tc xp:

+B mt ring:

K hiu - SR, n v - m2

/g.

Mt s vt liu xp thng dng:


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Mt s vt liu xp thng dng:

Diatomit: SR = 5 - 10 m2/g.Al2O3 hot tnh,silicagel: SR = 200 - 400 m2/g.Zolit: SR =700 - 800 m2/g.Than hot tnh: SR = 800 -

1000 m2

/g. Vi cu trc ht ca vt liu xp: c ht s

cp v ht th cp, b mt ring l tng bmt cc ht s cp trong 1 gam vt liu.

B mt ring c xc nh bng phngphp hp ph, tnh theo BET.

+ xp:

K hiu - n v - cm3/g


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K hiu , n v cm /g.

L khong trng gia cc ht s cp v th cptrong vt liu xp.

Ph thuc vo s sp xp (s phi tr) ca ccht s cp v th cp, xp c th thay i t

0,1 n 1cm3

/g.Thng s phi tr gim th xp tng v bn c hc ca vt liu gim.

c tnh t khi lng ring tht v khilng ring biu kin theo cng thc:

= 1/bk - 1/tht . ( 5 . 1 )

+Phn b l xp theo kch thc:


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L xp c kch thc khc nhau do ssp xp ca cc ht s cp v th cp,c th din t theo m hnh 5.1, phnchia thnh l xp b ( ng knh di

0,2 nm ), l xp trung (t 0,2 n 50 nm)v l xp ln (trn 50 nm).

Phn b l xp theo kch thc cxc nh bng phng php hp ph,ngng t mao qun v np thu ngndi p sut ( n 250 MPa ).


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L xpln

L xptrung

L xpb Ht xc tc

Ht th cp

Ht s cp

H.5.1-M hnh cu trc xp nhiu dng.Ht xc tc c cha cc tinh th zolit l ht th cp

+H s khuch tn hiu dng:

xp v phn b l xp theo kch thc l nhng yu t quantrng nh hng n qu trnh khuch tn trong xc tc xp


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trng nh hng n qu trnh khuch tn trong xc tc xp.

H s khuch tn hiu dng De trong ht xc tc c th din ttheo cng thc sau:

De = D . / ( 5 . 2 )

Trong D - H s khuch tn trong pha kh.

- xp ca xc tc. - H s ni ln trng thi hnh hc ca l xp:

quanh co, g gh ca b mt l xp, s thay i ng knh cal xp...

Thng t s / c gi tr t 1/10 n 1/20 v h s khuch tnhiu dng c th c xc nh bng thc nghim.

b/Hiu sut s dng b mt bn trongca xc tc :


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ca xc tc : Ngoi h s khuch tn hiu dng De, cn ph thuc vo vn tc phn

ng b mt (tc hot tnh ca xc tc)

, vo kch thc v hnh dng ca htxc tc.

Gn ng vi ht xc tc c hnhdng khc nhau, ta c:

= th / ( 5 . 3 )

Trong . - Modun Thiele


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= Vh/Sh Rs/De ( 5 . 4 ).Vh - Th tch ca ht xc tc.

.Sh - B mt ngoi ca ht xc tc.

.Vh/Sh - c trng cho kch thcht xc tc:

Vi ht hnh tr bn knh R, Vh/Sh = R/2.

Vi ht hnh cu bn knh R, Vh/Sh = R/3.

.RS - Vn tc phn ng b mt.

.Th - Hm tang hyperbolic. S ph thuc vo theo pt (4.3) cdin t hnh 5.2.


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T (5.3) v (5.4), khi kch thc ca htxc tc ln (V /S ln) vn tc phn ng


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xc tc ln (Vh/Sh ln), vn tc phn ng

b mt ln v h s khuch tn hiu dngb th ln, do hiu sut s dng bmt xc tc c gi tr thp. T c trngca hm tang hyperbolic ta c:

-Khi < 1 th th= , do = 1.

-Khi 3 th th= 1 , do = 1/.

Do vy vi tng qu trnh phn ng yu


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Do vy, vi tng qu trnh phn ng yu

cu xc tc c b mt ring, kch thcht v cu trc xp thch hp c tch sSR. cc i. Hot tnh ca mt n vkhi lng xc tc Ag c xc nh

Ag = As . SR . ( 5 . 5 )

As Hot tnh ca mt n v b mt xc

tc

V.2-Lp xc tc: a/Lp tnh:


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a/Lp tnh:

Hay s dng v n gin, cho xc tclu mt hot tnh ( > 6 thng ).

Ht xc tc khng c qu b,

thng kch thc t 3 n 10 mm, trlc ca lu th qua lp khng qu ln, vilng xc tc ln hay FV lndng lu th

chuyn ng theo hng xuyn tm thaycho hng trc ng.

Do kch thc ht xc tc ln nn


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Do kch thc ht xc tc ln nn

hiu sut s dng b mt xc tcthng thp.

TBP thng gp:

-Thit b ng chm, xc tc t trong ng,cht ti nhit i gia cc ng.

-Thit b c lp xc tc on nhit vitrao i nhit trung gian hay cho thmnguyn liu lnh vo iu chnh nhit phn ng.

b/Lp chuyn ng:

Dng cho xc tc mau mt hot tnh v c


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Dng cho xc tc mau mt hot tnh v c

th lin tc ly xc tc lm vic ra em iti sinh v b sung xc tc c hot tnh cao voh thng TBP m khng phi ngng qutrnh.

thc hin lp xc tc chuyn ngthit b c cu to phc tp hn, tuy nhin hot ca lp xc tc hu nh khng thay itrong sut qu trnh lm vic, l mt uim c bn i vi qu trnh lm vic lin tc.Xc tc cn c bn c hc cao chng mimn v v vn trong khi chuyn ng.

Lp xc tc chuyn ng c 3 dng vi nhng c im ring: -Lp ri: + Ht xc tc chuyn ng nh trng lc. + Kch thc ht tng ng vi lp tnh.


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g g p +Cn c c cu phn phi ht xc tc ng u trong

khp tit din ca thit b nh sng phn phi hnh 5.2.

H.5.3-a phn phi ht rn

+H thng TBP xc tc chuyn ng lp ri c th din t hnh5.4.


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TBTS

TBP

TBP TBTS

Kh trKh tr

H.5.3-H thng TBP v TBTS xc tc lp ri

TBTS

TBP

TBP TBTS

Kh trKh tr

H.5.3-H thng TBP v TBTS xc tc lp ri

-Lp si:


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*iu kin to lp si: Dng kh i t di lnngc vi chiu trng lc, khi vn tc dng kh ln lc nng ca dng kh cn bng vitrng lng ca ht rn, ht rn bt u llng, th tch ca lp n rng ra, ta c trngthi nh cht lng.

Tr lc ca dng kh hu nh khng thay ikhi bt u to tng si nh hnh 5.4 .

P


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P Vn tc kh,VLp siL. tnh

Dng kh

Lp htXc tc rn

Vf

H.5.5-nh hng vn tc kh n trng thi ca lp

xc tc rn

*Tnh cht ca lp si:

Nh cht lng: Mt thong nm ngang, chy


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c t g: Mt thong nm ngang, chynh cht lng, tun theo nh lut Acsimt ...

Hin tng to bt:

.Do ht rn chuyn ng to s khc nhau vmt ht trong lp v to thnh cc bt kh.

.Bt nh ni ln nhanh trong lp to nn nhngphn t kh c TGL thp nht.

.Bt i ln ko theo ui l cc ht rn, khi lnn mt lp bt v ra v cc ht rn li chm xung

trong lp huyn ph ko theo cc phn t kh bhp ph trong xc tc, to nn nhng phn t cTGL ln.

Do to bt v chuyn ng hn lon caht xc tc lp si c ch chuyn


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ht xc tc lp si c ch chuyn

ng ca dng gn vi m hnh KLT( hnh 5.5 )

Hu qu TGL ca cht phn ng trong

lp xc tc khng ng u cn ctnh n khi tin hnh cc phn ng cc trng ni tip to sn phm ph nh

crcking xc tc, cc phn ng oxi ho ...


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H.5.6-Quan h gia bt v chuyn ng ca ht rn

H s trao i nhit gia lp si v thnh ln:Vai tr chnh ca qu trnh trao i nhit gia


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lp si vi thnh l do s chuyn ng hnlon ca cc ht rn mang nhit.

Vn tc trao i nhit c th tng hng chcln so vi trng hp ch c mi trng kh.

H s trao i nhit K ph thuc vo vn tcdng ca kh nh hnh 5.6.

Nh tnh cht ny lp si ngy cng c s

dng cho cc qu trnh phn ng xc tc tonhiu nhit.

cK

K


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edb

am

Kmax

Vn tc kh,VVf Vt

H.5.7-nh hng ca vn tc kh n h s trao

i nhit K ca lp si vi thnh.ab-Lp tnh; bcd-Lp si; me-Khng c lp ht

-Lp cun theo:

Khi vn tc dng kh ln, cc ht rn cunth d kh t th h l th T thi


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theo dng kh to thnh lp cun theo. Trng thi

ny c s dng rng ri vn chuyn cc vtliu ri dng ht nn cn gi l lp vn chuyn.

Ph thuc vo c trng ca ht, vn tc khc th gp 2 n 5 ln vn tc ht

Ch dng chy gn vi m hnh LT, thigian tip xc ca pha kh vi xc tc ng u vtng i ngn.

Nh vy, lp cun theo c s dng thay th

cho lp si trong thit b crcking xc tc hin idng ng ng vi xc tc cha zolit c hot tnhcao.

VI-MT S LOI TBP TRONG CNGNGHIP HO DU VI 1-L ng:


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VI.1 L ng:

C b mt trao i nhit ln v nhit cao ( > 300 0C )bng cch t nhin liu lng hay kh.

Thng gp trong cng nghip ho du vi cc chc nngsau:

-TBP cho cc qu trnh pha kh, kh-rn, thu nhit nhit cao ( n 10000C ) nh nhit phn hydrocacbon c mt hi nc( steam cracking ) sn xut etylen v olefin, chuyn ho kh tnhin vi hi nc ( steam refoming ) sn xut hi dro v kh tnghp (synthesis gases, thnh phn gm CO v H2), dehydroclo hodicloetan sn xut vinyl clorua.

-Thit b gia nhit hn hp nguyn liu cho cc h thngchng luyn, cho cc h thng TBP nh crcking xc tc, refoming

xc tc, dehidro ho xc tc...

Cu to ca l ng gm 2 phn:

-Zn i lu: Nhit thp hn, c ch


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Zn i lu: Nhit thp hn, c ch

trao i nhit l do tip xc cng bc gia khchy vi ng trc khi i ra ng khi.

-Zn bc x: Nhit cao ca bung la,ni t nhin liu.C ch trao i nhit ch yul do bc x nhit ca bung la n thnhng.

Nguyn liu i ngc chiu t cui zn ilu n zn bc x, y t nhit cn thitcho phn ng, sau c lm lnh nhanh

trnh phn hu sn phm v thu hi dng hinc p sut cao ( hnh 6.1 ).

Nguyn

Hinc


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g yliu

Nc mm

Nhin liu

Khng kh

Zn bc x Zn iluH.6.1.1-S l ng

L c cng sut nhit khc nhau

c bit l ng cho qu trnh cracking hi


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c bit l ng cho qu trnh cracking hisn xut etylen c cng sut ln v hini, vi yu cu a nhanh nhit nguynliu t zn i lu 450 - 5000C ln nnhit phn ng ( 800 - 9000C ) nn ng

trong zn bc x c t vo gia c bc x t hai pha ( hnh 6.2 ).

Mt s thng s ca l ng:


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-S dn ng: 1 n 100.

-Chiu di 1 dn: 20 n 100 m.

-ng knh ng: 30 n 80 mm. -Thi gian lu trong zn bc x: 1 - 2 s .


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H.6.1.2-L ng hin i sn xut etylen, bc x t hai pha.

H.6.1.3-L


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ng choqu trnhRefominghi nc

trong s cng nghsn xutNH3 t kh

t nhin.

H.6.1.4-L ng


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L ng

trongcngnghsn xut

NH3 tkh tnhin.

VI.2- H thng thit b crcking xc tc: c trng ca phn ng: -Phn ng phc tp: song song ni tip theo m hnh


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Phn ng phc tp: song song, ni tip theo m hnh

sau ( hnh 6.3 )

VR/SLO

LCO Xng VGO/HCO

Cc KhH.6.2.1-S phn ng crcking

.VR(Vacuum Residue ):Cn chn khng

.SLO(Slurry Oil):Du c

.VGO(Vacuum Gazoil):Gazoil chn khng

.HCO(Heavy cycle Oil ):Du tun hon nng

.LCO(Light Cycle Oil ):Du tun hon nh

.Kh:Hydrocacbon t C 1 n C4

Trong xng l sn phm chnh, khv cc l sn phm ph to thnh theo


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phn ng ni tip chuyn ho snphm chnh, do vy m bo hiusut xng cao cn c TGL ng u vngn, chuyn ho X khng qu cao.

-Phn ng tin hnh nhit tngi cao ( 450 - 5000C ) v thu nhit ctmch C-C theo phn ng:

CnH2n+2 CqH2q+2 + CpH2p - Q( n = q + p )

c trng ca xc tc:

Xc tc nhanh chng mt hot


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-Xc tc nhanh chng mt hottnh do cc to thnh che ph b mt,xc tc c ti sinh bng cch t ccb mt vi khng kh 650 - 7200C theophn ng to nhit:

C + O2 (khng kh) COX + Q

Do vy cn tin hnh vi lp xc tc

chuyn ng lin tc ly ra i ti sinhv b sung xc tc.

-Xc tc crcking hin nay cha zolit c hottnh cao c s dng dng ht vi cu vi


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tnh cao c s dng dng ht vi cu vi

ng knh c 0,05 n 0,1 mm.

Vi tnh nng ca xc tc nh vy cng vi ctrng ni tip ca phn ng yu cu TGL ng

u v ngn nn lp cun theo vi ngng c s dng cho qu trnh crcking

Th h c ca TBP crcking xc tc sdng lp tnh lm vic theo chu k, lp ri vlp si.

Do xc tc nhanh mt hot tnh nnlng xc tc tun hon ln v xc tc


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lng xc tc tun hon ln v xc tc

sau ti sinh c nhit cao nn n clm cht mang nhit cho phn ngcrcking ( nh hnh 4.4 )

Qu trnh phn ng trong ng ng lon nhit v t l xc tc/nguyn liuquyt nh nhit phn ng v

chuyn ho X.

T cc c trng ni trn h thng TBP crckingxc tc hin i gm:

Thit b h ki d i l


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Thit b phn ng crcking dng ng ng vi lp xctc cun theo, lm vic theo ch on nhit

Nguyn liu c un nng s b tip xc vi xctc nng, ho hi v phn ng trong lp cun theo.

Do phn ng crcking tng th tch nn cho vn tc lp

cun theo n nh, ng knh ca ng phn ng cm rng dn v pha cui ( theo chuyn ho X ), sau xc tc c tch nhanh khi hn hp sn phm.

Vn tip xc gia xc tc nng vi nguyn liu ducng nh phn ly xc tc khi hn hp kh phn ngc nh hng ln n vic to cc v kh nn c nhiuhng quan tm.

Cn qu trnh ti sinh t cc to nhiu nhitc thc hin trong lp si.

t cc trit hn trong lp si vi s khuy


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t cc trit hn trong lp si vi s khuytrn mnh ca lp ht, nht l vi xu hng sdng nguyn liu cho qu trnh crcking ngycng nng hn, hai tng si ni tip c sdng vi tng u nhit thp hn (630 -6500C) v tng sau nhit cao hn (n

7200C). Vic chia thnh 2 tng ti sinh xc tc nh vy

ngoi vic t cc trit do TGL ca ht trongthit b ng u cn gip cho xc tc trnh tip

xc vi hi nc nhit cao v thnh phncc tng ti sinh sau hu nh khng cn hidro.

Hnh 6.4 trnh by mt dng h thngRCC ca UOP.


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Nguynliu v

pha phntn

Pha phntn

React ngng

Sn phmphn ng

Khng khth cp

Khng kh

th cp

Khng khs cp

Thit b honnguyn xc tc2 giai on

Kh x sau honnguyn

H.6.2.3-H thng FCC ca ABB Lummus Global Inc.(Hydrocarbon Processing, November 2000, trang 107).


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H.6.2.4-H thng FCC ca Kellogg Brown &Root, Inc.( nt , trang 108 )


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H.6.2.5-H thng FCC ca Shell Global SolutionsInternational B.V. ( nt , trang 108 )


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H.6.2.6-H thng FCC ca Stone & Webster Inc., aShaw Group Co/Institut Francais du Ptrole. ( nt ,trang 110 )


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H.6.2.7-H thng FCC ca UOP LLC.( nt , trang 110 )


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H.6.2.8-H thng RCC(Resid Catalytic cracking) ca Stone &Webter Inc., a Shaw Group Co. and Institut Francais du Ptrole.

( nt , trang 139 )


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H.6.2.8-H thng MSCC ca UOP LLC . 1-TB ti sinh; 2-TBP3-ng dn xc tc nng ti sinh; 4-B phn to lp ri; 5-Bphn nh hp ph sn phm khi b mt xc tc. ( nt , trang94 ).


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VI.3-H thng TBP sn xut xng alkyl ho:

Xng alkyl ho c sn xut bng


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phn ng alkyl ho izobutan vi ccolefin C3, C4 vi s c mt ca xc tcl axit mnh, ngha l t cc nguyn liukh chuyn thnh sn phm lng, xng

alkylat c tr s octan cao, khng chabenzen nn c ch hin nay.

c trng ca phn ng v xc tc:

-S phn ng c th trnh by nh sau:

C4H8 + H+ C4H9

+ i-C4H10i B t


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n-C4H10i-C4H9

+C4H8

i-C8H17+ i-C4H10

i-C8H18C4H8

i-C12H26

i-C16

H34

Butenizo-Butan

Buten Tert-Butylcacbenium cation

izo-OctanButen

izo-Butanizo-Octyl cacbenium

cation

H+

H+

-T nguyn liu ban u buten v i-butan tothnh xng vi thnh phn ch yu l i-octanv hp cht ny trong iu kin phn ng vi


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v hp cht ny trong iu kin phn ng vixc tc axit c th tip tc phn ng vi butenthnh hydrocacbon nng C12, C16 ...l sn phmph v c nhit si cao trn 2000C, nmngoi thnh phn ca xng.

hn ch phn ng ph ny v qu trnhpolime ho olefin trong cng nghip dng t si-butan/buten ln (t 10 dn 20) v h thng

TBP gn vi LT , c thi gian lu ngu.

-Qu trnh alkyl ho tin hnh pha lng vixc tc l axit mnh H2SO4, HF v gn ythm axit rn.


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Vi xc tc H2SO4 v HF l h d th lng-lngcn to nh tng m bo b mt tipxc pha, c bit vi H2SO4 l axit c nhtv khi lng ring ln nn TBP cn c

khuy trn mnh. Do vy vi xc tc ny trongcng nghip s dng h thng nhiu thit bkhuy ni tip.

Vi xc tc HF c nht v khi lng ringthp, d to nh tng vi hydrocacbon nndng thit b ng ng c cu khuy tnh ttrong ng ng.

-Phn ng alkyl ho to nhit, khong 90Kj/mol, gii nhit phn ng bng cc gii phpsau:


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sau:

+Vi xc tc H2SO4, nhit phn ng t 0n 100C, bng cch bay hi i-butan ( Tsi =-110C ) trc tip t hn hp phn ng nh

cng ngh ca Kellogg - Exxon ( hnh 6.5 ) haytrao i nhit qua thnh vi alkylat nh cngngh ca Stratco ( hnh 6.6 ).

+Vi xc tc HF, nhit phn ng cao

hn ( 30 n 500

C ) c th lm lnh qua thnhbng nc ( hnh 6.8 v 6.9 ).

H thng TBP:


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-Cng ngh Kellogg-Exxon vi xc tcH2SO4 gm 5 thng c khuy ni tip ttrong thit b dng ng nm ngang dip sut t 0,3 n 0,7 MPa ph thuc

vo thnh phn ca nguyn liu, hi i-butan to thnh do gii nhit phn ng t5 thng c tp trung v i ra nh

thit b dng ng n h thng nn vlm lnh ho lng.

H.6.3.1-H thng TBP sn xut xng alkyl

ho ca Kellog-Exxon


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H.6.3.2-H thng TBP alkyl ho nmngang, lm lnh qua thnh ca Stratco.


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TBP nm ngang ng thi tm thit b lch v pha di trnh lng ng H2SO4, to nh tt hn. Hnh 6.7 trnh by thit

b ny


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H.6.3.4-TBP alkyl ho vi xc tc HF caPhillips.


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H.6.3.5- TBP alkyl ho vi xc tc HF caUOP.


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-Vi xc tc rn-cng ngh mi ang nghincu pht trin, c th a ra TBP ca cng


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ngh Akylene ca UOP hnh 6.10. * Xc tc mau mt hot tnh do polime bm

vo b mt xc tc nn s dng lp xc tcchuyn ng lin tc thay th v ti sinh xc

tc: dng lp cun theo TGL ng u theoyu cu ca phn ng.

* Ti sinh xc tc bng cch dng i-butanbo ho hidro kh ni i gim kh nng

hp ph ca cc hp cht nng khng no lnb mt xc tc v ra sch b mt.

H.6.3.6-TBP alkyl ho vi xc tc rn theo cng ngh Alkylene

ca UOP.


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VI.4-H thng TBP sn xut MTBE(Metyl Tert-ButylEther) Qu trnh tin hnh pha lng, nhit 50 - 1000C, p sut 1,5

MPa


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MPa

T s MeOH/i-B= = 1,1 gim phn ng polime ho olefin.

Xc tc l axit, trong cng nghip dng axit rn l nha cationitc nhm axit mnh -SO3H, dng ht hnh cu ng knh c 0,8- 1mm.

+H thng TBP: Ph thuc vo nguyn liu i-B= :

-Rafinat-1(phn on C4 ca crcking hi sau khi tchButadien) cha 45 - 50% i-Buten.

-Phn on C4 crcking xc tc cha 15 - 20% i-Buten.

-Hn hp sn phm dehidro ho i-Butan cha 40 - 50% i-Buten.

Nhng nguyn liu c hm lng i-B= thp nh trnthng dng TBP on nhit, cn nguyn liu cha i-B= cao hn nh sn phm dehydrat ho TBA, i khi dng


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TBP ng chm lm lnh bng nc - phn ng tin hnh n cng, chuyn ho ht i-

B= thng dng cng ngh CD(Catalytic Distillation)ca hng CD Tech hay cng ngh RWD(Reaction withDistillation) ca UOP.

y l cng ngh phn ng c nhiu u im tin hnhcc qu trnh thun nghch to nhit pha lng c nhit phn ng xp x nhit si ca hn hp ( t chuynho cao, tn dng nhit phn ng tch v tinh ch snphm phn ng, gim s thit b trong cng ngh).

-H thng TBP c trnh by hnh 6.4.

E-6

Hn hp


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E 6

Hn hp C4 i thuhi metanol d

Hn hpi-B=+MeOH

1 2

1-TBP on nhit cTN trung gian

2- Thp CD

MTBE

VI.5-H thng TBP Refoming xc tc:

Qu trnh Re foming xc tc tin hnh


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ng thi cc phn ng ng phnho, ng vng, thm ho H/C trnxc tc lng chc: hidro ho -dehydroho v xc tc axit-base.

Ngoi mc ch tng tr s octan choxng chng trc tip n nay cng nghRe foming xc tc cn sn xut H/C

thm cho Tng hp ho du.

-c trng ca phn ng v xc tc:


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Thu nhit, nhit phn ng 400 -5000C, thng tin hnh lp onnhit xen k vi l ng gia nhit.

B mt xc tc cng b cc tothnh che ph lm gim hot tnh, nhngchm hn nhiu so vi xc tc crcking

Cng ngh c tin hnh vi lp xc tc tnh di p sut hidrokhong 2 n 3 MPa hn ch phn ng to cc, tui th ca

xc tc c th n hng nm ( Hnh 6.5.1 )


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Cng ngh mi tin hnh vi xc tc lp ri, lin tc ti sinh xctc, p sut thp khong 0,5 MPa, t phn ng ph hn v

hot tnh lp xc tc cao v n nh


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H.6.5.2-Hthng TBPrefoming xctc lp ri

ca IFP. 1, 2, 3, 4-

TBP xctc lp ri ;5- h thngti sinh xctc

H.6.5.3-H thng TBP refoming xc tc lp ri ca UOP LLC.


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Dng kh phnng i qua lp xctc theo hngxuyn tm v saukhi i qua mi lp

CCR Stacked Platforming Reactors3 Reactor System

Catalyst In


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xc tc c unnng tr li nnhit phn ng l ng 5

Cc lp xc tc 2,3, 4 chng ln

nhau, nh vy xctc ri t trnxung .

ng i ca khphn ng v xctc trong thit b

nh hnh 6.5.3

Reactor No. 1 Feed

Catalyst Transfer Pipes

Scallops or Outer Screen

Reactor No. 1

Catalyst Reduction Zone

Reactor No. 2 Feed

Reactor No. 1 Effluent

Reactor No. 2

Reactor No. 3

Reactor No. 3 Feed

Reactor No. 2 Effluent

Catalyst OutCatalyst Out

Reactor No. 3 EffluentUOP 3046-108

VI.6-H thng TBP sn xut phnol:

-Phng php ch yu sn xut phnol


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hin nay l i t nguyn liu cumen( trn 90%).

-c trng ca phn ng: C hai giaion phn ng: oxi ho cumen thnhhidroperoxit (H/P) v chuyn ho H/Pthnh phnol c mt xc tc axit c

tin hnh ring bit nn ta xem xt tnggiai on.

S phn ng chnh v ph ca giai on oxi ho:


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C6H5CH(CH3)2 C6H5C(CH3)2OO*

C6H5C(CH3)2OOH C6H5C*(CH3)2O2

C6H5COCH3 + CH3O*

+R*

C6H5C(CH3)2OH + C6H5C(CH3)2O*+RH

C6H5C(CH3)2OH + R*

CH3OH

HCOOH

Axetophenon

Dimetyl phenyl cacbinol

Nh vy, phn ng ph l chuyn ho tipcumen H/P hay gc cumen H/P thnh dimetylphenyl cacbinol, axetophenon v mt s hp


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p y p pcht khc.

hn ch phn ng ph mt mt duy tr chuyn ho thp (X = 20 - 25 %) gim

nng H/P v gc H/P trong hn hp phnng v gim dn nhit phn ng theomc tch t H/P trong sn phm ( ni cchkhc, theo mc

TB Pu HDau

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