thiet bi phan ung trong cong nghiỆp hoa dau pt

7/30/2019 THIET BI PHAN UNG TRONG CONG NGHIP HOA DAU pt

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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 sn phmca mt qu trnh sn xut,do quyt nh nng sut (do vn tc phnng r ) v hiu qu ( chuyn ho X v chn lc S) ca sn xut.

-Vn tc phn ng chuyn ho cht i:

Ri= dN

i/ 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 C i = Ni / V, do pt (1.1) thnh:

Ri = dCi / dt . ( 1 . 2 )

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- chuyn ho ca cht i:

Xi= (C

0- C

1) / C

0= 1 - C

1/C

0. ( 1 . 3 )

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

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 itrong hn hp phn ng .

Cj -tng nng cc sn phmtrong hn hp phn ng

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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 phmc th gm mt s lng ln ccthit b thc hin cc qu trnh chuyn khi v truyn nhit nh

chng luyn, hp th,hp ph, trch ly, un nng, lm lnh,ngng t ...m sinh vin lm quen trong mn hc "Qu trnhv thit b ho hc ".

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I.2-c im :

-a dngDo 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 nhnggii 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 2-3MPa gim th tch TBP, tng cng vn tc phn ng v hs 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 trngthi 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, xc tc v hthng 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 nh hngln 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, un nng, tin

hnh phn ng, lm ngui v tho sn phm, cthc hin trong mt thit b.

Do cc thng snh nng , nhit , p sut...thay i theo thi gian.

V d: tin hnh phn ng trong thit b loi thng c

khuy nng cht phn ng thay i theo thi giannh hnh 1.2.

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

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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 l cht

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

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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 mtthi 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 trungbnhtheo 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 tintheo th t trc saunh chuyn ng capit-tng trong xi lanhV do nng chtphn ng thay i t t,bt u u vo l CA0n u ra l CALnh hnh 1.4

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*M hnh khuyl tng :

-L m hnh

dng chy trongthit b c

khuy trn mnh,cht phn ng ivo c trn ln

ng u ngaytc khc trongthit b, do nng cht

phn ng thay i

t ngt ti uvoca thit bnh hnh 1.5

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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 sut TBPtheo m hnh khuy ltng thp hn y ltng.

Ni mt cch khc, m bo chuyn ho X

nh nhau thit b theo m hnh khuy l tng cnc th tch VR ln hn nhiu so vi m hnh y ltng, c bit khi X yu cu cao.

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* m bo nng sut thit b cao vi m hnhkhuy l tng h thng nhiu thit b khuyni tipc s dng nh hnh 1.6.

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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 b thigian lu ngu hn, nhit cth iuchnh khc nhau tng thit b v cht phn ng

th hai c th cho vo t t theo tng thit b theoyu cu

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3/Theo trng th pha:

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

*H d th: i vi h ny cn ch to b mt tip xc pha ln tng cng vn tc phn ng.-H d th lng - lng: Cn c khuy trn tt, to nh tng cb mt tip xc ln.

-H d th kh - lng: m bo b mt tip xc pha ca h nycn khuy trn hoc si bt hoc dng m rnc b mt ringln.

-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 liu xpc 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 c

hiu 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 xctc 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 trao inhit phn ng vi nguyn liu vo t nhit mphn ng c th tin hnh .

*Ch nhit theo qui hoch:-Thng dng t ch nhit ti u cho qu

trnh phn ng .

-Gp thit b loi ng , thit b c nhiu ngn onnhit v h thng nhiu thit b khuy ni tip .

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Gi s ti thi im t1nng 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 vo cth rt khc nhau. Do vy cn nghin cuphn b thi gian luca chng trong

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 thigian 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)

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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 tvn 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 )

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II.2-Gi tr ca hm phn b TGL trong cc mhnh l tng:

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

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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 raca 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 mira khi thit b nn TGL ca nhng phn t ny

bng . Ngha l TGL ca cht phn ng trong m

hnh 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 TGL ldt, nh vy c th vit:

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

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Nh vy, do khuy TGL ca cc phn tdngvo 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 ni tip).

t chuyn ho nh nhau so vi mhnh LT thit b loi ny cn c th tchln 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 nv 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 tnh cnbng 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).

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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) cah thng tng t nh m hnh LT:TGL ca cc phn t ca dng ng u vbng 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 tinsn 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 ta ctrng thi trung gian gia hai m hnh

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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* nghin cu hin tng trn ln ca dng chy trongng gy nh hng n TGL trong thit b v chuynho, 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 Drc 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 caht trong ng.

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a/ng rng, chy dng:

-Gradien vn tc dng theo tit din ngang lnguyn nhn ch yugy 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 thuc vokhong cch r n tm ng l vr(hnh 2.6)

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

vr= ( 2FV/.r02 )[ 1 - ( r / r0 )2 ] (2.21)Trong r0l 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 lt tm(r=0) = tTB / 2

Thay r = r0c 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 lr+dr (hnh vnh khn 2r.dr ) c vn tc nhnhau l vr, t nh ngha ca hm F(t) ta c:

dF(r) = vr . 2r.dr / FV ( 2.23)Thay vrt 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 phn pt(2.22) ta rt ra:

rdr = r02 . tTB.dt / 4 t

2

Thay vo pt (2.23) ,c:

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

T xc nh F(t) cho trng hp ng rngchy 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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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 nh ngu .

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

Vrt nh,c th b qua,ta c ch

dng chy ta m hnh LT .

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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 c haihng trc ng v ng knh ng

c bit theo hng ng knh

theo m hnh nh hnh 2.9.

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Theo m hnh ny khi dng chy qua mtlpht 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 /dht10 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 khng nguv do nh hng n chuyn ho X .

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

-Dng LTvi vn tc di l v .

-Dng khuch tn dcvi h s khuch tn theohng trc l Dlv 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 cht choVR = S .l:

A vo = A ra + A phn ng + A tch t.(2.26)

H.2.10-M hnh ng phn ng c khuch tndc .

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v (CA/l) - Dl (2CA/l

2 ) + 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/L

2).2CA/z2

T pt (2.27) thnh:

(v/L).CA/z - (Dl/L2 ).2CA/z

2 + kCAn = 0. ( 2 . 28 )

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

(Dl/L.v).2C

A/z2 - C

A/z - k.t

TB.C

A

n = 0.(2.29)

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(Dl/L.v) . 2CA/z

2 - CA/z - k . tTB . CAn = 0. ( 2 . 29 )

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

- ti u ra z = 1 , CA = CAL

c CALv t tnh chuyn ho XA, . .XAph 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 =

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H.2.11-Biu din VRthc/VRLTph thuc vo Pel, k.tTBv X caphn ng bc 1.

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H.2.12-Biu din VRthc/VRLT vo Pel, k.tTB.C0v X caphn ng bc 2

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

Hnh sau l kt qu thc nghim ca mts tc giv s ph thuc ca Dl/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.

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H 2 13-S liu thc nghim s ph thuc Dl/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 PHN B TGL THITB 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 theothi 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 dng hnh 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 .

C ch th(z=0) { ( 2 . 31 )

= C0 khi t > 0 .

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

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X l h F(t) Gi h th


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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 ti thiim ny l FV.Ct. T nh ngha ca hm F(t)

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

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

-Tn hiu cho ch th vo dngxung:

= 0 khi t 0 .

Cch th (z=0)=C0 khi 0


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Xc lp hm E(t):

Gi s ton b lng ch th cho vo thit

b dng xung vung l q0, ti thi im tnng ch th dng ra l Ct, trongkhonh khc dt th tch dng ra l FV.dtv lng ch th i ra l Ct.FV.dt, chnh

l 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)

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Bi tp 2.4 - Cho tn hiu vo dng xung, nng ch thra 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 l gitr ca tch phn, cn din tch cc ch nht l gi tr catng.

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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 ng nht vTGLv nh vy, nng ca cht phn ng

dng ra ( t tnh chuyn ho X = 1 - Cra/C0) l nng trung bnh ca cc phn t ca dngvi TGL trong thit b khc nhau . Do c thvit:

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

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Trong :C nguyn tc 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 ) .

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

trong TBP c hm phn b TGL E(t) nh bitp 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 )

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Lp bng tnh phng trnh (2.39):

t , ph E(t) k.t e-kt e-kt . E(t) .t

5 0,03 1,53 0,2154 0,0323 100,05 3,07 0,0464 0,0116 15

0,05 4,60 0,0100 0,0025 20

0,04 6,14 0,0021 0,0004

25 0,02 7,68 0,0005 0,0000530 0,01 9,21 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 2.5trong thit b phn ng c TGL trung bnh nh bi tp

2.4 nhng dng chy theo ch LT.Li gii:-Thi gian lu trung bnh ca thit b bi tp 2.4 cxc 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 bng tTB.Vi phn ng bc 1 t pt dCA/dt=kCA, c:

CA / CA0 = e-kt

TB = e-0,3.15

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

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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 thitb lm vic lin tc phn b t 0 n .

- Khuy tng cng h s trao i nhit gia mi

trng 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 tp haylm vic nhit v p sut caoc th ly t 0,15 n 0,20.

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Thit b lm vic gin on

- H s y . Thng vi thit b phn ng c khuy ly gitr t 0,75 n 0,80 ; trng hp mi trng phn ng to btc 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 un nhit phn ng + t lm ngui sn phm + t tho snphm + 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.

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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 b th i( hnh 3.1 ):

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

( 3 . 3 )

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Thit b lm vic lin tc

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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 i ra.

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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trng thi dng khng c tch t ta c:


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trng thi dng khng c tch t, ta c:

C i-1.F V(i-1) = Ci.F Vi + Ri.VRi hay

C i-1 = C i . F Vi/F V(i-1) + Ri . VRi /F V(i-1) .

pha lng coi nh hn hp phn ng khng thay i thtch, ngha l lu lng dng khng i: F V(i-1) = F Vi = FV, c:

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

Trong tTBil TGL trung bnh thit b th i, tTBi = VRi/FV. Phng trnh (3.4) c gi lpt c trngca TBP loi

thng c khuy.

T pt ny c th tnh chuyn ho X v th tch thit b VRi.

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*Vi phn ng chuyn ho cht Asn 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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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 2.5 v2.6 thy rng cng mt iu kin , thit b

theo ch KLT c chuyn ho thp nht.

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Phn ng bc 0:

i = 1 C0 - C1 = k1 . tTB1

i = 2 C1 - C2 = k2 . tTB2

i = 3 C2 - C3 = k3 . t TB3

... ...... 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 )

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Phn ng bc 2:


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Gii pt (3.4)''' c:

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

i = 1 C1= (- 1 ) / 2k1.tTB1

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

C2 = [-1 ]/2k2.tTB2 .

( 3 . 7 )

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*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 hthng 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 hthng n thit b ni tip theo cng thc (3.6) thnh:

Xn1

= 1 - Cn/C0 = 1 - ( 1 + k . tTB )

-n

. ( 3 . 6 )'

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b/Ph h d h h


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b/Phng php dng hnh:

Phng trnh (3.4) c th vit li dng:

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

V trica 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 l nghim

ca phng trnh (3.8).

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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 C0vi h s gc l -1/tTB1giao im ca ng thng ny vi ngcong R c to ng vi C1 v R1, l nng cht phn ng v vn tc phn ng

thit b 1.

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Phng php dng hnh


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Phng php dng hnh

T C1trn trc honh v ng thng vi h s

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

Bi ton c gii, ta tm c s thit b nca 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 dinRi = ki. f (C) tng ng .

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Phng php dng hnh


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Li gii:


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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 bng nnnng 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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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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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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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/m3v RA1 = 8,9 kmol/m3.h

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T CA1tip 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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Nh vy CA4

< CA yu cu

, ta c s thit btrong h thng n = 4 .

Cng cn lu rng th tch cc thit btrong 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?)

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IV/C TRNG NHIT TRONG THIT B PHN


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IV/C TRNG NHIT TRONG THIT B PHNNG

Phn ng ho hc lun c hiu ng nhit,thng kh ln c th thay i nhit ca qu trnh.

Nhit l thng s cng tnh, nh hngmnh n vn tc phn ng. Do khngch nhit phn ng lin quan n lm vican ton ca thit b v qu trnh sn xut.

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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 khitho 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.

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IV.1-Ch nhit ti u:


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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 cng nh chuyn ho X ca qu trnh phn ng.

Ch nhit ti u ph thuc vo c

trng nhit ng, ng hc ca phn ngcng nh tnh nng ca xc tc

Cho nn ch nhit ti u a dng, tu

tng trng hp c th.

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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/RT

2 - k20.CA0.XA.e-E2/RT.E2/RT

2 = 0 .

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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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Nh h hit ti t h h


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

chuyn ho XAnh cng thc (4.2) Lc u khi XAcn thp nn tin hnh phn

ng nhit cao m bo vn tc ln,sau khi XAtng ln cn gim dn nht .

Trong thc tin sn xut, phn ng c c trngny ( nh SO2 + 1/2O2 SO3+ q ) c thchin trong thit b trong TBP c nhiu ngn xc

tc on nhit c trao i nhit trung giangia cc ngn nh hnh 4.1 .

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IV 2 C ii h d h


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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 ngoi vng xontrong thit b.

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

pha kh to nhiu nhit.

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

do b ph cc b mt nn phi dng lpxc tc chuyn ng v tun hon giathit b phn ng v thit b ti sinh bngphn ng t cc to nhit nh hnh 4.4

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d/iu chnh nhit


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d/iu chnh nhit ban u T0i vi

cc qu trnh onnhit. e/Bay hi cu t c

nhit si xp x

nhit phn ng vih thng lm lnhngng t v cho chtlng quay v TBP (

hnh 4.6 ).

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IV 3 t h t hit


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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 T0 nnhit lm vic ca thit b.

C kh nng trao i nhit gia sn phm haykhi phn ng vi nguyn liui vo. iu kin

ny ph thuc vo c trng v kt cu ca thitb phn ng.

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-Cc loi thit b c kh nng t nhit:

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

ln vi khi phn ng nng ctrong thit b, nu phn ng to nhit ln th qu trnh tin hnh trong loithit b ny c kh nng t nhit sau

khi c "mi" (hnh 4.8a ).

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


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b/Thit b phn ng dng ngn la: Thit b dngny thng c s dng tin hnh cc phnng to nhit cao nh qu trnh clo ho ( t

hydro trong mi trng clo sn xut HCl ) hayqu trnh oxy ho mt phn hydrocacbon snxut axtylen.

Trong khng gian cht hp ca ngn la nhit

thay i t T0rt thp n nhit cao cth t bc chy, ngha l ta c gradien v nhit, v nng , v th ho hc rt ln(c tht hng vn n v ), nh qu trnh truynnht, chuyn cht v hot ho xy ra mnhm trn mt ca ngn la n c th t nhit(hnh 4.8b ).

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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 T1theo thi gian lm vic ca xctc bng cch iu chnh thng s .

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- qu trnh c th tin hnh t nhit, khi

m my cn phi un nng hn hp mtanol -khng kh n 300 - 4000C.Sau , khi lp xctc l cc tinh th bc t trn li ng nngn nhit phn ng th dng un nng, ktthc qu trnh "mi " phn ng.

-TBP cn c mng phng n v thit b lmlnh ng chm t st ngay sau lp xc tc lm lnh nhanh hn hp sn phm t nhit phn ng xung 200 - 3000C, gim phn hufomalehit to thnh nhit cao theo phn

ng: CH2O CO + H2

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


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*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 QStheo T, nghim ca pt l giaoim ca hai ng biu din.

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+Kho st QR

:

Gi h b 1 h h ht A t hit kh


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Gi s c phn ng bc 1 chuyn ho cht A, to nhit, khngthun nghch, tin hnh trong TBP loi thng c khuy th tchV

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 / ( eE/RT + k0.tTB) . Do :

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

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ng biu din ca QRph thuc vo T theopt (4 4) c dng ch S ngoi ra khi cc thng


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pt (4.4) c dng ch S, ngoi ra khi cc thngs khc nh VR, FVthay i ta c mt h

ng cong S trong mt phng vi trc to QR - T nh hnh 4.11.

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+Kho st Q :


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+Kho st QS:

QSc hai thnh phn:

- Trao i nhit vi nguyn liu vo, nhit nguyn liu t T 0 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.

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ng biu din ca QStheo T l ng


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g QS gthng, khi F thay i ta c cc ng thngvi h s gc thay i ( hnh 4.12a ), khi T0

thay i c cc ng thng song song(hnh 4.12b).

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Nh vy, c th c 3 giao imgia


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y, g gng QRdng ch S v ng thng

QS, ngha l c 3 trng thi dng ca hthng( hnh 4.13 ).Trng thi dng 1 cT1thp v QR1,QS1u thp, phn nghu nh khng xy ra. Trng thi dng 3 nhit cao T3do c chuyn hocao , l iu ta mong i. Trng thi dng2 khng n nh, s c kho st mc

3/.

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b/Trng thi dng ph thuc vo chiu


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g g p thay i ca cc thng s v bin php

mi phn ng: a/Gi s FVthay i, cc thng s

khc c nh:

QRl cc ng cong A, B, C, D v E,QScoi nh khng i,c cc trng thidng t 1 n 9, ng vi nhit dng

t T1n T9nh hnh 4.14.

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T rt ra cch mi phn ng:

Khi m my ban u dng lu lngdng FVthp phn ng tin hnh trng thi dng c chuyn ho v nhit

dng cao ( T9 ), sau tng dn FVn nng sut thit k( nhng khng thtng qu ng D ).

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b/Gi s T0thay i, cc thng s khc c nh:


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QRkhng i, QSl cc ng thng song

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

- Khi T0tng t thp ln, QSln lt l cc ngthng A, B, C, D v E, cc trng thi dng l 1, 2,3, 4 v 9.

-Ngc li, khi T0t cao gim xung, QSln ltl cc ng thng E, D, C, B v A, cc trng thidng l 9, 8, 7, 6 v 1.

Nh vy, trng thi dng ph thuc vo

chiu thay i ca T0nh hnh 4.16b.

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T rt ra bin php mi phn ng:


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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 ivo trng thi dng 8, 7, 6 (ng vi cc ng

QSl D, C v B) l cc trng thi dng c nhit v chuyn ho cao, m khng b ri votrng thi dng c nhit v chuyn hothp( tuy rng vi cc ng QS l B, C v D tac th gp cc trng thi dng 2, 3 v 4).

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IV.5-c trng nhit ca thit b loi


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ng:

-Do khng c khuy trn nn nhit ca thit b loi ny thay it im nyn 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 khng gianthit b theo hai thng s l v r.

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a/Phn b nhit trong ng phn ng:


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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 knh ng

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

).

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

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


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-Nhit thay itheo c hai hng l v r.

-C ch truyn nhit trong lp ht c 3 thnh phn:

+Dn nhit ca lp ht: b v ht xp v dintip xc nh.

+Dn nhit ca lu th: pha kh thng nh.+Truyn nhit do dng lu th chuyn ng

mang nhit: thng ng vai tr ch yu.

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Cn tnh thm thnh phn khuch tn theo hng ng knhvi h s khuch tn ngang l Dr v a vo pt mt th nguyn


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

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

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A = Lng ra - Lng vo


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= -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.

2

C/r

2

.dr + C/r.dr +

2

C/r

2

.dr

2

).S hng 2C/r2.dr2qu b, b qua v chia tipcho rdr:

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

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


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(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.

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Nh vy:


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

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 hngng knh ng.

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

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b/im nng Tmax:


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-Vi phn ng to nhit, c trao inhit qua thnh, tn ti Tmaxtrn trctm ng nh hnh (4.18).

-V tr v gi tr Tmax -i vi mt qu

trnh phn ng, kch thc ng v iukin trao i nhit xc nh-ph thuc vohot tnh xc tc(ngha l ph thuc vovn tc phn ng ) nh hnh 4.20

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-Xc nh cc thng s nh hng n Tmax: Dng FV 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)

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

dQ = -K ( T - TC ) 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 )

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Phng trnh cn bng vt cht trong S.dl vi vn tcphn ng l R:


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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 )

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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 nhit

v khong nhit thch hp khc nhau cn thitk kch thc ng phn ng ph hp:

-Khi nhit phn ng trong khong 50 n200 Kj/mol, ng knh ng phn ng c th t

20 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 ng

ca thit b ng chm th s ng s tng mnh vdo gi thnh thit b tng vt ln.

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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 trng hpcn thit c th s dng xc tc c hot tnh thp(hay xc tc c) u ng phn ng gim gitr ca Tmax .

Tmax t l nghch vi h s trao i nhit K nncn lu vic s dng tc nhn trao i nhitcng nh lm sch b mt trao i nhit mbo K c gi tr ln.

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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 phi lmlnh bng du nng 260-2700C (hnh 4.22).

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c/ nhy nhitcaTBP loi ng:


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y g

- nhy thng sl 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, khng

thun nghch, to nhit tm thy nhy canhit im nng Tmax i vi cc thng sTC, K/RTL vT on nhit.

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

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


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trong ng phn ng t thay i theo chiu ding, c th ni lp xc tc b ngui v vn tcphn ng thp, do chuyn ho X khngt yu cu.

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

din nhit ti tm ng c cc i cao hn T0n 80 K (Tmax=420 K).

Khi tng TCln mt t na (TC=342,5 K) nhit im nng vt nhit ban u n 100 K.

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. d/ n nhca TBP loi ng:

Khc vi TBP loi thng c khuy phn


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-Khc vi TBP loi thng c khuy, phn

ng tin hnh trong thit b loi ng lun lun nnh. Do c im ca thit b loi ny mt s rilon no ti mt th tch nh ca ng phnng khng lan rng ra qui m khp thit b m sb dng chy ko ra khi zn phn ng, trngthi dng ca h thng quay v trng thi c.

- TBP loi ng c trao i nhit gia snphm i ra v nguyn liu vo ( hay c trao inhit gia lp xc tc vi nguyn liu vo ) cth c trng thi dng khng n nh

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V-TBP XC TC KH-RN ( XC TC RN ).

-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 ucho 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, lpcun theo dng kh.

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V.1-B mt tip xc pha v hiu sut s dng b mt bntrong 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 mt bntrong rt pht trin, t 1-2 m2/g n gn 1000m2/g.

a/Cc c trng ca xc tc xp:+B mt ring:

K hiu - SR, n v - m2/g.

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Mt s vt liu xp thng dng:


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Mt s vt liu xp thng dng:

Diatomit: SR = 5 - 10 m2/g.Al2O3hot 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 b mt ccht s cp trong 1 gam vt liu.

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

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+ xp:

K hiu - , n v - cm3/g.


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u , c /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 cc hts cp v th cp, xp c th thay i t 0,1 n1cm3/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 khi lngring biu kin theo cng thc:

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

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+Phn b l xp theo kch thc:

L k h th kh h d


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L xp c kch thc khc nhau do s

sp xp ca cc ht s cp v th cp, cth din t theo m hnh 5.1, phn chiathnh 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 c xcnh bng phng php hp ph, ngng t

mao qun v np thu ngn di p sut (n 250 MPa ).

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+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 ca lxp...

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.

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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 ngb mt (tc hot tnh ca xc tc) , vokch thc v hnh dng ca ht xc

tc.

Gn ng vi ht xc tc c hnh dngkhc nhau, ta c:

= th / ( 5 . 3 )

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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 thc ht

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.

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


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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 thc htv cu trc xp thch hp c tch s SR.cc i. Hot tnh ca mt n v khi lngxc tc Ag c xc nh

Ag = As . SR . ( 5 . 5 )

As Hot tnh ca mt n v b mt xctc

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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 thchuyn ng theo hng xuyn tm thay cho

hng trc ng.

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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 cth lin tc ly xc tc lm vic ra em i tisinh v b sung xc tc c hot tnh cao vo hthng TBP m khng phi ngng qu trnh.

thc hin lp xc tc chuyn ng thitb c cu to phc tp hn, tuy nhin hot

ca lp xc tc hu nh khng thay i trongsut qu trnh lm vic, l mt u im cbn i vi qu trnh lm vic lin tc. Xctc cn c bn c hc caochng mi mn vv vn trong khi chuyn ng.

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+H thng TBP xc tc chuyn ng lp ri c th din t hnh 5.4.


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-Lp si:

*i ki t l i D kh i t d i l


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*iu kin to lp si: Dng kh i t di ln

ngc vi chiu trng lc, khi vn tc dng kh ln lc nng ca dng kh cn bng vi trnglng ca ht rn, ht rn bt u l lng, thtch ca lp n rng ra, ta c trng thi nh cht

lng. Tr lc ca dng kh hu nh khng thay i khi

bt u to tng si nh hnh 5.4 .

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*Tnh cht ca lp si:

Nh cht lng: Mt thong nm ngang, chy


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g g g g ynh 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 xungtrong lp huyn ph ko theo cc phn t kh b hp

ph trong xc tc, to nn nhng phn t c TGLln.

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


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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 ucn ctnh n khi tin hnh cc phn ng cc trng ni tip to sn phm ph nhcrcking xc tc, cc phn ng oxi ho ...

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


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si vi thnh l do s chuyn ng hn lonca cc ht rn mang nhit.

Vn tc trao i nhit c th tng hng chc lnso 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 sdng cho cc qu trnh phn ng xc tc tonhiu nhit.

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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 ( > 3000C ) bngcch 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 t nhinvi hi nc ( steam refoming ) sn xut hi dro v kh tng hp(synthesis gases, thnh phn gm CO v H2), dehydroclo ho dicloetan sn xut vinyl clorua.

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

xc tc, dehidro ho xc tc...

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Cu to ca l ng gm 2 phn:

-Zn i lu: Nhit thp hn, c ch traoi hi l d i b i kh h


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i nhit l do tip xc cng bc gia kh chyvi 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 thnh ng.

Nguyn liu i ngc chiu t cui zn i lun zn bc x, y t nhit cn thit chophn ng, sau c lm lnh nhanh trnhphn hu sn phm v thu hi dng hi ncp sut cao ( hnh 6.1 ).

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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 hi

sn xut etylen c cng sut ln v hin i,vi yu cu a nhanh nhit nguyn liut zn i lu 450 - 5000C ln n nhit phn ng ( 800 - 9000C ) nn ng trong zn

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

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H.6.1.3-L


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ng cho qutrnhRefoming

hi nctrong s cng nghsn xutNH3t kht nhin.

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H.6.1.4-L ng


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L ng

trongcng

nghsn xut

NH3 tkh tnhin.

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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 )

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c trng ca xc tc:

-Xc tc nhanh chng mt hot tnh


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Xc tc nhanh chng mt hot tnh

do cc to thnh che ph b mt, xc tcc ti sinh bng cch t cc b mtvi khng kh 650 - 7200C theo phnng to nhit:

C + O2 (khng kh) COX + Q

Do vy cn tin hnh vi lp xc tcchuyn ng lin tc ly ra i ti sinh

v b sung xc tc.

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-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 ngu v ngn nn lp cun theo vi ng ngc s dng cho qu trnh crcking

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

si.

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


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g

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.

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T cc c trng ni trn h thng TBP crcking xctc hin i gm:

Thit b phn ng crcking dng ng ng vi lp xc tc


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Thit b phn ng crcking dng ng ng vi lp xc tccun theo, lm vic theo ch on nhit

Nguyn liu c un nng s b tip xc vi xc tcnng, 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 ng c

nh hng ln n vic to cc v kh nn c nhiu hngquan tm.

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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 s dngvi tng u nhit thp hn (630 - 6500C) vtng sau nhit cao hn (n 7200C).

Vic chia thnh 2 tng ti sinh xc tc nh vyngoi vic t cc trit do TGL ca ht trongthit b ng u cn gip cho xc tc trnh tipxc vi hi nc nhit cao v thnh phn cc tng ti sinh sau hu nh khng cn hidro.

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Hnh 6.4 trnh by mt dng h thngRCC ca UOP.


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Nguyn liuvpha phn tn

Pha phn tn

React ngng

Sn phm phnng

Khng kh thcp

Khng kh thcp

Khng kh scp

Thit b honnguyn xc tc 2 giai

on

Kh x sau honnguyn

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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 Solutions International B.V. ( nt , trang 108 )


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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 , trang 94).


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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 tc laxit mnh, ngha l t cc nguyn liu khchuyn thnh sn phm lng, xng alkylat

c tr s octan cao, khng cha benzennn c ch hin nay.

c trng ca phn ng v xc tc:

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-S phn ng c th trnh by nh sau:


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


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hp cht ny trong iu kin phn ng vi xctc axit c th tip tc phn ng vi buten thnhhydrocacbon nng C12, C16...l sn phm ph vc nhit si cao trn 2000C, nm ngoi thnhphn ca xng.

hn ch phn ng ph ny v qu trnhpolime ho olefin trong cng nghip dng t s i-butan/buten ln (t 10 dn 20) v h thngTBP gn vi LT , c thi gian lu ngu.

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


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+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 cngngh ca Kellogg - Exxon ( hnh 6.5 ) hay traoi nhit qua thnh vi alkylat nh cng ngh

ca Stratco ( hnh 6.6 ). +Vi xc tc HF, nhit phn ng cao hn

( 30 n 500C ) c th lm lnh qua thnh bngnc ( hnh 6.8 v 6.9 ).

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


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


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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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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 cationit cnhm 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 tch Butadien)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.

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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 cho xngchng trc tip n nay cng ngh Refoming xc tc cn sn xut H/C thm

cho Tng hp ho du.

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-c trng ca phn ng v xc tc:

Thu nhit, nhit phn ng 400 -


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p g5000C, thng tin hnh lp onnhit xen k vi l ng gia nhit.

B mt xc tc cng b cc to thnh

che ph lm gim hot tnh, nhng chmhn nhiu so vi xc tc crcking

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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 xc tclp ri ; 5- hthng ti sinhxc tc

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Dng kh phn ngi qua lp xc tctheo hng xuyntm v sau khi iqua mi lp xc tc


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c un nng trli n nhit phn ng l ng5

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

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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 ( trn90%).

-c trng ca phn ng: C hai giai on

phn ng: oxi ho cumen thnh hidroperoxit(H/P) v chuyn ho H/P thnh phnol cmt xc tc axit c tin hnh ring bitnn ta xem xt tng giai on.

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


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H.6.6.2-H thng TBP phn hu H/P cumen thnh phnol v axton


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H.6.6.3-H TBP phn hu H/P theo m hnh dng tun hon.


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7/H thng TBP tng hp 1,2 dicloetan (1,2 DCE):

Phn ng:

CH CH Cl CH Cl CH Cl


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CH2=CH2 + Cl2 CH2Cl - CH2Cl

Kh Kh Lng( Ts = 850C )H = - 220 Kj/mol

Thc t l phn ng kh-lng: sc etylen v clovo 1,2DCE lng ho tan FeCl30,5% vi vai trxc tc. Phn ng xy ra trong pha lng vi vntc cao ngay nhit thng.

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Nh vy, qu trnh c c trng d th kh-


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lng, to nhiu nhit, tin hnh nhit thp c nhit si ca pha lng.

nhit hi cao ( 80 - 1000C ) phn ngph clo ho tip thnh tricloetan c ctng cng, v vy cng ngh c tin hnhclo ho nhit thp v gii nhit phnng bng nc.

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n nay phn ng ph theo c ch chui gc b hn ch bng cch dng xc tc FeCl3 vclo c cha oxi, nn c th tin hnh nhit


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cao hn v s dng nhit phn ng ( kh ln) tch sn phm dng hikhng b ln vixc tc.

Cng ngh tin b nht dng TBP dngCD, kt hp vi thp chng, tn dng nhit phnng tinh ch DCE khng ch t thit b clo hom t cc b phn khc ca h thng lin hp,v d t cng on oxiclo ho hay cng onnhit phn DCE v tch VC.

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H.6.7.H TBP tng hp 1, 2 DCE nhit cao, tn dng nhitphn ng tinh ch sn phm.


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8/TBP sn xut Fomaldehit:


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-Fomaldehit c sn xut t mtanolbng qu trnh oxi-dehidro ho trn xctc bc v oxi ho trn xc tc oxit Fe-Mo.

-y l cc qu trnh pha hi nn cntrnh to hn hp n metanol - khng khc thnh phn t 6 n 40% th tch

metanol.

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a/Cng ngh xc tc bc:

Nhit phn ng: 650 - 7200C.

Lm vic trn gii hn n, do thiu oxi cho phn ng oxi


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ho (1) nn c thm phn ng chnh dhidro ho mtanol

CH3OH + 1/2O2 CH2O + H2O ( 1 ) H = - 156 Kj/mol

CH3OH CH2O + H2 ( 2 ) H = + 85 Kj/mol

Tng nhit phn ng ca qu trnh l to nhit, qu trnhnm phm vi khuch tn ngoi, on nhit v t nhit,iu chnh nhit phn ng bng cch pha thm hi ncvo hn hp phn ng.

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TBP nh hnh 4.10


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b/Cng ngh xc tc oxit Fe-Mo:

Nhit phn ng: 290 - 3100C.


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Lm vic di gii hn n ( hn hpkhong 6% th tch mtanol ), d oxi nnch c phn ng oxi ho ( 1 ) to nhiu

nhit l phn ng chnh. Xc tc lp tnh, TBP dng ng chm (

hnh 4.22 ), ng knh ng tng i

nh, c 20 - 25 mm.

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Kt qu ti u ho TBP ca vin xc tc Nvxibic cho thy rnghiu sut phn ng v kinh t hn nu kt hp thit b ng chm vi

lp xc tc on nhit nh hnh 6.8.


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Kt qu nghin cu cho thy, trn xc tc nyngoi phn ng chnh ( 1 ) vi nng lng hotho E1= 40,67 Kj/molcn c phn ng ph oxi

ho fomaldehit thnh CO v H2O i l


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ho fomaldehit thnh CO v H2O vi nng lnghot ho thp hn:

CH2O + 1/2O2 CO + H2O (3)

H=-215 Kj/mol E3 = 28,22 Kj/mol

Do vy, khi nhit xung thp tc tngi ca phn ng ph s tng ln, hin tng nyb hn ch bng lp xc tc on nhit cuiTBP.

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Qua mt s loi TBP ca cc cng ngh ho du thy rng,TBP c s dng trong cng nghip ca mt s hng c thkhc nhau do nhng gii php k thut m h nghin cu v pdng

Nh l l t ht t h i


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Nhng v nguyn l l xut pht t nhng c im v nghc v nhit ng ca phn ng, t nhng c trng vhot tnh, nhit lm vic v nhy ca cc tnh nngca xc tc vo nhit , thi gian sng ca xc tc.

Nhng ci tin, hon thin v TBP l kt qu ca nhng

thnh tu trong nhiu lnh vc khoa hc v cng ngh, trong c xc tc, ng hc, ho hc v l thuyt v TBP, m hnhho, ti u ho .

thiet bi phan ung trong cong nghiỆp hoa dau pt

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