Đồ án chưng cất tháp đĩa rượu etylic - nước

0

LI M AU

Trong cng nghip ha cht v thc phm cng nh nhiu qu trnh

sn xut u da trn c s s tip xc gia cc pha v s di chuyn vt

cht t pha ny sang pha khc. Cc qu trnh vt cht di chuyn t pha

ny sang pha khc khi c s tip xc gia cc pha gi l qu trnh truyn

cht. Vic nghin cu cc qu trnh truyn cht cng nh quy trnh cng

ngh v thit b ca cc qu trnh l cng vic cn thit ca mt k s

ngnh cng ngh ha hc. Chnh v vy, cc k s tng lai ca lp Lin

Thng HCNH3 K III trng i hc Cng nghip H Ni chng em,

c cc thy c trong khoa giao cho v hng dn lm n mn hc

Qu trnh & Thit b.

n mn Qu trnh & Thit b l mn hc quan trng ca cc sinh

vin ngnh cng ngh ho. Qua qu trnh lm n mn hc, gip em

bc u c vn dng cc kin thc cc thy c dy vo gii quyt

cc vn k thut thc t m cc k s tng lai cn bit.

Qua y, em xin gi li cm n chn thnh ti thy gio hng dn

cng cc bn sinh vin gip em hon thnh n ny. Do thi gian c

hn v s non nt v kinh nghim ca bn thn nn khng th khng cn

thiu st em rt mong c s gp ca cc thy, cc c em c th rt

ra c nhiu kinh nghim qu bu cho qu trnh hc tp, cng tc v sau.

LI M AU

Trong cng nghip ha cht v thc phm cng nh nhiu qu trnh

sn xut u da trn c s s tip xc gia cc pha v s di chuyn vt

cht t pha ny sang pha khc. Cc qu trnh vt cht di chuyn t pha

ny sang pha khc khi c s tip xc gia cc pha gi l qu trnh truyn

cht. Vic nghin cu cc qu trnh truyn cht cng nh quy trnh cng

ngh v thit b ca cc qu trnh l cng vic cn thit ca mt k s

ngnh cng ngh ha hc. Chnh v vy, cc k s tng lai ca lp Lin

Thng HCNH3 K III trng i hc Cng nghip H Ni chng em,

c cc thy c trong khoa giao cho v hng dn lm n mn hc

Qu trnh & Thit b.

n mn Qu trnh & Thit b l mn hc quan trng ca cc sinh

vin ngnh cng ngh ho. Qua qu trnh lm n mn hc, gip em

bc u c vn dng cc kin thc cc thy c dy vo gii quyt

cc vn k thut thc t m cc k s tng lai cn bit.

Qua y, em xin gi li cm n chn thnh ti thy gio hng dn

cng cc bn sinh vin gip em hon thnh n ny. Do thi gian c

hn v s non nt v kinh nghim ca bn thn nn khng th khng cn

thiu st em rt mong c s gp ca cc thy, cc c em c th rt

ra c nhiu kinh nghim qu bu cho qu trnh hc tp, cng tc v sau.Em xin chn thnh cm n!

TRNG I HC CNG NGHIP H NI N MN: QTTB

MC LC:

PHN I:TNG QUAN.

CHNG I: GII THIU CHUNG.I S LC V CHNG LUYN:

II THIT B CHNG LUYN:

III HN HP ETYLIC NC:

CHNG II: THUYT MINH DY CHUYN CNG NGH

I S CNG NGH:

II THUYT MINH LU TRNH:

PHN II: TNH TON THP CHNG LUYN.

CHNG I: CN BNG VT LIU.

I - CN BNG VT LIU TON THIT B.

II XC NH CH S HI LU THCH HP & S A L

THUYT.

CHNG II: TNH NG KNH THP.

I NG KNH ON LUYN:

II NG KNH ON CHNG:

CHNG III: TNH S A THC T THEO NG CONG

NG HC.

I. H S CP KHI:

II H S CP KHI TRONG PHA LNG:

III H S CHUYN KHI TRONG PHA HI:

IV - XC NH S N V CHUYN KHI I VI MI A

TRONG PHA HI:

CHNG IV: TNH TR LC CA THP.

I TR LC CA A KH:

SV: L Vn NLp: C-H Ha 3 K3

GVHD: Th.S Nguyn Th Hu

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II TR LC DO SC CNG B MT:

III TR LC THY TNH DO CHT LNG TRN A TO RA:

IV TNG TR LC THU LC CA THP:

CHNG V: TNH CN BNG NHIT LNG

I - CN BNG NHIT LNG THIT B UN NNG HN HP

U:

II - CN BNG NHIT LNG CA THP CHNG LUYN:

III - CN BNG NHIT LNG CA THIT B NGNG T:

IV - CN BNG NHIT LNG CA THIT B LM LNH:

PHN III: TNH TON C KH.

I CHN VT LIU LM THN THP

II TNH NG KNH CC NG DN:

III TNH Y V NP THIT B:

IV KHI LNG Y V NP THIT B:

V CHN BCH GHP:

VI TNH GI V TAI TREO:

PHN IV : TNH THIT B PH.

CHNG I: THIT B GIA NHIT HN HP U.

I B MT TRUYN NHT CA THIT B GIA NHIT HN

HP U:

II S NG CA THIT B GIA NHIT:

III NG KNH THIT B GIA NHIT:

IV S NGN CA THIT B GIA NHIT:

CHNG II: THIT B GIA NHIT HN HP Y

I B MT TRUYN NHT CA THIT B GIA NHIT HNHP Y:

II S NG CA THIT B GIA NHIT:

III NG KNH THIT B GIA NHIT:



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IV S NGN CA THIT B GIA NHIT:

CHNG III: TNH BM V THNG CAO V.

I - CC TR LC CA QU TRNH CP LIU:

II CHIU CAO THNG CAO V SO VI A TIP LIU:III TNH BM:



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PHN I:TNG QUAN.

CHNG I: GII THIU CHUNG.I S LC V CHNG LUYN:

Trong cng ngh ha hc c nhiu phng php phn ring hn hp hai

hay nhiu cu t tan mt phn hay hon ton vo nhau nh : hp th, hp ph, li

tm, trch li, chng...Mi phng php u c nhng c th ring v nhng u

nhc im nht nh. Vic la chn phng php v thit b cho ph hp tu

thuc vo hn hp ban u, yu cu sn phm v iu kin kinh t.

i vi hn hp Etylic v Nc l hai cu t tan hon ton vo nhau theo

bt k t l no c nhit si khc bit nhau th phng n ti u tch hn

hp trn l chng ct.

Chng l qu trnh dng tch cc cu t ca mt hn hp lng cng nh

hn hp kh lng thnh cc cu t ring bit da vo bay hi khc nhau ca cc

cu t trong hn hp (ngha l khi cng mt nhit , p sut hi bo ha ca

cc cu t khc nhau). Thay v a vo trong hn hp mt pha mi to nn s

tip xc gia hai pha nh trong qu trnh hp thu hoc nh kh, trong qu trnh

chng ct pha mi c to nn bng s bc hi hoc ngng t.

Trong trng hp n gin nht, chng ct v c c khng khc g nhau,

tuy nhin gia hai qu trnh ny c mt ranh gii c bn l trong qu trnh chng

ct dung mi v cht tan u bay hi (ngha l cc cu t u hin din trong c

hai pha nhng vi t l khc nhau), cn trong qu trnh c c th ch c dung mi

bay hi cn cht tan khng bay hi.

Khi chng ta thu c nhiu cu t v thng th bao nhiu cu t s thu

c by nhiu sn phm. Nu xt h n gin ch c 2 cu t th ta thu c 2

sn phm:

Sn phm nh ch yu gm cu t c bay hi ln v mt phn

rt t cc cu t c bay hi nh.

Sn phm y ch yu gm cu t c bay hi nh v mt phn rt

t cu t c bay hi ln.

i vi h Etylic - Nc th:



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Sn phm nh ch yu l Etylic.

Sn phm y ch yu l Nc.

C nhiu cch phn loi cc phng php chng:

Theo p sut lm vic:

- p sut thp.

- p sut thng.

- p sut cao.

Theo nguyn l lm vic:

- Chng gin on.

- Chng bng hi nc trc tip.

- Chng lin tc.

Theo phng php cp nhit y thp:

- Cp nhit trc tip.

- Cp nhit gin tip.

II THIT B CHNG LUYN:

Trong sn xut thng s dng rt nhiu loi thp nhng chng u c mt

yu cu c bn l din tch b mt tip xc pha phi ln ,iu ny ph thuc vo

phn tn ca lu cht ny va lu cht kia .

Thp chng luyn rt phong ph v kch c v ng dng, cc thp ln nht

thng c ng dng trong cng nghip lc ho du. Kch thc ca thp :

ng knh thp v chiu cao thp tu thuc lu lng pha lng, pha kh ca thp

v tinh khit ca sn phm. Trong chng luyn, hai loi thp c dng ch

yu l thp m v thp a.

Thp a: thn thp hnh tr, thng ng pha trong c gn cc a

chia thn thp thnh nhng on bng nhau, trn a pha lng v pha hi c cho

tip xc vi nhau. Ty theo cu to ca a, ta c:

* Thp a c ng chy chuyn v khng c ng chy chuyn.

* Thp a l(li), thp chp, thp supap v mt s dng khc.



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Thp m: thp hnh tr, gm nhiu on ni vi nhau bng mt bch

hay hn. m c cho vo thp theo mt trong hai phng php: xp ngu nhin

hay xp th t.

III HN HP ETYLIC NC:

1. Etylic:

- Etylic l cht lng linh ng khng mu v c th ha tan v hn trongnc. Nhit si ca n l 78.40C, Etylic c ng dng rt nhiu trong cngnghip, i sng hng ngy. Etylic l sn phm ca qu trnh ln men hoc qutrnh tng hp khc.2. Nc:

Nc l cht lng linh ng, khi lng phn t 18 (kg/kmol), nhit si1000C, khi lng ring 1000(kg/m3).

Hai cht lng ny c bay hi khc nhau. y Etylic bay hi trc do

nhit si ca n thp hn ca Nc. Hn hp n mn yu nn trong qu trnh

la chn thit b chng luyn th ta nn s dng loi thp ccbon thng

trnh lng ph.

*KT LUN: Vi hn hp Etylic Nc, trong bi ny ta s dng

phng php chng luyn lin tc p sut thng s dng thp a l c ng

chy chuyn.

Hn hp u c nng sut: F = 9,0 ( tn/h) = 9000 (kg/h).

Nng hn hp u : aF = 0,32 phn khi lng.

Nng sn phm nh : aP = 0,83 phn khi lng.

Nng sn phm y : aW = 0,005 phn khi lng.

Loi thp: Thp a l c ng chy chuyn

- Thp lm vic p sut thng l 1 (atm)

- Hn hp u c gia nhit n nhit si

- Tng tit din ca l hoc rnh chim t 8 15% tit din thp.

- ng knh l t 3 8 mm

- a c lp cn bng cng c th lp a xin mt gc vi dc 1/45

1/50



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Thp a l c ng chy chuyn hot ng vi hiu sut cao n nh. Khc phc

c nhc im ca cc loi thp khc.



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CHNG II: THUYT MINH DY CHUYN CNG NGH

I. S CNG NGH:



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H thng thit b cng ngh chng luyn lin tc thp a l c ng chy chuyn

tng qut gm c :

(1) : Bm ly tm.

(2) : Thng cao v.

(3) : Thit b un nng hn hp u dng a hn hp u ti nhit

lm vic. S dng thit b loi ng chm, dng hi nc bo ho un

nng v n c h s cp nhit ln, n nhit ngng t cao. Hi nc bo ho i

ngoi ng, lng i trong ng.

(4) : Lu lng k.

(5): Thp chng luyn: gm c 2 phn : phn trn gm t trn a tip liu

tr ln nh gi l on luyn, phn di gm t a tip liu tr xung gi l

on chng.

(6): Thit b ngng t hon ton sn phm nh, nc lnh i trong ng.

(7) : Thit b lm lnh sn phm nh.

(8) : Thng cha sn phm nh.

(9) : Thit b gia nhit hn hp y. B phn un bc hi y thp, c th t

trong hay ngoi thp.

(10) : Thit b tch nc ngng.

(11) : Thng cha hn hp u.

(12) : B phn phn phi lng.

(13) : Van x kh khng ngng.

(14) : Thng cha sn phm y.

II THUYT MINH LU TRNH:

Hn hp Etylic, Nc t thng cha (11) c bm vo thng cao v (2)

nh bm (1) ri dn xung thit b un nng (3). S c mt ca thng cao v m

bo cho lng hn hp u vo thp khng dao ng, trong trng hp cng sut

bm qu ln hn hp u s theo ng tun hon trn v b cha hn hp u.

(3) dung dch c un nng n nhit si bng hi nc bo ho. Ra khi

thit b un nng, dung dch i vo thp chng luyn (1) v tr a tip liu. Do

c un nng n nhit si nn ti y Etylic thc hin qu trnh chuyn



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khi t pha lng sang pha hi v tin v nh thp. Nc l cu t kh bay hi

nhit ny n vn ang th lng v phn phi xung di. Nh vy trong thp,

hi Etylic i t di ln gp lng Nc i t trn xung. V nhit cng ln cng

thp nn khi hi Etylic i t di ln c mang theo mt phn cu t Nc, cu t

c nhit si cao s ngng t li v cui cng trn nh ta thu c hn hp

gm hu ht cu t Etylic d bay hi. Hi Etylic vo thit b ngng t (6) c

ngng t li. Mt phn cht lng ngng i qua thit b lm lnh (7) n nhit

cn thit ri i vo thng cha sn phm nh (8). Mt phn khc hi lu v thp

a trn cng tng mc tch.

Tng t qu trnh dch chuyn ca Etylic, Nc s ko theo 1 phn cu t

v cng xung thp nhit ca thp cng tng khi cht lng Nc i t trn

xung gp hi Etylic c nhit cao hn, mt phn cu t c nhit si thp

c bc hi v do nng Nc kh bay hi trong cht lng ngy cng tng.

Cui cng y thp ta thu c hn hp lng gm hu ht l Nc kh bay hi.

Cht lng y thp khi ra khi thp c lm lnh ri a vo thng cha sn

phm y (14). tit kim hi t ngi ta c th dng hi nh thp un

nng hn hp ban u. Thp chng luyn lm vic ch lin tc, hn hp u

vo v sn phm c ly ra lin tc.

III. CH THY NG CA THP A L C NG CHY CHUYN:Ty thuc vo vn tc dng kh, trong thp a l c ng chy chuyn c cc ch chuyn ng sau y: vn tc kh b, kh qua lng dng tng bong bng ring l nn thp lm vic ch si bong bng. Lc ny cht lng va i qua ng chy chuyn va cng btqua l a. Nu tng vn tc ln th kh i qua lng thnh tia lin tc. Khi thp lm vic ch dng, cht lng khng lt qua l a c. ch ny thp lm vic un.Tip tc tng vn tc kh ln na, thp chuyn sang ch bt, tc l kh ha vilng to thnh bt. Lc ny lp cht lng trn a khng cn na, m ch c btlinh ng v xoy mnh. V vy, ch ny a lm vic tt nht. Nu tip tctng vn tc ln, trong thp c hin tng bn cht lng. i vi loi a ny ta thng cho thp lm vic ch dng hoc bt.



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u im ca ca thp a l c ng chy chuyn:- Hiu sut cao v b mt tip xc pha ln- Cu to thp n gin, d lp t sa cha.- Tr lc trong thp khng ln lm

- Gii hn lm vic tng i rngNhc im :

Yu cu lp t cao: : mm lp phi rt phng , i vi nhng thp cng knh qu ln (>2.4m) t dng mm xuyn l v khi cht lng phnphi khng u trn mm



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PHN II: TNH TON THP CHNG LUYN.

CHNG I:CN BNG VT LIU.

I - CN BNG VT LIU TON THIT B:

1. Thng s ban u :

Gi :

F : Lu lng hn hp u; kg/h, kmol/h.P : Lu lng sn phm nh; kg/h, kmol/h.W : Lu lng sn phm y; kg/h, kmol/h.aF : Nng hn hp u, % khi lng.aP : Nng sn phm nh, % khi lng.aW : Nng sn phm y, % khi lng.xF : Nng hn hp u, % mol. xP : Nng hn hp nh, % mol.xW : Nng hn hp u, % mol. M : Khi lng phn t, kg/kmol.

thun tin trong qu trnh tnh ton ta k hiu :

Etylic : A, v c MA = 46 (kg/kmol).

Nc : B, v c MB = 18 (kg/kmol).

2. Tnh cn bng vt liu:

Phng trnh cn bng vt vit cho ton thp:

F = P + W (1)

Phng trnh cn bng vt liu vit cho cu t nh:

F.aF = P.a P + W.aW (2)

T (1) v (2) suy ra:

Tnh nng phn mol ca cu t Etylic:

Thnh phn mol trong hn hp u:

xF =

N

F

E

F

E

F

Ma

Ma

Ma

+1 =

1832,01

4632,0

4632,0

+ = 0,1555 phn mol



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P W F W P F

F P Wa a a a a a

= =


Thnh phn mol trong sn phm nh:

xP =

1883,01

4683,0

4683,0

1 +

=

+N

P

E

P

E

P

Ma

Ma

Ma

= 0,6564 phn mol

Thnh phn mol trong sn phm y:

xW =

N

W

E

W

E

W

Ma

Ma

Ma

+1 =

18005,01

46005,0

46005,0

+ = 0,00196 phn mol

H phng trnh cn bng vt liu:Phng trnh cn bng vt liu cho ton thp : F = P + W ; (1)i vi cu t d bay hi (etylic) : FaF = Pap + Waw ( 2) Thay (1) vo (2) rt ra : Lng sn phm nh :

P = F . = 9000 . 005.083.0005.032.0

= 3436.3636 (kg/h)

T suy ra lng sn phm y :W = F P = 9000 3436.3636 = 5563.6364 ( kg/h)Xc nh thnh phn mol:

GF = F . ( B

F

A

F

Ma

Ma

+1

)

= 9000 . ( 1832.01

4632.0

+ ) = 402.609 ( kmol/ h )

GP = GF . ( wP

WF

xxxx

) = 402.609 . ( 00196,06564,000196,01555,0

)

= 94.457 ( kmol/ h)

GW = GF - GP = 402.609 94.457 = 308.152 ( kmol / h )

Nh vy ta c bng tng kt thnh phn sn phm nh sau:



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II XC NH CH S HI LU THCH HP & S A L THUYT:

1.Phng trnh ng lm vic ca on chng, on luyn

n gin cho vic thit lp ng lm vic ca on chng on luyn.

Ta gi thit:

- Dng mol pha hi i t di ln khng i trn ton b chiu cao ca thp.

Dng mol pha lng i t trn xung khng i trong on chng v on

luyn. Tc tha mn iu kin sau:

+Nhit ha hi mol ca cc cu t bng nhau theo cng thc kinh

nghim ca Trouton: r/ T = 21 kcal/ kmol = const.

+ Khng c nhit ha tan Q= 0.

+ S sai khc v nhit lng ring ca nhit si trn cc tit din

khc nhau ca thp c b qua.

+ Hn hp u vo thp nhit si.

- Cht lng i ra khi thit b ngng t c thnh phn bng thnh phn i ra

khi nh thp.

- Hi bc ln t y thp c nng bng nng sn phm y.

a) Phng trnh ng nng lm vic ca on luyn :

- Phng trnh cn bng vt liu:

D0 = L0 + P

Trong : Do l lng hi trn on luyn.

Lo l lng lng trn on luyn.

- Phng trnh cn bng vt liu cho cu t d bay hi l.

Doy = Lox + Pxp



- -

Nng

phn

khi lng

Nng

phn mol

Lu lng

(kg/h)

Lu lng

(kmol/h)

Hn hp u 0,32 0,1555 9000 402,609

Sn phm nh 0,83 0,6564 3436,3636 94,457

Sn phm y 0,005 0,00196 5563,6364 308,152

15


(Lo + P) y = Lox + Pxp

y = xLo

L Po ++ xp P

PL P+

t Lo/ P = R( ch s hi lu)

=>

b) Phng trnh ng nng lm vic ca on chng.

- Phng trnh cn bng vt liu:

Du = Lu W

Trong : Du l lng hi trn on chng.

Lu l lng lng trn on chng.

- Phng trnh cn bng vt liu cho cu t d bay hi:

Du y = Lu x- Wxw

=> ( Lu W)y ' = Lu x ' - W xw

M Lu = Lo + F = Lo + P+ W

v W= F P

=> ( Lo+P) y ' = ( F + Lo ) x ' - ( F P) xw

=> y ' = WxPLPFx

PLFL

+

+

+

00

0

t : R= Lo/ P, f= F/P Lo = R.P

=>

3. Xc nh ch s hi lu Rmin :

Dng ng cn bng theo s liu ng cn bng sau:Theo bng IX.2a S tayII Trang 147

X 0 5 10 20 30 40 50 60 70 80 90 100

Y 0 33,2 44,2 53,1 57,6 61,4 65,4 69,9 75,3 81,8 89,8 100



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y = R

R 1+ x +

1R 1+

xp

y ' = R fR 1

+

+x -

f 1R 1

+xw


T0c 100 90,5 86,5 83,2 81,7 80,8 80 79,4 79 78,6 78,4 78,4

* T ng cn bng lng (x) hi (y) vi xF = 0.1555 y*F = 0.50 phn mol

FF

Fp

xyyx

R

= *

*

min [III-81]

454,01555.050.0

50.06564.0*

*

min =

=

=

FF

Fp

xyyx

R

Trong :

xP _nng phn mol ca Etylic trong pha lng sn phm nh

xF_ nng phn mol ca Etylic trong pha lng hn hp u.

y*F _nng phn mol ca Etylic trong pha hi nm cn bng pha lng

hn hp u.

3. Xc nh ch s hi lu thch hp Rth:Xc nh ch s hi lu thch hp da vo iu kin th tch thp nh nht

tc l tng ng vi Nlt(Rx+1) nh nht (Nlt : S bc thay i nng l thuyt)

RX = Rmin vi b l h s = 1,2 2,5

Vn chn ch s hi lu thch hp rt quan trng, nu lng hi lu qu

b th thp s v cng cao, iu ny rt kh thc hin, nu lng hi lu ln th



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thp c thp i nhng ng knh li ln, sn phm nh thu c chng bao

nhiu.

Xc nh RX thch hp theo s bc thay i nng c tin hnh nh sau :

Cho nhiu gi tr RX ln hn gi tr Rmin vi RX = (1,2 2,5) Rmin . Vi mi gi

tr trn, ta xc nh c ng nng lm vic ca on chng, on luyn hay

ni n gin l xc nh c tung ca ng nng lm vic ca on luyn

vi trc tung c to l B, vi:

P

X

xBR 1

=

+

T , ta xc nh ng lm vic ca on chng v on luyn v v th

ta s xc nh c s a l thuyt ng vi tng gi tr , Rx.

Vi RX1= 0,5448Phng trnh on luyn : y = 0.3527x + 0.4249Phng trnh on chng: y = 3.1116x 0.004C 10 a trong on luyn 5, on chng 6.

Vi RX2= 0.6356Phng trnh on luyn : y = 0.3886x + 0.401Phng trnh on chng: y = 2.994x 0.004

C 9 a trong on luyn 4, on chng 6. Vi RX3 = 0.73

Phng trnh on luyn : y = 0.422x + 0.38Phng trnh on chng: y = 2.89x -0.0037





C 7 a trong on luyn 3, on chng 5. Vi RX6 = 1,091



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C 6 a trong on luyn 3, on chng 5. Vi RX7 = 1,098


C 6 a trong on luyn 3, on chng 4.

T cc phng trnh lm vic trn ta v th biu din quan h x y tmc s a l thuyt tng ng vi tng gi tr Rx.

* th xc nh s a l thuyt :



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- -22


* S a l thuyt Nlt v ch s hi lu Rx c tng hp trong bng sau :

1,2 1,4 1,6 1,7 1,8 1,9 2,0 2,2 2,4Rx 0.5448 0.6356 0.73 0,772 0.8172 0.8626 0.908 1,091 1,098Nlt 10 9 8 8 7 7 7 6 6

Nlt(Rx+1) 15,5448 14,7204 13,84 14,176 12,2704 13,0382

13,356 12,656 12,858

T cc gi tr tm c bng trn, ta xy dng th biu din mi quan h giaRx Nlt(Rx+1)

V th mi lin h gia Rx&Nlt(Rx+1):

xc nh ch s hi lu thch hp ngi ta xc nh ch s hi lu t iu kinth tch thp nh nht ( khng tnh n cc ch tiu kinh t vn hnh). Mt khc ddng nhn thy th tch lm vic ca thp t l vi tch s N lt(Rx+1).T th trn,im cc tiu ca ng v c s cho ta gi tr th tch thit b b nht v ngvi im s c ch s hi lu thch hp.



- -23


M Nlt(Rx+1) = 12,2704 l gi tr nh nht, ng vi gi tr nh nht th Rx=0.8172Vy ch s hi lu thch hp Rth= 0.8172 v s a l thuyt Nlt= 7

V th ng lm vic ng vi Rth:

4. Phng trnh ng nng lm vic SV: L Vn NLp: C-H Ha 3 K3


- -24


4.1.on luyn

11 ++

+=

th

P

th

th

Rxx

RRy [III- 78]

*Thay s :

361.045.018172.0

6564.018172.0

8172.0+=

++

+= xxy

Trong :y: L nng phn mol ca cu t d bay hi trong pha hi i t di ln.x: L nng phn mol ca cu t d bay hi trong pha lng chy t trn xung.Rth: L ch s hi lu thch hp.4.2. on chng - Lng sn phm u trn mt n v sn phm nh :

262.4457.94609.402

===

P

F

GGf ( kmol h2 u / kmol sn phm nh );

+Phng trnh ng nng lm vic ca on chng :

0035.0795.200196.018172.0

1262.418172.0262.48172.0

11

1=

+

+

+=

+

+

+= xxx

Rfx

RfRy w

thth

th

* Nhit si - Nhit si ca hn hp u toF : t xF = 0.1555 theo bng t-x, ni suy toF =84.67oC- Nhit hi ti nh thp toP : t xp = 0.6564 theo bng t y ni suy toP = 79.18- Nhit si ca sn phm y tow : T xw= 0.00196 theo bng t- x ni suy tow =99.63oC



- -25


CHNG II:

TNH NG KNH THP.

i vi thp a l c ng chy chuyn, khi cc cht lng khng c cc tp

cht c hc v khng ng cn trn b mt th tc hi c tnh theo cng thc:

xy

y

0,05. (m / s) = cng thc [122-tnh ton...tp2]

y chnh l tc gii hn trn ca hi i trong thp a l c ng chy

chuyn.

ng knh thp a l c ng chy chuync tnh theo cng thc: - ng knh ca thp ( D) l mt i lng c trung cho nng sut ca

thp. D = hay D = 0,0188. (m) [181- 2].

Trong : g :lng hi trung bnh i trong thp (Kg/h) .

:khi lng ring trung bnh ca hn hp kh hi (Kg/m) .

:vn tc trung bnh ca dng hi i trong thp (m/s) .

V :lu lng trung bnh ca pha hi i trong thp (m/h)

V lng hi v lng lng thay i theo chiu cao thp v khc nhau trong

mi on nn lng hi trung bnh trong tng on khc nhau v do ng

knh on chng v on luyn ca thp c th khc nhau.



- -26


n'

2 2 1

1 2 2

g'1,y'1=yw

F,xF

n

G1,x1=xF

G'1,x'1P,xP

GRx

1.Lng hi trung bnh i trong thp1.1.Lng hi trung bnh i trong on luyn Lng hi trung bnh i trong on luyn c tnh gn ng bng trungbnh cng ca lng hi i ra khi a trn cng ca thp v lng hi i vo adi cng ca on luyn.

2

1ggg tb+

= , kg/h (STQTTB T2-181)

Trong : + gtb :lng hi trung bnh ca on luyn (kg/h)+ g : lng hi i ra khi a trn cng ca on luyn (kg/h);+ g1 : lng hi i vo a di cng ca on luyn ( kg/h);

* Lng hi i ra khi nh thp : g = GR + GP = GP(Rth + 1) (STQTTB T2-181)Trong

GP : lng sn phm nh ( kg/h) GR : lng lng hi lu nh thp (kg/h) Rth : ch s hi lu thch hp

g = 3436.3636( 0.8172 + 1) = 6244.56 (kg/h)* Lng hi i vo on luyn: Lng hi g1, hm lng hi y1, lng lng ca a th nht ca on luyn G1c xc nh theo h phng trnh cn bng vt liu v nhit lng cho onluyn :



- -27


=

+=

+=

dd

PP

P

rgrgaGaGy

GG

11

1111

11

.gg

(STQTTB T2-182)

Trong 32.01 == Faa phn khi lng

r1 : n nhit ha hi ca hn hp hi a th nht ca on luyn (kJ/kg) ;r : n nhit ha hi ca hn hp hi ra khi nh thp (kJ/kg) r1 = rAy1 + (1- y1)rB; STQTTB T2-

182) r = rAy + (1- y)rB;

Trong : y : hm lng hi sn phm nh = aP = 0.83 phn khi lng rA ; rB : ln lt l n nhit ha hi ca Etylic v Nc nguyn cht (kJ/kg)T xF= 0.1555 tra bng IX2a [STQTTB T2-148] suy ra nhit tF = 84.670C.

. Tra bng I.212 [STQTTB T1-254] vi nhit hn hp u 0Ft ta c: rA = 200.132 ( Kcal/kg) = 837.913 (KJ/kg) rB = 554.33 ( Kcal/kg) = 2320.87(KJ/kg ) r1 = 837.913 y1 + 2320.87 ( 1-y1) = 2320.87 1482.957y1(*)

T xP = 0.6564 tra bng IX2a [STQTTB T2-148] suy ra nhit ca hnhp nh tP= 79.180C, ni suy t bng I.212 [STQTTB T1 -254] vi nhit hnhp nh tP ta c:

rA = 202.328(Kcal/kg) = 847.107 (KJ/kg)rB = 559.82(Kcal/kg) = 2343.85 (KJ/kg)r = aPrA + (1- aP )rB == 0.83 x 847.107 + (1- 0.83) x 2343.85 = 1011.55 (KJ/kg) (**)

Thay (*)&(**) vo h trn ta c :

( )

=

+=

+=

55,1011.56,6244957,148287,2320

83,0.3636,343632,0.g3636,3436g

11

111

11

yg

GyG

Gii h ta c: g1 = 4857,88( kg/h ); y1 = 0,681 phn khi lng G1 = 1421,52 ( kg/h ) lng hi trung bnh ca on luyn :



- -28


)/(22,55512

56,624488,48572

1 hkgggg dtb =+

=

+=

Thay y1vo (*) ta c : r1 = 2320,87 1482.957.0,681= 1310,976 (KJ/kg) 1.2. Lng hi trung bnh i trong on chng

2

''1' n

tbggg += [STQTTB T2-182]

Trong :+ gtb: lng hi trung bnh ca on chng ( kg/h)+ gn: lng hi i ra khi on chng ( kg/h) + g1 : lng hi i vo on chng (kg/h)

V lng hi i ra khi on chng bng lng hi i vo on luyn nn tac : gn = g1

2

1'1' ggg tb+

= [STQTTB T2-182]

Ta c h phng trnh cn bng vt liu v nhit lng cho on chng :

==

+=

+=

''11

'1

'1

'1

'1

'1

'1

'1

...

...

nn

www

w

rgrgrgxGygxG

GgG

[STQTTB T2-182]

Trong :- G1: lng lng i vo on chng(kg/h)- g1 : lng hi i vo on chng (kg/h)- x1 : hm lng lng(phn khi lng)- rn : n nhit ha hi ca hn hp i vo a trn cng ca on chng

(KJ/kg)- r1 : n nhit ha hi ca hn hp hi i vo a th nht ca on chng ( KJ/kg)- xw : thnh phn cu t d bay hi (Etylic ) trong sn phm y ;- r1 : n nhit ha hi ca hn hp hi i vo a trn cng ca on chng(KJ/kg)- rA : n nhit ha hi ca Etylic (KJ/kg)- rB : n nhit ha hi ca Nc (KJ/kg)

wyy =1' l nng cn bng ng vi wx . Da theo th ng cn bnglng hi trn ta c:ng vi 00196.0=wx ta c 0021.0=wy ta i sang phn khi lng



- -29


( ) 0053.0189979.0460021.0460021.0

0021.010021.00021.0* =

+

=

+

=

BA

Aw MM

My

Mt khc ta c:

BnnAn

BA

ryyrrryyrr)1(

)1('''

'1

'1

'1

+=

+=

T 00196.0=wx ni suy t ( Bng IX. 2a_ 148_STQTTB tp II ) suy ra toC=99.63oC.ni suy t bng I.212 [STQTTB T1 -254] vi nhit hn hp y tW ta c:

RA = 194.148(Kcal/kg) = 812.859(KJ/kg); RB = 539.37 (Kcal/kg) = 2258.23 (KJ/kg);

)/(569,225023.2258)0053.01(0053.0859.812'1 kgKJr =+=

Thay vo h phng trnh ta c :

=

+=

+=

976,1310.88,4857569,2250.00196,0.6364,55630053,0.

6364,5563

'1

'1

'1

'1

'1

'1

ggxG

gG

)/(757,2829'

0031.0);/(3934,8393

1

'1

'1

hKggx

hKgG

=

=

=

lng hi trung bnh ca on chng :

)/(8185,38432

88,4857757,28292

''1' hKgggg ntb =

+=

+=

1.3 Lng lng trung bnh i trong on luyn

G = = 2

3636,3436.8172,052,1421 + = 2114,858(Kg/h)

1.4 Lng lng trung bnh i trong on chng:

G = G = = 4567,490723934,839352,1421

=

+ (Kg/h).

2. Tnh khi lng ring trung bnha) i vi pha hi :*on luyn :



- -30


[ ] ( )[ ] [ ] );/(1614,1

)27318.79.(4,22273.18)5557,01(465557.0

.4,22273.)1(.

1832.4,22

273.)1(.

3

`

`

mkgT

MyMy

STQTTT

MyMy

L

BtbAAtbAtbl

L

BtbAAtbAtbl

=

+

+=

+=

+=

-Vi 2

1yyy ptbA+

=

+) y1= 0.681 phn khi lng ta i sang phn mol:

( ) 455.018

681.0146681.0

46681.0

1 =

+=y (phn mol)

+) yp=0.6564(phn mol)

5557.02

455.06564.0=

+=tbAy (phn mol)

- T bng IX 2a(STQTTBT2-148) ta c totbL= 79.18oC*on chng:

[ ] ( )[ ] [ ] );/(7994.0

)27363.99.(4,22273.18)23,01(4623.0

.4,22273.)1(.

1832.4,22

273.)1(.

3

`

`

mkgT

MyMy

STQTTT

MyMy

C

BtbAAtbAtbC

C

BtbAAtbAtbC

=

+

+=

+=

+=

-Vi 2

11' yyytbA

+=

+) y1= 0.681 phn khi lng ta i sang phn mol:

( ) 455.018

681.0146681.0

46681.0

1 =

+=y (phn mol)

+) y1=yw=0.0053 (phn mol)

23.02

455.00053.0=

+=tbAy (phn mol)

- T bng IX 2a(STQTTBT2-148) ta c totbC= 99.63oCb) i vi pha lng :

xtbBtbA

xtbA

tbA

xtb

aa

+=11

` [STQTTBT2-183]

Trong :



- -31


- tba : phn khi lng trung bnh ca cu t Etylic - xtbA : khi lng ring ca cu t Etylic (kg/m3)- xtbB : khi lng ring ca cu t Nc (kg/m3)

* on luyn :

xtbB

tbAL

xtbA

tbAL

xtL

aa

+=11

575.02

83.032.02

=

+=

+=

pFtbAL

aaa (phn khi lng)

406.02

1555.06564.02

=

+=

+=

wptbAL

xxx (phn mol)

Ni suy t bng IX2a(STQTTBT2-148) nht trung bnh Ct OotbLX 18.79=Theo bng I.2 trong (STQTTBT1-9) ta ni suy c:

xtbA = 735.779 (kg/m3)xtbB = 972.451 (kg/m3)

)/(66.820451.972575,01

779.735575,01 3mkgxtbL

xtbL

=

+=

*on chng:

xtbB

tbAC

xtbA

tbAC

xtbc

aa

+=11

2

wFtbAC

aaa +=

aw=0.005(phn khi lng)

1625.02

005.032.0=

+= tbACa (phn khi lng)

079.02

00196.01555.02

=

+=

+=

wFtbAC

xxx (phn mol)

Ni suy t bng IX2a(STQTTBT2-148) nht trung bnh Ct OotbCX 63.99=Theo bng I.2 trong (STQTTBT1-9) ta ni suy c:

xtbA = 716.3515 (kg/m3)xtbB = 958.259(kg/m3)

);/(41.908259.9581625.01

3515.7161625.01 3mkgxtbC

xtbC

=

+=

3.Tnh tc ca kh v hi i trong thp



- -32


- Tc hi i trong thp ph thuc cu to thp ,ch lm vic , khi lng

ring

- Vi thp a l c ng chy chuyn : trnh tao bt

= ( 0,8 0,9 ) [ 117- 2] .

Trong : : vn tc hi i trong thp .(m/s)

: vn tc gii hn (m/s)

Vi thp a l c ng chy chuyn :

= 0,05 . (m/s) [186-2].

3.1.Tc hi i trong on luyn Ta c : = 820.66 (kg/m ).

= 1.1614(kg/m ).

Vy : = 0,05 . 1614,166,820

= 1,329 (m/s).

trnh to bt : = 0,8 = 1,063 (m/s).

3.2. Tc ca hi i trong on chngTa c : =908,41 (kg/m )

= 0,7994 (kg/m )

Vy = 0,05 . 7994.0

41.908 = 1,685 (m/s) .

trnh to bt : = 0,8 = 1,348 (m/s).

4. ng knh ca thp* ng knh on Luyn:

)(206,1063,11614,122,55510188,0

..0188,0 mgD

LytbL

tbL =

==

* ng knh on chng:

)(123,1348,1.7994,08185,38430188,0

..0188,0 mgD

CytbC

tbC ===

Da vo kt qu nhn c chn ng knh theo bng quy chun cho thp (bng6.27-sch TTQTTBT2-115)Ta c DT = DL= DC=1,2 (m)Khi vn tc ca hi i trong thp s l:



- -33


=

on Chng : = 1,1802 (m/s) .

on Luyn : = 1,1732 (m/s).

Lc ny gi tr ca vn tc lm vic

= 0,88

= 0,7

CHNG III:



- -34


TNH S A THC T

THEO NG CONG NG HC.

- V ng cong cn bng y*=f(x) v xy dng ng nng lm vic ng vi

Rth = 0,8172 xc nh c s a l thuyt Nlt = 7 a.

- Dng cc ng thng A1C1, A2C2,ti cc im c honh tu song song

vi trc tung trn th.

AiCi = *y y

I. H S CP KHI TRONG PHA HI:

H s cp khi trong pha hi tnh cho 1m2 din tch lm vic ca a:

y = y y

2

D .(0,79Re 11000) kmolkmol22,4 m .s.kmol

+ [II 164]

Trong :

Dy - H s khuch tn trung bnh trong pha hi, (m2/s).

yRe - chun s Reynol i vi pha hi.

1.H s khuch tn trung bnh trong pha hi:

T s tay tp II trang 127 ta c:

BABA

y MMvvP

TD 11.

.

.10.0043.02

31

31

23

4

+

+

=

(m2/s) [STQTTBT2-127]

Trong : Dy : l h s khuch tn ca pha hi (m2/s); P : l p sut tuyt i ca h 2 cu t Etylic Nc = 1atm = 1.033at ;

BA ; : ln lt l th tch mol ca hi Etylic Nc (cm3/mol); TC : l nhit tuyt i ca on chng = 99.63oC TL : l nhit tuyt i ca on luyn = 79.18oC MA , MB : l khi lng mol phn t ca 2 cu t Etylic Nc ( vC)

MA = 46 (kg/kmol) MB = 18 (kg/kmol).



- -35


+ A = 6.3,7 + 2.14,8+1.7,4 = 59.2 (cm3/mol);

+ B = 2.3,7 +1.7,4 = 14.8(cm3/mol);

(Tra bng II-127 ta c th tch nguyn t ca :C = 14,8; H = 3,7; O = 7,4)*Trong on chng:

Theo phn trn c tytbC= 99,63 C = 99,63 + 273 = 372,63 K

Thay vo ta c :

( )sm

MMvvP

TDBA

BAC

CyC

/10.063,2181

461.

8,142,59.033,1

)63,372.(10.0043,0

11.

.

.10.0043,0

252

31

31

23

4

2

31

31

23

4

=+

+

=

=+

+

=

* Trong on luyn :

Theo phn trn TytbL = 79,18 C = 79,18 + 273 = 352,18 K.

Thay vo ta c:

( )sm

MMvvP

TDBA

BAL

LyL

/10.895,1181

461.

8,142,59.033,1

)18,352.(10.0043,0

11.

.

.10.0043,0

252

31

31

23

4

2

31

31

23

4

=+

+

=

=+

+

=

2. Chun s Reynol pha hi:

y yy

y

.h.Re

=

[STQTTBTII -164]

Trong : y tc hi tnh cho mt ct t do ca thp, (m/s).

h kch thc di, chp nhn h=1m.

y khi lng ring trung bnh ca hi, (kg/m3).

y nht trung bnh ca hi, (Ns/m2).

2.1. Tc hi tnh cho mt ct t do ca thp:



- -36


i vi on chng: yC = C= 1,1802 (m/s)

i vi on luyn: yL = L = 1,1732 (m/s)

2.2. Khi lng ring trung bnh ca hi:

Theo phn trc ta c:

i vi on chng: )3/(7994,0 mkgytbCyC ==

i vi on lun:)3/(1614,1 mkgytbLyL ==

2.3. nht trung bnh ca hi:

nht trung bnh ca hi c tnh bng biu thc:

hhy

tb A tb B

A B

My .M (1 y ).M =

+

(Ns/m2).

Trong : Mhh, MA, MB : khi lng mol ca hn hp, Etylic v Nc.

,hh A B , : nht ca hn hp,Etylic v Nc; (Ns/m2).Tra nht ca hi ph thuc vo nhit da vo ton hnh I.35[Tr

117.ST-I]

*i vi on chng:

C t0ytb = 99.630C.

=>

=

=

23

23

/.10.013,0/.10.011,0msNmsN

B

A

Khi lng mol trung bnh ca hn hp:

MhhC = ytbC . MA + ( 1 ytbC).MB = 0,23 . 46 + ( 1 0,23) . 18

= 24,44 ( kg/kmol)

Suy ra: )/(10.21,1

10.013,018).23,01(

10.011,046.23,0

44,24 25

33

mNsyC

=

+=

*i vi on luyn: C t0ytb = 79,180C.



- -37


=>

=

=

)/(10.012,0)/(10.01,0

23

23

mNsmNs

B

A

Khi lng mol trung bnh ca hn hp:

MhhL = ytbL . MA + ( 1 ytbL).MB = 0,5557 . 46 + ( 1 0,5557) . 18

= 33,56 ( kg/kmol)

Vy: )/(10.041,1

10.012,018).5557,01(

10.01,046.5557,0

56,33 25

33

mNsyL

=

+=

Suy ra chun s Reynol pha hi:

on chng:

Rey = 23,7797110.21,11.7994,0.1802,1

5 =

on luyn:

Rey = 999,13088810.041,11614,1.1.1732,1

5 =

H S CP KHI TRONG PHA HI TNG ON:

y = y y

2

D .(0,79Re 11000) kmolkmol22,4 m .s.kmol

+ [II 164]

on chng:

yC = 067,04,22)1100023,77971.79,0.(10.066,2 5

=

+

kmolkmolsm

kmol

..2

on luyn:

yL = 097,04,22)11000999,130888.79,0.(10.895,1 5

=

+

kmolkmolsm

kmol

..2

II H S CP KHI TRONG PHA LNG:

c xc nh theo cng thc sau:



- -38


x =0,62x xx

2x

3800. .D kmolPr kmolM .h m .skmol

[II 165]

Trong :

x khi lng ring trung bnh ca lng, (kg/m3).

Dx - h s khuch tn trung bnh trong pha lng, (m2/s).

Mx khi lng mol trung bnh ca lng, (kg/kmol).

h = 1 kch thc di.

Prx = x

x xD

- chun s Prant i vi pha lng.

x nht trung bnh ca lng, (Ns/m2).

1. Khi lng ring trung bnh ca lng:

on chng:

xC = xtbC = 908,41 (Kg/m3).

on luyn:

xL = xtbL= 820,66 (Kg/m3).

2. H s khuch tn trung bnh trong pha lng:

c xc nh nhit trung bnh theo cng thc sau :

Dx = D20 [ ]1 b(t - 20)+ [II 134]Trong :

b = B

B

0,23

- h s nhit 20oC.

D20 - H s khuch tn pha lng nhit 200C.

B - nht dung mi Nc 20C, (cP hoc 10-3 N.s/m2).

B - Khi lng ring ca dung mi Nc 20C, (kg/m2).

Ti nhit 20C :

Nc = 1.005 (cP ). ( Bng I.102 94, STQTTBT1)

Nc20 = 998 (kg/m ). (Tra bng I.2 STQTTBT1- 9)



- -39


Vy: b = 02,0998

005,12,03

=

H s khuch tn pha lng c tnh theo :

D20 = A B

B A B

1 161.10M M

1/ 3 1/ 3 2A.B. ( )

+

+ , (m2/s) [II 133]

A, B h s lin hp ph thuc vo tnh cht ca cht tan v dung mi

ta c: A =1,24 ( Bng VIII.6 133, STQTTBT2)

B = 4.7 (Bng VIII.7 134, STQTTBT2)

VA ,VB : th tch mol ca Etylic v Nc

Theo bng VIII.2 [ 127-2] c th tch nguyn t : C = 14,8 ; H = 3,7 ; O

= 7,4

Vy : VA = 2.14,8 + 6.3,7 + 1.7,4 = 59.2 (cm/mol)

VB = 2.3,7 + 1.7,4 = 14.8 (cm/mol).

H s khuch tn 20 C l:

( )smBA

MMD

BA

BA

/10.179,1

)8,14()2,59(.005,1.7,4.24,1

181

461.10

...

11.10

29

2

31

31

6

2

31

31

20

6

20

=

=

+

+=

+

+

=

H s khuch tn trong on chng:

t = 99,63 C

D = 1,179. 10 . [1 + 0,02.(99,63 - 20)] = 3,057 . 10 (m/s).

H s khch tn trong on luyn:

t = 79,18 C

D = 1,179. 10 .[ 1 + 0,02.(79,18 - 20 )] = 2,574. 10 (m/s)

3. Khi lng mol trung bnh ca lng:

on chng: C xtbC = 0,079

xC tbC A tbC BM x M (1 x )M= +

= 0,079 . 46 + ( 1 0,079) . 18



- -40


= 20,212 (kg/kmol).on luyn: C xtbL = 0,406

xL tbL A tbL BM x M (1 x )M= +

= 0,406 .46 + ( 1 0,406).18

= 29,368 ( kg/kmol)

4. nht trung bnh ca lng:

nht ca lng c tnh:

lg x = xtb.lg A + (1-xtb).lg Bon chng:

C xtbC = 0,079 & TtbC = 99,630C

Tra bng I.101[Tr92.ST-I]kt hp ni suy ta c:

=

=

)(2853.0)(328.0

cPcP

B

A

Vy:

lg xC = 0,079.lg 0,328 + (1 - 0,079).lg 0,2853

=> xC = 0,2885 (cP)= 0,2885.10-3(Ns/m2).

on luyn:

C xtbL = 0,406 & TtbL =79,18oC.

Tra bng I.101[Tr92.ST-I]kt hp ni suy ta c:

=

=

)(362,0)(441,0

cPcPA

B

Vy:

lg xL = 0, 406.lg 0,441 + (1 - 0, 406).lg 0,362

=> xL = 0,3922 (cP) = 0,3922.10-3(Ns/m2).

5. Chun s Prant i vi pha lng:

on chng:

PrxC = 889,10310.057,3.41,90810.2885,0

9

3

=

on luyn:



- -41


PrxL = 667,18510.574,2.66,82010.3922,0

9

3

=

H S CP KHI TRONG PHA LNG:

on chng:

xC = 093,0889,103.1.212,2010.057,3.41,908.38000 62,09

=

kmolkmolsm

kmol

..2

on luyn:

xL = 0697,0667,185.368,2910.574,2.66,820.38000 62,09

=

kmolkmolsm

kmol

..2

III H S CHUYN KHI TRONG PHA HI:

- Ti mi gi tr ca x tm tg ca gc nghing ca ng cn bng v cc i lng

y , ycb , x, xcb c tm trn th.

m = tg = cb

cb

xxyy

- H s chuyn khi trong pha hi:

y

y x

1K 1 m=+

(Kmol/m.s) IX.33[Tr162.ST-II]



- -42


BNG S LIU V NG CONG NG HC

x 0,05 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9

ycb 0,332 0,442 0,531 0,576 0,614 0,654 0,699 0,753 0,818 0,898

y 0,136 0,276 0,451 0,496 0,541 0,586 0,631 0,676 0,721 0,766

xcb 0,015 0,035 0,11 0,152 0,22 0,31 0,495 0,55 0,64 0,715

m 5,6 2,554 0,89 0,541 0,406 0,358 0,648 0,513 0,606 0,714

x 0,093 0,0697

y 0,067 0,097

Ky 0,013 0,024 0,044 0,056 0,062 0,065 0,051 0,057 0,053 0,049

IV - XC NH S N V CHUYN KHI I VI MI A TRONG

PHA HI:

yyT

y

K .fm

G= IX.65a[Tr173.ST-II]

Trong :

yG : lng hi i trung bnh trong thp

f : din tch lm vic ca a (m2).i vi thp a l c ng chy chuyn:

chf F m.f=

F mt ct t do ca thit b (m2).

V D = 1,2 < 2 => chn m = 1.

fch mt ct ngang ca ng chy chuyn (m2).

- xc nh Cy theo cng thc sau ng vi mi gi tr ca x:

yTy

mC e=



- -43


- xc nh BC : BC = y

ACC vi AC = ycb - y;

1. Lng hi i trung bnh trong thp:

on chng:

GyC = 3600.2

''1 ngg +

Trong :

)/(73,156)/(757,2829'1 hkmolhkgg == ( MW = 18,055 kmol/h)

)/(563,161)/(88,48571' hkmolhkgggn ===

MyF = )/(068,30)431,01(18431,0.46)1.(. 262 kmolkgyMyM FOHFOHC =+=+

GyC = 3600.2

''1 ngg + = )/(044,0

3600.2563,16173,156 skmol=+

on luyn:

GyL = 3600.21 dgg +

Trong : gd = 6244,56 (kg/h) = 171,648 (kmol/h)

( v MP = 36,38 kg/kmol)

GyL = 3600.21 dgg + = )/(0463,0

3600.2648,171563,161 skmol=+

2. Din tch lm vic ca a:

Ta c: F = 2

2D (m )4

pi

Xc nh ng knh ng chy chuyn bi biu thc:

Xch

X c

4.Gd (m)3600. . . .m

=

pi

Trong :

Gx lu lng lng trung bnh i trong thp (kg/h).

x - khi lng ring ca lng trong ng chy chuyn(kg/m3).

c tc cht lng trong ng chy chuyn(m/s). chn = 0,1 0,2 (m/s).

m s ng chy chuyn trn mi a.



- -44


- Lu lng lng trung bnh i trong thp:

on chng:

GxC = '

1 1G G2+ = 4907,4567 (kg/h)

on luyn:

GxL = 1 RG G

2+

= 2114,858 (kg/h)

Ta c:

==

==

)/(66,820)/(41,908

3

3

mkgmkg

xtbLxL

xtbCxC

Chn tc trong ng chy chuyn: c = 0,15 (m/s)

Thay cc gi tr vo biu thc tnh dch,ta c:

on chng:

)(113,01.15,0.41,908.14159,3.3600

4567,4907.4 md Cch ==

on luyn:

)(074,01.15,0.41,908.14159,3.3600

858,2114.4 md Lch ==

Din tch lm vic ca a:

on chng:

fC =2D

4pi -

C 2chd

4pi = )(042,1

4113,0.14159,3

42,1.14159,3 2 m=

on luyn:

fL =2D

4pi -

L 2chd

4pi = )(073,1

4074,0.14159,3

42,1.14159,3 2 m=



- -45


Ta lp c bng sau:

Ky 0,013 0,024 0,044 0,056 0,062 0,065 0,051 0,057 0,053 0,049f 1,042 1,073

Gy 0,044 0,0463myT 0,308 0,568 1,02 1,298 1,437 1,506 1,182 1,321 1,228 1,136Cy 1,361 1,765 2,773 3,662 4,208 4,509 3,261 3,747 3,414 3,114AC 0,196 0,166 0,08 0,08 0,073 0,068 0,068 0,077 0,097 0,132

BC 0,144 0,094 0,029 0,022 0,017 0,015 0,021 0,021 0,0280,042

Qua th ta xc nh c ng cong ph:

Xc nh c s a thc t l 17 a. Trong :

S a on chng l 12 a.

S a on luyn l 5 a.

T ng knh l D = 1,2 m , tra bng IX.4a [II-169] ta chn:

Khong cch gia cc a l H = 0,35 (m)

Chiu dy a 005,0= (m)



- -46


Chiu cao thp c tnh bi cng thc sau:

H = Ntt (H + ) + ( 0,8 1) [II-169]Vy chiu cao on chng l:

HC = 12.(0,35 + 0,005) = 4,26(m)

Vy chiu cao on luyn l:

HL = 5.(0,35 + 0,005) = 1,745 (m) Ta c chiu cao ca thp l:

H = HC + HL = (4,26 + 1,745) + ( 0,8 1) = 6,805(m)

Quy chun li chiu cao lm vic ca thp l:

H =6,8 (m)

CHNG IV:TNH TR LC CA THP.



- -47


*Cu to a l:

Chn thp a l c ng chy chuyn vi:

Tit din t do bng 10% din tch a.

ng knh l: dl = 3mm = 0,003 (m).

Chiu cao ca chy trn: hc = 50mm = 0,05 (m).

Din tch ca 2 bn nguyt bng 20% din tch a.

L b tr theo hnh lc gic u.

Khong cch gia 2 tm l bng 7mm.

S l trn 1 mm:

N = 16000003,02,110,010,0%.10

22

=

=

=

lolo

dia

dD

SS

Gi a l s hnh lc gic.

p dng cng thc (V.139), trang 48, [ST-II]: N = 3a(a+1) + 1

Gii phng trnh bc 2 a = 74 N = 16651 (l)

S l trn ng cho: b = 2a + 1 = 149 (l)

I TR LC CA A KH:

p dng cng thc (IX.140), [tr 194.ST-II]:2

yk

' .P

2

= (N/m2).

i vi a c tit din t do bng 10% din tch a th = 1,821. on chng:

Vn tc hi qua l: C = 802,111,01802,1

1,0==

C (m/s)

Nn: PkC = 1,82. )/(325,10127994,0.802,11 22 mN=

2. on luyn:

Vn tc hi qua l: oL = )/(732,111,01732,1

1,0smL ==

Nn: PkL = 1,82. )/(468,14521614,1.732,11 22 mN=

II TR LC DO SC CNG B MT:



- -48


V a c ng knh l > 1mm: p dng cng thc IX.142 [II-194]:

s 2lo lo

4P1,3d 0,08d

=+

(N/m2)

1. on chng:

Ti nhit trung bnh ca pha lng trong phn chng TtbC = 99,63oC

Tra bng I.242, [trang 300.ST-I]kt hp ni suy ta c:

=

=

)/(10.443,34)/(10.831,17

3

3

mNmN

B

A

p dng cng thc (I.76), [trang 299. ST-I]:

BA

BA

BA

+=+=

.111

Nn: => )/(0117,010.443,3410.831,17

10.443,34.10.831,1733

33

mNC =+

=

Vy: )/(998,11003,0.08,0003,0.3,10117,0.4 2

2 mNP C =+=

2. on luyn:

Ti nhit trung bnh ca pha lng trong phn luyn TtbL = 79,18oC

Tra bng I.242, [trang 300.ST-I]kt hp ni suy ta c:

=

=

)/(10.7943,36)/(10.5697,19

3

3

mNmN

B

A

p dng cng thc (I.76), [trang 299. ST-I]:

BA

BA

BA

+=+=

.111

Nn: => )/(0128,010.7942,3610.5697,19

10.7942,36.10.5697,1933

3.3

mNL =+

=

Vy: )/(13,13003,0.08,0003,0.3,10128,0.4 2

2 mNP L =+=

III TR LC THY TNH DO CHT LNG TRN A TO RA:

p dng cng thc [tr 194.ST-II]:



- -49


Pb = 1,32

X3c

c

GK.h Km.L

+ x .g

Trong :

hc chiu cao ca chy trn (m).

K : t s gia khi lng ring cht lng bt v khi lng ring ca

cht lng, ly gn bng 0,5.

Gx: lu lng th tch ca pha lng, (kg/h).

Lc : chiu di ca g chy trn,(m).

m h s lu lng qua ca chy trn.

x khi lng ring ca lng (kg/m3).

g gia tc trng trng ly bng 9,81 (m/s2).

Tnh chiu di g chy trn:

Ta c: Squt - S = Sbn nguyt

22

R2%20

2cosR

2sinR

21.2

2R

pi=

- sin = 0,2pi

Dng php lp : = 1,627 (Rad)

Nn: Lc = Dsin 2

= 1,2. sin2627,1

= 0,872 (m)

1. on chng:

Theo phn trc ta c:

===

=

)/(402,5)/(41,908

4567,4907)/(4567,4907

)/(41,90833

3

hmhmhkgG

mkg

xC

xtbC



- -50

R

Lc


Lp t s: 5195,6872,0402,5

>==C

xC

LG

=> m = 10000

Vy: PbC = 1,3 =

+ 81,9.41,9085.0.)

872,0.10000402,5(05,0.5,0 3 2 331,72(N/m2)

2. on luyn:

Theo phn trc ta c:

===

=

)/(58,266,820858,2114)/(858,2114

)/(66,8203

3

hmhkgG

mkg

xL

xtbL

Lp t s: 596,2872,058,2

m = 6400

Vy: PbL = 1,3 =

+ 81,9.66.8205.0.)

872,0.640096,2(05,0.5,0 3 2 295,935(N/m2)

IV TNG TR LC THU LC CA THP:

Tng tr lc ca 1 a trong on chng ca thp l:

PC = PkC + PC + PbC

= 101,325 + 11,998 + 331,72 = 445,043(N/m2)

Tng tr lc ca 1 a trong on luyn ca thp l:

PL = PkL + PL + PbL

= 145,468 + 13,13 + 295,935 = 454,533(N/m2)

Kt lun:

Tng tr lc thy lc ca thp:

P=nttC.PC+nttL.PL=13.445,043+ 5. 454,533= 8058,224(N/m2)

CHNG V:

TNH CN BNG NHIT LNG



- -51


I - CN BNG NHIT LNG THIT B UN NNG HN HP U:

Theo nh lut bo ton nng lng ta c: Tng nhit lng i vo thit b

un nng bng tng nhit lng i ra khi thit b un nng.

Ta c:

D f F ng11 xq1Q Q Q Q Q+ = + + (J/h) [II-196]

Trong :

D 1Q : nhit lng do hi t mang vo, (J/h).

fQ : nhit lng do hn hp u mang vo, (J/h).

FQ : nhit lng do hn hp u mang ra, (J/h).

ng1Q : nhit lng do nc ngng mang ra, (J/h).

xq1Q :nhit mt mt ra mi trng xung quanh ly bng 5% nhit tiutn, (J/h).

1. Nhit lng do hi t mang vo D 1Q :

D 1Q = 1 1 1 1 1 1D . D (r C ) = + (J/h) [II-196]

Trong :

D1- lng hi t,(Kg/h).

r1 - n nhit ha hi ,(J/Kg).

1 - hm nhit ca hi t,(J/Kg).

1 - nhit nc ngng,(0C).C1 nhit dung ring ca nc ngng,(0C).

Ta coi thit b lm vic p sut 2at.

Tra STI-tr254,ti P=2 at kt hp ni suy ta c:

r1 527 (Kcal/kg) 2208.103 (J/Kg)

2. Nhit lng do hn hp u mang vo: fQ



- -52


f f fQ F .C .t= (J/h) [II-196]

Trong :

F lng hn hp u, (Kg/h).

tf nhit u ca hn hp, (0C).

Cf nhit dung ring ca hn hp u, (J/Kg.).

3. Nhit lng do hn hp u mang ra: FQ

F F FQ F .C .t= (J/h) [II-196]

Trong :

CF nhit dung ring ca hn hp khi i ra, (J/Kg.).

tF - nhit hn hp khi ra khi thit b un nng, (0C).

4. Nhit lng do nc ngng mang ra: ng1Q

.ng1 ng1 1 1 1 1 1Q G .C D .C .= = (J/h) [II-196]Trong :

Gng1 lng nc ngng bng hi t, (Kg/h).

5. Nhit mt mt ra mi trng xung quanh ly 5% nhit tiu tn: xq1Q

xq1 1 1Q 0,05.D .r= (J/h) [II-196]

Vy lng hi t ( lng hi nc) cn thit un si dung dch u n

nhit si l:

F ng1 xq1 f F f1

1 1

Q Q Q Q Q QDr 0,95.r

+ +

= =

F F f f1

1

F(C .t C .t )D0,95.r

= (Kg/h)

tf nhit hn hp u vo bng nhit mi trng;

tf = 250C

Tra bng I.153 v bng I.154 ( STTI 172) nhit dung ring CP(J/Kg.)



- -53


Ni suy ti tf = 250C :

=

=

)./(75,4178)./(5,2652

kgJCkgJC

B

A

Nhit dung ring ca hn hp l:

)./(35,369075,4178).32,01(5,2652.32,0

).1(.

kgJ

CaCaC BFAFf

=

+=

+=

tF nhit hn hp khi i ra khi thit b un nng:

tF = 84,670C

Tra bng I.153 v I.154 trang 172 STTI nhit dung ring CP(J/Kg.)

Ni suy ti tF = 84,670C

=

=

)./(66,4220)./(95,3449

dokgJCdokgJC

B

A

Nhit dung ring ca hn hp l:

)./(033,397466,4220).32,01(95,3449.32,0

).1(.

kgJ

CaCaC BFAFf

=

+=

+=

Lng hi nc cn thit l:

D1 = )/(866,104710.2208.95,0)25.35,369067,84.033,3974.(9000

.95,0...(

31

) hkgr

tCtCF ffFF=

=

II - CN BNG NHIT LNG CA THP CHNG LUYN:

Tng nhit lng mang vo thp bng tng nhit lng mang ra khi thp, ta

c s cn bng nhit :



- -54

H2O

H2O

H2O

H2O

Qy

QF

QD1

QD2

Qxq2

Qng1

Qxq1

QW

Qf

QP

QR

Qng2


T s trn ta c:

F D2 R y w xq ng2Q Q Q Q Q Q Q+ + = + + + [II-196]

Trong :

D2Q : nhit lng do hi t mang vo thp, (J/h),

RQ : nhit lng do lng lng hi lu mang vo thp, (J/h),

yQ : nhit lng do hi t mang ra nh thp, (J/h),

wQ : nhit lng do sn phm y mang i, (J/h),

xqQ : nhit lng mt mt ra mi trng ( ly 5% nhit tiu tn y

thp), (J/h),

ng2Q : nhit lng do nc ngng mang ra ngoi, (J/h)

1. Nhit lng do hn hp u mang ra: FQ

Ta c:

)/(10.028,367,84.9000.033,3974.. 9 hJtCFQ FFF ===

2. Nhit lng do hi t mang vo thp: 2DQ

D 2Q = 2 2 2 2 2 2D . D (r .C ) = + , (J/h) [II-197]

Trong :

D2 lng hi t cn un si dung dch y thp,(Kg/h)

r2 - n nhit ha hi,(J/Kg).SV: L Vn NLp: C-H Ha 3 K3


- -55


22 ,C - nhit v nhit dung ring ca nc ngng (0C).

3. Nhit lng do lng lng hi lu mang vo thp: RQ

R R R RQ G .C .t= (J/h) [II-197]GR lng lng hi lu.

GR =GP.Rth=3436,3636.0,8172=2808,196 (Kg/h)

tR nhit ca cht lng hi lu

tR = tP = 79,180C

CR nhit dung ring ca lng lng hi lu ti tR = tP = 79,180C

Tra bng nhit dung ring ca cht lng ta c:

=

=

)./(4190)./(25,2980

dokgJCdokgJC

B

A

=> RC = aP.CA + (1-aP).CB

= 0,83.2980,25 + (1-0,83). 4190

= 3185,91 (J/Kg.)

Vy : QR = 2808,196.3185,91.79,18 = 4,723.108 (J/h)

4. Nhit lng do hi t mang ra nh thp: yQ

y th dQ P(1 R )= + (J/h) [II-197]

d : nhit lng ring ca hi nh thp.

d 1 P 2 P.a .(1 a ) = +

21 , ; nhit lng ring ca tng cu t t0 = tP = 79,180C.

Trong : Ctr P .+=

Tra bng I.212 [ST-I. Tr254] ta c:

=

=

)/(10.854,2343)/(10.107,847

32

31

kgJrkgJr

Tra bng nhit dung ring ca cht lng ti tP = 79,180C ta c:

=

=

)./(4190)./(75,3209

dokgJCdokgJC

B

A

Nhit lng ring ca Etylic l:



- -56


)/(10.101,175,3209.18,7910.107,847. 6311 kgJCtr AP =+=+=

Nhit lng ring ca Nc:

)/(10.676,24190.18,7910.854,2343. 6322 kgJCtr BP =+=+=

)/(10.369,1

10.676,2).83,01(10.101,1.83,06

66

kgJd

=

+=

Vy nhit lng do hi t mang ra nh thp:

)/(10.324,210.354,1).8172,01.(457,94

).1.(

8

6

hJ

RPQ dthy

=

+=

+=

5. Nhit lng do sn phm y mang i: wQ

w w w wQ G .C .t= (J/h) [II-197]

Gw = 5563,6364 ( kg/h)

Cw: nhit dung ring ca sn phm y tw = 99,630C

w w A w BC a .C (1 a ).C= + (J/Kg.)

Tra bng I.153 (ST-I. tr171) kt hp phng php ni suy ta c:

=

=

)./(26,4229)./(45,3514

dokgJCdokgJC

B

A

Thay s ta c:

CW = 0,005.3514,45 + ( 1 0,005).4229,26

=4225,686 ( J/kg.)

Vy nhit lng sn phm y mang i:

QW = 5563,6364.4225,686.99,63 = 2,432.109 ( J/h)

6. Nhit lng do nc ngng mang ra ngoi 2ngQ

ng2 ng2 2 2 2 2 2Q G .C . D .C .= = (J / h) [II-198]

Trong :

Gng2 lng nc ngng, bng lng hi t , (Kg/h).

2 - nhit nc ngng (0C)

C2 nhit dung ring ca nc ngng(J/Kg.)

7. Nhit lng mt mt ra mi trng: xqQ



- -57


xqQ = 0,05.D2.r2 (J / h) [II-198]

Vy lng hi t cn thit un si dung dch y thp l:

y w ng2 xq2 F R2

2

y w F R

1

Q Q Q Q Q QD

Q Q Q Q0,95.r

+ + + =

+

=

Thay s ta c:

D2 = )/(67,87510.2208.95,010.732,410.028,310.432,210.324,2

3

8998

hkg=+

III - CN BNG NHIT LNG CA THIT B NGNG T:

1. Nu ch ngng t lng hi lu:

th nl n 2 1P.R .r G .C .(t t )= [II-198]

lng nc lnh cn tiu tn l:

thnl

n 2 1

P.R .rGC (t t )

=

Trong :

r - n nhit ngng t, (J/Kg).

Cn nhit dung ring ca nc nhit trung bnh

1 2tb

t tt2+

=

t2,t1 nhit ra v vo ca nc lm lnh.

Ta chn: t1 = 250C

t2 = 600C

Vy nhit trung bnh: 0tb25 60t 40 C

2+

= =

Ti ttb = 400C tra bng nhit dung ring I.147 [I-165] kt hp ni suy:

Cn = 4781,31529 (J/Kg.)

Ta ly r tP = 79,18oC ta c:

Tra bng I.212 [Tr254. ST-I], kt hp ni suy ta c :



- -58


==

==

)/(2343850)/(33,554)/(847107)/(328,202

kgJkgKcalrkgJkgKcalr

B

A

Suy ra :

r =847107.0,6564 + 2343850.(1- 0,6564)

= 1361387,895 (J/kg) .

Vy lng nc lnh cn tiu tn

)/(14534,22845)2560.(31529,4781

895,1361387.8172,0.3636,34361 hkgGnl =

=

2. Nu ngng t hon ton :

P. (Rth + 1).r = Gnl2.Cn.(t2 - t1) [II-198]

Vy lng nc lnh cn tiu tn

)/(565,50800)2560.(31529,4781

895,1361387).8172,01.(3636,34362 hkgGnl =

+=

IV - CN BNG NHIT LNG CA THIT B LM LNH:

1. Nu trong thit b ngng t ch ngng t lng hi lu:

p dng cng thc:

( )P 1 2 n3 n 2 1' 'P r C t t G .C (t t ) + = [II-198] lng nc lnh cn tiu tn l:

( )P 1 2n3

n 2 1

' 'P r C t tG

C (t t )

+ =

Trong : Gn3 : lng nc lm lnh tiu tn (kg/h).

CP: nhit dung ring ca sn phm nh ngng t,(J/Kg.).

t1, t2 : nhit u v cui ca sn phm nh ngng t.

Cn nhit dung ring ca nc nhit trung bnh,(J/Kg.).

Ta chn : t2 = 250C

t1 = tP = 79,180C

Cttttb0

'2

'1' 09,52

218,7925

2=

+=

+=

Tra bng nhit dung ring I.153 [I-171]kt hp ni suy:



- -59


=

=

)./(9325,4180)./(83,2812

dokgJCdokgJC

B

A

=> CP = aP.CA + (1-aP).CB = 0,83.2812,83 +(1- 0,83).4180,9325

= 3045,407 (J/Kg.)

Ti ttb = 52,090C tra bng nhit dung ring I.147[I-165] kt hp ni suy:

Cn = 4184,383 (J/Kg.)

Ta ly r ttb = 52,09oC ta c:

Tra bng I.212 [II-254], kt hp ni suy ta c :

==

==

)/(518,2141891)/(528,511)/(945,2428297)/(989,579

kgJkgkcalrkgJkgkcalr

B

A

Suy ra : r =2428297,945.0,6564 +2141891,518.(1- 0,6564)

= 2329888,697 (J/kg) .

Vy lng nc lnh cn dng l:

[ ] )/(747,58539)2560.(383,4184

)2518,79.(407,3045697,2329888.3636,34363 hkgGn =

+=

2. Nu ngng t hon ton:

p dng cng thc:

( )P 1 2 n4 n 2 1' 'P.C t t G .C (t t ) = [II-198] lng nc lnh cn tiu tn l:

( )P 1 2n4

n 2 1

' 'P.C t tG

C (t t )

=

Trong :

Gn4 : lng nc lm lnh tiu tn (kg/h).

CP: nhit dung ring ca sn phm nh ngng t,(J/Kg.).

t1, t2 : nhit u v cui ca sn phm nh ngng t.

Cn nhit dung ring ca nc nhit trung bnh,(J/Kg.).

Vy lng nc lnh cn dng l:SV: L Vn NLp: C-H Ha 3 K3


- -60


Gn4 = )/(542,3871)2560.(383,4184)2518,79.(407,3045.3636,3436 hkg=

PHN III:

TNH TON C KH.

I CHN VT LIU LM THN THP:



- -61


Thn hnh tr l b phn ch yu to thnh thit b ha cht. Ty theo iu

kin lm vic m ngi ta la chn vt liu v phng php ch to. Do iu kin

n l thp lm vic p sut thng v nhit khng cao lm. Dung dch

cha Etylic v Nc do ta chn vt liu l thp cacbon vi tn thp l C T3 lm

thn thp , l mt vt liu bn chu nhit. Thn thp c ch to bng cch

cun tm vt liu vi kch thc nh sau hn gip mi li.

c trng c hc ca mc thp ny tra t bng XII.4,7 [STQTTBT2-310] - Chiu dy tm thp (mm): 4-20mm- Gii hn bn (N/m2):

k=380.106(N/m2) ch= 240.106(N/m2)- nht va p ak .106= 0,08 (J/m2)- %=15%- H s gin khi ko nhit 20-100oC at.106= 11.0(1/oC)- Khi lng ring = 7.85103(kg/m3)- H s dn nhit 20-100oC =50(w/m.)

Cch ch to :T Dt = 1200 ( mm ) hn h quang tay vi kiu hn gip mi 2bn : (Tra bng XIII.8 STQTTBT2-362 ) Khi ch to loi ny cn ch :

p sut v nhit lm vic nh hng n chiu dy vt liu.

Chn vt liu ph thuc vo mi trng lm vic.

m bo ng hn cng ngn cng tt.

Ch hn gip mi.

B tr ng hn dc ( cc ng thn ring bit ln cn) cch nhau t

nht l 100 (mm).

Khng khoan l qua mi hn.

Quan h gia chiu cao v ng knh ca thp (ng knh trong) i

vi thit b t thng ng theo yu cu sn xut ho cht: H/D30.

Chiu dy thn thp c xc nh theo cng thc sau:

[ ]tD .Ps C

2. . P= +

, (m)

Trong :



- -62


Dt ng knh trong ca thn hnh tr (m) - h s bn ca thnh hnh tr theo phng dc;[ ] - ng sut cho php.C s b xung do n mn, bo mn v dung sai v chiu dy (m)

P p sut trong ca thit b (N/m2)

Do mi trng lm vic l hn hp lng-hi nn:

P = Plv + Ptt (N/m2)

Trong : Plv p sut lm vic , N/m2

Ptt p sut thy tnh ca ct cht lng , N/m2

Ptt = x .g.HH - chiu cao max ca ct cht lng (m) : khi lng ring ca ct cht lng (Kg/m3)g : gia tc trng trng , g = 9,8 m/s2

Ta c chiu cao ct cht lng H = Hthp = 6,8 (m)

Khi lng ring ca cht lng trong thp:

)/(535,8642

66,82041,9082

3mkgxtbLxtbCtb =+

=

+=

Do : Ptt= x .g.H =864,535.9,8. 6,8=57612,6124 (N/m2))/(5894,1589666124,5761210.81,9.03317,1 24 mNP =+=

tnh ton sc bn ca thit b th trc ht phi xc nh ng sut cho

php ca vt liu l thp CT3. ng sut cho php ca thp theo gii hn bn khi

ko v khi chy c xc nh theo cng thc di y:

[ ] kkk

.n

= ( N/m2)

[ ] ccc

.n

= (N/m2)

Trong : nk. nc: l h s an ton theo gii hn bn v khi chy

k c, : gii hn bn khi ko v khi chy, N/m2



- -63


: h s hiu chnh, y l thit b loi II khng b t nng trc tip nn : 1 =

Tra bng XIII.3 [II-356] ta c nk = 2,6 ; nc = 1,5

Tra gii hn bn khi chy v khi ko [II-309] ta c:6

k6

c

380.10

240.10

=

= (N/m2)

Do : [ ]6

6kk

k

380.10. .1 146,15385.10n 2,6

= = = (N/m2)

[ ]6

6cc

c

240.10. .1 160.10n 1,5

= = = (N/m2)

So snh hai kt qu ng sut cho php theo gii hn bn khi ko v gii hn

bn khi chy ta chn [ ] theo gi tr nh hn, nn [ ] 6146,15385.10 = (N/m2)Tnh : ta thit k chn hn theo phng php hn bng tay bng h quang in, kiu hn gip mi hai bn thnh c l nhng c gia c hon ton khi

h 0,95 = = tra [II-362]

Ta lp t s [ ] 504298,873

5894,15896695,0.10.15385,146. 6

>==P

do vy ta c th b qua

P di mu ca cng thc tnh chiu dy thp.

Tnh C: C= C1 + C2 + C3

C1: b sung do n mn, xut pht t iu kin n mn ca vt liu ca mi

trng v thi gian lm vic ca thit b, (m). Chn C1 =1 mm= 0,001m: do tc

n mn ca thp l 0,05- 0,1mm/nm.

C2 b sung do hao mn C2 = 0 ( v nguyn liu u l lng hi khng phi l

rn nn c th b qua C2.

C3 b sung theo dung sai chiu dy, chn C3 = 0,5.10-3 m/nm

Khi chiu dy ca thp l:

S= [ ] [ ] )(10.469,310.5,195,0.10.15385,146.25894,158966.2,1

..2. 33

6 mmCPDt

=+=+



- -64


Chun ha chn s = 5 (mm).

Kim tra ng sut cho thnh thit b theo p sut th.

Theo [II-366] ng sut tnh theo p sut th phi tha mn iu kin:

[ ]t 0 cD (s C) .P2.(s C). 1,2+

=

, (N/m2)

Trong : P0 l p sut th tnh ton:

P0 = Ptt + P1, (N/m2)

Vi Ptt- p sut th thy lc. V P = 158966,5894 [ ] 60,07 0,5 .10 (N/m2) tra bng XIII.5 nn:

Ptt =1,5.P = 1,5. 158966,5894=238449,884 (N/m2) [II-358]

Tnh P1 p sut thy tnh ca ct cht lng.

P1 = .g.H - khi lng ring ca cht lng nhit trung bnh,(Kg/m3) .g gia tc trng trng g = 9,8 m/s2.

Nng trung bnh ca hn hp trong thp :

Xtb = 2713,0300196,06564,01555,0

3=

++=

++ WPF XXX

Da vo th T x,y: Cttb 01305,82=

Ni suy khi lng ring ca Etylic v Nc theo [I-9]:

=

=

5087,970976,732

B

A

(kg/m3)

Vy khi lng trung bnh ca hn hp l:

)/(322,8795087,970

32,01976,73232,011 3mkgaa

B

F

A

F=

+=

+=

)/(02,585988,6.8,9.322,879 21 mNP ==

0P = 238449,884 + 58598,02 = 297047,904 (N/m2)

Do : [ ] )/(3,53758970

95,0.10).5,15.(2904,297047.10).5,15(2,1 2

3

3

mN=

+=

Ta thy 66

10.2002,110.2403,53758970 == tha mn iu kin



- -65


Vy chiu dy thp l 5 (mm).

II TNH NG KNH CC NG DN:

ng knh ng dn c tnh theo cng thc :

d = (m). [STQTTB-I-369].

trong : V : lu lng cc cht chuyn ng trong ng (m/s).

V = .

G ; lu lng dng pha (kg/s)

: khi lng ring ca dng pha .

: vn tc ca cc cht trong ng.

1) ng knh ng dn sn phm nh.

ng dn sn phm nh l ng ni gia lp thp v thit b ngng t .

- nng cu t d bay hi : y = 0,6564 (phn mol).

- nhit nng hn hp l : t = t = 79,18 C

hn hp i ra khi nh thp l hn hp hi c khi lng mol phn t trung bnh

l :

M = y . 46 + (1- y ).18 = 36,38 (kg/kmol).

- khi lng hn hp ra khi nh thp :

g = 171,648 (kmol/h)= 0,04768 (kmol/s)= 1,7346 (kg/s).

- khi lng ring ca hn hp l :

= [ ]

)273.(4,22273.)1(.

tMyMy BA+

+ = )18,79273.(4,22

273.38,36+

= 1,259 (kg/m ).

- thp lm vic p sut thng vi hi qu nhit ,chn = 30 50 (m/s).

Chn vn tc hi l : = 40 (m/s).

- ng knh ng dn sn phm nh :

d= )(209,0259,1.40.3600.785,0

38,36.648,171 m=

theo STQTTB-II-434 ta chn d = 200(mm) khi chiu di ng hi bn

ngoi l l = 130(mm).

- tnh li vn tc ta c : = 43,88(m/s).

2) ng knh ng hi lu sn phm nh :



- -66


- hn hp trong ng l hn hp lng c nng : x= x = 0,6564(phn mol)

Nhit ca hn hp l : t= t = 79,18 C ,khi lng phn t ca hp

cht l : M = 36,38(kg/kmol).

khi lng ring ca hn hp lng c tnh theo cng thc :

B

P

A

P aa

+=11

[STQTTB-I-5].

Vi : :khi lng ring ca hn hp (kg/m ).

A ,B : khi lng ring ca Etylic, nc Ta c : a = 0,83 (phn khi lng).

Ta c : t = 79,18 C tra s tay QTTB-I-5 ta c:

A = 735,779 (kg/m ).

B = 972,451 (kg/m ).

Thay vo cng thc trn ta c: = 767,535 (kg/m )- lng sn phm nh hi lu l:

V = = 535,767.36008172,0.38,36.457,94

= 1,02.10 (m/s).

- ta coi cht lng t chy nn = 0,1 0,5 (m/s) chn = 0,3(m/s).

Ta c: d= 3,0.785,0

10.02,1 3 = 0,066 (m)= 66 (mm).Theo STQTTB-II-434

- ta chn d = 70 (mm),khi chiu di ng l = 110 (mm) .

Tnh li vn tc hi ta c = 0,265 (m/s).

3 )ng knh ng tho sn phm y:

Hn hp c tho l hn hp c nng x= x = 0,00196(phn mol).

Nhit hn hp l : t = t = 99,63 C .

Khi lng ring ca hn hp lng c tnh theo cng thc :

= A

Wa + B

Wa

1 [STQTTB-I-5].

Vi : : khi lng ring ca hn hp .

A ,B : Khi lng ring ca Etylic v Nc. Ta c : a = 0,05 (phn khi lng).



- -67


A = 716,3515(kg/m ).

B = 958,259 (kg/m ).

Do : =942,348 (kg/m ).khi lng sn phm y l :

G = 5563,6364 (kg/h).

Chn vn tc : = 0,3 (m/s).

d = )(08345,03,0.348,942.3600.785,0

6364,5563 m=

Chn d = 80 (mm),khi chiu di ng bn ngoi : l = 110 (mm).-

STQTTB-434.

Tnh li vn tc : = 0,33 (m/s).

4) ng knh ng dn nguyn liu u .

- ng dn nguyn liu u l ng ni gia thit b gia nhit hn hp u v thp

ti a tip liu.

- lng hn hp u : F= 9000(kg/h).

Nng hn hp u ;x= x = 0,1555 (phn mol).-khi lng ring ca

hn hp :

a = 0,32 (phn khi lng).

- nhit ca hn hp l : t=t = 84,67C .

A = 730,563 (kg/m ).

B = 968,731(kg/m ).

Thay vo cng thc : = A

Fa + B

Fa1

= 877,218 (kg/m ).

Chn vn tc = 0,4 (m/s).

Ta c: d = )(0953,04,0.218,877.3600.785,0

9000 m=

- chn d = 100 (mm) ,khi chiu di ng bn ngoi l = 120(mm)

5) ng knh ng hi lu sn phm y :

hn hp c hi lu l hn hp vi nng cu t d bay hi :



- -68


y = 0,0053 (phn mol).

-nhit ca hn hp : t=t = 99,63 C

Ta c khi lng ring ca hn hp l :

[ ] [ ] )/(594,0)63,99273.(4,22

273.18).0053,01(46.0053,0)273.(4,22

273.).1( 3. mkgtMyMy BWAW

=

+

+=

+

+=

Khi lng mol ca hn hp hi l:

M = y.MA +(1-y ).MB = 0,0053.46 + (1- 0,0053).18

= 18,15 ( kg/kmol)

- lng sn phm y hi lu l:

g= 2829,757 (kg/h).

- vn tc hi l : = 40 (m/s).

Do : d = )(2053,040.594,0.3600.785,0

757,2829 m=

Chn d = 200 (mm),khi chiu di ng ni bn ngoi l = 130(mm).

III TNH Y V NP THIT B:

y v np thit b l b phn quan trng ca thit b v thng c cng

loi vt liu vi thn thp. V thp lm vic p suy thng v thn tr hn nn

ta chn y v np thit b hnh elip c g i vi thit b thng ng c P> 7.104

Chiu dy ca np v y thit b c xc nh theo cng thc sau:

[ ]t t

h b

D .P DS . C3,8. . .k P 2.h

= +

Trong : P= 158966,5894(N/m2).

h : h s bn mi hn;

k: h s hiu chnh .

hb: chiu cao np



- -69


C: h s hiu chnh C= 1,5.10-3 m V c tng thm mt cht

ty thuc chiu dy:

Thm 2mm khi S - C < 10mm

Thm 1mm khi 10 < S C ==Pk h

+ np: [ ] 30858,727

5894,15896695,0.8333,0.10.15385,146.. 6

>==Pk h

Nn ta c th b qua P mu s ca cng thc tnh chiu dy ca y v np.

1. Chiu dy ca np:



- -70


[ ]t n t

nh b

D .P DS . C3,8. . .k 2.h

= +

C

CSn

+=

+=

4

6

10.68,83,0.2

2,1.83333,0.95,0).10.15385,146.(8,3

5894,158966.2,1

=> S - C = 8,68.10-4 = 0,868 (mm).

Ta thy: S - C


Ta c: 66

6 10.2002,110.240

2,110.153,141 ==


Theo chn bch lin bng thp ni thit b P = 0,1.106 (N/m2) ; D = 1,2 (m). Tra theo bng XIII.27 Ta c bng sau :

VI TNH GI V TAI TREO:

Thng thng ngi ta t thp trn gi gi l chn thp. xc nh

c chn ta phi xc nh c ton thit b c ti trng bao nhiu t nn n.

Ta c:

Mtb = Mnp + My + Ma + Mc.lng + Mt + Mch + Mbb (Kg)

1. Khi lng np v y tra bng XIII.11 [II-384] :

Mnp = My = 64,2 (Kg).

2. Khi lng ca a:

Mia = Ntt.ma = Ntt. a.Va (Kg)Trong :

a - khi lng ring ca vt liu lm a: do a lm bng vt liuthp cacbon CT3 tra bng XII.7 [II-313]. Ta c a = 7850 (Kg/m3).

Va = .FtdTrong : Ftd din tch a bng 90% din tch mt ct thp.

Ftd = )(02,14

9,0.2,1.4

9,0.. 222 mD == pipi



- -

Py.10-6

(N/m2)

Dt(mm)

D Db D1 D0

Bu lng

(mm)

(mm) db Z(ci) h

0,1 1200 1340 1290 1260 1213 M20 32 25

73


: chiu dy a ta chn l 5 (mm)Khi lng a ton thp l:

Ma = Ntt. .Va = 16.7850. 5.10-3 .1,02 = 640,56 (Kg)3. Khi lng ca cht lng trn a Mc.lngGi s thp cha ton b cht lng nh hn hp u:

Mc.lng = Fia.(V V )

Trong :

F : khi lng ring trung bnh ca cht lng trong thp

)/(535,8642

66,82041,9082

3mkgxtbLxtbCF =+

=

+=

V: th tch ton thp

V = )(691,78,6.4

2,1..4. 322 mHD == pipi

Va: bng th tch ca a l.

Va = 2

tt.DN . .90%.4

pi = 16.0,005.0,9. =4

2,1. 2pi 0,0814 ( m3)

Vy khi lng ca cht lng trn a:

Mc.lng = Fia.(V V ) = 864,535.(7,6910,0814) = 6578,77 (Kg)

4. Khi lng thn thp:

Mt =ML + MC = t t.V (m)

Trong : Vt = 2 2t n tH . .(D D )4pi

Dn: ng knh ngoi cng v thit b

Dn = Dt + 2. = 1,2 + 2.0,005 = 1,21 (m)

=>Vt = 6,8 )2,121,1(4

pi=0,053(m3)

Vy: Mt = 7850. 0,053 = 416,05 (Kg)

5. Khi lng ca ng chy chuyn:



- -74


Do ng knh ng chy chuyn ca on chng ln hn on luyn nn ta

tnh theo on chng.

Ta c: Mch = n t

2 2tt ch ch c tN . .(d d ).H .4

pi

Trong :

chtd - ng knh trong ca ng chy chuyn.

chtd =0,113(m)

chnd - ng knh ngoi ca ng chy chuyn.

chnd = chtd +2.0,003 = 0,113 + 0,006 = 0,119(m)

Hc chiu di ng chy chuyn = 0,3 (m).

Vy: Mch =16. = 7850.3,0).113,0119,0(422pi 41,195 (kg)

6. Khi lng bch, chi tit ph nh bu lng, c vt, thanh , bch, kp a

Ta ly khi lng bch, cc chi tit ph: bu lng, c vt, thanh , l

khong: Mbb = 1000 (Kg).

*Vy khi lng ton thp l:

Mtb = 64,2+ 64,2 +640,56 + 6578,77+ 416,05 +41,195+ 1000 = 8804,975 (Kg)

Trng lng thp :

P = m.g = 8804,975.9,8 = 86288,755 (N)

Chn chn 4 chn mi chn chu ti trng l

G = )(189,215724

755,862884

NP ==



- -75


m bo an ton cho thit b, ta chn: Gc = 2,5.104 (N).

Truc

thiet

b

Theo ay thiet b

Tra bng XIII.35 [II - 437] ti trng 1 chn cho php l 2,5.104 (N). Ta c bng

sau:

Ti trng cho php trn 1chn G.10-4 (N/m2) 2,5

B mt F.104 (m2)444

Ti trng cho php ln bmt q.10-6 (N/m2)

0,56

L mm 250B 180B1 215B2 290H 350H 185S 16L 90



- -76


d 27

*Chn tai treo: tai treo c gn trn thn thp gi cho thp khi b dao

ng trong iu kin ngoi cnh

Ta chn 4 tai treo. Ti trng mi tai treo l:

G = )(189,215724

755,862884

NP ==

Ti trng 1 tai treo cho php l 2,5.104 (N)

c tra theo bng XIII.36 [II-438]

Ti trng cho php trn 1 taitreo G.10-4 (N/m2)

2,5

B mt F.104 (m2) 1,73

Ti trng cho php ln b mt q.10-6 (N/m2)

1,45

L

mm

150B 120B1 130H 215S 8l 60a 20d 30

Khi lng 1 tai treo 34,8



- -77


PHN IV :

TNH THIT B PH.

CHNG I:

THIT B GIA NHIT HN HP U.

un nng hn hp u gm 0,32(phn mol/phn mol) Etylic v 0,83

(phn mol/phn mol) Nc theo phn khi lng vi nng sut 9000 (kg/h). Ta gi

thit dung dch u c nhit u l 250C, cn un nng ti nhit si ca hn

hp l tF = 84,67 0C ; nhit si tra theo bng ng cn bng x,y-t tng ng

vi xF = 0,1555 (phn mol). un nng hn hp u ta dng thit b gia nhit

loi ng chm loi ng, dng hi nc bo ha un si hp u.

Chn thit b trao i nhit loi ng chm thng ng vi cc thng s:

Chiu cao ng 2m.

ng knh ngoi ng: dn = 29 mm = 0,029 m.

B dy ng: t = 2 mm = 0,002 m.

ng knh trong ca ng: dtr = 0,025 m.

Dung dch i trong ng, hi t i ngoi ng.

Chn vt liu ch to ng l thp khng g.

Chn hi t l hi nc bo ha P = 2atm, c ts = t = 119,62 C

[STQTTB-I-230].

Ngoi ra ta cn xc nh cc thng s sau:

B mt truyn nhit ca thit b gia nhit hn hp u (m2).

S ng (n) ca thit b gia nhit.

S ngn (m) trong thit b.

ng knh thit b.

I B MT TRUYN NHT CA THIT B GIA NHIT HN HP U:

2

tb

QF (m )q

=



- -78


Trong :

Q lng nhit cn dng un nng hn hp u n nhit cn thit (W).

qtb lng nhit ti t hi nc bo ho qua thnh ng trao i nhit n dung

dch (W/ m2).

1. Lng nhit cn dng un nng hn hp u n nhit cn thit:

P F fQ F.C (t t )=

Trong : F = 9000(Kg/h) = 5,236009000

= (Kg/s)

CP nhit dung ring ca dung dch nhit trung bnh (J/ kg.)

Xc nh nhit trung bnh ca dung dch:

Trong thit b gia nhit hn hp u ta cho 2 lu th i ngc chiu nhau, khi

hiu s nhit trung bnh ca dung dch c xc nh theo cng thc:

1 2tb

1

2

t tt tlnt

=

nu 12

tt

>2

1 2tb

t tt2

+ = nu 12

tt

>2

Hi t l hi nc bo ha nn nhit khng thay i v nhit p

sut chn (2atm) : tbh = 119,62 0C.

d bh ft t t = = 119,62 25 = 94,620C

c bh Ft t t = = 119,62 84,67 = 34,950C

Khi : 7074,295,3462,94

==

c

d

tt

> 2

Nn hiu s nhit trung bnh c xc nh:

Cttb091,59

95,3462,94ln

95,3462,94=

=

Nhit trung bnh ca dung dch l:

ttb = tbh - tbt =119,62 59,91 = 59,710C

Vi ttb = 59,710C tra bng I.153 [ST-I .tr171], kt hp vi ni suy ta c:



- -79


=

=

78,418923,2966

B

A

CC

(J/kg.)

Nng hn hp u l: aF = 0,32

=> CP = aF.CA + (1-aF).CB

= 0,32.2966,23 + (1- 0,32) .4189,78 = 3798,244 (J/Kg.)

Vy : Q = 2,5.3798,244.( 84,67 25) = 566603,049 ( W)

2. Nhit ti trung bnh:

Chn thit b loi ng chm lm bng thp khng g, ch chy trong ng

ca dung dich l ch chy xoy.

K hiu:

ttb1 nhit hi nc bo ha 1atm: ttb1 = tbh = 119,620C.

tT1 nhit mt ngoi ng

tT2 nhit mt trong ng

ttb2 nhit dung dch.

1t :hiu s gia hi t v mt ngoi ng: 1t = ttb1 tT1

2t :hiu nhit gia mt trong ng v dung dch: 2 T2 tb2t t t =

Tt : hiu nhit gia mt ngoi ng v mt trong ng :

Tt = tT1 tT2

: chiu dy thnh ng (m)tm : nhit mng nc ngng: tm = 0,5(ttb1 + tT1)

q1 : nhit ti ring pha hi ngng t

q2 : nhit ti ring pha dung dch

:1 h s cp nhit pha hi ngng t

:2 h s cp nhit pha dung dch

2.1. Nhit ti ring bn hi nc bo ha:

q1 = 11. t (W/m2)

H s cp nhit pha hi nc bo ha:

411

r2,04.A.t .H

=

(W/m2.)



- -80


Trong :

A ph thuc nhit mng nc ngng tm

r - n nhit ha hi ly theo nhit hi bo ha th (J/Kg)

t = 119,62 C P=2at: r=2203290,94 (J/kg) [STQTTB-I-414].

H- chiu cao ng (m) : H = 2m

Gi thuyt Cttt Tbh 011 2==

Th: tT1 = tbh - 1t = 119,62 2 = 117,620C

=> tm = Ctt bhT 01 62,118)62,11762,119.(21)(

21

=+=+

Theo bng s liu A tm [ST-II. tr29] kt hp ni suy ta c:

A = 187,379

=> 332,104132.210.2203.379,187.04,2

25,03

1 =

= (W/m2.)

Vy nhit ti ring bn hi nc bo ha :

q1 = 11. t = 10413,332.2 = 20826,664 (W/m2)

2.2. Nhit ti ring v pha dung dch:

2 2 2q . t= (W/m2)

H s cp nhit pha dung dch

xc nh h s cp nhit v pha dung dch th cn phi da vo ch

chy ca dung dch v cu to thit b: chn ch chy xoy Re = 104.

Ta c: 2 tr.dNu = 2 tr

Nu.d

=> = (W/m2.)

Trong : Nu chun s nuyxen.

dtr - ng knh trong ca ng dtr = 0,025 (m)

- h s dn nhit ca dung dch (W/m.)H s dn nhit ca dung dch c xc nh:

P3A.C . .

M = (W/m.do)

Đồ án chưng cất tháp đĩa rượu etylic - nước

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