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PHN I

N QTTB SV: L Th Lng LT Ho3-K3

M U I. GII THIU V HN HP C CHNG LUYN Trong qu trnh ch bin mt sn phm t sn phm th ban u n khi n tr thnh mt sn phm c ch c ng dng rng ri trong i sng th cng nghip ha hc ng gp khng nh trong s pht trin kinh t nc nh, cc phng php khc nhau nh: lng, lc, hp th, kt tinh, Chng ct.. nhng c bit l phng php chng luyn n c ng dng rng ri trong rt nhiu ngnh, lnh vc c bit l ngnh cng ngh ln men, cng ngh tng hp hu c, lc - ha du, cng ngh sinh hc. Vic la chn phng php v thit b cho ph hp ty thuc vo hn hp ban u, yu cu sn xut v iu kin kinh t. Chng l phng php dng tch ring cc hn hp lng cng nh cc hn hp ha lng thnh cc cu t ring bit da vo nhit si khc nhau ca cc cu t trong hn hp.

Khi chng ta thu c nhiu sn phm v thng c bao nhiu cu t th c by nhiu sn phm. Ring i vi phng php chng luyn hai cu t th sn phm nh gm ch yu l cu t d bay hi cn sn phm y gm ch yu l cu t kh bay hi.

Trong sn xut ta thng gp cc phng php chng khc nhau nh : chng n gin, chng bng hi nc trc tip, chng chn khng v c bit hn l chng luyn.

Chng luyn l phng php thng dng dng tch hon ton hn hp cc cu t d bay hi c tnh cht ha tan mt phn hoc ha tan hon ton vo nhau.Chng luyn p sut thp dng cho cc hn hp d b phn hy nhit nhit cao, cc cu t d bay hi v ngc li.Chng n gin dng tch cc hn hp gm cc cu t c bay hi rt khc nhau .Phng php ny thng dng tch s b v lm sch cu t khi tp cht.

Chng bng hi nc trc tip dng tch cc hn hp gm cc cht kh bay hi v tp cht khng bay hi ,thng dng trong trng hp cht c tch khng tan vo nc

Chng chn khng dng trong trng hp cn h thp nhit si cu t .V d nh trng hp cc cu t trong hn hp d b phn hy nhit cao hay trng hp cc cu t c nhit si qu cao.Chng nht luyn l phng php ph bin nht tch hon ton hn hp cc cu t d bay hi c tnh cht ha tan mt phn hay hon ton vo nhau theo bt k t l no bng cc thit b khc nhau nh: thp m, thp chp, thp ai vi hn hp Ru etylic- nc l hn hp 2 cu t ha tan hon ton vo nhau v c nhit si khc bit nhau cng iu kin p sut. Do phng php ti u tch cc hn hp trn l chng, ta nn dng thit b chng luyn loi thp m. Phng php ny da vo bay hi khc nhau gia cc cu t bng cch thc hin qu trnh chuyn pha v trao i nhit gia 2 pha lng- kh. Sn phm nh thu c gm cc cu t c bay hi ln v mt phn cu t c bay hi thp. Cn sn phm y thu c ch yu l cu t kh bay hi v mt phn cu t d bay hi. Ngy nay thp m c s dng rng ri v n c nhiu u im: Hiu sut cao v b mt tip xc ln, gii hn lm vic tng i rng, thit b n gin, gn nh d tho ri sa cha, tr lc khng cao nhng lm vic n nh v chng c sn phm i hi c tinh khit cao.Thp m c th lm vic p sut thng v p sut chn khng, lm vic lin tc hoc gin on.

Tuy nhin thp m cng c nhng hn ch sau: Kh lm t u m,Nu thp cao th phn phi cht lng khng u.

khc phc nhc im trn, ngi ta chia m ra nhiu tng v c t thm a phn phi cht lng i vi mi tng m nu thp qu cao.

Trong thp m, cht lng chy t trn xung phn b u trn b mt m, Kh i t di ln v tip xc vi cht lng v qu trnh chuyn khi xy ra.

Vt liu gia cng l thp khng g bi v hn hp cn tch l h n mn mnh, mt khc tuy gi thnh sn xut cn cao nhng p ng c nhng tiu chun c bn ca thit b ha cht l: chng n mn, bn nhit, c tnh tt, tui th lm vic lu di Bi n ny bao gm 6 ni dung chnh: s cng ngh sn xut trong thc t, tnh ton thit b chnh: Tnh ton cn bng vt liu ton thit b, chiu cao, ng knh thp, tr lc ca thp m, tnh cn bng nhit lng, tnh v chn thit b ph: thit b gia nhit hn hp u, bm.Tnh ton c kh v la chn: Tnh b dy, ng knh ng dn, Tnh y v np, chn bch ghp, tnh gi tai treo II. GII THIU V RU ETYLIC V NC.

1 . Nc:

-Nc l cht lng khng mu,khng mi,khng v.

- Nhit si p sut 760mmHg l 1000 C

- Ho lng 00C

- Khi lng ring l 997,08 kg/m3 250C

- nht bng 0,8937.10 N.s/ m2 = 893,7 Cp 250C

- Nhit dung ring Cp= 0,99892Kcal/kg. 250C

- Nhit ho hi p sut kh quyn r = 540 Kcal/kg

- Nc c cng thc phn t H2O ,cng thc cu to H-O-H

-Nc c hp cht phn cc mnh,c th ho tan nhiu cht rn,lng,kh

-Nc cn thit cho sinh hot hng ngy,sn xut nng nghip,cng nghip,xy dng,giao thng vn ti

- Nc dng iu ch oxy

2. Etylic : Ru etylic c cng thc C2H5OH, c nhit si l 78,40C, khi lng phn t mol M= 46.Ru Etylic c iu ch t nhiu ngun gc khc nhau. Ph bin nht l phng php ln men. Nguyn liu l go,ng, khoai, sn, ng ca qu chn.Ngy nay ngi ta s dng xenlulozo c trong g, nh v bo, mn ca,lm nguyn liu iu ch etylic. Ngoi ra trong cng nghip ngi ta iu ch bng cch hydrat ha anken v thy phn dn xut halogen trong dung dch kim.Trong thc t ru etylic c ng dng rng ri, n lm nguyn liu sn xut cao su tng hp, iu ch mt s hp cht hu c nh axit axetic.Ru etylic l mt dung mi ha tan nhiu hp cht hu c nh axit axetic.Ru etylic l mt dung mi ha tan nhiu hp cht hu c, n cn c s dng trong y t lm cn etylic, ngoi ra n cn s dng lm thc phm. PHN I: GII THIU CHUNG:I. S DY CHUYN CNG NGH :1. S dy chuyn cng ngh sn xut ( hnh 1):

Hnh 1.

Ch thch:

1: thng cao v

2: b cha dung dch u

3: thit b gia nhit hn hp u

4: lu lng k

5: thp chng luyn

6: thit b ngng t

7: thit b lm lnh

8: b cha sn phm nh

9: b cha sn phm y

10: thit b un si y thp

11: cc tho nc ngng12: bm pt tong

2 . Thuyt minh dy chuyn sn xut:Nguyn liu u c cha trong thng cha (2) v c bm (12) bm ln thng cao v (1). Mc cht lng cao nht c c khng ch bi ng chy trn tr li b cha dung dch u (2). Hn hp u t thng cao v (1) t chy xung thit b un si hn hp u (3). Lu lng c khng ch bng cch iu chnh h thng van v lu lng k (4) hi nc bo ha t ni hi vo un si hn hp u n nhit si sau khi t ti nhit si hn hp ny c a vo a tip liu ca thp chng luyn (5) loi m.Trong thp hi i t di ln tip xc trc tip vi lng chy t trn xung, ti y xy ra qu trnh bc hi v ngng t nhiu ln. Theo chiu cao ca thp, cng ln cao th nhit cng thp nn khi hi i qua cc tng m t di ln, cu t c nhit si cao s ngng t. Qu trnh tip xc lng - hi trong thp din ra lin tc lm cho trong pha hi cng giu cu t d bay hi. Cui cng trn nh thp ta s thu c hu ht l cu t d bay hi (c th y l Ru Etylic) v mt phn cu t kh bay hi ( Nc ). Hn hp hi ny c a vo thit b ngng t (6) v ti y n c ngng t hon ton (tc nhn l nc lnh ). Mt phn cht lng sau khi ngng t c a hi lu tr v thp chng luyn v cng c khng ch bng lu lng k, phn cn li t yu cu s c a vo thit b lm lnh (7) lm lnh n nhit cn thit sau c a vo thng cha sn phm nh (8). Cht lng hi lu i t trn xung di, gp hi c nhit cao i t di ln, mt phn cu t c nhit cao tip tc ngng t thnh lng i xung. Do nng cu t kh bay hi trong pha lng ngy cng nhiu , cui cng y thp ta thu c hn hp lng gm hu ht l cu t kh bay hi (Nc ) v mt phn rt t cu t d bay hi (R.Etylic ) ,hn hp lng c a ra khi y thp qua thit b phn dng, mt phn c a ra thng cha sn phm y (9), mt phn c a vo thit b un si y thp (10) v mt phn c hi lu tr li y thp. Thit b ny c tc dng un si tun hon v bc hi sn phm y ( to dng hi i t di ln trong thp). Nc ngng ca thit b gia nhit c tho qua thit b tho nc ngng ( 11). Thp chng luyn lm vic ch lin tc, hn hp u vo v sn phm c ly ra lin tc.II. CH THY NG CA THP M Trong thp m c 3 ch thy ng l ch chy dng, ch qu v ch chy xoy.Khi vn tc kh b lc ht phn t ln hn v vt lc . Lc ny qu trnh chuyn khi c xc nh bng dng khuch tn phn t. Tng vn tc lc l tr ln cn bng vi lc ht phn t. Qu trnh chuyn khi lc ny khng ch c quyt nh bng khuch tn phn t m c bng c khuch tn i lu. Ch thy ng ny gi l ch qu . Nu ta tip tc tng vn tc kh ln na th ch qu chuyn sang ch chy xoy. Trong giai on ny qu trnh khuch tn s c quyt nh bng khuch tn i lu. Nu ta tng vn tc kh ln n mt gii hn no th s xy ra hin tng o pha. Lc ny cht lng s chim ton b chiu cao thp v tr thnh pha lin tc, cn pha kh khuch tn vo trong pha lng v tr thnh pha phn tn. Vn tc kh ng vi thi im ny gi l vn tc o pha. Kh sc vo lng v to thnh bt kh v th trong giai on ny ch lm vic trong thp gi l ch si bt. ch ny vn tc chuyn khi nhanh ng thi tr lc cng tng nhanh. Trong thc t, ta thng cho thp m lm vic ch mng c vn tc nh hn vn tc o pha mt t v qu trnh chuyn khi trong giai on si bt l mnh nht nhng v giai on kh khng ch qu trnh lm vic. u im ca thp m:

-Hiu sut cao v b mt tip xc pha ln .

-Cu to thp n gin .

-Tr lc trong thp khng ln lm .

- Gii hn lm vic tng i rng .Nhc im :

-Kh lm t u m .-Thp cao qu th phn phi cht lng khng u .III. BNG K CC K HIU THNG DNG TRONG BN N NY:

*. Ru Etylic : K hiu E ME = 46 (kg/kmol)

*. Nc : K hiu N MN = 18(kg/kmol)

GF : Lu lng hn hp u (kmol/h)

GP : Lu lng sn phm nh (kmol/h)

GW: Lu lng sn phm y(kmol/h)

aF : Nng phn khi lng cu t d bay hi trong hn hp u (% khi lng).

aP : Nng phn khi lng cu t d bay hi trong sn phm nh (% khi lng).

aW : Nng phn khi lng cu t d bay hi trong sn phm y (% khi lng).

xF : Nng phn mol cu t d bay hi trong hn hp u (% mol)

xP : Nng phn mol cu t d bay hi trong sn phm nh (% mol)

xW : Nng phn mol cu t d bay hi trong sn phm y (% mol)

F: lu lng hn hp u, kg/h.

P: lu lng sn phm nh, kg/h.

W: lng sn phm y, kg/h.

M: khi lng mol phn t, kg/ kmol.

: nht ng lc, N.s/m2.

( : khi lng ring, kg/m3.PHN II:TNH TON THIT B CHNH Cc s liu ban u:-Nng sut tnh theo hn hp u F = 7,5 tn/gi = 7500 kg/h-Nng cu t d bay hi trong:+ Hn hp u : aF = 0,34 phn khi lng+ Sn phm nh: aP = 0,86 phn khi lng+ Sn phm y : aW = 0,005 phn khi lng - Thp lm vic p sut thng. - Hn hp u c gia nhit n nhit si

I. TNH CN BNG VT LIU TON THIT B:1. Tnh ton cn bng vt liu cho ton thp:

1.1. H phng trnh cn bng vt liu

Phng trnh cn bng vt liu cho ton thp :

F = P + W (1) IX.16 [II-144] i vi cu t d bay hi (Ru Etylic) :

FaF = Pap + Waw ( 2) IX.17 [II-144] Thay (1) vo (2) rt ra : Lng sn phm y : T suy ra lng sn phm nh :

P = F W = 7500-4561,40=2938,6 ( kg/h)1.2. i t phn khi lng sang phn molp dng cng thc:

VIII.1 [II-126] Trong : (Kg/Kmol) 18 ( Kg/Kmol)Thay s liu vo ta c:

( phn mol ) Nng phn mol trong sn phm nh :

( phn mol)Nng phn mol trong sn phm y :

(phn mol)Khi lng phn t hn hp u :

M=x.ME+(1-x).MN =0,1678.46 + (1-0,1678).18 = 22,6984 (kg/kmol )

- Phng trnh cn bng vt liu cho ton thp :

GF = GP + GW ( IX.16 Tp II tr 144)

Tnh theo kmol/h:

- Lng hn hp u vo tnh theo kmol/h:

-Lng sn phm nh tnh theo kmol/h : GP = GF.

-Lng sn phm y tnh theo kmol/h:

GW =G F- GP = 330,4198 77,8091 = 252,6107 (kmol/h).Tnh theo kg/h:

-Lng sn phm nh tnh theo kg/h:

-Lng sn phm y tnh theo kg/h:

W= F-P = 7500- 2938,5964 = 4561,404 (kg/h) Bng Tng kt thnh phn nh sau:

P.khi lngPhn molLu lng

(kg/h)Lu lng

(kmol/h)

Hn hp u0,340,16787500330,4198

Sn phm nh0,860,70622938,596477,8091

Sn phm y0,0050,0019624561,404252,6107

1.3.Ch s hi lu ti thiu: (Rmin) :Tra bng IX 2a (tr 148)c bng cn bng lng hi ca Ru Etylic Nc.

Dng ng cn bng theo s liu ng cn bng sau:X05102030405060708090100HH.P

Y033,244,253,157,661,465,469,975,381,889,810089,4

T0C10090,586,583,281,780,88079,47978,678,478,478,15

T s liu trong bng trn ta v th ng cn bng lng (x) hi (y).

Hnh 1: th ng cn bng lng hi:

+) xc nh ch s hi lu ti thiu: th ng cn bng cn bng lng(x)- hi (y)Da vo th x y t gii tr xF = 0,1678 ta dng ln ng cn bng ca th cn bng lng hi (x) ( y ) ta tm c y*F = 0,5101 do nng cu t d bay hi trong pha hi cn bng vi nng trong pha lng xF ca hn hp u.

Ch s hi lu ti thiu c xc nh theo cng thc.

IX.24 [II.158]

Phng trnh ng lm vic ca on luyn:

( II 144 )Vi Rx ch s hi lu thch hp.

ng nng lm vic ca on luyn ct trc tung ti im c tung B = (0, ) vi Rmin = 0,5729 th Bmax = = 0,449 Phng trnh ng lm vic ca on chng:

( II 144 )

1.4. Xc nh ch s hi lu thch hp: Rth H s hiu chnh:

Vn chn ch s hi lu thch hp l rt quan trng, v khi ch s hi lu b th s bc ca thp ln nhng tiu tn t hi t, ngc li khi ch s hi lu ln th s bc ca thp c t hn nhng tiu tn hi t li rt ln.Vi mi gi tr ca Rx > Rmin t th cn bng lng hi ca hn hp Etylic Nc ta xc nh c mt gi tr ca Nlt tng ng. Cng d nhn thy c th tch lm vic ca thp t l vi tch s NLT(RX+1).Trong NLT: s n v chuyn khi hay s a l thuyt

Thc t gi tr Rth ng vi NLT(RX+1) nh nht, ta ln lt ly cc gi tr ca nm trong khong 1,2 2,5 tm ra gi tr Rth.

2. Phng trnh ng nng lm vic

2.1. on luyn

IX.20 [II-144]

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.

2.2.on chng

IX.22 [II.158]

3. th xc nh s a l thuyt ng vi tng gi tr Rx:

a. Trng hp = 1,2 => Rx = 1,2 .0,5729 = 0,6875 Phng trnh ng nng lm vic on luyn.

y = 0,4074. x + 0,4185 Phng trnh ng nng lm vic on chng.

y = 2,9239. x 0,0038

N lt = 11Ta v c th sau:

b. Trng hp = 1,4 => Rx = 1,4 .0,5729 = 0,8021 Phng trnh ng nng lm vic on luyn.

y = 0,4451. x + 0,3919

Phng trnh ng nng lm vic on chng.

y =2,8015. x 0,0035

N lt = 10Ta c th sau :

c. Trng hp = 1,6 => Rx = 1,6 .0,5729 = 0,9166 Phng trnh ng nng lm vic on luyn: y = 0,4782. x + 0,3685 Phng trnh ng nng lm vic on chng.

y =2,6939. x 0,00333

N lt = 9Ta c th sau:

d. Trng hp = 1,8 => Rx = 1,8 .0,5729 = 1,0312 Phng trnh ng nng lm vic on luyn:

y = 0,5077. x + 0,3477


y =2,5983. x 0,0031


e. Trng hp = 1,9 => Rx = 1,9 .0,5729 = 1,0885 Phng trnh ng nng lm vic on luyn:

y = 0,5211. x + 0,3381 Phng trnh ng nng lm vic on chng.

y =2,55441. x 0,0030


f. Trng hp = 2,0 => Rx = 2,0 .0,5729 = 1,1458 Phng trnh ng nng lm vic on luyn:

y = 0,5334. x + 0,3291


y =2,513. x 0,0030


g. Trng hp = 2,1 => Rx = 2,1 .0,5729 = 1,2031

Phng trnh ng nng lm vic on luyn:

y = 0,5461. x + 0,321


y =2,4736. x 0,0029


h. Trng hp = 2,2 => Rx = 2,2.0,5729 = 1,2634


y = 0,5582. x + 0,312


y =2,4343. x 0,0028


i. Trng hp = 2,4 => Rx = 2,4.0,5729 = 1,375


y = 0,5789. x + 0,2973


y =2,367. x 0,0027


k. Trng hp = 2,5 => Rx = 2,5.0,5729 = 1,4323


y = 0,5889. x + 0,2903


y =2,3347. x 0,0026


T cc gi tr ca Rx thay vo phng trnh on chng, on luyn ta v cc ng nng lm vic on chng v on luyn, sau tm s bc thay i nng NLT bng cch m s tam gic gia ng cn bng lng hi v ng lm vic on chng, on luyn t ta c bng sau:1,21,41,61,81,9

B0,41850,39190,36850,34770,3381

Rx0,68750,80210,91661,03121,0885

NLT1110988

NLT(Rx+1)18,562518,02117,249416,249616,708

2,02,12,22,42,5

B0,32910,3210,3120,29730,2903

Rx1,14581,20311,26341,3751,4323

NLT88877

NLT(Rx+1)17,166417,624818,107216,62517,0261

- Bng cch v th Rx v NLT(Rx+1) ta xc nh c Rth v NLT chnh xc ng vi im NLT(Rx+1) nh nht khi Rth = 1,0312 v NLT = 8 a.

Khi : =

- Phng trnh ng nng lm vic on luyn l:

y = =

=>Y = 0,5077 x + 0,3477 Phng trnh ng nng lm vic on chng l:

y = =

=>Y = 2,5983 x 0,00313Ta c th T X, Y

Vy ch s hi lu thch hp Rth= 1,0312 v s a l thuyt Nlt= 83. Tnh lng hn hp u F, lng sn phm nh P, lng sn phm y W theo n v kmol/h

EMBED Equation.3 - Lng sn phm u tnh theo lng sn phm nh :

( kmol h2 u / kmol sn phm nh )II. TNH NG KNH THP CHNG LUYN: ng knh thp c xc nh theo cng thc:

( m )

IX.90 [II.181]

Trong : gtb l lng hi trung bnh i trong thp (kg/h)

(y.y)tb : tc hi trung bnh i trong thp ( kg/m.s2)

V lng hi v lng thay i theo chiu cao ca thp v khc nhau trong mi giai on nn ta phi tnh lng hi trung bnh cho tng on.

1. Xc nh lu lng cc dng pha i trong thp:1.1. Lng hi trung bnh i trong on Luyn: Lng hi trung bnh i trong on luyn c tnh gn ng bng trung bnh cng ca lng hi i ra khi a trn cng ca thp v lng hi i vo a di cng ca on luyn.

,( kmo/h) [II181] + gtb :lng hi trung bnh ca on luyn, kmol/h + g : lng hi i ra khi a trn cng ca on luyn, kmol/h + g1 : lng hi i vo a di cng ca on luyn, (kmol/h) * Lng hi i ra khi nh thp :g g = GR + GP = GP(Rth + 1) [II181] Vi:

GP : lng sn phm nh, kmol/h =77,8091 GR : lng lng hi lu nh thp, kmol/h GR= Rx .GP= 77,8091.1,0312=80,236 (kmol/h)

Rth : ch s hi lu thch hp g = 77,8091 (1,0312+1)=158,045 ( kmol/h)* Lng hi i vo a di cng ca on Luyn: Lng hi g1, hm lng hi y1, lng lng ca a th nht ca on luyn G1 c xc nh theo h phng trnh cn bng vt liu v nhit lng cho on luyn : (kmol/h) [II-182]

Trong :

=0,1678 phn mol. r1 : n nhit ha hi ca hn hp hi a th nht ca on luyn,( kcal/kmol) r : n nhit ha hi ca hn hp hi ra khi nh thp, kcal/kmol r1 = rEy1 + (1- y1)rN [II-182]

r = rEy + (1- y)rN [II-182]

Trong :

y : hm lng hi sn phm nh = xP = 0,7062 phn mol. rE; rN : ln lt l n nhit ha hi ca Etylic v Nc nguyn cht, kcal/kmol. *) T xF= 0,1678 tra bng IX.2a [II-145] suy ra nhit tF = 84,260C. Tra bng I.212 [I-254] vi nhit hn hp u ta c: RE =200,296 ( Kcal/kg) = 200,96 . ME kcal/kmol = 200,96. 46 = 9244,16 (kcal/kmol) RN = 554,74 ( Kcal/kg) = 554,74. MN = 554,74. 18 =9985,32 (kcal/kmol)- n nhit ha hi ca hn hp u i vo a luyn th nht :

r1 = 9244,16 y1 + 9985,32 ( 1-y1) (*) *) T xP = 0,7062 tra bng IX.2a [II-145] suy ra nhit ca hn hp nh tP= 78,980C , ni suy t bng I.212 [I-254] vi nhit hn hp nh tP ta c: rE = 202,408 (Kcal/kg) = 202,408. M E = 202,408. 46 = 9310,768 (kcal/kmol)

rN = 560,02 (Kcal/kg) = 560,02. MN = 560,02. 18=10 080,36 (kcal/kmol)

- n nhit ha hi ca hn hp i ra khi nh thp, r (ng vi xp = 0,7062) r = yrE + (1- y )rN ( II- 182)

r = 0,7062.9310,768 +(1-0,7062).10 080,36 =9536,874 (kcal/kmol) (**) Thay (*)&(**) vo h trn ta c :

EMBED Equation.3 (kmol/h)

r1 = 9244,16. 0,4363 + 9985,32 ( 1-0,4363) =9661,9518Vy vi y1= 0,4363 (phn mol) r1 = 9661,9518(kcal/kmol)

Lng hi trung bnh ca on Luyn :

( II - 181)1.2. Lng hi trung bnh ca on chng [II-182] Trong :

+ gn: lng hi i ra khi on chng,( kmol/h). + g1 : lng hi i vo on chng, (kmol/h). V lng hi i ra khi on chng bng lng hi i vo on luyn nn ta c :

gn = g1 ;

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

[II-182] Trong : - 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. - rE: n nhit ha hi ca Etylic, (kcal/kmol) - rN: n nhit ha hi ca Nc, (kcal/kmol) l nng hi i vo on chng.

Lng sn phm y: Gw= 252,6107(kmol/h)

l nng cn bng ng vi . Ni suy theo bng s liu ng cn bng [II-145] ta c: ng vi ta c (phn mol)

Mt khc ta c:

[II-182] Ti ni suy t bng IX.2a [II-145] suy ra . Ni suy theo bng I.212 [I-254] ta c :

rE = 194,1492 Kcal/kg = 194,1492.46 = 8930,725 (kcal/kmol) rN = 539,073 Kcal/kg = 539,073. 18 = 9703,314 (kcal/kmol)

n nhit ha hi ca hn hp hi i vo a chng th nht:

r1 = 8930,725. 0,0197 + (1 0,0197).9703,314 = 9688,094 (kcal/kmol)

Lng hi trung bnh ca on Chng :

(kmol/h)2. Tnh khi lng ring trung bnh:2.1. Pha hi * on Chng Khi lng ring trung bnh ca pha hi i vi on chng c tnh theo cng thc: ( kg/m3) IX.102 [II-183]

Trong : T: Nhit lm vic trung bnh ca thp (0K) Ta c:

Vi:

Nng pha hi u on chng: ydC = y1 = 0,0197( phn mol) Nng pha hi cui on chng: ycC = y1 = 0,4363(phn mol)Nng trung bnh ca pha hi trong on chng l :

EMBED Equation.3 (phn mol) Vi ytbC = 0,228 phn mol.Ni suy theo bng IX.2a [II-145] ta c t0tbC = 90,520C Vy TC = 90,52 + 273 =363,52(0K)Khi lng mol trung bnh ca pha hi trong on chng:

[II-183] Vy khi lng ring trung bnh ca pha hi i vi on Chng l:

* on Luyn Khi lng ring trung bnh i vi pha hi i vi on luyn c tnh theo cng thc:

IX.102 [II-183]

Ta c:

Vi:

Nng pha hi u on luyn l: ydL = y1 = 0,4363 (phn mol) Nng pha hi cui on luyn l: ycL = yp = xp = 0,7062 (phn mol) Vy:

(phn mol) Vi ytbL = 0,571 P.mol. Ni suy theo bng IX.2a [II-145] ta c t0tbL= 81,86 0CVy: TL = 81,86+273 = 354,86 0KKhi lng mol trung bnh ca hi trong on luyn :

Vy khi lng ring trung bnh ca pha hi i vi on Luyn l:

=

2.2.Pha lng Khi lng ring trung bnh ca pha lng c tnh theo cng thc:

` IX.104a [II-183] Trong :

- : phn khi lng trung bnh. -: khi lng ring ca Etylic (kg/m3) -: khi lng ring ca nc (kg/m3) * on Chng Khi lng ring trung bnh ca pha lng i vi on chng c tnh theo cng thc sau:

phn khi lng (phn mol) Vi xtbC = 0,085(phn mol).Ni suy theo bng IX.2a [II-145] ta c t0tbC = 87,7 0C Ni suy t bng I.2 [I-9] nhit trung bnh

Ta suy ra :

= 727,685 (kg/m3)

=966,61 (kg/m3)

Khi lng mol trung bnh ca lng trong on chng:

* on Luyn Khi lng ring trung bnh ca pha lng i vi on luyn c tnh theo cng thc sau:

(phn mol)

Khi lng mol trung bnh ca lng trong on luyn:

Vi xtbL = 0,437. Ni suy t bng IX.2a [II-145] ta c t0tbL = 80,504 0C

Ni suy t bng I.2[I-9] nhit trung bnh: ta suy ra :

= 734,521 (kg/m3)

= 971,647 (kg/m3)

3. Vn tc hi i trong thp:Vn tc hi trong thp m c tnh theo cng thc: w = (0,8 0,9) wS [II-187]

Vi ws l tc sc, c tnh theo cng thc: Y = 1,2.e-4X IX.114 [II-187]

Trong :

Y = [II-187] X = [II-187]Trong : : B mt ring ca m (m2/ m3) Vd: Th tch t do ca m (m3/ m3)

G: Gia tc trng trng (m2/s) : khi lng ring trung bnh ca pha lng v pha hi (kg/m3)

: nht ca pha lng theo nhit trung bnh v nht ca nc 20 0C (N.s/m2)

- nht ca nc 20 0C tra bng I.102 trong ( I-94) ta c: = 1,005 .10-3 N.s/m2

* Chn m vng loi Rasiga loi ln xn kch thc: 20 x 20 x 2,2 tra t bng IX.8 [II- 193] Kch thc m, mmB mt ring,

, m2/m3Th tch t do, Vd, m3/m3S m trong 1m3Khi lng ring xp,, kg/m3

20 x 20 x 2,22400,7346.103600

3.1. Vn tc hi i trong on Chng

* Lng lng trung bnh i trong on Chng GxC =

* Lng hi trung bnh i trong on Chng Gy= (kmol/h)

* nht hn hp lng ca on Chng I.12 [I-84]

Trong :

: nht ca hn hp lng ca on chng

EMBED Equation.3 : nht ca nc 20oC

xtbc : nng phn mol ca Etylic on chng

E ; N: ln lt l nht ca 2 cu t Etylic v Nc nhit

T bng I.101 [I-91] ni suy vi

E = 0,393.10-3 (N.s/m2)

N = 0,329.10-3 (N.s/m2)

C phn mol -Do : nht ca pha lng tnh theo nhit trung bnh ca on chng l :

T bng I.102 [I-94] tm c n = 1,005.10-3 (N.s/m2 ) ( 200C) Vy: X =

Y = 1,2.e-4X = 1,2.e-4. 0,46 = 0,192 Vi loi m vng ta chn nh trn nn:

Y = 0,192 =

ws = 2,1 (m/s). Vn tc hi i trong on Chng l:

WC = 0,85.ws = 0,85.2,1 = 1,8 ( m/s).3.2. Vn tc hi i trong on Luyn

* Lng lng trung bnh i trong on Luyn GxL = kmol/h Vy GxL= GtbL.= 79,213.30,236=2395,084 (kg/h)

GyL= gtbL.= 33,988.157,0225=5336,88(kg/h) * Lng hi trung bnh i trong on Luyn GyL=gtbL = (kmol/h) * nht ca hn hp lng on Luyn

Mt khc:

EMBED Equation.3

T bng I.101 [I-91] ni suy ta c :

Vy:

Vy: X =

Y = 1,2.e-4X = 1,2.e-4.0,371 = 2,27 Vi loi m vng ta chn nh trn nn:

Y = 2,27 =

ws = 1,8 (m/s) Vn tc hi i trong on Luyn l:

WL = 0,8. ws = 0,8.1,8 = 1,49 (m/s)Lng hi trung bnh trong thp l :

Gtbc=GyC=Gtbc.=155,79.24,384=3798,783(kg/h)4. ng knh thp chng luyn:4.1. ng knh on Chng

Chn DC=1m. Theo bng Quy chun cho thp chng (Bng 6.27-tnh ton QT& TB II trang 115 )ta ly ng knh l: DC = 1m.Khi vn tc hi thch hp i trong thp l :

* Th li iu kin lm vic thc t

Tc hi thc t i trong on chng:

, m/s. T s gia tc thc t vi tc sc:

Vy chn ng knh l 1(m) c th chp nhn c.

4.2. ng knh on Luyn:Lng hi trung bnh i trong thp (kg/h)

gtbL=gyL=

Theo bng Quy chun cho thp chng (Bng 6.27-tnh ton QT& TB II ltrang 115 ) ta ly ng knh l: DL = 1m.

* Th li iu kin lm vic thc t

Tc hi thc t i trong on luyn:

(m/s). T s gia tc thc t vi tc sc:

Vy chn ng knh l 1m c th chp nhn c. Quy chun ta ly ng knh thp chng l DL = DC = D = 1m. III. CHIU CAO THP CHNG LUYN: * Chiu cao lm vic ca thp m i vi thp m, chiu cao lm vic ca thp hay chiu cao lp m c tnh theo cng thc:

H = hv . my , m IX.68 [II-175] Trong :

- hv : chiu cao ca mt n v chuyn khi, m

- my : s n v chuyn khi tnh theo pha hi Ta c:

vi chiu cao ca mt n v chuyn khi c tnh theo cng thc sau:

IX.75 [II- 177] Vi:

- h1: l chiu cao ca 1 n v chuyn khi i vi pha hi,( m) - h2: l chiu cao ca 1 n v chuyn khi i vi pha lng, (m) - m: l h s phn b trung bnh iu kin cn bng pha

- Gx, Gy: l lng lng v lng hi trung bnh i trong thp,(kg/h) * Chiu cao ca mt n v chuyn khi IX.76 [II-177]

IX.77 [II-177]

Trong :

a: h s ph thuc vo dng m, vi m vng th a = 0,123

(x: nht ca pha lng, Ns/m2Vd: th tch t do ca m, m3/m3(x: khi lng ring ca lng, kg/m3(: H s thm t ca m, n ph thuc vo t s gia mt ti thc t ln tit din ngang ca thp v mt ti thch hp, xc nh bng th (hnh IX.16) [II .178]

1. Xc nh chun s Reynolt1.1. on Chng

* Chun s Reynolt ca pha hi on Chng Theo [II-178]:

T bng I.101 [I- 91] nhit

E= 0,378.10-3 N.s/m2 N= 0,318.10-3 N.s/m2 + Tnh yC:

* Chun s Reynolt ca pha lng on Chng Ta c:

XC = (phn mol)

T bng I.101 [I- 91] nhit

E= 0,393.10-3 N.s/m2 N= 0,329.10-3 N.s/m2 + Tnh xC:

1.2. on Luyn * Chun s Reynolt ca pha hi ca on Luyn Theo II-178 :

(phn mol)

T bng I.101 [I-91] nhit

E= 0,424.10-3 N.s/m2 N= 0,35.10-3 N.s/m2 + Tnh yL:

* Chun s Reynolt ca pha lng on Luyn

Ta c:

xL = (phn mol)

T bng I.101 [I-91] vi x= 0,437 ni suy t bng IX.2a (II-145) c nhit . nhit ny : E= 0,432.10-3 N.s/m2 N= 0,355.10-3 N.s/m2 + Tnh xL:

2. Xc nh h s khuch tn2.1. on Chng

* H s khuych tn ca pha hi on Chng , m2/s. VIII.5 [II-127]

Trong : DyC : l h s khuch tn ca pha hi, m2/s. PC : l p sut tuyt i ca h 2 cu t Etylic nc = 1at : ln lt l th tch mol ca Etylic nc, cm3/mol. TC : l nhit tuyt i ca on chng ME , MN : l khi lng mol phn t ca 2 cu t Etylic nc ( vC)

Tra bng [II -127] ta c th tch nguyn t ca C = 14,8: H = 3,7:O=7,4 + = 2.14,8 + 6.3,7+7,4 = 59,2 (cm3/ng.t). + = 7,4 + 2.3,7 = 14,8 (cm3/ng.t). M T = 363,520K

Thay vo ta c :

+ Tnh h s khuch tn ca hn hp lng 20o C (m2/s) [II- 133]

Trong :

- A, B : l h s ph thuc vo bn cht dung mi v cht tan , l h s lin hp k n nh hng ca ru Etylic va nc.Tra bng VIII.7(II-134) Ta c A = ru Etylic =1,24

B = Nc =4,7

- DxC20 : l h s khuch tn ca pha lng 20o C, m2/s. - ME ; MN : l khi lng phn t ca etylic-nc - : l th tch mol ca Etylic-nc, cm3/mol.ln lt bng :14,8 v 59,2 (cm3/ng.t) - : nht ca dung mi Nc 200C

T bng I.102 [I-94]

* H s khuch tn ca hn hp lng on Chng VIII.15 [II-134]

: b l h s hiu chnh ;

VIII.16 [II-135]

: nht ca dung mi nc 200C (cP).Tra bng 1.101.(92-I) ta c:

Tra bng I.2 (9-I)ta c:

2.2. on Luyn * H s khuych tn ca pha hi trong on Luyn

, m2/s VIII.5 [II-127] Trong :

DyL : l h s khuch tn ca pha hi, m2/s. PL: l p sut tuyt i ca h 2 cu t Etylic-Nc= 1at : ln lt l th tch mol ca hi Etylic-Nc, cm3/mol. TL: l nhit tuyt i ca on luyn.

ME , MN : l khi lng mol phn t ca 2 cu t Etylic-Nc ( vC)

Ta c: TL= 81,6 + 273 = 354,860KThay vo ta c :

* H s khuych tn ca pha lng on Luyn

VIII.15 [II-134] b l h s hiu chnh VIII.16 [II-135]Trong :: nht ca dung mi nc 200C.

: Khi lng ring dung mi nc 200C.

Ni suy t bng I.2 [I-9] ta c:

3. Xc nh chun s Prand3.1. on Chng

* Chun s Prand trong pha hi on Chng [ II- 178] Thay s ta c :

* Chun s Prand trong pha lng on Chng

3.2. on Luyn

* Chun s Prand trong pha hi on Luyn

* Chun s Prand trong pha lng on Luyn 4. Tnh h s thm t 4.1. on Chng

* Mt ti thch hp Uth = .B, (m3/m2.h) B : Hng s, B = 0,065, (m3/m.h). H s ny c cho trong bng IX.6 [II-177]

: b mt ring ca m, m2/m3. Vi loi m vng Rasiga loi: 20 x 20 x 2,2 th:

- Vd = 0,73, m3/m3 - = 240, m2/m3 - a = 0,123

Uth = 240.0,065 = 15,6 (m3/m2.h) * Mt ti thc t (m3/m2.h) [ II- 177] Trong :

VxC : lu lng th tch ca pha lng trong on chng, m3/s. Ft : tit din ngang ca thp (on chng) M:

Ft = m2

T hnh IX.16 [II-178 ] ni suy c: C = 0,46

+ Chiu cao ca mt n v chuyn khi i vi pha hi on chng:

( CT II-177) + Chiu cao ca mt n v chuyn khi i vi pha lng on Chng:

4.2. on Luyn

* Mt ti thch hp Uth = .B (m3/m2.h) B : Hng s, B = 0,065, m3/m.h. H s ny c cho trong bng IX.6 [II-177]

: b mt ring ca m, m2/m3 Vi loi m vng Rasiga loi: 20 x 20 x 2,2 th:

- Vd = 0,73 m3/m3 - = 240 m2/m3 - a = 0,123

Uth = 240.0,065 = 15,6 (m3/m2.h)

* Mt ti thc t (m3/m2.h) [ II- 177]

Trong :

VxL : lu lng th tch ca pha lng trong on luyn, m3/s.

Ft : tit din ngang ca thp (on luyn)

M: Ft = m2

T hnh IX.16 [II-178 ] ni suy c: L = 0,28 + Chiu cao ca mt n v chuyn khi i vi pha hi on Luyn:

+ Chiu cao ca mt n v chuyn khi i vi pha lng on Luyn:

5. Tnh s n v chuyn khi5.1. on Chng

S n v chuyn khi c xc nh theo cng thc:

IX.74 [II-176] Vi mi gi tr ca y trong khong (0,0197;0,5101) tm c y*.T xy dng th .

Ta chia on [0,0197;0,5101] thnh 20 on bng nhau, c chiu di: HC =

* Kt qu c tng hp bng sau:

yy*y*-y

0,01970,09550,075813,1926

0,044220,17840,134187,4526

0,068780,23670,16795,9559

0,093260,28120,187945,3208

0,117780,31760,199825,0045

0,14230,34840,20614,8520

0,166820,37460,207784,8127

0,191340,39670,205364,8694

0,215860,41560,199745,0065

0,240830,43170,190875,2391

0,26490,44570,18085,5309

0,289420,45570,166286,0139

0,313940,46880,154866,4574

0,338460,47870,140247,1306

0,362980,48770,124728,0179

0,38750,4960,10859,2165

0,412020,50370,0916810,9075

0,436540,51510,0785612,7291

0,461060,54380,0827412,086

0,485580,56490,0793212,6071

0,51010,58350,073412,6239

T bng s liu trn ta v th , t th ta tnh din tch gii hn bi ng cong, trc honh v hai trc tung v yF chnh l s n v chuyn khi cn tm ( v yF l nng u v cui ca hi).

+p dng cng thc SIMSON :

t:

F==

=3,8 Vy

5.2. on Luyn

S n v chuyn khi c xc nh theo cng thc:

IX.74 [II-176] Vi mi gi tr ca y trong khong (0,7556 0,5101) tm c y* .T xy dng th .


Ta chia on [0,7556; 0,5101] ra thnh 10 on bng nhau, c di:

HL =


yy*y*-y

0,51010,58350,073413,6239

0,534650,60180,0671514,892

0,55920,62040,061216,3398

0,583750,63960,0558517,9051

0,60830,65960,051319,4931

0,632850,68080,0479520,855

0,65740,70700,049620,1612

0,681950,72950,0475521,0304

0,70650,75690,050419,8412

0,731050,78630,0552518,0995

0,75560,78660,03132,258

T bng s liu trn ta v th , t th ta tnh din tch gii hn bi ng cong, trc honh v hai trc tung v yP chnh l s n v chuyn khi cn tm ( v yP l nng u v cui ca hi).

+p dng cng thc SIMSON :

-t F== = 4,65Vy :

6. Tnh mm l gi tr trung bnh ca tag gc nghing v ng cn bng vi mt phng ngang.i lng ny cho php tnh gn ng ta c th nhn bng cch chia ng cn bng thnh nhiu on v tm gi tr tag gc nghing ng vi nhng on chia, sau ly gi tr trung bnh.

m= ( II-168)

m=tg= (II- 169)6.1. Xc nh cho on chng :mC Chia ng cn bng thnh 5 phn t xW xF X0,051660,068240,084820,10140,11798

X

0,011650,01690,023560,03180,0421

Y0,093260,117780,14230,16680,19134

Y

0,28120,31760,34840,37460,3967

m

EMBED Equation.3 4,6973,8923,8872,9852,706

6.2. Xc nh cho on luyn :mLChia ng cn bng thnh 7 phn t

X0,221640,275480,329320,383160,437

Xcb (X)0,11400,14030,17460,22070,2833

Y0,534650,55920,583750,60830,63285

Ycb (Y)0,60180,62040,63960,65960,6808

m

0,62610,45270,36090,31570,3119

7. Xc nh chiu cao thp

7.1. on Chng

Chiu cao ca mt n v chuyn khi:

Vy chiu cao lp m on Chng (hoc chiu cao lm vic ca on chng):

7.2. on Luyn

Chiu cao ca mt n v chuyn khi:

Vy chiu cao lp m on Luyn l:

7.3. Chiu cao ton thp Chn:

- Khong cch gia cc ngn m l 200(mm) = 0,2(m) + on Chng c HC = 1,06 m, ta chia lp m on Chng ra thnh cc ngn m nh c chiu cao l: 0,4 m. Vy on Chng ta s chia thnh 3 ngn m, vy s khong cch gia cc ngn m l: 3 1 = 2 (khong cch).do :Chiu cao ton b lp m on Chng l: HC = 3.0,4+(3-1).0,2=1,6 (m) + on Luyn c HL = 4,8 (m), ta chia lp m on Luyn thnh cc ngn m nh c chiu cao l: 0,4 m, Vy on Luyn ta s chia thnh 12 ngn m, vy s khong cch gia cc ngn m l: 12 1 = 11 (khong cch).do : Chiu cao ton b lp m on Luyn l:

HL = 12.0,4+(12-1).0,2= 7 (m) Chiu cao ca lp m:

H = HC + HL = 1,6 + 7 = 8,6 (m)

Vi ng knh thp D = 1 (m) - Khong khng gian gia hai lp m on chng v luyn l: 0,6 (m)

- Khong cch t ngn m trn cng n np Zn = 0,6 (m)

- Khong cch t m n y Z = 0,8 (m)

Chiu cao ton thp: HT = 8,6 + 0,6 + 0,8 + 0,6 = 10,6 (m) IV. TR LC CA THP M: Tr lc i vi thp m c th c xc nh theo cng thc:

(N/m2) IX.118 [II-189] Trong : Tn tht p sut ti im o pha c tc ca kh bng tc ca kh khi i qua m kh, N/m2. Gx, Gy: Lu lng ca lng v kh, kg/h Khi lng ring ca lng v kh, kg/m3

nht ca lng v kh, N.s/m2 Tn tht p sut ca m kh, N/m2

khi Rey > 40

A; m; n ; c : l hng s Tra t bng IX.7 [II-189] ta c : A= 5,15 ; m = 0,342; n = 0,19 ; c = 0,038 Khi Rey < 40 th

1. Tr lc on Chng

* Chun s cimet

[II-188]

* Chun s Reynolt

IX.117 [II-188]

* Tc hi thch hp

Vy :

Vi:

do Rey < 40

= 203,59

2. Tr lc on Luyn * Chun s Acsimet

[II-188]

* Chun s Reynolds

IX.117 [II-188]

* Tc hi thch hp

Vy :

Vi:

do Rey > 40

EMBED Equation.3

Vy tr lc ca ton thp :

V. TNH CN BNG NHIT LNG :

1. Cn bng nhit lng ca thit b un si hn hp u

Phng trnh cn bng nhit lng ca thit b un nng hn hp u: QD1 + Qf = QF + Qng1 + Qxq1 (J/h) IX.149 [II-196]Trong :

QD1 : Nhit lng do hi t mang vo, J/h.Qf : Nhit lng do hn hp u mang vo, J/h.QF : Nhit lng do hn hp u mang ra, J/h.

: Nhit lng do nc ngng mang ra, J/h.

: Nhit lng mt mt ra mi trng xung quanh, J/h. Chn hi t l hi nc bo ho p sut 2 at, c to si = 119,62 oC

1.1.Nhit lng do hi t mang vo

IX.150 [II-196] Trong :

- D1 : lng hi t, kg/h. - r1 : n nhit ha hi, J/h. - : hm nhit ca hi t, J/kg.( nhit lng ring ca hi t (J.kg)). - : nhit nc ngng, o C. - C1: nhit dung ring ca nc ngng, J/kg .1.2. Nhit lng do hn hp u mang vo

Qf = F.Cf.tf (J/h) IX.151 [II-196] Trong :

- F: lng hn hp u, kg/h = 7500 kg/h - tf: nhit ca hn hp u, o C. Hn hp vo nhit thng: tf = 200C

- Cf : nhit dung ring ca hn hp u, J/kg . Ly tf = 20oC v ni suy t bng I.153 [I-171] ta c :

CE = 2480 (J/kg ) CT = 4180 (J/kg )

(J/kg ).

Vy: Qf = F.Cf.tf = 7500.3602.20 = 540 300 000 (J/h)

1.3. Nhit lng do hn hp u mang ra QF = F.CF.tF (J/h) IX.152 [II-196] - F: lng hn hp u, kg/h = 7500, kg/h. - tF : nhit si ca hn hp, o C. - CF: nhit dung ring ca hn hp i ra khi thit b un si, J/kg .

- CE ; CN : ln lt l nhit dung ring ca Etylic, nc nhit toF = 84,26 oC

Ni suy t bng I.153 [I-171] ta c :

CE = 3283,9 (J/kg )

CN= 4198,5 (J/kg )

EMBED Equation.3 (J/kg )

Vy:

QF = F.CF.tF = 7500.3887,536.84,26= 2 456728375 ( J/h )

1.4. Nhit lng do nc ngng mang ra Qng1 = Gng1.C1. = IX.153 [II-197] Trong :

Gng1: Lng nc ngng, bng lng hi t, kg/h

Do Gng1 = D1 (kg/h) 1.5. Nhit lng mt mt ra mi trng xung quanh Lng nhit mt ra mi trng ly bng 5% nhit tiu tn

Qxq1 = 0,05D1.C1r1 (J/h) IX.154 [II-197]1.6. Lng hi t cn thit un nng dung dch u n nhit si Theo IX.155 [II-197]:

Chn p = 2 at suy ra tso = 119,62oC tra bng I.212 [I-254] ta c :

2. Cn bng nhit lng ca thp chng luyn Phng trnh cn bng nhit lng ca thp chng luyn: Tng lng nhit mang vo thp bng tng lng nhit mang ra:

QF + QD2 + QR = Qy + QW + Qxq + Qngt IX.156 [II-197] Trong :

QF : nhit lng do hn hp u mang vo thp, J/h. QD2 : Nhit lng do hi t mang vo thp, J/h.

(J/h) IX.157 [II-197] QR: Nhit lng do lng lng hi lu sn phm nh mang vo thp :

QR = GR.CR.tR (J/h) IX.158 [II-197] :Nhit do hi mang ra nh thp :

IX.159 [II-197] Qw :Nhit lng do sn phm y mang ra :

(J/h) IX.160 [II-197] :Nhit do nc ngng mang ra :

IX.161 [II-198] :Nhit lng mt mt ra mi trng xung quanh :Ly bng 5% nhit tiu tn y thp

IX.162 [II-198]Chn hi t l hi nc bo ho p sut 2 at, c t0 si = 119,620C

* Vy lng hi t cn thit un si sn phm y :

(kg/h) IX.163 [II-198]2.1. Nhit lng do lng lng hi lu mang vo thp

QR = GR.CR.tR = P.Rth.CR.tR * Tnh CR:

T bng I.153( I-171) nhit toP= 78,98oC ta ni suy c : CE = 3207,25 (J/kg )

CN = 4208,98 (J/kg )

EMBED Equation.3 (J/ kg )

* Tnh GR:

GR = P . Rx = 2938,596.1,0312= 3030,28 (J/h )

Vy: QR = GR.CR.tR = 3030,28. 3347,4922.78,98 = 801160377 ( J/h )

2.2. Nhit lng do hi mang ra nh thp

Tra bng I.212 [I-254] ni suy nhit toP = 78,98oC c:

2.3. Nhit lng do sn phm y mang ra(hoc nhit lng do hi mang ra nh thp) QW = CW.W.tW ( J/h ) T ni suy theo bng IX.2a [II-145] ta c

Ni suy t bng I.153 [I-171] c CE = 3514,405 (J/ kg )

CN = 4229,25 ( J/kg )

EMBED Equation.3

EMBED Equation.3 2.4. Nhit lng do nc ngng mang ra

T bng I.149 [I-168] ta c

( J/kg )

Nhit tn tht ra mi trng :

Qxq = 0,05.D2r2 = 0,05.762952,1904.D2

Lng hi t cn thit un si lng sn phm y :

EMBED Equation.3 3. Cn bng nhit lng cho thit b ngng t

Phng trnh cn bng nhit lng cho thit b ngng t ( ngng t hon ton ): P(Rth +1)r = GnlCn(t2 t1) [II-198] Trong :

r: n nhit ngng t, J/kg

Nhit ca hi nh thp l: t0d = 57,60C

Tra bng I.212 [I-254] ni suy nhit toP = 78,98oC c :

Cn: Nhit dung ring ca nc nhit trung bnh (J/kg. )

Gnl: Lng nc lnh tiu tn cn thit, kg/h. t1, t2: Nhit vo v ra ca nc lm lnh, 0C

Nhit vo ca nc lnh ly l nhit thng: t1 = 250C.Nhit ra ca nc lnh chn l: t2 = 450C trnh hin tng ng cn li trn b mt truyn nhit v trnh s kt ta ca cc mui.

Tra t bng I.153 [I-171] ta c Cn = 4176,25(J/kg )

Vy lng nc lnh cn thit ngng t hon ton sn phm nh :

4. Cn bng nhit lng ti thit b lm lnh Coi lm lnh sau khi ngng t hon ton th ta c phng trnh cn bng nhit lng ca thit b lm lnh:

IX.167 [II-198] Trong :

Gn4: Lng nc lnh tiu tn, kg/h. t1, t2: Nhit u v cui ca sn phm nh ngng t, 0C

Chn t1 = 25oC ,t2 = 45o C

* Tnh t1

+ hiu s nhit trung bnh:

Nhit cui ca sn phm nh ngng t ly l: t2 =250C

Vy CP ti

Ni suy theo bng I.153 [I-171] ti 30,39oC ta c

CE = 2590,52 (J/kg )

CN = 4177,4 (J/kg )

(J/kg )

Lng nc lnh tiu tn :

PHN III:TNH TON V CHN THIT B PH I. THIT B GIA NHIT HN HP U:i vi qu trnh chng luyn, nng cao hiu qu lm vic th hn hp u thng c a vo thp trng thi lng si ( xt n nh hng ca trng thi nhit ng) nhm to ra s tip xc tt gia 2 pha lng - hi. iu ny thc hin nh thit b gia nhit hn hp u.

Da trn cc tiu chun kinh t, k thut ta chn thit b truyn nhit ng chm kiu ng. Tc nhn un nng l hi nc bo ho 2at v nhit 119,620C un nng hn hp u t nhit 200C ti nhit si ca hn hp l 99,900Cv n c h s cp nhit ln v n nhit ngng t cao.

Trong thit b 2 lu th i ngc chiu nhau, hi t i t trn xung, truyn n nhit ho hi cho hn hp lng i t di ln v ngng t thnh lng i ra khi thit bThit b trao i nhit loi ng chm thng ng vi cc thng s:

- Chiu cao ng: h0 = 1,5 (m)

- ng knh ng: d = 30 (mm)= 0,03 m - Chiu dy thnh ng: = 2,0 (mm)

Vy ng knh trong ca ng l: d0 = 26 (mm)= 0,026 m - Dung dch i trong ng, hi t i ngoi ng

- Chn vt liu ch to ng l thp khng g CT3Vy h s dn nhit ca vt liu l: ( = 50 (W/m. ) [II-313]

Chn hi t l hi nc bo ho p sut 2 at, c t0 si = 119,620C

1. Tnh hiu s nhit trung bnh:Nhit vo ca dung dch l: td =250C

Nhit ra ca dung dch l: tc = tF = 84,26 0C

Hi t l hi nc bo ho nn nhit khng thay i v l nhit si p sut chn ( 2 at ): 119,620C [I-314] Hiu s nhit gia cht ti nhit v lu th

Vy nhit trung bnh ca hai lu th: 60,2

Suy ra nhit trung bnh hn hp u ttb2 =119,62 60,2oC=59,42 0C2.Tnh lng nhit trao i dng un nng hn hp u n nhit i siQ = FCP(tF tf) [II-46]

T bng I.153 [I-171] ni suy ti 59,42 oC ta c :

CE = 2717,54(J/kg )

CN = 4175,4 (J/kg )

(J/kg )

Q = FCP(tF tf) = 7500.3679,727(84,26 20)= 1773444428(J/h) = = 492623,45 (J/s)3. Din tch trao i nhit

K hiu:th : Nhit hi t hi nc bo ho 2 at (oC) : th = ttb1 = 119,62 oC

tT1 : Nhit mt ngoi ng (oC)

tT2 : Nhit mt trong ng (oC)

tdd : Nhit dung dch (oC) : tdd = ttb2 = 59,42 oC

(t1 : Hiu nhit gia hi t v mt ngoi ng (oC) : (t1 = th tT1

(t2 : Hiu nhit gia mt trong ng v dung dch (oC) : (t1 = tT2 tdd(tT : Hiu nhit gia mt ngoi ng v mt trong ng (oC) : (tT = tT1 tT2( : Chiu dy thnh ng (m)

tm : Nhit mng nc ngng (oC) : tm = 0,5.( th + tT1)

q1 : Nhit ti ring pha hi ngng t (W/m2)

q2 : Nhit ti ring pha dung dch (W/m2)

(1 : H s cp nhit pha hi ngng t (W/m2.)

(2 : H s cp nhit pha dung dch (W/m2. )

3.1. H s cp nhit pha hi ngng t

(W/m. ) V.101 [II-28]

Trong : - r: n nhit ngng t ly theo tobh (J/kg) - : chnh lch nhit gia hi bo ha v thnh ngoi ng - H: chiu cao ng = 1,5 m

- A: h s tra theo tom ( nhit mng nc ngng)

Vi ni suy theo I.212 [I-254]

Gi s: (t1 = 4,6 oCKhi : =117,32

T tom = 116,6 oC tra theo [II-29] ta c A=186,6Vy:

= =6941,5 (W/m2. )3.2. Tnh h s cp nhit pha hn hp chy xoy (V-40 .15-II)Chn d = 30x2 mm , H= 1,5 m

* Re : Chun s Reynolt: qu trnh truyn nhit t hiu qu, dung dch phi ch chy xoy

Gi s Re = 10500

* Chun s Pr:

Chun s Prand ca dng tnh theo nhit dng:

V.35 [II-12]

Trong :

+CP : Nhit dung ring ca dung dch ttb, (J/kg. ) = 3679,727 (J/kg. ) +: nht ca dung dch totb , (N.s/m2) Ti totb = 59,42 oC ni suy theo I.101[I-91]

+: H s dn nhit ca hn hp, (W/m. )

Cng thc tnh h s dn nhit: I.32 [I-123] Trong : +A: H s ph thuc mc lin kt ca cht lng Ru v nc.V Ru , nc l hai cht lng lin kt nn A = 3,58.10-8 [I-123]

+( : Khi lng ring ca dung dch (kg/m3) Tra t bng I.2 [I-9]ti ttb= 59,42oC ta c:

Nng khi lng ca dung dch: aF = 34 %

( Theo IX .104a [II-183], ta c:

EMBED Equation.3 +M: Khi lng mol phn t ca dung dch, (kg/kmol)

Nng phn mol ca dung dch l: xF = 0,1678 (kmol/kmol)

Vy

(W/m )

Vy:

* Chun s Prt:

Chun s Prand tnh theo nhit tng:

- Nhit ti ring v pha hi ngng t :

q1=(t1. =6941,5.4,6=31930,9 -hiu s nhit hai pha thnh ng :

[II-3]

: b dy ca thnh =2 mm

: h s dn nhit ca vt liu lm ng (W/m )

Theo bng [II-313] vi thp CT3 th = 50(W/m )

Tra bng V.1 trong (II-4) ta c:

r1: Nhit tr do lp cn b bm bn ngoi thnh ng :

=0,232.10-3 (m2. /W)

r2: Nhit tr do lp cn b bm bn trong thnh ng :

=0,387.10-3 (m2/W)

Do (tt= =31930,9.0,659.10-3= 21,04 (oC)( tT1 = th - (t1 = 119,62 4,6 = 114,6(oC)

( tT2= tT1-(tt=114,621,04=93,56(oC)(t2= tT2- ttb2 =93,5659,42=34,14(oC) * Tnh CPt:

Ti tT2 = 93,56oC ni suy theo I.153 [I-171]

CE = 3423,5(J/kg )

CN = 4217,12(J/kg )

* Tnh :

Ti tT2 = 93,56 oC ni suy theo I.2[I-9]

* Tnh t: Ti tT2 = 93,56oC ni suy theo bng I.101[I-91]

* Tnh: Theo I-123 : , (W/m. )

= 0.4148Vy:

Vy chn l ph hp ( Kt qu chp nhn c

( Nhit ti ring trung bnh:

(W/m2)

4. B mt truyn nhit

5. S ng truyn nhit (ng)

Quy chun ta c n = 127 (ng ) t bng V.11 [II-48]

Chn cch sp xp ng theo hnh lc gic, gi a l s ng trn mt cnh hnh lc gic.(Tng s ng [II-48] l: nO = 3a(a-1)+1 (ng)

Chn a = 7 (ng) ( nO = 3a(a-1)+1 = 3.7.(7-1) + 1 = 127 (ng) S ng trn ng cho hnh lc gic: b = 2a 1 = 13 ng

6. ng knh trong ca thit b trao i nhit: ng knh ngoi ca ng: dn = 0,03m

t l bc ng, t = (1,2 ( 1,5).dn

Chn bc ng l: t = 1,5.dn = 1,5.0,03 = 0,045m S ng xuyn tm b= 13 (Theo [II-49] ng knh trong ca thit b l:

Dt = t.(b - 1) + 4.dn = 0,045.(13 1) + 4.0,03 = 0,66 m = 660mm 7. Tnh li vn tc v chia ngnTc chy thc t ca thit b gia nhit c xc nh theo cng thc sau:

GF: Lng hn hp u (kg/ h)

: khi lng ring trung bnh ca hn hp u ( kg/ m3)

d: ng knh trong ca ng ( m )

Thay s vo ta c:

Vn tc gi thit

S ngn ngnQuy chun, ta chia thit b lm: m= 5 ngnS ng l 127 ng chia lm 5 ngn

B tr theo hnh lc gic, s ng trn ng cho l 13 ng

Chiu cao ng l h0= 1,5 (m), ng knh ngoi ca ng l 0,03 (m), ng knh trong ca ng l 660 mm.

ng knh ng: d = 30 (mm)= 0,03 m

Chiu dy thnh ng: = 2,0 (mm)

ng knh trong ca ng l: d0 = 26 (mm)= 0,026 m

Vy cc thng s ca thit b gia nhit hn hp u l:

F = 15,5 m2 L = 1,5 m

dn =30 mm

D = 660 mm

S ng : n = 127 ng

S ngn : m = 5 ngn II. TNH TON C KH V LA CHN:1. Tnh ton thn thp

* Chn vt liu lm thn thp:

Thp chng luyn c thn hnh tr t thng ng lm vic khong nhit t = 25 1000C v p sut thng nn ta chn vt liu lm thn hnh tr thp bng thp cacbon c k hiu CT3. Thp ny rt bn nhit. Thit b thuc nhm 2 loi II( H s iu chnh [II-356] l: ( = 1,0

c trng c hc ca mc thp ny tra t bng XII.4 [II-309] v [II-313] Vt liuGii hn bn ko (k (N/m2)Gii hn bn chy (c (N/m2)H s gin khi ko 200-

1000C at (1/oC)Khi lng ring ( (kg/m3)H s dn nhit W/m.

CT3380.106240.10611.1067,85.10350,0

* Cch ch to:

T Dt = 10600 (mm) Hn tay bng h quang in vi kiu hn gip mi 1 bn c tm lt khp chu vi. Theo bng XIII.8 [II-362] ta c h s bn mi hn = 0,9

* Tnh ton sc bn vt liu :

- Cng thc tnh ng sut cho php :

do tT < 470o t bng XIII.4-337 dng cng thc XIII.1 XIII.2

[II-355]

Trong :

- k : l ng sut ko ( N/m2)

- nb, nC : h s an ton theo gii hn ko , gii hn chy ;

Tra t bng XIII.2, XIII.3 [II-356] =1,0, nk = 2,6 , nC = 1,5

( ng sut gii hn bn ko l:

( ng sut gii hn bn chy l:

Chn ng sut sao cho t min = =146.106 ( N/m2)

2.Tnh chiu dy thn thp

Theo II-360 : [II-360]

- Dt : ng knh trong thn thp (m). Theo tnh ton ban u Dt = 0,66m

- [k] : ng sut cho php vi loi vt liu chn, N/m2.

- ( : H s bn hn ca mi hn dc

Trn thn hnh tr c 2 l ng knh 150mm= 0,15(m) lp knh quan st cc v tr quan st phn phi cht lng v cht hi lu.

V cc l c b tr trn mt ng sinh nn gi tr h s bn ca thn hnh tr c khot l c tnh theo cng thc sau:

XIII.16 [II-362] L: chiu cao thn hnh tr, m

L = Ht = 10,6 (m)

- P : p sut lm vic trong thp (N/m2)

(N/m2) - Pmt:p sut ca hi trong thp.Thp lm vic p sut thng nn Pmt = 1 at =9,81.104 (N/m2)

- (l : Khi lng ring ca cht lng trong thp, kg/m3.

Ta ly theo khi lng ring ln nht l khi lng ring trung bnh pha lng on chng: (l = kg/m3 - Hl : Chiu cao ct cht lng trong thp,( m).Ta ly chiu cao ln nht l chiu cao thp: Hl = HT = 10,6 (m)

- Pl : p sut thy tnh ca ct cht lng = l.g.H1 - g: Gia tc trng trng: g = 9,81 (m/s2)

P = 9,81.104 + 9,81.864,622.10,6 = 182914(N/m2)

- C: S b sung do n mn, bo mn v dung sai v chiu dy, (m)

C = C1 + C2 + C3 (m)

- C1 : L h s hiu chnh do n mn. B sung do n mn xut pht t iu kin n mn vt liu ca (m)

Chn C1 = 1 (mm) = 10-3 (m)

- C2 : H s hiu chnh do bo mn c hc ( m)

Thp chng luyn ch cha lng v hi nn t bo mn C2 =0

- C3 : H s hiu chnh do dung sai (m)

Tra t bng XIII.9 [II-364] C3 = 0,8.10-3 (m);

Vy: C = C1 + C2 + C3 = 10-3 + 0 + 0,8.10-3 = 1,8.10-3 (m)

V gi tr nn ta c th b qua i lng P mu s ca cng thc tnh chiu dy thit b.

S= 6,45.10 + 1,8.10= 2,445.10

Chun ha S= 2(mm)

* Kim tra ng sut thnh thit b theo cng thc XIII.26 [II-365]

(N/m2)

p sut th:

Po = Pth + Pl (N/m2) [II-366]

Trong :

Pl : p sut ct cht lng trong thp (N/m2)

Pl = g . (l . Hl

(1: Khi lng ring ca nc nhit trung bnh ca thp, kg/m3

Vi nhit trung bnh ca thp nh trn ta ni suy theo bng I.2 [I-9] c:

(1 = = 963,334( kg/m3) ( Pl = g . (l . Hl = 9,81 .983,334.10 = 102252,9693 (N/m2)

Pth : p sut thu lc (N/m2)

Theo bng p sut thu lc khi th [II-358]:

( Pth = 1,5.P1 = 1,5.102252,9693 = 153379,454 (N/m2) ( Po = 141754,59 +94503,06 = 255632,4233 (N/m2)

Vy:

Vy St = 2 mm l chp nhn c .

3. Tnh chiu dy y v np thit b

* Chn y v np dng elip c g lp vi thn thit b bng cch ghp bch, tm c c l ly sn phm y v sn phm nh. Vt liu lm y v np bng thp CT3.

*Chi tit cu to:

- ng knh: Dt = 1 (m)

- Chiu cao g, tra bng XIII.12 [II-385]: h = 25 (mm)

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

(m) [II 385]

Trong :

- [] : ng sut cho php (N/m2)

-: H s bn mi hn = 0,9

- hb: Chiu cao phn li ca y v np, (m). Tra bng XIII.10 [II-382] ta c:

hb = 250 (mm)= 0,25(m) - k: H s khng th nguyn

XIII.48 [II-385]

i vi np thp c ng knh ng dn sn phm nh d = 150mm= 0,15(m)

i vi y thp c ng knh ng dn sn phm y d = 80mm

- C: h s hiu chnh (m)

3.1. Chiu dy np

[II 385]

V gi tr nn ta c th b qua i lng P mu s ca cng thc tnh chiu dy np.

V vy ta thm vo C: 2mm sao vi i lng b sung phn tnh chiu dy thn thp, ta c: C= 2 + 1,8 =3,8 (mm)

Vy Sn = 0,861+ 3,8 = 4,661 (mm)

Quy chun ta ly Sn = 5 (mm)

* Kim tra ng sut np bng nc :

Vy Sn = 5 (mm) l chp nhn c.

3.2. Chiu dy y thit b

V gi tr nn ta c th b qua i lng P mu s ca cng thc tnh chiu dy np.

V vy ta thm vo C: 2mm so vi i lng b sung tnh phn tnh chiu dy thn thp, ta c: C = 2 + 1,8 = 3,8 (mm)

Vy: Sd = 0,796 + 3,8 = 4,596 (mm)

Chun ha ta c: S = 5 (mm)

* Kim tra ng sut bng nc :

Vy S = 5 (mm) l chp nhn c .

Vy Sn = Sd = 5 (mm)

Tra bng XIII.11 [II-384], vi ng knh Dt = 600mm, S = 5mm, ta tra c khi lng ca y v np: m = 17,5 (kg)

Vy ta c thn thp vi nhng thng s nh sau:

- ng knh thp: Dt = 1000mm

- Chiu cao phn li ca y v np: hb = 250mm=0,25(m) - Chiu cao g: h= 25mm=0,025(m) - Chiu dy y v np: S = 5mm

- Khi lng ca y v np: m = 17,5kg

4. Tnh ng knh ng dn

ng knh cc ng dn v ca ra vo ca thit b c xc nh t cng thc sau:

(I- 369)

Trong : V : lu lng th tch (m3/s)

: tc trung bnh (m/s)

4.1. ng knh ng dn sn phm nh

Lng hi nh thp l: gd = 104,404 (kmol/h) = 4802,6088 (kg/h)= 1,02 (kg/s) Nhit ca hi nh thp l: tP = 78,98 0C

(Lu lng th tch ca hi nh thp l:

: Khi lng ring ca hi nh thp, kg/m3

EMBED Equation.3 Theo bng II.2 [I-370] i vi hi bo ho trong ng dn khi p = 1at ta chn tc hi nh thp: h = 20 (m/s) [ II 74 ]

ng knh ca ng dn hi nh thp l:

EMBED Equation.3 Quy chun: D = 200mmTra bng XIII.32 ta c chiu di on ng ni l=140 (mm)

( Tc thc t ca hi nh thp l:

4.2. ng knh ng dn hi lu sn phm nh

Lng hi ngng t hi lu l: GR = P.R = 2938,5964.1.0312 = 3030,28 (kg/h) = 0,842(kg/s) Nng khi lng ca hi ngng t hi lu: aR = aP = 86 %

(Khi lng ring ca hi ngng t hi lu l:

Ni suy t bng I.2 [I-9] nhit tPo = 78,98 oC

(Lu lng th tch ca hi ngng t hi lu l:

Do sn phm nh t chy v hi lu. Tra bng vn tc II.2 [I-370]

w = 0,1 0,5 m/s. Nn ta chn tc hi ngng t hi lu l: wR = 0,2 (m/s) ( ng knh ca ng dn hi ngng t l:

Quy chun: Dy = 100mm (Tc thc t ca hi ngng t hi lu l:

4.3. ng knh ng dn hn hp u

Lng hn hp u vo thp l: F = 7500 (kg/h) = 2,083 (kg/s)

Nhit ca hn hp u: tF = 84,26 oC

( Khi lng ring ca Etylic v nc bng I.2 [I-9] theo t = tF :

=720,047(kg/m3)

=960,982(kg/m3)

Nng khi lng ca hn hp u: aF = 34%

( Khi lng ring ca hn hp u l:

( Lu lng th tch ca hn hp u l:

(m3/s)

Chn tc hn hp u l: wF = 0,3 (m/s) ( ng knh ca ng dn hn hp u l: Quy chun ta c : Dt = 125(mm)4.4.ng knh ng dn sn phm y Lng sn phm y l: W = 4561,404 (kg/h) = 1,267 (kg/s)Nhit ca hn hp y: = 108,66(0C)Khi lng ring ca Etylic v nc c ni suy t bng I.2 [I-9] nhit =

( Lu lng th tch ca sn phm y l:

Chn tc sn phm y l: w = 0,2 (m/s)

( ng knh ca ng dn sn phm y l:

Quy chun Dt = 125(mm)4.5. ng knh ng dn hi lu sn phm y

Lng hi sn phm y hi lu l: gtbC = 155,79 (kmol/h) = 0,779 (kg/s)

Nhit ca hi sn phm y hi lu: = 99,6270C

Chn tc hi sn phm y hi lu l: w = 20 (m/s)

Quy chun ta c :Dt = 200(mm)

5. Tra bch

5.1. Chn bch lin bng thp ni thit b

Do khng th ch to c than thp vi chiu di ln nn ta buc phi dung bch ni cc phn li vi nhau. Vi thn thp hnh tr lm vic iu kin thng ta chn mt bch lin bng thp CT3 ni thn vi y v np thit b. Theo bng XIII.27 [II-419]

i vi thn thp ta v chiu cao ca thp ln nn ta phi s dng bch ni cc on thn thp li vi nhau. Ta s dng loi bch lin c ch to bng thp CT3 vi khong cch gia cc bch chn l: 1,88m. Vy vi thp c chiu cao:H = 10,6(m) ta phi s dng s bch l: Z = (bch trn cng ni thn thit b vi np) = 7 (ci Kt qu tng hp bng sau:Py.106DtDDbD1D0dbhZ

N/m2MmCi

0,110001140109010601013M202024

5.2. Chn bch lin bng thp kim loi en ni cc b phn ca thit b v ng.

Tn ng dnDy Dn DD(

D1D0hZ

MmCi

Sn phm nh300325435395365M202212

Hi lu nh100108205170148M16144

ng dn liu125133235200178M16144

Sn phm y125133235200178M16144

Hi lu y200219290255232M16228

5.3. Kch thc chiu di on ng ni

Da vo ng knh ca cc ng ta tra c s liu sau:

Theo bng XIII.32 [II-434]

Tn ng dnDyPy < 2,5.10-6 N/m2

mm

Sn phm nh200130

Hi lu nh100120

ng dn liu125120

Sn phm y125120

Hi lu y200130

6. Tnh li m , dm m , a phn phi cht lng

6.1. a phn phi cht lng

Chn kiu a loi 2 vi cc thng s nh sau:

Tra t bng IX.22 [II-230] kt qu c tng hp ti bng sau:

ng knh thpng knh a, Ddng knh ngBc ng, tS ng

MmChic

100060044,5x2,57040

6.2 Li m Chn ng knh li l: D1 = 980mm

Theo bng IX.22 [II-230] chiu rng ca bc b vi m (25 x 25) = 20,5mm

6.3. Dm m

Chn dm m hnh ch nht c chiu cao bng 2 ln chiu rng. Dm c lm bng vt liu thp CT3, hai u thanh dm c hn vo thn thit b.

Tnh bn ca dm:

, kN/m2

Trong :

: ng sut php, kN/m2

: ng sut tip, kN/m2 - Dm s chu tc dng phn b gy nn bi khi lng ca lp m ca cht lng v ca thnh thit b.

- m bo bn cho thanh dm, ta coi cht lng chon y thp.

-V khi lng ring ca Nc ln hn ca Etylic nn ta coi thp cha ton Nc. ng thi coi c thp l mt khi tc dng ln mt thanh dm chung.

* Th tch ca mt ngn m:

Vi Hn: Chiu cao mt ngn m, chn Hn = 0,4m

Theo bng IX.8 [II-193] th khi lng ring ca xp: = 650 kg/m3 * Khi lng ca mt ngn m:

* on Chng c HC = 1,06(m), vi khong cch mi ngn m l 0,2(m) vy ta phi s dng 5 dm m trong dm trn cng c chiu cao lp m l: HC= 1,06 4.0,2 = 0,26(m).

Th tch ngn m tng ng vi chiu cao m HC l:

Vy khi lng ngn m trn cng on Chng l:

Vy khi lng ca lp m on Chng l:

mC = 2.md1 + mdC = 2.153,075+81,64 = 387,79 (kg)

* on Luyn c HL =4,8(m), vi khong cch mi ngn m l 0,2m vy ta phi s dng 24 dm m trong dm trn cng c chiu cao lp m l:

HL= 4,8 23.0,2 = 0,2 m.

Th tch ngn m tng ng vi chiu cao m HL l:

Vy khi lng ngn m trn cng on Luyn l:

Vy khi lng ca lp m on Luyn l:

mL = 10.md1 + mdL = 10.153,075 + 156 =1686,75 (kg)

Vy khi lng m trong thp chng l: md = mC + mL = 387,79+1686,75= 2074,54 (kg)* Khi lng ca cht lng trong thp:

Vi T =960,982 (kg/m3) tra nhit tFo = 84,26o C

EMBED Equation.3

* Din tch bao quanh thp:

* Lc phn b tc dng ln thanh dm ca thnh thit b l:

EMBED Equation.3 = 7850.33,284 = 261290,3837(N/m) - Khi lng ca np bng khi lng ca y: m = 47,9(kg) * Lc phn b tc dng ln dm ca thit b:

* Momen i vi trc x, Mx Mx = MA.MB =

* Lc ct ngang ti hai u A v B:

Qy = RA = RB =

EMBED Equation.3 - Biu ni lc :

Mt ct nguy him ti B,C,A

V trng thi ng sut l n nn ta c:

- : ng sut cho php ca thp CT3

- Wx: Kch thc dm o, m2 Wx

M:

Vy ta chn: b = 50mm

Vy: h = 2b = 100mm

Vi h l chiu cao, b l chiu rng

*Kim tra bn ca thanh dm c b = 50mm, h = 100mm. Theo cng thc:

, ti v tr 1, 2, 3

Ta c: (max = (min =

Ti v tr (1):(kN/m2)

( < [(] => Tho mn iu kin bn

Ti v tr (2):

m vy

kN/m2

( < [(] => tho mn iu kin bn

Ti v tr (3): kN/m2

( < [(] => tho mn iu kin bn

Vy chn thanh dm vi thng s: b = 50mm, h = 100mm l t c yu cu thit k.

7. Tnh chn tai treo v chn

* Tai treo

Mun chn c loi tai treo thch hp vi thit b ta phi tnh c trng lng ton thit b.

- Chiu cao ca ton thp: HT = 10,6(m)

-Tng khi lng ca ton thp:

M = mt + mnap + mdem + mday + mchatlong + mbx Vi:

mbx: H s b sung bao gm khi lng cc chi tit ph ca thp nh bch, bulng

-Khi lng thp:

mt = (t. Vt

Thn thit b lm bng thp CT3 vi (t = 7850kg/m3 [II.313]

-Th tch thp:

- Dt, Dn: ng knh trong v ngoi ca thp

- Dn = 1 + 2.0,003 = 1,006m

Dt = 1m

Mbx: Khi lng cc b phn khc, gi thit chn: mbx = 500kg

M = 10455,088 + 500 =10955,088(kg)

-Trng lng ca thp l:

P = M.g = 10955,088.9,81 = 107469,41 (N)

- Ta s dng 4 tai treo lm bng thp CT3, ti trng ca mi tai treo l:

Theo bng XIII.36 [II-438], ta chn tai treo c ti trng cho php l:

G = 2,5.104 N.

B mt l: F = 173.10-4 m2, ti trng cho php ln b mt l:

Q = 1,45.106 N/m2.

LBB1HSladKL 1 tai treo

Mmkg

15012013021586020303,48

*Chn Tra bng XIII.35 [II-437], ta chn chn c ti trng cho php l:

G = 2,5.104 N

B mt l: F = 444.10-4 m2, ti trng cho php ln b mt l:

Q = 0,56.106 N/m2.

LBB1B2Hhsld

Mm

250180215290350185169027

III. TNH THNG CAO V:1. Cc tr lc trong qu trnh tip liu

p sut ton phn cn thit khc phc tt c sc cn thy lc trong h thng k c ng dn vo thit b khi dng chy ng nhit.

[I-376]

Trong :

- : p sut ng lc hc cn thit to tc cho dng chy ra khi ng dn, N/m2.

[I-377]

- : khi lng ring ca cht lng trong ng, kg/m3.

- : tc dng chy, m/s.

- : p sut khc phc tr lc ma st khi dng chy n nh trong ng thng, N/m2.

[I-377]

- : h s ma st

- L : chiu di ng ni (m)

- dt : ng knh tng ng ng dn (m)

- : tn tht p sut cc b (N/m2)

[I-377]

Vi:

- : h s tr lc cc b

- : p sut cn thit nng cht lng ln cao hoc khc phc p sut thy tnh .

[I-377]

- H: Chiu cao cht lng, m

- : p sut cn thit khc phc tr lc trong thit b

- : p sut b sung cui ng dn trong nhiu trng hp

Trong tnh ton thit b chng luyn vi thp m th:, = 0

1.1. Tr lc ca on ng t thit b gia nhit hn hp u n thp

*Tnh p sut ng hc:

Khi lng ring ca dung dch ti tos = tF = 84,26 oC trong phn tnh ton c kh tnh = 629,326 (kg/m3)

d = 125 mm ( ng knh ng dn liu, tnh trong phn tnh ton c kh )

Tc trung bnh ca lu th:

*Tnh p sut khc phc tr lc ma st:

(Pm ==.(Pd, N/m2 Chn chiu di ng dn: L = 2,5m

ng knh tng ng ca ng: dtd = d = 80mm = 0,08m

-Chun s Re ca lu th :

nht ca hn hp ti toF = 84,26 oC.

Ni suy theo bng I.101 [I-91] ta c:

Vy dng chy trong ng l dng chy xoy

-Chn ng lm bng ng trng km mi bnh thng. Tra bng II.15 [I-381] ta c

mm Chn =0,1mm

Vy:

Ta thy Regh < Re < Ren v vy h s ma st c tnh theo cng thc sau:

[I-380]

* Tnh tr lc cc b:

, N/m2 - : H s tr lc cc b

Trn ng c :

+ Mt van chn tiu chun, m hon ton. Theo bng II.26 [I-397] ta c:

+ Tr lc ca 3 khuu 90o do 3 khuu 30o to thnh. Chn a/b =1. T bng II.26 [I-394]

+Tr lc ra khi ca ng, t m:

Tng tr lc cc b ca h thng ng dn:

Vy p sut tng cng l:

Chiu cao ct cht lng tng ng vi l:

1.2. Tr lc ca ng dn t thng cao v n thit b gia nhit hn hp u

* Tnh p sut ng hc:

Khi lng ring ca hn hp:

Ni suy t bng I.2 [I-9] 25oC ta c:

Tc ca hn hp chy trong ng dn l:

*Tnh :

(Pm ==.(Pd, N/m2 - Chun s Reynolt ca lu th

nht ca dung dch 250C Ni suy t bng I.101 [I-91] 25oC ta c :


[I-380]

Chn chiu di on ng l: L = 10m

*Tnh tr lc cc b:

, N/m2 - : H s tr lc cc b

Tr lc cc b ca cc on ng :

+ Tr lc vo ng thng : chn =0,5

+ c 2 khuu 90o trong c 3 khuu 30o to thnh

Chn a/b=1 [I-394]

+ 1 van tiu chun [I-397]


1.3. Tr lc ca thit b gia nhit hn hp u

*Tnh p sut ng hc:

- Tc lu th i trong ng l:

: Th tch ca hn hp

, m3/s

: Tit din b mt truyn nhit

, m2 Vi:

n: S ng ca thit b gia nhit, n = 127 ng

m: S ngn ca thit b gia nhit, m = 5 ngn

-Theo phn tnh ton thit b gia nhit hn hp u c :

Vy:

*Tnh :

(Pm ==.(Pd, N/m2 Chun s Reynolt ca lu th:

Theo phn tnh ton thit b un si hn hp u nhit trung bnh t = 59,42 0C ta tnh c :

Vy:

Vy hn hp ch chy xoy

Ta c:


[I-380]

Chiu di on ng l:

L = m.h = 5.1,5 = 7,5(m)

*Tnh tr lc cc b:

, N/m2 : H s tr lc cc b

Khi vo thit b, dng chy qua nhiu ch ngot v t thu t m

+ Tit din ca vo bng tit din ca ra ca ng :

+ Tit din khong trng 2 u thit b i vi mi ngn l:

+ Tit din ng truyn nhit ca mt ngn l :

+ Khi dng chy vo thit b gia nhit tc l t m :

+ Khi dng chy t cc khong trng vo cc ngn , tc l t thu 2 ln :

(Ni suy theo bng II.16 [I-389])

+ Khi dng chy t cc ngn ra khong trng , tc l t m 2 ln :

Tra bng I16 [I-388] ta c :

* Tr lc thy tnh ca thit b:


2. Tnh chiu cao ca thng cao v so vi a tip liu

Cc k hiu:

H0: Chiu cao tnh t mt thong b cha dung dch n mt thong thng cao v, m

H1: Chiu cao tnh t ay thp n a tip liu, m.

H2: Chiu cao tnh t ni t bm n y thp, m.

Z: Chiu cao tnh t a tip liu n mt thong thng cao v, m.

Bm lm vic lin tc trong qu trnh chng luyn, a dung dch t b cha ln thng cao v, mc cht lng trong thng cao v c gi mc khng i nh ng chy trn duy tr p sut khng i cho qu trnh cp liu.

( Lu lng bm: GB = GF = 7500(kg/h)

Vit phng trnh Becnuli cho hai mt ct 1-1 v 2-2 so vi mt tiu chun 2-2.

Trong :

- P1, P2: p sut ti mt ct 1-1 v 2-2 N/m2 P1 = Pa = 9,81.104 N/m2 P2 = P1 +

- w1: Vn tc dung dch ti mt ct 1-1 m/s. Coi w1 = 0 v tit din thng cao v rt ln so vi tit din ng.

- w2: Vn tc dung dch ti mt ct 2-2 m/s. w2 = 0,27 (s)

- (1: Khi lng ring ca dung dch trc khi gia nhit, kg/m3

200C

(1 = 907,176(kg/m3)

- (2: Khi lng ring ca dung dch sau khi gia nhit, kg/m3 (2 = 872,4115(kg/m3)

Hm = H1 + H2 + H3 = 0,022+0,043+1,53=1,6(m)

IV. TNH TON BM tnh v bm trong vic a hn hp u ln thng cao v m bo yu cu cng ngh th ta phi tnh cc tr lc ca ng ng dn liu ca thit b gia nhit hn hp u t tnh ra chiu cao thng cao v so vi v tr a tip liu vo thp. Cui cng tnh cng sut v p sut ton phn ca bm. Ta chn bm ly tm.

Bm ly tm lm vic p sut thng, vi s vng quay ca bm l n = 60 vng /pht, chiu cao ht ca bm nhit thng t = 250C l 5,75(m) theo bng II.31 [I-439]. chiu cao ny bm lm vic tun hon m bo v khng xy ra hin tng xm thc.

- Chiu cao y ca bm l:

Hd = Hc + Z + Hdy + Hb= 1,06+3,19+1,5+0,25 = 6 (m)

Chiu cao lm vic ca bm l:

HF = Hh + Hd = 5,75 + 6 =11,75(m)

* Tn tht p sut trn ng ng:

* Tnh :

Trong phn tn tht p sut t thng cao v n thit b gia nhit hn hp u ta tnh c:

Re = 28257,899

=0,025 - Chiu di ng:

L = H = 11,75(m)

* Tnh :

-Trn ng ng c 2 khuu mi khuu do 3 khuu 30o to thnh

-C 2 van tiu chun nn

Vy:

(P = (Pd + (Pm + (Pc = 33,0666+350,50596+97,133 = 480,70556( N/m2 )

- Chiu cao ct cht lng tng ng l:

Hm=

Do p sut ton phn ca bm l:

H = HF + Hm = 11,54+0,054=11,594

*Nng sut ca bm :

*Cng sut trn trc bm :

[I-439] Trong :

- Q: Nng sut ca bm, m3/s

- (: Khi lng ring ca cht lng, kg/m3 - G: Gia tc trng trng, m/s2 - H: p sut ton phn ca bm, m

- (: Hiu sut chung ca bm

( = (o . (tl . (ck Theo bng II.32 [I-439] i vi bm ly tm ta c cc thng s sau:

- Hiu sut th tch: (o = 0,85 0,96

- Hiu sut thu lc: (tl = 0,8 0,85

- Hiu sut c kh: (ck = 0,92 0,96

Ta chn:

- Hiu sut th tch: (o = 0,9

- Hiu sut thu lc: (tl = 0,85

- Hiu sut c kh: (ck = 0,96

( = 0,9 . 0,85 . 0,96 = 0,7344

Vy cng sut ca bm l:

*Cng sut ca ng c in :

[I-439]

Trong :

- : hiu sut ca ng c in = 0,8

- : hiu sut truyn ng = 1

Trong thc t thng ta phi chn cng sut ca ng c ln hn cng sut tnh ton

[I-439]

Trong :

l h s d tr cng sut t bng I I.33[I-139] vi

chung luyen thap dem.doc

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