phn I

n mn hc Nguyn c TonPhn II: Qu trnh v thit b truyn khi.Tnh cc thng s c bn ca thit b (thp m) chuyn khi lm vic p sut kh quyn (760 mm Hg) chng luyn hn hp hai cu t Axeton- Nc; m bo cc yu cu v nng sut tnh theo hn hp u v thnh phn nguyn liu, nh v y. Cc yu cu c th nh sau:GF=2450kg/h, aF=20% , aw= 2% , ap=97%1- Xc nh nng phn mol v lu lng (kmol/ h) ca nguyn liu, sn phm nh v sn phm y.2- V ng cn bng trn th x y.3- Xc nh ch s hi lu thch hp, ng lm vic v s a l thuyt ca thp.4- Xc nh ng knh, s a thc t v chiu cao c bn ca thp.5- Xc nh nhit nh, y v v tr a tip liu ng vi trng hp nguyn liu vo thp trng thi si.

Tnh ton thit b chnh

I. Tnh cn bng vt liu:1. Tnh ton cn bng vt liu:- Phng trnh cn bng vt liu chung cho ton thp.F = P + W[II 144]- i vi cu t d bay hiFaF = Pap+ Waw[II 144]- Lng sn phm nh l:

Trong :F: nng sut tnh theo hn hp u, kg/s hoc kg/haF, ap, aw: ln lt l nng cu t d bay hi trong hn hp u, sn phm nh, sn phm y, phn khi lngu bi cho F = 2450kg/h.Vy ta c lng sn phm nh l:

kg/h- Lng sn phm y l:W = F - P = 2450464,21 =1985,79 kg/h2. i nng phn khi lng sang nng phn mol ca aF, ap, aw:p dng cng thc

[II 126]Trong :aA, aB: nng phn khi lng ca axeton v ncMa, MB: khi lng mol phn t ca axeton v ncVi Ma= MC3H6O=58 kg/kmolMb= MH20=18kg/kmolThay s liu vo ta c:

phn mol

phn mol

phn mol3. Tnh khi lng phn t trung bnh ca hn hp u, sn phm nh, sn phm y:Theo cng thc: M = x.MA + (1 - x)MBTrong :M: Khi lng phn t trung bnh, kg/kmolx: Nng phn mol - Khi lng phn t trung bnh ca sn phm nh:Mp = xp.MA + (1 - xp)MBMp = 0,909.58 +(1- 0,909).18Mp = 54,36 kg/kmol- Khi lng phn t trung bnh ca hn hp u:MF = xF.MA + (1 - xF)MBMF = 0,072.58 +(1- 0,072).18MF = 20,88 kg/kmol- Khi lng phn t trung bnh ca sn phm y:Mw = xw.MA + (1 - xw)MBMw = 0,006.58 +(1- 0,006).18Mw = 18,24 kg/kmol4. i n v ca F, P, W t kg/h sang kmol/h:

5. Lng hn hp u trn mt n v sn phm nh:

II. Xc nh s bc thay i nng :1. Xc nh Rmin da trn th y x:Dng ng cn bng theo s liu ng cn bng tra trong bng phn cn bng lng hi v nhit si ca hai cu t 760mmHg (phn trm s mol) ca axeton v axit axetic ta c bng sau: [II 145]x05102030405060708090100

y060,37280,382,784,285,586,988,290,494,3100

toC10077,969,664,562,661,660,759,85958,257,556,9

Bng 1 - T s liu trong bng trn ta v th ng cn bng lng (x) hi (y) ( c th k v th v vit l xem phn ph lc, in th ri ghp vo)

(hnh 1), vi gi tr ca xF = 0,072 ta dng ln ng cn bng v tm c gi tr y*F = 0,654- Rmin: lng hi lu ti thiu c tnh theo cng thc

[II 158]y*F: nng cu t d bay hi trong pha hi cn bng vi nng trong pha lng xF ca hn hp.

=>2. Tnh ch s hi lu thch hp:- Ch s hi lu lm vic thng c xc nh qua ch s hi lu ti thiu theo cng thc (IX.25 [4]):Rth = b.Rmintrong b l h s d, T ta c bng s liu sau.

b1,41,61,82,02.22,4

Rth0,6130,7010,7880,8760,9641,051

B0,5630,5340,5080,4850,4630,443

Nlt8,47,57,16,86,66,5

Nlt(Rth + 1)13,54912,75812,69512,75712,96213,332

Xy dng th quan h gia Rx Nlt(Rx+1). ( c th k v th v vit l xem phn ph lc, in th ri ghp vo)

Qua th ta thy, vi Rx = 0,788 th Nlt(Rx + 1) l nh nht hay th tch thp nh nht. Vy ta c Rth = 0.788 , Nlt= 7,1 on chng 3,4 a, on luyn 3,7 a.3. Phng trnh ng nng lm vic ca on luyn:

[II 148]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 a xung.Rx : ch s hi lu.Thay s liu vo ta c.

yL= 0,441x + 0.5084. Phng trnh ng nng lm vic ca on chng:

[II.158]Trong :

: lng hn hp u tnh cho 1kmol sn phm nh.Thay s liu vo ta c.

yc = 8,125x 0,0428III. Tnh ng knh thp chng luyn:ng knh thp c xc nh theo cng thc

, m [II - 181]Trong :gtb: lng hi trung bnh i trong thp, kg/h.(y.y)tb: tc hi trung bnh i trong thp, kg/m2.sV lng hi v lng lng thay i theo chiu cao ca thp v khc nhau trong mi on nn ta phi tnh lng hi trung bnh cho tng on.1. ng knh on luyn:a. Xc nh lng hi trung bnh i trong on luyn:Lng hi trung bnh i trong on luyn 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.

, kg/h [II - 181]Trong :gtb: lng hi trung bnh i trong on luyn, kg/h.g: lng hi i ra khi a trn cng ca thp, kg/h.gl: lng hi i vo a di cng ca thp, kg/h.* Lng hi ra khi nh thp:g = GR + Gp = Gp(Rx+1)[II 181]g = 464,21(0,788 + 1)g = 830,01 kg/h* Lng hi i vo on luyn: Lng hi g1, hm lng hi y1 v lng lng G1 i vi a th nht ca on luyn c xc nh theo h phng trnh.

g1 = G1 + Gp(1)g1.y1 = G1.x1 + Gp.xp (2)[II - 182]g1.r1 = g.r(3)Trong :y1: hm lng hi i vo a 1 ca on luyn, phn khi lng.G1: lng lng i vi a th nht ca on luyn.r1: n nhit ho hi ca hn hp hi i vo a.r: n nhit ho hi ca hn hp hi i ra khi nh thp.x1 = xF = 0,072 phn mol tng ng vi 0,2 phn khi lngr1 = ra.y1 + (1-y1).rb[II - 182]Vi ra, rb: n nhit ho hi ca cc cu t nguyn cht l Axeton va Nc t01 = tF . ( c th k v th v vit l xem phn ph lc, in th ri ghp vo)

T x1= xF = 0,072 tra th trn ta c t01 = tf = 74,20CVi t01 = 74,20 C ni suy theo bng I.212 trong [I 254] ta c;

kJ/kg.

kJ/kg. r1 = 502,81.y1 + (1 - y1).2364,70 r1 = -1861,89y1 + 2364,70 kJ/kg.r = ra.y + (1 y).rb [II - 182]Vi ra, rb: n nhit ho hi ca cc cu t nguyn cht l Axeton va Nc t02 = tp. T xp = 0,909 tra th lng hi trn ta c tp = 57,4y: hm lng hi i ra khi nh thp, phn khi lng.y = yp = 0,948phn mol.(ngoi suy ng cn bng xp=0,909 phn mol)i ra phn khi lng:

Vi t02 = tP = 57,40C ni suy theo bng I.212 trong [I 254] ta c:

kJ/kg.

kJ/kg. r = 521,51.0,983 + (1 - 0,983).2425,52 r = 553,88kJ/kg.Thay cc gi tr tnh c vo h phng trnh trn ta cg1 = G1 + 464,21g1.y1 = 0,2G1 + 464,21.0,97g1 (-1861,89y1 + 2364,70) = 830,01.553,88=459724,42Gii h phng trnh ta c: g1 =564,79kg/hG1 = 100,58 kg/hy1 = 0,833phn khi lng Thay y1 = 0,833 vo r1 ta c: r1 = -1861,89.0,833 + 2364,70 = 813,98 kJ/kgVy lng hi trung bnh i trong on luyn l:

kg/h.b. Tnh khi lng ring trung bnh* Khi lng ring trung bnh i vi pha hi c tnh theo

, kg/m3. [II - 183]Trong :MA MB: khi lng phn mol ca cu t Axeton va NcT: nhit lm vic trung bnh ca thp, 0K.ytb1: nng phn mol ca cu t 1 ly theo gi tr trung bnh

[II - 183]

Vi : nng lm vic ti 2 u mi on thp, phn mol.

= yp = 0,948 phn mol

= y1 = 0,833 phn khi lng i sang phn mol ta c:

phn mol

phn molNhit trung bnh on luyn.T= (TF + TP)/2 = (74,2 + 57,4)/2 = 65,8 0C = 338,8 KVy khi lng ring trung bnh ca pha hi i vi on luyn l:

kg/m3.* Khi lng ring trung bnh i vi pha lng

, kg/m3 [II - 183]Trong :

: khi lng ring trung bnh ca lng, kg/m3.

: khi lng ring trung bnh ca cu t 1 v 2 ca pha lng ly theo nhit trung bnh, kg/m3.

: phn khi lng trung bnh ca cu t 1 trong pha lng.

phn khi lng

: nhit trung bnh ca on luyn theo pha lng ng vi t0 = 65,80C. Ni suy theo bng I.2 trong [I-9] ta c:

kg/m3.

kg/m3.Vy khi lng ring trung bnh ca lng trong on luyn l:

kg/m3 Tnh ytb Chn thp a li, lm vic u n (ng knh l 2,5mm, chiu cao ng chy chuyn trn a 10 12 mm, thit din t do ca a 12,8%). Thng ly tc lm vic ytb khong 80 90% tc gii hn gh: (IX.111 [4])

[m/s]trong ytb, xtb l khi lng ring trung bnh ca pha hi v pha lng ca on luyn.Vy,

Tc lm vic ytb c ly bng 0,9.gh, nn:ytb = 0,9.1,08= 0,971 [m/s]T cc thng s gtb, ytb, ytbtnh c, ta c ng knh on luyn bng:

Quy chun ng knh. Do = 0,38 m2. ng knh on chng:a. Lng hi trung bnh i trong thp

[II - 182]Trong :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 (gn= g1) nn ta c th vit:

[II - 182]Lng hi i vo on chng gl, lng lng G1 v hm lng lng xl c xc nh theo h phng trnh cn bng vt liu v cn bng nhit lng sau:G1 = g1 + GwG1.x1 = g1.yw + Gw.xw[II - 182] g1.r1 = g1.r1Trong :r1: n nhit ho hi ca hn hp hi i vo a th nht ca on chng.xw: thnh phn cu t d bay hi trong sn phm y.r1: n nhit ho hi ca hn hp hi i vo a trn cng ca on chng.Ta c:Gw = W = 1985,79 kg/h.xw = 0,006 phn mol tng ng vi 0,02 phn khi lng y1 = yw xc nh theo ng cn bng ng vi xw = 0,006 phn mol. yw = 0,0724 phn mol.i y1 = yw = 0,0724 phn mol ra phn khi lng ta c:

phn khi lngr1 = ra. y1 + (1- y1).rb [II 182]Vi ra, rb: n nhit ho hi ca cc cu t nguyn cht t0 = tw. Vi xw = 0,006 tra th hnh 9 ta c tw = 97,30C. T t0 =tw =97,30C ni suy theo bng I.212 trong [I 254] ta c.

kJ/kg.

kJ/kg.=>r1 = 476,22.0,201 + (1 - 0,201).2267,99r1 = 1907,84 kJ/kgThay vo h phng trnh trn ta c:G1 = g1 + 1985,79G1.x1 = g1.0,201 + 1985,79.0,02g1.1907,84 = 564,79.813,98Gii h phng trnh trn ta c:g1 = 240,97 kg/hG1 = 2226,76 kg/hx1 = 0,0396 phn khi lngVy lng hi trung bnh i trong on chng l:

kg/h.b. Tnh khi lng ring trung bnh:* Khi lng ring trung bnh i vi pha hi c tnh theo:

, kg/m3.[II - 183]Trong :MA MB: khi lng phn mol ca cu t Axeton va NcT: nhit lm vic trung bnh ca thp, 0K.ytbc: nng phn mol ca cu t 1 ly theo gi tr trung bnh

[II - 183]

Vi : nng lm vic ti 2 u mi on thp, phn mol.

= y1 = yw = 0,0724 phn mol

= y1 = 0,608 phn mol

phn mol nhit trung bnh on chng.T =(TF + TW)/2 = (74,2 + 97,3)/2 = 85,80C = 358,8 KVy khi lng ring trung bnh ca pha hi i vi on chng l:

kg/m3.* Khi lng ring trung bnh i vi pha lng

, kg/m3[II - 183]Trong :

: khi lng ring trung bnh ca lng, kg/m3.

: khi lng ring trung bnh ca cu t 1 v 2 ca pha lng ly theo nhit trung bnh, kg/m3.

: phn khi lng trung bnh ca cu t 1 trong pha lng.

Vi a1: nng phn khi lng ca pha lng a di cng ca on chng.Ta c: a1 = x1 = 0,0396 phn khi lng

phn khi lng

: nhit trung bnh ca on chng theo pha lng ng vi t0 = 85,80C. Ni suy theo bng I.2 trong [I-9] ta c:

kg/m3.

kg/m3.Vy khi lng ring trung bnh ca lng trong on chng l:

kg/m3

Vy,

Tc lm vic ytb c ly bng 0,9.gh, nn:ytb= 0,9.1,469= 1,323 [m/s]T cc thng s gtb, ytb, ytbtnh c, ta c ng knh on chng bng:

Quy chun ng knh : DU= 0,32 mb. S ngn thc t Ntt :S ngn thc t c tnh theo cng thc: (IX.59 [4])

trong :Nlt s bc thay i nng hay s a l thuyttb hiu sut trung bnh ca thit b:

(IX.60 [4]) l bay hi tng i ca hn hp:

(IX.61 [4]) hh nht ca hn hp lng [N.s/m2]:

A nht ca cu t A ph thuc nhit xtB nht ca cu t B ph thuc nhit xt Ti nh thp:xP = 0,909 [phn mol]; yP* = 0,948[phn mol]; tP = 57,4 [oC];suy ra: = 1,825A = 0,23416 .10-3 Ns/m2 (ni suy bng I.101 [3])B = 0,4898 .10-3 Ns/m2 (ni suy bng I.101 [3])

Vy, ta c:. = 0,45P=60%( th hnh IX.11 [4])

Ti v tr tip liu:xF = 0,072 [phn mol]; yF* = 0,654 [phn mol]; tF = 74,2 [oC];suy ra: = 24,36A = 0,2087 [10-3 Ns/m2] (ngoi suy bng I.110 [3])B = 0,38948 [10-3 Ns/m2] (ngoi suy bng I.101 [3])

Vy, ta c:. = 9,1F = 30%(ngoi suy th hnh IX.11 [4])

Ti y thp:xW = 0,006 [phn mol]; yW* = 0,0724 [phn mol]; tW = 97,3 [oC];suy ra:

= 12,48A = 0,17405[10-3 Ns/m2] (ngoi suy bng I.110 [3])B = 0,293855 [10-3 Ns/m2] (ngoi suy bng I.101 [3])

Vy, ta c:. =3,7W = 35%(ngoi suy th hnh IX.11 [4])

T cc gi tr P, F, Wtm c, tbbng:

Suy ra s ngn thc t ca thp :

S a on luyn : 3,7/0,4167 =9 [mm]S a on chng : 3.4/0,4167 = 8 [mm]

III. Chiu cao c bn ca thp:

(IX.54 [4])trong Hd khong cch gia cc mm chiu dy ca mm, chn = 4 [mm] = 0,004 [m]Hd = 200 [mm] chn theo bng IX.4a [4] ng vi ng knh on luynDO = 380[mm] v DU = 320 [mm]Vy,

IV:Nhit nh, y v v tr tip liu ng vi trng hp nguyn liu vo trng thi si:T th t-x ta c th xc nh d dng nhit nh,y v v tr a tip liu ca thp:Tp=57,40C , TF= 74,20C , TW= 97,30C - 18