LI M U

LI M U

Ngy nay, trong bi cnh cc nn kinh t trn th gii u pht trin khng ngng, khoa hc k thut ngy cng th hin vai tr quan trng trong mi lnh vc ca i sng ni chung v kinh t ni ring. Mt trong nhng ngnh c ng gp v cng to ln l ngnh cng nghip ho hc, c bit l ngnh sn xut cc ho cht c bn.Hin nay, cc ngnh cng nghip cn s dng rt nhiu ho cht c tinh khit cao. Nhu cu ny t ra cho cc nh sn xut ho cht phi chn la s dng nhiu phng php nng cao tinh khit ca sn phm nh : trch ly, chng ct, c c, hp thu Tu theo c tnh yu cu ca sn phm m chn phng php cho ph hp. i vi h Clorofom - Benzen l h 2 cu t tan ln vo nhau, ta chn phng php chng ct nng cao tinh khit cho sn phm. n mn hc Qu trnh v Thit b l mt mn hc mang tnh tng hp trong qu trnh hc tp ca cc k s Cng ngh Ha hc tng lai. Mn hc gip sinh vin gii quyt nhim v tnh ton c th v: quy trnh cng ngh, kt cu, gi thnh ca mt thit b trong lnh vc sn xut ha cht - thc phm. y l bc u tin sinh vin vn dng nhng kin thc hc ca nhiu mn hc vo gii quyt nhng vn k thut thc t mt cch tng hp.Nhim v ca n ny l thit k h thng chng ct hn hp Clorofom - Benzen vi nng sut nhp liu l 1500 kg/h, nng nhp liu l 40% (kmol Clorofom/kmol hn hp), nng sn phm nh l 99% (kmol Clorofom/kmol hn hp), nng sn phm y l 5% (kmol Clorofom/kmol hn hp).

MC LC

TNG QUAN6I.L THUYT V CHNG CT :61.Khi nim :62.Phng php chng ct :63.Thit b chng ct :6II.GII THIU V NGUYN LIU :71.Benzen:72.Clorofom :83.Hn hp Clorofom - Benzen:9QUY TRNH CNG NGH10CN BNG VT CHT V NNG LNG13I.CC THNG S BAN U :13II.XC NH SUT LNG SN PHM NH V SN PHM Y:13III.XC NH CH S HI LU THCH HP :14IV.CN BNG NHIT LNG:151.Cn bng nhit cho ton thp162.Thit b ngng t sn phm nh163.Thit b lm ngui sn phm nh174.Thit b un si dng nhp liu175.Thit b lm ngui sn phm y17TNH TON THP CHNG CT18I.TNH S A L THUYT V S A THC181.Phng trnh ng lm vic:182.Xc nh s a l thuyt:183.Tnh ton s a thc:20II.TNH NG KNH THP: (theo trang 181-184, [2])221.Phn ct (luyn):232.Phn chng :26III.TNH CHIU CAO THP:28IV.TNH TON CHP:28V.TNH TR LC THP: (xem trang 192, [2])33TNH TON KT CU THIT B CHNH37I.TNH TON B DY THP, NP, Y, BCH GHP, M371.B dy thn thp372.y v np383.Chn bch ghp v m cho thn, y v np39II.TNH TON NG DN401.ng dn dng nhp liu412.ng dn hi nh thp413.ng dn dng hon lu424.ng dn hi y thp435.ng dn lng vo ni un44III.TNH TON CHN , TAI TREO THP461.Tnh ton trng lng ca ton thp462.Chn v tnh ton chn 473.Chn v tnh ton tai treo47TNH TON THIT B PH49I.THIT B NGNG T SN PHM NH491.Hiu s nhit trung bnh492.H s truyn nhit493.Tnh ton h s cp nhit cc dng494.Xc inh h s truyn nhit525.Xc nh b mt truyn nhit526.Cu to thit b52II.THIT B LM NGUI SN PHM NH531.Hiu s nhit trung bnh532.H s truyn nhit533.Tnh ton h s cp nhit cc dng534.Xc nh h s truyn nhit575.Xc nh b mt truyn nhit576.Cu to thit b57III.THIT B UN SI Y THP581.Hiu s nhit trung bnh582.H s truyn nhit583.Tnh ton h s cp nhit cc dng594.Xc nh h s truyn nhit615.Xc nh b mt truyn nhit616.Cu to thit b truyn nhit61IV.THIT B GII NHIT SN PHM Y621.Hiu s nhit trung bnh622.H s truyn nhit623.Tnh ton h s cp nhit cc dng624.Xc nh h s truyn nhit665.Xc nh b mt truyn nhit666.Cu to thit b66V.THIT B UN SI DNG NHP LIU661.Hiu s nhit trung bnh672.H s truyn nhit673.Tnh ton h s cp nhit cc dng674.Xc nh h s truyn nhit705.Xc nh b mt truyn nhit706.Cu to thit b70VI.BN CAO V711.Tn tht ng ng dn712.Xc nh tng h s tn tht cc b723.Tn tht ng ng dn trong thit b un si dng nhp liu724.Chiu cao bn cao v73VII.BM741.Nng sut742.Ct p743.Cng sut76TNH KINH T78LI KT79TI LIU THAM KHO80

MC LC HNH1. Hnh 1. th cn bng lng hi ca h Clorofom BenzenTrang 112. Hnh 2. Xc nh gi tr yF* trn th x-yTrang 183. Hnh 3. th xc nh s a l thuytTrang 224. Hnh 4. Hnh phng to s a l thuyt on chngTrang 225. Hnh 5. Hnh phng to s a l thuyt on ctTrang 236. Hnh 6. th xc nh hiu sut theo tch Trang 247. Hnh 7. S cc dng pha trong thpTrang 258. Hnh 8. M t g chy trn trn aTrang 339. Hnh 9. y (np) elipTrang 4010. Hnh 10. Bch ghp thnTrang 4111. Hnh 11. Bch ghp ng dnTrang 4212. Hnh 12. Chn Trang 4913. Hnh 13. Tai treoTrang 5014. Hnh 14. Minh ha cch b tr ng trong thit b ngng t sn phm nhTrang 5415. Hnh 15. Ni un KettleTrang 60

MC LC BNG1. Bng 1. Thnh phn cn bng lng hi v nhit si ca hn hpTrang102. Bng 2. Gi tr tnh ton dng nhp liu, sn phm nh v sn phm yTrang 153. Bng 3. Tnh cc tnh cht ha l ca hn hp theo nhit Trang 184. Bng 4. Kt qu tnh ton hiu sut ti cc v trTrang 235. Bng 5. Gi tr ca bch ghp thnTrang 406. Bng 6. Gi tr ca bch ng dn dng nhp liuTrang 427. Bng 7. Gi tr ca bch cho ng dn hi nh thpTrang 438. Bng 8. Gi tr ca bch cho ng dn hon luTrang 449. Bng 9. Gi tr ca bch cho ng dn hi y thpTrang 4510. Bng 10. Gi tr ca bch cho ng dn lng vo ni unTrang 4511. Bng 11. Gi tr ca bch cho ng dn ra khi ni unTrang 4612. Bng 12. Gi tr ca chn Trang 48

Chng I:TNG QUAN

I. L THUYT V CHNG CT :Khi nim : Chng ct l qu trnh dng tch cc cu t ca 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 nhit , p sut hi bo ho 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. Chng ct v c c kh ging nhau, tuy nhin s khc nhau cn bn nht ca 2 qu trnh ny 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 ct 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 s thu c 2 sn phm : Sn phm nh ch yu gm cu t c bay hi ln (nhit si nh) Sn phm nh ch yu gm cu t c bay hi nh (nhit si ln) i vi h Clorofom Benzen: Sn phm nh ch yu gm Clorofom v mt t Benzen. Sn phm y ch yu l Benzen v mt t Clorofom.Phng php chng ct :Cc phng php chng ct c phn loi theo : p sut lm vic : p sut thp p sut thng p sut cao Nguyn l lm vic : Chng n gin (mt bc) Chng li cun theo hi nc Chng ct Cp nhit y thp : Cp nhit trc tip Cp nhit gin tipVy : i vi h Clororom - Benzen, ta chn phng php chng ct lin tc, cp nhit gin tip bng ni un p sut thng.Thit b chng ct :Trong sn xut, ngi ta thng dng nhiu loi thit b khc nhau tin hnh chng ct. Tuy nhin, yu cu c bn chung ca cc thit b vn ging nhau ngha l din tch tip xc pha phi ln. iu ny ph thuc vo mc phn tn ca mt lu cht ny vo lu cht kia. Nu pha kh phn tn vo pha lng ta c cc loi thp mm, nu pha lng phn tn vo pha kh ta c thp chm, thp phun, y ta kho st 2 loi thng dng l thp mm v thp chm. Thp mm : thn thp hnh tr, thng ng pha trong c gn cc mm c cu to khc nhau, trn pha lng v pha hi c cho tip xc vi nhau. Tu theo cu to ca a, ta c : Thp mm chp : trn mm b tr c chp dng trn, xupap, . Thp mm xuyn l : trn mm c nhiu l hay rnh. Thp chm (thp m) : thp hnh tr, gm nhiu bc ni vi nhau bng mt bch hay hn. Vt chm c cho vo thp theo mt trong hai phng php sau : xp ngu nhin hay xp th t.So snh u, nhc im ca cc loi thp :

Thp chmThp mm xuyn lThp chp

u im- Cu to kh n gin.- Tr lc thp.- Lm vic c vi cht lng bn nu dng m cu c ca cht lng.- Tr lc tng i thp.- Hiu sut kh cao.- Kh n nh.- Hiu sut cao.

Nhc im- Do c hiu ng thnh hiu sut truyn khi thp.- n nh khng cao, kh vn hnh.- Do c hiu ng thnh khi tng nng sut th hiu ng thnh tng kh tng nng sut.- Thit b kh nng n.- Khng lm vic c vi cht lng bn.- Kt cu kh phc tp.- C tr lc ln.- Tiu tn nhiu vt t, kt cu phc tp.

Vy: S dng thp mm chp chng ct h Clorofom Benzen.II. GII THIU V NGUYN LIU :1. Benzen:L mt hp cht mch vng, dng lng khng mu v c mi thm nh. Cng thc phn t l C6H6. Benzen khng phn cc,v vy tan tt trong cc dung mi hu c khng phn cc v tan rt t trong nc. Trc y ngi ta thng s dng benzen lm dung mi. Tuy nhin sau ngi ta pht hin ra rng nng benzen trong khng kh ch cn thp khong 1ppm cng c kh nng gy ra bnh bch cu, nn ngy nay benzen c s dng hn ch hn.a) Cc tnh cht vt l ca benzen: Khi lng phn t: 78,11 (g/mol) T trng (): 0,879 Nhit si: 80,6 oC Nhit nng chy: 5,5 oC

b) ng dng: Ngy nay mt lng ln benzen ch yu : Sn xutstyrencho tng hppolymer. Sn xutcumencho vic sn xut cng lcaxetonvphenol. Sn xut cyclohexan tng hp t nilon. Lm dung mi, sn xut dc liu.

c) iu ch :Thng thng cc hidrocacbon t c iu ch trong phng th nghim, v c th thu c lng ln n bng phng php chng ct than , du m.Tuy nhin c th iu ch Benzen theo cc phng php sau: ng vng v dehydro ha ankaneCc ankane c th tham gia ng vng v dehydro ha to thnh hidrocacbon thm nhit cao v c mt xc tc nh Cr2O3, hay cc kim loi chuyn tip nh Pd, Pt:

CH3(CH2)4CH3 C6H6 Dehydro ha cc cycloankane Cc cycloankane c th b dehydro ha nhit cao vi s c mt ca cc xc tc kim loi chuyn tip to thnh benzen hay cc dn xut ca benzen:

C6H12 C6H6 i t acetylen: un acetylene vi s c mt ca xc tc l than hot tnh hay phc ca niken nh Ni(CO)[(C6H5)P] s thu c benzen:

3C2H2 C6H61. Clorofom :L cht lng linh ng khng mu, v hi ngt; ts= 61,15oC;; chit sut vi tia D 20oC: 1,4455. Khng tan trong nc, tan trong hu ht cc dung mi hu c. Clorofom khng chy v cng khng to hn hp n vi khng kh. L cht u tng hp freon; dng lm dung mi. Clorofom c tc dng lm bun ng v gy m, nhng v c nn khng c dng trong gii phu. Clorofom c i vi s trao i cht v ni tng, c bit l gan.a) Cc tnh cht vt l ca clorofom : Khi lng phn t: 119,5 (g/mol) Khi lng ring: 1,488 g/cm3 Nhit si: 61,15oC Nhit nng chy: -63,5 oC

b) ng dng:Trc y, Clorofom cs dng ch yu tng hp cht lm lnhR-22cho my iu ha khng kh. Tuy nhin, v R-22 gy ra ssuy gim ozonnn clorofom gn nh t s dng cho mc ch ny. Hin nay, Clorofom l mt dung mi ph bin v n kh tr, trn hp vi hu ht cc cht lng hu c, v d bay hi. Dung mi sn xutthuc nhumvthuc tr su. Clorofom chadtri(hydro nng),CDCl3, l dung mi ph bin chophng php o ph cng hng t ht nhn (NMR).c) iu ch :Trong cng nghip, ngi ta iu ch clorofom bng t nng hn hp clov clomtan hay mtan. nhit 400-500C, phn ng halogen ha gc t dodin ra, chuynmtanhayclomtandn dn thnh cc hp cht clo ha.CH4+ Cl2 CH3Cl +HClCH3Cl + Cl2CH2Cl2+ HClCH2Cl2+ Cl2 CHCl3+ HClTip tc phn ng clo ha, clorofom chuyn thnhCCl4:CHCl3+ Cl2 CCl4+ HClHn hp sn phm sau phn ng gm 4 cht: clomtan, diclomtan, clorofom (triclomtan), v cacbon tetraclorua, tch chng bngchng ct.

1. Hn hp Clorofom - Benzen:Ta c bng thnh phn cn bng lng (x) hi (y) v nhit si ca hn hp Clorofom - Benzen 760 mmHg.(Tham kho [2])

Bng 1.Thnh phn cn bng lng hi v nhit si ca hn hpx (% phn mol)05102030405060708090100

y (% phn mol)06.512.627.54154.66674.68390.596.2100

t (oC)80.680.179.678.477.275.974.573.17168.765.761.5

Hnh 1. th cn bng lng hi ca h Clorofom Benzen

Chng II:QUY TRNH CNG NGH

Ch thch cc k hiu trong qui trnh:1. Bn cha nguyn liu2. Bm3. Bn cao v4. Van5. Lu lng k6. By hi7. Thp chng ct8. Thit b ngng t sn phm nh9. B phn chnh dng10. Thit b lm ngui sn phm nh11. Bn cha sn phm nh12. Thit b un si y thp ( Ni un Kettle)13. Thit b lm ngui sn phm y14. Bn cha sn phm y15. p k16. Nhit k17. Thit b un si dng nhp liu

Hn hp Clorofom - Benzen c nng Clorofom l 40% (phn mol), nhit nguyn liu lc u l 300C ti bn cha nguyn liu (1), c bm (2) bm ln bn cao v (3). Dng nhp liu c gia nhit ti nhit si trong thit b truyn nhit ng chm. Sau hn hp c a vo thp chng ct (7) a nhp liu v bt u qu trnh chng ct. Lu lng dng nhp liu c kim sot qua lu lng k (5).Trn a nhp liu, cht lng c trn vi phn lng t on ct (luyn) ca thp chy xung. Trong thp, hi i di ln gp lng i t trn xung. y c s tip xc v trao i gia hai pha vi nhau. Pha lng chuyn ng trong phn chng cng xung pha di cng gim nng cc cu t d bay hi v b pha hi to nn t ni un (12) li cun cu t d bay hi. Nhit cng ln trn cng thp, nn khi hi i qua cc a t di ln th cu t c nhit si cao l Benzen s ngng t li, cui cng trn nh thp ta thu c hn hp c cu t Clorofom chim nhiu nht (nng 99% phn mol). Hi ny i vo thit b ngng t (8) c ngng t hon ton. Mt phn cht lng ngng t i qua thit b lm ngui sn phm nh (10), ri c a qua bn cha sn phm nh (11). Phn cn li ca cht lng ngng t c hon lu v thp a trn cng vi t s hon lu thch hp v c kim sot bng lu lng k. Cui cng y thp ta thu c hn hp lng hu ht l cu t kh bay hi (Benzen). Hn hp lng y c nng Clorofom l 5% phn mol, cn li l Benzene. Dung dch lng y i ra khi thp vo ni un (12). Trong ni un dung dch lng mt phn s bc hi cung cp li cho thp tip tc lm vic, phn cn li ra khi ni un c cho qua thit b lm ngui sn phm y (13) ri sau vo bn cha sn phm y (14). H thng lm vic lin tc cho ra sn phm nh l Clorofom, sn phm y l Benzen.

Chng ct Clorofom Benzen GVHD: Trn Tn Vit

SVTH: V Vn ThcTrang 48

Chng III:CN BNG VT CHT V NNG LNG

I. CC THNG S BAN U :Chn loi thp l thp mm chp.

Hn hp: Khi chng luyn hn hp Clorofom Benzen th cu t d bay hi l Clorofom. Nng sut nhp liu :GF = 1500 kg/h Nng nhp liu :xF = 40 % (kmol Clorofom / kmol hn hp) Nng sn phm nh :xD = 99 % (kmol Clorofom / kmol hn hp) Nng sn phm y: xW = 5 % (kmol Clorofom / kmol hn hp) Nhp liu trng thi lng si Chn: Nhit nhp liu: tFV = 30 oC i vi thit b un si y thp: p sut hi t : Ph = 1,6 at i vi thit b ngng t sn phm nh: Nhit dng nc lnh i vo: tV = 30 oC Nhit dng nc lnh i ra: tR = 50 oC i vi thit b lm ngui sn phm nh: Nhit sn phm nh sau khi lm ngui:tDR = 35 oC Nhit dng nc lnh i vo: tV = 30 oC Nhit dng nc lnh i ra: tR = 50 oC i vi thit b lm ngui sn phm y: Nhit sn phm y sau khi lm ngui: tWR = 40 oC Nhit dng nc lnh i vo: tV = 30 oC Nhit dng nc lnh i ra: tR = 50 oC Cc k hiu: GF , F : sut lng nhp liu tnh theo kg/h, kmol/h. GD, D : sut lng sn phm nh tnh theo kg/h, kmol/h. GW, W: sut lng sn phm y tnh theo kg/h, kmol/h. , xi : nng phn khi lng, phn mol ca cu t i.

II. XC NH SUT LNG SN PHM NH V SN PHM Y:Khi lng mol trung bnh ca hn hp: M = xMC + (1-x) MB

Phng trnh cn bng vt cht: Vi cc gi tr xF, xD, xW bit, gii h trn ta c D v W, t suy ra GF, GD, GW. Cc gi tr tnh ton c th hin trong bng di y:

Bng 2. Gi tr tnh ton ca cc dng nhp liu, sn phm nh v sn phm y

Nhp liu FSn phm nh DSn phm y W

kg/hkmol/hPhn klPhn molMFkg/hkmol/hPhn klPhn molMDkg/hkmol/hPhn klPhn molMW

150015,85650,52940,094,60703,0685,90499,34599,0119,085796,9329,9527,4625,080,075

V ti nh thp, nng phn mol ca Clorofom trong pha lng v pha hi bng nhau nn khi lng mol ca pha hi v pha lng ti nh thp l bng nhau:MxD = MyD = xDMC + (1 xD)MB = 0,99 x 119,5 + (1 0,99) x 78 = 119,085 kg/kmolTng t ti y thp: MxW = MyW = xWMC + (1 xW)MB = 0,05 x 119,5 + (1 0,05) x 78 = 80,075 kg/kmol

III. XC NH CH S HI LU THCH HP :* Ch s hi lu ti thiu :T s hon lu ti thiu l ch lm vic m ti ng vi s mm l thuyt l v cc. Do , chi ph c nh l v cc nhng chi ph iu hnh (nhin liu, nc v bm) l ti thiu. tm Rmin ta cn xc nh gi tr nng pha hi cn bng vi nng pha lng ca dng nhp liu, hay yF*.

Hnh 2. Xc nh gi tr yF* trn th x-yDa vo hnh trn ta c xF = 0,40 yF* = 0,546

T s hon lu ti thiu : T s hon lu lm vic tnh theo cng thc kinh nghim: R = 1,3Rmin + 0,3 = 1,3 x 3,041+ 0,3 = 4,253

IV. CN BNG NHIT LNG:- S dng bng I251, trang 314, [1] tra chn hi nc bo ha cn s dng un nng hn hp u.- Chn hi t l hi nc 1,6at:+ Nhit ha hi:rh = 2227 x103 J/kg+ Nhit si:th = 112,7 oC- S dng th T- x,y ca hn hp Clorofom - Benzen ta c:Dng sn phm y c nhit : Trc khi vo ni un ( lng): tS1 = 80,10 oC Sau khi c un si ( hi): tS2 = 80,22 oCNi suy cc gi tr Nhit dung ring (Tra bng I.153, trang 171, [1]), nhit ha hi ( Tra bng I.212, trang 254, [1]). S dng cc cng thc tnh Enthanpy v nhit ha hi sau: h = Ct r = xrC + (1 -x)rB C =xCC + (1 -x)CB

Bng 3. Tnh cc tnh cht ha l ca hn hp theo nhit f (hh u)DFW

Nhit (oC)3061,9275,9080,10

Cu tClorofomBenzenClorofomBenzenClorofomBenzenClorofomBenzen

Cp J/kg.1037,001777,501083,781940,081104,062013,481110,152035,43

r kcal/kg61,88102,3858,9197,1657,5594,7257,1493,98

x %kl50,5349,4799,350,6550,5349,477,4692,54

Cp kJ/kg.1,4031,0891,5541,966

r kcal/kg (L)81,91159,16375,93791,233

r kJ/kg (L)342,945247,705317,934381,976

h42,10067,455117,945157,507

1. Cn bng nhit cho ton thpQ + GFhFS = (R+1) GDrD + GDhDS + GWhWS + QtGi s lng nhit mt mt (tn tht) bng a = 5% nhit hu ch: Qt = 0,05Q Nhit lng cn cung cp:

= 958876,329 kJ/h = 266354,536 W

Nu dng hi nc bo ha ( khng cha m) cp nhit th: Q = Gh1rh

Vy lng hi nc cn dng l: Gh1 = 0,120 kg/s4. Thit b ngng t sn phm nhQnt = (R + 1)GDrD = GnCn(tR tV) Trong : Cn : nhit dung ring ca nc nhit trung bnh (tR + tV)/2.tV = 30 oC, tR = 50 oC Tra bng I.249, trang 310, [1] c Cn = 4178 J/kg. Lng nhit trao i: Qnt = (4,253+1) x 703,068 x 247,705 = 914902,481 kJ/h = 254139,578 W Lng nc cn dng: Gn = 3,041 kg/s5. Thit b lm ngui sn phm nhQln1 = GD(hDS hDR) = Gn1Cn(tR tV)Trong : hDR = CDRtDR, tDR = 35 oC CDR = 1,049 kJ/kg. Lng nhit trao i: Qln1 = GD(hDS hDR) = 703,068 (67,455 - 1,049 x 35) = 21612,310 kJ/h = 6003,420 W Lng nc lnh cn dng l:

Gn1 = 0,072 kg/s6. Thit b un si dng nhp liuQF = GF(hFS hf) = Gh2rh (1-a) Lng nhit trao i: QF = GF(hFS hFV) = 1500(117,945 - 42,100) = 113768,030 kJ/h = 31602,231 W Lng hi t s dng:

0,015 kg/s7. Thit b lm ngui sn phm yQln2 = GW(hWS hWR) = Gn2Cn(tR tV)Trong : hWR = CWRtWR, tWR = 40 oC CWR = 1,767 kJ/kg. Lng nhit trao i: Qln2 = GW(hWS - hWR) = 796,932(157,507 - 1,767 x 40) = 69195,215 kJ/h = 19220,893 W Lng nc lnh cn dng l: Gn2 = 0,230 kg/s

Chng IV:TNH TON THP CHNG CT

I. TNH S A L THUYT V S A THC1. Phng trnh ng lm vic:- Phn ct:

- Phn chng: Sut lng mol tng i ca dng nhp liu ( h s tip liu) l:

Phng trnh on chng:

Xc nh s a l thuyt:V ng lm vic cho phn chng v phn ct ca thp trn th x-y: T im (xD; yD) k ng thng i qua im (0; 0,188), ct ng thng x = xF ti im A. Ni A vi im (xW; yW) ta c ng lm vic on ct.

Hnh 3. th xc nh s a l thuyt

Hnh 4. Hnh phng to s a l thuyt on chng

Hnh 5. Hnh phng to s a l thuyt on ctT ta c c s a l thuyt trong qu trnh chng ct hn hp Clorofom - Benzen l 31 a gm: 10 a ct v 21 a chng.

Tnh ton s a thc:S a thc t tnh theo hiu sut trung bnh :

Trong : Ntt : s mm thc tNlt : s mm l thuyttb : hiu sut trung bnh ca a, l mt hm s ca bay hi tng i v nht ca hn hp lng : = f(). Xc nh hiu sut trung bnh ca thp tb : theo mc 23, trang 170-171, [2]: Tnh hiu sut a ti 3 v tr : a nh , a nhp liu, a y. ng vi mi v tr, xc nh bay hi tng i ca cu t d bay hi v nht ng hc N.s/m2) ca cht lng: bay hi tng i:

Vi : x : phn mol ca Clorofom trong pha lng.y* : phn mol ca Clorofom trong pha hi cn bng vi pha lng. nht ng hc N.s/m2) ca cht lng: Ni suy bng I.101, trang 91,[1]: (tng ng vi nhit ti v tr xt) xc nh nht ca Clorofom v Benzen, sau tnh nht ca hn hp lng theo cng thc I.12, trang 84, [1]:

Trong : nht ng hc ca Clorofom v Benzen, (cP)

: Phn mol ca Clorofom v Benzen trong hn hp. Xc nh tch s , tra trn hnh IX. 11- trang 171, [2], xc nh c .

Hnh 6. th xc nh hiu sut theo tch

- Kt qu tnh ton c th trong bng sau:Bng 4. Kt qu tnh ton hiu sut ti cc v tra nha nhp liua y

y (phn mol)0,9960,5460,065

x (phn mol)0,9900,4000,050

2,6481,8041,321

t (oC)61,9275,9080,10

.103 Ns/m2 (C)0,38430,34230,3298

lgC-3,415-3,466-3,482

.103 Ns/m2 (B)0,38300,33120,3157

lgB-3,4168-3,4799-3,5007

lg (hn hp)-3,4153-3,4742-3,4997

(hn hp) Ns/m20,00038430,00033560,0003164

(hn hp) cP0,38430,33560,3164

1,01760,60540,4179

%495662

Hiu sut a trung bnh: Cho on ct (luyn) : Cho on chng: Cho ton thp: S a thc t: S a ct: ; lm trn l 19 a S a chng: ; lm trn l 36 a V tr nhp liu:a 20 (tnh t nh) Tng s a l 56 a.

II. TNH NG KNH THP: (theo trang 181-184, [2])

(m) gtb : lng hi trung bnh i trong thp (kg/h). yy) tb: tc hi trung bnh i trong thp (kg/m2.s).Lng hi trung bnh i trong on chng v on ct khc nhau, lm cho ng knhon chng v on ct cng khc nhau, do ta s tnh ring cho 2 on.

1. Phn ct (luyn):a. Lng hi trung bnh i trong on ct:

(kg/h)gd : lng hi ra khi a trn cngca thp (kg/h).g1 : lng hi i vo a di cngca on ct (kg/h). Xc nh gd :gd = GD + GL = GD(R + 1)= 703,068 x (4,253 + 1) = 3693,515 (kg/h) Xc nh g1 : T h phng trnh :

Hnh 7. S cc dng pha trong thpTrong : G1 : lng lng a th nht ca on ct.

: hm lng hi a th nht ca on ct.r1 : n nhit ho hi ca hn hp hi i vo a th nht ca on ct rd : n nhit ho hi ca hn hp hi i ra nh thp .

Vi:

rc, rb : n nhit ha hi ca Clorofom, Benzen. Tnh rd : tD = 61,92 oC , tra bng I.212, trang 254, [1] n nhit ho hi ca Clorofom : rcd = 58,91 (kcal/kg) . n nhit ho hi ca Benzen : rbd = 97,16 (kcal/kg) .

yD = xD = 0,99 = 0,9935

= 59,163 (kcal/kg) t1 = tF = 75,90 oC n nhit ho hi ca Clorofom : rc1 = 57,55 (kcal/kg) n nhit ho hi ca Benzen : rb1 = 94,72 (kcal/kg)T h pt (I), c th tnh c g1 theo cng thc sau:

T tnh c: G1 = 2342,570 (kg/h) ; = 0,6180 y1 = 0,5136

= 3369,577 (kg/h)b. Tc hi trung bnh i trong on ct:

xtb, ytb : khi lng ring trung bnh ca pha lng v pha hi, (kg/m3) : h s tnh n sc cng b mt h: Khong cch gia cc a, chn h = 0,3 (m)

ytb : tnh theo cng thc (IX.102), trang 183, [2]

+ Nng phn mol trung bnh pha hi:

+ Nhit trung bnh on ct : oC

= 3,892 (kg/m3) xtb : tnh theo cng thc (IX.104a), trang 183, [2]

+ Nng phn khi lng trung bnh pha lng:

+ T ttbL = 68,91 oC, tra bng I.2, trang 9, [1] Khi lng ring ca Clorofom:xtbC = 1397,190 (kg/m3) Khi lng ring ca Benzen : xtbB = 826,645 (kg/m3)

= 1191,140 (kg/m3) : H s tnh n sc cng b mt+ Khi < 20 dyn/cm th = 0,8+ Khi > 20 dyn/cm th = 1Trong , c tnh theo cng thc (I.76), trang 299, [1]:

C, B : Sc cng b mt ca Clorofom v Benzen nhit lm vic (dyn/cm)+ T ttbL = 68,91 oC , tra bng I.242 trang 300, [1] Sc cng b mt ca Clorofom : C = 20,50 (dyn/cm). Sc cng b mt ca Benzen : B = 22,63 (dyn/cm). =10,76 dyn/cm < 20 dyn/cm = 0,8

ng knh on ct :

= 0,784 (m).1. Phn chng :a. Lng hi trung bnh i trong on chng:

(kg/h)gn : lng hi ra khi on chng (kg/h).g1 : lng hi i vo on chng (kg/h). Xc nh gn : gn = g1 = 3045,638 (kg/h) Xc nh g1 : T h phng trnh :

Vi : G1 : lng lng a th nht ca on chng .r1 : n nhit ho hi ca hn hp hi i vo a th nht ca on chng. Tnh r1 : tW = 80,10 oC, tra bng I.212, trang 254, [1]: n nhit ho hi ca Clorofom : rcw = 57,14 (kcal/kg) . n nhit ho hi ca Benzen : rbw = 93,98 (kcal/kg) .

yW = xW = 0,05 = 0,0746

(kcal/kg) Tnh r1: = 0,6180

(kcal/kg).

= 2395,183 (kg/h)

GW = 796,932 (kg/h) (kg/h)

= 2720,410 (kg/h)

b. Tc hi trung bnh i trong on chng:Tnh ton tng t phn ct, kt qu th hin trong bng di y:yW (%mol)y1 (%mol)y'tb (%mol)t (oC)T (K)'ytb (kg/m3)xtb1 (%kl)'xtbC'xtbB'xtb (kg/m3)

5,0051,3628,1878,00351,003,11429,001383,100817,100927,106

C.103 (N/m)B.103 (N/m)hh.103 (N/m)hh (dyn/cm)h (m)(yy)tbC

19,2721,5410,1710,170,80,31,530

ng knh on chng :

(m)Kt lun :V hai ng knh on ct v on chng khng chnh lch nhau qu ln nn ta chn ng knh ca ton thp l : 0,8 (m) = 800 (mm).Nh vy vn tc pha hi trong phn chng v phn ct ca thp l:

+ Phn ct : = 0,478 (m/s).

+ Phn chng : = 0,482(m/s).

III. TNH CHIU CAO THP:- Chiu cao thn thp (tnh theo cng thc (IX.54), trang 169, [2]):Hthn = Ntt (h + ) + 0,8 = 56 x (0,3 + 0,005) + 0,8 = 17,88 (m)Trong : : chiu dy ca mm, chn = 5 (mm) = 0,005 (m)h : khong cch gia cc a, chn h = 0,3 (m)

- Chn y (np) ellip tiu chun c ht = 0,25 x 0,8 = 0,2 (m)- Chn chiu cao g: hg = 50 (mm) = 0,05 (m)- Chiu cao y (np): Hn = ht + hg = 0,2 + 0,05 = 0,25 (m)Kt lun: Chiu cao ton thp: H = Hthn + 2Hn = 17,88 + 2 x 0,25 = 18,38 (m).

IV. TNH TON CHP:

ng knh ng hi ca chp (chn):dh = 50 (mm) = 0,05 (m)

S chp phn b trn a: ; ( : ng knh trong ca thp) chn n= 27 (chp) Chiu cao chp pha trn ng dn hi:h2 = 0,25dh = 0,25 x 50 = 12,5 (mm) ng knh chp:, ch : chiu dy chp, chn ch = 2 (mm)

dch = 74 (mm) Khong cch t mt a n chn chp :Chn S = 12,5 (mm) Chiu cao mc cht lng trn khe chp :Chn h1 = 30 (mm) Tit din thp :(m2) Lu lng hi i trong thp:

Chiu cao khe chp: :H s tr lc ca a chp:Chn = 2 y := 4,554 (m/s) y :Khi lng ring trung bnh ca pha hi:

(kg/m3) x:Khi lng ring trung bnh ca pha lng:

= 1059,123 (kg/m3) g :gia tc trng trng:9,81 (m/s2)

(m) Chn b = 20 (mm) S lng khe h ca mi chp :

,c: khong cch gia cc khe: Chn c = 3 (mm)

Chn i = 44 (khe) Chiu rng khe chp: Chn a = 2 (mm)

Lu lng lng trung bnh i trong thp (kg/h): Phn ct: Phn chng: Chn s knh chy chuyn: z = 1 Chiu cao lp cht lng trn mm:hm = h1 + S + hsr + b = 30 + 12,5 + 5 + 20 = 67,5 (mm)(hsr : khong cch t mp di ca chp n mp di ca khe chp) Tit din ng hi : Tit din hnh vnh khn :

Tng din tch cc khe chp: S3 = iab = 44 x 0,002 x 0,02 = 0,0018 m2 Tit din l m trn ng hi : Nhn xt : S1 S2 S3 S4 (hp l) Bc ti thiu ca chp trn a:tmin = dch + 2ch + l2 l2 : khong cch nh nht gia cc chpl2 = 12,5 + 0,25dch = 12,5 + 0,25 x 74 = 31 (mm) Chn: l2 = 35 (mm) tmin = 74 + 2 x 2 + 35 = 113 (mm)

m l chp hs : Theo cng thc (5.2), trang 108, [3]:

SS = nS3: tng din tch cc l chp trn mi mm (m2)

hs > hso Pha kh s thi qua mp di ca chp (thit k kh hp l). Chiu cao cht lng trn g chy trn: Theo cng thc (7.41), trang 56, [8]Lw : chiu di g chy trn (m)Chn Lw = 0,7 = 0,7 x 0,8= 0,56 (m)B dy g chn trn = 2 mm = 0,002 m Phn ct:

Phn chng:

Gradient chiu cao mc cht lng trn mm : = Cg'nh , theo cng thc (5.5), trang 111, [3]- Din tch knh chy chuyn hnh vin phn:

Hnh 8. M t g chy trn trn a- Khong cch gia 2 g chy trn:

- Din tch gia 2 g chy trn :

- Chiu rng trung bnh ca mm Bm :

- H s iu chnh tc pha kh Cg : Phn ct:

+

+ Tra th 5.10, trang 111, [3] c : Cg = 0,63hm = 67,5 ; hsc = 12,5 mm ; khong cch gia 2 chp (l2) bng khong 50% dch nn tra hnh 5.14, trang 112, [3], c 4 = 3,5 = 0,875.S hng chp nh = 5, do L = 0,63 x 0,875 x 5 = 2,77 (mm)- Chiu cao g chy trn hw: Do hm = hw + how + 0,5 hwL = hm - howL 0,5L = 67,5 7,11 0,5 x 2,77 = 59,005 (mm) Phn chng: tnh tng t phn ct, ta c bng sau:XC

Cg4ChwC(mm)

5,5010,700,6243,157,320

Vy chn chiu cao g chy trn chung cho ton thp l hw = 58 (mm)- gim p do ma st v bin i vn tc pha kh thi qua chp khi khng c cht lng:

, (mm cht lng) E : h s, ph thuc t s Saj/Srj , tra trn hnh 5.16, trang 115, [3] y ta c Saj/Srj 1 nn E = 0,65 Sr : Tng din tch ng hi ca mi mm, (m2)Sr = n x Srj = 27 x 0,00196 = 0,05292 (m2)

- Kim tra s n nh ca mm:

Vy mm n nh.Do lu lng kh v lng ca phn chng u ln hn so vi phn ct nn trong cc tnh ton tip theo, cc s liu ly theo phn chng (nu cc gi tr c tnh ring bit).- gim p ca pha kh qua mt mm ht : ht = hfv + hs + hss + how + 0.5 ; (mm cht lng)+ Chiu cao thy tnh lp cht lng trn l chp n g chy trn hss:hss = hw - (hsc + hsr + b) = 58 (12,5 + 5 + 20) = 20,5 (mm) ht = 13,24 + 20,78 + 20,5 + 8,63 + 0,5 x 3,1 = 64,7 (mm)- Tn tht thy lc do dng lng chy t knh chy chuyn vo mm hd c xc nh theo biu thc (5.10), trang 114, [3]:

, (mm cht lng)Khong cch gia hai g chy trn l = 571 < 1500 nn khong cch gia mp trn ca g chy trn v mp di ca knh chy chuyn c chn l 12,5 mm.Do khong cch gia mp di ca knh chy chuyn v mm :S5 = 58 12,5 = 45,5 (mm) = 0,0455 (m)Nn tit din gia ng chy chuyn v mm di:Sd = 0,0455 x Lw = 0,0455 x 0,56 = 0,0255 (m2)

(mm)- Chiu cao mc cht lng trong knh chy chuyn c tnh theo biu thc (5.9), trang 115, [3]: hd = hw + how + + h'd +ht = 58 + 8,63 + 3,1 + 0,13 + 64,7 = 134,56 (mm)- Chiu cao hd dng kim tra mm: m bo iu kin thp khng b ngp lt khi hot ng, ta c:hd = 134,56 (mm) < 0,5h = 150 (mm)- Cht lng chy vo knh chy chuyn:+ Khong cch ri t do trong knh chy chuyn dtw :ho = h + hw - hd = 300 + 58 134,56 = 223,44 (mm)

+ (mm)- i lng ny kim tra cht lng chy vo thp c u khng v cht lng khng va p vo thnh:

Trong : dw = 0,5 x ( l) = 0,5 x (800 571) = 114,5 (mm)V. TNH TR LC THP: (xem trang 192, [2])1. 2. 3. 4. 5. 6. 1. Tr lc a kh Pk : Cng thc (IX.137), trang 192, [2]

:h s tr lc a kh, thng = 4,5 5 , chn = 5 y :khi lng ring ca pha hi (kg/m3) o : vn tc hi qua rnh chp (m/s): (Vy: lu lng pha hi trung bnh)a. Phn ct:VyL : lu lng pha hi trung bnh i trong phn ct

(m3/h)

(m/s)

b. Phn chng:Tnh ton tng t nh phn luyn ta c bng kt qu sau:VyC (m3/h)oC (m/s)PkC (N/m2)

871,5685,173209,062

2. Tr lc do sc cng b mt Ps: Cng thc (IX.138), trang 192, [2]

: sc cng b mt trung bnh ca hn hp (N/m2) dtd : ng knh tng ng ca khe rnh chp (m)

fx : din tch tit din t do ca rnh H : chu vi rnh a. Phn ct:L =10,76 dyn/cm = 10,76 10-3 (N/m2) ( tnh trn)

b. Phn chng:C =10,17 dyn/cm = 10,17 10-3 (N/m2) ( tnh trn)

3. Tr lc ca lp cht lng trn a ( Tr lc thy tnh Pt ): Cng thc (IX.139), trang 194, [2]

b : khi lng ring ca bt, thng b = (0,4 0,6)x hb:chiu cao lp bt trn a (m)( Cng thc (IX.110), trang 185, [2])

hx : Chiu cao lp cht lng ( khng ln bt ) trn a (m)hx = S + 0,5b = 12,5 + 0,5 x 20 = 22,5 (mm) = 0,0225 (m) F : Phn din tch b mt a c gn chp (m2) f : Tng din tch cc chp trn a:

(m2) hch : Chiu cao ca chp hch = hw + how + ch (mm)a. Phn ct:bL = 0,5xL = 0,5 x 1191,140 = 595,570 (kg/m3)hchL = hw + howL + ch = 58 + 7,11 + 2 = 67,11 (mm) = 0,067 (m)

b. Phn chng:Tnh ton tng t phn ct, ta c bng sau:

xCbChowChchChbCPtC (N/m2)

927,106463,5530,008630,068630,083331,964

Chn chiu cao chp hch = 0,068 (m) cho c thp. Tng tr lc qua mt a ct:PL = PkL + PsL +PtL = 240,261 + 11,824 + 408,978 = 661,063 (N/m2)Tng tr lc qua mt a chng:PC = PkC + PsC +PtC = 209,062 + 11,176 + 331,964 = 552,202 (N/m2) Tng tr lc phn ct:PL = Nttct x PL = 20 x 661,063 = 13221,26 (N/m2)Tng tr lc phn chng:PC = Nttchng x PC = 36 x 552,202 = 19879,272 (N/m2) Tng tr lc ca ton thp :P = PL + PC = 13221,26 + 19879,272 = 33100,532 (N/m2) = 0,33 (atm) Kim tra li khong cch mm h = 0,3 (m) bo m cho iu kin hot ng bnh thng ca thp: V PL > PC nn ta ly PL kim tra :

Vy khong cch a chn tha mn iu kin trn.

Chng V:TNH TON KT CU THIT B CHNH

I. TNH TON B DY THP, NP, Y, BCH GHP, M1. B dy thn thpV thp hot ng p sut thng nn ta thit k thn hnh tr bng phng php hn h quang in, kiu hn gip mi 2 pha. Thn thp c ghp vi nhau bng cc mi ghp bch. m bo cht lng ca sn phm v kh nng chng n mn, ta chn vt liu lm thn thp l thp khng g m X18H10T. Cc thng s cn tra v chn phc v cho qu trnh tnh ton Nhit tnh ton: t = tW = 80,1 oC (hng dn trang 9,[5]) p sut tnh ton: v thp hot ng p sut thng nn: P = Pmt + Pthy tnh + P (trang 9-10,[5]) Khi lng ring trung bnh ca pha lng trong ton thp:x = 1059,123 kg/m3 ( tnh trn) P = 105 + xgH + P = 105 + 1059,123 x 9,81 x 18,38 + 33100,532 = 324068,67 N/m2 0,32 N/mm2Do khi lng ngp y thp (H =18,83 m) th khng xt ti P, nn chn P = 0,3 N/mm2 (hng dn trang 10,[5]) H s b sung do n mn ha hc ca mi trngThi gian s dng thit b l 15 nm, vi mi trng c n mn t 0,05 mm/nm n 0,1 mm/nm Ca = 0,1 x 15 = 1,5 (mm) ng sut cho php tiu chunTra hnh 1.2, trang 16, [5].V vt liu l X18H10T []* = 144 N/mm2 H s hiu chnhV thit b c cha cht c hi = 0,9 (trang 17, [5]) ng sut cho php [] = []* = 0,9 x 144 = 129,6 N/mm2 H s bn mi hnTra bng XIII.8, trang 362, [2]. V s dng phng php hn h quang in, kiu hn gip mi 2 pha h = 0,95 Tnh b dy

Ta c: p dng cng thc 5-3, trang 96, [5]

S = B dy thc ca thn : S = S + Ca + Co = 0,98 + 1,5 + 1,52 = 4 mmTra bng 5-1, trang 94, [5]: B dy ti thiu: Smin = 3 mm B dy S tha iu kin. Kim tra bnp dng cng thc 5-10, trang 97,[5]

iu kin: (tha)p dng cng thc 5-11, trang 97, [5]

= 0,77 > P = 0,3 (tha)Kt lun: S = 4 mm1. y v np

Hnh 9. y (np) elip Chn y v np c dng hnh elip tiu chun, c g, lm bng thp X18H10T Chn b dy y v np bng vi b dy thn thp: S = 4 mm.p dng cng thc 6-10, 6-11, trang 126, [5] Kim tra tha iu kin bn : ( tnh trn)

V y v np c hnh ellip tiu chun vi Rt = Dt iu kin trn c tha nh kim tra phn thn thp.

Kt lun: Kch thc ca y v np: ng knh trong: Dt = 800 mm ht = 200 mm Chiu cao g: hg = 50 mm B dy: S = 2 mm Din tch b mt trong - Tra bng XIII.10, trang 382, [2]: Fb mt = 0,82 m2 1. Chn bch ghp v m cho thn, y v np Mt bch l b phn quan trng dng ni cc phn ca thit b cng nh ni cc b phn khc vi thit b. Cc loi mt bch thng s dng: Bch lin: l b phn ni lin vi thit b (hn, c v rn). Loi bch ny ch yu dng thit b lm vic vi p sut thp v p sut trung bnh. Bch t do: ch yu dng ni ng dn lm vic nhit cao, ni cc b bng kim loi mu v hp kim ca chng, c bit l khi cn lm mt bch bng vt liu bn hn thit b. Bch ren: ch yu dng cho thit b lm vic p sut cao.Chn bch c ghp thn, y v np lm bng thp CT3, cu to ca bch l bch phng hn (khng c).

Hnh 10. Bch ghp thn Chn bch ghp thn, y v npTra bng XIII.27, trang 420, [2], ng vi Dt = 800 mm v p sut tnh ton P = 0,3 N/mm2 chn bch c cc thng s sau: Bng 5. Gi tr ca bch ghp thnDtDDbD1D0hBu lng

dbZ

(mm)(ci)

80093088085081122M2024

Tra bng IX.5, trang 170, [2], vi h = 300 mm khong cch gia 2 mt bch l 1200 mm v s mm gia 2 mt bch l 4.

Ta c: S mt bch cn dng ghp l: (15+1) x 2 = 32 bch Chn m kn ca mi ghp bch ch yu do vt m quyt nh. m lm bng cc vt liu mm hn so vi vt liu bch. Khi xit bu lng, m b bin dng v in y ln cc ch g gh trn b mt ca bch.Vy, m bo kn cho thit b ta chn m l ch, c b dy l 3 mm.(tra bng 7-2, trang 146, [5])

II. TNH TON NG DNTnh chi tit ng dn

Hnh 11. Bch ghp ng dnng dn thng c ni vi thit b bng mi ghp tho c hoc khng tho c. Trong thit b l thp chng ct ny, s dng mi ghp tho c.i vi mi ghp tho c, ngi ta lm on ng ni, l on ng ngn c mt bch hay ren ni vi ng dn: Loi c mt bch thng dng vi ng c ng knh d > 10 mm. Loi ren ch yu dng vi ng c ng knh d 10 mm, i khi c th dng vi d 32 mm.ng dn c lm bng thp X18H10T.Bch c lm bng thp CT3 , cu to ca bch l bch lin khng c.1. ng dn dng nhp liu

, (m)Nhit ca cht lng nhp liu vo thp l tF = 75,9 oCTi nhit ny th: Khi lng ring ca Clorofom : C = 1386,355 kg/m3 Khi lng ring ca Benzen: B = 819,305 kg/m3

Nn: F = 1032,753 kg/m3 Chn loi ng ni cm su vo thit b Cht lng t chy t bn cao v, chn vn tc cht lng trong ng ni l F = 0,2 m/s.ng knh trong ca ng ni:

m = 51 mm Chn ng theo tiu chun c Dy = 50 mm.Tra bng XIII.33, trang 434, [2]: Chiu di on ng ni: l = 85 mmVn tc thc ca dng nhp liu l:

m/sTra bng XIII.26, trang 409, [2]: Cc thng s ca bch ng vi P = 0,3 N/mm2 l:Bng 6. Gi tr ca bch ng dn dng nhp liuDyDnDDD1hBu lng

dbZ

(mm)(ci)

50571401109012M124

1. ng dn hi nh thpNhit ca pha hi ti nh thp l tyD = 61,92 oC.Khi lng ring ca pha hi ti nh thp:

= 4,333 kg/m3Tra bng II.2, trang 370,[1]:Chn vn tc hi ra khi nh thp l yD = 20 m/s.GyD = gd = 3693,515 kg/hng knh trong ca ng ni:

DD = = 0,123 m = 123 mm Chn ng c Dy = 125 mm.Vn tc thc ca dng hi l:

m/s Chn loi ng ni cm su vo thit b Suy t bng XIII.33, trang 434, [2] Chiu di on ng ni l = 100 mm.Tra bng XIII.26, trang 409, [2] Cc thng s ca bch ng vi P = 0,3 N/mm2 l:Bng 7. Gi tr ca bch cho ng dn hi nh thpDyDnDDD1hBu lng

dbZ

(mm)(ci)

12513323520017814M168

1. ng dn dng hon luNhit ca cht lng hon lu l tD = 61,92 oCTi nhit ny th: Khi lng ring ca Clorofom: C = 1408,024 kg/m3 Khi lng ring ca Benzen: B = 833,984 kg/m3

Nn: xL =xD = 1401,752 kg/m3 Chn loi ng ni cm su vo thit b. Chn vn tc cht lng trong ng ni l L = 0,3 m/s. GL = RDMyD = 4,253 x 5,904 x 119,085 = 2990,447 kg/hng knh trong ca ng ni:

DL = = 0,050 m = 50 mm Chn ng c Dy = 50 mm. Vn tc thc ca dng hon lu l:

m/sTra bng XIII.33, trang 434, [2] Chiu di ng ni l = 85 mmTra bng XIII.26, trang 409, [2] Cc thng s ca bch ng vi P = 0,3 N/mm2 l:Bng 8. Gi tr ca bch cho ng dn hon luDyDnDDD1hBu lng

dbZ

(mm)(ci)

50571401109012M124

1. ng dn hi y thpNhit ca pha hi ti y thp l tyW = 80,22 oC.Khi lng ring ca pha hi ti nh thp:

= 2,763 (kg/m3)Chn vn tc hi vo y thp l yW = 20 m/s.GyW = g1 = 2395,183 kg/hng knh trong ca ng ni:

DW = = 0,124 m = 124 mm Chn ng c Dy = 125 mm.

19,62 m/sSuy t bng XIII.33, trang 434, [2] Chiu di on ng ni l = 100 mm.Tra bng XIII.26, trang 409, [2] Cc thng s ca bch ng vi P = 0,3 N/mm2 l:Bng 9. Gi tr ca bch cho ng dn hi y thpDyDnDDD1hBu lng

dbZ

(mm)(ci)

12513323520017814M168

1. ng dn lng vo ni unNhit ca cht lng ti y thp l txW = 80,10 oCTi nhit ny th: Khi lng ring ca Clorofom: C = 1379,730 kg/m3 Khi lng ring ca Benzen: B = 814,890 kg/m3

Nn: xW = 840,567 kg/m3Chn vn tc cht lng trong ng ni l xW = 0,5 m/s.GxW = G1 = 3192,115 kg/hng knh trong ca ng ni:

DxW = = 0,052 m = 52 mm Chn ng c Dy = 50 mm.Vn tc thc ca dng lng l:

LW =m/sTra bng XIII.32, trang 434, [2] Chiu di on ng ni l = 100 mm.Tra bng XIII.26, trang 409, [2] Cc thng s ca bch ng vi P = 0,3 N/mm2 l:

Bng 10. Gi tr ca bch cho ng dn lng vo ni unDyDnDDD1hBu lng

dbZ

(mm)(ci)

50571401109012M124

Khi lng mt bch ghp ng dn lng:

mbch ghp ng lng = = 1,265 kgKhi lng mt bch ghp ng dn hi:

mbch ghp ng hi = = 3,904 kg

III. TNH TON CHN , TAI TREO THP1. Tnh ton trng lng ca ton thpTra bng XII.7, trang 313, [2] Khi lng ring ca CT3 l: CT3 = 7850 kg/m3 Khi lng ca mt bch ghp thn:

mbichghepthan = = 30,51 kg Khi lng ca mt a:

ma =

= 16,11 kg Khi lng chp trn mt a:

mchp

= 7,501 (kg) Khi lng ng hi trn mt a:

moh = Khi lng ca thn thp:

mthn = (D2ng D2t) Hthn X18H10T == 1427,12 kg Khi lng ca y (np) thp:my(np) = Fb mt S X18H10T = 0,82 x 0,004 x 7900 = 25,91 kg. Khi lng g chy trn trn mt a thp X18H10T : hcc = h S5 + hw + mm = 0,3 0,0455 + 0,058 + 0,005 = 0,3175 (m)mtrn = Lw(hcc + hw)trn = 0,56 x (0,3175 + 0,058 ) x 0,002 x 7900 = 3,32 (kg). Khi lng ca ton thp kh: m = 32mbch ghp thn + 56ma + mthn + 2my(np) + 56(mtrn + mchp + moh ) = 4350,36 kg Khi lng dung dch:Gi s lng Benzen cha y thp trong trng hp nguy him nht:B = 836 kg/m3

Khi lng ton thp: mthp = 4350,36 + 7702,6 = 12052,96 kg

1. Chn v tnh ton chn Chn chn : thp c trn bn chn. Vt liu lm chn thp l thp CT3.

Hnh 12. Chn Ti trng cho php trn mt chn: Gc = = 2,96 x 104 N m bo an ton cho thit b, ta chn: Gc = 4 x 104 N

Tra bng XIII.35, trang 437, [2] chn chn c cc thng s sau:Bng 12. Gi tr ca chn LBB1B2Hhsld

2602002253304002251610027

Th tch mt chn : V1chn = [2(H s)sB2 + LsB] x 10-9 = [2 x (400 16) x 16 x 330 + 260 x 16 x 200] x 10-9 = 4,89 x 10-3 m3 Khi lng mt chn : mchn = Vchn CT3 = 4,89 x 10-3 x 7850 = 38,38 kg

1. Chn v tnh ton tai treoChn tai treo: tai treo c gn trn thn thp gi cho thp khi b dao ng trong iu kin ngoi cnh. Chn vt liu lm tai treo l thp CT3.Ta chn bn tai treo, ti trng cho php trn mt tai treo: Gt = Gc = 4 x 104 N.

Hnh 13. Tai treo

Tra bng XIII.36, trang 438, [2] chn tai treo c cc thng s sau:Bng 13. Gi tr ca tai treoLBB1HSlaD

1501201302158602030

Khi lng mt tai treo: mtai treo = 3,48 kg (trong bng)Tra bng XIII.37, trang 439, [2] Chn tm lt tai treo bng thp CT3 c cc thng s sau: Chiu di tm lt: H = 260 mm. Chiu rng tm lt: B = 140 mm. B dy tm lt: SH = 6 mm.Th tch mt tm lt tai treo: Vtm lt =BSHH= 140 x 6 x 260 x 10-9 = 2,184 x 10-4 m3Khi lng mt tm lt tai treo: mtm lt = Vtm lt. CT3 =2,184 x 10-4 x 7850 = 1,71 kg

Chng VI:TNH TON THIT B PH

I. THIT B NGNG T SN PHM NHChn thit b ngng t v - ng loi TH, t nm ngang.Chn theo tiu chun trang 175, [6]ng truyn nhit c lm bng thp X18H10T, kch thc ng 25 x 21. ng knh ngoi: dn = 25 mm = 0,025 m1. B dy ng: t = 2 mm = 0,002 m1. ng knh trong: dtr = 0,021 mChn: Nc lm lnh i trong ng vi nhit vo tV = 30 oC v nhit ra tR = 50 oC. Dng hi ti nh i ngoi ng vi nhit ngng t tngng = 61,92 oC1. Hiu s nhit trung bnhChn kiu truyn nhit ngc chiu, theo cng thc (V.8), trang 5, [1]

= 20,304 K1. H s truyn nhitH s truyn nhit K c tnh theo cng thc nh i vi tng phng:

W/(m2.K) Vi: 1. n : h s cp nhit ca dng nc lnh (W/(m2.K)).1. ngng : h s cp nhit ca dng hi ngng t (W/(m2.K)).1. rt: nhit tr qua thnh ng v lp cu.1. Tnh ton h s cp nhit cc dng1. Xc nh h s cp nhit ca nc i trong ngNhit trung bnh ca dng nc trong ng: tf = ( tV +tR) = (30+50) = 40 oCTi nhit ny:Tra bng I.249, trang 310, [1]1. Khi lng ring ca nc: n = 992,2 kg/m3 1. nht ca nc: n = 6,59 x 10-7 m2/s 1. H s dn nhit ca nc: n = 0,634 W/(mK)1. Chun Prandtl: Prn = 4,31Chn vn tc nc i trong ng: vn = 0,5 m/s

S ng: 17,70Tra bng V.II, trang 48, [2] chn n = 19 ng Vn tc thc t ca nc trong ng:

= 0,466 m/s.Chun s Reynolds :

= 14849 > 104 : ch chy rip dng cng thc 3.27, trang 110, [4] cng thc xc nh chun s Nusselt:

Trong : 1 h s tnh n nh hng ca h s cp nhit theo t l gia chiu di L v ng knh d ca ng.Tra bng 3.1, trang 110, [4] chn 1 = 1

H s cp nhit ca nc i trong ng trong: n = 1. Nhit ti qua thnh ng v lp cu

, W/m2.Trong :1. tw1 : nhit ca vch tip xc vi hi ngng t, oC1. tw2 : nhit ca vch tip xc vi nc lnh, oC

1. B dy thnh ng: t = 0,002 m1. Tra bng XII.7, trang 313, [2]: H s dn nhit ca thp khng g: t = 16,3 W/mK1. Nhit tr lp bn trong ng: r1 = 1/5500 m2 K/W (chn)1. Nhit tr lp cu ngoi ng: r2 = 1/5800 m2 K/W (chn)Nn: rt = 4,769 x 10-4 m2 K/W

1. Xc inh h s cp nhit ca hi ngng t ngoi ngiu kin:1. Ngng t hi bo ha.1. Khng cha khng kh khng ngng.1. Hi ngng t mt ngoi ng.1. Mng cht ngng t chy tng.1. ng nm ngang.p dng cng thc (3.65), trang 120, [4] i vi ng n chic nm ngang th:

Tra bng V.II, trang 48, [2] vi s ng n = 19 th s ng trn ng cho ca hnh 6 cnh l b = 5Tra hnh V.20, trang 30, [2] h s ph thuc vo cch b tr ng v s ng trong mi dy thng ng l tb = 0,7 (v xp xen k v s ng trong mi dy thng ng l 5) H s cp nhit trung bnh ca chm ng: ngng = tb1 = 0,71Dng php lp: chn tW1 = 52,3 oCNhit trung bnh ca mng cht ngng t: tm = (tngng + tW1) = 57,11 oCTi nhit ny th:1. Khi lng ring ca Clorofom : C = 1416,636 kg/m31. Khi lng ring ca Benzen : B = 839,179 kg/m3

Nn: = 1410,279 kg/m31. nht ca Clorofom : C = 4,004 x 10-4 N.s/m2 1. nht ca Benzen: B = 4,033 x 10-4 N.s/m2Nn: lg = xDlgC + (1 xD)lgB = 0,99lg(4,004 x 10-4) + (1 0,99)lg(4,033 x 10-4)= -3,397 = 4,004 x 10-4 N.s/m21. H s dn nhit ca Clorofom : C = 0,1137 W/(mK) 1. H s dn nhit ca Benzen : B = 0,1366 W/(mK)Nn theo cng thc (I.33), trang 124, [1]: = CxD + B(1 -xD) 0,72xD(1 -xD)(B - C) = 0,1138 (W/mK)1. Nhit ngng t ca dng hi: r = rD = 247,705 kJ/kg = 247705 J/kg Nn: 1 = 2125,467 W/(m2K) ngng = tb1 = 0,71 = 1478,827 W/m2K qngng = ngng (tngng tW1) = 14312,894 W/m2 qt = qngng = 14312,894 W/m2 (xem nhit ti mt mt l khng ng k) tw2 = tw1 - qtrt = 45,474 oC Prw2 = 3,893 (tra bng I.249, trang 310, [1]) Nun = 87,754

n = = 2649,333 W/m2K qn = n (tW2 tf) = 2649,333 x (45,474 - 40) = 14501,739 W/m2Kim tra sai s:

= 100% = 1,32 % < 5% (tha)Kt lun: tw1 = 52,3 oC v tw2 = 45,474 oC1. Xc inh h s truyn nhit

= 655,091 W/(m2K)1. Xc nh b mt truyn nhit1. B mt truyn nhit c xc nh theo phng trnh truyn nhit:

F = 19,107 m2H s an ton c cho l 20 % (theo trang 164, [4]), din tch b mt truyn nhit l: Ftt = F x 1,2 = 19,107 x 1,2 = 22,928 m21. Cu to thit bS ng truyn nhit: n = 19 ng.

Chiu di ng truyn nhit: L = = 16,701 m chn L = 16 m, dng tm ngn kiu r qut chia lm 4 pass, mi pass 4 m.Khi s ng l: 19 x 4 = 76 (ng) b tr ng theo hnh lc gic u (gm 5 hnh - kiu dng cho s ng l 91).S ng trn ng cho: b = 11 ngBc ng: t = (1,2 1,5)d = 35 mmp dng cng thc V.140, trang 49, [2]: ng knh trong ca thit b:

Hnh 14. Minh ha cch b tr ng.D = t(b-1) + 4dn = 0,450 mII. THIT B LM NGUI SN PHM NHChn thit b lm ngui sn phm nh l thit b truyn nhit ng lng ng. ng truyn nhit c lm bng thp X18H10T, chn theo tiu chun trang 175, [6] Kch thc ng trong: 16 x 1,6 mm Kch thc ng ngoi: 38 x 3 mmChn: Nc lm lnh i trong ng vi nhit vo tV = 30 oC v nhit ra tR = 50 oC. Sn phm nh trong ng ngoi vi nhit vo tDV = 61,92 oC v nhit ra tDR = 35 oC.1. Hiu s nhit trung bnhChn kiu truyn nhit l ngc chiu, nn:

= 7,965 K1. H s truyn nhitH s truyn nhit K c tnh theo cng thc:

,W/(m2.K) Vi: n : h s cp nhit ca dng nc lnh (W/m2.K). D : h s cp nhit ca dng sn phm nh (W/m2.K). rt : nhit tr qua thnh ng v lp cu.1. Tnh ton h s cp nhit cc dng1. Xc nh h s cp nhit ca nc trong ngKch thc ca ng trong: ng knh ngoi: dn = 16 mm = 0,016 m B dy ng: t = 1,6 mm = 0,0016 m ng knh trong: dtr = 0,0128 mNhit trung bnh ca dng nc trong ng: tF = (tV + tR) = 40 oC.Ti nhit ny th:Tra bng I.249, trang 310, [1] Khi lng ring ca nc: n = 992,2 kg/m3 nht ca nc: n = 6,59 x 10-7 m2/s H s dn nhit ca nc: n = 0,634 W/(mK) Chun s Prandtl: Prn = 4,31Vn tc nc i trong ng:

= 0,564 m/s.Chun s Reynolds :

= 10953 > 104 : ch chy rip dng cng thc (3.27), trang 110, [4] cng thc xc nh chun s Nusselt:

Trong : 1 h s tnh n nh hng ca h s cp nhit theo t l gia chiu di L v ng knh d ca ng.Tra bng 3.1, trang 110, [4] chn 1 = 1

H s cp nhit ca nc trong ng: n = 1. Nhit ti qua thnh ng v lp cu

,W/m2Trong : tw1 : nhit ca vch tip xc vi sn phm nh (ngoi ng trong), oC tw2 : nhit ca vch tip xc vi nc lnh (trong ng trong), oC

B dy thnh ng: t = 0,0016 m H s dn nhit ca thp khng g: t = 16,3W/(mK) Nhit tr lp bn trong ng: r1 = 1/5500 m2.K/W Nhit tr lp cu ngoi ng: r2 =1/5800 m2.K/WNn: rt = 4,524 x 10-4 m2.K/W1. Xc nh h s cp nhit ca dng sn phm nh ngoi ngKch thc ca ng ngoi: ng knh ngoi: Dn = 38 mm = 0,038 m B dy ng: t = 3 mm = 0,003 m ng knh trong: Dtr = 0,032 mNhit trung bnh ca dng sn phm nh ngoi ng: tD = (tDV + tDR) = 48,46 oC.Ti nhit ny th:Tra bng I.2, trang 9, [1] Khi lng ring ca Clorofom : C = 1433,503 kg/m3 Khi lng ring ca Benzen : B = 848,694 kg/m3

Nn: = 1427,062 kg/m3Tra bng I.101, trang 91, [1]1. nht ca Clorofom : C = 4,322 x 10-4 N.s/m2 1. nht ca Benzen : B = 4,446 x 10-4 N.s/m2Nn: lg = xDlgC + (1 xD)lgB = 0,99 x lg(4,322 x 10-4) + (1 0,99) x lg(4,446 x 10-4)= -3,364 = 4,323 x 10-4 N.s/m21. H s dn nhit ca Clorofom : C = 0,1128 W/(mK) 1. H s dn nhit ca Benzen : B = 0,1383 W/(mK)Nn theo cng thc (I.33), trang 124, [1]: = CxD + B(1 -xD) 0,72xD(1 -xD)(B - C) = 0,1129 (W/mK)Tra bng I.153, trang 171, [1]:1. Nhit dung ring ca Clorofom : CC = 1063,690 J/(kg.K)1. Nhit dung ring ca Benzen : CB = 1869,415 J/(kg.K)

Vy: c = cC + cB(1 - ) = 1068,967 J/(kg.K)

p dng cng thc V.35, trang 12, [2]: = 4,093Vn tc ca dng sn phm nh ngoi ng:

= 0,227 m/sng knh tng ng: dt = Dtr dn = 0,032 0,016 = 0,016 mChun s Reynolds :

= 11983 > 104 : ch chy rip dng cng thc 3.27, trang 110, [4] cng thc xc nh chun s Nusselt:

Trong : 1 h s tnh n nh hng ca h s cp nhit theo t l gia chiu di L v ng knh d ca ng.Tra bng 3.1, trang 110, [4] chn 1 = 1

H s cp nhit ca dng sn phm nh ngoi ng: D = Dng php lp: chn tW1 = 42,3 oCTi nhit ny th:Tra bng I.101, trang 91, [1]1. nht ca Clorofom : C = 4,568 x 10-4 N.s/m2 1. nht ca Benzen : B = 4,791 x 10-4 N.s/m2Nn: lg = xDlgC + (1 xD)lgB = 0,99lg(4,568 x 10-4) + (1 0,99)lg(4,791 x 10-4)= -3,340 = 4,570 x 10-4 N.s/m21. H s dn nhit ca Clorofom : C = 0,1122 W/(mK) 1. H s dn nhit ca Benzen : B = 0,1395 W/(mK)Nn theo cng thc (I.33), trang 124, [1]: = CxD + B(1 -xD) 0,72xD(1 -xD)(B - C) = 0,1123 (W/mK)Tra bng I.153, trang 171, [1]:1. Nhit dung ring ca Clorofom : CC = 1054,450 J/(kg.K)1. Nhit dung ring ca Benzen : CB = 1837,075 J/(kg.K)

Vy: c = cC + cB (1 - ) = 1059,576 J/(kg.K)

p dng cng thc V.35, trang 12, [2]: = 4,313Nn: NuD = 69,600 D = 491,087 W/(m2K) qD = D (tD tW1) = 491,087 x (48,46 42,3) = 3025,094 W/m2 qt = qD = 3025,094 W/m2 (xem nhit ti mt mt l khng ng k) tw2 = tw1 - qtrt = 42,3 3025,094 x 4,524 x 10-4 = 40,93 oC PrW2 = 4,239 (tra bng I.249, trang 310, [1]) Nun = 67,372 n = 3337,032 (W/m2K) qn = n (tW2 tf) = 3337,032 x (40,93 - 40) = 3108,356 W/m2Kim tra sai s:

= 100% = 2,72 % < 5 % (tha)Kt lun: tw1 = 42,3 oC v tw2 = 40,93 oC

1. Xc nh h s truyn nhit

= 358,634 W/(m2K)

1. Xc nh b mt truyn nhitB mt truyn nhit c xc nh theo phng trnh truyn nhit:

F = = 2,102 m2i vi trng hp truyn nhit, h s an ton thng c chn l 20% nn din tch b mt truyn nhit cn thit l Ftt = F x 1,2 = 2,522 m2 1. Cu to thit b

Chiu di ng truyn nhit: L = = 55,747 m chn L = 56 m

Kim tra: > 50 l = 1: thaKt lun: Thit b lm ngui sn phm nh l thit b truyn nhit ng lng ng vi chiu di ng truyn nhit L = 56 m.Tin hnh b tr thnh 14 dy, mi dy di 4m. ng hng trn c ni vi hnh di bng mt bch, cc chi tit bch v ng, ng v thnh ng c hn vi nhau.

III. THIT B UN SI Y THPChn thit b un si y thp l ni un Kettle.

Hnh 15. Ni un Kettle

Chn theo tiu chun trang 175, [6]ng truyn nhit c lm bng thp X18H10T, kch thc ng 25 x 21. ng knh ngoi: dn = 25 mm = 0,025 m1. B dy ng: t = 2 mm = 0,002 m1. ng knh trong: dtr = 0,021 mHi t l hi nc 1,6 at.Tra bng 1.251, trang 314, [1]: 1. Nhit ha hi: = rn = 2227 x 103 J/kg1. Nhit si: = tn = 112,7 oCDng sn phm ti y c nhit :1. Trc khi vo ni un (lng): tS1 = 80,10 oC1. Sau khi c un si (hi): tS2 = 80,22 oC1. Hiu s nhit trung bnhChn kiu truyn nhit ngc chiu, nn:

= 32,540 K.1. H s truyn nhitH s truyn nhit K c tnh theo cng thc nh i vi tng phng:

W/(m2.K) Vi: 1. n : h s cp nhit ca hi t (W/(m2.K)).1. S : h s cp nhit ca sn phm y (W/(m2.K)).1. rt: nhit tr qua thnh ng v lp cu.1. Tnh ton h s cp nhit cc dng1. Xc nh h s cp nhit ca dng sn phm y ngoi ngp dng cng thc V.89, trang 26, [2]

Nhit si trung bnh ca dng sn phm ngoi ng:

oCTs = 80,16 + 273 = 353,16 KTi nhit trung bnh th:1. Khi lng ring ca pha hi trong dng sn phm ngoi ng:

2,763 kg/m31. Khi lng ring ca pha lng:Tra bng I.2, trang 9, [1]: C = 1379,568 kg/m3Tra bng I.2, trang 9, [1]: B = 814,824 kg/m3

Nn: : = 840,498 kg/m3Tra bng I.101, trang 91, [1]1. nht ca Clorofom : C = 3,297 x 10-4 N.s/m2 1. nht ca Benzen: B = 3,156 x 10-4 N.s/m2Nn: lg = xWlgC + (1 - xW)lgB = 0,05 x lg(3,297 x 10-4) + (1 0,05)lg(3,156 x 10-4)= -3,500 = 3,163 x 10-4 N.s/m21. H s dn nhit ca Clorofom : C = 0,1022 W/(mK) 1. H s dn nhit ca Benzen : B = 0,1300 W/(mK)Nn theo cng thc (I.33), trang 124, [1]: = CxW + B(1 -xW) 0,72xW(1 -xW)(B - C) = 0,1265 (W/mK)Tra bng I.153, trang 171, [1]:1. Nhit dung ring ca Clorofom : CC = 1110,240 J/(kg.K)1. Nhit dung ring ca Benzen : CB = 2035,680 J/(kg.K)

Vy: c = cC + cB(1 -) = 1179,294 J/(kg.K)1. Sc cng b mt:Tra bng I.242, trang 300, [1]1. Sc cng b mt ca Clorofom : C = 0,01898 N/m1. Sc cng b mt ca Benzen : B = 0,02128 N/m

Nn: = 0,01003 N/m1. Nhit ha hi: Tra bng I.213, trang 256, [1]1. Nhit ha hi ca Clorofom : rC = 239210,306 J/kg 1. Nhit ha hi ca Benzen: rB = 393441,970 J/kg

Nn: r = rC + rB (1 - )= 239210,306 x 0,0746 + (1- 0,0746) x 393441,970 = 381933,581 J/kg1. Nhit ti qua thnh ng v lp cu

W/m2Trong :1. tw1 : nhit ca vch tip xc vi hi t (trong ng), oC1. tw2 : nhit ca vch tip xc vi sn phm y (ngoi ng), oC

1. B dy thnh ng: t = 0,002 m1. H s dn nhit ca thp khng g:Tra bng XII.7, trang 313, [2]: t = 16,3 W/mK1. Nhit tr lp bn,cu:Bng 31, trang 419, [4]:1. Lp bn trong ng: r1 = 1/4800 m2.K/W1. Lp cu ngoi ng: r2 = 1/5800 m2.K/WVy: rt = 5,034 x10-4 m2.K/W1. Xc nh h s cp nhit ca hi t trong ng

p dng cng thc 3.65, trang 120,[4]: Dng php lp: chn tW1 = 111,06 oCNhit trung bnh ca mng nc ngng t: tm = (tn + tW1) = 111,880 oCTi nhit ny th:1. Khi lng ring ca nc: n = 949,515 kg/m3 1. nht ca nc: n = 2,555 x 10-4 N.s/m2 1. H s dn nhit ca nc: n = 68,438 x 10-2 W/(mK)

Nn: = 21113,700 W/m2K qn = n (tn tW1) = 34626,468 W/m2 qt = qn = 34626,468 W/m2 (xem nhit ti mt mt l khng ng k) tw2 = tw1 - qtrt =111,06 - 34626,468 x 5,034 x 10-4 = 93,627 oC

= 2565,926 W/(m2K) (vi q = qt) qS = S (tW2 tS) = 2565,926 x (93,627 80,16) = 34556,419 W/m2Kim tra sai s:

= 100% = 0,20 % < 5 % (tha)Kt lun: tw1 = 111,06 oC v tw2 = 93,627 oC1. Xc nh h s truyn nhit

= 1063,228 W/(m2K)1. Xc nh b mt truyn nhitB mt truyn nhit c xc nh theo phng trnh truyn nhit:

F = = 7,699 m2H s an ton ca thit b c cho l 20 % cho nn din tch cn l :Ftt = 1,2F = 9,238 m21. Cu to thit b truyn nhitChn s ng truyn nhit: n = 61 ng. ng c b tr theo hnh lc gic u.

Chiu di ng truyn nhit: L = = 2,096 m chn L = 2 mTra bng V.II, trang 48, [2] S ng trn ng cho: b = 9 ngBc ng: t = (1,2 1,5)d = 35 mmp dng cng thc V.140, trang 49, [2]: ng knh trong ca thit b: D = t(b-1) + 4dn = 0,380 mIV. THIT B GII NHIT SN PHM YChn thit b lm ngui sn phm y l thit b truyn nhit ng lng ng. ng truyn nhit c lm bng thp X18H10T, chn theo tiu chun trang 175, [6]1. Kch thc ng trong: 25 x 2 mm1. Kch thc ng ngoi: 48 x 3 mmChn: 1. Nc lnh i trong ng trong vi nhit vo tV = 30 oC v nhit ra tR = 50 oC.1. Sn phm y i ngoi ng trong vi nhit vo tWS = 80,1 oC v nhit ra tWR = 40 oC.1. Hiu s nhit trung bnhChn kiu truyn nhit trong ng lng ng l ngc chiu, nn:

= 18,241 K1. H s truyn nhitH s truyn nhit K c tnh theo cng thc:

,W/(m2.K) Vi: 1. n : h s cp nhit ca dng nc lnh (W/m2.K).1. W : h s cp nhit ca dng sn phm y (W/m2.K).1. rt : nhit tr qua thnh ng v lp cu.1. Tnh ton h s cp nhit cc dng1. Xc nh h s cp nhit ca nc trong ng trongKch thc ca ng trong: 1. ng knh ngoi: dn = 25 mm = 0,025 m1. B dy ng: t = 2 mm = 0,002 m1. ng knh trong: dtr = 0,021 mNhit trung bnh ca dng nc: tF = (tV + tR) = 40 oC.Ti nhit ny th:Tra bng I.249, trang 310, [1] Khi lng ring ca nc: n = 992,2 kg/m3 nht ca nc: n = 6,59x10-7 m2/s H s dn nhit ca nc: n = 0,634 W/(mK) Chun s Prandtl: Prn = 4,31Vn tc nc i trong ng:

= 0,669 m/s.Chun s Reynolds :

= 21327 > 104 : ch chy rip dng cng thc (3.27), trang 110, [4] cng thc xc nh chun s Nusselt:

Trong : 1 h s tnh n nh hng ca h s cp nhit theo t l gia chiu di L v ng knh d ca ng.Tra bng 3.1, trang 110, [4] chn 1 = 1

H s cp nhit ca nc trong ng: n = 1. Nhit ti qua thnh ng v lp cu

,W/m2.Trong : tw1 : nhit ca vch tip xc vi sn phm y (ngoi ng trong), oC tw2 : nhit ca vch tip xc vi nc lnh (trong ng trong),oC

B dy thnh ng: t = 0,002 m H s dn nhit ca thp khng g: t = 16,3 W/(mK) Nhit tr lp bn trong ng: r1 = 1/5500 m2.K/W Nhit tr lp cu ngoi ng: r2 =1/5800 m2.K/W Nn: rt = 4,769 x10-4 m2.K/W1. Xc nh h s cp nhit ca dng sn phm y ngoi ng trongKch thc ca ng ngoi: ng knh ngoi: Dn = 48 mm = 0,048 m B dy ng: t = 3 mm = 0,003 m ng knh trong: Dtr = 0,042 mNhit trung bnh ca dng sn phm y ngoi ng: tW = (tWS + tWR) = 60,05 oC.Ti nhit ny th:Tra bng I.2, trang 9, [1] Khi lng ring ca Clorofom : C = 1410,923 kg/m3 Khi lng ring ca Benzen : B = 835,948 kg/m3

Nn: = 862,164 kg/m3Tra bng I.101, trang 91, [1] nht ca Clorofom : C = 3,899 x 10-4 N.s/m2 nht ca Benzen: B = 3,898 x 10-4 N.s/m2Nn: lg = xWlgC + (1 - xW)lgB = 0,05 x lg(3,899 x 10-4) + (1 0,05)lg(3,898 x 10-4)= -3,409 = 3,898 x 10-4 N.s/m21. H s dn nhit ca Clorofom : C = 0,1140 W/(mK) 1. H s dn nhit ca Benzen : B = 0,1360 W/(mK)Nn theo cng thc (I.33), trang 124, [1]: = C xxW + B.(1 -xW) 0,72xW.(1 -xW)(B - C) = 0,1332 (W/mK)Tra bng I.153, trang 171, [1]:1. Nhit dung ring ca Clorofom : CC = 1081,073 J/(kg.K)1. Nhit dung ring ca Benzen : CB = 1930,263 J/(kg.K)

Vy: c = cC + cB(1 -) = 1866,898 J/(kg.K)

p dng cng thc V.35, trang 12, [2]: = 5,462Vn tc ca dng sn phm y ngoi ng trong:

= 0,287 m/sng knh tng ng: dt = Dtr dn = 0,042 0,025 = 0,017 mChun s Reynolds :

= 10791 > 104 : ch chy rip dng cng thc 3.27, trang 110, [4] cng thc xc nh chun s Nusselt:

Trong : 1 h s tnh n nh hng ca h s cp nhit theo t l gia chiu di L v ng knh d ca ng.Tra bng 3.1, trang 110, [4] chn 1 = 1

H s cp nhit ca dng sn phm y ngoi ng: W = Dng php lp: chn tW1 = 46 oCTi nhit ny th:Tra bng I.101, trang 91, [1]1. nht ca Clorofom : C = 4,420 x 10-4 N.s/m2 1. nht ca Benzen : B = 4,584 x 10-4 N.s/m2Nn: lg = xWlgC + (1 xW)lgB = 0,05 x lg(4,420 x 10-4) + (1 0,05)lg(4,584 x 10-4)= -3,340 = 4,576 x 10-4 N.s/m21. H s dn nhit ca Clorofom : C = 0,1126 W/(mK) 1. H s dn nhit ca Benzen : B = 0,1388 W/(mK)Nn theo cng thc (I.33), trang 124, [1]: = C.xW + B.(1 -xW) 0,72xW.(1 -xW)(B - C) = 0,1355 (W/mK)Tra bng I.153, trang 171, [1]:1. Nhit dung ring ca Clorofom : CC = 1060,000 J/(kg.K)1. Nhit dung ring ca Benzen : CB = 1856,500 J/(kg.K)

Vy: c = cC + cB (1 -) = 1797,067 J/(kg.K)

p dng cng thc V.35, trang 12, [2]: = 6,067Nn: NuW = 71,505 W = 560,464 W/(m2K) qW = W (tW tW1) = 560,464 x (60,05 - 46) = 7874,525 W/m2 qt = qW = 7874,525 W/m2 (xem nhit ti mt mt l khng ng k) tw2 = tw1 - qtrt = 46 7874,525 x 4,769 x 10-4 = 42,244 oC PrW2 = 4,139 Nun = 115,498 n = 3486,952 W/(m2K) qn = n (tW2 tf) = 3486,952 x (42,244 - 40) = 7826,087 W/m2Kim tra sai s:

= 100% = 0,61 % < 5 % (tha)Kt lun: tw1 = 46 oC v tw2 = 42,27 oC1. Xc nh h s truyn nhit

= 392,472 W/(m2K)

1. Xc nh b mt truyn nhitB mt truyn nhit c xc nh theo phng trnh truyn nhit:

F = = 2,685 m2i vi trng hp truyn nhit, h s an ton thng c chn l 20% nn din tch b mt truyn nhit cn thit l Ftt = F x 1,2 = 3,222 m2 1. Cu to thit b

Chiu di ng truyn nhit: L = = 44,59 m chn L = 48 m

Kim tra: > 50 l = 1: thaKt lun: Thit b lm ngui sn phm y l thit b truyn nhit ng lng ng vi chiu di ng truyn nhit L = 48 m.Tin hnh b tr thnh 12 dy, mi dy di 4m. ng hng trn c ni vi hnh di bng mt bch, cc chi tit bch v ng, ng v thnh ng c hn vi nhau.

V. THIT B UN SI DNG NHP LIUChn thit b gia nhit nhp liu l thit b trao i nhit kiu ng chm, t nm ngang.ng truyn nhit c lm bng thp X18H10T, kch thc ng 20 x 2 mm ng knh ngoi: dn = 20 mm = 0,020 m B dy ng: t = 2 mm = 0,002 m ng knh trong: dtr = 16 mm = 0,016 m Chn s ng n = 7, xp theo hnh lc gic.Hi t l hi nc 1,6 at.Tra bng 1.251, trang 314, [1] Nhit ha hi: rH2O= rn = 2227000 J/kg Nhit si: t H2O= tn = 112,7 oCDng nhp liu c nhit :Trc khi vo ni un (lng): tF = 30 oCSau khi c un (lng si): tFS = 61,92 oC1. Hiu s nhit trung bnhChn kiu truyn nhit ngc chiu, nn:

= 65,448 K1. H s truyn nhitH s truyn nhit K c tnh theo cng thc nh i vi tng phng:

W/(m2.K)Vi: n : h s cp nhit ca hi t W/(m2.K) F : h s cp nhit dng nhp liu W/(m2.K) rt : nhit tr qua thnh ng v lp cu.1. Tnh ton h s cp nhit cc dng1. Xc nh h s cp nhit ca dng nhp liu trong ngTi nhit trung bnh th ca dng nhp liu tF = (tV + tR) = (30+61,92) = 45,96 oCTra bng I.2, trang 9, [1]1. Khi lng ring ca Clorofom : C = 1438,378 kg/m31. Khi lng ring ca Benzen : B = 851,444 kg/m3

Nn: = 1072,595 kg/m3Tra bng I.101, trang 91, [1]1. nht ca Clorofom : C = 4,422 x 10-4 N.s/m21. nht ca Benzen: B = 4,586 x 10-4 N.s/m2Nn: lg = xFlgC + (1 xF)lgB = 0,4 x lg(4,422 x 10-4) + (1 0,4) x lg(4,586 x 10-4)= -3,345 = 4,520 x 10-4 N.s/m21. H s dn nhit ca Clorofom : C = 0,1126 W/(mK) 1. H s dn nhit ca Benzen : B = 0,1388 W/(mK)Nn theo cng thc (I.33), trang 124, [1]: = C xxF + B.(1 -xF) 0,72xF.(1 -xF)(B - C) = 0,1208 (W/mK)Tra bng I.153, trang 171, [1]:1. Nhit dung ring ca Clorofom : CC = 1059,940 J/(kg.K)1. Nhit dung ring ca Benzen : CB = 1856,290 J/(kg.K)

Vy: c = cC + cB(1 -) = 1462,324 J/(kg.K)

p dng cng thc V.35, trang 12, [2] : = 5,469Vn tc ca dng nhp liu i trong ng:

= 0,276 m/s.Chun s Reynolds :

= 10480 > 104 : ch chy rip dng cng thc 3.27, trang 110, [4] cng thc xc nh chun s Nusselt:

Trong : 1 h s tnh n nh hng ca h s cp nhit theo t l gia chiu di L v ng knh d ca ng.Tra bng 3.1, trang 110, [4] chn 1 = 11. Nhit ti qua thnh ng v lp cu

, W/m2Trong : tW1 : nhit ca vch tip xc vi hi t pha v, oC tW2 : nhit ca vch tip xc vi nhp liu trong ng, oC

B dy thnh ng: t = 0,002 m H s dn nhit ca thp khng g:Tra bng XII.7, trang 313, [2]: t = 16,3 W/mKTra bng 31, trang 419, [4] Nhit tr lp bn trong ng: r1 = 1/5800 m2.K/W Nhit tr lp cu ngoi ng: r2 = 1/5500 m2.K/WNn: rt = 4,769 x 10-4 m2.K/W1. Xc nh h s cp nhit ca hi ngng t ngoi ngiu kin: Ngng t hi bo ha. Khng cha khng kh khng ngng. Hi ngng t mt ngoi ng. Mng cht ngng t chy tng. ng nm ngang.p dng cng thc (3.65), trang 120, [4] i vi ng n chic nm ngang th:

Tra bng V.II, trang 48, [2] vi s ng n = 7 th s ng trn ng cho ca hnh 6 cnh l b = 3Tra hnh V.20, trang 30, [2] h s ph thuc vo cch b tr ng v s ng trong mi dy thng ng l tb = 0,78 (v xp xen k) H s cp nhit trung bnh ca chm ng: ngng = tb1 = 0,781Dng php lp: chn tW1 = 111,2 oCNhit trung bnh ca mng nc ngng t: tm = (tn + tW1) = 111,95 oCTi nhit ny th: Khi lng ring ca nc: n = 949,460 kg/m3 nht ca nc: n = 2,547 x 10-4 N.s/m2 H s dn nhit ca nc: n = 68,439 x 10-2 W/(mK)Nn: n = ngng = 0,781 = 17049,202 W/(m2K) qn = n (tn tW1) = 17049,202 x (112,7 111,2) = 25573,804 W/m2 qt = qn = 25573,804 W/m2 (xem nhit ti mt mt l khng ng k) tw2 = tw1 - qtrt = 111,2 25573,804 x 4,769 x 10-4 = 99,00 oCTi nhit ny th:Tra bng I.101, trang 91, [1] nht ca Clorofom : C = 2,920 x 10-4 N.s/m2 nht ca Benzen: B = 2,637 x 10-4 N.s/m2Nn: lg = xFlgC + (1 xF)lgB = 0,4 x lg(2,962 x 10-4) + (1 0,4) x lg(2,695 x 10-4)= -3,561 = 2,747 x 10-4 N.s/m21. H s dn nhit ca Clorofom : C = 0,0922 W/(mK) 1. H s dn nhit ca Benzen : B = 0,1286 W/(mK)Nn theo cng thc (I.33), trang 124, [1]: = C xxF + B.(1 -xF) 0,72xF.(1 -xF)(B - C) = 0,1037 (W/mK)Tra bng I.153, trang 171, [1]:1. Nhit dung ring ca Clorofom : CC = 1138,505 J/(kg.K)1. Nhit dung ring ca Benzen : CB = 2115,763 J/(kg.K)

Vy: c = cC + cB(1 -) = 1632,299 J/(kg.K)

p dng cng thc (V.35), trang 12, [2]: = 4,325Nn: NuF = 72,293 aF = 468,406 W/(m2K) qF = aF (tW2 - tF) = 468,406 (99 45,96) = 24845,694 (W/m2)Kim tra sai s:

= 100% = 2,85% < 5 % (tha)Kt lun: tw1 = 111,2 oC v tw2 = 99,0 oC1. Xc nh h s truyn nhit

= 374,464 W/m2K1. Xc nh b mt truyn nhitB mt truyn nhit c xc nh theo phng trnh truyn nhit:

F = = 1,289 m2Din tch truyn nhit c tnh vi h s an ton 20 % nn gi tr din tch thc t l: Ftt =1,289 x 1,2 = 1,547 m2.1. Cu to thit bChn s ng: n = 7 ng, c b tr theo hnh lc gic u

Chiu di ng truyn nhit: = 3,909 m chn L = 4 m, dng tm ngn chia lm 2 pass, mi pass 2 m.

Kim tra: > 50 l = 1 (tha)Khi s ng l: 7 x 2 = 14 (ng) b tr ng theo 2 hnh lc gic u (kiu dng cho s ng l 19).S ng trn ng cho: b = 5 ngBc ng: t = (1,2 1,5)d = 25 mmp dng cng thc V.140, trang 49, [2]: ng knh trong ca thit b: D = t(b-1) + 4dn = 0,180 mKt lun: Thit b trao i nhit dng nhp liu l thit b truyn nhit ng chm c chiu di L = 2 m, D = 0,180 m.

VI. BN CAO V1. Tn tht ng ng dnChn ng dn c ng knh trong l dtr = 50 mmTra bng II.15, trang 381, [1] nhm ca ng: = 0,2 mm = 0,0002 m (n mn t)Tn tht ng ng dn:

, mTrong : 1 : h s ma st trong ng ng. l1 : chiu di ng ng dn, chn l1 = 30 m. d1 : ng knh ng dn, d1 = dtr = 0,05 m. 1 : tng h s tn tht cc b. vF : vn tc dng nhp liu trong ng dn0. Xc nh vn tc dng nhp liu trong ng dnCc tnh cht l hc ca dng nhp liu c tra nhit trung bnh:

tF = = 45,96 oCTi nhit ny th:1. Khi lng ring ca Clorofom : C = 1438,378 kg/m31. Khi lng ring ca Benzen : B = 851,444 kg/m3

Nn: = 1072,595 kg/m31. nht ca Clorofom : C = 4,422 x 10-4 N.s/m21. nht ca Benzen: B = 4,586 x 10-4 N.s/m2Nn: lg = xFlgC + (1 xF)lgB = 0,4 x lg(4,422 x 10-4) + (1 0,4) x lg(4,586 x 10-4)= -3,345 = 4,520 x 10-4 N.s/m2Vn tc ca dng nhp liu i trong ng:

= 0,198 m/s

0. Xc nh h s ma st trong ng ngChun s Reynolds :

= 23475 > 4000 ch chy ri (trang 378,[1])p dng cng thc II.60, trang 378, [1]Chun s Reynolds ti hn: Regh = 6(d1/)8/7 = 3301p dng cng thc II.62, trang 379, [1]Chun s Reynolds khi bt u xut hin vng nhm: Ren = 220(d1/)9/8 = 109674Mc c, trang 379, [1]V Regh < ReF < Ren ch chy ri ng vi khu vc qu .

p dng cng thc (II.64), trang 379, [1]: H s ma st = 0,0321. Xc nh tng h s tn tht cc b Ch un cong:Tra bng 9.5, trang 110, [7]: Chn dng ng un cong 90 o c bn knh R vi R/d = 2 th u1 (1 ch) = 0,9.ng ng c 8 ch un u1 = 0,9 x 8 = 7,2 Van :Tra bng 9.5, trang 110, [7]: Chn van cu vi m hon ton th van (1 ci) = 10ng ng c 3 van cu van = 10 x 3 = 30Ni suy mc 27, bng II.16, trang 399, [1]: Tn tht van mt chiu: vanmtchiu = 1,333 Lu lng k : l1 = 0 (coi nh khng ng k). Vo thp: thp = 1 Nn: 1 = u1 + van + thp + vanmtchiu = 7,2 +30+1+1,333 = 39,533

Vy: = 0,117 m1. Tn tht ng ng dn trong thit b un si dng nhp liu

, mTrong : 2 : h s ma st trong ng ng. l2 : chiu di ng ng dn, l2 = 3,5 m. d2 : ng knh ng dn, d2 = dtr = 0,016 m. 2 : tng h s tn tht cc b. v2 : vn tc dng nhp liu trong ng truyn nhit1. Vn tc dng nhp liu trong ng truyn nhit: v2 = 0,276 m/s1. Xc nh h s ma st trong ng ngChun s Reynolds : Re2 = 10480 > 4000: ch chy ri nhm: = 0,0002Chun s Reynolds gii hn: Regh = 6(d1/)8/7 = 898Chun s Reynolds khi bt u xut hin vng nhm: Ren = 220(d1/)9/8 = 30437V Regh < Re2 < Ren ng vi khu vc qu

Tra bng II.13, trang 379, [1]: = 0,0391. Xc nh tng h s tn tht cc b t thu:

Tra bng II.16, trang 382, [1]: Khi = 0,1024 th t thu 2 (1ch) = 0,470 t m:

Tra bng II.16, trang 382, [1]: Khi = 0,1024 th t m 2 (1ch) = 0,806

Nn: 2 = t thu 2 + t m 2 = 1,275

Vy: = 0,038 m1. Chiu cao bn cao vChn : Mt ct (1-1) l mt thong cht lng trong bn cao v. Mt ct (2-2) l mt ct ti v tr nhp liu thp.p dng phng trnh Bernoulli cho (1-1) v (2-2):

z1 = z2 + +hf1-2Trong : z1: cao mt thong (1-1) so vi mt t, hay xem nh l chiu cao bn cao v Hcv = z1 z2: cao mt thong (2-2) so vi mt t, hay xem nh l chiu cao t mt t n v tr nhp liu: ( nhp liu a th 37 t di ln )z2 = hchn + hy + (nttC 1)h + 0,5 = 0,4 + 0,25 + 36 x 0,3 + 0,5 = 11,95 (m) P1 : p sut ti mt thong (1-1), chn P1 = 1 at = 9,81 x 104 N/m2 P2 : p sut ti mt thong (2-2)Xem P = P2 P1 = nttL x PL = 13221,26 N/m2 v1 : vn tc ti mt thong (1-1), xem v1 = 0 m/s. v2 : vn tc ti v tr nhp liu, v2 = vF = 0,198 m/s. hf1-2 : tng tn tht trong ng t (1-1) n (2-2):hf1-2 = h1 + h2 = 0,117 + 0,038 = 0,155

Vy: Chiu cao bn cao v: Hcv = z2 + +hf1-2

= 11,95 ++ 0,155= 13,36 mChn Hcv = 15 m.VII. BM1. Nng sutNhit dng nhp liu l tF = 30 oC.Ti nhit ny th:1. Khi lng ring ca Clorofom : C = 1469,500 kg/m31. Khi lng ring ca Benzen : B = 868,500 kg/m3

Nn: = 1094,729 kg/m31. nht ca Clorofom : C = 5,100 x 10-4 N.s/m21. nht ca Benzen: B = 5,600 x 10-4 N.s/m2Nn: lg = xFlgC + (1 xF)lgB = -3,268 = 5,394 x 10-4 N.s/m2Sut lng th tch ca dng nhp liu i trong ng:

= 1,370 m3/hVy: chn bm c nng sut Qb = 2 m3/h1. Ct pChn : Mt ct (1-1) l mt thong cht lng trong bn cha nguyn liu. Mt ct (2-2) l mt thong cht lng trong bn cao v.p dng phng trnh Bernoulli cho (1-1) v (2-2):

Trong : z1: cao mt thong (1-1) so vi mt t, chn z1 = 1 m. z2: cao mt thong (2-2) so vi mt t, z2 = Hcv = 15 m. P1 : p sut ti mt thong (1-1), chn P1 = 1 at. P2 : p sut ti mt thong (2-2), chn P2 = 1 at. v1,v2 : vn tc ti mt thong (1-1) v(2-2), xem v1= v2 = 0 m/s. hf1-2 : tng tn tht trong ng t (1-1) n (2-2). Hb : ct p ca bm.

Tnh tng tr lc trong ngChn ng knh trong ca ng ht v ng y bng nhau: dtr = 50 mmTra bng II.15, trang 381, [1] nhm ca ng: = 0,2 mm = 0,0002 m (n mn t)Tng tr lc trong ng ht v ng y

hf1-2 =Trong : lh : chiu di ng ht.Chiu cao ht ca bm: Tra bng II.34, trang 441, [1] hh = 4 m Chn chiu cao ht thp hn 1,5 m loi tr kh nng dao ng trong bm, trnh xm thc: hh = 2,5 m Chn lh = 3 m. l : chiu di ng y, chn l = 3 m. h : tng tn tht cc b trong ng ht. : tng tn tht cc b trong ng y. : h s ma st trong ng ht v ng y. vF : vn tc dng nhp liu trong ng ht v ng y (m/s).

= 0,283 m/s Xc nh h s ma st trong ng ht v ng y :Chun s Reynolds :

= 28710 > 4000 : ch chy riChun s Reynolds ti hn: Regh = 6(dtr/)8/7 = 3301Chun s Reynolds khi bt u xut hin vng nhm: Ren = 220(dtr/)9/8 = 109674V Regh < ReF < Ren ch chy ri ng vi khu vc qu .

p dng cng thc II.64, trang 379, [1]: == 0,031 Xc nh tng tn tht cc b trong ng ht.Ch un cong:Tra bng 9.5, trang 110, [7]: Chn dng ng un cong 90o c bn knh R vi R/d = 2 th u1 (1 ch) = 0,9ng ht c 2 ch un u1 = 0,9 x 2 = 1,8VanTra bng 9.5, trang 94, [7]: Chn van cu vi m hon ton th v1 (1 ci) = 10.ng ht c 1 van cu v1 = 10Nn: h = u1 + v1 = 11,8 Xc nh tng tn tht cc b trong ng y Ch un cong:Tra bng 9.5, trang 110, [7]: Chn dng ng un cong 90o c bn knh R vi R/d = 2 th u2 (1 ch) = 0,9ng y c 4 ch un u2 = 0,9 x 4 = 3,6Van:Tra bng 9.5, trang 94, [1]: Chn van cu vi m hon ton th v2 (1 ci) = 10ng y c 1 van cu v2 = 10Vo bn cao v : cv = 1Nn: = u1 + v1 + cv = 14,6

Vy: hf1-2 == 0,158 mTnh ct p ca bmHb = (z2 z1) + hf1-2 = (15 1) + 0,158 = 14,158 m1. Cng sutChn hiu sut ca bm: b = 0,8.

Cng sut thc t ca bm: Nb == 105,391W = 0,142 Hp.Kt lun: m bo thp hot ng lin tc ta chn 2 bm li tm loi B (theo bng (II.36), trang 444,[1]), c: Nng sut: Qb = 2 m3/h Ct p: Hb = 15 m Cng sut: Nb = 0,142 Hp

Chng VIITNH KINH T

Lng ng truyn nhit cn dng:

Vng = (0,0252 0,0212)xx(16x19) + (0,0162 0,01282)xx56 + (0,0382 0,0322)xx56 + (0,0252 0,0212)xx(2x61) +(0,0252 0,0212)xx48 +(0,0482 0,0422)xx48 +(0,0202 0,0162)xx(4x7) + (0,0162 0,01282)xx(4x7) = 0,117 (m3)mng = VngX18H10T = 0,117 x 7900 = 139 kg

= ( 42,22 36,662)xx(60 + 48) x10-6 x7900 = 925 kg Lng thp X18H10T cn dng:M1 = 56(mmm + mtrn + moh + mchp ) + mthn + 2my(np) = 3374,04 kg Lng thp CT3 cn dng:

M2 = 32mbchghpthn + mbchghpnglng + mbchghpnghi + 4mchn + 4mtai treo + 4mtm lt = 1250 kg S bulng cn dngN = 16 x 24 + 6 x 4 + 56 x 5= 688 ci

STTTn thit bKch thcS lngn giThnh tinGhi ch

1ng truyn nhit925 kg70.0006.4750.000

2X18H10Ttm 4 mm3375 kg70.000236.250.000

3CT31250 kg10.00012.500.000

4BulngM16, M12688 ci5.0003.440.000

5p k t ng2 ci800.0001.600.000

6Nhit kin tr3 ci200.000600.000

7Lu lng k2 ci500.0001.000.000

TNG CHI PH MUA VT LIU, THIT B320.140.000

Tin gia cng ch to v lp t ( ly bng 300 % tin vt t), do gi thnh thp chng l: 1.280.560.000 VN

LI KT

Vi h thng chng ct Clorofom - Benzen dng thp mm chp nh thit k, ta thy bn cnh nhng u im cng cn c nhiu nhc im. Thit b c u im l nng sut v hiu sut kh cao nhng thit b cn rt cng knh, i hi phi c s vn hnh vi chnh xc cao. Bn cnh , khi vn hnh thit b ny ta cng phi ht sc ch n vn an ton lao ng trnh mi ri ro c th xy ra, gy thit hi v ngi v ca. n ny gip sinh vin t tng hp li tt c cc kin thc mt cch linh ng, chnh xc v hp l. ng thi n cng gip sinh vin bc u tip cn vi cng vic thit k, nghin cu su hn v cc kin thc c bn, bit cch tnh ton, a ra cc gii php v la chn cc thng s cn thit ph hp vi yu cu ban u.Do thi gian hn hp nn ni dung n ch cp n vic thit k cc phn chnh quan trng ca thit b, cha ci thin c quy trnh thit b t mc ti u nht.Vi kin thc hn hp cng vi hn ch l t tip xc vi thc t nn vic thit k khng trnh khi thiu st, rt mong s ng gp kin ca qu thy c v cc bn em c thm nhiu kinh nghim hn trong vic thit k sau ny.

TI LIU THAM KHO

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