Trng H BKHNVin KH & CN Mi TrngLI M U

Hin nay, do cc ngnh cng nghip sn xut ra cc sn phm phc v con ngi, ng thi cng sn xut ra mt lng cht thi ln lm hng h cn bng sinh thi, gy nhim mi trng nghim trng.Trong cc loi nhim, nhim khng kh l nh hng trc tip n con ngi v cc loi ng thc vt. V vy, trong nhng nm gn y nhim khng kh t cc ngnh cng nghip nc ta ang l vn quan tm khng ch nh nc m ca ton b x hi bi mc nguy hi ca n ln mc bo ng.CO2 l mt trong nhng cht kh c sn sinh nhiu trong cc ngnh cng nghip v sinh hot. Vic x l CO2 c nhiu phng php khc nhau. V vy, n mn hc vi nhim v tnh ton thit k thp m hp th CO2 l mt phng n gp phn x l kh thi nhim.Trong n ny s i kho st phng n: Dng H2O hp th CO2 trong hn hp kh CO2 Khng kh. Vi ni dung kho st thit k thp m x l CO2 vi nng sut tnh theo hn hp kh vo thit b 10000 m3/h ,nng CO2 trong hn hp kh u vo l 4,5% th tch. p sut lm vic trong thp 15at ,nhit lm vic 30oC.Sau 15 tun tm hiu, tnh ton v nhn c s hng dn nhit tnh, to mi iu kin thun li em hon thnh n mn hc ny ca c gio hng dn Nguyn Lan Phng v cc thy c b mn n 1 v k thut kim sot nhim khng kh nhng do cha c nhiu kinh nghim tnh ton, nn khng th trnh khi nhng sai st, em rt mong nhn c kin ca cc thy, c n sau c kt qu tt hn.Em xin chn thnh cm n!

Phn 1:M UI. C s l thuyt ca qu trnh hp th.1. Khi nim qu trnh hp th:Hp th l qu trnh xy ra khi mt cu t ca pha kh khuch tn vo pha lng do s tip xc gia hai pha kh v lng. Nu qu trnh xy ra ngc li, ngha l cn s truyn vt cht t pha lng vo pha kh, ta c qu trnh nh kh. Nguyn l ca hai qu trnh l ging nhau.Qu trnh hp th tch b mt hay nhiu cht nhim ra khi dng kh thi(pha kh) bng cch x l vi cht lng( pha lng). khi ny hn hp kh c tip xc vi cht lng nhm mc ch ha tan chn la mt hay nhiu cu t ca hn hp kh to nn mt dung dch cc cu t trong cht lng.- Kh c hp th gi l cht b hp th.- Cht lng dng hp th gi l dung mi ( hay cht hp th ).- Kh khng b hp th gi l kh tr.Qu trnh hp th dng :- Thu hi cc cu t qu.- Lm sch khng kh.- Tch hn hp kh thnh tng cu t ring bit.- To thnh mt dung dch sn phmTrong trng hp th nht v th 3, bt buc ta phi tin hnh qu trnh nh sau khi hp th tch cu t c hp th ra khi dung mi. i vi trng hp khc, qu trnh nh l khng cn thit, tr khi phi dng li dung mi (dung mi qu).Qu trnh hp th ph thuc ch yu vo dung mi, do cn chn dung mi theo nhng tnh cht sau y:1. C tnh cht ha tan chn lc, ngha l ch ha tan cu t cn tch v ha tan khng ng k cc cu t cn li, cn nhng cu t khng c kh nng ha tan . y l iu kin quan trng nht.2. nht ca dung mi cng b th tr lc qu trnh cng nh. Tng tc hp th, v tng h s chuyn khi.3. Nhit dung ring b, tit kim nhit nng khi hon nguyn dung mi.4. C nhit si khc xa nhit si ca cu t ha tan, d dng tch ring cc cu t ra khi dung mi qua chng luyn.5. C nhit ng rn thp, trnh hin tng ng rn lm tc thit b.6. Khng to thnh kt ta khi ha tan, trnh tc thit b v d thu hi.7. t bay hi trnh tn tht, r tin,d kim v khng c hi vi ngi.8. Khng c v n mn thit b.Tuy nhin, trong thc t khng c dung mi no t c cc tiu chun nu.V vy khi chn ta phi da vo cc iu kin c th ca sn xut.Trong sn xut, ngi ta dng nhiu loi thit b khc nhau thc hin chng luyn v hp th. Cc thit b thng dng trong sn xut l :- Thit b b mt.- Thit b la mng.- Thit b loi phun.- Thit b loi m (thp m).- Thit b loi a (thp a).2. Cc yu t nh hng n qu trnh hp th:* nh hng ca nhit .Khi cc iu kin khc khng i m nhit thp tng th h s Henry s tng. Kt qu l nh hng ng cn bng chuyn dch v pha trc tung. Nu ng lm vic AB khng i th ng lc trung bnh s gim, s a l thuyt s tng v chiu cao thit b s tng. Thm ch c khi thp khng lm vic c v nhit tng qu s vi yu cu k thut. Nhng nhit tng cng c li l lm cho nht c hai pha kh v lng gim.* nh hng ca p sut.Nu cc iu kin khc gi nguyn m ch tng p sut trong thp th h s cn bng s tng v cn bng s dch chuyn v pha trc honh. Khi ng lm vic AB khng i dn n ng lc trung bnh tng qu trnh truyn khi s tt hn v th s a l thuyt s gim lm chiu cao thp thp hn.Tuy nhin, vic tng p sut thng km theo s tng nhit . Mt khc, s tng p sut cng gy kh khn trong vic ch to v vn hnh ca thp hp th.* Cc yu t khc.Tnh cht ca dung mi, loi thit b, cu to thit b, chnh xc ca dng c o, ch vn hnh thpu nh hng nhiu n hiu sut hp th.II. Gii thiu v thp m :Thp m c s dng cho qu trnh hp th, hp ph, chng luyn v cc qu trnh khc. Thp m hnh tr, bn trong c y m. m c nhiu loi, ph bin nht l cc loi m sau y: - m vng, kch thc t 10100mm. - m ht.kch thc t 20100mm. - m xon, kch thc t 0,31mm, ng knh vng xon c 38mm, chiu di dy khng qu 25mm - m li bng g.Tt c cc loi m trn u c yu cu chung: - Cc b mt ring ln(k hiu , th nguyn m2/m3) - Th tch t do ln( k hiu Vtd, th nguyn m2/m3) - Khi lng ring b. - Bn ha hc.Trong thc t khng c loi m no c ht cc yu cu trn nn ty theo iu kin c th m chn loi m cho ph hp.Thp m c nhng u im:- C b mt tip xc pha ln, hiu sut cao.- Cu to n gin.- Tr lc trong thp khng ln lm.- Gii hn lm vic tng i rng.Tuy nhin, thp m c nhc im l kh lm t u m. Do , nu thp cao qu th cht lng phn b khng u. V vy, ngi ta phi chia tng v mi tng t thm b phn phn phi cht lng. III. Tng quan v cht b hp th:1. Tng quan v Cacbon ixt (CO2)* Tnh cht ha l:Cacbon ixt(CO2)l mt kh khng mu m khi ht phi nng cao to ra v chua trong ming v cm gic nhi mi v c hng(nguy c b ngt th rt ln ). T trng ring ca n 25o C l 1,98kg.m-3, khong 1,5 ln nng hn khng kh. Phn t Cacbon ixt (O=C=O) cha hai lin kt li i v c hnh dng tuyn tnh. N khng c lng cc in ,khng chy. nhit di -78o C, Cacbon ixt ngng t li thnh tinh th mu trng gi l bng kh . Cacbon ixt lng ch c khi to ra di p sut trn 5,1 baz, iu kin p sut kh quyn n chuyn trc tip t cc pha kh sang rn hay ngc li theo mt qu trnh gi l thng hoa.

Bng thuc tnh Cacbon ixt(CO2)

T trng v pha1,98 kg/m3 298 K1,6 g/m3( rn)

ha tan trong nc1,45 kg/m3

im nng chy-57o C ( 216 K-p lc)

im si-78o C (195 K-thng hoa)

pKa6,35

pKb10,33

nht0,07 cP -78o C

Phn t gam44,01 g/mol

* Tnh cht ha hc: Kh Cacbon ixt (CO2) khng chy v khng duy tr s chy nhng mt s kim loi c tnh kh mnh c th chy c trong CO2VD: CO2 +2Mg 2MgO + C - CO2 l oxit axit :1. Tc dng vi nc:Tan tt trong nc ,nht l nhit thp. CO2 l anhydrit ca axit cacbonic.Nc s hp th mt lng nht nh cacbon dioxit, v nhiu hn lng ny khi kh b nn. Khong 1% cacbon dioxit ha tan chuyn thnh axit cacbonic. Axit cacbonic phn ly mt phn thnh cc ion bicacbonat ( HCO3-) v cacbonat( CO3).CO2 + H2O H2CO3 ; ( k1 = 4,01.10-7 ; k2 = 5,2.10-11)2. Tc dng vi baz:CO2 + 2 NaOH Na2CO3+ H2OCO2 + NaOH Na + HCO33. Tc dng vi oxit baz:CO2 + CaO CaCO34. Phn ng vi mui trung ha to mui axit.K2CO3 +CO2 + H2O 2KHCO3K2SO3 +CO2 +H2O KHCO3 +KHSO3Ngoi ra trong CO2 cacbon ha tr 4 nn c th b kh to COCO2 + O2 H2CO3Ngun gc ca kh Cacbon ixt (CO2): CO2 sinh ra do qu trnh t chy nguyn liu.Nguyn nhin liu t chy hon ton CO2 + hi nc +cc cht khc.(bi , NOx)Nguyn nhin liu t chy khng hon ton CO2 + hi nc + CO + cc cht khc.CO2 sinh ra do qu trnh sn xut phn Ure: bao gm kh thi cha CO, CO2, NH3, SO2 ch yu khu kh ha than.CO2 sinh ra t ngnh cng nghip sn xut gang thp: bi , CO, CO2, hydrocacbon, phenol, benzene, SO3, NH3, Tc ng n mi trng & con ngi .Hin nay, theo s liu ca cc nh khoa hc, hm lng CO2 trong kh quyn ngy cng tng. iu ny lm tng hiu ng nh knh ca Tri t v tng nhit trung bnh ca Tri t , gy hiu qu xu n mi trng v con ngi.Hm lng cacbon dioxin trong khng kh trong lnh l khong 0,004% v trong khng kh b thi ra t s th l 4,5%. Khi th trong khng kh vi nng cao ( khong 5% theo th tch )gy ngt th . Nng cao dn n t vong.Phng php hp th :Do ha tan trong nc ca CO2 trong nc cao cn lu lng nc khng ln CO2 + H2O H+ + HCO3-

Phn 2:Tnh ton Cng Ngh1. Chn qu trnh cng ngh :Chn ngun x l kh t hn hp kh CO2khng khChn dung mi l nc v nc l dung mi hp th r tin, d tm, khng n mn thit b.Chn thp hp th l thp m nn dung mi hp th l nc sch khng to ra cn lng lm cn tr dng kh v lng.Thuyt minh quy trnh cng ngh:Dng hn hp kh CO2khng kh c x l s b trc khi c qut thi qua lu lng k o lu lng v i vo thp m thc hin qu trnh hp th. Thp hp th lm vic nghch dng.Dung mi hp th l nc. Nc sch t b cha c bm ln bn cao v . Sau i qua lu lng k o lu lng dng chy v i vo thp hp th, nc c chy t trn xung .Kh CO2 c thi t y thp ln, qu trnh hp th c thc hin.Hn hp kh CO2khng kh sau khi hp th t TCVN c thi pht tn ra mi trng. Dung dch nc sau khi hp th c cho chy vo b cha thc hin qu trnh trung ha v thi ra mi trng2. Tnh ton cng nghCc s liu ban u:Hn hp kh cn tch: CO2 - khng khDung mi:H2OLu lng kh thi vo thp (Nm3/h):10000Nng kh thi vo thp (phn th tch) 4,5%Hiu sut hp th:80%Thit b hp th:Thp mNhit :30oCp sut:15 atTnh cn bng vt cht :Phng trnh cn bng ca dung dch hp th CO2 bng H2O c biu din theo nh lut henry.y* = m.x Trong : m = y*: Nng phn mol ca CO2 trong dng kh iu kin cn bng .x :Nng phn mol kh ha tan trong pha lng.Pt: p sut tng ca h hp th ;m: h s phn b.H: H s henry 30oC: H = 1,4.10-6 mmHg ( II - 139 )

Y* = x = x = 1228,1xTnh theo nng phn mol tng i ta c: y = ; x = Thay vo ta c: = m Y* = = ( phng trnh cn bng )

Nng th tch ban u ca dng kh.y = 4,5% = 0,045Nng u ca pha kh theo t s mol:Y = = = 0,047 ( Kmol CO2/Kmol khng kh )Hiu sut H = 80% = 0,8Nng cui ca pha kh theo t s mol :Yc = Y( 1 0,8) = 0,0094 ( Kmol CO2/Kmol khng kh) Ytb = = = 0,0282 ( Kmol CO2/Kmol khng kh)Lng hn hp kh vo thp hp th l:Gy = = = 446,42 ( Kmol/h )Lng mol kh tr:Gtr = Gy( 1 y ) = 446,42 ( 1 0,045 ) = 426,34 ( Kmol/h )Lng kh CO2 u vo:GCO2 vo = Gy Gtr = 446,42 426,34 = 20,06 ( Kmol/h )Lng kh CO2 u ra:GCO2 ra = Gy ( 1 yH ) = 446,42 ( 1 0,045.0,8 ) = 430,35 ( Kmol/h )Hp th CO2 bng nc, chn dung mi sch khi vo thp nn : X = 0 Vi X : nng u ca pha lng, Kmol CO2/Kmol H2OLng dung mi ti thiu cn thit hp th khi gi thit nng cui ca dung mi t n nng cn bng :Gxmin = Gtr. M Xc = Xcbc = = = 3,66.10-4 (Kmol CO2/Kmol H2O) Gxmin = Gtr. = 426,34. = 43798,56 ( Kmol/h )Lng dung mi thc t thng ln hn dung mi ti thiu l 20% Gx = Gmin.1,2 = 52558,63 ( Kmol/h )Thit lp phng trnh ng lm vic:Gtr(Y Yc) = Gx(X X) Y = 123,29X + 0,0094 Xc thc t = Gtr. = 3,05.10-4 (Kmol CO2/Kmol H2O)T phng trnh ng cn bng v ng lm ta c:XYcbYlv

000.0094

0.000050.00620.0156

0.00010.01240.0217

0.000150.01860.0279

0.00020.0250.0341

0.000250.03170.0394

0.00030.03820.0464

0.0003050.03890.047

th ng cn bng v ng lm vicI. Tnh kch thc thp hp thI.1. Tnh ng knh1. Tnh khi lng ringi vi pha lng = + Trong l khi lng ring trung bnh ca hn hp lng l phn khi lng ca CO2 trong pha lng, l khi lng ring ca CO2 v H2O 30oCTra bng I.5 v th I.1: = 995,68 (kg/m3) = 720 (kg/m3) Ta c xtb = = 1,525.10-4 (Kmol CO2/Kmol H2O)Khi lng ring trung bnh ca pha lng:Mxtb = xtb.MCO2 + ( 1 xtb )MH2OMxtb = 1,525.10-4.44 + ( 1 - 1,525.10-4 ).18 = 18 ( kg/kmol)p dng cng thc:aCO2 = = = 0,37.10-4 (kg/kg) = = = 995,57 (kg/m3)i vi pha khTnh MytbKhi lng mol phn t trung bnh ca hn hp kh l:Mytb = ytb.MCO2 + (1-ytb).Mkkytb = = = 0,0274 Mytb = 29,411 (kg/kmol) = = = = 17,74 (kg/m3)2. Tnh lng kh trung bnh i trong thp:Vytb = (m3/h)V = = = 740,09 (m3/h)Vc = .( 1 + Yc) = = 713,45 (m3/h) Vytb = = 726,77 (m3/h)3. Tnh nht x, y: i vi pha lng x p dng cng thc lg x = xtb.lgCO2 + ( 1 xtb ).lgH2O ( I 84 )Trong :CO2, H2O : nht trung bnh ca pha lng ca CO2 v ca nc 30oCTra bng: CO2 ( 30o ) = 0,16.10-3 ( Ns/m2 ) ( I 101 ) H2O ( 30o ) = 0,8007.10-3 ( Ns/m2 )( I 102 ) lg x = 1,525.lg(0,16.10-3) + ( 1 1,525 ).lg(0,8007.10-3) = -3,097 x = 7,998.10-4 ( Ns/m2 )i vi pha kh y = + Trong : CO2, kk : nht trung bnh ca pha kh ca CO2 v ca khng kh 30oCTra bng: CO2 = 0,024.10-3 ( Ns/m2 )( th I 35 )kk = 0,019.10-3 ( Ns/m2 )( Bng I 114 ) y = = = 1,92.10-5 ( Ns/m2 )4. Tnh ng knh thpD = (m)Tnh wtb l vn tc kh i trong thp (m/s)p dng cng thc:lg[ ] = A 1,75..( (*)Trong :, : lng lng v lng hi trung bnh (kg/s) = = = 259 (kg/s) = = = 3,58 (kg/s)g: gia tc trng trng, g = 9,81 (m/s2)x: nht ca pha lng theo nhit trung bnh, x = 7,998.10-4 ( Ns/m2 )n: nht ca nc 20oC, n = 1,005.10-3 ( Ns/m2 ): l b mt ring ca m,m2/m3Vd: th tch t do ca m,m3/ m3Chn loi m c = 165 m2/m3, Vd = 0,76 m3/ m3Gi tr A = 0,022 i vi kh hp th

Thay vo phng trnh (*) ta c:lg[ = 0,022 1,75. ..( 2lgws = 0,18 3,07 = -2,89 ws = 0,036 (m/s) wtb = ( 0,8 : 0,9 )ws = 0,9.0,036 = 0,0324 (m/s) D = = = 2,8 (m)Kim tra:+ < Ta c: dtd = = = 0,0184 (m) = = 6,6.10-3 < ( tha mn iu kin )Kim tra theo mt ti: U = Vi V1: l lu lng th tch cht lng, m3/hF: thit din thp,m2F = = = 6,15 (m2)V1 = = = 950,16 U = 154,5 (m3/m2h)Mt ti ti hn: Uth = d.b ( hng s b chn b = 0,158 ) Uth = 165.0,158 = 26,07 (m3/m2h) = = 5,93 > 1Vy m thm t rt tt

I.2 Tnh ton chiu cao thpChiu cao thp c xc nh theo phng php s n v chuyn khi:H = hdv.my (m)Trong : H: l chiu cao thp, mHdv: chiu cao mt n v chuyn khi, mmy: s n v chuyn khi Xc nh chiu cao mt n v chuyn khiHdv = h1 + h2 (m)Trong : h1, h2: chiu cao 1 n v chuyn khi ng vi pha kh v pha lngm: gi tr trung bnh ca tg gc nghing ng cn bng Ycb = f(x) vi mt phng ngangTnh h1 v h2:Tnhh1 = .. (m) Trong : a: h s ph thuc vo dng m, vi m vng a = 0,123: h s thm t ca m, do > 1 nn = 1 Re: chun s Renoyd i vi pha khRey = Chun s Pran: Pry = Dy :h s pht tn phn t kh CO2 trong pha kh, m2/sDy = . (m2/s)Trong : T: nhit lm vic tuyt i T = 303KP: p sut lm vic P = 15 atvCO2: th tch mol ca CO2, vCO2 = 44,8 cm3/molvkk: th tch mol ca khng kh, vkk = 29,9 cm3/mol Dy = . = 1,2.10-5 (m2/s) Pry = = = 0,09 h1 = .. = .0,25. 0,092/3 = 0,023 (m)Tnh h2 = 256.. (m)Rex = = = 12,76Prx = Dx: h s khuch tn ca CO2 vo nc 30oCDt = D25.[ 1 + 0,02( t 25 )] (m2/s)Trong D25 l h s khuch tn ca CO2 vo nc 25oCD25 = (m2/s)A: h s i vi cht kh tan trong nc, A = 1B: h s, dung mi l nc, B = 4,7vCO2, vH2O: th tch mol ca CO2 v H2O 25oCvCO2 = 44,8 cm3/mol, vH2O = 18,9 cm3/molH2O: nht ca nc 25oC, H2O = 1,005.10-3 Ns/m2 = 1,005 cP ( bng I-102 ) D25 = 1,466.10-9 (m2/s) Dx = 1,466.10-9.[ 1 + 0,02( 30 -25 )] = 1,6126.10-9 Prx = = 498,18 h2 = 256.. = 256.. = 0,933 (m)Tnh m: T phng trnh ng cn bng ta c m= 128,59 hdv = h1 + h2 = 0,023 + .0,933 = 1,68 (m)Xc nh s n v chuyn khi my:Da vo ng cn bng v ng lm vic v ga tr xc = 3,05.10-4XYcbYY - Ycb1/( Y - Ycb)

000.00940.0094106,38

0.000050.00620.01560.0094106,38

0.00010.01240.02170.0093107,73

0.000150.01860.02790.0093107,73

0.00020.0250.03410.0091109,89

0.000250.03170.03940.0077129,87

0.00030.03820.04640.0082121,95

0.0003050.03890.0470.0081123,46

Cng thc tnh s n v chuyn khi:my = (II.175)Ta xc nh s n v chuyn khi theo phng php tch phn th. Ta v th vi h trc ta v Y. Gi tr ca tch phn bng din tch hnh thang cong gii hn bi th v 2 ng Y = 0,0094, Yc = 0,047

Hnh thang cong

Y1/( Y - Ycb)S

0.0094106,38

0.0156106,380,659

0.0217107,730,653

0.0279107,730,668

0.0341109,890,675

0.0394129,870,635

0.0464121,950,881

0.047123,460,074

my =4,245

H = hdv.my = 1,68.4,245 = 7,1316 (m)Quy chun H = 7,2 (m)y thc cht l chiu cao ca lp m. Ngoi chiu cao cao ca lp m thp cn tnh n chiu cao t mt trn ca m n nh thp v chiu cao t di lp m xung y thp v khong cch gia 2 lp m. Ta c cng thc:Hthp = Hm + Hm lp + Hm m + Hm yHm lp = Hm y = 1,5 mHm m = 0,8 mVy chiu cao thp Hthp = 7,2 + 1,5 + 0,8 + 1,5 = 11 mI.3. Tnh tr lcp dng cng thc:P = Pu + PkTrong : Pu: Tn tht p sut ca m t ti im o pha c tc ca kh bng tc ca kh i qua m kh (N/m2)Pk: Tn tht ca m kh (N/m2)Tn tht p sut ca m kh tnh theo cng thc:

[II-189]Rey = 80,64 > 40 => ch xoy v m l m vng ln xn = = = 6,65Tnh tr lc m kh:

= N/m2Tn tht p sut ca m t tnh theo cng thc:Pu = Pk[ 1 + A]Trong :Gx, Gy: lu lng ca lng v ca kh (kg/h)

: khi lng ring ca lng v ca kh (kg/m3)

: nht ca lng v kh (Ns/m2)A1: h sTra bng: A = 10; m = 0,945; n = 0,525; c = 0,105 Pu = 71,83.[ 1 + 10.] Pu = 7774,95 N/m2 P = Pu + Pk = 71,83 + 7774,95 = 7846,88 N/m2II. TNH TON CC THIT B PH

Trong vic hp th CO2 bng nc s dng thp m cn c cc thit b ph gip cho qu trnh vn chuyn cht lng v cung cp kh vo thp theo ch lm vic ca thp gip vic hp th t c hiu sut mong mun.Trong cc thit b ph th bm v my nn kh l hai thit b quan trng nht.II.1. BMTrong cng nghip, bm ly tm c s dng rng ri v c nhiu loi khc nhau v cu to cng nh cch vn hnh.Bm ly tm c phn loi theo nhiu cch khc nhau nh theo s bc, theo cch t bm, theo iu kin vn chuyn ca cht lng t gung ra thn bm v theo 1 s c trng khcTheo dy chuyn cng ngh trong bi ta chn bm ly tm 1 cp nm ngang.I.1.1. Nguyn tc lm vic ca bm ly tmNguyn tc hot ng: Bm ly tm lm vic theo nguyn tc ly tm. Cht lng c ht v y cng nh nhn thm nng lng l nh tc dng ca lc ly tm khi cnh gung quay. B phn chnh ca bm l cnh gung trn c gn nhng cnh c hnh dng nht nh, bnh gung c t trong thn bm v quay vi tc ln. Cht lng theo ng ht vo tm gung theo phng thng gc ri vo rnh gia cc gung v cng chuyn ng vi gung. Di tc dng ca lc ly tm, p sut ca cht lng tng ln v vng ra vo thn bm, vo ng y theo phng tip tuyn. Khi tm gung to nn p sut thp. Nh p lc mt thong b cha, cht lng dng ln trong ng ht vo bm. Khi gung quay cht lng c ht lin tc, do cht lng c chuyn ng u n. u ng ht c lp li lc ngn khng cho rc v vt rn theo cht lng vo bm gy tc bm v ng ng. Trn ng ht c van mt chiu gi cho cht lng trn ng ng ht khi bm ngng lm vic. Trong ng y c lp van mt chiu trnh cht lng bt ng dn vo bm gy ra va p thy lc lm hng bmI.1.2. Cc thng s c trng ca bmp sut mt thong P1= 9,81.104 N/m2p sut lm vic P = 15 atm=15.1,013.105=1519500 N/m2Gia tc trng trng g = 9,81 m/s 30oC: nc = 995,68 kg/m3nc = 8,007.10-3 Ns/m2p sut ton phn ca bm H(m):p dng phng trnh becnulli ta c

Mt ct 1-1 v 1-1:

Mt ct 1-1 v 2-2

Trong :P1: p sut b mt nc khng gian htP2: p sut khng gian y: khi lng ring ca ncPv: p sut trong ng ht lc vo bmPr: p sut ca cht lng trong ng y lc ra khi bmHh, Hd: chiu cao ng ht v ng yhmh, hmd: tn tht p sut do tr lc gy ra trong ng ht v ng yhmh + hmd =

: p sut ton phn cn thit khc phc sc cn thy lc trong h thng, p sut ton phn ca bm l hiu p sut gia hai giai on ht v y1: vn tc nc b cha, 1=02: vn tc nc khi vo thp hay trong ng yv: vn tc nc khi vo bmr: vn tc nc khi ra khi bmThc t: 2 = r

Xc nh tn tht p sut do tr lc gy ra trn ng ng ht ca bm

hmh =

Trong :

: p sut ng lc hc cn thit to tc cho dng chy ra khi ng

: p sut khc phc tr lc ma st khi chy n nh trong ng thng

: p sut cn thit khc phc tr lc cc b

=

+ Tnh ng knh ng ht: Trong : V l lu lng th tch cht lng i trong ng, m3/sV = = = 0,28 (m3/s)Vn tc cht lng trong ng ht: wh = 1 2 m/s. Chn wh = 2 m/s dh = 0,42 m

+ Tnh : Chun s Re ca cht lng trong ng ht:Re = = = 1045333,33 > 105 Dng ch chy xoy nn h s ma st c tnh: = -2 lg[ (I-380)Trong : nhm tuyt i. Chn vt liu lm ng l thp ni khng hn

m: nhm tng i, c xc nh theo cng thc:

= = = 1,66.10-4 = -2 lg[= -2 lg[= 8.356 = 0,0143+ H s tr lc cc b: Cht lng vo ng thng, u ng ht c lp li chn an bng kim loi = 1 + cVi c = o.. Chn = 0,9 tr lc ca ng c lp li chn an bng kim loi l = 1 + 0,13.1 = 1,13Trn ng ht cn lp 1 van 1 chiu. Theo I-399 => van = 2,1 2,5Chn van = 2,4 => h = 2,4 + 1,13 = 3,53+ Tnh L: chiu cao ht ca bm ly tm 30oCTra bng II-34 (I 441) s ph thuc chiu cao ht ca bm ly tm vo nhit . nhit lm vic T=30oC th chiu cao ht ca bm khong 4m th m bo khng xy ra hin tng xm thc. => L = 4m p lc ton phn cn thit khc phc sc cn thy lc l:

= .( 1 + 3,53 + 0,0143. ) = 9292,065 (N/m2)Tn tht p sut do tr lc gy ra trn ng ng ht ca bm:

hmh = = = 0,95 (m)

Xc nh tn tht p sut do tr lc gy ra trn ng ng y:+ Tnh ng knh ng y: dd = Vi Vn tc cht lng trong ng y wd = 2 (m/s) dd = = 0,42 m+ Tnh : Chun s Re ca cht lng trong ng y:Re = = = 1045333,33 > 105 Dng ch chy xoy nn h s ma st c tnh: = -2 lg[ (I-380)Trong : nhm tuyt i. Chn vt liu lm ng l thp ni khng hn

m: nhm tng i, c xc nh theo cng thc:

= = = 1,66.10-4 = -2 lg[= -2 lg[= 8.356 = 0,0143+ H s tr lc cc b: Theo bng II.16(I-393), i vi thnh nhn Re > 2.105 th b qua tn tht ma st cong=A.B.C

Gc

Chn:

H s tr lc cc b ca ton ng y:

Chn chiu di ng y l 12m p lc ton phn cn thit khc phc sc cn thy lc l:

Pd = .( 1 + 3,3475 + 0,0143. )

= 9471,05 (N/m2)

Hmd = = = 0,97 (m)

Hm = hmd + hmh = 0,97 + 0,95 = 1,92 (m)

Vy tn tht p sut ton phn ca bm:

= 10 + 1,92 + + = 157,54 mCng sut ca bm:Cng sut yu cu trn trc bm:

p dng cng thc: (kW) I-439Trong :: khi lng ring ca nc, kg/m3N: hiu sut ca bm, kWQ: nng sut ca bm(m3/s); Q= 0,28 ,m3/sg: gia tc trng trng(m/s2)H: p sut ton phn ca bm tnh bng mt ct ct cht lng bm: hiu sut ca bm

(I-439)

Vi : hiu sut th tch tnh n s hao ht cht lng chy t vng p sut cao n vng p sut thp v do cht lng r t cc ch h ca bm

: hiu sut thy lc

: hiu sut c khHiu sut ton phn ph thuc vo loi bm v nng sut. Khi thay i ch lm vic ca bm th hiu sut cng thay i

i vi bm ly tm:

Chn: ;;

Vy cng sut yu cu trn trc bm:

= = 561,75 (kW)Cng sut ng c in Ndc(kW)

Vi: : hiu sut truyn ng

: hiu sut ng c in

= = 695,66 (kW)Thng thng ng c in c chn c cng sut ln hn so vi cng sut tnh ton. Chn =1,15 = .Ndc = 1,15.695,66 = 800 (kW)Chn cng sut ng c in l 800 kW

II.2. My nn ly tmThp lm vic iu kin P=15 at, T=30oC.Ta chn my nn ly tmMy nn ly tm l mt loi my nn v y kh nh tc dng ca lc ly tm do bnh gung sinh ra. Dng my nn ly tm khi p sut y t 10-20 at. nn ca my ly tm nh nn my c nhiu cp thng t 3-7 cp nn trong mt cp t 1,2-1,5 khi tc vng nh hn 200m/sng knh bnh gung t 700-1400 mm. Cnh gung c th cong ra hoc hng tmCng ca my nn ly tmp dng cng thc:

(I-465)Trong : PA, PB: p sut trc v sau khi nn, atT1: nhit u ca kh, K

1AABB22

T1=30+273=303Km: ch s a bin, m=1,21,62. Chn m=1,4

R: hng s kh, R

p dng phng trnh becnulli cho mt ct 1-1 v mt ct A-A. chn mt ct 1-1 lm chun

Do ng nm ngang nn ZA=0.

Chn vn tc kh trong b cha tnh: =0

Phng trnh becnulli cho mt ct 2-2 v B-B. Chn mt ct B-B lm chun

Vn tc kh trong ng y:

Vi:P1 = Pa: p sut kh quyn, P1 = 9,81.104 (N/m2)P2: p sut cui ng y, N/m2.

ZB : chiu cao ng y

Khi lng ring ca hn hp kh thi iu kin u vo ca kh

= = 1,18 (kg/m3)hmh, hmd : tr lc trn ng ng ht v ng yXc nh p sut trc khi nn:

Trong :P1 : p sut kh quyn

(nh bm ly tm)

* ng knh ng ht.

[I-369].V: Lu lng th tch u vo ca kh thiV = = = 0,2 (m3/s)Kh trong ng dn P > 1at th = 15 25m3/h II.2(I-370)Chn vn tc ht h = 20m3/h

= = 0,11Chun s ReynolRe = = = 135208,33> 4000 Dng ch chy xoy nn h s ma st c tnh: = -2 lg[ (I-380)Trong : nhm tuyt i. Chn vt liu lm ng l thp ni khng hn

m: nhm tng i, c xc nh theo cng thc:

= = = 6,36.10-4 = -2 lg[= -2 lg[= 7,0246 = 0,02H s tr lc cc b trong ng ht:

Trong :

: H s tr lc ca ng thng, on ng thng c u li ra pha trc c

: h s tr lc ca vanChn van 1 chiu.Theo II.16 [I-399] ta c dh =0,11 m => 2 = 1,5-1,9 chn 2 = 1,6tr lc cc b ca ng ht = 1 + 2 = 1,6 + 0,5 = 2,1Chn chiu di ng ht Hh =Lh =5 (m)

= .( 1 + 2,1 + 0,02. ) = 946,145 (N/m2)

hmh = = = 81,73 (m) PA = P1 - Ph = 9,81.104 946,145 = 97153,855 (N/m2)

Xc nh p sut trc khi sau khi nn:PB = P2 + gLdTrong : P2: p sut cui ng ng yP2 = Ph + Pd + P

ng knh ng .

[I-369].V: Lu lng th tch u vo ca kh thiV = = = 0,2 (m3/s)Kh trong ng dn P > 1at th d = 15 25m3/h II.2(I-370)Chn vn tc ht d = 20m3/h

= = 0,11Chun s ReynolRe = = = 135208,33> 4000 Dng ch chy xoy nn h s ma st c tnh: = -2 lg[ (I-380)Trong : nhm tuyt i. Chn vt liu lm ng l thp ni khng hn

m: nhm tng i, c xc nh theo cng thc:

= = = 6,36.10-4 = -2 lg[= -2 lg[= 7,0246 = 0,02H s tr lc cc b trong ng ht: = 1 + 2 + 3Trong :

: H s tr lc ca ng thng, on ng thng c u li ra pha trc c

: h s tr lc ca van, Chn van 1 chiu.Theo II.16 [I-399] ta c dh =0,11 m => 2 = 1,5-1,9 chn 2 = 1,63 : h s tr lc cc b ca ng ng cong3 = A.B.CGc = 900 => A =1

Chn =2 => B =0,15

= 0,5 => C =1,45

= 10,151,45 = 0,2175H s tr lc cc b trn ng ng y = 1 + 2 + 3 = 1,6 + 0,5 + 0,2175 = 2,3175 (m)Chn chiu di ng y Hd =Ld =5 (m)

= .( 1 + 2,3175 + 0,02. ) = 997,475 (N/m2)

hmd = = = 86,17 (m) p sut cui ng ng y:P2 = 946,145 + 997,475 + 1519500 = 1521443,62 (N/m2)p sut sau khi nn:PB = P2 + gLd = 1521443,62 + 1,18.9,81.5 = 1521501,5 (N/m2)Do

= 281,16.303. 1]= 356223,6 (J/kg)

Cng sut ca my nn Cng sut l thuyt

[I-466]Trong :G: Nng sut ca my nn, kg/sG = = = 3,65 (kg/s)L: Cng nn 1 kg khNlt: Cng sut l thuyt ,kWNlt = = 1300,216 (kW)Cng sut thc t ca my nn

(kw) [I-466]Trong :Ndn: cng sut ng nhit, kwNdn= 1300,216 kW

: hiu sut ng nhit thng bng 0,65 0,75.

Chn = 0,75Ntt = = 1733,621 (kW)Cng sut trn trc ca my nn

Trong :Nhd: cng sut hiu dng,kw

: hiu sut c kh ca my nn. i vi my nn ly tm =0,96 0,98. Chn = 0,98Nhd = = 1769 (kW)II.3 Cng sut ca ng c in

[I-466]Trong :

h s d tr cng sut thng ly bng 1,11,15.Chn =1,1

hiu sut truyn ng ( 0,96 0,99 ) = 0,99

hiu sut ng c in =0,95Ndc = 1,1. = 2069Nh vy ta chn ng c in c cng sut 2100 kW

PHN III: TNH TON C KHI. Chiu dy thn thpThit b lm vic p sut kh quyn, dng hp th kh CO2, thn thp hnh tr, c ch to bng cch un tm vt liu vi kch thc nh sn, hn ghp mi, thp c t thng ngChn thn thp lm bng vt liu X18H10T( C < 0,1%, Cr khong 18%, Ni khong 10%, Ti khng qu 1 1,5%).Chn thp khng g, bn nhit v chu nhit.Thng s gii hn bn ko v gii hn bn chy ca thp loi X18H10T:k = 540.106(N/m2)c = 220.106 (N/m2) (II-310) gin tng i: = 38% nht va p : ak = 2.106J/m2

Chiu dy thn thp hnh tr, lm vic vi p sut bn trong c xc nh bng cng thc:, m Bng XIII.8-II-360

Trong :Dt.: ng knh trong thp, m: h s bn ca thnh thn tr theo phng dc, vi thn hay c l gia c hon ton th = h i vi mi hn c.Vi hn tay bng h quang in, thp khng g ta c: = h = 0,95 [Bng XIII.8-II-362]C : h s b sung do n mn, bo mn v dung sai v chiu dy, m[k]: ng sut cho php ca loi thp X18H10T.P: p sut trong thit b, N/m2.P: p sut trong thit b ng vi s chnh lch p sut ln nht bn trong v ben ngoi thp, N/m2P= Pmt+ PttTrong :Pmt : p sut lm vic, Pmt= 15.1,013.105 = 1519500 N/m2Ptt : p sut thu tnh ca ct cht lngPtt = x.g.H (N/m2) [II- 360 ]Vi:x: khi lng ring ca nc, kg/m3g: gia tc trng trng, g= 9,81 m/s2H: chiu cao ct cht lng, Ht= 10 m=> Ptt = x.g.H = 995,68.9,81.10 = 97676,21 ( N/m2)=> P = Pmt+ Ptt = 1519500+ 97676,21 = 1617176,21 ( N/m2)* Tnh CC ph thuc vo n mn, bo mn v dung sai ca chiu dy. i lng C c xc nh theo cng thc:C = C1+ C2+ C3 ,m [II-363]Trong :C1: h s b sung do n mn. i vi vt liu l thp X18H10T c bn 0,05 0,1mm/nm th ly C1= 1mm.C2: i lng b sung do hao mn, C2= 0C3: i lng b sung do dung sai ca chiu dy,Chon C3 = 0,8mm*Tnh [k]:Theo bng XIII-4 ,ta c th chn gi tr nh nht, tnh theo cng thc sau:

Theo gii hn bn khi ko th ta c:: h s hiu chnh, =1k= 540.106(N/m2)k: h s an ton bn, k= 2,6 [II- 356]

= = 207,692.106

Theo gii hn bn chy:Trong :: h s hiu chnh, =1c = 220.106 (N/m2)c: h s an ton theo gii hn chy, c= 1,5 [II-356]

== 146,667.106Ta ly gi tr b hn trong hai gi tr va tnh c:[k] = 146,667.106( N/m2)Do = = 86,158 > 50 b dy thn thp c tnh theo cng thc sau:

= + C, mC = C1+ C2+ C3 = 1+0+0,8= 1,8 mmS= 16,25 +1,8= 18,05 mmQuy trn S = 18 mmKim tra ng sut ca thnh theo p sut th:

, N/m2Trong :P0: p sut th, c xc nh theo cng thcP0 = Pth+ PttPth: p sut thu lc ly theo bng XIII.5Chn Pth = 1,5P = 1,5 1617175,21 = 2425764,32 (N/m2)Ptt: p sut thu tnh, Ptt= 97676,21 (N/m2)=> P0 = 2425764,32 + 97676,21 = 2523440,53 (N/m2)Vy = = 230,88.106 (N/m2) = = 183,33.106Nh vy l cha tho mn.Chn S = 25mm = = 161,62 .106 (N/m2)Tho mn S = 25 mm

II. Chiu dy np v y thit bNp v y cng l nhng b phn quan trng ca thit b, c ch to cng loi vt liu vi thn thit b l thp X18H10T. Thit b t thng ng.p sut trong l P = 1617176,21 > 0,7.105 N/m2 ngi ta thng dng np elip c g.

p sut tnh ton P = 1617176,21 N/m2Chiu dy ca np v y thit b c xc nh theo cng thc:

Trong :P : p sut trong ca thit b.hb : chiu cao phn li ca y v np, hb= 0,25Dt=0,252,8=0,7 m[k] : ng sut cho php ca thit b , [k] = 146,667.106(N/m2)h : h s bn ca mi hn hng tm, vi mi hn tay bng h quang in, vt liu thp cacbon khng g chn h= 0,95C: i lng b sung, Cn = 1,8 ( mm )k : h s khng th nguyn, chn k = 1V: = = 86,158 > 39 B dy np v y c tnh theo cng thc:

= . + CS = 0,0171+ C ( m ) S = 17,1 + 1,8 = 18,9 mmQuy trn S = 20 mmKim tra ng sut ca y thit b theo p sut th:

Ta c: N/m2 = = 215,78.106 Khng tha mnChn S =25 mm = = 169,426 .106tha mn iu kin. Vy S = 25 mmChiu cao g 50mm ( Bng XIII.11)3. ng knh ng dna. ng knh ng dn khp dng cng thc

[I-369].V: Lu lng th tch ca kh thiV = = = 0,2 (m3/s)Kh trong ng dn P > 1at th = 15 25m3/h II.2(I-370)Chn vn tc kh k = 20m3/h

= = 0,11Quy chun dk = 110 mmVt liu l nha PVCng dn kh c hn vo thit b bn ngoi c lp mt bch. Theo bng XIII.32 trang 434 th on ng ni di 150mm m bo cho kh c phn b u trong thp ta s dng a c l ( l c ng knh 20mm, bc l 20mm, a c l dy 5mm)b. ng knh ng dn cht lng

Trong : V l lu lng th tch cht lng i trong ng, m3/sV = = = 0,28 (m3/s)Vn tc cht lng trong ng : wl = 1 2 m/s. Chn wl = 2 m/s dl = 0,42 mQuy chun dl = 420 mmVt liu l nha PVCng dn lng c hn vo thit b bn ngoi c lp mt bch. Theo bng XIII.32 trang 434 th on ng ni di 150mm4. Mt bcha. Bch ni thit b: ni thit b ( thn, np v y) ta c th dng bch lin kiu I ch to bng thp CT3.Ghi ch:D ng knh ngoi ca bchDb ng knh vng bu-lngDI ng knh trong ca bchDn ng knh quy c ca bchDy ng knh trong ca ngh Chiu dy ca bchdb ng knh bu-lng

Bch lin kiu I

Vi ng knh ca thp Dt = 2800mm v p sut tnh ton P = 1617176,21 N/m2 , tra bng XIII.27( S tay 2-423) ta c cc thng s ca bch nh sau:

py.10-6DtKch thc nih

N/m2DDbDID0Bu lng

dbz

1,6mmcimm

2800303029602900M426845

Kch thc b mt m bt kn dng cho loi bch trn:(Bng XIII.31- T2).

Dtpy.10-6, N/m2

0,11,6

D2D4

280028592835

c. Bch ni ng dn vi cc b phn khc trong thit bChn bch lin bng kim loi en, kiu I

hDnDyD

Dt

D1

db

Tra bng XIII.26, s tay 2-414ng vi ng dn kh c ng knh Dy=110 mm ta c cc thng s sauy.10-6N/m2DyngKiu bchh

DnDDbD1Bu lng

dbz

mmcimm

1,6110108215180158M16826

ng vi ng dn lng c ng knh Dy=420 mm ta c cc thng s sauy.10-6N/m2DyngKiu bchh

DnDDbD1Bu lng

dbz

mmcimm

1,6420426580525490M271638

Tnh cc thit b ph khc ng tho m Chn ng tho m theo bng XIII.Vt liu l thp khng r X18H10Tng tho m c hn vo thn thit b, bn ngoi c lp mt bchTra theo bng chiu di on ng ni l 140mm ng nhp liu:Ta chn kch thc ng nhp liu ging nh ng tho m.S dng knh quan st theo di qu trnh vn hnhng knh ng quan st d = 250 mmChiu di ng ni l = 140mm Li mCc thng s li m

ng knh thp (mm)ng knh li D1 (mm)Chiu rng bc b (mm)m 50 x 50

2800272050

Li m c cu to l 2 na v thp khng r ni li vi nhau bn trn c hn cc l tay c th d dng cm nm khi tho lp. B mt li c cu to bi cc thanh thp khng r c kch thc b x h = 15 x 60

B phn phn phi lngDa no cc thng s bng IX.22 ( II 230 )

ng knh thpa phn phi loi 2

ng knh a Ddng dn cht lng

d x StS lng l (loi2)

mmChic

2800176570 x 5110154

+B dy ng: 5mm+ng knh l: 70 mm+Bc l (khong cch gia cc l) = 110 mmChn , v , tai treoThng thng ngi ta khng t trc tip thit b ln b mt m phi c tai treo hay chn , v thit b thit b c n nh khi vn hnh. Mun chn c chn , v hay tai treo thch hp ta phi tnh trng ti ca thp. Trng ti ca thp:Pthp = Pthn + Py, np + Pcht lng + Pbch + Pm (N)Khi lng thn thit bMth = V. = S.H. = (/4).(Dn2 Dt2). H.Trong :Mth: khi lng ca thn thit b, kgDn, Dt: ng knh ngoi v trong ca thit b, mH: chiu cao ca thp, m: khi lng ring ca thp, = 7,9.103 kg/m3 Mthn = (/4).[(Dt + 2.S)2 Dt2]. H.= (/4).[ ( 2,8 + 20,025)2 2,8 ]107,9.103 = 6134,84(kg)Khi lng ca y v np thpKhi lng ca np: S= 25 mm; Dt=2800mm Tra bng XIII-11( s tay 2- 384) ta c chiu cao g h= 50 mm v Mnp=1789 kgKhi lng y: S= 25 mm; Dt=2800 mm Tra bng XIII-11( s tay 2- 384) ta c chiu cao g h= 50 mm v Mnp=1789 kgVy Mnp- y= 3578 (kg)Khi lng ca mm l m vng Rasig ln xn: m bng s kch thc 30303,5Tra bng thng s k thut IX.8( s tay 2-193)m = 570 kg/m3Mm = Hm .(/4).D2.m= 7,2.(/4)2,82570= 25270,47 (Kg)Khi lng ca cht lng:

Trong : : khi lng ring cht lng 30oCD: ng knh thp, mH: chiu cao thp trc khi b nc chon ht, mVy: Ml = 995,68..10 = 61309,21 (kg)Khi lng bchBch ni thn thit b:

p dng cng thc: Trong :

: khi lng ring ca thp lm bch(CT3), = 7,85.103(kg/m3)Db: ng knh vng bulong, Db= 2960 mm = 2,96 mDI: ng knh trong ca bch, DI= 2900 mm= 2,9 mh: chiu di ca bch, h=40 mm= 0,04 m Mbt = . (2,962 2,92).0,04. 7,85.103 = 86,71 (kg)C 4 bch loi 1: Mb1 = 4.86,71 = 346,84 (kg)Bch ni ng dn kh:Mbk = = . (0,182 0,1582).0,026. 7,85.103 = 1,192 (kg)C 4 bch dn kh: Mbk = 4.1,192 = 4,768 (kg)Bch ni ng dn lng:Mbl = = . (0,5252 0,4902).0,038. 7,85.103 = 5,7 (kg)C 4 bch dn lng: Mbl = 4.5,7 = 22,8 (kg)

Bch ni ng ng vi l y v npDo Mb = Mbt+Mbk+Mbl = 346,84 + 4,768 + 22,8 = 374,41 (kg)Trng ti ca thp l:Khi lng ca ton thpMtt= Mm + Mbch + Mcht lng + Mthn + Mnp,y=25270,47 +374,41 + 61309,21 + 6134,84 + 3578 = 96666,93( kg )Trng ti ca thp:P = Mtt g = 96666,93 9,81 = 948302,58(N )Ta chn chn gm 4 chn Ti trng trn mt chn: 948302,58/4 = 237075,645 (N)Theo bng XIII.35 trang 437- S tay 2Cc thng s v chn c trnh by bng sau:LBB1B2HhSld

mm

90065078010601500110060500100

PHN KT LUN:Trn y ton b phn thit k h thng thp m hp th cu t CO2 vi dung mi l nc, trong qu trnh tnh ton do rt di v c nhiu cng thc phc tp cng nh phi lm trn s hoc tra s liu t cc ti liu tham kho nn bn thit k khng trnh khi sai st trong tnh ton, rt mong c s gp v thng cm ca thy c v bn c.Tuy nhin bn thit k gip em h thng li ton b kin thc hc t nhng mn c s ca ngnh mi trng. ng thi qua vic thit k cng gip em c c ci nhn hon chnh v s ca mt h thng x l kh thi.

TI LIU THAM KHO1. TS Trn Xoa, PGS.TS Nguyn Trng Khung, TS Phm Xun Ton- S tay qu trnh v thit b cng ngh ha tp 1- NXB KHKT H Ni2. TS Trn Xoa, PGS.TS Nguyn Trng Khung, TS Phm Xun Ton- S tay qu trnh v thit b cng ngh ha tp 2- NXB KHKT H Ni3. GS.TSKH Nguyn Bin - C s cc qu trnh v thit b cng ngh ha cht tp 4- NXB KHKT H Ni

MC LC

Phn 1: M U 2I. C s l thuyt ca qu trnh hp th 2 1. Khi nim qu trnh hp th ..2 2. Cn bng vt cht v ng lm vic ca thp ..3 3. Cc yu t nh hng n qu trnh hp th ..6 II. Gii thiu v thp m ............6 III. Tng quan v cht b hp th ....................7 1. Tng quan v Cacbon ixt(CO2) ..7 2. Quy Trnh cng ngh...10Phn 2: TNH TON CNG NGH ..11 I. Cc thng s ban u...11 II. Tnh cn bng nhit lng...15 III. Tnh kch thc thp hp th..171. Tnh khi lng ring:172. Lng kh trung bnh i trong thp:19

3. nht :204. Tnh ng knh thp21II.2 Tnh ton chiu cao thp:22II.3. Tr lc lp m28PHN 2: TNH TON CC THIT B PH29I. BM29I.1. Nguyn tc lm vic ca bm ly tm29I.2. Cc thng s c trng ca bm30II. My nn kh37II. Cng ca my nn ly tm38II.2 Cng sut ca my nn44II.3 Cng sut ca ng c in45PHN III: TNH TON C KH46

GVHD: Nguyn Lan PhngSinh vin: Nguyn Trng QuytPage 53