thuyet minh tblbd
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Ti liu tham kho
n tt nghip Thit b lc bi tnh in
Mc lc
Phn m u .
Phn I : Gii thiu cng ngh sn xut kh.
I . Lu trnh cng ngh sn xut kh .......................................................
II. cc cng v sn xut chnh ...............................................................
II.1. Cng v to kh .............................................................................
II.1.1. Nguyn l qu trnh kh ho than m .........................................
II.1.2. Lu trnh cng ngh...................................................................
1. Vn chuyn v cung cp nguyn liu ...........................................
2. Thuyt minh lu trnh...................................................................
3. iu chnh t l H2/N2 trong kh than m.....................................
4. Cc thit b chnh trong dy truyn .............................................
Phn II : Gii thiu v thit b lc bi in ...........................................
I. Vi nt v thit b lc bi in ti phn xng ..................................
I.1. C s chn thit b lc bi in ......................................................
I.2. So snh vi cc thit b khc............................................................
I.3. Nguyn l qu trnh .........................................................................
II.C s l thuyt ca qu trnh..............................................................
II.1.C s ..............................................................................................
II.2. Qu trnh vt l sy ra trong thit b .............................................
II.3. Dng in trong cht kh ...............................................................
II.4. Qung sng......................................................................................
III. Cc yu t nh hng trong thit b lc bi in .............................
IV. Cu to v nguyn l hot ng.......................................................
Phn III : Tnh cng ngh .....................................................................
I. Th tch lm vic ca thit b ................................................
II. S ng in cc .....................................................................
III. Vn tc chuyn ng ca kh trong ng ...............................
IV. in p ti hn v cng dng in ................................
V. Cng sut tiu th ca thit b ..............................................
VI. Cc thng s hnh hc ..........................................................
VII. Nhng s c v bin php s l............................................
Phn IV : Tnh c kh..............................................................................
I.Thn thp .............................................................................................
I.1. Tnh chiu dy v ............................................................................
I.2. Kim tra bn v ...............................................................................
II. Tnh chn y v np thit b ...........................................................
III. Xc nh cc ca trn thit b .........................................................
IV . Chn dng bch v m ................................................................
V . ng sut do trng lng v m men gi .........................................
VI . V ng .............................................................................................
Phn kt lun ..........................................................................................
Phn m u
Trong cng nghp ho chtvic s l bi trong kh l 1 vn ht sc quan trng , vic s l gim c nh hng c hi ca kh i vi mi trng , gim c s ph hu ng k i vi cc thit b ho cht . Chnh v th thit b lc bi in rt cn thit cho cc nh my ho cht .
Trong bn n ny em dc giao nhim v thit k thit b lc bi tnh in vi lu lng kh 28.800 m3/h , v hiu sut thu bi t 98% .
Thit b lc bi in c s dng rt rng di do tnh cht lm vic ca thit b lm vic vi hiu st thu bi rt cao vi nhng ht bi c ng knh nh .
Bn n ny c cp n .
Phn 1 : Tng quan v cng ngh to kh .
Phn 2 : Gii thiu thit b lc bi in .
Phn 3 : Tnh cng ngh .
Phn 4 : Tnh c kh .
Bn n ny c hon thnh vi s gip nhit tnh ca cc thy c gio trong b mn My v thit b ho cht , c bit em xin chn trng cm n K s T Hng c tn tnh gip em hon thnh n ny .
Sinh vin thit k
Nguyn V Thnh
phn i :
gii thiu cng ngh sn xut kh than
I. Lu trnh cng ngh ca xng to kh:
- Nhim v chnh ca phn xng to kh l ch kh lm nguyn liu cho qu trnh tng hp NH3 v thu hi kh CO2 tng hp Urea.
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S lu trnh cng ngh
- Than cc t kho than c a ln bunke c dung tch 50 tn/l ri t than c a vo l kh ho UGI theo chu k. Hin ti xng c 10 l to kh. Than c chia ra lm 3 loi: >12; 20 ( 40 v >40 mi loi dng cho tng l khc nhau c th p dng cng ngh mt cch thch hp cho tng c ht than. Than cc, hi nc 5 at, khng kh phi hp phn ng trong l kh ho nhit T ( 1100oC to thnh hn hp kh than m c thnh phn gm cc kh CO, C, H2S, H2, N2, CH4. Kh than m ra khi l kh ho c i qua l t tr nhit nhm gia nhit cho hn hp hi nc v khng kh giai on ch kh thi xung. Kh than m sau i qua ni hi nhit tha nhm thu hi nhit lng sn xut hi nc, lm ngui kh v tch mt phn bi. Sau kh than m i qua van ba ng vo thu phong ti ra lm lnh v ra s b kh than trc khi vo thp ra. Sau khi i qua cc thp ra kh than c a vao kt kh. Kt kh ngoi tc dng cha kh n cn c tc dng trn kh than ca cc l vi nhau. Sau , kh than a qua lc bi in tch nhng ht bi nh cn li. Kh than m t lc bi in c a qua qut tng p nng p t > 100 mm H2O ln n 900 ( 1800 mm H2O p lc ny cn thit thng tr lc ca thp hp th bng keo tananh. Sau , kh than c a vo thp hp th H2S thp p bng dung dch tananh. Dch tananh sau khi hp th c ti sinh. Dch tananh sau ti sinh c bm a i hp th tun hon cn lu hunh c tch ra di dng bt qua mng dn i thu hi c lu hunh rn. Kh than m sau khi hp th H2S thp p c a vo on I ca my nn 6 on. Sau khi ra on III ca my nn kh than m c p sut l 20 at i qua thit b trao i nhit trung gian ri vo l chuyn ho CO thnh CO2. Hn hp kh than sau chuyn ho CO cn mt lng kh H2S 12 mm; 20 ( 40 mm v >40 mm mi loi dng cho tng l khc nhau c th p dng cng ngh mt cch thch hp cho tng c ht than.
2. Thuyt minh lu trnh:
Khi to kh than m bng phng php gin on, cn phi thi xen k khng kh v hi nc vo l kh ho. Trong khong t ln thi khng kh vo ln trc n thi khng kh vo l ln sau c gi l mt vng tun hon lm vic. Mt tun hon kh nh vy mt khong 3 pht. Trong thc t sn xut, m bo an ton v nng cao sn lng kh than th qu trnh ch kh khng th n c tin hnh thi gi v tun hon ch kh, m phi tri qua mt tun hon lm vic gm 5 giai on:
- Giai on thi gi.
- Giai on ch kh thi ln ln 1.
- Giai on ch kh thi xung.
- Giai on ch kh thi ln ln 2.
- Giai on thi sch.
+ Giai on thi gi:
Khng kh ngoi tri c qut ht a vo ng khng kh chung qua van tm chn cho qut cp khng kh vo y l kh ho, qua tng than c nhit cao xy ra phn ng to nhit. Kh thi gi i ra t nh l kh ho theo ng xin sang l t. Trong kh thi gi cn mt lng CO, ngi ta a thm mt lng khng kh ln 2 vo l t chy kh CO trong l nhm mc ch thu hi nhit v trnh nhim mi trng. Nhit c tch lu trong gch tch nhit ca l t, kh tip tc i sang l hi nhit tha, kh i trong ng chm ca ca l hi nhit tha truyn nhit cho nc i ngoi ng sn xut hi nc 5at. Sau khi i ra t y ni hi nhit tha qua van ng khi phng khng ra ngoi.
Giai on thi gi ko di khong 42s. Mc ch l tng nhit cho tng than. Khng kh c thi vo y l t chy nguyn liu, nhit lng to ra c gi trong tng nhin liu cung cp nhit cho phn ng gia ccbon v hi nc trong giai on ch kh.Thnh phn chnh ca kh thi gi bao gm: CO2 = 15,5% ; O2 = 0,3% ; CO = 8%.
+ Giai on ch kh thi ln ln 1:
Hn hp hi nc 5at v khng kh qua van hi nc v khng kh vo ng ng thi ln, vo y l kh ho qua tng than (sau bc thi gi, tng nhin liu c nhit rt cao ) xy ra phn ng ch kh. Kh than m sinh thnh qua ca nh l sang y l t ri sang l hi nhit tha. Kh ngui ra t y ni hi nhit tha, lc ny van ng khi ng, kh qua van 3 ng vo thu phong ti ra. Sau kh ra ng ng kh than chung qua thp ra lm lnh. Kh than i t di ln, nc tun hon i t trn xung lm lnh kh than v gim mt phn bi. Cui cng kh tip tc qua van thu phong v kt kh ri qua lc in tip tc sang cng on tinh ch kh. Bi lng trong l t i vo thu phong y l. Ti y, bi b dng nc cun theo ra khi thit b thu phong cng vi bn thi thu phong ti ra cng nh thp ra tp trung vo 2 b lng ri a v h mi trng.
Giai on ch kh thi ln ln mt ko di khong 49s . Mc ch ca vic cho khng kh v hi nc giai on ch kh l thu c kh than m t l (H2+CO)/N2 = 3,1 - 3,2. Ngoi ra phn ng gia khng kh vi C cng cn cung cp mt lng nhit cho phn ng gia hi nc v ccbon. Thnh phn ca hn hp kh trong giai on ny l: CO2 = 7,5% ; O2 = 0,3% ; CO = 34,8% ; H2 = 52,6% ; N2 = 4,8%.
+ Giai on ch kh thi xung:
Hn hp hi nc v khng kh qua b van hi nc v khng kh thi xung i vo nh l t ly nhit, theo ng xin sang nh kh ho qua tng than c nhit cao. Kh than m sinh thnh i ra t y l qua van 3 ng v tr thi xung, qua ti ra ra ng ng kh than chung v sang thp ra ri v kt kh.
Giai on ch kh thi xung ko di 69s. Thnh phn kh bao gm: CO2 = 5,5% ; O2 = 0,3% ; CO = 34,6% ; H2 = 56,5% ; N2 = 3,1%. Trong giai on ch kh, nu nh hi nc v khng kh ch i qua tng nhin liu t di ln trn th do nhit cht kh ho tng i thp v phn ng kh ho thu lng nhit ln, s lm nhit nhin liu phn di tng kh ho tt xung, thm ch c th lm tt l. V vy tng kh ho s mng i, cn phn trn tng nhin liu khng ngng c kh than nng gia nhit lm cho tng kh ho di chuyn ln pha trn do nhit phn trn l kh than s tng cao, lng tn tht nhit do kh than mang theo ra ngoi s rt ln v gy nn kt tng, bm vo vch l. trnh hin tng k trn th sau giai on ch kh thi ln cn phi thay i hng i ca dng kh, tc l cho hi nc v khng kh vo l t tn dng nhit l t ri a vo nh l pht sinh. Kh than m to ra ra khi t pha y l.
+ Giai on ch kh thi ln ln 2:
Cng ging nh chu trnh thi ln ln 1. Giai on ny ko di khong 15s. Sau giai on ch kh thi xung, nhit tng nhin liu gim i nhiu nn cn phi thi khng kh li nng nhit l. Nhng ngay sau khi kt thc ch kh thi xung th phn di ca l v trong tng nhin liu cn st li kh nhiu kh than m, nu thi gi ngay khng kh v than m s gp nhau phn di l d dng gy ra s c n. V vy, sau giai on ch kh thi xung th hi nc v khng kh li phi thay i phng hng mt ln na: Thi qua tng nhin liu theo chiu t di ln tin hnh ch kh ln 2, thi sch kh than m y l nhm to iu kin an ton cho bc thi khng kh vo l. Gai on thi kh ln 2 cng sinh ra kh than m, nhng mc ch chnh vn l trnh n.
+ Giai on thi sch:
Cng ging nh chu trnh thi gi nhng khc l van ng khi th ng cn van 3 ng th m. Kh qua thp ra ri v kt kh. Giai on ny chim khong 5s. Mc ch ca n nh sau: Sau khi thi ln ln 2 trong khong trng trn nh l v trong cc ng ng c cha y kh than m, nu chuyn sang giai on thi gi ri sau cho phng khng th khng nhng gy tn tht kh than m m cn dn ti hin tng nguy him l khi kh than m thi ti ming ng khi trn ln vi khng kh s bng n nu gp phi tia la. V vy, trc khi chuyn sang giai on thi gi ta phi thi kh vo phn y l. Kh than kh sinh ra v kh than m cn st li cng c a vo kt kh tng lng thu hi.
Thnh phn (%) kh than m thu c nh sau:
H2 CO CO2 N2 CH4 O2
37,0 33,3 6,6 22,4 0,3 0,3
3. iu chnh t l H2/N2 trong kh than m:
Yu cu i vi l kh ho phi t c sn lng kh cao, cht lng kh tt. Gi cht lng kh tt ch yu l t l H2/N2 phi ph hp vi yu cu v t l ny cn phi n nh. Ni chung phng php iu chnh t l H2/N2 c p dng ch yu l tng gim khng kh (thm N2) , cn tng gim thi gian thi sch v ch kh than t l b tr. Phng php b tr c u im l hiu qu nhanh,nhng thng gy ra thnh phn kh khng n nh, dao ng ln.
4. Cc thit b chnh trong dy chuyn:
a. L kh ho UGI :
( = 2.745mm & 3.000mm.
L bao gm 4 phn chnh nh sau:
+ Ni hi v kp :
H = 2961mm; FTN = 13m2; Lng nc cha: 12m3.
Tc dng: Chng hin tng nhit tng nhin liu qu cao lm cho x chy ra bm dnh vo thnh l gy hin tng treo liu, ng thi sn xut ra hi nc thp p 0,5 - 0,8 t
+ M gi: Bng gang, cao: 1400mm
Tc dng: Phn phi kh u cho tng than
ng knh vnh ln nht ( = 1200 mm, gn y tng ln ( = 1400 mm. M gi tng trn cng khoan 20 l ( = 20 mm. Din tch thng gi: 0,9m2
+ Ghi l: Tc quay: 0,6 - 1 v/f
+ Mm tro: ton b trng lng tng tro x v tng nhin liu. trn mm tro c nh 4 gt tro hnh cong li lim gi l g y tro
Tc dng: y tro x ra khi mm tro
b. L t :
H =10.880mm. ( = 3.354mm.
Tc dng: -Tr nhit nhm gia nhit cho hn hp hi nc v khng kh giai on ch kh thi xung.
- Loi b mt phn bi trong kh thi gi v kh than m thi ln.
Cu to: - Chp trn, chp di v phn hnh tr trn.
- V l lm bng thp cun dy 8 mm, phn hnh tr trn c xp gch chu nhit, phn chp v hnh tr di c xy lt bng gch chu la.
L t khng ch nhit : Tmax = 850OC; Tmin = 600OC.
c. Ni hi nhit tha :
Tc dng: -Thu hi nhit lng ca kh thi gi v kh than m thi ln sn xut hi nc.
Lm ngui kh thi gi truc khi phng khng, lm ngui kh than m thi ln trc khi a vo thu phong ti ra.
- Tch mt lng bi trong kh thi gi v kh than m giai on ch kh thi ln.
Cu to: Hnh tr trn, 2 mt trn v di c gn hai mt sng lng ng chm.
(ng= 76 x 3 , (l= 2300
Hng= 6000
Hl= 11714 , FTN = 480m2
d. Thu phong ti ra:
Tc dng:- Khng cho kh than m sau thu phong ti ra i ngc tr li l kh ho gy n
- Lm lnh v ra s b kh than ra l trc khi vo thp ra
Cu to: - Trn hnh trn, di hnh chp nn
- ng ng kh vo cm su ngp trong nc 70 mm
- ng knh ti ra: 3000mm
- Dung tch: 15m3
PLV = 700mmH2O TLV = 80oC
e. Kt kh :
Tc dng: - Cha Kh than m
- Trn kh than cc l vi nhau
- C tc dng cn bng ph ti h thng sn xut, gip cc cng v sau n nh ph ti mt cch lin tc
Cu to : Gm 2 tng hnh tr trn, 1 chp chung
+Tng di cha nc c ng knh:
( = 27928 x 14;
H = 11312
+Tng Trn cha kh c ng knh:
( = 27016 x 6;
H = 95
+ Chp chung: Dc = 26100 x 4; H = 9585
+ Chp an ton, van phng khng
phn II
Gii thiu v thit b lc inI. vi nt v thit b lc bi in ti phn xng to kh :
I.1. C s chn thit b lc bi in :
Phn xng lin hp c 2 b lc bi in kiu ng 5,8m2. Th kh vo don 1 my nn l kh than m, v trong kh than m c hm lng bi nht nh d gy ra mi mn, xi lanh, xc mng v tc van ca vo, nh hng n chu k vn hnh v tui th ca my nn. Chnh v vy phi tin hnh kh sch bi bn trc khi vo my nn. Do vy trnh nhng hu qu trn ta cn phi s dng b lc in m bo cht lng kh than p ng c yu cu sn xut.
Trong qu trnh sn xut phi phng hm lng O2 trong kh than qu cao gy ra nguy him v n lc bi in nn phi lp thm b cnh bo t ng v hm lng.
II..2 So snh thit b lc b in vi cc thit b khc:
- Phng php lc kh bng c hc, phng lng bi th cng knh km hiu qu v khng th tch c cc phn c phn t c kch thc b. Dng xyclon c gn gng hn nhng tiu tn nng lng v bn thn thit b nhanh b bo mn vi yu cu th kh vi hm lng bi t, chnh v vy nh my dng phng php lm sch kh bng lc in vi phng php ny th kh thu c vi sch 98% nng lng tiu hao nh.
II.3. Nguyn l qu trnh:
Di tc dng ca in th cc phn t kh phn ly thnh ion , cc in t tch in dng v cc in t tch in m. Cc ion ny di tc dng ca lc in trng bt u chuyn ng v cc in cc tri du. Do tc chuyn ng v ng nng ca cc ion v in t v cc in cc s tng ln khi tng in th in trng.
Khi tc ca ion v in t bt u vt gi tr ti hn nh lc sn c, chng c kh nng ln lt phn ly cc phn t trung ho gp trn ng
V vy xy ra s ion ho ton b th kh gia cc in cc. ng thi vi s to thnh mt s ng k cc ion trong trng hp ng nht gia cc bn song song th cng in tng nhanh v phng ra tia la in.
Cng vi s ion ho bng va chm cng cn nhn thy s chuyn ng mnh lit ca khng kh do cc phn t khng kh nhn c xung lng t cc ion chuyn ng theo hng xc nh. Trong k thut lc kh bng in ta thc hin ion ho bng cch ch to mt in th cao trn hai in cc. Khi ion ho bng cch ny ch nn mt lp khng kh b xuyn thng mt phn khong cch gia hai in cc, cn mt phn lp khng kh b xuyn thng c cng dng nh cht cch in chng s ngn mch do tia la in hay h quan sinh ra gia cc in cc.
Thc t to c lp khng kh ny bng cch chn dng in cc v khong cch gia cc in cc tng ng vi cc in th xc nh. Trong trng hp ny dng cc in cc c hai dng bn song song khng c li , bi v 1 im bt k ca trng gia cc in cc c in th nh nhau, ngha l trng ng nht.
Khi hiu in th gia 2 in cc phng ln n mc t gi tr in th xuyn thng, do tnh ng nht ca trng, ton b lp kh gia cc in cc s b xuyn thng v phng tia la in trong trng hp ny khng xy ra s ion ho khng kh.
Gia cc in cc c dng hnh tr ng tm hay hnh tr v bn phng th to thnh lung khng ng nht. Gn dy dn th in th ca trng ln n ni cc ion c kh nng ion ho cc phn t trung ho, nhng cng xa dy dn th in th trng gim xung v do tc chuyn ng ca cc ion gim xung n mc khng xy ra c s ion ho bng va chm.
Du hiu bn ngoi ca s tch in ion l hin tng pht sng nht quanh b mt dy dn qung sng Khc vi vng to thnh ion c hai u, hin tng ny gi l tch in qung.
Khi lc kh bng in ta ch dng qung m, qung ny t hn qung dng nhng th hiu cao hn.
C cu lng bi trong my lc in rt phc tp ch c mt mt phn bi rt nh hay m sau khi ri trong min qung sng s bm trn dy dn c qung sng. Cn phn ln cc phn l lng trong khng kh sa khi tch in m s chuyn ng v pha cc in cc lng v cui cng s b kh trn in cc ny khi lp bi dn in tt bm trn in cc lng, chng s tch in cng du v b y vo khng kh phn bi ny c th ly ra khi my lc bi in. Nu bi khng dn in th ht v in cc do lc ca in trng v to thnh 1 lp cht trn in cc cc lp bi tch in m bm trn in cc s y cc phn t cng du gn n, ngha l tc dng ngc li vi in trng chnh. in th trong cc l hng ca lp bi bm trn in cc c th vt qu in th ti hn v gy ra qung sng, khng kh trong cc l hng to thnh cc ion dng, cc ion ny s trung ho cc phn t bi in tch m. Hin tng ny gi l qung sng nghch v lm gim t ngt hiu qu lng bi. loi tr nh hng c hi ca bi bm trn in cc, ngi ta lc in cc hoc tng tnh dn in ca bi bng cch tng m cho bi nh phun nc vo dng kh nng trc khi cho kh vo thit b lc in.
II. C s l thuyt ca qu trnh :
II. 1. C s:
Ta ly mt ngun in mc vo 2 bn kim loi. t song song vi nhau. Cch nhau mt lp khng kh. Trn mt t mt in th. Ta thy kim in th khng lch i chng t mch khng c dng in chy qua, v vy khng kh nm gia 2 bn kim loi khng dn in.
Nu ta tng hiu s in th gia 2 in cc th thy kim in k lch i, iu chng t trong mch c dng in chy qua v khng kh gia 2 bn b ion ho.
S ion ho khng kh c th xy ra theo 2 trng hp :
1.T ion ho: xy ra khi tng hiu s in th gia 2 in cc n mt gi tr vt qu gi tr ng vi hng s in mi ca khng kh.
2.Thit b ion ho: xy ra di tc dng ca cc nhn t nh tia phng x tia rnghen...
Hin nay trong k thut lc in ngi ta thng s dng
3.Phng php t ion ho:
Di tc dng ca in th, cc phn t kh phn chia thnh cc ion v cc in t t do, cc ion v cc in t ny chuyn ng n cc in cc tri du. Trn ng i n cc in cc, cc ion v in t s va p vo cc phn t kh trung ho, cc ht bi l lng, ion ho lun cc phn t . ( Ngi ta ng dng hin tng ny lm sch h thng kh khng ng nht).
Tc chuyn ng ng nng ca cc ion v in t cng tng khi in th gia 2 in cc cng tng. Nhng nu ta tng hiu s in th gia hai in cc n mt gi tr ti hn no gi l in th xuyn thng kh.
Khi c cng dng in tng rt nhanh, gia 2 bn cc xut hin tia la in, hin tng ny gi l t phng in. S xut hin tia la in l do gia 2 in cc song song c mt in trng ng nht. Cho nn khi c in th cao th s ion v in t t do to thnh ng thi rt ln( Kh nng ion ho cc im gia hai cc l nh nhau) do xut hin tia la in gia 2 cc. Hin tng ny ging nh l chp mch v sau khng tip tc ion ho cc phn t kh na. v th trong lc in ngi ta ht sc trnh xut hin tia la in bng cch lm cho in trng gia 2 cc khng ng nht vi nhau: mt in cc l tm phng hoc ng, cn in cc kia l dy dn.
in cc dy l m
in cc ng l dng
Hnh v
T hnh v ta thy rng: chung quanh in cc dy mt n v ng sc rt ln, ngha l in trng mnh, do kh nng ion ho mnh, cng xa in cc dy dn in th ca in trng cng gim, nn kh nng ion ho do va p cng yu. V chung quanh in cc m c kh nng ion ho mnh, nn xut hin qung sng v nghe thy ting n lp bp nh ting dn nt, v th ngi ta gi l in cc qung. in cc tm v ng l in cc dng bi s lng trn in cc , nn ngi ta gi l in cc lng.
Trong thc t lm sch bi nh tnh in cn phi c 2 iu kin:
1. Kh cha bi phi t gi tr 108 ion trn 1 cm3 kh.
2. Cn c in trng mt chiu mnh.
II. 2. Qu trnh vt l xy ra trong thit b lc in:
Qu trnh lm sch kh nh lc in trng c th chia lm ba giai on:
1. Tch in cho cc ht bi l lng trong lng kh
2. Dch chuyn cc ht c tch in n cc in cc di tc dng ca in trng.
3. Lng trn cc in cc cc ht mang in tch chuyn ng
Nh vy khi cho dng kh c cha cc ht bi i qua trng in cc th mi ht bi s chu tc dng ca cc lc sau:
+. Lc li cun ca dng kh chuyn ng: Theo kt qu nghin cu ca th tc li cun cc ht bi ca dng kh thng vo khong V = 0.5(3m/s.
+ Lc trng trng:
Vn tc ca cc ht chuyn ng di s tc dng ca trng trng l khng ln lm , m thi gian lu li ca dng kh trong thit b lc in khong 10- 15 (s) , trong cng thi ht c ng knh nh ri xung di tc dng ca trng lc l khng ng k nn b qua nh hng ca lc trng trng trng ln chuyn ng ca ht .
+ Lc cm ng t ca cc in tch c cm ng:
Cng thc tnh :
7
1
1, L pht sinh.
2, L t.
3, Van ba ng.
4, TB thu phong.
5, Ni hi nhit tha.
6,
ng khi.
7, Thp ra.
8, Kt kh.
9, TB lc bi in.
s lu trnh sn xut kh than
7
8
9
9
4
3
2
5
6
Nh ta bit trong thit b lc in, in th in trng nhng im khc nhau th khng bng nhau v in th ch thay i ln nht l vng in cc qung nn ti thy r tc dng ca lc cm ng t ln ht . Cn i b phn vng tch in bn ngoi cng in trng thay i rt t v thc t coi nh khng i (
0
dx
dE
). Do nh hng ca Fu ln ht vng ngoi l khng ng k.
+ Lc tc dng ca lung gi in:
Tc dng c hc ca dng ion ln cc phn t kh theo phng n in cc lng gi l gi in, gi in sinh ra khi phng in cc mi nhn v bt u t cc im ca qung.
+ Lc cu-lng ( hay lc in trng: u l lc ch yu tc dng ln cc ht trong in trng).
+ Tr lc ca mi trng: Khi chuyn ng trong trng in lc, cc ht tch in gp phi tr lc ca mi trng, tr lc ny ph thuc vo kch thc ht, tc chuyn ng ca ht, nht ca mi trng.
II. 3. Dng in trong cht kh S in ho kh :
S chuyn dch ca cc in tch c hng gi l dng in. Cc c chia ra : vt dn in v khng dn in . vt dn in c cc in tch chuyn ng t do , cn vt khng dn in th khng c tnh cht trn .
Cc kh iu kin trung ho , ngha l iu kin bnh thng chng gm cc nguyn t v cc phn t trung ho, chng l ngun in ho v ch nn dn in. Nu kh c in ho s to thnh cc in dng v in m (electrn). Trong kh c th to thnh cc in m v mt s phn t kh trung ho kt hp vi cc in t t do.
Kh c in ho di nh hng cc tc ng bn ngoi khc nhau : Nung nng rt mnh, tc dng ca tia rnghen , tia bc x v do va p phn t kh vi tc nhanh ca electrn v in.
Ngay iu kin bnh thng , kh chu tc dng ca tia v tr nn trong kh bao gi cng tn ti mt s lng in t v ion. Qu trnh in ho di tc dng ca cc in v electrn gi l s in ho va p
Nu trong kh cha cc in t t do ( cc electrn v in) c t gia hai cc ( mt cc l b mt vt tr cn cc kia l dy dn t gia ) dc ni vi ngun in c hiu in th nht nh v to nn mt in trng khng u th di tc dng ca in trng cc in tch t do trong kh bt u chuyn ng theo hng ng sc t ca in trng. chiu chuyn ng ca mi in tch c xc nh bng du ca chng. Cc in v electron chuyn ng trong kh to nn dng in.
Khi tng th hiu gia cc in cc (tng cng in trng) , cng dng in tng hu nh khng ng k nh t l vi th hiu
Tip tc tng hiu in th th s tng cng chm li . Dng in cc i ng vi cng in ho cho gi l dng in bo ho.
Trn hnh 9.5 cho thy vng I v vng II ng vi s tng cng dng in hu nh t l vi s tng hiu din th, vng III ng vi dng in bo ho v vng IV ng vi iu kin to thnh qung sng . in tch qung sng ch pht sinh trong in trng khng u vi iu kin l hnh dng khong cch in cc xc nh . Khi in cc ngoi l mt khi tr rng hoc phin cn in cc trung tm l dy dn s to nn in trng gia hai cc khng u . Gn dy dn trung tm c cng in trng rt ln nn xy ra s va p in . Mu xanh da tri l c im mu sc ca in cc qung sng cnh dy dn . b mt phin hoc b mt khi tr rng khng c s in ho . in tch qung ph thuc vo du in tch trn dy dn c th c in in tch qung sng dng hoc m.
II.4. Qung sng trong thit b lc bi in :
-Thc cht ca qu trnh lc in l s np in (tch in) cho cc ht bi cha trong kh . Cc ht ny s c tch khi kh di tc dng ca in trng . Qu trnh ny c th xy ra trong cc thit b lc bi in gm mt vng hoc hai vng, trong mt vng tch in cho ht bi vng in ho v vng lng bi.
- tch in cho ht bi th dng in c to nn bi qung sng trong in trng khng u gm hai h thng in cc : dy dn trung tm ng tr rng hoc nhiu dy dn t gia hai phin lo bi c qung sng bao quanh gi l in cc qung sng , cn in cc th hai lng bi di tc dng ca in trng gi l in cc lng . in tch qung sng ch pht sinh in trng xc nh . iu kin ph thuc vo hnh dng , v tr in cc , thnh phn , p sut , nhit kh .
Tng p sut tronh thit b lc c th cho php thit b lc lm vic vi cng in trng cao , tng nhit kh cho kt qu ngc li
III. Cc yu t nh hng trong thit b lc bi in:
Khi thit b lc in lm vic s c rt nhiu yu t ng thi tc dng, nh hng n c tuyn vn ampe cng nh s tch in ca cc ht. thit b lm vic vi hiu sut ln, thi gian lm sch l ngn nht, qu trnh n gin d iu chnh, th phi chn c tuyn ca thit b thch hp, ngha l phi ch n cc yu t : Kch thc, tnh cht vt l ca ht, nhit , khi lng ring ca kh v thnh phn kh bi.
Hiu qu thc t trong thit b lc bi in ph thuc vo nhiu nhn t : Cc tnh cht vt l ca kh lm sch (thnh phn ho hc, nhit v p sut kh ) tnh cht ca bi (thnh phn ho hc , cc tnh cht in , phn tn ht bi) lp trn cc lng , hm lng bi ban u trong kh , bn ca bi trn in cc lng v in cc qung , cc thng s in ca thit b (in p , cng din trng , tc kh , phn tn bi trong in trng).
Sau y xt nh hng cc nhn t :
III.1. nh hng cc tnh cht kh cn lm sch:
Cng in trng ph thuc vo th hiu cp cho in cc qung . Nh bit , in tch ht bi , tc chuyn ng ca chng (sau khi tch in) n in cc lng ph thuc vo cng in trng . Do vy duy tr th hiu cc i cho php trn in cc qung l iu kin quan trng nht thit b lm vic c hiu qu v t mc thu bi cc i .
Nh xt , khi tng nhit kh th hiu in cc qung s gim , iu duy tr c s xuyn thng. Th hiu xuyn thng ca thit b lc in t l nghch vi nhit tuyt i ca kh . Ngoi ra , nhit kh nh hng n tnh cht lp bi trn in cc lng .
nh hng ca m kh n hiu in th trong thit b lc in th ngc li so vi nh hng ca nhit : tng m to kh nng th hiu xuyn thng , ngoi ra th hiu nh hng n c tnh ca lp bi trn in cc lng.Th hiu trong thit b cn ph thuc vo thnh phn ho hc ca kh , thng l cc tp cht kh mang in m cng nh SO3 b hp th trong lp bi lm thay i tnh cht ca lp .
III.2. nh hng cc tnh cht ca bi trn in cc lng:
nh hng kch thc ht bi ( phn tn ht ) n cng tc ca thit b lc l tr s in tch m ht bi nhn c , l tc chuyn ng ca ht bi n cc lng sau khi chng c tch in. in tch cc ht ln cng nh tc chuyn ng ca chng n cc lng ln hn so vi cc ht nh . V vy hiu qu thu cc ht ln hiu qu cao hn , thi gian chuyn ng n cc ht ln t hn . Ngoi ra , phn tn ca ht bi c lin quan n hin tng gi l s kn qung sng v lin quan n cu to lp bi trn in cc lng .
Thc nghim chng minh rng : in tch m ht bi tch t l bc nht vi bn knh ht ( i vi ht c ng knh nh hn 10-4cm ) v t l vi bnh phng bn knh ht (i vi ht c ng knh ln hn 10-4cm). Nh vy ht cng to th s chuyn ng n in cc lng cng nhanh , cng qua nghin cu ngi ta thy rng tr s in tch cn ph thuc vo cng in trng .
Tnh cht vt l ca ht nh : dn in , h s cch in ,trng lng ring ... ca ht ng vai tr ch yu
Thnh phn ho hc m lin quan n n l cc tnh cht v in ca ht bi n nh hng trc ht n tr sut ( dn in)ca lp bi trn in cc lng , do nh hng n hiu qu lm vic ca thit b lc in, nh hng
III. Cu to, nguyn l lm vic thit b lc in :
- H thng lm sch kh nh lc in trng bao gm hai phn ch yu.
+ Phn in( bin p, chnh lu, cc b phn ph khc)
+ Phn thit b lm sch : gm 1 phng bn trong t cc in cc lng v in cc qung kh cn lm sch s i qua phng ny.
S nguyn l lm vic ca h thng:
Dng in xoay chiu cng nghip (220-380v) sau khi qua b iu chnh in p ri qua bin th cao p, in p c tng t 220-380V n 220V-10KV, sau qua chnh lu cao p 3 bin thnh dng 1 chiu v a vo in cc ca thit b lc in.
Thit b lc in c chia lm 2 loi:
-Loi 1 ngn: bao gm c in cc qung v in cc lng do s ion ho ct bi cng nh s lng u c xy ra.
-Loi 2 ngn: s ion ho v s lng sy ra trong 2 ngn ngn th nht gm c cc in cc qung gi l ngn ion ho, ngn th 2 gm cacs in cc lng v gi l phng lng trong thc t, lm sch kh cng nghip th loi thit b lc in 1 ngn c s dng rng ri hn nhiu.
Cc b phn ch yu ca thit b lc in:
1) Cc b phn a kh vo, phn phi kh v dn kh ra khi thit b:
- l cc yu t quan trng ca mt thit b lc in vic phn phi kh u vo cc ngn c thit b, cng nh phn phi u theo khp tit din ca thit b l cc iu kin bt buc thit b lm vic c hiu qu. iu chnh lng kh i qua cc ngn ca thit b lc ta cn t cc van iu chnh trn ng ng dn kh vo v ra khi thit b.
-Cc b phn a kh vo v dn kh ra khi thit b phi c cu trc sao cho dng kh i qua thit b mt cch t t, u n v ng hng, thng u vo ngi ta t cc ng loa hay li phn phi kh, cn u ra th lm cc ng tht dn t tit din ca thit b lc in n tit din ng dn kh.
2) V ca thit b lc in
-V ca thit b lc in l b phn cha cc in cc lng, in cc qung v phn bi c lng. Ph thuc vo iu kin cng ngh m c th lm v thit b t thp( thp thng hay thp c bit) hoc gch, b tng ct thp..... v c lp lt hay lp bo n.
3) Cc loi in cc:
Nh trn ta nu trong thit b c 2 loi in cc:
+ in cc qung.
+ in cc lng.
2 loi in cc ny cn m bo 2 iu kin sau:
-C bn v cng c hi c tui th cao trong iu kin rung ng do s bi lng bng c kh hay nh hng ca lc in trng v ca dng kh chuyn ng.
-Phi c tnh bn trong mi trng xm thc cng nh trong mi trng kh nhit cao.
Cc in cc qung:
i vi cc in cc qung, ngoi 2 iu kin trn cn phi m bo hnh dng in cc tht chnh xc to nn mt in trng ng nht v m bo cng cn thit. Theo nguyn tc chung c th chia in cc quang thnh 2 nhm ch yu nh sau:
-Nhm 1: Gm cc in cc qung khng c im tp trung in tch m in tch phn b dc theo in cc theo cc khong cch hng tch khc nhau ph thuc vo ch lm vic ca thit b v trng thi b mt ca in cc.
Nhm 1 gm 4 loi ch yu sau:
+ Loi 1 : c tit din hnh trn ng knh 2(4mm c khi dng xon l xo vi ng knh vng xon 20mm, khong cch gia cc vng xon l: 30mm.
+ Loi 2: tit din hnh vung vi cc cnh 3(4mm.
+ Loi 3: Dng hnh sao vi s nh khc nhau.
+ Loi 4: Hnh det.
i vi cc in cc nhm ny phi c tnh ton sao cho s din tch tnh ton sao cho din tch tit din ngang ca in cc phi m bo bn c hc , ng thi phi nghin cu cng ca cc b mt sao cho m bo cng dng in qung.
Nhm 2: gm
+ Kiu dy thp gai.
+ Kiu rng ca.
+ Kiu gn thm cc gai.
Nhm ny gm cc in cc in hnh cc im tp trung in tch theo chiu di in cc.
in cc qung nhm 1 dng c i vi c lc in tm v lc in ng.
in cc qung nhm 2: dng cho lc in tm c nhiu u vit hn cc in cc qung thng c gia cng t kim loi nh thp cu trc, thp nh hnh, ng nhm bc...... sao cho m bo lm vic tt trong mi trng cho.
Cc in cc lng.
i vi thit b lc in loi ng:
Ta dng cc ng hnh lc gic hoc hnh trn, hn ni tip nhau tu thuc vo ng knh ca thit b
i vi thit b lc in loi tm
+ Dng tm. + Dng thanh.
+ Dng li
chng III
tnh cng ngh
Cc thng s ban u:
- Nng sut ca thit b Vs = 28.600m3/h.
- Nhit kh bi vo TB: < 450C
- Mc lm sch: 95%.
- Hm lng lm sch: 95%
- Hm lng ca bi vo thit b: S4 = 200mg/m3
- Hm lng ca bi ra thit b: S4 = 5mg/m3
- p sut kh than bi vo thit b: Pb = 300 ( 400mmH20
- Kch thc di ht bi khng kh:
D((m)
60
% khi lng
8
6
16
33
25
12
Tnh ton:
I.Th tch lm vic ca thit b.
a. Th tch lm vic ca thit b ta c:
VTB = Vs.(l
+ Vs = 28.800m2/h. Nng sut lm vic ca thit b
+ (l: Thi gian lu ca ht bi trong thit b.
- Theo yu cu k thut v trn thc t ta chn mt s thng s sau:
+ ng knh ng trn ca in cc lng: D = 325mm
+ ng knh cc in cc qung: d = 3mm
(lm bng thp bc ch).
trnh s xut hin tia la in gia 2 in cc th phi m bo t l gia bn knh ng v bn knh dy dn: eq \f(R,r) ( 2, 72
Ta c: eq \f(R,r) = eq \f(D,d) = eq \f(325,3) = 108,33 >> 2,72
Vy chn ng knh nh trn l tho mn.
+ Dng in lm vic ca b lc: I ( 200mA
+ in p lm vic ca b lc: U = 50 ( 70kv.
Hiu s in th gia cc in cc:
V = E.R* (KV)
F: gradien in th (KV/cm)
Chn E = 4KV/cm (khi nhit thng < 400C).
R*: Khong cch gia cc in cc khc tn (cm).
R* = R r
V = F (R r) = 4. (
)
2
3
,
0
5
,
32
-
= 65,6K2/cm
V ( 6,5.1012 n v in t.
Thi gian lu li ca ht bi trong th c tnh theo CT: (2 22) (I - 88b)
(l =
A
r
R
2
2
2
-
Theo (I 88b) ta c cng thc tnh h s A.
A =
EMBED Equation.3
r
R
d
bui
d
e
v
lg
.
.
.
16
,
2
.
81
,
0
1
(
.
min
min
m
l
+
Trong :
- v: Hiu s in th v = 6,5.1012 v in t
- e: in tch in t e = 1,6.10-19 v in t.
db: iu kin nh nht ca bi d = 5(m = 5.10-4cm.
(: nht ca kh ly: ( = 2,32.10-5N.S/m2.
(+: Qung ng t o trung bnh ca nguyn t kh
( = (273
273
40
273
10
.
12
,
1
3
2
273
30
+
=
+
+
-
t
= 1,28.10-3 (cm)
vi t=400C: nhit kh
(2730K = 1,2.10-3 (cm) (ng i trung bnh ti 00C)
Thay s vo ta c :
A =
7
,
62
10
.
098
,
5
10
.
196
,
3
5
,
1
5
,
162
lg
.
10
.
43
,
2
.
10
.
5
.
16
,
2
)
10
.
5
10
.
28
,
1
.
81
,
0
1
(
10
.
6
,
1
.
10
.
5
,
6
8
6
5
4
4
3
19
12
=
=
+
-
-
-
-
-
-
-
Vy thi gian lu ca ht bi trong thit b:
(l = eq \f(R2-r2,2A) =
)
(
1
,
2
7
,
62
.
2
)
15
,
0
25
,
16
(
2
2
s
=
-
Thi gian lu li ca kh trong thit b lc in thng ly khong (3 ( 3,5)s. i vi cc kh n mn ho hc in cc lm bng ch (chng n mn) lc thi gian lu khong: 6s.
i vi cc kh trong nh my phn m H Bc th cc kh thng
(02 + CO2 + N2 + H2 + S02)
Vy ta chn thi gian lu: (l = 3,5(s)
Th tch lm vic ca thit b:
VIV = VS . (l =
3
28
3600
5
,
3
.
800
.
28
m
=
II.Tnh s ng in cc:
- Ta ly chiu di mi ng L = 4.400(mm)
Th tch mi ng s l:
V0 = eq \f((D2,4) .L =
368
,
0
4
4
,
4
.
325
,
0
.
14
,
3
2
=
(m3)
Vy s ng cn thit ca thit b lc in:
Nl = eq \f(Vcv,V0) = eq \f(28,2.368) = 76,08 ng
Vy chn s ng n = 80 ng
Vy thi gian lu li thc t ca kh bi trong thit b lc in.
(l = (l . eq \f(n,n1) = 3,5. eq \f(80,76.08) = 3,68 (s)
- Tit din ngang ca mi ng.
f = eq \f(( (D2-d2),4) =
2
2
2
08
,
0
4
)
03
,
0
325
,
0
(
m
=
-
p
III. Vn tc chuyn ng ca kh trong ng:
( = eq \f(Vs,n.f) = eq \f(28.800,80.0.08.36000) = 1,25 (m/s).
Ta thy tc nh vy l t yu cu cho c 4 ma trong nm v tho mn yu cu k thut v m bo an ton cho thit b.
IV.in p ti hn cho v cng dng in.
- Cng ti hn ca in trng m ti bt u xut hin s phng in t cc ion ho c th xc nh theo cng thc gn ng ca pik. Mt cch cht ch dng cho khng kh v dy in cc tit din trn.
E0 = 3,04 ((+0,0311)
)
/
(
10
6
m
v
r
b
Theo( VII 197)
r = 1,5(mm) = 0,0015m
(: t s gia khi lng n v ca kh trong iu kin lm vic v trong iu kin tiu chun.
(t = 200C, P = 1,013.105 N/m2)
( = eq \f(PKq +PK,1.013.105) . ( eq \f(273 + 20,273+t) )
Pkq: p sut kh quyn Pkq = 1,013.105 N/m2
PK: p sut (d) tng i ca kh trong thit b lc PK = 300Pa = 300 N/m2
( =
938
,
0
)
40
273
20
273
(
10
.
013
,
1
)
300
10
.
013
,
1
(
5
5
=
+
+
+
Cng ti hn E0 ca in trng.
E0 =
)
/
(
10
.
21
,
5
10
).
0015
,
0
938
,
0
0311
,
0
938
,
0
(
04
,
3
6
6
m
v
=
+
a. in p ti hn U0:
U0 = E0.r.ln eq \f(R,r) = 5,21.106 . 0,0015.Ln
0015
,
0
325
,
0
= 42.031,8v ( 42,03KV.
b. Cng dng in I0:
- Cng n v ca dng I0 nu cc m (cc ion ho c cp in vi in p V = 50KV.
- Hng s z ca thit b lc in:
Z =
r
R
R
R
ln
.
10
.
9
2
2
9
(VII 198).
Khi nhn hot ng ca ion k = 2,1.10-4m2/vs
Z =
13
2
9
4
10
.
77
,
3
0015
,
0
1625
,
0
.
)
2
325
,
0
.(
10
.
9
10
.
1
,
2
.
2
-
-
=
Ln
Cng dng in:
I0 = Z (U U0) U = 3,77.10-13 (50.000 42032).50.000
= 1,5.10-4 A/m = 0,15mA/m.
V.Cng sut tiu th ca thit b:
- Theo 1-42 (VIII 29)
N = eq \f(v.n.L.i,1000.() + 0,5KW.
Trong :
- v: hiu s in th gia cc in cc.
v =6,5.1012 n v in tnh
n: S ng: n = 80
L: chiu di ng; L = 4,4m
I: mt dng in.
I = 1.5.10-4A/m.
- (: hiu sut ng c ,( = 0,8
N =
)
(
79
,
4
5
,
0
10
.
8
,
0
.
1000
10
.
5
,
1
.
4
,
4
.
80
.
10
.
5
,
6
6
4
12
Kw
=
+
-
-
VI.Thng s hnh hc:
- Ta c in tch ct ngang lm ca thit b:
S = 80.f = 80.0,08 = 6,4m2
Thng thng trong thc th ngi ta b tr gia cc ng phi c khong cch. Do vy din tch ct ngang ca thit b s phi cng thm phn .
Ta ly din tch ca khong chng S = 7m2
Din tch ct ngang ca thit b.
S = 6,4 + 7 = 13,4m2
ng knh trong ca thit b
Dt = 2
m
S
13
,
4
14
,
3
4
,
13
.
2
14
,
3
=
=
Vy ta ly ng knh ca thit bi l:
Dt = 4,6m
Chiu cao ca thit b
H = HLV + HLK + Hc + Hs
HLv: chiu cao lm vic ca thit b H = 4,4 m
HLK: chiu cao lc kh
H = 4,5m
Hc: chiu cao H = 3,8m
Hs : Chiu cao s cch in t trn mi
Tng chiu cao H = 14 m
VII,Nhng s c sy ra trong thit b lc in trong qu trnh lm vic v cc bin php s l:
a, Thit b lc bi khng duy tr c in p , khi ng in th h thng bo v ct mch :
- Nguyn nhn : +c vt l trong hp cch in .
+ S cch in b hng .
+ C s cong vnh ca khung in cc qung sng .
+ Cc sc cch in b ph lp hi m ngng t .
+Dy nung nng trong hp cch in b hng , khng c in vo
+ in cc b t .
Bin php sa cha :
+Kim tra loi b vt l trong hp cch in .
+Thay th s cch in , un phng khung treo cc in cc qung sng
+Loi b in cc b hng .
b, Khng thy xut hin dng in qung sng :
Nguyn nhn : Cc in cc qung sng bc kn bi bi ro rung lc yu . Thanh dn in cao p khng dn in .
Bin php sa cha : Tm v loi b s h hng , sa cha v thay th dy dn cao p .
c, in p lm vic dng in qung sng b hng:
Nguyn nhn : D in s cch in , c s c vi phm v cng ngh v d nh hm lng m kh cao ,nn nh hng ti s cch in .
Bin php sa cha : Lm sch s cch in , thng bo vi i iu khin kim tra li iu kin cng ngh .
d, Hiu sut kh bi gim :
Nguyn nhn : + B kh bi qu ti .
+ in p khng tng
S l : Gim lu lng vo , kim tra li b cch in
Phn 3:
Tnh c kh
I, Tnh thn thp :
- Loi thit b ny lm vic iu kin p sut thp khi ch to loi ny cn ch :
m bo ng hn cng ngn cng tt .
Ch hn gip mi .
B ch cc ng hn dc ( cc on tr ring bit ln cn )cch nhau t nht l 100mm.
B ch mi hn v tr d quan st.
Khng khoan l qua cc mi hn .
1, Tnh chiu dy v:
Cng thc tnh chiu dy thn thp chu p sut trong :
S =
[
]
P
P
Dt
-
j
s
.
2
.
+ C ,m Theo( IX-360)
Trong :
+ Dt : ng knh trong ca thit b Dt =4.6m
+( : H s bn ca thnh hnh tr theo phng dc i vi thnh kn hay c l c gia c hon ton . Chn ( =(h ( h s hn bn ) , ly ( =(h = 0.95.
+ C: H s b xung do n mn , bo mn dung sai v chiu dy .
C= C1 + C2 + C3 . (m)
C1 : H s b do n mn ho hc , chn C1 = 1 mm.
C2 : H s b xung do bo mn , trong trng hp ny khng c ht rn chuyn ng vi tc ln nn chn C2 = 0.
C3 : H s b xung do dung sai ca chiu dy ,
chn C3 =0.8.10-3 (m).
Do vy :
C= C1 + C2 + C3= 10-3 +0 + 0.8.10-3 = 1.8.10-3 (m).
+ P: p sut trong lm vic trong ca thit b
P = 300 mm H2O = 300. 9,81 = 2943(N/m2).
Vi iu lm vic ca thit b ta chn vt liu lm thit b l loi chng n mn
Do vt liu c chn l thp khng r X18H10T.
Khi lng ring ( = 7900 kg/m3
+
[
]
s
: ng sut cho php ca vt liu .
Ta c :
s
k*
[
]
s
k =
k
k
n
h
s
.
*
( N/m2) ( Theo IX . 355).
[
]
s
c =
c
c
n
h
s
.
*
( N/m2)
Trong :
h
: H s iu chnh
h
= 0.9 (Theo IX- 356)
nb ,nc : H s an ton theo gii hn bn v gii hn chy .
nb= 2.6 Tra bng (XIII. 3) trong (IX-356).
nc= 1.5
s
k*.,
s
c*.: Gii hn bn ko , gii hn chy .
s
k*.= 550.106 (N/m2). (Theo IX-310)
s
c*.=220.106 (N/m-2).
Do :
EMBED Equation.3
[
]
s
k =
5
,
2
9
,
0
.
10
.
550
6
= 190,385.106 (N/m-2).
[
]
s
c =
5
,
1
9
,
0
.
10
.
220
6
= 132.106 (N/m-2).
m bo iu kin bn ta tnh ton vi gi tr ng sut b hn .
[
]
s
k =
[
]
s
c =132.106 (N/m-2).
T ta suy ra :
[
]
j
s
.
P
=
EMBED Equation.3
95
.
0
.
2943
10
.
132
6
= 42385 > 50
Do ta c th b qua P mu s trong cng thc tnh S:
S =
[
]
p
P
Dt
-
j
s
2
.
+ C =
95
,
0
10
.
132
.
2
2943
.
6
,
4
6
+ 1,8. 10-3 = 5,63. 10-3 (m).
m bo cho thit b ta chn S = 8.10-3 (m).
Tc l S = 8 mm.
* Kim tra bn ca v thit b :
Kim tra v bng cch kim tra ng sut theo p sut th :
s
=
[
]
j
)
(
2
.
)
(
C
S
Po
C
S
Dt
-
-
+
s
c*./ 1.2 (N/m-2).
Vi :
+
2
.
1
s
=
.
2
,
1
10
.
220
6
= 183033106 (N/m-2).
+ Po : p sut th
Po =Pth + Pl .
Trong :
Pth : p sut th thu lc .
Theo (IX-358) ta c Pth = 1,25.P = 1,5.2943 = 4414,5 (N/m-2).
Pl : p sut ca nc .
Pl = g.
r
H2O. H = 9,81.996,5.12,7=1240244 (N/m-2).
Do :
Po =Pth + Pl =4414,5 + 1240024,4 = 128493 (N/m-2).
Thay cc thng s vo ta c :
s
=
95
,
0
).
10
.
8
,
1
10
.
8
.(
2
128439
).
10
.
8
,
1
10
.
8
6
,
4
(
3
3
3
3
-
-
-
-
-
-
+
= 50,2.106 (N/m-2). <
s
c /1,2 (Tho mn).
Vy thn thp vi chiu dy S = 8mm l tho mn iu kin bn cho v thit b .
2, Tnh chn y v np thit b :
Np v y cng l nhng b phn quan trng ca thit b v thng c ch to cng loi vt liu vi v thit b .
Ta tnh tnh ton vi y np hnh nn bi v loi ny thng dng cho cht lng nht, huyn ph , vt liu bt m d dnh v loi ht
Loi y , np hnh nn ny khng c g c ni vi thn thit b bng cch hn cht li vi nhau.
hnh v
+ Do loi thit b lm vic trong iu kin nc bn nn ta chn gc dy 2
a
= 1200
B dy ti thiu ca y S ph thuc vo gc nh nn
a
, vo ng sut cho php
[
]
s
, vo p sut P v h s
j
Vi na gc nh
a
=600 .
Ta c :
[
]
j
s
.
P
=
EMBED Equation.3
95
.
0
.
2943
10
.
132
6
= 42385 > 50.
Theo ( X 174) ta c :
S =
[
]
j
s
a
.
cos
2
.
P
Dt
=
95
,
0
.
10
.
132
.
60
cos
.
2
2943
.
6
,
4
6
= 2,2.10-3 (m).
ph hp vi v thit b m vn m bo iu kin bn,
ta chn chiu dy ca y S = 8mm.
3, Xc nh cc ca vo ra c lp trn thit b
a, Xc nh ca vo dng kh :
ng ng dn kh tnh theo cng thc :
D =
w
Vs
.
785
,
0
(m).
Trong :
Vs : Lu lng kh , Vs = 28800 m3/h.
W : Vn tc chuyn ng kh , w = 1,25 m/s .
Do vy :
D =
3600
.
785
,
0
.
25
,
1
28800
= 0,95 (m).
b, Ca nhn , ca ngi :
Cc ca ny dng xem xt bn trong thit b , ra , lm sch , tho lp cc b phn bn trong thit b .
Trn thn ta b tr cc c :
+ Ca nhn : dng quan st trng thi bn trong ca thit b , n c b tr sao cho thun tin quan st ng thi ti v tr nh sng c th chiu sng c .
+ Ca ngi :
Tt c cc thit b c ng knh >800 mm u phi c ca ngi . Hnh dng ca ngi cn sao cho d ch to v chu vi bt kn l b nht , ng thi vic la chn ca ngi cn ph thuc vo t0 , P thng ca ngi c hnh dng trn hoc van .
Kch thc :
van : 400x500 (mm).
Trn : 500(mm).
i vi loi thit b lc bi in ta chn dng hnh trn.
c, Ca phng n :
Do thit b lc bi in lm vic vi dng in c cng v hiu in th cao nn n rt nguy him trong vic chy n . Trong trng hp nh vy th vic c ca phng n l rt cn thit .
Kch thc ca phng n bng vi kch thc ca ngi D = 500mm.
4, Chn dng bch v vt liu m :
a, Chn loi m :
Vng m c tc dng gi kn khng cho cht lng hoc cht kh lt qua khe h . Vic chn vt liu m tu thuc vo nhit v p sut tnh n mn ca mi trng , chi tit tnh , chuyn ng tng i . Yu cu i vi vng m l :
+ C tnh cht n hi bt kn cc khe h ti b mt p cht trong qu trnh lm vic .
+ Khng b bin i tnh n hi theo thi gian lm vic .
+ Khng lm h hng b mt p cht , ng thi khng lm tr ngi vic tho thp bch .
Do trong hn hp kh c CO2 , SO2 , H2S ,O2 ...nn vt liu bt kn l :
+ m mm : Aming chu axt
+ Loi m : m phng.
b, Tnh bn bch :
Chn vt liu ging vi vt liu ch to v : 1X18H10T ,
c
EMBED Equation.3
[
]
s
= 132.106 (N/m2).
Chn vt liu lm bu lng l thp 45.
[
]
s
b = k
[
]
s
*b
Trong :
+ K : H s ph thuc vo ng knh bu lng v nhit lm vic ca bch .
+
[
]
s
*b : ng sut cho php ca vt liu lm bu lng
[
]
s
*b = 143.106 (N/m2).
Tnh bn bch :
Phn sau y s trnh by tnh bch theo phng php ti trng ti hn , phng php ny mang tnh kim tra . Ta a ra 1 bn v phc tho ca bch ( chn bch quy chun ) c y cc kch thc sau tnh kim tra xem chiu dy vnh bch c tho mn iu kin bn khng .
Chn dng bch Phng hn (hnh v) : v loi ny d ch to v d thay th
Trong :
h : Chiu dy vnh bch.
D: ng knh ngoi ca m
Dt : ng knh qua tm bu lng.
D1: ng knh c
Dy: ng knh trong ca ng
Dn: ng knh ngoi ca ng.
Db: ng knh Bu lng .
Tnh ton bch cho cc ca lp :
1 , Bch lp ca tho nc :
325
f
Vi Py = 0,002493.10-6 (N/m2)
Dn = 325 mm.
S = 8 mm
Dy = 317 mm.
Tra trong (IX - 415 ) ta c :
D = 435 mm.
D1 = 365 mm.
Dt = 395 mm.
h = 22 mm
Db : Dng bu lng M20 tng 12 chic.
Kim tra bch v bn :
Chiu dy bch phng c tnh theo cng thc :
h =
[
]
b
y
t
n
D
D
D
S
D
M
.
2
.
75
,
0
.
.
.
275
,
1
2
-
-
-
l
s
(Theo XI - 153).
Trong :
[
]
s
: ng sut ca vt liu
[
]
s
= 132.106 (N/m2).
l
: H s
l
= 1 i vi vt liu c kim loi .
S : Chiu dy ng S = Dn Dy = 8 mm.
M : M men thc t .
M =
[
]
=
+
+
-
-
2
2
.
4
3
).
(
).
2
(
.
.
4
S
S
D
S
D
D
D
y
b
y
t
s
p
=
=
+
+
-
-
2
2
6
6
8
.
4
3
).
8
317
(
8
).
20
.
2
317
395
(
.
10
10
.
132
.
4
p
= 1,8.106 (N/mm2).
Do vy:
h =
20
.
2
317
395
8
.
75
,
0
).
8
325
(
1
.
10
.
8
,
1
.
132
275
,
1
2
6
-
-
+
-
=
h = 21,6 mm.
Vy ta thy h > h nn bch tho mn iu kin bn.
2 , Bch lp ca tho nc :
500
f
Vi Py = 0,002493.10-6 (N/m2)
Dn = 500 mm.
S = 8 mm
Dy = 516 mm.
Tra trong (IX - 415 ) ta c :
D = 630 mm.
D1 = 550 mm.
Dt = 580 mm.
h = 22 mm
Db : Dng bu lng M20 tng 16 chic.
Kim tra bch v bn :
Chiu dy bch phng c tnh theo cng thc :
h =
[
]
b
y
t
n
D
D
D
S
D
M
.
2
.
75
,
0
.
.
.
275
,
1
2
-
-
-
l
s
(Theo XI - 153).
Trong :
[
]
s
: ng sut ca vt liu
[
]
s
= 132.106 (N/m2).
l
: H s
l
= 1 i vi vt liu c kim loi .
S : Chiu dy ng S = Dn Dy = 8 mm.
M : M men thc t .
M =
[
]
=
+
+
-
-
2
2
.
4
3
).
(
).
2
(
.
.
4
S
S
D
S
D
D
D
y
b
y
t
s
p
=
=
+
+
-
-
2
2
6
6
8
.
4
3
).
8
500
(
8
).
20
.
2
500
580
(
.
10
10
.
132
.
4
p
= 2,8.106 (N/mm2).
Do vy:
h =
20
.
2
500
580
8
.
75
,
0
.
516
1
.
10
.
8
,
2
.
132
275
,
1
2
6
-
-
-
=
h = 15,6 mm.
Vy ta thy h > h nn bch tho mn iu kin bn.
3 , Bch lp ca tho nc :
1000
f
Vi Py = 0,002493.10-6 (N/m2)
Dy = 1000 mm.
S = 10 mm
Dn = 1020 mm.
Tra trong (IX - 415 ) ta c :
D = 1140 mm.
D1 = 1060 mm.
Dt = 1090 mm.
h = 22 mm
Db : Dng bu lng M20 tng 24 chic.
Kim tra bch v bn :
Chiu dy bch phng c tnh theo cng thc :
h =
[
]
b
y
t
n
D
D
D
S
D
M
.
2
.
75
,
0
.
.
.
275
,
1
2
-
-
-
l
s
(Theo XI - 153).
Trong :
[
]
s
: ng sut ca vt liu
[
]
s
= 132.106 (N/m2).
l
: H s
l
= 1 i vi vt liu c kim loi .
S : Chiu dy ng S = Dn Dy = 8 mm.
M : M men thc t .
M =
[
]
=
+
+
-
-
2
2
.
4
3
).
(
).
2
(
.
.
4
S
S
D
S
D
D
D
y
b
y
t
s
p
=
=
+
+
-
-
100
.
4
3
).
10
100
(
100
).
20
.
2
1000
1090
(
.
10
10
.
132
.
4
6
6
p
= 8,44.106 (N/mm2).
Do vy:
h =
20
.
2
1000
1090
10
.
75
,
0
.
1020
1
.
10
.
44
,
8
.
132
275
,
1
2
6
-
-
-
=
h = 17,6 mm.
Vy ta thy h > h nn bch tho mn iu kin bn.
4, Nhng iu ch khi thit k mt bch :
Cc kch thc lp ca bt k loi bch no trong h thng thit b v ng ng u ph hp vi quy phm . Cc kch thc khc c th t thit k sao cho khong cch gia cc vng m v tm anh c , kch thc hnh hc v khi lng ca bch u nh nht
S l anh c ca vnh bch hnh trn nn l bi s ca 4.
Vng m phi t pha di cc anh c .
p sut cng cao , vng m cng phi hp anh to.
Phi thit k sao cho anh c khi xit cht ch chu lc ko .
anh c phi tin nhn v to iu kin tng sc n hi , gim ng sut tp trung . Dng anh ln t tn hn l dng anh nh .
II, Tnh ng sut do trng lng v m men gi Mg :
- Thit b t ngoi tri b tc dng ca ti trng gi nn cn tnh kim tra n theo bn v n nh . c bit l xc nh cc kch thc ca cc chi tit quan trng nht ca thit b l chn v cc bu lng bt chn vi b
Cc thng s hnh hc ca thp : H =14m.
Dt = 4,6 m
1, Tnh lc tc ng ca gi :
Ta c cng thc sau :
P =
F
P
g
.
.
.
j
x
( Theo XI - 129 ).
Trong :
+ Pg : p sut gi trong lc a .
Tra bng 22 trong (XI129) ta c Pg = 480 (N/m2)
+ F : L b mt trc din ca thp trc gi
F = H.D = 14.4,6 = 64,4 (m2 ).
+
x
: H s tr l pha tr ca v thp .
x
= f (Re). Ta xc nh
x
theo th 29.
- Xc nh chun s Reynol : Re =
n
D
V
.
v : Vn tc gi v = 4
Pg
= 4
480
= 87,6 (m/s).
n
: nht ng hc ca khng kh :
300C c
n
= 15,16.10-6 (m2/s).
Do ta c :
Re =
6
10
.
16
,
15
6
,
4
.
6
,
87
-
= 26,5.10-6 .
Tra bng 6 ta c
x
= 0,6.
+
j
: H s tnh n nh hng mch nhp ca gi .
1
.
1
C
m
j
+
=
Vi : C1: H s mch nhp ca gi .
C1 = 0,35 tra bng 22.
m
: H s tu thuc chu k lc l ca thit b .
T =
g
J
F
H
G
H
.
.
.
max
.
.
79
,
1
+Gmax : Khi lng ln nht ca thp .
Gmax = Gv + Gc + Gtin
Trong :
+ Gv =
.
r
Vv.
Vt liu lm v l thp 1X18H10T c
.
r
= 7900 (kg/m3).
Vv = V1- V2
V1 = H1.
p
.
2
2
n
D
= 12,7.3,14.
2
2
6
,
4
= 211(m3)
V1 = H1.
p
.
2
2
t
D
= 12,7.3,14.
2
3
2
10
.
8
,
2
6
,
4
-
-
= 209 (m3)
Vv = V1- V2 = 211 209 = 2 (m3)
Gv =
.
r
Vv.= 2.7900 = 15800 (kg).
Gtin = 50.5 = 250 Kg
Gc = 80.12.5 = 950 kg.
Vy khi lng ln nht a thp khong.
Gmax = 17000 Kg.
+ J : Mmen qun tnh ca thp :
J =
(
)
4
4
64
t
n
D
D
-
p
= 0,4.Dn3.S = 0,4.(4,6)3.8.10-3 = 0,31 (m4).
Do vy chu k lc l ca thit b l :
T = 1,79.14.
81
,
9
.
31
,
0
.
4
,
64
14
.
17000
= 873,6
Tra th 30 theo T ta c
4
,
3
=
m
Ta c
19
,
2
35
,
0
.
4
,
3
1
.
1
1
=
+
=
+
=
C
m
j
.
Vy lc tc ng ca gi :
P =
F
Pg
.
..
.
j
x
= 0,6.2,19.480.64,4 = 40618 (N).
M men gi do P to ra l :
Mg = P.
2
H
= 40618.
2
14
= 284326 (Nm) .
2 , ng sut trn gi :
- ng sut ln nht trn mt chn ca gi :
Wo
Mg
Fo
G
max
0
+
=
s
(Theo XI-141)
Trong :
- Fo : Din tch mt chn ca gi , hnh vnh khn
Fo =
4
p
(D22 - D21).
Vi : D2 = 1,08-1,18.Dn =1,15.Dn = 1,15.4,6 = 5,29 (m)
D1= 0,95.Dn = 0,95.4,6 = 4,37 (m).
Fo =
4
p
(5,292 - 4,372) = 6,97 (m2)
- Wo : M men chng un ca chn :
Wo =
(
)
2
4
1
2
4
.
32
D
D
D
-
p
=
(
)
29
,
5
37
,
4
29
,
5
.
32
4
4
-
p
.
Wo = 7,76 (m3).
Do vy :
)
/
(
7
,
60566
76
,
7
284326
97
,
6
81
,
9
.
17000
2
0
m
N
=
+
=
s
Gi tr
o
s
khng cho php vt qu ng sut cho php ca thp lm chn , ng thi cng phi nh hn ng sut cho php ca b tng .
Nu ly h s an ton cho b tng l 2-3 th
[
]
bt
s
= 250
350.103 (N/m2).
3, B dy mt chn :
B dy mt chn SD c xc nh da trn c s ti trng
o
s
tc dng ln 1 phn t c b rng 1cm v b di b ca mt chn .
Ti trng bng :
q =
2
.
2
2
0
b
q
b
s
=
=
2
15
,
0
.
7
,
60556
2
= 681 (N).
ng sut phn t :
[
]
s
s
s
s
=
=
=
p
p
M
S
b
S
W
M
2
2
0
.
.
3
.
vi
[
]
s
= 132.106 (N/m2).
V th b dy mt chn :
Sp =
[
]
[
]
s
s
s
s
0
.
3
.
b
M
=
=
009
,
0
10
.
132
7
,
60556
.
3
.
15
,
0
6
=
(m).
m bo bn ta ly Sp = 0.02(m) = 2 (cm)
ng knh anh c:
ng sut ti thiu trn mt chn s l :
0
0
min
min
0
W
Mg
F
G
-
=
s
.
T iu kin
s
0min = 0 , c th rt ra h s n nh Ks.
Ks =
Mg
P
G
Mg
M
G
G
.
min
=
=
15
,
1
284326
932
,
1
.
17000
=
MG : M men ca thp vi tay n PG = 0,42.Dn = 0,42.4,6 = 1,932 (m).
Ta thy Ks = 1,15 < 1,5 , nn thp cn c anh b gi cht , mi anh s chu lc ko P m trn i th ta c th tnh c nh sau :
P =
n
Fo
.
0
s
.
Fo : Din tch hnh vnh khn :
F1 =
EMBED Equation.3
(
)
4
.
2
2
b
D
n
+
p
=
(
)
37
4
15
,
0
.
2
6
,
4
14
,
3
2
=
+
(m2)
F2 =
(
)
4
05
,
0
.
2
2
-
n
D
p
=
(
)
16
4
05
,
0
.
2
6
,
4
14
,
3
2
=
-
(m2)
F0 = F1 - F2 = 37 16 = 21 (m2)
n: S anh b n = 40.
Do :
P =
n
Fo
.
0
s
.= 60556,7.
40
21
= 31792 (N).
T ta tnh ra ng knh chn ren bu lng .
d =
[
]
C
P
d
+
s
p
.
.
4
=
3
6
10
.
8
,
1
10
.
132
.
31792
.
4
-
+
p
= 0,019 (m) = 19 mm.
Vi :
C : h s b xung do n mn , bo mn , dung sai , C=1,8.10-3 m.
[
]
s
: ng sut cho php ca vt liu ,
[
]
s
= 132.106(N/m2).
Thng ta ly ng knh bu lng nn ln hn hoc bng M24.
Vy ta dng bu lng nn M24 , tng s bu lng n = 40.
III, V ng :
V ng l mt b phn tng i quan trng trong cu to ca thit b lc bi in , n c nhim v l gi cht v nh v cho cc ng in cc lng u nhau trong trnh lm vic khng nh hng n hiu sut thu hi bi ca thit b .
V ng c ch to t phi tm , ngoi ra c th c . Ch vt liu lm v ng phi bn v cng hn vt liu lm ng
Ta b tr 2 v ng nh v cho gin ng , v ng c lin kt vi v thit b nh gi c hn c nh vi v thit b
Chn chiu dy v ng S = 15mm
IV. Dy treo in cc qung:
Trong qu trnh lm vic th dy in cc qung c treo l lng trong ng in cc lng , to ra in trng u v n nh th dy in cc qung lun trong trng thi ung a , chnh v vy h thng dy in cc s c treo trn dy c ni vi t cch in .
Tnh ng knh dy treo :
Khi lng ton b c cu cng dy : M = 150 Kg .
Thit b lm vic trong 2 na , mi na c 3 dy treo cch u nhau .
Lc tc dng ln mi dy treo l :
q =
n
Q
=
6
10
.
150
= 250 (N).
Chn vt liu lm dy treo l thp CT40 , c ng sut ko cho php
[
]
5
10
.
140
=
k
s
(N/m2).
Tit din ngang ca dy treo l :
F =
[
]
k
q
s
=
5
10
.
140
250
= 1,78.10-5 (m2).
Do vy ng knh dy treo l :
D =
p
F
4
=
14
,
3
10
.
78
,
1
.
4
5
-
= 0,005 (m).
tho mn iu kin bn ta chn
D = 10 (mm).
Phn kt lun
c s hng dn tn tnh ca thy gio T Hng c cng cc thy c gio trong b mn My v thit b ho cht v du kh v tham kho nhiu ti liu lin quan n thit b lc bi in em chn ra c phng cch tnh ton thit k thit b lc bi in mt cch thch hp nht .
Trong khun kh mt bn n tt nghip i hc , vi s c gng n lc ca bn thn em xin chn thnh cm n thy gio T Hng c v cc thy c gio trong b mn tn tnh hng dn v ch dy gip em hon thnh n ny.
Tuy bn n ny hon tt nhng khng trnh khi s sai st xy ra . Em mong c s ch dy ca cc thy c gio trong hi ng bo v n tt nghip . Em hy vng s tip thu c nhiu kin thc qu bu qua s gp ca cc thy c .
Sinh vin
Nguyn V Thnh
Ti liu tham kho
[
]
I
. H Th An .
Cc qu trnh v thit b thu c - 1977.
[
]
II
. B mn my v thit b ho cht .
Bi tp my v thit b ho cht - HBK 1969.
[
]
III
. Hong Kim C.
K thut lc bi v lm sch kh Nxb gio dc-1999
[
]
IV
. B mn t ng ho.
Thu bi v lm sh kh-1974.
[
]
V
. ISO 1400.
[
]
VI
. Cc tiu chun Vit Nam v kh thi.
[
]
VII
. Trn Ngc Chn.
nhim khng kh v s l kh thi - Nxb KHKT.
[
]
VIII
. H L Vin, Nguyn Minh Tuyn
My v thit b ho cht Nxb KHKT 1971
[
]
IX
. S tay qu trnh v thit b ho cht - Tp 2.
[
]
X
. H L Vin.
Thit k tnh ton thit b ho cht .
[
]
XI
. H Hu Phng.
Gio trnh c s tnh ton thit b ho cht.
L to kh
Kt kh
Lc bi in
Kh H2S trung p
ra on 3
(19 - 21 atm)
Than
Hi nc
Khng kh
vo on 1
Bin i CO2
My nn 667
Kh H2S thp p
on 4 / 667
Kh
tinh ch
Phn xng urea
CO2
Hp th CO2
Than
Kh than than
Tc nhn kh
1
2
3
4
5
6
EMBED AutoCAD.Drawing.14
2
1
a
2
2
1
b
c
1
2
d
1
b: in trng u.
a, c, d: in trng khng u.
2: in cc qung(-).
1: in cc lng(+).
i
Dng bo ho
Xuyn thng
Qung sng
u
iv
iii
i
ii
in cc dn - khi tr rng.
thay i th hiu trong h thng cc
Hnh 9.5. S thay i dng in khi
khi tr lng.
thng cc in cc lm dy dn
Hnh 9.4. Cc ng sc trong h
6
9
10
4
5
380V
7
8
20--100KV
3
2
1
Kh sch
220 --
Kh bi
1: B iu chnh in p.
2: Bin cao p .
3: Chnh lu p
4: Dy cao p ..
5: Tip t .
6: Cch in .
7: Phng lc .
8: in cc lng (+).
9: in cc qung (-).
10: C cu cng in cc qung.
Bi
V
Gi
in cc lng
V ng
Bch ni
B tng
Sd
b
PAGE
62
Nguyn V Thnh Lp My v Thit b HC-Du kh K42