AMH XUN HUY

GVHD: THS.NG

TRNG I HC CNG NGHIP THNH PH H CH MINH

N MN HC X L NC TI : X L NC CP L HIGIO VIN HNG DN NHM SV THC HIN LP : THS NG XUN HUY : NHM 4 : CMT 11 TH

NHM THC HIN:NHM 04.

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GVHD: THS.NG

DANH SCH SINH VIN NHM 4

STT 1 2 3 4 5 6 7 8

H v tn Dng Th Nga L Th Thu Ngn ng Thi phng Nguyn Th Tho Bi Thu trang Trng Vn Truyn Hong Th Vn L Th Vn

MSSV 09011293 09027693 09012553 09023843 09018953 09018663 09026683 09027903

Ghi ch

NHM THC HIN:NHM 04.

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AMH XUN HUY

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NHN XT CA GIO VIN HNG DN

......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...........................................................................................................................................

THANH HA NGY 17 THNG 06 NM 2011

NHM THC HIN:NHM 04.

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LI CM NTrong sut thi gian hc va qua, chng em c cc thy c khoa Mi Trng tn tnh ch dy, truyn t nhng kin thc qu bu, n Mn Hc l dp chng em tng hp li nhng kin tc hc, ng thi rt ra nhng kinh nghim cho bn thn cng nh trong cc phn hc tip theo. hon tt n ny, chng em xin chn thnh cm n Thy NG Xun Huy tn tnh hng dn, cung cp cho chng ti nhng kin thc qu bu, nhng kinh nghim trong qu trnh hon thnh n. Xin chn thnh cm n cc thy c Khoa Mi Trng ging dy, ch dn to mi iu kin thun li cho chng ti trong sut thi gian va qua. Vi ln u lm n, kin thc cn hn ch, kinh nghim thc t cha c nn trong n ny cn nhiu thiu st, chng em rt mong nhn c s gp ca cc thy c v cc bn nhm rt ra nhng kinh nghim cho cc n trong hc k sp ti.

thanh ho ngy 18 thng 6 nm 2011 Nhom Sinh vin thc hin nhm : 04

NHM THC HIN:NHM 04.

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AMH XUN HUY

GVHD: THS.NG

MC LC

trang

PHN I : TNG QUAN ............................................,61. S CN THIT CA TI ..................................................6 2. TIU CHUN NC U VO..............................................7 3. C TNH CA NC.............................................................7 4. CC PHNG PHP LM MM NC.............................11 5. CNG TRNH THU V TRM BM.....................................11

PHN II: XC INH CC THNG S1.P LC TN THT....................................................................12 2.HIN TNG KH XM THC...............................................13

PHN III : TNH TON THIT K CC CNG TRNH N V1.B HO PHN.................................................................................15 2.B TRN C KH............................................................................17 3.B LNG NGANG............................................................................19 4.B LC NHANH TRNG LC......................................................26 5.TNH TON MNG THU NC....................................................27 6. TNH TON H THNG RA LC.............................................30 7.TNH TON NG LC....................................................................30 8.TNH TON P LC TN THT KHI RA LC......................32 9.TNH TON CHON BM RA LC.............................................33 10.B CHA NC SCH..................................................................35

PHN IV : TNH TON CAO TRNH CNG NGH.........37 PHN V : TNH TON MT BNG X L........................39 PHN VI : TNH KINH T.....................................................41NHM THC HIN:NHM 04. LP:CDMT11TH

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AMH XUN HUY

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PHN I: TNG QUANS CN THIT CA TITrong cng nghip nng lng ca hi nc c s dng rng ri . Ni hi c cu to khc nhau ,ch yu l do cc thit b ph dng t cc nhin liu khc nhau (kh,madut,than, i khi c ci).V vy s lm vic chc chn v n nh ca l hi ph thuc rt nhiu vo cht lng nc cp cho l hi sinh hi. Trong nc sng ca Sng M c ho tan nhng tp cht,m c bit l cc loi mui canxi v magi v mt s mui cng khc.Trong qu trnh lm vic ca l hi,khi nc si v bc hi,cc mui ny ny s tch ra pha cng di dng bn hoc cu tinh th bm vo vch ng ca l hi.Cc cu v bn ny c h s dn nhit rt thp,thp hn so vi kim loi hng trm ln,do khi bm vo vch ng, s lm gim kh nng truyn nhit t khi n mi cht trong ng, lm cho mi cht nhn nhit t hn v tn tht nhit do khi thi tng ln, hiu xut l gim xung, lng tiu hao nhin liu ca l tng ln. Khi cu bm trn cc ng sinh hi, cc ng ca b qu nhit s lm tng nhit ca vch ln qu mc cho php c th lm n ng. Khi cu bm ln vch ng s tng tc n mn kim loi ng, gy ra hin tng n mn cc b. Ngoi nhng cht sinh cu, trong nc cn c nhng cht kh ho tan nh xy v nht l hm nc. Nh trnh by trn khng th dng nc thin cung cp nc ngay cho l hi m cn phi x l loi b cc tp cht sinh ra cu.

2. TIU CHUN V NC U VOCht lng nc cung cp cho l hi c ngha rt quan trng i vi vic m bo s an ton khi l vn hnh.Ngun nc cung cp cho l thng ln cc tp cht tan v khng tan trong nc. Nhng cht tan trong nc :thng dng lng cc v c th phn hy thnh ion nh:Ca2+,Mg2+,Na+,k+,Hco3-,Cl-. Nhng cht khng tan lm cho nc b c.Nhng hy nh c kch thcNHM THC HIN:NHM 04. LP:CDMT11TH

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AMH XUN HUY

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< 0,0001mm hu nh khng lng ng m l lng trong nc

3. C TNH CA NC: PH:L mt trong nhng ch tiu quan trng,c biu th th cht kim hoc acid ca nc. PH < 5,5 l nc c tnh acid mnh PH = 5,5 -6,5 l nc c tnh acid yu PH =6,5-7,5 l nc trung tnh PH =7,5-8,5 l nc c tnh kim yu PH >8,5 l nc c tnh kim mnh V ti chn nc u vo l nc sng M nn yu cu PH = 6,5 -7,5 l nc trung tnh. Ty theo cp phn ly ca acid trong nc c PH khc nhau, c th gip ta kho st qu trnh hnh thnh cu cn trong lo hi, v cc anion c th lk vi cc ion KL hnh thnh cc cht c ha tan khc nhau. Ngoi PH cn c cc ch tiu sau: cng, kim, kh kt,

CNGL ch tiu ht sc quan trng,n biu th nng cc ion Ca+ v Mg+ c trong nc v cng l kh nng bm cu cn trn b mt truyn nhit,thng o bng cng c hoc cng miligam ng lng. Da theo cng,c th chia nc thnh cc loi: Nc rt mm,c cng < 40H Nc mm c cng bng 4- 80H Nc binh thng c cng 8-160H Nc cng c cng 16-300H Nc rt cng c cng >3000 Ta chn nc rt mm ,c cng < 40H La chn cng ngh y kh cng cp noc sinh l hoat ta s dng pp ho cht. lam mm noc bng vi v s a pp c hiu qa i vi thnh phn ion bt ki ca noc. khi cho vi vo noc kh ocNHM THC HIN:NHM 04.

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AMH XUN HUY

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cng can xi v ma gie mc tong ong vi hm lng ion hirocacbonnat trong nc. nu cho vi vo noc sau khi a chuyn tt c CO2 v ion hirocacbonnat thanh ion cacbonat v lng xung doi CACO3 th tu trong noc c to cn khong tan Mg(OH)2 lm gim cng magie, nhng tng cng lc khong gim v CA2+ ca vi mi cho vo thay ion Mg2+ kt hp vi anion ca cc axit to thnh mui can xi ca cc axit mnh tan trong noc. cng ton phn : Ca2+ Co = 20,04 iu kin: Ca2+ + 20,04 HCO3= 61,02 Ca2+ 20,04 => cng cacbonat: HCO3Ck = 61,02 cng canxi: Ca2+ CCa = 20,04 cng can xi: Mg2+NHM THC HIN:NHM 04. LP:CDMT11TH

Mg2+ + 12,16 =

250 + 20,04

50 =16,6(meq/l) 12,16

Mg2+ = 16,6 (meq/l) 12,16 150 = 2,46 (meq/l) 61,02 Mg2+ + 12,16 > 61,02 HCO3-

= 2,46 (meq/l)

= 12,5 (meq/l)

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AMH XUN HUY

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CMg = 12,16 cng magi Mg2+ CMg =

= 4,1(meq/l)

= 4,1(meq/l)

12,16Trong trng hp nay, cacion ca2+ v magie 2+ cn d nm trong dng kt hp vi anion ca axit manh, ngoi vi phi cho thm vo noc ho cht co cha ion CO32- chuyn lng ion d Ca2+ ca vi thnh hp cht khong tan CaCO3 khi cho Na2CO3 vo nc ion CA2+ cn d s chuyn thnh cn theo phn ng: CaSO4 + Na2CO3 CaCl2 + Na2CO3 CaCO3 + CaCO3 + Mg(HO)2 + Na2SO4 2NaCl CaSO4

cn magi chuyn thnh cn do cho thm vi vo theo p: MgSO4 + Ca(OH)2 s

NHM THC HIN:NHM 04.

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AMH XUN HUY

GVHD: THS.NG

vi v soda

phn

Clo

gin ma

b trn c kh

lng tip xc

ti cacbonic

lc

bn b nn bn kh nc

b cha

4. cc phng php lm mm ncphng php ho hc: lm mm nc bng ho cht bng cch pha cc hp cht khc nhau vo nc kt hp vi ion Ca2+ to thnh hp cht khng tan trong nc phng php nhit :un nng hoc chng ct phng php trao i ion : lc nc cn lm mm qua lp cation

c kh nng trao i ion Na+ hoc H+.c trong tp ca cc ht cation kl vi ion ca2+ v mg2+ ho tan trong noc chung b gi li trn lp vt liu lc phng php tng hp :l phi hp 2 hay 3 pp trn

kimBiu th tng hm lng v mui ca acid yu. kim c nh hng xu n cht lng ca hi v tui th ca cc b mt truyn nhit.

kh kt:L tng hm lng cc vt cht cn li sau khi chng ct nc ,c o bng mg/ lNHM THC HIN:NHM 04. LP:CDMT11TH

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AMH XUN HUY

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V vy , m bo s an ton cc thit b l hi cn: Ngn nga to cu bm trn cc b mt t. Duy tr sch ca l hi mc cn thit. Ngn nga qu trnh n mn trong ng ng nc v hi.

5. Cng trnh thu v trm bm kt hp t trong lng sng, lng hTrch dn hinh minh ha H2.1 trang 25-ttct x l &pp nc-Trnh Xun Lai 1.My bm chm t thp hn mc nc thp nht song H10,5m 2.Ming ht ca my bm v tr cao hn y song H2 1m. Do mc nc cht ca bi l 1m nn ta phi o su xung t 1,5m-2m 3. ngn nga vt ni trong lng sng v ma l (g,bo,lc bnh,xc xc vt )phi bc li B40 xung quanh cc tr sn trm bm. ngn nga rong ru,rc,ti bong i vo ming ng ht ca my bm,t mt lng (kiu lng chim)li chn ngoi my bm. Lng lm bng khung thp,ngoi qun li ng,ng knh si dy ng l 1mm.Mt li 2 2mm. Lng c ng knh ln hn ng knh my bm 50mm,chiu cao ln hn chiu cao my bm 0,3- 0,4m. t ph t nh my bm ko di su xung di ming ht.Qua kinh ngim cho thy li ng ngoi nhim v ngn rc cn c tc dng chng ru,h,c,bm vo my bm,n mn my bm.

NHM THC HIN:NHM 04.

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GVHD: THS.NG

PHN II: Xc nh Cc Thng S1. p lc cng tc:H1 = 1,2m , H3 = 0,5m , H3 = 11m H1 l tn tht p lc trn ng ng ht Ta c :

Trong :

H5 l tn tht p lc trn ng ng y H = H3 + Htt = 11+ (0,5+1,2)=12,7m Cng sut tn tht : N = N= (hp) (kw)

Q = 0,1157m3/ s, H= 12,7m , Y=1000kg/m3. N= N= = = =2,4 (kw)

=3,23 ( hp)

H2 = 2m, V1=1,2m/s, h1 =1,2m,

=5,5NPSH: Tac:

=

-(

+ H2+H1) +2+1,2 ) = 2,227

=5,5 (

Ta c: 600 vng, C=700, Q= 0,1157m3/s > 10 )^ = 10 ( = 1,92 (m)

2. Hin tng kh xm thc:NPSH b = 2,227 > Hbh +

h =(1,92+0,24)12LP:CDMT11TH

NHM THC HIN:NHM 04.

AMH XUN HUY

GVHD: THS.NG

Tnh ton sc va thy lc khi my bm ngng hot ng. D= 300mm, L=1000m, Hd=30m, Zd=2m

Zk=1m,

Hckmax=5m,

V=1,2m/s,

q=1162m/s ( Hd+ Zd) ( 30+2) = 0,93 m/s

Tnh ton vn tc d :

D d =Vo D d = 1,2 -

Kim tra nc trong ng b t dng khng Z = Hck K x V2 d . Ly Hck = 8m Ta c: Ta c: = , k= = 6,1m Z = 2,7m > Zd = 2m

= 142,3 =

(dao ng p lc khi ng van)

Ta c :

> 2( Hd + Zd) 142,3 > 64

p lc ln nht khi va H max = 2Hd+Zk+hck max + =60 + 1+5m = 66m.Mt khc , H max = Hd + Zk +

= 173,2m

Vy p lc cho php ca van v ng khi va Hx = 80m,vy phi t thit b chng va

Lng nc cn x gim p lc 63 80m Hva = Hmax -

=

vc=

Lu lng cn x : Q =F.Vc= Chn thit b chng va c c tnh m van khi p lc p=80m, Qmin = 60 l/s p lc va ln nht :Hmax= 173 m , Qmax= 100 l/s ,iu kin ng dn D=300mm, ng dn bng gang m/s,V0=1.2 m/s Tn tht p lc trong ng ng : H= a=1162

= i.l Vi i =

Chn i = 1.5 ng vi d = 300 mmNHM THC HIN:NHM 04.

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AMH XUN HUY

GVHD: THS.NG

H = 1.5

III. TNH TON THIT K CC CNG TRNH N V1) B ho phnC nhim v ho tan phn cc v lng cn bn. Trm c cng sut kh ln 41000 m3/ng Dng b ho phn khuy trn bng cch sc kh nn. B xy dng bng b tng. Sn phn gm cc thanh g xp cch nhau 10-15 mm, sn cch y b 0,5 m. Bn di sn t mt dn ng phn phi kh nn. Dung tch b ho trn: Wh= Trong : + Q : cng sut trm, Q=10000 m3/ng = 416.67 m3/h + LP : liu lng phn cho vo nc. LP = 25,298 (g/m3) + bh : nng dung dch trong b ho, bh=10%. + :khi lng ring ca dung dch ( y l nc). =1 T/m3 + n : s gi gia hai ln pha ch, ph thuc Q. Q=10000m3/ng n=4 gi. Wh=416,67.7.25,298 = 0.738 (m3) 10000.10.1 Q.n.L P 10000.bh .

(m3)

Chn hai b ho trn, dung tch mi b : Wh = 1 m3. Kch thc mi b : 1 x1x1 m Dung tch b tiu th: +bh=10% +btt:nng dung dch trong b tiu th, btt=52.10 =4 wtt= 5 NHM THC HIN:NHM 04.

W = tt

+Wh: dung tch b ho trn, Wh=2 m3

W .bh h btt

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B ha ph n

ng cp nu c ng cp kh

B tiu th

dung d ch ghi ph n cc ng ph n phi gi b trn ng

ng x c n

ng x c n

2.

B TRN C KH Thit k b trn c kh cho nh my nc cng sut 10000 m3 / ngy

Hnh 7.3: S cu to b phn ng c kh 1.ng dn nc t b trn sang 2. Ngn phn phi nc 3. Cc my khuy trc ngNHM THC HIN:NHM 04. LP:CDMT11TH

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4. Cc vch ngn c l 5. Ngn thu nc 6. Mng phn phi nc vo b lng Thi gian khuy trn 3s Cng khuy trn :g = 1000/s Nhit nc t = 200c Th tch b trn V = 3. 0,1157 m3/ s = 0,347m3 Chn b trn vung : a.a.h = 0,34. 0,34.3 ng dn nc vo ca y b , dung dich phn cho vo ca ng dn vo b, nc i t di len trn qua mng trn l mt pha ca thnh b dn sang ngn phn ng. Dng my khuy tua bin 4 cnh nghing gc 45o hng ln trn a nc t di ln ng knh cnh khuy D < chiu rng b Trong b t 4 tm chn ngn chuyn ng xoy ca nc, chiu cao tm chn3m , chiu rng 3cm,=0.03m=1/10 ng knh b Trch dn hinh minh ha H9.5 trang 166-ttct x l &pp nc-Trnh Xun Lai

My khuy t cch y H = D ng knh my khuy Chiu rng cnh khuy = 1/5 k my khuy Chiu di cnh khuy = Nng lng cn chuyn vo nc : P = G2V Hiu sut ng c Cng sut ng c : P = N =( =( = 0,5 = 13.2 (vng/s) 0,001 =350 l/s = 0,3 kw

Tnh ton b to bng c kh Dng cnh khu tuabin trc ng 4 cnh nghing gc 45 o qut nc xung y b xoay v ti cn lng ng y b khi phi ngng hot ng T=200NHM THC HIN:NHM 04.

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GVHD: THS.NG

Thi gian keo t 20 pht. Cng khuy 3 bc G1 = 70 , G2 = 50 ,G = 30 Lu lng nc nc x l Qmax= 0.1157 m3/s

Thit k 2 b to bng dung tch 1 b V = (Chn h = 4m F = B chia lm 3 ngn bi cc tm chn khoan l D = 150 mm Vn tc nc qua l trn vch ngn V = 91 m/s Cch b tr tm chn, my v kch thc Th tch nc khuy trn ca 1 my V = 1.7 Cng sut tiu th cn thit ca my my khuy bc 1 P1=G2v =702 Chn my khuy iu kin D = 0.5 m tuabin 4 cnh nghing gc 450, hng di vng quay ca ng c: n=((

m3

xung

= 1.2 (vng/s)

Hiu sut ng c 75% => Pc = = 0.086 kw

3) B lng ngang:B lng ngang c nhim v lm sch s b trc khi a nc vo b lc hon thnh qu trnh lm trong b. Ta tnh vi b lng ngang dng lng cn c keo t. Ta chn tc lng tnh ton u0=1,625 m/h t hiu qu lng R=70%, tiu chun nc ra c c 2NTU a) Tnh ton kch thc b S b lng : chn 2 hoc 3 b => chn 2 b d n gin, r tin. S mt 2 b lng : F = Chiu di l = T s = = = 256 m2 , chn chiu rng b B=4

4mg/l.

D tnh thit k hai b lng vi cng sut ca nh my l Q=10000 m3/ngay = 0,1157m3/s

NHM THC HIN:NHM 04.

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GVHD: THS.NG

Chon h = 2m , => V ngang = V0 =

= =

= 16 >1 5 ( t/m) = 7,2 ( mm/s) < 16,3 mm/s

V xoy cn nht ng hc To = 20oc V= 1,01

m2 /s

Re =

=

=14319 < 20.000 (t/m)

Tnh Fr m bo iu kin n nh dng v khng to ra vng nc cht trong b

Fr = Dng n nh .

=

=2,1

>

Vy ta s xy 2 b lng.Mi b kch thc rng 4m mi b t 1 my ca cn c nh = dy xch chiu di b = 32m Vng phn phi nc vo : hiu qu lng ph thuc rt nhiu vo kt qu lm vic ca b to bng cn mng. Hoc ng dn nc t b to bng cn sang b lng lm sao khng ph v bng cn ,ng thi khng bng cn lng xung y. Mng dn v phn phi cng b lng cng tt Vn tc trong mg : v = 0,3 /s m bo phn phi u nc vo 2 b lng mi b t 3 ca ly nc t mng dnchung vo ca ly nc t van bm vo iu chnh lu lng v tn tht p lc qua ca. Tn tht p lc qua ca thu: Chn h 0,01 m m ba phn phi qua u 6 ca Theo nguyn tc phn phi tr lc ln: Ca thu : 600mm. vn tc qua ca

NHM THC HIN:NHM 04.

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AMH XUN HUY

GVHD: THS.NG

V=

=

=0,07 m/s =1 = 3,6.10-3

p lc tn tht H =

b) Tnh ton tm phn phi nc vo b : khng gy ra hin tng ngn dng, khng to ra cc vng nc cht v cc vng xoy nh lm gim hiu qu lng th mt iu kin quan trng cn c t ra l nc i vo b lng cn c phn phi u trn ton b mt ct ngang ca b. Bin php hiu qu nht to ra s phn phi u vn tc l dng cc tm phn phi khoan l. a) Tnh ton kch thc b : - S b lng ngang chn l N=2 b - Din tch mt bng ca mi b : F=Q 416,67 = = 109,65m2. (u0 = 0,53 mm/s = 1,9 m/h) N .u 0 2.1,9

- Ta chn phng php co cn c kh vi h thng co t trn dm cu chy, kch thc b chn B x L = 9 x 50 m, m bo t sL 50 = = 5,6 >5. B 9 L 50 = = 16,67 >15. H 3

- Chn chiu cao vng lng H =3m, m bo t s - Vn tc dng chy ngang trong b : v0 =

Q 416,67 = =10,716 m/h = 5,35 mm/s. N .B.H 2.9.3

v0H1 =( H11,5 )1,5 = 2,17.10-5

= 4,2.10-4 m3/s

)1,5=

C tng chiu di mng 1 b = L/2 = 28/2 = 14m t mi b 4 mng thu nc, mi mng di 11,2m. - Kim tra ti trng thu nc trn 1m di mp mng : q=Q 115,7 1,35 l/s.m di = 2.L 2.90

q < 3 l/s.m m bo yu cu - Khong cch gia cc tm mng l : 1,8m. - 4 mng phi ti mt lu lng lng l0,1157 = 0,05785 m3/s, mi mng phi ti mt lu 2

0,05785 = 0,0145 m3/s, ly vn tc nc dn trong mi mng l 0,3m/s ta s c tit 4 0,0145 = 0,048 m2, chn kch thc tit din ngang ca mng l bxh = 3

din mi mng l 20x10cm.

- Cc rnh thu nc mi bn mp mng ta b tr dng rng ca, vi gc ca ch V l 900 . Chiu di mp mng ca mi mng l 20m, ti trng nc thu 1m di mp mng l q=0,0145 = 0,000725 m3/s. Ta chn khong cch gia cc nh rng trn 1m di l 20cm, tc 20

trn 1m di s c 5 rng thu nc, lu lng nc thu trn mi rng l q 0=0,000725 = 0,000145 m, chiu cao mi rng l hR=4,49 cm, ta chn hR=5 cm. 5 NHM THC HIN:NHM 04.

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AMH XUN HUY

GVHD: THS.NG

50

90 100 200 200 200

200

m ng thu nu c500

B l n nag g n g

3000

i = 2%

120 10

500 00

4.) B lc nhanh trng lc:- B lc c tnh ton vi 2 ch lm vic l bnh thng v tng cng. - Dng vt liu lc l ct thch anh vi cc thng s tnh ton:

dmax = 1,6 (mm) dmin = 0,7 (mm) dtng ng =0,8 1,0 (mm)- H s dn n tng i e = 20%, h s khng ng nht k = 2,0. - Chiu dy lp vt liu lc = 1,2 (m)NHM THC HIN:NHM 04. LP:CDMT11TH

23

9000

2000

150

AMH XUN HUY

GVHD: THS.NG

- H thng phn phi nc lc l h thng phn phi tr lc ln bng chp lc u c khe h. Tng din tch phn phi ly bng 0,8% din tch cng tc ca b lc (theo quy phm l 0,8 1,0 m). Tng din tch b lc ca trm x l : F = T .v 3,6.W .t at .v bt 1 2 bt Trong : Q =416,67m3/h = 10000m3/ng .Cng sut ca TXL. T : Thi gian lm vic ca 1 trm trong mt ngy m (gi). T = 24h vbt : Vn tc lc tnh ton ch lm vic bnh thng (m/h) - Tra bng vi b lc nhanh 1 lp vt liu lc vi c ht khc nhau, dt = 0,8 1mm, vbt = 7m/h. a : S ln ra mi b trong 1ng ch lm vic bnh thng, ly a = 2 ln. W : Cng ra lc (l/s_m2).Tra bng :W = 10(l/s_m2) t1 : Thi gian ra lc (gi). t1 = 6 ' = 0,1 gi t2 : Thi gian ngng b lc ra ,t2 = 0,35 gi F =10000 63 m 2 24 .7 3,6.10 .0,1 2.0,35 .7

Q

- S b lc cn thit: N = 0,5F

= 0,5

6 3

= 4(b)

- Din tch mi b lc : f=F 63 = 16 m 2 N 4

- Chn kch thc mi b : LxB = 4x4 m - Kim tra li tc lc tng cng khi ng mt b ra hoc sa cha. vtc = vbt. N N = 7 4 1 9 (m/h) 1NHM THC HIN:NHM 04.N 4

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+ N1 : S b lc ngng sa cha :N1 = 1 vtc = 9m/h < vtccf = 8 10m/h H = h + hv + hn + hp (m) Trong : h :Chiu cao lp si (m).Tra bng h = 0,4 m (ra bng gi nc kt hp). hv :Chiu dy lp vt liu lc. hv = 1,2 m hn :Chiu cao lp nc trn lp vt liu lc (m):hn 2 m.Ly hn=2m hP :Chiu cao ph k n vic dng nc khi ng 1 b ra. hP = 0,5m H = 0,4 + 1,2 + 2 + 0,5 = 4,1 m 5. TNH TON MNG THU NC RA LC Chn dc y mng theo chiu nc chy i = 0,01. Mi b b tr 2 mng thu. Khong cch gia cc tm mng l 2 (m) < 2,2 (m) Khong cch t tm mng n tng l 1 (m) < 1,1 (m) Lu lng nc ra mt b lc l: qr = F1b W (l/s) Trong : Chiu cao ton phn ca b lc nhanh :

- W: Cng nc ra lc, W = 10 (l/s.m2) - F1b: Din tch ca mt b: F1b = 16 (m2) qr = 10 16 = 160 (l/s) = 0,160 (m3/s)0,1602

Do mt b b tr 2 mng thu nn lu lng nc i vo mi mng l: q1m = =0,8 (m3/s)

Chn mng hnh tam gic, ta i tnh ton mng dng ny. Chiu rng ca mng c tnh theo cng thc:NHM THC HIN:NHM 04.

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GVHD: THS.NG

Bm = K Trong :

5

qm2 (1,57 a) 3 +

+ qm : tnh ton trn = 0,8 (m3/s) + a: T s gia chiu cao hnh ch nht v mt na chiu rng mng, a = 1,5 (quy phm l 1 1,5) + K: h s ph thuc vo hnh dng ca mng, vi mng c tit din y hnh tam gic ta ly K = 2,1 Bm= 2,15

(0,8) 2 0,06 (m) (1,57 + 1,5) 2

Chiu cao ca phn mng ch nht H1 =1,5 Bm 1,5 0,06 = = 0,045 (m) 2 2 1 0,06 = 0,03 (m) 2

Chiu cao ca y mng tam gic H2 = 0,5 Bm = Chiu cao ton b mng Hm = H 1 + H 2 + Trong : m m

(m)m

l chiu dy y mng, ly

= 0,1 (m)

Do Hm = 0,045 + 0,03 + 0,1 = 0,175 (m) Kim tra khong cch t b mt lp vt liu lc ti mp trn ca mng thu nc c xc nh theo cng thc: h= Trong : + e : trng n ca vt liu lc khi ra, e = 20% + H : Chiu cao lp vt liu lc (m) => h =1,2 20 + 0,25 (m) = 0,49 (m) 100 H e + 0,25 (m) 100

Theo quy phm, khong cch gia y di cng ca mng dn nc ra phi nm cao hn lp vt liu lc ti thiu l 0,07 (m).NHM THC HIN:NHM 04.

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AMH XUN HUY

GVHD: THS.NG

Chiu cao ton phn ca mng thu nc ra l: Hm = 0,175 (m) . V mng dc v pha mng tp trung 0,01, mng di 8 (m) nn chiu cao mng pha mng tp trung l: Hm = 0,175 + 0,08 = 0,255 (m) Do khong cch gia mp trn lp vt liu lc n mp di cng ca mng thu Hm phi ly bng: Hm = 0,255 + 0,07 = 0,325 (m)

Mng thu kiu y hnh tam gic.Khong cch t y mng thu ti y mng tp trung nc :

hm= 1,75 Trong :

3

q2m + 0,2 (m) g B2m

- qm : Lu lng nc chy vo mng tp trung nc; qm =qr = 0,23328 ( m3/s) - Bttm: Chiu rng ca mng tp trung , Theo quy phm, chn Bttm = 0,7 (m) - g : Gia tc trng trng, g = 9,81 m/ s2Vy: hm = 1,753 2 0,23328 2 9,81 0,7

+ 0,2 (m)

hm = 0,59 (m) Chn vn tc nc chy trong mng khi ra lc l 0,8 (m /s) Tit din t ca mng khi ra l: Fmng = Fmng =

qr ( m2 ) Vk0,23328 = 0,29 ( m2) 0,8

Chiu cao nc trong mng tp trung khi ra l: h = B = 0,7 = 0,42 (m) mF

0,29

NHM THC HIN:NHM 04.

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GVHD: THS.NG

Theo TTVN 33.85 y ng thu nc sch t nht phi cch mc nc trong mng khi ra l 0,3m, vy ta phi b tr ng thu nc sch c ct y ng cch y mng mt khong 0,75 (m)

6. TNH TON H THNG RA LCB c s dng h thng phn phi nc tr lc ln l sn chp lc. Ra lc bng gi v nc kt hp. Quy trnh ra b: u tin, ngng cp nc vo b. Khi ng my sc kh nn, vi cng 18 (l/s.m2), cho kh nn sc trong vng 2 pht. Cung cp nc ra lc vi cng 2,5 (l/s.m2), kt hp vi sc kh trong vng 5 pht. Kt thc sc kh, ra nc vi cng 8 (l/s.m2) trong vng 5 pht. Cung cp nc vo b tip tc qu trnh lc v x nc lc u 7. TNH TON S TRC LC S dng loi chp lc ui di, loi chp lc ny c khe rng 1 (mm). S b chn 50 chp lc trn 1 (m2) sn cng tc, tng s chp lc trong mt b l: N = 50 F1b = 50 29,16 = 1458 (ci) B tr 40 hng chp lc, mi hng 40 ci tng cng 1600 Sau pha ra gi nc ng thi, cng ra nc thun tu l W = 8 (l/s.m2) Lu lng nc i qua mt chp lc l: q =W 8 = = 0,16 (l/s) = 1,6 10-4 (m3/s) N 50

Ly din tch khe chp lc bng 0,8% tng din tch sn cng tc, tng din tch khe chp lc trong mt b = 0,829,16 = 0,233 (m2) 1000,233 = 1,07 10-4 (m2) 1600

Din tch khe h ca mt chp lc l: f1 khe =

Vn tc nc qua khe ca mt chp lc khi l: V= f 1k e hq

1,6 10 -4 = = 1,5 (m/s) m bo theo quy phm. 1,07 10-4

Vy chn 50 chp lc trong 1m2 b, khong cch gia tm cc chp lc theo chiu ngang v chiu dc b u l 13,17 (cm).NHM THC HIN:NHM 04.

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GVHD: THS.NG

Tn tht qua h thng phn phi bng chp lc l: hPP = Trong :V 2K 2 g 2

(m)

(Theo 6.114 TCVN 33.85)

-

VK : Vn tc nc qua chp lc; VK = 1,5 (m/s)

g

: H s lu lng ca chp lc, v dng chp lc c khe h nn =0,5 : Gia tc trng trng, g = 9,81 (m/s2) hPP =2 = 0,459 (m) 2 9,81 0,5 2 1,5

Tnh ton cc ng ng k thut ng ng dn nc ra lc Lu lng nc cn thit ra 1 b lc: Qr = 3,6.f.W (m3/h) f: din tch 1 b, f = 29,16m2 W: cng nc ra, W = 8l/s.m2 Qr = 3,6.29,16.8 = 839,808(m3/h) = 233,28(l/s) Ta chn ng knh ng l 450mm, v = 1,5m/s H thng cp kh Cng ra gi thun tu l: W = 18 (l/s.m2) Vn tc ca gi trong ng l: V = 20 (m/s) (quy phm l 15 20 m/s) Lu lng gi cung cp cho mt b l: qgi = W F1b = 18 29,16= 524,88 (l/s) = 0,525 (m3/s) ng knh ng dn gi chnh: dd =4 qgi

V

=

4 0,525 = 0,183 (m) 3,14 20

Chn ng dn gi c ng knh l: 200mm ng ng thu nc sch ti b cha : S dng 2 ng ng thu nc t 4 b lc v b cha. ng ng c t trn cao trong khi b lc v xung thp khi ra khi khi b lc.NHM THC HIN:NHM 04.

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ng t mt b ra ng thu nc sch chung ti 1 lu lng l 52,08l/s chn ng knh ng 250mm. ng ng chung s nhn lu lng tng dn, do ng knh ng cng tng dn. C th: ti b lc u tin, ng ti lu lng 1 b = 52,08l/s chn ng knh 250mm. n b lc th 2, ng nhn lu lng = 52,08.2 = 104,16l/s chn ng knh 300mm n b lc th 3, ng nhn lu lng = 52,08.3 = 156,24l/s chn ng knh 400mm n b lc th 4, ng nhn lu lng = 52,08.4 = 208,32l/s chn ng knh 450mm. ng ng x kit Ly ng knh ng l D100 (mm). ng ng x ra lc Lng nc x chnh bng lng nc cp cho ra lc. Qx = 233,28l/s Ly ng knh ng bng ng knh ng cp nc ra lc l D450 (mm). 8. TNH TON P LC TN THT KHI RA LC Tng tn tht qua sn chp lc Theo tnh ton trn l: 0,459 (m) Tng tn tht qua lp vt liu h = 0,22 L W (m) Trong :

-

L :Chiu dy lp si dy = 0,2 (m) W : Cng nc ra lc = 8 (l/s.m2) h = 0,22 0,2 8 = 0,352 (m) hVLL = ( a+ b W) hL

Vy

Tn tht p lc bn trong lp vt liu lc Trong :

- a,b l cc thng s ph thuc ng knh tng ng ca lp vt liu lc, vi d t= 0,8-1 (mm) => a = 0,76; b = 0,017NHM THC HIN:NHM 04.

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GVHD: THS.NG

- hL : Chiu cao lp vt liu lc = 1,2 (m)Vy hVLL = ( 0,76+ 0,017 8) 1,2 = 1,08 (m) Tng tn tht trn ng ng dn nc ra lc Tnh vi 1 ng: h = hdd + hCB Trong :

- hdd: Tn tht trn chiu di ng t trm bm nc ra n b cha. S b chn bng100 (m). Theo tnh ton trn ta c lu lng nc chy trong ng 0,4167/2 = 0,20836 (m3/s), ng knh ng Dchung = 450 (mm). Tra bng ta c 1000 i = 5,13 hdd = i L = 5,13100 = 0,513 (m) 1000

qr =

- hCB : Tn tht p lc cc b trn van kho, s b chn hCB = 0,3 (m)Vy h = 0,513 + 0,3 = 0,813 (m) TINH TON CHN BM RA LC hB = h +hr+hdt + hdl (m) Trong : V c 2 ng ng nn h = 2.0,813 = 1,626m 9. p lc cn thit ca bm ra lc c tnh theo cng thc:

- h : chnh lch hnh hc gia mc nc thp nht trong b cha nc sch ticao mng thu nc, c tnh theo cng thc: h = h1 + hK + hS + h + hl + Hm + Hm Vi ly

h1 : chnh gia ct M ti trm x l v cao MNTN trong b cha,hK hS h hl : Chiu cao hm phn phi nc: hK = 1 (m) : Chiu dy sn chp lc, hS = 0,1 (m) : Chiu cao lp vt liu ; h =0,3 (m) : Chiu cao lp vt liu lc; hl = 1,2 (m)

h1 = -0,5m

NHM THC HIN:NHM 04.

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AMH XUN HUY

GVHD: THS.NG

Hm : Khong cch t mp di ca mng phn phi n lp vt liu lc, Hm = 0,724 (m)Hm : Chiu cao mng thu nc ra lc; Hm = 0,6 (m) => h = -0,5 + 1 + 0,1 + 0,3 + 1,2 + 0,724+ 0,6 = 3,424 (m)

- hr : Tng tn tht p lc khi ra lc:hr = hPP + hVLL+ h (m) Theo tnh ton trn ta c: hr = 0,459 + 1,08 + 0,352 = 1,891 (m)

- h: Tng tn tht trn ng ng dn nc ra lc:h = 1,626 (m)

- hdt : p lc d tr ph v kt cu ban u ca ht vt liu lc, ly hdt = 2 (m)Tm li: hB= 3,424 + 1,891 +1,626 + 2 = 8,941 (m) tin cho tnh ton, ly hB = 9 (m). Vy chn bm nc ra lc c Qr = 0,23328 ( m3/s) v p lc Hr = 9 (m) Chiu cao xy dng b lc Chiu cao xy dng b lc c xc nh theo cng thc: Hxd = hk + hS + hd + hl +hn + hBV Trong :

- hk , hS , hd , hl : l cc h s c trnh by trn hBV = 0,5 (m)

- hn : chiu cao lp nc trn vt liu lc 2 (m)Hxd = 1+ 0,1+ 0,3 + 1,2 + 2 + 0,5

Hxd = 5,1 (m)

5400 Lp nc trn vt liu lc

Lp vt liu lc 760

NHM THC HIN:NHM 04.

Lp vt liu

32

LP:CDMT11TH

Sn chp lc Hm thu nc

AMH XUN HUY

GVHD: THS.NG

500

2000 5100 1200 300 100

1000

10. b cha nc schThit k b cha nc sch c dung tch = 20% Qtrm do dung tch b: Wb =20 36000 = 7200 (m3) 1003600

Thit k 2 b vung, mi b c dung tch 3600 (m3/ng) vi chiu cao mi b l 4,5 (m) Din tch mt b l: F1b = 4,5 = 800 (m2) Vy kch thc 1 b l 28,3 28,3 800 (m2)NHM THC HIN:NHM 04.

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- Tng chiu cao ca b l : HB = HNB + Hbv Trong : Hbv :Chiu cao bo v ,ly Hbv = 0,5m HB = 4,5 + 0,5 = 5m

Tnh ton kho chun b cloa) Tnh lng Clo cn dng: - Lng clo cn thit trong mt gi xc nh theo cng thc : qCl2 = Trong : Q :Cng sut trm ; Q = 36000m3/ng = 1500m3/h LCl :Lng clo cn thit kh trng LCl = LClS b + LClKh trng LClS b =18,608 mg/l = 18,608g/m3 LClKh trng :Lng clo dng kh trng nc trc khi dn nc vo b cha nc sch.Ly theo tiu chun: Vi nc mt LClKh trng = 3mg/l = 3g/m3 LCl = 18,608 + 3 = 21,608g/m3 - Lng clo kh trng trong 1 gi l: qClo =1500 .21 ,608 1000

Q.LCl 1000

(kg/h)

= 32,412 (kg/h)

lng clo tiu th trong 1 ng l 777,888kg/ng, trong 1 thng l 23,34T/thng. Tnh s Clorat - Clorat dng nh lng clo hi vo nc - Chn loi clorat chn khng - Lng clo tiu th trong mt ngy l : C = qClo.24 = 32,412.24 = 777,888 (kg/ng) - Dng bnh Clo lng c dung tch 500 (l) ,nng sut bc hi ca Clo l S = 5 kg/h_m2 trong iu kin bnh thng. - S lng bnh lm vic ng thi l :NHM THC HIN:NHM 04.32 ,412 5

= 7 bnhLP:CDMT11TH

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GVHD: THS.NG

- S bnh Clorat d tr l 2 bnh. Mi bnh c t ln mt bn cn.

- S trm clo c th hin trn hnh.Phn IV

Tnh ton cao trnh cng nghTnh ton cao trnh cng ngh da vo tn tht ca tng cng trnh v tn tht trn ng ng dn nc ti cng trnh . m bo nc trong trm l t chy Ly ct mt t ti y b lc bng 0.00m

1 - B lc nhanh:

- T ct mt t ti b lc = 0, chiu cao ton b b lc = 5,1m. Mc nc trong bcch thnh trn b 0,5m Ct mc nc cao nht trong b lc: Z2 = 5,1 - 0,5 = 4,6m. 2 - B cha nc sch: Mc nc cao nht trong b cha nc sch: Z1 = Z2 - hngB.lc-BC - hB.lc Trong : Z1 :Ct mc nc cao nht trong b cha nc sch hngB.lc-BC :Tn tht p lc trn ng ng dn t b lc n b cha nc sch . C th ly s b hngB.lc-BC = 1m hB.lc :Tn tht p lc trong ni b b lc. hB.lc = 2,5m Vy Z1 = 4,6- 1- 2,5 = 1,1 m 3- B lng ngang: - Mc nc cao nht trong b lng ngang l : Z3 = Z2 + hngB.lng-B.lc + hB.lng Trong : hngB.lng-B.lc :Tn tht p lc trn ng ng dn t b lng n b lc . Ly s b hngB.lng-B.lc = 0,7m (quy phm 0,5 1m)NHM THC HIN:NHM 04.

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GVHD: THS.NG

hB.lng :Tn tht p lc trong ni b b lng hB.lng = 0,6m Vy Z3 = 4,6+ 0,7 + 0,6 = 5,9m 4 - B phn ng zc zc ngang: - Ct mc nc cao nht trong b phn ng l Z4 = Z3 + hngB.p-B.lng + hP Trong : hngB.p-B.lng :Tn tht p lc trn ng ng dn t b phn ng n b lng. V b phn ng lion vi b lng nn hngB.p-B.lng = 0m hP :Tn tht p lc trong ni b b phn ng hP = 0,4m Vy Z4 = 5,9 + 0,4 = 6,3 m 5- B trn: Ct mc nc cao nht trong b trn l: Z5 = Z4 + hBT_Bf + hBT hBT_Bf : tn tht t b trn n b phn ng, hBT_Bf = 0,4m hBT: tn tht trong b trn, hBT = 0,3m Z5 = 6,3 + 0,4 +0,3 = 7m.

NHM THC HIN:NHM 04.

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GVHD: THS.NG

NHM THC HIN:NHM 04.

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GVHD: THS.NG

Phn VTnh ton mt bng trm x l1. Din tch trm kh trng : - Trm kh trng c t cui hng gi. - Din tch trm kh trng ly theo tiu chun l : - 3m2 cho mt Clorat *Trm c 7 Clorat lm vic v 2 Clorat d tr - 4m2 cho mt cn bn (c 7 cn bn) Vy tng din tch ca trm l : F = 9.3 + 7.4 = 55 m2 - Trm c chia lm 2 gian : *Mt gian cha Clorat c din tch :f1 = 27m2

*Mt gian cha bnh Clo lng c din tch :f2 = 28 m22. Din tch sn phi ct - Din tch sn phi phi m bo phi ton b lng ct trong cc b lc ,chiu dy lp ct phi bng 0,2m. - Th tch khi ct 1 b lc bng din tch b nhn vi chiu dy lp vt liu lc: VB = FB.HB = 29,16.1,2 = 35m3 - Din tch sn phi ct bng th tch khi ct chia chiu dy lp ct phi: S =35 0,2

= 175 m2

0,2 :Chiu dy lp ct phi (m). Chn 2 sn phi ct b tr 2 bn b lc175 2

- S1sn =- Kch thc sn l : 3. Din tch trm bm cp II

= 87,5 m2

vi cng sut trm l Q = 36000m3/ng ta ly STBII = 150m2NHM THC HIN:NHM 04. LP:CDMT11TH

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4. Trm bin th - Ly theo quy phm .Kch thc (4 x 4)m = 16m2 5. Phng bo v - Trm c Q =36000m3/ng ly Sbv = 10m2 .Kch thc (5 x 2)m 6.Nh hnh chnh S = 120m2 kch thc (6 x 20)m 7. Nh c kh - kho - Ly theo quy phm S = 30m2.Kch thc l (6 x 5)m. Mt bng quy hoch trm x l c th hin trn hnh v. 8. Phng th nghim ho nc Din tch 42m2. Kch thc l (6 x 7)m. Tnh Ton x l nc sau lc Mi ngy ra 8 b .Chu k ra 1 b 24gi Lng cn dc d li 2 b W=9000.(12-3)= 81000 000 mg Nng cn trong nc ra : C=W/V Lu lng nc ra : V=2.6.12.60.33 Mi ln ra ln lt 2 b = 285120 l C= 81000 000/285120 =284mg/l Do C=284 < C0=1900 mg/l .V vy a nc quay li b trn Dung tch b iu ho Chiu su lp nc 3 m Din tch b F= 285,120/3 =10x10 m Tnh Ton Sn Phi Bn Dung tch cn trong mt ngy 700 m3 Thi gian tch cn trong sn l 7ngy Chiu su sn phI bn 3 m Din tch sn phi :f=NHM THC HIN:NHM 04.700 .7 = 2450 m2 =50x50 m 3

lng nc ra 2b

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PHN VI TNH KINH T Kinh ph xy dng tram x lon v n gi vn d

Stt A. I. II. 1/.

Vt t thit b chi phi trc tip cng trinh thu v trm bm cm x l gin ma than cc b tng

s lng

thnh tin 4,6600

7.939,155 m3 m3 75,6 6,2 5.200.000 393,12 1.000.000 6,2 8,9

my khuy tua bin 4 cnh nghieng gc 45 3/. P=4,1124kw b lng b tng gung cnh khuy p = 161,675 kw 4/. b ti cacbonic b tng my thi kh nn p = 132,5 w ci 2 16.000.00 32 0 m3 ci m3 ci 3,9 2 1.000.000 3,9 45.000.00 90 0 400 2 1.000.000 400 5.000.000 10

5/.

b lc b tng m3 131,72 1.000.000 131,72LP:CDMT11TH

NHM THC HIN:NHM 04.

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bm gi ci bm nc ra lc ci ct lc than angthraxit si - 16 32 - 8 16 - 4 8 6/. - 2 4 i nc ra lc b tng - Q=87m3/h; H=22m. 7/. b cha v trm bm cp 2 b tng - bm Q= 300m3/h; 8/. 9/. 10/. 11/. 12/. 13/. 14/. 15/. H=22m. my pht in phng th nghim nh hnh chnh kho sn nn v ng ni b nh xe cng v hng ro bo v nh ho cht b tng m3 ci b m3 m3 m3 m3 m3

5

90.000.00 450 0

5

65.000.00 325 0

68,75 68,75 55 13,75 13,75 13,75

500.000 34,375 2.000.000 137,5

250.000

24,07

m3 m3 Ci

1.000.000 15,97 15,97 2 100.000.0 200 00

423 3 2

1.000.000 423 150.000.0 450 00 533.000.0 1.066 00 700 1.000 200 900 60 700

m3

87,4 41

1.000.000 87,4LP:CDMT11TH

NHM THC HIN:NHM 04.

AMH XUN HUY

GVHD: THS.NG

nh lng phn vi Q=1m3/h; H=30m. III. mng li ng ng

ci

3

30.000.00 90 0 5.000

NHM THC HIN:NHM 04.

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AMH XUN HUY

GVHD: THS.NG

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