Đồ án xử lý nước cấp - nước mặt

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N MN HC XL NC CP THIT KTRM XL NC CP

GVHD: THS. NGUYN THNH MUSVTH : NGUYN XUN NHT 11M Page 1

LI NI U

Sau qu trnh hc tp v rn luyn ti trng Kin Trc H Ni, khoa th, di sdybo ca cc thy c, tng bc em c tip thu nhng kin thc p ng cho nghnghip

trong tng lai. Vi nhng kin thc c c, em iu kin nhn n mn hc XL

Nc Cp. Em c nghin cu v tm hiu vhthng cp nc thnh phA. Do vy trong

n ny em nhn ti THIT KDY CHUYN CNG NGHHTHNG CP

NC THNH PHA. pht trin kinh t, thnh ph cn c mt h thng c sh tng

vng chc, trong hthng cp nc ng vai tr v cng quan trng, nh hng n spht

trin ca cc ngnh khc. Di spht trin ca thnh ph, hthng cp nc c ca thnh phang dn qu ti v nhiu khu vc ca thnh phhin vn cha c cung cp nc. V vy nhu

cu bc thit ca thnh phA hin nay cn phi xy dng mt hthng cp nc mi p ng

nhu cu pht trin ca thnh phhin ti v trong tng lai.

Vi nhng ti liu v nhng thng tin c svthnh phA, sau mt thi gian lm vic,

em hon thnh n. Trong n ny, hthng cp nc thnh phA sc thit kcho

giai on 2012 - 2030, m bo cho spht trin ca thnh phtrong vng 18 nm ti sc

nc cho cc nhu cu.

Mc d c cgng, song do cn cha c kinh nghim trong thit kv khi lng

n tng i ln nn n vn khng thtrnh khi mc nhiu li. Em knh mong nhn c s

chbo ca cc Thy, C em hon thin hn na kin thc ca mnh.

Em xin c gi li cm n n thy gio THS. Nguyn Thnh Mu, c gio Ths.

Minh Hng v cc thy c trong bmn Cp thot nc, khoa Kthut Mi trng tn tnh

hng dn em trong sut thi gian va qua.

Knh chc cc Thy, C mnh khe v Thnh cng !

H Ni Ngy 11thng 11 nm 2014

Sinh vin :Nguyn Xun Nht

Lp :2011M

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A.

TNG QUAN VN:

I.

KHI LNG THC HIN:

1.Phn bn v -A1:

+ S cao trnh trm x l (phng n chn).

+Mt bng trm x l (phng n chn).

+ Chi tit mt cng trnh n v (phng n chn).

2.Phn thuyt minh - A4:

+Phn tch, xut, la chn dy chuyn cng ngh x l nc.

+ Tnh ton cc cng trnh ca phng n chn.

II. TI LIU THAM KHO:

- Tiu chun xy dng Vit Nam 33-2006.-X l nc cp cho sinh hot v cng nghip Ts.Trnh Xun

Lai.

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B.CHNG I:

NH GI CHT LNG NC NGUN V LA CHN DY

CHUYN CNG NGH

I. CHT LNG NC NGUN V YU CU CHT LNG

NC SAU X L:

1. Cht lng nc ngun.

STT CH TIU GI TR

1 Q Cng sut (m3/ng) 20.000

2 pH pH 7.8

3 C Hm lng cn (mg/l) 800

4 M mu (Pt-Co) 55

5 ton Nhit (toC) 25

6 Hm lng Ca+(mg/l) 8

7 P Hm lng mui (mg/l) 360

8 Ki kim (mgl/l) 3.7

9 Vi khun E.coli (con/100mlnc)

120

2.Yu cu cht lng nc sau khi x l

.Bng tiu chun vsinh nc n ung

(Ban hnh km theo Quyt nh ca Btrng BY t

s01/ 2009/ BYT / Q ngy 17 / 6 /2009)

STT Tn chtiu

n

v

tnh

Gii hn

ti aPhng phpth

Mc

gim st

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I Chtiu cm quan v thnh phn v c

1 Mu sc (a) TCU 15TCVN 6185-1996

(ISO 7887-1985) A

2 Mi v(a) Khng cmi, vl

Cm quanA

3 c (a) NTU 2(ISO 7027 - 1990)

TCVN 6184- 1996A

4 pH (a) 6,5-8,5 AOAC hoc SMEWW A

5 cng (a) mg/l 300 TCVN 6224 - 1996 A

6Tng cht rn ho tan

(TDS) (a)

mg/l 1000TCVN 60531995

(ISO 96961992)

B

7 Hm lng nhm (a) mg/l 0,2 ISO 120201997 B

8Hm lng Amoni, tnh

theo NH4+ (a)mg/l 1,5

TCVN 59881995

(ISO 5664 1984)B

9 Hm lng Antimon mg/l 0,005 AOAC hoc SMEWW C

10 Hm lng Asen mg/l 0,01TCVN 61821996

(ISO 65951982)B

11 Hm lng Bari mg/l 0,7 AOAC hoc SMEWW C

12Hm lng Bo tnh chung

cho cBorat v Axit boric mg/l 0,3ISO 9390 - 1990

C

13 Hm lng Cadimi mg/l 0,003TCVN6197 - 1996

(ISO 5961-1994)C

14 Hm lng Clorua (a) mg/l 250TCVN6194 - 1996

(ISO 9297- 1989)A

15 Hm lng Crom mg/l 0,05TCVN 6222 - 1996

(ISO 9174 - 1990)C

16 Hm lng ng (Cu) (a) mg/l 2(ISO 8288 - 1986)

TCVN 6193- 1996C

17 Hm lng Xianua mg/l 0,07TCVN6181 - 1996

(ISO 6703/1-1984)C

18 Hm lng Florua mg/l 0,71,5TCVN 6195- 1996

(ISO10359/1-1992)B

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19Hm lng Hydro sunfua

(a)mg/l 0,05 ISO10530-1992 B

20 Hm lng St (a) mg/l 0,5TCVN 6177-1996 (ISO

6332-1988)

A

21 Hm lng Ch mg/l 0,01TCVN 6193- 1996 (ISO

8286-1986)B

22 Hm lng Mangan mg/l 0,5TCVN 6002- 1995

(ISO 6333 - 1986) A

23 Hm lng Thungn. mg/l 0,001

TCVN 5991-1995 (ISO

5666/1-1983 ISO

5666/3 -1983)

B

24 Hm lng Molybden mg/l 0,07 AOAC hoc SMEWW C

25 Hm lng Niken mg/l 0,02TCVN 6180 -1996

(ISO8288-1986) C

26 Hm lng Nitrat mg/l 50 (b)TCVN 6180- 1996

(ISO 7890-1988) A

27 Hm lng Nitrit mg/l 3 (b)TCVN 6178-1996 (ISO

6777-1984)

A

28 Hm lng Selen mg/l 0,01TCVN 6183-1996 (ISO

9964-1-1993)C

29 Hm lng Natri mg/l 200TCVN 6196-1996 (ISO

9964/1-1993)B

30 Hm lng Sunpht (a) mg/l 250TCVN 6200 -1996

(ISO9280 -1990)A

31 Hm lng km (a) mg/l 3TCVN 6193 -1996

(ISO8288-1989)C

32 xy ho mg/l 2 Chun bng KMnO4

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2.Tnh ton liu lng ha cht a vo .

a)Xc nh liulng phn dng keo t:

+Hm lng phn xc nh theo mu: Cn c vo mu ca nc ngun M

= 620th theo TCVN 33-2006 ta c cng thc xc nh lng phn nhm nh sau:

Pp= 4= 4= 29.66 (mg/l)+Hm lng phn xc nh theo hm lng cn nc ngun:

Bng 6.3 TCXDVN 33-2006

Hm lng cn khng tan ca nc

ngun(mg/l)

Liu lng phn khng cha nc dng

x l nc c (mg/l)

n100 25 35

101200 30 40

201400 35 45

401600 45 50

601800 50 60

8011000 60 70

10011500 70 80

Cn c vo hm lng cn ca nc ngun C = 800 mg/l v theo Bng 6.3

TCXDVN 33-2006 th lng phn nhm cn thit keo t l 60 (mg/l).

So snh gia liu lng phn nhm tnh theo hm lng cn v theo mu

chn liu lng phn tnh ton PP= 60 (mg/l).

b)

Xc nh mc kim ha.

Trong qu trnh keo t nc bng phn nhm th kim trong nc gim,

trong nc s xut hin cc ion H+, cc ion ny s c kh bng kim t nhin

ca nc. Nu nh kim t nhin ca nc nh khng trung ha ta phi

tin hnh kim ha nc bng vi CaO.

Kim tra kim ca nc theo yu cu keo t xc nh theo cng thc 6-2

TCXDVN 33-2006:

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1 ( / )PK tP

D K K mg le

Trong :

- e : ng lng ca phn khng cha nc. i vi Al2(SO3), e = 57- DK: Liu lng ho cht kim ho (mg/l).

- PP : Liu lng phn dng keo t PP = 60 (mg/l).

- K : H s i vi vi (theo CaO) K = 28.

- Kt : kim nh nht ca nc ngun Kt= 3,7 (mg/l).

Dk= 28 (- 3,7 + 1) = - 46,13 < 0

=> khng cn phi kim ha.

3.Kim tra n nh ca nc sau khi x l.

Sau khi cho phn nhm vo keo t th pH ca nc gim, do kh nng

nc c tnh xm thc. Cn phi kim tra n nh ca nc.

a)Kimtra kim ca nc sau khi keo t

* ( / )p

i io

PK K mg le

Trong :

- :K*i kim ca nc sau khi keo t.

- Kio: kim ban u ca nc ngun, Kio = 3,7 (mgl/l).

- PP : Liulng phn dng keo t, PP = 60 (mg/l).

- e : ng lng ca phn khng cha nc. /v Al2(SO3)e = 57

Ki*= 3,7 - = 2,647 (mg/l)

b)Kim tra n nh ca nc sau khi keo t

n nh ca nc c nh gi bng ch s J.

Theo TCXDVN 33-2006, Nu J < - 0,5 Nc c tnh xm thc.

J > 0,5 Nc c tnh lng ng.

Ch s J c xc nh nh sau:

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J = pH*- pHs

Trong :

- pH*: pH ca nc sau khi kheo t.

- pHs: pH ca nc trng thi bo ho CaCO3sau khi keo t.

pHs: c tnh theo cng thc sau:

pHs= f1(t) - f2(Ca2+) - f3(Ki

*) + f4(P)

Trong : f1(t0), f2(Ca

2+), f3(K1), f4(p) l nhng trsphthuc vo nhit ,

nng canxi, kim, tng hm lng mui trong nc, xc nh theo th

trn hnh H-6.1 TCXDVN 33:2006.

-Xc nh lng CO2ca nc sau khi keo t:

)/(4402*2 lmg

e

PCOCO

p

p

Trong :

- CO*2: Lng CO2 ca nc sau khi keo t

- CO02: Lng CO2 ca nc ngun.(= 5.2 )

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- ep : ng lng phn nhm ep= 57 mgl/l

CO2*= CO2

0+44 = 5,2+ 44= 51,52 (mg/l)Tra biu Langlier hnh H-6.2 trong TCXDVN 33:2006

Vi t = 250C f1(250C) = 2

Ca2+= 8 f2(8) = 0,9

= 3,7 (mg/l) f3(3,7) = 1,57P = 360 (mg/l) f4(360) = 8,82

- Vi:

CO2*= 51.52 (mg/l)

= 3,7 (mg/l)t0= 250C

P = 360 (mg/l)

Tra biu Langlier

ta c:

pH*= 6,8

pHs= f1(t) - f2(Ca2+) - f3(Ki

*) + f4(P) = 20.91,57 + 8,82 = 8,35

Nh vy: J = pH*- pHs= 6,88,35 = -1,55

Vi ch s J va tnh = - 1,55 > 0,5

Kt lun:Nc c tnh xm thc=> to lp bo v bng cacbonat mt trongthnh ng phi kim ha nc.

+ V pH* hm lng vi a vo kim ha xc nh theo cng thc

bng 6.20 TCVN 33-2006:

DK= ek Ki*(mg/l))

Trong :

- ek: ng lng ca ho cht a vo kim ho, dng vi ek= 28.

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- : Hsphthuc n nh ca J v pH ca nc vi J = 1,55 , pH*=

6,8. Tra biu Hnh 6-4 trong TCXDVN 33-2006 ta c = 0,4.

DK= 28 0,4 3,7 = 41,44(mg/l)

c)Hm lng cn ln nht trong nc sau khi a ho cht vo kim ho

v keo t

Ta c cng thc tnh nh sau:

Cmax=oCmax+ K Pp+ 0,25M + DK(mg/l)

Trong :

- oCmax: Hm lng cn ln nht ca nc ngun = 800 (mg/l)

- K: H s ph thuc vo tinh khit ca phn s dng. i vi phn nhm

khng sch K = 1.

- Pp: Lng phn a vo keo t Pp= 60 (mg/l)

- M: mu ca nc ngun theothang Platin - Coban. M = 55.

- DK: Liu lng vi a vo kim ho DK= 41,44 (mg/l).Cmax= 800 + 160 + 0,2555 + 41,44 = 915,19(mg/l).

4.La chn dy chuyn cng ngh.

a) xut cc phng n dy chuyn cng ngh x l.

Da vo bng phn tch mu nc v so snh vi tiu chun cht lng nc mt

dng lm ngun cp nc TCVN 33-2006 v tiu chun 01/2009/BYT/Q nc ngun

c cht lng kh tt m bo cc ch tiu v v sinh i vi nc n ung, sinh hot.

Cn c vo kt qu cht lng nc sau khi x l s b:

- Cng sut Q = 20000 ( /ng)- Hm lng cn max Cmax= 800 (mg/l).

- kim ca nc ngun = 3,7 (mgdl/l)- mu M = 55(pt/ Co)

- Tiu chun cht lng nc sinh hot TXDVN 33-2006.

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C th a ra hai phng n dy chuyn cng ngh sau:

Phng n I:

Phng n II:

b)nh gi v la chn dy chuyn cng ngh.

-Skhc nhau c bn gia 2 dy chuyn cng ngh l lng v lc.

+phng n 1 sdng cng nghlng dng tm lamen.u im ca cng

ngh ny l:

Kt cu chu ti nng ca b, thng bn v b tm nghing cng nh cac cu co. Nu cn thit c th b sung h nng co.

nh v cc cng cp liu c thit k c bit t c t l ti u

gia din tch lm trong nc v din tch c c.

Khng gian gia cc tm nghing rng cho php x l bn cp liu nng

c v ht rn th.

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Khng c on mch v chy ri bmt

C cu to kt bng tch hp vi tc khuy c th thay i c.

V vic s dng b lc aquazur V em li hiu qu kh cao, u im ca loi b

ny:

B lc Aquazur V tp hp tt c cc nguyn l lm vic tt ca mt thit b lc

v ra hiu qu:

- Nc lc c cp lin tc tng phn hoc ton b vo b lc trong c thi

gian ra m bo qut nc b mt. Cc b lc khc khng chu s tng lulng v vn tc lc trong thi gian ra b.

- Chiu su lp nc trn b mt b lc 1-1,2 m nn tit kim chiu cao xy

dng cng trnh, gim kch thc xy dng cng trnh.

- B lc ph hp vi tc lc cao. thc hin iu ngi ta ct lc c

chiu cao t 1,5-2 m thng thng l 1,2m.

- N gi mt p sut dng trn tt c b dy ca ct v ko di trong tt c chutrnh lc.

- Vic ra lc vi tc dng ca dng nc qut b mt lm tng hiu qu ra lc

cng nh tit kim nc ra.

- Tn tht p lc khi ra lc l ti thiu ( do c dng nc qut trn b mt) nn

p lc bm ra lc khng cn cao, tiu hao nng lng t.

+phng n 2 sdng cng nghblng ngang v blc nhanh trng lc. 2

loi bny ra i kh lu do hiu qukhng cao nh 2 loi btrn.

xut chn dy chuyn 1 lm dy chuyn x l.

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C.CHNG II

TNH TON CC CNG TRNH TRONG DY CHUYN CNG NGH

PHNG N CHN.

II.1.Thit khthng pha chnh lng dtrho cht.

II.1.1.Tnh ton thit bpha chphn.

a) Bha tan phn

Cng sut trm xl kh ln l 62000 (m3/ng) nn ta sdngBha tan phncc khuy trn bng kh nn.

S cu to chung:

Xc nh cu to ca bha tan:

Dung tch b ha tan hu ch c xc nh bng cng thc 6-3(TCVN 33-

2006):

W =

(m3)

Trong :

- Q: Cng sut trm xl(m3/h), Q = 62000 (m3/ng)=2583.33 (m3/h)

- ng d n k h ng k hnn

- n g a n c v o

17

2

3 45

6

1

7

2 - g h i ph n c c

3 - h t h n g p h n ph i g i t r n

4 - h t h n g p h n ph i g i d i

5 - ng d n dun g dc h ph n s ang b t i u t h

6 - ng x cn

B ha t an ph n c c k hu y t r n bng k hnn.

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Dung tch btiu thtnh theo cng thc 6-4(TCVN 33-2006):

Wtt=t

h

b

bW(m3)

Trong :

- W: Dung tch bha trn W=2 3.82=7.64 (m3)

- bh : Nng phn trong bho trn 10%

- bt : Nng phn trong btiu th5%

Wtt=t

h

b

bW=

= 15.28(m

3)

Chn 2 btiu thvi dung tch 1 bl 15.28 (m3).

W1= b l h = 2,1 2,1 1,8 = 7,94 (m3)

- y btiu thc dc 5% vpha ng x. Chn ng knh ng xD =

125 (mm). Mt trong b c p gch men chu axit. Dung dch phn bo ha

c dn bng ng tchy sang btiu thsau dng bm nh lng a dung

dch phn 5% tbtiu thvo nc xl.

c)Chn my qut gi v tnh ton ng dn kh nn.

- Lu lng kh nn cn a vo bha tan:

Qh= 0,06 Wh Fh(

/pht)

Trong :

12

3

4

5

6

- bm nh l ng6

- n g phn ph i gi5

- n g x c n4

- n g dn kh3

- n g d n n c2

1 - n g dn du ng d ch

B t i u t h ph n kh uy t r n bng k hnn.

ph n t b h a t r n

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- Qh: Lu lng kh nn cn a vo b(/pht)- Wh: Cng sc kh trong bha trn ly W = 10 (l/s.m2)

- Fh: Din tch mt bng ca bpha phn. F = 2(1,5 1,5) = 4,5 (m2)

Qh= 0,06 10 4,5 = 2,7 (/pht)= 0,045 (m3/s)- Lu lng kh nn cn a vo btiu th:

Qt= 0,06 W F (/pht)Trong :

- Qt: Lu lng kh nn cn a vo btiu th(/pht)- Wt: Cng sc kh trong btiu thly W = 5 (l/s.m2)

- Ft: Din tch mt bng ca bpha phn. Ft= 2(2,1 2,1) = 8,82 (m2)

Qt = 0,06 5 8,82 = 2,646 (/pht)= 0,044 (m3/s)- Tng lu lng kh nn cn a vo bha trn v tiu thl:

Qg= Qh+ Qt= 2,7 + 2,646 = 5,346 (/pht) =0,0891 (m3/s)- ng knh ng gi chnh:

Dc= (m)

Theo TCXD 33-2006,tc gi trong ng phi ly bng 10-15 (m/s )

=>chn v=15 (m/s)

Dc= = = 0,087(m)Chn Dc = 90(mm)

Thli tc : v= = = 14 (m/s)Nm trong phm vi tc cho php.

-ng knh ng dn gi n thng ha trn:

Dh= = = 0,063(m)Chn Dh= 65 (mm)

Thv= 13,5(m/s) nm trong gii hn tc cho php.

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-ng knh ng dn gi n ythng ha trn:

Dh= == 0,045(m) = 45(mm)-ng knh ng nhnh vo thng ha trn: thit k3 ng

Qnh= = 0,0075(m

3/s)

Tnh slkhoan trn gin ng gi bha trn:

Theo mc 6.22 TCXDVN 33-2006

Tc khng kh qua l 20-30 (m/s)

ng knh l 3-4 (mm)

Chiu di ng nhnh :

ln=1,5m ; chn 3mm ; vl=25m/s

din tch l: fl=

=

= 7.(m2)Tng din tch cc ltrn 1 ng nhnh:

Fl = =

= 3.

(m2)

Sltrn 1 nhnh : n = = 43 (l)

Nu khoan 1 hng lth khong cch cc l:

l= = 35 mm

Nu khoan 2 hng th l=70 mm.

d)Tnh ton lng ha cht dtrtring kho.

- Tnh ton lng phn dtr.i vi trm x l nc th lng phn phi d tr s dng trong 15

ngy.

Lng phn khi tiu th trong 15 ngy c tnh nh sau:

Pl= n (tn)

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Trong :

- n = 15 ngy, sngy sdng.

- Q : Cng sut trm xl Q = 62000 (m3/ng).

PP : Lng phn cho vo nc tnh theo sn phm tinh khit PP= 48

mg/l.

b : Tltinh khit trong phn kh Pk= 50% (Phn Vit Nam sn xut)

Pl= 15

= 89.28(tn)

Din tch mt bng ngn cha phn dtrc xc nh theo cng thc :

F =0

p

Ghb10000

TPQ

Trong :

- : Hstnh n din tch i li v thao tc trong kho, ly = 1,3.

- b : Tltinh khit trong phn kh Pk= 50% (Phn Vit Nam sn xut).

- h : Chiu cao cho php ca lp ha cht.Vi phn nhm cc ta c h=2 m.

- T : Thi gian giha cht. T = 15 ngy.

- Go : Khi l

ng ring ca ho ch

t. Thng l

y Go= 1,1 (tn/m

3).

- Q : Cng sut trm xl Q = 62000 (m3/ng).

- PP : Lng phn cho vo nc ngun tnh theo sn phm tinh khit. PP=

48 (mg/l)

F =

=

= 53(m2)

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- Tnh ton lng vi dtr.

Lng vi cc tiu thtrong 1 ngy:

G =b10000

DQ k

Trong :

- Dk : Lng vi cn a vo nc kim ha. Theo phn trn c Dk=

29.68 (mg/l).

- b : Tlvi tinh khit b = 70% .- Q : Cng sut trm xl Q = 62000 (m3/ng).

- : Trng lng ring ca dung dch phn = 1 (T/m3)

G==2.63(T/ngy)

Khi lng vi cn dtrcho 15 ngy th chn bin php dtrt trong kho vti bng my.

Tng lng vi cha trong kho : M = 152.63 = 39.45 (T)

Din tch mt bng cha vi:

F =

=

= 31.06(m2)

Chn F = 35 (m2). Kch thc kho cha BL = 57 (m).

Chiu cao ca vi cc cha trong kho l 1,5m. Kho vi xy tng bao quanh, c

mi che v c ca thng sang phng pha dung dch. Vi cc c a sangphng

ti vi bng xe y.

II.1.2.Tnh ton thit bpha chvi sa.

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a). S cu to thit bpha chvi sa.

Vi cc c ti trong cc bthnh vi st, sau dng gu ngom vn chuyn

bng cu palng a vo cc bc y hnh cn v c lp my khuy c kh

pha long thnh vi sa. Sau mi ln pha, mvan xy hnh cn v cho cn

cha ti chy vo rthp, ri dng palng a rny ra ngoi xng.

b) Tnh ton bv thit bpha chvi sa.

Bti vi.

Dung tch bti vi xc nh theo cng thc:

Wvt= Mq (m3)

Trong :

- M : Lng vi cc tiu thtrong 15 ngy. M = 39.45 (tn).

- q : Lng nc cn thit ti 1 ln vi cc. Theo quy phm th q= 35

m3/tn. Chn q = 4 m3/tn.

Wvt= 39.454 = 157.8 (m3)

- m y b m n h l n g6

- g u x c v i t i5

- m y k h u y4

- b pha v i sa3

- l n g n g v i c c2

7

1

- ng dn v i sa

- b t i v i

h t h ng pha vi sa

- ng x cn9

8 - n g d n n c s c h n

10 - mo n o r a y

4

6

2

1 8

9

7

5

3

10

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Ly chiu cao bti vi l h =2,5 (m)Din tch bti vi F = 63.12 (m2).

Kch thc bti vi BL = 8x8=64 (m2).

Bpha vi sa.

Dung tch bpha vi sa 5% chn ph hp vi thi gian sdng ht lng vi

ti 1 ln l 8 gi. (theo mc 6.34 TCXDVN 33-2006 l t612 h)

Dung tch bpha vi sa xc nh theo cng thc:

Wvs= b10000

LnQ

v

v

(m3

)

Trong :

- Q : Cng sut trm xl Q = 62000 m3/ ng= 2583.33 m3/h.

- n :Thi gian gia hai ln pha ch, n = 10 ting.

- Lv : Liu lng vi tinh khit cho vo nc xl, Lv= 29.68 (mg/l).

- bv: Nng vi khi ha, b = 5%.

- : Trng lng ring ca dung dch vi, ly = 1 (T/m3)

Wvs=

=

= 12.27 (m3)

Bpha vi sa c thit kxy kiu hnh trn, y c xy dc vpha tm

b, ng knh bly bng chiu cao cng tc ca bd = h. y blp ng xcn

D150mm di ming xc t rst hng cn. gicho vi khng blng

v c nng u 5% phi lin tc khuy trn bng my khuy.

Mt khc ta c:

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Wv=

=

>d = = = 3.95

Vy d h = 4 (m) .

Tnh ton la chn my khuy.

Dng my khuy pha vi ti thnh vi sa v gicho dung dch khng b

lng xung y b. Chn my khuy kiu cnh phng c cc chtiu:

Svng quay ca trc ng c : n = 40 (vng/pht)

(theo mc 6.36 TCXDVN 33-2006 : tc khuy bng my khng nhhn 40

vng/pht)

Chiu di cnh khuy theo quy phm l t(0,4 0,5)d.

Chn lck = 0,4d =1,6 (m).

Chiu di tnh ton ton phn ca cnh qut l : 1.62 = 3.2 (m).Din tch bn cnh hu ch ly 0,15 m2cho 1 m3b. (Quy phm = 0,1 0,2 m2).

Din tch hu ch cnh khuy: 0,1512.27 = 1.84(m2)

Chiu rng mi cnh khuy:

ck

1 1.84b 0,92

2 2

(m)

Cng sut ca ng c cho quay cnh qut ly l 3,0KW.

m bo cho cng trnh lm vic n nh v an ton ta chn 2 bpha vi,

trong 1 blm vic v 1 bdphng.

II.2. Tnh ton thit kbtrn c kh.

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II.2.1.Cu to btrn c kh.

Hnh II.6.s cu to thit btrn c kh

II.2.2.Tnh ton btrn c kh

II.2.2.1.Xc nh kch thc ca btrn c kh:

V = tQ (m3)Trong :

- t: thi gian khuy trn. Theo 6.58 TCXDVN 33-2006 t = 45 90(s) chn t

= 80s.

- Q: cng sut trm x l, Q = 62000 (m3/ng) = 0.717 (m3/s).

=> V = 80 x 0,717 = 57.36 (m3).

Thit k 2 b trn,th tch 1 b l : v=V/2=28.68 (m3).

vung kch A xB = 2.4 2.4m. H = 5 m

Chn chiu cao bo v Hbv= 0,5m Hxd=5,5m.

II.2.2.2.Xc nh kch thc cnh khuy v lng lng cn thit cho my

khuy:

D

A

H

B

A

h

mt c t 1-1mt b n g

I

b

b I

b

h

m

m

Hbv

m

m

ho cht

n c th

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+ Dng my khuy tua bin su cnh nghing gc 450 hng xung di

a nc t trn xung. ng knh my khuy D 1/2xchiu rng b-> chn D =

1 m. Trong b t bn tm chn ngn chuyn ng xoy ca nc, chiu cao

ca tm chn 4,6m, chiu rng 0,24 mbng 1/10 ng knh b.

- ng knh my khuy ly :D = 1 m.

- My khuy t cch y mt khong:h = D=1 m.

- Chiu rng bn cnh khuy :B= 15

D = 1 0,8 0,16( )5

m

- Chiu di bn cnh khuy: A= 14

D = 1 1 0,25( )4

m

-

-chiu di vng trn: L = 1,5D = 1,5m

y b t dc 2 % v mt pha v t ng ng x kit thau ra, x cn

khi cn thit.

Nng lng cn thit truyn vo nc:

P = G2 .V. (kW)

Trong :

- G : Cng khuy trn. Theo iu 6.58 TCXDVN 33 -2006

G = 500 1500s-1Chn G = 500s-1

- - V : Th tch vng trn:2 2

33,14 1 1,5 1,184 4

DV L m

- : h s nht ca cht lng, = 0,0001P = 10002 0.001 1.18 = 1180(J/s) = 1.18 (kW).

S vng quay ca my khuy:

n =1/3

5. .

P

K D

(vng/pht)

Trong :

- P: Nng lng cn thit truyn vo nc , P = 1180 (W).

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Hnh II.7. S cu to b phn ng c kh

II.3.2.Tnh ton b phn ng c kh.

II.3.2.1.Xc nh kch thc ca b phn ng c kh:

- Th tch b b phn ng c kh c tnh theo cng thc :

V = t.Q (m3)

Trong :

- t: thi lu nc li trong b phn ng ( Mc 6.80 TCXDVN 33- 2006)t=

1030 pht, chn t= 20 pht = 1200s.

- Q: cng sut trm x l, Q = 62000 (m3/ng) = 0.717 (m3/s).V = 1200 x 0,717= 860.4 (m3).

Thit k 2 b to bng cn c kh, dung tch mi b:

V1b= V/2 = 430.2 (m3).

Mi b chia lm 3 ngn, c ngn bi cc tm chn khoan l lp ng nha

PVC-D150mm, vn tc nc qua l trn vch ngn v = 0,1m/s.

D h

H

H

bv

I

mt c t 1-1

L L L

L L L

A

A

mt bn g

I

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Chn chiu cao ca bphn ng h = 4.3 m, chiu rng b = 5m, tit din ngang30 m2

Tit din ngang ca mt b l: f =BxH = 5x4,3=21.5 (m2).

Chiu cao bo v: Hbv= 0,5 m. Hxd= 4,8m.Chiu di 1 b :

L= = 20.1(m)

Chiu di mi bung: l=

= 6.7 (m)Dung tch mi bung l : A x B x H = 5 x 6.7x 4,3 = 144.05 (m3).

- Tng din tch cc l lp ng nha PVC D150mm vi tc nc chy qua l

Vl= 0,1 m/s s l:

= =

= 7.17 ()Vi ng knh l D = 150 mm th din tch ca mi l s l:

=

=

= 0,0176 (

)

Tng sltrn thnh mng sl:= = 407 l.L D = 150 c b ch hng 16 hng theo chiu cao b, 17 hng theo chiu

rng b.(sai)

+ Xc nh tng s l vch ngn sang vng n nh, trc b lng v=0,5 (theo

iu 6.77 TCXDVN 33-2006).

f= = = 1.435 ()n=

=81 l

L vch ngn sang vng n nh b ch hng 8 hng theo chiu cao b, 7 hng

theo chiu rng b.

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II.3.2.2.Xc nh kch thc cnh khuy v lng lng cn thit cho my

khuy:

- Chn b phn ng to bng c kh dng cnh khuy tuabin trc ng,

4 cnh nghing 450, qut nc xung y b xi v ti cn lng ng y khi

ng c phi ngng hot ng.

+Mi ngn t 1 my khuy, tng s my khuy n = 6.

+ Th tch nc khuy trn ca 1 my:

V1m= V1b/3 =430.2 /3 = 143.4 (m3).

+ ngknh my khuy ly: D = 1,5m.

+My khuy t cch y mt khong :h = D = 1,5m.

+ Chiu rng bn cnh khuy = 15

D = 1 1.5 0,3( )5

m

+Chiu di bn cnh khuy = 14

D = 1 1,5 0,375( )4

m

y b t dc 2 % v mt pha v t ng ng x kit thau ra, x cn

khi cn thit.

- Cng khuy trn xc nh theo iu 6.83 TCXDVN 33 -2006nh sau: G1=70s-1; G2=50s

-1; G3=30s-1.

* Nng lng cn cho my khuy bc 1

- Cng sut tiu th cn thit ca my khuy bc 1:

P1= G2

1.V. (kW)

Trong :

+ G1: Cng khuy trn ca my khuy bc 1. G1= 70s-1+ V : Th tch b trn (m3), V = 92,6 (m3).

+ : h s nht ca cht lng, = 0,0001

P1= 702 92,6 0.001= 454 (J/s) = 0,454 (kW).

Vi hiu sut ng c =0,8. Cng sut ng c N1= P1/ 0,8 = 0,567(KW).

- S vng quay ca my khuy bc 1:

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n1=1/3

1

5. .

P

K D

(vng/pht)

Trong :

+ P1: Nng lng cn thit truyn vo nc, P1 = 454 (W).

+ K: h s sc cn ca nc, vi cnh khuy nghing 450, K = 1,08.

+ : khi lng ring ca cht lng, = 1000 (Kg/G/m3).

+ D: ng knh cnh khuy, D =1,5m.

n1= (

)

= 0,57 (vng/s)= 34 (vng/pht).



P2= G2

2.V. (kW)

P2= 502 92,6 0.001= 232 (J/s) = 0,232(kW).

Cng sut ng c N2= P2/ 0,8 = 0,29 (KW).


n2=1/3

2

5. .

P

K D

(vng/pht).

n2= ( )

= 0,45 (vng/s)= 27 (vng/pht).



P3= G2

3.V. (kW)

P3= 302 92,6 0.001= 83 (J/s) = 0,083 (kW).

Cng sut ng c N3= P3/ 0,8 = 0,1 (KW).


n3=1/3

3

5. .

P

K D

(vng/pht).

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n3= ( )

= 0,32 (vng/s)= 19 (vng/pht).

Kt lun: B phn ng c kh c 2 b mi b c 3 ngn cc thng s tht k ca 1

b nh sau:

b (m) L (m) H (m) Hbv(m) Hxd(m)

5 20.1 4,3 0,5 4,8

B phn ng c trang b: 2 my khuy to bng bc 1 c cng sut ng c

0,56 KW v svng quay ca my khuy l 34 (vng/pht); 2 my khuy to bng

bc 2 c cng sut ng c 0,29 KW v s vng quay ca my khuy l 27

(vng/pht). 2 my khuy to bng bc 3 c cng sut ng c 0,1 KW v s vng

quay ca my khuy l 19 (vng/pht). ng knh cnh khuy 1,5m.

II.4. Tnh ton thit kblng Lamen.

II.4.1.Cu to blng lamen.

Hnh 5.7 Cu to blng Lamella

Ch thch:

1- Vng phn phi nc vo b lng.

2 - Vng thu cn.

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3-H thng gt cn

4 - Tm lng mng (lamen)

5-H thng ng thu nc sau lng.

6- Vng lng.

7- ng dn nc sau lng sang b lc.

II.4.2.tnh ton b lng lamen.

(Vic tnh ton blng da trn cun Xl nc cp cho sinh hot v cng

nghip ca tc gi- TS Trnh Xun Lai (NXB Xy Dng, 2004; Chng 6, mc

6.3.4).

II.4.2.1.C s tnh ton:

- B lng lamen cng ging nh b lng thng v cng gm 3 vng:

+Vng phn phi nc

+Vng lng

+Vng tp trung v cha cn

- c im ca b lng lamen:

+Vng lng c chia thnh nhiu lp mng vi khong khng gian nh

hp, nhcc tm c t nghing. Khi dng cc tm ln sng hoc tm phng

th tin lp rp v qun l hn. Dng cc ng th chc chn hn v m bo kch

thc c ng u hn v tc dng chy c thtng hn nhng li chng b

lng cn, tng khi lng cng tc ty ra. y dng cc tm c dng na lc

gic v khi ghp cc tm li th sto thnh khi ng c mt ct ngang nh nhng

ng lc gic ghp li. Nh vy sva m bo c tnh linh ng trong thi cng

cng nh bn xy dng khi hp khi cc tm.

+Khu vc lng c lp cc m-un dng khi hp chnht. Cc m un

ny to nn bi s lp ghp ca cc tm Lamella nghing ( 60o ). Nhng tm

Lamella ny bng nha PVC cht lng cao. Hai tm Lamel ghp li vi nhau s

cho ra nhng ng hnh lc gic ( dng ging nh tong ).

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-Tc dng v c ch ca qu trnh lng:

+Nc tbphn ng vo blng schuyn ng gia cc bn vch ngn

nghing theo hng tdi ln v cn lng xung n bmt bn vch ngn

nghing strt xung theo chiu ngc li v dng tp hp ln tp trung vh

thu cn, t theo chu kxi. Cht ni c tp trung vkhoang trng gia cc

tng v dn i theo mng chm.

+Khi gim chiu cao lng th gim chy ri ca dng chy tdo, gim

c dao ng ca thnh phn tc thng ng ca dng nc. Kt qul tng

hssdng dung tch v gim c thi gian lng( chcn mt vi pht).

II.4.2.1Tnh ton cc kch thc ca blng.

Theo s tnh ton, trong khong thi gian lng T, ht cn chuyn ng tA

n B. Quo AB c thphn tch thnh chuyn ng tA n C vi tc vtb

ca dng nc v tC n D vi tc ri cn u0. C thxc lp cc tng quan:

Hnh 5.10 S tnh ton ng lng

Tvtg

hHAC tb .

sin

0

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Tuh

CD .cos

0

Do : cos.cotcot0

0ggh

H

u

vtb

Hay )cos(cot 00

h

Hg

u

vtb

Nu gi din tch b mt lng l F v lu lng nc x l l Q, tc ca dng

nc i ln theo phng thng ng v0l:

sin0

tbv

F

Qv

vtb - Tc trung bnh ca dng nc i ln

theo vch ngn nghing.

Vy :sinF

Qvtb

Thay vtbvo cng thc trn ta c:

F

Q

hH

hu .

)cos(cos 00

Trong h c gi tr bng 0,05 - 0,15m v

H0=1-1,5(m). T phng trnh trn cho thy

cng vi lu lng nc x l v vng tc lng cn u0, b lng lamella vi

dng chy ngc chiu s c din tch b mt b hn so vi b lng ngang.

Khi tnh ton b lng Lamella cng da trn 2 ch tiu c bn ban u l tc

lng cn u0v gc nghing ca cc vch ngn song song (thng ly t 450-600).

m bo khng gian phn phi nc u vo cc lng, khong cch phn

di cc vch ngn ly l 1,0-1,2(m). Chiu cao vng cha cn thng ly t 1,0-

1,5(m). Lp nc trn b mt tnh t mp cc vch ngn nghing ly ln hn

0,5(m)( m bo thu nc u. Nc b lng c din tch mt ln cn phi thit

k h thng thu nc b mt bng cc mng hoc ng.

Hnh 5.11 Kch thcng

lng(mm)

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Tnh ton kch thc cng trnh:

- Chn cc thng sc bn:

+Tm mng: Chn loi tm nha, c phn ln sng hnh lc gic, khi

ghp cc tm li vi nhau thnh khi sto thnh cc hnh ng. Vi chiu cao h=

52(mm), d=60(mm). Chiu di mi tm L =1(m).

+Tit din hnh ng:

f=5230 + 5215=2340 (mm2) = 0,00234(m2)

+Chu vi t: c= 630 = 180 (mm) = 0,18(m)+Chiu di ng: Lo= 1(m)

+Gc nghing : chn = 600.

+Vn tc lng Uochn theo bng 6.9 (TCXDVN33-2006) chn Uo= 0,45

(mm/s).

+ Chiu cao khi tr lng: H= L.sin 600= 10,867 = 0,867(m)

Theo , ta c:

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Cng sut nc i vo b lng:

LQ Q

Trong :

QL- Cng sut nc vo b lng

Q - Cng sut thit k. Q = 62000 (m3/ng).

- H s d phng. Chn = 1,05

Vy ta c QL = 1,05 62000 = 65100 (m3/ng) = 0,753 (m3/s)

Din tch mt bng b lng:

2

h

H.cos + h.coso

Qu

F

(Cng thc 6.28 sch X l nc cp cho sinh hot v cng nghip ca tc

gi - TS Trnh Xun Lai)

Trong :

uo: Tc lng ca ht cn; uo= 0,5(mm/s) = 5.10-4 (m/s).

h: Kch thc tit din ng lng.

H: Chiu cao khi tr lng= 60o; cos = 0,5;

Ta c :

F=

=

= 175.4(m2)

- Chn s b lng l 2 n nguyn.Din tch mt bng 1 b l F1= 87.7 (m2).

-Chn chiu rng 1 b l B1= 5 (m) => chiu di 1 b l L1= 87.7/5 = 17.54

(m), lm trn 17.6(m).

- Din tch thc ca 1 b : F1= 5 17.6 =88 (m2).- Chiu di phn phn phi nc u bv khu vc btr my gt cn cui b,

chn theo cu to v kch thc my gt cn. Chn chiu di phn phn phi L2 =

2m; phn cui bL3= 3m (vng thu cn). Tng chiu di xy dng ca blng

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L = L1+L2+L3= 17.6 + 2 + 3 = 22.6 (m)

- Chiu di vng n nh Lchnh l khong cch gia vch ngn c l n

tng ng lng.

+Tc thc t ca ht cn:

u0=

=

= 4,98.(m/s)

+Vn tc nc chy trong cc ng lng:

v0= = = 4,9.(m/s).+Bn knh thy lc. R=

=

= 0,013 (m).

Vi: f tit din ng lng.

cchu vi t ca ng lng.

+S Reynon : Re =

Trong :

v0vn tc nc chy trong ng lng.

R bn knh thy lc.

h s nht dng hc ca nc. t0= 250C, = 0,9110-6m2/s .

Re =

= 70 < 200.

Vy nc trong ng lng ch chy tng.

Chun s Froude:

Fr =

=

= 1,9.> Dng chy trong ng lng l chuyn ng n nh.

X cn: Ta dng phng php gt cn bng c kh v x cn bng thu lc.

Th tch vng cha cn ca 1 b c xc nh theo cng thc:

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Wc=N

mCQT

.

).(. ma x

(m3).

Trong :

T: Chu kgia hai ln xcn, ly T = 8(h).Q: lu lng nc vo blng (m3/h). Q = 62000(m3/ng) = 2583,33 (m3/h).

Cmax= 613,18 (mg/l). y l hm lng cn ca nc ngun sau khi choho cht vo.

m: Lng cn cn li sau blng. Theo quy phm ly m = 10 (mg/l).

: Nng trung bnh ca cn khi c gt vhthu cn.

Vi T = 24h ta ly = 60000 (g/m3).N: Slng blng. N=2.

Wc=8 2583,33 (613.8 10)

60000.2

= 311.9(m3).

Ti mi blng sbtr 02 hnh chp thu cn. Thtch hu ch mi hnhchp Wc= 623.9/2 = 155.95(m

3). Kch thc y mi hnh chp chn l 66 m(dhp khi vi blng).

- Chiu cao vng cha cn:

Hc=155.95

2 (6 6) 2 (6 6)

Wc

= 2.2(m).

Vi hthu cn y c kch thc 3317.3 m.

Tnh chiu cao b lng:

+Chiu cao phn nc trong trn cc ng lng: h1= 1,2 m;+Chiu cao t tm lng nghing: h2= 0,867 ~ 0.9 m

+Chiu cao phn khng gian phn phi nc di cc ng lng nghing:

h3= 2 m (lp t thit b gt cn c kh).

- Phn cui b:

HXD= h1+ h2+ h3+ Hc+ HDT= 1,2 + 0,9 + 2 + 2,0+0,5 = 6,6 (m)

Trong :

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h1: Chiu cao phn nc trong trn cc khi tm lng lamella.

h2: Chiu cao khi tm lng lamella.

h3: Chiu cao vng lng di cc tm lamella.

Hc : Chiu cao vng cha cn blng

HDT : chiu cao dtr, ly bng 0,5 (m).

- Phn u b:

HXD= h1+ h2+ h3+ HDT= 1,2 + 0,9 + 2,0 +0,5 = 4,6 (m)

Tnh ton mng thu nc lng:

Thu nc lng dng mng thu t trn sut chiu di b lng. Ti mi b dnghai mng thu, t cch thnh 1,0m. Khong cch gia hai mng thu l 1,4m.

Lu lng nc ti mi mng thu l: 30,324 0,081 /2 2

q m s

Tc nc ti cui mng thu V = 0,7m/s

Tit din t ca mng cn 20,081

0,1150,7

qf m

V

Chn chiu rng B = 0,4m, H = 0,4m

Dc hai bn thnh mng t tm thp c rng ca thu nc lng, chiu

cao ca rng ca l 100mm.

Lu lng nc thu qua mng ch V: 5/ 20 1,4q h

Vi chiu cao nc trung bnh qua mng ch V l h = 0,05m th lu lng

qua mi mng:5/ 2 4 3

0 1,4 0,05 7,83 10 /q m s

S lng rng ca trn mt thnh mng:

4

0

0,08194,72

2 2 7,83 10

qn

q

B tr 95 rng ca trn mi thnh mng. Khong cch gia cc rng ca

trung bnh l 189mm. Trn mi mng trn ch V c cc l gn bulng vo thnh

mng. Mng thu nc c th iu chnh cao khi cn thit.

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Mng thu nc chung cui b lng tnh vi tc nc l V = 0,7m/s.

Tit din t ca mng: F = f 2 2 = 0,46 m2.

Kch thc mng: B H = 0,7 0,7 m.

H thng thu nc sau lng:

Phn thu nc sau lng dng h thng mng x khe ch V t theo sut chiu

di b lng (l=13m), ti trng thu ly q=2 l/s.m (mc 6, phn 6.2.2 x l nc

cp cng nghip v sinh hot-Trnh Xun Lai)di mp mng, mng thu c 2 pha,

tng chiu di mng l:

L =

=

= 115 (m).

Cng sut lng:Q = 40000 (/ng)= 0,46 (/s).Khong cch tim mng a=1,5 m(v chiu rng mi b lng l 4,2 m)

Tim mng cch tng dc a=1 m.

m bo thu u nc trn ton b chiu di mng, cu to pha ngoi

thnh mng phi gn cc tm iu chnh chiu cao mp mng bng thp khng g,

x khe hnh ch V, gc y 90

0

. Chiu cao hnh ch V l 5 cm, y ch V l 10cm, mi mt di c 5 khe ch V, khong cch gia cc nh l 20 cm.

Lu lng nc qua 1 khe ch V:

2

5

0 .4,1 hq

trong : h- chiu cao mc nc trong khe ch V(m)

X cn:

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Ta dng phng php gt cn bng c kh v x cn bng thu lc. Th tch

vng cha cn ca 1 b c xc nh theo cng thc:

Wc=N

mCQT

.

).(. max

(m3).

(Cng thc 6.10-mc 6.68-TCXDVN 33-2006)

Trong :

T- Chu k gia hai ln x cn, ly T = 8(h).

Q- lu lng nc vo b lng (m3/h). Q = 62000 (m3/ng) = 2583.33

(m3/h).

Cmax= 613,18 (mg/l).y l hm lng cn ca nc ngun sau khi cho ha cht vo.

m- Lng cn cn li sau b lng. Theo quy phm ly m = 10 (mg/l).

-Nng trung bnh ca cn khi c gt v h thu cn. Tra bng 6.8

TCXDVN 33-2006.

Vi T = 8h ta ly = 35000 (g/m3).

N- S lng b lng. N=2.Wc=

= 178 ().

Ti mi b lng s b tr 02 hnh chp thu cn. Th tch hu ch mi hnh

chp Wc= 178/2 = 89(m3). Kch thc y mi hnh chp chn l 33m(d hp

khi vi b lng).

- S lng bn tch li b lng sau mt ngy tnh theo cng thc sau:

1000

)( 1ma x CCQG

= 36190.8(kg).

Trong :

- G : Trng lng cn kh (kg).

- Q: lng nc x l, Q= 62000(m3/ng).

- CMax: Hm lng cn sau b trn. Cmax= 613.8 (mg/l)

- C1: Hm lng cn yu cu sau b lng (mg/l). Theo iu 6.61

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TCXDVN 33 - 2006 th C = 10 (mg/l).

- Chiu cao vng cha cn:

Hc=

=

= 4.9 (m).

Chiu cao xy dng b lng:

-Phn cui b:

HXD= h1+ h2+ h3+ Hc+ HDT

= 0,8 + 0,9 + 1.5 + 4,9+0,5 = 8,6 (m)

Chn HXD= 8,6 (m).

Trong :h1- Chiu cao phn nc trong trn cc khi tm lng lamella.

h2- Chiu cao khi tm lng lamella.

h3- Chiu cao vng lng di cc tm lamella.

Hc- Chiu cao vng cha cn b lng

HDT - chiu cao d tr, ly bng 0,5 (m).

-

Phn u b:HXD= h1+ h2+ h3+ HDT= 0,8 + 0,9 + 1.5 +0,5 = 3.7 (m).

Chn HXD= 3.7 (m).

Tnh ton lu lng nc x cn b lng:

- Lng nc dng cho vic x cn 1 b lng tnh bng phn trm lu lng

nc x l, c xc nh theo cng thc:

P cK W N

P 100%q T

(Cng thc 6.12-mc 6.68-TCXDVN 33-2006)

Trong :

+ Wc -Th tch vng cha v nn cn. Wc= 178 (m3).

+ KP - H s pha long cn. Gt cn bng c kh chn Kp= 1,2.

+ N - s lng b lng. N = 2.

+ T- thi gian gia 2 ln x cn (h). T= 8 (h).

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+ q- Lu lng nc tnh ton (m3/h). q = 2583.33 (m3/h).

P =100 = 2.06%

- Vy lu lng nc dng cho vic x cn 2 b lng, tnh theo th tch ncgia cc ln x (8h) l:

= 2.06% 60000 = 1236 ().Vy lu lng nc dng cho vic x cn 2 b lng trong mt ngy m l:

= =3708 ().Chn thi gian x cn ca b lng l t = 20 pht (mc 6.74- TCXDVN 33-2006).

Lu lng mt ln x l:Q1ngnX= =

= 0,515 ().

ng knh ng x cn l :

Dx=

== 0,43 (m)

(Chnvn tc x Vx=1,5m/s.mc 6.74 TCXDVN 33-2006)Chn Dx= 450(mm). m bo yu cu D>150mm

Vn tc x thc t l : Vx =2

14

D

Q N

=1,45(m/s).

II.5. Tnh ton thit kblc AquazurV.

II.5.1.Cu to blc AquazurV c 2 mng thu nc ra lc.

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Hnh 5.12: S cu to b lc nhanh Aquazur- V

Ch thch:

1 -Mng ch V phn phi nc lc v nc qut b mt khi ra lc.

2 -Lp ct lc.

3 -Mng thu nc ra lc

4 -Lp si lc.

5 -Lp chp lc

6 - ng cp nc ra lc.

II.5.2.Tnh ton blc AquazurV c 2 mng thu nc ra lc.

II.5.2.1.La chn mt sthng s.

- B lc mt lp vt liu lc l ct thch anh.

- dmin0,7 (mm); dmax1,6 (mm).dtd0,75 (mm). ( Tra bng 6.11 TCXDVN

33-06)

- n tng i: e 30%.( Tra bng 6.13 TCXDVN 33-06)

- Chiu dy lp vt liu lc l: l 1,3 (m).( Tra bng 6.11-TCXDVN 33-2006

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- H s khng ng nht K=d60/d10; d60 v d10 l ng knh mt sang (mm)

c 60% v 10% lng ct lt qua K= 1,5 .( Tra bng 6.16 TCXDVN 33-2006)

- Vn tc lc: V = 7 (m/h) (Bnh thng)

Tc lc khi lm vic bnh thng: vbt7 (m/h) ( Do b lc Aquazur c h

thng qut b mt nn khi 1 b ra lc cc b khc khng phi lm vic tng

cng, do vy khi tnh ton b lc ta ch tnh vi vn tc lm vic bnh thng)

-Phng php ra lc: Ra lc bng nc v gi kt hp. (Theo mc 6.123

TCXDVN 33-2006) .Cng gi ra lc l: Wgio60 (m3/m2h) (16,6 l/s), trong

2 pht. Ra kt hp nc, gi trong thi gian 5 pht vi cng gi 60 (m3/m2h)

(16,6 l/s), nc 10 (m3/m2h) (2,8 l/s). Sau , Cng nc ra thun tu l: Wn

24 (m3/m2h), cng qut b mt 7 (m3/m2h), trong 4 pht.

II.5.2.2.Xc nh din tch mt bng v slng bcng tc.

- Tng din tch mt bng ca b lc tnh nh sau:

F bt

vT

Q

. (m2)

Trong :

Q - Cng sut trm. Q = 62000 m3/ng.

T-Thi gian lm vic trong ngy ca b. T = 24 h.

vbt- Tc lm vic bnh thng ca b. vbt= 7 m/h.

Ta c: F = 369(m

2)

S b lc: n= = =9.6Chn s b lc l n= 10b.

- Kim tra vn tc lc tng cng: vtc= 9,5 (m/h) ( Tra bng 6.11 TCXDVN 33-

06).

- Kim tra tc lc tng cng:

=

=

=8 < 9,5 (m/h).

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Nh vy s b lc n=10l hp l.

- Din tch mi b l Fi

= 36.9 (m2).

- Chn b c 2 ngn lc,b rng mi ngn l2.5 m.- Chiu di mt ngn l:

L =

=

= 7.38 (m).Tnh n kch thc mng thu nc gia b lc (0,6m) (tnh phn f), th

chiu rng ca mi b lc l:

B = 2

2.5 + 0,6 = 5.6 (m).

Chn kch thc mi b: LB = 7.4 5.6 = 41.44Dintch mi ngn lc:

= 6,2 3 = 18.6 ().Din tch lc ca 1 b:

= 2 18.6 = 37.2 (). Chu k lc:

- Chiu dy lp vt liu lc l 1,3(m).

- Th tch ct lc bng:

= 1,3 37.2 =48.36 ().- rng ca lp ct lc bng 30% .

Th tchgi cn trong lp vt liu lc l:

Vcn= 0,3 Vct= 0,3 48.36 = 14.508 (m3).

+Vi vn tc lc : vtb= 7(m/h)cn cha c 1/4 th tch cc l rng (theo tng

kt ca phng th nghim cng ty T vn cp thot nc s 2).

+Th tch cha cn ca 1m3 ct lc:V= 0,31= 0,075 (m3)

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Hnh 5 5. Chp lc di ui

+Khi lng cn chim 3-4%(cn nh 3%):

G =30kg/m30,075 = 2,25(kg/m3)

+Tc lc 7m3

/h, lp ct dy 1,2m, hm lng cn sau b lng theo iu 6.61TCXDVN 332006 l12g/m3, mi khi ct trong mt gi s phi gi li c

khi lng cn l g= 712 = 84g = 0,084kg.

- Chu k lc m bo cht lng: TL= 2,25/0,084 = 27h.

- cht m bo cht lng nc, Chn chu k lc l T = 24h

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Tng din tch khe h ca chp lc mt b:

fchp lc= 1.482 0,00012 = 0,177 m2

.

=

= 0,6%

Nm trong khong 0,60,8% Theo (mc 3.4.1- X L Nc Cp-Ts.Nguyn

Ngc Dung)

Nh vy m bo tng din tch khe h theo quy phm.

II.5.2.3.Tnh ton hthng phn phi nc lc.

Chiu rng mng tp trung nc vo bchn =0,8 (m). Vn tc nc trongmng =0,4-0,6 m/s. Chn =0,6 m/s.

Chiu cao lp nc trong mng tp trung:

= =

=1.49 (m).

Ca tmng tp trung sang mng phn phi:

= ==0,149 ().

Thit kca c kch thc: B H=0,4 0,4 m.Chiu rng mng phn phi nc vo b, chn =0,4 (m). Vn tc nctrong mng =0,3-0,4 (m/s). Chn =0,4 (m/s).

Chiu cao lp nc trong mng phn phi l:

= =

=0,56 (m). Thit kmng cao 0,6m.

Ca tmng phn phi vo mng chV.

= = =0,06 (). (=0,6-0,8 m/s ).Thit kca c kch thc: B H=0,3 0,3 m.

ng thu nc sch ti bcha.

Sdng mt ng chung thu nc tcc blc vbcha. ng ng c

t trn cao trong khi blc v thp xung khi ra blc.

ng knh ng t1 bra ng thu nc sch chung l 400 mm.

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ng knh ng chung l.

=(m).

Trong :Q:Lu lng ton trm. Q=0,717 (/s).:Vn tc nc chy trong ng. =1,1 (m/s).Vy =0,89 (m).Chn ng knh ng l 800 (mm). Kim tra li tc nc chy :

= =1,02 (m/s).II.5.2.4.Tnh ton hthng phn phi nc ra lc.

Quy trnh ra lc.(Theo mc 6.123 TCXDVN 33-2006).

Bc 1: ng van dn nc lc ng thi m van x nc ra h mc

nc trong b lc.

Bc 2: Khi ng my bm gi, m van dn gi xo trn vt liu v r

cn. Lu lng bm gi 60m

3

/m

2

/h. Nc th vn tip tc cho vo lc to tcdng qut b mt v gim ti cho cc ngn lc khc vi lu lng 6m3/m2/h. Qu

trnh ny ko di 2 pht.

Bc 3: Gi nguyn my gi ang vn hnh, khi ng 1 my bm nc

ra vi lu lng 10 m3/m2/h. Lu lng ny c thit lp bng cch ng van

dn nc chnh trnng ng ra v ch m van by-pass. Qu trnh ra vi gi

nc kt hp ko di 5 pht.

Bc 4: ng van gi, tt my bm gi, ng thi m van dn nc

chnh t my bm ra bt u qu trnh vt cn vi lu lng 24 m3/m2/h. Qu

trnh ny ko di t 4pht. Sau , kha van dn nc ra, tt my bm ra, ng

van x nc ra tch nc cho chu k lc mi.

- Tng lng nc ra lc theo chu trnh ra lc nh trn l:

W = ( 606 11 + 5 +4 ) 41.44= 146 (m3).

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Vn tc ca gi qua khe h:

Wg= 20 l/s.m2= 0,02 m3/s.m2

= = = 2,22 (m/s).Vn tc ca hn hp gi + nc qua khe h:

Ta c :

60 (m3/m2h)= 16,6 (l/s.m2).

10 (m3/m2h)= 2,8 (l/s.m2).

Wg+n= Wn2+ Wg= 16,6 + 2,8 = 19,4 (l / s. m2

)= 0,0194 (m3

/s. m2

).

= =

= 3,28 (m/s).Vn tc ca nc thun tu qua khe h:

Ta c :

24 (m3/m2h)= 6,67(l/s.m2).

Wn2= 6,67(l / s. m2)= 0,00667(m3/s. m2).

Qn2= Wn2F = 6,67 30 = 200 (l/s).

= =

= 1,1 (m/s).

Mng thu nc ra lc.

- H thng mng thu nc ra lc gm 1 mng chnh (mng tp chung) chy

theo chiu dc b, lng mng c dc 1% v pha mng tp trung.

- Phn b lc thnh 2 ngn. 2 mng chy theo chiu ngang ni vi mng chnh.

- nh cc mng c cng cao , m bo thu c nc u theo b mt b

lc.

+Chiu rng ca mng thu nc ra lc xc nh theo cng thc:

53

2m

m)a57,1(

q.KB

(m)

Trong :

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K - h s ph thuc vo hnh dng ca mng, k 2,1

qm - lu lng nc ra tho theo mng (m3/s).

+Lu lng x nc ra lc tnh ton bng lu lng nc ra lc vi cng

ra lc 6,67( l/s.m2):

Qn2= Wn2F = 6,67 30 = 200 (l/s)=0,2(m3/s).

a : t s gia chiu cao phn ch nht vi mt na chiu rng mng, a 1,2

= 2,1

= 0,6 (m)

Chiu cao phn ch nht ca mng : H1= 0,75 Bm 0,75 0,6 = 0,45(m).

Chiu cao phn tam gic ca mng : H2= 0,5 Bm 0,5 0,6 = 0,3(m).

Chiu cao hu ch ca mng : H= H1+ H20,45 + 0,3 = 0,75 (m).

Chiu cao ton phn ca mng : Hm = H + = 0,75 + 0,1 = 0,76 (m).

: chiu dy y mng, = 0,1 (m)

Thit k mng c dc i= 0,01 v pha mng tp trung

Khong cch t b mt lp vt liu lc n mp trn mng thu nc:

Hm L . e 0,25 (m).

L : chiu dy lp vt liu lc, L1,3 (m)

e : dn n tng i ca vt liu lc, e 30%

Hm1,3 . 30% 0,25 0,64 (m).

Chiu cao mng pha mng tp trung : Hcm= 0,76 + 0,013 = 0,79 (m).

ng dn nc ra lc.- Lu lng nc cn ra 1 b lc vi cng ra nc bng 5,56 (l/s.m2)

l: q = 166,8 (l/s).

- Theo 6.120 TCXDVN 33-2006, vn tc trong ng phn phi nc ra lc cho

php t: 1,5 2,0 (m/s) s b chn Vr1,7 (m/s).

- ng knh ng tnh theo cng thc sau:

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drr

rq4

v.

.

0,387 (m)

Chn ng dn nc ra lc bng gang xm DN 400 ng x nc ra lc.

- ng x nc ra lc c ni t mng thu nc ra lc trong b xung

mng thot nc chung pha di. Vn tc trong ng dn v thot nc ra lc

theo 6.120 TCXDVN 33-2006bng 1,5 2,0 m/s.

- Lu lng x nc ra lc tnh ton bng lu lng nc ra lc vi cng

ra lc20 (m3/m2h) = 6,67 (l/s.m2).

- Lu lng nc qut b mt vi cng bng 1,67 l/s.m l: 1,67 30 = 50,1

(l/s).

+Tng lu lng nc x l: q = 200 + 50,1 = 250,1( l/s).

+Chn vn tc qua ng x bng 1,7 m/s.

+ng knh ng thot nc ra lc l:

drr

rq4v..

0,432 (m).

- Chn ng thot nc ra lc l ng gang xm ng knh DN 400 mm.

ng x kit:

Chn ng knh ng x kit D = 150 mmtheo cu to.

y b c dc 0,0005 v pha ng x kit.

Tnh ton h thng cp gi ra lc.- Cng ra gi thun tu : Wg60 (m

3/m2h) = 0,01667(m3/m2.s)

- Chn vn tc gi trong ng gi l: Vg20 (m/s).(Mc 6.122 TCXDVN 33-

2006).

Lu lng gi cn cung cp:

qgWg. Fb0,01667 30 0,5(m3/s)

ng knh ng dn gi l:

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dg

0,17(m3/s)

Chn dg= 150mm.

II.5.2.5.Tnh ton chiu cao blc.

- Chiu cao xy dng b lc:

HXD= h1+ h2h3+ h4 + h5+ h6(m)

(Cng thc 4-54 X l nc cp TS.Nguyn Ngc Dung).

Trong :

+ h1: chiu cao t chp lc n y b, h1 = 1,0(m)+ h2: chiu dy lp vt si

h2= 0,30 (m)

(bng 4-7 X l nc cp TS.Nguyn Ngc Dung).

+ h3: chiu dy lp vt liu lc, h3= 1,3(m)

+ h4 : chiu cao lp nc trn mt vt liu lc, h4= 2,0 (m)

+ h5 : chiu cao bo v b, h5= 0,5 (m).HXD=1,0 + 0,3 + 1,3 + 2,0 + 0,5 = 5,1(m).

II.5.2.6.Thit biu khin tc lc.

Xi phng ng tm iu chnh tc

lc

Cu to: xem hnh v.

1. Xi phng ng tm2. Van gi

3. Ct lc

4. Hm thu nc lc

5. ng thu nc lc

6. Mng tp trung nc lc

C ch hot ng:

Hnh 56. Xi phng ng tm

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Bt u chu k lc: tn tht p lc qua lp vt liu lc (VLL) cn nh, mc nc

trong b lc thp, phao h xung lm van gi m, p sut ti nh xiphng bng p sut

kh quyn, xiphng lm vic nh mt ng trn thun tu.

Trong qu trnh lc: tn tht p lc qua lp VLL tng dn, lu lng lc c xuhng gim, nc trong b lc c xu hng dng ln, phao cng b nng ln lm cho van

gi b ng li dn dn. Khi p sut ti nh xiphng nh hn p sut kh quyn v

xut hin chn khng. ln ca p sut chn khng bng mc tng ca tn tht p lc

trong lp VLL, kt qu l tng tn tht p lc vn nh ban u v lu lng lc c

duy tr n nh.

Cui chu k lc: tn tht p lc qua lp VLL rt ln v p sut chn khng trong

xiphng cng rt ln dn n vic xut hin chn khng ngay trong lp VLL. ng h

o chn khng ti nh xiphng s ch n mt gi tr gii hn no gip ta quyt nh

ngng chu k lc vtin hnh thi ra.

II.5.2.7.Tnh ton sn phi vt liu lc.

- Th tch vt liu lc trong 1 b:

V = Fb . hVLL = 41.44 1,3 = 53.9 ~ 54 (m3).

- Thit k sn phi vt liu lc vi chiu cao phi bng 0,25 m.

Din tch sn phi : Fs= V / 0,25 = 54/ 0,25 = 216 (m2).

- Thit k 2 sn phi din tch 1 sn : F1s = Fs/ 2 = 108(m2).

- Kch thc mi sn : a b = 30 3.6 (m).

II.5.2.8.Tnh ton hthng bm ra lc.

- Vi lu lng nc ra lc qr= 250,2 (l/s).- p lc cn thit ca bm ra lc :

dtdtthhrb hhhhH (m)

Trong :

- hd: Tng tn tht (k c tn tht p lc theo chiu di v p lc cc b )

trn ng ng dn t trm bm n b lc, s b ly hd= 2 (m).

- htt: Tng tn tht p lcqua b lc khi ra :

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htt= h + hlc+ hs

Trong :

+ h : Tn tht qua h thng phn phi bng an lc s b ly h = 1,03 (m).

+ hsd: Tn tht p lc qua lp si hsd= 0,22 h Wn

= 0,22 0,30 10 = 0,66 (m).

(Cng thc 4-46 X l nc cp TS.Nguyn Ngc Dung).

+ hlc: Tn tht p lc qua lp vt liu lc:

hlc= (a + bW) Le =(Cng thc 4-47 X l nc cp TS.Nguyn Ngc Dung).

Vi dtd= 0,9a = 0,76, b = 0,017

htt= 1,03 + 0,66 + 1,157 = 2,847 (m).

II.5.2.9.Tnh ton sn phi bn.

Lng nc ra mt blc nhanh:

= 146(m3)- Hm lng cn tblc nhanh

Hm lng cn c trong nc trc khi vo b lc nhanh l : C1= 12

(mg/l).

Hm lng cn c trong nc sau khi ra khi b lc nhanh l : C2= 2

(mg/l).

Hm lng cn b gi li trong b lc nhanh l : CGL = 12 2 = 10

(mg/l).

Lng cn tch lytrong 1 ngy ca blc nhanh :

GGL=n

CQT GLb

Trong :

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- GGL : Hm lng cn tch ly trong 1 ngy ca 1 blc (mg)

- T : Thi gian lm vic ca blc T = 24gi.

- Qb : Lu lng nc ca mt bQb= 146(m3

/h)- CGL : Hm lng cn bgili trong b(mg/l)

- n : Sln ra btrong 1 ngy, n = 1.

GGL=n

CQT GLb =

= 35.04 (kg/ngy)

C 10 bl

c n

n t

ng hm lng c

n trong 1 ngy l 350.4 kg.

Tng lng cn kh trung bnh xra trong mt ngy:

G = GLng + Gloc= 36190.8 + 350.4 = 36514.2 ( kg/ngy)

Lng bn kh to thnh sau 15 ngay:

Gnn=36514.2 15= 547713 (kg).

Trong thc tcn to thnh a ra sn phi nm trong hn hp vi nc u c

m 96% v sau khi phi m gim xung cn 80%. Vic tnh ton chiu caobn trong sn phi ta chn theo m trung bnh 88%

Vi bn m 80%, trong 100kg hn hp c 88kg nc, 12 kg bn.

Khi lng bn sinh ra trong 15 ngay l:

Gbn= Gnn+ Gnc = Gn+ nG

W=547713 +

= 4199133 (kg)

Vbn=bun

bunG

= = 3817 (m3).

: Khi lng ring ca bn long =1.1 (T/m3).Chiu cao bn trong sn mi sn phi l: 2,5 m

Ta c din tch cn thit ca sn phi bn:

F = 1526(m2).

Chn 2 hhnh chnht kch thc mi hl B =35; L = 45m.

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- Chiu su sn phi: H = Hsi+Hchacn+hdtr=1,6 m.

Vi Hsi- Chiu cao lp si = 0,3m

Hchacn = 1 m.

hdtr- Chiu cao d tr =0,3m

- Thng xuyn tho lp nc trong trn mt lp bn lng ra khi hra h

thng thot nc ca thnh ph.

Khi h cha y bn cn, em bm chm di ng, t vo htp trung

nc u ra, bm ht nc lm kh lp cn cha trong h. Khi nng bn

kh t khong 25%, ttrng bn = 1,2t/m3xc bn kh ra ngoi, sau chnhsa li cc lp si v hng rt nc y h, ri cho htrli lm vic.

- Bn kh vt thln t chuyn i chn lp.

- Thng xuyn tho lp nc trong trn mt lp bn lng ra khi hra h

thng thot nc ca thnh ph.

II.6. Tnh ton thit kcc cng trnh phtr.

II.6.1.Tnh ton thit ktrm khtrng.

Lng clo a vo khtrng, Lclo= 2 mg/l (t2-3 mg/l - theo iu 6.162

TCXDVN 33-2006)

+ Liu lng clo dng trong 1 gi:

= = 5 (kg/h)

+ Lng clo dng trong 1 ngy :

Qngclo= Q1h

clo24 = 5 24 = 120 (kg/ngy)

+ Lng clo dng trong 1 thng:

Qngclo= Qng

clo30 = 120 30 = 3600 (kg/thng)

Vcl= 3600 / 1,43 = 2517,4 (l)

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+ 1.43trng lng ring ca clo

- Lng nc tnh ton cho Clorat lm vic ly bng 0,6 m3/kg Clo (theo iu

6.169 TCXDVN 33-2006)

Lu lng nc cp cho trm clo trong 1 gi:

Qclo= 0,6 . L1h

clo= 0,6 . 5= 3 (m3/h)

- Chn 4 bnh clo loi 1000 (l), 3 bnh hot ng v 1 bnh dtr.

Xc nh dung tch bcha

Dung tch iu ha bcha ly s bbng 20%Qng= 0,260000 = 12000 m3- Ct mt t ti v tr xy dng trm x l: Ztrm= 0,00 m

- B cha nc phn mng li xc nh xy 2 b vi kch thc 1 b 40

30 5,0 (m). Trong chiu cao lp nc trong b 4,8 (m).

B tr b cha theo kiu na ni na chm, 4,4(m) chm di t 0,4 (m) ni.

- Ct mc nc cao nht trong b cha l:ZB.cha

max= ZM+ 0,4 = 0+ 0,4 = 0,4 (m)

- Ct y b cha:

Zy bc= ZM- 4,4 = - 4,4(m).

II.6.2.Kch thc cc cng trnh phtrkhc.

- Nh hnh chnh : 830 = 180 (m2)

- Phng bo v: 54 = 20 (m2)

- Phng th nghim : 612 = 60 (m2)

- Trm bin p v my pht in: 612 = 72 (m2)

- Kho cha vt liu v xng c kh v sa cha : 920 = 180 (m2)

-Nh t - xe my : 525 = 90 (m2)

II.6.3.Tnh ton cc cng trnh xl bn cn.

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II.6.3.1.Slng v hm lng nc thi ra lc.

a) S lng nc ra lc.

Theo tnh ton trm x l c ton b l 10b lc nhanh, 2 b lng. Mi ngy 1ln ra b lc, khi ra lc th ra lun phin tng b (mi ln ra mt b).

Lng nc ra mt b lc vi chu trnh ra lc nh nu trn l:

b111111 F)

606060(

tWtWtWWn (m

3)

Trong :

- W1, W2, W3: Cng cc pha ra lc ln lt l : 6 (m3/m2.h), 10

(m3/m2.h), 24 (m3/m2.h).

- t1, t2, t3: Thi gian ra lc ca cc pha : t1 = 4 pht, t2 = 6 pht, t3 = 4 pht

- Fb: Din tch mt b lc, Fb= 37.2 (m2)

W = (60

611 +

5 +

4 ) 37.2= 131.44(m

3).

Lng nc cn ra 10b l : 131.44 10 = 1314.4 (m3).

II.6.3.2.Slng v hm lng nc thi tblng.

Lu lng nc dng cho vic x cn 2 b lng trong mt ngy m l:

= = 3708 ().II.6.4.Tnh ton xl nc ra lc v bn cn tblng.

II.6.4.1.Xc nh dung tch biu ha lu lng.

- Th tch b iu ha c xc nh theo cng thc :

V = Q - qth.t (m3)

Trong :

Q : Lu lng nc t b lc.

Q = 106 10 = 1060 (m3/ng)

qth

: Lu lng bm tun hon (m

3

/ng).

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Lu lng nc tun hon tnh m bo khi bm lm vic khng b

gin on khng nh hng n ch lm vic ca cc cng trnh x l.

qth 5%.Q TXL=5%

=129.2 (m3/h).

- m bo bm ht nc ra lc trong 24 h

qth24

W (m3/h) =

= 44.2 (m3/h)

Chn qth= 50 m3/h.

- Vy dung tch b iu ha l :

V = 106050 = 1010 (m3).

Chn 2 b iu ha lu lng nn dung tch mt b l 505 (m3). Chn b c

dng hnh tr kch thc mi b HxD = 5,9 x 11m.Chiu cao bo v 0,4 m

HXD= 6,3m.

Xc nh hm lng ha cht s dng

tng hiu qu qu trnh keo t dng thm ha cht l phn nhm.

Theo TCXDVN 33-2006 th lng phn nhm cn thit keo t Pp= 130

(mg/l)

II.6.4.2.Tnh ton thit kbkeo tvblng.

* S cu to b

D

D

pu

1

2

3

h

h

h

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Hnh II.17. S cu to b keo t v b lng.

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Xc nh kch thc ca b :

- Din tch mt ngn phn ng xoy l :

nH

tQF

f

th

60

(m2)

Trong :

t : Thi gian nc lu li trong b. Chn t = 18 pht (theo TCXDVN 33-

2006, t = 1520).

Q : Lu lng nc x l 32,67 (m3/h).

h2: Chiu cao b phn ng ly bng 0,9 chiu cao vng lng ca b lng

theo TCXDVN 33-2006 chiu cao vng lng 2,65 (m). Chnchiu cao vnglng l 3,5 (m) th h2= 0,93,5 = 3,15 (m).

- n : S b phn ng, n = 2.

= = 1,55 (m2).- ng knh ca ngn phn ng:

= = = 1,4 (m)Xc nh tit din b lng :

- Th tch cng tc ca b lng c xc nh bng cng thc:

)( 3mktQW h

Trong :

t : Thi gian nc lu li trong b ly t = 0,75 h.kh: H s khng iu ha gi. Ly kh= 1.

W= Q t kh = 32,670,751 =24,5(m3).-Din tch tit din ngang ca vng lng c xc nh theo cng thc sau:

NV

QF

h

6,3. (m2)

Trong :

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Q : Lu lng tnh ton, Q =32,67 (m3/h).

Vh: Tc tnh ton ca dng nc i ln. Theo TCXDVN 33-2006

chn Vh= 0,6 (mm)/s.

N : S b lng ng, N =2

: H s s dng dung tch b chn = 1,5.

F =1,5 = 11,3(m2)- ng knh b lng xc nh theo cng thc :

= = = 4 (m)

Xc nh ng knh ng dn nc vo mi b :

Vi Q= 32,67 m3/h = 0,00907(m3/s).

- Vn tc trong ng l v = 0,9 (m/s) (theo TCXDVN 33-2006 )

= = 80(mm)II.7.Xc nh cao trnh trm xl.

II.7.1.Cao trnh bcha nc sch.

- Ct mt t ti v tr xy dng trm x l: Ztrm= 0.0 m

- B cha nc phn mng li xc nh xy 2 b vi kch thc 1 b 30

30 4,8 (m). Trong chiu cao lp nc trong b 4,8 (m).

B tr b cha theo kiu na ni na chm, 4,4(m) chm di t- 0,4 (m) ni.

- Ct mc nc cao nht trong b cha l:

ZB.chamax= ZM+ 0,4 = 0+ 0,4 = 0,4(m)

- Ct y b cha:

II.7.2.Cao trnh blc Aquazurv.

- Ct mc nc trong blc Aquazurvl :

ZnB.lc= ZB.chamax+ hB.lc + hngB.lc - B.cha(m)

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Trong :

hB.lc: Tn tht p lc trong blc. Theo iu 6.355 TCXDVN 33-2006

hB.lc= 3,4 m.

hngB.lc - B.cha: Tn tht p lc trn ng ng dn tblc n bcha,

hngB.lc - B.cha= 0,8 m.

ZnB.lc= 0,4+ 0,8 +3,4 =4,60 (m).

II.7.3. Cao trnh b lng Lamen.

-Ct mc nc trong blc blng Lamen l:ZnB.lng= Z

nB.lc+ hB. lng + hng

B.lng - B.lc(m)

Trong :

hB. lng: Tn tht p lc trong blng. Theo iu 6.355 TCXDVN 33-2006

hB. lng= 0,5 (m).

hngB.lng - B.lc: Tn tht p lc trn ng ng dn tblng Lamen n b

lc hngB.lng - B.lc= 0,5m.

ZnB.lng= 4,60 + 0,50 +0,50 = 5,60 (m).

II.7.4. Cao trnh b phn ng c kh

-Ct mc nc trong bphn ng c kh l:

ZnB.phn ng= Zn

B.lng+ hB. phn ng+ hngB.phn ng - B.lng(m)

Trong :

hB. phn ng: Tn tht p lc trong bphn ng c kh. Theo iu 6.355

TCXDVN 33-2006 hB. phn ng= 0,2 (m).

hngB.phn ng - B.lng: Tn tht p lc trn ng ng dn tbphn ng c kh

n blng Lamen. V bphn ng lin vi blng nn hngB.phn ng - B.lng= 0

ZnB.phn ng= 5,60 + 0,2 + 0 = 5,80 (m).

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II.7.5. Cao trnh b trn c kh

Ct mc nc trong b trn c kh l:

ZnB.trn= Zn

B. trn+ hB. trn+ hmngB.trn - B.phn ng (m)

Trong :

hB. trn: Tn tht p lc trong ni bbtrn c kh. Theo iu 6.355

TCXDVN 33-2006 hB. phn ng= 0,1 m.

hmngB.trn - B.phn ng: Tn tht p lc trong mng dn tbn bphn ng

c kh. hmngB.trn - B.phn ng= 0,1m

ZnB.trn= 5,80 + 0,1 +0,1= 6,0 (m).

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MC LC:

CHNG I: NH GI CHT LNG NGUN NC V LA CHN

DY CHUYN CNG NGH.

I.1.Cht lng nc ngun v yu cu cht lng nc sau khi xl.

I.1.1.Cht lng nc ngun ....................................................................3

I.1.2.Yu cu cht lng nc sau khi xl.............................................4

I.2.La chn dy chuyn cng ngh.

I.2.1.Cc chtiu xl .............................................................................7

I.2.2.Tnh ton liu lng ha cht a vo .............................................7

I.2.2.1.Xc nh lng phn dng keo t........................................7

I.2.2.2.Xc nh mc kim ha .......................................................8

I.2.3.Kim tra n nh ca nc ..........................................................9

I.2.3.1.Kim tra kim ca nc sau khi xk.................................9

I.2.3.2.Kim tra kim ca nc sau khi keo t................................9

I.2.3.3.Hm lng c

n l

n nh

t sau khi a ha cht vo ki

m ha v

keo t..................................................................................13

I.2.4.La chn dy chuyn cng ngh.....................................................13

I.2.4.1.xut cc phng n dy chuyn cng nghxl..............13

I.2.4.2.nh gi v la chn dy chuyn cng ngh...........................14

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II.4.2.1.C s tnh ton.......................................................................35

II.4.2.2.Tnh ton cc kch thc ca blng....................................36

II.5. Tnh ton thit kblc AquazurV.

II.5.1.Cu to blc AquazurV c 2 mng thu nc ra lc.................47

II.5.2.Tnh ton blc AquazurV c 2 mng thu nc ra lc..............48

II.5.2.1.La chn mt sthng s......................................................48

II.5.2.2.Xc

nh din tch m

t b

ng v s

l

ng bcng tc...........49

II.5.2.3.Tnh ton hthng phn phi nc lc.................................51

II.5.2.4.Tnh ton hthng phn phi nc ra lc...........................53

II.5.2.5.Tnh ton chiu cao blc.....................................................57

II.5.2.6.Thit biu khin tc lc................................................57

II.5.2.7.Tnh ton sn phi vt liu lc...............................................58

II.5.2.8.Tnh ton hthng bm ra lc.............................................58

II.5.2.9.Tnh ton sn phi bn..........................................................59

II.6. Tnh ton thit kcc cng trnh phtr.

II.6.1.Tnh ton thit ktrm khtrng.................................................61

II.6.2.Kch thc cc cng trnh phtrkhc........................................62

II.6.3.Tnh ton cc cng trnh xl bn cn.........................................63

II.6.3.1.Slng v hm lng nc thi ra lc..............................63

II.6.3.2.Slng v hm lng nc thi tblng.........................63

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II.6.4.Tnh ton xl nc ra lc v bn cn tblng.....................63

II.6.4.1.Xc nh dung tch biu ha lu lng.............................63

II.6.4.2.Tnh ton thit kbkeo tv blng..................................64

II.7.Xc nh cao trnh trm x l..........................................................67

II.7.1.Cao trnh bcha nc sch.....................................................67

II.7.2.Cao trnh blc Aquazurv........................................................68

II.7.3. Cao trnh b lngLamen..........................................................68

II.7.4. Cao trnh b phn ng c kh...................................................68

II.7.5. Cao trnh b trn c kh............................................................69

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