tinh toan thiet ke he thong hap thu khi so2 dai hoc bkhn

7/27/2019 Tinh Toan Thiet Ke He Thong Hap Thu Khi So2 Dai Hoc Bkhn

1/41

TRNG I HC BCH KHOA H NIVIN KHOA HC V CNG NGH MI TRNG

______________________OOO_____________________

N MN HCCC QU TRNH C BN TRONG CNG NGH MI TRNG

bi:Tnh ton thit k h thng hp th kh SO2

GVHD: TS. V c ThoSVTH: Mai Th Hin

Lp: K thut mi trng K52SHSV: 20071073

H Ni, thng 12/2010


2/41

NHIM V THIT K N MN HCH v tn Sinh vin: Mai Th HinLp: K thut mi trngKha: 52

I. u thit k:Thit k h thng hp th kh SO2II. Cc s liu ban u:- Hn hp kh cn tch: SO2 trong khng kh- Dung mi: nc- Lu lng kh vo thp: 25000 m3/h- Nng SO2: y = 0,028( mol/mol)- Hiu sut yu cu:= 84%- Nhit p sut v lng dung mi: m phng theo mt s iu kin- Loi thit b: Thp m

III. Cc phn thuyt minh v tnh ton:1. M u2. Tnh ton thit k thp hp th (ng knh, chiu cao, tr lc)3. Tnh ton thit b ph- Tnh bm- Tnh my nn kh4. Tnh ton c kh5. Kt lunIV. Cc bn v:1. Bn v s dy chuyn kh A3 hoc A42. Bn v thp hp th kh A1V. Gio vin hng dn:TS. V c ThoVI. Ngy giao nhim v: Ngy 6 thng 9 nm 2010VII. Ngy hon thnh nhim v: Ngy thng nm 2010 Gio vin hng dn

Ch nhim khoa ( H tn v ch k)( H tn v ch k)

nh gi kt qu Ngy thng nm 2010- im thit k Cn b bo v- im bo v ( H tn v ch k)- im tng hp


3/41

PHN M U

Vn x l cc cht nhim khng kh v ang nhn c s quan tmca ton nhn loi ni chung v ca Vit Nam ni ring. Vi mc ch vic

thc hin n mn hc thc s cn thit, trong qu trnh lm n em hiuc nhng phng php, cch tnh ton, la chn thit b c kh nng ngdng vo thc tin c th x l cc cht thi gy nhim

Sau 15 tun tm hiu, tnh ton v nhn c s hng dn nhit tnh ca cthy c trong Vin, nhng do hn ch v ti liu v kinh nghim tnh ton, nnkhng th trnh khi nhng sai st, em rt mong nhn c kin ca cc thyc n sau c kt qu tt hn

Em xin chn thnh cm n!H Ni, ngy 3 thng 12 nm 2010

Sinh vin thc hin Mai Th Hin


4/41

PHN NI DUNG

I.Gii thiu chung1. S lc v kh SO2

Trong s nhng cht gy nhim khng kh th SO2 l mt cht gy nhim

kh in hnh. Sulfuro l sn phm ch yu ca qu trnh t chy cc nguynnhin liu c cha S. Cc nh my in thng l ngun pht sinh ra nhiu SO2trong kh thi, ngoi ra cn phi k n cc qu trnh tinh ch du m, luykim, tinh luyn qung ng, sn xut ximang v giao thng vn ti cng lnhng ni pht sinh nhiu kh SO2

Kh SO2 l cht kh khng mu, c mi hng cay khi nng trong kh quynl 1 ppm. Kh SO2 l kh tng i nng nn thng gn mt t ngang tmsinh hot ca con ngi, n cn c kh nng ha tan trong nc nn d gy phn ng vi c quan h hp ca ngi v ng vt. Khi hm lng thp, S2

lm sng nim mc, khi nng cao> 0,5 mg/m3

, SO2 s gy tc th, ho, vimlot ng h hpSO2 lm thit hi ma mng, lm nhim c cy trng. Ma axit c ngun

gc t kh SO2 lm thay i pH ca t, nc, hy hoi cc cng trnh kin trc,n mn kim loi. Ngoi ra nhim SO2 cn lin quan n hin tng m quangha

Chnh v nhng tc ng tiu cc trn m vic gim ti lng cng nh nn pht thi SO2 vo mi trng l vn rt c quan tm2. Phng php x l SO2

Kh SO2 thng c x l bng phng php hp th, tc nhn s dng hp th thng l sa vi, sa vi kt hp vi MgSO4 hoc dung dch kim...Trong phm vi n ny, vi nhim v c giao l hp th kh SO2 bng nc.y l phng php hp th vt l nn hiu sut hp th khng cao. Do t phi chn iu kin lm vic ca thp hp th nhit thp v p sut caonng cao hiu sut hp th3. Thp m

Thp m c s dng rng ri trong cng nghip ha cht v c im dthit k, gia cng, ch to v vn hnh n gin. Thp m c s dng trocc qu trnh hp th, chng luyn, hp ph v mt s qu trnh khc. Thp cdng hnh tr, trong c cha m, ty vo mc ch thit k m m c th xp hay ln xn. Thng thng lp m di thng c sp xp, khot lp 3 tr i, m c ln xn

Thp m c nhng u im sau:- Cu to n gin

- B mt tip xc pha ln, hiu sut cao- Tr lc trong thp khng qu ln


5/41

- Gii hn lm vic tng i rngTuy nhin, thp c nhc im l kh thm t u m lm gim kh nng hth...II.Thit k n mn hc1. u thit k:

Thit k h thng hp th kh thi p dng trong cng nghip2. Cc s liu ban u- Hn hp kh cn tch: SO2 trong khng kh- Dung mi: nc- Lu lng kh vo thp: 25000 m3/h- Nng SO2: y = 0,028( mol/mol)- Hiu sut yu cu:= 84%- Nhit p sut v lng dung mi: m phng theo mt s iu kin- Loi thit b: Thp m3. Phng php hp th x l SO2S ca h thng

2

1

3

4

5


6/41

1. B cha dung mi

2. Bm cht lng

3. Thp hp th

4. My nn kh

5. Van an ton

Thuyt minh dy chuyn:

- Hn hp kh cn x l cha SO2 v khng kh c my nn kh a vot pha di y thp. Nc t b cha c bm li tm a vo thp hpth, trn ng ng c van iu chnh lu lng v ng h o lu lng Nc c bm vo thp vi lu lng thch hp, ti t trn xungdi theo chiu cao thp hp th

- Hn hp kh sau khi i qua lp m xy ra qu trnh hp th s i ln nhthp v ra ngoi theo ng ng thot kh. Kh sau khi ra khi thp cnng kh SO2 gim, mc gim ty thuc vo hiu sut hp th cathp hp th

- Nc sau khi hp th SO2 i xung y thp i v ra ngoi theo ngng thot cht lng. Nc sau khi hp th nu nng SO2 cao s cx l v ti s dng.

Gi:

Gy: lu lng hn hp kh vo thp( kmol/h)

Gx: lu lng nc vo thp( kmol/h)

Gtr : lu lng kh tr( kmol/h)Y: nng phn mol tng i ca SO2 trong kh i vo thp ( kmol SO2/kmolkk)

Yc: nng phn mol tng i ca SO2 trong kh i ra khi thp ( kmolSO2/kmol kk)

X: nng phn mol tng i ca SO2 trong nc i vo thp( kmolSO2/kmol dm)


7/41

Xc: nng phn mol tng i ca SO2 trong nc i ra khi thp( kmolSO2/kmol dm)

Theo bi: y = 0,028 (mol/mol) Y =0,028

1 1 0,028d

d

y y

= = 0,0288 (kmol

SO2/kmol kk)Bit hiu sut hp th l:= 84%

Do : Yc = Y( 1-) =0,0288.( 1-0,84)= 4,608.10-3 (kmol SO2/kmol kk)

yc =3

34,608 10

1 1 4,608 10c

c

Y Y

=+ + =4,587.10

-3 (kmol/kmol)

ytb= 2d c y y+

=

30, 028 4,587 102

+ = 0,0163 kmol/kmol

Dung mi ban u l nc X = 0

Gi s iu kin lm vic ca thp l T =250CT =298K

P =1atm = 760mmHg

P=1 atm = 1,0326 at

Ta coi hn hp kh l l tng. Theo phng trnh trng thi kh l tng ta c:

Gy = n =1 25000

0,082 298 PV RT

= = 1023,08( kmol/h)

Gtr =1023,08

1 1 0,0288 y

d

GY

=+ + = 994,44( kmol/h)

Thit lp phng trnh ng cn bng:

Theo nh lut Henry ta c: ycb= mx

Ycb= 1 (1 )mX

m X +

Ta c m= P

250C vi kh SO2 th =0,031.106 mmHg

m =6

0,031.10760 = 40,79

Gx, XdYc

XcGy, Yd


8/41

Ycb =40,79

1 39,79 X

X

Thit lp phng trnh ng lm vic:

Phng trnh cn bng vt liu cho thit b:

Gtr . Y + Gx. X = Gtr . Yc + Gx. X

Gtr ( Y- Yc) = Gx( X- X)

Do X = 0 nn pt tr thnh: Gtr .( Y- Yc) = Gx. X

Y= xtro

GG .X+Yc

Gi thit Xc= Xcbc th lng dung mi ti thiu cn hp th l:

Gxmin = Gtro. d ccbc d

Y Y X X

T phng trnh ng cn bng Ycb= 1 (1 )mX

m X +

Xcb= (1 )Y

m m Y

Y = 0.0288(kmol SO2/kmol kk)

Xcbc =0,0288

40,79 39,79 0,0288+ =6,868.10-4 (kmol SO2/kmol nc)

Gxmin = 994,44.3

40,0288 4,608 10

6,868 10

= 35028,38 kmol/h

Lng dung mi cn thit hp th : Gx = . GxminThng thng = 1,21,5. Chn = 1,2

Gx = 1,2.35028,38 = 42034,056 kmol/h xtro

GG =

42034,056994,44 =42,269

Y = 42,269X+ 4,608.10-3


9/41

khi Yd = 0.0288 th Xc =34,608 10

42,269d Y =

30,0288 4,608 1042,269

= 5,723.10-4

(kmol SO2/kmol H2O) xc = 1c

c

X X + = 5,72.10

-4( kmol/kmol)

xtb= 2d c x x+ =

40 5,72 102

+ =2,86.10-4( kmol/kmol)

X Y Ycb0 4,608.10-3 00,0001 8,835.10-3 4,095.10-30,0002 0,013 8,223.10-30,0003 0,0173 0,0124

0,0004 0,0215 0,01660,0005 0,0257 0,0210,0005723 0,0288 0,02390,0006868 0,0288

V th ng cn bng v ng lm vic

th ng cn bng, ng lm vic

y = 41.697x

R 2

= 0.9999

y = 42.269x + 0.0046

R 2 = 1

0

0.005

0.010.015

0.02

0.025

0.03

0.035

0 0.0002 0.0004 0.0006 0.0008

X

Y ng lm vic

ng cn bng

II.1 Tnh ng knh thp:


10/41

1. Tnh khi lng ring: i vi pha lng:

p dng cng thc:O H

SO

SO

SO

xtb

aa

2

2

2

211

+=

Trong : xtb : Khi lng ring trung bnh ca hn hp lng, kg/m3. :2SOa Phn khi lng ca SO2 trong pha lng O H SO 22 , : Khi lng ring ca SO2 v H2O 250C, kg/m3.

- Tra bng I.5 250C c: O H 2 = 997,08 (kg/m3)- Tra bng I.2 250C c: 2SO (200C) = 1383 (kg/m3)

2SO (400C) = 1327(kg/m3)Dng phng php ni suy => 2SO (250C) = 1369 (kg/m3)- Tnh :2SOa

p dng cng thc 222 2

.. (1 )

SO tb

SOSO tb H O tb

M xa

M x M x=

+ Trong

:2SOa Phn khi lng trung bnh ca SO2 trong hn hp.xtb: Nng phn mol trung bnh ca SO2 trong pha lng, (kmol

SO2/kmol H2O)xtb = 2,86.10-4 (kmol SO2/kmol H2O)

2SOa =4

4 464 2,86.10

64 2,86.10 (1 2,86.10 ) 18

+ = 1,016.10-3

- Tnh khi lng phn t ca hn hp lng Mx x M = xtb. 2SO M +(1xtb). O H M 2 = 2,86.10-464+(1-2,86.10-4)18

=18,013156Lm trn Mx =18

i vi pha kh:- Tnh Myp dng cng thc: My = ytb. 2SO M + (1 - ytb). KK M Trong :

My: Phn t lng trung bnh ca hn hp kh, (kg/kmol) 2SO M , KK M : Khi lng phn t ca SO2 v khng kh, (kg/kmol)

ytb: Phn mol trung bnh ca SO2 trong hn hp(kmol SO2/kmol hn hp kh) My = 0,016364 + (1-0,0163)29 = 29,5705


11/41

- Tnh ytb = 2[ (1 ) ] 273

22,4tb SO tb kk y M y M

T +

=

[0,0163 64 (1 0, 0163) 29] 27322, 4 298

+

ytb

= 1,209 kg/m3

- Tnh xtb : 2 22 2

11 SO SO xtb SO H O

a a

= +

2 22 2

3 3

1 1 997,361 1, 016.10 1 1, 016.101369 997,08

xtbSO SO

SO H O

a a

= = = ++ (kg/m3)

2. Lng kh trung bnh i trong thp:

Vytb = 2cd V V +

(m3

/ h) (II.183)vi: V: Lu lng hn hp u iu kin lm vic (m3/ h)

Vc: Lu lng kh thi i ra khi thp (m3 / h): Vc = Vtr * (1 + cY )(II.183)

V = y ytb ytb

G M

=1023,08 29,5705 25023,15

1,209 = m3/h

vi: Mytb: Khi lng mol phn t trung bnh ca hn hp kh (kg /kmol)

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

Tng t: Vtr = tr ytb ytb

G M

=994,44 29,5705 24322,65

1,209 = m3/h

Vc = Vtr (1+ cY ) = 24322,65(1+4,608.10-3 )=24434,73 m3/h

Vytb = 24728,94 m3

/h

3. nht , x y :i vi pha lng:p dung cng thc: 2 2lg lg (1 ) lg x tb SO tb H O x x = + I-84Trong :

O H SO 22, : nht ca SO2 v H2O 250C, Ns/m2

Tra bng I-101 s tay I: 2SO (200C)= 0,304.10-3 Ns/m2 2SO (300C)= 0,279.10-3 Ns/m2


12/41

2SO (250C) =0,2915.10-3 Ns/m2

Tra bng I-102 s tay I:2 H O (250C)= 0,8937.10-3 Ns/m2

xtb: Nng phn mol trung bnh ca SO2 trong pha lng, (kmol SO2/kmol H2O)

xtb = 2.86*10-4 (kmol SO2/kmol H2O)lg x = 2,86.10-4lg(0,2915.10-3)+(1-2,86.10-4)lg(0,8937.10-3)= -3,0489 x = 8,935.10-4 Ns/m2

i vi pha kh:

p dng cng thc: KK

KK tb

SO

SOtb

y

y M y M y M

).1(.

2

2 +=

Trong 2, , y SO kk : nht trung bnh ca pha kh, ca SO2 v ca khng kh

iu kin lm vic 250

C, Ns/m2

2, , y SO kk M M M : khi lng phn t ca pha kh, ca SO2 v ca khngkh iu kin lm vic 250C v P=1atmTra th I-35 ta c : 2SO (250C)=0,0125.10-3, Ns/m2

kk (250C)=0,018.10-3, Ns/m2

2

2

5 2

3 3

29,5705 1,77.10 ( / ). 0,0163 64 (1 0,0163) 29(1 ).0,0125.10 0,018.10

y y

tb SO tb KK

SO KK

M Ns m y M y M

= = = ++

4. Tnh vn tc o pha:p dng cng thc:

Y =1,2.e-4X ( II-187 )

Vi0,162

3

. .. .

s d ytb x

d xtb n

Y g V

=

1 14 8

ytb x

y xtb

G X

G

= s : tc o pha, m/s

V: th tch t do ca m, m3/m3

d : b mt ring ca m, m2/m3

Thp hp th SO2 mang tnh axit nn ta chn m vng Rasig ln xn: m bng s kch thc 25253.0

V= 0,75 m3/m3

d = 195 m3/m3

g: gia tc trng trng, g=9,81m/s2


13/41

Gx, Gy l lng lng v lng hi trung bnh( kg/s)

Gx = 2 * *2 2 2SO bihapthu xd xc tro d

xd xd GG G G Y G G + = + = + = 42034,056+994,44 0,0288 0,842

Gx = 42046,08 kmol/hGx=42046,08 18=756829,44 kg/h Gx=756829,44 210,23043600 = kg/s

Gy = 2 1023.08 1023,08 994,44 0,0288 0,842 2 2 yd yd SO bihapthu yd yc G G GG G + + + = =

Gy = 1011,05 kmol/hGy=1011,05 29,5705=29897,25 kg/h Gy=

29897,25 8,3053600

= kg/s

1 1 1 14 8 4 8210, 2304 1, 209 0,969

8,305 997,36 ytb x

y xtb

G X G

= = =

T phng trnh ca Y ta c:

4 3 4*0,969 3

0,16 0,164

3

1,2. 1,2. 9,81 0,75 997,36 0,6668,935.10195*1,2091,005.10

X d xtb

s

xd ytb

n

e g V e

= = =

m/s

Thng thng: (0,8 0,9)tb s = Chn tb =0,85 s tb =0,850,666=0,5661 m/s

5. Tnh ng knh thp

ng knh thp:

Cng thc: D=43600

ytb

tb

V

=

4 24728,94 3,933600 0,5661 = m Quy trn D=3,9 m

Kim tra:

+ 150td d

D1

m thm t rt tt

II.2 Tnh ton chiu cao thp:Chiu cao thp c xc nh theo phng php s n v chuyn khi:

H = hdv.my (m)Trong : H: chiu cao thp, m

hdv: chiu cao mt n v chuyn khi, mmy: s n v chuyn khi

Xc nh chiu cao mt n v chuyn khi:

hdv = h1 + 2'. y

x

m Gh

G (m)

Trong : h1: chiu cao 1 n v chuyn khi ng vi pha kh

h2: chiu cao 1 n v chuyn khi ng vi pha lng


15/41

m: gi tr trung bnh ca tg gc nghing ng cn bng Y*=f(X)vi mt phng ngang

Tnh h1 v h2:

0.25 2/31 Re Pr

d y y

d

V h a = , m

Trong : a : h s ph thuc vo dng m, vi m vng a=0,123

: h s thm t ca m, do1th

U U

> nn =1

Rey: chun s Renoyd i vi pha kh

0,4

Re y s y y d

=

50,4 1,209 0,666Re 93,3

1, 77.10 195 y = =

Pr y: chun s Pran:Pr y y y y D

=

Dy =kkSO

23 / 1

kk

3 / 1

SO

5,14

M1

M1

)vv(P T 10.0043,0

22

++

(m2/s)

Trong : T: nhit lm vic tuyt i T=298K

P: p sut lm vic P=1atm

2SOv : th tch mol ca SO2, 2SOv =44,8 cm3/mol

kk v : th tch mol ca khng kh,kk v =29,9 cm3/mol

Dy =4 1,5

51/3 1/3 2

0,0043.10 298 1 1 1,0826.101,0326 (44,8 29,9 ) 64 29

+ =

+ m2/s

Vy5

51,77.10

1,209 1,0826. 0Pr

1 y

= =1,3523

0,25 2/31 93,3 *1,35230,123 1 1950,75h = = 0,119 m

2/3 0.25 0.52 256 ( ) Re Pr x x x

xh

=


16/41

Trong : Gx=42046,0818=756829,44 kg/h Gx=756829,44 210,2304

3600= kg/s

Rex l chun s Renoyd i vi pha lng: 0,04Re x xt d x

G F

= = 40,04 210,2304

11,95 195 8,935.10

= 4,03886

Pr x l chun s Pran i vi pha lng:Pr x x x x D

= Dx: h s khuch tn ca SO2 vo nc nhit 250CDt = D20[1+b(t-20)] (m2/s)Trong : D20: h s khuch tn ca SO2 vo nc 200C

D20=2 22 2 2

6

1/3 1/3 2

10 1 1( ) SO H O H O SO H O M M AB v v

+

+ (m2/s)

A: h s, i vi cht kh tan trong nc A=1B: h s, dung mi l nc B=4,7

2SOv : th tch mol ca SO2 200C, 2SOv =44,8 cm3/mol2 H Ov : th tch mol ca H2O 200C, 2 H Ov =18,9 cm3/mol2 H O

: nht ca nc 200C, 2 H O =1,005*10-3 Ns/m2=1,005 cP(bng I.102-94)

D20=6

9

1/3 1/3 2

10 1 1 1,466.1064 181 4,7 1,005(44,8 18,9 )

+ =

+ (m2/s)

b =2

23

0,2 H O H O

2 H O

: khi lng ring ca nc 200C, 2 H O = 998,23 kg/m3 bng I.5-11

b =3

0,2 1,005 0,02998,23

;

Dx = 1,466.10-9[1+0,02(25-20)]= 1,6126.10-9 m2/s

94

Pr 555,5997,36 1,6126.10

8,935.10 x

= =

2/3 0,524 0,25256 ( ) 4,03886 555,5 08,935.10

997, 95

,367h m

= =

Tnh m: T phng trnh ng cn bng ta c: m=41,697Gy=1011,05 29,5705=29897,25 kg/h Gy=

29897,25 8,3053600

= kg/sVy ta xc nh c chiu cao ca mt n v chuyn khi:hdv = 0,119 +

41,697 8,305 0,795210,2304

=1,428 mXc nh s n v chuyn khi:

Da vo gi tr Xcbc= 6.868*10-4 (kmol SO2/kmol nc)


17/41

X Y Ycb Y-Ycb 1/(Y-Ycb)0 4,608*10-3 0 4,608*10-3 2170,0001 8,835*10-3 4,095*10-3 4,74*10-3 210,970,0002 0,013 8,223*10-3 4,777*10-3 209,3360,0003 0,0173 0,0124 4,9*10-3 204,080,0004 0,0215 0,0166 4,9*10-3 204,080,0005 0,0257 0,021 4,7*10-3 212,770,0005723 0,0288 0,0239 4,9*10-3 204,080,0006 0,03 0,025 5*10-3 2000,0006868 0,0336 0,0288 4,8*10-3 208,33

Cng thc tnh s n v chuyn khi:

my =d

c

Y

cbY

dY Y Y (II.175)

V hnh thang cong quan h gia Y v f(Y)=1

cbY Y

Y 1/(Y-Ycb) S my4,608.10-3 217 0,904514595 6,024726748,835.10-3 210,97 0,875287245

0,013 209,336 0,88884440,0173 204,08 0,8571360,0215 204,08 0,8753850,0257 212,77 0,64611750,0288 204,08 0,242448 0,03 200 0,734994

0,0336 208,33


18/41

Hnh thang cong

0.00E+00

5.00E+01

1.00E+02

1.50E+02

2.00E+02

2.50E+02

1 2 3 4 5 6 7 8

Y

1 / ( Y

- Y c

b )

Din tch hnh thang cong chnh bng s n v chuyn khi l my =6,025 H=1,42856,025=8,61 m. Quy chun H=8,6 my thc cht l chiu cao lp m. Chiu cao ca thp ngoi chiu cao ca lm cn tnh n chiu cao t mt trn ca m n nh thp v t mt dm ti y thp v khong cch gia hai lp mp dng cng thc:Hthp = H + Hm- np+ Hm-m+ Hm- y

Hm-np = 1 m Hm-m = 0,5 m do tch lp m lm i

Hm-y = 1 mVy chiu cao thp Hthp = 8,6+1+0,5+1=11,1 mII.3. Tr lcp dng cng thc

k u P P P +=

Trong : k P :Tn tht m kh

u P

:Tn tht m tThp hp th t hiu sut cao nht khi vn tc ca kh bng vn tc io pha


19/41

=> Tr lc ca thp m i vi h kh-lng di im o pha c th xc nc bng cng thc sau:

0,405 0,0450,225

1(1 ) 1 . . y x x

u k k y x y

G P K P P A

G

= + = + [II-190] ( * )

Trong : Pu: tn tht p sut khi m t ti im o pha c tc ca kh bn

tc ca kh i qua m kh(N/m2) PK : tn tht ca m kh (N/m2) Gx, Gy: lu lng ca lng v ca kh (kg/h) y x , : khi lng ring ca lng v ca kh (kg/m3)

y x ,

: nht ca lng v kh (Ns/m2

) A1: h s (ng vi im tc lm vic bng 0.85 tc o pha)=> A1 = 5,1

* Tn tht p sut ca m kh tnh theo cng thc:

2

...

42

...

2

3

'2' y y

d

d t y

td

K V H

d H

P == [II-189]

Rey = 93,3 => ch xoy v m l m vng ln xn

=>'

0,2

16,0Re y = = 0,2

1693,3 =6,46

Tnh tr lc m kh:2

3'

4 2 y yd

k d

H P V

= =2

36,46

0,758, 6 195 0,5661 1, 209

4 2 =1243,66 N/m2

0,0450,405 0,225 4

5

210,2304 1,209 8,935.101243,66 1 5,1. . . 7424,98,305 997,36 1,77.10u

P

= + = N/m2

=> 21243,66 7424,9 8668,56( / )k u P P P N m = + = + =II.4. M phngBng m phng 1 s iu kin: nh kmDa vo bng m phng km theo ta c cc nhn xt nh sau:nh hng khi thay i nhit :

Nhit tng khng c li cho qu trnh hp th. Nhit tng lm gim hisut hp th v t c yu cu phi tng thm kch thc thit b, tnng knh v chiu cao.

nh hng ca p sut:


20/41

p sut c nh hng ti hiu sut hp th, lm tng hiu sut hp th. Nhnu p sut tng th chi ph kinh t cng tng theo nh l phi lp t thm mnn, chi ph nng lng tng do cng sut hot ng ca my tng rt mnh.

T bng m phng ta chn P=3 atm, T=250C D=2 m; H=9,2 m

Hthp= 9,2+1+0,5+1=11,7 m; Gx=14000 kmol/h

PHN 2: TNH TON CC THIT B PH

Trong vic hp th SO2 bng nc s dng thp m cn c cc thit b phgip cho qu trnh vn chuyn cht lng v cung cp kh vo thp theo ch lm vic ca thp gip vic hp th t c hiu sut mong mun.Trong ccthit b ph th bm v my nn kh l hai thit b quan trng nht.I. BMTrong cng nghip, bm ly tm c s dng rng ri v c nhiu loi khcnhau v cu to cng nh cch vn hnh.Bm ly tm c phn loi theo nhiu cch khc nhau nh theo s bc, theo

cch t bm, theo iu kin vn chuyn ca cht lng t gung ra thn bm vtheo 1 s c trng khcTheo dy chuyn cng ngh trong bi ta chn bm ly tm 1 cp nm ngang.

I.1. Nguyn tc lm vic ca bm ly tm Nguyn tc hot ng: Bm ly tm lm vic theo nguyn tc ly tm. Cht lngc ht v y cng nh nhn thm nng lng l nh tc dng ca lc ly tmkhi cnh gung quay. B phn chnh ca bm l cnh gung trn c gn nhngcnh c hnh dng nht nh, bnh gung c t trong thn bm v quay vitc ln. Cht lng theo ng ht vo tm gung theo phng thng gc rivo rnh gia cc gung v cng chuyn ng vi gung. Di tc dng ca lly tm, p sut ca cht lng tng ln v vng ra vo thn bm, vo ng y theo phng tip tuyn. Khi tm gung to nn p sut thp. Nh p lc mtthong b cha, cht lng dng ln trong ng ht vo bm. Khi gung quay chlng c ht lin tc, do cht lng c chuyn ng u n. u ng hc lp li lc ngn khng cho rc v vt rn theo cht lng vo bm gy t bm v ng ng. Trn ng ht c van mt chiu gi cho cht lng trn nng ht khi bm ngng lm vic. Trong ng y c lp van mt chiu trnhcht lng bt ng dn vo bm gy ra va p thy lc lm hng bmI.2. Cc thng s c trng ca bmp sut mt thong P1= 9,81.104 N/m2p sut lm vic P= 3 atm=31,013.105=303900 N/m2Gia tc trng trng g=9,81 m/s2 250C: nc=997,08 kg/m3

nc=0,8937.10-3

Ns/m2

p sut ton phn ca bm H(m):p dng phng trnh becnulli ta c


21/41

Mt ct 1-1 v 1-1:

)1(.2..2.

2211

mhhvv h H g g

P

g g

P +++=+

Mt ct 1-1 v 2-2

)2(.2..2.

222

2

md d

r r

h H g g

P

g g

P

+++=+

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

hmh + hmd= g P .

P : p sut ton phn cn thit khc phc sc cn thy lc trong h thnp sut ton phn ca bm l hiu p sut gia hai giai on ht v y1: vn tc nc b cha, 1=02: vn tc nc khi vo thp hay trong ng y1: vn tc nc khi vo bm2: vn tc nc khi ra khi bmThc t: 2 = 2

g g

P P h H H m .2.

'112

0

+

++=

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

hmh = g P h.

Trong : cmd h P P P P ++=d P : p sut ng lc hc cn thit to tc cho dng chy ra khi ng

2

2

1

1

11

h H P1

2


22/41

2. 2h

d P

=

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

2.

.2h

hm d

L P

=

c P : p sut cn thit khc phc tr lc cc b

2.

.2

hc P =

cmd h P P P P ++= =2.

.2.

..2. 222

hh

h

h

d L

++

++=

h

h

d L

.12. 2

ng knh ng ht: 0,785h hV d

= Trong : V l lu lng th tch cht lng i trong ng, m3/s

2

2

d 3. 14000 18 0,07( / ).3600 3600 997,08

x H O

H O

G M V m s

= = =

Theo bng II.2(I-370) cht lng trong ng ht ca bm c h=0,8-2,0 (m/s).

Chn h = 1,5 (m/s) 0,07 0,2438( )

0,785 1,5hd m= =

Quy chun dh=0.244m 20,07 1,5( / )

0, 244 0, 785hm s = =

Chun s Re ca cht lng trong ng ht

2

2

4

. . 1,5 0,244 997,08Re 408337,56 40008,937.10

h h H O

H O

d

= = = >

Dng ch chy xoy nn h s ma st c tnh nh sau

0,91 6,812.lg

Re 3,7

= +

(I-380)

Trong :: nhm tuyt i. Chn vt liu lm ng l thp ni khng hn

310.07,0 = : nhm tng i, c xc nh theo cng thc:

340, 07 10 2,87 10

0,244hd = = =

0,9 41 6,81 2,87 10

2.lg 408337,56 3,7

= +

0,0165 =


23/41

H s tr lc cc b:Cht lng vo ng thng, u ng ht c lp li chn an bng kim loi

c += 1

Vi .0=c

Chn 0 0,9T

F F = 0

0,131,0

== Bng II.16(I-382,384)

tr lc ca ng c lp li chn an bng kim loi l1 0,13 1 1,13ong = + =

Trn ng ht cn lp 1 van 1 chiu. Theo I-3991,9 2,1van = Chn 2 = 1,13 2 3,13h = + =Tra bng II-34(I-441) s ph thuc chiu cao ht ca bm ly tm vo nhit nhit lm vic T=250C th chiu cao ht ca bm khong 4,5m th m bo khng xy ra hin tng xm thc. Tuy nhin loi tr kh nng dao ntrong bm nn gim chiu cao ht khong 11,5m so vi gi tr trong bng. Vychn chiu cao ht l 3,5m p lc ton phn cn thit khc phc sc cn thy lc l:

2 2997,08 3,51,5 1 3,13 0,0165* 4898,17( / )2 0,244h P N m = + + =

hmh = g P h.

=

4898,17 0,5( )997,08 9,81

m=

Xc nh tn tht p sut do tr lc gy ra trn ng ng y:ng knh ng y:Theo bng II.2( I-370) vn tc cht lng trong ng y ca bm ld= 1,5-2,5 m/s. Chn d = 2,0 m/s

=> 0,07 0,211( )0,785 2d d m= =

Quy chun dd = 21cm

Vn tc ca ng y l 2

0,07 2( / )0, 21 0, 785d

m s = =Chun s Re ca cht lng trong ng y

2

2

4

. . 2 0,21 997,08Re 468584,1 40008,937.10

d d H O

H O

d

= = = >

Dng ch chy xoy nn h s ma st c tnh nh sau

0,91 6,812.lg Re 3,7

= + (I-380)

Trong :


24/41

: nhm tuyt i. Chn vt liu lm ng l thp ni khng hn 310.07,0 =

: nhm tng i, c xc nh theo cng thc:3

40, 07 10 3,33 100,21

d d = = =

0,9 41 6,81 3,33 102.lg

468584,1 3,7

= + 0,0167 =

Theo bng II.16(I-393), i vi thnh nhn Re > 2.105 th b qua tn tht ma stcong=A.B.CGc 1900 == A

Chn:15,02 == B

d

R

d 45,15,0 == C

ba

=> 2175,045,115,01 ==cong H s tr lc cc b ca ton ng y: 0,2175 3,13 3,3475cong thang = + = + =Chn chiu di ng y l Hd=12m. p lc ton phn cn thit khc phc sc cn thy lc l:

2 2997,08 122 1 3,3475 0,0167 10572,61( / )2 0,21d P N m = + + =

hmd = .d P

g =

10572,61 1, 08( )997,08 9,81

m=

hm= hmh + hmd=0,5+1,08=1,58m2

1 98100 98100 4898,17 93201,83( / )h P P N m= = =2

2 10572, 61 303900 314472,61( / )d P P P N m= + = + =

Vy p sut ton phn ca bm:2313472,61 93201,83 1,53,5 12 1,58 39,8( )997,08 9,81 2 9,81

H m= + + + + = Cng sut ca bm:Cng sut yu cu trn trc bm:p dng cng thc:

.10

...3

H g Q N = (kW) I-439

Trong :: khi lng ring ca nc, kg/m3 N: hiu sut ca bm, kW

Q: nng sut ca bm(m3/s); Q= 22

..3600

x H O

H O

G M ,m

3/s


25/41

Q=14000 18

997,08 3600

=0,07 m3/s

g: gia tc trng trng(m/s2)H: p sut ton phn ca bm tnh bng mt ct ct cht lng bm: hiu sut ca bm

tk tl .0= (I-439)Vi 0 : hiu sut th tch tnh n s hao ht cht lng chy t vng p sut can vng p sut thp v do cht lng r t cc ch h ca bm

tl : hiu sut thy lctk : hiu sut c kh

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

i vi bm ly tm:96,092,0

85,08,0

96,085,00

=

=

=

ck

tl

Chn: 0 0,95 = ; 0,85tl = ; 0,95ck = 0. 0,95.0,85.0,95 0,767tl tk = = =

Vy cng sut yu cu trn trc bm:0, 07 997, 08 9,81 39,8 35,5( )

1000 0,767 N kW = =

Cng sut ng c in Ndc(kW)

dctr dc

N N

.=

Vi: 85,0=tr : hiu sut truyn ng 0,95dc = : hiu sut ng c in

35,5 44( ). 0,85 0,95dc tr dc

N N kW

= = =Thng thng ng c in c chn c cng sut ln hn so vi cng suttnh ton. Chn =1,15 . 1,15 44 50,6( )cdc dc N N kW = = =Chn cng sut ng c in l 51 kWII. My nn khThp lm vic iu kin P=3atm, T=250C.Ta chn my nn ly tmMy nn ly tm l mt loi my nn v y kh nh tc dng ca lc ly tm do bnh gung sinh ra. Dng my nn ly tm khi p sut y t 2-10 at. nn camy ly tm nh nn my c nhiu cp thng t 3-7 cp nn trong mt cp t 1,2-1,5 khi tc vng nh hn 200m/sng knh bnh gung t 700-1400 mm. Cnh gung c th cong ra hochng tmCc iu kin ca kh u vo T=250C, P=1atmII. Cng ca my nn ly tmp dng cng thc


26/41

1

21

1

. . . 1 ( / )1

m

m

db

P m L RT J kg m P

= (I-465)

Trong : PA, PB: p sut trc v sau khi nn, at

T1: nhit u ca kh, K T1=25+273=298K m: ch s a bin, m=1,21,62. Chn m=1,4

R: hng s kh, R 8314 8314 281,16

29,5705 y M = = =

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

mh A A A h Z g g

P

g g

P +++=+

.2..2.

2211

Do ng nm ngang nn ZA=0.Chn vn tc kh trong b cha tnh:1 =0

mh A A h g g

P

g

P =

.2..

21

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

md B B B h Z

g g P

g g P

+++=+.2..2.

222

2

Vn tc kh trong ng y: 2 = B

md B B h Z g P

g P ++=

..2

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

d B

h A

P P P

P P P +=

=

2

1

ZB : chiu cao ng y : Khi lng ring ca hn hp kh thi iu kin u vo ca kh

a P

1 A

A

BB

2

2


27/41

= 3.63 kg/m3

hmh, hmd: tr lc trn ng ng ht v ng y Xc nh p sut trc khi nn:

h A P P P = 1

Trong : P1 : p sut kh quyn cmd h P P P P ++= (nh bm ly tm)

++=

h

hh

d L

P .12. 2

* ng knh ng ht.

h

h

V d

.785,0= [I-369].

V: Lu lng th tch u vo ca kh thi3. 1023,08 29,5705 2,315( / )

.3600 3600 3,63 y y

y

G M V m s

= = =

Kh trong ng dn P > 1at th = 15 25m/s II.2(I-370)Chn vn tc ht h = 25m/s.

2,315 0,34( )0, 785 25h

d m= =Chun s Reynol

5

. . 25 0,34 3,63Re 1743220,34 40001,77.10

h hd

= = = >

=> Dng ch chy xoy nn h s ma st c tnh nh sau:0,91 6,812.lg

Re 3,7 = +

[I-380]

Trong :

: nhm tuyt i. Chn vt liu lm ng l thp ni khng hn 310.07,0 = : nhm tng i, c xc nh theo cng thc:

340, 07 10 2, 06 10

0,34hd = = =

0,9 41 6,81 2,06 102.lg

1743220,34 3,7

= + 0,0144 =

* H s tr lc cc b trong ng ht :


28/41

21 +=hTrong :

1 : H s tr lc ca ng thng, on ng thng c u li ra pha trc 5,01 =

2 : h s tr lc ca vanChn van 1 chiu.Theo II.16 [I-399] ta c dh =0,34 m =>1 2,1 2,5 = chn 5,21 = tr lc cc b ca ng ht 1 2 0,5 2,5 3 = + = + =Chn chiu di ng ht Hh =Lh =5 (m)

2 23,63 525 1 3 0,0144 4777,72( / )2 0,34h P N m = + + =

4777,72

134,17( ). 3, 63 9,81h

mh

P h m g

= = =

4 21 9,81.10 4777,72 93322,28( / ) A h P P P N m= = =* Xc nh p sut sau nn:

p dng cng thc: d B L g P P ..2 +=Trong :

P2: p sut cui ng ng y P P P P d h ++=2

Vi:

++=

d

d d d d

L P .1

2. 2

* ng knh ng yTheo II.2(I-370) kh trong ng y ca my nn smd /2515 = . Chn vn tcy )/(20 smd =

2,315 0,384( )0, 785 20d d m= = Quy chun dd = 38,5 cm

Chun s Re trn ng ng y:5

. . 20 0,385 3,63Re 1579152,54 40001,77.10

d d d

d

= = = >>

Dng ch chy xoy nn h s ma st tnh theo cng thc:0,91 6,812.lg

Re 3,7 = +

[I-380]

Trong : : nhm tuyt i, chn vt liu lm ng l thp ni khng hn

310.07,0 =

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


29/41

340,07.10 1,82.10

0,385d d = = =

Thay vo cng thc:

0,9 41 6,81 1,82.102.lg

1579152,54 3,7

= +

0,0142 =* H s tr lc cc b trn ng ng yp dng cng thc:

321 ++=d

Trong : 1 : h s tr lc ca ng thng;1 =0,5

2

: h s tr lc ca van, chn van 1 chiu. Theo bng II.16 ta cdd = 0,385 m c 2 = 2,1 2,5 => chn2 =2,5 3 : h s tr lc cc b ca ng ng cong;3 =A.B.C

Gc = 900 => A =1

Chnd d

R =2 => B =0,15

ba = 0,5 => C =1,45

3 = 10,151,45 = 0,2175H s tr lc cc b trn ng ng y

321 ++=d = 0,5+ 2,5+ 0,2175= 3,2175Chn chiu di ng y Hd=Ld =5 (m)vy tr lc trong ng y:

2 2. 3,63 20 51 . 1 3,2175 0,01422 2 0,385

d d d

d

L P d

= + + = + +

23195,79( / . ) N m=Tn tht p sut trn ng ng y:

3195,79 89,74( )3,63 9,81

d md

P h m

g = = =

p sut cui ng ng y P P P P d h ++=2 = 4777,72+3195,79+303900 = 311873,51(N/m2)

Vy p sut sau khi nn l:d B L g P P ..2 += = 311873,51+ 3,63 9,815 = 312051,56(N/m2)

Thay cc s liu vo cng thc tnh cng ca my nn ta c:


30/41

1 1,4 11,4

21

1

1, 4 312051, 56. . . 1 281,16 298 1 120769,8( / )1 1,4 1 93322,28

m

m

db

P m L RT J kg m P

= = =

II.2Cng sut ca my nn*Cng sut l thuyt

. ( W)1000lt G L N k = [I-466]

Trong : G: Cng sut ca my nn, kg/s

1023,08 29,5705 8,4( / )3600

G kg s= =

L: Cng nn 1 kg kh Nlt: Cng sut l thuyt ,kW 8,4 120769,8 1014,5( W)1000lt N k

= =

Cng sut thc t ca my nn

dn

dntt

N N

= (kw) [I-466]

Trong :

Ndn: cng sut ng nhit, kw Ndn= 1014,5 kW

dn : hiu sut ng nhit thng bng 0,65 0,75.Chn dn = 0,7

1014,5 1449,3( )

0,7dn

tt dn

N N kW

= = =

Cng sut trn trc ca my nn

ck

tt hd

N N =

Trong : Nhd: cng sut hiu dng,kw ck : hiu sut c kh ca my nn. i vi my nn ly tmck =0,96

0,98. Chn ck = 0,97

1449,3 1494,1( W)0,97

tt hd

ck

N N k

= = =

II.3Cng sut ca ng c in


31/41

dctr

hd dc

N N

..= [I-466]

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

:tr hiu sut truyn ng ( 0,96 0,99 ) tr = 0,98 :dc hiu sut ng c indc =0,95

1494,1. 1,15 1604,8( W). 0,98 0,95hd

dctr dc

N N k

= = = Nh vy ta chn ng c in c cng sut 1600 kWPHN III: TNH TON C KHI. Chiu dy thn thpThit b lm vic p sut kh quyn, dng hp th kh S02, thn thp hnhtr, c ch to bng cch un tm vt liu vi kch thc nh sn, hghp mi, thp c t thng ng

Chn thn thp lm bng vt liu X18H10T( C < 0,1%, Cr khong 18%, Ni khong 10%, Ti khng qu 1 1,5%).Chn thp khng g, bn nhit v chu nhit.Thng s gii hn bn ko v gii hn bn chy ca thp loi X18H10T:k = 540.106(N/m2)c = 220.106 (N/m2)

gin tng i: = 38% nht va p : ak = 2.106J/m2

Chiu dy thn thp hnh tr, lm vic vi p sut bn trong c xc nh bn

cng thc:.

2.[ ].t

k

D P S C P

= + , m Bng XIII.8-II-360

Trong :

Dt.: ng knh trong thp, m : h s bn ca thnh thn tr theo phng dc, vi thn hay c l gia c

hon ton th = h i vi mi hn c.Vi hn tay bng h quang in,thp khng g ta c: = h = 0,95 [Bng XIII.8-II-362]

C : h s b sung do n mn, bo mn v dung sai v chiu dy, m [k ]: ng sut cho php ca loi thp X18H10T.

P: p sut trong thit b, N/m2.

P: p sut trong thit b ng vi s chnh lch p sut ln nht bn trong v bengoi thp, N/m2


32/41

P= Pmt+ PttTrong :

Pmt : p sut lm vic, Pmt= 3 x 1,013.105 = 303900 N/m2 Ptt : p sut thu tnh ca ct cht lng

Ptt = x.g.H (N/m2) [II- 360 ]Vi:

x: khi lng ring ca nc, kg/m3 g: gia tc trng trng, g= 9,81 m/s2 H: chiu cao ct cht lng, Ht= 11,7 m

=> Ptt = x.g.H = 997,08 x 9,81 x11,7 =114441,9 ( N/m2)=> P = Pmt+ Ptt = 303900+ 114441,9 = 418341,9( N/m2)

* Tnh C C ph thuc vo n mn, bo mn v dung sai ca chiu dy. i lng Cc xc nh theo cng thc:

C = C1+ C2+ C3 ,m [II-363]Trong :

C1: h s b sung do n mn. i vi vt liu l thp X18H10T c bn0,05 0,1mm/nm th ly C1= 1mm.

C2: i lng b sung do hao mn, C2= 0 C3: i lng b sung do dung sai ca chiu dy*Tnh [k ]:Theo bng XIII-4 ,ta c th chn gi tr nh nht, tnh theo cng thc sau:

[ ] .k k k n

=

Theo gii hn bn khi ko th ta c:

: h s hiu chnh, =1 k = 540.106(N/m2) k : h s an ton bn, k = 2,6 [II- 356]

[ ] .k k k n

= =

6540.10 .12,6 = 207,692.10

6

Theo gii hn bn chy:[ ] .ck cn

=

Trong : : h s hiu chnh, =1


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

[ ] .ck cn

= =

6220.10 .11,5 = 146,667.10

6

Ta ly gi tr b hn trong hai gi tr va tnh c:[k ] = 146,667.106( N/m2)

Do [ ]6. 146,667.10 0,9

418341,5

9k h

P = =333,0617>50

b dy thn thp c tnh theo cng thc sau:.

2.[ ].t

k

D P S C

= + = 62

2 146,667.10418341,

09

,95C + , m

C = C1+ C2+ C3 = 1+0+0,18= 1,18 mmS= 3,0025+1,18=4,1825 mmQuy trn S = 5 mm Kim tra ng sut ca thnh theo p sut th:

0[ ( )]2( ) 1,2t c D S C P

S C

+ = P0 = 627512,85 + 114441,9 = 741954,75(N/m2)

Vy3

63

[2 (5 1,18).10 ] 741954,75204,842.102 (5 1,18).10 0,95

+ = = N/m2

66220.10 183,33.10

1,2 1,2

c = = N/m2

Nh vy l khng tho mn.Chn chiu dy thn thp S = 6 mm

36

3[2 (6 1,18).10 ] 741954, 75162,425.10

2 (6 1,18).10 0,95

+ = =

Tho mn iu kin.Vy S = 6 mmII. Chiu dy np v y thit b


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Np v y cng l nhng b phn quan trng ca thit b, c ch to cnloi vt liu vi thn thit b l thp X18H10T. Thit b t thng ng.p sut trong l P = 418341,9 > 0,7.105 N/m2 ngi ta thng dng np elip cg.

p sut tnh ton P=418341,9 N/m2

Chiu dy ca np v y thit b c xc nh theo cng thc:.

3,8[ ] 2k h b DP DS C

k P h = +

Trong :

P : p sut trong ca thit b. h b : chiu cao phn li ca y v np , h b= 0,25Dt=0,252=0,5 m [k ] : ng sut cho php ca thit b , [k ] = 146,667.106(N/m2) h : h s bn ca mi hn hng tm, vi mi hn tay bng h quang

in, vt liu thp cacbon khng g chn h= 0,95 C: i lng b sung, Cn = 1,18 ( mm ) k : h s khng th nguyn, chn k = 1

V:6[ ] 146,667.10. . .1.0,95 333,06 30

418341,9k

hk P

= = >

B dy np v y c tnh theo cng thc:

.3,8[ ] 2k h b

DP DS C k h = + = 62 418341,9 2. 0,003163,8 146,667.10 1 0,95 2 0,5C C + = +

S = 0,00316 + C ( m )xt thy S = 0,00316 + C< 10 mm tng thm 2 mm so vi gi tr C C = 1,18+2=3,18 mm S = 6,34 mmQuy trn S = 7mm Kim tra ng sut ca y thit b theo p sut th :

20[ 2 ( )]

7,6 ( ) 1,2t b c

h b

D h S C P k h S C

+ =


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Ta c:6

6220.10 183,33.101,2 1,2

c = = N/m2

2 320

3

2 2 0,5 (7 3,18).10[ 2 ( )]7,6 ( ) 7,6 1 0,95 0,5 (7 3,18).10

741954,75t b

h b

D h S C P k h S C

+ + = =

=215,418.106

N/m2

Khng tha mnChn S =8 mm

2 32

03

2 2 0,5 (8 3,18).10[ 2 ( )]7,6 ( ) 7,6 1 0,95 0,5 (8 3,18).10

741954,75t b

h b

D h S C P k h S C

+ + = =

=170,768.106 N/m2tha mn iu kin. Vy S = 8 mm3. ng knh ng dna. ng knh ng dn kh p dng cng thc

.785,0V

d = [I-369]

Trong : V: lu lng th tch kh trong ng, m/s

3. 1023,08 29,5705 2,315( / ).3600 3600 3,63 y y

y

G M V m s

= = =

: tc trung bnh kh trong ng, m/s P = 3atm, = 1525 m/s. Ta chn = 25m/s

d: ng knh ng dn kh, m. Thay vo cng thc tnh ng knh ng dn kh ta c:

2,315 0,34( )0, 785 25k d m= =

Quy chun d= 350 mm

22,315 24,07( / )

0,35 0, 785k m s = =

b. ng knh ng dn cht lng

.785,0V

d =

Trong :V: lu lng th tch nc trong ng

2

2

d 3. 14000 18

0,07( / ).3600 3600 997,08 x H O

H O

G M V m s

= = =

Theo bng II.2 [I-370] vn tc cht lng trong ng y ca bm


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l = 1,5-2,5 m/s. Chn l = 2,0 m/s

0,07 0,211( )0, 785 2l d m= =

Quy chun dl = 200 mm

20,07 2,23( / )0,2 0,785l m s = =4. Mt bcha. Bch ni thit b: ni thit b ( thn, np v y) ta c th dng bch lin kiu I (hnh 8. 2) chto bng thp CT3.

Hnh 8. 2. Bch lin kiu IVi ng knh ca thp Dt = 2000mm v p sut tnh ton p = 418341,9 N/m2 ,tra bng XIII.27( S tay 2-423) ta c cc thng s ca bch nh sau:

y.10-6 Dt

Kch thc nih

N/m2 D D b DI D0 Bu lng

Ghi ch:D ng knh ngoi ca bchDb ng knh vng bu-lngDI ng knh trong ca bchDn ng knh quy c ca bchDy ng knh trong ca ngh Chiu dy ca bchd b ng knh bu-lng


37/41

d b z

0,4mm ci mm

2000 2170 2100 2060 2015 M30 48 50

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

Dt

y.10-6, N/m20,11,6

D2 D4mm

2000

2054 2030

b. Bch ni ng ng vi l y v np: ta dng kiu bch t do bng thpTra bng XIII-28( s tay 2-425) ta c

y.10-62/ m N

Dy Kiu bch

5

D0 h h1 D bmm

0,25 0,6 350 283 28 20 -

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

Tra bng XIII.26, s tay 2-414ng vi ng c ng knh Dy=200 mm ta c cc thng s sau

d b

D1

D t

h

Dn

Dy

D


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y.10-6

N/m2Dy

ng Kiu bchhDn D D D1 Bu lngd b z

mm ci mm0,4 200 219 290 255 232 M16 8 22

5. Chn , v , tai treoThng thng ngi ta khng t trc tip thit b ln b mt m phi c tai trehay chn , v thit b thit b c n nh khi vn hnh. Munchn c chn , v hay tai treo thch hp ta phi tnh trng ti ca thp.Trng ti ca thp:Pthp = Pthn + Py, np+ Pcht lng+ P bch + Pm(N)

Khi lng thn thit bMth = V. = S.H. = (/4).(Dn2 D2t). H.

Trong : Mth: khi lng ca thn thit b, kg Dn, Dt: ng knh ngoi v trong ca thit b, m H: chiu cao ca thp, m : khi lng ring ca thp, = 7,9.103 kg/m3

Mthn = (/4).[ ( 2,0 + 20,006)2 2,02 ]11,77,9.103 = 3493(kg)Khi lng ca y v np thp

- Khi lng ca np: S= 8mm; Dt=2000mm Tra bng XIII-11( s tay 2- 384) ta c chiu cao g h= 25 mm vMnp=283 kg- Khi lng y: S=8mm; Dt=2000 mm Tra bng XIII-11( s tay 2- 384) ta c chiu cao g h= 25 mm vMnp=283 kg

Vy Mnp- y= 566 (kg)Khi lng ca m

m l m vng Rasig ln xn: m bng s kch thc 25253.0

Tra bng thng s k thut IX.8( s tay 2-193)

m= 600 kg/m3

Mm= Hm.(/4).D2.m= 9,2.(/4)2,02600= 17332,8 (Kg)

Khi lng ca cht lng:


39/41

H D M l .4.

.2

=

Trong : : khi lng ring cht lng 250CD: ng knh thp, mH: chiu cao thp trc khi b nc chon ht, m

Vy:23,14 2,0997,08 11,7 36630,725( )

4l M kg = =

Khi lng bcho Bch ni thn thit b:

p dng cng thc: ( )2 2. . .4b b I thep M D D h

=

Trong :

thep

: khi lng ring ca thp lm bch(CT3),thep

= 7,85.103

(kg/m3

) D b: ng knh vng bulong, D b= 2100 mm = 2,1 m DI: ng knh trong ca bch, DI= 2060 mm= 2,06 m h: chiu di ca bch, h=50mm=0,05m ( )2 2 32,1 2,06 0,05 7,85.10 51,27( )4b M kg

= =

1 3 51,267 153,8( )b M kg = =o Bch ni ng dn:

2 22 4 (0,255 0,232 ) 0,22 7850 6,07( )4b M kg = =

o Bch ni ng ng vi l y v np2 2

3 4 (0,445 0,415 ) 0,26 7850 16,53( )4b M kg = =

Do M b = M b1+M b2+M b3 = 153,8+6,07+16,53= 176,4 kgTrng ti ca thp l:

Khi lng ca ton thpMtt= Mm+ M bch+ Mcht lng+ Mthn+ Mnp,y=17332,8 +176,4 + 36630,725 + 3493+ 566 = 58198,44( kg )

Trng ti ca thp:P = Mtt g = 58198,44 9,81 =570926,7(N )Ta s dng v vi cc kch thc trnh by trong bn v. Vt liu v lthp CT3 Nhng iu kin cn ch khi ch to:

m bo ng hn cng ngn cng tt Ch hn gip m


40/41

B tr cc ng hn dc B tr mi hn v tr d quan st Khng khoan l qua mi hn

PHN KT LUN

Trn y ton b phn thit k h thng thp m hp th cu t SO2 vdung mi l nc, trong qu trnh tnh ton do rt di v c nhiu cng thc phc tp cng nh phi lm trn s hoc tra s liu t cc ti liu tham kho n bn thit k khng trnh khi sai st trong tnh ton.

Tuy nhin bn thit k gip em h thng li ton b kin thc hc tcc mn Cc qu trnh c bn I,II,III,nhng mn c s ca ngnh mi trng vha. ng thi qua vic thit k cng gip em c c ci nhn hon chnh vs ca mt h thng x l kh thi.


41/41

TI LIU THAM KHO

1. TS Trn Xoa, PGS.TS Nguyn Trng Khung, TS Phm Xun Ton- Stay qu trnh v thit b cng ngh ha tp 1- NXB KHKT H Ni

2. TS Trn Xoa, PGS.TS Nguyn Trng Khung, TS Phm Xun Ton- Stay qu trnh v thit b cng ngh ha tp 2- NXB KHKT H Ni

3. GS.TSKH Nguyn Bin- C s cc qu trnh v thit b cng ngh hacht tp 4- NXB KHKT H Ni

tinh toan thiet ke he thong hap thu khi so2 dai hoc bkhn

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