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D liu cho trc : Cng sut ca thp chng ct :Qv = 6,5 triu tn/nm

= 21666,67 tn/ngy (Nh my hot ng 300 ngy mt nm )Gi s:

- Thp chng ct kh quyn c 50 a

- Sn phm nh: Kh + naptha

- Kerosen ly ra a 15

- LGO ly ra a 25

- HGO ly ra a 40

- Vng stripping y thp c 5 a

- p sut ti nh: 1,5 atm, tn tht p sut trn tng a l 8mmHg

- Nguyn liu u vo l du th Bch H.

* Yu cu tnh ton:

- Tnh nhit , lu lng Q ti cc vng ca thp

- Tnh ton cu trc thp

- Thit k bn v PFD cho phn xng CDU

- Xy dng bn v P&ID cho mt thit b

A. TNH NHIT , LU LNG THPCHNG CT KH QUYN

Bng A.1. c trng du th Bch H

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STT TINH CHT C BAN CA DU TH KQ PHN TICH1 Ty trong, 0APIT trng, d60600FKhi lng ring 150C, g/ml

40,20,82410,8236

2 Ham lng kh (C1-C4) ha tan trong du, %KL -3 Ham lng lu huynh, %KL 0,0414 im chy, 0C 365 Hng s c trng K UOP 12,396 nht ng hc 500C, cSt

700

C, cSt

5,286

3,4347 Trong lng phn t 240,58 Ham lng parafin, %KL 27,239 Ham lng nc, %TT 0,20

10 Ham lng tro, %KL 0,01411 Ham lng Nit, %KL 0,03212 Ap sut hi bo ha Reid, psi 3,7913 Ham lng tp cht c hc, %KL 0,00514 Chi s axit, mgKOH/g 0,037

15 Cok Conradson, %KL 0,8216 Ham lng mui NaCl, mg/kg 51,3217 Ham lng nha, %KL 1,8818 Ham lng Asphanten, %KL 0,5019 Ham lng Vanadi, ppm

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S liu ly t trang Web: Congnghedaukhi.com

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T d liu ca du th Bch H ta v cc th sau:

0

1 0 0

2 0 0

3 0 0

4 0 0

5 0 0

6 0 0

0 2 0 4 0 6 0 8 0 1 0 0

%V

o

C

nh i t

Hnh 1: ng cong TBP ca du th Bch H

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0

0 ,1

0 ,2

0 ,3

0 ,4

0 ,5

0 ,6

0 ,7

0 ,8

0 ,9

1

0 2 0 4 0 6 0 8 0 1 0 0 1 2 0

%

d d

Hnh 2: th biu din mi quan h: t trng -%V ca du th Bch H

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0

1 0 0

2 0 0

3 0 0

4 0 0

5 0 0

6 0 0

0 2 0 4 0 6 0 8 0 1 0 0

% V

t

0

0 ,1

0 ,2

0 ,3

0 ,4

0 ,5

0 ,6

0 ,7

0 ,8

0 ,9

1

d

Hnh 3: Cc ng c trng ca du th

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Cn c vo ng TBP chia du th thnh 5 phn on sau bng thp chng ctkh quyn:

- Phn on kh 4,5%- Phn on Naphta 17,5%

- Phn on kerosen 6,8%

- Phn on LGO 15,2%

- Phn on HGO 11,0%

- Phn on AR 45,0%

Phn t lng trung bnh ca du m v sn phm c th tm c nhHnh 3.13[1] vi KUOP = 12,39

Chng ta s dng hi nc stripping cn AR, cc phn on sn ly ra t nhthp. Ton b hi nc dng stripping b ngng t thit b lm lnh dng hi bayra t nh thp chng ct v chy ra ngoi y t bnh hi lu.Chng ta dng hi nc qu nhit 220oC, p sut 3atm. Entanpy cua hi nc cho Hnh 3.14 [1] .Lng hi nc cn dng c tm theo Hnh 3.15 [1] .

Bng 1 c trng cc phn on nh sau:( coi t khi bng khi lng ring )

Phnon

%V Th tch T khi Khi lng Phnt

lng

Skmol/h[m3/h] d [tn/h]

Kh 4,5 49,42 0,5828,665

64 447,89

Naphta 17,5 192,2 0,736

141,457

138 1025,05

Kerosen 6,8 74,7 0,778

58,103

170 341,78

LGO 15,2 166,9 0,802

133,884

225 595,04

HGO 11 120,8 0,81598,460

280 351,64

55 604,0 0,762

460,286

188 2448,33

AR 45 494,2 0,895

442,494

470 941,48

Du th 100 1098,3 0,823

902,78

240,5 3753,76

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I, TNH TON NHIT VNG NP LIU Y THP:I.1: S vng np liu y thp :

Hnh I.1: S dng vng np liu y thp

- Va: dng hi bay ln t du th

- La: dng lng chy xung t du th

- Vo: dng hi sinh ra nh stripping- Lo: dng hi lu lng

- Wo: dng hi nc stripping y thp,

Theo iu kin tm thy bng 1v hnh I.1 ta c 2 phng trnh sau:

- Tng cc phn on gm GO v nh hn :

Va + Vo Lo = 55

- Phn on AR :

La + Lo Vo = 45Trong cc s hng u dc tnh bng phn trm th tch so vi du th . CnVa + L a = 100

rm nhit ti a np liu ta phi da vo ng bay hi flash ca duti vng . Nhit ti a Ta s l nhit ng vi bay hi Va % trn ngflash . Mun th cn tnh Va v p sut hi ring phn ca du ti a np liu,tc p sut gy ra bi dng hi (Va + Vo) .Nh vy cn bit lu lng Va ,VoWo. y l mt bi ton kh khn , ch c gii da vo nhng s liu rt ra tkinh nghim ca cc nh vn hnh.

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c trng phn on A

0

10 0

20 0

30 0

40 0

50 0

60 0

0 20 40 60 80 100 120

%V

t

0,83

0,84

0,85

0,86

0,87

0,88

0,89

0,9

0,91

0,92

d

I.2 Lu lng cc dng

Kinh nghim ch ra rng cn lm bay hi 3 -5 % th tch (so vi duth) bng cch stripping lng y thp. Theo H3.15 [1] th stripping 3% AR(6,6% du th) cn dng 1,2 lb hi nc cho 1 gallon AR, tc khong 0.14 kghi nc cho 1lt AR. Vy lng hi nc cn dng stripping l :

Wo =494,2.0,14.10

3

= 69188(kg/h) = 3843,78(kmol/h)Thng thng phi lm bay hi 1 lng du th vt qu tng lng

cc phn on distillat (cc phn on ly ra pha trn a np liu) mtlng 3 5%. Ta dng Vo = 3%; gi s Lo = 6%. Theo h phng trnh lp ra I.1 ta c:

55 = Va + 3 - 6 = Va 3

45 = La + 6 3 = La + 3

Va = 58%, La = 4 2%

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I.3 nng cc dng vng np liu : bit c lu lng mol ca cc dng cn bit nng, y l t khi

ca cc dng nu chng th lng , cng nh phn t lng trung bnh ca ccdng .

Theo H3.17 [1], t khi Va l 0,775 ;ca dng La l 0,887

Phn t lng ca phn on t 55 -58 % l 322 (theo H3.13 [1]) phn t lngca 0% - 55% l 168 .Do phn t lng trung bnh ca Va l :

MVa =

.. += 176.

Khi lng ca dng Va : 0 ,775.58%. 1098,3=493 ,69 tn /hLu l n g mo l d n g Va : (4 93 ,69 .103)/ 176 = 2805 kmol/h

Vic nh gi t khi ca dng Vo, Lo c th thc hin bng cch da vo

nhng lun im sau :- V nguyn liu c np vo vng gn y thp nn thc t thp chng ct hotng gn nh khng c vng chng ct. Vng y thp chng ct vi 5 a ngvai tr l stripping lm tng kh nng phn tch. Trong 2 dng La v Lo th dng Lacha nhiu hp cht nh hn dng Lo, v La vn l pha lng ca dng du th vanp vo y thp, n nm cn bng Flash vi dng hi Va, cn dng Lo l dnghi lu lng c trao i cht kh tt vi dng hi trong vng pha trn anp liu . Vy qu tnh strippinh ch yu l xy ra vi dng La. Bn cht ca dngLo thay i khng nhiu trong vng np liu.- Nh vy mc d nh vy dng Lo vn nng hn dng La v nh hn dng ARvi d = 0,895

0,880< dLo < 0,895Ly t khi ca dng Lo l 0,892 nh vy theo H I.1 ta c phng trnh cn bngkhi lng m :

mAR= mLa+ mLo mVothay s ta c :

442,494= 42%.1098,3.0,887 + 6%.1098,3.0,892 3%.1098,3. Vo Vo = 0,773 tn/m3 , nn t khi ca Vo= 0,773. t bng 3.13[1] th phn tlng trung bnh va Vo = 168Lu lng mol dng Vo l : 3%.0,773.1098,3/168=151,6 kmol/h

I.4 :Nhit ti a np liu :Nu ta coi t trng bng khi lng ring thi ta v c ng c trng TBP caphn on cn nh sau :p sut ti a np liu Pa : 1,5.760 + 50.8 =1540 mmHgp sut hi ring phn ca du ti a np liu Pa :

Pa = WoVoVaVoVa

+++)(

Pa = ,,,

+++

.1540=669,5 mmHg

Nh vy phi v ng flash 669,5 mmHg ly T 58 trn ng . chnh lnhit ti a np liu Ta. Trc ht cn c vo TBP v ng flash ta lm

nh sau :

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Theo hnh 3.11[2] ta c : B =

TT =

= 5,43(oC/%)

A= 3,5 (oC/%) T50 = T50DRL T50FRL = 24 ,

M T50DRL = 335oC nn T50FRL = 311

oC ,

Nh vy ta v c ng flash ti 760mmHg.Sau tnh tin n v pha di mt on ng vi 13oC ta s c ng flash ti

669,5 mmHg .Ly T58 trn ng ny ta c Ta = 350oC

I.5 : Nhit ti y thp :Phng php tnh nhit y thp chng ct da trn cn bng nhit hay cngi l da trn cn bng entalpy ca vng np liu- y thp. Tng entalpy dngvo phi bng tng entalpy cc dng ra khi vng ang xt.S liu Entalpy ca du ly trang 83[2] , entalpy ca nc ly hnh 3.14[1] sliu xut pht tm entalpy cho bng sau :

Bng 2 : S liu lin quan n vng y thp

Dng Nhit [oC]

d Th tch[m3/h]

Khi lng[kg/h]

Entalpy

Kcal/kg Kcal/h

VoLa (lng)

Lo (lng)Wo

350

350220

0,887

0,892

461,3

65,9

409,2.103

58,8.103

69188

215

214715

87978.103

12583,2.103

49469,4.103

RaARVoWo

Td350350

0,8940,773

494,232,95

4424942547069188

x255765

442494.x6494,85.103

52,93.106

Ev = Er : 87978.103 +12583,2.103+49469,4.103

= 442494.x+6494,85.103+52,93.106 x=205

Cn c vo biu trang 83[2] ta c T= 335 oCII : TNH TON VNG A LY HGO :

Ta quyt nh stripping 6% so vi HGO ly ra. Theo H3.15 [1] l cn dng

35 kg hi nc cho 1 m3 HGO).

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Hnh II.1: S dng HGO

c trng phn on HGO

290

300

310

320

330

340

350

360

370

0 20 40 60 80 100 120

%V

t

0,815

0,82

0,825

0,83

0,835

0,84

0,845

d

Do vy tng lng hi nc cn dng l:

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W1= 35.120,8= 4228 kg/h hay 234,9 kmol/h

Trong s Hnh II.1 th

V1(tng cc phn on hi)= 49,42+192,2+74,7+166,9+ 120,8 =604,0 m3/h

S1(dng hi b stripping t dng lng): S1= L1 L1C th coi S1 c t khi bng t khi ca phn on gn n nht (LGO) d=0,802 ,

S1 =94

100.

100

6.120,8 = 7,71 m3/h hay 6183,4 kg/h

Cc s liu v dng hi lu R1 ti a ly HGO c xc nh da trn cn bngkhi lng cc dng u vo v ra khi thp stripping :

120,8.0,815 +0,802.7,71 =(120,8+7,71). L1

L1 = 0,8142

khi lng ca L1= (120,8+7,71).0,8142 = 103981,5 kg/hTheo phng php tnh trn cn bng nhit ta tnh c cc s liu bng sau:

Bng 3: S liu lin quan ti vng HGO

(Nhit gi nh T1 = 300 oC)Dng Nhit

[oC]d Th

tch[m3/h]

Khi lng[kg/h]

Entalpy

kcal/kg kcal/h

VoVa (hi)Vo (hi)R1(lng)

Wo

350350296350

0,7750,802

0,8142

63732,95

493675

26426R169188

260255180770

128,35.106

6,738.106

226.R153,27.106

RaV1(hi)Lo(lng)R1 (hi)

Wo

300350300

300

0,7620,8920,8142

604,065,9

46024858783

R1

69188

240220230

740

110,46.106

12,93.106

230R151,2.106

Ev = Er R1 = 275320 kg/hLu lng mol ca L1 = 351,64 kmol/hLu lng mol dng S1= 7,71.0,802.103/225=27,48 kmol/h

Phn t lng dng L1 = 48,2764,351.103981,5

+=274

Lu lng mol dng R1=275320/274=1004,8 kmol/h

Tng s mol hi qua a ly HGO l :

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n(V1+R1+Wo)= 2448,33+1004,8+3843,78= 7298,92 kmol/h

S mol hi HGO ti a ly HGO l:

n(R1+L1) =1004,8 +351,64 =1356,44 kmol/hp sut hi P1 ca HGO :

P1 = 1,5.760+40.8=1460 mmHgp sut hi ring phn P1 ca HGO :

P1 = 146092,729844,1356 = 271mmHg

Nh vy phi v ng flash ca HGO 271mmHg ri tnh tin xung pha di35oC theo trang 72 [2]

V ng flash ti 271mmHg

Ta c : B=60

1070 TT =60306343 =0,62(oC/%)

A=0,1(oC/%) , C 0M T50DRL =335TFRL= 333

Tnh tin xung di 35oC ta c Tflash 271mmHg l 298oC , nhit gi nhl chp nhn c . Vy nhit ly HGO l 298 oC.III, TNH NHIT TI A LY LGO :S dng nh hnh sau :

S III.1 :S dngLGO

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c trng phn on LGO

%V

1

,

,

,

,

,

,

,

d

Vn stripping 6% so vi LGO ly ra vy lng nc W2 cn dng l :W2= 35.166,9=5841,5 kg/h hay 324,53 kmol/h .Nhit gi nh l 240oC

Trong s III.1 th :V2 (tng phn on hi t a LGO tr ln)=49,42+192,2+74,7+166,9

= 483,22 m3/hKhi lng ring ca V2=( 4,5.0,58+17,5.0,736+6,8.0,778+15,2.0,802)/44

=0,749 .Ta c th coi S2 c t khi bng t khi ca kerosen bng 0,778

S2 =94

100.

100

6.166,9=10,65 m3/h hay 8285,7kg/h

Cc s liu v dng hi lu ni R2 ti a ly LGO tm c bng cch da vo cnbng khi lng cc dng du vo va ra thp stripping :

166,9.0,802 + 10,65 .0,778 =(166,9+10,65) L2 L2 = 0,800Theo khi lng ring ca L2 = 0,800. vy t khi ca dng R2= 0,800Khi lng dng L2 = 142040 kg/h

Bng 4: S liu lin quan ti vng LGO

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Dng Nhit

[o

C]

d Thtch

[m3

/h]

Khilng

[kg/h]

Entalpy

Kcal/kg Kcal/h

VoV1(hi)R1(hi)S1(hi)R2(hi)Wo+W1

298298298240298

0,7620,814

20,8020,800

604,0

7,71

4602862753206183,4

R273416

238228229140738

109,55.106

62,77.106

1,416.106

140.R254,18.106

Ra

V2(hi)R1+L1(lng)R2 (hi)Wo + W1

240298240240

0,7490,814

20,800

361931,8379301,5

R273416

198180190705

71,66.106

68,27.106

190R251,76.106

Theo cn bng Entalpy ta c :Ev = Er R2= 724520 kg/h

Lu lng mol dng lng L2 :595,04 kmol/hLu lng mol dng S2 : 8285,7/170 = 48,74 kmol/h

Phn t lng dng L2: 142040/(595,04+48,74) = 221.Lu lng mol dng R2: 724520/221=3278,4 kmol/h .Tng s mol qua a ly LGO :n = nV2 + nR2 + n(Wo + W1) = 2409,76 + 3278,4+4078,7 =9766,86 kmol/hTng s mol hi qua a ly LGO :n = nR2 + nL2= 3278,4 + 595,04 =3873,44 kmol/hp sut ti a ly LGO l :

P2= 1,5.760 + 25.8 = 1340 mmHgp sut hi ring phn ca a ly LGO:

P2 = 44,387386,9766

.1340= 515 mmHgNh vy ta cn v ng flash ca LGO 515mmHg .

V ng flash ca LGO ti 760mmHg.T hnh v c trng phn on LGO ta c :

B=60

1070 TT =

60

246285 = 0,65 (oC/%)

A= 0,12(oC/%) ,C 0M T50DRL = 265TFRL= 260.Tnh tin xng di 20 oC ta c Tflash = 240 oC

Vy gi nh l chnh xc .Nhit ti a ly LGO l 240oC.

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IV. TNH NHIT TI A LY KEROSEN :

S dng vng Kerosen

c trng phn on karosen

0

50

100

150

200

250

300

0 20 40 60 80 100 120

%V

t

0,765

0,77

0,775

0,78

0,785

0,79

0,795

0,8

d

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Trong hnh trn ta ly stripping 5%. Lng hi nc W3 cn dng l :W3= 38.74,7=2838,6 kg/h hay 157,7 kmol/h

Nhit gi nh l 200oCV3 l tng phn on hi bay qua a ly Kerosen, ngha l:

V3 =49,42+192,2+74,7= 316,32 m3/hV c t trng l :

dV3= 28,86,8.0,778)17,5.0,7364,5.0,58( ++

=0,722

S3 c t khi coi bng t khi ca Naphta ,v bng 0,736

S3= 7,7495

5 = 3,93 m3/h hay 2893,64 kg/h

Cc s liu dng hi lu ni R3 c th tm c nh cn bng khi lng vo v rathp stripping : 3,93.0,722 + 74,7.0,778= (3,93+74,7) L3

L3= 0,775.Vy t khi ca dng R3 =0,775 khi lng dng L3 = 60,94 kg/h

Bng 5: s liu lin quan ti vng Kerosen

Dng Nhit [oC]

d Th tch[m3/h]

Khi lng[kg/h]

Entalpy

Kcal/kg Kcal/h

VoV2(hi)

R2 (hi)S2(hi)R3 (lng)

Wo+W1+W2

240

240240200240

0,749

0,8000,7780,775

483,22

10,65

361931,8724520

R379257,5

198

190195112705

71,66.106

129,49.106

1,62.106

112R355,88.106

RaV3 (hi)

R2+L2(lng)R3 (hi)

Wo+W1+W2

200240200

200

0,7220,8000,775

316,32 228383823560

R3

79257,5

180140170

685

41,1.106

115,3.106

170R354,29.106

Theo cn bng Entalpy ta c :Ev = Er R3= 821724.14 kg/hLu lng mol L3 =341,78 kmol/hLu lng mol dng S3: 2893,64 /138 = 20,97 kmol/hPhn t lng dng L3= 60,94 /(341,78 +20,97 ) = 168Lu lng mol dng R3 : 821724.14 /168=4891,22 kmol/hp sut ring phn ca hi kerosen P3

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P3 = )8.15760.5,1(79257,5/181814,724891,22341,784891,22

+++

+= 581 mmHg

Tng t nh i vi phn on HGO,v LGO ta v ng flash 760mmHg ritnh tin xung 15oC . Ta c ng flash ti 581 mmHg .Nhit To trn ng

flash tm c l 198oC . Nhit gi nh l chp nhn c. Vy nhit a lyKerosen l 200oCV .TNH NHIT TI NH THP CHNG CT

S dng vng nh thpTrong hnh trn ta vn stripping 5 % , lng hi nuc cn stripping l :

W4= 42.192,2= 8072,4 kg/h hay 448,47kmol/hNhit gi nh l 120oC.

V4 l tng phn on hi i qua a ly napthaV4= 49,42+192,2= 241,62 m3/h v c khi lng l :mV4 = 170,122 tn/h v t khi dV4= (4,5.0,58+17,5.0,736)/22=0,704.Cch tm nhit ti nh thp cng dc thc hin bng cch gi nh . sau kim tra li xem gi nh c chp nhn c khng. Vic kim tra c thchin bng cch thit lp ng flah ca sn phm nh ,sau ly T100 trn ngflash .Gi s chng ta lm vic vi dng hi lu lnh L c nhit l 30 oC.ML = 130Dng L tip xc vi dng hi ang bay ln lm ngng t mt lng lng R4. R4ng vai tr l hi lu nng trong nh thp.Dng hi nc W3 = 42.74,7=3137,4 kg/h

Dng S3 : S3= 7,7495

5 = 3,93 m3/h hay 2893,64 kg/h

Dng L3 = 58103+2893,64= 60996,64kg/h

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c trng phn on Naphta

0

50

100

150

200

250

0 20 40 60 80 100 120

%V

t

0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

0,9

d

Ta c bng s liu sau:

Bng 6: S liu lin quan ti vng nh thp

Dng Nhit [oC]

d Th tch[m3/h]

Khi lng[kg/h]

Entalpy

Kcal/kg Kcal/h

VoV3(hi)R3 (hi)S3 (hi)L(lng)

W0 3

20020020030

200

0,7220,7750,7040,736

316,32 228383821724.142893,64

L82394.9

18017017315

685

41,1.106

139,69.106

0,5.10

6

15L54,29.106

RaV4 (hi)R3+ L3(lng)L (hi)

W0 3

120200120

120

0,7040,7750,736

241,62 170122882720,78

L

82394.9

140112138

644

23,82.106

98,86.106

138L53,06.106

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Theo cn bng Entalpy ta c :Ev = Er L= 486504 kg/h hay 3742.3 kmol/hp sut hi ring phn ca HC nh thp :

P4 = +++

++

82394.9/181025,05447,893,3742

1025,05447,893,37421,5.760= 607 mmHg

V ng flash ti 760mmHg ri suy ra ng flash ti 607mmHgV ng flafh ca phn on nh ,ri tnh tin 12oC ta c T100 ca n l119oC .sai khc 1 oC nn c th chp nhn c nhit si nh thp l 120oC .C th tnh dng hi luuw nng ti nh thp nh phng trnh :

R4(138-65) = L(138-15) = 486504(138-15)R4= 819725,92 kg/h

VI.XC NH PHN TCH CA PHP CHNG CT DU TH:

Bng 7: Tm phn tch gia cc phn on

i lng LGO HGO Kerosen- LGO Naphta - kerosen

S a NiDng hi lu ni Ri,

[m3/h]Dng hi ly ra Vi,

[m3/h]

di hi lu= Ri/ ViF= NihiT50,TBP phn on nngT50,TBP , phn on nh

T50,TBP phn tch

15R2= 724,52/0,8

= 905,65.

V2= 483,22

1,87428,11335195140+25

10R3=821,7/0,775

= 1060,29

V3= 316,32

3,35233,52265150115+20

15R4=819,726/0,73

6=1113,76

V4= 241,62

4,61069,1520214062

+10

Gi tr phn tch ly c nh biu Hnh 3.3[1]Kt qu cho thy php chng ct tt, cc phn on t ln vo nhau.

VII.HI LU VNG :Phng php hi lu vng ngi ta trch mt phn dng lng ti a no em rangoi lm lnh ,ri bm tr v thp chng ct v tr cao hn vi ba a.Hi lu vng l mt bin php c s dng rt ph bin trong chng ct ti ccnh my lc ha du.Trc khi hi lu: (c biu din bi cc mi tn lin nt )Lu lng hi ti a ly phn on sn l: Ri + ViLu lng lng ti i ly phn on sn l: Ri + Li

1. Ti a ly HGO :R1 + L1 = 379301,5 kg/h

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R1 + V1 = 460286+275320 = 735606 kg/h

2. Ti a ly LGO:

R2 + L2 =823560 kg/hR2 + V2 =361931,8+681520 =104345,18 kg/h

3. Ti a ly kerosen:R3 + L3 = 882720,78 kg/hR3 + V3=228383+821724.14= 1050107 kg/h

4. Ti nh thp:R4 =81972=5,92 kg/hR4 + V4 =819725,92 + 170122=989848 kg/hL + V4 = 486504+ 170122= 656626kg/h

5. Ti a np liu:Lo + La = 409,2.10

3 +58,8.103 = 468000 kg/hVo + Va = 25470+493690=519160 kg/h

6. Ti y thp :AR = 442494 kg/h

Ly hi lu vng ti v tr a ly HGO:Gi s ta ly 200 tn/ h t a ly HGO thc hin hi lu vng ,lng nhitcn ly bi dng du th ti thit b trao it nhit l :

200000.(180- 110)= 14.106 kcal/h

y 112 l Entalphy ca dng lng ti 170o

C.Lu lng ca cc dng trong thp thay i nh sau:- Dng hi lu lnh L:L= L - 14.106/(138- 15)= 476504 14.106 (138- 15)=412617,4 kg/h- Dng hi ra khi nh thp chng ct :L+V4=412617,4 +170122=582739,4 kg/h- Dng hi lu nng R4:R4 = R4- 14.106/(138-60 ) = 819725,92 179487,2= 640265,74 kg/h- Dng bay ln a trn cng :R4+V4 = 640265,74 +170122= 810387,7 kg/h

- Dng hi lu nng :R3 = R3 - 14.106/(170- 110) = 821724.14 233333,3 = 588390,8 kg/h- Dng hi lu nng chy vo a ly KerosenR3 +L3= 588390,8 +60996,64=649387,44 kg/h- Dng hi bay ln t a kerosen:R3 + V3= 588390,8 +228383= 816773,8 kg/h

- Dng hi lu R2:R2= R2 - 14.106/(190 100)= 724520 -15555=468964 kg/h- Dng hi lu nng chy vo a ly LGO

R2 + L2 =468964 +142040= 611004 kg/h

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- Dng hi nay ln t a LGO:R2+V2= 468964 + 361931,8=8300895,8 kg/h- Dng hi lu nng chy vo a HGO :

(R1+L1)+200000=379301,5 +200000= 579301,5 kg/hCc gi tr thu c biu din bi mi tn t ntB.TNH TON CC THNG S CA THP CHNG CT KH QUYN:T tnh ton phn A ta thy lu lng cc dng vng Kerosen l ln nht .Ta cth ly s liu vng ny tnh ton .Theo kt qu ca phn A ta c cc d liu sau:

Khi lng dng lng : 649387,44 kg/hKhi lng ring dng lng : 775 kg/m3 20oCLu lng th tch dng lng: 649387,44.106/(775.3600 )=232755,4

cm3/s 20oC.Khi lng dng hi: R3 + V3+ W0+W1+W2 =816773,8 + 79257,5= 896031,3 kg/hPhn t lng trung bnh R3 tm coi bng 170Phn t lng trung bnh hi bng :

M v = 79257,5/1872,1814/170588390,8896031,3

++ =92,5

Khi lng ring dng hi 200oC , 1260 mmHg

v =)273200(082,0.760

5,92.1260

+= 3,95 kg/m3

Lu lng th tch dng hi 200oC , 1260 mmHg :Qv= 896031,3/3,95=226613,9 m3/h hay = 62,95.106 cm3/s

Khi lng ring dng lng 200oC c th c tnh theo cng thc (4-10)[1]

L = o a(t-20)

Trong :

o: khi lng ring dng lng 20oC (kg/m3)

t: nhit , oC

a: h s ph thuc o. Theo bng 4/5 [1] a= 0,83 L = 775 0,83(200-20)= 625,6 kg/m3

Nn lu lng th tch 200oC l :QL=649387,44.106 /625,6.3600=288339,8 cm3/s

I.TNH NG KNH A :Lu lng a kh ln nn bi th ta chn a c ng knh ngoi ln ,chn a s6 trong bng 4/2 [1]. Khong cch l gia hai a c la chn l 0,37 ln ngknh ngoi chp, bng 0,37.15,48= 5,73cm

Theo cng thc (4-6)[1] ta c

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qv=30,122/3

3

2

21

21

2

2/1

4,01 Ha

aa

aa

a

H

A

v

vL

+

+

Trong :A: tng din tch cc khe trn 1 chp, A = 93,61 (cm3)H: chiu cao khe chp, H = 3,81 (cm)

L, V: khi lng ring dngv kh, kg/m3 a1, a2: chiu rng y khe, nh khe chp a1= 0,84cm, a2 = 0,42cmThay s ta c :

qvmax= 30,122/1

95,3

95,36,625

.

+

+ 42,0

42,084,04,01

42,084,0

42,0.93,61.3,811/2

=32204,1 cm3/s

Nu chn m khe chp h3/H = 0,9 th theo cng thc (4-7) [1], lu lng hi qua1 chp l :

qv= )21(4,02/3)21(4,02

aaa

Hhaaa

++

.2/3

3

H

h.qvmax

= 9,0)42,084,0(4,042,0

9,0)42,084,0(4,042,0

++ 3/2.32204,1 = 26710,7 cm3/s

S chp trn mt a :

qv

Qv

= 26710,762,95.10 6

=2357 chpTng tit din cc khe chp

2357 93,61= 220610 cm2Vi s lng chp rt ln nh trn th a t nht cng phi l loi 2 dng. Khi ,dng chy ca pha lng 2 a lin tip s khng ging nhau (do khc nhau bi di b chn ng chy chuyn). iu ny dn ti s khc bit v din tch vng mgia 2 a .V th m bo l trnh dng lng vng m lun ln, ta cntng vng ln so vi a 1 dng. Ni cch khc, t l din tch vng cha chps b i.

Cn c vo bng 4/4 [1] ta chn phn trm cc vng theo din tch iac l Vng kt ni : 4%

Vng m :10%

Vng ng chy chuyn :16%

Vng cha cc chp : 70%Din tch vng chc cc chp, theo bng 4/3[1] l :

220610/0,24= 919208 cm2

Din tch a : A =919208/0,7=1313154,7 cm2

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ng knh a: D= 2.

A = 1293,4 cm

Ch l nu dng chp b hn th a s b hn

C th tinhd ng knh a bng cng thc (4-9)[1] vi khong cch gia cc al 60cm sc cng b mt l 20dyn/cm.

D =( )vLvv

v

K

m

4

=)92,36,625(92,3196.14,3.84,0

896031,3.4

=11,85m

=1185cmII. S SP XP MT A : phn trc ta chn loi chp s 6 cho thp. Khi chiu cao lp lng trn a

phi vo khong > 10 cm (ln hn chiu cao chp l 9,52 cm).Nu dng a 1 dng: Vi tit din ng chy chuyn bng 8% din tch a th di b chn L ca ng chy chuyn l 68% D hay 879,5cm .T s QL/L= 288339,8 /879,5= 327,85 cm2/sTheo biu H 4.19[1] th gradien mt lng i ng vi1dy chp khong10 mm.Khi l/d = 0,37 th s chp ti a trn 1 dy l

d

D

37,1=

48,15.37,1

4,1293= 61 chp

S dy chp l: 2357 /61= 39 dyVy gradien mt lng i= 39 = 39.10 =390 >>25cm .Kt qu tnh s b cho thy phi dng a 2 dng.

B tr chp;Trn mt a, cc chp c sp xp theoquy lut tam gic u nh hnh bn.Nu la chn a 2 dng 1 bc :Phn trc ta chn a c 16 % dintch l ng chy chuyn. Trong ngchy chuyn gia chim 8%; 2 ng chychuyn 2 bn mi ng 4% .Theo biu H 4.30 [1] th chiu di b chn ng chy

chuyn bn l : 52% D hay 672,6 cm .ng chychuyn gia chim 8% din tcha hay 96972 cm2 . Do b rng ngchy chuyn khong 75cm .

Mt a nhn t trn xung mi bc c 1179 chp c xp thnh 19 hay 20 dy khi dng chy t 2 bn vo gia. Cn khi dng chy t gia ra 2 bn th s dy c th nhiu hn, l trnhdng lng qua vng m s ln hn.

dy trung bnh h ca lp lng trn a ch c th nh gi 1 cch gn ng :

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++= 65 hhh

h5: chnh gia mt cht lng trn b chn ng chy chuyn vi mt b chn .Theo cng thc (4-15)[1] ta c :

0

2/3

5 15,0 FL

Qh L

=

H s Fo c tra theo biu H 4.23 [1].

Vi5,2L

QL = 2,5672,6

288339,8=0,0246

L/D=0,68 th F0 =1,04Thay s ta c :

h5=0,15.3/2

672,6

288339,8

.1,04= 8,77 cm

Ly h1 = 2,5 cm, theo bng 4/2 [1] th vi chp s 6, chiu cao ng hi : h 6 = 8,25(cm ) = 3cmKhi h = 8,87+8,25+3=20,12 cmVi h =20,12 cm ,Ql/L= 215 , h1 = 2,5 cm , =0,37 th i=3,8 mm = 3,8.20=76mm >25mm

Nh vy vic la chn a 2 dng 1 bc khng ph hp .Ta s dng a 4 dng 1 bc:

.

Hng dng lng trn mt a

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i vi a 4 dng 1 bc, lp lng trn mt a c chia lm 4 vng 1-2, 2-3,3-4, 4-5. Mi vng c 9 hay 10 dy. Do 2 a lin tip c s ng chy chuyn lkhc nhau nn gradien mt lng cc vng cng khc nhau.Ta c qui tc sau :1-2 = 4-5, 2-3 = 3-4Q2 = Q3 = Q4 = 2Q1 = 2Q5S2= S3 = S4 = 2S1 = 2S5S1=2% A L1/D = 0,42 L1 = 543 cmS2=4% A L2/D = 0,52 L2 = 672 cm

S3=4% A S3= 36768 cm2

Theo hnh 4.30 [1] th ng chy chuyn gia chim 4% din tch a hay bng

36768 cm2 .Do vy b rng ng chy chuyn 28,5 cm

III.TNH CHIU CAO THP:p dng cng thc tnh chiu cao thp :

H=( N- 2).h + 2a + bTrong

H: chiu cao thpN: s a thc t ca thph: khong cch gia hai a l 0,8ma:chiu cao nh thp ti a trn cng (3m)

b:khong cch gia 2 a tip liu (2m)

Thay s ta c : H =(50-2).0,8+2.3+2 = 46mIII.GRADIEN MT LNG :1.Gradien mt lng quanh a :

Vic tnh gradien mt lng c tnh nh biu 4.19[1]. Vi h5: chnh giamt cht lng trn b chn ng chy chuyn vi mt b chn .Theo cng thc (4-15)[1] ta c :

0

2/3

5 15,0 FL

Qh L

=

H s Fo c tra theo biu H 4.23 [1].

Vi 5,21L

QL

= 2,5543/4288339,8

=0,01L1/D=0,42 F0 = 1,0

Thay s ta c :

h5=0,15.3/2

543

288339,8/4

.1,0= 3,9cm

Ly h1 = 2,5 cm, theo bng 4/2 [1] th vi chp s 6, chiu cao ng hi : h 6 = 8,25(cm )V ly = 2,3Khi : dy trung bnh ca lp lng trn mt a : 3,9+8,25+2,3=14,45 cm

i= 1,8mm

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Chng ta cn tnh n nh hng ca dng hi nh hnh 4.20[1]

i lng vou =.0,841313154,7

62,95.106

. 00395,0 = 3,6

Theo hnh 4.20 th h s iu chnh bng 0,96.vy Gradien mt lng ng vi 1 dychp l 0,96.1,8=1,73mm , trong vng l 1,73.9= 15,57 mm2.Gradien mt lng vng trung tm a :L2= 672cmL2/D= 0,52Theo cng thc (4-15)[1] ta c :

0

2/3

5 15,0 FL

Qh L

=

H s Fo c tra theo biu H 4.23 [1].

Vi5,2

2L

QL = 2,5672

/4288339,8=0,0062

L1/D=0,42 F0 = 1,015Thay s ta c :

h5=0,15.3/2

672

288339,8/4

.1,015= 3,4cm

Ly h1 = 2,5 cm, theo bng 4/2 [1] th vi chp s 6, chiu cao ng hi : h 6 = 8,25(cm )V ly = 3,2Khi : dy trung bnh ca lp lng trn mt a : 3,4+8,25+3,2=14,85 cm

i= 1,7mm

Chng ta cn tnh n nh hng ca dng hi nh hnh 4.20[1]

i lng vou =.0,841313154,7

62,95.106

. 00395,0 = 3,6 ,2

2

L

QL=107

Theo hnh 4.20 th h s iu chnh bng 0,95.vy Gradien mt lng ng vi 1 dychp l 0,95.1,7=1,62mm , trong vng l 1,62.9= 14,54 mmIV.TNH P SUT A:H thng ng hi chp cng nh s tn ti ca lp lng trn mt a to ra 1 trlc i vi dng hi i qua a .Do p sut pha trn a phi b hn p sut

pha di a, ngha l mi a gy ra 1 gim p sut. Ta gi ngn gn l p sut

a.p sut a b gy ra bi lp cht lng trn a, bi gim p khi dng hi i quah thng ng hi-chp v bi sc cng b mt. Trong nhiu trng hp c th bqua gim p gy bi sc cng b mt. gim p gy bi s chui lun ca dng hi qua h thng ng hi-chp ho c thtnh theo cng thc (4-11) [1]

2

0

=o

v

vL

v

oA

Qkh

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H s ko c gi tr ph thuc t s tit din hnh vnh khn so vi ng hi. Theobng 4/2 [1]:Tit din hnh vnh khn =90,30 (cm2 )Tit din ng hi = 76,00 (cm2 )

tit din vnh khn/tit din ng hi = 90,30/76,00 = 1,19Theo H 4.22 [1] th ko = 0,00145

Vy2

6

2357.76

10.95,62

95,36,625

95,300145,0

=oh = 1,138cm

Theo cng thc (4-13) [1] th gim p hl gy ra bi lp lng hot ng trn aphi bng chiu cao lp lng t mt lp lng n v tr thp nht khe chp mdng hi chui qua :

hl = h3 + h4 + h5 + /2 = 0,9H + (h6 h1 h2 H) + h5 + /2

= 0,9.3,81 + (8,25 - 2,5 - 0,65 - 3,81) +3,4 + 3,2/2 = 11,03(cm)

Vy theo (4-14) [1] :h = ho + hl = 1,138+11,03=12,169 cmVy p sut a tnh theo chiu cao ct lng tng ng l 12,169cmV.NG CHY CHUYN :

1. ng chy chuyn trung tm(3)

Theo H 4.5 v cng thc (4-16) [1] th chiu cao ca lp cht lng trong ngchy chuyn l: hc = (h6 + h5 + D) + h + hk

hkl gim p gy ra bi s chy ca cht lng t chn ng chy chuyn qua

khe ng chy chuyn trn qua mt a.

C th tnh hk theo cng thc (4-16a) [1] :2

210.65,1

= =

k

Lk

A

Qh

Ak l tit din khe ng chy chuyn , Ak= L.hk

Khe ng chy chuyn phi nh sao cho mp di ca tm chn phi ngptrong lp lng trn mt a1khong su ngn khng cho dng hi t adi chui qua ng chy chuyn ln a trn. ng thi khe ng chy chuyn

phi ln tc chy ca dng lng trong ng chy chuyn ra a di lva phi ,trnh gy hin tng dn .ngp lt .

Nh vy cao khe phi nh hn mc cht lng trn a:'kh < ++= 65 hhh = 15 1

Ly hk = 15 cmAk= L.hk= .1293,4.15= 19401 cm

2

2

2

4.19401

8,28833910.65,1

= kh = 0,228 cm < 25mm

Vy thay s vo ta c:hc= 14,85+12,169+0,228= 27,247 cm

Thi gian lu cht lng trong ng trung tm l :

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t=L

c

Q

hA .04,0.=

288339,8/4

70,04.27,241313154,7.= 16,8s

Tc chy xung ca cht lng trong ng chy chuyn l 27,247/16,8=1,62 cm/sKt qur tnh cho thy kch thc ng chy chuyn l hon ton thch hp2.ng chy chuyn ngoi cng:

Ak= 543.15=8145cm2

hk = 1,65.10-2

2

8145.8

8,288339

= 0,32cmhc =14,85+12,169+0,32=27,39 cm

t =8/8,288339

39,27.02,0.7,1313154= 17,7 s

Tc chy xung ca cht lng trong ng chy chuyn l 27,39/17,7=1,54 cm/s.VI.HIN TNG NGP LT :

Hin tng ngp lt (flooding) l hin tng trn ngp cht lng trong ton bkhng gian gia cc a. Khi b ngp lt, thp chng ct b tc dn ti ngng hotng v cht lng khng chy kp, hi khng bay ln c. Ngoi ra ,n cn cth gy n thp chng ct do s tng p sut qu mc (dn , tch t qu nhiu

pha hi).Hin tng ngp lt xy ra do cht lng trong ng chy chuyn khng chy kp

hoc do cht lng trn mt a b li cun qu d di di tc ng ca dnghi bay ln vi tc qu ln. Ni chung l s phn b khng hp l vngcha chp v vng ng chy chuyn.Do tc hi ca hin tng ngp,lt chng ta phi xem xt, nh gi kh nngkhng b ngpngp lt.Ta c th nh gi kh nng theo cng thc (4-19) [1] .

2

2

2

2

1

1

8,12,1

+

=

l

ll

S

S

S

Sf

S1l: din tch l thuyt ca vng cha cc chp

S1: din tch thc ca vng cha cc chpS2l: din tch ca vng ng chy chuynS2: din tch thc ca vng cha ng chy chuynf: i lng nh gi kh nng khng b ngp ltyu cu f < 1,2

Theo cng thc (4-20) [1] :

( ) vvLl

v

lK

QS

/1

=

Chn khong cch gia cc a l T =80 (cm)

Theo biu H.4.24 [1] th K1 =250

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Do

( ) 95,3/95,36,625250

226613,91

=lS = 72,25 m2

Theo cng thc (4-21) [1]:S2l= K2.(QL+ eS1)

Trong :

K2: h s tra theo biu hnh 4.26 [1]

Vi L - V = 621,65 kg/m3 ; T = 80 cm th K2= 0,15

e : lng lng b li cun t 1 m2 mt a trong 1 pht v ctnh theobiu H 4.25 [1].

Ta c:

S1 = 70% A = 0,7.1313154,7 = 919208,3 cm2

lS1 /s1 = 0,79 e =0,0101

S2l = 0,15.(288339,8+0,0101.91,9208)=43251 m2.Vy thay s ta c :

22

92,91.8,1

92,91

25,72.2,1

25,72

+

=f =1,003

f< 1,2.Nh vy khng sy ra hin tng ngp lt trong thp .v thit k thp nh vy lhp l .