Do an Cong Nghe Loc Dau

7/29/2019 Do an Cong Nghe Loc Dau

1/39

CC D LIU CHO TRC

* Tnh ton thp chng ct kh quyn c cu trc nh sau:- Thp c 48 a

- Sn phm nh: Kh + naptha

- LGO ly ra a 26

- HGO ly ra a 38

- Vng stripping y thp c 6 a

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

- Nguyn liu u vo l du th m Bch H,

- Qv = 19744 tn/ngy

* Yu cu tnh ton (tnh ton thp chng ct hon chnh)

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

- Tnh ton cu trc thp

Gentle - 1 - 29/03/2013

n Cng Ngh Lc Du Lc Ha Du K50


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n Cng Ngh Lc Du Lc Ha Du - K50

A. TNH NHIT , LU LNG THP

CHNG CT KH QUYN

T d liu v tnh cht du Bch H ta v c cc ng c trng ca du

th:

Bng A.1. c trng du th Bch H

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ng TBP ca du th

0

100

200

300

400

500

600

700

0 20 40 60 80 100%V

oC

t

HNH A.1.ng TBP ca du th

c tnh %V-d ca du th

0,0000

0,1000

0,2000

0,3000

0,4000

0,5000

0,6000

0,7000

0,8000

0,9000

1,0000

0 20 40 60 80 100 120

%V

d

d

HNH A.2.ng c trng V-d ca du th

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

kh quyn:- Phn on kh + naphtha 21,95% th tch

- Phn on kerosene 6,58%

- Phn on LGO 15%

- Phn on HGO 10,96%

- Phn on AR 45,51%

Gi s, coi s chung ct l l tng. Khi ng TBP ca nguyn liu xem

nh trng vi ng TBP ca tng phn on. T ta c c trng ca tng phn

on ,nh sau :

gentle - 4 - 3/29/2013


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Bng A.2.S liu %V-t-d ca phn on kh + naphtha

c tnh%V-t ca sp nh

0

20

40

60

80

100

120

140

160

180

200

0,00 20,00 40,00 60,00 80,00 100,00 120,00

%V

oC

t

HNH A.3.ng cong chng ct phn on kh + naphtha

gentle - 5 - 3/29/2013

%V t d

16,54 70 0,6440

20,77 80 0,6846

25,56 90 0,6981

31,03 100 0,7057

37,13 110 0,7135

43,96 120 0,7248

51,85 130 0,7297

60,27 140 0,7367

68,47 150 0,743

76,49 160 0,7482

84,37 170 0,7531

92,30 180 0,7583100,00 190 0,7633


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c tnh %V_d ca sp nh

0,6200

0,6400

0,6600

0,6800

0,70000,7200

0,7400

0,7600

0,7800

0,00 20,00 40,00 60,00 80,00 100,00 120,00

%V

dd

HNH A.4 ng c trng V- d phn on kh + naphtha

Bng A.3.S liu %V-t-d ca phn on Kerosen

c tnh %V-t ca Kerosen

0

50

100

150

200

250

0 20 40 60 80 100 120

%V

oC

t

HNH.A.5 ng cong chng ct phn on kerosen

gentle - 6 - 3/29/2013

%V t

0 188,4

5 196,4

10 198.0

20 199,7

30 201,6

40 203,6

50 204,760 206,7

70 209,1

80 212,1

90 216,9

100 228.0

%V t d

0 190 0,7633

24,77 200 0,768

49,09 210 0,7725

73,71 220 0,7767

100,00 230 0,7809


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c tnh %V-d ca kerosen

0,762

0,764

0,766

0,768

0,77

0,772

0,774

0,776

0,778

0,78

0,782

0 20 40 60 80 100 120%V

d d

HNH.A.6 ng c trng V- d phn on kerosen

Bng A.4 S liu %V-t-d phn on LGO

%V t D

0 230 0,7809

12,47 240 0,7852

25,93 250 0,7894

40,33 260 0,7934

55,67 270 0,7971

71,27 280 0,8005

86,40 290 0,8038

100,00 300 0,8066

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c tnh %V-t ca LGO

0

50

100

150

200

250

300

350

- 20,00 40,00 60,00 80,00 100,00 120,00%V

oC

t

HNH.A.7.ng cong chng ct phn on LGO

c tnh %V-d ca LGO

0,78

0,785

0,79

0,795

0,8

0,805

0,81

- 20,00 40,00 60,00 80,00 100,00 120,00

%V

dd

HNH.A.8 ng c trng V- d phn on LGO

Bng A.5 S liu %V-t-d phn on HGO

%V t d

0 300 0.8066

16,33 310 0,809

31,75 320 0,8109

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47,99 330 0,8127

65,60 340 0,814

83,58 350 0,8172

100,00 360 0,8273

c tnh %V-t ca HGO

300

310

320

330

340

350

360

370

- 20,00 40,00 60,00 80,00 100,00 120,00

%V

oC

t

HNH.A.9 ng cong chng ct phn on HGO

c tnh %V-d ca HGO

0,805

0,81

0,815

0,82

0,825

0,83

- 20,00 40,00 60,00 80,00 100,00 120,00

%V

dd

HNH.A.10 ng c trng V- d phn on HGO

Bng A.6 S liu %V-t-d phn on AR

%V t D

0 360 0.8273

3,49 370 0,8391

6,68 380 0,8487

9,76 390 0,855213,18 400 0,8581

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17,42 410 0,8598

22,90 420 0,8626

29,11 430 0,8658

35,66 440 0,8674

41,53 450 0,86846,39 460 0,8685

50,54 470 0,869

53,89 480 0,8713

57,55 490 0,8754

60,58 500 0,8796

63,33 510 0,8836

65,90 520 0,8873

68,18 530 0,8904

69,92 540 0,892871,48 550 0,8948

73,21 560 0,8969

75,13 570 0,8994

77,21 580 0,9023

79,04 590 0,9054

80,44 600 0,9081

c tnh %V-t ca cn

0

100

200

300

400

500

600

700

- 20,00 40,00 60,00 80,00 100,00

%V

oC

T

HNH.A.11 ng cong chng ct phn on AR

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c tnh %V-d ca cn

0,83

0,84

0,85

0,86

0,87

0,88

0,89

0,9

0,91

0,92

- 20,00 40,00 60,00 80,00 100,00

%V

d Series1

HNH.A.12 ng c trng V- d phn on AR

i vi thp chng ct ny, chng ta dng hi nc qu nhit 260 oC ,P= 3atm

stripping cn AR, cc phn on sn ly ra t sn ca thp chng ct. Ton b

hi nc dng stripping b ngng t thit b lm lnh dng bay hi ra t nh thp

chng ct v chy ra ngoi t y bnh hi lu. Entanpy ca hi nc cho H3.14

[1], lng hi nc cn dng c tm theo H3.15 [1]. Entanpy ca du xc nh nh

[2].

Bng A.7 .c trng cc phn on

(coi s o t khi bng s o khi lng ring)

Phn

on%V

Th tch

(m3/h)T khi d

Khi

lng

(tn/h)

Phn t

lng M

S

Kmol/h

Kh +naphta

21,95 220,15 0,7175 157,952 118 1338,58

Kerosen 6,58 66 0,7728 51,005 171 298,27LGO 15 150,45 0,8098 121,837 217 561,46HGO 10,96 109,9 0,8137 89,424 279 320,52 54,47 546,5 0,7689 420,218 167 2518,83AR 45,51 456,5 0,8816 402,449 475 847,24Du th 100 1003 0,8203 822,667 244,4 3366,07

I. Tnh ton iu kin vng np liu-y thp

I.1 S dng vng np liu-y thpTheo iu kin tm thy bng A.7 v hnh I.1 ta c h 2 phng trnh sau:

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Va + Vo Lo = 54,49

La + Lo Vo = 45,51

Va + La = 100

HNH.I.1: S dng vng np liu-y thp

Trong :

- 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

tm c nhit ti a np liu ta da vo ng bay hi Flash ca du ti

vng . Nhit ti a np liu Ta s l nhit ng vi bay hi Va % trn ng

Flash . Mun th cn bit Va v p sut hi ring phn ca hi du ti a np liu

(bit Va, Vo, Wo, P ti a np liu). Do khng th bit chnh xc Vo, Lo nn bi ton ch

c gii quyt da vo nhng s liu rt ra t kinh nghim.I.2 Lu lng cc dng

Kinh nghim ch ra rng cn lm bay hi 3 -5 % th tch (so vi du th) bng

cch stripping lng y thp. Theo H3.15 [1] th stripping 3% AR (6,6% du th)

cn dng 1,4 lb hi nc cho 1 gallon AR, tc khong 0.168 kg hi nc cho 1l AR.

Vy lng hi nc cn dng stripping l :

Wo = 0,168. 465,5.103 = 76692 (kg/h) = 4260,67 (kmol/h)

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Wo

Lo

Du th

Va

La

Vo

Ta

Hi nc Wo

AR= 456,5 m3/h


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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) mt lng 3 5%. Ta

dng Vo = 3%; gi s Lo = 6%. Theo h phng trnh lp ra I.1 ta c:

54,49 = Va + 3 - 6 = Va -3

45,51 = La + 6 -3 = La +3 Va = 57,49%

La = 42,51%

I.3 nng cc dng vng np liu

Theo H3.17 [1], t khi Va l 0,777 ;ca dng La l 0,878

Phn t lng ca phn on t 54,4957,49 % l 322 (theo H3.13 [1])

Do phn t lng trung bnh ca Va l :

17549,57

332249,54167

=+

=VaM

Khi lng dng Va l: 0,777.57,49% .1003 = 448,018 (tn/h)

Lu lng mol dng Va l: 448.018/175 = 2550,33 (kmol/h)

Theo bi, vng stripping y thp gm 6 a ( a 42-48) nn thc t thp

chng ct hot ng theo kiu gn nh khng c vng chng. Kh nng phn tch

vng ny khng cao, m ch yu ch xy ra s phn tch nhng hp phn nh

nht xt li khi cc dng lng ang chy xung. Trong s 2 dng La v Lo th dng La

cha nhiu cc hp phn nh hn so vi dng Lo. Nguyn nhn l do dng La vn lpha lng ca dng du th va np vo y thp, n nm cn bng Flash vi dng hi

Va, cn dng Lo l dng hi lu lng c trao i cht kh tt vi dng hi trong

vng pha trn a np liu. Vy qu trnh stripping xy ra ch yu vi dng L a. Bn

cht dng Lo thay i khng nhiu trong vng np liu-y thp.

Nh vy mc d dng La sinh ra ti a np liu cng dng Lo chy t trn

xung qua a np liu nhng dng Lo vn phi nng hn dng La nhng khng th

nng hn AR.

0,878 < dLo < 0,8816Ta coi t khi dng Lo l 0,881. Nh vy theo H I.1 ta c phng trnh cn bng

khi lng sau:

mAR = mLa + mLo mVo

hay: 402,449 = 42,51%.1003.0,878 + 6%.1003.0,881 3%.1003.Vo

Vo = 0,81 (tn/m3) hay t khi ca dng Vo l 0,81 (= dLGO)Nh vy phn t lng trung bnh ca dng hi Vo l 217.

Lu lng mol dng Vo l: 3%.1003.0,81/217 = 112,32

I.4 Nhit ti a np liu

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p sut ti a np liu Pa:

Pa = 1,5.760 + 42.8 = 1476 mmHg

p sut hi ring phn ca du ti a np liu Pa

( ) ( )

mmHg

WVV

VVP

ooa

oaa 5681476.

67,426032,11233,2550

32,11233,2550' =++

+=

++

+=

V ng Flash, (coi ng FRL trng ng Flash)

Theo H3.11 [2] ta c:

Vi B = 5,24 (oC/%) th:

A = 3,43 (oC/%)

T50 = T50DRL T50FRL = 22

M T50DRL = 336 oC nn t50FRL = 314 oC

Nh vy ng Flash (ti1atm) c xy dng vi nghing A = 3,43 (oC/%),

t50 =314 oC.

Tnh tin ng Flash (ti 1 atm) xung pha di 14 oC ta s c ng Flash

ti 568 mmHg.

Ly t57,49 trn ng Flash ta c Ta = 340 oC

I.5 Nhit ti y thp chng ct

Nhit y thp chng ct l nhit dng AR.

Phng php tnh T da trn cn bng Entanpy ca vng np liu-y thp. S

liu tnh ton cho bng I.1, Entanpy ca cc dng du ly t biu trang 83 ti liu[2], entanpy ca hi nc cho H3.14 [1]

Bng I.1 S liu lin quan n vng np liu-y thp

Dng Nhit dTh tch

(m3/h)

Khi lng

(kg/h)

Entanpy

Kcal/kg Kcal/h

VoLa (lng) 340 0,878 426,4 374.379 213 79742727

Lo (lng) 340 0,881 60,2 53036 212 11243632Wo 220 76692 716 54911472Ra

AR (lng) T 0,8816 402449 x 402449.xVo(hi) 340 0,81 30,1 24381 251 6119631

Wo 340 76692 766 58746072Ta c cn bng Entanpy:

Ev = Er

145897831 = 64865703+ 402449.x

x = 201Cn c vo biu trang 83 [2] ta c T = 332 oC

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II. Tnh nhit ti a ly HGO

HNH II.1.S dng vng HGO

Ta quyt nh stripping 6% so vi HGO ly ra. Theo H3.15 [1] cn dng 0.3

pounds hi nc stripping 1 gallon HGO (tng ng l 35 kg hi nc cho 1 m3

HGO).

Do vy tng lng hi nc cn dng l:

W1 = 35.109,9 = 3846,5 (kg/h)

Hay: 213,7 kmol/hTrong s , H II.1 c:

V1 (tng cc phn on hi) = 220,15+66+156,45+109,9

(tng ng 2518,83 kmol.h) = 546,5 (m3/h)

S1 (dng hi b stripping t dng lng L1) :

S1 = L1 L1

C th coi S1 c t khi bng t khi ca phn on pha trn gn phn on

HGO nht (tc l phn on LGO) :

dS1= 0,8098

015,79,109.94.100

100.61 ==S (m3/h) hay 5680,66 (kg/h)

Cc s liu v dng hi lu R1 ti a ly HGO c xc nh da trn cn bng

khi lng cc dng u vo v ra khi thp stripping :

109,9.0,8137+1,015.0,8098=116,915 L1

=> L1 = 0,8135

Khi lng dng L1 l 95110 kg/hVy t khi dng R1 l 0,8135.

gentle - 15 - 3/29/2013

Nguyn liu

Hi nc

a ly HGO

Ta

= 340 oC

V1

R1 W

0

R1

W1

S1

L1

W1

HGO

L1 = 109,9 m3/h

Va Vo Wo Lo

Thpstrip

ping


16/39


S dng phng php cn bng Entanpy tnh nhit a ly HGO.

Bng II.1 S liu lin quan n vng ly HGO

(Nhit gi nh T1 = 296)

Dng Nhit (oC) T khi Th tch(m3/h)

Khi

lng(kg/h)

Entanpy

Kcal/kg Kcal/h

VoVa (hi) 340 0,777 576.6 448019 264 117825312Vo (hi) 340 0,81 30,1 24381 251 6119531R1(lng) 294 0,8135 R 1 181 181.R 1

Wo 340 76692 766 58746072Ra

V1 (hi) 296 0,7689 546,5 420218 236 99171448

Lo(lng) 340 0,881 60,2 53036 212 11243632R1 (hi) 296 9,8135 R 1 228 228.R 1Wo 296 76692 737 56522004

T bng s liu trn, cn bng Entanpy ta c :

R1 = 335190,02 (kg/h)

Lu lng mol dng L1=320,52 (kmol/h)

Lu lng mol dng S1 = 5680,66/217 = 26,18 (kmol/h)

phn t lng dng 27418,2652,320

951101 =+=L

Lu lng mol dng 32,1223274

02,3351901 ==R (kmol/h)

Tng s mol hi qua a ly HGO l:

n=nV1 + nwo+ nR1

= 2518,83 +4260,67 + 1223,32 = 8002,82 (kmol/h)

S mol hi HGO ti a ly HGO l :

n = nR1 + nL = 1223,32 + 320,52 = 1543,84(kmol/h)

p sut ti a ly HGO l:

P1 = 1,5.760 + 38.8 = 1444 (mmHg)

p sut hi ring phn ca hi HGO l :

2791444.82,8002

84,1543. 1

'1 === P

n

nP (mmHg)

V ng Flash

Theo H3.11 [2], vi B = 0,6 (oC/%) th

A = 0,1 (oC/%)

T50 = T50DRL T50FRL = 0M T50DRL = 333 oC

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Nn T50FRL = 333 oC ToFRL = 328 oC

c ng Flash 279 mmHg ta tnh tin ng Flash 760 mmHg xung

pha di 33 oC (T0 = 295 oC).Vy nhit ti a ly HGO l T1= 296 oC

III. Tnh nhit ti a ly LGO

Hnh III.1.S dng vng LGO

Quyt nh stripping 6% so vi LGO ly ra. Khi tng lng hi nc cn dng l:

W2 = 35.150,45 = 5264,7 (kg/h)Trong s HIII.1 ta c:

V2 = 220,15 + 66 + 15045 =436,6 (m3/h 0

V2: tng cc phn on hi t a ly LGO tr ln

=2198,31(kmol/h) = 330794 (kg/h)

Nn V2 = 0,7577 (tn/m3)

Hay t khi dng hi V2 l 0,7577

S2 (dng hi stripping t dng lng L2) :

S2 = L2 L2Ta coi rng: S2 c t khi bng t khi phn on kerosen

dS2 = 0,7728

6032,945,150.100.94

100.62 ==S (m3/h)

mS2 =7421,4 (kg/h)

Ta c: mL2 = 150,45.0,8098 +0,7728.9,6032 = 160,0532L2

L2 = 0,8076 (tn/m3)

ML2 = 129259 (kg/h)

gentle - 17 - 3/29/2013

a ly LGO

T1

= 296 oC

V2 R

2W

0

R2

W2

S2

L2

W2

LGO

L2 = 150,45 m3/h

V1

R1

Wo

R1

W1

Thpstrip

ping

L1

S1

W1

hi nc


18/39


S dng phng php cn bng Entanpy tnh nhit ti a ly LGO

Bng III.1: S liu lin quan ti vng ly LGO

(Nhit gi nh T2 = 241 oC)

DngNhit

(oC) T khiTh tch

(m3/h)

Khi lng

(kg/h)

Entanpy

Kcal/kg Kcal/h

VoV1(hi) 296 0,7689 546,5 420218 236 99171448R1 (hi) 296 0,8135 335190 228 76423320S1(hi) 296 0,8088 7,015 5680,66 229 1300871,14

R2 (lng) 241 0,8076 R 2 139 139R 2Wo + W1 296 80538,5 737 59356874,5

RaV2 (hi) 241 0,7577 436,6 330794 199 65928006

R1 + L1(lng) 296 0,8135 430300 181 77884300R2 (hi) 241 0,8076 R 2 195 195.R 2Wo + W1 241 80538,5 706 58860181

Cn bng Entanpy cho bng s liu trn, ta thu c:

R2 = 599643,333 (kg/h)

Lu lng mol dng L2 = 561,46 (kmol/h)


4,74212 ==S (kmol/h)

phn t lng dng 2144,4346,561129259

2 =+=L

Lu lng mol dng R2 = 2802,072 (kmol/h)

Tng s mol hi qua a ly LGO :

n = nV2 + nR2 + n Wo + W1 = 2198,31 + 2802,072 + 4474,36 = 9474,742

S mol hi LGO ti y l :

n = nR2 + nL2 = 2802,072 +561.46 = 3363.532 (mmHg)

p sut ti a ly LGO l :

P2 = 1,5 .760 + 26.8 = 1348 (mmHg)p sut hi ring phn ca hi LGO l:

4791348.742,9474

532,3363. 2

'2 === P

n

nP (mmHg)

V ng Flash

Theo H3.11 [2] vi B = 0,67 (oC/%) th :

A = 0,12 (oC/%) ; T50 = 0

Mt khc T50DRL = 263 oC

Nn T50FRL = 263 oC ToFRL = 257 oC

gentle - 18 - 3/29/2013


19/39


ng Flash ca LGO ti 479 mmHg c xc nh bng cch tnh tin ng

FRL 1 atm xung pha di 16 oC. Khi ToFRL = 241 oC.

Vy nhit ti a ly LGO l T2 = 241 oC.

IV. Tnh nhit ti a ly Kerosen.

HNH IV.1: S dng vng Kerosen

Ta stripping 5% so vi lng Kerosen ly ra, Theo H3.15 [1] cn 38 kg hi

nc stripping 1m3 Kerosen. Vy lng hi nc cn dng sripping l:

W3 = 38.66 = 2508 (kg/h) = 139,3 (kmol/h)Trong s HIV.1:

V3 (tng lng hi nc cc phn on t Kerosen tr ln) :

V3 = 220,15+66=286,15 (m3/h)

nV3 = 1338,58 + 289,27= 1536,85 (kmol/h)

mV3 = 157952 + 51005 = 208957 (kg/h)

Nn V3 =730,2 (kg/m3)

Vy : t khi dng hi V3 l 0,7302

S3 (dng hi stripping t dng lng L3) :S3 = L3 L3

Ta coi rng: S3 c t khi bng t khi phn on nh :

dS3 = 0,7175

4737,366.100.95

100.53 ==SV (m3/h)

mS3 = 3,4737.0,7175= 2492,4 (kg/h)

Ta c: mL3 = 66.0,7728 + 0,7175.3,4737 = 69,4737L3

L3 = 0,77 (tn/m3) mL3 = 53485 (kg/h)

gentle - 19 - 3/29/2013

Thpstrip

ping

W0

W1

W2

a lyKerosen

T2

= 241 oC

V3

R3

R3

W3

S3

L3

W3

L3 = 66 m3/h

V2 R2

Wo

R2W1

L2

S2

W2

hi nc

Kerosen


20/39


S dng phng php cn bng Entanpy tnh nhit ti a ly Kerosen.

Bng III.1: S liu lin quan ti vng ly Kerosen


DngNhit

(oC) T khiTh tch

(m3/h)

Khi lng

(kg/h)

Entanpy

Kcal/kg Kcal/h

VoV2(hi) 241 0,7577 436,6 330794 199 65928006R2 (hi) 241 0,8076 599643,333 195 116930450S2(hi) 241 0,7728 9,6032 7421,4 196 1454594,4R3 (lng) 191 0,77 R 3 109 109R 3Wo+W1+ W2 241 85803,2 706 60577059,2RaV3 (hi) 191 0,7302 286,15 208957 173 36149561

R2+ L2(lng) 241 0,8076 728902,333 139 101317424,3R3 (hi) 191 0,77 R 3 168 168.R 3Wo+W1+W2 191 85803,2 677 58088766,4

Cn bng Entanpy cho bng s liu trn, ta thu c:

R3 = 836176 (kg/h)

Lu lng mol dng L3 = 298,27 (kmol/h)


4,24923 ==S (kmol/h)

phn t lng dng 167122,2127,298

534953 =+=L

Lu lng mol dng R3 = 836176/167 = 5007 (kmol/h)

p sut hi ring phn ca hi Kerosen l :

591)8.15760.5,1.(18/2,8580385,15365007

27,2985007. 3

2033

'33'3 =+++

+=

+++

=

Pnnn

nnP

WVR

LR(mmHg)

V ng Flash

Theo H3.11 [2] vi B = 0,185 (oC/%) th : A = 0,05 (oC/%)

T50 = 0

Mt khc T50DRL = 204,7 oC

Nn T50FRL = 204,7 oC ToFRL = 202,2 oCng Flash ca Kerosen ti 591 mmHg c xc nh bng cch tnh tin

ng FRL 1 atm xung pha di 13,5 oC. Khi : ToFRL = 202,2-13,5 = 188,7 oC.

Vy nhit ti a ly LGO l T3 = 191 oC.

V. Tnh nhit ti nh thp chng ct

gentle - 20 - 3/29/2013

R3

T3

= 191oC

T4

R4

R4

V3

R3

Wo

W1 W

2

W3

S3

L3

Kh = 10,03 m3/hE

L

V4+L+W


21/39


HNH V.1.S dng vng nh thp

Trong s Hnh V.1:

V4: phn on bay hi ln nh thp

L :dng hi lu lnh nh thp.Dng L tip xc vi dng hi ang bay ln lm

ngng t mt lng lng R4.Gi s chng ta lm vic vi dng L c nhit l 30 oC.

R4 ng vai tr l dng hi lu nng nh thp.

Ti nh thp chng ct, dng hi V4 qua thit b lm lnh E v bnh ngng tth c chia lm 2 dng l dng kh nh bay ln nh thp v dng lng chy ra

y thp cha naphtha (c trng cho bng V.1).

Bng V.1: c trng dng hi phn on nh

Phn

on%V

Th tch

(m3/h)T khi

Khi lng

(kg/h)

Phn t

lng

S

kmol/hKh 1 10,03 0,5741 5758 71 81LN 4,61 46,23 0,678 31344 100 313,44HN 16,34 163,89 0,7374 120850 128 944,14V4 21,95 220,15 0,7175 157952 118 1338,58

Dng hi lu lnh L coi l c tnh cht nh HN :

dL = 0,7374 ; ML = 128

Bng V.2: S liu lin quan ti phn on nh


DngNhit

(oC)T khi

Th tch

(m3/h)

Khi lng

(kg/h)

Entanpy

Kcal/g Kcal/h

VoV3(hi) 191 0,7203 286,15 208957 173 36149561

gentle - 21 - 3/29/2013

Naphtha = 210,12 m3/h

Nc lng

W 0,1,2,3

Bnh hilu


22/39


R3 (hi) 191 0,77 836176 168 140477568S3(hi) 191 0,7175 3,4737 2492,4 174 433677,6L (lng) 30 0,7374 L 16 16.LWo+W1+W2+W3 191 88311,2 677 59786682,4Ra

V4 (hi) 128 0,7175 220,15 157952 144 22745088R3 + L3 (lng) 191 0,77 889671 109 96974139L (hi) 128 0,7374 L 142 142.LWo+W1+W2+W3 128 88311,2 654 57755524,8

Cn bng Entanpy phn on nh theo bng s liu trn, ta thu c :

L = 471212,2 (kg/h) hay 3681,345 (kmol/h)

p sut hi ring phn ca hydrocarbon ti nh thp l :

597760.5,1.18,4906345,368158,1338

345,368158,1338. 4

304

4'4 =++

+=

+++

=

Pnnn

nnP

WLV

LV(mmHg)

V ng Flash

Theo H3.11 [2] vi B = 1,4 (oC/%) th:

A = 0,45 (oC/%)

T50 = 7 oC

Mt khc T50DRL = 127 oC

Nn T50FRL = 120 oC

ToFRL = 142,5 oC

ng Flash ca Kerosen ti 591 mmHg c xc nh bng cch tnh tinng FRL 1 atm xung pha di 14 oC. Khi T100 FRL = 128 oC.

Vy nhit ly ti phn on nh l T4 = 128 oC.

Tnh dng hi lu nng R4 :

R4(142-70) = L(142-16) = 471212,2.126

R4 = 824621,35 (kg/h)

VI. Xc nh phn tch ca php chng ct du th

Chng ta s nh gi phn tch nh biu Packie H3.3 [1] vi trng

hp dng hi nc:

i lng LGO-HGO Kerosen-LGO HN-KerosenS a Ni 12 11 15Dng hi lu ni

Ri (m3/h)R2 = 742,5 1085,9 118,3

D n g h i l y r a

Vi (m3/h)V2 = 436,6 V3 = 286,15 V4 = 220,15

hi lu hi=Ri/Vi 1,7 3,8 5,1

F=Ni.hi 20,4 263 204,7T50,TBPp nng (oC) 333 263 204,7T50,TBPcc p nh (oC) 189 146 127

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23/39

W(h i)1,5atm

128 oC

W(l ng)1,5atm

30 oC

W(h i)1,5atm

112 oC

W(l ng)1,5atm

112oC


T50,TBP(oC) 144 117 77,7 phn tch +23 +18 +21

Gap Gap GapKt qu v phn tch cho thy php chng ct rt tt, cc phn on rt t ln

vo nhau.Cht lng sn phm tt.

VII. Tnh cn bng nhit cho ton thp

Vi s chnh lch khng ln, ta coi nhit ly cc sn phm cc phn on

sn chnh bng nhit ti a ly phn on

Da vo cc s liu cho ta lp c bng cn bng nhit sau :

DngNhit

( oC)T khi

Th tch

(m3/h)

Khi lng

(kg/h)

Entanpy

Kcal/kg Kcal/h

Vo

La (lng) 340 0,678 426,4 374379 213 79742727V4 (hi) 340 0,777 576,6 448018 264 117825312L (lng) 30 0,7374 471212,2 16 7539395,2W (hi) 220 88311,2 716 63230819,2RaAR (lng) 332 0,8816 456,5 402449 201 80892249L (hi) 128 0,7374 471212,2 142 66912132,4L1 (lng) 296 0,8137 109,9 89424 181 16185744L2 (lng) 241 0,8098 150,45 121837 139 16935343L3 (lng) 191 0,7728 66 51005 109 5559545Kh (hi) 30 0,5741 10,03 5758 114 656412

Naphtha(lng) 30 0,7243 210,12 152194 18 2739492W (lng) 30 88311,2 654 54475116,96PA X

Lu :ti nh thp chng ct, dng hi nc i ra khi nh thp 128 oC qua

thit b lm lnh ngng t thu dng nc ngng 30 oC.

Do hi nc c ngng t l hi nc qu nhit (128 oC; 1,5 at) v qu trnh

ny ngng t nc xung di nhit bo ha (30 oC) Bi vy ,qu trnh phi xy ra

theo 3 giai on sau:- Giai on 1: Lm ngui hi nc qu nhit ti nhit bo ha

- Giai on 2: ngng t hi nc bo ha nhit khng i

- Giai on 3: lm ngui cht lng ngng t n nhit cn thit

Qu trnh lm lnh ngng t hi nc c th c m t theo s sau:

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Q1

Q

Q3


24/39


Q = Q1 + Q2 + Q3

Q1 = mCp(128 - 112) = mCp.6

Q2 = m .rngngt

Q3 = mCn(112 - 30) = m.Cn.82

Trong :

-m: khi lng ca nc ngng t

-Cp: nhit dung ring trung bnh ca hi nc qu nhit ,Cp=0,503(kcal/kg.)[3]

-Cn: nhit dung ring ca nc ngng ,Cn = 0,998 (kcal/kg.) [3]- rngngt :n nhit ngng t ca hi nc bo ha

rngngt = 532 kcal/kg [3]

Vy:

Q = 88311,2(6.0,503 + 532 + 82.0,998) = 54465116,96 (kcal/h)

Cn bng nhit cho bng s liu VII.1 ,ta thy tng lng nhit dng cho hi lu

vng l:

X = 23982219,04 (kcal/h)

VIII. Tnh ton hi lu vng cho thp chng ct

tng phn tch cho qu trnh chng ct du th ngi ta thng sdng

phng php hi lu vng. Theo phng php ny, ngi ta trch mt dng lng ti

mt a no , a ra ngoi lm lnh ri hi lu n tr li thp v tr cao hn 3 a

so vi v tr ly n ra.

Kh quay li thp chng ct, dng hi lu vng c nhit thp hn nhiu so

vi nhit ti a m n a vo (do trao i nhit vi dng du th) lm ngng tmt lng hi ng k. Kt qu l lu lng hi bay ln t a gim i nn ng

thi lu lng lng chy xung a cng nh bt. Kt qu l lu lng lng v hi

ti cc a pha trn u gim xung.

VIII.1 Trc khi hi lu.

Lu lng hi ti a ly phn on sn l: Ri + Vi

Lu lng lng ti i ly phn on sn l: Ri + Li

1.1 Ti a ly HGO

R1 + L1 = 335190,02 + 95110 = 430300,02 (kg/h)

R1 + V1 = 335190,02 + 420218 = 755408 (kg/h)

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Q2


25/39


1.2 Ti a ly LGO

R2 + L2 = 599643,333 + 129259 = 728902,333 (kg/h)

R2 + V2 = 599643,333 + 330794 = 930437,333 (kg/h)

1.3 Ti a ly Kerosen

R3 + L3 = 836174 + 53495 = 889669 (kg/h)R3 + V3 = 836174 + 208957 = 1045131 (kg/h)

1.4 Ti nh thp

R4 = 824621,35 (kg/h)

R4 + V4 = 824621,35 + 157952 = 982573,35 (kg/h)

L + V4 = 471212,2 + 157952 = 629164 (kg/h)

1.5 Ti a np liu

Lo + La = 53036 + 374379 = 427415 (kg/h)

Vo + Va = 24381 + 448018 = 472399 (kg/h)

1.6 Ti y thp

AR = 402449 (kg/h)

VIII.2 nh lng hi lu vng

Quyt nh hi lu vng ti 3 v tr trn sn thp chng ct.

2.1.ti a ly HGO

Trch ra 220 tn/h du th thc hin hi lu vng, dng hi lu khi quay

li thp a th 35 ch cn 200o

C sau khi trao i nhit vi dng du th.Nh vy lng nhit hi lu vng trao i vi dng du th l :

Q1 = 220000(181 - 113) = 14960000 (kcal/h)

y 113 kcal/kg l entanpy ca HGO ti 200 oC

Lu lng cc dng trong thp chng ct s thay i nh sau :

- Dng hi lu lnh L

L = L 14960000/(142 - 16) = 471212,2 118730,16 = 352482,04 (kg/h)

- Dng hi ra khi thp chng ct

L + V4 = 352482,04 + 157952 = 510434,04 (kg/h)- Dng hi lu nng R4

R4 = R414960000/(142 - 70) = 824621,35 207777,78 = 616843,57 (kg/h)

- Dng hi bay ln vo a trn cng

R4 + V4 = 616843,57 + 157952 = 774795,57 (kg/h)

- Dng hi lu nng R3

R3 = R3 14960000(168 - 109) = 836174 253559 = 582615 (kg/h)

- Dng hi lu nng chy vo a ly phn on kerosen

R3 + L3= 582615 + 53495 = 636110 (kg/h)

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26/39


- Dng hi bay ln t a ly Kerosen

R3 + V3= 582615 + 208957 = 791572 (kg/h)

- Dng hi lu nng R2

R2 = R2 14960000(195 - 139) = 59964,333 267142,857 = 332500,476 (kg/h)- Dng hi lu nng chy vo a ly LGO

R2 + L2= 332500,476 + 129259 = 461579,476 (kg/h)

- Dng hi bay ln t a ly Kerosen

R2 + V2= 332500,476 + 330794 = 663294,276 (kg/h)

- Dng hi lu nng chy vo a ly HGO

(R1 + L1) + 220000 = 430300,02 + 220000 = 650300,02(kg/h)

2.2 Ti a ly LGO

Tin hnh ly ra 140 tn LGO trong 1h, hi lu vng trao i nhit vi dng du

th v quay tr li thp a s 23.Khi dng PA c nhit l 150 oC.

Vy lng nhit PA chuyn cho du th:

Q2 = 140000(139 - 82) = 7980000 (kcal/h)

Lu lng cc dng trong thp s thay i nh sau:

- L = L 7980000/(142 - 16) = 352482,04 63333,33 = 289148,71 (kg/h)

L + V4 = 289148,71 + 157952 = 447100,71 (kg/h)

- R4 = R4 7980000/(142 - 70) = 616843,57 110833,33 = 506010,24 (kg/h)R4 + V4 = 506010,24 + 157952 = 663962,42 (kg/h)

- R3 = R3 7980000/(168 - 109) = 582615 135254 = 447361 (kg/h)

R3 + L3 = 447361 + 53495 = 500856 (kg/h)

- R3 + V3 = 447361 + 208957 = 656318 (kg/h)

(R2 + L2) + 140000 = 461579,476 + 140000 = 601579,476 (kg/h)

2.3 Ti a ly Kerosen

Trch ra 18,5 tn/h LGO, hi lu vng trao i nhit vi dng du th v quay

tr li thp a s 12. Khi nhit dng h xung cn 100 oC.Vy lng nhit trao i vi dng du th:

Q3 = 18500(109 - 53) = 1036000 (kcal/h)

Lu lng cc dng trong thp s thay i nh sau:

- L = L 1036000/(142 - 16) = 289148,71 8222,22 = 280926,49 (kg/h)

L + V4 = 280926,49 + 157952 = 438876,49 (kg/h)

- R4 = R4 1036000/(142 - 70) = 506010,24 14388,89 = 491621,35 (kg/h)

R4 + V4 = 491621,35 + 157952 = 649573,35 (kg/h)

(R3 + L3) + 18500 = 500856 + 18500 = 519356 (kg/h)

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27/39


Cc kt qu thu c c th hin qua s dng di y :

HNH.VIII.2.3.1.S phn b cc dng ti cc a

Nh vy ,sau hi lu vng ,lu lng hi trong thp tng i n nh.iu ny

to iu kin thit k thp c ng knh ng u trong ton thp .

phn B ta s i tnh ng knh thp theo nhng s liu hi lu ny.

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983


28/39


B . TNH TON CC THNG S CNG NGH CA

THP CHNG CT KH QUYN

Kt qu tnh ton phn A cho thy lu lng cc dng pha vng HGO l

ln nht. Ta s dng cc s liu v cc dng pha a ly phn on HGO tnh ccthng s a ca thp chng ct.

Theo kt qu tnh ton c t phn A, ta c d liu nh sau:

- Khi lng dng lng: 650300,02 kg/h

- Khi lng ring dng lng: 813,5 kg/m3 20 oC

- Lu lng th tch dng lng: 5,2220513600.5,813

10.02,650300 6= (cm3/s)

- Khi lng dng hi: R1 + V1 + Wo = 668000 + 76692 = 744692 (kg/h)- Phn t lng trung bnh hi l:

9767.426083,2518274/247782

744692=

++=VM

- Khi lng ring dng hi 296 oC; 1444 mmHg

RTMPV .=

Trong :

: khi lng ring dng hi, kg/m3

P: p sut tng, atm

M: phn t lng dng hi

R: hng s kh l tng (l.atm/mol.oK)

T: nhit (oK)

Vy: 95,3)273296(082.0.760

97.1444=

+=V (kg/m3)

- Lu lng th tch dng hi: 18853095,3

744692==VQ (m3/h) = 52369339 (cm3/s)

- Khi lng ring dng lng 296 oC c tnh theo cng thc(4-1)trang 107 [1].

L = o a(t-20)

Trong :

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

t: nhit , oC

a: h s ph thuc o. Theo bng 4/5 [1] a = 0,76

L = 813,5 0,76(296 - 20) = 603,74 kg/m3

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29/39


- Lu lng th tch dng lng 296 oC

12,107774,60302,650300

===L

L

mQ

(m3/h) = 299200 (cm3/s)

I. Tnh ng knh a

Do lu lng hi l ln nn a chc chn phi to, bi vy ta chn loi chp ln Chn loi chp dng trong thp l chp s 6 ( theo bng 4/2 [1] )

Theo cng thc (4-6) [1] ,lu lng hi cc i qua 1 chp l :

2/3

2

21

21

2

2/1

max 4,0112,30 Ha

aa

aa

a

H

Aq

v

vLv

+

+

=

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 = 0,42cm

2/32/1

max 81,342,0

42,084,04,01

84,042,0

42,0

81,3

61,93

95,3

95,374,60312,30

+

+

=vq

=31648,1(cm3/s)

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

qua 1 chp l :

( )

( )

262501,31648.9,0.)42,084,0(4,042,0

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

.4,0

/4,0

2/3

max

2/3

3

212

3212

=+

+=

++

= vv qh

h

aaa

Hhaaaq

S chp trn 1 a:

199526250

52369339 ==v

v

qQ (chp )

Tng tit din tt c cc khe chp: 93,61.1995 = 186752 (cm2).

Vi 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 m

gia 2 a .V th m bo l trnh dng lng vng m lun ln, ta cn tng

vng ln so vi a 1 dng. Ni cch khc, t l din tch vng cha chp s b

i.iu ny c nh hng c v vn cng ngh cng nh vn kinh t ;v vy cnc tnh ton k lng .

gentle - 29 - 3/29/2013

(Cm3/s )


30/39


Ta chn phn trm cc vng theo din tch a nh sau :

- Vng cha cc chp 70%

- Vng kt ni 4%

- Vng m 10%

- Vng ng chy chuyn 16%Chn l/d = 0,37 hay l = 0,37.15,48 = 5,73 (cm )

Theo bng 4/3 [1] th =0,24.Nh vy din tch vng cha cc chp l :

186752/0,24 = 778133 (cm2)

Din tch a :

A = 778133/0,7 = 1111619 (cm2)

Nn ng knh a l :1191 (cm)

Ch : ng knh a cn c th xc inh qua cng thc (4-9) [1] nh sau :

( )VLvv

v

K

mD

=

4(m)

Trong :

mv: lu lng khi lng dng hi trong thp chng ct (kg/h)

: phn din tch khng b chim bi ng chy chuynKv: h s ph thuc khong cch gia cc a v sc cng b mt

Vi khong cch gia cc a l 60cm, sc cng b mt 20 dyn/cm th theo biu H 4.17 [1] , Kv =196

( )88,10

95,374,60395,3196.84,0

744692.4=

=

D (m)

II. Sp xp mt a

II.1 Chn la cu trc ng chy chuyn.

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 1dng vi tit din ng chy chuyn bng 8% din tch a th di b chn L ca ng

chy chuyn l 68% D hay 809 cm).

T s 370809

299200==

L

QL

= l/d =0,37

Theo 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 :

5648,15.37,1

119037,1

==d

D(chp)

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31/39


S dy chp ti thiu l: 1995/56 = 36 (dy)

Vy gradien mt lng:

= 36.i = 360 mm >> 25mm(25mm l gradien mt lng ti a cho php).

Nh vy kt qu tnh s b cng cho thy t nht cng phi dng a 2 dng. II.2 B tr chp

Trn mt a, cc chp c sp xp theo

quy lut tam gic u nh hnh bn.

Phn trc ta chn a c 16 % din tch

l ng chy chuyn. Trong ng chy chuyn

gia chim 8%; 2 ng chy chuyn 2 bn mi ng

4%.

Theo biu H 4.30 [1] th chiu di b chn ng chy

chuyn bn l : 52% D hay 619cm . ng chy chuyn gia

chim 8% din tch a hay 88929,52 cm . Do b

rng ng chy chuyn gia l khong 75cm. HNH B.II.2.1.B tr mt

a

Nhn t trn xung

mi bc c 998 chp c xp thnh 22 hay 23 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 :

++= 65 hhh

h5: chnh gia mt cht lng trn b chn ng chy chuyn vi mt

b chn

Theo cng thc (4-15) [1]:

oL F

LQh ..15,0

3/2

5 = (cm)

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

Vi 0157,0619

2/2992005,25,2

==L

QL

L/D = 0,52

Th Fo = 1,023

Nn: 95,5023,1.619.2

299200.15,0

3/2

5 =

=h (cm)

Ly h1 = 2,5 cm, theo bng 4/2 [1] th vi chp s 6, chiu cao ng hi :

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32/39


h6 = 8,25 (cm )

Mt cch gn ng ta ly i = 1,8 (cm)

Khi 168,125,895,5 =++=h (cm)

Vi 16=h cm, QL/L = 242 cm, h1 = 2,5cm, = 0,37 th i = 4,1 mm

S b ta tnh c gradien mt lng ng vi 1 bc a l :

= 4,1.22 = 90,2 mm >> 25 mmNh vy a 2 dng 1 bc cng khng th p ng c yu cu k thut ca thp.

Ta quyt nh s dng a 4 dng 1 bc.

Hnh B.II.2.2.Hng dng lng trn ai 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 l khc

nhau nn gradien mt lng cc vng cng khc nhau.

gentle - 32 - 3/29/2013

1 2 3 4 5


33/39


Ta c quy tc sau:

1-2 = 4-5, 2-3 = 3-4Q2 = Q3 = Q4 = 2Q1 = 2Q5

S2 = S3 = S4 = 2S1 = 2S5

Theo hnh 4.30 [1] th:

S1 = 2% A nn L1/d = 0,42 L1 = L5 = 500 (cm)

S2 = 4% A nn L2/d = 0,52 L2 = L4 = 619 (cm)

S3 = 4% A = 44464,76 (cm2 )

Do b rng ng chy chuyn gia khong 37 (cm).

III. Gradien mt lng

III.1 Gradien mt lng xung quanh a.Tra biu H 4.23 [1] vi :

L1/D = 0,42

0134,0500

4/2992005,25,2

2

==L

QL

th Fo = 1,031.

36,4031,1.500.4

299200.15,0..15,0

3/23/2

15 =

=

= o

L FL

Qh (cm)

Ly h1 = 2,5 cm th h6 = 8,25 cm

Gi s = 2,39 cm

Khi 1639,225,836,465 =++=++= hhh (cm)

Theo biu H 4.19 [1] th i = 2,6 (mm)

C 525,310.95,384,0.1111619

52369339 3 == VoU

Theo H 4.20 [1] th h s hiu chnh do nh hng ca dng hi l 0,97

1-2 = 4-5 = 0,97.2,6.9 = 22,7 (mm).III.2 Gradien mt lng gia a

Theo biu H 4.23 [1] th F = 1,016

Khi 0078,0619

4/2992005,25,2

2

==L

QL

L2/D = 0,52

46,3016,1.

619.4

299200.15,0..15,0

3/23/2

2

5 =

=

= o

L F

L

Qh (cm)

Ly h1 = 2,5 cm th h6 = 8,25 cm

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Ly gradien mt lng = 3,21 cm

Khi 1521,325,846,365 =++=++= hhh (cm)

Theo biu H 4.19 [1] th i = 2,7 (mm)Theo H 4.20 [1] ,h s hiu chnh do tc ng ca hi vi:

525,3=VoU ; 121619.4299200

2

==L

QL

h s hiu chnh l 0,96.Vy Gradien mt lng trong vng l :

2-3 = 3-4 = 0,96.2,7.9 = 23,33 (mm)IV. p sut a.

H thng ng hi chp cng nh s tn ti ca lp lng trn mt a to ra 1

tr lc 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

qua h thng ng hi-chp v bi sc cng b mt. Trong nhiu trng hp c th b

qua 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

th tnh theo cng thc (4-11) [1]2

.

=o

V

VL

Voo

A

Qkh

H s ko c gi tr ph thuc t s tit din hnh vnh khn so vi ng hi.

Theo bng 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 = 1,19

Theo H 4.22 [1] th ko = 0,00145

Vy 14,11995.76

52369339.

95,374,603

95,300145,0

2

=

=oh (cm)

Theo cng thc (4-13) [1] th gim p hl gy ra bi lp lng hot ng trn

a phi bng chiu cao lp lng t mt lp lng n v tr thp nht khe chp m

dng 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) +4,36 + 2,27/2 = 10,214 (cm)Vy theo (4-14) [1] :

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h = ho + hl = 1,14 + 10,214 = 11354 (cm)

V. Thi gian lu ca cht lng trong ng chy chuyn

V.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 ng

chy chuyn l:hc = (h6 + h5 + ) + 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

2.10.65,1

=

h

Lk

A

Qh

Ak l tit din khe ng chy chuyn , Ak = L.hkKhe ng chy chuyn phi nh sao cho mp di ca tm chn phi ngp

trong lp lng trn mt a1khong su ngn khng cho dng hi t a di

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 l va phi ,trnh gy

hin tng dn .ngp lt .

Nh vy cao khe phi nh hn mc cht lng trn a:

161565' =++=< hhhhk cm

Ly hk = 10 cm

Ak = L.hk = 1190.10 = 11900 (cm2 )

65,011900.4

299200.10.65,1

22 =

= kh (cm )

hc = 15 + 11,354 + 0,65 = 27,004 (cm)

Thi gian lu ca cht lng trong ng chy chuyn trung tm l :

164/299200

004,27.04,0.1111619.04,0.

=== Lc

Q

hA

t (s)

Tc chy xung ca cht lng trong ng chy chuyn l :

27,004/16 = 1,69 (cm/s)

V.2 Hai ng chy chuyn quanh trc (2 v 4)

Tng t nh trn ta cng tnh c cc gi tr:

Ak = L.hk = 619.10 = 6190 (cm2 )

41,26190.4

299200.10.65,1

22 =

= k

h (cm )

hc = 15 +11,354 + 2,41 = 28,764 (cm)

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1,174/299200

764,28.04,0.1111619.04,0.===

L

c

Q

hAt (s)

Tc chy xung ca cht lng trong ng chy chuyn l:

28,764/17,1 = 1,68 (cm/s). V.3.Hai ng chy chuyn ngoi cng (1 v 5):

Ak = L.hk = 500.10 = 5000 (cm2 )

92,05000.8

299200.10.65,1

22 =

= kh (cm)

hc = 16 +11,354 + 0,92 = 28,274 (cm)

8,16

8/299200

274,28.02,0.1111619.02,0.===

L

c

Q

hAt 1 (s)

Tc chy xung ca cht lng trong ng chy chuyn l:

28,274/16,81 = 1,682 (cm/s).

Nh vy trong cc ng chy chuyn, thi gian lu ca cht lng l khng qu

b (> 5s), trnh xy ra hin tng cc bng hi khng kp thot ht ra khi khi lng

bay ln a trn, gy hin tng bng kh nh hng n hiu sut tch v cu trc

a, thp.

VI. Hin tng ngp lt

Hin tng ngp lt (flooding) l hin tng trn ngp cht lng trong ton b

khng gian gia cc a. Khi b ngp lt, thp chng ct b tc dn ti ngng hot ng

v cht lng khng chy kp, hi khng bay ln c. Ngoi ra ,n cn c th 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 dng hi bay

ln vi tc qu ln. Ni chung l s phn b khng hp l vng cha chp v

vng ng chy chuyn.Do tc hi ca hin tng ngp,lt chng ta phi xem xt, nh gi kh nng

khng 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

+

=

S

S

S

Sf ll

Trong :

S1l: din tch l thuyt ca vng cha cc chp

S1: din tch thc ca vng cha cc chp

S2l: din tch ca vng ng chy chuyn

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S2: din tch thc ca vng cha ng chy chuyn

f: i lng nh gi kh nng khng b ngp lt

yu cu f < 1,2

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

( ) VVL

Vl

K

QS

/11

= (m2)

Chn khong cch gia cc a l T =60 (cm)

Theo biu H.4.24 [1] th K1 =230

Do :

( )

54353,6995,3/95,374,603220

1885301 =

=lS (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 = 600 kg/m3 ; T = 60 cm th K2 = 0,21e : lng lng b li cun t 1 m2 mt a trong 1 pht v c tnh theo

biu H 4.25 [1].

Ta c:S1 = 70% A = 0,7.111,619 = 77,81333 (m2 )

S2= 16% A = 0,16.11,619 = 17,785904 (m2 )

Vi S1l/S1 = 0,894 th e = 0,024

S2l = 0,21(1077,12/60 + 0,024.7781333) = 4,162099 (m2 )

Vy 76,0785904,178,1

162099,4

81333,772,1

54353,6922

=

+

=f

Nh vy f < 1,2 nn khng c hin tng ngp lt song thit k nh vy chat hiu qu tht cao v din tch vng cha chp v ng chy chuyn hi ln .

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TI LIU THAM KHO

[1]. Ts. Phan T Bng, Gio trnh cng ngh lc du, nh xut bn xy dng HNi, 2002.

[2]. Ts. Phan T Bng, Gio trnh ha hc du m v kh t nhin, nh xut bn

giao thng vn ti H Ni, 1999.

[3]. S tay qu trnh v thit b cng ngh ha cht, tp 1, nh xut bn khoa hc

v k thut, H Ni, 2004.

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MC LCA. TNH NHIT , LU LNG THP............................2

CHNG CT KH QUYN....................................................2

I. Tnh ton iu kin vng np liu-y thp......................................................................11I.1 S dng vng np liu-y thp..................................................................................11

I.2 Lu lng cc dng.........................................................................................................12

I.3 nng cc dng vng np liu...................................................................................13

I.4 Nhit ti a np liu............................................................................................13

I.5 Nhit ti y thp chng ct.................................................................................14

II. Tnh nhit ti a ly HGO......................................................................................15

III. Tnh nhit ti a ly LGO..............................................................................................17

IV. Tnh nhit ti a ly Kerosen.................................................................................19

V. Tnh nhit ti nh thp chng ct.....................................................................................20VI. Xc nh phn tch ca php chng ct du th.......................................................22

VII. Tnh cn bng nhit cho ton thp..............................................................................23

24

VIII. Tnh ton hi lu vng cho thp chng ct................................................................24

VIII.1 Trc khi hi lu........................................................................................................24

VIII.2 nh lng hi lu vng.....................................................................................25

B . TNH TON CC THNG S CNG NGH CA ......................................28

THP CHNG CT KH QUYN.........................................................................28

I. Tnh ng knh a.........................................................................................................29II. Sp xp mt a.............................................................................................................30

II.1 Chn la cu trc ng chy chuyn................................................................................30

II.2 B tr chp.............................................................................................................31

III. Gradien mt lng..........................................................................................................33

III.1 Gradien mt lng xung quanh a..................................................................................33

III.2 Gradien mt lng gia a.................................................................................33

IV. p sut a...........................................................................................................................34

V. Thi gian lu ca cht lng trong ng chy chuyn......................................................35

V.1 ng chy chuyn trung tm(3).......................................................................................35V.2 Hai ng chy chuyn quanh trc (2 v 4)...............................................................35

V.3.Hai ng chy chuyn ngoi cng (1 v 5):.............................................................36

VI. Hin tng ngp lt.............................................................................................................36

TI LIU THAM KHO......................................................38

TI LIU THAM KHO..........................................................................................37

MC LC..................................................................................................................38

n Cng Ngh Lc Du Lc Ha Du K50

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