do an cong nghe loc dau
TRANSCRIPT
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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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n Cng Ngh Lc Du Lc Ha Du - K50
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n Cng Ngh Lc Du Lc Ha Du - K50
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 :
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n Cng Ngh Lc Du Lc Ha Du - K50
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
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%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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n Cng Ngh Lc Du Lc Ha Du - K50
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
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%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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n Cng Ngh Lc Du Lc Ha Du - K50
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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n Cng Ngh Lc Du Lc Ha Du - K50
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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n Cng Ngh Lc Du Lc Ha Du - K50
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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n Cng Ngh Lc Du Lc Ha Du - K50
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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n Cng Ngh Lc Du Lc Ha Du - K50
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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n Cng Ngh Lc Du Lc Ha Du - K50
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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n Cng Ngh Lc Du Lc Ha Du - K50
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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n Cng Ngh Lc Du Lc Ha Du - K50
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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n Cng Ngh Lc Du Lc Ha Du - K50
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.
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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
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n Cng Ngh Lc Du Lc Ha Du - K50
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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n Cng Ngh Lc Du Lc Ha Du - K50
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)
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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
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n Cng Ngh Lc Du Lc Ha Du - K50
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)
Lu lng mol dng 4,43171
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
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n Cng Ngh Lc Du Lc Ha Du - K50
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
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n Cng Ngh Lc Du Lc Ha Du - K50
S dng phng php cn bng Entanpy tnh nhit ti a ly Kerosen.
Bng III.1: S liu lin quan ti vng ly Kerosen
(Nhit gi nh T3 = 191 oC)
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)
Lu lng mol dng 122,21118
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
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n Cng Ngh Lc Du Lc Ha Du - K50
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
(Nhit gi nh T4 = 128 oC)
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
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Naphtha = 210,12 m3/h
Nc lng
W 0,1,2,3
Bnh hilu
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n Cng Ngh Lc Du Lc Ha Du - K50
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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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
n Cng Ngh Lc Du Lc Ha Du - K50
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:
gentle - 23 - 3/29/2013
Q1
Q
Q3
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n Cng Ngh Lc Du Lc Ha Du - K50
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
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n Cng Ngh Lc Du Lc Ha Du - K50
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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n Cng Ngh Lc Du Lc Ha Du - K50
- 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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n Cng Ngh Lc Du Lc Ha Du - K50
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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n Cng Ngh Lc Du Lc Ha Du - K50
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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n Cng Ngh Lc Du Lc Ha Du - K50
- 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 .
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(Cm3/s )
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n Cng Ngh Lc Du Lc Ha Du - K50
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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n Cng Ngh Lc Du Lc Ha Du - K50
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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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.
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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
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