Đồ án chưng cất Lê Trung Hiếu

8/13/2019 n chng ct L Trung Hiu

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1. Tng quan........................................................................................................................... 31.1. Gii thiu v nguyn liu............................................................................................. 31.2. L thuyt v chng ct................................................................................................. 32. Quy trnh cng ngh chng ct hn hp methanol nc............................................. 43. Tnh cn bng vt cht v nng lng.............................................................................. 53.1. Cc thng s ban u................................................................................................... 53.2.

Cn bng vt cht......................................................................................................... 6

3.3. Cn bng nng lng.................................................................................................. 94. Tnh ton thp chng ct................................................................................................. 124.1. ng knh thp........................................................................................................ 124.2. Trlc ca thp......................................................................................................... 144.3. Chiu cao mc cht lng khng bt trong ng chy chuyn.................................. 165. Tnh ton c kh thp chng ct...................................................................................... 175.1. B dy mm................................................................................................................ 175.2. Chiu cao thp mm xuyn l................................................................................... 185.3. B dy thn thp......................................................................................................... 185.4. B dy y v np c g............................................................................................. 19

5.5.

Bch ghp thny v np...................................................................................... 20

5.6. Lp cch nhit............................................................................................................ 205.7. Ca ni ng dn vi thit b....................................................................................... 215.8. Chn thp.............................................................................................................. 235.9. Tai tr eo thp................................................................................................................ 246. Tnh ton thit b ph...................................................................................................... 256.1. Thit b un si y thp............................................................................................ 256.2. Thit b lm ngui sn phm y.............................................................................. 276.3. Thit b ngng t sn phm nh.............................................................................. 306.4. Thit b un si dng nhp liu................................................................................. 326.5. Thit b lm ngui sn phm nh............................................................................. 356.6. Bn cao v................................................................................................................... 376.7. Bm............................................................................................................................. 40Kt lun........................................................................................................................................ 42TI LIU THAM KHO........................................................................................................... 43


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Li m uCng ngh ha hc l mt trong nhng ngnh ng gp rt ln trong s pht trin

ca nn cng nghip hin i nc ta.Trong ngnh sn xut ha cht cng nh s dngsn phm ha hc, nhu cu s dng nguyn liu c tinh khit cao phi ph hp vi quitrnh sn xut hoc nhu cu s dng.

V th, cc phng php nng cao tinh khit lun lun c ci tin v i mi

ngy cng han thin hn, nh l: c c, hp th, chng ct, trch ly, Ty theo ctnh yu cu ca sn phm m ta c s la chn phng php ph hp. i vi hMethanolNc l hai cu t tan ln vo nhau hon ton v c s khc bit nhit si ln,ta phi dng phng php chng ct nng cao tinh khit ca Methanol.

n mn hc Qu trnh v Thit b l mt mn hc mang tnh tng hp qutrnh hc tp cacc k s cng ngh ha thc phm tng lai. Mn hc gip sinh vingii quyt nhim v tnh ton c th v:yu cu cng ngh, kt cu ca mt thit b trongsn xut ha cht - thc phm. y l bc u tin sinh vin vn dng nhng kinthc hc ca nhiu mn hc vo gii quyt nhng vn k thut thc t mt cchtng hp.

Em chn thnh cm n thy Nguyn Ngc Anh Tuncng cc thy c trong bmn Qu trnh & Thit b hng dn em tn tnh trong thi gian qua. Tuy nhin, trongqu trnh hnh thnh n khng th trnh nhng sai st, em rt mong nhn c kinnh gi thy c v ni dung ca n.

Tp HCM, ngy 7 thng 12 nm 2013Sinh vin

L Trung Hiu


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1. Tng quan1.1. Gii thiu vnguyn liu1.1.1. Methanol

Methanol cn gi l methyl alcohol, alcohol g hoc ru mnh g l mt hp chtha hc vi cng thc phn t l CH3OH. y l cht lng khng mu, d bay hi v rtc. nhit phng, methanol l mt cht lng phn cc. Cc thng s ca methanol:

- Phn t lng: 32,04 g/mol.- Khi lng ring: 0,7918 g/cm3.- Nhit nng chy: -97oC (176K).- Nhit si: 64,5oC ( 337,8K).- nht: 0,59 Ns/m2 20oC.

Methanol c dng lm cht chng ng, lm dung mi, lm nhin liu cho ng ct trong, nhng ng dng ln nht l lm nguyn liu sn xut cc ha cht khc.

-Dng methanol trong sn xut nha Urea- formaldehyd v nha phenol-formaldehyd. Nhng cht ny l nguyn liu cho ngnh cng nghip carton th.

1.1.2.Nc

Trong iu kin bnh thng, nc l cht lng khng mu, khng mi, khng vnhng khi nc dy s c mu xanh nht. Tnh cht vt l:

- Khi lng phn t : 18 g / mol- Khi lng ring d4

oC : 1 g / ml

- Nhit nng chy : 00C- Nhit si : 1000C

1.1.3. Hn hp Methanol-ncBng 1.1 Cn bng lng-hi cho hn hp methanol-nc 1 atm[1, 39, 47]

toC 100 92,3 87,7 81,7 78 75,3 73,1 71,2 69,3 67,5 66 64,5

x 0 5 10 20 30 40 50 60 70 80 90 100y 0 26,8 41,8 57,9 66,5 72,9 77,9 82,5 87 91,5 95,8 100 y: x l thnh phn lng- y l thnh phn hi- toC l nhit .

1.2. L thuyt vchng ct1.2.1. Khi nim

Chng ct l qu trnh dng tch cc cu t ca mt hn hp lng (cng nh hnhp kh lng) thnh cc cu t ring bit da vo bay hi khc nhau ca cc cu ttrong hn hp (ngha l khi cng mt nhit , p sut hi bo ha ca cc cu t khcnhau).

Thay v a votrong hn hp mt pha mi to nn s tip xc gia hai pha nh

trong qu trnh hp thu hoc nh kh, trong qu trnh chng ct pha mi c to nnbng s bc hi hoc ngng t.Khi chng ct ta thu c nhiu cu t v thng th h c bao nhiu cu t s thu

c by nhiu sn phm. Nu xt h n gin ch c 2 cu t th ta thu c 2 snphm:

Sn phm nh ch yu gm cu t c bay hi ln v mt phn rt t cc cu t c bay hi b.


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Sn phm y ch yu gm cu t c bay hi b v mt phn rt t cu t c bayhi ln.

Kt lun: Khi chng cth methanol - nc th snphm nh ch yu l methanolv sn phm y ch yu l nc.

1.2.2. Cc phng php chng ct Phn loi theo p sut lm vic:p sut thp, p sut thng, p sut cao.

Phn loi theo nguyn l lm vic: chng ct n gin, chng bng hi nc trc tip,chng ctng kh.

Phn loi theo phng php cp nhit y thp:cp nhit trc tip, cp nhit gin tip.Phn loi theo s cu t c trong h:h hai cu t, h ba cu t hoc t hn mi cu t, h

nhiu cu t ( trn mi cu t).Kt lun:i vi h methanol - nc, chn phng php chng ct lin tc cp nhit

gin tip.1.2.3. Thit b chng ct

Trong sn xut thng dng nhiu loi thit bchng ctkhc nhau.Nu pha kh phntn vo pha lng ta c cc loi thp mm, nu pha lng phn tn vo pha kh ta c thp

chm, thp phun, y ta kho st 2 loi thng dng l thp mm v thp chm.- Thp mm: thn thp hnh tr, thng ng pha trong c gn cc mm ccu to khcnhau, trn pha lng v pha hi c cho tip xc vi nhau. Ty theo cu to ca a, tac:

- Thp mm chp: trn mm b tr c chp dng trn, xupap, ch s- Thp mm xuyn l: trn mm c nhiu l hay rnh

- Thp chm : thp hnh tr, gm nhiu bc ni vi nhau bng mt bch hay hn. Vtchm c cho vo thp theo haiphng php: xp ngu nhin hay xp th t.

2. Quy trnh cng ngh chng ct hn hp methanol ncHn hp methanol - nc c nng nhp liu methanol 12% (theo phn khi lng),

nhit khong 30

0

C ti bnh cha nguyn liu (1) cbm (2) bm ln bn cao v (3).Sau , hn hp c gia nhit n nhit si trong thit b un si dng nhp liu (4 ),ri c a vo thp chng ct (5) a nhp liu.

Trn a nhp liu, cht lng c trn vi phn lng t on luyn ca thp chyxung. Trong thp, hi i t di ln gp cht lng t trn xung. y, c s tip xcv trao i gia hai pha vi nhau. Pha lng chuyn ng trong phn chng cng xungdi cng gim nng cc cu t d bay hi v b pha hi to nn t hi nc ccp trc tip vo y thp li cun cu t d bay hi. Nhit cng ln trn cng thp,nn khi hi i qua cc a t di ln th cu t c nhit si cao l nc s ngng tli, cui cng trn nh thp ta thu c hn hp c cu t methanol chim nhiu nht (cnng 98% phn khi lng). Hiny i vo thit b ngng t (6) v c ngng thon ton. Mt phn ca cht lng ngng t c hon lu v thp a trn cng. Phncn li c lm ngui ti thit b lm ngui sn phm nh (7) ri a v bnh cha sn

phm nh(8).Mt phn cu t c nhit si thp c bc hi, cn li cu t c nhit si cao

trong cht lng ngy cng tng. Cui cng, y thp ta thu c hn hp lng hu ht lcc cu t kh bay hi (nc). Hn hp lng y c nng methanol l 1% phn khilng, cn li l nc. Dung dch lng y i ra khi thp i vo thit b ni un(9), ri


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c a qua thit b lm ngui sn phm y (10) v cui cng i vo bn cha snphm y (11). H thng lm vic lin tc chora sn phm nh l methanol, sn phmy l nc.Ch thch cc k hiu trong qui trnh:

1. Bn cha nguyn liu2. Bm

3. Bn cao v4. Thit b un si nhp liu.

5. Thp chng ct6. Thit b ngng t7. Thit b lm ngui sn phm nh8. Bn cha sn phm nh9. Thit b ni un10. Thit b lm ngui sn phm y

11. Bn cha sn phm y12. Lu lng k

13. By hi14. Bnh phn phi3. Tnh cn bng vt cht v nng lng

3.1.Cc thng sban uChn loi thp chng ct l thp mm xuyn l v hot ng theo ch lin tc.Khi chng ct h methanolnc th cu t d bay hi l methanol v thu nhn snphm nh.Hn hp:

Methanol: CH3OH, Mm= 32 g/mol. Nc: H2O, Mn= 18 g/mol.

Nhp liu vo thp trng thi lng si. Nng sut nhp liu: 1500l/h Nng dng nhp liu: 12% khi lng Nng dng sn phm nh: 98% khi lng Nng dng sn phm y: 1% khi lng Chn Nhit nhp liu ban u: tF= 30

0C. Nhit sn phm nh sau khi lm ngui: tD1= 40

0C Nhit sn phm y sau khi lm ngui: tW1= 50

0C Nhit nc ban u: tn= 30

0C

Nhit nc sau khi lm nguisn phm nh: tn1= 500C. K hiu GF, F: sut lng nhp liu tnh theo kg/h, kmol/h. GD, D: sut lng sn phm nh tnh theo kg/h, kmol/h. GW, W: sut lng sn phm y tnh theo kg/h, kmol/h. Gn: sut lng khi lng ca nc lm lnh, kg/h tn, tn1: nhit vo v ra ca nc lm lnh ti thit b ngng t ,

0C


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tsF: nhit nhp liu lng si,0C

L : sut lng dng hon lu, kmol/h. xi, ix : nng phn mol, phn khi lng ca cu t i.

3.2. Cn bng vt cht3.2.1.Nng phn mol ca methanol trong thp

0712,018/)12,01(32/12,0

32/12,0

/)1(/

/

nFmFmFF

MxMx

Mxx

965,018/)98,01(32/98,0

32/98,0

/)1(/

/

nDmD

mDD

MxMx

Mxx

00565,018/)01,01(32/01,0

32/01,0

/)1(/

/

nWmW

mWW

MxMx

Mxx

3.2.2. Sut lng dng nhp liu, sn phm y v sn phm nh

T s liu ca bng 1.1, xy dng th t-x,y cho h Methnol- nc:

Hnh 3.1. th t, x y h methanol ncT hnh 3.1, ti xF = 0,0712, nhp liu vo thp chng ctnhit l: tF= 90,1

0C.Tra [1, 9, 1.2]khi lng ring ca methanol ti 90,10C: m= 724,89 kg/m

3

Tra [1, 311, 1.249]khi lng ring ca nc ti 90,10C: n= 965,23 kg/m3

Khi lng ring ca nhp liu: 310.077,123,965

12,01

89,724

12,011

n

F

m

F

F

xx

(3.1)

F= 928,3 kg/m3

Khi lng mol ca nhp liu: molgMxMxM nFmFF /997,1818)0712,01(320712,0).1(. Sut lng nhp liu: GF= VF F= 1,5 928,3 = 1392,45 kg/h

Sut lng mol nhp liu: F ==

= 73,3 kmol/h

- Phng trnh cn bng vt cht cho ton thpF = D + WFxF= DxD+ WxW

60

70

80

90

100

0 10 20 30 40 50 60 70 80 90 100

Nhit

(0C)

Phn mol x-y


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- Gii h phng trnh sau:

0712,03,7300565,0965,0

3,73

WD

WD

D = 5,01 kmol/h v W= 68,29 kmol/hKhi lng mol ca sn phm y:

kmolkgMxMxM nwmww /08,1818)00565,01(3200565,0).1(. Sut lngkhi lng sn phm y: Gw= Mw W = 1234,68 kg/hSut lng khilng sn phm nh: GD= GFGw= 1392,451234,68 = 157,77 kg/h

Kt lun:Dng nhp liu Sn phm y Sn phm nh

Sut lng khi lng (kg/h) 1392,45 1234,68 157,77

Sut lng mol (kmol/h) 73,3 68,29 5,01

Phn mol methanol 0,0712 0,00565 0,965

Phn khi lng methanol 0,12 0,01 0,98

3.2.3. Tm t s hon lu v dng hon lu- T bng 3.1, xy dng th cn bng pha ca h Methanol-nc p sut 1atm.

Hnh 3.2. ng cn bng pha h methanol nc- Vi xF= 0,0712, xc nh

*

Fy t hnh 2 c*

Fy = 0,341

T s hon lu ti thiu:

Rmin= *Fy

*

Fy =

= 2,313

T s hon lu:R = 1,3xRmin+ 0,3 = 1,32,313 +0,3 = 3,31

0

10

20

30

40

50

60

70

80

90

100

0 20 40 60 80 100

y(%)

x(%)


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3.2.4. Cc phng trnh lm vic3.2.4.1.Phng trnh lm vic phn ct

y =

= Phng trnh lm vic phn ct: y = 0,768x + 0,224

3.2.4.2.Phng trnh lm vic phn chngSut lng mol tng i ca dng nhp liu

64,1400565,00712,0

00565,0965,0

WF

WD

xx

xxf

Phng trnh lm vic phn chng:

01788,0.165,400565,0131,3

64,141.

131,3

64,1431,3

1

1

1

xxxR

fx

R

fRy W

y = 4,165x0,017883.2.5. Tnh lu lng dng hon lu

R == = 3,31

Sut lng mol dng hon lu: L = 16,58kmol/h.Khi lng mol ca sn phm nh:molgMxMxM nDmDD /51,3118)965,01(32965,0).1(.

Sut lng khi lng dng hon lu: GL= MD L = 31,5116,58 = 522,44 kg/h.3.2.6. Tm s mm thc t

Hnh 3.3. S mm l thuyt ca thp chng ct.

Thp chng ct c 11 mm l thuyt. V tr nh:

- Vi xD= 0,965, xc nh*

Dy t hnh 3.2 c*

Dy = 0,985T hnh 3.1, ti xD= 0,965, sn phm ra khi thp chng ctc nhit l: tD= 65

0C.

bay hi tng i: =*

Dy *Dy =

= 2,382

Ni suy t [1, 95, 1.102] nht ca nc 650C: n= 0,4355 cP


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Dng ton [1, 90, 1.18] nht ca methanol 650C: m= 0,336 cP nht ca sn phm nh: lgD= (1- xD)lg n + xDlg m (3.2)

lgD= lg0,4355 (1 - 0,965) + lg 0,336 0,965=> D= 0,339 cP => D= 0,3392,382 = 0,81

Tra [2, 171, IX.11] Hiu sut trung bnh ti v tr nh: Ed = 51,5% V tr nhp liu :

- Vi xF= 0,0712, xc nh*Fy t th 2 c

*Fy = 0,341

T hnh 3.1, ti xF= 0,0712nhp liu i vo thp chng ctc nhit l: tF= 90,10C.

bay hi tng i: =

*

Fy *

Fy =

= 6,75

Ni suy t [1, 95, 1.102] nht ca nc 90,10C: n= 0,316 cPDng ton t [1, 90, 1.18] nht ca methanol 90,10C: m= 0,265 cP nht ca dng nhp liu theo (3.2): => F= 0,312 cP

=> F= 0,312 6,75 = 2,106Tra [2, 171, IX.11] Hiu sut trung bnh ti v tr nhp liu: EF = 41,5% V tr y:

- Vi xW= 0,00565, xc nh*

Wy t hnh 3.2 c*

Wy = 0,0368T hnh 3.1, ti xW= 0,00565, sn phm y i ra thp chng ctc nhit l:tw= 99

0C.

bay hi tng i: =

*

Wy *

Wy =

= 6,724

Ni suy t [1, 95, 1.102] nht ca nc 990C: n= 0,2868 cPNi suy t [1, 90, 1.18] nht ca methanol 990C: m= 0,243 cP nht ca sn phm y theo (3.2): => w= 0,287 cP

w= 0,287 x 6,724 = 1,927Tra [2, 171, IX.11] Hiu sut trung bnh ti v tr y: Ed = 42,5%.

Hiu sut trung bnh: = = = 45,17 S mm thc t:

S mm thc t:Ntt= =

= 24,35.

S mm phn ct:NtL=

= 13S mm phn chng:NtC=

= 11

Kt lun:Thp chng ct h methanol nc c 24 mm v 1 ni un.

3.3. Cn bng nng lng3.3.1.Nhit lng ti thit b ngng tD D DQ = G (1 + R) r , (J/h) (IX.159, [2]) (3.3)

Ni suy t [1, 257, 1.213]Nhit ha hi ca methanol ti 650C: rm= 1119,97 kJ/kg.Tra t [1, 312, 1.250] Nhit ha hi ca nc ti 650C: rn= 2345,2 kJ/kg.

Nhit ha hi ca sn phm nhti 650C: rD = rm + (1 - )rnrD= 0,981119,97 + (10,98)2345,2 = 1144,5 kJ/kg.

Nhit lng ti thit b ngng t:


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Qnt= GD(R + 1)rD = 157,77(3,31 + 1)1144,5= 7,78105kJ/h. Sut lng nc lm lnh cn dng:

Cn bng nhit: Qnt= (R + 1)GDrD= GnCn(tn1tn)Ni suy t [1, 165, 1.147]Nhit dung ring nclnh ti 400C (= :

Cn= 0,99869 Kcal/kg.0C = 4,181 kJ/kg.0C.

Gn= = = 9304 kg/h3.3.2.Nhit lng trao i thit b lm ngui sn phm nh.

QlnD= GDCD(tD1tD) = , (J/h) (IX.166, [2]) (3.4)[1, 165, 1.147]Nhit dung ring ca nc ti 52,50C ( t =

):

Cn= 0,99939 Kcal/kg.0C = 4,184kJ/kg.0C

[1, 172, 1.154]Nhitdung ring ca methanol ti52,50C: Cm= 2726,25 J/kg.0C

Nhit dung ring ca sn phm nh: CD= Cm + (1 - )Cn (3.5)CD= 0,982,72625 + (10,98)4,184 = 2,755 kJ//kg.0C

Nhit lng trao i nhit ti thit b lm ngui sn phm nh:

QlnD= GDCD(tD1tD) = 157,772,755(6540) = 10868kJ/h. Sut lng nc lm lnh cn dng ti thit b lm nguisn phm nh:[1, 165, 1.147]Nhit dung ring ca nc ti 400C ( t =

):

Cn= 0,99869 Kcal/kg.0C = 4,181 kJ/kg.0C

GnlnD=

=

= 129,97 kg/h3.3.3.Nhit lng ti thitb un si nhp liu.

Cn bng nhit lng ca thit b un si nhp liu:QD1+ Qf = QF+ Qng1+ Qxq1, (J/h) (IX.149, [2]) (3.6)

QD1: nhit lng do hi t mang vo- Qf: nhit lng do nhp liu mang vo- QF: nhit

lng do nhp liu mang ra- Qng1: nhit lng do nc ngng mang ra- Qxq1: nhit lngmt ra mi trng xung quanh ly bng 5%nhit tiu tn Nhit lng do hi t mang vo QD1: QD1= D1(r1+ C1t1) (IX.150, [2]) (3.7)

D1: lng hi t kg/hr1: nhit ha hi ca nc 2at, 2208kJ/kg.C1: nhit dung ring ca nc ngng, 1,014 (kcal/kg) = 4,245(kJ/kg)t1: nhit nc ngng 2at, t1= 119,62

0C [1, 166, 1.148]. QD1= D1(2208 +4,245 119,62) = 2715,8 D1 (kJ/h).

Nhit lng do nhp liu mang vo Qf: Qf= GFCftF, (J/h) (IX.151. [2]) (3.8)[1, 172, 1.154]Nhit dung ring methanol ca nhp liu ti 300C: Cm= 2,620 kJ/kg.

0C

[1, 165, 1.249] Nhit dung ring ca nc ti 300

C:Cn= 0,99866 Kcal/kg.0C = 4,181 kJ/kg.0C.

Nhit dung ring ca nhp liu(3.5):Cf= 0,122,620 + (10,12)4,181 = 3,994 kJ//kg.0CNhit lng nhp liu ti thit b gia nhit:

Qf= GFCftF= 1392,453,99430= 1,67105kJ/h. Nhit lng do nhp liu mang ra:QF= GFCF tsF, (J/h) (IX.152, [2]) (3.9)

[1, 172, 1.154]Nhit dung ring methanol ca nhp liu ti 90,10C: Cm= 2,913 kJ/kg.0C

[1, 165, 1.249] Nhit dung ring ca nc ti 90,10C:


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Cn= 1,00504 Kcal/kg.0C = 4,208 kJ/kg.0C.

Nhit dung ring ca nhp liu(3.5):CF= 0,122,913+(10,12)4,208 = 4,053 kJ//kg.0CNhit lng nhp liu ti thit b gia nhit:

QF= GFCFtsF= 1392,454,05390,1= 5,08105kJ/h. Nhit lng do nc ngng mang ra Qng1:Qng1= D1 C1 t1, (IX.153, [2]) (3.10)

Qng1= D1

4,245

119,62 = 507,8

D1kJ/h.

Nhit lng mt ra mi trng xung quanh:Qxq1= 0,05 D1r1= 0,05 D12208 = 110,4 D1kJ/h.

Lng hi t cn thit un nng dng nhp liu:2715,8 D1+ 1,67105= 5,08105+ 507,8 D1+ 110,4 D1 => D1= 163 kg/h.

3.3.4.Nhit lng trao i thit b lm ngui snphmy

[1, 165, 1.249]Nhit dung ring ca nc ti 720C ( t =

:Cn= 1,0016 Kcal/kg.

0C = 4,193 kJ/kg.0C.[1, 172, 1.154]Nhit dung ring methanol ca sn phm y720C: Cm= 2,82 kJ/kg.

0CNhit dung ring ca sn phm y:Cw= 0,012,82 + (10,01)4,193= 4,179 kJ//kg.0CNhit lng trao i ti thit b lm ngui sn phm y:Qw= GwCw(tw2tw1) = 1234,684,179(9945) = 2,79105kJ/h.

Sut lng nc lm lnh cn dng ti thit b lm nguisn phm y:

[1, 165, 1.147]Nhit dung ring ca nc ti 400C ( t =

):Cn= 0,99869 Kcal/kg.

0C = 4,181 kJ/kg.0C

GnlnD=

=

= 3336,52 kg/h3.3.5.Nhit lng cung cp cho y thp

Tng lng nhit mang vo thp bng tng nhit lng mang ra:QF,ra + QD2 + QR = Qnt+ QW,ra + Qxq + Qng2, (IX.156, [2]) (3.11)

QF,ra: nhit lng nhp liu vo thp - QD2: nhit lng do hi t mang vo thp- QR:nhit lng do dng hon lu mang vo thp. - Qy: nhit lng do hi mang ra nhthp. - Qw,ra: nhit lng do sn phm y mang ra.- Qxq: nhit lng tn tht bng 5%nhit tiu tn y thp.- Qng2: nhit lng do nc ngng mang ra. - D2: lng hi tcn thit un si dung dch trong thp.

Nhit lng do dng hon lu mang vo thp[1, 172, 1.154]Nhit dung ring methanol ca dng hon lu 650C: Cm= 2,785 kJ/kg.

0C[1, 165, 1.249]Nhit dung ring ca nc ti 650C:

Cn= 1,00065 Kcal/kg.0C = 4,19 kJ/kg.0C.

Nhit dung ring ca dng hon lu:CL= 0,982,785+ (10,98)4,19 = 2,813 kJ//kg.0CQR= GLCLtR= 522,442,81365 = 0,95510

5

kJ/h. Nhit lng do sn phm y mang ra:

Qw,ra= GwCw,ratw, (J/h) (IX.160, [2]) (3.12)[1, 172, 1.154]Nhit dung ring methanol ca sn phm y990C: Cm= 2,960 kJ/kg.

0C[1, 165, 1.249]Nhit dungring ca nc ti 990C:

Cn= 1,00734 Kcal/kg.0C =4,218 kJ/kg.0C.

Nhit dung ring ca sn phm: Cw,ra= 0,012,960 + (10,01)4,218 = 4,205 kJ//kg.0CQw,ra= GwCw,ratw= 1234,684,20599 = 5,14 105kJ/h.


12/43

12

Nhit lng do nc ngng mang ra:Qng2= D2C1t2, (J/h), (IX.161, [2]) (3.13) Qng2= D24,245 119,62 = 507,8D2kJ/h.

Nhit lng tn tht ra mi trng xung quanh ly bng 5%nhit tiu tn y thp:Qxq= 0,05D2r1, (J/h), (IX.162, [2]) (3.14)

Qxq= 0,05D2r1= 0,05D2 2208 = 110,4D2 kJ/h. Nhit lng do hi hi t mang vo thp:

QD2= D2(r1+ C2t2), (J/h), (IX. 157, [2]) (3.15)QD2= D2(r1+ C2t2) = D2(2208 + 4,245119,62) = 2715,8D2 kJ/h. Lng hi cn thit un si dung dch y thp:

5,08 105 + 2715,8D2+ 0,955105= 7,78105 + 5,14 105+ 507,8D2+ 110,4D2 D2= 328,23 kg/h.

4. Tnh ton thp chng ct4.1. ng knh thp4.1.1. Phn ct

4.1.1.1. Khi lng ring trung bnh ca pha lng trong phn ct:Phn mol trungbnh ca pha lng trong phn ct: xL=

=

= 0,518

T hnh 3.1, xc nh nhit si ca pha lng trong phn ct: tL= 72,70C.

Phn khi lng trung bnh ca pha lng trong phn ct: = = = 0,55[1, 9, 1.2]Khi lng ring methanol ti 72,70C: m= 743,3 kg/m

3[1, 311, 1.249]Khi lng ring nc ti 72,70C: n= 976,18 kg/m

3Khi lng ring trung bnh pha lng trong phn cttheo (3.1):L = 832,69 kg/m

34.1.1.2.Khi lng ring trung bnh ca pha hi trong phn ct:

Phn mol trung bnh ca pha hi trong phn ct : yL= 0,768xL + 0,224 yL= 0,768 0,518 + 0,224 = 0,622

Nhit trung bnh ca pha hi trong phn ct: thL= 800C

Khi lng mol trung bnh ca pha hi trong phn ct:MhL= yLMm+ (1yL)Mn= 0,62232 + (10,622)18 = 26,708 kg/kmolKhi lng ring trung bnh pha hi trong phn ct:

3/922,0

)27380(273

4,22

708,261mkg

RT

PM

hL

hL

hL

Chn khong cch 2 mm h = 300 mm

[3, 256, 6.2] => C = 0,03 Vn tc pha hi trong thp:

smChL

L

/902,0922,0

69,832

03,0

( 6.18, [3]) (4.1)Lu lng pha hi i trong phn chng ca thp:

sm

TM

tRGQ

oD

hLD

V /174,0360027351,31

)27380(4,22)31,31(77,157

3600

4,22)1( 3

ng knh thp chng ct


13/43

13

mQ

D Vt 496,0902,0785,0

174,0

.785,0

(6.16, [3]) (4.2)

4.1.2. Phn chng4.1.2.1. Khi lng ring trung bnh ca pha lng trong phn chng:

Phn mol trung bnh ca pha lng trong phn chng:

xC= = = 0,0384T hnh 3.1, xc nh nhit si ca pha lng trong phn chng: tc= 93,80C. Phn khi lng trung bnh ca pha lng trong phn chng:

= =

= 0,065[1, 9, 1.2]Khi lng ring methanol ti 93,80C: m= 720,82 kg/m

3[1, 311, 1.249]Khi lng ring nc ti 93,80C: n= 962,64 kg/m

3Khi lng ring trung bnh pha lng trong phn chngtheo (3.1): c = 942,1 kg/m

34.1.2.2. Khi lng ring trung bnh ca pha hi trong phn chng:

Phn mol trung bnh ca pha hitrong phn chng: yC= 4,165xC0,01788 yC= 4,1650,03840,01788 = 0,142 Nhit trung bnh ca pha hi trong phn chng: thC= 960C

Khi lng mol trung bnh ca pha hi trong phn chng:MhC= yCMm+ (1yC)Mn= 0,14232 + (10,142)18 = 19,988 kg/kmol

Khi lng ring trung bnh pha hi trong phn chng:

3/66,0

)27396(273

4,22

988,191mkg

RT

PM

hC

hChC

Chn khong cch 2 mm h = 300 mm[3, 256, 6.2] => C = 0,03

Vn tc pha hi trong thp:

smChCv

c /13,166,0

1,94203,0

( 6.18, [3]) Lu lng pha hi i trong phn chngca thp:

smTM

tRGQ

oD

hCD

V /182,0360027351,31

)27396(4,22)31,31(77,157

3600

4,22)1( 3

ng knh thp chng ct

mQ

D Vt 453,013,1785,0

182,0

.785,0

(6.16, [3])

Chn ng knh thp l Dt= 500 mm. Vn tc pha hi trong thp theo thc t:

smD

Q

C

VC /927,0

5,014,3

182,04

..14,3

.422

smD

Q

L

V

L /887,05,014,3

174,04

..14,3

.422


14/43

14

4.2.Trlc ca thpCu to mm lChn thp mm xuyn l c ng chy chuyn vi:

Tit din t do bng 8% din tch mm. ng knh l dl= 4 mm = 0,004 m Chiu cao g chy trn: hg= 25mm = 0,025 m

Din tch ca 2 bn nguyt bng 20% din tch mm. L b tr theo hnh lc gic u. Khong cch gia 2 tm l bng 8 mm. Mm c lm bng thp khng g X18H10T. S l trn 1 mm:

N =22

004,0

5,0.08,008,0

%8

l

t

l

m

d

D

S

S= 1250 l

Gi a l s hnh lc gic.p dng cng thc [2, 48, V.139]:N = 3a(a-1) +1 => a = 20,91 21, N = 1261 l

S l trn ng cho: b = 2a - 1 = 41 l.4.2.1. Tr lc ca mm kh PkChn h s tr lc i vi tit din t do ca l bng 8% din tch mm: = 1,82.

4.2.1.1. Phn ctVn tc hi qua l: 088,11

08,0

887,0

%8' LL

m/s

p dng cng thc:

PkL=

= 1,82

= 103,15 N/m2. (IX.140, [2]) (4.3)4.2.1.2. Phn chng

Vn tc hi qua l: 588,1108,0927,0

%8' C

C

m/s

PkC=

= 1,82

= 80,65 N/m2(IX.140, [2])

4.2.2. Tr lc do sc cng b mt Ps4.2.2.1. Phn ct

[1, 301, 1.242]Sc cng b mt methanol 72,70C: m= 0,0182 N/m.[1, 311, 1.249]Sc cng b mt nc 72,70C: n= 0,6391 N/m.p dng cng thc :

=

+

=

+

= 56,5098 ([I.76, [1]) (4.4) hhL= 0,0177 N/m.Mm c ng knh l ln hn 1mm:

PsL=

=

= 13,61 N//m2. (IX.142, [2])

4.2.2.2. Phn chng[1, 301, 1.242]Sc cng b mt methanol 93,80C: m= 0,0163 N/m.[1, 311, 1.249]Sc cng b mt nc 93,80C: n= 0,6006 N/m.p dng cng thc theo (4.4): hhC= 0,0159 N/m.


15/43

15

Mm c ng knh l ln hn 1mm:

PsC=

=

= 12,23 N//m2.(IX.142, [2])

4.2.3. Tr lc thy tnh do cht lng trn mm to raTr lc thy lc do lp cht lng bt trn mm:Pb= 1,3hbkbgChiu cao lp cht bt: hb= hg+ h

h chiu cao lp cht lng trn g chy trn: h = vi Qv,l: lu lng cht lng, m2/s; P: chu vi ca g; k = b/L: t sgia khi lng cht lng bt v khi lng ring ca cht lng, ly gn bng 0,5.

Chiu di ca g chy trn: Squt- S= Sbn nguyt

22

2

%20

2cos

2sin

2

1.2

2RRR

R

- sin= 0,2 => = 1,627 rad = 93,32o

Nn Lg= mDt 364,0)2

32,93sin(.5,0)

2sin(.

4.2.3.1. Phn ctLu lng cht lng trong phn ctca thp:Qv,l=

=

= 1,4010-4m3/s.Vi MmL= 0,51832 + (10,518)18 = 25,252 kg/kmol.khi lng mol trung bnhcht lng trong phn luyn.

h = =

= 5,5710-3, m.Chiu cao lp cht lng bt trn mm:hb= hg+ h = 0,025 + 5,5710-3= 0,03057 m.Tr lc qua lp cht lng bt: Pb= 1,3

hb

k

L

g

PbL= 1,30,030570,59,81 = 162,32N/m2.4.2.3.2. Phn chng

Lu lng cht lng trong phn chngca thp:

Qv,c=

=

Vi MmC= 0,0384 32 + (10,0384)18 = 18,54 kg/kmol - khi lng mol trung bnhcht lng trong phn chng.


16/43

16

Qv,c = 4,2810-4m3/s.h = =

= 0,012 m.

Chiu cao lp cht lng bt trn mm:hb= hg+ h = 0,025 + 0,012 = 0,037 m.Tr lc qua lp cht lng bt:Pbc= 1,3hbkCg

Pbc= 1,3

0,037

0,5

942,1

9,81 = 222,27 N/m2.

Kt lun:Tng tr lc thy lc ca 1 mm trong phn ctca thp:PL= PkL+ PsL+ PbL= 103,15+ 13,61 + 162,32 = 279,08N/m

2. (IX.135, [2]) (4.5)Tng tr lc thy lc ca 1 mm trong phn chng ca thp:

PC= PkC+ PsC+ PbC= 80,65+ 12,23 + 222,27= 315,15 N/m2. (IX.135, [2])

Tng tr lc thy lc ca thp:P = NtL.PL+ NtCPC= 13279,08 + 11315,15= 7094,69 (N/m2)

Kim tra li khong cch mm h = 300 mm m bo u kin cho thp hot ng bnhthng:

h > 1,8

vi cc mm trong phn chng tr lc thy lc qua mt mm ln hn tr lc thy lc quamt mm trong phn luyn:

== 0,061m (h = 0,3 m > 0,061 m)

Kt qu l iu kin c tha.Kim tra tnh ng nht ca hot ng mm. Tnh vn tc ti thiu qua l ca pha hivl,min cho cc mm hot ng bnh thng:

vl,min= 0,67 = 0,67

= 11,304 m/s < 11,588 m/sKt lun:Mm hot ng bnh thng

4.3. Chiu cao mc cht lng khng bt trong ng chy chuynB qua s to bt trong ng chy chuyn, chiu cao mc cht lng trong ng chy chuynca mm xuyn l c xc nh theo biu thc: hd= hg+ hb+ P + hd

mm cht lng4.3.1. Tn tht thy lc do dng chy t ng chy chuyn vo mm

Sd: tit din gia ng chy chuyn v mm.

Sd= 0,8 Smm= 0,84

0,52= 0,16 m2 thp khng b ngp lt khi hot ng th: hd

2

1 h= 150 mm (5.11, [4])

4.3.1.1. Phnctp dng cng thc:hdL

= 0,128(

= 0,128

= 1,2710-4mm cht lng(5.10, [4])(4.6)PbL=

=

= 0,0342 m cht lng = 34,2 mm cht lng

Chiu cao mc cht lng khng bt trong n chy chuyn phn luyn:=> hdL= 25 + 30,57 + 34,2+1,2710-4=89,77 mm cht lng < 150 mm


17/43

17

4.3.1.2. Phn chngp dng cng thc (4.6): hdC

= 1,1910-3mm cht lngPbc=

=

= 0,0341 m cht lng = 34,1 mm cht lng

Chiu cao mc cht lng khng bt trong n chy chuyn phn chng:=> hdC= 25 + 37+ 34,1 + 1,19

10-3=96,1 mm cht lng < 150 mm.

Kt lun:Khi thp hot ng th cc mm trong thp u khng b ngp lt5. Tnh ton c kh thp chng ct

5.1.Bdy mm5.1.1. Cc thng s cn tra v chn phc v cho qu trnh tnh ton

Nhit tnh ton: t = tmax= 100 (oC)

p sut tnh ton: P = Pthy tnh+ PgChn b dy g chy trn l 3mm.Th tch ca g chy trn: V = 0,3640,0250,003= 2,7310-5m3[2, 313, XII.7] Khi lng ring ca thp X18H10T l: X18H10T= 7900 kg/m

3 Khi lng g chy trn:m = VX18H10T= 2,7310-57900 = 0,2157 kgp sut do g chy trn tc dng ln mm trn

78,10

4

5,014,3

81,92157,0

4

22

t

gD

mgP

N/m2

Khi lng ring ca cht lng ti sn phm y thp:VixW= 0,00565 suy ra tW= 99

0C[1, 311, 1.249]Khi lng ring ca nc 99oC: n= 959,08 kg/m

3[1, 9, 1.2] Khi lng ring ca metanol 99oC: m= 715,1 kg/m

3[1, 5, 1.2] Khi lng ring ca sn phm ytheo (3.1): LW= 955,82kg/m

3p sut thy tnh:Pthy tnh= LW g(hg + hlC) = 955,829,81(0,025 + 0,012) = 346,934 N/m2 P = 346,934 + 10,78 + = 357,714 N/m2

H s b sung do n mn ha hc ca mi trng:Thi gian s dng thit b l trong 20 nm: Ca= 1 mm

ng sut cho php tiu chun:V vt liu l X18H10T: []*= 146,67 N/mm2

H s hiu chnh:= 1ng sut cho php: [] = []*= 146,67 N/mm2

Mun n hi[6, 45, 2.12]:E = 2.105N/cm2H s Poisson[2, 313, XII.7]: = 0,33

H s iu chnh:b=10

410 t

dt = 0,6

5.1.2. Tnh b dyi vi bn trn c ngm kp cht theo chu vi:

ng sut cc i vng chu vi:2

max16

3

S

DP

(6.36, [7]) (5.1)


18/43

18

i vi bn c c l: ][16

3 2

maxmax

S

DP

bb

l

S 437,06,067,14616

10357,7143500

][16

3 6

b

t

PD

mm

Nn: S + Ca= 1,437 mm => Chn S = 2 mmKim tra iu kin bn:

vng cc i tm:T

oD

PRW

64

4

(6.35, [7]) (5.2)

i vi bn c c l:Tbb

olo

D

PRWW

64

4

Vi:)1(12 2

3

ESDT

3

24

3

24 )1(.

16

3

64

)1(12

ES

PR

ES

PRWW

bbb

olo

m bo iu kin bn th: Wlo< S

243,022000006,0

)33,01(25010357,714163

3

246

loW < 2S 1

B dy S chn tha iu kin.Kt lun:Smm= 2 mm

5.2.Chiu cao thp mm xuyn lChiu cao thp: H = Ntt (H+ Smm m (IX.54, [2]) (5.3)

H = 24 ( 0,3 + 0,002) + 0,8 = 8,048 m

Chn y (np) tiu chun c 25,0t

t

D

hsuy ra ht= 0,25Dt= 0,250,5 = 0,125 m

Vi htl chiu cao ca y (np) li ra- Chn chiu cao g: hg= 25 mm = 0,025 m- Chiu cao y (np): Hn= ht + hg= 0,125 + 0,025 = 0,15 m

Kt lun: Chiu cao ton thp: Htt = H + 2Hn= 8,348 m m5.3. Bdy thn thp

V thp hot ng p sut thng nn ta thit k thn hnh tr bng phng php hn h quang in, kiu hn gip mi 1pha. Thn thp c ghp vi nhau bng cc mi ghp

bch. m bo cht lng ca sn phm ta chn thit b thn thp l thp khng g mX18H10T.

5.3.1. Cc thng s cn tra v chn phc v cho qu trnh tnh tonNhit tnh ton: tmax= tnc= 100

0C

[2, 310, XII.4]ng sut ko ca X18H10T ng vi b dy (1-3 mm):k1= 540106N/m2.[2, 310,XII.4]ng sut ko ca X18H10T ng vi b dy (4-25mm):k2= 550106N/m2.[2, 310, XII.4] ng sut chy ca X18H10T: c= 220106N/m2.[2, 356, XIII.2] H s iu chnh: ; v thp chng ct khng b t nng v ccch ly vi ngun t nng trc tip.[2, 356, XIII.3] H s an ton bn: nk= 2,6; nc= 1,5Cs b sung do n mn,bo mn v dung sai theo chiu dy: C = 1mm tnh thi gianlm vic l 20 nm


19/43

19

5.3.2. Tnh tonb dy thn thpGii hn bn cc ng sut:

[k1] =

m

2

[k2] =

m2

[c] = m2

- h s bn ca thnh thp theo phng dc: = h = 0,95.p sut mi trng Pmi trng= 1 at = 10

5N/m2

Khi lng ring trung bnh ca pha lng:

tL =

=

= 887,4 kg/m3

p sut lm vic ca thp chng ct:P = tLgHtt + P + Pmi trng = 887,49,818,348+ 7094,69 + 105= 0,18106N/m2.

V

=

0,95 = 774,092 > 50, do b qua ilng P mu s ca cng

thc:

S =

+ C =

+ 0,001 = 1,3210-3 m (XIII.8, [2]) (5.4)Chn thp X18H10T theo tiu chun [2, 364, XIII.9]: S =2mmKim tra ng sut ca thnh theo p sut th (dng nc). Ly p sut th pth= 1,5P = 0,181061,5 < 0,2 106p sut th tnh ton p0:p0= pth + tLgHtt+ P = 0,2 106+ 887,49,818,348 + 7094,69= 0,352106N/m2 =

=

= 92,82106 <

= 183,31065.4. Bdy y v np c g

Chiu dy S c xc nh theo cng thc sau:

S =

+ C, m (XIII.47, [2]) (5.5)

Trong ht-chiu cao phn li ca y(np),m;hh s bn ca mi hn hng tmkh s khng th nguyn, i vi vi y c l c tng cng hon ton th k = 1

Bi v =

= 774,09> 30 nn i lng P mu s c thb qua. Chiu dy np c tnh theo cng thc:


20/43

20

S =

+ C =

+ 0,001=> S =1,3410-3m = 1,34 mmSC = 1,151 = 0,34 => thm 2mm vo S. => S =3,34 mmChn chiu dy theo tiu chun 4 mm.Kim tra ng sut thnh ca np thit b theo p sut th thy lc:

= = = 27,18106


21/43

21

Nhit lng tn tht ra mi trng xung quanh:Qm= 0,05.Q= 328,23 110,4= 36236,592 (kJ/h) = 10065,72 (W)

Nhit ti mt mt ring:

qm = va

a2v1v

a

a

tb

m t.)tt.(f

Q

(W/m2)

Trong :tv1 : nhit ca lp cch nhit tip xc vi b mt ngi ca thp.tv2 : nhit ca lp cch nhit tip xc vi khng kh.tv: hiu s nhit gia hai b mt ca lp cch nhit. an tan ta ly tv= tmax= ty - tkkChn tkk= 30

oC tv= tmax= 99,830 = 69,8 (K)ftb: din tch b mt trung bnh ca thp (k c lp cch nhit), m

2.

ftb= DtbH = HSDD

HDD athnttnt

2

22

2

= (Dt + Sthn + a)H

Ta c phng trnh:

8,69151,0

348,8)004,05,0(

72,10065

aa

aa

1

)504,0(

415,36

36,415a= 0,504 + a

a= 0,01423 (m) = 14,23 (mm)Kt lun: chn a= 15 (mm).Th tch vt liu cch nhit cn dng:V = (Dt + 2Sthn + a).a.H = .(0,5 + 20,004 + 0,015)0,0158,348 = 0,206 (m

3).5.7. Ca ni ng dn vi thit b5.7.1. ng nhp liu

Khi lng ring ca hn hp: F= 928,3 kg/m3Chn loi ng ni cm su vo thit b.Chn vn tc cht lng trong ng ni l vF= 1 m/sng knh trong ca ng ni:

Dy= 023,01928,33600

45,13924

3600

.4

FF

F

v

Gm

Chn ng theo tiu chun [2, 454, XIII.32]c Dy= 25 mm[2, 434, XIII.32]=> Chiu di on ng ni l = 90 mm

[2, 409, XII.26]=>Cc thng s ca bch ng vi P = 0,18106N/m2l:Dy Dn D Dd Dl h

Bu lngdb Z

(mm) (ci)25 32 100 75 60 12 M10 4

5.7.2. ng hi nh thpChn vn tc hi ra khi nh thp l vHD= 100 m/s-Nng trung bnh ca pha hi nh thp: yD= xD= 0,965

=>Nhit trung bnh ca pha hi nh thp: THD= 65,8oC


22/43

22

- Khi lng ring trung bnh ca pha hi trong phn luyn:

3/133,1

)2738,65(273

4,22

51,311mkg

RT

PM

HD

DhD

ng knh trong ca ng ni:

Dy= 046,010014,3133,136001)(3,31157,774

3600

)1(.4

HDhD

D

v

RG

m

=> Chn ng c Dy= 50 mm[2, 434, XIII.32]=> Chiu di on ng ni l = 100 mm[2, 409, XIII.26]=>Cc thng s ca bch ng vi P = 0,18106N/m2l:

Dy Dn D Dd Dl hBu lng

db Z(mm) (ci)

50 57 140 110 90 12 M12 45.7.3. ng hon lu

Chn loi ng ni cm su vo thit b.Chn vn tc cht lng trong ng ni l vLD= 1 m/s[1, 311, 1.249]Khi lng ring ca nc 65oC: n= 980,5 kg/m

3[1, 9, 1.2]Khi lng ring ca metanol 65oC: m= 751,0 kg/m

3[1, 5, 1.2] Khi lng ring ca dng hon lu

310.33,15,980

98,01

0,751

98,011

n

D

m

D

L

xx

L= 754,53 kg/m3

ng knh trong ca ng ni:

Dy= 0156,01754,533600522,444

3600

.4

LDL

L

v

G

m

Chn ng c Dy= 20 mm[2, 434, XIII.32]=> Chiu di on ng ni l = 80 mm[2, 409, XIII.26]=>Cc thng s ca bch ng vi P = 0,18106N/m2l:


db Z(mm) (ci)

20 25 90 65 50 12 M10 45.7.4. ng hi y thp

-Nng trung bnh ca pha hi y thp: yW= xW= 0,00565=>Nhit trung bnh ca pha hi y thp: THW= 99,8oC

- Khi lng ring trung bnh ca pha hi trong phn chng:

3/591,0

)2738,99(273

4,22

08,181mkg

RT

PM

HW

WhW

Chn vn tc hi vo y thp l vHW= 120 m/sng knh trong ca ng ni:


23/43

23

Dy= 078,0120.591,03600

1234,684

3600

.4

HWhW

W

v

Gm

Chn ng c Dy= 80 mm[2, 434, XIII.32]=> Chiu di on ng ni l = 110 mm[2, 409, XIII.26]=>Cc thng s ca bch ng vi P = 0,18106(N/mm2):

Dy Dn D Dd Dl h Bu lngdb Z(mm) (ci)

80 89 185 150 128 14 M16 45.7.5. ng dn lng ra khi y thp

Chn vn tc cht lng trong ng ni l vLW= 0,5 m/sng knh trong ca ng ni:

Dy= 03,05,0955,823600

68,12344

3600

.4

LWLW

LW

v

Gm

=> Chn ng c Dy= 32 mm

[2, 434, XIII.32]=> Chiu di on ng ni l = 90 mm[2, 409, XIII.26]=>Cc thng s ca bch ng vi P = 0,18106N/m2l:


db Z(mm) (ci)

32 38 120 90 70 12 M12 45.8. Chn thp

5.8.1. Tnh trng lng c tonthp[2, 313, XII.7]Khi lng ring ca thp CT3 l: CT3= 7850 (kg/m

3)Khi lng ca mt bch ghp thn:

mbch ghp thn= 785002,05,063,044

22

3

22

CTt hDD = 18,11 (kg)

Khi lng ca dung dch trong thp:mdd= 0,4.(D

2t/4).Htt .tL= 0,40,5

28,348/4887,4 = 581,82 (kg)Khi lng ca mt mm:

mmm= THXmmt SD 10182

%)10%8%100(4

=4

0,520,0020,827900 = 2,54 (kg)

Khi lng ca thn thp:

mthn=4

.(D2ngD2t).Hthn. X18H10T= 7900348,85,0508,0

4

22 = 417,69 (kg)

Khi lng ca y (np) thp:my(np)= Sb mt .y . X18H10T = 0,310,004 7900 = 9,796 (kg)Khi lng ring lp cch nhit [3, 416, 28]: cch nhit= 600kg/m

3Khi lng lp cch nhit:mcch nhit= cch nhit V = 600 0,206 = 123,6 (kg)Khi lng ng nhp liu:mng nhp liu= (Dn

2/4Dt2/4) .l.X1810T+ (D

2/4Dn2/4).h.CT3

= (0,0322/40,0252/4). 0,09.7900 + (0,12/40,0322/4).0,012.7850 = 0,887 (kg)


24/43

24

Khi lng ng hi nh thp:mng hi nh thp = (Dn

2/4Dt2/4) .l.X1810T+ (D

2/4Dn2/4).h.CT3

= (0,0572/40,052/4). 0,1.7900 + (0,142/40,0572/4).0,012.7850 = 1,67 (kg)Khi lng ng hon lu:mng hon lu = (Dn

2/4Dt2/4) .l.X1810T+ (D

2/4Dn2/4).h.CT3

= (0,0252/40,022/4). 0,08.7900 + (0,092/40,0252/4).0,012.7850 = 0,665 (kg)

Khi lng ng hi y thp:mng hi y thp = (Dn

2/4Dt2/4) .l.X1810T+ (D

2/4Dn2/4).h.CT3

= (0,0892/40,082/4). 0,11.7900 + (0,1852/40,0892/4).0,014.7850 = 3,31 (kg)Khi lng ng lng y thp:mng lng y thp = (Dn

2/4Dt2/4) .l.X1810T+ (D

2/4Dn2/4).h.CT3

= (0,0382/40,0322/4). 0,09.7900 + (0,1202/40,0382/4).0,012.7850 = 1,19 (kg)Khi lng ca tan thp:m = 10 mbch ghp thn + 24 mmm + mthn + 2 my(np)mdd+ 10mbch thn+ mbch ng = 1392,4 (kg)

5.8.2. Tnh chn thpChn chn : thp c trn bn chn.

Vt liu lm chn thp l thp CT3.Theo aythiet b

Truc

thiet

b

Ti trng cho php trn mt chn: Gc=

4

81,91392,4

44

mgP = 3414,86 (N)

m bo an tan cho thit b, ta chn: Gc = 0,5.104(N)

[2, 437, XIII.35]chn chn ccc thng s sau:L B B1 B2 H h s l d

160 110 135 195 240 145 10 55 235.9. Tai tr eo thp

Chn tai treo: tai treo c gn trn thn thp gi cho thp khi b dao ng trong iukin ngoi cnh.Chn vt liu lm tai treo l thp CT3.Ta chn bn tai treo, ti trng cho php trn mt tai treo: Gt= Gc= 0,5.10

4 (N).

[2, 438, XIII.36]chn tai treo ccc thng s sau:


25/43

25

L B B1 H S l a d100 75 85 155 6 40 15 18

[2, 439, XIII.37]Chn tm lt tai treo bng thp CT3 c cc thng s sau:Chiu dy tm lt

(H)B (mm) SH(mm)

260 140 6Th tch mt tm lt tai treo: Vtm lt = 260140610-9= 2,18410-4(m3)Khi lng mt tm lt tai treo: mtm lt = Vtm lt. CT3 = 2,18410

-4 7850 = 1,71(kg)6. Tnh ton thit b ph

6.1.Thit bun si y thpChn thit b un si y thp l ni un Kettle. ng truyn nhit c lm bng thpX18H10T, kch thc ng 38 x 3. ng knh ngoi: dn= 38 mm = 0,038 m. B dy ng:t = 3 mm = 0,003 m. ng knh trong: dt= 0,032 m.

Nhit sn phm y khi vo ni un (lng): 990C v sau khi ra ni un vo thp(hi): 99,80C.

Nhit hi bo ha 2at vo ni un [1, 314, 1.251]: 119,60C6.1.1. chnh lch nhit trung bnh logarith

tlog=

=

=20,2

6.1.2. H s truyn nhit, K

21

11

1

tr

K

1: h s cp nhit ca sn phm y2: h s cp nhit ca hi t 2 at

rt: nhit tr cathnh ng vlp cu.B dy thnh ng: t = 0,003 m[2, 313, XII.7]H s dn nhit ca thp X18H10T: t= 16,3 W/mK[3, 419, 31]Nhit tr lp bn trong ng: r1= 1/5800 m

2.K/W[3, 419, 31]Nhit tr lp cu ngoi ng: r2 =1/5800 m

2.K/WNn: rt= 5,289.10

-4 m2.K/W6.1.3. Xc nh h s cp nhit ca sn phm y

p dng cng thc:

1= 7,77 . 10-2.

0,37

s

0.11745,0

0,775,00,3330,033

h

h

T.c.

q...

r.

(V.89, [2]) (6.1)

Nhit trung bnh ca sn phm y:ts=

=

= 99,4 (0C)

Khi lng ring trung bnh pha hi sn phm y:3

/592,0

)2734,99(273

4,22

08,181mkg

RT

PM

hC

hChC

[1, 310, 1.249]Khi lng ring ca nc: n= 958,81 kg/m

3.


26/43

26

[1, 9, 1.2]Khi lng ring ca methanol: m= 714,66 kg/m3.

Khi lng ring ca sn phm y theo (3.1)=> = 955,55 kg/m3.[1, 310, 1.249] Sc cng b mt ca nc: n= 0,5897 N/m[1,301,1.242] Sc cng b mt ca methanol: m= 0,01576 N/mSc cng b mt sn phm y theo (4.4): = 0,0153N/m.[1, 310, 1.249]H s dn nhit ca nc: n= 0,68188 W/mK

[1, 134, 1.130]H s dn nhit ca methanol: m= 0,19909 W/mK = )).(1.(.72,0)1.(. mnWWWnWm xxxx = 0,674 W/mK (6.3)

[1, 95, 1.102] nht ca nc: n= 2,85610-4N.s/m2[8, 16, 9] nht ca methanol: m= 2,415.10

-4N.s/m2p dng cng thc(3.2): = 2,853.10-4 N.s/m2[1, 310, 1.249]Nhit dung ring ca nc: Cn= 4219,52 J/kgK[1, 172, 1.154]Nhit dung ring ca methanol: Cm= 2961,85 J/kgK=> Nhit dung ring ca sn phm ytheo : C = Cm Wx + Cn. (1 - Wx ) = 4206.94 J/kgK[1, 312, 1.250]Nhit ha hi ca nc:rn= 2261,56 kJ/kg

[8, 38, 45]Nhit ha hi ca methanol:rm= 1015,35 kJ/kg=> Nhit ha hi ca sn phm y:r = rm Wx + rn.(1 - Wx ) = 2249,1 kJ/kg6.1.4. Xc nh h s cp nhit ca hi t trong ng

p dng cng thc:

4

ngW1n

32

n

2).dt-.(t

.g..r725,0

n

nn

( 3.65, [3]) (6.4)

Dng php lp:chn tW1= 118,26oC

Nhit trung bnh ca mng nc ngng t: tm=

2

tt W1n 118,93oC

Ti nhit ny th:- Khi lng ring ca nc: n= 943,9453 kg/m3

- nht ca nc: n= 2,34568.10-4N.s/m2

- H s dn nhit ca nc: n= 0,685786 W/mKNn: 2=20334,76W/m

2Kqn= 2(tntW1) = 27248,58W/m

2qt= qn= 27248,58 W/m

2 (xem nhit ti mt mt l khng ng k)tw2= tw1 - qtrt= 103,85

oC1= 6126,025 W/m

2K (vi q = qt)qS= 1(tW2tS) = 27249,96W/m

2

Kim tra sai s:

=n

Sn

q

qq 100% = 5,05. 10-3% < 5% (tha)

Kt lun: tw1 = 118,26oC v tw2 = 103,85

oC


27/43

27

6.1.5. Xc nh h s truyn nhit

025,6126

110.289,5

76,20334

1

1

4

K = 1348,95 W/m2K

6.1.6. B mt truyn nhitB mt truyn nhit c xc nh theo phng trnh truyn nhit:

F =2,02.95,1348.3600

1000.23,328.8,2715

. log

tK

Q = 9,087 m2

6.1.7. Cu to thit bChn chiu di ng l 1,5m. ng c b tr theo hnh lc gic u.

S ng truyn nhit: n=

2

tn ddL

F

= 61 (ng).

6.2. Thit blm ngui sn phm yChn thit b lm ngui sn phm y l thit b truyn nhit ng lng ng.

ng truyn nhit c lm bng thp X18H10T: Kch thc ng trong: 38 x 3 Kch thc ng ngoi: 57 x 3

Chn:Nc lm lnh i trong ng trong vi nhit vo tV= 30

oC v nhit ra tR= 50oC.

Sn phm y i trong ng ngoi vi nhit vo tWS= 99oC v nhit ra tWR= 45

oC.6.2.1. chnh lch nhit trung bnh logarith

tlog=

=

= 28,7

6.2.2.Nhit ti qua thnh ng v lp cu:

21

t

tt rrr

B dy thnh ng: t = 0,003 m[2, 313, XII.7]H s dn nhit ca thp X18H10T: t= 16,3 W/mK[3, 419, 31]Nhit tr lp bn trong ng: r1= 1/5800 m


2.K/WNn: rt= 5,289.10

-4 m2.K/W6.2.3. Xc nh h s cp nhit ca nc

Tnh h sRe

Re =

v: vn tc nc lm ngui, m/s

: nht ng lc ca nc lm ngui, m2/s.d: ng knh ng nh, m.

Nhit trung bnh dng nc trong ng: = 0,5(30 + 50) = 40 (0C)[1, 310, 1.249] Khi lng ring ca nc 400C: = 992,2 kg/m3

[1, 310, 1.249] nht ng lc ca nc 400C: = 0,659 10-6m2/s[1, 310, 1.249] H s dn nhit ca nc: = 0,634W/m.K


28/43

28

Chun s Prn= 3,54 Vn tc dng nc lm ngui:

v = =

= 1,16 m/s.

=> Ren= =

= 56398,4> 10

4=> ch chy ri

p dngcng thc xc nh chun s Nusselt: 25,0

2w

n43,0

n

8,0

nlnPr

Pr.PrRe..021,0Nu

( 3.27, [4]) (6.5)

Trong : 1h s tnh n nh hng ca h s cp nhit theo t l gia chiu di L vng knh d ca ng.chn 1= 1 [3, 110, 3.1]

H s cp nhit ca nc trong ng: 2=tr

nn

d

.Nu

(6.6)

6.2.4. Xc nh h s cp nhit ca dng sn phm y ngoi ngp dng cng thc cho sn phm y chy ngoi ng nh

25,0

1

36,06,0

Pr

Pr.PrRe..28,0

w

w

wwlwNu

(6.14, [9]) (6.7)

Trong : 1h s tnh n nh hng ca h s cp nhit theo t l gia chiu di L vng knh d ca ng.chn 1= 1 [3, 110, 3.1]

H s cp nhit ca dng sn phm y ngoi ng: 1=t

W

d

.Nu

Nhit trung bnh ca dng sn phm y ngoi ng: tW= (tWS+ tWR) = 72 (oC).

Ti nhit ny th:

Khi lng ring ca nc: n= 976,6 (kg/m

3

) Khi lng ring ca methanol: m= 744 (kg/m3)

= 973,56 (kg/m3)[1,310, 1.249] nht ng hc ca nc: n= 0,39110-3N.s/m2[8, 16, 9] nht ca methanol: m= 3,144.10

-4N.s/m2 nht ca sn phm y 720C theo (3.2) = 3,905.10-4 N.s/m2

H s dn nhit ca nc: n= 0,669 (W/mK) H s dn nhit ca methanol: m= 0,203 (W/mK)

Nn: = m.xW+ n.(1 - xW)0,72 xW.(1 - xW)(n- m) = 0,204 (W/mK) Nhit dung ring ca nc: cn= 4188,6 (J/kgK)

Nhit dung ring ca methanol: cm= 2820 (J/kgK)Nn: c = cm Wx + cn. (1 - Wx ) = 4174,91 (J/kgK)ng knh tng ng

Dtd=

= Dd (6.8)D: ng knh ng lnD: ng knh ng nh


29/43

29

=> Dtd= 0,0510,038 = 0,013 m.- Vn tc dng sn phm y:

)038,0051,0(56,9733600

68,12344

)(3600

42222

dD

Gv W = 0,388 m/s

Chun s Re

410.905,356,973013,0388,0Re

t

w

vd

= 12575,25> 104=> ch chy ri

204,0

10905,391,4174Pr

4

cw = 7,99


Ti nhit ny th: nht ca nc: n= 4,4696.10

-4N.s/m2 nht ca methanol: m= 3,4124.10

-4N.s/m2p dng cng thc (3.2)W1= 4,463.10

-4N.s/m2 H s dn nhit ca nc: n= 0,66156 W/mK

[1, 134, 1.130]H s dn nhit ca methanol: m= 0,20494 W/mKNn: ))(1.(72,0)1(. mnwwwnwmw xxxx = 0,65374 W/mK

Nhit dung ring ca nc: cn= 4183,0 J/kgK[1, 172, 1.154]Nhit dung ring ca methanol: cm= 2776 J/kgK

Nn: cw= cm wx + cn. (1 - wx ) = 4168,93 J/kgK

=>1

11

1Prw

www

c

= 2,85

Nn Nuw= 220,6531= 3462,56 W/m

2K

qw= 1(twtw1) = 30470,517W/m2qt= qw= 30470,517W/m

2(xem nhit ti mt mt l khng ng k)tw2= tw1 - qtrt= 47,08

0CPrw2= 3,765Nun= 225,22352= 4462,241 W/m

2Kqn= 2(tw2) = 31592,663W/m2

Kim tra sai s:

=W

nW

q

qq 100% = 3,683% < 5% (tha)

Kt lun: tw1 = 63,2oC v tw2 = 47,08

oC6.2.5. Xc nh h s truyn nhit

241,4462

110.289,5

56,3462

1

1

4

K = 960 W/m2K



30/43

30

F =960.28,7.3600

1000.102,79

.

5

log

tK

Q= 2,813 m2

6.2.7. Cu to thit bChn s ng truynnhit: n = 19ng. ng c b tr theo hnh lc gic u.

- Chiu di ng truyn nhit: L =

2tn ddn

F

= 1,35 m chn L = 1,5 m

[2, 48, VII]S ng trn ng cho: b = 5ng- Bc ng: t = 1,2dn= 0,0456 m- ng knh trong ca thit b (p dng [2, 49, V.140]):D = t.(b-1) +4dn= 0,3344 m

6.3. Thit bngng tsn phm nhChn thit b ngng t v ng loi TH, t nm ngang.ng truyn nhit c lm bng thp X18H10T, kch thc ng 38 x 3: ng knh ngoi: dn= 38 (mm) = 0,038 (m) B dy ng: t = 3 (mm) = 0,003 (m)

ng knh trong: dtr= 0,032 (m)Chn:Nc lm lnh i trong ng vi nhit vo tV= 30


Dng hi ti nh i ngoi ng vi nhit ngng t tngng= 65oC v nhit hi l

65,80C6.3.1. chnh lch nhit trung bnh logarith

tlog=

=

= 24,14

6.3.2. Xc nh h s cp nhit ca nc i trong ngNhit trung bnh ca dng nc trong ng: tf= (tV+ tR) = 40 (

oC).

Ti nhit ny th: Khi lng ring ca nc: n= 992,2 (kg/m

3) nht ca nc: n= 0,659 10-6(m2/s) H s dn nhit ca nc: n= 0,634 (W/mK) Chun s Prandtl: Prn= 4,31

[2, 48, V.II]chn n = 127 (ng)Vn tc thc t ca nc trong ng:

22 032,01272,9923600

93044

3600

4

trn

n

ndn

Gv = 0,03 (m/s).

Chun s Reynolds :

6

.

10.659,0

032,003,0Re

n

trn

n

dv

= 1456,753< 2320: ch chy tng

[3, 110, 3.27]cng thc xc nh chun s Nusselt:25,0

2

43,01,033,0

Pr

Pr.Pr..Re..15,0

w

ln GrNu

(6.9)

[3, 110, 3.1]chn 1= 1


31/43

31

Chun s Grashof

2

32

.

tlgGr

= h s dn n th tchH s dn nhit ca nc: n= 3,87.10

-41/Kt = chnh lch nhit gia thnh ng v dng nc

t = tw2tf, oC

H s cp nhit ca nc i trong ng trong: n=tr

nn

d

.Nu

6.3.3.Nhit ti qua thnh ng v lp cu

t

2w1wt

r

ttq

, (W/m2).

Trong : tw1: nhit ca vch tip xc vi hi ngng t,

oC tw2: nhit ca vch tip xc vi nc lnh,

oC

21

t

tt rrr

B dy thnh ng: t = 0,003 (m) H s dn nhit ca thp khng g: t= 16,3 (W/mK) Nhit tr lp bn trong ng: r1= 1/5000 (m

2.K/W) Nhit tr lp cu ngoi ng: r2 =1/5800 (m

2.K/W)Nn: rt= 5,565.10

-4(m2.K/W)6.3.4. Xc nh h s cp nhit ca hi ngng t ngoi ng

iu kin:- Ngng t hi bo ha.- Khng cha khng kh khng ngng.- Hi ngng t mt ngoi ng.- Mng cht ngng t chy tng.- ng nm ngang.

[3, 120, 3.65]i vi ng n chic nm ngang th:

4

nW1ngng

32

1).dt-.(t

.g.r.725,0

(6.10)

[2, 48, V.II]vi s ng n = 127 th s ng trn ng cho ca hnh 6 cnh l: b = 13[2, 30, V.20]h s ph thuc vo cch b tr ng v s ng trong mi dy thng ng

l tb= 0,6H s cp nhit trung bnh ca chm ng: ngng= tb1= 0,421Dng php lp: chn tW1= 62.35 (

oC)Nhit trung bnh ca mng cht ngng t: tm= (tngng+ tW1) = 64,175 (

oC)Ti nhit ny th: Khi lng ring ca nc: n= 980,95 (kg/m

3) [1, 9, 1.2]Khi lng ring ca methanol: m= 751,83 (kg/m

3)p dng cng thc (3.1)= 755,36 (kg/m3)


32/43

32

nht ca nc: n= 4,3994.10-4(N.s/m2)

nht ca methanol: m= 3,383.10-4(N.s/m2)

p dng cng thc (3.2)= 3,414.10-4(N.s/m2) H s dn nhit ca nc: n= 0,66234 (W/mK) [1, 134, 1.130]H s dn nhit ca methanol: m= 0,2048 (W/mK)

Nn: = m

.xD+

n.(1 - x

D)0,72 x

D.(1 - x

D)(

n-

m) = 0,2097 (W/mK)

Nhit ngng t ca dng hi: r = rD= 2208 (kJ/kg)Nn: 1= 5501,20201 (W/m

2K)ngng= 2310,50485 (W/m

2K)qngng= ngng (tngngtW1) = 6122,83784(W/m

2)qt= qngng= 6122,83784(W/m

2) (xem nhit ti mt mt l khng ng k)tw2= tw1 - qtrt= 58,94 (

oC)Gr = 5409817Prw2= 2,879Nun= 16,21665

n= 321,29227 (W/m2

K)qn= n (tW2tf) = 6086,124(W/m2)

Kim tra sai s:

=ngng

nngng

q

qq 100% = 0,6% < 5% (tha)

Kt lun: tw1 = 62,35oC v tw2 = 58,94


29227,321

110.565,5

50485,2310

1

1

4

K = 243,8 (W/m2K)


F =14,248,2433600

1000107,78

.

5

log

tK

Qnt = 36,72 (m2)

6.3.7. Cu to thit bS ng truyn nhit: n = 127 (ng). ng c b tr theo hnh lc gic u.

Chiu di ng truyn nhit: L =

2

ddn

F

trn= 2,63 (m) chn L = 3(m)

S ng trn ng cho: b = 13(ng)Tra bng trang 21, [3] Bc ng: t = 0,0456(m)[2, 49, V.140]ng knh trong ca thit b: D = t(b-1) + 4dn= 0,7 (m)

6.4. Thit bun si dng nhp l iuChn thit b ngng t v ng, t ngang.ng truyn nhit c lm bng thp hp kim X18H10T, kch thc ng 38 x 3:


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33

ng knh ngoi: dn= 38 mm = 0,038 m, b dy ng: dt = 3 mm = 0,003 m, ng knhtrong: dtr= 0,032 m, nhit tr lp bn trong ng: r1= 1/5000 m

2.K/W, nhit tr lp cungoi ng: r2 =1/5800 m

2.K/WChn:

Nhp liu i trong ng vi nhit vo tV= 30,0oC v nhit ratR= 90,1

oC.Dng hi ngng t i ngoi ng vi nhit ngng t t ngng= 119,6

oC, c p sut 2at, n

nhit ha hi r = 2280kJ/kg.6.4.1. Hiu s nhit trung bnh

Chn kiu truyn nhit ngc chiu, nn:

1,54

1,906,119

306,119

)1,906,119()306,119(log

Ln

t

6.4.2.Nhit ti qua thnh ng v lpcuB dy thnh ng: t = 0,003 mH s dn nhit ca thp khng g: t= 16,3 W/mK

Nhit tr lp cu trong ng: r1= 1/5800 m2.K/W

Nhit tr lp cu ngoi ng: r2 =1/5800 m2.K/WNn: rt= 5,289.10

-4m2.K/W6.4.3. Xc nh h s cp nhit ca dng nhp liui trong ng

Nhit trung bnh ca dng nhp liu trong ng: tf= (tV+ tR) = 60,05oC

Ti nhit ny th:Khi lng ring ca nc: n= 983,17kg/m

3[1, 9, 1.2]Khi lng ring ca methanol: m= 755,95kg/m

3p dng cng thc (3.1)=>= 948,94 kg/m3

nht ca nc: n= 4,696.10-4N.s/m2

nht ca methanol: m= 3,5.10-4N.s/m2

p dng cng thc (3.2) => = 4,6.10-4N.s/m2 H s dn nhit ca nc: n= 0,65896 W/mK

[1, 134, 1.130]H s dn nhit ca methanol: m= 0,20549 W/mKNn: = )).(1.(.72,0)1.(. mnFFFnFm xxxx = 0,57 W/mK

Nhit dung ring ca nc: cn= 4183 J/kgK[1, 172, 1.154]Nhit dung ring ca methanol: cm= 2760,25 J/kgK

Nn: c = cm.F+ cn. (1F) = 4012,27 J/kgK[2, 12, V.35]=>

cPr = 3,24

Chn thit b c s ng l 127 ng

Vn tc thc t ca dng nhp liu trong ng:

004,0032,0..127.94,948.3600

1392,45.4

3600

422

trF

FF

dn

Gv m/s

Chun s Reynolds :

05,26410.6,4

94,948.032,0.004,0.Re

4

.

F

FtrFF

dv

< 2320 : ch chy tng

p dngcng thc xc nh chun s Nusselt(6.9)


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34

chn e1= 16.4.4. Xc nh h s cp nhit ca hingng t ngoi ng

p dng cng thc (6.10)cho hi ngng ng nm ngangDng php lp: chn tW1= 119,06 (

oC)Nhit trung bnh ca mng nc ngng t: tm= (tn+ tW1) = 119,33 (

oC)Ti nhit ny th:

Khi lng ring ca nc: n= 943,63 (kg/m3) nht ca nc: n= 2,385.10

-4(N.s/m2) H s dn nhit ca nc: n= 0,685866 (W/mK)

Nn: 1= 24345,51222 (W/m2K)

qn= 1(tntW1) = 13146,5766(W/m2)

qt= qn= 13146,5766(W/m2) (xem nhit ti mt mt l khng ng k)

tw2= tw1 - qtrt= 111,74 (oC)

Ti nhit ny th: Khi lng ring ca nc: 949,63 Khi lng ring ca methanol: 699,912 = 910,64 nht ca nc: n= 2,552.10

-4(N.s/m2) nht ca methanol: m= 2,2239.10

-4(N.s/m2)p dng cng thc (3.2)W2= 2,53.10

-4(N.s/m2) H s dn nhit ca nc: n= 0,684348 (W/mK)

[1, 134, 1.130]H s dn nhit ca methanol: m= 0,1978 (W/mK)Nn: W2= n.xF+ m.(1 - xF)0,72 xF.(1 - xF)(n- m) = 0,589 (W/mK)

Nhit dung ring ca nc: cn= 4235,96 (J/kgK)[1, 172, 1.154]Nhit dung ring ca methanol: cm= 3023,7 (J/kgK)

Nn: cW2

= cn Fx + c

m. (1 -

F

x ) = 4090,49 (J/kgK)

=>2W

2W2W2W

cPr

= 1,76

Nn: NuF= 13,88F= 247,24(W/m

2K)qF= F(tW2tf) = 12870,523(W/m

2)Kim tra sai s:

=n

Fn

q

qq 100% = 2,85% < 5% (tha)

Kt lun: tw1

= 119,06oC v tw2

= 111,74oC6.4.5. Xc nh h s truyn nhit

24,247

110.289,5

51222,24345

1

1

4

K = 216,7 (W/m2K)



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35

F =1,54311,02613600

10002208163

. log

tK

Q= 8,53 (m2)

6.4.7. Cu to thit b

Chiu di ng truyn nhit: L =

2

ddn

F

trn= 0,61 (m) chn L = 1(m)

S ng trn ng cho: b = 13 (ng)[2, 49, V.141]Bc ng: t = 0,0456 (m)[2, 49, V.140]ng knh trong ca thit b: D = t(b-1) + 4dn= 0,7 (m)

6.5.Thit blm ngui sn phm nhChnthit b lm ngui sn phm nhl thit b truyn nhit ng lng ng.ng truyn nhit c lm bng thp X18H10T: kch thc ng trong: 38 x 3 v kchthc ng ngoi: 57 x 3.Chn:Nc lm lnh i trong ng trong vi nhit vo tV= 30


Sn phm nhi trong ng ngoi vi nhit vo tDS

= 65oC v nhit ra tDR

= 40oC.6.5.1. chnh lch nhit trung bnh logarith

tlog=

=

= 12,33

6.5.2.Nhit ti qua thnh ng v lp cuB dy thnh ng: t = 0,003 m[2, 313, XII.7] H s dn nhit ca thp X18H10T: t= 16,3 W/mK[3, 419, 31]Nhit tr lp bn trong ng: r1= 1/5800 m


2.K/WNn: rt= 5,289.10

-4 m2.K/W

6.5.3. Xc nh h s cp nhit cancNhit trung bnh nc chy trong thit b: = 0,5(30 + 50) = 40 (0C)[1, 310, 1.249] Khi lng ring ca nc 400C: = 992,2 kg/m3

[1, 310, 1.249] nht ng lc ca nc 400C: = 0,659 10-6m2/sH s dn nhit ca nc: = 0,634Chun s Prn= 3,54 Vn tc dng nc lm ngui:

v = =

= 0,045 m/s.

=> Ren=

=

= 2196,93 < 2320 => ch chy tng

p dngcng thc xc nh chun s Nusselt(6.9), chn 1= 1 [3, 110, 3.1]

H s cp nhit ca nc trong ng: n=tr

nn

d

.Nu

6.5.4. Xc nh h s cp nhit ca dng sn phm nhngoi ngNhit trung bnh ca dng sn phm nhngoi ng: tD= (tDS+ tDR) = 52,5 (

oC).Ti nhit ny th:

Khi lng ring ca nc: n= 986,85 (kg/m3)


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36

[1, 9, 1.2]Khi lng ring ca methanol: m= 762,75 (kg/m3)

p dng cng thc (3.1)= 766,23 (kg/m3)[1,310, 1.249] nht ng hc ca nc: n= 0,52810-3(N.s/m2)[8, 16, 9] nht ca methanol: m= 3,8475.10

-4(N.s/m2) nht ca sn phm nh 52,50C (3.2): = 3,89.10-4 N.s/m2 H s dn nhit ca nc: n= 0,65 (W/mK)

[1, 134, 1.130]H s dn nhit ca methanol: m= 0,206 (W/mK)Nn: = n.xD+ m.(1 - xD)0,72 xD.(1 - xD)(n - m) = 0,209 (W/mK)Nhit dung ring ca nc: cn= 4178 (J/kgK)[1, 172, 1.154]Nhit dung ring ca methanol: cm= 2726,25 (J/kgK)

Nn: c = cm Dx + cn. (1 - Dx ) = 2755,285 (J/kgK)ng knh tng ng (6.8)=> Dtd= 0,0510,038 = 0,013 m. Vn tc dng sn phm nh:

)038,0051,0(23,7663600

157,774

)(3600

42222

dD

Gv D = 0,063 m/s

Chun s Re

410.89,3

23,766013,0063,0Re

tD

vd= 1613,22< 2320 => ch chy tng

cD Pr = 5,13

p dng cng thc cho sn phm nhchy ngoi ng nh (6.9)chn 1= 1 [3, 110, 3.1]

H scp nhit ca dng sn phm nhngoi ng: D=t

D

d

Nu .


Ti nhit ny th: nht ca nc: n= 5,86.10

-4N.s/m2 nht ca methanol: m= 4,165.10

-4N.s/m2W1= 4,22.10

-4N.s/m2 H s dn nhit ca nc: n= 0,6425 W/mK

[1, 134, 1.130]H s dn nhit ca methanol: m= 0,207 W/mKNn: ))(1.(72,0)1(.1 mnDDDnDmw xxxx = 0,2096 W/mK

Nhit dung ring ca nc: cn= 4178,0 J/kgK[1, 172, 1.154]Nhit dung ring ca methanol: cm= 2697,95 J/kgK

Nn: cw1= cm Dx + cn. (1 - Dx ) = 2727,55 J/kgK=>

1

11

1Prw

www

c

= 5,49

Nn NuD= 14.37D= 225,50685 W/m

2Kqw= D(tDtw1) = 1418,4381W/m

2qt= qw= 1418,4381W/m

2(xem nhit ti mt mt l khng ng k)


37/43

37

tw2= tw1 - qtrt= 45,460C

Prw2= 3,832Nun= 13,359184n= 264,67884 W/m

2Kqn= n(tw2) = 1445,0904W/m2

Kim tra sai s:

=W

nW

q

qq 100% = 1,88% < 5% (tha)

Kt lun: tw1 = 46,21oC v tw2 = 45,46


67884,264

110.289,5

50685,225

1

1

4

K = 114,4 W/m2K


F =3992,2.12,3.3600

1000.10868. log

tKQ = 0,247 m2

6.5.7. Cu to thit bChn s ng truynnhit: n = 7ng. ng c b tr theo hnh lc gic u.

- Chiu di ng truyn nhit: L =

2

tn ddn

F

= 0,32 m chn L = 0,5m

[2, 48, VII]S ng trn ng cho: b = 3ng- Bc ng: t = 1,2dn= 0,0456 m- ng knh trong ca thit b : D = t.(b-1) +4dn= 0,2432 m

6.6.Bn cao v6.6.1. Tn tht ng ng dn

Chn ng dn c ng knh trong l dtr= 25 mm[1, 381, II.15]=> nhm ca ng: e = 0,2 mm = 0,0002 mTn tht ng ng dn:

g

v

d

lh F

2.

2

1

1

111

m

Trong :1: h s ma st trong ng ng.

l1: chiu di ng ng dn, chn l1= 40 md1: ng knh ng dn, d1= dtr= 0,025m1: tng h s tn tht cc b.vF: vn tc dng nhp liu trong ng dn

6.6.2. Xc nh vn tc dng nhp liu trong ng dn

Nhit trung bnhdng nhp liu: tF=2

1,9030

2

FSFV tt = 60,05oC

Ti nhit ny th:


38/43

38

Khi lng ring ca nc: n= 983,17 kg/m3

Khi lng ring ca methanol: m= 755,95 kg/m3

=>F= 948,94kg/m3

nht ca nc: n= 4,688.10-4N.s/m2

nht ca methanol: m= 3,510.10-4N.s/m2

=>F= 3,58.10-4N.s/m2

Vn tc ca dng nhp liu i trong ng:

22 025,0..94,948.3600

1392,45.4

3600

4

trF

F

Fd

Gv = 0,83m/s

6.6.3. Xc nh h s ma st trong ng ngChun s Reynolds :

410.58,3

94,948.025,0.83,0Re

F

FtrF

F

dv

= 55025,52 > 4000 : ch chy ri

Chun s Reynolds ti hn: Regh= 6(d1/e)8/7= 1494,93

Chun s Reynolds khi bt u xut hin vng nhm:Ren= 220.(d1/e)9/8= 50285

V ReF> Ren=> ch chy ri ng vi khu vc qu . [1, 379, II.13]h s = 0,0366.6.4. Xc nh tng h s tn tht cc b

Ch un cong :[1, 382, II.6]Chn dng ng un cong 90oc bn knh R vi R/d = 2 th u1(1 ch)= 0,15.ng ng c 4 ch un =>u1= 0,15. 4 = 0,6 Van :Chn van cu vi m hon ton th van(1 ci)= 4.ng ng c 2 van cu =>van= 4. 2 = 8 Lu lng k : l1= 0 (coi nh khng ng k). Ti ming ra ca bn cao v[1, 385, 10]: = 1 nn: 1= 9,6

Vy:81,9.2

83,0.6,9

025,0

40.036,0

2

1

h = 2,36 (m)

6.6.5. Tn tht ng ng dn trong thit b un si dng nhp liu

g

v

d

lh

2.

2

2

2

2

2

22

,m

Trong :h2: h s ma st trong ng ng.l2: chiu di ng ng dn, l2= 1m.d2: ng knh ng dn, d3= dtr= 0,032m.

2: tng h s tn tht cc b.v2: vn tc dng nhp liu trong ng dn

6.6.6. Vn tc dng nhp liu trong ng dn

Nhit trung bnh dng nhp liu: tF=2

1,9030

2

FSFV tt = 60,05oC

Ti nhit ny th: Khi lng ring ca nc: n= 983,17 kg/m

3 Khi lng ring ca ru: m= 755,95 kg/m

3


39/43

39

=>F= 948,94kg/m3

nht ca nc: n= 4,688.10-4N.s/m2

nht ca ru: m= 3,510.10-4N.s/m2

=>F= 3,58.10-4N.s/m2

Vn tc ca dng nhp liu i trong ng:

22 032,0.127.94,948.3600

1392,45.4

3600

4

trF

F

F dn

G

v = 0,004m/s6.6.7. Xc nh h s ma st trong ng ng

Chun s Reynolds : ReF= 339,286 ch chy tng

p dng cng thc [1, 377, II.58]: 2=FRe

64= 0,189

6.6.8. Xc nh tng h s tn tht cc bt thu:

[1, 388, II.16]Khi2

2

1 032,0

025,0

F

Fo = 0,610 th t thu 3 (1ch) = 0,463

C 1 ch t thu =>t thu 3 = 0,463t m:

[1, 382, II.16]Khi2

2

1 025,0

032,0

F

Fo = 0,61 th xt m 3 (1ch) = 0,507

C 1 ch t m =>t m 3 = 0,507 nn: 2= t thu 3 + t m 3 = 0,97

Vy: 42

2 10.62,781,9.2

004,0.97,0

032,0

1.198,0.127

h m

6.6.9. Chiu cao bn cao vChn :Mt ct (1-1) l mt thong cht lng trong bn cao v.Mt ct (2-2) l mt ct ti v tr nhp liu thp.Ap dng phng trnh Bernoulli cho (1-1) v (2-2):

z1 +g

P

F.

1

+

g

v

.2

2

1 = z2 +g

P

F.

2

+

g

v

.2

2

2 +hf1-2

Trong :z1: cao mt thong (1-1) so vi mt t, hay xem nh l chiu cao bn cao v Hcv= z1.z2: cao mt thong (2-2) so vi mt t, hay xem nh l chiu cao t mt t n v trnhp liu:z2= hchn + hnhp liu = 0,120 + 3,85 = 3,97P1: p sut ti mt thong (1-1), chn P1= 1 at = 9,81.10

4N/m2P2: p sut ti mt thong (2-2)Xem P = P2P1= nttL .PL = 13.279,08 = 3628,04 N/m

2


40/43

40

v1 : vn tc ti mt thong (1-1), xem v1 = 0 m/sv2 : vn tc ti v tr nhp liu, v2 = vF= 0,08 m/shf1-2: tng tn tht trong ng t (1-1) n (2-2):hf1-2= h1 + h2 = 2,3608 mVy:Chiu cao bn cao v:

Hcv= z2+ gvv

g

PP

F .2.

2

1

2

212

+hf1-2 = 3,97 + 81,9.2083,0

94,948.81,9

3628,04 2

+ 2,3608 = 6,76 m

Chn Hcv= 7 m6.7. Bm6.7.1.Nng sut

Nhit dng nhp liu l tF= 30oC.

Ti nhit ny th:Tra [1, 9, 1.2]Khi lng ring ca methanol ti 300C: m= 783 kg/m

3

Tra [1, 311, 1.249]Khi lng ring ca nc ti 300C: n= 995,7 kg/m3

Khi lng ring ca nhp liu: F= 964,27 kg/m3

nht ca nc: n= 8,007.10-4N.s/m2

nht ca methanol: m= 0,51.10-3N.s/m2=> F= 7,75.10

-4N.s/m2Sut lng th tch ca dng nhp liu i trong ng:

27,964

1392,45

F

FF

GQ

= 1,44 m3/h

Vy: chn bm c nng sut Qb= 2 m3/h

6.7.2. Ct pChn :Mt ct (1-1) l mt thong cht lng trong bn cha nguyn liu.Mt ct (2-2) l mt thong cht lng trong bn cao v.p dng phng trnh Bernoulli cho (1-1) v (2-2):

z1 +g

P

F.

1

+

g

v

.2

2

1 + Hb = z2 +g

P

F.

2

+

g

v

.2

2

2 +hf1-2

Trong :z1: cao mt thong (1-1) so vi mt t, chn z1= 1m.z2: cao mt thong (2-2) so vi mt t, z2= Hcv= 7m.P1: p sut ti mt thong (1-1), chn P1= 1 at.P2: p sut ti mt thong (2-2), chn P2= 1 at.v1,v2 : vn tc ti mt thong (1-1) v(2-2), xem v1= v2 = 0m/s

hf1-2 : tng tn tht trong ng t (1-1) n (2-2).Hb: ct p ca bm.6.7.3. Tnh tng tr lc trong ng

Chn ng knh trong ca ng ht v ng y bng nhau: dtr= 50 mm[1, 381, II.15]=> nhm ca ng: e = 0,2 mm = 0,0002Tng tr lc trong ng ht v ng y

hf1-2=g

v

d

ll Fh

tr

h

2.

2


41/43

41

Trong :lh: chiu di ng ht.Chiu cao ht ca bm[1, 441, II.34]=> hh= 4 mChn lh= 6 ml: chiu di ng y, chn l= 10 mh : tng tn tht cc b trong ng ht.

: tng tn tht cc b trong ng y.: h s ma st trong ng ht v ng y.vF: vn tc dng nhp liu trong ng ht v ng y m/s

283,0050,0..3600

2.4

3600

422

tr

b

Fd

Qv m/s

6.7.4. Xc nh h s ma st trong ng ht v ng yChun s Reynolds :

410.75,7

27,964.05,0.283,0Re

F

FtrF

F

dv

= 17605,7> 4000 : ch chy ri

Chun s Reynolds ti hn: Regh= 6(dtr/e)

8/7

= 3301,065Chun s Reynolds khi bt u xut hin vng nhm:Ren= 220(dtr/e)

9/8= 109674,381V Regh< ReF< Ren=> ch chy ri ng vi khu vc qu .

p dng cng thc [1, 379, II.64]: =25,0

Re

100.46,1.1,0

Ftrd

= 0,033

6.7.5.Xc nh tng tn tht cc b trong ng htCh un cong:

[1, 382, II.16]Chn dng ng un cong 90oc bn knh R vi R/d = 2 th u1(1 ch)= 0,15.ng ht c 2 ch un =>u1= 0,3

Van:Chn van cu vi m hon ton th v1(1 ci)= 4.ng ht c 1 van cu =>v1= 4 nn: h= u1 + v1 = 4,3

6.7.6.Xc nh tng tn tht cc b trong ng y Ch un cong:

[1, 382, II.16]Chn dng ng un cong 90oc bn knh R vi R/d = 2 th xu2(1 ch)= 0,15.ng y c 4 ch un =>u2 = 0,15. 4 = 0,6

Van:[1, 94, 9.5] Chn van cu vi m hon ton th v2(1 ci)= 4.ng y c 1 van cu =>v2= 4

Vo bn cao v: cv= 1 nn: = u2+ v2+ cv= 5,6Vy: hf1-2=

81,9.2

283,0.6,53,4

05,0

106033,0

2

= 0,084m

6.7.8. Tnh ct p ca bmHb= (z2z1) + hf1-2= (71) + 0,084 = 6,084 m

6.7.9. Cng sutChn hiu sut ca bm: hb= 0,8.


42/43

42

Cng sut thc t ca bm: Nb=8,0.3600

81,9.27,964.084,6.2

.3600

.

b

Fbb gHQ

= 40 W

Kt lun: m bo thp hot ng lin tc ta chn 2 bm li tm, c:Nng sut: Qb= 2 m

3/hCt p: Hb= 6,084 mCng sut: Nb= 40 W.

Kt lunSau khi hon thnh n Qu trnh & Thit, em tng hp li nhng kin thc

hc va qua v tm hiu thm cc vn khc nh: Bit cch thit k c thp chng ct mm xuyn l Methanol Nc khi bit trc lu

lng, nng nhp liu, nng sn phm nh v sn phm y. Tnh ton c iu kin lm vic ca thit b v kh nng chu bn ca thit b v tnh

n mn c hc v ha hc. Vi h thng chng ct metanol - nc dng thp mm xuyn l nh thit k, ta thy

bn cnh nhng u im cng cn c nhiu nhc im. Thit b c u im l nng sutv hiu sut cao nhng thit b cn rt cng knh, i hi phi c s vn hnh vi

chnh xc cao. Bn cnh , khi vn hnh thit b ny ta cng phi ht sc ch n vn an ton lao ng trnh mi ri ro c th xy ra, gy thit hi v ngi v ca.


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TI LIU THAM KHO[1]. S tay Qu trnh v thit b cng ngh ha cht Tp 1 , NXB Khoa hc v K thut.[2]. S tay Qu trnh v thit b cng ngh ha cht Tp , NXB Khoa hc v K thut.[3]. Phm Vn Bn, V B Minh, Hong Minh Nam, Qu trnh v thit b cng nghha hc thc phm Tp1, NXB HQG Tp.HCM.[4]. V Vn Ban, V B Minh, io trnh Qu trnh v thit b cng ngh ha hc thc

phm Tp , NXB HQG Tp.HCM.[5]. Trnh Vn Dng , Qu trnh v Thit b trong Cng Ngh Ha Hc Bi tp Truynkhi, Nh xut bn i hc Quc gia TpHCM,2004.[6]. H L Vin, Thit k v Tnh ton cc thit b ha cht, Nh xut bn Khoa hc vK thut, H Ni, 1978.[7].Nguyn Minh Tuyn, C s Tnh ton Myv Thit b Ha cht Thc phm,

Nh xut bn Khoa hc v K thut, H Ni, 1984.[8].Bng tra cuQu trnh c hc truyn nhit-truyn khi, Nh xut bn i hc Qucgia TP.HCM.[9]. PGS-TS Hong nh Tn, PGS-Ts Bi Hi, Bi tp nhit ng hc k thut v

truyn nhit, NXB HQG Tp.HCM

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