Mam-Chop Diendandaihoc.vn 05472324042011

8/2/2019 Mam-Chop Diendandaihoc.vn 05472324042011

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N MN HC GVHD: L XUN HI

SVTH: TRN VN PHC Trang 1

MC LCLi mu....................................................................................................... 5

CHNG 1: GII THIU V ACETONE V QUI TRNH CNG NGH........ 6

1. Gii thiu s b .......................................................................................... 7

2. Sn xut Aceton.......................................................................................... 8

3. Cng ngh chng ct hn hp Acetone Nc ..................................... 9

4. Chn loi thp chng ct v phng php chng ct .......................... 9

5. S qui trnh cng ngh v thuyt minh qui trnh cng ngh.......... 10

CHNG 2: CN BNG VT CHT CN BNG NNG LNG .......... 14

1. Cn bng vt cht ..................................................................................... 15

1.1 th cn bng Acetone Nc ........................................................ 16

1.2 Xc nh chs hi lu thch hp ........................................................ 17

1.3 Vng lm vic ............................................................................... 19

1.4 Xc nh s mm l thuyt v s mm thc t .................................... 19

2. Cn bng nng lng .............................................................................. 21

2.1 Cn bng nhit lng ca thp chng ct .......................................... 21

2.2 Cn bng nhit lng ca thit b ngng t ........................................ 23

2.3 Cn bng nhit lng ca thit b lm lnh sn phm nh ................ 23

2.4 Cn bng nhit lng ca thit b lm ngui sn phm y (trao inhit vi nhp liu ban u) .................................................................. 24

CHNG 3: TNH TON THIT B CHNH .................................................. 25

I. Kch thc thp .................................................................................... 26

1. ng knh on ct ........................................................................... 26

2. ng knh on luyn ......................................................................... 28

3. Chiu cao thp ..................................................................................... 30

II. Tnh ton chp v ng chy chuyn .................................................... 31

A. Tnh cho phn ct .................................................................................. 31


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B. Tnh cho phn chng ............................................................................ 32

III. Tnh chi tit ng dn ............................................................................. 37

1. ng knh ng dn hi vo thit b ngng t ..................................... 37

2. ng dn dng chy hon lu ............................................................... 38

3. ng dn dng nhp liu ........................................................................ 38

4. ng dn dng sn phm y ................................................................ 38

5. ng dn t ni un qua thp ................................................................ 39

IV. Tnh tr lc thp .................................................................................... 39

A. Tng tr lc phn ct ............................................................................ 39

1. Tr lc a kh Pk .............................................................................. 39

2. Tr lc do sc cng b mt.................................................................. 40

3. Tr lc ca lp cht lng trn a ( Tr lc thy tnh Pt )................... 40

B. Tng tr lc phn chng ...................................................................... 41

1. Tr lc a kh Pk ............................................................................. 41

2. Tr lc do sc cng b mt ................................................................. 41

3. Tr lc ca lp cht lng trn a ( Tr lc thy tnh Pt ) .................. 42

CHNG 4: TNH C KH ........................................................................... 44

1. Tnh b

dy thn tr

c

a thp ................................................................. 452. Tnh - chn b dy y v np thit b ................................................... 47

3. Chn bch v vng m ........................................................................... 48

4. Tnh mm ................................................................................................... 49

5. Chn v tai treo thit b ...................................................................... 50

6. Tnh bo n ............................................................................................... 53

CH

NG 5: TNH THI

T BPH

.................................................................. 58

I. Thit b gia nhit hn hp u hay thit b lm ngui sn phm y 59

1. iu kin nhit ca qu trnh .............................................................. 59

2. Nhit ti .................................................................................................... 60

3. Chn thit b ............................................................................................. 60


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II. Thit b lm ngui sn phm nh ....................................................... 64

1. iu kin nhit ca qu trnh ............................................................. 64

2. Nhit ti .................................................................................................... 65

3. Chn thit b ............................................................................................. 65

III. Thit b ngng t hi lu ....................................................................... 69

1. iu kin nhit ca qu trnh .............................................................. 69

2. Nhit ti .................................................................................................... 70

3. Chn thit b ............................................................................................. 70

4. Xc nh h s cp nhit t dng sn phm nh n thnh ng ........... 70

5. Xc nh h s cp nhit t thnh ng n nc ................................... 71

6. Nhit ti ring ........................................................................................... 72

7. H s truyn nhit .................................................................................... 728. B mt truyn nhit .................................................................................. 72

9. Chiu di mi ng .................................................................................... 72

IV. Thit b ni un ....................................................................................... 73

1. iu kin nhit ca qu trnh ............................................................. 73

2. Nhit ti .................................................................................................... 74

3. Chn thit b ............................................................................................. 74

4. Xc nh h s cp nhit t dng sn phm nh n thnh ng ........... 745. Tnh h s cp nhit ca sn phm y ni ............................................ 75

6. H s truyn nhit .................................................................................... 75

7. B mt truyn nhit .................................................................................. 75

8. Chiu di mi ng .................................................................................... 75

V. Tnh bn cao v- Bm ............................................................................. 76

1. Tnh bn cao v......................................................................................... 76

2. Tnh bm .................................................................................................. 79

CHNG 6: TNH GI THNH THIT B ..................................................... 81

TI LIU THAM KHO .................................................................................. 83


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LI MUMt trong nhng ngnh c sng gp to ln n ngnh cng nghip

nc ta ni ring v th gii ni chung l ngnh cng ngh ho hc.cbit l ngnh ha cht c bn.

Trong thc t, chng ta s dng rt nhiu dng ho cht khc nhau: hnhp nhiu cht hay n cht tinh khit. M nhu cu v mt loi ho cht tinhkhit cng rt ln. Qu trnh c thp ng phn no tinh khit theo yucu l chng ct: la qu trnh tch cc cu t trong hn hp lng lng, hayhn hp lng kh thnh cc cu t ring bit da vo bay hi khc nhauca chng.

V i vi h acetone nc, do khmg c im ng ph nn c thtc bt k tinh khit theo yu cu nh qu trnh chng ct.

Nhim v thit k: tnh ton h thng chng luyn lin tc tch hn hphai cu t : acetone nc vi cc s liu sau y:


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Nng sut sn phm nh : 1500 Kg/h

Nng sn phm nh : 98% theo khi lng

Nng nhp liu : 30%

Ap sut lm vic : p sut thng.

CHNG 1

GII THIU V

ACETONE V QUI TRNHCNG NGH


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1. Gii thiu sb :Acetone c cng thc phn t : CH3COCH3 .Khi lng phn t bng

58.079 vCL mt cht lng khng mu, d lu ng v d chy, vi mt cch m du

v c mi thm.N ha tan v hn trong nc v mt s hp cht hu c nh : eter,

metanol, etanol, diacetone alcoholng dng : Acetone c ng dng nhiu lm dung mi cho cng nghip,

v d cho vecni, sn, sn mi, cellulose acetate, nha, cao su N ho tantt t acetate, nitroxenluloz, nha phenol focmandehyt, cht bo, dung mipha sn, mc in ng ng. Acetone l nguyn liu tng hp thy tinh huc.T Acetone c th tng hp ceten, sumfonat (thuc ng), cc holofom.c tm thy u tin vo nm 1595 bi Libavius, bng chng ct khan

ng, v n nm 1805 Trommsdorff tin hnh sn xut Acetone bng cchchng ct Acetat ca b tt v soa : l mt phn on lng nm gia phnon ru v eter.

Mt s thng s vt l v nhit ng ca Acetone : Nhit nng chy : -94.6 0C ; Nhit si : 56.9 0C ;

T trng :4

20d ; Nhit dung ring Cp : 22 Kcal/mol (chun 102 0C) nht : 0.316 cp ( 250C)


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Nhit tr : 0.5176 cal/g ( 200C)Tnh cht ho hc :

Cng hp vi natri bisunfit:OH

CH3COCH3 + H2O CH3 - C - SO3Na

CH3( 1-metyl-1-hydroxi etan sunfont natri )

Cng hp axit HCN:OH

CH3CO + HCN CH3-C-CNCH3

( pH= 4-8 )Phn ng ngng t :

OH OCH3-CO-CH3 + HCH2C=O CH3-C-CH3-C-CH3

CH3 CH3( 4-oxy-4-mtyll-2-pentanon)

Acetone kh b oxi ha bi thuc th Pheling, Tluen, HNO3, KMnO4 ,Chb oxi ha bi hn hp KMnO4 + H2SO4, Sunfcrmic K2Cr2O7 + H2SO4B gy mch cacbon.

CH3-C-CH3 CH3-C-CH2-OH CH3-C-CH=O CH3COOH +HCOOH O O O

Phn ng kh ho :

CH3COCH3 + H2 CH3CHOH-CH3

iu ch :Oxy ha ru bc hai:

CH3CHOH-CH3 CH3COCH3 + H2OTheo phng php Piria : nhit phn mui canxi ca axit cacboxylic:

(CH3COO)2Ca CH3COCH3 + CaCO3T dn xut c magi :

O OCH3-C-Cl + CH3-MgBr CH3-C-CH3 + Mg-Br

Cl2. Sn xut Acetone :

Trong thi k chin tranh th gii ln th nht, do nhu cu v ngunAcetone rt ln, tong khi c s gii hn trong vic thu dc Acetone t schng ct g, nn b sung ngun Acetone Hoa K p dng phngphp chng ct khan Ca(CH3COO)2 thu c bng cch ln men ru cmt xc tc vi khun chuyn carbohydrate thnh Acetone v ButylAlcohol.Cng ngh ny c ng dng ch yu trong sut chin tranh th giiln th nht v nhng nm 20 .


8/79



Tuy nhin, n gia nhng nm 20 v cho n nay cng ngh trn cthay bng cng ngh c hiu qu hn (chim khong phng php snxut Acetone ca Hoa K) : Dehydro Isopropyl Alcol.

Ngoi ra, cn mt s qa trnh sn xut Acetone khc :- Oxi ha Cumene Hydro Peroxide thnh Phenol v Acetone.- Oxi ha trc tip Butan Propan.- Ln men Carbo hydrate bi vi khun c bit.- Cng ty Shell s dng n nh mt sn phm ph.

Tng hp Acetone bng cch Dehydro Isopropyl Alcol c xc tc: CH3CHOHCH3 + 15.9 Kcal ( 327

0C ) xuctac CH3COCH3 + H2 Xc tc s dng y : ng v hp kim ca n, oxit kim loi v

mui. nhit khong 325 0C , hiu sut khong 97%. Dng kh nng sau phn ng gm c : Acetone, lng Isopropyl Alcol

cha phn ng, H2 v mt phn nh sn phm ph ( nh Propylene,

diisopropyl eter ). Hn hp ny c lm lnh v kh khng ngngc lc bi nc . Dung dch lng c em i chng ct phnon, thu c Acetone nh v hn hp ca nc, Isopropyl Alcol( t ) y.

3. Cng ngh chng ct hn hp Acetone Nc :Ta c Acetone l mt cht lng tan v hn trong nc v nhit si ca

Acetone ( 56.9 0C 760 mmHg) v Nc ( 100 0C 760 mmHg) : l khcch xa nhau nn phng php hiu qu nht thu c Acetone tinh khitl chng ct phn on da vo bay hi khc nhau ca cc cu t tronghn hp.

Trong trng hp ny ta khng th s dng phng php c c v cccu tu c kh nng bay hi, v khng s dng phng php trch ly cngnh hp th do phi a vo mt pha mi tch chng, c th lm cho qutrnh phc tp hn, hay qu trng tch khng c hon ton.4. Chn loi thp chng ct v phng php chng ct :

Chng ct l qu trnh phn tch cc hn hp lng thnh cc cu tring bit da vo s khc nhau v bay hi ca chng ( hay nhit si ), bng cch lp i lp li nhiu ln qu trnh bay hi ngngt, trong vt cht i t pha lng vo pha hi hoc ngc li.

i vi chng ct ta c hai phng php thc hin :

- Chng ct n gin (dng thit b hot ng theo chu k):Phng php ny s dng trong cc trng hp sau :+ Khi nhit si ca cc cu t khc xa nhau .+ Khi khng i hi sn phm c tinh khit cao .+ Tch hn hp lng ra khi tp cht khng bay hi .+ Tch s b hn hp nhiu cu t .


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- Chng ct lin tc hn hp hai cu t (dng thit b hot ng lintc): l qu trnh c thc hin lin tc, nghch dng, nhiu on.Ngoi ra cn c thit b hot ng bn lin tc .

Trong trng hp ny, do sn phm l Acetone vi yu cu c tinh khit cao khi s dng , cng vi hn hp Acetone Nc l hn

hp khng c im ng ph nn chn phng php chng ct lintc l hiu qu nht.

Chn loi thp chng ct :C rt nhiu loi thp c s dng, nhng u c chung mt yu cu c

bn l din tch b mt tip xc pha phi ln, iu ny ph thuc vo phntn ca mt lu cht ny vo lu cht kia .

Ta kho st hai loi thp chng ct thng dng l thp mm v thpchm:

- Thp mm gm thn thp hnh tr, thng ng, pha trong c gn ccmm c cu to khc nhau, trn pha lng v pha hi c cho tip

xc vi nhau.Gm c : mm chp, mm xuyn l , mm van. Thng s dng mmchp .

- Thp chm l mt thp hnh tr, gm nhiu on ni vi nhau bngmt bch hay hn . Vt chm c y trong thp theo mt hay haiphng php : xp ngu nhin hay xp th t .

Chn loi mm chp thc hin qu trnh chng ct v nhng u imsau:

- D dng lm v sinh thng qua cc ca sa cha .- Vi cng mt chc nng, tng khi lng thp mm thng nh hn

so vi thp chm.- Hiu sut mm khng i i vi mt khong vn tc dng lng hockh.

- C th lp t ng xon gii nhit trn mm khi cn thit .- Thp mm thch hp trong trng hp c s mm l thuyt hoc sn v truyn khi ln .

- Thp c thit k c th gic mt lng lng nht nh trnmm.

- Chi ph thp mm c ng knh ln r hn so vi thp m .- D dng a vo hoc loi b cc dng bn .

- Tnh n nh cao.5. S qui trnh cng ngh v thuyt minh qui trnh cng ngh :a. S qui trnh cng ngh (xem trang sau)


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11/79



13

San Pham ay

N c Long

Ng uye n Lie u2

1

San Ph am ay

T

P T

7

T

T

6

4

N c

12

Sa n P ham nh

Hi Nc

11

5

Nc

9

10

3

P

14

T

15

N c

8

Hi khong ngngN c


12/79

O AN MON HOC GVHD: LE XUAN HAI

SVTH: TRAN VAN PHUC Trang 12

Ch thch :1. Bn cha nguyn liu .2. Bm.3. Bn cao v .4. By hi .

5. Lu lng k .6. Van .7. Thp chng ct .8. Thit b ngng t sn phm nh .9. B phn chnh dng .10. Thit b lm ngui sn phm nh .11. Bn cha sn phm nh .12. Ni un .13. un si nhp liu bng sn phm y .14. Ap k .

15. Nhit k .b. Thuyt minh qui trnh cng ngh :Hn hp Acetone- Nc c nng Acetone 30% ( theo khi lng) ,

nhit khong 27 0C ti bnh cha nguyn liu (1) c bm (2) bmln bn cao v (3). T c a n thit b gia nhit (3) ( trao i nhitvi sn phm y). y, hn hp c un si n nhit si . Sau, hn hp c a vo thp chng ct (7) a nhp liu.

Trn a nhp liu, cht lng c trn vi phn lng ton ct cathp chy xung. Trong thp hi, i t di ln gp cht lng t trnxung . y, c s tip xc v trao i gia hai pha vi nhau. Pha lng

chuyn ng trong phn chng cng xung di cng gim nng cccu t d bay hi v b pha hi to nn t ni un (12) li cun cu td bay hi . Nhit cng ln trn cng thp, nn khi hi i qua cc a tdi ln th cu t c nhit si cao l nc s ngng t li, cui cngtrn nh thp ta thu c hn hp c cu t Acetone chim nhiu nht (c nng 98% theo khi lng ). Hi ny i vo thit b ngng t (8) vc ngng t mt phn ( ch ngng t hi lu). Mt phn cht lngngng i qua thit b lm ngui sn phm nh (10), c lm ngui n30 0C , ri c a qua bn cha sn phm nh (11). Phn cn li cacht lng ngng c hi lu v thp a trn cng vi t s hon luti u . Mt phn cu t c nhiot si thp c bc hi, cn li cu tc nhit si cao trong cht lng ngy cng tng . Cui cng, y thpta thu c hn hp lng gm hu ht l cu t kh bay hi ( nc). Hnhp lng y c nng Acetone l 2% theo khi lng, cn li lnc. Dung dch lng y i ra khi thp, mt phn dc un, bc hi ni un (12) cung cp li cho thp tip tc lm vic, phn cn li ctrao i nhit vi nhp liu ( sau khi qua bn cao v ).


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H thng lm vic lin tc cho ra sn phm nh l Acetone, snphm y sau khi trao i nhit vi nhp liu c thi b.

CHNG 2


14/79



CN BNG VT CHT

CN BNG NNG LNG

1. Cn bng vt cht Cc s liu ban u :

Nng sut sn phm nh thu c :1500 ( Kg/h )Sn phm c nng Acetone : 98% theo khi lng.Nhp liu c nng Acetone : 30% theo khi lng .Thit b hot ng lin tc.

Cc k hiu :F : lng nhp liu ban u ( Kmol/h )D : lng sn phm nh ( Kmol/h )W : lng sn phm y ( Kmol/h )xF :nng mol Acetone trong nhp liuxD :nng mol Acetone trong sn phm nhxW : nng mol Acetone trong sn phm y


15/79



Phng trnh cn bng vt cht cho ton b thp chng ct :F = D + W ( 1 )F * xF = D * xD + W * xW ( 2 )

Chuyn t phn khi lng sang phn mol:

xF =

21

1

)1(

M

x

M

x

M

x

FF

F

=

18

)30.01(

58

30.05830.0

= 0.117(phn

molAcetone )

x D =

21

1

)1(

M

x

M

x

M

x

DD

D

=

18

)98.01(

58

98.058

98.0

= 0.938

(phnmolAcetone)

xW =

21

1

)1(

M

x

M

x

M

x

ww

w

=

18

)02.01(

58

02.058

02.0

= 0.006 ( phn

molAcetone) (Chn sn phm y c nng khi lng ca Acetonel 2% )

Tnh Mtb :Mtb

F

= xF * M1 + (1- xF ) * M2= 0.117 * 58 + (1 0.117 ) * 18= 22.68 ( Kg/Kmol)

MtbD

= xD * M1 + (1- xD ) * M2= 0.938* 58 + (1 0.938) * 18= 55.52 ( Kg/Kmol)

MtbW

= xW * M1 + (1- xW ) * M2= 0.006 * 58 + (1 0.006 ) * 18= 18.24 ( Kg/Kmol)

Sut lng sn phm nh :

D =Dtb

MD = 52,55

1500 = 27.017 ( Kmol/h )

( 1 ) v ( 2 ) ta c h phng trnh :

WDx*Wx*Dx*F

WDF

F


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006.0*938.0*017.27117.0*

017.27

WF

WF

)/(83.199

)/(84.226

hKmolW

hKmolF

hay

)/(86.3644

)/(86.5144

hKgW

hKgF

1.1 th cn bng Acetone Nc :

Thnh phn cn bng lng (x), hi (y) tnh bng %mol v nhit sica hn hp hai cu t 760 mmHg ( Acetone nc ):

x 0 5 10 20 30 40 50 60 70 80 90 100y 0 60.3 72 80.3 82.7 84.2 85.5 86.9 88.2 90.4 94.3 100t 100 77.9 69.6 64.5 62.6 61.6 60.7 59.8 59 58.2 57.5 56.9

ACETONE-NC

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x = phan mol acetone trong pha long

y=phanmolacetonetro

ngpha


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1.2 Xc nh ch s hi lu thch hp :

a. Ch s hi lu ti thiu :Do nhp liu trng thi lng bo ha, nn Rminc xc nh nh

sau:

Rmin =FF

FD

xyyx

*

*

xF = 0.117 y *F = 0.74 ( Xc nh tng cn bng )

Rmin = 0.3178b. Ch s hi lu thch hp :Cho cc gi tr Rxi > Rxmin tm cc gi tr tung Bi tng ng v

v cc ng nng lm vic ca on luyn ng vi cc gi tr Bi :

Bi =1i

D

Rx

x

Tm cc im a( y= x= xD ), b( y= x= xw ) v ng x = xF ( song songvi trc tung ). C mi gi tr Bi ta vc ng nng lm vic caon luyn v on chng .

Nh vy ng vi mi gi tr Rxi ta c s n v chuyn khi chungtng ng l mxi.

Ta c bng sau :Rx Bi mx mx* (Rx + 1)0.8 0.5211 6.86 12.3480.9 0.4936 6.00 11.4000.92 0.4885 5.92 11.366

0.95 0.4810 5.91 11.5241.0 0.4690 5.89 11.7801.1 0.4466 5.82 12.2221.2 0.4263 5.79 12.7381.3 0.4078 5.74 13.2021.4 0.3908 5.64 13.536

Th tch thp l V = f * Hf : tit din thp, m2H : chiu cao lm vic ca thp, m

Ta bit tit din ca thp t l thun vi lng hi i trong thp, m

lng hi li t l thun vi lng lng hi lu trong thp, nh vy tit dinthp t l vi lng hi lu .Tc l f ( Rx + 1 ) * GDTrong mt iu kin lm vic nht nh th GD l khng i,nn f ( Rx + 1).Cn chiu cao thp t l vi sn v chuyn khi H mx , nn cui

cng ta c th vit V = f*H mx ( Rx + 1)


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T ta s lp c s ph thuc gia Rx _ mx * ( Rx + 1 ) . Mi quanh ny s cho ta tm c mt gi tr Rx m th tch ca thit b chng ctng vi n l ti u Rxth .

V th quan h gia (mxi*(Rxi + 1) _ Rxi ) tm Rxth .

( Xem trang sau)

1.3 V

nglm vic :

Phng trnh ng lm vic lm ct :

y =1x

x

R

Rx +

1xD

R

x

=0.479*x + 0.4885Phng trnh ng lm vic phn chng:

y =1

x

x

R

FRx -

1

1

xR

L x

Vi L = DL0 = 8.396

y = 4.8521*x 0.0231.4 Xc nh s mm l thuyt v s mm thc t :

Do iu kin nhp liu l lng bo ha, ta c ng nhp liu lng :

x = xF = 0.117

Chsohoi lu thch hp

1111.5

1212.5

1313.5

14

0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

Rxi

mxi*

(Rxi+1

)


19/79



k cc ng lm vic ca phn ct v phn chng trn cng thc s bc thang l 5.92 , tng ng vi s mm l thuyt l 6 ( kc ni un ) Xc nh hiu sut trung bnh ca thp tb :

tb = f ( , )

=y

y

1 xx1 : bay hi tng i

x, y : nng phn mol ca cu t d bay hi trong pha lng, pha hi nht ca hn hp lng : tra theo nhit

tb = (1 + 2 + 3) / 31 , 2 , 3 : ln lt l hiu sut mm nh, mm y, mm nhp

liu.T gin x-y, t-x,y : tm nhit ti cc v tr v nng pha

hi cn bng vi pha lng :V tr mm nh :

xD = 0.938yD = 0.962 Dy = 0.988tD = 57.55 O C

V tr mm nhp liu :xF = 0.117yF = 0.74 Fy = 0.902

tF = 68.5 O CV tr mm y :

xW = 0.006

yW = 0.09 wy = 0.242tW = 96.00 C

Xc nh nht, bay hi tng i, hiu sut ti cc v tr :V tr mm nh :

tD = 57.55 C nc = 0.49 * 10-3 Ns/m2

acetone = 0.2386 * 10-3 Ns/m2

( Tra bng I.102 v I.101 S tay tp mt ) lg hh = xD* lg acetone + ( 1 xD ) * lg nchh = 0.2477*10

-3 Ns/m2

= 1.6733* = 0.4145*10-3

1 = 0.63 ( Hnh IX.11- S tay tp 2 )V tr mm nhp liu :

tF = 68.5 C nc = 0.4145 * 10-3 Ns/m2

acetone = 0.2175 * 10-3 Ns/m2

( Tra bng I.102 v I.101 S tay tp mt )


20/79



lg hh = xF * lg acetone + ( 1 xF ) * lg nchh = 0.3844 * 10

-3 Ns/m2 = 21.48* = 8.258 * 10-3

2 = 0.31 ( Hnh IX.11- S tay tp 2 )V tr mm y :

tW = 96 C nc = 0.2962 * 10-3 Ns/m2

acetone = 0.1759 * 10-3 Ns/m2

( Tra bng I.102 v I.101 S tay tp mt ) lg hh = xW * lg acetone + ( 1 xW ) * lg nchh = 0.2953 * 10

-3 Ns/m2 = 16.385* = 4.838 * 10-3

3 = 0.33 ( Hnh IX.11- S tay tp 2 )

hh = ( 1 + 2 + 3 )/3= ( 0.64 + 0.3 + 0.33 )/3= 0.432

Ntt =432.0

92.5 = 13 ( mm )

S mm thc t cho phn ct : 10S mm thc t cho phn chng : 3

V nhp liu mm s : 102. Cn bng nng lng

2.1 Cn bng nhit lng ca thp chng ct

Phng trnh cn bng nng lng :

QF + QD2+ QR = Qy + Qw + Qxq

2+ Qng

2

Nhit lng do hn hp u mang vo QF (j/h)QF = . .F FF C t

F= 5144.86 (Kg/h)tF = 68.5

oC : nhit i vo ca hn hp u ( trng thi lng si )CF : nhit dung ring :

tF = 68.5oC Cnc = 4190 (J/Kg. )

Cacetone = 2332.62 (J/Kg.)CF = Fx .Cacetone + ( 1- Fx ).Cnc

= 0.3* 2332.62 + ( 1- 0.3 )*4190 = 3632.78 (J/Kg. ) QF = 5144.86 * 3632.78 68.5 = 1.28010

9 (J/h) = 355.63 (KW) Nhit lng do hi t mang vo thp QD

2(J/h):

QD 2 = D2 *2 = D2 * (r2 + C2* t2)


21/79



Dng hi nc p sut 2at , r2 = 2173 (Kj/Kg), to= 119.6 oC

2 : nhit lng ring ca hi t ( J/Kg)r2 : n nhit ha hi ( J/Kg)t2 , C2 : nhit

oC v nhit dung ring ca nc ngng (J/Kg.) Nhit lng do lu lng lng hi lu mang vo :

QR = GR * CR * tRCR = CD :nhit dung ring ca sn phm nh :tD = 57.3

oC Cnc = 4187.96 ( J/Kg.)Cacetone = 2296.06 (J/Kg. )

CD= CR = Dx * Cacetone + ( 1 - Dx ) Cnc= 0.98*2296.06 + ( 1-0.98 )*4187.94 = 2333.89 ( J/Kg. )

GR = D *R = 1500*0.92 = 1380( Kg/h)tR = tD = 57.3

oC QR = 1380*2333.89*57.25 = 1.844*10

8 = 51.22(KW)

Nhit lng do hi mang ra nh thp Qy :Qy = D .( 1+ R).DNhit lng ring ca hi nh thp D :

D = acetone. Dy + nc ( 1 - Dy )

ViD

y = 0.988 (phn khi lng )acetone , nc :: nhit lng ring ca acetone, nc :

acetone = racetone + tD . Cacetonenc = rnc + tD . Cnc

rnc, racetone , Cacetone , C nc tra bng I.212 v bng I.153 (S taytp mt ) tD=57.3

oCCacetone = 2296.06 ( J/Kg. )Cnc = 4187.94 (J/Kg. )

racetone = 521.46 (Kj/Kg)nc = 2425.6 ( Kj/Kg)

nc = 2665.36*103(J/Kg)acetone = 652.91 *103(J/Kg)

D = 677.06 ( Kj/Kg) Qy= 1.95*10

9(J/h) = 541.65 (KW) * Nhit lng do sn phm y mang ra Qw :

Qw=. .

w wW C t

W= 3644.86 (Kg/h)tw = 96

oC

wx = 0.02 ( phn khi lng )

nhit 96 oC Cacetone = 2422(J/Kg.)Cnc = 4222( J/Kg.)

Cw= 4186 (J/Kg. )


22/79



Qw = 1.465*109 ( J/h) =406.86 (KW)

Nhit lng tn tht ra mi trng xung quanh Qxq2:Ly Qxq2 = 5%QD

2

Nhit lng do nc ngng mang ra Qng2 (J/h)Qng2 = Gng2 . C2 .t2

Vy lng hi t cn thit un si dung dch y thp :

D2 =2

2 20.95*

y w xq F R y w F RQ Q Q Q Q Q Q Q Q

r

= 930 (Kg/h)

2.2 Cn bng nhit lng ca thit bngng t :

Chngng t hi lu :

1 1 2 1. . . .( ) x D n nD R r G C t t

Gn1 = 1 2 1

. .

.( )

x D

n

D R r

C t t Chn nhit vo, ra ca nc lm lnh t1 = 27

oC , t2 = 400C

t = 1 2( ) 27 40

2 2

t t =33.5 0C

Nhit dung ring ca nc nhit trung bnh Cn = 4180.94(J/Kg. )

yD = 0.962 tD (hi) = 57.30C

An nhit ha hi r057.3 C

acetone = 521.46 * 103 (J/Kg)

r057.3 C

nuoc = 2425.60*103 ( J/Kg)

rD = 544.31 *103 ( J/Kg)Suy ra lng nc lnh cn tiu tn Gn = 13820 (Kg/h) =3.839( Kg/s)

2.3 Cn bng nhit lng ca thit blm lnh sn phmnh :

Phng trnh cn bng nng lng :' '

1 2 3 2 1.( .( )) . .( ) D D n nD r C t t G C t t

Nhit vo ca sn phm nh t '1

=57.3 0C

Nhit ra ca sn phm nh t'

2 = 30 0CNc lm ngui c nhit vo, ra l :t1 = 27 0C, t2 = 40 0CNhit trung bnh ca nc lm lnh ttb = (27+ 40 )/2 = 33.5 0CNhit dung ring ca nc nhit ttb l Cn = 4176.6 (J/Kg. )Nhit trung bnh ca sn phm nh : t '

tb= (57.3 + 30 )/2 = 43.65

0C t '

tb= 43.65 0C Cacetone = 2251.86 ( J/Kg. )

Cnc = 4177.74 (J/Kg. )


23/79



CD = 2290.38 ( J/Kg. )An nhit ha hi rD = 544.31 *10

3 ( J/Kg)Suy ra lng nc cn dng :

Gn3 =' '

1 2

2 1

.( ( ))

.( )

D D

n

D r C t t

C t t

= 16764.76 ( Kg/h)

2.4 Cn bng nhit lng ca thit blm ngui sn phmy ( traoi nhit vi nhp liu banu ):

Phng trnh cn bng nhit lng :' ' ' '

1 1 2. .( ) . .( )F F f xq wQ F C t t Q W C t t

tf = 270C

tF = 68.50C

047.752

f Ft t

t C

Cacetone = 2265.19 ( J/Kg. )Cnc = 4180.81 ( J/Kg. )

Cf = . (1 ).F acetone F nuoc x C x C = 0.3*2265.19 + ( 1- 0.3)* 4180.81= 3606.12 ( J/Kg. )

t '1=tw = 960C

Chn t '2

= 70 0C

' '

' 01 2 832

t tt C

Cacetone = 2379.75 ( J/Kg. )Cnc = 4196 ( J/Kg. )

Cw = . (1 ).w acetone w nuoc x C x C = 0.02 * 2379.75 + ( 1- 0.02 )*4196 = 4159.67 (J/Kg.)

Chn Qxq = 5% Q 'w

0.95* ' ' '1 2. .( ) . .( )w F F f W C t t F C t t

'' '

1 2

. .( ) 1500*3606.12*(68.5 27)2184.86( / )

0.95* * ( ) 0.95*4159.67* (96 70)

F F f

w

F C t t W Kg h

C t t

Cn 3644.85 - 2184.86 = 1460 (Kg/h) thi ra ngoi.


24/79



CHNG 3

TNH TON

THIT B CHNH

I. Kch thc thpng knh thp c xc nh theo cng thc sau :

D = 0.0188tbyy

tbg

)*(

Trong gtb

: lng hi trung bnh i trong thp ( Kg/h )


25/79



(tb * y )tb : tc hi trung bnh i trong thp ( Kg/h )V rng lng hi v lng lng thay i theo chiu cao ca thp v

khc nhau trong mi on cho nn ta phi tnh ng knh trung bnhring cho tng on : on chng v on ct .1. ng knhon ct :

Nng trung bnh ca pha lng :xm = ( xF + xD )/2 = (0.117 + 0.938 )/2 = 0.5257

Nng trung bnh ca pha hi theo phng trnh ng lm vic :ym = 0.479 * x

m + 0.4885

= 0.479 * 0.5275 + 0.4885= 0.7412

Nhit trung bnh ca pha hi, pha lng t gin t-x,y:xm = 0.5275 t

x = 60.25

OCym = 0.7412 t

y = 68.3

OC Khi lng mol trung bnh v khi lng ring pha hi :

Mm = ym * Macetone + ( 1 ym ) * Mnc= 0.7412 * 58 + ( 1 0.7412 ) * 18= 47.648 ( Kg/ Kmol )

y = ''

*4.22

*

y

Om

T

TM=

)3.68273(*4.22

273*648.47

= 1.701 ( Kg/m3 )

Khi lng ring pha lng :

xm = 0.5275 xm =

18*)5275.01(58*5275.0

58*5275.0

= 0.7825 ( phn khi lng )

tx = 60.25

O

C acetone = 746 ( Kg/m

3

)nc = 983 ( Kg/m

3 )( Bng I.2 S tay tp mt )

'

1

x

='

'

acetone

mx

+

'

'

)1(

nuoc

mx

=746

7825.0 +983

)7825.01(

= 1.270 * 10-3

= 787.28 (Kg/m3 )

Lng hi trung bnh i trong on luyn :tb = ( g + g1 )/2 ( Kg/h )g : lng hi ra khi a trn cng

g= GR + GD = GD ( Rx + 1)= D ( Rx +1 ) = 1500 ( 0.92 + 1)= 2880 ( Kg/h )

g1 : lng hi i vo a di cng ca on ct


26/79



g1 = G1 + GD = G1 + D = G1 + 1500 (Kg/h )

Lng hi g1, hm lng y 1 , lng lng G1c xc nh theo hphngtrnh sau :

dd

D

rgrg

xDxGyg

DGg

**

***

11

1111

11

( x1 = Fx )

r1 = racetone * y 1 + ( 1 - y 1 ) * rncr= racetone * Dy + ( 1 - Dy ) * rnc

Ti v tr nhp liu :tF = 68.5

OC ra = racetone = 509.38 ( Kj/Kg )rb = rnc = 2388.57 ( Kj/ Kg )

( Bng I.212-S tay tp mt ) r1 = ra * y 1 + ( 1 - y 1 ) * rb

= 2388.57 - 1879.19 * y1

Ti v tr nh thp :tD = 57.25

OC ra = racetone = 521.46 ( Kj/Kg )rb = rnc = 2425.60 ( Kj/ Kg )

( Bng I.212-S tay tp mt )yD = 0.962 ( phn mol ) Dy = 0.988 ( phn khi lng )

r= ra * Dy + ( 1 - Dy ) rb

= 544.31 ( Kj/ Kg ) g1 * r1 = g* r= 1567612.8

8.1567612*

98.0*150030.0**

1500

11

111

11

rg

Gyg

Gg

( r1 = 2388.57 1879.19* y 1)

448.409

448.1909

834.0

1

1

1

G

g

y

gtb = ( g1 + g) /2

= ( 1909.448 + 2880 )/2 = 2394.72 ( Kg/h) Vn tc hi i trong thp :

( y * y )tb = 0.065 * [] *tbtb yx

h ** (Kg/m2*s )

xtb

: khi lng ring trung bnh ca pha lng:

x = 787.28 ( Kg/m3 )

y tb : khi lng ring trung bnh ca pha hi:


27/79



y = 1.701( Kg/ m3)

h : khong cch mm ( m ), chn h = 0.3[] : h s tnh n sc cng b mt

hh

1 =acetone

1 +nuoc

1 (acetone , nuoc tra bng I.242- S tay tp mt

)

= (604.17

1 +706.64

1 )

hh = 13.84 < 20 dyn/cmNn theo s tay tp hai : [] = 0.8 ( y * y )tb = 0.065 * 0.8 * 701.1*28.787*3.0

= 1.042 (Kg/m2*s )ng knh on ct :

Dct = 0.0188 *042.1

72.2394= 0.901 ( m )

Chn Dct theo tiu chun : 0.9 ( m)2. ng knhon luyn :

Nng trung bnh ca pha lng :xm = ( xF + xW )/2 = (0.117 + 0.006 )/2 = 0.0615

Nng trung bnh ca pha hi theo phng trnh ng lm vic :ym = 4.8521 * x

m - 0.023

= 4.8521 * 0.0615 0.023= 0.2754

Nhit trung bnh ca pha hi, pha lng t gin t-x,y:

xm = 0.0615 tx = 73.75 OCym = 0.2754 t

y = 89.91

OC Khi lng mol trung bnh v khi lng ring pha hi :

Mm = ym * Macetone + ( 1 y

m ) * Mnc

= 0.2754 * 58 + ( 1 0.2754 ) * 18= 29.016 ( Kg/ Kmol )

y = ""

*4.22

*

y

Om

T

TM=

)91.89273(*4.22

273*016.29

= 0.974 ( Kg/m3 )

Khi lng ring pha lng :

x

m = 0.0615 x

m = 0.1743 ( phn khi lng )tx = 73.75OC acetone = 727.44 ( Kg/m

3 )nc = 975.44 ( Kg/m

3 )( Bng I.2 S tay tp mt )

"

1

x=

"

"

acetone

mx

+

''

''

)1(

nuoc

mx


28/79



=44.727

1743.0 +44.975

)1743.01(

x = 920.73 (Kg/m3 )

Lng hi trung bnh i trong on chng :gtb = ( g

n + g

1 )/2 ( Kg/h )

g1: lng hi i vo on chnggn : lng hi ra khi on chngV lng hi ra khi on chng bng lng hi i vo on luyn

nn gn = g1Hay gtb = ( g1 + g

1 )/2

Lng hi i vo on chng g1, lng lng G1 v hm lng lng

x1c xc nh theo h phng trnh cn bng vt cht v cn bngnng lng :

g

rgrgrg

xWygxG

WgG

nn

WW

11

'''

1

'

1

'

1

'

1

'

1

'

1

'

1

******

Vi W= 3644.86 ( Kg/h )Wx = 0.02

xW = 0.006 tng cn bng yW = 0.09 ( phn mol )

Wy = 0.242 ( phn khi lng )

gi * r1 = 1567612.8r1 = ra * Wy + ( 1 - Wy ) rb

Ti y thp :tW = 96OC ra = 477.71 ( Kj/Kg )

rb = 2273.43 ( Kj/Kg )( Bng I.212- S tay tp mt )

r1 = 1838.86 ( Kg/Kg )

062.0

)/(35.4497

)/(49.852

'

1

'

1

'

1

x

hKgG

hKgg

Lng hi trung bnh i trong on chng :

gtb = ( g1 + g

1 )/2 = ( 1909.448 + 760.58 ) /2 = 1380.97 (Kg/h) Vn tc hi i trong thp :

( y * y )tb = 0.065 * [] *tbtb yx

h ** (Kg/m2*s )

xtb

: khi lng ring trung bnh ca pha lng = x = 920.73 (Kg/m

3)

ytb

: khi lng ring trung bnh ca pha hi = y = 0.974 ( Kg/ m

3 )h : khong cch mm ( m ), chn h = 0.3


29/79



[] : h s tnh n sc cng b mt

hh

1 =acetone

1 +nuoc

1

(acetone , nuoc tra bng I.242 nhit ty = 89.91

O C- S tay tp mt )

= ( 151

+ 75.601

)hh = 12.03 < 20 dyn/cm

Nn theo s tay tp hai : [] = 0.8 ( y * y )tb = 0.065 * 0.8 * 974.0*73.920*3.0

= 0.853 (Kg/m2*s )ng knh on chng :

Dchng = 0.0188 *853.0

97.1380= 0.756 ( m )

Chn Dchng theo tiu chun = 0.8 ( m)

Do chn Dthp = 0.8 ( m )3. Chiu cao thp :

Chiu cao thp c xc nh theo cng thc sau :H = Ntt * ( H+ ) + ( 0.8 1.0 ) ( m )

Vi Ntt : sa thc t = 13 : chiu dy ca mm, chn = 4 ( mm ) = 0.004 ( m )H: khong cch gia cc mm ( m ) , chn theo bng IX.4a-

S tay tp hai, H= 0.3 ( m )( 0.8 1.0 ) : khong cch cho php nh v y thp

H = 13 * ( 0.3 + 0.004 ) + ( 0.8 1.0 ) = 5 ( m )

II. Tnh ton chp vng chy chuyn Chn ng knh ng hi dh = 47 ( mm ) = 0.047 ( m ) S chp phn b trn a :

N = 0.1 *2

2

hd

D = 0.1 *2

2

047.0

8.0= 29 ( chp )

( D : ng knh trong ca thp ) Chiu cao chp pha trn ng dn hi :

h2 = 0.25 * dh = 0.01175 ( m ) ng knh chp:

dch = 22 )*2( chhh dd

ch : chiu dy chp, chn bng 2 ( mm )

dch = 22 )2*247(47 = 69.39 ( mm )

Chn dch = 70 (mm)


30/79



Khong cch t mt a n chn thp :S = 0 25 ( mm ), chn S = 12.5 ( mm )

Chiu cao mc cht lng trn khe chp :h1 = 15 40 ( mm ), chn h1 = 30 ( mm )

Tit din thp :F = * D2/4 = 3.1416 * 0.82/4 =0.5062 ( m2 )

Bc ti thiu ca chp trn mm :tmin = dch + 2*ch + l2l2 : khang cch nh nht gia cc chp

l2 = 12.5 + 0.25*dch = 12.5 + 0.25*71 = 36.25 (mm)chn l2 = 35 (mm)

tmin =70 + 2*2 + 35 = 109 (mm)A. Tnh cho phn ct :

Chiu cao khe chp :

b = ( *

2

y * y)/ (g * x ) : h s tr lc ca a chp = 1.5 2 , chn = 2y = ( 4* Vy )/ ( 3600 * * d

2h * n )

Vy : lu lng hi i trong thpVy = gtb /

tby= 2193.16/1.701 = 1407.83 ( m3/h)

y = ( 4* 1407.83) / ( 3600 * * 0.0472 * 29 )

= 7.772 ( m/s) b = ( 2 * 7.7722 * 1.701 ) / ( 9.81 * 787.28 )

= 26.6 * 10-3 ( m )Chn b = 20 ( mm ) ( 10 50 mm )

Chiu rng khe chp : a = 2 : 7 mm ; chn a = 2 ( mm ) S lng khe h ca mi chp :

i = /c * ( dch b

dh

*4

2

)

c = 3 4 mm ( khong cch gia cc khe ) , chn c = 3 ( mm )

i = 3.1416/3 * ( 70 -20*4

472) = 44.4 ( khe )

Chn i = 45 ( khe )Gx : lu lng lng trung bnh i trong thp ( Kg/h )

Gx = 21

* ( G1 + GF )

=2

1 * (409.448 + 5144.86 )

= 2777.15 ( Kg/h)z : sng chy chuyn , chn z = 1c : tc cht lng trong ng chy chuyn , c = 0.1 0.2 ( m/s )Chn c = 0.15 ( m/s )


31/79



x = 787.28 ( Kg/m3 )

B. Tnh cho phn chng : Chiu cao khe chp :

b = ( * 2y * y)/ (g * x ) : h s tr lc ca a chp = 1.5 2 , chn = 2y = ( 4* Vy )/ ( 3600 * * d2h * n )

Vy : lu lng hi i trong thpVy = gtb / '

tby= 1380.97 / 0.984 = 1407.83 ( m3/h)

y = ( 4* 1417.83) / (3600 * 3.1416 * 0.0472 * 29 )

= 7.828 ( m/s) b = ( 2 * 7.828 * 0.974 ) / ( 9.81 * 920.73 )

= 0.0132 ( m )Chn b = 20 ( mm ) ( 10 50 mm )

Chiu rng khe chp : a = 2 : 7 mm ; chn a = 2 ( mm )

S lng khe h ca mi chp :i = /c * ( dch

b

dh

*4

2

)

c = 3 : 4 mm ( khong cch gia cc khe ) , chn c = 3 ( mm )

i = 3.1416/3 * ( 70 -20*4

472) = 44.4 ( khe )

Chn i= 45 ( khe )Gx : lu lng lng trung bnh i trong thp ( Kg/h )

Gx =2

1 * ( G1 + GF )

=2

1 * ( 4405.44 + 5144.86 )

= 4775.15 ( Kg/h)z : sng chy chuyn , chn z = 1c : tc cht lng trong ng chy chuyn , c = 0.1 0.2 ( m/s )Chn c = 0.15 ( m/s )x = 920.73 ( Kg/m

3 )l1 : khong cch nh nht gia chp v ng chy chuyn

Chn l1 = 75 ( mm )c : b dy ng chy chuyn, chn c = 2 ( mm )

Chiu cao lp cht lng trn mm :hm = h1 + ( S + hsr + b )

= 30 + 12.5 + 5 + 20= 67.5 (mm)

hsr : khong cch t mp di ca chp n mp di ca khe chpchn hsr = 5 mm

Tit din ng hi :


32/79



Srj = S1= * 2,h td /4 = 3.1416*0.0472/4 = 0.001735 m

2

Tit din hnh vnh khn :Saj = S2 = *( d

2ch,t - d

2h,n )/4 = 3.1416*(0.07

2- 0.051

2)= 0.001805 (m2 )

Tng din tch cc khe chp :S3 = i.a.b = 45*0.002*0.02 =0.0018 m2

Tit din l m trn ng hi :S4 = .dhi.h2 = 3.1416 * 0.047*0.01175

= 0.001735 m2Nn ta c S1 S2 S3 S4 ( hp l )

L tho lng :Tit din ct ngang ca thp F = 0.5026 m2C 1 m2 chn 4 cm2 l tho lng . Do tng din tch l tho lng

trn mt mm l:0.5026 *4 /1 = 2.0104 cm2

Chn ng knh l tho lng l 5mm = 0.5cmNn s l tho lng cn thit trn mt mm l :

2

2.0104

*

4

d 10 l

** Kim tra khong cch gia cc mm :

hmin = 23300*2

,

**

* *

y y

x ch t

F

n d

y , x : khi lng ring trung bnh ca pha hi, pha lng

y= ('

y+ "

y)/2 = 1.3375 ( Kg/m3)

x = ('x+ "x)/2 = 854.005 (Kg/m3)y : vn tc hi trung bnh i trong thp

y =

'

' "

* *

2

y y y ytb tb

y y

= 0.7441 (m/s)

hmin = 23300*2

1.3775 0.5026*0.7441*

854.005 29* *0.071

= 0.12 (m) < 0.3 (m)

Vy khong cch gia hai mm l 0.3 m l hp l. m l chp hs :

hs = 7.55*1/ 3 2/ 3

2/ 3* *y G

s

x y s

VH

S

Hs = hso = b = 20 (mm)VG = ( Vy + V

'y) /2 = 0.3924 (m

3/s)


33/79



Ss = n * S3= 29*0.0018 = 0.0522 (m2)

hs = 7.55*1/ 3 2/3

2 /31.3375 0.3924* 20 *854.005 1.3375 0.0522

= 24.805 (mm)

Nn ta c 1.24s

so

h

h

1 : kh hp l

Chiu cao mc cht lng trn g chy trn :

hwo = 2.84 * K*2/ 3

L

w

Qh

L

Vi K : h s hiu chnh cho g chy trn , ph thuc vo 2 gi tr :

x = 0.226*2.5 2.5

4.7720.226* 6.685

0.48

L

w

Q

L

Trong :QL :lu lng pha lng trung bnh trong thp c tnh nh

sau : + Lu lng cht lng trong phn ct ca thp :' '

' 3

'

* * * * 1500*0.92*47.6480.00042( / )

3600* * 3600*55.52*787.28 3600*55.52*787.28

D x m x mL

D x

G R M D R M Q m s

M

+ Lu lng cht lng trong phn chng ca thp :

" ""

" "

3

* *( )* ( )*

3600* 3600*

1500*0.92 5144.86 29.016( )* 0.00220( / )

55.52 22.68 3600*920.73

D x m x mFL

D F x D F x

G R M D R M G FQ

M M M M

m s

+ Lu lng cht lng trung bnh trong thp :QL = ( Q

'L + Q

"L) /2 = 0.0013 ( m

3/s)= 4.772 ( m3/h)

Lw : chiu di g chy trn = 0.6*D = 0.48 (m)

0.6wL

D

Tra th hnh IX22 trang 186 S tay tp hai, c K = 1.12

Do how = 2.84 * 1.12 *2 /3

4.72214.60( )

0.48mm

Gradient chiu cao mc cht lng trn mm : = Cg *

' *nh-Chiu rng trung bnh ca mm Bm :

+ Chiu rng ca ng chy chuyn : dw = 0.08* D= 0.064 (m)+ Din tch ca ng chy chuyn Sd = 0.04*F = 0.02 (m

2)Khong cch gia hai g chy trn l = D - 2.dw= D ( 1 -

2*0.08)=0.84D


34/79



+ Din tch gia hai g chy trn :A = F - 2*Sd = F(1 -2*0.04 ) = 0.92F

Chiu rng trung bnh : Bm =0.92*

0.860.84*

A FD

l D = 0.688(m)

- H siu chnh tc pha kh Cg ph thuc hai gi tr :

+ x = 1.34 * 4.7221.34* 9.1970.688

L

m

Q

B

+ 0.82* v* G = 0.82*0.7806*1.3375 = 0.74

Vi v=2 2

4* 4*0.39240.7806( / )

* 3.1416 *0.8

GV

m sD

Tra th hnh 5.10 trang 80 Tp 3 (K thut phn ring ) cCg = 0.71

- Gi tr 4* ' tra t hnh 5.14a trang 81 Tp 3 vi:x = 9.179

hsc = 12.5hm = 67.5c 4. ' = 6.5 hay ' = 6.5/4 = 1.625- S hng chp nh = 5Khi = 0.71 * 1.625 * 5 = 5.77 (mm)

o Chiu cao g chy trn hw :Do hm = hw + how + 0.5 =Suy ra hw = hm - how - 0.5* = 67.5 - 14.6 - 0.5*5.77 = 50.015 (mm)

o Kim tra sn nh ca mm : < 0.5 ( hfv + hs )

gim p do ma st v bin i vn tc pha kh thi quachp khi khng c cht lng, hfv :

hfv =274* E *2

*y G

x y r

V

S

Saj/Srj1 , nn theo hnh 5.16 trang 83 Tp 3 c E= 0.65Sr = n*Srj = 29*0.001735 = 0.050837 (m

2)

Nn hfv = 274* 0.65 *2

1.3375 0.3924*

850.01 1.3375 0.050837

= 16.645(mm)

Do 0.5*(hfv + hs ) = 0.5 * (16.645 + 24.805 ) = 20.725 > ( = 5.77

) Vy mm n nh . gim p ca pha kh qua mt mm ht :

- Chiu cao thy tnh lp cht lng trn l chp g chy trn hss :hss = hw - (hsc + hsr + Hs )

= 50.015 - ( 12.5 + 5 +20 )= 12.915 (mm)


35/79



- gim p ca pha kh qua mt mm :ht = hfv + hs + hss + how + 0.5*

= 16.645 + 24.805 + 12.515 + 15.71 + 0.5*5.77 = 72.56 (mm) Chiu cao lp cht lng khng bt trn ng chy chuyn :

hd = hw + how + + h'd +ht

- Tn tht thy lc do dng lng chy t ng chy chuyn vomm h'dc xc nh theo biu thc sau :

h'd = 0.128*2

100*

L

d

Q

S

, mmcht lng

Khong cch gia hai g chy trn l = 0.84D = 0.84*800 =672


36/79



d =v

Qy

*

*4

Qy : lu lng hi i ra khi nh thp ( m3/s)

Qy =*3600 y

yG

=701.1*3600

2880 = 0.470 ( m3/s)

v : vn tc hi i qua ng, chn v = 30 ( m/s )

d1 =4*0.470

3.1416*30= 0.141 ( m ) = 141 ( mm )

Chn d1 = 150 ( mm )Theo s tay tp hai Bng XIII-32 trang 434 , chn l1 = 130 ( mm ) (

chiu di on ni ng )2. ng dn dng chy hon lu :

d =

v

Q

*

*4

Lng hon lu G = GD * R = 1500 * 0.92 = 1380 ( Kg/h)

Q =x

G

*3600=

28.787*3600

1380 = 4.869 * 10-4 ( m3 /h )

(x : khi lng ring pha lng trong on ct = x = 787.28 Kg/m

3 )Chn v = 0.6 ( m/s )

d2 =44*4.869*10

3.1416*0.5

= 0.035 ( m ) = 35 ( mm )

Chn d2 = 40Theo s tay tp hai Bng XIII-32 trang 434 , chn l2 = 100 ( mm )

3. ng dn dng nhp liu :

d =v

Q

*

*4

Q =F

FG

*3600=

F*3600

86.5144

tF = 68.5OC ( bng I.2 S tay tp mt ) acetone = 734.52 ( Kg/m

3)nc = 978.32 ( Kg/m

3)F = 0.117 * 734.52 + ( 1 0.117 ) * 978.32

= 949.80 ( Kg/m3)

Q = 1.505 * 10-3 ( m3/s )Chn v = 0.5 ( m/s )

d3 =34*1.505*10

3.1416*0.5

= 0.062 ( m ) = 62 ( mm )

Chn d3 = 80 ( mm )l3 = 110 ( mm )

4. ng dn dng sn phmy :


37/79



d =v

Q

*

*4

Q =W

WG

*3600=

W*3600

86.3644

tW = 96O

C ( bng I.2 S tay tp mt ) acetone = 698.2 ( Kg/m3

)nc = 960.8 ( Kg/m

3)W = 0.006*698.2 + (1 0.006 ) * 960.8

= 959.22 ( Kg/m3) Q = 1.056 * 10-3 ( m3/s )

Chn v = 1.0 ( m/s )

d4 =34*1.056*10

3.1416*1.0

= 0.0366 ( m ) = 36.6 ( mm )

Chn d3 = 40 ( mm )l3 = 100 ( mm )

5. ng dn tniun qua thp :

d =v

Q

*

*4

Q ="*3600 y

WG

=

974.0*3600

86.3644 = 1.039 ( m3/s )

Chn v = 35 ( m/s )

d5 =4*1.039

3.1416*35= 0.194 ( m ) = 194 ( mm )

Chn d5 = 200 ( mm )l5 = 130 ( mm )

IV. Tnh trlc thpTr lc thp chp c xc nh theo cng thc :

P = Ntt * P ( N/m2)

Ntt : s mm thc ca thpP : tng tr lc qua mt mm

phn chng v phn ct, tr lc qua cc a khng ng u . Do chnh xc , tr lc sc tnh ring cho tng phn .A. Tng trlc phn ct :

Tng tr lc qua mt da :P = Pk + Ps +Pt

1. Tr lc a kh Pk :Pk = * y*

20 ( N/m

2) : h s tr lc a kh, = 4.5 : 5 , chn = 5y =

y = 1.701 ( Kg/m

3)o : vn tc hi qua rnh chp ( m/s )


38/79



o =bain

Vy

***

'

Vy : lu lng pha hi trung bnh i trong phn ct

Vy = 1407.83 (m3/h) = 0.3911 (m3/s)

O =02.0*002.0*45*29

3911.0 = 7.492 (m/s)

Pk = 5*1.701* 7.4922 /2 = 238.67 ( N/m2)

2. Tr lc do sc cng b mt :

Ps =tdd

*4

: sc cng b mt trung bnh ca hn hp tx = 60.25

OC acetone = 18.6*10-3 (N/m)

nc = 66.2*10-3 (N/m)

( Tra bng I-242 S tay tp mt )

hh

1 =acetone

1 +nuoc

1

hh = 14.52*10-3 (N/m)

dtd : ng knh tng ng ca khe rnh :

dtd =H

fx

.4=

).(2

..4

ba

ba

fx : din tch tit din t do ca rnhH : chu vi rnh

dtd = 4.2.20 / 2(2+20) = 3.64 ( mm ) = 3.64*10-3 (m)

Ps = 4. / dtd = 4*14.52 10-3/ ( 3.6410-3 ) = 15.972 ( N/m2)3. Tr lc ca lp cht lng trn a ( Tr lc thy tnh Pt )

Pt = b.g.( hb hr /2) (N/m2)

b : khi lng ring ca bt, thng b = ( 0.4 0.6 ). xChn b = 0.5*x = 0.5*787.28 ( Kg/m

3) = 393.64 ( Kg/m3)hr : chiu cao ca khe chp (m), hr = b = 20 (mm) = 0.02 (m)hb : chiu cao lp bt trn a (m)

hb =b

bxchbxxc

F

fhhfhfFhh

.

.).(..).).(( x

hc : chiu cao on ng chy chuyn nh ln trn ahc = 50.015 (mm) = 0.05 (m)hx : chiu cao lp cht lng ( khng ln bt ) trn a

hx = S + 0.5*b = 0.0125 + 0.5* 0.02 = 0.0225 (m) = 22.5 (mm)F : phn din tch b mt a c gn chp ( ngha l tr hai phn

din tch a b tr ng chy chuyn )Fo = F 2*Sd = 0.92 * F = 0.4624 (m

2)f : tng din tch cc chp trn a


39/79



f = 0.785* dch2 *n

= 0.785 * (70/100)2*29 = 0.115 (m2)hch : chiu cao ca chp

hch = hc + = 0.05 + 0.0157 = 0.0657 (m)

hb = 64.393*4624.0115.0*64.393*0225.0*28.787*)115.04624.0(*)0225.00157.005.0(

+

64.393*4624.0

64.393*115.0*)0225.00657.0(

= 0.0786 (m) = 78.6 (mm)Pt = 393.64 * 9.81 * ( 0.0786 0.02/2 ) = 275.14 (N/m

2)o Tng tr lc qua mt a :

P = Pk + Ps +Pt= 238.67 + 15.972 + 275.14 = 529.79 (N/m2)

o Tng tr lc phn ct:Pct = Ntt.ct * P.ct = 10* 529.79 = 2 (N/m

2)B. Tng trlc phn chng :

Tng tr lc qua mt da :P = Pk + Ps +Pt

1. Tr lc a kh Pk :Pk = *

y*

20 ( N/m

2) : h s tr lc a kh, = 4.5 5 , chn = 5y =

y = 0.974 ( Kg/m

3)o : vn tc hi qua rnh chp ( m/s )

o =bain

Vy

***

"

Vy : lu lng pha hi trung bnh i trong phn chng :Vy =1417.83 (m

3/h)= 0.3938 (m3/s)

O =02.0*002.0*45*29

3938.0 = 7.545 (m/s)

Pk = 5* 0.974 * 7.5452 /2 = 138.61 ( N/m2)

2. Tr lc do sc cng b mt :

Ps =td

d

*4

: sc cng b mt trung bnh ca hn hp tx = 73.75

OC acetone = 16.95*10-3 (N/m)

nc = 63.72*10-3 (N/m)

( Tra bng I-242 S tay tp mt )

hh

1 =acetone

1 +nuoc

1

hh = 13.39*10-3 (N/m)


40/79



dtd : ng knh tng ng ca khe rnh :

dtd =H

fx

.4=

).(2

..4

ba

ba

fx : din tch tit din t do ca rnhH : chu vi rnh

dtd = 4.2.20 / 2(2+20) = 3.64 ( mm ) = 3.64*10 -3 (m)Ps = 4. / dtd = 4*13.39* 10

-3/ ( 3.6410

-3 ) = 14.71 ( N/m2)3. Tr lc ca lp cht lng trn a ( Tr lc thy tnh Pt )

Pt = b.g.( hb hr /2) (N/m

2)b : khi lng ring ca bt, thng

b = ( 0.4 0.6 ).

x

Chn b = 0.5*x = 0.5* 920.73 ( Kg/m

3) = 460.365 ( Kg/m3)hr : chiu cao ca khe chp (m), hr = b = 20 (mm) = 0.02 (m)hb : chiu cao lp bt trn a (m)

hb =b

bxchbxxc

F

fhhfhfFhh

.

.).(..).).(( x

: chiu cao ca cht lng trn ng chy chuyn = h = 15.71(mm)

= 0.0157 (m)hc : chiu cao on ng chy chuyn nh ln trn a

hc = 50.015 (mm) = 0.05(m)hx : chiu cao lp cht lng ( khng ln bt ) trn a

hx = S + 0.5*b = 0.0125 + 0.5* 0.02 = 0.0225 (m) = 22.5 (mm)F : phn din tch b mt a c gn chp ( ngha l tr hai phn

din tch a b tr ng chy chuyn )

Fo = F 2*Sd = 0.92 * F = 0.4624 (m

2

)b = 393.64 (Kg/m3)

f : tng din tch cc chp trn af = 0.785* dch

2 *n= 0.785 * (70/100)2* 29 = 0.115 (m2)

hch : chiu cao ca chphch = hc + = 0.05 + 0.0157 = 0.0657 (m)

hb =365.460*4624.0

115.0*365.460*0225.0*73.920*)115.04624.0(*)0225.00157.005.0(

+

365.460*4624.0

365.460*115.0*)0225.00657.0(

= 0.08125 (m) = 81.25 (mm)Pt = 460.365 * 9.81 * ( 0.08125 0.02/2 ) = 321.78 (N/m

2)

o Tng tr lc qua mt a :P = Pk + Ps +Pt

= 138.61 + 8.03 + 321.78 = 4576.11 (N/m2)


41/79



o Tng tr lc phn chng :Pchng = Ntt.chng * P.chng = 3*457.11 = 1425.32 (N/m

2)Tng tr lc ca thp P :

P= Pct + Pchng= 5297.9 + 1425.32 = 6723.22 (N/m2) = 0.068 (at)

** Kim tra li khong cch mm h = 0.3 m m bo iu kin hotng bnh thng ca thp : h > 1.8*

.

d

x

P

g

V P-ct > P-chng nn ta ly P-ct kim tra :

1.8* 529.79920.73*9.81

=0 0.106 < 0.3 tha

Vy chn h= 0.3 l hp l .

CHNG 4


42/79



TNH TON C KH

1. Tnh b dy thn tr ca thpThn ca thp c ch to bng phng php hn h quang. Thn

thp c ghp t nhiu on bng mi ghp bch.Tra bng IX.5 ta chn vi ng knh trong ca thp D = 800(mm),

khong cch gia cc a H =300 (mm), chn khong cch gia hai mt

ni bch 1200(mm), sa gia hai mt bch n = 4Chn vt liu lm thn l thp khng gX18H10T. nhit lm vict = 0C .

Tc n mn ca thp 0.1 mm/nm .Da vo bng XII.4 v bng XII.7 ( Tnh cht vt l ca kim loi en v

hp kim ca chng ), cc thng sc trng ca X18H10T ( vi chiu dytm thp 4 25 mm):


43/79



Gii hn bn ko : k = 550.106 N/m2

Gii hn bn chy : ch = 550.106 N/m2H sdn khi ko nhit 20 100 0C l 16.6*10-6 1/0CKhi lng ring = 7.9*103 ( Kg/m3)H s an ton bn ko : nk = 2.6H s an ton bn chy : nch = 2.6Nhit nng chy : t = 1400 0CM un n hi : E = 2.1*105 N/mm2H s Poatxng = 0.33

iu kin lm vic ca thp chng ct :Ap sut bn trong thp ( tnh ti y thp ) vi mi trng lm vic lng

-kh:P = Ph + PL + P

Ap sut hi trong thp : Ph = 1 at = 9.81*104 ( N/m2)

Ap sut thy tnh ca ct cht lng :PL = L . g .HL = 854.01 (Kg/m3)H= 1.1 H0 = 4.347 (m) ( C kn ct cht lng y, np )PL = 854.01 *9.81 * 4.347 = 3.6420 *10

4 (N/m2)Tng tr lc ca thp : P = 7780.45 ( N/m2)Do p sut ti y thp : P = 1.423 * 105 (N/m2) < 0.25*106 (

N/m2)Theo bng XIII.8 : gi tr bn hn ca thn hnh tr, hn h quang in,

Dt = 800 (mm), thp hp kim h = 0.95Do trn thn c khot l nn h s bn ca thnh hnh tr theo phng

dc c tnh theo cng thc :0

0

5 6*0.24. 0.95*( ) 0.68

5h

H d

H

(Xem tng ng 6 l quan st ng knh = 240(mm), cha tnhn mt s l ni ng dn hi)ng sut cho php [k] ca vt liu c tnh :

[k] =6

6 2550.10. .0.95 200.95*10 ( / 2.6

k

k

N mn

)

[ch] =6

6 2220.10. *0.95 139.33*10 ( / )1.5

ch

ch

N mn

Chn []= Min ( [k], [ch]) = 139.33 *106 ( N/m2) B dy thp c tnh theo cng thc :

S=P

DP t

].[2

.

Do5

6

10*423.1

68.0*10*33.139].[

P

=665.80 > 25 nn :


44/79



S=].[2

.t

DP=

68.0*10*33.139*2

8.0*10*423.16

5

= 6.01*10-4( m)

= 0.601 (mm) B dy thc t ca thn thp :

S = S + CTrong C = Ca + Cb + CoChn thit b lm vic trong 15 nm : H s b sung do n mn :

Ca = 15*0.1 = 1.5 (mm)H s b sung do bo mn c hc ca mi trng Cb = 0H s b sung do sai lch khi ch to v lp rp : C0 = 0.4 (mm)H s quy trn bng 1.499 (mm)

Do C = 1.5 + 0.4 + 1.499 = 3.399(mm)Khi S = S +C = 4 (mm)

Kim tra b dy ca thn :- Kim tra iu kin :

1.0003.0800

5.14

1.0

D

CSa

( tha )

- Kim tra p sut tnh ton bn trong thit b :

[P] = )/(10*90.5)5.14(800

)5.14(*68.0*10*33.139*2

)(

).(]..[2 256

mNCSD

CS

a

a

Nh vy [P] > P (hp l)Nn chiu dy ca thn S = 4(mm)

2. Tnh - chn b dyy v np thit by v np cng l mt b phn quan trng thng c ch to cngloi vt liu vi thn thit b . S dng thp khng gX18H10T .

- Chn loi y np hnh elip c g- Tnh b dy y v np ging nhau :

Cc thng sy v np :y- np elip c :

25.0t

t

D

h

ht = 0.25 * Dt = 0.25 * 0.8 = 0.2 (m) = 200(mm)

Chn chiu cao g h = 25 (mm)Nn din tch b mt trong 0.76(m2 ) tra bng XIII.10 trang 382 Stay tp 2


45/79



D = 800

25

h

=

200

Bn knh cong bn trong y- np thp : Rt =Dt = 800(mm)Chiu dy y v np elip ca thit b chu p sut trong :

S =b

t

hk

t

h

D

Pk

PD

.2*

.].[*8.3

.

H s khng th nguyn :

k = 1 -tD

d = 1-8.0

2.0 = 0.75

(ng knh l np : d = 0.2 m)Do :

S =2.0*2

8.0*

10*423.195.0*75.0*10*95.200*8.3

10*423.1*8.056

5

= 0.419*10-3 (m) = 0.419 (mm)C = Ca + Cb + C0 = 1.5 + 0.4 + 0 +C0 + 2 = 4.581 (mm)(i vi y np dp th i lng b sung tng thm 2 mmkhi S < 10 mm Theo s tay tp 2 )

Kim tra p sut d cho php tnh ton :. B dy y np cn tha biu thc sau :

125.0

t

a

D

CS

125.00041.0800

5.15

(tha)

. Do p sut d cho php tnh theo cng thc :

[P] = )/(106*66.1)5.154(800

)5.15(*95.0*10*96.200*2

).(

).(]..[2 256

mNCSR

CS

at

a

P = 1.423*105

(N/mm2

) (tha)Vy b dy y np thit b l 5 (mm)3. Chn bch v vngma. Bch vm ni v bt kn thit b:

Mt bch l b phn dng ni cc phn ca thit b cng nh ni ccb phn khc vi thit b .

Chn loi bch lin khng c bng thp CT3 .


46/79



Bng XIII-27 trang 417 S tay tp hai . Cho cc kiu bch lin bngthp CT3 (Kiu I )vi thit by np nh sau :

ng knh bn trong ca thit b Dt = 800 (mm)ng knh bn ngoi ca thit b Dn = 808 (mm)ng knh tm bu lng Db = 880 (mm)

ng knh mp vt D1 = 850 (mm)ng knh bch D = 930 (mm)Chiu cao b1ch h = 22 (mm)ng knh bu lng db = M20 (mm)S bu lng z = 24 (ci)

Theo bng XIII-31 _ Tng ng vi bng XIII-27 : kch thc b mtm bt kn :

Dt = 800 (mm)H = h = 22 (mm)D1 = 850 (mm)

D2 = 847 (mm)D4 = 827 (mm)V do Dt < 1000 (mm) nn D3 = D2 +1 = 848(mm)

D5 = D4 1 = 826 (mm)

b. Bch ni ccng dn ( Bng XIII-26 trang 409 S tay tp hai ) Chn vt liu l thp CT3 , chn kiu 1

Ta c bng sau :STT Loingdn

Dy(mm)

Kch thc ni h(mm)

l(mm)Dw

(mm)D(mm)

Db(mm)

D1(mm)

Bulngdb(mm)

z(con)

1 Vo TBNT 150 159 260 225 202 M16 8 16 1302 Hon lu 40 45 130 100 80 M12 4 12 1003 Nhp liu 80 89 185 150 128 M16 4 14 1104 Dngspy 40 45 130 100 80 M12 4 12 1005 Hivoy 200 219 290 255 232 M16 8 16 130

Theo bng XIII.30 tng ng vi bng XIII-26 : kch thc b mt mbt kn :D1 tra theo bng XIII-26 .z : s rnh

Ta c bng sau :STT D

(mm)D1(mm)

D2(mm)

D3(mm)

D4(mm)

D5(mm)

b(mm)

b1(mm)

z(rnh)

f(mm)


47/79



1 150 202 191 192 171 170 5 11 3 4.52 40 80 69 70 55 54 4 1 2 43 80 128 115 116 101 100 5 1 3 44 40 80 69 70 55 54 5 1 3 45 200 232 249 250 229 228 5 1 3 4.5

4. Tnh mmChn b dy mm bng b dy thn S = 4 (mm) v vt liu l thp

X18H10T Chiu cao mc cht lng ln nht trn mm khi mm b ngp lt : hd

= 145.49 (mm)Pl = l .g . hd = 854.01 * 9.81 * 0.1455 = 1218.98 (N/mm

2 ) Nu mm khng c l :

max = 222

24.0*98.1218*

004.0*8

)33.03(3**

.8

)3(3 tRPS

= 15.222*106 (N/mm2)

= 0.33 : h s Poat xng gim sc bn do cc l trn mm gy ra th hin bng h s hiu

chnh: b

b = 758.0194

47194

t

dth

dh : ng knh ng hi bng 47 (mm)t : ( hnh v chp ) = 2*112*cos30 = 194. ng sut cc i ca mm c c l :

max =758.0

10*222.15 6max b

= 20.1 *106 (N/mm2 )

< [] = 139.33(N/mm2 )Vy b dy mm l 4 (mm)

5. Chnv tai treo thit ba. Tnh sb khi lng thp :

Khi lng np bng khi lng y ( Gi sng ng dn vo npv y gn nh nhau ) ;

Vi np y elip c Dt = 800(mm), chiu dy S = 5(mm), chiu cao g

h = 25 (mm) .Tra bng XIII.11 trang 384 S tay tp hai , ta cGnp = Gy = 1.01*30.2 (Kg) = 30.504 (Kg) Gnp y = 2*30.502 = 61 (Kg)

Khi lng mm :ng knh trong ca thp Dt = 0.8 (m)B dy mm m = 0.004(m)ng knh ng hi dh = 0.047 (m)Sng hi n = 29 (ng )


48/79



Din tch ng chy chuyn hnh vin phn Sd = 0.04*FSng chy chuyn trn mi mm z = 1S mm Nt = 13 mm

Mm = Nt .(F z.Sd n..d2

h/4) * *= 13 ( 0.5026*1 0.04*0.5026

29**0.0472/4)*0.004*7.93*103 (Kg/m3)= 177.514 (Kg)

Khi lng chp trn mm ca ton thp :Mchp = Nt *n *( * d

2ch* hch *ch * + *d

2ch*ch i * ch b*a )7.9*10

3= 13*29 (3.1416*0.070*0.0567*0.002 +

3.1416*0.072*0.002/4 -- 45* 0.002*0.02*0.002 )*7.0*103

= 98.26 (Kg)

Khi lng mm :

M mm = * D* Hthn * thn * = 3.1416* 0.8 (5 2( 0.2 + 0.025) 0.130) 0.004*7.9*103= 351.03(Kg)

Khi lng ng hi :Mng hi = *dh*hhi * h *n * Ntt *

= 3.1416*0.047*0.054*0.002*29*13*7.9*103= 47.49(Kg)

Vi hh = hchp h2 - chp +hi= 0.0657 0.01175 +2 2 = 0.054 (m)

v h2 : chiu cao chp pha trn ng hi

Khi lng mng chy chuyn :Mm = lw*hcc*cc **Nt

hcc = h S1 + hw +mm= 0.3 0.0375 + 0.05 + 0.004 = 0.316 (m)

Mmm = 0.48*0.3165*0.002*0.004 * 7.9*103 = 31.20( Kg)

Khi lng bch ni thn :ng knh bn ngoi ca thp Dn = 0.808 (m)ng knh mt bch ca thn Db = 0.930(m)Chiu cao bch h = 0.022(m)Chia thp lm 4 on , nn s mt bch l 8CT3 = 7.85*103 (Kg/m3) Mbch = 0.785*(0.93

2 0.8082)*8*0.0222*7.85*103 = 229.96(Kg)

Khi lng bch ni cc ng dn :Mb = 0.785* [(0.26

2 0.15

2)*0.016 + 2*(0.132 0.042)*0.012 +(0.1852 0.082)*0.016 + ( 0.292 0.22* 0.016

]*7.85*103


49/79



= 12.56 (Kg) Khi lng dung dch trong thp ( xem Vdung dch = 0.85 Vthp )

Mdd = 0.85 ( * D2t Ho/4 + Vy ) * xtb

= 0.85*( 3.1416*0.82 *5/4 + 2*79.6*10-3 )*854.005= 1939.96(Kg)

Vy tng khi lng ca ton thp :Mthp = 2949(Kg)

b. Chn tai treo : Chn vt liu lm tai treo l thp CT3 . Tm lt l vt liu lm thn:

[CT3]= 130 *106( N/m2)

Chn s tai treo : n = 4 Ti trng ln mt tai treo ( 4 tai treo v 4 chn ):

Q0 =8

Q =8

81.9*2949 = 0.413*104 (N)

Chn ti trng cho php ln mt tai treo l 0.5 *104 N

Theo bng XIII.36 S tay tp hai : tai treo thit b thng ng .B mt F = 72.5*104(m2)Kch thc tai treo : cho bng sau :

Khilngtaitreo(Kg)

Titrngchophplnbmtq.106(N/m2)

L(mm)

B(mm)

B1(mm)

H(mm)

S(mm)

l(mm)

a(mm)

d(mm)

1.23 0,69 100 75 85 155 6 40 15 18


50/79



20

40

18

75

15

155

15

BOPHAN TAI TREO

c. Chn chn:

Chn vt liu lm tai treo l thp CT3 .Ti trng cho php ln mt chn:0.5*104 N

Theo bng XIII.35 S tay tp hai. Chn thp i vi thit b thng ng:

BmtF.10-4m2

Titrngchophplnbmtq.106(N/m2)

L B B1 B2 H h S l d

172 0.29 160 110 135 195 240 145 10 55 23


51/79



55

23

145

160 280

10

110

135

Truc thiet b

10

10

195

236

10

BOPHAN CHAN

6. Tnh bo n

Vi vn tc hi trung bnh trong c 2 phn ca thp : chng v ct :

y = 0.7441 < 10 (m/s) v mng nc ngng chuyn ng dng, hs cp nhit trong thp :

= 1.15*Ht

gr

..

... 32

( W/m2.)

Nhit ngng t ( n nhit ha hi ) ca hi trong thp:( nhit cao nht t = tw = 96

0C)r = racetone *

acetonex + rnc * nuocr

Thnh phn khi lng tnh gi tr trung bnh ca thp v nh thp :

5.02

02.098.0

acetonex (phn trm khi lng)

acetonex = 0.273 ( phn mol) .Chn nhit ngng tnh l nhit cao nht trong thp

tngng = 960C

racetone = 114.1 (Kcal /Kg)rnc = 543 (Kcal /Kg)Do r = 0.5*114.1 + 0.5*543 = 328.55 (Kcal/Kg)


52/79



xc nh cng nhit khi cha c lp cch nhit, chn nhit tng bn trong thp l tw1 = 80

0C .T cng thc trn :

= 1.15 * 432

4

32

*

**61.46

**5

8.4186***81.9*55.328

tt

(W/m2.)

Cc gi tr (Kg/m3), (W/m.), ( Ns/m2) tra nhit trung bngca mng :

Tm = 882

8096

2

1

wng tt 0C

Mhh = xacetone * 58 + ( 1-xacetone)*18 = 27.48 (Kg/Kmol)

hh = 66.817

4.966

5.0

6.708

5.0

11

nuoc

nuoc

acetone

acetone xx

( Kg/Kmol)

= arclog ( xacetone * log acetone + (1- xacetone )log nuoc )

= arclog ( 0.273* log 0.188 + ( 1- 0.273 )*log 0.328)*10-3= 0.287*10-3 (Ns/m2)

Cphh = 0.5 * Cacetone + ( 1- 0.5) * Cnc= 0.5 * 2396 + 0.5* 4206 = 3301 ( J/Kg.)

hh = 3.58*10-8 *Chh * hh * 3

hh

hh

M

(W/m2.)

= 3.58*10-8 * 3301*817.66 * 348.27

66.817

= 0.311 ( Kcal/W.h.0C )

t1 = tng tw1 = 96 80 = 160

C = 46.61* 24.2132

16*10*287.0

311.0*66.8173

32

( W/m2.)

. qng = ng * t1 = 2132.24*16= 35115.78 (Kcal/m2.h)

Tng nhit tr ca thnh ng v lp cn :

rw = rcn1 + w

w

rcn2

Cng nhit trung bnh :

qtb = 23.7511211.214436.12878

2

kkngqq

(Kcal/m2.h.)Sai s gia qtb v qng :

%5%4242.036.12878

3.751136.12878

. Do ta phi tnh lp.

o Tnh lp ln 1 :Chn t1 = 6

0C tw1 = 900C


53/79



tm = 930C

hh = 812.97 (Kg/m3)

= 0.272 *10-3 ( Ns/m2)Cphh = 3314.12 ( J/Kg.

0C)hh = 0.309 9 ( Kcal/m.h.

0C) t1 = 6 0C

ng = 2738.83 ( Kcal/m2.h.)

qng = 16432.96 ( Kcal/m2.h)

tw = 13.61 (0C )

tw2 = 82.4 (0C )

t2 = 53.4 (0C)

qkk = 1976.2 ( Kcal/m2.h)

qtb = 9204.56 (( Kcal/m2.h)

Sai s gia qtb v qng = 0.44 > 0.05 . Do ta phi tnh lpln 2.o Tnh lp ln 2 :

Chn t1 =40C tw1 = 92

0C tm = 94

0Chh = 810.95 (Kg/m

3) = 0.269 *10-3 ( Ns/m2)Cphh = 3316.75 ( J/Kg.

0C)hh = 0.308 ( Kcal/m.h.

0C) t1 = 4

0Cng = 2737.89 ( Kcal/m

2.h.)

qng = 10951.56 ( Kcal/m

2

.h)tw = 9.068 (0C )

tw2 = 86.93 (0C )

t2 = 57.93 (0C)

qkk = 2187.93 ( Kcal/m2.h)

qtb = 6569.74 (( Kcal/m2.h)

Sai s gia qtb v qng = 0.4 > 0.05 . Do ta phi tnh lp ln3 :

o Tnh lp ln 3 :V th qua qkk v qng xc nh c tw1 = 95.5

0C

t1 = 96 96.5 = 0.5

0

Ctm = 95.75 0Chh = 809.27 (Kg/m

3) = 0.265 *10-3 ( Ns/m2)Cphh = 3320.69 ( J/Kg.

0C)hh = 0.3075 ( Kcal/m.h.

0C)o t1 = 0.5

0C


54/79



ng = 5103.35 ( Kcal/m2.h.)

qng = 2551.68 ( Kcal/m2.h)

tw = 2.11 (0C )

tw2 = 93.89 (0C )

t2 = 64.89 (0C)

qkk = 2521.31 ( Kcal/m2.h)qtb = 2536.495 (( Kcal/m

2.h)Sai s gia qtb v qng = 0.006 < 0.05 ( Tha)

Khi c lp cch nhit, nhit lng ta ra mi trng :qtt = 5%qtb = 0.05*2536.495 = 126.28 (Kcal/m

2.h)- H s cp nhit ca hi :

ng = 5103.35 ( Kcal/m2.h.0C)

kk = 2521.31 ( Kcal/m2.h.0C)

Do qtt = ng (tw1 tw2 ) = )(1

)( 21322

2

ww

wi

wwtt

rtt

- Nhit ca thnh pha trong thit b :

tw1= tng -35.5103

82.12696

ng

ttq

= 95.975 0C

- Nhit tr ca thnh thit b v lp cu :

0004705.0785.14

004.0

5000

111

w

wcauw

rr

(m2.h.0C/Kcal)

- Nhit lp tip xc gia thnh thit b v tm cch nhit tw2 :

qtt = )(1

21 ww

wi

ttr

tw2 = tw1 qtt* 1wr = 95.975 126.82*0.0004705 = 95.9150C

- Nhit lp cu tip xc vi khng kh :

tw4= tkk +31.2521

82.12629

kk

ttq

= 29.050 0C

- Nhit tw3 :

qtt = )(1

43

2

ww

cau

ttr

tw3 = qtt * rcu 2 + tw4

= 126.82*2800

1 + 29.050 = 29.0956 0C

Do b dy lp cch nhit 2 tnh theo cng thc :

)( 322

2 ww

tt

ttq

Chn lp cch nhit l aming cactng c h s dn nhit l = 0.144 (W/m.0C) = 0.124 (Kcal/m.h.0C)


55/79



)(65)(065.0)0956.29915.95(82.126

124.02

mmm

CHNG 5

TNH THIT B PH


56/79



I. Thit bgia nhit hn hpu hay thit blm ngui sn phmy1. iu kin nhit ca qu trnh :

San pham ay

Nguyen lieu

Nguyen lieu

San pham ay

960C

700C

68.50C

270C


57/79



Chnh lch nhit u nh :tn = 96 - 68.5 = 27.5

0CChnh lch nhit u ln :

tL = 70 -27 = 430C

T s

43

1.5627.5

L

n

t

t

< 2 nn c th tnh hiu s nhit

trung bnh : Ltbn

tt

t

=35.25 0C

Nhit trung bnh ca tng lu th :Do T1 - T2 < t2 - t1 nn :

Ttb = ( T1 + T2)/2 = 830C

ttb = Ttb - tb = 83 - 35.25 = 47.750C

2. Nhit ti :Nhit lng cn thit un nng nhp liu ( tnh phn cn bng

nhit ) : ).(. 12''

TTCWQ ww = 2184.86*4159.67*( 96 - 70)= 2.363*108 (J/h) =65.64(KW)

3. Chn thit b:a. Chn loi thit bng chm, t nm ngang vt liu l ng thau, h s

dn nhit = 93 W/h..Thit b gm 91 ng, xp thnh 5 hnh su cnh, sng vng ngoi

cng l 31 ng.Chn ng knh ngoi ca ng dh = 0.025m, loi ng l 25x2mm.ng knh trong ca thit b :

Dtr =T.(b 1) +4.dht : bc ng, chn t =1.2dh =1.2*0.025 = 0.03 (m)b = 2.a 1 =2.6 1 =11(a =6 :sng trn mt cnh ca hnh su cnh ngoi cng Dtr = 0.03*(11-1) + 4*0.025 = 0.4(m)

b. Cc lu cht u trng thi lng . V sn phm y c bn caohn nn cho chy bn trong ng, cn nhp liu cho chy ngoi ng

c. Xc nh h s cp nhit t thnh ng n dng nhp liu :- Tit din ngang ca khong ngoi ng l :

S = 0.785*(D2tr n.d2)

= 0.785*(0.42

91.0.0252

) = 0.081 m2

- Tc chy ca dng nhp liu :

w =S

F

F **3600

nhit trung bnh cu dng nhp liu ttb = 47.750C, ta tra cc

thng s :acetone =759.48 (Kg/m

3)


58/79



nc =988.51 (Kg/m3)

F =919.80 ( Kg/m3)

acetone = 0.251*10-3(Ns.m2)

nc = 0.573*10-3(Ns.m2)

F = 0.520*10-3(Ns.m2)

w = 0.019 (m/s)- ng knh tng ng khong ngoi ng :

dt =dnD

dnD

.

. 22

=0025*914.0

0025*914.0 22

= 0.038 (m)

- Tnh chun s Re :

Re =F

tdd

..= 1277.11

- Tnh chun s Pr :+ Chn t1 = 4

0C tt1 = ttb + t1 = 47.75 + 4 = 51.750C

Tra hnh V.12 trang 12 S tay tp hai :Pr47.75 = 3.55Pr51.75 = 3.45

007.145.3

55.3

Pr

Pr25.025.0

75.51

75.46

- Chun s Gr :

2

1

23 ....

tdgGr td

: h s gin n th tch, = 0.574*10-3 (1/) (tra bng 33 trang

420 tp 10)

23

323

)10*052.0(

4*10*574.0*80.919*038.0*81.9

Gr = 3866978.92

(Gr)0.1 = 4.56- Chun s Nu :

Nu = 0.15* 1* Re033 * Pr0.43 *Gr0.1(Pr/Prt )

0.25

1 : h s hiu chnh tra theo bng V.2 trang 15 , s taytp hai

(l/d > 50 nn chn 1 = 1) Nu = 0.15* 1* 1277.110.33*3.6560.43*4.56*1.007 = 11.00

- H s cp nhit 1 = 08.171038.0

591.0*00.11..

tdd

Nu (W/m2. )

( tra 47.75 0C bng 0.591 W/m.)d. Xc nh h cp nhit t dng sn phm y n thnh ng :

Ttb = 830C , tra cc thng s :

acetone =571 (Kg/m3)


59/79



nc =969.9 (Kg/m3)

w =964.8 ( Kg/m3)

acetone = 0.20*10-3(Ns.m2)

nc = 0.35*10-3(Ns.m2)

w = 0.349*10-3(Ns.m2)

- Vn tc trong dng ng :

w =22

'

021.0*785.0*91*86.960*3600

86.2184

*785.0**

dn

W

w

w = 0.02 (m/s)- Tnh chun s Re :

Re =w

d

.. = 1163.97

10 < Re 50 nn chn 2 = 1)

Nu = 0.15* 1* 1163.970.33*2.60.43*4.27*0.995 = 9.875

- H s cp nhit 2 = 47.316021.0

673.0*875.9..

d

Nu (W/m2. )

( tra 47.75 0C bng 0.673 W/m.)e. Nhit ti ring :


60/79



q1 = 1* t1 = 171.08*4 = 648.32 (W/m2)

q2 = 2* t2 = 361.47*4 = 1265.88 ( W/m2)

Sai s gia q1 v q2 :

495.088.1265

32.68488.1265

2

12

q

qq> 0.05 ( 5% ) , Nn ta phi tnh lp

o Tnh lp ln 1 :Chn t1 = 4.(1 + 0.459) = 6

0C tt1 = 47.75 + 6 = 53.75

0C Pr53.75 = 3.35Pr47.75 = 3.55

015.135.3

55.3

Pr

Pr25.025.0

75.53

75.47

- (Gr)0.1 = 4.75- Nu = 13.2- 1 = 205.29 (W/m2.)- q1 = 205.26*6 = 1231.72 (W/m

2)

- 0265.088.1265

123188.1265

2

'

12

q

qq< 0.05 (Hp l )

f. H s truyn nhit tng qut :

K = [2

21

1

11

rr ]-1

= Cu = 93 (W/m2.) = 2 (mm)= 0.002 (m)

1 = 205.29 (W/m2. )2= 316.47 ( W/m2.)Xem h s cu bn ca nc bn r1 = 0.8455*10

-3 (W/m2.)Xem h s cu bn ca nc thng r2 = 0.5945*10

-3 (W/m2.)

K = [47.316

110*5945.0

93002.0

10*8455.029.205

1 33 ]-1

= 105.11g. B mt truyn nhit :

F = 72.1725.35*11.105

1000*64.65

.log

tKQ (m2)

h. Chiu di mi ng :L = 48.2

91*025.0*1416.3

72.17

..

hdn

F

(m)

Chn L = 2.5 (m)


61/79



II. Thit b lm ngui sn phmnh1. iu kin nhit ca qu trnh :

N c

N c

S an pham nh

S an pham nh

Chnh lch nhit u nh :tn = 30 27= 3

0CChnh lch nhit u ln :

tL = 57.3 40 = 17.30C

Hiu s nhit trung bnh :16.8

3

3.17ln

33.17log

t 0C

Nhit trung bnh ca tng lu th :t2tb = ( 27 +40)/2 = 33.5

0Ct1tb = t2tb + tlog = 33.5 + 8.16 = 41.66

0C

57.30C

300C400C

270C


62/79



2. Nhit ti :Nhit lng cn thit lm ngui sn phm nh ( tnh phn cn

bng nhit ):

Q1N = 85.252)2740(6.4176*3666*1000

76.16764 (KW)

3. Chn thit b:a. Chn gi tr K : tra bng 4.4 Sch Tnh ton qu trnh v thit b,

Nguyn Bin , c K ( chi lu cng bc ) = 300 (Kcal/m2.)

hay K = 300*4.1868*1000/3600 = 348.9 (W/m2.)- B mt truyn nhit :

F = 812.8816.8*9.348

1000*85.252

.log

tKQ (m2)

- F = .d.n.LChn L = 4 (m)

d = 0.032 (m)

n =Ld

F

..= 220.86 (ng )

Chn theo tiu chun : thit bng chm, t nm ngang, vt liu lng thau, h s dn nhit = 93 W/h. .

Thit b gm 241 ng, xp thnh 8 hnh su cnh, sng vng ngoicng l 50 ng. Chn ng knh ngoi ca ng dh = 0.032m, loi ng l32x3mm.Chiu di

L = 3.66 (m)+ ng knh trong ca thit b :

Dtr =t.(b 1) +4.dht : bc ng, chn t =1.5dh =1.5*0.032 = 0.048 (m)b = 2.a 1 =2.9 1 =17(a = 9 :sng trn mt cnh ca hnh su cnh ngoi cng

Dtr = 0.048* (17-1) + 4*0.032 = 0.896(m)Chn D = 900(mm) = 0.9 (m)

b. Cc lu cht u trng thi lng . V sn phm y c bn caohn nn cho chy bn trong ng, cn nhp liu cho chy ngoi ng .

c. Xc nh h s cp nhit t dng sn phm nh n thnh ng- Tit din ngang ca khong ngoi ng l :

S = 0.785*(D2tr n.d2)= 0.785*(0.92 241*0.0322) = 0.442 m2

- Tc chy ca dng sn phm nh :

w =S

D

D **3600


63/79



nhit trung bnh cu dng sn phm nh t1tb = 41.660C, ta tra

cc thng s :acetone =766.17 (Kg/m

3)nc =991.25 (Kg/m

3)D =770.67 ( Kg/m

3)acetone = 0.26*10-3(Ns.m2)nc = 0.63*10

-3(Ns.m2) D = 0.27*10

-3(Ns.m2) w = 1.223*10-3 (m/s)

- ng knh tng ng khong ngoi ng :

dt =dnD

dnD

.

. 22

=032.0*2419.0

032.0*2419.0 22

= 0.065 (m)

- Tnh chun s Re :

Re =F

tdd

..= 226.9

10 < Re < 2000 : ch chy dng- Tnh chun s Pr :

+ Chn t1 = 60C tt1 = t1tb - t1 = 41.66 - 6 = 35.66

0CTra hnh V.12 trang 12 S tay tp hai :

Pr35.66 = 3.81Pr41.66 = 3.60

986.081.3

6.3

Pr

Pr25.025.0

66.35

66.41

- Chun s Gr :

2

1

23 ....

tdgGr td

: h s gin n th tch, = 1.53*10-3 (1/)(tra bng 33 trang 420 tp 10)

23

323

)10*27.0(

6*10*53.1*67.770*065.0*81.9

Gr = 201493544.7

(Gr)0.1 = 6.767- Chun s Nu :

Nu = 0.15* 1* Re033 * Pr0.43 *Gr0.1(Pr/Prt )

0.25

1 : h s hiu chnh tra theo bng V.2 trang 15 , s tay tp hai(l/d > 50 nn chn 1 = 1) Nu = 0.15* 1* 226.90.33* 3.60.43* 6.767* 0.986 = 10.4

- H s cp nhit 1 = 4.26065.0

165.0*4.10..

tdd

Nu (W/m2. )

( tra 41.66 0C bng 0.165 W/m.)d. Xc nh h s cp nhit t thnh ng n nc :


64/79



ttb2 = 33.50C , tra cc thng s :

nc =993.95 (Kg/m3)

nc = 0.75*10-3(Ns.m2)

w = 0.349*10-3(Ns.m2)

- Vn tc nc trong ng :

w =22 026.0*785.0*241*95.993*3600

76.16764

*785.0**

dn

G

n

n

w = 0.0366 (m/s)- Tnh chun s Re :

Re =

..d= 1261.12

10 < Re 50 nn chn 2 = 1) Nu = 15.71

- H s cp nhit 2 = 85.378..

d

Nu (W/m2. )

( tra 33.50

C bng 0.627 W/m2

.)e. .Nhit ti ring :- q1 = 1* t1 = 26.4*6 = 158.4 (W/m2)- q2 = 2* t2 = 378.85*1 = 378.85 (W/m2)Sai s gia q1 v q2 :

58.02

12

q

qq> 0.05 ( 5% ) . Do ta phi tnh lp :


65/79



o Tnh lp ln 1 :- Tnh chun s Pr :

+ Chn t1 = 110C tt1 = t1tb - t1 = 41.66 - 11 = 30.66

0CTra hnh V.12 trang 12 S tay tp hai :

Pr30.66 = 3.9

Pr41.66 = 3.6

980.09.3

6.3

Pr

Pr25.025.0

66.30

66.41

- Chun s Gr :

2

1

23 ....

tdgGr td

Gr = 369404832 (Gr)0.1 = 7.19-Chun s Nu :

Nu = 0.15* 1* Re033 * Pr0.43 *Gr0.1(Pr/Prt )0.25 Nu = 10.98

- H s cp nhit 1 = 88.27..

tdd

Nu (W/m2. )

- q1 = 1* t1 = 27.88*11 = 306.69 (W/m2)

Sai s gia q1 v q2 :

19.02

'

12

q

qq> 0.05 ( 5% )

o Tnh lp ln 2 :

Chn t1 = 13 tt1 = 41.66 -13 = 28.860C

Pr28.86 = 3.85Pr41.66 = 3.60

983.085.3

6.3

Pr

Pr25.025.0

66.28

66.41

- Gr = 436569346.9- (Gr)0.1 = 7.31- Nu = 11.2- 1 = 28.43 (W/m2.)- q1 = 369.58 (W/m

2)

- 024.02

'

12

q

qq< 0.05 (Hp l )

Vy cc thng s chn l ph hp .III. Thit bngng t hi lu


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1. iu kin nhit ca qu trnh

Nc

NcSan pham nh

San pham nh

Thng thng ngi ta chn loi thit b ngng t vi cht lm lnhbng nc ngng t hi.

Dng nng 57.30

C (hi ) 57.30

C (lng )Dng lnh 27 0C (lng ) 40 0C (lng )Chnh lch nhit u nh :

tn = 57.3 40 = 17.30C

Chnh lch nhit u ln :tL = 57.3 27 = 30.3

0CHiu s nhit trung bnh :

20.23

3.17

3.30ln

3.173.30log

t 0C

Nhit trung bnh ca tng lu th :tt1 = 57.3 = 33.50C

tt2 = 57.3 23.2 = 34.10C

2. Nhit tiNhit lng cn thit ngng t sn phm ra nh thp ( tip theo

phn cn bng nhit ):Q = 1500 *0.92 *544.31 *1000/3600 = 208652.17(W)

57.30C57.30C

270C

400C


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3. Chn thit bChn loi thit bng chm, t nm ngang vt liu l ng thau, h s

dn nhit = 93 W/h. .

Thit b gm 187 ng, xp thnh 7 hnh su cnh, sng vng ngoi

cng l 43 ng.Chn ng knh ngoi ca ng dh = 0.032m, loi ng l 32x3mm.ng knh trong ca thit b :

Dtr =t.(b 1) +4.dht : bc ng, chn t =1.5dh =1.5*0.032 = 0.048 (m)b = 2.a 1 =2.8 1 =15(a = 8 :sng trn mt cnh ca hnh su cnh ngoi cng )

Dtr = 0.048*(15-1) + 4*0.032 = 0.8(m)4. Xc nh h s cp nhit t dng sn phm nh n thnh ng nhit trung bnh cu dng nng tt1 = 57.3

0C, ta tra cc thng s :

=748.97 (Kg/m3) = 0.23*10-3(Ns.m2)r = 521.42 *10-3 : n nhit ngng t ca Acetonec = 2296.22 J/Kg : nhit dung ring ca Acetone = 0.165 W/m. : h s cp nhit ca Acetone

dt =dnD

dnD

.

. 22

=032.0*1878.0

032.0*1878.0 22

= 0.066 (m)

+ Chn t = 3 0C tw1 = 57.3 - 3 = 54.30C

- H s cp nhit 1 =

.28* 43

323

4

32

066.0*3*10*23.0

165.0*97.784*10*42.521*28.1

**

**

tddt

r

= 2966.56 (W/m2. )- H s cp nhit trung bnh ca chm ng :

ch = * 1Tra th V.18 trang 29 S tay tp 2 : = 0.90

ch = 0.90*2966.56 = 2669.90 ( W/m2. )

5. Xcnh h s cp nhit t thnhngn nc

- Nhit trung bnh ca dng lnh ttb2 = (40 + 27)/2 = 33.50

C , tra ccthng s :nc =993.95 (Kg/m

3)nc = 0.75*10

-3(Ns.m2)- Vn tc nc trong ng :

w =22 032.0*785.0*187*95.993

893.3

*785.0**

dn

G

n

n


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Trong Gn = 3.893 (Kg/s) w = 0.039 (m/s)

- Tnh chun s Re :

Re =

..d= 1343.46

10 < Re 50 nn chn

2= 1)

Nu = 22.69

- H s cp nhit 2 = 18.547..

d

Nu (W/m2. )

( nc tra 33.50C bng 0.627 W/m2.)

6. Nhit ti ring- q1 = ch* t1 = 2669.90*3 = 8009.7 (W/m

2)- q2 = 2* t2 = 547.18*15 = 8207.67 (W/m

2)Sai s gia q1 v q2 :

024.0

2

12

q

qq< 0.05 ( 5% )

7. H s truyn nhit

K = [2

21

1

11

rr ]-1

= Cu = 93 (W/m2.) = 3 (mm)= 0.003 (m)1 = 2966.56 (W/m2. )


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2= 547.18 ( W/m2.)

Xem h s cu bn ca nc bn r1 = 0.387*10-3 (W/m2.)

Xem h s cu bn ca nc thng r2 = 0.464*10-3 (W/m2.)

K = [18.547

110*464.0

93

003.010*387.0

56.2966

1 33 ]-1

= 328.09 (W/m2. )8. B mt truyn nhit

F = 41.2720.23*09.328

17.208652

. log

tKQ (m2)

9. Chiu di ming

L = 489.1187*032.0*1416.3

41.27

..

hdn

F

(m)

Chn L = 1.5 (m)Tm li : Thit b ngng t hi lu Thit b chm ng c :

ng knh thit b D = 0.8 (marketing) = 800(mm)Sng n = 187 ngChiu di ca ng L = 1.5 (m), ng knh d = 32(mm) = 0.032 (m)Nc i trong ng, dng sn phm nh i ngoi ng , thit bt nm

ngang.IV. Thit bniun

1. iu kin nhit ca qu trnh

Hi Nc

NcSan pham ay

ay thap (hi)

ay thap


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Ta dng hi nc bo ha c p sut tuyt i l 2at, nhit si 119.60C, cp nhit cho dung dch.

Dng nng 119.6 0C (hi ) 119.6 0C (hi )Dng lnh 96 0C (lng ) 106.5 0C (hi )

(Tra nhit si ca Acetone p sut 2at. Bng I.207 trang 246 S taytp 1)

Chnh lch nhit u vo :tn = 23.6

0CChnh lch nhit u ra :

tL = 13.10C

Hiu s nhit trung bnh :

84.17

1.13

6.23ln

1.136.23log

t 0C

2. Nhit ti

Nhit lng hi nc cn cung cp (tip theo phn cn bng nhit ):QD2 = Qy + Qw + Qxq2 + Qng2 QF - QR = 208652.17(W)Hay QD2 = 1.05 ( Qy + Qw QF - QR )

= 1.05* ( 1.95*109 + 1.465*109 1.280*109 1.844*108 )= 2.048*109 (J/h) = 5,689*105 (J/s)

3. Chn thit bt ni un Kette ring bit vi thp . Ni un tip nhn dng lng i ra

bn di thp. Nh hi nc bo ha c nhit cao hn nhit dnglng trn s gip ha hi mt phn lng y thp mc ch to hi chophn ny c iu kin i ln nh thp.

ng dng ttrong ni c kch thc : ng knh ngoi ca ng dh =0.032m, loi ng l 32x3mm.

Thit b gm 187 ng, xp thnh 7 hnh su cnh, sng vng ngoicng l 43 ng.4. Xcnh h s cp nhit t dng sn phmnhn thnhng- Xc nh chun s Re :

Khi ngng hi mt ngoi ng nm ngang :

119.60C 119.60C

106.50C

90.60C


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Rem =r

Mam-Chop Diendandaihoc.vn 05472324042011

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