Đồ án công nghệ thực phẩm 1_2

8/7/2019 n cng ngh thc phm 1_2

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PHN M U

Trong cng nghip,vic phn tch cc cu t t hn hp ban u l rt cn thit nhm mc ch hon thin,khaithc,ch bin...

C rt nhiu phng php phn tch cc cu t trong cng nghip,trong c phng php chng luyn l mttrong nhng phng php hay c s dng.

Chng l phng php tch cc cu t t hn hp ban u da vo bay hi khc nhau ca chng trong hnhp.Hn hp ny c th l cht lng hoc cht kh.Thng khi chng mt hn hp c bao nhiu cu t ta s thuc by nhiu sn phm.Vi hn hp c hai cu t ta s thu c hai sn phm l sn phm nh gm phnln l cu t d bay hi v sn phm y cha phn ln l cu t kh bay hi.

Trong thc t c nhiu phng php chng khc nhau nh : chng bng hi nc trc tip,chng ngin,chng luyn...Chng luyn l phng php chng ph bin nht dng tch hn hp cc cu t d bayhi c tnh cht ha tan hon ton hoc mt phn vo nhau.

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V V THUYT MINH DY CHUYN SN XUT

I. Thuyt minh dy chuyn sn xut :

Hn hp u t thng cha 1 c bm 2 bm lin tc ln thng cao v 3.Mc cht lng cao nht thng cao

v c khng ch nh ng chy trn. T thng cao v, hn hp u qua thit b un nng 4 . Ti y,dung dchc gia nhit bng hi nc bo ha n nhit si .Sau ,dung dch c a vo thp chng luyn quaa tip liu.Thp chng luyn gm 2 phn : phn t a tip liu tr ln trn l on luyn ,cn t a tip liu tr xung lon chng.

Nh vy, trong thp,pha lng i t trn xung tip xc vi pha hi i t di ln. Hi bc t a di lnqua cc l a trn v tip xc vi pha lng ca a trn,ngng t mt phn,v th nng cu t d bay hitrong pha lng tng dn theo chiu cao ca thp.V nng cu t d bay hi trong lng tng nn nng can trong hi do lng bc ln cng tng.Cu t d bay hi c nhit si thp hn cu t kh bay hi nn khinng ca n tng nn th nhit si ca dung dch gim. Tm li theo chiu cao thp nng cu t d bayhi ( c pha lng v pha hi) tng dn ,nng cu t kh bay hi ( c pha lng v pha hi) gim dn v nhit gim dn. Cui cng nh thp ta s thu c hn hp hi c thnh phn hu ht l cu t d bay hi cn y thp ta s thu c hn hp lng c thnh phn cu t kh bay hi chim t l ln. duy tr pha lngtrong cc a trong on luyn,ta b sung bng dng hi lu c ngng t t hi nh thp . Hi nh thpc ngng t nh thit b ngng t hon ton 6,dung dch lng thu c sau khi ngng t mt phn c dnti hi lu tr li a luyn trn cng duy tr pha lng trong cc a on luyn,phn cn li c a quathit b lm lnh 7 i vo b cha sn phm nh 8. Cht lng y thp c tho ra y thp,sau mt

phn c un si bng thit b gia nhit y thp 9 v hi lu v y thp, phn cht lng cn li c a vob cha sn phm y 10.Nc ngng ca cc thit b gia nhit c tho qua thit b tho nc ngng 11.

Nh vy, thit b lm vic lin tc (hn hp u a vo lin tc v sn phm cng c ly ra lin tc).

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II.S dy chuyn

Ch thch1-Thng cha hn hp u 5-Thp chng luyn2-Bm 6-Thit b ngng t hi lu

3-Thng cao v 7-Thit b lm lnh sn phm nh4-Thit b gia nhit hn hp u 8-Thng cha sn phm nh9-Thit b gia nhit y 10-Thng cha sn phm y11-Thit b tho nc ngng

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TNH TON THIT B CHNH

Gi thit:

- S mol pha hi i t di ln l bng nhau trong tt c mi thit din ca thp- S mol cht lng khng thay i theo chiu cao ca on chng v luyn- Hn hp u i vo thp nhit si- Cht lng ngng t trong thit b ngng t c thnh phn bng nhau ca phn hi i ra nh thp- Cp nhit y thp bng hi t gin tip

Yu cu thit b

F : nng sut thit b thit b tnh theo hn hp u = 2,5kg\sThit b lm vic p sut thng p= 1atLoi thp :thp m

iu kin

aF : nng metylic trong hn hp u =25% khi lng.

aP : nng metylen trong sn phm nh =96% khi lng.

aW : nng metylen trong sn phm y =1,5% khi lng.

M1 = 32 kg\kmol,khi lng phn t metylen.

M2 = 18 kg\kmol,khi lng phn t nc.

TNH CN BNG VT LIU

1.tnh cn bng vt liu:Theo phng trnh cn bng ca ton thp:

F=W+P (1)V phng trnh cn bng vt liu cho ring cu t d bay hi

F.aF=P.aP+W.aF (2)

T (1) v (2) ta suy ra lng sn phm y l

lng sn phm nh l

P=F-W= 0.621693122 kg/s

W=( )F P

W P

F a a

a a

=1.878306878 kg/s

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Tnh lng hn hp u F,lng sn phm nh P,lng sn phm y W theo n v kmol/saF,M1

F=( 1FaM + 1 2

FaM ).F= 0.25 1 0.25 *2.532 18 + = 0.124 kmol/s=446.4 kmol/h

P= (1

Pa

M+

2

1P

a

M

).P=

0.96 1 0.96*2.5

32 18

+

=0.081 kmol/s=291.6 kmol/h

W=F- P= 0.043 kmol/s= 154.8 kmol/h

2.tnh ch s hi lu thch hp,s a l thuyt

2.1 i nng t phn khi lng sang phn mol:

xF =1

1 2

0.2532 0.15789

1 0.25 1 0.2532 18

F

F F

a

M

a a

M M

= = ++

xP =1

1 2

0.96

32

1 0.96 1 0.96

32 18

P

P P

aM

a a

M M

= =

++

= 0.931

xW =

W

1

1 2

0.01532 0.0085

1 0.015 1 0.01532 18

W W

a

M

a a

M M

= = ++

Da vo ng cn bng lng hi ca h methanol- nc trong bng IX.2a trang 149 ni suyta c

yF = 0.5112, yW = 0.0085

th ng cn bng lng hi h methanol-nc:

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2.2 Ch s hi lu ti thiu:

Rmin=*

*

0.931 0.5112

0.5112 0.15789P F

F F

x y

y x

=

= 1,188305

2.3 Tnh ch s hi lu thch hp:

p dng cng thc : Rth= .RminTrong :

+ 1,2 =< =< 2.5 l h s hi lu+Rmin=1,188305

a) Phng trnh on luyn cho bi cng thc:

y=1

R

R +x +

0.931

1R +b)Phng trnh on chng cho bi cng thc:

y=1

R f

R

++

x -1

1

f

R

+

xW

trong

'

'

Ff

P=

= 6,174913 v wx

= 0.0085

Rth ng luyn y=f(x) m=y(0) N N(R+1)

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1.2 1.425966 0.5878x+0.3838 0.3838 12 29.112

1.25 1.485381 0.5976x+0.3746 0.3746 11 27.339

1.5 1.782458 0.6406x+0.3346 0.3346 10 27.825

2 2.37661 0.7038x+0.2757 0.2757 8 27.013

2.5 2.970763 0.7482x+0.2345 0.2345 7 27.793

=1.2

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1.25 =

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1.5 =

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2 =

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2.5 =

Vy s a l thuyt l NLT= 8 v Rth= 2.37661

II.Tnh ng knh thp:Cng thc tnh ng knh thp m:

D=0.0188 ( . )tb

y y tb

g

(m)

Vi gtb lng hi trung bnh i trong thp ( . ) y y tb : tc hi trung bnh i trong thp kg\m2.s2.1 Lng hi trung bnh i trong thp:

2.1.1: lng hi trung bnh i qua on luyn:

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gtbl= 2d l

g g+l lng hi trung bnh i trong on luyn,kg\h hoc kmol\h

gd :lng hi i ra khi a trn cng ca thp,kg\h hay kmol\h

gl : lng hi i vo a di cng ca on luyn,kg\h hay kmol\h

gd=GR+GP=GP(Rx+1)trong GR= lng cht lng hi lu = GP*Rx= 291.6*2.37661= 693.1kmol/h

GP =P= 291.6 kmol/h lu lng sn phm nh,kmol/h

Vy gd= 291.6(2.37661+1)= 984.62 kmol/hLng hi g1,hm lng hi y1,lng lng G1 i vi a di cng ca on

luyn c xc nh theo h phng trnh cn bng vt liu v cn bng nhitlng sau:

(1 )

(1 )

l l P

l l l F P P

l l d r

l a l l b

d a d d b

g G G

g y G x G x

g r g r

r r y y r

r r y y r

= + = + =

= +

= +

Trong x1 = xFV r1 : n nhit ha hi ca hn hp hi i vo a di cng ca on luyn, rd ln nhit ha hi ca hn hp hi i ra khi nh thpV ra,rb l n nhit ha hi ca cc cu t A,B nguyn chtNi suy theo bng I.213 i vi methanol v bng I.250 i vi nc t s tay qutrnh thit b cng ngh ha cht tp 1

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Ni suy ta c vi xF = 0.15789 th y*F = 0.5112, tF = 84.20C.Vi xP = 0.931 th tP= 65.5oCVi xW = 0.0085 th tw= 99.70C

Gi tr 084.2Ft C= 065.5Pt C=099.7

Wt C=

Ar (kcal/kg) 256.48 267.25 246.21

Ar (kcal/kmol) 8207.36 8552 7878.72

Br (kcal/kg) 548.68 559.41 539.58

Br (kcal/kmol) 9876.24 10069.38 9712.44

Vi cc gi tr n nhit ha hi nhit u vo ta tnh c r1 .Vi cc gi tr n nhit hahi nhit nh ta tnh c rd.Vi yd l nng hi tng ng vi nng pha lng xPtheo phng trnh ng nng lm vic on luyn :

Ta c h pt[ ]

291.6

* *0.15789 0.931*291.6

(9876.24 1368.88 ) 984.62* 8207.36*0.931 (1 0.931)*9876.24 8752111

l l

l l l

l l

g G

g y G

g y

= +

= + = + =

Gii ra ta c gl = 937.95kmol/h, Gl=646.35kmol/h, yl=0.3982 kmol/kmolVy lng hi trung bnh i trong on luyn l :

. 937.95961.29 /

2 2d l

tbl

g g g kmol h

+ += = =

984 62

Lng lng trung bnh i trong on luyn l :

G tbl =. 646.35

669.74 /2 2

R lG G

kmol h+ +

= =693 1

.

2.1.2 Lng hi trung bnh i qua on chng:

Lng hi trung bnh i qua on chng c xc nh gn ng bng cng thc tnh:

' 'n l

tbc

g gg

+=

Vi gtbc l lng hi trung bnh i quan on chng,kg/h hay kmol/h.'ng l lng hi i ra khi on chng , 'ng =gl

lg l lng hi i vo on chng ,kg/h hay kmol/h

Lng hi i vo on chng gl,lng lng G1 v hm lng lng x1 c xc nh theo h phtrnh sau:

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Gl= gl + GWGl*x1 =gl*yW + GW*xW

lg *r1 = gn*rn= g1*r1

Trong y1= yW = 0.0085 theo ng cn bng ng vi xw

' ' (1 ')l a l l br r y y r = + n nhit ha hi ca hn hp hi i vo a th nht ca on chng

' ' (1 ')n a n n br r y y r = + n nhit ha hi ca hn hp hi i vo a trn cng ca on chng

Thay s v tnh tng t phn luyn ta c hGl= gl+ 154.8

Gl*x1 =0.0085 gl + 13.1589712.44gl =8752111

Gi h ta c g1 = 901.1 kmol/h , Gl = 1055.9 kmol/h, x1= 0.0197 kmol/hVy lng hi trung bnh i qua on chng l' 'n l

tbc

g gg

+= =

937.95 901.1

2

+=919.53 kmol/h

Lng lng trung bnh i trong on chng l :

Gtbl =' 646.35+1055.9

851.13 /2 2

l lG G

kmol h+

= =

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