Nhung Khai Niem Chung Ve Truyen Khoi

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truyen khoi

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Slide 1

Mn hcQU TRNH THITB TRUYN KHI

GII THIU MN HCCHNG I: NHNGKHINIM CHUNG V TRUYN KHI CHNG II: HP THCHNG III. HP PH CHNG IV. CHNG CT CHNG V. TRCH LY CHNG VI. SY

MC TIU MN HC

Trnh by c nh ngha v phnloi c cc qu trnh truyn khi. Trnh by cc biu din thnh phnpha Trnh by c qu trnh khuch tn, ng lc khuch tn. Lm cc bi tp lin quan nchng.

Chng 1: Nhng Kin Thc CBn Ca Qu Trnh Truyn KhiI. nh ngha v phn loi cc qu trnhtruyn khiII. Cc biu din thnh phn phaIII. Cn bng phaIV. Qu trnh khuch tnV. ng lc khuch tnVI. Phng php tnh thit b truynkhi

I. nh Ngha & Phn Loi1. nh ngha:Qu trnh di chuyn vt cht t pha ny sang pha khc khi hai pha tip xctrc tip vi nhau gi l qu trnh truyn khi hay qu trnh khuch tn.

I. nh Ngha & Phn Loi2. Phn loi:

Hp thuChngHp phTrch ly

Kt tinhSyHa tanTrao i ion

II. Cc Biu Din Thnh Phn Pha 1.Cc loi nng thnh phn

a. Thnh phn phn mol (x,y)b. Thnh phn phn khi lng (yc. Thnh phn t s mol (X,Y)

, x)

d. Thnh phn t s khi lng (Y , X)e. Nng mol (Gy , Lx)

f. Nng khi lng Gy

, Lx

II. Cc Biu Din Thnh Phn PhaGiG, L : sut lng mol ca pha y (pha kh), pha x (pha lng), ,kmol/hGi , Li sut lng mol ca cu t ang xt trong pha y, pha x, kmol/hy, x :nng phn mol ca cu t ang xt trong pha y, pha x Y, X : nng t s mol ca cu t ang xt trong pha y, pha x Vx, VY : lu lng th tch pha x, pha y, m3/h

II. Cc Biu Din Thnh Phn Pha

G , L : sut lng k/lng ca pha y (pha kh), pha x (pha lng), ,kg/h

G:i

, Li

:sut lng k/lng ca cu t ang xt trong pha y, pha x

kmol/h

y, x

: nng phn k/lng ca cu t ang xt trong pha y,

pha x

Y , X

: nng t s k/lng ca cu t ang xt trong pha y,

pha xi: cu t bt kz ca hn hp

II. Cc Biu Din Thnh Phn Pha

Thnh phn khi lng

x Li

yGi

XLi

YGi

Thnh phn phn molLGixiy LGLGNng t s molXLiYGiL LiG GiNng t s khi lng L LiG GiNng molLGCx iCiVyVxyNng khi lngLiGCCixVyVxy

II. Cc Biu Din Thnh Phn Pha

2. Quan h gia cc nng thnh phn pha

II. Cc Biu Din Thnh Phn Pha2. Quan h gia cc nng thnh phn pha

II. Cc Biu Din Thnh Phn Pha

2. Quan h gia cc nng thnh phn pha

II. Cc Biu Din Thnh Phn Pha

2. Quan h gia cc nng thnh phn pha

x Cxy CyC xxxM (1 x)MABC yyyM (1 y)MAB

x Cxy CyC xM A yxM (1 x)MABC yM A yyM (1 y)MAB

II. Cc Biu Din Thnh Phn Pha BI TPBi 1. Hn hp dung dch bao gm Etanol v nc, trong etanol chim 30% th tch, nhit lm vic 20 C. Xc nh:a. Thnh phn phn khi lng.b. Thnh phn t s molc. Thnh phn t s khi lngd. Nng mole. Nng khi lng

II. Cc Biu Din Thnh Phn Pha BI TPTa c: Thnh phn thnh phn mol:

XC2 H5OH

= 0.3

a, Thnh phn phn khi lng:

x x * Me tan ol

x * Me tan ol

(1

x)Mnuoc

0.3* 460.3* 46 (1 0.3)18

0.523kg / kg

II. Cc Biu Din Thnh Phn Pha BI TPb, Thnh phn t s mol

X x1 x

0.31 0.3

0.43mol /

mol

c, Thnh phn t s khi lng

X x * Me tan ol

0.3* 46

1.1kg / kg

(1

x)Mnuoc

0.7 *18

II. Cc Biu Din Thnh Phn Pha BI TPd, Nng mol:

T nhit 200 C

tra bng ta c:

e tan ol

789kg / m3

nuoc

998kg / m3

1x

1 x

924kg / m3

Cx

e tan ol

nuocx

10.7kmol / m3

xM e tan ol

(1

x)Mnuoc

II. Cc Biu Din Thnh Phn Pha BI TPe, Nng khi lng

Cx

xM e tan ol

492,4kmol / m3

xM e tan ol

(1

x)Mnuoc

II. Cc Biu Din Thnh Phn Pha BI TP2. Chn n v ng ng vi loi thnh phnphn mola. kmol CO2 / kmol hn hp khb. mol CO2 / mol ncc. kg NaOH / kg hn hpd. kg NaOH / kg nc

II. Cc Biu Din Thnh Phn Pha BI TPBi 3 (BTVN)Mt hn hp Etanol - Nc, trong t s gia s mol Etanol trn s mol Nc bng 52%. Xc nh t s khi lng, nng phn mol, nng khi lng ca Etanol trong hn hp.Ly kt qu t s khi lng, xc nh nng phn khi lng v t s mol, nng phn mol, ca etanol trong hn hp (xem nh d liu trn cha c).

III. Cn Bng Pha 1.3.1. Khi nim v cn bng pha:

PhaxPha y

y M

x My

x

III. Cn Bng Pha

AMMONIAC + KHNG KH

- Gi:x ,yln lt l pha lng v pha kh-x ,y l nng ca ammoniac trong pha lng v pha kh

-vt , vn

l vn tc

NHIT V P SUT KHNG I

NC

ca pha kh volng, v ca lng vo khBan u, x = 0, y>0

III. Cn Bng Pha

AMMONIAC+ KHNG KH

Qa trnh truynkhi xy ra

vt vn

NHIT V P SUT KHNG I

NC +AMMONIAC

- y gim dn, x tng dn

III. Cn Bng Pha

Qa trnh cn bng ng

AMMONIAC+ KHNG KH

vt

vn

y ycbx xcb

NHIT V P SUT KHNG I

NC +AMMONIAC

Ti trng thi cn bng pha, qu trnh di chuyn vt cht gia hai pha l bng nhau

III. Cn Bng Pha 1.3.1. Khi nim v cn bng pha: Ti mi iu kin xc nh s tn ti mt mi quan h cn bng gia nng ca cu t trong hai pha v c biu din bng ng cn bng Khi cn bng th s khuch tn tng cng ca haipha bng 0 Khi cha cn bng, s xy ra qu trnh khuch tn ca cu t gia hai pha a h v trng thi cn bng Gii hn ca qu trnh truyn khi l khi h ttrng thi cn bng

III. Cn Bng Pha

Chiu khuch tn ca cu t s tun theo quy lut:

Nu nh y < ycb

Nu nh y > ycb

vt cht chuyn t pha x vo pha yvt cht chuyn t pha y vopha

xCht phn b s i vo pha no cnng lm vic thp hn nng cn bng

III. Cn Bng PhaBi 1: Trng hp no sau y vt chtch yu chuyn t pha x sang pha y?

a. x < xcbb. x = xcbc. x > xcbd.Khng xc nh

III. Cn Bng Pha 1.3.2.Quy tc phaQuy tc pha cho php xc nh c th thay i baonhiu yu t m cn bng khng b ph hy.C = k -+ n Trong :C: bc t do: s pha trong hk: s cu t c lp trong hn: s yu t nh hng n cn bng ca h

1.3. Cn Bng Pha1.3.3. Cc nh lut v cn bng pha

nh lut Henry: i vi dung dch l tng p sut ring phn p ca kh trn cht lng t l vi phn mol x ca n trong dung dchp = H.x

H l hng s Herry, c n v l p sut, phthuc vo tnh cht ca kh, lng, nhit

1.3. Cn Bng Pha1.3.3. Cc nh lut v cn bng pha

nh lut Raoult: p sut ring phn ca mtcu t trn dung dch bng p sut hi bo ha ca cu t ( cng nhit ) nhn vi nng phn mol ca cu t trong dung dchp = Pbhi.x

III. Cn Bng Pha trng thi cn bng, ta c: nh lut Henry: nh lut Raoult:p* = Po.x Theo Clapeyron v Dalton, ta c:p* = P.y* Phng trnh cn bng: y*=(H/P)x y*=(Po/P)x

III. BI TP1. Mt dung dch l tng tun theo nh lut Henry c hng s Henry l 950 mmHg, p sut lm vic ca h 860 mmHg, nng pha lng bng 0,28 phn mol. Xc nh nng cn bng ca pha kh.Ta c:H = 950 mmHg = 1.25 atm P=860 mmHg = 1.13atm x= 0.28mol/mol Nng cn bng pha kh: ycb

III. BI TPThng qua phng trnh ng cn bng ta c nng cn bng ca pha kh.

y =(H/P)x= (1.25/1.13)0.28= 0.31 mol/mol

III. BI TP2(BTVN). Mt dung dch l tng tun theo nh lut Raoult c p sut hi bo ha bng 680 mmHg, p sut lm vic ca h 735mmHg. Xc nh nng cn bng ca pha kh khi nng pha lng bng 0,33 phn mol.Ta c:

pbh

= 680 mmHg = 0.89 atm

P = 735 mm Hg = 0.97 atmx = 0.33 mol/mol

III. BI TPThng qua phng trnh ng cn bng ta cnng cn bng ca pha kh.

ycb

pbh P

x 0.890.97

0.33

0.3mol

/ mol

IV. Qu Trnh Khuch Tn1. Khuch tn phn t Xy ra trong lp mng ch chuynng dng ng lc l chnh lch nng gia hai b mt tip xc. Khuch tn t ni nng cao n ni nng thp trong lp mng Xy ra rt chm.-Phthucvobmt,thigian,nng

IV. Qu Trnh Khuch Tn2. Khuch tn i lu Xy ra trong nhn pha ch chuynng xoy Xy ra l nh s xo trn ca cc phn ttrong dngng lc ca qu trnh khuch tn i lu l s chnh lch nng trong nhn v nng b mt tip xc.Khuch tn phn t quyt nh tc cho c qutrnh khuch tn

IV. Qu Trnh Khuch TnBTVNQu trnh truyn khi xuyn pha l g?

V. ng Lc Khuch TnNu tnh theo pha yy = y* yhayy = y y*Nu tnh theo pha xx = x* xhayx = x x*

V. ng Lc Khuch Tn 1.Phng trnh truyn khi

Vn tc ca mt qu trnh no cng t l thun vi ng lc v t l nghch vi tr lc.Trong qu trnh truyn khi ng lc l hiu s nng v tr lc l s cn tr cht khuch tn chuyn ng qua lu th

Phng trnh truyn khi:

G ky F ytb

kx F xtb

Trong : G: lu lng mol ca cu t truyn t phany sang pha kia kmol/h ???? ky, kx: h s truyn khi tnh theo nng pha kh v pha lng mol/s.m2 (n v ng lc)

ytb , xtb

: ng lc trung bnh ca qu trnh

F: b mt tip xc pha, m2: thi gian truyn khi

2.ng lc trung bnh- ng lc ca qu trnh thay i t u n cui nn khi tnh ton phi dng ng lc trung bnh.-Khi ng cn bng l ng cong th tnh theo ng lc trung bnh tchphn.

y

yd yc

tbyddy

y yc

ycb

2.ng lc trung bnh- Khi ng cn bng l ng thng

tbth tnh theo ng lc trung bnh logarit.

y

yc yd

y

yc yd

tb2,3lg

ycyd

ln ycyd

2.ng lc trung bnhBI TP1. Qu trnh truyn khi ngc chiu c ng lm vic l ng thng y=0,35.x+0,6; nng phn mol ca dng lng vo v ra khi thit b ln lt l: 0,1mol/mol v 0,6mol/mol. Xc nh ng lc trung bnh ca qu trnh truyn khi theo pha kh, bit phng trnh ng cn bng y=1,1.x.

2.ng lc trung bnhBI TPTa c:Phng trnh ng lm vic: y=0,35.x+0,6Phng trnh ng cn bng: y=1,1.x

xd= 0.1 mol/molxc= 0.6 mol/mol

Tnh ng lc trung bnh ca qu trnh

truyn khi

ytb

ytb

ycln yd ycyd

2.ng lc trung bnhBI TPV y l qu trnh truyn khi ngc chiu nnTh

vhvxdo pxcng trnh lm vic ta c:

yd=0,35* xc

+0,6

= 0.35*0.6 +0.6= 0.81 mol/molyc=0,35* xd +0,6= 0.35*0.1 +0.6= 0.635 mol/mol

2.ng lc trung bnhBI TPThvvo phng trnh cn bng ta c nngxdxc pha hi cn bng:

*yd 1,1.xc

*yc 1,1.xd

1.1* 0.6 1.1* 0.1

0.660.11

ng lc truyn khi:

yd

yd

y*d

0.81