Do an Co Dac 1 KNO3word2007

bi: thit k h thng 2 ni c c xui chiu tun hon cng bc c c dung dch KNO3 vi nng sut 12000 kg/h

Trng HCN Vit Tri n Mn Hc QTTB CN Ha Hc

( N MN HC QU TRNH THIT B( H v tn : Phng Th Hng Lp : CH1D11

Khoa : Cng Ngh Ha Hc Gio vin hng dn : Nguyn Thi HinNI DUNG BI:Tinh toan thit k h thng c c lin tuc 2 ni xui chiu tun hon cng bc c c dung dch KNO3 vi nng sut 12000 kg/h.Chiu cao ng gia nhit: 2,5 m

Nng u vo ca dung dch: 8%

Nng cui ca dung dch: 38%

p sut hi t ni 1: 3,0 at

p sut hi ngng t: 0,2 atNHN XET CUA GIAO VIN

(((((

Phu Tho, Ngay Thang Nm 2013 Ngi nhn xet

MC LC

5LI M U

PHN I 7GII THIU CHUNG

81. Phn loi thit b c c:

92. C c nhiu ni:

103.Gii thiu v dung dch KNO3:

104. S dy chuyn sn xut :

114.1. S dy chuyn h thng c c 2 ni xui chiu tun hon cng bc.

124.2. Nguyn l lm vic ca h thng

PHN II 14TNH TON THIT B CHNH

141. S liu ban u :

142.Tnh cn bng vt liu :

142.1. Xc nh lng nc bc hi ( lng hi th ) ton b h thng v trong tng ni:

142.1.1. Xc nh lng hi th bc ra trong ton b h thng:

142.1.2.Xc nh lng hi th bc ra t mi ni :

152.2. Xc nh nng cui ca dung dch ti tng ni

153.Tnh cn bng nhit lng :

153.1.Xc nh p sut v nhit trong mi ni:

153.1.1 Xc nh p sut v nhit hi t trong mi ni.

163.1.2 Xc nh nhit v p sut hi th mi ni.

173.2.Xc nh tn tht nhit :

3.2.1. Tn tht nhit do nng : 17

3.2.2 Tn tht do p sut thu tnh:18

3.2.3 Tn tht do ng ng 19

193.3.Tnh hiu s nhit hu ch ca h h thng v tng ni

193.3.1 H s nhit h ch trong h thng c xc nh :

203.3.2 Xc nh nhit si ca tng ni

203.3.3 Xc nh nhit hu ch mi ni;

203.4.Lp phng trnh cn bng nhit lng:

213.4.1 Nhit lng vo gm c:

213.4.2 Nhit lng mang ra:

213.4.3 H phng trnh cn bng nhit:

254.Tnh h s cp nhit , nhit lng trung bnh tng ni:

4.1.Tnh h s cp nhit khi ngng t hi.25

264.2. Xc nh nhit ti ring v pha hi ngng t:

4.3.Tnh h s cp nhit t b mt t n cht lng si W/m2 :27

294.3.1 Khi lng ring :

294.3.2 Nhit dung ring :

294.3.3 H s dn nhit:

314.3.4 nht :

324.4.Nhit ti ring v pha dung dch :

324.5.So snh q2i v q1i :

335. Xc nh h s truyn nhit cho tng ni

346..Hiu s nhit hu ch

346.1. Xc nh t s sau :

346.2.Xc nh nhit hu ch mi ni :

357. So snh (Ti', (Ti tnh c theo gi thit phn phi p sut

358. Tnh b mt truyn nhit (F)

PHN III 36TNH TON C KH V LA CHN

361. Bung t

361.1 .Xc nh s ng trong bung t:

371.2. ng knh ca bung t :

371.3 Chiu dy bung t :

401.4.Chiu dy li ng :

421.5 .Chiu dy y bung t :

451.6.Tra bch lp y vo thn bung t :

452.Bung bc

452.1 Th tch bung bc hi :

462.2. Chiu cao bung bc :

472.3. Chiu dy bung bc:

482.4 .Chn chiu dy np bung bc ( nh y bung t ):

502.5. Tra bch lp thn bung bc :

513. Chiu dy ng c g bng thp CT3

524. Tnh ton mt s chi tit khc

524.1. Tnh ng knh cc ng ni dn hi , dung dch vo, ra thit b :

524.1.1 ng dn hi t vo :

534.1.2 ng dn dung dch vo :

534.1.3. ng dn hi th ra :

544.1.4. ng dn dung dch ra:

544.1.5. ng tho nc ngng :

544.1.6 ng tun hon:

584.2. Tnh v chn tai treo gi :

584.2.1. Tnh Gnk :

624.2.2.Tnh Gnd. :

644.3. Chn knh quan st :

654.4.Tnh b dy lp cch nhit :

PHN IV 66TNH TON THIT B PH

661.Gia nhit hn hp u :

661.1.Nhit lng trao i :( Q)

661.2.Hiu s nhit hu ch:

711.3.B mt truyn nhit:

711.4.S ng truyn nhit:

721.5.ng knh trong ca thit b un nng :

731.6.Tnh vn tc v chia ngn:

732.Chiu cao thng cao v:

823.Bm

823.1.Xc nh p sut ton phn do bm to ra:

853.2.Nng sut trn trc bm:

863.3.Cng sut ng c in:

864.Thit b ngng t baromet:

884.1.Lng nc lnh cn thit ngng t:

894.2.ng knh thit b

894.3.Knh thc tm ngn:

904.4. Chiu cao thit b ngng t:

914.5.Cc kch thc ca ng baroomet:

924.6.Lng khng kh cn ht ra khi thit b:

934.7.Tnh ton bm chn khng:

PHN V 96KT LUN

97Ti liu tham kho:

97Chuyn i n v thng gp:

LI M U bc u lm quen vi cng vic ca mt k s ha cht l thit k mt thit b hay h thng thc hin mt nhim v trong sn xut, em c nhn n mn hc: Qu trnh v thit b Cng ngh Ha hc vi bi l: thit k h thng thit b c c hai ni tun hon cng bc .Vic thc hin n l iu rt c ch cho mi sinh vin trong vic tng bc tip cn vi vic thc tin sau khi hon thnh khi lng kin thc ca gio trnh C s cc qu trnh v thit b Cng ngh Ha hc trn c s lng kin thc v kin thc ca mt s mn khoa hc khc c lin quan, mi sinh vin s t thit k mt thit b, h thng thit b thc hin mt nhim v k thut c gii hn trong qu trnh cng ngh .Qua vic lm n mn hc ny, mi sinh vin phi bit cch s dng ti liu trong vic tra cu ,vn dng ng nhng kin thc,quy nh trong tnh ton v thit k,t nng cao k nng trnh by bn thit k theo vn bn khoa hc v nhn nhn vn mt cch c h thng.

n ca em trnh by v thit b c c tun hon cng bc . Thit b c c tun hon cng bc c nhng u im nh:

H s cp nhit ln hn trong tun hon t nhin ti 3 n 4 ln v c th lm vic c iu kin hiu s nhit hu ch nh (3-5C) v cng tun hon khng ph thuc vo hiu s nhit hu ch m ph thuc vo nng sut ca bm. C c tun hon cng bc cng trch c hin tng bm cn trn b mt truyn nhit v c th c c nhng dung dch c nht ln m tun hon t nhin kh thc hin.Tuy nhin khuyt im ca thit b ny l tn nng lng bm, thng ng dng khi cng bay hi ln.

Trong n mn hc ny ca em c chia thnh 5 ni dung chnh:Phn 1: Gii thiu chung

Phn 2: Tnh ton thit b chnh

Phn 3: Tnh ton c khPhn 4: Tnh ton thit b phPhn 5: Kt lun

Do hn ch v thi gian, chiu su v kin thc, hn ch v ti liu, kinh nghim thc t v nhiu mt khc nn khng trnh khi nhng thiu st trong qu trnh thit k. Em rt mong nhn c s ng gp kin, xem xt v ch dn thm ca thy c gio v cc bn n c hon thin hn.

Em xin chn thnh cm n thy hng dn em hon thnh n ny.PHN I

GII THIU CHUNGTrong cng nghip sn xut ha cht v thc phm v cc ngnh cng nghip khc ni chung thng phi lm vic vi cc h dung dch lng cha cht tan khng bay hi, lm tng nng ca cht tan ngi ta thng lm bay hi mt phn dung mi da trn nguyn l truyn nhit, nhit si, phng php ny gi l phng php c c.

C c l mt phng php quan trng trong cng nghip sn xut ha cht, n lm tng nng cht tan, tch cht rn ha tan dng tinh th, thu dung mi dng nguyn cht. dung dch c chuyn i khng mt nhiu cng sc m vn m bo c yu cu. thit b dung c c gm nhiu loi nh: thit b c c c ng tun hon trung tm, thit b c c bung t treo, thit b c c loi mng, thit b c c c vnh dn cht lng, thit b c c phng t ngoi, thit b c c tun hon cng bc, thit b c c ng tun hon trung tm..

Ty tng sn phm nng sut khc nhau m ngi ta thit k thit b c c ph hp vi iu kin cho nng sut c cao, v to ra c sn phm nh mong mun,gim tn tht trong qu trnh sn xut.Qu trnh c c ca dung dch m gia cc cu t c chnh lch nhit si rt cao th thng c tin hnh bng cch tch mt phn dung mi. Tuy nhin, ty theo tnh cht ca cu t kh bay hi ( hay khng bay hi trong qu trnh ) m ta c th tch mt phn dung mi (hay cu t kh bay hi) bng phng php nhit hay phng php lnh.

- Phng php nhit: Di tc dng ca nhit (do un nng) dung mi chuyn t trng thi lng sang trng thi hi khi dung dch si. c c cc dung dch khng chu c nhit ( nh dung dch ng) i hi c c nhit thp, thng l chn khng. l phng php c c chn khng.

- Phng php lnh: Khi h nhit n mt mc yu cu no th mt cu t s tch ra di dng tinh th n cht tinh khit thng l kt tinh dung mi tng nng cht tan. Ty theo tnh cht ca cc cu t - nht l kt tinh dung mi, v iu kin bn ngoi tc dng ln dung dch m qu trnh kt tinh c th xy ra nhit cao hay thp v c khi phi dng n my lnh.

1. Phn loi thit b c c: Cc thit b c c rt phong ph v a dng. Tuy nhin ta c th phn loi theo 1 s c im sau:

- Theo nguyn l lm vic: C 2 loi thit b c c lm vic theo chu k v lm vic lin tc.

- Theo p sut lm vic bn trong thit b: Chia ra 3 loi: Thit b lm vic Pd, Pck- Theo ngun cp nhit:

Ngun ca phn ng chy nhin liu.

Ngun in.

Ngun hi nc: Nay l ngun cp nhit thng gp nht.

Ngun nc nng, du nng hoc hn hp iphenyl cho thit b chu k c cng sut nh.

Cu trc ca mt thit b c c thng c 3 b phn chnh sau:

- B phn nhn nhit: thit b t nng bng hi nc, b phn nhn nhit l dn ng gm nhiu ng nh trong hi nc ngng t bn ngoi cc ng, truyn nhit cho dung dch chuyn ng bn trong cc ng.

- Khng gian phn ly: Hi dung mi to ra cn cha c dung dch nn phi c khng gian ln tch cc dung dch ri tr li b phn nhit.

- B phn phn ly: tc cc git dung dch cn li trong hi.

Cu to ca mt thit b c c cn t cc yu cu sau:

- Thch ng c cc tnh cht c bit ca dung dch cn c c nh: nht cao, kh nng to bt ln, tnh n mn kim loi.

- C h s truyn nhit ln.

- Tch ly hi th tt.

- Bo m tch cc kh khng ngng cn li sau khi ngng t hi t.2. C c nhiu ni:

C c nhiu ni l qu trnh s dng hi th thay hi t, do n c ngha kinh t cao v s dng nhit.

Ngyn tc c c nhiu ni c th tm tt nh sau:

Ni th nht dung dch c un bng hi t, hi th ca ni ny a vo un ni th hai, hi th ni th hai c a vo un ni th ba,hi th ni cui cng i vo thit b ngng t. Dung dch i vo ln lt t ni n sang ni kia, qua mi ni u bc hi mt phn, nng tng dn ln.

iu kin cn thit truyn nhit trong cc ni l phi c chnh lch nhit gia hi t v dung dch si, hay ni cch khc l chnh lch p sut gia hi t v hi th trong cc ni ngha l p sut lm vic trong cc ni phi gim dn v hi th ca ni trc l hi t ca ni sau. Thng thng th ni u lm vic p sut d cn ni cui lm vic p sut thp hn p sut kh quyn (chn khng).

C c nhiu ni c hiu qu kinh t cao v s dng hi t so vi mt ni. Lng hi t dng bc hi 1 kg hi th trong h thng c c nhiu ni s tng. Di y l s liu v lng tiu hao hi t theo 1 kg hi th:Trong h thng c c 1 ni: 1,1 kg/ kg

Trong h thng c c 2 ni: 0,57 kg/ kg




Qua s liu ny cho thy, lng hi t gim i theo s ni tng nhng khng gim theo t l bc 1 m t ni 1 ln ni 2 gim 50%, cn t ni 4 ln ni 5 gim i 10%, thc t t ni 10 ln ni 11 gim i khng qu 1% ngha l xt v mt hi t h thng c c nhiu ni khng th qu 10 ni. Mt khc s ni tng th hiu s nhit c ch gim i rt nhanh do b mt un nng ca cc ni s tng.V vy, cn la chn s ni thch hp cho h thng c c nhiu ni.

3.Gii thiu v dung dch KNO3: Kali nitrat hay cn gi l dim tiu kali l cht lng dng nhng tinh th lp phng, nng chy 334C. Khng ht m, tan trong nc v tan tng nhanh theo nhit nn rt d kt tinh li. N kh tan trong ru v ete 400C, KNO3 phn hu thnh kali nitrit v oxi:It is an odorless, colorless, nontoxic solution, which is used extensively in various industries and applications around the A naturally occurring compound, Liquid Calcium Chloride CaCl2 can be found most often in sea water and mineral springs. KNO3 = KNO2+ O2A large natural underground brine deposit in northern Alberta has provided Ward Chemical with a static high quality concentration of calcium chloride (CaCl2) since 1985, enabling us to become one of the largest producers of premium liquid calcium chlorideProperties of Liquid Calcium ChlorCalcium chloride (CaCl2) , in the liquid form , is a highly soluble hygroscopic solution that is also exothermIts ability to draw in moisture from its surroundings, resist evaporation, and release heat in a chemical reaction makes it the perfect substance for road construction and maintenance, including ice and dust control and base stabilization.Do nhit nng chy KNO3 l cht oxi ho mnh, nng s oxi ho ca Mn, Cr ln s oxi ho cao hn.

Hn hp ca KNO3 v cc hp cht hu c s chy d dng v mnh lit. Hn hp gm 75% KNO3, 10% S, 15% than l thuc sng en.

Dim tiu kali cn dc dng lm phn bn, cht bo qun tht v dng trong cng nghip thu tinh. nc ta nhn dn thng khai thc dim tiu t phn di hay ng hn t t trong cc hang c di . Phn di trong cc hang lu ngy b phn hu gii phng kh NH3. Di tc dng ca mt s vi khun, kh NH3 b oxi ho thnh nitr v axit nitric. Axit ny tc dng ln vi to thnh Ca(NO3)2, mui ny mt phn bm vo thnh hang, mt phn tan chy ngm vo t trong hang. Ngi ta ly t hang ny trn k vi tro ci ri dng nc si di nhiu ln qua hn hp tch ra KNO3

Ca(NO3)2 + K2CO3 2KNO3 + CaCO3

Phng php ny cho php chng ta sn xut c mt lng dim tiu tuy t i nhng tho mn kp thi yu cu ca quc phng trong cuc khng chin chng Php trc y.4. S dy chuyn sn xut : 4.1. S dy chuyn h thng c c 2 ni xui chiu tun hon cng bc.Trong s gm nhng thit b chnh sau (nh hnh v)

1. Thng cha dung dch u 6,6. 2 ni c c

2. Bm 7. Thit b ngng t ( Baromet )

3. Thng cao v 8. Thit b thu hi bt

4. Lu lng k 9. Thng cha sn phm

5. Thit b gia nhit hn hp u 10,11. Thng cha nc ngng4.2. Nguyn l lm vic ca h thng

Dung dich u Na2SO4 cha trong thng (1) c bm (2) a vo thng cao v (3) t thng cha thng cao v c thit k c g chy trn n nh mc cht lng trong thng, sau chy qua lu lng k (4) vo thit b trao i nhit (5) (thit b ng chm). thit b trao i nhit dung dich c un nng s b n nhit si bng hi nc bo ha cung cp t ngoi vo, ri i vo ni (6). ni ny dung dich tip tc c dung nng bng thit b un nng kiu ng chm, dung dch chy trong cc ng truyn nhit hi t c a vo bung t un nng dung dch. Mt phn kh khng ngng c a qua ca tho kh khng ngng.Nc nng c a ra khi phng t bng ca tho nc ngng. Dung dch si , dung mi bc ln trong phng bc gi l hi th. Di tc dng ca hi t bung t hi th s bc ln v c dn sang buong t ca thit b (7). Dung dch t ni (6) t di chuyn qua ni th (7) do s chnh lch p sut lm vic gia cc ni, p sut ni sau < p sut ni trc. Nhit ca ni trc ln hn ca ni sau do dung dch i vo ni th (7) c nhit cao hn nhit si, kt qu l dung dch s c lm lnh i v lng nhit ny s lm bc hi mt lng nc gi l qu trnh t bc hi. Dung dch sn phm ca ni (7) c a vo thng cha sn phm (9) qua thit b bm (2). Hi th bc ra khi ni (7) c a vo thit b ngng t Baromet (10). Trong thit b ngng t , nc lm lnh t trn i xung, y hi th c ngng t li thnh lng chy qua ng Baromet vo thng cha cn kh khng ngng i qua thit b tch bt (11) hi s c bm chn khng (12) ht ra ngoi cn hi th ngng t chy vo thng cha nc ngng.H thng c c xui chiu ( hi t v dung dch i cng chiu vi nhau t ni n sang ni kia ) c dng kh ph bin trong cng nghip ha cht.Loi ny c u im l dung dch t chy t ni trc sang ni sau nh s chnh lch p sut gia cc ni. Nhit si ca ni trc sang ni sau , do , dung dch i vo mi ni ( tr ni 1 ) u c nhit cao hn nhit si , kt qu l dung dch i vo s c lm lnh i v lng nhit s bc hi thm mt lng hi nc gi l qu trnh t bc hi.Nhng khi dung dch va ni u c nhit thp hn nhit si ca dung dch , th cn phi un nongd dung dch , do tiu tn thm mt lng hi t .V vy , khi c c xui chiu , dung dch trc khi vo ni nu cn c un nng s b bng hi ph hoc nc ngng t.Uses of Calcium ChlorThe innate properties of liquid calcium chloride CaCl2 lend this substance to a variety of applications. PHN IITNH TON THIT B CHNH1. S liu ban u : Thit k h thng 2 ni c c xui chiu tun hon cng bc c c dung dch KNO3 vi nng sut 12000 kg/h.

Chiu cao ng gia nhit: 2,5 m

Nng u vo ca dung dch: 8%

Nng cui ca dung dch: 38%

p sut hi t ni 1: 3,0 at

p sut hi ngng t: 0,2 at2.Tnh cn bng vt liu : 2.1. Xc nh lng nc bc hi ( lng hi th ) ton b h thng v trong tng ni:

2.1.1. Xc nh lng hi th bc ra trong ton b h thng:

p dng cng thc (VI.1/ST 2 T 55)

W = Gd Gc = Gd (1 - )

W = 12000 (1 -) = 9473,684(kg/h)

2.1.2.Xc nh lng hi th bc ra t mi ni :

W1 : Lng hi th bc ra t ni 1

W2 : Lng hi th bc ra t ni 2

Chn t l phn phi hi th hai ni nh sau:

M ta c: W1 + W2 = 10266,667

(kg/h)2.2. Xc nh nng cui ca dung dch ti tng ni

x1:nng cui ca dung dich ti ni 1

x2:nng cui ca dung dich ti ni 2

p dng cng thc :

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3 W1 = Gd ( 1- ) x 1 =

x 1 =

x2 = 45%

3.Tnh cn bng nhit lng : 3.1.Xc nh p sut v nhit trong mi ni: 3.1.1 Xc nh p sut v nhit hi t trong mi ni. - chnh lch p sut gia hi t ni 1 v thit b ngng t l:

- Chn t l chnh lch p sut hi t 2 ni l:

m:

* Vy p sut hi t tng ni l:

* Xc nh nhit hi t 2 ni:

Tra bng (I.251/ST1-T316):

3.1.2 Xc nh nhit v p sut hi th mi ni.

NX: khi hi th i t ni 1 sang ni 2 ,v hi th t ni 2 i sang thit b ngng t th s chu tn tht v nhit l ,v khi n s tr thnh hi t cho ni 2: chn

Gi nhit v p sut ca hi th ni 1 v ni 2 ln lt l:

Ta c:

Tra bng (I.250/ST1-T312), ng vi mi nhit hi th ca mi ni s cho p hi th tng ng:

Kt qu tnh c cho ta bng di y:

Bng 1:LoiNi INi 2Hi ngng t

p sutNhit (0C)p sutNhit (0C)p sutNhit (0C)

Hi tP1= 4t1=142,9P2=1,443t2=109,56Png=0,4tng=75,4

Hi thP1=1,490t1=110,56P2=0,418t2 =76,4

3.2.Xc nh tn tht nhit : 3.2.1. Tn tht nhit do nng :

p dng cng thc:

(CT.VI.10/ST2 T59)

(o : tn tht nhit do nhit si ca dung dch ln hn nhit si ca dung mi p sut thng

T: nhit si ca dung mi nguyn cht p sut cho

EMBED Equation.3 r: n nhit ho hi ca dung mi nguyn cht p sut lm vic J/kg

* Tra bng (VI.2/ST2 T63)

* Xc nh nhit Ti

* Xc nh ri:

Tra bng (I.250/ST1 T312)

Vy:

3.2.2 Tn tht do p sut thu tnh:

- p dng cng thc VI.13

: p sut hi th trn b mt thong (at)

:chiu cao ca lp dung dch si k t ming trn ca ng truyn nhit (m)

: chiu cao ca ng truyn nhit (m)

: khi lng ring ca dung dch khi si (kg/m3 ). Ly gn ng bng khi lng ring ca dung dch 20C

: gia tc tng trng m/s2- Khi lng ring ca dung dch KNO3 20C ng vi mi nng c xc nh theo bng (I.46/ST1 T42)

Vy khi lng ring ca dung dch si l

x- chn h1 = 0,5 m v h2 = 2,5 m ( ra )

Tra bng (I.251/ST1- T314)

Vy:

3.2.3 Tn tht do ng ng

- Nh ni trn ta chn tn tht nhit do ng ng l :1oC

Vy:

tng tn tht nhit l:

3.3.Tnh hiu s nhit hu ch ca h h thng v tng ni

3.3.1 H s nhit h ch trong h thng c xc nh :

EMBED Equation.3 (CT VI.16/ST2 T67)

Hiu s nhit chung gia hiu s nhit hi t ni 1 v nhit ngng thit b ngng t.

(CT VI.16/ST2 T67)

Vy :

3.3.2 Xc nh nhit si ca tng ni

nhit hi th ca tng ni

3.3.3 Xc nh nhit hu ch mi ni;

3.4.Lp phng trnh cn bng nhit lng: GdCdtso W1i1 W1.i2 W2i3 W

Qm1 Qm2 D.i

W1Cn1

DCn1 (Gd-W1)C1ts1 (Gd-W1-W2 )C2.ts2

D:Lng hi t vo kg/h

:hm nhit ca hi t v hi th J/kg

: Nhit nc ngng ni 1, ni 2

Cd, C1,Cn1,Cn2,C2: nhit dung ring ca dung dch u ,cui v nc ngng.

Qm1,Qm2 : nhit lng mt mt ni 1 v ni 2

Gd : lng hn hp u i vo thit b

W1 , W2 : lng hi th bc ln t ni 1, ni 2

3.4.1 Nhit lng vo gm c:

- Ni 1: Nhit do hi t mang vo : D.i

Nhit do dung dch mang vo : G

- Ni 2: Nhit do hi th mang vo : W1.i2 Nhit do dung dch t ni 1 chuyn sang : (Gd W1)C1ts1

3.4.2 Nhit lng mang ra:

- Ni 1:

- Hi th mang ra : W1i1

- Nc ngng :D..Cn1

- Dung dch mang ra : (Gd W1)C1ts1

- Nhit mt mt : Qm1=0,05D(i - C1)

- Ni 2 :

- Hi th : W2i3

- Nc ngng : W1..Cn2

- Do dung dch mang ra : (Gd W1 W2)C2.ts2

- Nhit mt mt: Qm2 = 0,05W1(i2 Cn2)

3.4.3 H phng trnh cn bng nhit:

c thnh lp da trn nguyn tc :

Tng nhit i vo = Tng nhit i ra

- Ni 1 :

(1)

- Ni 2 :

(2)

m ta li c: (3)

Kt hp pt (1),(2),(3) ta c

(4)

(5)

- Nhit nc ngng ly bng nhit hi t

-Nhit si ca dung dch:

Tra bng I204/ Tr 236/ ST1 x0 = 12% : ts0 =100,8863C tnh c : ts1 = 114,06oC

ts2 = 86,897oC

- Nhit dung ring ca nc ngng tng ni tra theo bng

(I.249/ST1 T310)

(1 = 142,9 oC ( Cn1 = 4294,25 (J/kg )

(2 = 109,56 oC ( Cn2 = 4232,428 (J/kg )

- Nhit dung ring ca hi t vo ni 1 ,ni 2, ra khi ni 2 :

- Dung dch vo ni 1 c nng xd = 12%

p dng cng thc I.41 /ST1 T152 ta c:

Cd = 4186 (1- x) = 4186 (1- 0,12) = 3683,68 (J/kg )

- Dung dch trong ni 1 c nng x1 = 18,95 %

Cng p dng cng thc trn ta c:

C1 = 4186 (1- x) = 4186 (1- 0,1895) =3392,753 (J/kg )

- Dung dch trong ni 2 c nng xc = 45 %

p dng cng thc I.44/ST1 T152 ta c:

C2 = Cht.x + 4186 (1- x)

Vi Chr l nhit dung ring ca KNO3 c xc nh theo cng thc I.41/ST1 T152:

M.Cht = n1.c1 + n2.c2 + n3.c3

trong :M : KLPT ca KNO3 : M1 = 101

n1 : S nguyn t K : n1 = 1

n2 : S nguyn t N : n2 = 1

n3 : S nguyn t O : n3 = 3

c1 , c2 c3 : Nhit dung ring ca nguyn t K, N, O .

Tra t bng I.141 /ST1 T152

c1 = 26000 J/kg.nguyn t.



Vy :

(J/kg )

Vy: C2 = 1013,8614.0,45 + 4186.(1- 0,45) = 2758,53763 (J/kg )

- Xc nh hm nhit hi t v hi th;

Tra bng ( I.250/ST1 312 )

Thay cc kt qu ta tnh ton c vo pt (1) v pt (2) ta c kt qu sau :

Ta c bng s liu nh sauBng 3NiC

J/kg Cn

J/kg (, (CW , kg/hSai s

(

CBVCCBNL

13392,754294,25142,95133,3345093,4422,42

22758,537634232,428109,565133,3345173,2252,42

T l phn phi hi th 2 ni c th hin nh sau W1 : W2 = 1: 1,05

Sai s gia W c tnh t phn cn bng nhit lng v s gi thit trong cn bng vt cht < 5% ,vy tho mn.

4.Tnh h s cp nhit , nhit lng trung bnh tng ni:

4.1.Tnh h s cp nhit khi ngng t hi.

- Gi thit chnh lch nhit gia hi t v thnh ng truyn nhit ni 1 v ni 2 l :

- Vi iu kin lm vic ca phng phng t thng ng H = 2,5m ,hi ngng bn ngoi ng ,mng nc ngng chy dng nh vy h s cp nhit c tnh theo cng thc ( V.101/ST2 T28 ).

W/m2.

Trong :

: h s cp nhit khi ngng hi ni th i W/m2.

: hiu s gia nhit ngng v nhit pha mt tng tip xc vi hi ngng ca ni I ( o C ).

Gi thit:

ri: n nhit nhit ngng t tra theo nhit hi t:

(Tra bng I.250/ST1 T321)

ta c:

t1 = 142,9 oC r1 = 2315,5 .103 J/kg

t2 = 109,56oC r2 = 2235,232.103 J/kg

A: h s ph thuc nhit mng nc ngng

Vi tm c tnh:

tmi = 0,5(tTi +ti ) oC ( * )

ti: nhit hi t

tTi : nhit b mt tng

m ta li c:

( * * )

thay (**) vo (*) ta c :

Vi:t1 = 142,9 oC

tm1 = 142,9 0,5.2,97 = 141,415 oC

t2 = 109,56 oC tm2 = 109,56 0,5.2,74 = 108,245oC

Tra bng gi tr A ph thuc vo tm : (ST2 T 29 )

vi:t1 = 141,55oC A1 = 194,2123

t2 = 108,245 oC A2 = 182,7103Vy:

4.2. Xc nh nhit ti ring v pha hi ngng t:

( CT 4.14/QTTB1 T1 )

W/m2

q11 = 9175,0340.2,97 = 27249,8501 (W/m2)

q12 = 8908,4283.2,74 = 24409,0936 (W/m2)

Bng 4:

Ni

EMBED Equation.3

12,97141,415194,21239175,034027249,8501

22,74108,245182,710258908,428324409,0936

4.3.Tnh h s cp nhit t b mt t n cht lng si W/m2 : Ta xc nh h s ny theo cng thc:

(CT /QTTB1 T332)

(W/m2 )

Pi: p sut hi th at

Xem bng 1:

: hiu s nhit gia thnh ng vi dung dch si.

- Hiu s nhit gia 2 mt thnh ng truyn nhit

, oC

- Tng nhit tr ca thnh ng truyn nhit

m2 /W

r1 , r2 : nhit tr ca cn bn 2 pha tng ( bn ngoi cn bn ca nc ngng ,bn trong cn bn do dung dch.

- Tra theo bng ( V.I/ ST2 T4 )

r1 = 0,387.10-3 m2 /W

r2 = 0,232.10-3 m2 /W

- Tra bng ( VI.6/ST2 T80 ) ta chn b dy thnh ng truyn nhit l

- Chn vt liu lm ng truyn nhit l thp CT3, h s dn nhit ca n l:

W/m. ( bng PL. 14/ Bt T1/ 348 )

( m2 /W

(

Vy :

* : h s hiu chnh ,xc nh theo cng thc(VI.27/ST2 T71)

( dd:dung dch , nc: nc )

Trong :

: h s dn nhit , W/m.

:khi lng ring , kg/m3C: nhit dung ring , J/kg.

: nht , Cp

: ly theo nhit si ca dung dch.

ts1 = 114,06 oC

ts2 = 86,897 oC

4.3.1 Khi lng ring :

- Khi lng ring ca nc: tra bng (I.249/ST1 T310)

- Khi lng ring ca dung dch KNO3 :tra bng ( I.46 /ST1 42 )

4.3.2 Nhit dung ring :

- Nhit dung ring ca nc :tra bng ( I.249 /ST1 T 310 )

Cnc1 = 4239,902 J/kg.

Cnc2 = 4227,661 J/kg.

- Nhit dung ring ca dung dch KNO3:( theo bng 3 )

Cdd1 = 3392,75 J/kg.

Cdd2 = 2758,53763 J/kg.

4.3.3 H s dn nhit: - H s dn nhit ca nc: tra bng (I.129/ST1 T133 )

W/m.

W/m.

- H s dn nhit ca dung dch c xc nh theo cng thc

(I.32/ST1 T123 )

A:h s t l ph thuc hn hp cht lng :ta chn A = 3,58.10-8

M: khi lng mol ca hn hp lng. (hn hp ca chng ta l

KNO3 v H2O )

nn : M = 101.a +(1- a)18

ni 1 :x = 18,95 % khi lng

ni 2 : x = 45% khi lng.

Vy :

W/m.

W/m.

4.3.4 nht : - nht ca nc tra bng ;(I.104/ST1 96) v (I.102/ST1 95)

( Cp)

nht ca KNO3 ( bng I.107/ ST1- 101 )

Bng 5:

Ni

W/m.

W/m.

Kg/m3

Kg/m3M

10,46630,68551047,97947,792621,32

20,43090,67861241,018967,347128,566

Ni

J/kg.

J/kg.

Cp

Cp

13392,754239,9020,52580,2463

22758,53774227,6610,980,3280

Vy:

Vy h s cp nhit t b mt t n cht lng si hon ton xc nh nh sau:

(W/m2. )

(W/m3. )

4.4.Nhit ti ring v pha dung dch :

(W/m2)

EMBED Equation.3 (W/m2)

4.5.So snh q2i v q1i : - Chnh lch gia q21 , q11 v q22 , q12 ( )

Vy gi thit c chp nhn.

5. Xc nh h s truyn nhit cho tng ni

p dng cng thc:

N/m2.

Trong :

qtbi : nhit ti ring trung bnh ca tng ni (W/m2 )

:Hiu s nhit hu ch ca tng ni ( oC ) (xem bng 2)

Ta c:

(W/m2)

(W/m2)

Vy:

N/m2.

- Dung dch vo ni 1 nhit si nn lng nhit tiu tn ni 1 c tnh theo cng thc

Vi W1 = 3763,698

r1 = 2517,2.103 J/Kg - Ni 2 c hin tng qu nhit ca dung dch gi l hin tng t bay hi nn lng nhit cn thit ni 2 l:

6..Hiu s nhit hu ch

6.1. Xc nh t s sau :

6.2.Xc nh nhit hu ch mi ni : Cng thc:

oC

vy:

7. So snh (Ti', (Ti tnh c theo gi thit phn phi p sut

NX: Sai s ny nh hn 5% ,vy phn phi p sut nh trn l hp l

Bng 6.

ni oC

W/m2. W/m2

17,77410,57613741,77929088,9641

29,29350,58952903,36326978,9153

niKiN/m2. Qi , W,oC,oC

1970,91982206467,95329,821328,35494,92

2906,48582191142,02629,093330,36424,37

8. Tnh b mt truyn nhit (F)

Tnh b mt truyn nhit theo phng thc b mt truyn nhit gia cc ni bng nhau:

m2Vy:

Ta theo bng (VI.6-tr.80-T2) th Fchun ly bng 80 (m2 ).PHN IIITNH TON C KH V LA CHN1. Bung t

Thit b lm vic iu kin p sut thp ( fmin

- Bn di tc dng ca cc loi ng sut . tha mn yu cu ny cn kim tra mng ng theo gii hn bn un:

Trong :

- Pb : p sut lm vic .N/m2Pb = 0,4181.106 N/m2

- dn : ng knh ngoi ca ng truyn nhit , m

dn = 25 m = 0,025 mm

Nhn vo hnh v ta c :

Vy :

Vy tha mn iu kin nn ta chn chiu dy mng li ng l 17 mm

1.5 .Chiu dy y bung t :

y bung t l nhng b phn quan trng ca thit b thng c ch to cng vt liu vi thn thit b , y l thp CT3.

y ni vi thn thit b bng cch ghp bch .

y chn elip c g i vi cc thit b c thn hn thng ng p sut

trong >7.104 (N/m2 )

Chiu dy y phng y phng t c xc nh theo cng thc

XIII.47/ST2 T385

, m

Trong :

- hb : chiu cao phn li ca y ,m

Theo hnh XIII/ST2 T 381

hb = 0,25.Dtr = 0,25.1000 = 250 mm

- : h s bn ca mi hn hng tm

Xem bng XIII.8/ST2 T362 ta c :

- k : hng s khng th nguyn ( h s bn ca y ) , c xc nh theo cng thc XIII.48/ST2 T385

k = 1- (d/Dtr )

- Vi : d l ng knh ln nht ca l khng tng cng .y c mt l hnh trn cho dung dch vo c ng knh d , c tnh theo cng thc 1.19 /TTQTTB1 T 13

Trong :

+ V: lu lng dung dch vo ni 1 ,m3 /h

+ : vn tc thch hp ca dung dch trong ng , n gin ta chn

= 1 m/s

Quy chun d = 0,15 m (

- C : i lng b xung , c tnh theo cng thc XIII.17/ST2 T363.C tng thm mt t i vi y :

Thm 2 mm khi S C 20 mm

- P : p sut hi t 4at

- nn ta c th b i lng P mu

Vy:

( i lng b xung C khi S C = 1,86.10-3 m = 1,86 mm A = 194,2682 Thay s: 1= 2,04. 194,2682. = 7943,0329(W/m2.)

b) Nhit ti ring v pha hi ngng t:

q1 = 1.t1 [W/m2]

Thay s: q1 = 7943,0329.4,4 = 34949,34 (W/m2 )

Theo cng thc V.40-ST2/14 ta c: c) H s cp nhit pha hn hp chy xoy:Nu = 0,021.k.Re0,8.Pr0,43.()0,25M Nu = ( t = 0,021..k.Re0,8.Pr0,43.()0,25

*) k : H s hiu chnh tnh n nh hng ca t s gia chiu di L v ng knh d ca ng.Chn ng knh d = 38(2 mm; L = 2(m) Ta c: = 58,824 > 5m k= 1 (theo V.2- ST2/15)

*) Tnh chun s Pr: Pr =

EMBED Equation.3 (CT-V.35-ST2/12)+ Cp: Nhit dung ring ca hn hp t2tb = 79,25730C

Cp=C0= 3707,4 J/kg. + Tra bng (I .107- ST1/101)ta c nht dung dch: = 0,4104.10-3Ns/m2 + H s dn nhit ca hn hp =A.Cp.

- Tra bng I.46-ST1/42 : khi lng ring ca hn hp ttb =1037kg/m3

- M: khi lng mol ca hn hp lng. ( hn hp ca chng ta l KNO3 v H2O ) nn : M = 132.a +(1- a)18x = 13,3 % khi lng nn:

Vi A=3,58 .10-8 thay s:

=3,58.10-8.3707,4.1037. =0,511(W/m.)

Thay s: Pr = = 3,0281

+Hiu s nhit 2 pha thnh ng:

tt = tt- tt= q1.rt Trong : tt: Nhit thnh ng pha hn hp

rt: Tng nhit tr 2 bn ng truyn nhit

ng dn nhit lm bng lm thp CT3 c chiu dy = 2 (mm) nn: = 46,4 (W/m ) ( rt =0.6621.10-3 m2 /W( tnh trn)Thay s: tt =34949,3448.0,6621.10-3 = 23,140C

=>tt2 = tt1 tt = 143,988- 23,14=120,8480C

t2 = - t2tb=120,84879,2573= 41,59070C

*)Tnh chun s Pr=

EMBED Equation.3 t

Trong :

+ Cpt: Nhit dung ring ca hn hp tt=120,8480C Cpt =C1= 3457,1235J/kg.

+ t: nht ca hn hp tra bng I.107-ST1/101:

t = 0,2802.10-3Ns/m2 + t: h s dn nhit ca hn hp tt2t = A.Cp.

Vi: A = 3,85.10-8

: khi lng ring ca hn hp tt

Tra bng I.46-ST1/42 ta c: = 1005,2kg/m3

Thay vo cng thc ta c:

(t = 3,58.10-8.3507,1236.1005,2.

= 0,4638 ( W/m2.)

Thay s(25) : Pr= = 2,1188Hn hp chy xoychn Re = 10000 Thay s ta c h s cp nhit pha hn hp chy xoy :

t = 0,021..(10000)0,8. (3,0281)0,43.()0,25= 880,256d. Nhit ti ring v pha dung dch:

q2 = t.t2 =41,5807.880,256= 36610,444e. Kim tra sai s:

= = .100% = 4,75 %

Sai s nh hn 5% ta chp nhn gi thit chn chnh lch nhit

1.3.B mt truyn nhit:

Cng thc tnh:

F =

Trong : Nhit lng trao i : Q = (W)

q tb:Nhit ti ring trung bnh v pha dung dch

qtb =

Thay s:

F =

1.4.S ng truyn nhit:

Cng thc tnh: n=

Trong : F: B mt truyn nhit d: ng knh ng truyn nhit d = 0,034 m

H: Chiu cao ng truyn nhit H = 3 (m)

Thay s: n=

Qui chun n = 127 ng .Theo bng V.11-ST2/48 ta c:

S hnh 6 cnhSp xp ng theo hnh 6 cnh

S ng trn ng xuyn tm 6 cnhTng s ng khng k cc ng trong cc hnh vin phnTng ng trong tt c cc hnh vin phnTng ng trong thit b

Dy 1Dy 2Dy 3

613127127

1.5.ng knh trong ca thit b un nng :

D = t.(b 1) + 4.dn

(CT-V.50-ST2/49)

Trong : dn : ng knh ngoi ca ng truyn nhit

dn = 0,038 (m)

t: Bc ng. Ly t = 1,4 dn.

t = 1,4 .0,038 = 0,0532

b: s ng trn ng xuyn tm ca hnh 6 cnh

b= 13Thay s: D =0,0532.(13-1) + 4.0,038 = 0,7904 m

Qui chun: D = 0,8 m=800 (mm) (bng VIII.6-ST2/359 )1.6.Tnh vn tc v chia ngn: *) Vn tc thc:

G=12587 kg/h

n=127 ng d=0,034m =1007,8kg/m3

Thay s ta c:

(m/s)

*) Vn tc gi thit:

(m/s)

V > 5% nn ta cn chia ngn qu trnh cp nhit ch xoy.S ngn c xc nh nh sau:

S ngn cn thit:

EMBED Equation.3

(Quy chn 3 ngn2.Chiu cao thng cao v:

p sut ton phn cn khc phc sc cn thy lc trong h thng khi dng chy ng nhit:

(cng thc II.53-ST1/376)

Trong :

+: p sut cn thit to tc cho dng chy ra khi ng dn:

Vi:

: khi lng ring ca cht lng

w : vn tc ca lu th.

+: p sut khc phc tr lc khi dng chy n nh trong ng thng.

=

Vi: dtd; iu kin ca ng

L: chiu di ng dn

:h s ma st.

+: p sut cn thit khc phc tr lc cc b:

vi: : h s tr lc cc b

+: p sut cn thit khc phc tr lc trong thit b . =0

+: p sut b sung cui ng ng, =0

a)Tr lc ca on ng t thit b gia nhit hn hp u n ni c c :

+)p sut ng hc :

C 1007,8kg/m ( nhit ttb)

Chn ng knh ng dn liu l d= 70mm

Tc dng chy trong ng ca thit b gia nhit:Thay s ta c:

Vy : =409,8842(N/m)

+)p sut khc phc tr lc ma st:=

Chn chiu di ng dn l L=2m, dtd= 0,07m. Ch s Reynold:

Re = ;

: nht ca hn hp u nhit si( nhit cui khi gia nhit) c : = 0,4104.10-3(N.S/m)

=15,50327.>

Do nhit chy ca hn hp u trong ng l ch chy xoy.

Chn ng lm bng ng trng km mi bnh thng-Tra bng II.15-ST1/381):mm.Chn 0,1

C: 700

Ta c: Regh=

Ren= 220

Ta c Regh

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