Upload
phuonghuong911
View
19
Download
0
Embed Size (px)
Citation preview
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
Trong h thng c c 3 ni: 0,40 kg/ kg
Trong h thng c c 4 ni: 0,30 kg/ kg
Trong h thng c c 5 ni: 0,27 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.
c2 = 26000 J/kg.nguyn t.
c3 = 16800 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