co dac nacl

8/2/2019 co dac NaCl

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o an mon hoc May va Thiet b GVHD :Tran Van Nghe

MUC LUCLi noi au2

Chng I: Gii thieu tong quan

I.Nhiem vu cua o an3

II.Tnh chat nguyen lieu va san pham3

III.Co ac 4IV.Quy trnh cong nghe

5

Chng II :Thiet ke thiet b chnhA.Can bang vat chat va nang lng

8I. Can bang vat chat8II. Can bang nang lng9

B. Tnh thiet ke thiet b chnhI.He so truyen nhiet

14II.Be mat truyen nhiet va thi gian co ac

21III.Buong ot va ay23IV. Buong boc va nap24

C. Tnh c kh cho thiet b chnh27

I.Buong ot27

II.Buong boc28

III.ay30

IV.Nap32

V. Tnh cach nhiet cho than32

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VI.Moi ghep bch33

VII.V ong34

VIII.Khoi lng va tai treo35

IX.Cac ong dan ,ca37

Chng III :Tnh chi tiet thiet b phuI.Thiet b ngng tu Baromet39II.Bm 44

Chng IV : Tnh gia thanh thiet b

49

Ket luan51

Tai lieu tham khao52

Li noi au

Ngay nay, cong nghiep san xuat hoa chat la mot nganhcong nghiep quan trong anh hng en nhieu nganh khac.Mot trong nhng san pham c quan tam san xuat kha nhieula Natri Clorua (NaCl) do kha nang s dung rong rai cua no.NaCltinh khiet c s dung nhieu trong thc pham di dang muoi

an ,hay s dung nhieu trong nganh y te di dang dch truyen.

Nhiem vu cu the cua o an mon hoc la thiet ke he thongco ac chan khong gian oan NaCl t 10% len 27% ,nangsuat 1200kg /me ,s dung ong chum.

o an gom 4 chng :

Chng I:Gii thieu tong quan

Chng II :Thiet ke thiet b chnh

Chng III :Thiet ke cac chi tiet phu

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Chng IV: Tnh toan gia thanh thiet b

Co the noi thc hien o an mon hoc la mot c hoi tot chosinh vien on lai toan bo cac kien thc a hoc ve cac quatrnh va cong nghe hoa hoc. Ngoai ra ay con la dp ma sinh

vien co the tiep can vi thc te thong qua viec la chon, tnhtoan va thiet ke cac chi tiet cua mot thiet b vi cac solieu rat cu the va rat thc te.

o an nay c thc hien di s giup va hng dantrc tiep cua thay Tran Van Nghe , va cac thay co bo monMay va Thiet B khoa Cong nghe Hoa hoc va Dau kh trngai hoc Bach khoa thanh pho Ho Ch Minh. Em xin chan thanhcam n thay Tran Van Nghe va cac thay co khac cung nh

cac ban be a giup em trong qua trnh thc hien o an.

CHNG I. GII THIEU TONG QUAN

I. NHIEM VU CUA O ANNhiem vu cu the cua o an mon hoc nay la thiet ke he

thong co ac chan khong gian oan dung dch NaCl t nongo 10% en nong o 27%, nang suat 1200kg/me, s dung ongchum.

II. TNH CHAT NGUYEN LIEU VA SAN PHAM:

1 .Nguyen lieu :- NaCl la mot khoi tinh the mau trang, tan trong nc phan lythanh ion- La thanh phan chnh cua muoi an hang ngay- Khoi lng rieng dd 10% la 1073 (kg/m3)

- o nht la 1,07 *10-3 (Ns/m2) 200C (dung dch 10%).- o hoa tan 60oC la 27,1% , 20oC la 26,3%

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- Nguyen lieu em i co ac la dd NaCl 10% vi dung moi lanc .2. San Pham:

Khi ket thuc qua trnh co ac ,dung dch nhiet o t 75-80oC ,khi o o hoa tan cua dung dch khoang 27,5% .Nhng ohoa tan cua dung dch nhiet o thng chng 26,3%.V vay,qua trnh co ac NaCl nay la e tao dung dch bao hoa ,vakhi lam nguoi th se co NaCl tinh the ket tinh .Trong khi cacmuoi tap chat khac nh MgCl2 hay CaCl2 lai tan nhiet othng ,v vay qua trnh nay co the c s dung va thu dungdch muoi bao hoa va tach tap chat e san xuat muoi tinhkhiet khi ha nhiet o. Muoi tinh khiet thng c s dung

trong thc pham va y te.

3.Nhng bien oi cua nguyen lieu va san pham trongqua trnh co ac:

Trong qua trnh co ac, tnh chat c ban cua nguyen lieu vasan pham bien oi khong ngng. Thi gian co ac tang lamcho nong o dung dch tang dan en tnh chat dung dch thayoi:

Cac ai lng giam: he so dan nhiet, nhiet dung, he socap nhiet, he so truyen nhiet.

Cac ai lng tang: khoi lng rieng dung dch, o nht,ton that nhiet do nong o, nhiet o soi.

Yeu cau chat lng san pham :at nong o va o tinh khiet yeu cau.Thanh phan hoa hoc chu yeu khong thay oi.

III. CO AC:1.nh nghaCo ac la phng phap thng dung e lam tang nong o

mot cau t nao o trong dung dch hai hay nhieu cau t. Tuytheo tnh chat cua cau t kho bay hi (hay khong bay hi trongqua trnh o) ta co the tach mot phan dung moi (cau t debay hi hn) bang phng phap nhiet hay bang phng phap lamlanh ket tinh.

Co ac la qua trnh lam tang nong o cua chat ran hoa

tan trong dung dch bang cach tach bt mot phan dung moi quadang hi.

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2. Cac phng phap co ac:Phng phap nhiet (un nong): dung moi chuyen t trang thai

long sang trang thai hi di tac dung cua nhiet khi ap suatrieng phan cua no bang ap suat tac dung len mat thoang

chat long.Phng phap lanh: khi ha thap nhiet o en mot mc nao

o th mot cau t se tach ra dang tinh the n chat tinh khiet,thng la ket tinh dung moi e tang nong o chat tan.Tuytnh chat cau t va ap suat ben ngoai tac dung len matthoang ma qua trnh ket tinh o xay ra nhiet o cao haythap va oi khi phai dung en may lanh.

3.Phan loai va ng dung

a. Theo cau tao Nhom 1: dung dch oi lu t nhien (tuan hoan t nhien)dung co ac dung dch kha loang, o nht thap, am bao stuan hoan t nhien cua dung dch de dang qua be mat truyennhiet. Gom:

- Co buong ot trong (ong truc buong boc), co the coong tuan hoan tronghoac ngoai.

- Co buong ot ngoai ( khong ong truc buong boc). Nhom 2: dung dch oi lu cng bc, dung bm e tao vantoc dung dch t 1,5 - 3,5 m/s tai be mat truyen nhiet. Co uiem: tang cng he so truyen nhiet, dung cho dung dch acset, o nht cao, giam bam can, ket tinh tren be mat truyennhiet. Gom:

- Co buong ot trong, ong tuan hoan ngoai.- Co buong ot ngoai, ong tuan hoan ngoai.

Nhom 3: dung dch chay thanh mang mong, chay mot lan

tranh tiep xuc nhiet lau lam bien chat san pham. ac bietthch hp cho cac dung dch thc pham nh dung dch nc traicay,hoa qua epGom:

- Mang dung dch chay ngc, co buong ot trong hay ngoai:dung dch soi tao bot kho v.

- Mang dung dch chay xuoi, co buong ot trong hay ngoai:dung dch soi t tao bot va bot dev.

b. Theo phng phap thc hien qua trnhCo ac ap suat thng (thiet b h): co nhiet o soi, ap

suat khong oi. Thng dung co ac dung dch lien tuc e

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gi mc dung dch co nh e at nang suat cc ai va thigian co ac la ngan nhat.Tuy nhien, nong o dung dch atc la khong cao.

Co ac ap suat chan khong: Dung dch co nhiet o soi

di 100o

C, ap suat chan khong. Dung dch tuan hoan tot, ttao can, s bay hi nc lien tuc.

Co ac nhieu noi: Muc ch chnh la tiet kiem hi ot. Sonoi khong nen ln qua v se lam giam hieu qua tiet kiemhi. Co the co chan khong, co ap lc hay phoi hp ca haiphng phap. ac biet co the s dung hi th cho muc chkhac e nang cao hieu qua kinh te.

Co ac lien tuc: Cho ket qua tot hn co ac gian oan. Cothe ap dung ieu khien t ong, nhng cha co cam bien tin

cay.4.u iem va nhc iem cua co ac chan khong gianoan

u iem- Gi c chat lng, tnh chat san pham, hay cac cau

t de bay hi.- Nhap lieu va thao san pham n gian, khong can on

nh lu lng.

- Thao tac de dang.- Co the co ac en cac nong o khac nhau.- Khong can phai gia nhiet ban au cho dung dch.- Cau tao n gian, gia thanh thap.

Nhc iem- Qua trnh khong on nh, tnh chat hoa ly cua dung

dch thay oi lien tuc theo nong o, thi gian.- Nhiet o hi th thap, khong dung c cho muc ch

khac.- Kho gi c o chan khong trong thiet b.

IV. QUY TRNH CONG NGHE1. Thuyet minh quy trnh cong nghe- Khi ong bm chan khong en ap suat Pck = 0,7 at.- Sau o bm dung dch ban au co nong o 10% t be changuyen lieu vao noi co ac bang bm ly tam. Qua trnh nhaplieu dien ra trong vong 20 phut en khi nhap u 3240 kg thngng.

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- Khi a nhap lieu u 3240 kg th bat au cap hi ot (la hinc bao hoa ap suat 3 at) vao buong ot e gia nhietdung dch. Buong ot gom nhieu ong nho truyen nhiet (ongchum) va mot ong tuan hoan trung tam co ng knh ln

hn. Dung dch chay trong ong c gia nhiet bi hi ot i ngoaiong. Dung dch trong ong se soi va tuan hoan qua ong tuanhoan (do ong tuan hoan co ng knh ln hn cac ongtruyen nhiet nen dung dch trong ong tuan hoan se soi t hntrong ong truyen nhiet, khi o khoi lng rieng dung dch trongong tuan hoan se ln hn khoi lng rieng dung dch trong ongtruyen nhiet v vay tao ap lc ay dung dch t ong tuanhoan sang cac ong truyen nhiet). Dung moi la nc boc hiva thoat ra ngoai qua ong dan hi th sau khi qua buong boc

va thiet b tach giot. Hi th c dan qua thiet b ngng tubaromet va c ngng tu bang nc lanh, sau khi ngng tu thanhlong se chay ra ngoai bon cha. Phan khong ngng se cdan qua thiet b tach giot e ch con kh khong ngng cbm chan khong hut ra ngoai. Hi ot khi ngng tu chay rangoai qua ca thao nc ngng, qua bay hi roi c xa rangoai.- Qua trnh c tiep tuc en khi at nong o 27% (sau thi gianco ac a tnh: 121,3 phut) th ngng cap hi. M van thong ap,

sau o thao san pham ra bang cach m van thao lieu.

2. Cac thiet b c la chon trong quy trnh cong nghea Bm

Bm c s dung trong quy trnh cong nghe gom: bm lytam va bm chan khong.

+ Bm ly tam c cau tao gom vo bm, banh guong treno co cac canh hng dong. Banh guong c gan tren tructruyen ong. Ong hut va ong ay.

Bm ly tam c dung e bm dung dch NaCl t be changuyen lieu vao noi co ac.+ Bm chan khong c dung e tao o chan khong khi he

thong bat au lam viec.

b Thiet b co acay la thiet b chnh trong quy trnh cong nghe. Thiet b

gom ay, nap, buong boc va buong ot. Ben trong buongot gom nhieu ong truyen nhiet nho va mot ong tuan hoantrung tam co ng knh ln hn.

Tac dung cua buong ot la e gia nhiet dung dch, buongboc la e tach hon hp long hi thanh nhng giot long ri tr

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lai, hi c dan qua ong dan hi th. Ong tuan hoan c sdung e tao mot dong chay tuan hoan trong thiet b.

c Thiet b ngng tuThiet b ngng tu c s dung trong quy trnh cong nghe la

loai thiet b ngng tu trc tiep (thiet b ngng tu baromet). Chatlam lanh la nc c a vao ngan tren cung thiet b. Thietb thng lam viec ap suat chan khong nen no phai cat mot o cao can thiet e nc ngng co the t chay rangoai kh quyen ma khong can may bm.

d Thiet b tach longThiet b tach long c at sau thiet b ngng tu baromet

nham e tach cac cau t bay hi con sot lai, cha kp ngng

tu, khong cho chung i vao bm chan khong.e Cac thiet b phu tr khac

- Bay hi- Cac thiet b o ap suat, o nhiet o, cac loai van.

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CHNG II. THIET KE THIET B CHNH

A. CAN BANG VAT CHAT VA NANG LNGI. CAN BANG VAT CHATCac so lieu ban au: Dung dch NaCl co:

- Nhiet o au 25oC, nong o au 10%.- Nong o cuoi 27%.

Chon hi ot la hi nc bao hoa ap suat 3at. Ap suat ngng tu: Pck = 0,7 at.Co ac gian oan vi nang suat 1200 kg/me

1.Khoi lng rieng cua dung dch theo nong o

Nong o, % 10 15 20 27

Khoi lngrieng, kg/m3

1073 1110 1150 1205

2.Can bang vat chat cho cac giai oanG = Gc + WG.x = Gc.xc

Trong o

G , Gc : lng dung dch au va cuoi moi giai oan (kg)W : lng hi th boc len trong moi giai oan (kg)x , xc : nong o au va cuoi cua moi giai oanG.x, Gc.xc : khoi lng NaCl trong dung dch (kg)

Khoi lng dd au :G = 1200 *0,27 /0,1 = 3240 (kg)

a Giai oan 10% en 15%G = 3240 (kg)

x = 0,1 ; xc = 0,15 Lng san pham ( la dung dch NaCl 15% ):

Gc = G . 216015,0

1,0*3240 ==

c

x

x(kg)

Lng hi th:W = G - Gc = 3240 2160 = 1080 (kg)

b Giai oan 15% en 20%G = 2160(kg) ; x = 0,15 ; xc = 0,2

Gc = 16202,015,0*2160. ==

c

xxG (kg)

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W = G Gc = 2160 1620 = 540 (kg)

c Giai oan 20% en 27%G = 1620 (kg) ; x = 0,2 ; xc = 0,27

120027,0

2,0

*1620. === c

c x

x

GG (kg)

W = 1620 1200 = 420 (kg)

Tong lng hi th boc hiWt = 1080+540+420 = 2040 (kg)

Ta co bang tom tat ket qua can bang vat chat

Nong o dung dch, % 10 15 20 27

The tch dung dch trong

noi, m3 3,02 1,946 1,4 1

Khoi lng dung dch, kg 3240 2160 1620 1200

Lng hi th a bochi, kg

0 1080 1620 2040

Khoi lng rieng dungdch, kg/m3

1073 1110 1150 1205

II. CAN BANG NANG LNGAp suat lam viec trong buong boc thiet b co ac P1 = 0,3

at.Nhiet o hi th buong boc t1 = 68,7oC ( Bang I.251 trang314 Tai lieu [1] ).

ay cung la nhiet o soi cua dung moi (la nc) tren mat

thoang dung dch )P(sdm 1t = 68,7oC

Chon ton that nhiet o t noi co ac ve thiet b ngng tu

K1''' = .Nhiet o hi th thiet b ngng tu : t0 = 68,7 - 1 = 67,7oC.

1.Cac ton that nhiet o Nhiet o soi dung dcha Xac nh ton that nhiet o do nong o va nhiet o

soi dung dch NaCl theo nong o ap suat P1 = 0,3 at ( ') :

Theo tai lieu [3] trang 149:' = o' *f

Vi f tra tai lieu [3] trang 147 :

Vi )P(sdm 1t = 68,7oC ,f = 0,81

Nong o dung 10 15 20 27

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dch, %

Ton that o' , oC 1,9 3,25 4,85 8

Ton that ' ,oC 1,54 2,63 3,9 6,48

Nhiet o soi dd ,oC

70,2 71,3 72,6 75,2

b Ton that nhiet o do hieu ng thuy tnh '' . Nhiet osoi dung dch ap suat trung bnh

Tnh theo v du 4.8 trang 207 Tai lieu [4]

)P(sdm)P(sdm)P(sdd)P(sdd''

1tb1tbtttt ==

Vi: Ptb = P1 + 0,5 hh .g.Hop = P1 + P

P = 0,5 hh .g.Hop

ddhh 21 =

Trong o dd : Khoi lng rieng dung dch tnh theo nong o cuoi

nhiet o )PP(sdd 1t +

Hop : Chieu cao lp chat long soiTrong thiet b tuan hoan t nhien

Hop = ( )[ ] odmdd H.0014,026,0 +

Vi Ho : Chieu cao ong truyen nhiet

dm : Khoi lng rieng dung moi tsdm* Chon chieu cao ong truyen nhiet Ho = 1,5m

Tnh cho trng hp dung dch NaCl 10 %

Do trong khoang nhiet o nho, hieu so dmdd thay oikhong ang ke nen :

)/(999 3mkgdm =

)/(1073 3%)25( mkgdd =

[ ] 5454,05.1*)9991073(*0014.026.0 =+= opH (m) atmNHgP ophh 015,0/24,14355454,0*81,9*1073*

2

1*5,0***5,0 2 ====

atPPPtb 315,0015,03,01 =+=+=

Nhiet o soi cua H2O 0,315 at la 69,7 oC ( Bang I.251 trang314 Tai lieu [1] )

o tang nhiet o soi do cot thuy tnh

17,687,69)()(''

1=== PsdmPsdm tt tb

oC

nhiet o soi dung dch NaCl 10 % ap suat P1+ P2,7112,70)( 1 =+=+ PPsddt

oC

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Tnh tng t ta c

Nong o dung

dch, %

10 15 20 27

'' , oC 1 1,14 1,34 1,7

)2( 1 PPsddt + 72,2 73,6 75,3 78,6

2.Can bang nang lng cho cac giai oanTnh theo cong thc 2.4 trang 104 Tai lieu [4]:

* Phng trnh can bang nhiet

ctwcccD QQcDiWtcGtcGiDcD +++=++ ......).1.(...''

.

''

Vi

D : lng hi ot s dung (kg) %5= : t le nc ngng b cuon theo : nhiet o nc ngng (oC)

C : nhiet dung rieng nc ngng Co (J/kg o)

c, cc : nhiet dung rieng dung dch au va cuoi moi giaioan (J/kg o)

t, tc : nhiet o dung dch au va cuoi moi giai oan (oC)

''Di : entanpi cua hi ot (J/kg)

''

wi : entanpi cua hi th (J/kg)Qt : nhiet lng ton that (J)Qc : nhiet lng co ac (J)

* Nhiet lng do hi ot cung cap ( do co 5% hi nc ngngcuon theo )

QD = D.(1-).( .ci ''D ) = D.(1-).r =G(cctc - c t ) +W (iw- cctc) Q-

cd

r = .ci ''D : nhiet hoa hi cua nc ap P

* Nhiet dung rieng cua dung dchTnh theo cong thc 2.11 trang 106 Tai lieu [4]

cdd = 4190.(1-x) + c1.xTrong o

x: nong o dung dchc1: nhiet dung rieng NaCl khan (J/kg o)

Theo cong thc 2.12 trang 183 Tai lieu [4]

c1 = 89,885,58

10*2610*26 33=

+(J/kg o)

Vay nhiet dung rieng dung dch theo nong o

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Nong o dung dch. % 10 15 20 27

Nhiet dung rieng dung

dch, J/kg o

3771 3652 3370 3082,7

* Chon hi ot co ap suat PD =3 at tD =132,9oC* Nhiet hoa hi cua nc ap suat 3 at

r = 2171*103 J/kg o (Bang I.251 trang 314 Tai lieu [1] )* Entanpi cua hi th 73,05oC

''wi =2620*10

3 J/kg ( Bang I.250 trang 312 Tai lieu [1] )

* Ton that nhiet Qt = 0,05*QD* Xem nhiet co ac la khong ang ke

a Giai oan a dung dch 10% t 25oC en 72,2oCG = Gc = 3240 (kg)c = cc =3771 (J/kg o)t = 25oC ; tc =72,2oC ; W = 0 kg

Nhiet lng tieu ton cho qua trnh:Q1 =3240*3771*(72,2-25) =5,77*108 (J)

Nhiet lng can cung cap ( ke ca ton that ):

QD1 =81 10*07,6

95,0=

Q(J)

Lng hi ot s dung:

D1 = 3,29410*2171*)05,01(

10*07,63

8

=

(kg)

b Giai oan a dung dch t 10% en 15%:G = 3240 (kg) ; c =3771 (J/kg o ); t =72,2 (oC)Gc = 2160 (kg) ; cc = 3562 (J/kg o) ; tc = 73,6 (oC)W = 1080 (kg)

Nhiet lng tieu ton cho qua trnh:

Q2 = 2160*3562*73,6 3240*3771*72,2 + 1080*2620*103

Q2 =25,14*108 (J)Nhiet lng can cung cap ( ke ca ton that )

QD2 =8

82 10*46,26

95,0

10*14,25

95,0==

Q(J)

Lng hi ot s dung

D2 = 128310*2171*)05,01(

10*46,263

8

=

(kg)

c Giai oan a dung dch t 15% en 20%:G = 2160 (kg) ; c = 3562 (J/kg o) ; t = 72,2(oC)

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Gc = 1620 (kg) ; cc = 3370 (J/kg o ); tc = 75,3(oC)W = 540 (kg)

Nhiet lng tieu ton cho qua trnhQ3 = 1620*3370*75,3 2160*3370*72,2 + 540*2620*103

Q3 = 12,6*108

(J)Nhiet lng can cung cap ( ke ca ton that nhiet ):

QD3 =8

8

10*26,1395,0

10*6,12= (J)

Lng hi ot s dung

D3 = 64310*2171*)05,01(

10*26,133

8

=

(kg)

d Giai oan a dung dch t 20% en 27%:

G = 1620 (kg) ; c = 3370 (J/kg o ); t = 75,3(oC)Gc = 1200 (kg) ; cc = 3082,7 (J/kg o ); tc = 78,6 (oC)W = 420 (kg)

Nhiet lng tieu ton cho qua trnhQ4 = 1200*3082,7*78,6 1620*3370*75,3 + 420*2620*103

Q4 = 9,8*108 (J)Nhiet lng can cung cap ( ke ca ton that nhiet ):

QD4 =8

8

10*32,1095,0

10*8,9= (J)

Lng hi ot s dung:

D4 = 50010*2171*)05,01(

10*32,103

8

=

(kg)

* Tong nhiet lng:QD = 6,07*108 + 26,46*108 + 13,26*108 + 9,8*108 =56,11*108 (J)

* Tong lng hi ot:D = 294,3 + 1283 + 643 + 500 =2720,3 (kg)

* Lng hi ot rieng:

Drieng = 33,12040

3,2720 ==WD (kg/kg hi th)

* Tom tat can bang nang lng:

Nong o dung dch. %10(25o

C)10(72,2oC

)15 20 27

Nhiet lng hu ch,J*10-8

0 5,77 30,9 43,5 53,3

Tong nhiet lng cungcap, J*10-8

6,07 32,53 45,79 56,11

Lng hi ot s dung, 294,3 1577,3 2220,0 2720,3

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kg

B. TNH THIET KE THIET B CHNH

I. HE SO TRUYEN NHIET1.He so truyen nhiet trong qua trnh soi1.1 Cac k hieu va cong thc

1 : he so cap nhiet khi ngng tu hi (W/m2K)

2 : he so cap nhiet pha dung dch soi( W/m2K)

q1 : nhiet tai rieng pha hi ngng (W/m2)q2 : nhiet tai rieng pha dung dch soi(W/m2)qv : nhet tai rieng pha vach ong truyen nhiet (W/m2)

1vt : nhiet o trung bnh vach ngoai ong (oC)

2vt : nhiet o trung bnh vach trong ong (oC)

tD : nhiet o hi ngng, tD = 132,9(oC)tdd : nhiet o dung dch soi (oC)

1vD1ttt =

ddv2 ttt 2 =

21 vvvttt =

( )1

vDm tt2

1t += : nhiet o mang nc ngng (oC)

1.1.1 Pha hi ngng:

111 t.q = (1)Theo cong thc V.101 trang 28 Tai lieu [2]

4

11 H*t

r*A*04.2

= (2)

Vi A=25,0

32 *

phu thuoc vao nhiet o mang tm

tm , oC 40 60 80 100 120 140 160 180 200

A 139 155 169 179 188 194 197 199 199

: khoi lng rieng cua nc nhiet o tm (kg/m3) : he so cap nhiet cua nc nhiet o tm (W/mK) :o nht cua nc nhiet o tm (Pas)r : an nhiet ngng tu cua hi nhiet o tD

r = 2171*103 (J/kg)

H = 1,5 m: chieu cao ong truyen nhiet1.1.2 Pha dung dch

trang 15


16/52


q2 = 22 t. (3)Theo cong thc VI.27 trang 71 Tai lieu [2]

435,02565,0

2 ***

=

dd

n

n

dd

n

dd

n

ddn

c

c

(4)

Trong o

nnnn ,c,, : he so dan nhiet (W/mK), khoi lng rieng (kg/m3),

nhiet dung rieng (J/kg o), o nht (Pas) cua nc

dddddddd ,c,, : cac thong so cua dung dch theo nong o

n : he so cap nhiet tng ng cua nc (W/m2K)

15,07,0 **56,0 pqn = (5), (cong thc V.90 trang 26 Tai lieu [2])

Vi q : nhiet tai rieng (W/m2)

p : ap suat tuyet oi tren mat thoang (N/m2)p = p1 = 0,3 at = 29430 (N/m2)

* Cac thong so cua nc ( Bang I.249 trang 311 Tai lieu [2] ):tsdm = 68,7 (oC)

=n 978,5 (kg/m3)cn = 4186 (J/kg o)

n = 0,4*10-3 (Ns/m2)

n = 66,7*10-2 (W/mK)

* Cac thong so cua dung dch: dd tra bang I.107 trang 101 Tai lieu [1] ( 40

oC )

dd tnh theo cong thc I.32 trang 123 Tai lieu [1]

3

dd

dddddd

8

M**c*10*58.3

= , W/mK

Mdd =18

1

5,58

1

xx +

Vi x : nong o dung dch cdd va dd xac nh theo nong o

Nong o dungdch, %

10 15 20 27

tsdd, oC 72,2 73,6 75,3 78,6

dd , kg/m3 1073 1110 1150 1205

cdd , J/kg o 3771 3562 3370 3082,7

trang 16


17/52


dd , Ns/m2 0,71*10-3 0,78*10-3 0,89*10-3 1,08*10-3

Mdd 19,34 20,09 20,9 22,14

dd , W/mK 0,55 0,54 0,528 0,504

1.1.3 Pha vach ong truyen nhiet

Theo th du 19 trang 148 Tai lieu [4]

=v

vv r

tq (6)

vt = vv rq *

Trong o: ++=2v

v

1v r

1rr

1r

Lay3

1

10*464,01 =r

, (W/mK)-1: nhiet tr cua nc thng

3

2

10*389,01 =r

(W/mK )-1: nhiet tr cua cau ban

2v = mm: be day ong truyen nhiet

=v 17,5 W/mK: he so dan nhiet qua vach

33

3 10*389,05.17

10*210*464,0

++= vr =9,653*10-4, (W/mK)-1

1.1.4 He so truyen nhiet K

++=

2

v

1

1r

11K

, W/m2K

Do khong biet chnh xac nhiet o vach ong truyen nhietnen phai thc hien tnh lap nh sau

1 Chon1v

t (< tD ) 1t

2 Tnh 1 theo cong thc (2)

3 Tnh q1 theo cong thc (1)

4 Tnh vt theo cong thc (6) vi qv = q1 2v t,t 2 5 Tnh n theo cong thc (5) vi q = q1

6 Tnh 2 theo cong thc (4)

7 Tnh q2 theo cong thc (3)

8 Tnh qtb = ( )21 qq.21

+

9 Xac nh sai so ss =1

tb1

q

qq

Neu ss > 5% th chon lai 1vt va lap lai qua trnh tnh en khi atsai so nho

trang 17


18/52


10Tnh K theo cong thc (7)

1.2 Tnh K cho cac giai oan

a. Tmh nong o 10% :Chon

Kttv 9,4128 11 ==

Tnh 1 :

( ) 19145,1301289,132*2

1==+= ACt om

57,90835,1*9,4

10*2171*191*204

***04,2 4

3

4

1

1 ===

Ht

rA (W/m2K)

)/(5,445099,4*57,9083* 2111 mWtq ===

)(97,4210*653,9*5,44509*4

1 Crqt

o

vv ===

)(859,421282

Ct ov ==

)(8,122,72852 Cto==

525,470316*29430*5,44509*56,0**56,015,07,015,07,0

1 === pqn (W/m2K)435,0

3

32565,0

210*71,0

10*4,0*

4186

3771*

5,978

1073

667,0

55,0*525,4703

=

)/(56,3402 22 KmW=

76,435528,12*56,3402* 222 === tq (W/m2)

( ) ( ) )/(4403176,435525,445092

1

2

1 221 mWqqqtb =+=+=

=

=1

tb1

qqq

ss %101,05,44509

13,440315,44509==

(thoa)

Vay Ct ov 1281 =

K =)/(730

56,3402

110*653,957,9083

1

1 2

4

KmW=

++

b. Tnh nong o 15%:

Tnh tng t Ct ov 3,1281 =

K =)/(718

75,3141

110*653,9

18,9228

1

1 2

4

KmW=++

c. Tnh nong o 20%:

Tmh tng t5,128

1

=vt

(o

C)

trang 18


19/52


K =)/(42,704

75,2880

110*653,9

3,9331

1

1 2

4

KmW=++

d. Tnh nong o 27%:

Tnh tng t Ct ov 1291 =

K =)/(675

78,2425

110*653,9

99,9616

1

1 2

4

KmW=++

o Bang tom tat

Nong o dung dch,% 10 15 20 27tsdd, oC 72,2 73,6 75,3 78,6q1, W/m2 44509,5 42449,63 41057,72 37506,29

q2,W/m2 43552,76 43115,41 39083,09 34434,39qtb, W/m2 44031,13 42782,52 40070,4 35970,34

1 ,W/m2K 9083,57 9228,18 9331,3 9616,99

2 , W/m2K 3402,56 3141,75 2880,75 2425,78

K, W/m2K 730 718 704 675ss, % 1 0,78 2,4 4

2. He so truyen nhiet trong qua trnh gia nhiet dung dchban au t 25oC en 72,2oC:2.1 Cac k hieu va cong thc

Cac k hieu 1 , 2 , q1,q2,qv, 1vt , 2vt , tD, tdd, v21 t,t,t , tm nh muc1.12.1.1 Pha hi ngng:

111 t.q =

4

1

1*

**04,2Ht

rA

=

A xac nh theo tmr = 2171*103 J/kg

H = 1,5 m

2.1.2 Pha vach:

=v

vv r

tq

= 124 )/(10*653,9 KmWrv2.1.3 Pha dung dch:

222 * tq =

trang 19


20/52


l

NulNu dd

dd

**2

2 ==

Trong o

( ) nGrCNu Pr*=

dd

ddddc

*Pr=

2

2

3 ****

dd

dddd gtlGr

=

* C va n phu thuoc vao Pr va Gr nh sau Gr*Pr 310 th Nu = 0.5

Gr*Pr 5010 3 = th ( ) 125,0Pr.18,1 GrNu =

7

10*2500Pr* =Gr th ( )

25,0

Pr.54,0 GrNu = Gr*Pr 710*2> th ( ) 33,0Pr.135,0 GrNu =

* l : chieu cao ong truyen nhiet, l = 1,5 m

* dddddddddd c,,,, : khoi lng rieng ( kg/m3 ), he so dan n

the tch ( K-1 ), he so dan nhiet ( W/mK ), o nht ( Pa.s ), nhietdung rieng ( J/kg o ) cua dung dch NaCl lay nhiet o mang

+=

2vdd

_

m tt21

t

Vi )(6,48)252,72(21

_

Ct odd =+=

)/(1073 3mkgdd =

cdd = 3771 (J/kg o)310*78,0 =dd (Ns/m2)

mK/W577.0dd =

cua dung dch NaCl 25%

To

,o

C 0 20 40 60 80 100 120310. 0,425 0,455 0,48 0,505 0,535 0,57 0,605

2.1.4 He so truyen nhiet :

++=

2

v

1

1r

1

1K

, W/m2K

* Trnh t tnh lap

(1). Chon 1v tt 1 (2). Tnh 1

trang 20


21/52


(3). Tnh q1

(4). Tnh 2v1 ttt 2

(5). Tnh Nu2 2

(6). Tnh q2

(7). Tnh qtb = ( )21 qq21+

(8). Tnh ss =q

qq tb1 , tnh cho en sai so nho (va phai nho hn

5% )

2.2 Thc hien tnh lap

(1). Chon Ct ov 5,1241 =

4,85,1249,1321 == t (oC)

tm= ( ) Co7,1285,1249,132

2

1=+ 13.190A =

(2). )/(83,79215,1*4,8

10*217113,190*04,2 24

3

1 KmW==

(3). )/(35,665434,8*83,7921 2111 mWtq ===

(4). === 23,6410*653,9*35,66543. 41 vv rqt (oC)

)(27,6023,645,1242 Ct

o

v ==67,116,4827,602 == t (oC)

(5). Tnh 2 :

868,455,0

10*78,0*3771.Pr

3

===

dd

ddddc

( ) 3' 10*49,0435,546,4827,602

1 ==+= Ct om (K-1)

( )14

23

23

10*583,310*78,0

81,9*67,11*49,0*1073*5,1

== Gr

ta thay Gr*Pr > 2.107 14456)(Pr**135,0 33,0 == GrNu

5300*

2 ==l

Nu dd (W/m2K)

(6). )/(61851* 2222 mWtq ==

(7). qtb = 64197,25(W/m2)(8). ss=0.035=3,5% (thoa)Vay he so truyen nhiet giai oan nay

trang 21


22/52


)/(781

5300

110*653,9

83,7921

1

1 2

4

KmWK =++

=

II. BE MAT TRUYEN NHIET VA THI GIAN COAC

Phng trnh truyen nhiet cho khoang thi gian nho dTdQ= K*F(T-t)*dT

Gia s en cuoi qua trnh dung dch van ngap het be mattruyen nhiet F khong oi, T khong oi

d.F T=( )tTKdQ

Lay tch phan ta cF.T2 =

( ) Q

0 tTKdQ

(1)

T2 : thi gian co ac ( khong ke thi gian gia nhiet cho dungdch au en 83.48oC ), s

Q : nhiet lng tieu ton cho qua trnh nay, J

* Ta tnh tch phan (1) bang o th. Can xac nh Q,( )tTK1

tng thi iem.

Nong o dungdch, %

10 15 20 27

Q*10-8, J 0 26,46 39,72 50,04

t(tsdd), oC 72,2 73,6 75,3 78,6

K, W/m2K 730 718 704 675

T-t 60,7 59,3 57,6 54,3

510*

)tT.(K

1

2,257 2,349 2,466 2,728

Ve o th co : truc hoanh : Q*10-8 (x)

: truc tung :( )tT.K1

*105 (y)

trang 22


23/52

810.Q


0

1

1

2

2

3

3

0 20 40 60

x

y

T viec tnh tch phan o th ta co Giai oan 1 ( 10%15% ) : S1 = F. T1 = 60937 (m2s) Giai oan 2 ( 15%20% ) : S2 = F. T2= 31923(m2s) Giai oan 3 ( 20%27% ) : S3 = F. T3 = 26801 (m2s) Tong qua trnh co ac t 10% en 27%

S = F. T = 119661 (m2s)*Chon thi gian co ac la 80 phut

Be mat trao oi nhiet laF = 119661 / 4800 = 24,9 (m2) .

Chon F=25 (m2)Thi gian cua cac giai oan Giai oan 1: T1 = 60937 / 25 = 2438 s Giai oan 2 : T2= 31923/ 25 = 1277 s Giai oan 3 : T3 = 26801 / 25 = 1072 s

* Thi gian gia nhiet ban au.F.t.KQ = T

T =F.t.K

Q

Vi Q : nhiet lng dung cho gia nhiet, JK : he so truyen nhiet cho qua trnh gia nhiet, W/m2K

t : chenh lech nhiet o, K

( ) ( )Kt 89,74

48,839,132

259,132ln

48,839,132259,132=

=

T = 32,637925*82*781

10*07,68

= s (phut)

trang 23


24/52


* Chon thi gian nhap lieu 20 phutThi gian thao san pham 15 phut

* Tong thi gian co ac 1 me laTt = 20 + 6,3 + 80 + 15 = 121,3( phut)

III. BUONG OT VA AY:Dien tch be mat truyen nhiet : F = 32 (m2) (lay d 20%

e an toan )Chieu cao ong truyen nhiet : H = 1,5mChon ong truyen nhiet co ng knh : dng = 38mm

: dtr = 34 mm So ong can :

HdnF tr=

2005,1*034,0*

32===

Hd

Fn

tr

(ong)

Chon bc ong t = (1.2 5.1 ).dngChon t = 56 mm

Chon ong tuan hoan ng knh ong tuan hoan

Chon dtr (th) = 315 mmdng(th) = 325 mm

So ong truyen nhiet b chiem choGoi m : la so ong nam tren ng cheo ong tuan hoan

m=( dng(th)- dng)/t +1 = 1,4156

38*4325=+

Chon m=7co 5 ong tren ng cheo ong tuan hoan

a=(m +1)/2 = 4 ( cong thc V.139 Tai lieu [2] trang 48 )Tong so ong b chiem cho

371)14(*4*31)1.(3' =+=+= aan (cong thc V.139 Tai lieu [2]

trang 48) Xep ong theo hnh luc giac eu ( theo Tai lieu [2] trang 48 ) So hnh luc giac eu : 9 hnh So ong tren ng cheo : 19 ong

Tong so ong : 271 ong

So ong truyen nhiet con lai23437271 ==n (ong)

Nh vay ta co the chon so ong an toan la 234 ong . ng knh trong buong ot

trang 24


25/52


Dt = t.(b-1) + 4.dng= 56*(19-1) + 4*38 =1160 (mm)Vi b = 19 , so ong tren ung cheo luc giacChon ng knh buong ot Dt (b) = 1200 (mm)

ay :

Chon ay non tieu chuan co g, goc ay 60o

Tra bang XIII.21 trang 394 Tai lieu [2]Chieu cao g hg = 50 mmChieu cao phan non hn = 1087 (mm)Be mat trong :Ft= 2,608( m2)

The tch ay non Vay = 0,532 (m3)The tch truyen nhiet va ong tuan hoan

Vo = 436,015*4

315,0*5,1*

4

034,0**234

22

=+

(m3)

Cuoi qua trnh co ac Vdd = 1 > 0,532 +0,436dung dch van ngap het be mat truyen nhiet

IV. BUONG BOC VA NAP

1.ng knh Lu lng hi thTa tnh lu lng hi th trong giai oan au ( do lng hi th

trong giai oan nay la ln nhat )

1hi WV = /( 1 * T1) ( m3/s)Trong o

W1 : lng hi th trong giai au (kg)W1 = 1080 (kg)

1 : khoi lng rieng hi th ap suat P1 = 0,3 at

1 = 0,1876 (kg/m3 ) ( Bang I.251 trang 314 Tai lieu [1] )

T1 : thi gian gia nhiet giai oan au ( t 10% en 15% )T1= 2438 s

361,22438*1876,0

1080 == hiV (m3/s)

Van toc hi:

2

)(

2

)(

2

)(

0065,3

.

361,2*4

4

bbtrbbtrbbtr

hhi

DDD

V===

Van toc lang:Xac nh theo cong thc 5.14 trang 157 Tai lieu [3]

( )

h

lhl

o

dg

3

4

=

trang 25


26/52


27/52

h3

h2 =400 mm

h1 =50

h1= 50 mm : chieu caophan

g buong

boch2= 400 mm :chieu caophan

non


Utt : cng o boc hi the tch cho phep cua khoangkhong gian hi (m3/m2h)

h : khoi lng rieng hi th, 1876,0=h (kg/m3)

Ta co W =1530

80

60*2040

= (kg/h)Utt = f.(Utt (1at) )

Vi Utt (1at) : cng o boc hi the tch cho phep khi ap suatbang 1 at

f : he so hieu chnhChon Utt (1at) = 1650 (m3/m3h) (lay trung bnh gia 1600 va 1700)

f = 1,5 ( o th VI.3 trang 72 Tai lieu [2] ) Utt = 1,5*1650 =2475 (m3/m3h) The tch khong gian hi:

295,32475*1876,0

1530==kghV (m3)

Chieu cao phan khong gian hi trong tru boc

64,16,1*

295,3*4422

)(

=== bbtr

kghkgh

D

VH (m)

The tch dung dch trong buong boc trc khi co acVdd (bb) = Vdd Vo - Vay = 3,02 0,436 0,532 = 2,052 (m3)

Mat khac :Vdd(bb) = Vdd phan g + Vdd phan non + Vdd phan tru

= ( ) 32

)(

)()(

2

)(

2

)(2

1

2

)(*

4*

12

**

4h

DDDDD

hh

Dbbtr

btrbbtrbtrbbtrbtr

++++

3

222

2

*4

6,1)2,1*6,12,16,1(

12

4,005,0*

4

2,1*052,2 h ++++=

68,03 = h (m)

Chieu cao phan tru buong bocHb = Hkgh + h3 = 1,64 + 0,68 = 2,32 (m)

Chon chieu cao phan tru buong boc 2,4 (m)Chieu cao buong boc

2,4+ 0,4 + 0,05 = 2,85 (m)

Khi ket thuc co ac Vdd = 1 (m3

)The tch dung dch trong buong boc

trang 27


28/52


Vdd(bb) = 1 0,436 -0,532= 0,032 (m3)Tng t nh tren ta c chieu cao dung dch ngap phan buongboc la

h1 = 0,028 m = 28 (mm)

3.Nap- Chon nap elip tieu chuan co g, ng knh trong 1600 mm- Tra bang XIII.10 trang 382 Tai lieu [2]

Chieu cao g : hg = 50 (mm)Chieu cao phan Elip : ht = 400 (mm)Dien tch be mat trong : Ft = 3,03 (m2)

The tch nap :Vn= 0,637 (m3)Chieu cao cua thiet b:

0,45+2,85+1,5+1,087+0,05= 5,937 (m)

C. TNH C KH THIET B CHNHI. BUONG OT

ng knh trong : Dt = 1200 (mm) Chieu cao : H = 1500 (mm)

1.Cac thong so tra va chon1.1 Ap suat tnh toan

Buong ot chu ap suat trongPB = Phi ot Pa = 3 1 = 2 at = 0,2 (N/mm2)

1.2 Nhiet o tnh toanNhiet o hi ot tD = 132.9oCBuong ot c boc cach nhiet nen nhiet o tnh toan tB

= 132.9 + 20 C153o

1.3 Chon vat lieuVat lieu c chon la thep khong g X18H10T doNaCl co

tnh an monng suat cho phep tieu chuan 153oC

[ ] 115* =BD (N/mm2 ) ( hnh 1.2 trang 22 Tai lieu [7] )

ng suat cho phep:

[ ] [ ] 25,10995,0*115.* === BDBD (N/mm2)

Vi 95,0= la he so hieu chnh

He so ben moi han 95,0=h (bang 1-7 trang 24 Tai lieu [7] )

2.Tnh va chon be day tnh ben cho buong ot

Ta co [ ] 255192,0

95,0*25,109* >==BD

hBD

P

trang 28


29/52


Be day toi thieu than buong ot tnh theo cong thc

[ ]156,1

95,0*25,109*2

2,0*1200

*.2

*)(' ===hBD

BDBDtr

BD

PDS

(mm)

Be day nay qua nho. Tra bang 5-1 trang 128 Tai lieu [7] c

Smin = 3-4 mmDung dch an mon (NaCl ) nen Ca = 1Vay chon be day buong SB = 4 mm* Kiem tra ap suat tnh toan

1.00025,01200

14

)(


30/52


2.Tnh be day Tnh on nh cho buong bocBe day toi thieu c xac nh theo cong thc 5.14 trang 133

Tai lieu [7]4,0

)(

'

)(

'

***18,1

= bBtrB

BB

BB

bBtrBB D

l

E

P

DS

2850' =BBl (mm) : chieu dai tnh toan buong boc

87,81600

2850*

10*2

17,0*1600*18,1

4,0

5

' =

= BBS (mm)

Be day thc buong boc

SBB = 1013,0187,8' =++=++ oaBB CCS (mm)

* Kiem tra ieu kien :

( )( )a

)bB(tr

)bB(tr

'BB

)bB(tr

a

CS2D

Dl

DCS25.1

( )3

)bB(tr

a

)BB(c

BB

t

'BB

DCS2E

3.0Dl

The so ta c

>352,078,1

43.978,1159,0(thoa)

* Kiem tra ap suat cho phep

[ ])(

2

)(

'

)(**649,0

bBtr

aBB

bBtr

aBB

BB

bBtrBBBB

D

CS

D

CS

l

DEP

=

= 1729,01600

110

1600

110

2850

1600*10*2*649,0

2

5 =

(N.mm2)

PBB = 0,17 (N/mm2) < [ ] 1729,0=BBP (N/mm2) thoa

* Kiem tra lc nen chieu trucLc nen chieu truc ( trang 149 Tai lieu [7] )

( )BB

2BB)bB(tr

NCT P.4

S2DP

+=

( )35040417,0*

4

10*216002

=+

=

NCTP (N)

T so( )

( )250;2589,882

)( = aBB

bBtr

CS

D087,0= ck (bang trang 140 Tai

lieu [7] )

087,0*10*2

95,202*875*875 5

)(

== cBB

BBcc kEK

=0,077

trang 30


31/52


ieu kien

SBB CaBBc

NCT

EK

P

.*

9 7,210*2*077,0*350404

5 = (thoa )

Vay be day buong boc thoa lc nen chieu truc

* Kiem tra ong thi ap suat ngoai va lc nen chieu trucng suat cho phep khi nen ( cong thc 3.51 trang 140 Tai

lieu [7] )

[ ] 9,861600

110*10*2*077,0 5

)(

=

=

=bBtr

aBBcBBnD

CSEK (N/mm2)

ng suat khi nen ( cong thc 5.48 trang 145 Tai lieu [7] )

698,7)110(*)101600(*

350404

)).(.( )()( =+

=+

=

aBBbBtr

NCTBBn CSSD

P(N/mm2)

Kiem tra ieu kien ( cong thc 5.47 trang 145 Tai lieu [7] )

[ ] [ ]1

)( +BB

BB

BBn

BBn

P

P

(thoa )

Vay than buong boc thoa ong thi ieu kien ap suatngoai va lc nen chieu truc

Ket luan : Be day buong boc la SBB = 10 mm

III. AYTnh theo cong thc trang 178-179 Tai lieu [7]ay non chu cung ap suat ngoai vi buong boc

P = 0,17 (N/mm2)Chon s bo be day ay S = 10 mm

D : ng knh tnh toan cua ay non

1216

30cos

34*1,01200*9,0

cos

1,09,0' =+

=+

=

tt dDD (mm)

Vi dt = 32 mm la ng knh lo thao san pham

Xet( )

3

'

' 2**3,0

>

t

a

c D

CSE

D

l

( )35

1216

1102*

95,202

10*2*3,0

1216

1087

>

0,89 > 0,53

Xet :( )

( )a

bBtr

bBtr

BB

bBtr

aCS

D

D

l

D

CS

22

5,1)(

)(

'

)(

trang 31


32/52


0,180,898,22 (thoa)Vay ap suat cho phep tnh theo cong thc 5.19 trang 135 Tai

lieu [7]

[ ]5,2

''

'

**649,0

= D

CS

l

DEP a

DD

= 0,649* 68,01216

110*

1087

1216*10*2

5,2

5 =

(N/mm2) > 0,17 (N/mm2)

(thoa)* Kiem tra ieu kien on nh:

Lc nen chieu truc ( cong thc 6.26 trang 178 Tai lieu [7] )

DngDNCT PDP2

4

=

Vi Dng = Dtr + 2S = 1200 + 2*10 = 1220 mm20397417,0*1220*

4

2 ==

NCTP (N)

Lc nen chieu truc cho phep ( cong thc 6.27 trang 178 Tailieu [7] )

[ ] 22 cos)( acNCT CSEKP =

Xac nh Kc

67,66)110.(2

1200

).(2=

=

aDtrD

CS

D

067,0= ck tra bang trang 140 Tai lieu [7]

0595,0067,0*10*2

95,202*875**875

5=== ct

cc k

EK

[ ] 240405130cos.)110(*10*2*0595,0* 225 == NCTP ( N)

ieu kien on nh ( cong thc 6.30 trang 178 Tai lieu [7] )

[ ] [ ]133,0

68,0

17,0

2404051

203974=+=+

D

D

NCT

NCT

P

P

P

P(thoa )

Vay be day ay la 10 (mm)

IV. NAP ELIP- Nap elip tieu chuan co g

ng knh trong 1600 mmChieu cao g : hg = 50 mmChieu cao phan Elip : ht = 400 mmRt = Dt = 1600 mm

- Nap chu ap suat ngoai nh buong boc PN = 0,17 (N/mm2 )

- Vat lieu la thep khong g X18H10TEN = 2*105 (N/mm2)

trang 32


33/52


95,202=cN (N/mm2)

Chon be day nap SN = SBB = 10 mm* Kiem tra

1610

1600

S

R

N

t

==

17,21195.202*7,0

10*2*15,0

*

*15,0 5==

cN

NE

( vi 7.0= oi vi thep khong

g )

Ta thaycN

N

N

t E

S

R

*

*15,0< nen tnh ap suat cho phep theo cong

thc 6.7 trang 166 Tai lieu [7]

[ ]

t

aN

N R

CSP

*

)(*.2][

= ( kiem tra ieu kien 0,2 == che o chay roi

0178,0

156894

221,00032,0

227,0=+=

HH

h 32

2 10*51,103185,05,115.0

*0178,0181,9*2

5,0 +=

++=

- Tnh HH= 7 + 0,03185 + 1,51*10-3H

04,7= H m

8.Cac kch thc khac- Chieu day thanh thiet b 5 (mm)-

Lo hi vao 300 (mm)- Lo nc vao 100 (mm)- Hon hp kh va hi ra noi vi thiet b thu hoi 80 (mm)- ng knh ong noi t thiet b thu hoi en ong baromet 50(mm)- Khoang cach t tam thiet b ngng tu en tam thiet b thuhoi 675 (mm)- ng knh thiet b thu hoi 400 (mm)- Chieu cao thiet b thu hoi 1440 (mm)

- Hon hp kh va hi ra khoi thiet b thu hoi 50 (mm)- Ong thong kh 50 (mm)

trang 44


45/52


II. BM

1.Bm chan khongo Toc o hut cua bm chan khong

0oc va 760 mmHgSB = 17/80 = 0,2125 (m3/ph) = 12,75 (m3/h)

Cong suat bm chan khong

=

1**1

**

1

1

21

m

m

ck

kk

p

pp

m

m

t

VN

m : ch so a bien, thng m = 1,2 1,62. Lay m = 1,5p1 : ap suat trc khi nen. p1 = P Ph = 0,3 0,05 = 0,25 atPh = 0,05 ap suat hi nc trong hon hpp2 : ap suat sau khi nen. P2 = Pa = 1 at = 9.81*104 (N/m2)Vkk : the tch khong kh can hut (m3)t : thi gian co ac (s )

ck : he so hieu chnh, ck = 0.8

8361

25,0

1*10*81,9*25,0*

15,1

5,1

4800*8,0

79,74 5,115,1

4 =

=

N (W)

Chon bm chan khongHieu bm KBH-4

Toc o hut 0oC va 760 mmHg: 0,4 (m3/ph)Ap suat gii han: 110 (mmHg)Cong suat ong c 1,5 (kW)Khoi lng bm 38 (kg)

2.Bm nhap lieu

Cong suat bm

1000

*** gHQN=

Q : lu lng nhap lieu (m3/s)

0025167,01200

02,3==Q (m3/s)

H : cot ap cua bm (m)Phng trnh Bernoulli cho 2 mat cat 1-1 (mat thoang be cha

nguyen lieu) va 2-2 (mieng ong nhap lieu)

trang 45


46/52


Z1+ Hg

vp++

2

2

1.11

= Z2+ ++

g

vp

2

2

2.22

h1-2

Trong o Z1, Z2 : chieu cao hnh hoc cua mat cat so vi at. Chon Z1

= 2 m, Z2 = 6,5 m p1,p2 : ap suat tai 2 mat cat. p1 = p2 = 1 at v1,v2 : van toc dung dch tai 2 mat cat (m/s)

v1 = 0v2 = v : van toc dung dch i trong ong ( m/s)

h1-2 : tong ton that trong ong (m)

Ta co

+= dl

g2v

h2

21

: tong he so ton that cuc bo

88,415,0*219,1*25,0*2*2 90.. =+++=+++= ravanquanhkhucv

l, d : chieu dai, ng knh ong noi bm ( m) : he so ma sat

Xac nh Chon ng knh d = dhut = day = dnl = 50 (mm)Van toc chay trong ong

28,105,0*

0025167,0*4

.

.4

22 === d

Q

v (m/s)

Chuan so Re = 9672110*71,0

1073*05,0*28,1**3

==

dv

Vi : khoi lung rieng dung dch NaCl 10% (kg/m3) : o nht ong lc cua dung dch NaCl 10% (Pa.s)

Chon o nham ong thep 2,0= (mm)

07,33012,0

50*6*6Re

7

8

7

8

=

=

=

dgh

4,1096742,0

50*220,220Re

8

9

8

9

=

=

=

dn

Vay Regh < Re < Ren nen

029,096721

100

50

2,0*46,11,0

Re

100*46,11,0

25.025.0

=

+=

+=

d

Chieu dai ng ong t be len ca nhap lieu l =8 m Tong ton that ap suat:

8,088,405,07*029,0*

81,9*228,1

2

21 =

+=h (m)

trang 46


47/52


Chon 121 == Cot ap cua bm

H = (Z2 Z1) + 38,591,081,9*2

28,125,6

2

2

21

212 =++=++

hg

vpp

(m)

Cong suat bm

178,08,0*1000

0025167,0*81,9*1073*38,5==N (kW)

Chon bm theo bang 1.7 trang 35 Tai lieu [4]Hieu bm : X20/18Lu lng Q = 5,5*10-3 m3/sCot ap H = 10,5 mSo vong n = 48,3 v/phong c ien : Loai A02-31-2

Cong suat N = 3 kW

Hieu suat 83,0=

3.Bm vao thiet b ngng tu Cong suat bm

1000

*** gHQN n=

n : khoi lng rieng cua nc 25oC, n =996,9 (kg/m

3)

: hieu suat cua bm, = 0,8 Q : lu lng nhap lieu (m3/s)

310*626,74800*9,996

36491

.

===t

GQ

n

n

(m3/s)

t : thi gian co ac (s) H : cot ap cua bm (m)

Phng trnh Bernoulli cho 2 mat cat 1-1 (mat thoang bon chanc vao thiet b ngng tu) va 2-2 (mat thoang ca vao ongdan nc)

Z1+ Hg

vp++

2

2

1.11

= Z2+ ++g

vp

2

2

2.22

h1-2

Trong o Z1, Z2 : chieu cao hnh hoc cua mat cat so vi at. Chon Z1

= 2 m, Z2 = 12 m

p1 : ap suat tai mat cat 1-1, p1 = 1 at 10p1 =

mH2O

p2 : ap suat tai mat cat 2-2, p2 = 0,3 at 32 =

pmH2O

v1,v2 : van toc nc tai 2 mat cat, m/sv1 = 0

trang 47


48/52


v2 = v : van toc nc chay trong ong, m/s h1-2 : tong ton that trong ong, m

Ta co

+= d

lg2

vh

2

21

: tong he so ton that cuc bo88,415,0*219,1*25,0*2*2 90.. =+++=+++= ravanquanhkhucv

l, d : chieu dai, ng knh ong noi t be cha enthiet b ngng tu (m)

: he so ma satXac nh

Chon ng knh d = dhut = day = dnl =100 mmVan toc chay trong ong

11,0*10*626,7*4

*4

2

3

2 ===

dQv (m/s)

Chuan so Re = 11176010*892,0

9,996*1,0*1**3

==

dv

Vi : khoi lung rieng cua nc 25oC (kg/m3) : o nht ong lc cua nc 25oC, =0,892*10-3 (Pas)

Chon o nham ong thep 2,0= mm

34,7289

2.0

100*6,6Re

7

8

7

8

=

=

=

dgh

1,2392012,0

100*220*220Re

8

9

8

9

=

=

=

dn

Vay Regh < Re < Ren nen

025,0111760

100

100

2,0*46,11,0

Re

100*46,11,0

25,025,0

=

+=

+=

d

Chieu dai ng ong t be len ca nhap lieu l = 15 m Tong ton that ap suat

44,088,41,0

15*025,0*

81,9*2

12

21 =

+=h (m)

Chon 121 == Cot ap cua bm

H = (Z2 Z1) + 5,344,081,9*2

1)103(212

2

2

21

212 =+++=++

hg

vpp

(m) Cong suat bm

326,08,0*1000

81,9*9,996*5,3*10*626,7 3

==

N (kW)

trang 48


49/52


Chon bm theo bang 1.7 trang 35 Tai lieu [4]Hieu bm : X45/21Lu lng Q = 12,5*10-3 (m3/s)Cot ap H = 13,5 (m)

So vong n = 48,3 (v/ph)ong c ien : Loai A02-51-2

Cong suat N = 10 (kW)

Hieu suat 88,0=

CHNG IV. TNH GIA THANH THIETB

- Khoi lng thep lam thiet b khong ke ong truyen nhiet :2464 kg

Gia thep X18H10T : 50000 /kg

$thiet b = 50000*2472 = 123600000 ()- Ong truyen nhiet

Ong co d < 50 mm gia 50000 /mOng co d >50 mm gia 100000 /m

$ong = 234*1,5*50000 + 100000*1,5 = 17700000 ()- Bulong

Gia 1 bulong 3000 /cai

$bulong = (32*2 +40 )*3000 = 312000 ()- em :

Gia 250000

$em = 3*250000 = 750000 ()- Tai

Vat lieu CT3, gia 10000 /kg

Khoi lng 1 tai 3,48 kg $tai treo = 3,48*4*10000 = 139000 ()

- Ca quan sat : 300000 (/cai)Vay tong gia thanh thiet b chnh:

$thiet b chnh = (123,6 + 17,7 +0,312 + 0,75 + 0,139 +0, 3)*106 =142801000 ()- Nhiet ke gia: 150000 /cai

$nhiet ke = 150000*2 = 300000 ()-

Ap ke gia: 600000 /cai $ap ke = 60000*3 = 1800000 ()

trang 49


50/52


- Bm chan khong 1500000 /cai- Bm nhap lieu

Cong suat N = 0,23 kW = 0,3 HpChon bm 0,5 Hp

Gia bm 700000 /Hp $bm nhap lieu = 700000*0,5 = 350000 ()

- Bm vao thiet b ngng tu (bm nc)N = 0,39 kW = 0,52 HpChon bm 1 Hp

$bm nc = 700000*1= 700000 ()- Thiet b ngng tu baromet va bnh tach long gia: 15000000 ()- Ong dan lieu: 8m ong thep va 2 cut:

Gia 1 cut 2000 ()

1m ong d 50 la 20000 (/m)$nhap lieu= 20000*8+8*2000= 176000 ()

- Ong dan nc vao thiet b ngng tu : 15m ong thep va 2cut :

Gia 1m ong d100 la 30000 (/m)$ngng= 30000*15+ 10*2000= 470000 ()

Vay tong gia thanh thiet b ( cha ke tien gia cong lap at)$ = (142,801 + 0,3 + 1,8 +1,5 +0,35*2 +15+0,176+0,47)*106 =

162747000 ()Neu tnh gia gia cong bang 100% gia vat t th tong gia thanhthiet b la:

$tong = 2*162747000 = 325494000 ()

trang 50


51/52


KET LUAN

He thong co ac gian oan c thiet ke gom noi co acva thiet b ngng tu baromet kha n gian, khong phc tap,khong can thiet b gia nhiet ban au va bon cao v e on nhlu lng. Thi gian co ac tng oi ngan (121,3 phut ), he sotruyen nhiet at c trong qua trnh co ac la kha cao.

Thiet b tng oi nho gon, gia thanh khong qua cao co thechap nhan c.Tuy nhien ta kho co the ieu khien c qua trnh co ac

va thi gian co ac co the thay oi khong on nh.Qua trnh co ac la qua trnh can thiet trong cong nghe hoa

chat va thc pham nen can c nghien cu phat trien eco c hieu qua co ac cao, chi ph thap.

trang 51


52/52


TAI LIEU THAM KHAO

[1] Cac tac gia, So tay Qua trnh va thiet b Cong nghehoa chat, Tap 1, NXB Khoa hoc va Ky thuat, 1992

[2] Cac tac gia, So tay Qua trnh va thiet b Cong nghehoa chat, Tap 2, NXB Khoa hoc va Ky thuat, 1992

[3] Pham Van Bon (chu bien) Nguyen nh Tho, Giao trnhQT & TB CNHH tap 5 : Qua trnh va thiet b truyen nhiet, NXB HQuoc gia TP HCM, 2002

[4] Pham Van Bon Vu Ba Minh Hoang Minh Nam, QT & TB

CNHH tap 10 : V du va Bai tap, Trng H Bach Khoa TP HCM[5] Pham Van Bon : Hng dan o an mon hoc, Trng HBach Khoa TP HCM, 1993

[6] Tran Hung Dung Nguyen Van Luc Hoang Minh Nam Vu Ba Minh, QT & TB CNHH tap 1, quyen 2 : Phan rieng bang khong, lc ly tam, Bm, quat, may nen, Tnh he thong ngong, NXB H Quoc gia TP HCM, 1997

[7] Ho Le Vien, Thiet ke tnh toan cac chi tiet thiet bhoa chat, NXB Khoa hoc va Ky thuat, 1978

[8] Nguyen Van Lua, QT & TB CNHH tap 1, quyen 1: Khuay Lang Loc, NXB H Quoc gia TP HCM, 2002

co dac nacl

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