co dac naohi

8/3/2019 co dac Naohi

1/36

O AN MON HOC QTTB

I. M AU:Nganh cong nghiep san xuat NaOH la mot trong nhng

nganh cong nghiep san xuat hoa chat c ban. No ong vai trorat ln trong s phat trien cua cac nganh cong nghiep khac nh

det , tong hp t nhan tao, loc hoa dau, san xuat phen...NaOH la mot baz manh, co tnh an da, kha nang an monthiet b cao. V vay can lu y en viec an mon thiet b, ambao an toan lao ong trong qua trnh san xuat.

Trc ay trong cong nghiep NaOH thng c san xuatbang cach cho Ca(OH)2 tac dung vi dung dch Na2CO3 loang vanong . Ngay nay ngi ta dung phng phap hien ai la ien phandung dch NaCl bao hoa. Tuy nhien dung dch san pham thu cthng co nong o rat loang , kho khan trong viec van chuyeni xa. e thuan tien cho chuyen ch va s dung ngi ta phai coac dung dch en mot nong o nhat nh theo yeu cau.

Co ac la qua trnh lam tang nong o cua chat hoa tantrong dung dch bang cach tach bt mot phan dung moi qua danghi hoac dang ket tinh.

Trong khuon kho o an nay ta se tien hanh co ac theocach tach dung moi di dang hi. Qua trnh co ac thng tienhanh trang thai soi, ngha la ap suat hi rieng phan cua dungmoi tren mat thoang dung dch bng vi ap suat lam viec cuathiet b.

Qua trnh co ac thng c dung pho bien trong congnghiep vi muc ch lam tang nong o cac dung dch loang, hoace tach cac chat ran hoa tan.

Qua trnh co ac thng tien hanh cac ap suat khacnhau. Khi lam viec ap suat thng ( ap suat kh quyen) ta dungthiet b h , con khi lam viec ap suat khac ( vd ap suat chankhong ) ngi ta dung thiet b kn.

Qua trnh co ac co the tien hanh trong he thong co acmot noi hoac nhieu noi, co the lam viec lien tuc hoac gianoan.

Thiet ke o an mon hoc may va thiet b hoa chat giupsinh vien lam quen vi phng phap tnh toan may thiet b hoachat.

Tap o an nay thiet ke he thong co ac hai noi lam viec

lien tuc xuoi chieu co ac dung dch xut NaOH co nong o au15% en nong o cuoi 30%. Nang suat au vao la 1m3/hour.

Trang 1


2/36

O AN MON HOC QTTB

II. CHON QUI TRNH CONG NGHE :1. Qui trnh cong nghe :Nang suat cua qui trnh co ac la 1m3/h. ay la nang suat nhodo o ta chon qui trnh cong nghe nh sau.

2. Nguyen tac hoat ong cua he thong co ac:Dung dch t be cha nguyen lieu c bm len bon cao v,

t bon cao v dung dch chay xuong qua thiet b gia nhiet va cgia nhiet en nhiet o soi ng vi ap suat lam viec cua noi I.Dung dch sau o c a vao noi I. Do co s chenh lech ap suatgia noi I va noi II nen dung dch tiep tuc chay qua noi II roi cbm hut ra roi chuyen vao be cha san pham. Hi th trong noi Idung lam hi ot noi II e tan dung nhiet. Hi th noi II se ca qua thiet b ngng tu baromet va c chan khong hut ra ngoai.

Nguyen ly lam viec cua noi co ac : phan di cua thiet bla buong ot gom co cac ong truyen nhiet va mot ong tuanhoan trung tam. Dung dch i trong ong, hi ot se i trong khoangkhong gian pha ngoai ong. Nguyen tac hoat ong cua ong tuanhoan trung tam la : do ong tuan hoan co ng knh ln hn ratnhieu so vi cac ong truyen nhiet do o he so truyen nhietnho, dung dch se soi t hn so vi dung dch trong ong truyennhiet. Khi soi dung dch se co ds = 0.5 dd do o se tao ap lc

Trang 2


3/36

O AN MON HOC QTTB

ay dung dch t trong ong tuan hoan sang ong truyen nhiet. Ketqua la tao mot dong chuyen ong tuan hoan trong thiet b. eong tuan hoan trung tam hoat ong co hieu qua dung dch chnen cho vao khoang 0,4 0,7 chieu cao ong truyen nhiet. Phan phatren thiet b la buong boc e tach hi ra khoi dung dch, trong

buong boc con co bo phan tach bot e tach nhng giot long rakhoi hi th.

A.TNH THIET B CHNH

III.TNH CAN BANG VAT CHAT:1. Chuyen n v nang suat t (m3/h) sang (kg/h) :

Nang suat nhap lieu : GD =1 m3/h.Khoi lng rieng : NaOH= 1159 kg/m3 GD = GD

NaOH= 1159 kg/hNong o nhap lieu : xD = 15 %

Nong o cuoi cua san pham : xC = 30%Ap dung phng trnh can bang vat chat : GD xD = GC xC

Suy ra: GC=C

DD

x

xG =

30

151159 = 579.5 kg/h .

2. lng hi th boc len trong toan he thong :

Ap dung cong thc : )1(C

DD

x

xGW = kg/h

Trong o:W : Lng hi th cua toan he thong kg/hGD : Lng dung dch ban au kg/h

xD,xC : Nong o au,cuoi cua dung dch % khoi lngThay so vao ta co:

5.579)30

151.(1159)1( ===

C

DD

x

xGW kg/h.

3. Gia thiet phan phoi hi th trong cac noi :

Chon t so gia hi th boc len t noi I va II la : 1.1=II

I

W

W

Khi o ta co he phng trnh:

1.1=II

I

W

W

WI + WII = WGiai he tren co ket qua :

WI = 303.5 kg/hWII = 276 kg/h

4. Xac nh nong o dung dch tng noi :- Nong o cuoi cua dung dch ra khoi noi I :

Trang 3


4/36

O AN MON HOC QTTB

xC= 32.205.3031159

15.1159.=

=

IDDD

WG

xG%

- Nong o cuoi cua dung dch ra khoi noi II :

xC= 302765.3031159

15.1159. =

= IIIDDD

WWG

xG %

IV.CAN BANG NHIET LNG:1. Xac nh ap suat va nhiet o moi noi:Hieu so ap suat cua ca he thong co ac:Theo au bai ap suat ngng tu la: Png = 0.5 atChon ap suat cua hi ot vao noi I la : P1= 3.5 atKhi o hieu so ap suat cua ca he thong co ac la :

Pt =P1 Png = 3.5 0.5 = 3 at

Chon t so phan phoi ap suat gia cac noi la : 5.12

1 =PP

Ket hp vi phng trnh : P1 + P2 = Pt = 3 atSuy ra : P1 = 1.8 at

P2 = 1.2 atDa vao cac d kien tren va tra so tay qua trnh thiet b

tap I ta co bang sau ay :

Loai

Noi I Noi II Thap ngng tu Ap

suat(at)

Nhiet

o(0C)

Ap suat

(at)

Nhiet

o(0C)

Ap

suat(at)

Nhiet

o (

0

C)

Hiot

P1= 3.50 T1=137.9 P1=1.70 T2=114.5Png=0.5 tng=80.9

Hith

P1=1.76 t1 =115.5 P2=0.52 t2 =81.9

2. Xac nh nhiet o ton that :a. Ton that nhiet do nong o tang cao ( ):Ap dung cong thc cua Tiaxenko:

= o . f ay : o : Ton that nhiet o ap suat thng.f : he so hieu chnh v thiet b co ac lam viec ap

suat khac vi ap suat thng.

fi

i

r

t 2)'273(2.16

+=

Trang 4


5/36

O AN MON HOC QTTB

ti : nhiet o hi th cua noi th I

ri : an nhiet hoa hi cua hi nhiet o t i .T cac d kien tren ta lap c bang sau:

ailngNoi I

xC(%k.l)

o(0C )

t( 0C )

r.10-3(j/kg )

i(0 C )

Noi I 20.32 8.457 115.5 2218.7 9.33Noi II 30.00 17.0 81.9 2304.6 15.05

T ay ta co tong ton that nhiet do nong o tang cao : = I + II = 9.33 +15.05 = 24.38 0C

b. Ton that nhiet do ap suat thuy tnh ( ):Goi chenh lech ap suat t be mat dung dch en gia ong la

P (N/m2), ta co: P =

2

1 S.g.Hop N/m2

Trong o: s : khoi lng rieng cua dung dch khi soi , kg/m3

s =0.5 dd dd : Khoi lng rieng cua dung dch ,kg/m3

Hop: Chieu cao thch hp tnh theo knh quann sat mc chatlong ,m

Hop = [0.26+0.0014( dd- dm)].Ho

T P ta se tnh c ap suat trung bnh cua dung dch tngnoi thong qua cong thc:Ptbi= Pi+ Pi

( i ): noi th iTra so tay ta co c bang sau:

x C ,% t ,0C dd ,kg/m3

dm ,kg/m3

Noi I 20.32 115.5 1173.4 958Noi II 30.00 81.9 1276 958

Coi dd trong moi noi thay oi khong ang ke trong khoang nhieto t be mat en o sau trung bnh cua chat long.Chon chieu cao ong truyen nhiet la Ho=1.5 m.Noi I:

Hop1 = [0.26+0.0014( dd- dm)].Ho=[0.26+0.0014(1173.4-958)]*1.5=0.84234 ,mAp suat trung bnh:

Ptb1= P1+ P1=1.76+0,5.0,5.1173.4.10-4.0.84234=1.785 at

Trang 5


6/36

O AN MON HOC QTTB

Tra so tay tai Ptb1=1.785 (at) ta co t1=116.03 0C.Suy ra : 1=(t1+ 1) (t1+ 1)= 116.03 115.5 =0.53 0C

Noi II:Hop2 = [0.26+0.0014( dd- dm)].Ho=[0.26+0.0014(1276-

958)]*1.5=1.0578 ,m

Ap suat trung bnh:Ptb2= P2+ P2=0,52+0,5.0,5.1276.10-4.1,0578=0,554 at

Tra so tay tai Ptb2 = 0.554 (at) ta co t2= 83.37 0C.Suy ra : 2=(t2+ 2) (t2+ 2)= 83.37 81.9 =1.47 0CVat ton that nhiet cua hai noi la:

= 1+ 2 =0.53+1.47 = 2.00 0Cc. Ton that nhiet do tr lc thuy lc tren ng ong ( )Chap nhan ton that nhiet o tren cac oan ong dan hi th

t noi nay sang noi no va t noi cuoi en thiet b ngng tu la10C. Do o:

1=1.50C 2 =1.0 0C

d. Ton that chung trong toan he thong co ac: = + + =24.38+2.00+2.5=28.88 0C

3. Hieu so hu ch va nhiet o soi cua tng noi:Hieu so nhiet o hu ch moi noi:

Noi I: ti1=TI (T2+ 1) =137.9 (114.5+9.33+0.53+1.5)=12.04 0C

Noi II: ti2=T2 (tng + 2) =114.5 (80.9+15.05+1.47+1)=16.080CNhiet o soi thc te cua dung dch moi noi:

Noi I : ti1=TI tS1 suy ra tS1=T1 - ti1=137.9 12.04 = 125.86 0CNoi II : ti2=T2 tS2 suy ra tS2=T2 - ti2=114.5 16.08 = 98.42 0C

4. Can bang nhiet lng:a. Tnh nhiet dung rieng cua dung dch cac noi:Noi I:

Nong o au dung dch xD=15%20% nen ap dung cong thc: C1=C2=4186

( 4186 Cht)xC1

Cht : Nhiet dung rieng cua chat hoa tan ,j/kg.oM.Cht =n1.c1+ n2.c2+ n3.c3+. . . nn.cn (*)

Tra so tay tap I ta co:

Trang 6


7/36

O AN MON HOC QTTB

MNaOH =40n1=n2=n3 =1c1=cNa = 26 j/kg n.t.o

c2=cO = 16.8 j/kg n.t.o

c3=cH = 9.6 j/kg n.t.oThay vao (*) ta co: Cht= 131010.

40

6.98.1626 3 =++

j/kg.o

Nhiet dung rieng dung dch ra khoi noi II la:C2=C1 =4186 ( 4186 Cht )xC2 =4186 (4186 1310)0.3

=3323.2 j/kg.ob. Lap phng trnh can bang nhiet lng (CBNL):Noi I:

D.i+GD.CD.tD=W1.i1+(GD W1)C1.t1+D.Cng1. 1+Qxq1Noi II:

W1.i1+(GD W1)C1.t1=W2.i2+(GD W)C2.t2+W1.Cng2. 2+Qxq2Trong o:D: lng hi ot dung co he thong ,kg/hi,i1,i2: ham nhiet cua hi ot , hi th noi I va noi II ,j/kgtD, t1, t2: nhiet o soi ban au, rakhoi noi I va noi II cua

dung dch , 0CCD, C1, C2:nhiet dung rieng ban au, ra khoi noi I va noi II

cua dung dch , j/kg.o 1, 2:nhiet o nc ngng tu cua noi I va noi II ,0CCng1, Cng2: nhiet dung rieng cua nc ngng noi I va

noi II ,j/kg.o.

Qxq1,Qxq2 :nhiet mat mat ra moi trng xung quanh , JGD : lng dung dch luc ban au ,kg/h

Chon hi ot , hi th la hi bao hoa, nc ngng la long soi cung nhiet o, khi o ta co:

i- Cng1. 1=r ( 1) va i1- Cng2. 2=r( 2)tra so tay ta co bang cac thong so sau ay:

au vao au ra noi I au ra noi IIDung dch NaOH :+ tD=125.33 0C+ CD= 3558.1j/kg.o+ GD=1159 kg/hHi ot:+ 1=137.9 0C+ i= 2737000 j/kg+ Cng1=4290 j/kg.o

Dung dch NaOH :+ t1=125.86 0C+ C1= 3323.2 j/kg.oHi th :+ 2 =114.5 0C+ i1 =2706000 j/kg+Cng2 = 4290 j/kg.o+ W1=303.5 kg/h

Dung dch NaOH:+ t2=98.42 0C+ C2= 3323.2 j/kg.o+ G2=579.5 kg/hHi th :+ t2=81.9 0C+ i2=2643740 j/kg+ W2=276 kg/h

Trang 7


8/36

O AN MON HOC QTTB

Cho : Qxp1=0.05.D.(i Cng1. 1) =0.05.D.r( 1).Qxp1=0.05.W.(i1 Cng2. 2) =0.05.W1.r( 2).

Vay lng hi th boc len noi I la :

=+

+=

1121

11222

1 .)(.95.0

...).(.

tCir

tCGtCWGiWW DD

= =+

+86.125*2.332326437402156000*95.0

86.125*2.3323*115942.98*2.3323*5.5792643740*5.579289.9

kg/h

Lng hi th boc len noi II la:W2=W-W1=579.5 289.9 = 289.6 kg/h

Lng hi ot tieu ot chung la:

D= =+).(95.0

...).(.

111

11111

ng

DDDD

Ci

tCGtCWGiW

= )9.137*42902700600(*95.033.125*1.3558*115986.125*2.3323*)9.2891159(2706000*9.289

+

=314.6 kg/hc. Kiem tra lai gia thiet phan bo hi th cac noi:

C%(I) = %5%5.4%1005.303

9.2895.303


9/36

O AN MON HOC QTTB

r = rcau1 +

+ rcau2

Chon : rcau1 = rcau2 =1/5000 m2.h.o / Kcal=1/4300 m2

.o /W = 2 mm

Ong lam bang thep khong r ma hieu 40XH: =44 W/m.o r = 5.106*10-04 m2.o /W

Nhiet tai rieng trung bnh:

- Noi I : qtb1=2

21 qq +

Trong o :+ q1 : nhiet tai rieng pha hi ot cap cho thanh thiet b. Ta cocong thc tnh q1:

q1= 1. t1 (1)He so cap nhiet cua hi nc bao hoa ngng tu tren be matong thang ng c tnh theo cong thc cua Nusselt:

1=1.13.A.(1

. tH

r

).025 Kcal/ m2.h.o (*)

Chon : tT1 = 137.56 0C.Khi o : t1= T tT1 = 137.9 137.54 = 0.36 0C

TW = 0.5(T+tT1) =137.72 0Cr = r( 1)=2156 kj/kg =514.95 Kcal/kg an nhiet ngng tu cua

hi ot.H =1.5 m chieu cao be mat truyen

nhiet.A = 2362.02 tr so phu thuoc nhieto ngng tu cua nc.

Thay cac gia tr vao cong thc (*) ta co:

1=1.13*2362.05*(36.0*5.1

95.514)0.25 = 14832.17 Kcal/m2.h.o.

Thay 1 vao cong thc (1) ta co:q1 = 14832.17* 0.36 =5339.58 Kcal/m2.h = 6199.85 W/m2

+ q2 : nhiet tai pha dung dch soi. Ta co cong thc tnh q2:q2= 2. t2 (2)

He so cap nhiet t thanh thiet b en dung dch 2 c tnhbi cong thc:

2 = 1.6 . . p0.4.q20.7 kcal/m2.h.o (**)Trong o :

T = q1.r = 6199.85*5.106.10-4=3.16 0C = 0.76 tha so ke en tnh chat ly hoc cua NaOH.

Trang 9


10/36

O AN MON HOC QTTB

p = 1.76 at ap suat hi tren be mat thoang cua dungdch soi

Thay vao (**) ta co : 2= 1.6*0.76*(1.76)0.4.(5339.58) 0.7=620 Kcal/m2.h.o= 719.89

W/m2

.oThay vao (2)ta co :q2 = 719.89*(137.54 3.16 125.86) =6133.46 W/m2

+ kiem tra lai gia thiet t1:

co %5%07.1%10085.6199

46.613385.6199%100*

1

21=

=

q

qqthoa man ieu

kien sai so.

vay nhiet tai trung bnh noi I la:

qtb1= 221 qq +

= 246.613385.6199 +

=6166.65 W/m2

.

- Noi II :qtb2=2

21 qq +

Trong o :+ q1 : nhiet tai rieng pha hi ot cap cho thanh thiet b. Ta cocong thc tnh q1:

q1= 1. t1 (3)He so cap nhiet cua hi nc bao hoa ngng tu tren be matong thang ng c tnh theo cong thc cua Nusselt:

1=1.13.A.( 1. tH

r

)

.025

Kcal/ m

2

.h.o (*)Chon : tT1 = 114.18 0C.

Khi o : t1= T - tT1= 114.5 114.18 = 0.32 0CTW = 0.5(T+tT1) =114.34 0C

r = r( 1)=2221.5j/kg=530.6 kcal/kg an nhiet ngng tu cuahi ot.

H =1.5 m chieu cao be mat truyennhiet.

A = 2268.87 tr so phu thuoc nhieto ngng tu cua nc.

Thay cac gia tr vao cong thc (*) ta co:

1=1.13*2268.87(32.0*5.1

60.530)0.25 = 14783.24 Kcal/m2.h.o.

Thay 1 vao cong thc (3) ta co:q1 = 14783.24* 0.32 =4730.64 Kcal/m2.h =5492.79W/m2

+ q2 : nhiet tai pha dung dch soi. Ta co cong thc tnh q2:q2= 2. t2 (4)

Trang 10


11/36

O AN MON HOC QTTB

He so cap nhiet t thanh thiet b en dung dch 2 c tnhbi cong thc:

2 = 1.6 . . p0.4.q20.7 kcal/m2.h.o (**)Trong o :

T = q1.r =5492.79*5.106.10-4

=2.80

C = 0.76 tha so ke en tnh chat ly hoc cua NaOH.p = 0.52 at ap suat hi tren be mat thoang cua dung

dch soiThay vao (**) ta co :

2= 1.6*0.76*(0.52)0.4.(4730.64) 0.7=349.77 Kcal/m2.h.o= 406.12W/m2.oThay vao (4)ta co :

q2 = 406.12.(114.2 2.8 98.42) =5271.44 W/m2

+ kiem tra lai gia thiet t1:

co %5%03.4%10079.5492

44.527179.5492%100*

2

21=

=

q

qqthoa man ieu

kien sai so.vay nhiet tai trung bnh noi I la:

qtb1=2

21 qq + =2

44.527179.5492 +=5382.1 W/m2.

a. He so truyen nhiet moi noi:- Noi I :

K1= 04.12

6166.651

= iItb

t

q

=512.18 W/m2

.o- Noi II :

K2=08.16

5382.12 = iII

tb

t

q=334.71 W/m2.o

3. Hieu so nhiet o hu ch thc cua moi noi:- Cong thc chung:

=

i

i

i

m

mim

K

Q

t

K

Qt .' 0C

trong o : ti = tiI+ tiII = 12.04+16.08=28.12 0C.

i

i

K

Q= 32.90246.53486.367

71.334

178890

18.512

188410=+=+

- Tnh cho noi I:

Trang 11


12/36

O AN MON HOC QTTB

46.1132.902

12.28*86.367.' ==

=

i

i

i

I

I

iI

K

Q

t

K

Qt

0C

- Tnh cho noi II:

66.1632.902

12.28*46.534.' ==

=

i

i

i

II

IIiII

K

Q

t

K

Qt

0C

4. Kiem tra lai hieu so nhiet o hu ch :- Noi I :

%5%8.4%10004.12

46.1104.12%100*

'


13/36

O AN MON HOC QTTB

VI. TNH KCH THC BUONG BOC VA BUONG OT :1. Kch thc buong boc :

Do lng hi th boc len hai noi gan xap x bang nhau, nhieto noi hai nho hn nen khoi lng rieng cua hi noi II se nho

hn noi I suy ra the tch hi thoat ra noi II se ln hn noi I. Dovay ta ch can tnh ai dien noi II.Van toc hi ( hmax) cua hi th trong buong boc khong qua 70 80% van toc lang( 0).

0=h

hl dg

..3

)..(.4 m/s

l, h : khoi lng rieng cua giot long va hi th (kg/m3).d : ng knh giot long, chon d =0.0003 m : he so tr lc

0.2< Re < 500 = 6.0Re

5.18

500< Re


14/36

O AN MON HOC QTTB

Hb= 56.04.1*14.3

86.0*4

.

.422==

b

b

D

V

m

Do trong thiet b co hien tng dung dch soi tran ca len phanbuong boc do o oi hoi thiet b phai cao hn so vi tnh toan .Vay o chon Hb=1500 m (QTTB T5 trang 182 ).2. Kch thc buong ot:a. Xac nh so ong truyen nhiet :

So ong truyen nhiet c tnh theo cong thc : n=ld

F

..

F= 40 m2 : be mat truyen nhietl = 1.5m : chieu dai cua ong truyen nhietd : ng knh ong truyen nhiet

chon loai ong co ng knh : 38 x 2 mmdo 1> 2 nen lay d = dt = 34 mm.

Vay so ong truyen nhiet la :

n= ldF

.. = 2505.1*034.0*14.340

= ong.Chon so ong n= 271 ong ( STQTTB T2 trang 46 )

b. ng knh ong tuan hoan trung tam :

tth

fD

.4=

Chon ft = 0.3 FD =0.34

..2nd =0.3

4

271*034.0*14.32

=0.0738 m2.

Vay :

tth

fD

.4= =

14.3

0738.0*4 =0.307 m

Chon Dth=0.325 m = 325 mm (QTTB T5 trang 180 )c. ng knh buong ot :oi vi thiet b co ac tuan hoan trung tam va bo tr ong

ot theo hnh luc giac eu th ng knh trong cua buong ot cothe tnh theo cong thc :

Dt=l

dFdd nnth

.

..60sin..4.0).2(

022

++ m

Trong o :

=nd

t= 1.4 : He so, thng = 1.3 1.5.

t =1.4*dn : Bc ong , m ( thng t = 1.2 1.5dn)dn =0.038 m : ng knh ngoai cua ong truyen nhiet , m = 0.8 : He so s dung li ong, thng = 0.7 0.9

l =1.5 m : Chieu dai cua ong truyen nhiet mdth = 0.325 : ng knh ngoai cua ong tuan hoan trung tam.F = 40 m2 : Dien tch be mat truyen nhiet , m2

Thay vao ta co :

Trang 14


15/36

O AN MON HOC QTTB

Dt= 057.15.1*8.0

038.0*40*.60sin*4.1*4.0)038.0*4.1*2325.0(

022 =++ m

Chon Dt = 1200 mm (QTTB T5 trang 182 )Kiem tra dien tch truyen nhiet:

Dth t( b-1 )

b 1.71038.0*4.1

325.01 =+=+

t

Dth

Chon b= 9 ong ( STQTTB T2 trang 46 )Vay so ong truyen nhiet a b thay the bi ong tuan hoantrung tam la :

n = 61 ong( STQTTB T2 trang 46 )So ong truyen nhiet con lai la:

n = 271 61 = 210 ong.Be mat truyen nhiet F = 3.14*1.5*(210*0.034+0.325)=35.2 m2 >

32.05 m2 ( thoa man )

VII. TNH KCH THC CAC ONG DAN LIEU, THAO LIEU:ng knh cac ong c tnh theo cong thc tong quat sau ay:

d= ..

.4

v

Gm

Trong o :G : lu lng lu chat kg/sv : van toc lu chat m/s : khoi lng rieng cua lu chat kg/m3

1. Ong nhap lieu noi I :G= 1159 kg/h = 0.322 kg/s

Chon v= 0.4m/s = 1159 kg/m3.

d= ..

.4

v

G=

1159*4.0*14.3

322.0*4 =0.0297 m

Chon : d = 0.042m2. Ong thao lieu noi I ( nhap lieu noi II ):

G= 869.1 kg/h = 0.2414 kg/sChon v= 0.2m/s = 1159 kg/m3.

d= ...4

v

G

= 1159*2.0*14.32414.0*4

=0.0366 mChon : d = 0.042m

3. Ong thao lieu noi II:G1= 579.5 kg/h = 0.161 kg/sChon v= 0.1m/s = 1276 kg/m3.

Trang 15


16/36

O AN MON HOC QTTB

d= ..

.4

v

G=

1276*1.0*14.3

161.0*4 =0.04 m

Chon : d = 0.042m4. Ong dan hi ot noi I:

G= 314.6 kg/h = 0.0874 kg/s

Chon v= 20 m/s = 1.869 kg/m3.

d= ..

.4

v

G=

869.1*20*14.3

0874.0*4 =0.0546 m

Chon : d = 0.146m5. Ong dan hi th noi I:

G= 289.9 kg/h = 0.0806 kg/sChon v= 20m/s = 0.9558 kg/m3.

d= ..

.4

v

G=

9558.0*20*14.3

0806.0*4 =0.0733 m

Chon : d = 0.146m6. Ong dan hi th noi II:

G= 289.6 kg/h = 0. 0804kg/sChon v= 20m/s = 0.3158 kg/m3.

d= ..

.4

v

G=

3158.0*20*14.3

0804.0*4 =0.1273 m

Chon : d = 0.146m7. Ong dan nc ngng noi I:

G= 314.6 kg/h = 0.0847 kg/s

Chon v= 0.4 m/s = 1000 kg/m3.

d= ..

.4

v

G=

1000*4.0*14.3

0847.0*4 =0.016 m

Chon : d = 0.042m8. Ong dan nc ngng noi II:

G= 289.9 kg/h = 0.0806 kg/sChon v= 0.2m/s = 1000 kg/m3.

d= ..

.4

v

G=

1000*4.0*14.3

0806.0*4 =0.016 m

Chon : d = 0.042m

VIII. TNH C KH:

Trang 16


17/36

O AN MON HOC QTTB

e thuan tien trong qua trnh tnh toan va che tao, ta chon vatlieu che tao hai noi la nh nhau vi be day bang nhau. Chon vatlieu la thep CT.3 e che tao vo thiet b, ay va nap.

Tra so tay tap 2 co cac thong so : k=380.106 N/m2

ch=240.106

N/m2

He so an toan : nk=2.6nc=1.5

He so hieu chnh: =1.0ng suat cho phep theo gii han ben :

[ k] =

k

k

n= 0.1*

6.2

10.380 6=146*106 N/m2

ng suat cho phep theo gii han chay:

[ c] =

c

c

n= 0.1*

5.1

10.2406

=160*106 N/m2

Vay ng suat cho phep : [ ]=146*106

N/m2

.a. tnh than buong ot:Cong thc tnh chieu day than buong ot :

S= Cp

pDt +]..[2

.m

ta co cac thong so cua noi I nh sau :Dt : ng knh trong cua thiet b. Dt=1200 mm : he so ben thanh hnh tru theo phng doc.

Tren thanh thiet b co lo d=0.146 m. do o =l

dl=

0.903

C : so bo xung do an mon , bao mon va dung sai am ,mC=C1+C2+C3C1=1 mm : he so bo xung nhiet do an monC2=0 mm : he so bo xung do bao monC3=0.6 mm : dung sai am theo chieu day.

C=1+0+0.6 =1.6 mm =1.6*10-3 mp : ap suat trong thiet b ,N/m2. p =3.5 at =343350 N/m2.

Thay vao cong thc co :

S= CppD

t +]..[2.

= 36 10*6.1343350903.0*10*146*2343350*2.1

+ =3.16 mmChon S=6 mm.

Kiem tra : 1.00042.01200

161


18/36

O AN MON HOC QTTB

[p] = 66

1

1 10*099.1)001.0006.0(2.1

)001.0006.0(*907.0*10*146*2

)(

)(**][*2=

+

=+

CSD

CS

t

N/m2

Ta co : p = 0.343 *106 N/m2 < [p] = 1.099 *106 N/m2 ( thoa man).Vay chieu day buong ot la : S= 6 mm.Do trong buong ot noi II, ap suat hi th nho hn noi I nen

chac chan ieu kien se thoa.b. tnh than buong boc:* Noi I :

Chon be day than buong boc noi I la S = 6 mm.Ta co ap suat ben trong buong boc noi I la:

P = 1.76 at = 0.172*106 N/m2 < [p] = 1.099*106 N/m2.Vay chon be day than buong boc la : S=6 mm.

* Noi II :Do thiet b lam viec ap suat chan khong nen chu tac ongcua ap suat ngoai. V vay be day toi thieu cua than c tnhtheo cong thc:

S = 1.18.Dt.4.0

.

t

t

n

D

l

E

p

Ap suat lam viec trong buong boc : po = 0.52 at=0.051*106 N/m2.Tnh chieu cao dung dch trong buong boc :The tch cua cac ong truyen nhiet va ca ong tuan hoan trungtam la :

V1=0.25..H0 (dn2.n+Dth2)=0.25*3.14*1.5*(210*0.0342+0.3252)=0.41 m3.The tch cua phan ay :

Chon ay non co g ( Dt=1200 mm, h=40mm, 2 =900)

V2 =0.348 m3 ( STQTTB T2 trang 386 )

The tch dung dch trong noi :Van toc dung dch cung cap vao noi : v = 0.4 m/sVan toc dung dch trong ong tuan hoan trung tam :

v=00335.0

325.0

4.0*14.3*3600

4*4.0

.2

2

2

2

=

=thD

dv m/s

Thi gian lu cua dung dch trong thiet b :

=8.1700

00335.0

325.0*14.3*25.0

348.05,1

'

' 2=+=+

v

ll (s)

the tch dung dch trong thiet b :

V = 2.v. =2*3600

1*1700.8=0.945 m3 ( do dung dch trong

thiet b soi bot nen s=0.5 dd the tch dung dch phaitang hai lan )

Trang 18


19/36

O AN MON HOC QTTB

The tch phan buong boc b chiem cho :V3 = V V1 V2 = 0.945 0.41 0.348 =0.187 m3.

Chieu cao dung dch trong buong boc :

H=122.0

4.1*25.0*14.3

187.0

4.

22

3 ==bD

V

m

Chon H= 0.2 m.Khoi lng rieng cua dung dch : =1276 kg/m3.Ap suat tnh toan trong buong boc:

p=po+ .g.H = 0.051*106+ 1276*9.81*0.2 =0.0535*106 N/m2.Ap suat ngoai: pn=pkt p =0.1*106 0.0535*106=0.0465*106 N/m2

Moun an hoi cua vat lieu :0.2*1012 N/m2

Chieu dai tnh toan cua than: l=1,5 m.ng knh than : Dt=1.4 m.

S = 1.18*1.4*

4.0

12

6

4.1

5.1

10*2.0

10*0465.0

=3.8 mmChon : C= Ci =1.6 mmBe day thc cua than: S=S+ C =3.8+1.6 = 5.4 mmChon S = 6 mm.Kiem tra :

Ta co: 07.14.1

5.1==

tD

l

1.5.4.1

)001.0006.0(*2*5.1

)(*2 =

t

a

D

CS=0.127

=

= )001.0006.0(*2

4.1

)(*2 a

t

CS

D11.832

0.3* 151.04.1

005.0*2*

10*240

10*2.0*3.0

)(*2.

3

6

123

=

=

t

a

ct

t

D

CSE

Thoa ieu kien:

0.127< 07.1=tD

l 0.151Ap suat ngoai cho phep :

[pn]=0.649.Et.4.1

005.0.

4.1

005.0*

5.1

4.1*10*2.0*649.0..

2

12

2

=

t

a

t

at

D

CS

D

CS

l

D

=0.92*106N/m2

Ta co: pn = 0.0465*106 N/m2 < [pn] = 0.92*106 N/m2 ( thoa man )Vay chon be day than buong boc la : S =6 mm.

Trang 19


20/36

O AN MON HOC QTTB

c. tnh nap:Chon nap elip co g vi Dt =1400 mm.

Ta co : 0=t

t

D

h.25 ht=350 mm ( chieu sau cua elip o theo mat

trong )

Chieu cao g : h= 25 mm

Ban knh cong ben trong nh Rt= ==35.0*4

4.1

.4

22

t

t

h

D1.4 m

Neu lo co lap d= dhi ot =0,146 m.

=> Z=1- 9.04.1

146.01 ==

tD

d

* Noi I:Thiet b lam viec ap suat trong p =1,76 at =0.172 *106 N/m2.He so ben moi han : = 0.95Be day toi thieu cua lap :

S= =

= 66

6

10*172.09.0*95.0*10*146*2

4.1*10*172.0

.]..[2

.

pZ

Rp t

0.97 mm

Chon C=5.03 mmBe day thc cau nap thiet b : S=6 mm.

Kiem tra :

125.00036.04.1

001.0006.0


21/36

O AN MON HOC QTTB

ng knh ay : Dt= 1200 mm H= 675 mm

h = 40 mmRt= 180 mm.

Chieu cao cot chat long : H = H+h+ H1+H2.

Trong o :H1 : chieu cao cot chat long trong buong ot, H1= 1,5 mH2 : chieu cao cot chat long trong buong boc, H2 = 0.2 m H= 675+40+1500+200=2415 mm =2.415 m

Noi I :Ap suat trong buong ot: p0= 3.5 at = 0.343*106 N/m2.Ap suat tnh toan :

p= p0 + .g.H =( 0.343+1159*9.81*10-6*2.415)*106=0.37*106 N/m2.He so ben moi han: =0.95Be day toi thieu cua ay :

S = )10*37.095.0*10*146(*45cos*210*37.0*2.1

)]..([cos.2

.66

6

= ppDt

=2.3 mm

Chon C= 3.7 mm.Be day thc cua ay : S =S+ C = 2.3+3.7 = 6 mm.

Kiem tra : 354.045cos

25.00019.0

1200

3.2'=


22/36

O AN MON HOC QTTB

Dt D Db Di h db1200 1350 1300 1260 30 M24

+ bch noi buong boc va buong ot tng t bch noi ay

va buong ot.+ tnh bulong noi nap va buong boc:Lc nen truc sinh ra do xiet bulong: Q1=0.25..Dt2.p+Dtb.b0.m.p (*)Vi : Dt=1400 mm

P=0.172 N/mm2

ng knh trung bnh cua em : Dtb= Dt + 2(0.5b+2.5 )Chon chieu day thc em b= 10 mm , ta co :

Dtb=1400+2(0.5*10+2.5)=1415 mmChieu rong tnh toan cua em : b0=0.8 b =0.8*10 =8 mmChon em amiang co = 3 mm ; m =2Ap suat rieng can thiet e lam bien dang deo em, q0=10

N/mm2.

Thay vao (*) :Q1=0.25*3.14*(1400)2*0.172+3.14*1415*8*2*0.172=0.277*106 N.

Lc can thiet e ep chat em ban au :Q2=.Dtb.b0.q0 = 3.14*1415*8*10=0.356*106 N.

Chon khoang cach tng oi gia tam cac bulong la x =195 mm

Vay so bulong la : z= 24195

1490*14.3.==

x

Db con

Lc tac dung len mot bulong : qb= zQ

Vi: Q= max ( Q1,Q2) =Q2= 0.356*106 N

qb= 33.1483324

10*356.0 6==

z

QN

ng suat tac dung len bulong :

===

22 20*14.3*25.0

33.14833

.4

b

b

d

q

47.22 N/mm2.

Chon vat lieu lam bulong la thep CT.3 co [ ]100=86 N/mm2. =47.22 N/mm2< [ ]100=86 N/mm2 ( thoa man )

+ tnh bulong noi ay va buong ot:Lc nen truc sinh ra do xiet bulong: Q1=0.25. .Dt2.p+Dtb.b0.m.p (*)Vi : Dt=1200 mm

p=0.343 N/mm2

thay vao (*) co:Q1=0.25*3.14*(1200)2*0.343+3.14*1215*8*2*0.343=0.409*106 N.

Lc can thiet e ep chat em ban au :Q2=.Dtb.b0.q0 = 3.14*1215*8*10=0.305*106 N.

Trang 22


23/36

O AN MON HOC QTTB

Chon khoang cach tng oi gia tam cac bulong la x =170 mm

So bulong la : z= 24170

1300*14.3.==

x

Db con

Lc tac dung len mot bulong : qb=z

Q

Vi: Q= max ( Q1,Q2) =Q1= 0.409*106

N qb= 67.17041

24

10*409.0 6==

z

QN

ng suat tac dung len bulong :

===

22 24*14.3*25.0

67.17041

.4

b

b

d

q

37.67 N/mm2< [ ]100 =86 N/mm2

f. tnh v ong:Lay v ong lam bchChon be day v d =30 mmTa co ng kng ngoai cua ong truyen nhiet : dn=38 mm

l=2

3*38*4,1

2

3.=

t =46.07 mm

* Noi I:p=p1-p1=0.343-0.172= 0.171 N/mm2.Co :

174.0

07.46

30)

07.46

38*7.01(*6.3

171.0

)7.01(6.3

=

=

=

l

h

l

d

p

n

N/mm2


24/36

O AN MON HOC QTTB

+ Ong ot : V = =+ ).(.4

.).(.

4

2222

tnthtthn ddHn

DDH

= 0759.0)034.0038.0(*5.1*4

210*14.3)325.0331.0(*5.1*

4

14.3 2222 =+

m3

m5= 7850*1.01*0.0759=601.77 kg+ V ong :V =0.25*3.14*0.03*[1.352 210*0.0382-0.3312] =0.033 m3.m6=2*7850*1.01*0.033=526.69 kg+ Bch :

nap va buong boc : V = == )49.154.1(*03.0*4

14.3).(.

4

2222

tn DDH

0.00357 m3.m7=7850*0.00357*2=56.05 kgBuong ot va buong boc : V =

== )3.135.1(*3.0*4

14.3).(.

4

2222

tn DDH

0.0312 m3.

m8=7850*0.0312*2=49.02 kgkhoi lng bch : m =m7+m8=56.05+49.02=105.07 kg.MTB =m+m1+ m1 +m2 +m3 +m4 +m5 +m6 =

= 105.07+106+77+312.07+267.68+601.77+526.69 =1996.28 kg Tnh Mdd:The tch dung dch trong thiet b ( tnh phan tren ) : V=0.945 m3

Khoi lng dung dch trong thiet b la : Mdd =V. s=0.945*0.5*1276=602.91 kg.Vay khoi lng tong cong :

M=MTB+ Mdd=1996.28+602.91 =2599.19 kg.

Chon 4 tai treo. Trong tai cho moi tai treo se la :G= 5.6374

4

19.2599*81.9= N

D phong chon : G=1.0*104NTra so tay QTTB T2 trang 426 co :

F.104(m2)

L(mm)

B(mm)

B1(mm)

H(mm)

S(mm)

l(mm)

a(mm)

d(mm)

m(kg )

89.5 110 85 90 170 8 45 15 23 2.0

Trang 24


25/36

O AN MON HOC QTTB

B.TNH THIET B PHU1. Tnh thiet b ngng tu baromet:a. Lng nc lanh ti vao thiet b ngng tu :

Gn=).(

)..(

22

22

dcn

cn

ttC

tCiW

kg/s

W2 : lng hi i vao thiet b ngng tu , W2=289.6 kg/h =0.0804kg/si : ham nhiet cua hi ngng , i =2643.74 kj/kgt2C,t2D : nhiet o au ,cuoi cua nc lam nguoi, lay t2D=300C.

t2C=tng 10 =80.9 10 =70.9 0Ctng : nhiet o hi bao hoa ngng tu

Cn : nhiet dung rieng trung bnh cua nc, tra theo nhiet otrung bnh.

tng= 45.502

9.7030

2

22 =+

=+ Dc tt 0C

Cn = 4,18 kj/kg.o

Gn= ==

)309.70(*18.4

)9.70*18.474.2643(*0804.0

)(

).(

22

22

Dcn

cn

ttC

tCiW1.1 kg/s

b. The tch khong kh va kh khong ngng can hut ra khoithiet b :

Lng kh can hut ra khoi thiet b ngng tu baromet :

Gkk=25.10-6

.(Gn+W2)+10-2

.W2= 25.10-6*(1.1+0.0804)+10-2*0.0804=25.10-6.0,96+0.0804*10-2 =8.34*10-4 kg/s

The tch kh khong ngng can hut ra khoi thiet b :

Vkk=hng

kkkk

pp

tG

+ )273(.288

Vi tkk =t2D+4+0.1 (t2C t2D)=30+4+0.1*(70.9-30)=38.09 0C

png=0.5 at =49050 N/m2 : ap suat lam viec cua thiet b ngngtuph =0.0576 at =5652 N/m2 : ap suat rieng phan cua hi nctrong hon hp nhiet o tkk.

Vkk = =

+

565249050

)09.38273(10*34.8*2884

0.0017 m3/s

c. Cac kch thc chu yeu cua thiet b ngng tu baromet :+ ng knh trong :

Trang 25


26/36

O AN MON HOC QTTB

Dt=1.383hh

W

2

h : khoi lng rieng cua hi, h=0.3158 kg/m3

h : toc o hi trong thiet b ngng tu, h=

23.514.0*25.0*14.3

0804.02= m/s

Dt=1.383 =23.5*3158.0

3600/6.2890.22 m.

Chon ng knh trong thiet b ngng tu la Dt= 300 mm+ kch thc tam ngan :tam ngan co dang vien phan vi chieu rong la :

b=2

tD +50 =150+50=200 mm

tam ngan co uc nhieu lo , lay ng knh cua lo la dl=2mmchieu day tam ngan , chon =4 mmchieu cao g canh tam ngan , chon h0=40 mmcac lo xep theo hnh luc giac eu, ta co bc lo :

t=0.866dl tbo

f

f

tbf

f0 ty so gia tong dien tch tiet dien cac lo vi dien tch

tiet dien cua thiet b ngng tu ,

chontbf

f0 =0.08

t =0.866*2* 08.0 =0.49 mm 0.5 mm.+ chieu cao cua thiet b ngng tu :

mc o un nong nc : p= 804.0309.80

309.70

2

22 =

=

Dng

DC

tt

tt

tra bang VI-7 ( STQTTB T2 trang 80) ta co:so ngan n=8khoang cach trung bnh gia cac ngan : htb=400 mmchieu cao cua thiet b ngng tu :Hng=n.htb+800 = 8*400+800=4000 mm = 4 m+ kch thc ong Baromet:

ng knh trong cua ong baromet : dba=.

)(004.0 2WGn +

: van toc cua hon hp nc lanh va nc ngng chay trongong baromet,chon =0.7m/s

Trang 26


27/36

O AN MON HOC QTTB

dba=14.3*7.0

)0804.01.1(*004.0 +=0.046 m

lay ng knh ong baromet la dba= 50 mmchieu cao ong baromet : hba=h1+h2+h3 (m)

h1: chieu cao cot nc trong ong baromet can bang vi hieu soap suat kh quyen va ap suat trong thiet b ngng tu baromet.

h1=10.33760

0P m

Po : o chan khong trong thiet b ngng tuPo = Pa-Png=760 0.5 *735 =392.5 mm Hg h1=10.33*392.5/760 = 5.33 m

h2 : chieu cao cot nc trong ong baromet can thiet e khacphuc toan bo tr lc khi nc chay trong ong.

h2= + )(2

2

ba

ba

d

h

g(m)

1: he so tr lc cuc bo khi vao ong , 1=0.5 2 : he so tr lc cuc bo khi ra khoi ong , 2=1ta co ttb= 50.450C = 990 kg/m3

= 5.4 *10-4N.s/m2.

Chuan so Re : Re =4

10*4.5

990*05.0*7.0..=

ba

d=64167

Chon ong thep la CT.3 nen o nham = 0.2mm

tnh: Regh=6 33012.0

50*6

7/87/8

=

=

d

Ren = 220 1096742.0

50*220

8/98/9

=

=

d

Vay: Regh


28/36

O AN MON HOC QTTB

Chon loai thiet b ong chum thang ng, dung dch i trong ong, hiot i ngoai ong.

Dong nhap lieu ( dong lanh ): tD=30 0CtC= 125.330C

ttb= 0.5 (tD+tC) = 0.5*(30+125.33)=77.670

C.Dong nong : TD=TC=137.9 0C.Hieu nhiet o au vao : tvao=137.9 125.33 =12.570C

tra= 137.9 - 30 =107.9 0CHieu so nhiet o trung bnh :

ttb==

=

57.12

9.107ln

57.129.107

lnvao

ra

vaora

t

t

tt

44.34 0C.

chon t1= 0.83 0C tt1= 137.9-0.83=137.07 0C.

tm=2

9.13707.137 +=137.490C.

tra bang ta co : A = 2360r = 2156 kJ/kg= 514.95 Kcal/kg

chon : rcau1 =0.2*10-3 m2.o/W rcau2 =0.387*10-3 m2 .o /W r =6.32*10-4 m2.o/W thep= 2mm .H = 1.5 m.

Thep khong r 40XH co =44 W/m.o

1=1.13.A.(1. tH

r

).025 Kcal/ m2.h.o (*)

= 1.13*2360*

25.0

5.1*8.0

95.514

= 12137.7 Kcal/m2

.h.ota co : q1= 1. t1= 12137.7*0.83=10074.29 Kcal/m2.h=11697.37W/m2

co : T = q1.r = 11697.37*6.32.10-4=7.4 0C t2= tt1 - T ttb= 137.09 7.4 77.67 =52.02 0C.

ta co bang so lieu sau ay :

(W/m.o )

(kg/m3)

C( j/kg.o )

.103

( N.s/m2)ttb=77.670C 0.5735 1133.67 3785.75 1.018

tt2=129.740C0.5855 1082.55 3800.00 0.810

Chuan so Prandtl :

Pr=5735.0

75.3785*10*018.1. 3=

C =6.72

Trang 28


29/36

O AN MON HOC QTTB

PrW=5855.0

3800*10*81.0.3

=

C =5.257

Chon van toc dung dch trong ong truyen nhiet : v =0.01 m/sng knh ong truyen nhiet : d = 34/38 mm

Re = 310*018.1

034.0*01.0*67.1133..

=

dv=378.63

Tra bang: = 0.554*10-3 ( 0C 1)Chuan so Grashoft :

Gr = 23323

2

2

23

)10*018.1(*81.9

10*554.0*02.52*)81.9*67.1133(*034..0

.

...

=g

td

=13784810.1

Tra co = 1

Nu = 0.15. .Re0.33.Pr0.43.Gr0.1.25.0

Pr

Pr

W

=

=0.15*1*(378.63)0.33*(6.72)0.43*(13784810.1)0.1*(6.72/5.257)0.25=13.28

2= ==034.0

28.13*5735.0.

d

Nu224.07 W/m.o

q2= 2. t2=224.07*52.09=11671.76 W/m2.Kiem tra :

s = %100*37.11697

76.1167137.11697%100*

1

21 =

q

qq=0.22% < 5%

nhiet tai trung bnh :qtb= (11671.76+11697.37 )/2 =11684.56 W/m2.

He so truyen nhiet :

K= 34.44

56.11684=

tbtb

t

q

=263.52 W/m

2

.o.

Q = GD.C. t = = )3033.125(*75.3785*3600

67.1133113649 W

Be mat truyen nhiet :

F =34.44*52.263

113649

.=

tbtKQ

=9.7 m2.

Chon be mat truyen nhiet la F = 10m2.

So ong truyen nhiet la : n=5.1*034.0*14.3

10

..=

Hd

F

= 63 ong

Chon n = 91 ong.

Tra b = 11 ong.t =1,4.dn=1.4*0.038 =0.0532 mm D= t.(b 1)+4dn=0.0532*10+4*0.038= 0.684 m.

Chon D =800 m.Van toc chay trong ong :

v=3600*034.0*25.0*91*14.3

1

.25.0.. 22=

dn

G

=0.003 m/s

Trang 29


30/36

O AN MON HOC QTTB

do van toc dung dch chay trong ong cham nen thi gian truyennhiet la ln do o chon m=1 ( so pass pha vo ).

2.tnh bon cao v:Chieu cao bon cao v c at o cao sao cho thang ccac tr lc cua ng ong.

Phng trnh nang lng :Z1+ H

g

vp++

2

2

1.11

= Z2+ ++

g

vp

2

2

2.22

h1-2

p1=1,033 atp2=0,522 at =1327,4 kg/m3

=5,6.10-3 N.s/m2

Chieu cao t mat thoang noi xuong at la : Z=4 mng knh ong nhap lieu vao noi la : d = 80 mmVan toc dong chay trong ong:

v= 312,04,1327.08,0.14,3.3600

7500.4

..

.422

==d

GD m/s

chuan so Reynolds :

Re =

..dv= 33 10.92,510.6,5

4,1327.08,0.312,0=

chon ong thep CT.3 nen o nham = 0.2 mm

tnh: Regh=6 5,56482,0

80.6

7/87/8

=

=

d

Ren = 220 38/98/9

10.1,186

2,0

80.220 =

=

d

Vay: Regh


31/36

O AN MON HOC QTTB

h1-2= 017,045,880

15038,0

81,9.2

2,0

.2

22

=

+=

+

d

l

g

vm

chieu cao t mat thoang bon cao v en mat at :

Z1= 4 +81,9.4,1327

10.81,9).033,1522,0(4

+0.043=9.82 m

3. Tnh bm chan khong :Cong suat bm chan khong :

N=

1.110.

1

1

1

2

3

m

m

kkkk

CK p

pVp

m

m

CK : he so hieu chnh , CK=0.8m : ch so a bien , m=1.3p2 : ap suat kh quyen , p2=1.033 atap suat khong kh trong TBNT : pkk= p1=png - ph=0.5 0.0576

=0.4424atthe tch khong kh can hut khoi thiet b: Vkk= 0.0017m3/scong suat bm :

N=

1

4424.0

033.10017.0*10*81.9*4424.0

13.1

3.1

10*8.0

1 3..13.0

4

3 =0.9 kW

4. Tnh bm nc vao thiet b ngng tu:Cong suat cua bm :

N=

.1000

... HgQ(kW)

H : cot ap cua bm ( m) : hieu suat cua bm , chon =0.75 : khoi lng rieng cua nc 300C , = 997kg/m3.

Q : lu lng nc lanh ti vao Baromet : Gn= 1.1 kg/s

Q= 0011.0997

1.1 ==

nG m3/s

Phng trnh bernoulli cho hai mat cat 1 1 ( mat thoang be nc)va 2 2 ( mat thoang thiet b baromet )

Z1+ Hg

vp++

2

2

1.11

= Z2+ ++

g

vp

2

2

2.22

h1-2

Vi :

v1=v2=0 m/sp1=1,033 atp2= 0.5 at = 0.801*10-3 N.s/m2.

Chieu cao t mat thoang be nc xuong at la : Z1=2 mChieu cao t mat thoang thiet b baromet xuong at la Z2= 12

m

Trang 31


32/36

O AN MON HOC QTTB

Chon dhut=day=ng knh ca vao thiet b cua nc la d =100mm

Van toc dong chay trong ong:

v= 141.001.0*14.3

0011.0*4

.

.42

==d

Q

( m/s)

Chuan so Reynolds : Re = 310*801.0997*1.0*141.0.. =

dv =17553

Chon ong thep CT.3 nen o nham = 0.2 mmTnh Regh:

Regh=6. 72892.0

100*6

7/87/8

=

=

d

Ren = 220 2392022.0

100*220

8/98/9

=

=

d

Vay: Regh


33/36

O AN MON HOC QTTB

p2=1.033 at = 2*10-3 N.s/m2.

Chieu cao t mat thoang be cha nguyen lieu xuong at la :Z1=2 m

Chieu cao t mat thoang bon cao v xuong at la Z2= 10 m

Chon dhut=day=ng knh ong nhap lieu d =42 mmVan toc dong chay trong ong:

v=22

042.0*14.3

00028.0*4

.

.4 =d

Q

=0.202 m/s

Chuan so Reynolds : Re = 310*2

1159*042.0*202.0..=

dv=4919

Chon ong thep CT.3 nen o nham = 0.2 mmTnh Regh:

Regh=6. 27052.0

42*6

7/87/8

=

=

d

Ren = 220 =

=

8/98/9

2.042*220

d 90140

Vay: Regh


34/36

O AN MON HOC QTTB

v1=v= 091.0042.0*14.3

000126.0*4

.

.422==

d

Q

m/s

v2=0 m/sp1= 0.588 atp2=1.033 at

= 1.82*10-3

N.s/m2

.Chieu cao t ong thao lieu xuong at la : Z1=0.9 mChieu cao t mat thoang be thao lieu xuong at la Z2= 2 mChon dhut=day=ng knh ong thao lieu d =42 mm

Chuan so Reynolds : Re = 310*82.11276*042.0*091.0..

=dv

=2681

He so ma sat :

= =25.0Re

3164.025.0

2681

3164.0=0.044

tong he so ton that cuc bo : = vao+3. khuyu 90+2. van+ ra= 0.5+3*1.19+2*0.5+1 =5.57

chieu dai ong t ong thao lieu en be cha la : l=10 mtong ton that :

h1-2= =

+=

+ 57.5

042.0

10044.0

81.9*2

091.0

.2

22

d

l

g

v0.007 m

cot ap cua bm : H= (2-0.9)+1276

10000*)588.0033.1( +0.007=4.59 m

cong suat cua bm : N=75.0*1000

59.4*81.9*1276*000126.0=0.01 (kW)

7. Ca sa cha va knh quan sat :

8. Be day lp cach nhiet :

Be day lp cach nhiet buong ot noi I : = 5.11

3.12

35.12.1

2 ...8.2

q

td t

ng knh ngoai buong ot : d2=1212 mmChon lp cach nhiet amiang cactong, he so cap nhiet

=0.144 W/m.o.Nhiet o ngoai thanh buong ot , tt2= 0CDien tch buong ot: S= .d2.H =3.14*1.212 *1.5 =5.7 m2

Nhiet ton that: q1=104.5*0.98*5.7=584.9 W/m

= 5..1

3.135.12.1

5..1

1

3..12

35..12.1

2

9.5849.137*144.0.212.1*8.2

...8.2

=q

td t=11 mm

e thuan tien trong che tao chon chieu day lp cach nhiet chobuong boc noi I va noi II la 11 mm

Trang 34


35/36

O AN MON HOC QTTB

C.TNH GIA THANH Tnh thiet b chnh :+ khoi lng thiet b khong tnh bch va ong : MTB = 1289.44 *2=2578.88 kg.n gia thep vat t khong r =50.000 /kg

$thiet b = 2578.88*50.000 = 128.944.000 + ong : n gia ong thep khong r : d $ = 50.000/m

d> 50 mm => $ = 10.000/mso ong truyen nhiet : n =210 ong $ong = 210.1,5.50000+1,5.100000 =15.900.000

+tien bulong : n gia $ =3000/contong so bulong can dung : m= 2*(24*3+8*6)=240 con $bulong = 3000*240 =720.000

+ tien em : n gia $=250.000tong so em can dung : 6 cai $em=6*250000=1.500.000

+ tai : $tai=2*4*2*10000 =160.000 + Bch ghep : $bch=2*105.07*10000 =2.101.400 + ca sa cha : n gia $ =1.000.000 $cua=2*1.000.000 =2.000.000 + ca quan sat : knh thuy tinh day 5 mm , ng knh 120 mm

Trang 35


36/36

O AN MON HOC QTTB

$knh=2*0.12*120000 =28.800 Vay gia vat t cua thiet b chnh la : $ =151.354.200 .

Tnh thiet b phu :Gia thiet b baromet, bnh tach long, thiet b gia nhiet : $ =20.000.000

+ bm chan khong :cong suat : 0.9 kW=1.2 Hp $bm ck=7000000*1.2=8.400.000

+ bm nc vao TBNTcong suat:0.067 kW=0.09Hp $=7000000*0.09 =630.000

+ bm thao lieucong suat 0.01kW=0.013 Hp $=2000000*0.013=26.000

+ bm dung dchcong suat 0.034 kW=0.05 Hp $=2000000*0.05=100.000

tong tien bm la :$ =9.057.000 + tien ong : $em=50000* 100=5.000.000

+ap ke t ongtong so ap ke can dung : 4 cai $em=4*500000=2.000.000

+ lu lng ke : $=1.000.000

Gia thanh che tao bang 200% tien vat t:

tong gia tien = tien che tao + tien vat t= 3(151354200+20000000+2000000+1000000+9057000+5000000)

= 565.233.000 .

co dac naohi

Documents