do an sấy quặng mangandioxit

7/29/2019 do an sy qung mangandioxit

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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc

Mc Lc

Mc Lc ............................................................................................................1Li m u ........................................................................................................3

PHN I: GII THIU CHUNG .......................................................................4I.Gii thiu chung v vt liu sy ..................................................................41. Tnh cht ca nguyn liu .....................................................................4

1.1. Tnh cht vt l ...............................................................................41.2 Tnh cht ha hc ............................................................................4

2. ng dng ca MnO2 .............................................................................4II. Gii thiu chung v my sy thng quay .................................................5

1.nh ngha, phm vi ng dng v phn loi ..........................................52.Gii thiu v dy chuyn thit b sy thng quay ..................................7

3.La chn thit b .......................................................................................104.Thuyt minh quy trnh cng ngh ............................................................10PHN II : THIT K H THNG SY THNG QUAY ........................11I.Cc thng s ban u ................................................................................11II. Tnh ton v la chn nhin liu ...........................................................12

1. Nhit dung ring ca than ...................................................................122. Nhit tr cao ca than ...............................................................................133.Nhit tr thp ca than ..............................................................................134. Lng khng kh l thuyt t chy 1 kg nhin liu ..........................135. Entanpi ca hi nc trong hn hp khi ...............................................13

6. H s khng kh d bung t v bung trn theo l thuyt ...........14III. Tnh ton cc thit b chnh ...............................................................141. Lng m bay hi trong qu trnh sy ................................................142. Lng vt liu kh tuyt i ...............................................................153. Phng trnh cn bng nhit ................................................................154. Th tch thng sy ...............................................................................155. Chiu di ca thng .............................................................................156. Thi gian sy .......................................................................................167. S vng quay ca thng sy ................................................................17

8. Cng sut cn thit quay thit b ....................................................179. Cn bng l t than ............................................................................189.1.Nhit lng vo tnh khi t 1kg than ...............................................18a) Nhit lng vo bung t ..................................................................18

b)Nhit lng ra khi bung trn tnh khi t 1 kg than ........................19Qra = Q4 + Q5 + Qm (KJ) ..........................................................19

9.2.Phng trnh cn bng nhit l t than ............................................21Qvo = Qra .....................................................................21

10. Tnh h s truyn nhit .....................................................................25

b) Xc nh .............................................................................................2711. Cn bng nhit lng trong thit b sy ............................................3011.1. Nhit lng vo ..............................................................................30

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11.2. Nhit lng ra khi my sy ..........................................................3111.3. Phng trnh cn bng nhit ca thit b sy ..................................3311.4. Trng thi ca khi l vo my sy, i ra khi my sy v lulng kh ..................................................................................................33

PHN III : TNH TON C KH ..........................................................37I. Tnh ton h thng bnh rng dn ng ..............................................371. Chn ng c ......................................................................................372. T s truyn v s vng quay .............................................................383. Cng sut v momen xon trn trc ca bnh rng nh ....................384. Tnh ton b truyn bnh rng tr, rng thng ....................................39a) Xc nh khong cch trc .................................................................39

b) Xc nh thng s n khp .................................................................39c) Tnh rng v bn tip xc ...............................................................40

d) Kim tra bn un qua ti ...............................................................42Cc thng s kch thc b truyn : ........................................................43II. Xc nh ti trng ..............................................................................441. Trng lng vt liu nm trong thng .................................................442. Trng lng ca thng ........................................................................443. Trng lng ca vnh ai ....................................................................444. Trng lng ca bnh rng vng .......................................................455. Trng lng ca lp cch nhit ..........................................................456. Trng lng ca cnh mc nng ..................................................45

III. Kim tra bn cho thng quay ............................................................461. Khong cch gia hai vnh ai ...........................................................462.Mmen un do ti trng gy ra ...........................................................463.Mmen un do bnh rng vng gy ra ................................................464.Tng mmen un .................................................................................465.Mmen chng un ca thng ..............................................................46IV.Tnh vnh ai .....................................................................................461. Ti trng trn mt vnh ai .................................................................462.Phn lc con ln ..................................................................................473.B rng vnh ai ...................................................................................47V. Tnh con ln chn, con ln ............................................................481.Tnh con ln ....................................................................................482. Tnh con ln chn ...............................................................................49PHN IV : CC THIT B PH ...........................................................50I .Tnh ton l t ....................................................................................501.Th tch bung t ...............................................................................502.Din tch ghi l .....................................................................................51II .Qut thi vo my sy ........................................................................51Kt Lun ..................................................................................................53

Ti liu tham kho ...........................................................................................54Nhn xt ca trng khoa: ..............................................................................55

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Li m u

Sy l mt trong nhng cng on quan trng trong cng ngh sn xut.

Thc t ta thy nu khng c qu trnh sy th thnh phm sau khi sn xut xong c m rt cao, nh hng n qu trnh bo qun v lu tr. Cc qung nhn to sau

khi sn xut c thnh phm nu khng qua cng on sy d nh hng n cht

lng qung do nh hng ca iu kin mi trng. Nc ta l mt trong nhng

nc c iu kin thi tit kh m, chnh v vy cng on sy l mt cng on v

cng quan trng trong giai on sn xut qung, nng sn. trong n mn hc

ny, em s trnh by v quy trnh cng ngh v thit k thit b sy thng quay

sy qung mangandioxit nhn to vi nng sut 13 tn/gi c m u vo l

8,5% v m u ra l 0,5%.

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PHN I: GII THIU CHUNG

I.Gii thiu chung v vt liu sy

1. Tnh cht ca nguyn liu

1.1. Tnh cht vt l

Qung MnO2 l cht bt mu en c thnh phn khng hp thc, khi un nng s b

phn hy thnh oxit thp hn.

iu kin thng n l oxit bn nht trong cc oxit ca Mangan, khng tan trong

nc, tng i tr.

Khi lng ring: 5030 Kg/m3.

1.2 Tnh cht ha hc

Khi un nng, n tan trong axit v kim nh mt oxit lng tnh. Khi tan trong

dung dch axit n khng to nn mui km bn ca Mn+4 theo phn ng trao i m

tc dng nh cht oxi ha.

Khi tan trong dung dch KOH c to nn dung dch xanh lam cha cc ion Mn (III)

v Mn (V) v trong iu kin ny ion Mn (V) khng tn ti c.

Khi nu chy vi kim hoc oxit baz mnh to thnh mui Managanat. nhit cao, MnO2 c th b kh bi H2,CO2 hoc C to thnh kim loi.

Khi nu chy vi kim nu c mt cht oxi ha v d nh:KNO 3

,KClO3,O2..MnO2 b oxi ha thnh Mn theo phng trnh:

MnO2 + KNO3 +K2CO3 =K2MnO4 + KNO2 + CO2

2. ng dng ca MnO2

MnO2 tn ti trong t nhin di dng khong vt pirolusit l hp cht ca

Mangan c nhiu ng dng trong thc t.

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dng bt MnO2 l xc tc cho phn ng phn hy KClO2 v H2O2 ,phn ng oxi

ha NH3 thnh NO v axit Axetic thnh Axeton

c a vo nguyn liu nu thy tinh lm mt mu lc ca thy tinh,cho thy

tinh mu hng hoc en (vi lng ln).Trong cng nghip gm MnO2 to mu nu ,en cho men.

Trong cng nghip sn xut pin MnO2 c s dng lm mt in cc ca pin.

V d nh:

Mangan dioxit c xem l ng vin c nhiu u im lm nguyn liu ch to

in cc cho pin sc bi v chng c nhiu trong thin nhin v tng hp vi mi

trng . Tn dng nhng u im ca mangan dioxit nhiu phng php c hiu

qu c pht trin ci thin c tnh ca in cc MnO2)C nhm mc ch s

dng cho pin s cp.

in cc hn hp ca MnO2/Cacbon c ch to bng cch cho trc tip bt cacbon

vo trong dung dch Manganaxetat cng kt ta vi MnO2.nH2O trn b mt nn

cacbon . Hnh thi hc b mt v cu trc tinh th c xc nh bng phng php

hin vi in t qut ( SEM) v k thut nhiu x tia X (XRD) . Qut th vng tun

hon ( CV ) nh gi tnh cht in ha ca in cc c ch to.Kt qu chng minh bt cacbon c hiu qu lm tng in dung v ci thin tnh cht in

ha ca in cc Mangandioxit.

II. Gii thiu chung v my sy thng quay

1.nh ngha, phm vi ng dng v phn loi

Sy l qu trnh dng nhit nng lm bay hi nc ra khi vt liu.Qu

trnh ny c th tin hnh bay hi t nhin bng nng lng t nhin nh nnglng mt tri,nng lng gi.(gi l qu trnh phi hay sy t nhin).Dng

cc phng php ny ch tn nhit nng nhng khng ch ng iu chnh

c vn tc ca qu trnh theo yu cu k thut , nng sut thpbi vy trong

cc ngnh cng nghip ngi ta thng tin hnh qu trnh sy nhn to ( ngun

nng lng do con ngi to ra ).

Ty theo phng php truyn nhit trong k thut sy cng chia ra :

Sy i lu phng php sy cho tip xc trc tip vt liu vi khng kh

nng,khi l.(gi l tc nhn sy).

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Sy tip tip xc l phng php sy khng cho tc nhn sy tip xc trc tip

vi vt liu sy m tc nhn truyn nhit cho vt liu sy gin tip qua mt vch

ngn .

Sy bng tia hng ngoi l phng php sy dng nng lng ca tia hng ngoido ngun nhit pht ra truyn cho vt liu sy.

Sy bng dng in cao tn l phng php sy dng nng lng nhit trng

c tn s cao t nng trn ton b chiu di ca lp vt liu.

Sy thng hoa l phng php sy trong mi trng c chn khng chn rt

cao nhit rt thp nn m t do trong vt liu ng bng v bay hi t trng

thi rn khng qua trng thi lng.

Ba phng php cui c s dng trong cng nghip nn gi chung l phng

php sy c bit.

Trong cng nghip ha cht v thc phm, cng ngh v thit b sy i lu v

tip xc c dng nhiu hn c, nht l phng php sy i lu. N c nhiu

dng khc nhau v c th sy c hu ht cc dng vt liu sy.

Theo kt cu nhm thit b sy i lu c th gp cc dng sau:

Thit b sy bung( nng sut thp, lm vic khng thng xuyn)Thit b sy hm( nng sut cao, lm vic bn lin tc)

Thit b sy thp( sy vt liu dng ht nh thc, ng)

Thit b sy thng quay( nng sut khng cao, sy vt liu dng cc, ht v bt)

Thit b sy phun( sy vt liu dng huyn ph nh c ph tan, sa bt)

Thit b sy kh ng( sy vt liu dng b, nh v c cha m b mt)

Thit b sy tng si( nng sut cao)

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2.Gii thiu v dy chuyn thit b sy thng quay

H thng sy thng quay l h thng sy lm vic lin tc chuyn dng sy

vt liu ht, cc nh nh: ct, than , cc loi qung, mui v cc ha chtNaHCO3, BaCl2,

B

C

C

400

234.4

1

2

3

4

5 6 7 8 II 9 10 11 12 III V

13

14

15

16

17 18 19 20 21

1.Thng quay 2.Vnh i 3.Con Ln

4.Bnh rng 5.Phu hng sn phm 6.Qut ht

7.Thit b lc bi 8.L t 9.Con ln chn

10.M t qut chuyn ng 11.B tng 12.Bng ti13.Phu tip liu 14.Van diu chnh 12.Bng ti

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My sy thng quay gm 1thng hnh tr (1) t nghig vi mt phng nm

ngang 16o. Ton b trng lng ca thng c t trn 2 bnh ai (2).

Bnh ai c t trn bn con ln (3), khong cch gia 2 con ln cng1 b (11) c th thay i iu chnh cc gc nghing ca thng, ngha l iu

chnh thi gian thi lu vt liu trong thng .Thng quay c l nh c bnh rng

(4 ). Bnh rng (4) n khp vi vi bnh rng dn ng (12) nhn truyn ng ca

ng c (10) qua b gim tc.

Vt liu t c np lin tc vo u cao ca thng qua phu cha (14) v

c chuyn ng dc theo thng nh cc m ngn. Cc m ngn va c tc

dng phn b u theo tit din thng, o trn vt liu va lm tng b mt tip

xc gia vt liu sy v tc nhn sy. Cu to ca m ngn ph thuc vo kch

thc ca vt liu sy tnh cht v m ca n. Vn tc ca khi l hay khng kh

nng i trong my sy khong 23 m/s,thng quay 38 vng/pht. Vt liu kh

cui my sy c tho qua c cu tho sn phm (5) ri nh bng ti xch (13)vn

chuyn vo kho.

Khi l hay khng kh thi c qut (7) ht vo h thng tch bi, tch nhng ht bi b cun theo kh thi. Cc ht bi th c tch ra, hi lu tr li

bng ti xch (13). Kh sch thi ra ngoi.

Tc khi l hoc khng kh nng i trong thng khng c ln hn 3m/s bi

nu tc ln hn 3m/s th vt liu b cun nhanh ra khi thng.

Cc m ngn trong thng va c tc dng phn phi u vt liu theo tit din

thng , o trn vt liu va lm tng b mt tip xc gia vt liu sy v tc nhn

sy. Cu to ca m ngn ( cnh o trn ) ph thuc vo kch thc ca vt liu

sy v m ca n. Cc loi m ngn c dng ph bin trong my sy thng

quay gm cc loi:

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ebd

cba

S cu to cnh trong thit b sy thng quay

a. Cnh nng

b. Cnh nng chia khoang

c. Cnh phn b u(cnh phn phi ch thp)

d. Cnh hn hp

e. Cnh phn vng

i vi vt liu dng cc to nhng xp, nh trong thng sy c th b tr cnh nng

( Hnh a )

Ngc li vi vt liu sy dng cc to,nng th nn b tr cnh nng c chia khoang

( hnh b )

Khi sy vt liu dng ht hoc cc nh, nh ngi ta thng dng loi cnh phn

phi ch thp(hnh c)i vi vt liu sy c kch thc ht qu b c th to thnh bi th nn dng cnh

loi chia thnh khoang kn ( hnh e )

u im:

Qu trnh sy u n v mnh lit nh tip xc tt gia vt liu sy v tc nhn

sy. Cng sy ln c th t 100kg m bay hi/m 3h.

Thit b gn c th c kh ha v t ng ha ton b khu sy.

Nhc im:

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Vt liu b o trn nhiu nn d to bi do v vn phi nn trong nhiu trng hp

s lm gim cht lng sn phm.

3.La chn thit b

Theo s liu m ban u ca qung MnO2 nhn to l W1 = 8.5% , qu trnh

cn phi thc hin lin tc vi nng sut 13 tn/h ,vt liu dng bt, c th t chy

c nn chn cnh o trn kiu phn phi. Tc nhn khi l ( v nhit u ca

tc nhn sy 240oC) chuyn ng ca tc nhn sy trong thit b sy chn xui

chiu.

4.Thuyt minh quy trnh cng ngh

Vt liu sy l qung MnO2 nhn to v thng st bng h thng gu ti.

Qung MnO2 nhn to khi vo thng sy c m 8,5% , chuyn ng cng chiu

vi tc nhn sy.

Tc nhn sy s dng l khi l, to ra t nhin liu t than, sau khi qua

bung ha trn vi khng kh bn ngoi t nhit thch hp cho qu trnh sy.

Dng tc nhn sy c gia tc bng qut y t trc thit b v qut ht t

cui thit b.

Trn ng ng dn khi l vo bung ha trn v ng ng dn khng kh

t mi trng vo bung ha trn u c cc van, dng iu chnh lu lng cc

dng. t nhit k sau bung ha trn xc nh nhit ca tc nhn sy trc

khi vo thng sy, nu nhit qu cao ta s m van tho bt khi l ra, gim

lng khi l vo bung ha trn gim bt nhit , ngc li nu nhit cha

, ta kha bt van dn khng kh t mi trng vo bung ha trn.

Thng sy c dng hnh tr t nm nghing mt gc 30

so vi mt phng ngang,c t trn mt h thng con ln v con ln chn. Chuyn ng quay ca

thng c thc hin nh b truyn ng t ng c sang hp gim tc n bnh

rng gn trn thng. Bn trong thng c gn cc cnh kiu phn phi, dng o

trn vt liu sy mc ch l tng din tch tip xc gia vt liu sy v tc nhn

sy, do tng b mt truyn nhit v tng cng trao i nhit qu trnh sy

din ra trit .

Trong thng sy ht qung MnO2 c nng ln n cao nht nh sau

ri xung. Trong qu trnh vt liu tip xc vi tc nhn sy, thc hin cc qu

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trnh truyn nhit v truyn khi l bay hi m. Nh nghing ca thng m vt

liu s c vn chuyn i dc theo chiu di thng. Khi i ht chiu di thng sy,

vt liu s t c m cn thit cho qu trnh bo qun 0,5%.

Sn phm qung MnO2 nhn to sau khi sy c a vo bung tho liu, sau khiqua ca tho liu s c bao gi bo qun hay dng vo cc mc ch sn xut

khc.

Dng tc nhn sy khi qua bung sy cha nhiu bi do cn phi

a qua h thng lc bi trnh thi bi bn vo khng kh gy nhim mi

trng. y, ta s dng h thng lc bi bng nhm 4 xyclon. Khi l sau

khi lc bi s c thi vo mi trng. Phn bi lng s c thu hi quaca thu bi ca xyclon v c x l ring.

PHN II : THIT K H THNG SY THNG QUAY

I.Cc thng s ban u

Kiu thit b sy thng quay: phng thc sy xui chiu

Tc nhn sy: khi l

Nhit khi l vo thng sy : t1 = 240oC

Nhit khi l ra khi thng sy: t2 = 80oC

Vt liu l qung MnO2 nhn to dng bt mn.

Khi lng ring xp ca MnO2: 32,724=x ( Kg/m3 ) c tnh nh sau:

.22 MnOMnO tbx = ( Kg/m3)

Trong : : l h s in y ( chn = 0,15 do chn cnh khuy kiu phn phi )

221

2

+=MnOtb

Vi 1 : Khi lng ring ca vt liu trc khi sy, 1 = 5030 ( Kg/m3)

2 : Khi lng ring ca vt liu sau khi sy

12 = - lng m bay hi trong 1 m3 th tch

( ) ( ) 6,46275030.5,05,85030 12112 === WW ( Kg/m3 )

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8,48282

6,46275030

221

2 =+

=+

=

MnOtb ( Kg/m3 )

32,72415,0.8,4828.22 === MnOMnO tbx ( Kg/m3 )

m u vo ca vt liu: 8,5% m u ra ca vt liu: 0,5%

Nhit vt liu vo my sy: to = 25oC

Nhit vt liu ra khi my sy: t1 = 70oC

Nhit dung ring ca bn thnh phm trc khi vo my sy:C1 = 0,658(KJ/Kg)

Nhit dung ring ca bn thnh phm sau khi ra khi my sy: C2 = 0,68(KJ/Kg)

Cc thng s khng kh:

Nhit mi trng: t = 25oC

m tng i: o = 85%

Nng sut: 13 tn/h = 13000 Kg/h

II. Tnh ton v la chn nhin liu

Vt liu l than vi cc thnh phn nhin liu nh sau:

Nguyn t C H O N S A W

Hm lng 85,32 4,56 4,07 1,8 4,25 7,87 3T y ta c cc thng s lm vic nh sau:

( ) ( )( )%04,76

100

87,7310032,85

100

100=

+=

+=

AWCClv

( ) ( )( )%06,4

100

87,7310056,4

100

100=

+=

+=

AWHHlv

( ) ( )( )%63,3

100

87,7310007,4

100

100=

+=

+=

AWOO lv

( ) ( ) ( )%61,1100

87,731008,1100100 =+=+= AWNNlv

( ) ( )( )%79,3

100

87,7310025,4

100

100=

+=

+=

AWSSlv

1. Nhit dung ring ca than

Nhit dung ring ca than c xc nh theo cng thc:

Ct = 837 + 3,7 to + 625x

Trong : to: l nhit ca than , chn to = 25oC

x: l hm lng ca cht bc, x = 2,78%

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Suy ra: Ct = 837 + 3,7.25 + 625.0,0278

= 946875 ( J/Kg ) = 0.946875( KJ/Kg)

2. Nhit tr cao ca than

Theo cng thc Mendeleep:

Qclv = 339.Clv + 1256.Hlv 109.( Olv Slv) ( VII.37/ 110 STT2)

= 339.76,04 + 1256.4,06 + 109.( 3,63 3,79)

= 30894,36 ( KJ/Kg )

3.Nhit tr thp ca than

Qlvth = Qc 25( W + 9Hlv) = 30894,36 25.( 3 + 9.4,06) = 29905,86( KJ/Kg)

4. Lng khng kh l thuyt t chy 1 kg nhin liu

cung cp cho cc phn ng chy, thnh phn ca oxi trong khng kh l

21%Cc phn ng chy:

C + O2 CO2

2H2 + O2 2H2O

S + O2 SO2

p dng cng thc: ( VII/ 111 STT2)

Lo = 0,115Clv + 0,346Hlv + 0,043( Slv Olv)

= 0,115. 76,04 + 0.346.4,06 + 0,043.(3,79 3,63)

= 10,16 (Kgkkk/Kg than)

5. Entanpi ca hi nc trong hn hp khi

ih = ro + Ch.t ( QTTBT4 273)

Trong : ro: l nhit lng ring ca hi nc 0oC, ro = 2493 KJ/Kg

Ch: l nhit dung ring ca hi nc, Ch = 1,97 KJ/Kg

t: l nhit ca khi l vo, t =240o

CSuy ra : ih = 2493 + 1,97. 240 = 2965,8 (KJ/Kg)

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6. H s khng kh d bung t v bung trn theo l thuyt

p dng cng thc:( ) ( )[ ]

( ) ( )[ ]opkhohoopk

lv

h

lv

tt

lv

c

ttCiixL

tCWAHiWHCtQ

+++++

=1

1..91.9..

Trong :

Qclv: l nhit tr cao ca than, Qclv = 30894,36 KJ/Kg

: Hiu sut, chn = 0,9

tt: l nhit tr ca than , tt = 25oC

Ct: l nhit dung ring ca than, Ct = 0,946875 KJ/Kg

Cpk: l nhit dung ring ca khng kh, Cpk = 1,004 KJ/Kg

ih: l entanpi ca hi nc nhit t1 = 240oC

iho: l entanpi ca hi nc nhit mi trng, to =25oC

iho = ro + Ch.to

Vi: ro: nhit lng ring ca hi nc nhit 0oC, ro =2493 (KJ/Kg)

Ch: nhit dung ring ca hi nc, Ch = 1,97 KJ/Kg

to: nhit ca mi trng, to = 25oC

Nn: iho = 2493 + 1,97. 25 = 2542,25 ( KJ/Kg )

t1: nhit ca khi l, t1 = 240o

CLo : lng khng kh l thuyt t chy 1 kg nhin liu, L o = 10,16 Kgkkk/Kg

than

xo : hm m ca khng kh iu kin t =25oC.Tra th I-x [ 255 4] ta c :

xo = 0,0166( Kg m/ Kg kk )

Io = 16,33 Kcal/ Kgkkk = 68,37 ( KJ/Kgkkk )

( ) ( )[ ]

( ) ( )[ ]25240004,125,25428,29650166,0.16,10

240.004,1.03,00787,00406,0.918,296503,00406,0.9946875,0.259,0.36,30894

+

++++=

82,11=

III. Tnh ton cc thit b chnh

1. Lng m bay hi trong qu trnh sy

Lng m bay hi c tnh theo cng thc :

W = G1 2

21

100 W

WW

(QTTBT2/165)

Trong : G1: l lng vt liu vo my sy: G1 = 13000 Kg/h

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W1, W2: m u v m cui ca vt liu

W1 = 8,5%; W2 = 0,5%

Suy ra: W = 13000.5,0100

5,05,8

= 1045,226 (Kg/h)

2. Lng vt liu kh tuyt i

p dng cng thc :

Gk = G1 100100 1W

= 13000. 1005,8100

= 11895 ( Kg/h )

3. Phng trnh cn bng nhit

G1 =G2 + WSuy ra : G2 =G1 W

Trong : G2: lng vt liu kh ra khi thit b sy

G2 = 13000 1045,226 = 11954,774 ( Kg/h )

4. Th tch thng sy

Vth =A

W

Trong : A: cng bay hi m, chn A = 30 Kg/m3h

W: lng m bay hi, W = 1045,226 Kg/h

Suy ra: Vth =)(84,34

30

226,1045 3m=

5. Chiu di ca thng

Lt = 24

t

th

D

V

Trong : Vth: l th tch ca thng, Vth = 34,84 ( m3)

Dt: l ng knh thng sy

Thng t s gia chiu di thng sy v ng knh thng sy l:

Lt / Dt = 3,5 7. Chn Lt / Dt = 6

Suy ra: Lt = 6Dt 6Dt = 644 3

2

tht

t

th VDD

V =

Nn ng knh thng l:

15


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Dt = )(95,114,3.6

84,34.4

6

433 m

Vth ==

. Ly Dt = 2 m

Suy ra: Chiu di thng l:

Lt = )(096,112.14,384,34.44

22m

DV

t

th ==

. Chn chiu di thng l Lt = 12 m

Tnh li th tch thng v cng bay hi m

Th tch thng :

Vtt = )(68,374

12.2.14,3

4

322

mLD tt ==

Cng bay hi m :

A = )/(74,2768,37

226,1045 3hmKgVW

tt

== . Ly A = 28 Kg/m3h

6. Thi gian sy

Thi gian sy ca vt liu trong thng( Thi gian lu ca vt liu trong

thng) :

i vi my sy thng quay thi gian sy c xc nh theo CT :

)(200[)(120

21

21

WWA

WWxs +

=

(S tay -tp2-trang123)

Trong :

+ x : Khi lng ring xp trung bnh ca vt liu trong thng quay, vi x

=724,32 kg/m 3

+ W1,W2 : m u v cui ca vt liu, tnh bng % khi lng chung:

w1 = 8.5 % , w2 = 0,5 %

+ : H s cha vt liu ca thng (0,1-0,25),chn = 0,15+ A : Cng bay hi m, A = 28(kg m/ 3m h)

=s =+)]5,05,8(200[28

)5,05,8(32,72415,012019,503 (pht)

Kim tra li thi gian sy: (Theo 408- my v thit b sn xut ho cht)

Thi gian sy thc t s l:

= 1

...60

G

fV xtt

= 1300032,724.15,0.68,37.60

= 18,89 (pht)

Vtt : Th tch thc ca thng sy = 37,68 m2

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7. S vng quay ca thng sy

S vng quay ca thng xc nh theo cng thc:

,..

..

tgD

Lkmn

t

t= (S tay -tp2-trang122).

Trong :

+ Lt , Dt : Chiu di v ng knh ca thng sy (m)

Lt = 12 m, Dt = 2 m

+ : Gc nghing ca thng quay, Thng gc nghing ca thng di l

1 6o, chn = 3o. Suy ra: tg 052,03 == otg

+ m : H s ph thuc vo cu to cnh trong thng ; la chn cnh phnphi , m = 1

+ k : H s ph thuc vo phng thc sy v tnh cht ca vt liu,k=0,7.

V y l phng thc sy xui chiu

+ : Thi gian lu li ca vt liu trong thng quay hay cng chnh l thi

gian sy, pht

n= =

052,0.2.89,18

127,014,28 (Vng/pht )

8. Cng sut cn thit quay thit b

Ta c: xtt naLDN =3

0013,0 (VII.54-trang123-STT2)

Trong :

+ n : S vng quay ca thng, vng /pht

+ a : H s ph thuc vo dng cnh, a = 0,026 (theo bng VII.54-123)

+ x : Khi lng ring xp trung bnh, x = 724,32 kg/m3

+ Dt,Lt : ng knh v chiu di ca thng (m)

Dt = 2 m, Lt = 12 m

N=0,0013.23.12.0,026.4,28 724,32=10,059 (Kw)

Cng sut chn ng c:

Ndc =1,1.N = 1,1.10,059= 11,0649 (kw)

Chn chiu dy thng: Theo cng thc thc nghim ta c:

S = ( 0,007- 0,011 )Dt. Ly: S = 14mm = 0,014 m

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9. Cn bng l t than

9.1.Nhit lng vo tnh khi t 1kg than

a) Nhit lng vo bung t Qvo = Q1+Q2 + Q3 (KJ)

Trong :

Q1:Nhit lng do than mang vo(tnh theo 1kg than) ( KJ )

Q2:Nhit lng do khng kh mang vo bung t ( KJ )

Q3:Nhit lng do t chy 1kg than ( KJ )

Ta c:

Q1 = Gt Ct to =1. 0,946875.25= 23,67 (KJ)

Vi:

Ct: Nhit dung ring ca than Ct = 0,946875 KJ /kg

to: Nhit ca than vo l t to = 25 C

Gt: Khi lng ca than, Gt = 1 Kg

Q2 = Lo. .Io (KJ)

Trong :Lo: L lng kh kh l thuyt mang i t chy ht 1kg than

Lo = 10,16 (kg k3/kg than)

Io :Hm nhit khng kh trc khi vo my sy

Tnh Io:

Ni suy theo th I-x (QTTBT2-255) :

to: Nhit ca mi trng : to = 25 o C

o : m tng i ca khng kh : o = 85%

xo = 0.0166 ( Kg m/ Kg kkk )

=> Io = 16,33 (kcal/kgk3)

Io = 68,37 (KJ /kgk3)

Q2 = 10,16.68,37

= 694,64 ( KJ )

Q3 = Qclv

Trong : Q3: Nhit lng do t chy 1 kg than

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Q3 = 30894,36 (KJ )

Tng nhit lng vo bung t l:

Qvo = Q1 + Q2 + Q3

= 23,67 + 694,64+ 30894,36 = 694,64 + 30918,03 (KJ)

b)Nhit lng ra khi bung trn tnh khi t 1 kg than

Qra = Q4 + Q5 + Qm (KJ)

Trong :

Q4: Nhit do x mang ra ( KJ )

Q5:Nhit do khng kh mang ra khi bung t.( KJ )

Qm:Nhit mt mt ra mi trng. ( KJ )Ta c:

Q4 = Gx.Cx.tx (KJ)

Gx : Khi lng x khi t 1kg than

Gx = A = 7,87 %

Cx: Nhit dung ring ca x: Cx= 0,754 ( KJ /kg ) (Tra s tay T1- 162)

tx :Nhit do x mang ra tx = 260 o C

Thay s :

Q4 = 43,15260.754,0100

87,7= ( KJ )

Q5 = Gk.Ck.t1

Vi:

t1 : Nhit khi l vo bung trn t1 =240 o C

Gk : Khi lng ca cht kh trong lCk : Nhit dung ring ca khi l

Ck =K

OHOHONNCOCOSOSO

G

CGCGCGCGCG2222222222 0

. ++++(KJ /kg o C)

(S tay T2- VII.42-112)

Q5 = (GSO 2 .C SO 2 +G CO 2 .C CO 2 +G N 2 .C N 2+GO 2 .CO 2 +GH 2 O.CH 2 O)t1

Thnh phn khi lng ca cc cu t kh khi t 1 kg nhin liu 240 o C:

Khi lng SO2:

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GSO 2 = =100.2

lvS0,02.Slv = 0,02.3,79= 0,076 ( kg/kgthan )

Khi lng CO2

GCO 2 = 0,0367C lv = 0,0367.76,04 = 2,79 ( kg/kgthan )

Khi lng N2

GN 2= 0,769. .L0 + 0,01Nlv

= 0,769. .10,16 + 0,01.1,61

= 7,81. + 0,0161 ( kg/kgthan )

Khi lng m:

GH 2 O = mH 2 O + .Loxo

=100

9 WHLV ++ .Loxo

= 0166,0.16,10.100

306,4.9++

=0,395+0,17 (Kg/kg than)

Khi lng O2:

GO 2 = 0,231( - 1).Lo = 0,231( - 1 ).10,16

= 2,35.- 2,35 (kg/kgthan)

Tnh nhit dung ring ca cc kh 240 o C:

CSO 2 = 0,18 (kcal/kg) = 0,754 (KJ /kg)

CCO 2 = 0,222 + 43.106t1

=0,222 + 43.10 6 .240

= 0,2323 (kcal/kg) = 0,973 (KJ /kg)

CN 2 = 0,246 + 189.10-7.t1

= 0,246 + 189.10 7 .240

= 0,251 (kcal/kg) = 1,049 (KJ /kg)

CO 2 = 0,216 + 166.10-7.t1

= 0,216 + 166.10 7 .240

=0,22(kcal/kg) = 0,921 (KJ /kg)CH 2 O = 0,436 + 119.10-6.t1

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= 0,436 + 119.10 6 .240

=0,465(kcal/kg) =1,945 (KJ /kg)

Thay s vo Q5 ta c:

Q5 ={0,076.0,754 + 2,79.0,973 + [(7,81.+0,0161).1,049 + [(2,35.-2,35).0,921 ]+ [(0,17+0,395.).1,945]} 240 = 229,24 + 2670,08 (KJ)

C : Qm =10%Qvo = 0,1( 30918,03 + 694,64 )

= 3091,803 + 69,464 (kj)

Qra = Q4 + Q5 + Qm

= 15,43 + 229,24 + 2670,08 + 3091,803 + 69,464

= 3336,473 + 2739,544. ( KJ )

9.2.Phng trnh cn bng nhit l t than

Phng trnh cn bng l t

Qvo = Qra

30918,03 + 694,64 = 3336,473 + 2739,544.

27581,557 = 2044,904

= 13,49

: H s khng kh d thc t

Vy lng khng kh thc t cn cung cp cho l t l:

L = .Lo

= 13,49.10,16 = 137,06 ( kg k3/kg than)

Khi lng cc kh khi t 1 kg than:

Khi lng SO2: GSO 2 = 0,076 ( Kg/ Kg than )

Khi lng CO2: GCO 2 = 2,79 ( Kg/ Kg than )

Khi lng N2: GN 2 = 7,81+ 0,0161

= 7,81.13,49 + 0,0161 = 105,373 ( Kg/ Kg than )

Khi lng ca O2: GO 2 = 2,35 - 2,35

= 2,35.13,49 2,35 = 29,35 ( Kg/Kg than )

Khi lng ca H2O: GH 2 O= 0,395+ 0,17

= 0,395.13,49 + 0,17 = 5,499 (Kg/Kg than)

Qu trnh sy

21


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Vi cc thng s:

Chn nhit ca khng kh : to = 25 C

m tng i ca khng kh : o = 85%

Ni suy theo th I-x (QTTBT2-255) :

xo = 0.0166 (kg m/kgk3)

=> Io = 16,33 (kcal/kgk3)

Io = 68,37 (kj/kgk3)

Nhit ca khi l vo thng sy t1 = 240oC

Nhit ca khi l ra khi thng sy t2 = 80oC

A. Qu trnh sy l thuyt

*. Cc thng s i vo thit b sy

a) Hm m ca khi l

x1 =kk

hn

G

G(kg/kg k3)

+ Ghn : L lng hi nc trong khi

GH 2 O = 0,433 + 0,0638

= 0,433 + 0,0638.13,49 = 1,29 ( Kg/Kg than )

Trong :

: H s khng kh d =13,49

+ Gkk: Lng khng kh kh sau bung trn c xc nh :

Gkk= 1 + .Lo - GH 2 O

= 1 + 13,49.10,16 1,29 = 136,77 ( kg/kgthan )

Hm m ca khi l l:

x1 =77,136

29,1= 0,009 (kg/kg k3)

b) Entanpi ca khi l sau bung trn :

I1 =kk

nnooot

lv

c

G

iGILtCQ +++ ... [IV 304]

Trong :

Qclv

: Nhit tr cao ca than :Qclv

= 30894,36 (kj/kg than) : Hiu sut bung t : = 0,9

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23/56


Io : Entanpi ca hi nc : Io = 68,37 (kj/kg )

Ct : Nhit dung ring ca than : Ct= 0,946875 (kJ/kg C )

to : Nhit ca than khi vo l : to = 25oC

: H s khng kh d = 13,49Lo: L lng kh l thuyt mang t chy ht 1kg than:

Lo = 10,16 kg k3/kg than

Gkk: Lng khng kh kh sau bung trn :Gkk= 132,56 kg/kgthan

Gn: Lng hi nc trong khi, Gn = 1,29 Kg/Kg than

in: Entanpi ca hi nc nhit mi trng

in = ro + Ch.to= 2493 + 1,97.25 = 2542,25 ( KJ/Kg )

Vi: ro: Nhit lng ring ca hi nc 0oC, ro = 2493 ( KJ/Kg )

Ch: Nhit dung ring ca hi nc, Ch = 1,97 ( KJ/Kg )

Suy ra: I1 =56,132

25,2542.29,137,68.16,10.49,1325.946875,09,0.36,30894 +++

= 386,08 (kj/kgkkk)

c) p sut hi bo ha ti t1= 240 0C

Pbh = exp{1

1

59,233

17

t

t

+ - 5,093} = exp{ 24059,233240.17

+ - 5,093} = 33,85 (bar)

d) m tng i 1 ti x1.

1 = )622,0(.

1

1

xP

xP

bh += 009,0.

)009,0622,0(85,33

750

745

+= 4,19.10-4 (%)

*. Cc thng s i ra khi thit b sy

a) Hm nhit ca khi l:

Do qu trnh sy l thuyt nn : I1 = I2 = 386,08 (kj/kg k3)

b) Hm m ca khi l :

Lng m ca khi thi ra sau qu trnh sy l thuyt :

x2 =2

21

842,1 tr

tCI kk

+

Trong : Ckk Nhit dung ring ca khng kh Ckk=1,004 (kj/)

r : Nhit ha hi ca nc 80oC

r = 542,5(kcal/kg) = 2271,339 ( KJ/Kg)

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x2 =80.842,1339,2271

80.004,108,386

+

= 0,12 (kg m/kg k3)

c) p sut hi bo ha ti t2 = 80 oC

Pbh = exp{2

2

59,23317

tt+ - 5,093} = exp{ 8059,233

80.17+ - 5,093}=0,47(bar)

d) m tng i 2 ti d2.

2 = )622,0(.

2

2

xP

xP

bh += 12,0.

)12,0622,0(47,0

750

745

+= 0,34 (%)

*. Phng trnh cn bng nhit lng.

Phng trnh cn bng nhit lng trong qu trnh sy l thuyt :loIo + q = loI1 = loI2

Vi lo : Lng khng kh kh tiu hao cn thit bay hi 1 kg m

lo=oxx 2

1

=0166,012,0

1

= 9,67 (kg k3/kg m)

Lng kh tiu hao trong 1h l :

Lo= lo.W =9,67. 1045,226 = 10108,57 (kg k3/h)

Nhit lng ring :

qo = lo(I1 - Io) =9,67.(386,08 68,37) = 3072,26 (kj/kg m)

Nhit lng cung cp cho W kg hi m :

Qo= qo.W = 3072,26.1045,226 = 3211206,031 (kj/h) = 892,0017 (kw)

*) Lu lng th tch khng kh trc khi qu trnh sy c xc nh theo cng

thc : V1 = 1 .Lo

1 : L th tch khng kh m (m3

/kg)Tra bng (318-I.255 ST1) t = 240oC => 1 = 0,685 (kg/m

3)

=> 1 =1

1

=

685,0

1= 1,46 (m3/kgkk)

Vy V1 = 1 .Lo = 1,46.10108,57= 14758,51 (m3/h)

*) Lu lng th tch khng kh sau khi qu trnh sy c xc nh theo cng thc

V2 = 2 .Lo

2 : L th tch khng kh m (m3/kg)

24


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Tra bng (318-I.255 ST1) t = 80 oC => 1 = 1,000 (kg/m3)

=> 2 =2

1

=

000,1

1= 1 (m3/kg)

Vy V2 = 2 .Lo = 1.10108,57= 10108,57 (m3/h)

*) Lu lng trung bnh l:

Vtb =2

21 VV + =2

57,1010851,14758 += 12433,54 (m3/h)

Din tch t do ca thng sy

Ft = 0,785Dt2( 1 - ) ,m2

= 0,785.22.( 1 0,15 ) = 2,669 ( m2 )

Tc trung bnh ca dng kh trong thit b sy l:

Wtb =669,2

54,12433=

t

tb

F

V= 4658,5 ( m3/h ) = 1,29 ( m3/h )

10. Tnh h s truyn nhit

21

11

1

++

=K( s tay qu trnh v thit b T1 )

Trong :

1 :H s cp nhit i lu ca tc nhn sy n thnh thit b (W/ 2m C0 )

2 :H s cp nhit t thnh ngoi thit b sy n mi trng (W/ 2m C0 )

: H s dn nhit ca thnh thit b (W/ m C0 )

: Chiu dy ca thnh thit b (m)

a) Xc nh 1 :

Theo cng thc : 1 =k )(//

1

/

1 + (W/ 2m C0 )k: H s nhm k=(1,2 1,3) chn k = 1,25

/1 : H s cp nhit t tc nhn sy n thnh thit b sy do i lu cng

bc,ph thuc vo ch chuyn ng ca dng kh. (W/ m2 C0 )

//

1 : H s cp nhit t khng kh n thnh thit b do i lu t nhin. (W/ m2

C0 )

* Tnh /1 :

Tc ca dng kh vo Wtb = 1,29 m/sCh chuyn ng c c trng bi chun s Reynon :

25


26/56


Re =

ttb DW . (359-ST1)

: h s nht ng ca kh (m2/s)

Dt : mg knh trong ca thit b (m)Nhit trung bnh ca dng kh:

ttb =2

80240

2

21 +=+ tt

= 160oC

Ti t = 160oC

Ni suy theo bng I.255-318 s tay T1

h s dn nhit : = 3,64.10-2 (W/ m C0 )

nht ng hc : = 30,09.10-6 (m2/s)

Thay cc gi tr thc :

Re = 610.09,30

2.29,1 = 85743 > 104

=> Ch chy xoy

Nu = 0.018.Re0,8 . ( V.42.STT1/16 )

:H s ph thuc vo Re v6

2

12

==tt

D

L

Tra bng V.2-s tay T2 _ T 15

= 1,16

Nu = 0,018.(85743)0,8.1,16 = 184,62

m Nu =

tD./

1 (QTTBT1-T 196)

/

1 = 210.64,3.62,184. 2

=tu

D

N

= 3,36 (W/2

m C0

)

* Tnh //1 :

uN =25,0.47,0 Gr (QTTBT1-T 206)

V h s cp nhit ca thit b nm ngang ca dng kh l ch i lu t do

trong khng gian rng c tnh nh hng chuyn ng :

Trong chun s Grashof : Gr

Gr = 23 ...

tDg

26


27/56


g : Gia tc trng trng g = 9,81 m/s2

: H s dn n th tch (1/T) =T

1=

tbt+2731

=160273

1

+=2,31.10-3

t : Nhit chnh lch gia nhit trung bnh ca dng kh v thnh thit bsy.

Gi s nhit trung bnh ca thnh trong thit b sy l 130 C0

Nhit lp ngn cch l: tl =2

120160 += 140 0 C.

Ti 140oC. Tra bng (I.255-318-STT1)ta c:

W/ m C0

sm /2

Ta c:

201401601 === tt tb oC

Gr = 82633

2

3

10.91,46)10.8,27(

20.10.31,2.2.81,9...==

tDg

Vy 003,123)10.91,46.(47,0 25,08 ==Nu

Mt khc:

//1tu

DN = 15,2

2

10.49,3.003,123. 2//1 ===

t

u

D

(W/ 2m C0 )

Vy 8875,6)15,236,3(25,1)( //2/

11 =+=+= k (W/ 2m C0 )

b) Xc nh 2

//2/

22 += (T-394 s tay qu trnh v thit b T2)

/

2 : H s cp nhit t thnh thit b ra mi trng do i lu t nhin.

//2 : H s cp nhit t thnh thit b ra mi trng do bc x

* Tnh /2 : tmt C025=

Nhit lp bin gii gia khng kh v thnh thng sy ngoi cng l t ng :

chn 55=ngt C0

40

2

5525

2

=+

=+

= ngmtbgtt

t C0

t : Nhit gia thnh thit b ngoi v mi trng : t = 55 25 = 30 C0

27

6

2

10.80,27

10.49,3

=

=


28/56


T = 25 + 273 = 298 oK

40=bgt C0 tra bng I.255 (T-318 s tay qu trnh v thit b T1)

210.76,2 = W/ m C0

610.96,16 = m 2 /s

Chun s Grashof :

=

.

..2

3

tDgGr

ng

Trong :

Dng : ng knh ngoi ca thit b

Thng c cu to gm 3 lp :+ Lp v bo v v b dy thng.Vt liu lm thng l thp CT 5 c 50=

(W/ 2m C0 )

+

Tng nhit tr ca thnh thit b sy ca lp cch nhit v ca thnh bo

n

B dy ca thng b1=14mm = 0,014 m

Lp v bo v trong cng : b3=1 mm = 0,001 m

+ Chn vt liu cch nhit l bng thy tinh c h s dn nhit l 2 =0,0372 (W/

2m C0 ) (T-128 s tay qu trnh v thit b T1)

Nhit ca thnh thng: t2 = t1 = 120oC

Nhit lp v bo v: t4 = t3 = 55oC

Nhit lu ring : q1 = ( )thdkttt ttDqD = 1

Dt : ng knh trong ca thng, Dt = 2 m

1 : H s cp nhit t tc nhn sy n v thit b sy, 1 =6,8875 ( W/m2 )

tth: nhit trung bnh ca thnh thit b, tth = 120oC

tdk: nhit trung bnh ca dng kh, tdk= 240oC

Suy ra: q1 = ( ) ( ) 05,5193120240.8875.6.2.1 === thdkttt ttDqD ( W/m2 )

Theo cng thc n gin ta c :

q1 = ( )3222

ttDbtb

M Dtb = Dt + 2S + 0,5b2 =2 + 2.0,014 + 0,5.b2

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Vi b2 l b dy lp bo n

Suy ra: q1 = ( ) ( )55120.5,0014,0.220372,0

2

2

++ bb

5193,05b2 = 0,0372. .65 ( 2 + 2.0,014 + 0,5b2 ) 5189,25b2 = 15,41

b2 = 0,003 m2.( y l b dy ca lp cch nhit )

ng knh ngoi ca thit b

D ng = )(2 123 bbbDt +++

=2 + 2(0,014+0,003+0,001)= 2,036 (m)

T=25+273=298 K0

=> Gr =298.)10.96,16(

30.036,2.81,926

3

= 2,89.1010

M Chun s: Nu = 0,47.Gr0,25 = 0,47. (2,89.1010)0,25 = 193,92

036,2

10.76,2.92,193. 2/2

==ng

u

D

N = 2,63(W/ 2m C0 )

* Xc nh //2 theo cng thc:

//

2 =

21

42

41

0100100

..

TT

TCn

(W/ 2m C0 )

Trong :

Co :H s bc x ca vt en tuyt i 76,50 =C (W/ 2m K0 )

T1 ,T 2 : nhit tng i ca b mt ngoi thit b sy v mi trng xung

quanh

T1 =55+273=328 K0

T 2 =25+273=298 K0

n : Mc en tuyt i ca thit b n = 95,0

//2 =

298328

100

298

100

328.76,5.95,0

44

= 6,73 (W/ 2m C0 )

36,973,663,2//2/22 =+=+= (W/ 2m C0 )

H s truyn nhit chung ca tc nhn sy n mi trng xung quanh

29


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K =21

11

1

++ =

22

2

1

31

1

11

1

++

++

bbb

K =36,9

1

0372,0

003,0

50

001,0014,0

8875,6

1

1

+++

+ = 3,003 (W/moK)

Xc nh b mt trao i nhit

4/..2.. 2ntn DLDF +=

= ( ) 2036.2.785,0.212.036,2.14,3 + = 83,26 m2 (420-Thit k my ho cht T1)

+ Xc nh H s nhit trung bnh : tbt

tbt =n

n

1

1

lg3,2( Qu trnh v thit b tp 1- T 193)

21525240011 === tt C0

55258002 === ttn C0

tbt =36,117

55

215lg3,2

55215=

C0

Vy nhit tn tht ra ngoi mi trng l

Qmt =W

tFK tb...6,3 = =226,1045

36,117.26,83.003,3.6,3101,07 (Kj/kg m)

11. Cn bng nhit lng trong thit b sy

11.1. Nhit lng vo

Gi n l s kg than cn t trong 1hTng nhit lng : Qvo = Q1 + Q2 ( KJ/s )

*) Q1: Nhit lng do nhin liu sy mang vo my sy

Q1 = G1.C1.to

G1: Nguyn liu ca my sy G1 = 13000 ( Kg/h ) = 3,61 ( Kg/s )

C1: Nhit dung ring ca vt liu sy khi vo my sy.

C1= 0,1572 ( Kcal/Kg o C ) = 0,658 ( KJ/Kg ) (Tra bng I.143,STT1-160)

to: Nhit ca vt liu sy vo my sy to =25 C0

Q1 = 3,61.0,658.25 = 59,38 ( KJ/s ) = 59,38 ( KW )

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*) Q2 :L nhit lng do khi l mang vo khi sy khi t n kg than

Nhit do khi l mang vo my sy khi t 1 kg than chnh l nhit lng Q 5

tnh bung t :

Q2 = n.Q5

Vi Q5 = 229,24 + 2670,08 ( KJ/h )

M: = 13,49

Q2 =( 229,24 + 2670,08.13,49 )n = 36248,62.n ( Kg/h )

= 10,07.n ( KJ/s )

Vy tng nhit lng vo my sy: Qvo = 59,38 + 10,07.n ( KJ/h )

11.2. Nhit lng ra khi my syQra = Q ' 1 +Q ' 2 +Q ' 3 + Q ' 4 (kj)

Trong :

*) Q ' 1 : Nhit do vt liu sy mang ra khi my sy

Q ' 1 = G2.C'

1.t1

Trong : G2 : Khi lng ca vt liu ra khi my sy :

G2 = 11954,774 ( Kg/h ) = 3,32 ( Kg/s )

C '1 : Nhit dung ring ca vt liu ra khi my sy

C '1 = C1.(1 w2) + Cn .w2 ( KJ/Kg Co )

Cn : nhit dung ring ca nc 25 Co

Cn = 0,99875 ( Kcal/Kg Co ) = 4,18 ( KJ/Kg Co )

C '1 = 0,658.(1 - 0,005) + 4,18.0,005 = 0,68 ( KJ/Kg Co )

t '1 :Nhit sn phm ra khi my sy t'

1 = 70 Co

Q ' 1 = 3,32. 0,68. 70 = 158,032 ( KJ/s )

*) Q '2 :Nhit do bc hi nc v do hi nc mang ra ngoi:

Q '2 = w.[Cn1.(100 t0) + r + Cn2(t2 100)]

Trong :

Cn1 ,Cn2 : nhit dung ring ca nc to , t2

to =25 Co Cn1 = 0,999 ( Kcal/Kg Co ) = 4,183 ( KJ/Kg Co )

t2 =80 Co Cn2 = 1,003 ( Kcal/Kg Co ) =4,199 ( KJ/Kg Co )

(Tra bng I.147 STT1 165)

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r : nhit ho hi ca hi nc 80oC. Ni suy bng I.212/254

r = 542,5 ( Kcal/Kg ) = 2271,339 ( KJ/Kg )

W:

Vy : Q '2 =3600

226,1045 [4,183( 100 25 ) + 2271,339 + 4,199( 80 100 )]

= 725,314 ( KJ/s )

*) Q '3 : Nhit do khi l mang ra khi my sy nhit 80 Co

Q'

3 =(GSO 2 .CSO 2 +GCO 2 .CCO 2 +GN 2 .CN 2 +GO 2 .CO 2 +GH 2 O.CH 2 O)thh.n ( KJ/s )

Trong :

thh: Nhit hn hp kh: thh = 80 Co

Gi: Khi lng ca cc cu t trong hn hp kh t

Ci :Nhit dung ring ca cc cu t tng ng

Khi lng ca cc kh khi t 1kg than tnh cn bng nhit l t l:

GSO 2 = 0,076 ( Kg/Kgthan )

GCO 2 = 2,79 ( Kg/Kgthan)

GN2

= 105,373 ( Kg/Kgthan )G H 2 O = 5,499 ( Kg/Kgthan )

GO 2 = 29,35 ( Kg/Kgthan )

Nhit dung ring ca cc kh t2= 80 Co l:

+ CSO 2 = 0,167. ( Kcal/Kgo

C) = 0,699 ( KJ/Kgo

C )

+ CCO 2 = 0,222 + 43.106.t2

= 0,222 + 43.10 6 .80

= 0,225 ( Kcal/Kg o C ) = 0,944 ( Kj/Kg o C )

+ CN 2 = 0,246 + 189.10-7.t2

= 0,246 + 189.10 7 .80

= 0,248 ( Kcal/Kg o C) = 1,038 ( KJ/Kg o C )

+ CO 2 = 0,216 + 166.10-7.t2

= 0,216 + 166.10 7 .80

= 0,217 ( Kcal/Kg o C ) = 0,909 ( KJ/Kg o C )

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+ CH 2 O= 0,436 + 119.10-6.t2

= 0,436 + 119.10 6 .80

= 0,446 ( Kcal/Kg o C) = 1,867 ( KJ/Kg o C )

Vy :

Q '3 = (0,076.0,699+2,79.0,944+105,373.1,038+29,35.0,909 +5,499.1,867 ).80.3600

n

= 3,31.n ( KJ/s )

Q '4 : Nhit lng tn tht ra mi trng.

Q '4 = qmt.w = 101,07.3600

226,1045= 29,31( KJ/s)

Vy : Qra = 158,032 + 725,314 + 3,31n + 29,31

= 912,65 + 3,31.n ( KJ/s )

11.3. Phng trnh cn bng nhit ca thit b sy

Qra = Qvo

912,656 + 3,31.n = 59,38 + 10,07.n

n = 126,22 ( Kg )

Vy lng than cn t trong 1h l : n = 126,22 ( Kg )

11.4. Trng thi ca khi l vo my sy, i ra khi my sy v lu lng

kh

*) Hm m ca khng kh 200 o C tnh theo 1kg khng kh kh

x1 =kk

OH

m

m2 (kg/kg k3 ) (Theo QTTB T2 314)

Trong : m H 2 O: Lng hi nc trong khi l khi vo thng sy tnh theo 1 kg

than , m H 2 O = 1,29 ( Kg/Kg than )

mkk: Lng khng kh kh tnh theo 1kg than.

mkk= 1 + Lo - mH 2 O (STT2 VII.40 111)

= 1 +13,49.10,16 1,29

= 136,77 (kg/kg k3 )

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x1 =77,136

29,1= 0,009 (kg/kgkkk)

*) Tnh hm nhit ti t1 = 240 o C

I1 = (0,24 + 0,47x1)240 + 595x1 (CT10.4/T256/QTTBT2)Thay s ta c : I1 = (0,24 + 0,47.0,009)240 + 595.0,009

= 63,97 ( Kcal/Kgk3 ) = 267,83 ( KJ/Kgkkk )

=> Trng thi ca khi l trc khi vo my sy

t1 =240 o C

x1 = 0,009 (kg/kgk3)

I1 = 63,97 ( kcal/kgk3) = 267,83 ( KJ/Kgk3)

*) Nhit lng tiu hao thc t trong thit b sy:

= l .(I2 I1)

= 4,18.to qvl qmt ( KJ/Kg m ) (STT2 389)

qvl : Nhit tiu hao nung nng vt liu

to: Nhit m ca vt liu mang vo. to = 25oC

qmt :Nhit tn tht ra mi trng, qmt = 101,07 ( KJ/Kg m )

*) Nhit lng tiu hao nung nng vt liu

Theo (thit k my sy 219)

qvl =W

ttCG )( 0'

22.2

t '2 : Nhit vt liu trong bung t, chn t'

2 = 70 o C

C2 : Nhit dung ring ca vt liu sau sy C2 = 0,68 ( KJ/Kg o C )

G2 : Khi lng vt liu sau khi sy G2 = 11954,774 ( Kg/h )

Vy :

qvl =226,1045

)2570.(68,0.774,11954 = 349,99 ( Kg/h )

Thay s ta c:

= 4,18.25 349,99 101,07 = - 346,56 ( KJ/Kg m )

*) Tnh hm nhit ca khi l nhit t2 = 80 o C

I2 = t2 + ( 2493 + 1,97t2)x2 ( Kcal/Kg k3 )

Thay s :

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I2 = 80 + ( 2493 + 1,97. 80 )x2

= 80 + 2650,6x2

Mt khc: = 12

12

xx

II

(Theo CT10.8a-QTTBT2-258)

= 009,0

83,267806,2650

2

2

+

x

x= - 346,56 (kj/kg m)

2997,16x2 = 190,949

x2 = 0,064 ( Kg/Kg k3 )

T ta suy ra:

I2 = 80 + 2650,6.0,064 = 249,64 ( KJ/Kg k3

)Vy khi ra khi my sy khi l trng thi :

t2 = 80 o C

x2 = 0,064 ( Kg/Kg k3 )

I2 = 249,64 ( KJ/Kg k3 )

Lng khng kh tiu hao thc t bc hi 1 kg m l:

l = 12

1

xx = 009,0064,01

= 18,18 ( Kg/Kg k3

)

Tng lng kh kh cn thit :

L = W.l =1045,226.18,18 = 19002,21 ( Kg/Kg k3 )

Lng nhit tiu hao ring:

q = l(I1 I0) (s tay T2 VII.23-103)

=18,18.(267,83 68,37) ( Kg/Kg k3 )

= 3626,18 ( KJ/Kg m )Lng kh a vo my sy:

Lm

mmm

kk

kkOH.2

+= (QTTBT2-317)

=774,136

774,13629,1 +.19002,21 = 19181,43 ( Kg/h )

Lng kh i vo bung t 25 o C:

35


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V25 =kk

m

(m 3 /h)

= 185,1

19181,43

= 16186,86 ( m3

/h) = 4,49 ( m3

/s)

Vi:

m: Khi lng kh i vo my sy (kg/h)

kk : khi lng ring ca khng kh 25 o C:

Ni suy theo bng I.255-318-STT1) 25 o C ta c kk = 1,185 ( Kg/m 3 )

Lng kh i vo bung t 240 o C:

V200 =kk

m

(m 3 /h)

Ni suy theo bng I.255-318-STT1) 240 o C ta c :

kk = 0,685 ( Kg/m 3 )

Vy: V240 =685,0

43,19181= 28002,09 (m 3 /h) = 7,78 ( s)

Lng kh i ra khi my sy nhit 80 o C:

V80 =kk

OHmm

2+

(m 3 /h)

=000,1

29,143,19181 += 19182,72 (m 3 /h) = 5,34 (m 3 /s)

Vi: kk = 1,000 (Kg/m 3 )

Lu lng th tch trung bnh qu trnh sy:

Vtb= 0,5(V240 + V80) = 0,5.(7,87 + 5,34 ) = 6,605 (m 3 /s)

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PHN III : TNH TON C KH

I. Tnh ton h thng bnh rng dn ng

1. Chn ng c

Ta chn loi ng c nhn hiu 4A (do Lin x c ch to):

N c = 15 kw nn ta chn loi ng c c k khiu: 4A180M8Y3 c cc thng s

sau:

S vng quay n = 730 vng/pht, hiu sut = 0,875

(bng P.13-tnh ton thit k dn ng c kh t1-235)

Cng sut lm vic ca ng c: Nlv = Nd= 15.0,875 =13,125 ( KW )

Ta thy : Nlv > Nc do tha mn thng quay

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2. T s truyn v s vng quay

T s truyn ca ton b h dn ng l

28,4

730

== thungc

h n

N

n = 170,56

Chn hp gim tc bnh rng hnh tr 2 cp c t s truyn gia hp gim tc v

ng c l nhp= 40

Vy t s truyn ca cp bnh rng dn ng c l n hp =40

264,440

56,170===

hp

h

n

nu

S vng quay ca bnh rng nh n khp vi bnh rng gn trn thng quay

)/(25,1828,4.264,4. phtvngnuN thngbr ===

Ly Nbr = 19 ( vng/pht )

3. Cng sut v momen xon trn trc ca bnh rng nh

Cng sut trn trc ca bnh rng nh l

c

NN .1

=

: hiu sut truyn ng ca h dn ng tnh t ng c n bnh rng nh

= hp .ol.ms

Trong :

hp :bnh rng =0,9

01: hiu sut ln 0,99 0,995 chn ol =0,99

ms : hiu sut ca ma st 0,9 0,96 chn ms = 0,96

855,096,0.99,0.9,0 ==Cng sut trn trc bnh rng nh l

N1 = 0,855.15 = 12,75 (kw)

Momen xon tc dng ln trc ca bnh rng l:

38


39/56


56

1

6

1 10.08,11419

75,12.10.55,9.10.55,9 ===brN

N/mm

4. Tnh ton b truyn bnh rng tr, rng thnga) Xc nh khong cch trc

( )[ ]

32

1

..

.1.

baH

HPaw

u

KTuKa

+=

(Tnh ton h thng dn ng c kh)

T1 : Momen xon trn trc bnh rng 1 : T1 = 114,08.10 5N/mm

Ka:H s ph thuc vo vt liu ca bnh rng v loi rng.Vi rng thng v vt

liu bnh rng l thp ta c: Ka=49,5 Mpa

u: T s truyn ca cp bnh rng dn ng thng quay u= 4,264

[ ]H : ng sut tip xc cho php ca vt liu chn [ ]H =481,8MPa

Chn vt liu lm bnh rng nh l thp 45 c kch thc khng ln hn 60 mm c

nhm : 241 285

KHP:H s phn b khng u ti trng trn chiu rng v rng.Chn KHP=1,1 ba=0,3

bd=0,53. ba (u 1) = 0,53.0,3( 4,264 + 1 ) = 0,84

Vy aw ( ) mm86,13693,0.264,4.8,481

1,1.10.08,1141264,4.5,49

2

53 =+=

Chn aw = 1370 mm

b) Xc nh thng s n khp

Modun bnh rng

4,277,13).02,001,0( == wam mm

Chn modun theo bng IX-99 (Tnh ton h thng dn ng c kh)

m=15mm

S bnh rng nh: z 1 = ( )71,34

)1264,4(15

1370.2

1.

.2=

+

=

+um

aw

Ly z 1 =35

39


40/56


S bnh rng ln :

z 2 =u.z 1 = 4,264.34,71= 147,96

Ly z 2 =148

S rng tng l : z t=z1 +z 2 = 35 + 148 = 183

Tnh li khong cch trc

a w = mmzm t 1373

2

183.15

2

.==

c) Tnh rng v bn tip xc

ng xut tip xc trn mt rng phi tha mn iu kin sau:

( ) [ ]Hww

HmHH

dubuKTzzz += 2

1

..1..2...

(T-105-Tnh ton h thng dn ng c kh)

z H :H s hnh dng b mt tip xc

:Gc nghing ca rng hnh tr c s =0

w :Gc n khp.Theo tiu chun VN 1065-71 : 020=

w

Hz

2sincos.2= =

20.2sin0cos.2 =1,764

z m :H s tnh n vt liu ca bnh rng khp

274=mz MPa (Tnh ton h thng dn ng c kh)

z :H s tnh n s trng khp ca rng

3

4

=z (khi =0)

cos.11

2,388,121

+=zz

= 767,10cos.148

1

35

12,388,1 =

+

Ly z

= 1,8

856,03

8,14 == Z

40


41/56


ng knh bnh rng nh

[ ] bH

HPdw

u

uKTKd

2

1 )1( += = 33,68684,0.264,4.8,481

)1264,4(1,1.10.08,11477

2

5

=+

mm

Ly dw = 690mm

(T-96,Tnh ton h thng dn ng c kh)

+) T1 :Momen xon tc ng ln bnh rng 1. T 1 =114,08.10 5

+) KH : h s ti trng khi tnh v tip xc theo CT :

KH=KHP .K H .KHV

:HBK H s phn b khng u ti trng HBK =1,1

HK =1 ( bnh rng thng ) l h s phn b khng u ca ti trng cho cc i

rng ng thi n khp

HVK : H s ti trng ng :

HVK =

HH

wwH

KKT

db

...2

..1

1

+

Vi H=

u

avg wF ... 0

H =0,006 : H nh hng ca cc sai s n khp

Tra bng 6.15 (Tnh ton h thng dn ng c kh T 107)

0g : H s nh hng sai lch cc bnh rng 1 v 2

0g = 82

69,060000

19.690.

60000

.. 1 ===

ndw m/s < 2

Nn chn cp chnh xc l 9

09,6264,4

1373.69,0.82.006,0 ==H

wb :Chiu rng bnh rng nh

wb = 4121373.3,0. ==wba a mm

06,11.1,1.10.08,114.2

690.412.09,61

5=+=HVK

166,106,1.1.1,1 === HVHHPH KKKK

41


42/56


( )2

1

..

1..2...

ww

HmHH

dub

uKTzzz

+=

=

( )MPa29,169

690.264,4.412

1264,4.166,1.10.08,114.2.856,0.764,1.274

2

5

=+

Vy [ ]HH MPa = 29,169 =481,8 MPa

Nn bn tip xc c m bo

d) Kim tra bn un qua ti

m bo bn un cho rng th ng sut un sinh ra ti chn rng khng

c vt qu ng sut un cho php

[ ]1

11

..

.....2F

ww

FF

F

mdb

YYYKT

=

(Tnh ton h thng dn ng c kh)

Tnh KF

KF :H s ti trng :

KF=KFP.K F .KFV

KFP:H s phn b khng u KFP=1,24

K F :H s k n s khng u ti trng K F =1,05

KFV : H s k n ti trng xut hin trong vng n khp

KFV =FF P

wwF

KKT

dbV

...2

..1

1

+

Vi V F=u

avg wF ... 0

Tra bng 016,0=F

g 0 =82

V F= 24,16264,4

1373.69,0.82.016,0 =

KFV = 15,105,1.24,1.10.08,114.2690.412.24,16

1 5 =+

42


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KF=1,24.1,05.1,15 = 1,4973

Y :H s k n nghing ca rng

0= Y =1

Y :H s k n s trng khp ca rng

Y = 56,08,111

==

FY : H s dng rng vi h s dch chuyn x=0

1FY =4,08 2FY =3,6

Thay s vo ta c

31,1815.690.412

08,4.1.56,0.4973,1.10.08,114.2 51 ==F MPa

Nhn xt 1F [ ] 2361 =< F MPa

16,1608,4

6,3.31,18.

1

212

===F

FFFY

Y MPa

2F < [ ] 1852 =F MPa

Vy b truyn m bo v iu kin bn un

Cc thng s kch thc b truyn :

Khong cch trc

wa = 1373 mm

ng knh vng chia

52535.1511 === mzd

2220148.15. 22 === zmd

ng knh nh rng

225015.22220222 =+=+= mdda

Chiu rng vnh bnh rng nh

4121 == wbb mm

Chiu rng vnh bnh rng ln

2752,0.13732,0.2 === wab m

43

55515.2525211 =+=+= mdda


44/56


ng knh y rng

df1= d1 2m = 525 2.15 = 495mm

df2= d2 2m = 2220 2.15 = 2190 mm

II. Xc nh ti trng

1. Trng lng vt liu nm trong thng

Qvl=60

..1 gG =60

89,18.81,9.13000= 40150,695 (N)

(Hng dn thit k my ha T1-89)

G1 : Khi lng vt liu vo my sy G1=13000 kg/h

: Thi gian sy = 18,89 phtg : Gia tc trng trng g= 9,81 m/s2

2. Trng lng ca thng

=

t

tn

t LDD

Q4

)(22

g +

gLDD

tttnt

4

)( 22

Trong vt liu lm thng l thp CT5 c khi lng ring l 7850 (kg/ 3m )

Dn: ng knh ngoi ca thng

Dn = Dtt + 2(b1 + b2 + b3) = 2 + 2( 0,014 + 0,003 + 0,001 ) = 2,036 m

Dt: ng knh ngoi lp cch nhit ca thng

Dt = Dtt + 2( b1 + b2) = 2 + 2( 0,014 + 0,003 ) = 2,034 m

Dnt: ng knh trong lp cch nhit ca thng

Dnt= Dtt+2.b1 = 2 +2.0,014 = 2,028 (m)

Dtt: ng knh trong ca thng :Dtt = 2 m

= (D 2n Dt2 + D2

nt D2

tt ) .4

Lth. .g

= (2,0362 2,0342 + 2,0282 22).4

.12.7850.9,81 = 87765,19 (N)

3. Trng lng ca vnh ai

ng knh ca vnh ai chn s b

nv DD )2,11,1( = = 2,239 2,443 mChn Dv= 2,2 m

44


45/56


Trng lng ca vnh ai

gpbDD

Q vnv

v ...4

).(22

=

(Hng dn thit k my ha T1-251)bv: B rng ca vnh ai bv=0,18 m

=>4

)036,22,2.( 22 =

vQ . 12,756381,9.7850.18,0 = (N)

4. Trng lng ca bnh rng vng

Tnh trng lng ca tit din vnh khn

Qbr = 4 ( Da22- Dng2 )bw2..g ( N )

Trong : Da2 : ng knh nh ca bnh rng ln, Da2 = 2250 mm

Dng : ng knh ngoi ca thng sy, Dng = 2036 mm

Vt liu lm bnh rng l thp CT5 c = 7850 kg/m 3

bw2: b rng bnh rng ln, bw2 = 0,275m

Thay s vo ta c:

Qbr = 4 ( 2,252 2,0362 ) 0,275.7850.9,81 = 15255,53 ( N )

5. Trng lng ca lp cch nhit

Chn vt liu cch nhit l bng thy tinh c 200= kg/ 3m

Qbo=

4

).(2

tntDD

. gL bot ..

= )028,2034,2.(785,0 22 .12.200.9,81= 450,44 ( N )

6. Trng lng ca cnh mc nng

Qc = 1/2Qv = 7563,12/2 = 3781,56 ( N )

Chn Qc = 4000 N

Vy tng trng lng ca thng l

Q= Qvl + Qth+ Qv+ Qr+ Qc + Qbo= 40150,695+87765,19+7563,12+15255,53+4000+450,44 =155184,975 (N)

45


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III. Kim tra bn cho thng quay

1. Khong cch gia hai vnh ai

ld= 0,586.Lth=0,586.12 = 7,032 (m) = 703,2 (cm)

+ Ti trng 1 n v chiu di thng khng k n khi lng bnh rng vng

q=t

r

L

QQ =

1200

12,7563975,155184 = 123,02 ( N/cm )

2.Mmen un do ti trng gy ra

M1=8

..2

dlq =8

2,703.02,123 2= 7604024 (N.cm)

3.Mmen un do bnh rng vng gy ra

M2=4

. dr lQ =4

2,703.12,7563= 1329596,5 ( N.cm)

4.Tng mmen un

Mu = M1+ M2 = 7207813 + 1329596,5 = 8933620,5 (N.cm)

5.Mmen chng un ca thng

W= .4

.2

DS= 014,0.

4

2.2

=0,04399823 (m 3 ) = 43982,29 (cm 3 )

S chiu dy thng sy =14 mm

ng sut thn thng

12,20329,43982

5,8933620===

W

Mu (N/cm 2 )

Ta c < [ ]CT5 = 6.10 4 (N/cm 2 )

Vy vi chiu dy thng l S = 0,014 m th thng bn

IV.Tnh vnh ai

1. Ti trng trn mt vnh ai

97,776983cos.2

975,155184

cos2

/ ===o

QQ

N

Q: Ti trng ca thng

: Gc nghing ca thng =3 o

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2.Phn lc con ln

T =3

/Q(Hng dn thit k my ha T1)

52,448593

97,77698 ==T N

3.B rng vnh ai

Pr

Pr : Ti trng ring tnh cho mt n v chiu di vi thng quay chm

Pr=24000 N/cm

78,124000

52,44859= cm =>chn B= 180 mm

Chn ng knh vnh ai Dv =220 cm (Dv=1,1Dt 1,2Dt) th b rng con ln

phi: B > 130 mm

Chn b rng vnh ai B= 180 mm =18cm

Vi thng nng th b dy vnh ai l

23,696,2

180

6,2==

=h (mm)

Chn h = 7cm => ng knh ngoi ca vnh ai =220+ 7 = 227 cm\

* Kim tra

Mmen un : Mu = 2T.R.A = T.Dv.A (N.cm)

A: H s ph thuc cnh lpA = 0,080,09

Chn A=0,09

Mu = 44859,52.2200.0,09 = 8882184,96 (N/cm)

Vnh ai c cu to t thp c c ng sut cho php : [ ] 15600= N/ 2cmMmen chng un l :

[ ]37,569

1560096,8882184 ===

uMW ( 3cm )

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Mt khc

6

. 2hW= ( T-85 Hng dn thit k my ha T1)

78,1318

37,569.66==

B

Wh cm = 137,8 (mm)

Vy h = 7 cm , vnh ai bn

Chn vnh ai c tit din : B h = 180mm 70 mm

V. Tnh con ln chn, con ln

1.Tnh con ln

B rng con ln c tnh theo cng thc

Bc = 180+ 60 = 240(mm) = 24 (cm)

(T 250 - Hng dn thit k my ha T1)

Chn s b ng knh con ln theo cng thc

d ccB

T

)400300(

(T 250 - Hng dn thit k my ha T1)

23,624.300

52,44859=cd (cm)

Kim tra ng knh theo tiu chun

DdD c 33,025,0 (T 250 - Hng dn thit k my ha T1)

D: ng knh ngoi ca vnh ai D = 2,2 m=220cm

0,25.220 dc 0,33.22055 dc 72,6

Vy chn ng knh con ln dc = 60 (cm)

+ Lc tc dng ln 1 n v chiu di tip xc

19,249218

52,44859===

B

TP N/cm

+ ng sut tip xc tnh theo cng thc

Rr

rREP += ...418.0max (N/cm 2 )

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R: Bn knh ngoi ca vnh ai R = 1102

220= cm

r: Bn knh trong con ln r = 302

60

= cmE:Mmen n hi ca vt liu E = 1,75.10 7 (N/cm 2 )

30.110

3011010.75,1.19,2492.418.0 7max+

= = 17980,13 (N/cm 2 )

ng sut tip xc cho php ca CT5 l [ ] 5CT = 6.104 (N/cm 2 )

[ ] 6000013,17980 5max =


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Vy 67,65636

25,23640max ===l

PP (/cm)

ng sut khi vnh ai tip xc vi con ln:

R

EP..418.0max = (T 285 - Hng dn thit k my ha T1)

R : Bn knh con ln chn

R= 5,172

35= cm

E: Momen n hi ca vt liu, E= 710.75,1 (N/cm 2 )

P: lc tc dng ln mt n v chiu di tip xc

49,107115,17

10.75,1.67,656.418.0

7

max == (N/cm 2 )

m bo iu kin bn v [ ] 600005max =< CT (N/cm 2 )

PHN IV : CC THIT B PH

I .Tnh ton l t

- Lng than cn t : 126,22 Kg/h

- Nhit chy cao ca than : 30894,36( Kj/Kg than )

- Nhit chy thp ca than : 29905,86( Kj/Kg than )

- Lng khng kh thi vo l t : 14758,51 ( m3/h ) = 4,099 ( m3/s )

1.Th tch bung t

v

lv

th

q

nQV

.= ( m 3 ) ( T-109, L cng nghip)

q v :Cng nhit th tch ca bung t ph thuc vo tng loi l t

Theo bng II-L cng nghip ta ly q v =460.10 3 (kcal/m 3 h)

1868,4.10.460

22,126.86,29905

3

=V = 3,47 ( m 3 )

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2.Din tch ghi l

r

nQF t

..28,0= ( T-103, L cng nghip)

r=550.10 3 (W/m 2 )Vy din tch ghi l l :

3

10.550

22,126.86,29905.28,0=F = 1,92 ( m 2 )

Chn F=2,65m3

Chn loi ghi l c kch thc mt tm ghi (34045) mm kch thc tng ng

ca l thng gi (35) mm

Vy s tm ghi l:

49,125045,0.34,0

92,1

045,0.34,0===

F tm.

Ly = 126 tm

II .Qut thi vo my sy

Lng khng kh a vo my sy 240 C0

V240= 4,099 ( m 3 /s)

Cng sut ng c qut c xc nh theo cng thc:

3600..102

.

HQN =

Trong :

Q: Nng sut qut Q=14758,51 ( m3

/h) :Hiu sut thy lc =0,65

H: Tng tr lc cn khc phc ( mm H 2 O)

H i= = 1 + 2 + 3

1=m.2

.150

S

n( T-107, L cng nghip)

Ta c:S:Din tch ghi l F = 1,92( m 2 )

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n: Lng than t

m:H s ph thuc vo hm lng tro than v loi ghi l

theo T-106, L cng nghip chn m=40

1=40.2

92,1.150

22,126

= 7,683 ( mm H 2 O)

2 : Tr lc ca lp than.Chn tr lc ca lp than v tr lc ca ghi l l 120

mm H 2 O ( T-107, L cng nghip)

3 =gd

l

2

..1

. 2

++ ( mm H 2 O)

: H s ma st ph thuc vo chun s Re

l: Tng chiu di ca ng ng :10 m

d: ng knh trong ca ng

: H s tr lc cc b( c 1 van chn) = 0.32

: Khi lng ring ca khng kh 25 C0 185,1= Kg/m 3

: Vn tc kh i trong ng

Chn vn tc kh i trong ng l 25 m/s

457,03600.25.785,0

14758,51

.785,0.3600===

Qd m

+ Xc nh

Ta c chun s Reynol

Re =

..d

(T-197, QTTBT1)

D:ng knh trong ca ng

: nht ng lc ca kh 25 C0

Tra bng I.255,T-318, STT1) ta c

=18,4.10 6 ( Ns/m 2 )=1,185 Kg/m 3

Re= 84,73574910.4,18

185,1.457,0.256= >10 4

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Vy ch chy xoy

( ) ( )012,0

64,184,735749lg8,1

1

64,1Relg8,1

122=

=

= ( T-378,STT1)

Ta thit k h thng ng ng t qut n bung t c chiu di l 2m trn h

thng c t mt van chn tiu chun 32,0=

3 = 28,6281,9.2

185,1.25.132,0

457,0

12.012,0 2=

++ ( mm H 2 O)

Vy ta c ;

iH = = 1 + 2 + 3 = 7,683 + 120+ 62,28 =189,963 (mm H2O)

Cng sut ng c qut trc my sy l:

746,113600.65,0.102

963,189.51,14758==N (KW)

Kt Lun

Qua thi gian thc thc hin n ny di s gip tn tnh ca thy

gio ..............................cng nh cc thy c trong khoa.Em hon thnh n

vi ni dung tnh ton v thit k h thng sy thng quay lm vic vi nng sut

13000 kg/h.Cc s liu c tra cu nhiu ti liu khc nhau v d nh cc sch

tp 1,2,3 qu trnh thit b ,s tay tp 1 v tp 2, h dn ng c kh tp1 nncng thc tra cu ng theo quy nh, m bo vic tnh ton l chnh xc v hp

l.

Tuy nhin do ln u tin lm quen vi kiu tnh ton v thit k nh th ny

nn khng th trnh c sai st. Em rt mong c thy hng dn v cc thy

trong b mn chm chc cho nhng li m em gp phi. Vic lm n mn hc

ny thc s em li cho em hiu qu cho em ni ring v cho sinh vin trong

ngnh ni chung .Qua sinh vin c nng cao k nng tnh ton cng nh nhn

nhn vn thit k 1 cch h thng. c bit gip cho sinh vin bit cch s dng,

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tra cu ti liu. C th ni y l mt s chun b tt cho vic lm n sp ti.

Tuy nhin do hn ch v thi cng nh trnh nn bn thuyt minh ca em cn

nhiu thiu st. Em rt mong c s gip ca thy c v cc bn.

Mt ln na em xin chn thnh cm n thy gio Nguyn Th Hu v mt sthy c khc trong khoa gip ch bo tn tnh cho em trong thi gian qua.

Em xin chn thnh cm n!

Ti liu tham kho

1. S tay ho cng tp 1 NXBKHKT

2. S tay ha cng tp 2 NXBKHKT

3. Tnh ton qu trnh thit b tp 1,2.3.4

4.Cc my gia cng vt liu do - H L Vin

5.Hng dn thit k h thng my sy - Trn Vn Ph HBKHN

7.L cng nghip

8. Tnh ton h dn ng c kh T1 - Trnh Cht,L Uyn

9. Hng dn tnh ton v tht k thit b my ho cht

v mt s ti liu trn mng

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Nhn xt ca trng khoa:

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