do an sấy quặng mangandioxit
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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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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:
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
= 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
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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)
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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 :
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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
=
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
*) 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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
+ 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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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:
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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:
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
( )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 =+
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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)
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
Nhn xt ca trng khoa:
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Trng i Hc Cng Nghip H Ni Khoa: Cng ngh ho hc
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