Đồ Án Môn Học Sấy Băng Tải

n mn hc

Sy bng ti

GVHD: Nguyn Dn SVTH: Nguyn Th T Dip_02H

n mn hc: QTTB Sy bng ti

SVTH: Nguyn Th T Dip_02H 2 GVHD: Nguyn Dn

PHN 1 : M U 1.1 Li m u Trong ngnh cng nghip ni chung th vic bo qun cht lng sn phm l rt quan trng. cht lng sn phm c tt ta phi tin hnh sy tch m Vt liu sau khi sy c khi lng gim do gim cng chuyn ch, bn tng ln, cht lng sn phm c nng cao,thi gian bo qun ko di........ Qu trnh lm bc hi nc ra khi vt liu bng nhit gi l qu trnh sy. Ngi ta phn bit sy ra lm hai loi :sy t nhin v sy nhn to Sy t nhin dng nng lng mt tri lm bay hi nc trong vt liu nn n gin ,t tn km tuy nhin kh iu chnh c qu trnh sy v vt liu sau khi sy vn cn m cao .Trong cng nghip ho cht thng ngi ta dng sy nhn to,tc l phi cung cp nhit cho vt liu m .Phng php cung cp nhit c th bng dn nhit ,i lu ,bc x hoc bng nng lng in trng c tn s cao. i vi nc ta l nc nhit i nng m,do vic nghin cu cng ngh sy ch bin thc phm kh v lm kh nng sn c ngha rt c bit .Kt hp phi sy nhm tit kim nng lng,nghin cu nhng cng ngh sy v cc thit b sy ph hp cho tng loi thc phm ,nng sn ph hp vi iu kin kh hu v thc tin nc ta.T to ra hng ha phong ph c cht lng cao phc v cho xut khu v tiu dng trong nc. 1.2 BIN LUN TI



thc hin qu trnh sy, ngi ta s dng mt h thng gm nhiu thit b nh: thit b sy ( bung sy, hm sy, thit b sy kiu bng ti, my sy thng quay, sy phun, sy tng si, my sy trc ), thit b t nng tc nhn, qut, bm v mt s thit b ph khc, Trong n ny em s tnh ton v thit k thit b sy kiu bng ti. Thit b sy loi ny thng c dng sy cc loi rau qu, ng cc, cc loi nng sn khc, sy mt s sm phm ho hc Trong n ca mnh em s dng vt liu sy l ch vi tc nhn sy l hn hp khng kh nng. Ch l mt cy cng nghip lu nm, thch hp nht i vi kh hu nhit i. Ch khng n thun ch l th cy c dng gii kht m tr thnh mt sn phm c nhiu cng dng. Ch bin ch khng ch cung cp phc v nhu cu trong nc m cn xut khu, yu cu v u t thit b t tn km hn cc loi nng sn khc. Trong cng ngh sn xut ch th sy ch l mt khu rt quan trng. Ch sau khi thu hoch qua ch bin s c sy kh. Sau khi sy ch phi t c ti, kh nht nh theo yu cu m bo cht lng v tng thi gian bo qun. Vi cc yu cu v hnh thc, v sinh v cht lng sn phm ngi ta s dng thit b sy kiu bng ti vi nhiu bng ti lm vic lin tc vi tc nhn sy l khng kh nng c tun hon mt phn kh thi. Vt liu sy c cung cp nhit bng phng php i lu. u im ca phng thc sy ny l thit b n gin, r tin, sn phm c sy u, do c tun hon mt phn kh thi nn d dng iu chnh m ca tc nhn sy, tc ca khng kh i qua phng sy ln, nng sut kh cao, hiu qu PHN 2 : S CNG NGH & THUYT MINH 2.1 S CNG NGH CA QU TRNH



Vi cc thit b v phng thc sy nh chn, ta c s cng ngh ca qu trnh sy ch nh sau : Kh thi Hn hp kh sau khi sy Vt liu vo Hi nc Kh tun hon Vt liu ra Hi nc bo ho Khng kh Ch thch : 1 phng sy 2 - calorifer 3 - qut y 4 cyclon 5 qut ht 2.2 THUYT MINH LU TRNH

2

3

5

1

4



Do yu cu v kh ca ch nn dng tc nhn sy l hn hp khng kh nng. Khng kh ban u c a vo calorife, y khng kh nhn nhit gin tip t hi nc bo ho qua thnh ng trao i nhit. Hi nc i trong ng, khng kh i ngoi ng. Ti calorife, sau khi nhn c nhit sy cn thit khng kh nng i vo phng sy tip xc vi vt liu sy (ch) cp nhit cho hi nc trong ch bc hi ra ngoi. Trong qu trnh sy, khng kh chuyn ng vi vn tc ln nn c mt phn ch s b ko theo khng kh ra khi phng sy. thu hi kh thi v ch ngi ta t ng ng ra ca khng kh nng mt cyclon. Kh thi sau khi ra khi phng sy i vo cyclon tch ch cun theo v lm sch. Sau mt phn kh thi c qut ht ra ng ng dn kh thi ra ngoi khng kh. Mt phn kh cho tun hon tr li trn ln vi khng kh mi to thnh hn hp kh c qut y y vo calorife. Hn hp kh ny c nng nhit n nhit cn thit ri vo phng sy tip tc thc hin qu trnh sy. Qu trnh sy li c tip tc din ra. Vt liu sy ban u c m ln c a vo phng sy i qua cc bng ti nh thit b hng vt liu. Vt liu sy chuyn ng trn bng ti ngc chiu vi ciu chuyn ng ca khng kh nng v nhn nhit trc tip t hn hp khng kh nng thc hin qu trnh tch m. Vt liu kh sau khi sy c cho vo mng v c ly ra ngoi.



PHN 3 : CN BNG VT CHT 3.1 Cc k hiu

G1,G2: Lng vt liu trc khi vo v sau khi ra khi my sy, (Kg/h) Gk:Lng vt liu kh tuyt i i qua my sy , (Kg/h) W1, W2: m ca vt liu trc v sau khi sy, tnh theo % khi lng vt

liu t W: m c tch ra khi vt liu khi i qua my sy , (Kg/h) L:Lng khng kh kh tuyt i i qua my sy , (Kg/h) xo:Hm m ca khng kh trc khi vo caloripher si , (Kg/Kgkkk) x1,x2: Hm m ca khng kh trc khi vo my sy (sau khi i qua

caloripher si) v sau khi ra khi my sy, (Kg/Kgkkk) 3.2 Cc thng s ban u Thit k h thng sy bng ti sy ch vi nng sut khong 1400tn/ nm Gi thit mt nm nh my lm vic 350 ngy ,mi ngy lm 20 gi . Vy nng sut trung bnh trong mt gi l

G2= 20020*3501400000 = Kg/h

Ch sau khi thu hoch c s ch s b trc khi em vo phng sy. m ca ch lc ny t khong t (60-65)% .Chn m ca ch trc khi sy l W1=63%. sn phm ch sau khi sy t c kh,ti,xp theo yu cu m khng b gy vn,khng b m mc th ta khng ch m ra ca ch t khong W2=5% Theo kinh nghim ch kh c th chu c nhit trn di 1000C.Do ta chn nhit tc nhn sy vo thit b sy l t2=1000C. m bo tnh kinh t,gim tn tht nhit do tc nhn sy mang i ng thi m bo khng xy ra hin tng ng sng sau khi sy,ta chn t2 sao cho m tng i khng qu b nhng cng khng qu gn trng thi bo ho .Do nhit tc nhn ra khi bung sy c chn s b khong t2=700C Thng s khng kh ngoi tri c xc nh ti thnh ph Nng Nh vy,cc thng s ban u c xc nh l: Nng sut tnh theo sn phm : G2 =200 kg/h m vt liu vo : W1 = 63% m vt liu ra : W2 = 5% Nhit tc nhn sy vo : t1 = 1000C Nhit tc nhn sy ra : t2 = 700C Nhit khng kh ngoi tri : t0 = 260C ,Pobh =0.0343 at m mi trng : = 81% Hm m ca khng kh c tnh theo cng thc sau:



xo=0.622obhokq

obho

PP

P

*

*

{sch QTTBII_ trang 156}

thay s vo ta c

xo=0.622 0343.0*81.0033.10343.0*81.0

=0.0172(kg/kgkkk) -Nhit lng ring ca khng kh: Io=Ckkk*to+xo*ih , ( J/kgkkk ) {sch QTTBII- trang 156} Vi Ckkk: nhit dung ring ca khng kh ,J/kg Ckkk= 103 J/kg to: nhit ca khng kh to= 26oC ih: nhit lng ring ca hi nc nhit to , J/kg Nhit lng ring ih dc xc nh theo cng thc thc nghim ih=ro+Ch *to=(2493+1.97to)103 , J/kg {sch QTTBII _ trang 156} Trong : ro=2493*103 :nhit lng ring ca hi nc 0oC Ch= 1.97*103: nhit dung ring ca hi nc , J/kg T ta tnh c Io=69.76*103 J/kgkkk hay Io=69.76 (kJ/kgkkk) -Trng thi ca khng kh sau khi ra khi caloripher l: t1=100oC,P1bh=1.02 at Khi i qua caloripher si, khng kh ch thay i nhit cn hm m khng thay i. Do x1=xo nn ta c :

( ) bhkq

PxPx

11

1

1 622.0*+= = ( ) 02.1*0172.0622.0 033.1*0172.0 + =0.027=2.7%

-Nhit lng ring ca khng kh sau khi ra khi caloripher l: I1 = t1+(2493+1.97t1)103x1 , (J/Kgkkk) I1 = 100+(2493+1.97*100)*0.0172 = 146.268 ( KJ/Kgkkk ) -Trng thi ca khng kh sau khi ra khi phng sy: t2=70oC , P2bh=0.3177 at -Nu sy l thuyt th:I1=I2=146.268 (KJ/Kgkkk) Ta c I2=Ckkk*t2+x2*ih , J/Kgkkk T hm m ca khng kh

x2=h

kkk

itCI 22 * =

00

22

**tCrtCI

h

kkk

+ (Kg/Kgkkk)

x2= 26*10*97.110*249370*1010*268.146

33

33

+ =0.029 (Kg/Kgkkk)

( ) bhkq

PxPx

22

22 622.0

*+= = 3177.0)029.0622.0(

033.1*029.0+ = 0.1496=14.96%

3.3 Cn bng vt liu



3.3.1 Cn bng vt liu cho vt liu sy Trong qu trnh sy ta xem nh khng c hin tng mt mt vt liu,lng khng kh kh tuyt i coi nh khng b bin i trong sut qu trnh sy.Vy lng vt liu kh tuyt i i qua my sy l:

Gk=G1 100100 1W =G2 100

100 2W {sch QTTBII_trang 165} Trong : W1=63%, W2=5%; G2=200 ( Kg/h.)

Vy Gk = 200 1005100 = 190 (Kg/h)

Lng m tch ra khi vt liu trong qu trnh sy c tnh theo cng thc:

W=G21

21

W100WW

, (Kg/h) {sch QTTBII_ trang 165}

W=20063100563

=313.5 (Kg/h)

Lng vt liu trc khi vo phng sy G1=G2+W=200+313.5=513.5 (Kg/h) 3.3.2 Cn bng vt liu cho khng kh sy Cng nh vt liu kh ,coi nh lng khng kh kh tuyt i i qua my sy khng b mt mt trong sut qu trnh sy.Khi qua qu trnh lm vic n nh lng khng kh i vo my sy mang theo mt lng m l :Lx1 Sau khi sy xong , lng m bc ra khi vt liu l W do khng kh c thm mt lng m l W Nu lng m trong khng kh ra khi my sy l Lx2 th ta c phng trnh cn bng: Lx1+W=Lx2 {sch QTTBII_ trang 165}

L =12

Wxx (Kg/h)

Thay s L = 0172.0029.0

5.313 = 26567.8 ( Kg/h)

Vi L l lng khng kh kh cn thit lm bc hi W kg m trong vt liu. Ta c,ti t0=260C,ng vi 0 th 185.10 = kg/cm3 Lu lng th tch ca tc nhn sy trc khi vo calorifer l:

V= 08.22420185.1

8.26567

0

==L (m3/h)

Vy lng khng kh kh cn thit lm bc hi 1 Kg m trong vt liu l:

l = WL =

12

1xx (Kg/Kgm) {sch QTTBII_ trang 166}

Khi i qua calorifer si khng kh ch thay i nhit nhng khng thay i hm m, do xo=x1 nn ta c:

l = 12

1xx = 02

1xx



Thay s vo ta c

l = 0172.0029.0

1 = 84.745 (Kg/Kgm)

3.4 Qu trnh sy hi lu l thuyt Qu trnh hot ng ca h thng ny l: Tc nhn sy i ra khi bung sy c trang thi t2, 2,x2 c hi lu li vi lng lH v thi ra mi trng lt .Khi lng lH c ho trn vi khng kh mi c trng thi l t0, o ,x0 vi lng l0 Sau khi c ho trn,ta c lng khng kh l l ,c qut ht v y vo calorife gia nhit n trng thi I1,t1, 1 ri y vo bung sy Vt liu m c khi lng l G1 i vo bung sy v sn phm ra l G2 .Tc nhn i qua bung sy nhn hi nc bay hi t vt liu sy ng thi b mt nhit nn trng thi ca n l x2 ,t2, 2 Gi xM,IM l trng thi ca hn hp kh bung ho trn Ta c: l=lo+lH hoc L=Lo+LH -Chn t l hi lu l 50% vy l = 0.5(lo+lH) suy ra lH=lo Vy t s hi lu n : l s kg khng kh hi lu ho trn vi 1 kg khng kh ban u ( t mi trng)

n =o

H

ll ( sch k thut sy nng sn _trang 79)

Vy hm m ca hn hp kh c tnh theo cng thc sau:

xM= nnxxo

++

12 { sach QTTBII_ trang 176} (Kg/Kgkkk)

xM= 220 xx + =

2029.00172.0 + = 0.0231 (Kg/Kgkkk)

Nhit lng ring ca hn hp khng kh l:

IM= nnII

++

120 (KJ/Kgkkk)

IM= 11268.146*176.69

++ =108.03 (KJ/Kgkkk)

Ta c: IM=(103+1.97*103xM)tM + 2493*103xM

Suy ra tM=M

MM

xxI

33

3

10*97.11010*2493

+

Vi tM: Nhit ca hn hp kh

T : tM= 0231.0*10*97.1100231.0*10*2493*10*03.108

33

33

+ = 48.250C ,

Suy ra PMbh=0.11(at)



)622.0( += MMbh

kqM

M xPPx = )622.00231.0(11.0 033.1*0231.0 + = 0.336= 33.6 %

Lng khng kh kh lu chuyn trong thit b sy

5.1690231.0029.0

11

2

=== Mxxl Kg/Kg m

PHN 4 : CN BNG NHIT LNG & TNH TON THIT B CHNH

4.1 Tnh ton thit b chnh 4.1.1Th tch ca khng kh a/Th tch ring ca khng kh vo thit b sy:

v1=bhPP

RT

11

1

m3/Kgkkk ,{sch QTTB II- trang 157}

Vi R=287 (J/KgoK) T1=1000C+273=373K P=1.033(at) P1bh=1.02(at) 1=0.027 Thay s vo ta c:



v1= ( ) 410*81.9*02.1*027.0033.1373*287

=1.085 (m3/Kgkkk)

b/Th tch khng kh vo phng sy: V1=L*v1=26567.8*1.085=28826.1 (m3/h) c/ Th tch ring khng kh ra khi phng sy l:

v2=bhPP

RT

22

2

,vi T2=70+273=343K, 1496.02 = ,P2bh=0.3177at Thay s vo ta c :

v2 = ( ) 410*81.9*3177.0*1496.0033.1343*287

v2 = 1.018 (m3/Kgkkk) d/Th tch khng kh ra khi phng sy: V2=Lv2=26567.8*1.018=27046.0 (m3/h) e/Th tch trung bnh ca khng kh trong phng sy:

Vtb= 221 VV + =27936.05 (m3/h)

4.1.2 Thit b sy kiu bng ti Thit b sy kiu bng ti gm mt phng hnh ch nht trong c mt hay vi bng ti chuyn ng nh cc tang quay,cc bng ny ta trn cc con ln khi b vng xung.Bng ti lm bng si bng tm cao su,bn thp hay li kim loi,khng kh c t nng trong carolifer.Vt liu sy cha trong phu tip liu,c cun vo gia hai trc ln i vo bng ti trn cng.Nu thit b c mt bng ti th sy khng u v lp vt liu khng c xo trn do loi thit b c nhiu bng ti c s dng rng ri hn. loi ny vt liu t bng trn di chuyn n u thit b th ri xung bng di chuyn ng theo chiu ngc li.Khi n cui bng cui cng th vt liu kh c vo ngn tho. Khng kh nng i ngc vi chiu chuyn ng ca cc bng . qu trnh sy c tt,ngi ta cho khng kh di chuyn vi vn tc ln,khong 3m/s ,cn bng th di chuyn vi vn tc ( 0.3-0.6) m/ph Chn kch thc bng ti Gi Br : Chiu rng lp bng ti (m) h : Chiu dy lp tr (m) ,Ly h=0.1(m) : Vn tc bng ti , chn =0.4 m/ph : Khi lng ring ca ch , Chn 3320 m

Kg= -Nng sut ca qu trnh sy:

G1=Brh (Kg/h) suy ra Br= 601

hG =

60*4.0*320*1.05.513 =0.6686

(m) -Chiu rng thc t ca bng ti l :



Btt= rB , vi l hiu s hiu chnh

Chn =0.9 ,ta c Btt= 9.06686.0 = 0.7429 (m)

Gi Lb : Chiu di bng ti ,m (chiu di mt mt) ls: Chiu di ph thm, chn ls=1.2 (m) T: Thi gian sy, chn T=30 pht=0.5 gi

Lb= ***1hB

TG

tt

+ ls = 2.1320*1.0*7429.05.0*5.513 + =12(m)

Vy Lb=12(m) -Bng ti ch s dng mt dy chuyn nn ta chn chiu di ca mt bng ti l 4(m) suy ra s bng ti l 3 ng knh ca bng ti d=0.3m 4.1.3 Chn vt liu lm phng sy -Phng sy c xy bng gch -B dy tng 0.22 (m) c: +Chiu dy vin gch 0.2( m) +Hai lp va hai bn 0.01 (m) -Trn phng c lm bng btng ct thp c: +Chiu dy m02.01 = +Lp cch nhit dy m15.02 = -Ca phng sy c lm bng tm nhm mng,gia c lp cc nhit dy 0.01 m +Hai lp nhm mi lp dy 0.015 (m) -Chiu di lm vic ca phng sy: Lph = 4+2*0.6= 5.2 m -Chiu cao lm vic ca phng sy: Hph = 0.3+0.1*3+0.2*4 = 2 ( m ) -Chiu rng lm vic ca phng sy: Rph = 0.7429+0.66 = 1.4029. (m) Vy kch thc ca phng sy k c tng l: Lng = 5.2+2*0.22 = 5.64(m) Hng = 2.0+0.02+0.15 = 2.17 (m) Rng = 1.4029+0.22*2 = 1.8429 (m) 4.1.4Vn tc chuyn ng ca khng kh v ch chuyn ng ca khng kh trong phng sy a/Vn tc ca khng kh trong phng sy:

===3600*4029.1*2

05.27936

phph

tbkk RH

V 2.77 m/s b/Ch chuyn ng ca khng kh:

Re = tdkk l* {sch QTTB II _ trang 35}



Vi: Re: l hng s Reynol c trng cho ch chuyn ng ca dng lt ng knh tng ng

lt = phph

phph

RHRH

+**2

= 4029.12

4029.1*00.2*2+ =1.649( m)

Nhit trung bnh ca khng kh trong phng sy:

ttb = 221 tt + =

270100 + = 85oC

-T nhit trung bnh ny tra bng ph 9 trang 130 sch k thut sy nng sn ta c = 0.031 (W/moK) = 21.06*10 6 (m2/s) Vy Re = 610*06.21

649.1*77.2 = 15*10

4

Vy Re=15*10 4 suy ra ch ca khng kh trong phng sy l ch chuyn ng xoy 4.1.5 Hiu s nhit trung bnh gia tc nhn sy vi mi trng xung quanh

tb = 2

1

21

lntt

tt

Vi 1t : Hiu s nhit gia tc nhn sy vo phng sy vi khng kh bn ngoi = 1t 100-26=74oC 2t : Hiu s nhit gia tc nhn sy i ra khi phng sy vi khng kh bn ngoi 2t =70-26 = 44oC Vy tbt =

4474ln

4474 = 57.71oC

4.2 Tnh tn tht nhit 4.2.1 Tn tht qua tng 1 -Tng xy bng gch dy 0.22 (m) Tt1 -Chiu dy vin gch gach =0.2 (m) -Chiu dy mi lp va v = 0.01 (m) Tt2 Tra bng gach = 0.77( w/m) 2 v = 1.2 (w/m) 1 2 3 Lu th nng (khng kh nng) chuyn ng trong phng do i lu t nhin(v c s chnh lch nhit ) v do cng bc ( qut) .Khng kh chuyn ng theo ch chy xoy(do Re>104) Gi 1 l h s cp nhit t tc nhn sy n b mt trong ca tng phng sy



1 = k( //1/1 + ) Vi : //1 l h s cp nhit t tc nhn sy n thnh my sy do i lu t nhin ,W/m2 1/ l h s cp nhit t tc nhn sy n thnh my sy do i lu cng bc ,W/m2 k : h s iu chnh, k= 1.21.3 a/Tnh /1 Phng trnh chun Nuxen i vi cht kh: Nu = C l R0.8 = 0.018 l R0.8 Trong : l ph thuc vo t s

td

ph

lL

v Re

Ta c : td

ph

lL

= 649.1

2.5 =3.15

Re =15*10 4 Tra bng v tnh ton ta c l =1.205 {s tay QTTBII_ trang 15} Vy Nu = 0.018*1.205* (15*104)0.8 = 300

M Nu = phH1/ suy ra /1 =

phHNu = 65.4

2031.0*300 =

b/Tnh 1// Gi tT1l nhit trung bnh ca b mt thnh ng(tng) tip xc vi khng kh trong phng sy Chn tT1=70.0oC Gi ttbk l nhit trung bnh ca cht kh vo phng sy (tc nhn sy)

ttbk = 85270100 =+ oC

Gi ttbl nhit trung bnh gia tng trong phng sy vi nhit trung bnh ca tc nhn sy.

ttb = 5.7728570 =+ oC

Chun s Gratket : t trng cho tc dng tng h ca lc ma st phn t v lc nng do chnh lch khi lng ring cc im c nhit cao khc ca dng,k hiu Gr

Gr = T

tgH ph2

13

vi g l gia tc trng trng g=9.8(m/s2 ) Hph Chiu cao ca phng sy ,m 1t = ttbk-tT1= 85-77.5 = 7.5 , T=ttbk +273=358K Suy ra Gr=

358*10*09.215.7*2*8.9

122

3

=3.69*109



M chun s Nuxen l Nu = 0.47*Gr0.25 {s tay QTTB II_ trang 24} Suy ra Nu = 115.8

Hn na Nu = 1//H suy ra 1// =

phHNu =

2031.0*8.115 =1.74

T ( ) ( ) 58.774.165.42.1//11/1 =+=+= k c/Tnh 2 H s cp nhit ca b mt ngoi my sy n mi trng xung quanh //22/2 += Vi /2 H s cp nhit do i lu t nhin 2// H s cp nhit do bc x Ta c nhit ti ring ca khng kh t phng sy n mi trng xung quanh : q1= 11 * t =7.68*(85-70)=113.7 ,KJ/kg m Trong qu trnh truyn nhit n nh th:

q1= =

3

1

21

i i

i

TT tt

M 3

3

2

2

1

13

1

++=

=i ii (m2/w)

y : 321 ,, : b dy cc lp tng ,m 321 ,, : H s dn nhit tng ng , W/m m01.021 == _ B dy lp va c 2.121 == (w/m) m2.03 = _B dy ca vin gch c 77.03 = (w/m) Vy =

=

3

1i i

i

267.077.02.0

2.101.0

2.101.0 =++ (m2/w)

T

tT1-tT2=q1=

3

1i i

i

= 113.7*0.267 =31(oC)

tT2: Nhit tng ngoi phng sy ,0C tT2 = tT1-31=70-31= 39 ( 0C) Nhit lp bin gii gia tng ngoi phng sy v khng kh ngoi tri

Tbg = 2262 +Tt = 5.32

22639 =+ oC

Ti nhit Tbg ny tra bng ta tnh c : 210*67.2 = (W/mK) 610*024.16 = (m2/s) Nhit tng ngoi v nhit khng kh c lch l 2t =tT2-tkk = 39-26 = 13 (0C)



Chun s Gratkev l

Gr= ( ) 91223

22

3

10*74.122731310*024.16

13*17.2*81.9 =+=

TtgH ng

Chun s Nuxen l Nu = 0.47*Gr0.25 = 157.9

Suy ra 2/ = 11.217.2

0267.0*9.157 ==HngNu

H s cp nhit do bc x 2//

2// =

4

24

1

2 100100TT

ttC

kkT

on Vi n : en ca va ly n = 0.9 Co:H s bc x ca vt en tuyt i ,ly C0=5.67 T1 = tT2+273=39+273=312K T2 = tkk+273=26+273=299 K

T 91.5100299

100312

263976.5*9.0 44

2// =

= Nn 2//2/2 += = 2.11+5.91 = 8.02 Nhit ti ring t b mt ca tng ngoi n mi trng khng kh q2 = 3.10413*02.8* 22 ==t ,KJ/kg m So snh %808.0

7.1133.1047.113 ===

mazqq

Vy tn tht qua tng Qt=3.6*k* tbtF * M F = 2*L*H+2*R*H=2*5.2*2+2*1.4029*2=26.4(m)

k = 88.1276.0

11.81

58.71

111

13

121

=++

=++

=i ii

7.57ln

2

1

21 ==tt

ttttboC

T :QT = 3.6*1.88*26.4*57.7 = 10309.6 (KJ)

Vy qt= 89.325.3136.10309

W==TQ (KJ/Kgm)

4.2.2 Tn tht qua trn Trn c: Lp btng ct thp dy 55.1);(02.0 22 == m (W/m)



Lp cch nhit dy 058.0);(15.0 33 == m (W/m) tnh tn tht qua trn ta xc nh: 22 *3.1 =tr =1.3*8.11=10.543 ,W/m2K Do h s truyn nhit qua trn Ktr bng Ktr= 35.0

543.101

058.015.0

55.102.0

58.71

111

1

23

3

2

2

1

=+++

=+++

(W/m2K)

Vy tn tht qua trn: Qtr=3.6*Ktr*Ftr* t =3.6*0.35*(5.2*1.4029)(85-26)=542.3 (KJ/h) Nhit ti ring

qtr= WQtr =

5.3133.542 =1.73 ,KJ/kg m

4.2.3 Tn tht qua ca Hai u phng sy c ca lm bng thp dy 4 =5mm c h s dn nhit

4 =0.5W/mK Do h s dn nhit qua ca Kc bng :

Kc= 77.3

11.81

5.0005.0

58.71

111

1

24

4

1

=++

=++

,W/m2K

Ca pha tc nhn sy vo c chnh lch nhit (t1-t0) cn ca u kia c chnh lch nhit bng (t2-t0).Do : Qc= 3.6*Kc*Fc{(t1-t0)+(t2-t0)} Thay s ta c : Qc=3.6*3.77*(1.4029*2){(100-26)+(70-26)}=4493.5 (KJ/h)

qc= 5.3135.4493=

WQc =14.33 (KJ/kg m)

4.2.4 Tn tht nhit qua nn Nhit trung bnh ca tc nhn sy bng 850C v gi s tng phng sy cch tng bao che ca phn xng 2m.Theo bng 7.1 ca sch tnh ton & thit k h thng sy_trang 142.Ta c: q1=50W/m .Do tn tht qua nn bng: Qn=3.6*Fn*q1=3.6 (5.2*1.4029)50=1313.1 (KJ/h)

Suy ra qn= 5.3131.1313=

WQn =4.19 (KJ/kg m)

Nh vy tng tn tht nhit truyn qua kt cu bao che ra mi trng xung quanh bng: Qmt=Qt+Qc+Qtr+Qn=16658.5 (KJ/h)

qmt= 5.3135.16658=

WQmt =53.14(KJ/kg m)

4.2.5 Tn tht do vt liu sy mang i



Trong sy nng sn,nhit vt liu sy ra khi thit b sy ly thp hn nhit tc nhn sy tng ng t (510)0C.Trong h thng sy ny ,vt liu sy v tc nhn sy chuyn ng ngc chiu nn tV2=t1-(510)0C.V vy ta ly tV2=100-10=90 0C. Do nhit dung ring ca ch ra khi phng sy : CV2=Cvl * (1- 22 ) nC+ Vi Cvl : nhit dung ring ca ch ,ly Cvl=0.37(KJ/KgoK) C : nhit dung ring ca nc ,ly C=4.18 (KJ/Kg) Thay s ta c: CV2 =0.37*4.18(1-0.05)+4.18*0.05 CV2=1.68 (KJ/kgoK) Tn tht nhit do vt liu sy mang i l: QVL=G2*CV2(tV2-tV1)=200*1.68(90-26)=21481.8 (KJ/h)

qVl= WQVl =68.52 (KJ/kgm)

4.3 Qu trnh sy thc t c hi lu 4.3.1 Nhit lng b sung thc t = qqC vl1 Vi: 1 = 26oC nhit ca vt liu trc khi vo my sy(bng nhit mi tr 2 = 70oC nhit ca vt liu khi ra khi my sy -Vy nhit lng b sung thc t: 98.1214.5352.6818.4*26 == (KJ/Kgm) 4.3.2 Cc thng s ca qu trnh sy thc -Hm m ca tc nhn sy i ra khi my sy:

( )2211/

2 ***

tCrtCxI

xno

k

+++= , Kg/Kgkkk {s tayQTTBII_

trang105}

Thay s: ( ) 027.070*18.4249396.1270*10172.0*96.12268.146/

2 =+++=x (Kg/Kgkkk)

Vy : ( ) ( ) 0.141027.070*97.1249370** /222/2 =++=++= xtCrtI ho (KJ/Kgkkk) - m tng i ( ) ( ) 135.03177.0027.0622.0 027.0*033.1*622.0 * /2 2

/

2/ =+=+= bhPx

xP =13.5% -Lng khng kh kh lm bc hi 1 Kg m ht t ngoi vo:

1000172.0027.0

11/

2

/ === oo

xxl ,Kg/Kgm

-Qa trnh sy tun hon kh thi (n=1):

38.10511

0.141*176.691

/2/ =+

+=++=

nnIII oM (KJ/Kgkkk)

-Hm m ca hn hp khng kh:



022.011

027.0*10172.01

/2/ =+

+=++=

nnxxx oM (Kg/Kgkkk)

-Khi ra khi caloripher khng kh ch thay i nhit ch khng thay i hm m do : x/1 = x/M = 0.022(Kg/Kgkkk) t1 = 100oC -Vy nhit lng ring ca khng kh sy vo phng sy l: I/1 = t1 + (2493 + 1.97*t1)*x/1 = 100 + (2493+1.97*100)*0.022=159.18 ,KJ/Kgkkk -Lng hi lu thc t: l/H = l/o = 100( Kg/Kg m) -Nhit khi ho trn:

Cx

xItM

MMM

0/

/// 4.48

022.0*97.11022.0*249338.105

*97.11*2493 =+

=+=

th I - x biu din qu trnh sy l thuyt v sy thc ng AMB1C1 biu din qu trnh sy thc t c tun hon mt phn kh thi



4.4 Cn bng nhit lng = rV qq 4.4.1 Nhit lng vo Nhit do calorife si cung cp: qs=l(I1-IM)=100(159.18-105.38)=5380 ,KJ/Kg m Nhit lng do vy liu sy mang vo:

qvl= WCG vl 11 ** =

5.31326*18.4*37.0*5.513 =65.86 ,KJ/Kgm

Nhit lng do khng kh sy mang vo my sy: qkkv=l*IM=100*105.38=10538 ,KJ/Kgm Vy tng nhit lng vo: 86.159831053886.655380 =++=++= kkvvlsV qqqq ,KJ/Kgm 4.4.2 Nhit lng ra Nhit lng do vt liu sy mang ra:

qvlr= 06.695.31370*18.4*37.0*200** 22 ==

WCG vl ,KJ/Kgm

Nhit do tn tht ca phng sy: = 14.53q , KJ/Kgm Nhit do khng kh mang ra : qkkr=l*I2=100*141=14100 ,KJ/Kgm Nhit tn tht trong qu trnh sy: qt=l(I1-I2)= 100(159.18-141)=1818 , KJ/Kgm Vy tng nhit lng ra l:

38.1603918181410014.5306.69 =+++=+++= tkkrvlrr qqqqq ,KJ/Kgm So snh tng nhit lng vo v tng nhit lng ra:

%35.00035.038.16039

38.1603986.15983

max

====q

qq rv



PHN 5 : TNH TON THIT B PH 5.1 Calorifer Do yu cu v cht lng ca sn phm ch sau khi sy nn dng tc nhn sy l khng kh nng.Khng kh nng i qua calorifer si v nhn nhit trc tip t hi nc bo ho qua thnh ng.Khng kh dng sy c nhit theo yu cu l 1000C,cht truyn nhit l hi nc bo bo .

Thit b l loi ng chm,hi nc bo ho i trong ng,khng kh i ngoi ng.Hai lu th chuyn ng cho dng

5.1.1 Chn kch thc truyn nhit Chn ng truyn nhit bng ng,c gn nng h s truyn nhit,h s dn nhit ca ng l 385= W/m {sch QTTB tp I_ trang 125} yChn ng: -ng knh ngoi ca ng : dng = 0.03 (m) -ng knh trong ca ng : dtr = 0.025 (m)

-Chiu dy ca ng : = 2

trng dd = 0.0025 (m) -ng knh ca gn : Dg = 1.4dng = 0.042(m) -Bc gn : bg = 0.01 m

-Chiu cao ca gn : h = 2

ngg dD = 0.006 (m) -Chiu di ca ng : l = 1 (m)

-S gn trong trn mt ng : m =gbl = 100

-B dy bc gn : b = 0.002(m) -Tng chiu di ca gn : Lg=b*m=0.002*100=0.2(m) -Tng chiu di khng gn : Lkg = l-Lg = 1.0-0.2=0.8(m) -Lng khng kh cn thit cho qu trnh sy c hi lu (theo tnh ton thc t): l = 100( Kg/Kgm) L = 31350( Kg/h) -Nhit ca khng kh ban u khi hi lu : tM = 48.4oC -Nhit khng kh sau khi ra khi caloripher l t1=1000C -Th tch ring ca khng kh

v100oC = 057.1946.011

100

==o ,m

3/kg

v70oC = 97.0029.111

70

==o , m

3/kg



v48.4oC = 88.0125.111

4.48

==o , m

3/kg

v26oC = 8.0181.111

26

==o ,m

3/kg

vtb = =+210026 oo vv 0.93 ,m3/kg

-Lng khng kh kh i vo caloripher l: V=L*vtb = 31350*0.93=29155 , (m3/h) -H s cp nhit i lu 1 : +Nhit trung bnh ca khng kh trong caloripher ttb ttb = thnc- tbt M:

c

d

cdtb

tt

ttt

=

ln

+Chn nhit hi nc bo ho khi vo l : thn = 130oC +Chn nhit hi nc bo ho khi ra l thnc = 105oC Nn ta c: Cttt odhndd 10426130 === Ctttc ochnc 5100105 === Thay s vo ta c: Ct otb 5.32= Suy ra : ttb = 130-32.5 = 97.5oC ng vi gi tr ttb ta c: 935.0= (Kg/m3) 210*15.3 = (W/moC) 610*5.22 = (m2/s) 610*7.21= (Ns/m2) Pr = 0.69 5.1.2 Tnh ton Din tch b mt ca mt ng : (pha trong ca ng) Ftr = *dtr*l = 3.14*0.025*1.0 = 0.0785( m2) Din tch mt ngoi ca ng: Fng = * dng*l = 3.14*0.03*1.0= 0.0342 (m2) Din tch phn b mt ngoi ca mt ng Fbm = Fgn+Fkgn -Din tch phn c gn

Fgn = ngggg dDLD 22 *

4*

4** + = 0.02705(m2)

-Din tch phn khng gn



Fkgn = Lkg* ngd* = 0.8*3.14*0.03=0.07536 (m2) Vy : Fbm = 0.02705+0.07536 = 0.10241 (m2) Chn s ng xp hng ngang l: i = 20 Khong cch gia cc ng ny ng kia l 0.05(m) Khong cch gia ng ngoi cng dn caloripher l:x= 0.01 (m) Tit din t do ca mt phng vung gc vi phng chuyn ng ca khng kh: Ftd=Fng-Ftr=0.0942-0.0785=0.0157 ,m2 Chn tc dng kh qua calorifer l: 6=kk m /s Chun s Re:

Re = 260010*5.2201.0*6*

6 == gkk b

2300< Re< 104.Vy dng kh trong calorifer chy qu Chun s Nu(tnh cho trng hp lu th chy ngang qua bn ngoi chm ng c gn):

Nu = C 4.014.054.0

*** rn

egg

ng PRbh

bd

{sch QTTB_trang 226 }

Trong : dng : ng knh ngoi ca ng; dng = 0.03 (m) bg : bc ca gn ; bg = 0.01 (m) h :chiu cao gn ; hg = 0.006 (m) C,n : cc i lng ph thuc cch sp xp ng Chn cch sp xp ng l thng hng,nn ta c: C=0.116 , n=0.72

Vy Nu = 0.116 6.2169.0*2600*01.0006.0*

01.003.0 4.072.0

14.054.0

=

-H s cp nhit i lu:

04.68006.0

0315.0*6.21*2 ===

gbNu (W/m2)

H s cp nhit t hi nc bo ho n thnh ng 1 25.01 )*

(**04.2tH

rA = (W/m2)

Vi H=1.0 : chiu cao ng r : n nhit ho hi J/kg.Tra bng I250-s tay QTTB tp 1 r=2208*10 3 J/Kg -H s A c tr s ph thuc vo ttb Chn tT = 1100C:Nhit ti thnh ng truyn nhit

Vy ttb = Ctt ohndT 120

2130110

2=+=+

S dng phng php ngoi suy,tra bng QTTB trang 231 ta c



A = 188 t =130-120=100C Vy thay s vo ta tnh c: 8.2621 = (W/m2) q1 = t*1 =262.8*10 = 2628,KJ/Kgm Tnh h s cp nhit t mt ngoi ng n khng kh chuyn ng trong caloripher 2 -Lu th chy qua bn ngoi ng thnh ng c gn: 04.682 = (Do chn tc dng kh 6=kk m/s ) H s cp nhit i lu thc t: tt2 =32 W/m2 Vy k = 4.22

0785.010241.0

321

8.2621

1111

21

=++

=++

FtrFbm

tt

-Vy nhit lng ring: q2 = k*tTB =22.4*120 = 2688 ,KJ/Kgm

So snh %5%2.2100*2688

26882628100*21



Chn k = 1.2 Suy ra Ft = 1.2*28.3=34(m2) -B mt truyn nhit trung bnh:

Ftb= 09.020785.010241.0

2=+=+ trbm FF (m2)

-S ng truyn nhit trong caloripher:

n = 37809.0

34 ==tb

t

FF

ng

-S ng xp theo chiu ngang:

m = 1920

378 ==in ng

-Chn s ng xp theo hng ngang 19 ng S ng xp theo hng dc l 20 ng Vy kch thc caloripher: +Chiu di ca caloripher Lx = (i-1)*0.05+Dg*20+2x , (m) = (20-1)*0.05+0.042*20+2*0.01=1.81 ,m +Chiu rng caloripher: Bx = (m-1)*0.05+Dg+2x =(19-1)*0.05+0.042*19+2*0.01 = 1.718(m) +Chiu cao caloripher l: Hx =l+2a =1+2*0.1=1.2( m) a:b dy mi tm chn 5.2 Xyclon 5.2.1 Gii thiu v xyclon: Do yu cu v sch ca ch cng nh kh thi ngi ta s dng tc nhn sy l khng kh nng.Trong qu trnh sy khng kh chuyn ng vi vn tc ln nn mt phn ch s theo khng kh ra ngoi. thu hi kh thi v ch,ngi ta t ng ng ra ca khng kh nng mt xyclon tch sch hn. 5.1.2 Tnh ton - nhit 70oC th tch ring ca khng kh l: v70oC = 0.97 (m3/Kg) -Lu lng khng kh ra khi phng sy (vo xyclon) V2 = L*v70oC = 3040.95 (m3/h) -Gi P l tr lc ca cyclon th: 54010 4 vy khng kh chuyn ng theo ch chy xoy Do ng xp theo kiu hnh lang nn

( ) 26.023.0)96(

+= eRdsm {s tay QTTBI trang 404}

Vi s l khong cch gia cc ng theo phng ct ngang ca dng chuyn ng (theo chiu rng ca dng) S=0.005+0.006+0.03/2 =0.026 (m) m l s dy chm theo phng chuyn ng m = 19 d: ng knh ng : d = Dg = 0.049(m) Suy ra = 10.421 Vy tr lc do caloripher l:

9.107294.3*024.1*58.13

2**

22

2 === P (N/m2) 3.Tr lc do t m vo calorifer -Din tch ca mt ct ngang ca ng y

Fo = 22

,07065.023.0* m=

-Din tch ct ngang ca ng dn khng kh nng: Ft = H*Bx = 1.2*1.8=2.16( m2)

T s 032.016.2

07065.00 ==tF

F

Tra bng s tay QTTB I trang 387 ta c 95.0= Vy tr lc do t m vo caloripher l:

45.6526.11*024.1*95.0

2**

22

3 === P (N/m2) 4.Tr lc t thu t calorife ra ng dn khng kh nng -Khng kh nng c nhit t = 100oC 610*13.23 = (m2s) 916.0= (Kg/m3) -Din tch ct ngang ca ng dn khng kh nng



F2 = 07065.043.0*

2

= ,m2 -Vn tc ca khng kh nng trong ng

6.11**3600 2

,'

==F

Lkk (m/s)

-Chun s Reynol:

Re = 46 10*1510*13.233.0*6.11* ==

dkk Re>10 4 :Vy khng kh chuyn ng theo ch xoy

T s 032.01

2 =FF

Tra bng QTTB I trang 338 95.0= Vy tr lc do t thu caloripher l:

5.582

*2

4 == kkP (N/m2) 5.Tr lc ng ng dn khng kh t caloifer n phng sy +Chn ng ng di 2 (m) +ng knh ng d =0.3 (m) -Tnh ton ging ng t ming qut n caloripher ta c: Regh



Tra bng st QTTB I trang 385 ta c 48.0= Vy tr lc l 17.32

26.11*996.0*48.0

2**

22

7 === P ( N/m2) 8.Tr lc ca phng sy Ta c th chn 10008 =P (N/m2) 9.Tr lc ng ng dn kh t phng sy n xyclon ng ng ny c chia lm 2 on: + on 1:t phng sy i ra ,chn ng c ng knh 3(m),di l=1.5(m) Vn tc khng kh trong ng :

22

23.0**3600

=

Vkk = 9.11

23.0*14.3*3600

95.30402 =

(m/s2)

Chun s Reynon :

Re= 46 10*1810*02.203.0*9.11* ==

dkk Re>104 do khng kh chuyn ng theo ch chy xoy

7.53.0*2

9.11*996.0*5.1*0162.0*2

*** 229 === d

lP (N/m2) + on 2:t cui on 1 n xyclon,chn ng c ng knh 3(m),di l=8(m)

5.303.0*2

9.11*996.0*8*0162.0*2

*** 2210 === d

lP (N/m2) 10.Tr lc ng ng dn kh tun hon Chn ng ng c ng knh d=3m,di l=1.5m

7.53.0*2

9.11*996.0*5.1*0162.0*2

*** 2211 === d

lP (N/m2) 11.Tr lc ng ng dn kh t qut ht n xyclon Chn ng ng c ng knh d=3m,di l=4m

2.153.0*2

9.11*996.0*4*0162.0*2

*** 2212 === d

jP (N/m2) 12.Tr lc ti cc khuu T phng sy ra: 1.1,900 == (Sch TT& thit k h thng sy_trang 352) 6.77

29.11*996.0*1.1

2**

2

12 === P (N/m2) 13.Tr lc ti cc chc 3 Ta c F3/F2=1&V3/V2=1 suy ra A=0.6& 0.2'= (Sch St QTTB&CNHC tp1_tr390) Do 2.10.2*6.0'* === A 6.84

29.11996.0*2.1

2**

22

13 === P (N/m2) 14.Tr lc ca xyclon



Chn 59014 =P (N/m2) 15.Tr lc ca ton b h thng 141 ........... PPP ++= P = 2128.11 (N/m2) 5.4 Tnh cng sut ca qut v chn qut 5.4.1 Qut y hn hp kh vo xyclon Lu lng y vo: Q = V = L*vtM = 27588(m3/h) -p sut lm vic ton phn:

H = Hp* kMt *

8.760760*

293273 '+ {s tay QTTB I trang 463}

Vi : Hp : Tr lc tnh ton ca h thng Hp = P =2128.11 (N/m2) tM : Nhit lm vic ca hn hp kh tM = 48.4oC B = 760.8 mmHg : p sut ti ch t qut :Khi lng ring ca ca kh ktc 3/181.1 mKg= k : khi lng ring ca kh k lm vic 08.1=k ,Kg/m3 H = 2128.11* 8.2019

181.108.1*

8.760760*

2934.48273 =+ ,N/m2

-Cng sut trn trc ng c in :

N = trq

P gHQ

**1000*** , KW {st QTTB I trang 463}

Vi 95.0=tr : Truyn ng qua bnh ai q : Hiu sut ca qut ly 72% Q = 181.275 m3/h = 0.050359 m3/s) Q : Nng sut qut:Q=181.275 (m3/h)=0.05035(m3/s)

N = W,57.195.0*72.0*1000

08.1*81.9*72.2015*05035.0 K= -Cng sut thit lp i vi ng c in: Nc = N*k3 ,(KW) Vi k3 l h s d tr N = 1.57 chn k3 = 1.2 {bng II-48 trang 64 ST QTTB I} Suy ra N = 1.57*1.2=1.884,( KW) 5.4.2 Qut ht kh thi xyclon -Lu lng ht: Qh = L*v700C=30409.5 (m3/h) -Qu trnh tnh ton nh trn +p sut lm vic ton phn: H =1987.9,( N/m2) +Cng sut trn ng c in:



N = 1.45,KW +Cng sut thit lp i vi ng c in: Nc = 1.45*1.25=1.8125, KW 5.4.3 Chn qut C hai qut u s dng qut ly tm loi II4.70No4 {s tay QTTB I trang 482}vi cng mt hiu sut 72.0= PHN 6 : KT LUN Sau khi hon thnh xong n gip em tm hiu su hn v k thut sy,c bit l nguyn tc hot ng cng nh cch tnh ton thit k h thng sy.Mc ch cng nh tm quan trng ca thit b sy kiu bng ti trong quy trnh sn xut. V y l n mn hc u tin m em tip xc,phn ti liu tham kho cn hn ch v kin thc cn hn hp.Hn na cc cng thc tnh ton cn mang tnh tng i,nhiu h s t chn c th dn n sai lch kt qu. Tuy nhin,cng vi s gip ca bn b,c bit l s hng dn tn tnh ca thy gio gip em hon thnh n ny Em chn thnh cm n



PHN 7: TI LIU THAM KHO

1. C s cc QT&CNHC tp 2 _NXB H&TH Chuyn nghip 2. K thut sy nng sn _NXB KHKT H Ni _1991 3. S tay QTTB&CNHC tp1 4. S tay QTTB&CNHC tp 2 5. Tnh ton v thit k h thng sy_ Trn Vn Ph_NXB GD_2001



PHN 8 : MC LC

Phn 1: M u Trang 1 Phn 2: S cng ngh sy & thuyt minh Trang 3 Phn 3: Cn bng vt cht Trang 5 Phn 4: Cn bng nhit lng &tnh ton thit b chnh Trang 10 Phn 5: Tnh ton v chn thit b ph Trang 21 Phn 6: Kt lun Trang 33 Phn 7: Ti liu tham kho Trang 34 Phn 8: Mc lc Trang 35

Documents

Đồ Án Môn Học Sấy Băng Tải