N MN HC QT&TBCNHH

N QT&TBCNHH KHOA : HO K THUT

N MN HC QT&TBCNHH

KHOA:HO K THUT

TRNG I HOC BCH KHOA NNG.....(.......(.....( GVHD:PHM NH HA SVTH:TRNG VN THINCHNG 1

M U

I-GII THIU V I TNG SY Cng ngh ch bin thc phm l mt ngnh rng ln v bao gm nhiu qu trnh khc nhau, mi qu trnh ng mt vai tr ring, mt trong s c qu trnh sy. y l qu trnh quan trng trong ch bin v bo qun thc phm.

Sy : l qu trnh lm kh mt vt th bng phng php bay hi. Sy l mt qu trnh truyn cht m ng lc chnh ca n chnh l s chnh lch m trong vt liu. Mc ch chnh ca qu trnh sy l lm gim khi lng vt liu lm gim cng chuyn ch, tng bn v ko di c thi gian bo qun.Sy chia lm hai loi

Sy t nhin:L dng nng lng mt tri lm bay hi nc trong vt liu. Sy t nhin c u im l n gin t tn km nhng ta khng iu chnh c qu trnh sy v vt liu sau khi sy vn cn mt lng m kh cao

Sy nhn to: L qu trnh sy m ta phi cung cp nhit trong qu trnh sy cho vt liu m, phng php cung cp nhit c th bng dn nhit i lu, bc x hoc bng nng lng in trng

Sy thc phm lm cho m ca thc phm thp, b mt ngoi hp, hn ch s pht trin ca vi sinh vt hoc c th tiu dit mt s vi sinh vt trong qu trnh sy, m bo cht lng v v sinh cho thc phm . Nc ta, mt nc nhit i vi rt nhiu cy nng sn, trong s c cy ch.Cy ch c tn khoa hc l Camellia sinensis l loi cy m l v chi ca n c dng sn xut ch. N c ngun gc ng Nam nhng ngy nay n ph bin khp th gii. Ch l cy cng nghip lu nm, thch hp nht vi khng kh vng ni trung du.Cy ch chu c cc iu kin khc nghit v thi tit v th nhng cc vng t cao, nc ta cy Ch ch yu c trng cc tnh trung du pha Bc nh Ph Th ,Thi Nguyn v cc tnh ty nguyn nh Lm ng , Gia LaiCh khng nhng l mt thc ung thng thng m cn c rt nhiu cng dng khc cho ngnh dc phm.Mt khc cac ph liu ca ch dng sn xut cafein.Cc cht mu dung cho ngnh dc v dc phm v th ch xng ng c gi tr v mt thc phm,dc phm v cn xut khu.Vit Nam l mt trong nhng nc trng c ch,tuy sn lng ch cha cao, pht trin ch c c bit ch .Chng ta m thm nhiu nng trng trng ch v xy dng nhiu vng chuyn canh ch ln Vnh Ph,Ngha L,H Tin....

Bn vng trng ch ln Vit Nm:

-Vng thng du vi ch tuyt,ging cy ch mc trung bnh nhng nng xut cao,phm cht ch tt.

-Vng trung du vi ging ch ging to chu c hn v cc loi su bnh,nng xut cng cao v phm cht tng i tt.-Vng ng ng bng bc b,chuyn sn xut ch ung ti

-Vng Ty Nguyn.

II/CHN PHNG N SY V THIT B SY.

C nhiu phng n sy ch, mi phng n sy u c nhng u v nhc im ring ca n.Thit b sy c nhiu loi khc nhau n ph thuc vo tc nhn sy,khng kh nng hoc khi l,ph thuc vo phng thc lm vic,cch cung cp nhit,chiu chuyn ng ca tc nhn sy so vi chiu chuyn ng ca vt liu i vo v mt phn ph thuc vo vt liu un sy.C hai loi sy:-Sy gin on:C nng xut thp cng knh,thao tc nng nhc khng c b phn vn chuyn,nhiu khi khng m bo cht lng sn phm.Thit b sy gin on thng c s dng khi nng xut nh,sy cc loi sn phm c hnh dng khc nhau .-Sy lin tc:Cho cht lng tt hn,thao tc nh nhng hn.Mun sy ch dng toei xp,kch thc ng u c th chu c nhit sy t1=110oC v m cui W=4%c bit l cho nng sut cao th ta dng thit b sy bng ti lm vic lin tc c tun hon kh thi My sy bng ti vi tc nhn sy l khng kh nng.

III/CU TO THIT B SY BNG TI.

Thit b sy gm c mt hnh ch nht trong c hai bng ti v chuyn ng nh cc tang quay,cc bng ny ta trn cc con ln vng xung. Bng ti lm bng li kim loi dy v hai u hi cong m bo ch khng ri xung bng pha di v ri ra ngoi khi bng. Khng kh c t trong caloripher 5.Vt liu sy cha trong phiu tip liu b cun 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 khuy trn do thit b c nhiu bng ti c dng rng ri hn loi ny vt liu t bng ti trn di chuyn n u thit b th ri xung bng ti di chuyn ng theo hng ngc li khi n bng ti cui th vt liu kh c vo bng tho.Khng kh nng i ngc vi chiu chuyn ng ca bng ti hoc i t di ln trn xuyn qua lp vt liu.D qu trnh sy c tt ngi ta cho khng kh chuyn ng vi vn tc ln, khong 3m/s cn bng ti th chuyn ng vi vn tc 0.3-0.6 m/pht.

CHNG 2 PHN 2 : S CNG NGH & THUYT MINH

2.1 S CNG NGH CA QU TRNHVi 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

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.

III/CNG NGH SN XUT CH EN

CC S LIU BAN U

-Nng sut tnh theo sn phm : G = 200(kg/h)- m vt liu vo

: W1 = 62%

- m vt liu ra: W2 = 4.8%

-Nhit sy cho php: t1 = 106oC suy ra p1bh = 1.2778(at)-Nhit ra ca tc nhn sy: t2 = 72oC suy ra p2bh = 0.3478(at)-Cht ti nhit: Hi nc bo ha-Trng thi khng kh ngoi tri ni t thit b sy ta chn nhit l to = 25oC suy ra po = 0.0323(at ) m l

pkq = p = 1.033at

-Hm m ca khng kh c tnh theo cng thc sau:

xo = 0.622

{sch QTTBII trang 156}thay s vo ta c

xo = 0.622= 0.01596(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 Ckkk = 103 j/kg to : nhit ca khng kh to= 26oC ih : nhit lng ring ca hi nc nhit to ; (j/Kg)

ih = ro+ Ch *to = (2493+1.97to)103 ( j/Kg){sch QTTB 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.) Vy Io =103T ta tnh c : Io = 65574.31 (j/Kgkkk )

Hay : Io = 65.57431 (KJ/Kgkk)-Trng thi ca khng kh sau khi ra khi caloripher l:

t1 = 106oC ; pobh = 1.2778(at)Khi i qua caloripher si khng kh ch thay i nhit nhng khng lm thay

i hm m do

==0.02

= 2%-Nhit lng ring ca khng kh sau khi ra khi caloripher l:

I1 = 103t1+(2493+1.97t1) 103x1 (J/Kgkkk) I1 = 149121.05(J/Kgkkk)

= 149.12105 (kJ/Kgkkk) -Trng thi ca khng kh sau khi ra khi phng sy:

t2 = 72oC ; p2bh = 0.34782 (at)-Nu sy l thuyt th : I1 = I2 = 149.12105 (KJ/Kgkkk) Ta c : I2 = Ckkk*t2+x2*ih (J/Kgkkk)T hm m ca khng kh x2 = = (Kg/Kgkkk) x2==0.02926 (Kg/Kgkkk) = 0.1334 = 13.34%

CHNG4

CN BNG VT LIUI-CN BNG VT LIU CHO VT LIU SY

t mt s k hiu:

G1,G2 : Lng vt liu trc khi vo v sau khi ra khi my sy (Kg/s)Gk : Lng vt liu kh tuyt i i qua my sy (Kg/s)

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/s)L : Lng khng kh kh tuyt i i qua my sy (Kg/s)

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)

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 st qu trnh sy.Vy lng vt liu kh tuyt i i qua my sy l: Gk = G1= G2

{sch QTTBII trang 165}Trong : W1 = 62% ;W2 = 4.8% ; G2 = 200( Kg/h.)Vy Gk = 200 * = 190.4 (Kg/h)

Lng m tch ra khi vt liu W c tnh theo cng thc: W = G2 (Kg/h)

{sch QTTB trang 165} W = 200 = 301.05 (Kg/h)Vy phng trnh cn bng vt liu l: G1 = G2+W = 200 + 301.05 = 501.05 (Kg/h)II-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 L*x2 th c phng trnh cn bng:

L*x1 + W = L*x2

{sch QTTB trang 165} L = (Kg/h)Thay s : L = = 22635.34 ( Kg/h)Vi L l lng khng kh kh cn thit lm bc hi W Kg m trong vt liu.Vy lng khng kh kh cn thit lm bc hi 1 Kg m trong vt liu l:

l = = (Kg/Kgm) {sch QTTB trang 166}Khi i qua caloripher si khng kh ch thay i nhit nhng khng thay i hm m, do xo=x1 nn ta c:

l = = thay s vo ta c

l = = 75.188 (Kg/Kgm)III-QU TRNH SY HI LU L THUYT

Qu trnh hi lu c u im l: Tit kim nng lng, to ra ch sy du dng lm tng cht lng sn phm

C hai kiu sy hi lu:

+ Hi lu trc caloripher

+ Hi lu sau caloripher

y ta xt qua trnh hi lu trc caloripher lthi-S :

l lH lo

Qu trnh hot ng ca h thng ny l:Tc nhn sy i ra khi bung sy c trng thi t2, 2, x2 c hi lu li vi lng lH v thi ra mi trng lthi .Khi lng lH c ho trn vi khng kh mi c trng thi l t0,o ,x0, c qut ht v y vo caloripher gia nhit n trng thai t1,x1,1 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 = = 1

{sch k thut sy nng sn -trang79}Vy hm m ca hn hp kh c tnh theo cng thc sau:

xM = { sch QTTBII trang 176} (Kg/Kgkkk)

xM = = = 0.0223 (Kg/Kgkkk)Nhit lng ring ca hn hp khng kh l:

IM = (Kg/Kgkkk)

IM = = 107.98 (Kg/Kgkkk)Ta c:IM = (103+1.97*103xM)tM + 2493*103xMSuy ratM =

Vi tM : Nhit ca hn hp khT :tM = = 50.180C suy ra : pMbh=0.129(at)

= 0.2814 = 28.14%Lng khng kh kh lu chuyn trong thit b syVit cn bng cho 1 thit b sy ta c

LxM + G1W1 = Lx2 + G2W2

L===80672.196 (Kg/h)

l = =196.08 (Kg/Kgm)Lo = LH = = 40336.098 suy ra lo = lH = = 98.04 (Kg/Kg m) TH SY K THUYT C TUN HON KHNG KH THI

I

B

B1

t1=110oDD1C

Mt2=75otM=50.28o

Ato=26o

xo=0.0172 x1=xM= x2=0.0309 x 0.02405 XCHNG 5CN BNG NHIT LNG

I-CC THNG S V THIT B SY1/Th tch khng kh

a/Th tch ring ca khng kh vo thit b sy:

v1= ( m3/Kgkkk ){sch QTTB II trang 157}Vi R=287 (J/Kg.oK)

T1=106+273=379(oK)

P1bh =1.2778(at) ( 1at = 0.981*103 N/m2)

1=0.02Thay s vo ta c:

v1 = =1.153 (m3/Kgkkk)b/Th tch khng kh vo phng sy:

V1 = L*v1 = 22635.34*1.153 = 26098.55(m3/h)c/ Th tch ring khng kh ra khi phng sy l:

Vi R=287 (J/Kg.oK)

T2=72+273=345(oK)

P=1.033(at)

P2bh =0.3478(at)

2=0.1334 v2 = v2 =

v2 = 1.023(m3/Kgkkk)d/Th tch khng kh ra khi phng sy:

V2 = Lv2 = 22635.34*1.023 = 23155.95(m3/h)e/Th tch trung bnh ca khng kh trong phng sy:

Vtb = = 24627.25(m3/h)2/Chn kch thc ca bng ti

-GiBr:Chiu rng lp bng ti (m)

H:Chiu dy lp ch (m) Ly H=0.1(m)

:Vn tc bng ti ; = 0.4 m/pht

:Khi lng ring ca ch Chn

-Nng sut ca qu trnh sy:

G1=Brh60 suy ra Br===0.652 m-Thc t chiu rng ca bng ti:

Btt==0.72 (m) Chon =0.9 Gi Lb :Chiu di bng ti ta c:

ls:

Chiu di ph thm, chn ls=1.2 (m)

T:

Thi gian sy, chn T=30 pht=0.5 gi

Ta chn khong cch gia 2 trc ca hai xch =0.6 (m)

Lb= + ls = =12-Vy Lb=12(m)-Bng ti ch s dng mt dy chuyn nn ta chn chiu di ca mt bng ti l 6 m suy ra s bng ti l 2*ng knh ca bng ti d = 0.3m

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

+Lp cch nhit dy

-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 = 6 + 2*0.6 = 7.2 m

-Chiu cao lm vic ca phng sy:

Hph = 0.3*2 + 0.1*2+0.33*3 = 1.8 ( m )-Chiu rng lm vic ca phng sy:

Rph = 0.724 + 0.6 = 1.32 (m)Vy kch thc ca phng sy k c tng l:

Lng = 7.2 + 2*0.22 = 7.64 (m)

Hng = 1.8 + 0.02 + 0.15 = 1.97 (m)

Rng = 1. 32 + 0.22*2 = 1.76 (m)4- Vn tc ca khng kh v ch chuyn ng ca khng kh trong phng sy:

a/Vn tc ca khng kh trong phng sy:

b/Ch chuyn ng ca khng kh:

Re = {sch QTTB II trang 35} Vi: Re: l hng s Reynol c trng cho ch chuyn ng ca dng ltng knh tng ng

lt = = =1.52( m) Nhit trung bnh ca khng kh trong phng sy:

ttb = = = 89oC

-T nhit trung bnh ny tra bng ph 9 trang 130 sch k thut sy nng sn ta c: 0.03122 (W/moK) 21.46*10(m2/s) Vy Re = = 20.399*10

Vy Re = 22.399*10> 104 suy ra ch ca khng kh trong phng sy l ch chuyn ng xoy5-Hiu s nhit trung bnh gia tc nhn sy v mi trng xung quanh

=

Vi

: Hiu s nhit gia tc nhn sy vo phng sy vi khng kh bn ngoi

106-25=81oC

: Hiu s nhit gia tc nhn sy i ra khi phng sy vi khng kh bn ngoi

=72-25 = 47oC

Vy = 62.5 oCII-TNH TN THT1-Tn tht qua tng

-Tng xy bng gch dy 0.22 (m)-Chiu dy vin gch =0.2 (m)-Chiu dy mi lp va = 0.01 (m) Tra bng = 0.77( w/m) = 1.2 (w/m) Lu th nng chuyn ng thng phng do i lu t nhin(v c s chnh lch nhit ) v do cng bc ( quat) khng kh chuyn ng xoy do Re>10

Gi l h s cp nhit t tc nhn sy n b mt trong ca tng phng sy

= k()

Vi :

l h s cp nhit t tc nhn sy n thnh my sy do i lu t nhin v W/m2

: l h s cp nhit t tc nhn sy n thnh my sy do i lu cng bc n v W/m2

k : h s iu chnha/Tnh

Phng trnh chun Nuxen i vi cht kh:

Nu = CR0.8 = 0.018R0.8

Trong : ph thuc vo t s v Re

Ta c : = 4.74 Re = 20.399*10

Tra bng v tnh ton ta c := 1.14758

{s tay QTTB II tranh 15}

Vy Nu = 0.018*1.14758*(20.399*10)0.8 = 356.378

MNu =

suy ra = =

b/Tnh

Gi tT1l nhit trung bnh ca b mt thnh ng(tng) tip xc vi khng kh trong phng sy

Chn tT1 = 77 oC

Gi ttbk l nhit trung bnh ca cht kh vo phng sy tc tac nhn sy

ttbk =oC

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

ttb = oC

Ti nhit ny tra bng c : = 3.0742.10-2( w/m)

21.393*10(m2/s)

Chun s Gratkov : c 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 =

vi g l gia tc trng trng g=9.8(m/s2 )

Hph Chiu cao ca phng sy

Suy raGr==4.14. 109M chun s Nuxen l

Nu = 0.47*Gr0.25 {s tay QTTB II trang 24}

Suy ra Nu = 119.22

Hn naNu =

suy ra

== 2.04T

Suy ra

q1 = =10.686*(89-79) = 128.23Nhit ti ring ca khng kh t phng sy n mt trong ca tng l 128.23c/Tnh H s cp nhit ca b mt ngoi my sy n mi trng xung quanh

Vi

H s cp nhit do i lu t nhin

H s cp nhit do bc xTrong qu trnh truyn nhit n nh th:

q1=

M

(m2/w)

y :

: b mt dy cc lp tng

: H s dn nhit tng ng

B dy lp va c (w/m)

B dy ca vin gch co (w/m)

Vy

EMBED Equation.3 m2/wT

tT1-tT2=q1 = 128.23*0.276 = 35.44(oC)

tT2: Nhit tng ngoi phng sy

tT2 = tT1 32.79 = 77 35.44 = 41.56oC

Nhit lp bin gii gia tng ngoi phng sy v khng kh ngoi tri

tbg = oC

Ti nhit tbg ny tra bang ta tnh c (w/mk)

(m2/s)Nhit tng ngoi v nhit khng kh c lch l

= tT2 - tkk = 52.02 - 25= 16.56oCChun s Gratkev l

Gr=*1010Chun s Nuxen l

Nu = 0.47*Gr0.25 = 154.57Suy ra

=

H s cp nhit bc x

=

Vi

: en ca va ly 0.9

Co:H s bc x ca vt en tuytt i ly 5.67

T1 = tT2+273=325.02oK

T2 = tkk+273=298 oK

T :

Nn

= 2.09 + 5.86 = 7.95Nhit ti ring t b mt ca tng ngoi dn mi trng khng kh

q2 =

So snh

Cc gi thit trn c th chp nhn c*/Vy tn tht qua tng

Qt=3.6*k*

M

(m2)

(m2)

m2 k =

oC

T :QT = 3.6*2.06*26.4*57.7 = 14215.55 (KJ)

Vyqt= (KJ/Kgm)

2-Tn tht qua trn

Trn c:Lp btng ct dy thep (W/m)

Lp cch nhit dy (W/m)a/Cp nhit t tc nhn sy n mt di ca trn:

Cp nhit do i lu bc x

Nu=0.018*=356.378 Suy ra

Cp nhit do i lu t nhin

Chon nhit trn di l:

Gi ttb l nhit trung bnh gia trn phng sy vi nhit trung bnh ca tc nhn sy.

ttb = oC

tT1=96oC suy ra ( w/moC)

( m2/s)

T1=ttb+273 = 89+273 = 362oK

T2=299oC

Chun s Gratkov

Gr=*108

Chun s Nuxen l:

Nu = 0.47*Gr0.25 = 80.11

ly k =1.33Nhit ti ring

q1=

b/Cp nhit t bn ngoi n mi trng xung quanh

Trong qua trnh truyn nhit n nh th:

q1=

suy ra :tT1-tT2 = q1

suy ra :tT2 = tT1 q1 = 96 96.46*0.622 = 36oCHiu s nhit gia trn ngoi v khng kh:

oC

Nhit bin gii gia trn ngoi v khng kh

tbg = oC

Ti nhit ny tra bng c

EMBED Equation.3 w/moK

(m2/s)Cp nhit do i lu t nhin:

Nu=0.47*Gr0.25

Vi Gr l chun s Gratkov

Gr109Chun s Nuxen l:

Nu=0.47*Gr0.25 = 132.52Vy h s cp nhit do i lu t nhin l:

V b mt trn hng ln trn nn h s cp nhit i lu t nhin tng 30% vy ta c h s cp nhit i lu t nhin thc t l:

Cp nhit do bc x:

T1=tT2+273=36+273=309oK ; ; Co=5.67

T2=298oK

Suy ra

Vy =2.61 + 5.7 = 8.31Nhit ti ring q2 =

So snh

Vy cc gia thit nu trn c th chp nhn c*/Tn tht qua trn l:

Qtr = 3.6*k*

k : l h s truyn nhit

k =

Thay s vo ta c:

Qtr = 3.6*1.23*11.77*62.5 = 3257.35 KJNhit ti ring :

qtr = KJ/Kg m3-Tn tht qua nnNhit trung bnh cua tac nhn sy bng 850C va gia s tng phong sy cach tng bao che cua phn xng 2m

Qn = 3.6*qn*F

q1=53.89W/m

q1: Tn tht qua mt m2 F : Din tch nn Vy thay s vo ta c

Qn = 3.6*53.89*1.32*7.2= 1843.81 (KJ/h)

Nhit ti ring:

qn = (KJ/Kgm)4-Tn tht qua ca

Qc = 3.6*Kc*

Gi l h s cp nhit t khng kh nng n trong

Vi

h s cp nhit do cng bc ly = 6.18(ly gn bng h s cp nhit do i lu cng bc i vi tng)

cp nhit do i lu t nhin

Gi nhit ca trong l tT1=84.5oC

Nhi lp ngn cch

tbg=

Ti nhit ny tra bng c

(W/m)

(m2/s)+Chun s Gratkov

Gr = =

+Chun s Nuxen l:

Nu = 0.47*Gr0.25 = 100.52+H s cp nhit do t nhin l:

Vy h s cp nhit t trong khng kh nng n ca trong l :

ta chn k =1.25Nhit ti ring q1 = KJ/Kgm

Gi l h s cp nhit t c ngoi ra khng kh xung quanh

Ta c:

Trong qu trnh truyn nhit n nh th:

= suy ra

Vi tT2 l nhit ca ngoi ca phng sy

tT2 = tT1-

= tT2 - tkk = 37.73 25 = 12.73 oC

: lch nhit gia ca ngoi phng sy vi khng kh

T2 = 299oK

Gi tbg l nhit ngn cnh ca ca phng sy:

tbg =

Ti nhit ny tra bng c:

(W/m)

(m2/s)

l chnh lch nhit gia lp ngn cnh ca phng sy vi khng kh bn ngoi

Vy

Gr =

Vi t/2 = tbg + 273 = 31.37+273 = 304.37Chun s Nuxen l:

Nu = 0.47*Gr0.25 = 132.77

H s cp nhit do bc x :

Gi T/1 = tbg +273 = 31.73+273 = 304.37

T2 = 298oK

Suy ra

.

= 1.78+5.28 = 7.06

Nhit ti ringq2 =

So snh

Vy cc gi thit c th chp nhn cVy tn tht qua ca l:

Qc = 3.6*kc*

kc =

Qc =3.6*0.77*2.376*62.5 = 297.3

qc =

5-Tng tn tht ca phng sy

(KJ/Kgm)III-QU TRNH SY THC T C HI LU1-Ta phi tnh nhit lng b sung thc t

(KJ/Kgm)

ViCvl : nhit dung ring ca ch ly 1.18(KJ/KgoC)

C : nhit dung ring ca nc ly 4.182 (KJ/Kg)

= 25oC nhit ca khng kh ngoi mi trng ca vt liu trc khi vo phng sy

= 96oC nhit ca vt liu khi ra khi my sy-Vy nhit lng ring ca vt liu l:

qvl = (KJ/Kgm)-Vy

(KJ/Kgm)2-Cc thng s ca qu trnh sy-Hm m ca tc nhn sy i ra khi my sy:

Kg/Kgkkk {s tayQTTBII trang105}

Suy ra

(Kg/Kgkkk)Vy :

(KJ/Kgkkk)- m tng i

-Lnh khng kh kh lm bc hi 1 Kg m ht t ngoi vo:

-Qa trnh sy tun hon kh thi:

9KJ/Kgkkk)-Hm mm ca hn hp khng kh:

(Kg/Kgkkk)-Khi i qua caloripher khng kh ch thay i nhit ch khng thay i hm m do :

x/1 = x/M = 0.0218 (Kg/Kgkkk)

t1 = 100oC

-Vy nhit lng ring cua khng kh sy vo phng sy l:

I/1 = t1 + (2493 + 1.97*t1)*x/1 = 106 + (2493+1.97*106)*0.0218 = 164.9 (KJ/Kgkk)-Lng hi lu thc t:

l/H = l/o = 85.91( Kg/Kgm)-Nhit khi ho trn:

-Ta c th ca qu trnh sy nh sau:

TH SY THC T C HI LU KH THI

B

t1=110oCFe

E C

C1

t2=75oC tM

At1=26oC x/o x/M=x1 x/2 xIV-CN BNG NHIT LNG

Tng nhit ra bng tng nhit vo1-Nhit lng vo-Nhit do caloripher si cp:

qs = l/(I/1-I/M) (KJ/kgm)

qs = 85.91(164.9-107.34) = 4944.98( KJ/Kgm)-Nhit lng do vt liu sy mang vo:

qvls = (KJ/Kg m)

-Nhit lng do khng kh mang vo:

qkkv = lIM = 85.91*107.34 =9221.58 (KJ/Kgm)Vt tng lng nhit mang vo l : (KJ/Kg m)2-Nhit lng ra

-Nhit lng do vt liu mang ra :

qvlr = =56.44 (KJ/Kgm)-Nhit do tn tht ca phng sy:

KJ/Kgm-Nhit do khng kh mang ra:

qkkr = l*I2 = 85.91*149.11 = 12213.6(KJ/Kgm)-Nhit tn tht trong qu trnh sy:

Qt = l(I1-I2) = 85.91(164.9 149.11)

= 1356.52 (KJ/Kgm)Vy tng lnh nhit ra :

(KJ/Kgm)Ta so snh tng lng nhit vo v tng lng nhit ra

Vy cc gi thit v qu trnh tnh ton trn u c th chp nhn c

CHNG 6

TNH TON CC THIT B PHA-CALORIPHER.

Do yu cu v chnh xc ca ch nn phi dng tc nhn sy l khng kh nng . Khng kh nng i qua caloripher si v nhn nhit gin tip t khi l qua thnh ng.

Khng kh dng sy phi c nhit i theo yu cu l 110oC cht truyn nhit l khi l.

Thit b chn l loi ng chm.Khng kh nng i ngoi ng,khi l i trong ng.Hai lu th chuyn ng cho dng.

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 W/m {sach QTTB tp I trang 125}(Chn ng:

-ng knh ngoi ca ng:dng = 0.03 (m)-ng knh trong ca ng:dtr = 0.025(m)-Chiu dy ca ng:

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

-Chiu cao ca gn:h = = 0.006 (m)-Chiu di ca ng:l = 1.2 (m)-S gn trong trn mtj ng:m == 120-B dy bc gn :b = 0.002(m)-Tng chiu di ca gn :Lg=b*m=0.002*120=0.24(m)-Tng chiu di khng gn:Lkg = l-Lg = 1.2-0.24=0.96(m)-Lng khng kh cn thit cho qu trnh sy c hi lu (theo tnh ton thc t): l = 85.91( Kg/Kgm) L = 25863.21( Kg/h)-Nhit ca khng kh ban u khi hi lu :

tM = 50.81oC

-Nhit khng kh sau khi ra khi caloripher l t1=106oC

-Th tch ring ca khng kh

v106oC =

v72oC =

v50.8oC =

v25oC =

vtb =0.9585-Lng khng kh kh i vo caloripher l:

V=L*vtb = 25863.21*0.9585 = 24789.89 (m3/h)-H s cp nhit

+Nhit trung bnh ca khng kh trong caloripher ttb

ttb = thn-

M:

+Chn nhit hi nc khi vo l thnd = 130oC +Chn nhit hi nc khi ra l thnc = 115oCNn ta c:

Thay s vo ta c:

Suy ra :ttb = 130 39.07 = 90.93oC

ng vi gi tr ttb ta c:

(Kg/m3)

(W/moC)

(m2/s)

(Ns/m2)

Pr = 0.69 2-Tnh ton.

(Din tch b mt ca mt ng : (pha trong ca ng)

Ftr = *dtr*l = 3.14*0.025*1.2 = 0.0942( m2)(Din tch mt ngoi ca ng:

Fng = dng*l = 3.14*0.03*1.2= 0.11304 (m2)(Din tch phn b mt ngoi ca mt ng

Fbm = Fgn+Fkgn -Din tch phn c gn

Fgn =

= 0.027446(m2) -Din tch phn khng gn

Fkgn = Lkg*= 0.07536 (m2)Vy :Fbm = 0.027446+0.07536 = 0.1028 (m2)(Chn s ng xp hng l:

i = 30( Khong cch gia cc ng ny ng kia l 0.007(m)( Khong cch gia ng ngoi cng dn caloripher l 0.01 (m)(Din tch t do ca caloripher l Ftd

-Chiu di ca c caloripher l

Lx = 0.01*2+(30-1)*0.007+30*0.042 = 1.483(m) -Din tch tit din ca c caloripher l

Fx= Lx*hcao = 1.483*1.2 = 1.7796 (m2) -Din tch cn ca gn l:

Fcg = Dg*Lg*i = 0.042*0.24*30 = 0.3024 (m2) -Din tch cn ca ng l:

Fcng = dng*Lkg*I = 0.03*0.96*30 = 0.864( m2)Vy din tch phn t do:

Ftd = Fx-Fcng-Fcg = 1.7796-0.378-0.72 =0.6816 (m2) (Vn tc ca khng kh

=10.10 (m/s) ( H s cp nhit t hi nc bo ho n b mt ngang ca ng:

(W/m2) ViH=1.2 : chiu cao ng

r : n nhit ho hi J/kg {Tra bng I250-s tay QTTB tp 1 }

r=2179*10(J/Kg) = 520.1 (Kcal/Kg ) -H s A c tr s ph thuc vo tm

Chn tT = 129.6oC : Nhit thnh ng trong ca ng

Vy ttb =

Tra bng QTTB trang 231 ta c

A = 190.94

-:Hiu s nhit gia nhit hi ngng t v nhit thnh caloripher:

= thn - tT = 130-129.6 = 0.4oC

Vy thay s vo ta tnh c:

(W/m2)

q1 = 7191.86 (W/m2)(Tnh h s cp nhit t mt ngoi ng n khng kh chuyn ng trong caloripher

-Lu th chy qua bn ngoi ng chm ng c gn:

Nu = C {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 xp ng

i vi ng xp hng : C = 0.116 ; n =0.72

-Chun ss Reynol :

Re =

Re > 2300 : Khng kh chy theo ch qu

Pr = 0.69Vy Nu = 0.116

-H s cp nhit i lu:

-H s cp nhit lu thc t:

=55Vy k = -Vy nhit lng ring:

q2 = k* ttb = 7114.34So snh

Vy tt c cc gi thit trn c th chp nhn c3-Xc nh b mt truyn nhit -Lng nhit do caloripher cung cp:

{sch QTTB trang 168}

Trong :

(KJ/Kgm)

(J/Kgm)

Qs = qs*W = 6297.42*301.05 = 1895838.29 (J/Kg) -Hiu sut caloripher ly

-Lng nhit thc t do caloripher cp:

Qt = (J/Kg)

-Gi D l lng hi nc tiu tn trong 1h

Qt = D*r suy ra D = = = 966721.886 (J/Kg)

-Th tch nc tiu hao trong 1h:

(m3/h)

-Lng nhit thc t truyn t hi nc trong ng n thnh ng:

Qt = 3.6*k*F*

Suy raF =( m2)

-B mt truyn nhit thc:

Ft = k*F ; k = 1.2 n 1.5

Chn k = 1.2Suy ra

Ft = 129.7*1.2 = 155.64(m2)

-B mt truyn nhit trung bnh:

(m2)

-S ng truyn nhit trong caloripher:

n = ng

-S ng xp theo chiu ngang:

m = ngVy kch thc caloripher:

+Chiu di ca caloripher

Lx = 1.012 (m)

+Chiu rng caloripher:

Bx = (53-1)*0.007+0.01*2+53*0.042 = 2.61(m)

+Chiu cao caloripher l:

Hx = H+2*a = 1.2 + 0.02 = 1.22( m)B-CYCLON1-Gii thiu v cyclonDo yu cu v chnh xc ca ch,cng nh kh thi ngi ta dng tc nhn l khng kh nng.Trong qu trnh sy, khng kh nng 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 cyclon tch nhiu hn.

2-Tnh ton.

- nhit 72oC th tch ring ca khng kh l:

V72oC = 0.977 (m3/Kg)

-Lu lng khng kh ra khi phng syV2 = L*v72o = 25863.21*0.977 = 25268.36(m3/h)

--Gi l tr lc ca cyclon th:

54010Vy khng kh i trong ng theo ch chy xoy -Chuyn ng chy xoy da lm 3 vng

+Vng 1: Nhn thu lc hc: Khu vc ny nhm khng nh hng n h s ma st

Regh = 6= 6

=10: nhm tuyt i ca tn ( Bng II-15-s tay QTTB-trang 381)

+Vng 2: Khu vc nhm:Khu vc ny h s ma st ph thuc vo nhm m khng ph thuc vo Re

Ren = 220*=220*

Vy

Regh10:Vy khng kh chuyn ng theo ch xoy

T s

Tra bng QTTB I trang 388

Vy tr lc do t thu caloripher l:

5-Tr lc ng ng dn khng kh t caloripher n phng sy

+Chn ng ng di 1.5(m)

+ng knh ng d =0.38 (m)-Tnh ton ging ng t ming qut n caloripher ta c:

Regh