do an say

7/30/2019 do an say

1/49

Trng HCN H Ni Khoa CN Ha 2K3

B CNG THNGTRNG HCN H NI

CNG HO X HI CH NGHAVIT NAM

c lp T do Hnh phc-------o0o----------

N MN HC QU TRNH THIT B

H v tn SV : o Hng Qun

Lp : H Ha 2 Kha 3

Khoa : Cng Ngh Ho

Gio vin hng dn : V Minh Khi

NI DUNGThit k h thng sy thng quay lm vic xui chiu dng sy than vi

nng sut 3540 kg/gi

Cc s liu ban u:

m u ca vt liu: 8.5%

m cui ca vt liu: 2.5%

Nhit khi u vo : 210o

CNhit khi l ra: 80oC

Phn thuyt minh

1.Mi u.

2.V v thuyt minh dy chuyn.

3.Tnh ton thit b chnh.

4.Tnh ton v chn thit b ph

5.Tnh c kh

6.Kt lun

TT Tn bn v kh giy S lng1 v dy chuyn sn xut A4 12 v my sy thng quay A0 1

Ngy giao : 15/11/2010 Ngy hon thnh:15/01/2010

Trng khoa: Gio vin hng dn

Nguyn Xun Tng V Minh Khi

MC LC

1GVHD: V Minh Khi n Mn Hc

7/30/2019 do an say

2/49


Li m u:

Trong ngnh cng nghip ho cht, thc phm, vt liu xy dngth

sy l mt vn rt quan trng . Trong ngnh ha cht vt liu xy dng qu

trnh sy dng tch nc v hi nc ra khi nguyn liu v sn phm. trong

nghnh cng nghip v thc phm , sy l cng on quan trng sau thu hoch.

thc hin qu trnh sy ngi ta s dng cc thit b :bung sy, thng

sy ,hm sy

c thy V Minh Khi giao cho nhim v tnh ton thit k h thng sythng quay vi phng thc sy xui chiu ,sn phm sy l than. Mc d c

gng rt nhiu song vn con rt nhiu thiu st v y l ln u tin lm n

nn c ng cha c kinh nghim. Bn cnh trnh t nghin cu v kh

nng t duy cn b gii hn, nn n ca em khng th trnh nhiu thiu st.

Qua ln lm n ny em knh mong qu thy c ch bao em c th hon

thin tt hn n cng nhu bi tp ln m thy c gio cho em vo nhng lnsau

Em xin chn thnh cm n s gip tn tnh ca thy Gio hng dn,

cc thy c gio v bn b d gip em hon thin n ng thi hn.

Phn 1: Gii thiu chung v my sy thng quay1. 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

sy c th tin hnh t nhin bng nng lng t nhin nh nng lng 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 diu chnh c vn tc

ca qu trnh theo yu cu k thut sy, nng sut thp Bi vy trong ngnh

cng nghip ngi ta thng tin hnh qu trnh sy nhn to bng ngun nng

lng do con ngi to ra.


7/30/2019 do an say

3/49


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 sy vi

khng kh nng khi l (tc nhn sy).

-Sy tip xc: Phng php sy khng cho vt liu sy tip xc trc tip

vi tc nhn sy.

-sy bng tia hng ngoi: Phng php sy dng nng lng ca tia hng

ngoi, do ngun nhit pht ra truyn cho vt liu sy

-Sy bng dng in cao tn: Phng php sy dng nng lng in

trng c tn s cao t nng trn ton b chiu dy lp vt liu

-Sy thng hoa: phng php sy trong mi trng c chn khng rtcao, nhit thp nn m t do trong vt liu ng bng v bay hi t trng

thi rn thnh hi khng qua trng thi lng.

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

l cc phng php sy c bit.

Trong cng nghip ho cht v thc phm, cng ngh v tht b sy i lu v

tip xc c dng ph bin hn c, nht l phng php sy i lu. N cnhiu dng khc nhau v c th sy c hu ht cc dng vt liu sy.Theo kt

cu nhm tht b sy i lu c th gp cc dng sau:

-Thit b sy bung: Nng sut thp, lm vic khng thng xuyn.

-Tht b sy hm: Nng sut sy 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.

-Tht 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.

-Tht b sy tng si: nng sut cao.

2. Gii thiu chung v my sy thng quay


7/30/2019 do an say

4/49


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

My sy thng quay l 1 thng hnh tr t nghing 1- 6o, c 2 vnh ai

, vnh ai ny t vo con ln khi thng quay. Vt liu vo sy qua phu

np liu.Vt liu trong thng khng qu 20 25% th tch thng. Sau khi sy

xong ,thnh phm qua b phn tho sn phm ra ngoi.

Bn trong thng c lp cc cnh xo trn vt liu lm cho hiu sut sy

t c cao hn, pha cui thng c hp tho sn phm cn u thng cm vo

l t hoc ni vi ng to tc nhn sy.Gia thng quay, hp tho v l c c

cu bt kn khng kh nng v khi l khng thot ra ngoi. Ngoi ra cn cxyclone thu hi sn phm bay theo v thi kh sch ra mi trng.

Kh nng v vt liu c th i cng chiu hoc ngc chiu bn trong thng.

Pha u ch np liu bn trong thng sy c lp cc cnh xon 1 on khong

700 1000mm, chiu di ca on ny ph thuc vo ng knh ca thng.

Tc khi l hoc khng kh nng i trong thng khng c >3m/s

trnh vt liu b cun nhanh ra khi thng.Vn tc quay ca thng l 58vng /pht.

Cc m ngn trong thng va c tc dng phn phi va c tc dng

phn phi u cho 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 trn)

Ph thuc vo kch thc vt liu sy v m ca n.

Cc loi m ngn dng ph bin l:

-m ngn mi cho Nng v loi phi hp: Dng khi sy nhng vt liu

cc to, m , c xu hng ng vn. Loi ny c h s cht y vt liu khng

qu 0.1 - 0.2

-m ngn hnh qut c nhng khong thng vi nhau.

-m ngn phn phi hnh ch nht v kiu vt o c xp trn ton b

tit din ca thng c dng sy cc vt liu dng cc nh, xp, khi thng


7/30/2019 do an say

5/49


quay vt liu o trn nhiu ln, b mt tp xc gia vt liu sy v tc nhn sy

ln.

-m ngn kiu phn khu: sy cc vt liu c p nh, bi. Loi

ny ch cho php h s in y khong 0,15 - 0,25

-Nu nhit sy cn ln hn 200oC th dng khi l nhng khng dng

cho nhit >800oC.

*u v nhc im ca thng sy quay

-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/m3h.+thit b nh gn, c th c kh v t ng ha hon ton.

-Nhc im:

+Vt liu b o trn nhiu nn d to bi do v vn. Do trong nhiu

trng hp s lm gim cht lng sn phm sy.


7/30/2019 do an say

6/49


3.Nguyn l hot ng ca my sy thng quay:

15

8 1

2

3

4

5

6

910

11

12

14

3

7

13

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 ti

13.Phu tip liu 14.Van diu chnh 15.Qut thi

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

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

(2).


7/30/2019 do an say

7/49


Bnh ai c t trn bn con ln (3), khong cch gia 2 con ln

cng 1 b (11) c th thay i iu chnh cc gc nghing ca thng,

ngha l iu chnh thi gian 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 nt 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 ph

thuc vo kch thc ca vt liu sy tnh cht v m ca n. Vn tc cakhi l hay khng kh nng i trong my sy khong 2 3 m/s,thng quay 5 8

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.4. La chn thit b.

Theo s liu ban u ca than 25%, qu trnh sy cn thc hin lin tc

vi nng sut ln:3540 kg/h, vt liu dng cc c th t chy c nn la chn

cnh o trn kiu mi cho. Tc nhn sy l khi l v nhit u ca khi l

ln(210oC). Nn chiu chuyn ng ca tc nhn v vt liu sy l xui chiu.

Phn 2. Gii thiu v than

1. Gii thiu chung


7/30/2019 do an say

8/49


Than l mt loi nhin liu ha thch (tn ting Anh l anthracite)

c hnh thnh cc h sinh thi m ly ni xc thc vt c nc v bn

lu gi khng b xiha v phn hy sinh vt (biodegradation). Thnh phn

chnh ca than l cacbon , ngoi ra cn c cc nguyn t khc nh lu

hunh . Than , l sn phm ca qu trnh bin cht, l cc lp c mu en

hoc en nu c th t chy c. Than l ngun nhin liu sn xut in

nng ln nht th gii, cng nh l ngun thi kh carbon dioxide ln nht, c

xem

l nguyn nhn hng u gy nn hin tng nng ln ton cu. Than ckhai thc t cc m thanl thin hoc di lng t (hm l).

2. S dy chuyn

http://vi.wikipedia.org/wiki/Carbon_dioxidehttp://vi.wikipedia.org/wiki/N%C3%B3ng_l%C3%AAn_to%C3%A0n_c%E1%BA%A7uhttp://vi.wikipedia.org/w/index.php?title=M%E1%BB%8F_than&action=edit&redlink=1http://vi.wikipedia.org/w/index.php?title=L%E1%BB%99_thi%C3%AAn&action=edit&redlink=1http://vi.wikipedia.org/w/index.php?title=H%E1%BA%A7m_l%C3%B2&action=edit&redlink=1http://vi.wikipedia.org/wiki/Carbon_dioxidehttp://vi.wikipedia.org/wiki/N%C3%B3ng_l%C3%AAn_to%C3%A0n_c%E1%BA%A7uhttp://vi.wikipedia.org/w/index.php?title=M%E1%BB%8F_than&action=edit&redlink=1http://vi.wikipedia.org/w/index.php?title=L%E1%BB%99_thi%C3%AAn&action=edit&redlink=1http://vi.wikipedia.org/w/index.php?title=H%E1%BA%A7m_l%C3%B2&action=edit&redlink=1

7/30/2019 do an say

9/49


Than

Bi cha

Bng ti

L cao

Bi cha Khi l x l

Sy thng quay

2. Cc thng s ban u

Thit k h thng sy thng quay , phng thc sy xui chiu

Tc nhn sy khi l

Nhit khi l ban u t1 = 210oC

Nhit khi ra thng : t2 = 80o C


7/30/2019 do an say

10/49


Vt liu sy l than

m ban u vt liu khi vo my sy : w1 = 8.5 %

m ra vt liu khi my sy : w2 = 2.5%

Ni tin t thit my l H Ni chn cc thng s sau:

Nhit mi trng : to = 25oC

m tng i : = 80%

Phn 3. Tnh ton

3. Tnh ton v la chn

Trong thit b sy i lu (thng quay) khi l cng nh khng kh nng c

s dng nh l cht va mang nhit va mang m thi ra mi trng. Khi l cth l sn phm tn dng t cc loi l nung kim lai to ra t bung t. Trong

k thut sy ta xem khi l l hn hp l tng v hn th na xem n l mt

lai kh tng ng no . V vy chng ta tnh ton qu trnh chy to ra

khi l v xc nh entanpi v lng cha m ca n c th tnh ton thit b

sy dng khi l.

3.1. Nhit tr ring ca than:Ct = 837+3,7.to +625.x (CT1.48-STT1-153)

Trong :

x - hm lng cht bc x =3% =0,03

to - nhit ca than , chn to = 25 C

G = 837+3,7.25 +625.0,03 = 948,25 (J/kg C )

Thnh phn ca than (s tay tp 2)

C = 82% N= 1,8% W = 7%(thnh phn hi nc)

H = 4,56% S = 4,25% A= 7,6%(thnh phn tro)

O = 3,44%

T s liu trn ta tnh cc thng s lm vic

tr ca nguyn liu :


7/30/2019 do an say

11/49

7/30/2019 do an say

12/49


3.3.1. Entanpi ca hi nc :

ih = ro+ Cn.tl (QTTBT T4-273)

Trong :

tl L nhit ho hi

rn:Nhit lng ring ca hi nc 0 0 C = 2493 (kj/kg)

Ch: Nhit dng ring ca hi nc Ch=1,97 (kj/kg)

vy ih = 2493+1,97.210 = 2906,7 (kj/kg 0 C)

3.4.Tnh h s d khng kh bung t v bung ha

trn.

3.4.1.Nhit lng vo tnh khi t 1kg thanQvo = Q1+Q2 + Q3 (KJ)

Trong :

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

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

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

Ta c:Q1 = G Cn to = 0,96075.25.1= 24,019 (kj)

G: Khi lng 1Kg than

Cn:Nhit dung ring ca than Cn = 0,96075 (kj/kg )

Q2 = Lo.Io. (kj)

Io :Hm nhit khng kh trc khi vo my sy

to Nhit ca mi trng to = 25 o C

o : m tng i ca khng kh o = 80%

Tnh Io:

Hm m ca khng kh m

x =bh

bh

pp

p

.

622,0 (CT10.3a-256-QTTB T2)


7/30/2019 do an say

13/49


x o=76,23.8,06,735

76,23.8,0622,0

= 0,0256 (kg/kgkkk)

vi :


pbh : p sut hi nc bo ho 25 o C

pbh= 23,76torr (bng BL 32-365-QTTBT1)

P: p sut lm vic ca hn hp kh nc P = Pkq= 735,6torr

Io = (0,24+0,47.xo).25+595.xo

= (0,24+0,47.0,0265).25+595.0,0265

= 16,24 (kj/kgkkk) = 67,964 (kj/kgkkk)

Cch 2: Ni suy theo th I-x (QTTBT2-255) :

to Nhit ca mi trng to = 25 o C


x o=0,0265 (kg/kgkkk)Io= 67,964 (kj/kgkkk)

Q2 = 9,4877.67,964

=644,822 (kj/kgkkk)

Q3 = Qc lv .

Trong :

:Hiu sut l t = 0,95Q3 = 28884,168.0,95 = 27436,83 (kj/kgthan)

Tng nhit lng vo bung t l:

Qvo = Q1+ Q2 + Q3

= 24,019+644,822+27436,83 = 27460,849 + 644,822

Nhit lng ra khi bung trn

Qra = Q4 + Q5 + Qm


7/30/2019 do an say

14/49


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

G:khi lng kh khi t 1kg than

Gx = Trlv

Cx: Nhit dung ring ca x: Cx= 0,75kj/kg o C

(Tra s tay T1- 162)Tx :Nhit do x mang ra Tx = 100 o C

Thay s :

Q4 = 315,5100.75,0100086,7

= (kj)

C: Q5 = Gk.Ck.t1

Vi:

t1 : nhit khi l ra khi bung trn t =210 o C

Gk : khi lng ca cht kh trong l

Ck : nhit lng ring ca khi l

Ck=K

OHOHONNCOCOSOSO

G

CGCGCGCGCG2222222222

0. ++++(j/kg o C)

(S tay T2- VII.42-112)

Q5 = (GSO 2 .CSO 2 +GCO 2 .CCO 2 +GN 2 .CN 2 +GO 2 .CO 2 +GH 2 O.CH 2 O)t1

Thnh phn khi lng vo my sy tnh theo 1 kg nhin liu 210 o C:

Khi lng SO2:

GSO 2 = =100.2 S

0,02.S = 0,02.3,625= 0,073 kg/kgthan

Khi lng CO2

GCO 2 = 0,0367Clv = 0,03677.0,464 = 2,586( kg)


7/30/2019 do an say

15/49


khi lng N2:

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

= 0,769. .9,4877 + 0,01.1,547

=7,269.+0,01547 (kg/kgthan)

Khi lng m:

GH 2 O = mH 2 O + .Loxo

=100

9 WHLV ++ .Loxo

= 0265,0.4877,9.100

7918,3.9 ++

=0,423 + 0,251. (Kg/kg than)

Khi lng O2:

GO 2 .= 0,231( - 1).9,4877

= 2,192. - 2,192 (kg/kgthan)

Tnh nhit dung ring ca cc kh 210 o C:

CSO 2 = 0,205(kcal/kgo C) = 0,858 (kj/kg o C)

CCO 2 =0,222 + 43.106 t1

=0,222 + 43.10 6 .210

= 0,231 (kcal/kg o C) = 0,967 (kj/kg o C)

CN 2 = 0,246 + 189.t1

= 0,246 + 189.10-6.210= 0,285(kcal/kg o C) = 1,193 (kj/kg o C)

CO 2 = 0,216 + 166.t1

= 0,216 + 166.10-6 .210

=0,25(kcal/kg o C) = 1,047 (kj/kg o C)

CH 2 O = 0,436 + 119.t1

= 0,436 + 119.10 6 .210


7/30/2019 do an say

16/49


0,460(kcal/kg o C) = 1,926 (kj/kg o C)

Q5 = [(0,073.0,858+2,586.1,03+(7,296.+0,01547).1,193+(2,192.-

2,192).1,047+(0,423 + 0,157.).1,926].210

= 231,2+2411,34. (kj)

C : Qm =10%Qvo = 0,1(27460,849 + 644,822)

= 2746,0849 + 64,4822. (kj)

Qra = Q4 + Q5 + Qm

= 5,315+231,2+2411,34. + 2746,0849 + 64,4822.

= 2982,63 + 2475,822.

3.4.3.Phng trnh cn bng nhit l t than:Qvo = Qra

27460,849 + 644,822= 2982,63 + 2475,822.

24478,219 = 1831.

= 13,369

Vy khi lng khi t 1kg than l:

GSO 2 = 0,073 (kg/kgthan)

GCO 2 = 2,586 (kg/kgthan)

GN 2 = 97,2 (kg/kgthan)

G H 2 O = 3,78 (kg/kgthan)

GO 2 = 27,11 (kg/kgthan)

Vy lng khng kh thc t cn cung cp cho l t l:L = .Lo

= 13,369.9,4877 = 126,84 ( kg kkk/kg than)

4. CN BNG VT LIU

4.1. Lng m bc hi:

Lng m bay hi :


7/30/2019 do an say

17/49


W=2

212

100 w

wwG

(Kg/h) (7.27-QTTB T4-289)

= 5,8100

5,25,8

3540

= 232,131 (kg/h)

Lng vt liu kh:

Gk =

=100

100 22

WG 3540.

100

5,2100= 3451,5 (CT7.22- QTTB T4-289)

4.2. Phng trnh cn bng nhit:

G1= G2 +w

G2 :khi lng vt liu kh i ra khi thit b:

G1= 3540 + 232,131 = 3772,131 (kg/h)

4.3. TNH TON THIT B SY

4.3.1. Th tch thng sy :

Vth=A

W=

32

131,232=7,25 ( m 3 ) (CT10.1-T77 -TK my ho cht)

A : Cng bay hi m( V A =32 40) chn A=32

(Bng VII.3-122-STT2)

ng knh ca thng sy :

Chn ng knh thng Dt=1,2m4.3.2. Chiu di thng:

Xt t lt

t

D

L=(3,5 7) (T79-TK my ho cht)

chnt

t

D

L= 4

Chiu di thng sy Lt = 5m

Tnh li th tch thng sy v cng bay hi m


7/30/2019 do an say

18/49


64

52,1

4

22

=

=

= tt

tt

LDV (m 3 ) (CTVII.51-T121-STT2)

At=tt

V

W=

6

131,232=41,071 (kg/m 3 .h) (CT10.1-TK my ho cht)

4.3.3. Chn chiu dy thng

S=(0,007 0,01) Dt (CT10.15-T83-TKmy ha cht)

Chn S = 0,007.Dt = 0,007.1,2 = 0,0084m

Chn S = 0,01 (m)

Thi gian sy:

Ta c:)(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 =800 kg/m 3 (Bng 1.1-8-STT1)

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

+ : H s cha y, = 0,15 (Bng VII.5-122-STT2)

+ A : Cng bay hi m, A = 35(kg m/m3.h) (Bng VII.3-122-

STT2)

=s =+ )]5,25,8(200[35)5,25,8(80015,0120

8,707(pht)

Kim tra li thi gian sy:

Ta c th tch vt liu vo trong 1 pht l:

U= =60.

1

G

60.1350

131,3772= 0,0466 kg/pht (CT10.7-T81-TKmy ha cht)

(Theo 408- my v thit b sn xut ho cht)

Thi gian sy thc t s l:


7/30/2019 do an say

19/49


=U

Vtt . =0466,0

15,0.5= 18,21 (pht) (CT10.6-T81-TKmy ha cht)

4.3.4. 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 :

+ : Gc nghing ca thng quay, Thng gc nghing ca thng di l 2,5 3

cn thng ngn n 6, chn = 3

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

(Bng VII.4-T122-STT2)

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

(Bng VII.5-T122-STT2)

+ : Thi gian lu li ca vt liu trong thng quay, pht

n= =

otg3.21,18.2,1

57,01

3,6679 (Vng/pht).

V n = 3,6679tD

4(

tD

8 ) (Theo 10.3-T79-TK my ho cht)

vy chn Dt = 1.2 v Lt = 5m l hp l

4.3.5. Cng sut cn thit quay thit b:

Ta c: xtt naLDN = 321013,0 (VII.54-trang123-STT2)

Trong :

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

+ : H s ph thuc vo dng cnh,Chn loi cnh phn phi dng cnh nng)

Chn = 0,026 (theo bng VII.54-123/STT2)

+ x : Khi lng ring xp trung bnh, x = 800 kg/m 3

(Bng1.1-5-STT1)


7/30/2019 do an say

20/49


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

N= 028,1800.6679,3.026,0.52.11013,0 32 = (Kw)

Cng sut chn ng c:

Ndc =1,25.1,028 = 1,285 (kw)

4.3.6 . Nhit tn tht ra mi trng qua lp cch nhit:

W

tFKq tb

=

...6,3(Kj/kg m)

F: Din tch bao quanh thng sy m2

W: Lng m bc hi W= 232,131 ( kg/h)K: H s truyn nhit W/m2oC

21

11

1

++

=K(T3- s tay qu trnh v thit b T1 )

1 : H s cp nhit gia cc tc nhn sy n v thit b (W/m2 oC)

2

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

2 o

C) : dy thnh thit b

a. Xc nh 1 :

1 = )(//

1

/

1 +k (W/m2 oC)

k: H s nhm k = 1,2 1,3 chn k=1,2

/1 : H s cp nhit t tc nhn sy n thnh thit b,ph thuc vo ch

chuyn ng ca dng kh (W/m2)

//

1 :H s cp nhit t khng kh n thnh thit b (W/m2)

* Tm /1 :

Chn tc ca dng kh vo wtb = 2(m/s)

tDtbw .

Re = (CT V.36-13-STT2)


7/30/2019 do an say

21/49


: nht ng ca kh

Dt : mg knh ng thit b(m)

Nhit trung bnh ca dng kh:

=tbt 1452

80210

2

21 =+=+ ttoC

Ti t= 145`oC

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

h s dn nhit 210.53,3 = (W/ m oC)

nht ng hc 610.37,28 = (m2/oC)

4

610654,84596

10.37,28

2,1.2Re >==

Ch chy xoy

uN =0.018. .Re

8.0 (T4.31-QTTBT1-T 198)

:H s ph thuc vo 4=t

t

D

L

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

=1,15

Nu 073,18115,1.)64,84596.(018,08,0 ==

m

tu

DN

./1= (QTTBT1-T 196)

/1 = 33,52.1

10.53,3.073,181.2

==

t

u

DN (W/ 2m C0 )

Tnh //1 :

Ta c chun s

uN =25,0

.47,0 Gr (QTTBT1-T 206)

Gr: Chun s Grashof


7/30/2019 do an say

22/49


2

3 ...

tDgGr

= (CT V.39-T13-STT2)

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

Gi s nhit trung bnh ca thit b l 95 C0

Nhit lp ngn cch l

tl = 295145 +

= 120 0 C Ti 120 0 C Tra bng (I.255-318-STT1)ta c:

= 3.34.10-2 (w/moC)

= 25,45.10-2( sm /2 )

Ta c:251201451 === tt tb 0 C

T 418273145273 =+=+= tb K0

7

26

3

2

3

10.531,156418.)10.45,25(

25.2.1.81,9... ==

=

tDgGr

Vy Nu 47,93)10.531,156.(47,025,07 ==

Mt khc:

//1tu

DN = 6,2

2,1

10.34,3.47,93. 2//1 ==

=

tt

u

D

(Wm2 oC)

Vy 516,9)6,233,5(2,1)( //2/

11 =+=+= k (W/m2 oC)

b. Xc nh 2 :

//

2

/

12 += (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

tmt C025=

Nhit thnh thng ngoi cng chn tng =45 oC

=+=+

= 24525

2

ngmt

bg

ttt 35

o

C


7/30/2019 do an say

23/49


tbg = 35oC tra bng I.255 (T-318 s tay qu trnh v thit b T1)

210.715,2 = W/ m C0

610.48,16 = m 2 /s

=

.

..2

3

tDgGr

ng

ngD :ng knh ngoi ca thit b

Thng c cu to gm 3 lp :

+ Lp v bo v,lp cch nhit v b dy thng.Vt liu lm thng l thp CT5

c 50= (W/ 2m C0 )

+

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

n

Gi thit lp bo v dy b1=8mm = 0,008m

B mt lp cch nhit : b 2 =50 mm = 0,05m

Lp trong cng : b3=10 mm = 0,01m

Chn vt liu cch nhit l amiang c h s dn nhit l 2 =0,144(W/m2 oC)

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

ng knh ngoi ca thit b

D ng= Dt + 2(b1+b2 +b3)

=1,2+2(0,008+0,05+0,0112)=1,336 (m)

T=25+273=298 oK

202545 ==t oC

8

26

3

10.81,57298.)10.48,16(

20.336,1.81,9== Gr

M 25,0.47,0 GrNu =

6,129)10.81,57.(47,0 25,08 ==uN


7/30/2019 do an say

24/49


634,2336,1

10.715,2.6,129. 2/2 ===

ng

u

D

N (W/m2 oC)

+ Xc nh //2 theo cng thc:

//

2 =

21

4

2

4

10

100100..

TT

TCn

(W/ 2m C0 )

Co 76,5= (Wm2 oK) H s bc x ca vt en tuyt i

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

quanh

T1 =45+273=318oK

T 2 =25+273=298oK

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

//2 =

298318

100

298

100

318.76,5.95,0

44

=6,402(W/ 2m C0 )

036,9402,6634,2//2/

22 =+=+= (W/2m C0 )

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

21

11

1

++

=K=

22

2

1

31

1

11

1

++

++

bbb

K=036,9

1

144,0

05,0

50

01,0008,0

516,9

11

+++

+ =1,635 ( )2336,1.14,3.26.336,1.14,3 +

Xc nh b mt trao i nhit

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

=3.14.1,336.6+4

336,1.14,3.2 2=27,97 (m2) (420-Thit k my ho cht T1)


7/30/2019 do an say

25/49


+ Xc nh tbt

tbt =n

n

1

1

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

18525210011 === tt oC

= t1 to = 80 25 = 45 oC

tbt =3,107

55

185lg3,2

55185=

( oC)

Vy nhit tn tht ra ngoi mi trng l

W

tFKq tb

=

...6,3= 7,54

131,232

3,107.97,27.635,1.6,3= (Kj/kg m)

5. Cn bng nhit lng trongthit b sy5.1.Nhit lng vo.

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 = m = 3540(kg/h) = 0,98 (kg/s)

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

C1= 1,3 (kj/kg o C) (STT1-162)

to : Nhit ca mi trng =25oC

Q1 = 0,98.1,3.25 = 31,85 (kj/s)

Nhit lng mang vo thit b chnh l nhit lng ra khi bung t.Vy Q2 l

nhit lng do khi l mang vo khi sy :

Q2 = n.Q5


7/30/2019 do an say

26/49


Vi Q5 = 231,23 + 2411,34. (kj/h)

= 13,369

Q2 =( 231,23 + 2411,34.13,369)n = 32468,43.n (kg/h)

= 32468,43.n (kg/h) = 9,02.n (kj/s)

Vy tng nhit lng vo my sy: Qvo = 31,85 + 9,02.n (kj/h)

5.2. Nhit lng ra khi my sy:

Qra = Q ' 1 +Q ' 2 +Q ' 3 + Q ' 4

Trong :

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

Q ' 1 = G2.C'

1.t1

C '1 : Nhit dung ring ca vt liu ra khi my sy:1,09(KJ/Kg)

Q ' 1 =0,98.1,09.80=85,46 (kj/s)

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

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

Trong :

W: Lng m bc hi: 232,131(Kg/h)=0,0645(Kg/s)

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

to =25 oC Cn1 = 0,99892 (kcal/kg Co ) = 4,182 (kj/kgoC)

t2 =80oC Cn2 = 1,00294 (kcal/kg Co ) =4,1991 (kj/kgoC)

(Tra bng I.149 STT1 168)

r : nhit ho hi ca hi nc (ni suy theo bng I.212245STT1)c:

r = 559 (kcal/kg) = 2340,42 (kj/kg)Vy :

Q '2 = 0,0645.[4,182(100 25) + 2340,42 + 4,1991(80 100)]

= 167,222 (kj/s)

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

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

Trong :


7/30/2019 do an say

27/49


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 = 0,073 (kg/kgthan)

GCO 2 = 2,586 (kg/kgthan)

GN 2 = 97,2 (kg/kgthan)

G H 2 O = 3,78 (kg/kgthan)

GO2

= 27,11 (kg/kgthan)Nhit dung ring ca cc kh t2=80 Co l:

CSO 2 = 0,19(kcal/kgo

C) = 0,795 (kj/kgo

C)

CCO 2 =0,222 + 43.106 .t2

=0,222 + 43.10 6 .80

= 0,225 (kcal/kg o C) = 0,924 (kj/kg o C)

CN 2 = 0,246 + 198.t2

= 0,246 + 198.10-6.80

= 0,247(kcal/kg o C) = 1,034 (kj/kg o C)

CO 2 = 0,216 + 166.t2

= 0,216 + 166.10-6.80

=0,229(kcal/kg o C) = 0,958 (kj/kg o C)

CH 2 O = 0,436 + 119.t2

= 0,436 + 119.10 6 .80

= 0,445(kcal/kg o C) = 1,855 (kj/kg o C)

Vy :

Q '3 = (2,586.0,942+3,78.1,855+0,073.0,795+97,2.1,034+ 27,11.0,958).80.3600

n

=3,022.n (kj/kg o C)


7/30/2019 do an say

28/49


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

Q '4 = q.w = 54,7.0,0645 = 3,513 (kj/s)

Vy Qra = 85,46 + 167,22 + 3,022n + 3,513

= 256,193 + 3,022.n (kj/s)

5.3. Phng trnh cn bng nhit ca thit b sy.

Qra = Qvo

256,193 + 3,022.n = 31,85 + 9,02.n

224,343 = 5,998.n

n = 37,4 (kg)

Ly n = 37 (kg)

5.4. Trng thi ca khi l vo v ra khi my sy khi i vo v i ra khi

my sy.

* Hm m ca khng kh 210 o C tnh theo 1kg khng kh kh

x1 =kk

OH

m

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

Trong :

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

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

= 1 +13,369.9,4877 3,78

= 124,06 (kg/kg kkk)

x1 = 06,124

78,3

= 0,0305 (kg/kgkkk)

* Tnh hm nhit ti t1 = 210 o C

I1 = (0,24 + 0,47x1).t + 595x1 (CT10.4/T256/QTTBT2)

Thay s ta c : I1 = (0,24 + 0,47.0,0305) + 595 0,0305

= 71,56 (kcal/kgkkk)

Trng thi ca khi l trc khi vo my sy

t1 = 210 o C


7/30/2019 do an say

29/49


x1 = 0,0305 (kg/kgkkk)

I1 = 71,56 (kcal/kgkkk) = 299,61 (kj/kgkkk)

Nhit lng tiu hao thc t trong thit b sy:

= l .(I2 I1)= 4,18.to qvl q (kj/kg m) (STT2 389)

qvl : Nhit tiu hao nung nng vt liu

4,18.to Nhit m ca vt liu mang vo.

q :Nhit u vo ca vt liu sy

* 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 ra khi my sy chn t'

2 = 80 o C

C2 :nhit dung ring ca vt liu sau sy C2 = 1,3 (kj/kg o C )

G2:khi lng vt liu sau khi sy G2 = 3540 (kg/h)

Vy

qvl =131,232

)2580.(3,1.3540 = 1090,376 (kg/h)

Thay s ta c:

= 4,18.25 1090,376 54,47 = - 1040,576 (kj/kg m)

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

I2 = (0,24 + 0,47 x2).t2 + 595x2 (kcal/kgkkk)

Thay s :

I2 = (0,24 + 0,47 x2).80+ 595x2

=19,2 + 632,6x2 (kcal/kgkkk) =80,39 +2648,57x2 ( kj/kgkkk)

Mt khc:

=12

12

xx

II

(Theo CT10.8a-QTTBT2-258)


7/30/2019 do an say

30/49


= 0305,061,29957,264839,80

2

2

+

x

x= - 1040,576 (kj/kgm)

x2 = 0,0508 (kg/kgkkk)

T ta suy ra:I2 = 80,39+2648,57.0,0508 = 214,94 (kj/kgkkk)

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

t2 = 80 o C

x2 = 0,0508 (kg/kgkkk)

I2 = 214,94 (kj/kgkkk)

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

l =12

1

xx = 0305,00508,01

= 48,5(kg/kgkkk)

* Tng lng kh kh cn thit :

L = w.l = 232,131.48,5 = 11258,35 (kg/kgkkk)

Lng nhit tiu hao ring:

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

=48,5.(299,61 69,696) (kg/kgkkk)

= 11150,83 (kj/kg m)

* Lng kh a vo my sy:

Lm

mmm

kk

kkOH.2

+= (QTTBT2-317)

= 06,12406,12478,3 + .11150,83= 11490,59 (kg/h)

* Lng kh i vo bung t 25 o C:

V25 =kk

G

1

(m 3 /h)

=185,1

59,11490= 9922,79 ( m 3 /h) = 2,75 ( m 3 /s)

Vi:


7/30/2019 do an say

31/49


G1: Khi lng kh i vo my sy (kg/h) G1 = 11490,59 (kg)

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 210 o C:

V210 =kk

G

1

(m 3 /h)

Ni suy theo bng I.255-318-STT1) 210 o C ta c :

kk = 0,7316 (kg/m 3 )

Vy: V210 =7316,0

59,11490= 15706,11 (m 3 /h) = 4,363(m 3 /s)

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

V80 =kk

OHGG

21

+(m 3 /h)

=029,1

78,359,111490 += 11494,37 (m 3 /h) = 3,193(m 3 /s)

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

Nhit lng tiu hao ring :

q = l(I2-I0) = 48,5 .(219,94 69,696)

= 7044,33 (kj/kgm)

* Lu lng th tch trung bnh qu trnh sy:

Vtb = 0,5(V210 + V80) = 0,5.(4,363 +3,193) = 3,778 (m3

/s)* Tc ca khng kh trong thit b sy:

kk =t

tb

F

V= 961,0

778,3= 3,931 (m/s)

Trong : Ft = 0,0785.1.Dth 2 (1 - )

= 0,0785.1.2 2 (1 0,15) = 0,961(m 2 )


7/30/2019 do an say

32/49


PHN III: TNH TON C KH

I:Kim tra bn cho thng quay1: Trng lng vt liu nm trong thng.

Gvl=60

.. gGl =60

21,18.81,9.59,11490=34211,34 (N)

(Hng dn thit k my ha T1-89)

G: Khi lng vt liu vo my sy

: Thi gian syg : gia tc trng trng

2: Trng lng ca thng

=

ttn

t LDD

Q4

)( 22g +

gL

DDt

ttnt

4

)(22

=(D 2n D2th + D

2nt D

2

tt) . 4

Lth. .g

= 1,2162 1,22 + 1,3362 1,3162).4

.5.7850.9,81 = 27713,16 (N)

Trong vt liu lm thng l thp CT5 c khi lng ring l 7850 (kg/m3)

Dnt: ng knh ngoi ca thng, Dnt=1,336 m

Dth: ng knh trong cung ca thng , Dth=1,2 m

Dn: ng knh ngoi st lp cch nhit ca thng , Dn= Dth+2.b1

=1,2+2.0,008 =1,216 (m)


7/30/2019 do an say

33/49


Dtt: ng knh trong ca thng :Dtt = Dth+ 2b1 + 2b2

=1,2+2.0,008+2.0,05 = 1,316 (m)

3:Trng lng ca vnh ai:

ng knh ca vnh ai chn s b

ntv DD )2,11,1( = =1,4961,6032 (m)

Chn Dv=1,5m)

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=0,2

4

)216,15,1.( 22 =

vQ . 81,932581,9.7850.2,0 = (N)

4:Trng lng ca bnh rng vng :

rQ =

4

).( 22 nr DD rb . g.

Chn Dr=Dv

Qr=Qv=9325,81 (N)

5:Trng lng ca lp bo n

Chn vt liu cch nhit l bng thy tinh c 200= kg/m3

Qbo=4

).( 22 ntt DD . gL bot ..

= )616,1716,1.(785,0 22 .5.200.9,81=1949,86 (N)

6:Trng lng ca cnh mc nng :

Chn Qc =4000 N

Vy tng trng lng ca thng l

Q= Qvl + Qth+ Qv+ Qr+ Qc

Qbo = 34211,34+27713,16+9325,81+9325,81+4000+1949,86 =86525,98 (N)

+Khong cch gia hai vnh ai:


7/30/2019 do an say

34/49


ld= 0,586.Lth=0,586.5=2,93 (m) = 239 (cm)

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

q=t

r

L

QQ =

500

86,194998,86525 =169,15 (N/cm)

+Mmen un do ti trng gy ra

M1=8

.. 2dlq=

8

293.15,169 2= 1815169,8 (N/cm)

+ Mmen un do bnh rng vng gy ra

M2=4

. dr lQ =4

293.81,9325= 683115,6 ( N.cm)

+Tng mmen un

Mu = M1+ M2 = 1815169,8 +683115,6 = 1132054,2(N.cm)

+ Mmen chng un ca thng :

W= .4

. 2DS= 0112,0.

4

2,1. 2=0,0127 (m 3 ) = 12700 (cm 3 )

S chiu dy thng sy =0,0112 m

+ng sut thn thng

138,8912700

2,1132054===

W

Mu (N/cm 2 )

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

Vy vi chiu dy thng l S = 0,0112(m) th thng bn

II . Tnh vnh ai

+ Ti trng trn mt vnh ai

36,433223cos.2

98,86525

cos2

/ ===

QQ (N)

Q: Ti trng ca thng

: Gc nghing ca thng =3 o

+ Phn lc con ln


7/30/2019 do an say

35/49


T =cos2

/Q( T85 - Hng dn thit k my ha T1)

Thng thng chn 030

=

17,25012

2

3.2

36,43322==T

(N)

+ B rng vnh ai

Pr

Pr : Ti trng ring tnh cho mt n v chiu di vi thng quay chm Pr=2400N/cm

42,102400

17,25012= cm = 104,2 mm

ng knh vnh ai Dv =2cm

Chn b rng vnh ai B= 200 mm

Vi thng nng th b dy vnh ai l

h 92,766,2

200

6,2==

= (mm)

* Kim tra

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

A: H s ph thuc cnh lp

A = 0,08 0,09

Chn A=0,08

Mu = 25012,17.200.0,08 = 400194,72 (N/cm)

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

[ ]654,25

15600

72,400194===

uMW ( 3cm )

Mt khc


7/30/2019 do an say

36/49


6

. 2hW

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

h 774,220

654,25.66

== BW

cm =27,74 (mm)

Vy h=76,92 >27,74 (mm)

Vy vnh ai bn

Chn vnh ai c tit din : Bh = 200 100 mm

III : Tnh con ln chn con ln

1 .Tnh con ln

B rng con ln c tnh theo cng thc

Bc = B + 50 = 200 + 50 = 250(mm) = 25 (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)

335,325.300

17,25012=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 = Dv +2h = 1,5 + 2.0,07692 m =1,654m = 165,4 cm0,25.165,4 dc 0,33.165,4

41,35 dc 54,58

Vy chn ng knh con ln dc = 50 (cm)

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

6,1025

20

17,20512===

B

TP N/cm

+ ng sut tip xc tnh theo cng thc


7/30/2019 do an say

37/49


Rr

rREP

+= ...418.0max (N/cm

2 )

R: Bn knh ngoi ca vnh ai R = 7,822

4,165= cm

r: Bn knh con ln r = 252

50 = cm

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

25.7,82

257,8210.75,1.6,1025.418.0 7max

+= = 12781,13 (N/cm 2 )

ng sut tip xc cho php ca CT5 l [ ] 5CT = 6.10 6 (N/cm 2 )

[ ] 6000013,12781 5max =

7/30/2019 do an say

38/49


R

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

R : Bn knh con ln chn

R= 2020

40 = 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

87,709720

10.75,1.53,329.418.0

7

max == (N/cm2 )

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

(TK my ha cht)

IV) Tnh ton h thng bnh rng dn ng

1) Chn ng c

Ta chn loi ng c nhn hiu 4A100L8Y3 (do lien x c ch to):

N c = 1,5 kw nn ta chn

S vng quay n = 698 vng/pht, hiu sut = 0,74, cos = 0,65

(bng P.13-tnh ton thit k dn ng c kh t1-238)

2) T s truyn v s vng quay

* T s truyn ca ton b h dn ng l

6679,3

698==

thung

ch

n

NU = 190,3 190

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

v ng c l Uh=40

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

75,440

190===

hp

hdd

U

UU (CT323-48-TTHDCKT1)

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

4,176679,3.75,4. === thngddbr nUN (vng/pht) 18(vng/pht)


7/30/2019 do an say

39/49


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

* Cng sut trn trc ca bnh rng nh l

cNN .1 =

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

= hp .01.ms

hp bnh rng =0,9

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

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

85,096,0.99,0.9,0 ==

Cng sut trn trc bnh rng nh l

N1 = 0,85.1,5 = 1,3 (kw)

* Momen xon tc dng ln trc ca bnh rng l:

56

1

6

1 10.9,618

3,1.10.55,9.10.55,9 ===brn

(N/mm)

4) Tnh ton b truyn bnh rng tr rng thng

4.1 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 =6,9.10 5 (N/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

u: T s truyn ca cp bnh rng dn ng thng quay

[ ]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 rm : 241 285


7/30/2019 do an say

40/49


KHP:H s phn b khng u ti trng trn chiu rng v rng.Chn KHP=1,06

ba=0,3

bd=5,3. ba (u 1) = 0,53.0,3.(4,75+1) = 0,91Du (+) i vi rng n khp ngoi

Du(-) i vi rng n khp trong

Vy aw = 49,5.(4,75+1). 8,3703,0.75,4.8,481

06,1.10.9,62

53 = (mm)

4.2 Xc nh thng s n khp

* Modun bnh rng

m =(0,010,02)aw =3,708 7,416 (mm)Chn modun theo bng IX-99 (Tnh ton h thng dn ng c kh)

m=4 mm

*S bnh rng nh:

z1 =( )1.

.2

+um

aw=

( )25,33

1756,4.4

8,370.2=

+chn z1 = 33 (vng)

* S bnh rng ln :

z 2 =u.z1 =4.33 = 132 (vng)

S rng tng l : z t=z1 +z 2 =33+132=165 (vng)

* Tnh li khong cch trc

a w = 2642

132.4

2

.==t

zm(mm)

4.3 Tnh rng v bn tip xc

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

( )[ ]H

ww

H

mHHdub

uKTzzz

+=

2

1

..

1..2...

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


7/30/2019 do an say

41/49


zH :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=

ZHw

2sin

cos.2= =

20.2sin

0cos.2=1,76

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)

= [1,88 3,2(132

1

33

1+ )]cos = 76,10cos.

132

1

33

12,388,1 =

+

=1,76 1,8

856,03

8,14

== Z

* ng knh bnh rng nh

1

.2

+=u

ad ww = 83,91175,4

264.2=

+mm

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

T1 :Momen xon tc ng ln bnh rng 1. T 1 =24.10 5

KH : h s ti trng khi tnh v tip xc

KH=KHP.K H .KHV

:HBK H s phn b khng u ti trng HBK =1,06

HK =1 ( bnh rng thng )

HVK = HH

wwH

KKT

dbV

...2

..1

1

+


7/30/2019 do an say

42/49


Vi VH=u

avg wF ... 0

H =0,006 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 =100

077,060000

16.83,91.

60000

.. 1 === nd

V w m/s

344,075,4

264.077,0.100.006,0 ==HV

* wb :Chiu rng bnh rng nh

wb = 2,79264.3,0. ==wba a mm

01,106,1.10.24.2

83,91.2,79.344,01

5=+=HVK

01,1.1.06,1== HVHHPH KKKK = 1,071

( )2

1

..

1..2...

ww

HmHH

dub

uKTzzz +=

= 5,17983,91.75,4.2,79

071.1).175,4.(10.24.2.856,0.76,1.274

2

5

=+

MPa

Vy [ ]HH MPa = 5,179 =390 MPa

Nn bn tip xc c m bo

4.4 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

..

.....2

F

ww

FF

F

mdb

YYYKT

=

(Tnh ton h thng dn ng c kh)

Tnh KF


7/30/2019 do an say

43/49


KF:H s ti trng

KF=KFP.K F .KFV

KFP:H s phn b khng u KFP=1,14

K F :H s k n s khng u ti trng K F =1,22

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

KFV =FF P

wwF

KKT

dbV

...2

..1

1

+

Vi VF=

u

avg wF ... 0

Tra bng 016,0=F

G 0 = 100

VF= 92,075,4

264.077,0.100.016,0 =

KFV = 122,1.14,1.10.24.2

83,91.2,79.92,01

5=+

KF=1,14.1,22.1=1,39

Y :H s k n nghing ca rng

0= Y =1

Y :H s k n s trng khp ca rng

Y =17,1

856,0

11 ==Z

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

1FY = 4

2FY = 3,6

Thay s vo ta c

5,10964.83,91.2,79

1.4.39,1.17,1.10.24.2 5

1

==F

MPa


7/30/2019 do an say

44/49


Nhn xt 1F [ ] 2361 =< F MPa

758,114

6,3.064,13.

1

212 ===

F

FFF

Y

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 =264

ng knh vng chiad1 = m.z1 = 4.33 = 132 (mm)

d2 = m.z2 = 4.132 =528 (mm)

ng knh nh rng

da1 = d1 + 2.m =132 + 2.4 = 140 (mm)

da2 = d2 + 2.m = 528 +2.4 = 536 (mm)

Chiu rng vnh bnh rng nh

b1 = bw = 79,2 (mm)

Chiu rng vnh bnh rng ln

b2 = aw.0,2 =264.0,2 = 52,8 (mm)

ng knh y rng

df1 = d1 2m = 132 - 2.4 = 124 (mm)

df2 = d2 2m = 528 2.4 = 520 (mm)


7/30/2019 do an say

45/49


PHN V: CC THIT B PH

I ) TNH TON L T

*Nhit chy ca than

+ Nhit chy cao ca than.

Qcao=28884,168 (KJ/KG)

+ Nhit chy thp ca than.

Qthp=27790,118(KJ/KG) = 6637,6 (Kcal/KG)

* Lng khng kh cn thi vo l t

+V25 = 9922,79 ( m 3 /h)* Th tch bung t

v

lv

th

q

QV

.= ( 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 =260.10 3 (kcal/m 3 h)

310.260

37.6,6637=V =0,945 ( m 3 )

Din tch ghi l:

r

QF t

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

r=550.10 3 (W/m 2 )

Vy din tch ghi l l :

310.550

18,4.37.6,6637.28,0=F = 0,523 ( m 2 )

Chiu cao l:F

Vh = =

523,0

945,0= 1,8 (m)

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:


7/30/2019 do an say

46/49


183,34045,0.34,0

523,0

045,0.34,0===

Fn tm 34 tm

II ) Qut thi vo my sy

Lng khng kh a vo my sy 25 C0

V25=9922,79 ( m 3 /h)

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

.102.3600

.HQN =

Q: Nng sut qut Q=17181,52 ( m 3 /h)

:Hiu sut thy lc =0,6H: Tng tr lc cn khc phc ( mm H 2 O)

H i= = 1 + 2 + 3

1=m.2

.150

S

( T-107, L cng nghip)

S:Din tch ghi l F = 0,523 ( m2 )

: Lng than t :37 (Kg/h)

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

523,0.150

37

=8,897 ( 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 ngd:ng knh trong ca ng


7/30/2019 do an say

47/49


:H s tr lc cc b

:Khi lng ring ca khng kh 25 C0 2,1= Kg/m 3

:Vn tc kh i trong ng

Chn vn tc kh i trong ng l 20 m/s

42,020.785,0.3600

79,9922

.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 oC

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

=18,4.10 6 ( Ns/m 2 )

=1,2 Kg/m 3

Re= 087,54782610.4,18

2,1.42,0.206

= >10 4

Vy ch chy xoy

( ) ( )0131,0

64,1087,547826lg81,1

1

64,1Relg81,1

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 = 82,3381,9.22,1.20

.132,042,0

2.0131,0 2=

++ ( mm H 2 O)

Vy ta c ;

iH = = 1 + 2 + 3 =8,897+120+33,82=162,717 ( mm H 2 O)

Cng sut ng c qut trc my sy l:


7/30/2019 do an say

48/49


33,76,0.102.3600

717,162.79,9922==N ( KW)

Chn qut c k hiu No4 c cc thng s sau:

A = 5500 , Pd = 20mmH20, = 57%

Kt lun

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

gio V Minh Khi 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 3540 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

nn cng 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 nhng 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 d thc s em li cho em hiu qu cho em ni ring v cho

sinh vin trong nghnh 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, 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 gian cng nh trnh nn

bn thuyt minh ca em cn nhiu thiu st. Em rt mong c s gip cathy c v cc bn.

Mt ln na em xin chn thnh cm n thy gio V Minh Khi v mt

s thy c khc trong khoa gip ch bo tn tnh cho em trong thi gian

qua.

Em xin chn thnh cm n!


7/30/2019 do an say

49/49


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

5.L cng nghip

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

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

v mt s ti liu trn mng

do an say

Documents