113

Chng X

BM LY TM

10.1. KHI nim chung

Bm ly tm thuc loi bm cnh dn c s dng rng ri do c nhng u

im sau :

- Cu to n gin v chc chn, tho lp tin li, trng lng kch thc khng

ln khi c nng sut ln, din tch t my khng ln v nn my n gin.

- Ni trc tip vi ng c in hoc tuc bin hi thch ng vi kch thc ca

tt c trm bm v nng cao hiu sut ca lin hp bm.

- Khi ng bm nhanh v iu chnh n gin trong khong lu lng ln.

- Truyn nc u v lin tc. C th bm c nhiu loi cht lng khc nhau,

hn hp cht lng v cht rn.

- Gi thnh tng i r, s dng n gin, tin li.

Xt s kt cu cu mt bm ly tm n gin (Hnh 10-1).

32

1

6

7

8

4

6

114

Hnh 10-1

B phn c bn v quan trng nht l bnh cng tc 1 t trn trc 2.

Bnh cng tc gm c nhng bn l un cong t trong bung xon c nh.

u b phn dn hng vo 3 c ng ht 5 c lp van mt chiu 7 v li chn rc 8.

b phn dn hng ra ca bung xon 4 c ng y 6.

Trc khi khi ng bm, phi y cht lng vo bm (mi bm) khi bm

lm vic bnh cng tc quay, cc phn t cht lng trong rnh ca bnh cng tc

di nh hng ca lc ly tm b dn t trong ra ngoi, chuyn ng theo mng dn v

i vo ng y vi p sut cao hn. ng thi li vo ca bnh cng tc to nn mt

vng c chn khng v di tc dng ca p sut trong b cha ln hn p sut li

vo ca bm, cht lng b ht lin tc b y vo bm theo ng ht. Qu trnh ht v

y ca bm l qu trnh lin tc, to nn dng chy lin tc qua bm.

tin cho vic thit k, ch to v s dng, bm c phn loi nh sau:

Phn loi theo ct p ca bm:

- Bm ct p thp : H < 20m ct nc

- Bm ct p trung bnh : H = (2060)m ct nc

- Bm ct p cao : H > 60m ct nc.

Phn loi theo s lng bnh cng tc lp ni tip trong bm:

- Bm mt cp (mt bnh cng tc): ct p ca bm hn ch ( 10m ct nc)

bi s vng quay v sc bn vt liu cnh dn.

- Bm nhiu cp: gm 2 hay nhiu bnh cng tc mc ni tip nhm nng cao

ct p ca bm.

Phn loi theo cch dn cht lng vo bnh cng tc:

- Bm mt ming ht - bm c bnh cng tc ht cht lng t 1 pha (cn gi l

bm Congxon). Cch ht ny lu lng hn ch, pht sinh lc dc trc.

115

- Bm hai ming ht - Bm c bnh cng tc ht cht lng t hai pha, cch ht

ny lu lng bm tng gp i, khng gy lc hng trc, bm lm vic n

nh, b vng hn.

Ngoi ra cn phn loi theo s b tr ca trc bm: bm trc ngang, bm trc

ng. Theo iu kin chy ca cht lng vo bm: loi c c cu dn hng v khng

c c cu dn hng...

10.2. L thuyt c bn v bm ly tm

10.2.1. Phng trnh c bn ca bm ly tm

thit lp c phng trnh c bn ca bm, ta gi thit khi bm lm vic

khng c tn tht thu lc v bnh cng tc c v s bn l. iu cho php ta xem

dng chy trong bnh cng tc gm nhng dng nguyn t v tc ca nhng cht

im trn b mt dng nguyn t c xc nh nh nhau.

ng dng phng trnh m men ng lng ( cp n Chng IV) i

vi bm ta c biu thc m men quay ca trc (Hnh 10-2):

D1

D2

u2C2

2

2w2

2 w1

u1

C1o

1 90=

o

1 90=u1

1

C1

w1

Hnh 10-2

M = QT(c2 l2 - c1 l1) = QT(C2 R2 cos2 - C1 R1 cos1)

Cng sut trn trc ca bnh cng tc: N = M; - vn tc gc.

116

Cng sut thu lc: Ntl = QHT ; HT - Ct p l thuyt.

V b qua tn tht thu lc: N = Ntl , ngha l gQTHT = M

Suy ra: g

cosRCcosRCH 111222T

=

V: R1 = u1 ; R2 = u2 - vn tc theo chiu quay

cos1 = C1u ; C2cos2 = C2u - Thnh phn vn tc tuyt i chiu theo phng u, nn:

g

CuCuH u11u22T

= (10-1)

l phng trnh c bn ca bm ly tm hay cn gi l phng trnh le ca

bm.

Trong cc bm ly tm hin i 0C 90 uC u1o111 ==

( ct p

ca bm c li nht) nn ta c :

gCuH u22T = (10-2)

Xt phng trnh (10-2) ta thy ct p ca bm cng ln khi tc quay vnh

ngoi ca bnh cng tc cng ln v hnh chiu ca tc tuyt i trn phng tc

quay cng ln (2 nh, 2 ln). Ct p thc t ca bm ly tm nh hn ct p l thuyt do 2 nguyn nhn:

Do tn tht thu lc trong bm; Do c s bn l nht nh nn khng phi tt c

cc cht im ca cht lng u chuyn ng bng nhau.

N lm gim tc tuyt i hay hnh chiu ca tc tuyt i trn phng

tc quay.

Vy phng trnh tnh ct p thc t ca bm c dng.

gCu

KH u22t=

117

Trong : H - Ct p ton phn ca bm

t - Hiu sut thu lc ca bm, ph thuc vo cu to bm, cht lng ch to cc chi tit bm v kch thc bm.

K - H s c tnh n s bn l nht nh.

10.2.2. nh hng ca kt cu cnh dn n ct p ca bm ly tm

Kt cu ca bnh cng tc ni chung v cnh dn ni ring c nh hng quyt

nh n ct p ca bm ly tm. Hnh dng b tr kt cu ca cnh dn ch yu ph

thuc vo gc 1 (gc vo) v 2 (gc ra). - nh hng ca 1: Gc 1 biu th phng ca vn tc tng i li vo ca bnh cng tc. Nh

ta bit, trng hp c li nht v ct p ca bm l 1 = 90o, do 1 ch ph thuc

vo u1 v C1 (Hnh 11-2):

tgCu

1 11

=

Theo (11-2) th gc vo 1 khng nh hng trc tip n ct p ca bm. Nhng nu tr s gc 1 khng thch hp th s gy va p dng chy vi cnh dn li vo bnh cng tc, nh hng xu n hiu sut, ct p ca bm (thng 1 15 30o).

- nh hng ca 2: Gc 2 biu th phng ca vn tc li ra ca bnh cng tc (Hnh 10-2). L

thuyt v thc nghim chng t gc 2 c nh hng trc tip n phng v tr s cc thnh phn vn tc ca dng chy trong mng dn, do c nh hng quyt nh n

ct p ton phn HT v cc ct p thnh phn Ht, H ca bm.

Tu theo tr s ca 2, bnh cng tc c ba cch b tr cnh dn sau y : 2 < 90o - Cnh dn cong v pha sau (Hnh 10-3a)

118

2 = 90o - Cnh dn hng knh li ra (Hnh 10-3b) 2 > 90o - Cnh dn cong v pha trc (Hnh 10-3c)

W2

C2U2

Cu2C1

U1

W

a)

2

o

2 90

2 C2

Hnh 10-3

T (10-2) ta xc nh tr s ct p ng ca bm theo cng thc :

g2CCH

21

22

d

= (10-3)

trong : C1 v C2 - tc tuyt i ca cht lng vo v ra khi bnh cng tc.

Trong cu trc ca bm, thng thit k sao cho C1 = C2r .

Phng trnh (10-3) c vit li nh sau :

g2

Cg2CCH

2u2

2r2

22

d =

=

M ct p ton phn ca bm t (10-2) c dng :

gCu

H u22T =

- Nu 2 < 90o th C2u < u2 do C2u2 < u2 C2u

v gCu

21

g2C u22

2u2 <

Ngha l H < 1/2 HT (Ct p ng hc nh hn mt na ct p ton phn)

119

- Nu 2 = 90o th C2u = u2 , do

gCu

21

g2C u22

2u2

=

hay H = 1/2 HT (Ct p ng hc bng ct p tnh hc)

- Nu 2 > 90o th C2u > u2 do C2u2 > C2uu2

guC

21

g2C 2u2

2u2 >

H > 1/2 HT (Ct p ng hc ln hn mt na ct p ton phn)

Nhim v ca bm l to thnh ct p tnh hc truyn nc i xa hoc ln

cao. V vy bnh cng tc ca bm ly tm c cnh dn cong v pha sau (Nu 2< 90o) c li hn v ct p tnh hc. Tuy nhin tr s 2 xc nh trong mt gii hn nht nh (2 = 15 30o). 10.3. ng c tnh ca bm ly tm

Cc quan h H = 1(Q); N =2(Q); = 3(Q) biu th di dng th gi l cc ng c tnh ca bm.

C 2 phng php xy dng cc ng c tnh: bng l thuyt v thc nghim.

11.3.1. ng c tnh lm vic (n = const)

a - ng c tnh l thuyt :

T phng trnh c bn, ta c th xy dng ng c tnh l thuyt ca bm.

gCu

H u22T =

Xt tam gic vn tc (Hnh 10-4) ta c :

C2u = u2 - C2r cotg2

120

Mt khc : 222

Tu2 bD

QCpi

=

2 - H s tnh n s gii hn mt ct i ra do c cc cnh dn (2 < 1)

Do : T222

222

2T QbD

gcotug

uHpi

=

HT = A - B cotg2QT trong A, B l nhng hng s dng.

ng biu din phng trnh ny gi l ng c tnh c bn l thuyt, l

mt ng thng khng qua gc to , h s gc ca n ph thuc vo tr s gc 2 (Hnh 10-5)

Nu 2 < 90o th cotg2 > 0, ta c ng AD Nu 2 = 90o th cotg2 = 0, ta c ng AC Nu 2 < 90o th cotg2 < 0, ta c ng AB Nh phn tch trn i vi bm ly tm th 2 < 90o, do ng c tnh l

thuyt ca bm l ng nghch bin bc nht AD.

Cn ch ng c tnh l thuyt AD biu din phng trnh c bn (11-2)

trong cha k ti nh hng ca s cnh dn c hn v cc loi tn tht Khi k ti

nh hng do s cnh dn c hn, ng c tnh tr thnh on thng AD.

HT = KHT

trong K < 1 - h s k ti nh hng ca s cnh dn c hn.

Khi k ti cc loi tn tht thu lc ca dng chy qua bnh cng tc cc loi

tn tht ny u t l vi bnh phng ca vn tc, tc l vi bnh phng ca lu

lng, th ng c tnh l ng cong bc hai AD.

Khi k n cc loi tn tht c hc v lu lng th ng c tnh AD thp

hn mt cht so vi AD. ng AD chnh l ng c tnh c bn tnh ton ca

bm ly tm.

121

C2m

C 2u

C2

2

u2

W 2

2

QDIII

DII

DD'

C

O

A"

A'

A

HB)90

(Ho

2T

>

)90(HT o2 =

)90(H

o2T

122

Trnh t th nghim cng c th tin hnh ngc li t ch lm vic c Q ln

nht, sau gim dn (bng cch ng dn kho 5) cho n ch khng ti (Q =

0).

Ti mi im lm vic ng vi Q, H nht nh ta xc nh c cng sut thu

lc ca bm (Ntl) tng ng. So snh cng sut thu lc vi cng sut o c trn trc

bm ta xc nh c hiu sut ca bm .

Nh vy t cc s liu th nghim, ta c th xy dng c cc ng c tnh

thc nghim ca bm H - Q, N - Q, - Q (Hnh 10-7).

Cc ng c tnh thc nghim ni chung khng trng vi ng c tnh l

thuyt i vi mt bm c th. iu c th l gii l trong l thuyt tnh ton khng

th nh gi hon ton chnh xc cc loi tn tht trong bm so vi thc t.

2

54

1

3

CA

y

H,N

Q~H

Q~N

Q~

O Q

Hnh 10-6. Hnh 10-7.

Cc ng c tnh thc nghim trn thng c ghi trong cc ti liu k thut

ca bm. i vi bm ly tm, thng c thm ng biu din quan h chn khng

ht cho php vi lu lng [HcK] = (Q). 11.3.2. ng c tnh tng hp (n = var)

123

xc nh c nhanh chng cc thng s Q, , N ca bm thay i nh th no khi s vng quay lm vic ca bm thay i, ngi ta xy dng ng c tnh

tng hp ca bm (Hnh 10-8).

ng c tnh tng hp ca bm chnh l ng biu din quan h Q - H vi

cc s vng quay lm vic ca bm khc nhau, trn cc im lm vic cng hiu

sut c ni vi nhau thnh nhng ng cong, gi l ng cng hiu sut.

N H(KW)(m)

15 30

10 20

5 10

0 00

0

10

20 40 60

20

80

Q(l/s)Q h/m3

n=1980

1425

n =1230

n =2975

n=2520

N-n

=29

25

N-n=151

0

N-n=1425

N-n=2520

75%

79%

81%

83%

83%

81%

79%

75%

H50

(m)

40

30

20E

10

0807060

40

20

0 100 200 300 Q(l/s)

100 200 300 Q

n4=880n5=960n6=104

0

n3=800

n2=720

n1=640

60 70 75

78

75

7079

60

n6 =1040n5 =960

n4 =880

n3 =800

n2 =720

n1 =640

Hnh 10- 8 Hnh 10-9

xy dng ng c tnh tng hp, cn phi c cc ng c tnh lm vic

ng vi nhiu s vng quay khc nhau ca bm. Trn hnh 10- 9 biu th cc ng c

tnh H - Q, - Q ng vi nhiu s vng quay.

Cn c vo ng c tnh tng hp ca bm ly tm, ta c th chn ch lm

vic thch hp nht trong khi iu chnh bm, xc nh c kh nng lm vic ca

bm khi chn mua bm.

10.4. im lm vic, iu chnh bm

10.4.1. im lm vic

124

Bm bao gi cng lm vic trong mt h thng ng ng c th no y. Do

ta cn phi xc nh ch lm vic ca bm. Theo (10 - 7) c th xc nh c

ct p ca bm lm vic trong mt h thng ( cng chnh l ct p ca h thng):

H = Z + hw

trong :

Z - cao hnh hc ca bm; hw - Tn tht nng lng trong ng ht v ng y ca

bm. Nh bit trong thu lc hw ph thuc vo ng knh, chiu di ng dn,

lu lng truyn trong ng dn, vt liu v trng thi ng dn.

hw = AQ2

trong : A - h s tn tht ph thuc vo cc yu t ni trn.

Q - lu lng ca bm.

Do c th vit :

H = Z + AQ2 (10 - 4)

ng biu din quan h (10 - 4)

gi l ng c tnh ng dn (ng

c tnh li) trong h thng bm vi

iu kin cho trc. V th ng

c tnh ng dn ln cng th

ng c tnh lm vic ca bm

(Hnh 10-10). Giao im ca 2 ng

c tnh (ca bm v ca h thng

ng ng) chnh l im lm vic

ca bm ly tm (im A trn hnh 10-

10). Ti bm lm vic n nh v

ct p yca bm bng ct p

cn ca h thng (xem thm [1],

[5], [6]).

H,m

N,KW%

C

HO

O

H~Q

HA

NAN~Q

Q~

A

A

QAQ,l/s

Hnh 10-10

125

im A cho ta bit nhng thng s biu th ch lm vic ca bm (lu lng

QA, ct p HA , cng sut NA v hiu sut A ca bm).

10.4.2. iu chnh bm

iu chnh bm sang ch lm vic khc ngha l thay i im lm vic ca

bm theo mt yu cu no . C nhiu phng php iu chnh.

a - iu chnh bng kho

Ni dung ca phng php ny l to nn s thay i ng c tnh li bng

cch iu chnh kho ng y thay i lu lng Q ca h thng (khng iu

chnh kho ng ht v c th gy nn hin tng xm thc trong bm).

Trn hnh 10-11: Khi m kho hon ton c im lm vic A (HA,QA).

H

HBHA

BA

QQAQBQ Q QA QB

A

BHBHA

H

AB

Q

Hnh 10-11. Hnh 10-12.

Khi ng bt kho li th tn tht kho s tng ln (A B) lu lng ca h thng gim i, ngha l ng c tnh li s thay i (dc hn), trong khi ng

c tnh bm vn khng i v nh vy im lm vic ch mi l im B (HB,QB)

Phng php iu chnh ny n gin, thun tin nhng khng kinh t v gy thm tn

tht kho (hwK) khi iu chnh v ch iu chnh c trong phm vi hn ch.

b - iu chnh bng s vng quay trc bm

126

Ni dung phng php ny l lm thay i ch lm vic ca bm bng cch

thay i ng c tnh ca bm khi thay i s vng quay ca trc bm.

Trn hnh 10-12: im lm vic A (HA,QA) ng vi s vng quay nA.

Khi tng s vng quay n nB > nA th ng c tnh bm s thay i trong,

trong khi ng c tnh li khng thay i, im lm vic t im A chuyn n

im B (HB,QB).

So vi phng php iu chnh bng kho, phng php ny kinh t hn (v

khng mt tn tht nng lng v ch do kho) song phc tp hn v phi c thit b

thay i s vng quay.

Ngoi ra nu dng lu di vi nng sut nh hn ta c th iu chnh ng c

tnh lm vic ca bm bng cch ng dng nh lut tng t gt nh bt bnh cng

tc. Nhng phng php ny t c s dng v s ph hoi bnh cng tc khng phc

hi li c.

10.5. Ghp bm ly tm

Trong thc t c trng hp phi ghp nhiu bm lm vic trong cng mt h

thng, khi h thng c yu cu ct p hoc lu lng ln hn ct p v lu lng ca

mt bm. C hai cch ghp sau y :

10.5.1. Ghp song song

Dng trong trng hp h thng c yu cu lu lng ln hn lu lng ca

mt bm.

Trn hnh 10-13 biu th s ghp song song 2 bm. xc nh im lm

vic chung ca h thng bm ta v ng c tnh chung ca cc bm ghp bng cch

cng cc lu lng ca tng bm ghp vi cng mt ct p. Giao im ca ng c

tnh chung vi c tnh li l im lm vic ca cc bm ghp trong h thng.

127

H

1 2

Qc = Q1c + Q2c

Q2c

Q1c

B

H~Q(li)

H~QA

H2~QH1

~QQ'1Q'2

Qc < Q'1=Q'2

OH

BQ

Hnh 10-13

R rng tng lu lng ca 2 bm lm vic song song s ln hn lu lng mt

bm nhng nh hn lu lng ca hai bm lm vic ring r v tn tht trong ng dn

tng, lm tng ct p ton phn ca bm.

Sau khi nghin cu nguyn tc ghp bm song song ta thy:

- S iu chnh (thay i im lm vic) ca h thng bm ghp song song

tng i phc tp khi cc bm ghp c ng c tnh khc nhau nhiu. V vy ngi

ta thng ghp cc bm c ng c tnh gn ging nhau hoc nh nhau iu

chnh thun li.

- Cch ghp bm song song ch c hiu qu ln khi ng c tnh ca cc bm

ghp thoi (c dc nh) v ng c tnh li khng dc lm cch ghp ny ng

dng trong cc h thng bm cn c ct p H thay i t, khi lu lng Q thay i

nhiu.

- S lng bm ghp song song tng lu lng trong h thng c gii hn

nht nh, xc nh bi ng c tnh chung ca cc bm ghp v c tnh li.

10.5.2. Ghp ni tip

Dng trong trng hp h thng c yu cu ct p ln hn ct p ca mt bm .

iu kin cc bm ghp ni tip lm vic c bnh thng trong h thng l cc

bm ghp c cng mt lu lng .

128

Q1 = Q2 = ... = Qi

Ct p lm vic ca h thng c ghp ni tip bm khi Q = const bng tng ct

p cc bm ghp: Hc = H1 + H2 + ... + Hi

Xt hai bm 1,2 ghp ni tip (Hnh 10-14) lm vic trong mt h thng.

ng c tnh chung ca 2 bm

ghp (Hc - Q) c xy dng bng cch

cng cc ct p ca ring tng bm ghp

vi cng mt lu lng. Giao im ca

ng c tnh chung vi ng c tnh

li l im lm vic ca cc bm ghp

trong h thng (im A - Hnh 10-14).

T ta xc nh c lu lng Q v

ct p ca hai bm ghp (H1+H2).

O

1

2

H

Hc = H1+H2

H~Q

AH~Q

(li)

H2~QH1~QH2

H1

Q

Hnh 10-14

Ch :

- Khi ghp ni tip nn chn bm v h thng c ng c tnh dc nhiu mi

c hiu qu cao.

- Khi ghp ni tip, bm th 2 phi lm vic vi p sut cao hn bm 1, do

phi ch n bn ca bm v h thng ng ng.

10.6. ng dng lut tng t trong bm ly tm

10.6.1. S lin h gia lu lng, ct p vi s vng quay ca bm ly tm

Ta bit trong mt bm ly tm khi s vng quay lm vic (n) ca bm thay i

th cc thng s lm vic khc ca bm Q, H, N cng thay i theo. Thc nghim

chng t rng khi mt bm ly tm lm vic vi s vng quay n thay i t (di 50% so

vi n nh mc) th hiu sut ca bm thay i t c th xem = const. Mt khc cc vn tc u t l vi s vng quay nn cc tam gic vn tc s ng dng vi nhau. Do

129

cc ch lm vic khc nhau ca mt bm khi s vng quay n thay i t, xem

nh cc trng hp tng t.

ng dng cc cng thc xc nh lu lng Q v ct p H ca bm ly tm :

Q = pi D b C2rV

gcosCKuH t222 =

Gi H1 , Q1 , N1 l ct p, lu lng v cng sut ng vi s vng quay n1;

H2 , Q2 , N2 l ct p, lu lng v cng sut ng vi s vng quay n2;

Ta xc nh c cc quan h t l gia lu lng, ct p v cng sut vi s

vng quay ca bm ly tm nh sau :

2

1

2

1

n

n

QQ

= (11-5)

2

2

1

2

1

n

n

HH

= (11-6)

3

2

1

2

1

n

n

NN

= (11-7)

10.6.2. Xc nh im lm vic v ng c tnh khi ch lm vic ca

bm thay i

a) V ng c tnh mi ca bm khi s vng quay lm vic thay i: Bit

ng c tnh lm vic H = (Q) vi s vng quay n (Hnh 11-15) ng vi im lm vic A1(H1 , Q1), dng cng thc (11-5)(11-6) c th tnh c H1 , Q1 ng vi im

lm vic A1(H1 , Q1) khi s vng quay ca bm n n.

2

11n

'nH'H

=

130

=

n

'nQ'Q 11

Cng tng t nh vy i vi cc im khc A2 ,A3 ,...,Ai trn ng c tnh H

= (Q) ta c th tm ra cc im tng ng A2 ,A3 ,...,Ai. V ta c th v c ng c tnh mi ca bm H = (Q) ng vi s vng quay lm vic n ca bm.

H

H1

H'1

O Q'1 Q1 Q

A'1

A1A'

2

A'3

A2

A3

H=f(Q)

H=f'(Q)

H

H'1

H1

O Q1Q'

1 Q

H=f'(Q)

H=f(Q)

A1

A'1

n'

n

Hnh 10-15. Hnh 10-16.

b) V ng biu din nhng im lm vic tng t (ng cng hiu sut) t

cc quan h tng t (10-5)(10-6) ta c th vit:

2

1

1

1

1

'QQ

'HH

= hoc const

'Q'H

QH

21

12

1

1==

Nh vy cc im lm vic A1 ,A1 ,... ng vi cc cp tr s H1 , Q1 ; H1 , Q1

;... biu din nhng ch lm vic tng t. Cc im lm vic tng t biu th quan

h bc 2 gia lu lng v ct p - Quy lut Parabol: H = KQ2 (K - hng s). ng

cong ny l ng biu din cc im lm vic c hiu sut bng nhau gi l ng

cng hiu sut.

c) Xc nh s vng quay lm vic ca bm ng vi mt im lm vic cho

trc.

131

Chng hn cn xc nh s vng quay n c ng c tnh lm vic i qua im

A1(H1, Q1) khng nm trn ng c tnh cho trc H = (Q) ng vi s vng quay n.

Mun vy, trc ht ta phi xc nh ng cong cng hiu sut - biu din cc

im lm vic tng t vi im A1(H1 , Q1).

Nh vy c H1 = KQ12 hay 2

1

1

'Q'HK =

Khi bit K ri, v ng H = KQ2 i qua im cho trc A1 ct H = (Q) ti im A1 (Hnh 10-16) chnh l im lm vic tng t vi im A1. Khi bit c

im A1(H1 , Q1) t quan h tng t (10-5 ), hoc (10- 6 ) ta c th xc nh c s

vng quay n:

nQ'Q

' n1

1= hoc

1

1

H'H

n' n =

*Ch : bm c th lm vic vi s vng quay gim, nhng tng s vng quay

trong trng hp cn phi theo qui nh ca nh my ch to.

10.6.3. nh lut tng t

Hai my bm tng t phi tho mn cc tiu chun tng t sau y:

- Tng t hnh hc l hai my bm phi ng dng, ngha l cc gc b tr

cnh dn phi bng nhau v cc kch thc tng ng phi t l vi nhau.

- Tng t ng hc l cc tam gic vn tc tng ng ca cht lng chy trong

2 bnh cng tc ca 2 my bm phi ng dng, ngha l t l gia cc vn tc tng

ng phi bng nhau.

Ta thy rng nhng bnh cng tc tng t c ng knh D1 v D2 vi s vng

quay n1 v n2 to nn ct p H1v H2 .

T phng trnh c bn ca bm ly tm (xc nh ct p) v cng thc xc nh

lu lng ca bm, ta xc nh c cc quan h tng t ca lu lng ct p v cng

sut khi s vng quay v ng knh bnh cng tc ca 2 bm khc nhau :

132

322

311

2

1

DnDn

'QQ

= (10-8)

2

22

11

2

1

DnDn

HH

= (10-9)

5

2

13

2

1

2

1

DD

n

n

NN

= (10-10)

Cc quan h (10-8),(10-9),(10-10) th hin ni dung ca nh lut tng t ca

bm ly tm.

Da vo nh lut tng t ta c th tnh ton vi chnh xc cao nhng thng

s c bn ca bm thit k, nu bit thng s ca bm tng t v nghin cu bng

thc nghim loi bm mi trn m hnh nh lm gim nh cng vic v tng thm s

hon chnh cu to bm.

Tuy nhin nh lut tng t ch ng dng trong trng hp kch thc ca bm

khng khc nhau qu 2 n 3 ln v bm lm vic vi cng mt loi cht lng.

10.6.4. S vng quay c trng nS

S vng quay c trng l s vng quay ca bnh cng tc bm ly tm, c cc

chi tit my tng t hnh hc vi bnh cng tc ca bm nghin cu, c nng sut Q

= 75l/s v ct p H = 1m ct nc.

Dng nh lut tng t ta c:

3

3SS

22

2S

2S

nDDn

Q075,0

va DnDn

H1

==

DS v nS l ng knh bnh cng tc v s vng quay bm c nng sut

Q=75l/s , ct H = 1m ct nc.

T hai biu thc trn ta tnh c

133

23

3S

3

3

3S

2S

2

2

2S

Hn

n

DD

Hnn

DD

==

v : 2/32

S

2

2/33S

3S

Hnn

Hnnnn

Q075,0

==

Rt ra : 4/3S HQ n65,3n = (10-11)

Xt phng trnh (10-11) ta thy s vng quay c trng khng ph thuc vo

loi cht lng. Nh vy ta c th xc nh c s vng quay c trng ca bt k loi

bm no, nu bit c nng sut, ct p ca bm ng vi s vng quay n.

Dng khi nim v s vng quay c trng, ta c th :

- Thnh lp s phn loi thu lc bm ;

- So snh nhng loi bnh cng tc khc nhau ca bm ly tm ;

- Nghin cu nhng bm ln trn m hnh nh ;

- Chn bm tin li nht khi bit nng sut v ct p.

10.7. Mt s im ch trong kt cu v s dng bm ly tm

10.7.1. Lc hng trc trong bm ly tm

nhng bm ht mt pha, trong

thi gian lm vic bnh cng tc chu

tc dng ca lc hng trc v pha

ngc vi hng chuyn ng ca cht

lng i vo bm. Lc sinh ra lc ma

st ph lm tng thm tn tht nng

lng, gim nng sut v hiu sut ca

bm.Quan st s bnh cng tc

(Hnh 10-17),

P2

1 2P2 R2

R0R1

P2

P1

Hnh 10-17

134

khi cht lng i vo bnh cng tc c p sut p1 v khi i ra p sut tng ln p2 .

V p2 ln hn p1 nn cht lng i qua cc k h gia bnh cng tc v thn bm v cc

bung trng 1 v 2 v c th chy ngc v ng ht lm gim nng sut bm.

Lc hng trc l chnh p lc tc dng ln

hai mt trng 1 v 2 t hai pha, ngha l:

Pht = ( piR22 - piRo

2 ) p2 - (piR22 - piR1

2 ) p2 - (piR12 - piRo

2 ) p1

Sau khi bin i ta c:

Pht = pi (R12 - Ro

2 ) ( p2 - p1 )

Nu p2 - p1 = H (ct p ca bm) ta c:

Pht = pi H (R12 - Ro2) (10-12)

lm cn bng lc hng trc, thng dng cc bin php sau y :

- Vi bm c mt bnh cng tc: cho cht lng i vo t hai pha (bm c 2

ming ht) hoc trn a sau bnh cng tc c khoan nhng l, do lm gim

chnh p sut tc dng ln thnh ngoi ca bnh cng tc.

- Vi bm c nhiu bnh cng tc: c th b tr cc bnh cng tc i xng

ngc nhau; hoc dng pittong cn bng gn vo phn cui roto ca bm ; hoc dng

a cn bng gn vo trc bm cp cui cng.

10.7.2. Mt s im ch khi s dng bm

- Chn bm ng theo yu cu k thut, da vo ng c tnh ca h thng v

ng c tnh ca bm, trong c bit ch ng c tnh c bn (H - Q);

- Cc thit b v ng h o p sut, o chn khng, o in nn c y , phi

c van mt chiu ng ht d dng khi mi bm v khi ng bm;

- Trc khi khi ng bm phi kim tra tnh trng bm, du m bi trn, ng

c, cc mi ghp, h thng in ..., sau nc mi bm cho nc in y

135

vo trong ng ht v bnh cng tc ca bm, ng kho trn ng ng y li

trnh qu ti ng c khi khi ng bm;

- Khi khi ng bm, cho ng c quay n nh ri mi t t m kho ng

y cho n khi t nng sut cn thit;

- Trong khi bm lm vic cn theo di ng h o, ch nghe ting my

pht hin nhng hin tng bt thng v c bin php x l kp thi;

- Khi chun b tt my, lm th t ng tc ngc vi khi cho my chy: ng

kho trn ng y trc, tt my sau.

- Khi bm lm vic cht lng khng ln hoc ln t, cn dng my v kim tra

li:

+ Cc van hoc kho trn ng y;

+ ng ht khng kn hoc cha ui ht khng kh ra khi mi bm;

+ Li chn rc b lp kn hoc ming ng ht khng ng su cn thit

cch mt thong b ht;

+ Bnh cng tc quay ngc chiu (khi u dy ngc pha trong ng c in)

hoc s vng quay bnh cng tc tng, gim qu nh mc cho php...