- 1 -

Bai Tap:

Bai 1.Cho 1 bm nh hnh v

Cho biet lu lng cua bm la 25 l/s, ap suat chan khong pck = 6.87N/cm2, ng

knh ong hut d1 = 0.15m, ton that tren ng ong hut la 1m, gia toc trong trng g =

9,81 m/s2. Hay xac nh chieu cao at bm?

Giai:

-Ap dung cong thc :

hvHZ hckh g2

2

2

Vi: Hck = 6.87 N/cm2

= 6,87.104

N/m2

= 6,87 m H2O.

vh = Q

F .

Thay cac gia tr vao ta co cong thc ta tnh ra :

2

2

2h ck hg

vZ H h

Zh = 5,77m .

Ket luan: at bm cao khong qua 5,77m so vi mat thoang be hut.

Bai tap:

Mot may bm nc tieu hao tren truc N = 5,5kW.

Tnh cac thong so: cot ap, lu lng va hieu suat

cua bm. Biet ap suat d ca ra cua bm

mp

202

(cot nc) va ap suat chan khong tai

ca vao cua bm mpck 4

Van toc trong ng ong ay v = 4m/s, ng

knh ong ay d2 = 75 mm, ng knh ong hut d1 = 100mm

Bai giai:

Cong thc tnh chieu cao cot ap:

122

1

2

212

2ZZ

g

VVPPH

- 2 -

Co the bo qua chieu cao hnh hoc.

Vay:

g

VVPPH CKAK

2

2

1

2

2

Van toc V1, V2 c xac nh t phng trnh lien tuc:

Q = V1.S1 = V2.S2

Lu lng c xac nh:

2 2Q VS

Van toc tren ng ong ay:

1 2

1

.4

.

Q QV

S d

Vay chieu cao cot ap c xac nh: H =

Cong suat thuy lc: . . .

1000tl

g Q HN

Hieu suat cua bm la: tl

b

t

N

N

Bai tap bm Pittong.

B.1.

Xac nh cac kch thc c ban cua bm pittong tac dung n theo cac ieu

kien sau: Lu lng bm Qb = 5m3/h, ap suat chat long lam viec p = 20 at, so vong

quay cua truc ong c dan ong bm nc = 150 v/ph, ck = 0,95.

Giai:

Cac kch thc c ban cua bm Pittong la ng knh pittong, ng knh can

pittong, va hanh trnh pittong.

ng knh pittong cua bm co the xac nh t cong thc tnh lu lng bm

tac dung n theo phng trnh:

240

..

240

...

60

..

60

. .32 nKDnSDnSFnqQ S

Trong o F dien tch mat pittong

S hanh trnh pittong

n so vong quay cua truc ong c

D ng knh pittong

D

SK S he so ty le.

- 3 -

Chon KS tuy theo ap suat chat long lam viec va van toc quay cua truc ong c.

He so KS ln ap dung cho bm cao ap va co so vong quay nho, He so KS nho dung

cho bm thap ap va co so vong quay ln. Thong thng He so KS co gia tr trong

khoang 1,5 3, co the lay KS < 1,5 hoac KS > 3 tuy theo ieu kien lam viec cua

bm.

T cong thc tren, suy ra ng knh pittong cua bm.

33

...25,4

.240

Qs

b

s

ltb

nK

Q

nK

QD

Hieu suat Q trong khi tnh toan co the chon s bo theo lu lng cua bm:

Q < 15 m3/h chon Q = 0,85 0,9

Q = 15 60 m3/h chon Q = 0,90 0,95

Q > 60 m3/h chon Q = 0,95 0,98

Bm thiet ke co Q = 5m3/h, v vay co the chon Q = 0,85, bm co ap suat p =

20 at la loai bm thap ap, nen co the chon KS = 1,5.

ng knh pittong co gia tr la :

cmo

D 25,885,.150.5,1

139025,4 3

(Qb = 5m3/h = 1390 cm

3/s)

Lay tron ng knh tieu chuan D = 80 mm; can xac nh hanh trnh S sao cho

bao am lu lng cua bm:

2.

.4

D

qS

Lu lng rieng cua bm:

.b

Q

Qq

n

Vay 2

4.

.

qS

D

ng knh cua can pittong xac nh theo ty so

D

dK d phu thuoc vao ap

suat:

P < 15 at

D

dK d = 0,3 0,36

15 < p 50 at D

dK d = 0,5

50 < p 80 100 at D

dK d = 0,7

Chon

D

dK d = 0,5; vay d = 0,5.80 = 40 mm.

Cong suat cua bm:

.b bN pQ

Trong o p c tnh bang N/m2, Qb tnh bang m

3/s (1at = 0,981.10

5 Pa)

Cong suat ong c

- 4 -

.

bdc

Q ck

NN

B 2:

Xac nh cac thong so lam viec cua bm Pittong tac dung n, cho biet: ng

knh Pittong D = 145mm, hanh trnh Pittong S = 450mm, cong suat tren truc ong

c Ndc = 56,8 KW , so vong quay lam viec cua ong c n = 75v/ph. Hieu suat cua

bm Q = 0,98; ck = 0,96.

Bai lam:

Cac thong so lam viec c ban cua bm la lu lng Qb, ap suat cua chat long

lam viec p, cong suat cua bm Nb. T cac ieu kien lam viec cua bai toan:

Lu lng tc thi cua bm:

.ltq F S

Lu lng cua bm:

Qb = qlt.n.Q. =

Cong suat cua bm:

Nb = Nc. Q. ck

Ap suat cua chat long lam viec:

.b

b

Np at

Q

B 3:

Mt bm pittng tc dng n, c ng knh D = 120mm, Hy xc nh lu lng ca bm v cng sut ca ng c? bit hnh trnh ca pittng l 320mm, s vng

quay lm vic ca ng c ndc =120 vg/pht, hiu sut ca bm b = 0,94, hiu sut

c kh ck = 0,9. p sut cht lng khi lm vic o dc P = 40 bar. s: Qb = 6,8. 10

-3 m

3/s ; Nc = 30,2 KW

B.3

Xac nh cac kch thc c ban va cong suat cua bm Pittong tac dung n,

cho biet: Lu lng cua bm Qtb = 0,6m3/h, ap suat chat long lam viec p = 200 at,

so vong quay cua truc ong c n = 195v/ph, hieu suat thiet ke Q = 0,85; ck =

0,95. Lay Ks = 4; D

dK d = 0,75

S : chon Ks = 4; khi o D = 25mm.

D = 20mm; S = 123mm; Nb = 3,27kW; Nc = 4,04kW

B.4

Xac nh cac kch thc c ban va cong suat cua bm Pittong tac dung kep,

cho biet: Lu lng cua bm Qtb = 16m3/h, ap suat chat long lam viec p = 3 at, so

vong quay cua truc ong c n = 45v/ph, hieu suat thiet ke Q = 0,9; ck = 0,98.

Hng dan: cong thc tnh ng knh suy ra t cong thc tnh lu lng cua

bm:

QSd

b

nKK

QD

...2

1

36,32

- 5 -

S: D = 150mm; d = 50mm; S = 200mm; Nc = 1,73kW.

Bm Banh rang

Lu lng trung bnh cua bm banh rang tnh theo cong thc:

Q = q.n

Trong o: q la lu lng rieng cua bm trong 1 chu ky

n la so chu ky trong mot n v thi gian

Tnh lu lng rieng q:

The tch cua 1 ranh rang bang the tch cua 1 rang, goi a la the tch cua mot rang:

bht

a ..2

Trong o: t- bc rang;

Z

Dt

; D- ng knh vong lan

h- Chieu cao rang an khp; h = 2m ; m- modun cua banh rang

Z- la so rang; b- Chieu dai rang (chieu rong rang)

Vay: bmZ

Da .2.

2

Khi hai banh rang quay mot vong , the tch chat long c chuyen qua bm t

vung ay en vung hut la: 2.Z.a.

Vay lu lng cua bm vi so vong quay n trong mot n v thi gian la;

Q = 2.Z.a.n = 2Dmbn = 2.m2Z.b.n.

Bai tap:

B1. Xac nh cac kch thc c ban va cong suat cua bm banh rang, biet: Lu

lng cua bm Qb = 60l/ph, ap suat chat long lam viec p = 20 at, modun cua banh

rang m = 4mm, so vong quay cua truc bm n = 1450v/ph, so rang cua moi banh

rang Z = 14. Hieu suat thiet ke Q = 0,92, ck = 0,85.

Bai lam:

Cac kch thc c ban cua bm banh rang la: Modun cua banh rang m, ng

knh cua vong lan D, ng knh vong nh D2, ng knh vong tron c s D0 ,

chieu cao cua rang h, chieu rong cua rang b, khoang cach cua 2 tam banh rang L,

ng knh ong hut va ong ay d.

T cong thc tnh lu lng cua bm ta co:

Qb = 2Dmbn. Q = 2m2Zbn. Q

Vay be rong cua banh rang c xac nh:

Q

b

nZm

Qb

.....2 2

32mm

- 6 -

ng knh vong lan: D = m.Z =

ng knh vong tron c s: D0 = m.Z.cos0 =

ng knh vong nh rang : D2 = m(Z+2) =

Khoang cach gia 2 tam banh rang: L = mZ =4x14 = 56mm

Chieu cao rang: h = 2m = 2x 4 = 8mm

ng knh ong:

.

42

v

Qd b

oi vi ong hut Vh 1,5 2 m/s ta chon Vh = 2m/s dh = 25,3mm

oi vi ong ay V 3 5 m/s ta chon V = 4m/s d = 17,8mm Lay

tron dh = 25mm d = 20mm

Cong suat cua bm: .N pQ

Cong suat ong c dan ong bm: 2,5.

dc

Q CK

NN kW

Bai 2.

Xac nh cac thong so lam viec c ban cua bm banh rang, biet: Modun cua banh

rang m = 4,5mm, so rang cua moi banh rang Z = 13, chieu rong cua banh rang b =

50mm, goc an khp cua thc ren 0 = 200, ap suat cua chat long lam viec p = 3,3

at; So vong quay lam viec cua truc bm n = 1450v/ph; hieu suat cua bm Q =

0,65, ck = 0,645.

Bai lam

Cac thong so lam viec c ban cua bm la lu lng Q, ap suat cua chat long lam

viec p, cong suat N va hieu suat .

Lu lng trung bnh cua bm co the xac nh theo cong thc a hoc, con gia tr

chnh xac hn c xac nh theo cong thc sau:

Qlt = 2Dmbn = 2m2bn(Z+sin

20)

= 121 l/ph

(trong o sin20 = sin

220

0 = 0,117 )

Lu lng thc, trung bnh cua bm:

Qtb = Qlt.Q

Tnh cong suat cua bm: N = p.Q

.N pQ

Cong suat ong c dan ong bm: dcN

B.3 Bm Canh gat

- 7 -

Lu lng rieng cua bm c tnh: q = 2eb(2r-Z)

Trong o:

He so dao ong lu lng: 2min

max Q

Q

e: o lech tam gia stato va roto

b: Chieu rong cua canh gat.

r: ban knh mat lam viec cua bm

: Chieu day canh gat.

Z: so lng canh gat

Goi n la so vong quay cua bm trong mot phut, lu lng trung bnh ly thuyet cua

bm trong mot giay la:

60

.nqQlt

Zrneb

Qlt ..230

(m3/s)

Vay lu lng thc te cua bm la : Qtt = Qlt . Q Bai tap:

Xac nh cac kch thc c ban va cong suat cua bm canh gat tac dung n theo

cac so lieu sau:

Lu lng cua bm Q = 100 l/ph, ap suat chat long lam viec p = 30 at, so vong

quay cua truc bm n = 960 v/ph, hieu suat lu lng va hieu suat c kh Q = 0,90,

ck = 0,96.

- 8 -

Bai lam:

Cac kch thc c ban cua bm canh gat la ban knh roto r, ban knh trong cua

stato (ban knh vong hng) R, o lech tam gia stato va roto la e, chieu cao cua

canh gat h.

Ban knh trong cua stato xac nh t bieu thc tnh lu lng cua bm:

Q = 2ebn.Q (2R-Z) l/ph

Suy ra:

Q

Q

neb

ZebnQR

4

2

Neu bo qua chieu day canh gat , khong oi hoi o chnh xac cao, bieu thc tnh

R nh sau:

Qneb

QR

4

Neu trong ieu kien bai toan khong cho trc gia tr e, b, , Z, co the chon cac gia

tr o trong gii han sau:

e = 210 mm ; b = 10 40 mm ; = 2,0 2,5 mm ; Z = 7 16

Theo ieu kien bai toan va chon e = 4 mm; b = 25 mm; = 2,3mm; Z = 12, co the

tnh ban knh trong cua stato theo:

cmneb

ZebnQR

Q

Q7,9

9,0.5,2.4,0.960.14,3.4

9,0.960.12.23,0.5,2.4,0.210.100

4

2 3

Lay tron R = 100mm , khi o gia tr b bang:

cm

ZRen

Qb

Q

41,212.23,010.14,3.29,0.960.4,0.2

10.100

22

3

= 24,1mm

Ban knh roto: r = R e = 100 4 = 96 mm

Chieu cao phan lam viec cua canh gat: Hlv = 2e = 8 mm

Chieu cao cua canh gat: mmh

h lv 204,0

8

4,0

Cong suat cua bm : kWQp

N 9,460.1000

10.100.10.81,9.30

1000

. 34

Cong suat tren truc bm:

kWN

NCKQ

dc 66,596,0.9,0

9,4

- 9 -

Cho mt h thng bm nh hnh v. Bit p sut d ca ra mP

.152

, p sut

chn khng ti ca vo mP

.31

, ng knh ng y d2 = 75 mm, ng knh ng

ht d1 = 100 mm, tc trong ng ng y V2 = 4 m/s , chiu cao hnh hc t b ht n b cha l 12 m. Hy xc nh lu lng, ct p v hiu sut ca bm? Bit rng cng sut trn trc bm l 6,5 Kw, tn tht trn ng ng khng ng k, h

s cht lng 1 = 2 = 1.

Bi gii Cng thc tnh ct p ca bm c xc nh:

hhHg

VVPPH

2

2

1

2

212

hay hhCKAK H

g

VVPPH

2

2

1

2

2

Vn tc V1 , V2 c xc nh t phng trnh lin tc: Q = V1.S1 = V2.S2

Hay: Q = V2.S2

11

QV

S .

Th vo ta c 30,56.H m

Cng sut ca bm c xc nh: . .

1000.

. .

1000.

Q HN

Q H

N

Vy Q = 17,66 l/s; H = 30,56 m; = 0,81

Cu 3 (4im):

My bm ly tm bm nc t b A ln b B vi lu lng Q=24 m3/h, vi ng c tnh theo n1 = 950 v/ph nh sau:

Q ( l/s ) 4 6 8 10

H ( mH20 ) 9,1 8,15 6,75 5,0

- 10 -

( % ) 60 85 80 40

a- Hy dng ng dc tnh ca bm? b-Xc nh im lm vic, nu mc nc ti b ht cao hn tm my bm

mt on h1= 2 m v mc nuc b cha cao hn tm my bm mt on h2 = 3,2m.

Cho bit tng chiu di ng ng l = 28m, ng knh ng d = 75mm, = 0.03, =15,7 cho g = 10 m/s

2

c- Xc nh s vng quay n2 ca bm dim lm vic?

Bm ly tam

Banh cong tac bm ly tam co cac kch thc chnh: 2R1 = 100mm; 2R2 = 250mm;

b1 = 55mm; b2 = 23,7 mm; lu lng Q = 72,5 l/s; so vong quay cua truc n = 1450

v/ph.

Hay xay dng tam giac van toc ca ra va ca vao cua day canh cho dong

nguyen to trung bnh. Biet cac goc at canh 1 = 320; 2 = 23

0

.

Tnh cac van toc thanh phan C1U; C2U va cac goc 1; 2.

Bai lam:

e xay dng tam giac van toc, can xac nh cac van toc thanh phan:

+ Van toc vong: U = .R

Tai ca vao: 1 1 1

.. .

30

nU R R

Tai ca ra: 2 2.U R

+ Van toc hng knh:

F

QCR

Tai ca vao: 1

1 1 12R

Q QC

F R b

Tai ca ra: 2

2

R

QC

F

U1 C1U

C1R

C1 W1

1 1

- 11 -

Xac nh cac van toc thanh phan: C1U; C2U; va cac goc 1; 2.

C1U = U1 C1R.cotg1 =

11

1

R

U

Ctg

C 1 = 77

055

2tg 2 = 21040

Bai tap:

Bm ly tam hut nc t 2 be co mc nc chenh nhau theo o cao a = 1m len be

tren cung co o cao b = 5 m (so vi sat ben di), lu lng Qb = 10 l/s, theo cac

nhanh ng ong l = 5 m; d1 = 50mm; d2 = 75 mm.

Xac nh cot ap cua bm, biet he so ma sat ng ong = 0,03; khong tnh ton

that cuc bo. (hnh ve)

Bai lam:

Cot ap cua bm: H = a + b +hAB + he = b +hAB + hf (1)

T phng trnh (1): a + he = hf

2121

2

1

2

1 2.

2e

ee

e QKgS

Q

d

l

g

V

d

lh

Hay: 22

1 ffe QKQka (2)

Ma: Q = Qe + Qf hay Qe = Q Qf (3)

Thay vao (2): 212

1 ff QKQQKa

a + K1Q2 2K1QQf = 0

122 QK

aQQ f (4)

Hay 1 21 1

1

2

lK

d gS

C2R

2 2

C2U

C2 W2

U2

a Q

4l,d2 B

A

l,d2 f

l,d1

b

l,d1

e

- 12 -

Thay ai lng K1 va Q vao (4):

12 2f

Q aQ

QK

Qf = 6,26 l/s

Qe = Q Qf =

Cot ap H = b +hAB + hf

H = 9,2 m

Bai tap:

Mot bm nc tieu hao cong suat tren truc N = 70 kw, so vong quay n = 960

v/ph, cot ap H = 28 m.

Xac nh so vong quay n va lu lng Q sao cho lu lng nay cua bm giam i

25% so vi lu lng ban au. Biet hieu suat cua bm = 0,863.

s: Q = 0,219 m3/s ; Vay lu lng giam 25% la Q = 0,164 m3/s

So vong quay tng ng: n = 720 vg/ph

Bi tp: My bm ly tm bm nc t b di c cao 5m ln b trn c cao 14m vi lu lng Q = 7,3 l/s, ng c tnh nh bng di vi n = 1600 vg/ph theo ng ng ht l1= 10m; d1= 100mm, ng ng y l2=30m; d2 = 75mm. Bit h s ma st ca cc ng ng 1= 0,025; 2= 0,027 v h s tn tht cc b trong cc ng ng 1 = 2; 2 = 12.

a- Hy dng ng c tnh ca bm? b-Xc nh im lm vic? c- Xc nh s vng quay n2 ca bm im lm vic?

Phn Qut, my nn

Q (l/s) 0 2 4 6 8 10 12 14 16

H (m) 14,9 15,4 15,5 14,9 14 12,2 10,1 8 4,3

% 0 40 65 74 75 70 61 42 22

l2d2

l1d1

- 13 -

Bi 1:

Gung ng( bnh xe cng tc) ca qut ly tm c ng knh D1 = 480 mm; D2

= 600 mm; s vng quay n = 1500 v/ph; k = 1,2 kg/m3; W2 = 22 m/s; W1 = 25 m/s;

1 = 600, 2 = 120

0.

Hy dng tam gic vn tc cnh vo v ra ca cnh gung v xc nh chiu cao ct p ca qut.

Bi lm:

Cc vn tc vng: 11

60

D nU

22

60

D nU

Dng tam gic vn tc

Cc thnh phn Cu c tnh t tam gic vn tc: Cu1 =

Cu2 =

Chiu cao ct p l thuyt ca qut:

2 2 1 11

lt U UH U C U Cg

Ct khng kh

Tnh theo m ct nc nh sau:

kHk = ncHlt

2

K

K ltlt

H O

HH

ct nc

Bi tp: Cn nung nng khng kh vi lu lng 30000kg/h t 200C ln 1600C bng cch cho n chy qua thit b trao i nhit . Hy xc nh nng sut, p sut, cng sut ca qut khi lp trc v lp sau thit b trao i nhit. Tn tht thy lc ca dng khng kh chy trong h thng l 120 mm

H2O, iu kin nhit 200C v k = 1,2 kg/m

3. Hiu sut tng hai trng hp

nh nhau = 50% Bi lm:

iu kin nhit 200C th khi lng ring ca khng kh l k = 1,2 kg/m3,

Vy khi nhit 600C th khi lng ring ca khng kh k l;

kTk = kT

k

'

'

K KK

K

T

T

nhit 200C th chiu cao ct p tnh cho khng kh l Hk:

kHk = .H

U1

W1 C1

C1R

C1U

1 1 2

U2

C2

W2

CU2

2

- 14 -

.

K

HH

ct kh

Tn tht thu lc ph thuc vo khi lng ring v t l bc 2 vi vn tc lu lng, Ta c hai trng hp:

a. Qut lp trc thit b trao i nhit (qut y):

3600.

K

K

GQ

. . .

1000.

K Kg Q HN

b. Qut lp sau thit b trao i nhit (qut ht):

'

'3600.

K

K

GQ

K

K

KK H

Q

QH .

2''

'

'

KH ct khng kh ' ' '

' . . .

1000.

K Kg Q HN

So snh 2 trng hp, ta c:

'

N

N

'Q

Q

'

K

K

Nhn xt: Cng sut tiu th cho qut do ht khng kh b t nng tng ln bao nhiu ln

th khi lng ring khng kh gim i by nhiu ln so vi trc khi t nng. Cng sut gia tng l do qut phi ht th tch khng kh ln hn.

Do vy, i vi h thng qut ta nn lp qut v tr m khng kh c nhit thp nht trong h thng.

Bi tp:

My nn 1 cp ht lng khng kh 400m3/h p sut P1 = 1bar, nhit t1= 200C,

p sut nn l P2 = 7 bar vi s m a bin ca qu trnh nn n = 1,3. Xc nh: a- Cng sut l thuyt ca my nn, cng sut tiu th in nng ca my nn,

cho bit = 0,7

b- Xc nh s vng quay ca trc khuu nu bit h s np = 0,7, ng knh xi lanh d = 200 mm, hnh trnh S = 100 mm, s xi lanh Z = 2.

c- Lng nhit thi qua xi lanh my nn.

Bi lm

Cng sut l thuyt ca my nn:

- 15 -

11

1

1

2.1

nn

mnP

PVP

n

nN

j/h

Cng sut tiu th in nng ca my nn:

mnd ien

mn

NN

Vy Nin = 38,8kW S vng quay ca trc my nn:

ZnSdV

Vh ...4

. 2

n = 150 vg/ph Lng nhit thi qua xilanh: Q = [Qn] = [G.cn(t2 t1)]

1 1

1

.

.

P VG

R T ;

1

22 1

1

n

nPT T

P

;

1n v

n kc c

n

j/kg

0K N = 5,25 kW

Bai tp:

Trong mt bnh kn th tch V = 0,045m3 cha lng khong kh vi ap sut u

P1

= 2bar, nhit t1

= 250c, nhit va ap sut s thay i ra sao nu ta cp cho

khong kh lng nhit 16 kj.

Bai tp:

Tnh hiu sut th tch my nn khng kh khi bit cc thng s sau : ng knh pt tng 80 mm , hnh trnh 70 mm , nng sut 60 m3/gi. My nn mt cp 2 xy lanh

Bai tp:

Dng my nn kh nn kh metan n p sut 55 at vi nng sut 210 m3/gi (trong iu kin tiu chun), nhit 180C, p sut 1 at. Tnh s cp my nn v p sut ra mi cp v cng sut my nn vi hiu sut 0,7