Quan Day May Bien AP

7/28/2019 Quan Day May Bien AP

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2/39

3. V li s hon chnh MBA:

4. Tnh s vng dy qun cho mi vn:

(4.3)

Trong :

Tit din li thp c tnh bng m2

Nu tit din li thp c tnh bng cm2 v f = 50Hz th biu thc trn tr thnh.

(4.4)

5. Tnh s vng qun cho cun s cp v th cp:a. S vng qun cho cun s cp:

(4.5)

b. S vng qun cho cun th cp:

Khi my bin p mang ti th in p trn ti s st gim mt lng so vi lc khng ti.

m bo in p cung cp cho khi my vn hnh th phi tr hao lng st p ny khi tnh ton t

(5 15)%.

(4.6)

6. Tnh dng in pha s cp:

Tra bng chn hiu sut ca MBA v tnh ra dng in pha s cp

(4.7)

S2 ( VA ) 3 10 25 50 100 1000 (% ) 60 70 80 85 90 > 90

7. Tnh ng knh dy qun:

Chn mt dng in thch hp v tnh ng knh dy qun

Pha s cp:

(4.8)

nV =45

n2 = nV . (U2 + 5% 15%)

I1 =%.1

2

U

S

d1 = 1,13J

I1

tm ABf ...44,4

1n v =

n1 = nV . U1


3/39

Pha th cp:

(4.9)

Vi J l mt dng in (A / mm2); Chn ty vo ch lm vic ca MBA. MBA lm vic lin tc J = (2,5 5) A/mm2. MBA lm vic t J c th chn n 7A/mm2.

8. Tnh h s lp y (kl)

H s lp y cho bit b dy cun dy chim ch bao nhiu trong ca s ca li thp

(4.10)

Trong :

BD: B dy cun dy

C: B rng ca s c =2

a

Tnh b dy cun dy

- Cun s cp c b dy BD1 c tnh t s vng qun n1.

- Cun th cp c b dy BD2 c tnh t s vng qun n2.

- B dy c cun dy BD = BD1 + BD2 + (1 2)mm.

S vng dy qun cho 1 lp:

(4.11)

Trong :

hK: Chiu di h ca khun qun

d/ : ng knh dy k c cch in

S lp dy qun:

(4.12)

Trong :

n: S vng dy ca tng cun (s hoc th cp)

nVL: S vng dy qun cho 1 lp

d2 = 1,13J

I2

Kl =C

BD= 0,6 0,7; Ti a l 0,8

/d

hn Kvl =

VLn

nn L =


4/39

B dy cun dy s hoc th BD1(2) = nL1(2) . d/i

9. Tnh khi lng dy qun (W)

(4.13)

Vi: W1; W2 l khi lng ca cun s cp v th cp.

Khi lng ca tng cun dy c tnh theo biu thc.

(4.14)

Trong :LTB: L chiu di trung bnh ca mt vng dy (tnh bng dm).

n: S vng qun ca cun s cp hoc th cp.

d: ng knh dy qun cun s cp hoc th cp (tnh bng mm2).

W: L khi lng (tnh bng Kg).

4.2. Tnh ton qun mi my bin p.

La dang bai toan ma ngi th nhan c nhng yeu cau ky thuat can co chomot may bien ap cu the t khach hang nh ien ap nguon vao; ien ap ra can co;cong suat ngo ra; muc ch s dung ... Vi dang bai toan nay chung ta phai xac nh

c tiet dien loi thep; so vong day quan s cap, th cap va ng knh day quans cap, th cap ...

Co the tom tat bai toan nh sau:Biet trc: SLOI; U2; U1Can tm: SBA I2 I1; n1; n2; d1; d2 ...

4.2.1.Phng php tnh ton my bin p cm ng:

My bin p cm ng hay cn gi l my bin p hai dy qun, l loi my bin p c dy qun

s cp v th cp cch ly nhau. K hiu my bin p hai dy qun nh hnh 4.1. Trnh t tnh ton dy

qun v chn kch thc li thp c tin hnh theo cc bc sau:

U1

N1

N2

U2

Hnh 4.1: K hiu my bin p hai dy qun

W = W1 + W2

W1(2) = (1,2 1,3). 8,9. LTB. n.4

.2

d


5/39

Bc 1: Xc nh cc s liu yu cu:

- in p nh mc pha s cp U1 [ V ].

- in p nh mc pha th cp U2 [ V ].

- Dng in nh mc pha th cp I2 [ V ].

Trng hp nu khng bit r gi tr I2, ta cn xc nh c cng sut biu kin pha th cp

S2 :

S2 = U2 . I2 [ VA ] (4.15)

- Tn s f ngun in.

- Ch lm vic ngn hn hay di hn.

Bc 2: Xc nh tit din tnh ton cn dng cho li st (At ):

(4.16)

Trong :

At: l tit din tnh ton ca li thp [cm2]

S2: l cng sut biu kin cung cp ti pha th cp bin p [ VA ]

K: l h s hnh dng li thp.

Khi l thp dng EI (hnh 4.2) ta c K = 1 1,2

Khi l thp dng UI (hnh 4.3) ta c K = 0.75 0,85

Bm: l mt t thng s dng trong li thp. Ty theo hm lng silic nhiu hay t mchn Bm cao hay thp. Cng ty theo loi l thp c ch to theo dng dn t c nh hng hoc

khng nh hng m chn Bm cao hay thp.

i vi l thp dn t khng nh hng: Bm = (0,8 1,2)T

i vi l thp c dn t nh hng: Bm = (1,2 1,6)T.

m

tB

SKA

2..423,1=


6/39

Bc 3: Chn kch thc cho li thp, khi lng li thp.

Kch thc cho li thp:

Gi Ag l tit din tnh t kch thc thc s ca li thp, ta c:

(4.17)

Trong :

a: l b rng l thp [cm]

b: l b dy li thp [cm]

Nh vy gia Ag v At chnh lch nhau do:

Hnh 4.2 : Li thp dng E,I Hnh 4.3 : Li thp dng U,I

b

a a

b

a

Hnh 4.4: Cch o ly kch thcli thp dng E I

Hnh 4.5: Cch o ly kch thcli thp dng U I

baAg .=


7/39

B dy cch in trng trn l thp ( gim nh dng in Foucault chy qua cc l thp trong

li).

(4.18)

ba v c trn l thp do cng ngh dp nh hnh l thp gy nn.

chnh lch ny c xc nh bng h s ghp Kf, ta c:

Trong thit k tnh ton, tham kho gi tr Kf theo bng sau:

B dy l thp

(mm)

KfL thp t ba v L thp nhiu ba v

0,35

0,5

0,92

0,95

0,8

0,85

Ch :Nu o c b dy mi l thp v bit chnh xc s l thp ta tnh c A t v c th xem At =

Ag.

Da vo gi tr Ag, ta chn c kch thc a, b ca li thp.

d dng trong thi cng qun dy, thng gia a v b c mi quan h v kch thc nh sau:

b = a n b = 1,5a (4.19)

T , ta c quan h sau :

Ag = a.b = a2 (khi chn a = b).

Ag = a.b = a.1,5a = 1,5a2 (khi chn b = 1,5a).

Tm li: Khi bit trc gi tr Ag, ta c th xc nh dy gi tr a chn, bng cch tnh sau:

(4.20)

Phi hp gi tr a c sn trong thc t, chn gi tr a thch hp cho li thp, t tnh li chnh

xc gi tr b. Sau khi c kch thc l thp, ta chn khi lng li thp.

f

t

K

AAg =

g

gAa

Aaaaa == maxminmaxmin

5,1vvi


8/39

Khi lng li thp:

Trng hp li thp dng E I :(hnh 4.6)

Gi c l b rng ca s.

h l b cao ca s.

Ta c th tch li thp ( tr i khong khng gian trng ca 2 ca s) l:

(4.21)

Gi l khi lng ring ca thp k thut in = 7,8 kg/dm3.

Suy ra khi lng li thp l :

(4.22)

Hay:

(4.23)

Trong cng thc (4.9):

Wth: n v l [kg]

Cc kch thc a, b, c, h: n v l [dm]

Trng hp li thp E, I ng dng tiu chun, ta c quan h cc kch thc nh sau:

2ac = v

23ah =

Do cng thc (4.23) c th vit li thnh cng thc (4.24) cho l thp ng tiu chun:

(4.24)

Hnh 4.6: Cch o kch thc li thp dng E,I tnh ton

a/2

a/2 a/2a

h + a

bc c

h

h)c15,6ab(a

h)c7,8.2ab(aWth

++=

++=

V.Wth =

h)c(a2abV ++=

b46,8aW 2th =


9/39

Trng hp kt cu li thp dng UI : (hnh 4.7)

Th tch li thp tr i ca s l:

V = 2ab(2a + c + h) (4.25)

Suy ra khi lng li thp:

(4.26)

Trong :

Wth: n v l [kg]Cc kch thc a, b, c, h: n v l [dm]

Th d 1:

Xc nh khi lng li thp bin th c th dng ch to bin th vi cc yu cu theo hnh 4.8

GII:

Bc 1:

a ac

2a+c

a

a

2a+h

b

Hnh 4.7: Cch o kch thc li thp dng U,I tnh ton

h)c15,6ab(2aWth ++=

Hnh 4.8: Hnh th d 1

U1

= 110V U2

= 15V

I2

= 5A


10/39

Tham s ti th cp gm:U2 = 15V; I2 = 5A

Nn S2 = U2.I2 = 15.5= 75VA

Bc 2:

Chn dng li thp E, I ng tiu chun, mt t dng cho li thp chn l:

Bm = 1,2T, ta c:

At = 1,423(1 1,2)mB

S2 = 1,423(1 1,2)2,1

75= 10,269 cm2 12,32 cm2

Ta c:

At = 10,27 cm2 12,32 cm2

Bc 3:

Nu chn Kf = 0,95 (kh nng ghp st ti a), th tit din A g cn dng cho li thp so vi tit

din tnh ton At l:

Ag =95,0

32,1227,10 2cm= 10,81 cm2 12,97 cm2

Xc nh amin v amax theo khong Ag = 10,81 cm2 12,97 cm2.

5,1

81,10

5,1min==

gAa = 2,68 cm 2,7 cm

6,397,12max === Aga cm

Tm li, thc hin bin th c cng sut 75VA ta chn a trong khong t 2,7cm n 3,6cm.

p dng cng thca

Ab g= v Wth = 46,8a2b ta c th xc nh mt dy gi tr cho cc li thp c th

t c cng sut yu cu bi nh sau:

a (cm) 2,7 2,8 3 3,2 3,4 3,5 3,6

Ag (cm2) 10,81 12,97 10,81 12,97 10,81 12,97 10,81 12,97 10,81 12,97 10,81 12,97 10,81 12,97b (cm) 4 4,8cm 3,86 4,63 3,6 4,32 3,37 4,05 3,18 3,81 3,09 3,77 3 3,6

Wth (kg) 1,36 1,64 1,42 1,7 1,52 1,82 1,62 1,94 1,72 2,06 1,77 2,12 1,82 2,18Bng gi tr ny cho ta cc kch thc li thp c th to bin th ng theo yu cu trn, ta c

th chn mt trong cc kch thc ny tnh ton s b, sau nu cn ta s hiu chnh trong cc bc

tnh sau.

Gi s trong th d ny ta chn:


11/39

a = 3,2cm; b = 3,4cm; Wth = 1,63Kg

Ag = 10,88 cm2; At = 10,336 cm2 (Kf= 0,95)

Khi dng l thp E, I ng tiu chun, kch thc li thp cn dng ( to ra S 2 = 75VA) nh

hnh 4.9.

Ch :

Nu b dy mi l thp l 0,5mm v b = 34mm, tng s l thp ch E cn dng l

mm

mm

5,0

34= 68 (l).

Tm li:

B l thp gm 68 l thp ch E v 68 l thp ch I.

Khi lng li thp: Wth = 1,63kg

Bc 4:

Xc nh s vng to ra mt vn trong cun dy s cp v th cp.

(4.27)

Hnh 4.9: Kch thc li thp cn dng

32mm

a/2

34mm

64mm 16mm

48mm

96mm

Wth

=

1,63kg

32mm

mt

vBAf

n...44,4

1=


12/39

Trong :

nv: n v l [vng/vn]

f: n v l [Hz]

At: n v l [m2]

Bm: n v l [T]

Trng hp At dng n v l [cm2] v cc i lng khc c n v ging nh trn, ta c:

(4.28)

Khi f = 50Hz:

(4.29)

gn hn, (4.29) c ghi nhn ti mi mc gi tr ca Bm cho trc:

Vi (4.30)

Vi (4.31)

Vi (4.32)

Bc 5:

Xc nh st p pha th cp lc mang ti nh mc.Ta lun lun c U20 > U2.

Trong :

U20: l in p th cp khi khng ti.

U2: l in p th cp khi ti nh mc.

Thng ta t tham s U% vi nh ngha:

(4.33)

Tuy nhin, d tnh ton trong thit k, ta bin i nh sau:

(4.34)

Do :

(4.35)

mt

vBAf

n...44,4

104=

mt

vBA

n.

045,45=

t

vAn3,55

=TBm 8,0=

t

vA

n045,45

=

t

vA

n54,37

=

TBm 1=

TBm 2,1=

100.1100.U%2

20

2

220

=

=

U

U

U

UU

h

00

2

20 C100

1 =

+=U

U

U

2h20 .UCU =


13/39

giai on xc nh s b ban u, U% hay Ch c xc nh theo cc bng sau :

S2 (VA) 5 10 25 50 75 100 150 200 300U% 20 17 15 12 10 9 8 7,5 7

Hoc tham kho bng dng cho ph ti thun tr (h s cos = 1)

S2 (VA) 25 50 75 100 150 200 250 400 500 600 750 1000U% 8 6,5 6,1 6 5,9 5,2 5 4,3 4 3,9 3,8 3,75

Bng quan h: h s Ch theo S2

S2 (VA) Ch S2 (VA) Ch S2 (VA) Ch S2 (VA) Ch5

7,5

10

15

20

25

30

40

1,35

1,28

1,25

1,22

1,18

1,16

1,14

1,13

50

60

70

80

90

100

120

150

1,12

1.11

1,10

1,09

1,085

1,08

1,075

1,065

180

200

250

300

350

400

500

600

1,060

1,058

1,052

1,048

1,045

1,042

1,038

1,035

700

800

900

1000

1500

2000

3000

1,032

1,030

1,028

1,025

1,020

1,016

1,009

Bc 6:

Xc nh s vng dy qun ti s cp v th cp:

(4.36)

(4.37)

Th d 2:

Da vo kt qu tnh c trong th d 1 tnh ton s vng dy qun cho bin th (hnh 4.8).

Gii:

Trong th d 1, tm c a = 3,2cm, b = 3,4cm,

Ag = 10,88cm2, At = 10,336cm2.

Nu li thp c mt t l Bm = 1,2T, ta c:

Bc 4:

p dng (4.13) hay (4.18) ti tn s f = 50Hz:

nv=tA

54,37=

336,10

54,37=3,6319 3,632 vng/vn.

Bc 5:

ng vi S2 = 75VA, tra bng chn Ch = 1,1

Nn U20 = U.Ch = 15.1,1 = 16,5V

Bc 6:

v202 .nUN =

v11 .nUN =


14/39

Vi U1 = 110V, U20 = 16,5V v nv = 3,632 vng/vn.

Suy ra s vng pha s v th cp nh sau:

N1 = U1.nv = 110 . 3,632 = 399,52 vng 400 vng

N2 = U20 .nv = 16,5 . 3,632 = 59,928 vng 60 vng

Bc 7:c lng hiu sut ca my bin th, xc nh dng in pha s cp I1.

Trong thit k s b, hoc n gin ha, hiu sut c th tra bng theo S2. C th tham kho

mt s bng sau:

Theo Robert Kuhn:

S2 (VA) 3 10 25 50 100 1000 % 60 70 80 85 90 Ln hn 90

Theo Anton Hopp:

S2 (VA) 30 50 100 150 200 300 500 750 1000 % 86,4 87,6 89,6 90,9 91,3 93 93 95,3 94

Theo Walter Kehse:

S2 (VA) 10 20 30 50 100 150 300 500 % 80 80 85 90 91 92 92 92,5

Theo AEG (bin th ngun ca b chnh lu):

S2 (VA) 25 50 100 200 300 400 500 700 1000

% 76,5 84 85 86 88 90 90,5 91 92

Theo Newnes:

S2 (VA) 100 150 200 250 500 750 1000 1500 2000 2500 3500 5000 % 83,5 89,3 90,5 91,2 92,5 93,5 94,1 95 95,4 95,7 95,9 98,2

Theo Elektroteknik und Machinenbau (Vienne 16/8/1931):

S2 (VA) 150 250 500 1000 2000 3000 5000 % 88,5 89,6 91 92,8 94,2 94,9 95,7

Theo Nationnal Bureau of Atandard S.408 Westinghouse:

S2 (VA) 2,5 5 9 25 50 80 150 200 500 650 % 78 81,8 84,2 87,7 88,8 90,5 92,5 92,7 94,3 94,4

Theo Schindler:

S2 (VA) 100 200 300 500 % 92,5 93,5 94 94,5


15/39

Theo Transfor Matoren Fabrik Magnus

S2 (VA) 25 50 75 100 150 200 250 400 500 % 84,2 86,8 89 90 91 91,9 92 93,2 93,6

Sau khi tra bng, chn c % cho bin th, t xc nh c dng in pha s cp:

(4.38)

Bc 8:

Chn mt dng in J, suy ra tit din v ng knh dy dn pha s cp v th cp.

Chn J xc nh ng knh dy dn ph thuc vo cc yu t:

- Cp cch in vt liu.

- iu kin gii nhit dy qun.

- Ch lm vin (di hn hay ngn hn).Ta c th tham kho cc bng gi tr cho php ca J nh sau:

Bng quan h gia J theo S2, khi bin th vn hnh lin tc, iu kin gii nhit km (hoc cp

cch in thp).

S2 (VA) 0 50 50 100 100 200 200 500 500 1000J (A/mm2) 4 3,5 3 2,5 2

Trng hp vt liu cch in cp A (nhit ti a im nng nht cho php l 105 0C), my

lm vic ngn hn, ta c th chn J cao hn gi tr bng trn t (1,2 1,5)ln. C th ta c.

S2 (VA) 0 50 50 100 100 200 200 500 500 1000J(A/mm2) 5 6 4,5 5,5 4 5 3,5 4,5 3 4Ngoi ra ta cng c th chn J theo nhit pht nng cho php:

At (cm2)

J (A/mm2)

vi gia

nhit 400C

J (A/mm2)

vi gia

nhit 600C

At (cm2)

J (A/mm2)

vi gia nhit

400C

J (A/mm2)

vi gia nhit

600C1,0 4,6 5,5 6,0 2,3 2,81,4 4,0 4,9 6,5 2,25 2,72,0 3,5 4,3 7,0 2,2 2,62,4 3,3 4,0 7,5 2,15 2,62,8 3,1 3,7 8,0 2,1 2,5

3,0 3,0 3,6 9,0 1,9 2,43,5 2,8 3,4 10 1,8 2,34,0 2,7 3,3 15 1,6 1,94,5 2,6 3,2 20 1,4 1,85,0 2,4 3,0 30 1,25 1,55,5 2,35 2,8 40 1,15 1,4

Cn c theo cc s liu tham kho trn, chn J v suy ra ng knh dy qun s cp v th cp.

1

21

.%U

SI

=


16/39

Gi d1 v d2 l ng knh dy dn trn (khng tnh lp cch in bc bao quanh dy) ti s v

th cp. Ta c:

(4.39)

(4.40)

Th d 3:

Tnh s liu ng knh dy qun ca bin th kho st trong cc th d 1 v th d 2.

Gii:

Trong cc thnh phn tnh ton trc ta c:

I2 = 5A; S2 = 75VA; U1 = 110V.

Bc 7:Chn % = 88% ng vi S2 = 75VA

Dng in pha s cp l:

Bc 8:

Gi s bin th vn hnh 10 gi/ngy, cch in s dng cp A, chn J = 5,5 A/mm2 (ng vi S2= 75), suy ra ng knh dy qun s v th cp nh sau:

d1 = 1,13 5,5

775,0

= 0,424 mm, chn d1 = 0,45 mm.

d2 = 1,135,5

5= 1,07mm, chn d2 = 1,1mm

Chn dy emay c ng knh dy k c cch in l:

d1c = 0,5mm

d2c = 1,15mm

Bc 9:

Chn b dy cch in lm khun qun dy (ec) v b cao hiu dng qun dy (Hhd)

d thi cng qun dy, thng thng ta chn:

ak= a + (1 2)mm

bk= a + (1 2)mm

Hhd = H - [2ec + (1 2)mm]

Trong :

Hhd: l b cao hiu dng qun dy

ec: l b dy ba cch in, chn theo cp cng sut ca bin p.

J

Id 11 13,1=

J

Id 22

13,1=

A775,0110.88,0

75I 1 =


17/39

m bo bn c hc chn ec theo cp cng sut ca bin th nh bng sau:

S2 (VA) 1 10 10 200 200 500 500 1000 1000 3000ec 0,5 1 2 3 4

Bc 10:

Xc nh s vng cho mt lp dy qun s v th cp.

Gi: SV1 l s vng mt lp dy qun s cp.

SV2 l s vng mt lp dy qun th cp.

Ta c:

(4.41)

(4.42)

Trong :

Kq: l h s qun dy

+ Vi dy ng bc cotton: Kq = 0,9 0,93

+ Vi dy ng trng emay: Kq = 0,9 0,93

Bc 11:

Xc nh s lp cho mi phn dy qun s v th cp.

(4.43)

(4.44)

K.

d

HSV q

1cd

hd1 =

K.d

HSV q

2cd

hd2 =

1

11 SL

SV

N=

2

22 SL

SV

N=

Hnh 4.10: Chn kch thc cch in lm khun qun dy


18/39

Trong (4.43) v (4.44) lm trn s, sau xc nh b dy cch in gia mi lp dy qun bn

s v th cp, sau cng xc nh cch in gia cun dy s cp v th cp.

Ta c cng thc tng qut:

(4.45)

T cng thc tng qut (4.45) ta vit li cch tnh cho ec1 v ec2:

(4.46)

(4.47)

Trong :

ec1 v ec2: n v l [mm]SV1 v SV2: n v l [vng/lp]

nv: n v l [vng/vn]

Bc 12:

Xc nh b dy mi phn dy qun.

Khi bin p c li thp E I, cun dy s cp v th cp qun trn mt trc li (b tr ng

trc), ta xc nh b dy cun dy s v th cp nh sau:

Gi: BD1 l b dy cun dy s cp.

BD2 l b dy cun dy th cp.

BD l b dy tng ca c b dy.Ta c:

(4.48)

(4.49)

(4.50)

Trong :

ec3 l cch in gia s v th.

(4.51)

Cui cng, kim tra h s lp y kl1 theo b dy chon ch cun dy so vi b rng ca s li

thp, ta c:

(4.52)

B dy cch ingia 2 lp lin tip

nhau [mm]

Hiu in th gia hai lp

[V]1000

= 1,4

Vn

SV1cd1 0,0624e =

Vn

SV2cd2

0,0624e =

)e(dSLBD cd11cd11 +=

)e(dSLBD cd22cd22 +=

cd3c21 eeBDBDBD +++=

10001,4e 21cd3

UU +=

C

BDK ld1 =


19/39

Gi tr ti a cho php ca Kl1 b lt cun dy (k c cun dy s cp v th cp) vo ca s

l Kl1 = 0,7 0,8.

Nu Kl1 tnh tha mn gi tr ni trn th ta tnh tip cc bc cn li.

Nu khng tha mn gi tr ni trn ta phi tnh li, iu chnh li kt cu b lt dy qun.

Ch :

Cng c th kim tra bng cch tnh khc (ngay sau bc 8)

Gi kld2 l h s lp y, tnh theo tit din chon ch ca dy qun so vi tit din ca s mch

t.

(4.53)

Gi Sc1 v Sc2 l tit din dy qun s v th cp k c lp bc cch in, ta c:

(4.54)

Nu Kl2 = 0,4 0,46 th b dy b lt vo ca s, khong gi tr ny tng ng khong gi tr

Kl1 = 0,7 0,75.

Bc 13 :

Xc nh chiu di trung bnh cho mt vng dy qun s cp v th cp, suy ra tng b di cho

b dy s cp v th cp.

Trong bc ny, ty theo b dy s cp v th cp lp t theo dng no (cng mt tr hay hai

tr khc nhau) m ta c cch tnh khc.i vi bin th hai dy qun thng ta c mt s cch b tr dy qun nh sau (xem hnh 4.11)

Kl1 =

Tng din tch chon ch ca b

dy

Tit din ca s li thp

hcSNSN cdcd

...K 2211ld2 +=

S cp

Thcp

S cp

Thcp

S cp

S cp

Thcp

S cp Thcp

Hnh 4.11: Mt s cch b tr dy qun i vi bin th hai dy qun


20/39

Gi s kt cu dy qun s cp b tr bn trong v th cp bao bc quanh s cp, b di trung

bnh Ltb1 v Ltb2 cho b dy s v th cp xc nh nh sau (xem hnh 4.12).

t : a = a + 2ecb= b + 2ec

Ta c :

(4.55)

Tng t, suy ra:

(4.56)

Gi L1 v L2 l tng b di ca b dy qun s v th cp.

Hnh 4.12: Cch b tr dy qun s cp bn trong v thcp bn ngoi bao bc quanh s cp

1

//

tb1 BD.)b2(aL ++=

( )[ ]2cd1//

tb2 BDeBD2.)b2(aL ++++=


21/39

Ta c:

(4.57)

(4.58)

Bc 14:

Xc nh khi lng dy qun s cp v th cp:

(4.59)

Trong :

Kdp: l h s d phng do sai s trong thi cng thc t so vi tnh ton.

+ Vi dy emay: Kdp = 1,1 1,15.

+ Vi dy bc cotton: Kdp = 1,2 1,3.

Tng t, khi dy qun th cp c tnh:

(4.60)

Th d 4:

Xc nh khi lng s dng cho b dy bin th tnh trong cc th d 1, 2, 3.

Gii:

T cc th d 1, 2, 3 ta c cc kt qu sau:

a = 32mm, b = 34mm, c = 16mm, h = 48mm.

S vng dy s cp:N1 = 400 vng

d1/d1c = 0,45/0,5mm (dy trng emay)

S vng dy th cp:

N2 = 60 vng

d2/d2c = 1,1/1,15mm (dy trng emay)

Kim tra s b h s lp y Kl:

S1c =4

5,0. 2= 0,196 mm2 0,2mm2

S2c =415,1. 2 = 1,038 mm2 1,04mm2

Din tch ca s li thp:

Scs = c.h = 16 . 48 = 768mm2

Kl =768

4,142

768

04,1.60.2,0.400=

+= 0,185

8,9.10.4

.d.L.k=W

4-

2

1

1dp1

10.8,9.4

.d.L.K=W 4-

2

22dq2

.LNL tb111 =

.LNL tb222 =


22/39

Vi Kl tnh c qu thp so vi tiu chun cho php (0,7 0,8) ta c th iu chnh gim kch

thc li thp, gim khi lng dy. Tuy nhin mun duy tr cc tham s khc khng i ta phi

gi tit din li thp nh lc u tnh.

Ta th xt phng n iu chnh nh sau:

Chn Kl tng ln khong 0,36 v gi s s liu dy qun s v th cp khng i. Tng

din tch chon ch ca b dy khng i, vn bng 142,4mm2. Suy ra din tch ca s l:ca s l:

Scs =36,0

4,142= 395,55mm2

Cn c theo Scs tnh ra a:

V: c =2

av h =

2

3a

Nn: Scs =4

3 2a

Vy: a =3

55,395.43

4 =csS = 22,96 mm

i chiu theo th d 1 ta c th chn a ti mc thp nht l a = 24mm

Nu duy tr s vng nh c, theo th d 2 ta cn duy tr:

Ag = 10,88cm2 c At = 10,336cm2

Mun vy: b =4,2

88,10=

a

Ag= 4,5cm

Tm li, ta iu chnh li kch thc li thp gim khi lng thp v khi lng dy ng,

ng thi nng cao Kl, li dng ti a khong trng ca s li thp.

Ta chn:

a = 2,4cm; b = 4,5cm; Wth = 46,8a2b = 1,21kg 1,2kg

(Xem hnh 4.13)

Hnh 4.13: Kch thc li thp cn dng sau khi iu chnh

24mm

45mm

12mm 36mm


23/39

Nh vy, kt cu mi c iu chnh li c s liu nh sau:

Ag = 10,8cm2 vi Kf= 0,95 (kh nng ghp st)

At = Ag.Kf= 10,8 . 0,95 = 10,26cm2; Ssc = 432 cm2

Bm = 1,2T; nv = 3,66 vng/vn

N1 = 402 vng; N2 = 60 vng

d1/d1c = 0,45/0,5mm; d2/d2c = 1,1/1,15mm

H s lp y rnh (tnh theo tit din) l Kl = 0,53Cn c vo s liu mi, tnh li cc bc t bc 9 n bc 14

Bc 9:

B dy khun qun dy: ec = 1mm

Kch thc khun giy:

ak= a + 1mm = 25mm

bk= 45 + 1mm = 46mm

B cao hiu dng:

Hhd = h - (2ec + 1mm) = 36 - (2,1 + 1) = 33mm

Bc 10:S vng qun cho mi lp s cp v th cp:

SV1 = 627,6295,0.5,0

33.

1

==qcd

hd Kd

Hvng / lp

SV2 = 2726,2795,0.15,1

33= vng / lp

Bc 11:

S lp ca cun dy s v th cp:

SL1 = 748,662

402

1

1

==SV

N

lp

SL2 = 32,227

60

2

2==

SV

Nlp

Xc nh b dy cch in gia cc lp s cp vi nhau:

ec1 = 0,0624v

n

SV1 = 0,0624 66,3

620,25mm chn ec1 = 0,25mm


24/39

Xc nh b dy cch in gia cc lp th cp vi nhau:

ec2 = 0,0624 66,3

270,169mm chn ec2 = 0,2mm

Bc 12:

B dy cun dy s v th cp:BD1 = SL1 (d1c +ec1) = 7(0,5 + 0,25) = 5,25mm

BD2 = SL2 (d2c +ec2) = 3(1,15 + 0,2) = 4,05mm

B dy cch in gia cun s cp v th cp:

ec3 = 1,4.1000

21 UU + = 0,49mm

Chn ec3 = 0,5mm. Tng b dy b dy BD = BD1 +BD2 + ec3 = 9,8mm

Kim tra li h s lp y theo b dy ca s b chon ch

Kl =

c

BD=

12

8,9= 0,816

Ch : Nu cch in lp chn theo tiu chun k thut nh trn ta c th gim b dy cch in

tnh theo cng thc (4.31), (4.32), (4.33), (4.37) xung 0,5 ln.

Vi th d tnh ton trn nu hiu chnh:

ecd1 = ecd2 = 0,1mm v ec3 = 0,25mm, ta c:

BD1 = 4,2mm; BD2 = 3,75mm; BD= 8,2mm nn Kl = 0,683

Bc 13:

Chn cch b tr b dy ging nh hnh 4.12, ta c:

a = a + 2ec= 24 + 2.1 = 26mm.

b

= 45 + 2 = 47mm.B di trung bnh ca mt vng dy qun s cp:

Ltb1= 2 (a+b) + .BD1 = 2.(26 + 47) + 3,14 . 4,2 = 159,19 mm

Chn Ltb1= 159,2 mm = 1,592 dm .

B di trung bnh ca mt vng dy qun th cp:

Ltb2= 2(a+b) + .[2(BD1+ec3) + BD2] = 185,74 mm

Chn Ltb2= 186 mm = 1,86 dm.

Tng b di cun dy qun s cp:

L1= N1.Ltb1= 402.1,592 = 640 dm

Tng b di cun dy qun th cp:

L2= N2.Ltb2= 60.1,86 = 111,6 dm = 112 dm.

Bc 14:

Khi lng b dy qun s cp:

W1= 1,1 . 640 .4

45,0. 210-4.8,9

W1= 0,0996 kg 0,1 kg .


25/39

Khi lng b dy qun th cp:

W2= 1,1.112.4

1,1. 210-4.8,9

W2= 0,1042 kg 0,11 kg .

Tng khi lng b dy qun:W= W1 + W2 = 0,1 + 0,11 = 0,21 kg.

TM TT KT QU NH SAU :

Kch thc mch t:

S liu:

A = 24 mm; b = 45 mm; Ag = 10,8 cm2; At = 10,26 cm2

Bm = 1,2T; nv = 3,66vng/vn;N1 = 402 vng.

d1/d1c = 0,45mm/0,5mm; N2 = 60 vng Wth = 1,2kg

d2/d2c = 1,1mm/1,15mm; W2 = 0,11kg W1 = 0,1kg

H s lp y: Ll = 0,33 (tnh theo thit din chon ch)

Ll = 0,68 (tnh theo b dy chon ch)

B dy cch in gia cc lp s cp v th cp:

ec1 = ec2 = 0,1 mmB dy cch in khun: ec = 1 mm

B dy cch in gia s cp v th cp: ec3 = 0,25 mm

4.2.2.Phng php tnh ton my bin p t ngu:

a. Trnh t tnh ton my bin p t ngu:

Hnh 4.14: Kch thc mch t

24mm

45mm

12mm 36mm

U1

= 110V U2 = 15V

S2

=

75VA

I2

= 5A


26/39

i vi MBA t ngu 1 ng vo v 1 ng ra th 2 ng ny c th hon i cho nhau khi s dng,

do khi tnh ton phi m bo c 2 trng hp.

Gi SBA l dung lng ca bin p trong c 2 trng hp.

Cn c vo U1, U2 c suy ra I1, I2. Sau cn c vo s s tnh ra c dng

in chy trong on dy chung IC. Tip theo s dng cc biu thc phn 4.2.1 tm cc thng s cn li.

Th d:

Xc nh cc thng s ca MBA t ngu 110V / 220 V; S = 550VA.

Gii:

Gi Im l dng in ng vi in p thp ca my.

Khi ng ra l 110V

- Chn hiu sut = 90% th I1 c tnh

- Tnh dng in trong on dy chung

Khi ng ra l 220V:

I1 =220.9,0

550

. 1=

U

SBA

= 2,8A = 0,56 Im

IBC = IC = I2 I1 = 5 2,8 = 2,2 A = 0,44 Im

Hnh 4.15: My bin p t

ngu220V/110V

110

220

0A

B

C

110

220

I2

I1

IC

I2 =110

550

2

=U

SBA= 5A = Im

I2 =220

550

2

=U

SBA = 2,5A = 0,5 Im


27/39

- Chn hiu sut = 90% th I1 c tnh

- Tnh dng in trong on dy chung:

- Tm li ta c :

IBC (on 110 220) IAB (on 0 110)Ng ra l 110V 0,56 Im 0,44 ImNg ra l 220V 0,50 Im 0,60 Im

thun tin trong thi cng m vn m bo cng sut ca MBA c th chn mt loi dy

theo dng in 0,6 Im = 3A.

b. Qui trnh qun dy Adaptuer v Survoltuer:

S nguyn l:

IBC = IC = I1 - I2 = 5,6 - 2,5 = 3,1 A 0,6 Im

I1 =110.9,0

550

. 1=

U

SBA

= 5,6A = 1,1

Im

Hnh 4.16: S nguyn l Adaptuer

0

110

22

0

03

6

9

4,5

12

7,5

+

_

110

220

C+

Hnh 4.17: S nguyn l Survoltuer

7

110

220

5

3

6

9

160

8

80

4

21

0V

0

220

110

C

S

A

G1

G2 C

RA TI

NGUN VO


28/39

Tnh ton theo s

Da vo s nguyn l, tnh ton cc thng s ca adaptuer hoc survoltuer theo trnh t bi

ton ngc.

a. i vi adaptuer: Pha th cp c nhiu ng ra. Cn ch gi tr in p ca tng ng

tnh s vng qun cho chnh xc.b. i vi survoltuer: Cc u dy gallett G2 s iu chnh c in p ra trong khong

50V. Do c th chia u gi tr in p trong tng cp hoc thng tnh theo t l sau: T s 1 n

s 7 mi s cch nhau 5V; t s 7 n s 9 mi s cch nhau 10V.

Lm khun

Xc nh kch thc khun qun; k thut lm khun; bc cch in khun thc hin tng t

hon ton nh trng hp 4.2.1.

Qun dy

- Bc 1: Chun b cc vt t, thit b cn thit (bn qun, dy in t, gen cch in, ko,

dy ai... ).- Bc 2: G khun v m p ln bn qun v xit cht c nh v khun.

- Bc 3: Quay th v chnh b m v 0.

- Bc 4: Tin hnh qun dy.

Chn mt mt a trn khun qun t u dy ra u tin.

Kha cht vng dy u tin.

X ng gen 1mm vo u dy ra.

Tin hnh qun dy ri theo tng lp.

Lt cch lp (sau khi qun xong mi lp).

Cn khon 5 n 10 vng cui cng th t vo 1 on dy ai chun b kha vng

dy cui cng.

Kha vng dy cui, x gen vo u dy cui.

Tho khun ra khi bn qun, gi c nh cun dy.

Tho cun dy ra khi khun qun, bc thm mt lp cch in bn ngoi.


29/39

Cc im cn lu :

- Hai u dy ra phi c cch in bng gen v phi b tr mt a ca khun qun.

- Phi hn ch ti a cc mi dy b cc khi vut sa dy.

- Ch khng mt vng dy no nm ngoi giy cch in.

- Trc khi qun phi quan st v tr cc gallett G1 v G2 trn v hp chn hng ra cc u

dy ph hp.

- Cc mi ni, nu c th cng phi a ra s l hn ni bn ngoi.

- Cc u dy ra phi di a t v tr li thp n cc gallett m khng phi ni dy.

- Cc u dy ra phi c gen cch in, c bit l phn nm trong khun qun. Nn sp xp

ng th t thun tin trong vic lp rp.

- Khi qun xong phi bc cch in bn ngoi bng giy cch in dy t 1 n 2 lp. Sau

qun cun cm ng bn ngoi thun tin cho vic chnh ng h.

- i vi adaptuer do cun s cp dng dy rt nh nn khng cn qun xp lp, nhng cng

phi ri cho tng i u b mt. cun th cp th qun tng t nh survoltuer.

Lp rp

- Tho khun qun ra khi cun dy. X l cc mi ni, cch in cn thn; sau kim tra li

cc lp bc cch in.

- Lng xen k cc l thp vo lng cun dy.

- Co sch cc u dy ra, a vo ngun cho my vn hnh th o kim li cc cp in p

ng ra.

- Hn ni cc u dy vo ng s . Kim tra li s thay i in p ng ra khi iu chnh

gallett G2 .

Ch : Khi lp rp phi m bo cch in tuyt i gia cc phn mch in v v hp.

Cn chnh: i vi survoltuer

Chnh ng h:

- t G1 v tr 220V; G2 v tr s 1 cp ngun 220V cho my.

- o kim gi tr thc t ca ngun cung cp.

- Chnh cho ng h ch ng gi tr in p ngun bng cch thm hoc bt s vng qun

ca cun cm ng (Nu ng h ch cao hn th gim bt s vng qun ca cun cm ng v ngc

li).

Chnh chung:

- t G1 v tr 220V; iu chnh tng G2 sao cho ng h ch 240V.- Chm u dy C trong mch chung in vo cc u t s 9 v s 1 s c nhng v tr

chung reo.

- Gi nguyn u dy C ti v tr chung reo. H G2 t 1 n 2 s: nu chung khng cn

reo, th hn u dy C vo im .

- Cn nu khi h G2 nh trn m chung vn cn reo th phi tm im khc c in p

thp hn (di v hng c s cao hn trn gallett G2).


30/39

i vi Adaptuer:

- Kim tra cch lp cc diode trong cu chnh lu.

- Ch cc tnh ca t lc.

- Kim tra cc tnh in p ra, ghi k hiu cn thn.

- Cn ghi k hiu r rng trn cng tc chuyn i 110V/220V.

4.2.3.Phng php tnh ton my bin p dng cho b np c quy:

a. Tnh dng in v in p cn np cho c qui

Dng in np:

(4.61)

Trong :Q: Dung lng ca c quy [Ah].

t: Thi gian np (7 10)h. in p np:

(4.62)

b. Chn mch chnh lu, t tnh ra dng in v in p th cp MBA tng ng:

S dng chnh lu cu (4 diode):

(4.63)

Dng mch chnh lu sao 1 pha (bin th i xng _ 2 diode):

(4.64)

c. Tip tc tnh ton cc thng s cn li ca MBA tng t hon ton MBA cch ly.

d. Trng hp MBA np cho nhiu bnh c quy ng thi th trc tin phi chn phng

n np ri sau mi tnh cc bc t 1 n 3.

Trng hp np song song: Inptng ln; Unpkhng i. Trng hp np ni tip: Inp khng i; Unptng ln.

V d: Xc nh cc thng s ca MBA dng cho b np c qui. Bit rng ph ti l mt bnh

c qui c dung lng 45Ah; Ubnh = 12V.

Gii:

Dng in np v in p np cho c qui:

Inp =t

Q

I2 = 1,11 Inp; U2 = 1,11 Unp

I2 = 0,785 Inp; U2 = 1,11 Unp

Unp = ( 1,15 1,25 )Ubnh


31/39

Inp =t

Q=

10

45= 4,5A (chn t = 10h)

Unp = (1,15 1,25) Ubnh = 1,15.12 = 13,8 V; chn Unp = 15V.

Chn mch chnh lu cu:

I2 = 1,11 Inp = 1,11. 4,5 = 4,995A 5A;U2 = 1,11 Unp = 1,11. 15 = 16,65V 17V;

Vy dung lng bin p trong trng hp ny l SBA = U2. I2 = 17. 5 = 85VA.

Cn nu chn mch chnh lu sao 1 pha th:

I2 =0,785 Inp = 0,785. 4,5 = 3,5A;

U2 = 1,11 Unp = 1,11. 15 = 16,65V 17V;

Vy dung lng bin p trong trng hp ny l SBA = 2U2.I2 = 2.17. 3,5 = 119VA.


32/39

Gii thiu mt s s np in cho c quy:

4.2.4.K thut qun dy my bin p:a. Lm khun cch in:

Khun cch in nhm mc ch cch in gia cun dy v mch t, cn lm sn cng nh

giy cch in presspahn, php (fibre) hoc bng cht do chu nhit.

C 2 dng khun:

-Khun khng vch chn c s dng i vi my bin p ln (hnh 4.20a)

-Khun c vch chn thng s dng cc my bin p nh (hnh 4.20b)

Hnh 4.18: Mch np acquy 6V/12V _ DC; Ng vo chuyn i110V/220V

0110

220

_

+6 6

V

-

DC

- 12

+12

1CD 2CD- 6

+

12V

-

DC

G1

220V

_

6

24

12

Hnh 4.19: Mch np acquy 6V/ 12V/ 24V _DC;

Ng vo 220V iu chnh c

+

DC

G2


33/39

Ch :

Kch thc ca khun so vi kch thc ca li nh sau:

Cc h s d tr b, c v h c chn sao cho khng hp qu hoc rng qu, sau ny khi

lp vo mch t khng b cn d gy s chm masse. C th:- ak= ali cc l thp p cht vo nhau.

- ck< cli khong 0,5mm lp khun d lt vo ca s.

- hk< hli khong 1mm khe h mch t gia I vi ch E st kht nhau

- bk> bli khong 1mm d lp ch E vo khun.

- Gc tip gip gia ak, v bk theo chiu cao ca hk phi vung thnh, sc cnh khng un

ln khi lp l thp th mt trong ca akst kht vi mt l thp .

Nu c vt liu bng ba mica, baklt hoc cc tng chu nhit cng, b dy 0,5mm lm khun

qun dy rt tt.

Sau khi ly mu khun cun dy, thc hin khun nng cho kht khao vi khun cch in. Mcch l khi lp khun vo trc my qun dy lm sao cho tm ca khun trng vi tm trc my.

Khun nng lm bng g c kch thc nh hnh 4.21, gia mt phng akxbk khoan mt l c

ng knh bng ng knh trc my quay sut dc chiu di hk.

ng thi, gia cng thm 2 tm chn (m p) (hnh 4.22) bng g, vung, kch thc 15x15cm

(tt nht l g vn p), c b dy khong (3 5)mm p cht 2 u khun trn trc khi quay my

qun dy.

bk

= b + b

hk= h - h

ak

ck= c -

c bk = b +

b

Hnh 4.20: Kch thc khun

a). b).

ak

hk= h - h


34/39

b. K thut qun dy:

-Trc khi qun dy phi v s b tr cc dy ra v tr thc t sau ny khi ni mchkhng b vng v d phn bit (Hnh 4.23a).

-Trc khi qun dy c nh u dy khi u nh hnh v (Hnh 4.24a). Trong lc qun dy c

gng qun dy cho thng v song hng vi nhau. C ht mi lp dy phi lt giy cch in. i vi

dy qu b (d < 0,15) c th qun sut lun khng cn lt giy cch in gia cc lp. Ch lt cch

in k gia cun s cp v th cp m thi.

-Khi qun na chng mun a dy ra ngoi thc hin nh hnh (Hnh 4.24b). Dy a ra ngoi

ny phi c cch in bng ng gaine cch in. Vic ni dy gia chng cng phi a mi ni ra

ngoi cun dy (Hnh 4.24c).

-i vi loi khun khng c vch chn dy, gi cc lp dy khng b chi ra ngoi khun,dng bng vi hoc giy chn dy li c 2 pha u cun dy (Hnh 4.24d).

-Khi sp hon tt vic qun s vng dy, phi t dai vi hoc giy (H3-3c) sau y qun dy

chng ln bng vi, giy , cui cng ln dy qua v rt cht bng vi gia cho chc.

Hnh 4.22: tm chn (m p)

bk

ak

hk

Hnh 4.21: khun nng


35/39

c. Cch rp li l st ca mch t:

Hnh 4.23a: cch b tr cc dy ra v tr thc t Hnh 4.23b: cch ni cc mi ni dy

Hnh 4.24a: Cch c nh u dy khi u Hnh 4.24b: Cch ra u dy gia cun dy

Hnh 4.24c: Cch c nh u dy cui cun dy Hnh 4.24d: Cch gi cc lp dy

Hnh 4.23c: Cch un cc dy ra thc t(dng khoen kn)

Hnh 4.23d: Cch un cc dy ra thc t(dng khoen h)


36/39

-Tu theo dng l st ghp thnh mch t l dng EI hoc cc thanh ch I m ghp theo trt t

c tnh trc.

Cch ghp mch t vi l st EI:

Lp tng l st E sut dc chiu (b) ca khun, tr u i din nhau. Cc l st cui cng

thng rt kh lp phi dng ba st lt mt ming g ng dn dn, nh nhng cho l st p cht vo

li khun.

Sau khi lp cht cc l st ch E, v cc ch E tr u nn gia 2 gng t ch E c mt

khe h lp ch I. Cc l st ch I cng lp dn vo cc khe h c 2 pha ca khun

Ch :

Cc l st cng p cht, khi vn hnh MBA khi rung v khng pht ting .

Nu cc l st lng ngoi ting ku v rung, MBA cn b nng ln do t tr ln.

Cch ghp mch t vi l st ch I:

d. Hn 2 u dy vo-ra:

Cc u dy vo ra ca hai cun dy phi nm cng mt pha ca tai khun.

Hnh 4.25: Cch ghp mch t vi l st EI

Hnh 4.26: Cch ghp mch t vi l st ch I


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Vi nhng MBA dng c dy ng knh rt nh, cc u dy vo ra ngi ta khoan hai l st

nhau tai khun qun vi vng dy ca cc u ra phng dy qu nh rt d t.

Nhiu khi cc u ra ca cc loi dy qu nh, ngi ta gn mt ming tn st hoc tn ng

ri hn cc u dy ra ca cun dy v cc u dy ngun v ti. Dy ngun v ti s dng loi dy si

n, mm. Ty theo cng sut MBA m chn dy ngun, ti c tit din ph hp.

e. Th nghim:

S dng m k kim tra cch in gia 2 cun dy, gia cun dy vi li st. Nu 2 cun dy

chm nhau hoc chm li st phi tho ton b ri qun dy li.

u in ngun kim tra in p U2 c ng thit k khng.

f. Sy s b:

Thng trong iu kin mi trng m thp, lp may v ba cch in rt d ht m nn phi

sy s b cho kh hi m.

g. Tm sn cch in:

Thng cc MBA lm vic trong iu kin mi trng m thp phi tm sn cch in.

Sau khi sy s b phi tm sn cch in bng cch:- Nhng ton b MBA vo sn cch in n lc khng thy bt kh ni ln na mi ly MBA

ra.

- sn cch in t t vo cc cun dy.

h. Sy li v xut xng:

Sau khi tm sn phi sy li cho kh sn, kim tra cch in, U2 mt ln na ri cho xut xng.

4.2.5.Cc pan thng thng trong my bin p:

Pan chm masse:

- Trng hp ny gy hin tng in git, nu km s n cu ch, bc khi nh th do s chm

masse lm chp mch cun dy.-C th do b chm gia cc cc ni vi v st hoc c s c ni tt gia cc cc ni cc do

din. Dng n th hoc m k kim tra cc im cn lu xc nh ni b chm, chp mch... sau

sa cha li cho ht b chm masse.

-Nu my bin p vn vn hnh bnh thng, th ni b chm ch c 1 ch, c th ng dy ra

cc ni b trc lp cch in chm vo v bc my bin p hoc cc ni b lng lo chm b bc hoc

chm masse lp dy tip cn vi mch t. Trng hp sau cng ny, nu quan st khng thy c

ch chm masse.

-Nu my bin p vn vn hnh bnh thng m gy s git nh. Trng hp ny my bin p

khng b chm masse m do my bin p b m, in tr cch in b suy gim (nu dng bt th in

thy cch in bng M-gm k sao cho trn 1 M l tt. Nu khong t, lp cch in b lo ho cn

phi qun li ton b.

My bin p ang vn hnh b n cu ch:

-Nu my bin p b pht nhit thi qu, c th l do mch tiu th qu ln. Thay li dy ch

ng c v cho my bin p vn hnh khng ti, nu vn bnh thng chng t lc trc my bin p

lm vic qu ti.


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-Nu my bin p vn hnh khng ti m cu ch vn n th chc chn my bin p chp vng

trong cun dy, phi qun dy li.

-i vi my bin p c cng sut nh th s chp vng kh lm cu ch n ngay nhng c s

pht nhit rt nhanh.

-i vi my bin p np c quy, chnh lu ton k, lu diode b hng ni tt. Hoc mc nhm

2 cc (+) v cc (-) vo bnh c quy (Hnh 4.27)

-Nu my bin p b pht nhit thi qu, c th l do mch tiu th

My bin p vn hnh b rung ln, km s pht nhit:

-Do dng in tiu th qu ln, qu cng sut ca my nn my bin p rung ln pht ting r, lu pht nhit nhanh, chng chy my bin p. khc phc cn gim bt ti.

-Do mc khng ng vi in p ngun, nhm vo ngun c in p cao.

-Do mch t ghp khng cht. Phi sit cht li cc bulong p gia cc l st ca mch t v

tm verni vo cun dy v vo cc khe h chn cng cc l st li, dnh cht hn.

-Do bn cht l st ca mch t km phm cht, qu r st hoc qun thiu vng dy.

My bin p khng vn hnh:

-Nu n bo khng sng hoc khng cm thy my bin p rung nh nh do c dng in vo,

th lu ng dy vo b h mch, cc ni dy vo khng tip in, hoc tip xc xu o in.-Nu n bo sng, vn k hot ng m in p ly ra khng c, phi xem li cc ni dy ra b

tip in xu, t dy ra... Dng vn k hoc bt th in d tm xc nh ch pan khc phc.

-Nu b h mch bn trong cun dy, c th do mi ni dy cu th, khng hn ch nn tip

in xu sau mt thi gian s dng, hoc dy qun b gy t... Trng hp ny phi tho ra qun li.

Hnh 4.27: Cc s c i vi my bin p np c quy


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-i vi np c quy, c th diode chnh lu b hng t mch. Trng hp ny d pht hin khi

dng vn k o c in p xoay chiu U2, nhng khng c in p ra UDC ch cn thay mi diode m

thi.

My bin p lc vn hnh, lc khng:

-Nhn chung do ngun in cung cp vo my bin p lc c, lc khng hoc in p ra b t

qung, chnh l do tip xc xu. Nn kim tra li t ngun in cung cp n my bin p v t my

bin p n mch tiu th. Lu ni cu dao chnh, xit li cc c vt xit dy ch cho cht, co sch

ni tip in ht ten ng ti cu dao chnh, cc cc ni my bin p...

Mt s pan trong my bin p gia dng:

Ngoi s pan nu trn i vi my bin p gia dng c c mt s pan nh sau:

-Chung bo sm nhng in p ra vn khng cao do tc te iu khin chung b hng, nn thay

ci mi.

-Chung khng bo, mc d in p ra qu in p nh mc. Do tc te b hng lm h mch

chung, cun dy chung b chy.

-n bo khng sng nhng my bin p vn hot ng bnh thng. Do b t bng, mch n

b h mch.

-Vn k ch sai tr s in p. Hiu chnh li v i chiu vi vn k chun hoc thay vn k

mi.

-Khng tng c in p ra n in p nh mc. Do in p ngun xung qu thp ngoi

khong cho php ca my bin p hoc do qu ti (my bin p rung rn ln). Trng hp ny do s

thit k my bin p, cun s cp qun d vng nn c tr khng ln gy s st p ln bn trong cun

dy. V th khng th nng in p ln c, khi in p ngun b suy gim thi qu.

Mt s pan trong my bin p np c quy:

Ngoi s pan ni chung, cn ring i vi my bin p xc c quy c cc trng hp sau:

- My bin p pht nhit thi qu, n cu ch hoc cng tc bo v qu ti

(OVERLOAD) ca my xc ct mch. Cn phi xem li bnh c quy c b chm ni tt khng. Hoc

diode chnh lu ton k b ni ngn mch.

- My bin p mi vn hnh pht ting rung r v pht nhit. Cn ct

mch ngay, v do ni nhm cc cc (+) v cc (-) vo bnh c quy, gy ra dng in np ln trong mybin p. Nu lu c th lm hng diode, chy my bin p (trng hp khng c cng tc bo v

qu ti).

- My bin p np bnh yu. Do in p xc bnh thp hn in p ca c

quy. Lu 1 diode b hng t (chnh lu cu 4 diode), khng xc bnh c (chnh lu bn k).

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