n mn hc: thit k my bin p

1

LI NI U i vi chuyn ngnh Thit b in - in t , My bin p l mt lnh vc

rt quan trng. Chnh v vy, thit k mn hc My bin p c mc ch gip cho

sinh vin nm c nhng bc c bn nht trong vic tnh ton kt cu ca mt

My bin p.

Di s hng dn ca thy Bi c Hng - ging vin B mn Thit b

in - in t, em hon thnh n mn hc ca mnh vi ti Thit k My

bin p in lc ngm du. Trong qu trnh lm chc chn s khng trnh khi thiu

st, qua em mong thy gio v cc bn gp n c tt hn.

. SP = 250 kVA

. Um = 6,3/0,4 kV

. I0% = 2%

. P0 = 610 W

. UK% = 4

. Pk = 4100 W

. T ni dy Dy11

.iu chnh 2x2,5%

n mn hc: thit k my bin p

2

N THIT K MY BIN P A . Xc nh cc kch thc ch yu: I. Xc nh cc i lng in c bn:

1. Dung lng mt pha:

=3

250=S f 83,33 kVA

Dung lng trn mi tr:

=3

250=S t 83,33 kVA

2. Dng in dy nh mc: - Pha cao p:

=10.3,6.3

10.250=

U.3

S=I

3

3

1

pf1 22,9 A

- Pha h p:

=400.3

10.250=

U.3

S=I

3

2

pf2 360,8 A

3. in p pha nh mc: - Pha cao p: Ud1 = Uf1 = 6,3 kV - Pha h p:

3

400=

3

U=U 22f = 231 V

4. in p th nghim ca dy qun: - Dy qun cao p vi U1 = 6,3 kV Uth = 20 k V - Dy qun h p vi U2 = 0,4 kV Uth = 5 kV

II. Chn cc s liu xut pht v tnh cc kch thc ch yu: 1. Chiu rng quy i ca rnh t tn gia dy qun cao p v h p: vi Uth1 = 20 kV, ta c a12= 1,2 cm; 12= 4 mm. Trong rnh a12 ta dt ng cch in dy 12= 4 mm. Theo cng thc:

n mn hc: thit k my bin p

3

4 'P21 S.k=3

a+a

k l h s ph thuc dung lng my bin p, tra bng 13.1 ta c k = 0,45 thay vo ta c

44 'P21 33,83.45,0=S.k=3

a+a=1,51 cm

aR khong cch ph thuc kch thc hnh hc ca dy qun h p v cao p, vy:

71,2=51,1+2,1=3

a+a+a=a 2112R cm

2. H s quy i t trng: kR=0.95 3. Cc thnh phn in p ngn mch:

- in p ngn mch tc dng:

250.10

4100=

S.10

P=U

P

nnr =1,64%

- in p ngn mch phn khng: 64,1_4=U_U=U 222nr

2nnx =3,65%

4. Ta chn tn cn lnh m hiu 3405 c chiu dy 0,35mm. Ly mt t thng trong tr BT = 1,6T, h s kG = 1,02. p tr bng nm vi ng Baklt, p gng bng thp U khng dng bulng xuyn qua tr v gng. S dng li thp c mi ghp nghing 4 gc v ba mi p thng gia tr v gng. Theo bng 13.2 vi SP= 250 kVA ta chn tr c 6 bc, s bc thang ca gng ly nh hn tr 1 bc tc l ggng c 5 bc, h s in y kd=0,928; h s chm kn kc = 0.93, nn h s li dng li st kl = kd.kc = 0,928.0,93=0,86.

T cm trong gng BG = 1,6/1,02=1,57 T. Twf camr khe h khng kh mi ni thng Br = BT = 1,6 T, t cm mi ni xin BK= BT/ 2 =1,6/ 2 =1,13 T. Tra bng, ng vi tng gi tr mt t cm ta s c sut tn hao trong thp v tn hao t ho trong tr:

. Sut tn hao trong thp: -Trong tr pFeT = 1,230 W/kg -Trong gng pFeG = 1,17 /kg . Tn hao t ho -Trong tr qFeT = 1,602 VA/kg

n mn hc: thit k my bin p

4

-Trong gng qFeG = 1,486 VA/kg -Trong khe h khng kh +Ni thng qK= 1,92 VA/cm2 +Ni nghing qK= 0,272 VA/cm2 5. Cc khong cch cch in chnh . Gia tr v dy qun h p ao1= 5mm . Gia dy qun h p v dy qun cao p, a12 = 1,2 cm . ng cch in gia dy qun cao p v h p, 12= 0,4 cm . Gia cc dy qun h p, a22 = 2,2 cm . Tm chn cc pha 22= 0,2 cm . Gia dy qun cao p n gng, lo = 4 cm . Phn u tha ca ng cch in, ld2=3 cm 5. Cc hng s tnh ton , vi dy dn bng ng v in p dy qun cao p l

6,3 kV (bng 13.5 v 13.6) . a = d12/d = 1,36 . b = 2a2/d = 0,4 6. H s tn hao ph, vi cng sut 250 kVA ta chn kf = 0,94 (bng 13.7) 7. Chn vi di bin thin t 1,2 n 3,6, xc nh gi tr ti u ca ta

phi tnh cc s liu v cc c tnh c bn ca m.b.a:

+ 42l

2Tnx

RR'P

k.B.U.f

k.a.S=A

04,14=

88,0.6,1.65,3.50

95,0.71,2.33,8316= 4

22

+A1 = 5,66.10-2.a.A3.kl = 5,66.10-2.1,36.14,043.0,88 = 187 kg +A2 = 3,6.10-2.A2.kl.l0 = 3,6.10-2.14,042.0,88.4 = 25,1 kg +B1 = 2,4.10-2.kl.kG.A3.(a+b+e) = 2,4.10-2.0,88.1,02.14,043(1,36+0,4+0,411) = 129,43 kg e = 0,411 i vi thang nhiu bc + B2 = 2,4.10-2.klkG.A2.(a12+a22) = 2,4.10-2.0,88.1,02.14,042.(1,2+2,2) = 14,44 kg

+ C1 = 2nr

2T

2lf

2P

dqA.u.B.k.k

a.S.K

= 82,188=04,14.64,1.6,1.88,0.94,0

36,1.25010.46,2

222

22 kg

n mn hc: thit k my bin p

5

+ M = A.a

P.k.k.k.10.2453,0 nRf

2n

4-

trong :

kn = )e+1.(u

100.41,1 nxu

nru.-

n

= )e+1.(4

100.41,1 65,3

64,1.-

= 43,84

M = 04,14.36,1

4100.95,0.94,0.84,43.10.2543,0 24- = 9,04 MPa

+ Trng lng tn silic cc gc ca gng: Gg = 0,486.10-2.kl.kG.A3.x3 = 0,486.10-2.0,88.1,02.14,043.x3= 12,07x3 kg + Tit din tr li st: ST = 0,785.kl.kG.A2.x2 = 0,785.0,88.1,02.14,042.x2 = 139x2 + Tit din khe h vung gc: SK= ST = 139x2 + Tit din khe h cho SK = ST/ 2 =98,29x2 + Tn hao khng ti, theo cng thc 13-24: P0 = kf.( pT.GT + pG.GG ) = 1,25.( 1,23.GT + 1,17.GG ) = 1,538.GT + 1,463.GG kf h s tn hao ph trong st, tn cn ngui ly kf = 1,25 + Cng sut t ho, theo cng thc 13-26: Q0 = kf .(Qc + Qf + QK) Trong : . kf h s xt n s phc hi t tnh khng hon ton khi li l tn,

chn kf = 1,25 . Qc cng sut t ho chung ca tr v gng Qc = qT.GT + qG.GG = 1,602.GT + 1,486.GG . Qf Cng sut t ho i vi gc c mi ni vung gc

Qf = 40.qT.GG = 40.1,602.GG = 64,08.GG . QK cng sut t ho khe h khng kh ni gia cc l thp QK = 3,2.qK.SK = 3,2.0,272.139x2 = 120,9x2 Vy cng sut t ho tng l:

Q0 = kf .(Qc + Qf + QK)

n mn hc: thit k my bin p

6

= 1,25.( 1,602.GT + 1,486.GG + 64,08.Gg + 120,9x2 ) = 2.GT + 2,972.GG + 80,1.Gg + 151x2 Lp bng tnh gi tr cc tham s c bn ng vi tng gi tr ca bin thin t 1,2 n 3,6;t xc nh c ga tr ti u :

1,2 1,8 2,4 3,0 3,6 x= 4 1,047 1,16 1,25 1,32 1,38

x2=2

4 1,095 1,34 1,55 1,73 1,90

x3= 3

4 1,15 1,55 1,93 2,28 2,62

A1/x = 187/x 178,67 161,44 150,24 142,09 135,76A2.x2 = 25,1.x2 27,39 33,54 38,73 43,3 47,43GT = A1/x +A2/x2 206,05 194,99 188,97 185,39 183,19B1.x3 = 129,43x3 148,41 201,15 249,59 295,06 338,29B2.x2 = 14,44x2 15,82 19,37 22,37 25,01 27,40GG= B1.x3+ B2.x2 164,23 220,53 271,96 320,07 365,69GFe = GT + GG 370,28 415,51 460,93 505,46 548,89Gg = 12.07x3 13,84 18,76 23,27 27,51 31,55P0 = 1,538.GT + 1,463.GG

557,17 662,52 688,52 753,39 816,75

Q0 = 2.GT + 2,972.GG + 80,1.Gg + 151x2

2174,100 2750,43 3284,48 3787,81 4266,50

iox = Q0/(10.SP) 0,87 1,10 1,31 1,52 1,71Gdq= C1/x2 = 188,82/x2 172,37 140,74 121,88 109,01 9,52GCu = 1,66.Gdq 286,14 233,63 202,33 180,97 165,20KCuFe.GCu = 2,21.GCu 632,35 516,31 447,14 399,93 365,09Ctd = GFe + kCuFe.GCu 1002,6 931,82 908,07 905,39 931,97

J=dqG.4,2

4100.94,0 3,05 3,38 3,63 3,84 4,02

cp = M.x3 = 9,04.x3 10,36 14,05 17,43 20,61 23,63d = A.x3= 14,04.x 14,69 16,26 17,48 18,48 19,34d12 = a.d = 1,36.d 20,00 22,12 23,77 25,13 26,3l = .d12/ 52,29 38,58 31,10 26,30 22,942a2 = b.d = 0,4.d 5,88 6,50 6,99 7,39 7,74C= d12+ a12 + 2a2 + a22 29,26 32,02 34,16 35,92 37,44

n mn hc: thit k my bin p

7

Ta s thy gi thnh thp nht s ng vi 3,6 3,0; nhng vi gii hn P0 = 610 W, ta s ly gi tr 2,4 1,2, tung ng ng knh li st 16,26 d 14,69. Trong khong ny tt c cc tham s u t tiu chun.

Cn c vo ng knh li st tiu chun ta chn gi tr d = 16 cm, lc = 1,69; x = 4 = 4 69,1 = 1,14

Tit din li st s b: ST = 129.x2 = 139.1,142 = 181 cm2 ng knh trung bnh ca rnh du s b d12 = a.d = 1,36.16 = 21,76 cm Chiu cao dy qun s b

l = 69,1

76,21.=

.d12

=39,33 cm

Chiu cao tr li st s b lT = l + 2.l0 = 39,33 + 2.4 = 47,33 cm Khong cch gia cc tr li st s b C = d12+ a12 + 2a2 + a22 = 21,76 + 1,2 + 0,4.16 + 2,2 = 31,56 cm in p ca mt vng dy uv = 4,44.f.BT.ST.10-4 = 4,44.50.1,6.181.10-4 = 6,43 V Trng lng st s b : GFe = 407,18 kg Trng lng ng s b: Gdq = 145,29 kg Mt dng in s b: J = 3,32 A/mm2 ng sut trong dy qun: cp = 13,39 Mpa Tn hao khng ti P0 = 610,41 W Dng khng ti % I0% = 0,97

B. Tnh ton dy qun: I. Dy qun h p:

1. Sc in ng ca mt vng dy: uv = 4,44.f.BT.ST.10-4 = 4,44.50.1,6.181.10-4 = 6,43 V

n mn hc: thit k my bin p

8

2. S vng dy trong mt pha ca dy qun h p:

w1 = 43,6

231=

u

U

v

1f = 35,93 36 vng Nh vy in p mt vng dy

uv = 36

231=

w

U

1

1f = 6,42 V

3. Mt dng in trung bnh: Theo cng thc 13-18

Jcp = 76,21.250

42,6.4100.94,0.746,0=

d.S

u.P.k.746,0

12p

vnn

= 3,39 A/mm2 4. Tit din vng dy s b:

s1 = 39,3

84,360=

J

I

cp

1f = 106,44 mm2

Vi iu kin SP = 250 kVA,

If1 = 360,84 A Um = 400 V S1 = 106,44 mm2

Tra bng XV, ta chn kiu dy qun l kiu qun kp dy dn ch nht, do cng ngh ch to n gin, r, lm ngui tt thch hp vi my cng sut nh, nhng n cng c nhc im l bn c nh, ta s khc phc nhc im ny trong qu trnh chn dy v thng s dy.

5. S vng dy ca mt lp :

w11 = 2

36=

n

w1 =18 vng

n: S lp, v qun kp nn n = 2. 6. Chiu cao hng trc ca mi vng dy:

hv1 = 1+18

33,39=

1+w

l

11

1 = 2,07 cm = 20,7 mm

7. Cn c vo hv1 = 20,7 mm; tit din s1 = 106,44 mm2, theo bng VI.2, ta chn tit din mi vng dy bao gm 2 si chp song song, chia thnh 2 lp dy (qun ng kp ).

Kch thc dy h p nh sau:

PB 2x 52,79x5,20x6,4

02x1,4

8. Tit din thc ca mi vng dy

n mn hc: thit k my bin p

9

s1 = 2x79,52= 159,04 mm2 9. Mt dng in thc trong dy qun

J1 = 04,159

84,360 = 2,27 A/mm2

10. Chiu cao tnh ton ca dy qun h p l1 = hv1.(w11 + 1) + 1 = 2,05.(18+1) = 39,95 cm a01 a a1

b b a11 11. B dy ca dy qun h p a1 = 2.a+a11 a: b dy ca 1 si dy, a=4,6mm a11: khong cch gia hai lp ca dy qun ng kp, chn a11=5mm a1 = 2.4,6 +5 = 14,2 mm = 1,42 cm 12. ng knh trong dy qun h p: D1 = d+2a01 a01: 5mm = 0,5 cm; d= 16cm D1 = 16 +2.0,5 = 17 cm 13. ng knh ngoi ca dy qun h p: D1 = D1 + 2.a1 a1 = 1,42 cm, D1 = 17 cm D1 = 17+ 2.1,42 = 19,84 cm 14. B mt lm lnh ca dy qun h p: M1 = t.kK..(D1+D1).l1.10-4 t: s tr tc dng, t = 3 kK h s xt n s che khut b mt ca dy qun, kK = 0,75 thay vo: M1 = 3.0,75..(17+19,84).39,95.10-4 = 1,04 m2

n mn hc: thit k my bin p

10

15. Trng lng ng dy qun h p , theo cng thc 13-76a

GCu1 = 5-11"1

'1 10.s.w.

2

D+D.t.28

= 04,159.36.2

84,19+17.3.28 .10-5 = 88,6 kg

II. Dy qun cao p 1. Chn s iu chnh in p, cn c vo cng sut ca my bin p l

250 kVA ta b tr on dy iu chnh nm lp ngoi cng, mi lp iu chnh c b tr thnh hai nhm trn di dy qun ni tip vi nhau v phn b u trn ton b chiu cao dy qun nn khng xut hin lc chiu trc.

Cc u phn p c cc ca b i ni ba pha. Dng lm vic nh mc qua cc tip im chnh l dng nh mc ca dy qun cao p bng 22,9 A.

in p ln nht t ln hai u tip im ca b i ni l: - in p lm vic Ulv = 10%.Uf2 = 10%.6,3.103 = 630 V - in p th Uth = 2.Ulv S iu chnh in p

A x5 x4 x3 x2 x1 12 12 12 54 12 12 12 12 12 11 lp bnh 2 lp iu thng chnh