Do an Mon Hoc Thiet Ke May Bien AP 0009

n mn hc: thit k my bin p

LI NI Ui 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%

1


N THIT K MY BIN PA . Xc nh cc kch thc ch yu: I. Xc nh cc i lng in c bn:1. Dung lng mt pha: 250 Sf = = 83,33 kVA 3 Dung lng trn mi tr: 250 = 83,33 kVA St = 3 2. Dng in dy nh mc: - Pha cao p: Sp 250.10 3 = = 22,9 A I 1f = 3.6,3.10 3 3.U 1 - Pha h p: Sp 250.10 3 I 2f = = = 360,8 A 3.U 2 3.400 3. in p pha nh mc: - Pha cao p: Ud1 = Uf1 = 6,3 kV - Pha h p: U 400 U f2 = 2 = = 231 V 3 3 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: 2


a1 + a 2 ' = k.4 S P 3 k l h s ph thuc dung lng my bin p, tra bng 13.1 ta c k = 0,45 thay vo ta c a1 + a 2 ' = k.4 S P = 0,45.4 83,33 =1,51 cm 3 aR khong cch ph thuc kch thc hnh hc ca dy qun h p v cao p, vy:

a R = a 12 +

a1 + a 2 = 1,2 + 1,51 = 2,71 cm 3

2. H s quy i t trng: kR=0.95 3. Cc thnh phn in p ngn mch: - in p ngn mch tc dng: Pn 4100 =1,64% U nr = = 10.S P 10.250 - in p ngn mch phn khng:

U nx = U 2 _ U 2 = 4 2 _ 1,642 =3,65% n nr 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/kg3

n mn hc: thit k my bin p -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: + A=4

S 'P .a R .k R f .U nx .B 2 .k 2 T l 83,33.2,71.0,95 = 14,04 50.3,65.1,6 2 .0,88 2

= 16 4

+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 S P .a 2 + C1 = K dq . k f .k 2 .B 2 .u nr .A 2 l T = 2,46.10 2

250.1,36 2 = 188,82 kg 0,94.0,88 2 .1,6 2 .1,64.14,04 2

4

n mn hc: thit k my bin p + M = 0,2453.10 - 4 .k 2 .k f .k R . n trong :

Pn a.A

100 .(1 + e kn = 1,41. un100 .(1 + e = 1,41. 4

-.u nr u nx

)) = 43,84

- .1, 64 3, 65

4100 = 9,04 MPa 1,36.14,04 + Trng lng tn silic cc gc ca gng:M = 0,2543.10 -4 .43,84 2 .0,94.0,95. 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)

5

n mn hc: thit k my bin p = 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 : x= 4

x2= 4

2

1,2 1,047 1,095

1,8 1,16 1,34

2,4 1,25 1,55

3,0 1,32 1,73

3,6 1,38 1,90

3 1,15 1,55 1,93 2,28 2,62 x3= 4 A1/x = 187/x 178,67 161,44 150,24 142,09 135,76 2 2 A2.x = 25,1.x 27,39 33,54 38,73 43,3 47,43 2 GT = A1/x +A2/x 206,05 194,99 188,97 185,39 183,19 3 3 B1.x = 129,43x 148,41 201,15 249,59 295,06 338,29 2 2 B2.x = 14,44x 15,82 19,37 22,37 25,01 27,40 3 2 GG= B1.x + B2.x 164,23 220,53 271,96 320,07 365,69 GFe = GT + GG 370,28 415,51 460,93 505,46 548,89 3 Gg = 12.07x 13,84 18,76 23,27 27,51 31,55 P0 = 1,538.GT + 557,17 662,52 688,52 753,39 816,75 1,463.GG Q0 = 2.GT + 2,972.GG + 2174,100 2750,43 3284,48 3787,81 4266,50 80,1.Gg + 151x2 iox = Q0/(10.SP) 0,87 1,10 1,31 1,52 1,71 2 2 Gdq= C1/x = 188,82/x 172,37 140,74 121,88 109,01 9,52 GCu = 1,66.Gdq 286,14 233,63 202,33 180,97 165,20 KCuFe.GCu = 2,21.GCu 632,35 516,31 447,14 399,93 365,09 Ctd = GFe + kCuFe.GCu 1002,6 931,82 908,07 905,39 931,97 3,05 3,38 3,63 3,84 4,02 0,94.4100 J= 2,4.G dq

cp = M.x3 = 9,04.x3 d = A.x3= 14,04.x d12 = a.d = 1,36.d l = .d12/ 2a2 = b.d = 0,4.d C= d12+ a12 + 2a2 + a22

10,36 14,69 20,00 52,29 5,88 29,26

14,05 16,26 22,12 38,58 6,50 32,02

17,43 17,48 23,77 31,10 6,99 34,16

20,61 18,48 25,13 26,30 7,39 35,92

23,63 19,34 26,3 22,94 7,74 37,44

6

n mn hc: thit k my bin p Ta s thy gi thnh thp nht s ng vi 3,6 3,0; nhng vi gii hn P0 16,26 = 610 W, ta s ly gi tr 2,4 1,2, tung ng ng knh li st 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 1,69 = 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 .d 12 21,76 . = =39,33 cm l= 1,69 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

7

n mn hc: thit k my bin p 2. S vng dy trong mt pha ca dy qun h p: U 231 w1 = f 1 = = 35,93 36 vng u v 6,43 Nh vy in p mt vng dy U 231 uv = f 1 = = 6,42 V w1 36 3. Mt dng in trung bnh: Theo cng thc 13-18 P .u 4100.6,42 Jcp = 0,746.k n . n v = 0,746.0,94. S p .d 12 250.21,76 2 = 3,39 A/mm 4. Tit din vng dy s b: I 360,84 = 106,44 mm2 s1 = f 1 = J cp 3,39 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 : w 36 =18 vng w11 = 1 = n 2 n: S lp, v qun kp nn n = 2. 6. Chiu cao hng trc ca mi vng dy: l1 39,33 = = 2,07 cm = 20,7 mm hv1 = w 11 + 1 18 + 1 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: 4,1x20 x79,52 PB 2x 4,6 x20,5 8. Tit din thc ca mi vng dy

8

n mn hc: thit k my bin p s1 = 2x79,52= 159,04 mm2 9. Mt dng in thc trong dy qun 360,84 = 2,27 A/mm2 J1 = 159,04 10. Chiu cao tnh ton ca dy qun h p l1 = hv1.(w11 + 1) + 1 = 2,05.(18+1) = 39,95 cm

a01 a1 b a11

a

b

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

9

n mn hc: thit k my bin p 15. Trng lng ng dy qun h p , theo cng thc 13-76a ' " D1 + D1 GCu1 = 28.t. .w 1 .s1 .10 -5 2 17 + 19,84 = 28.3. .36.159,04 .10-5 = 88,6 kg 2

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 12 x4 x3 12 12 54 12 12 12 12 12 x2 x1

11 lp bnh thng

2 lp iu chnh

10

n mn hc: thit k my bin p 2. S vng dy cun cao p ng vi Um U f2 6,3.10 3 w2 = w 1 . = 36. 231 U f1 = 982 vng 3. in p mt cp iu chnh u = 2,5%.U2 = 2,5%.6,3.103 = 157,5 V 4. S vng dy cao p mt cp iu chnh: 157,5 24 vng wc = U = uv 6,42 5. S vng dy tng ng trn cc u phn p in p u dy 6615,0 V X1 6457,5V X2 6300,0V X3 6142,5V X4 5985,0V X5 Mc iu chnh + 5% +2,5% 0% -2,5% -5% S vng dy cao p 982+2.24 = 1030 vng 982+1.24 = 1006 vng 982 vng 982 1.24 = 958 vng 982-2.24=934 vng

6. Mt dng in trong cun cao p: J2 = 2Jcp J1 = 2.3,39 2,27 = 4,51 A/mm2 7. Tit din dy cao p s b: I 22,9 = 5,08 mm2 s2 = f'2 = 4,51 J2 8. Cn c vo cng sut my bin p 250 kVA, dng If2 = 22,9 A, U2=6,3 kV, s2=5,08 mm2 tra bng XV ta chn kt cu dy qun l kiu hnh ng nhiu lp tit din trn, u dim ca phng php ny l cng ngh ch to n gin, nhng nhc im l tn nhit km v bn c thp, nhng vi cng sut ca my nh th ta c th khc phc c nhng nhc im ny trong khi chn la thng s dy qun v kiu cch tn nhit. Theo bng VI.1 hn dy dn trn m hiu PTEV c quy cch nh sau:

2,36 ;4,36 2,46 9. Tit din thc ca mi vng dy, s2 = 2.4,36 = 8,72 mm2PTEV-2x 10.Mt dng in thc trong mi vng dy

11


I f 2 22,9 = = 2,62 A/mm2 s2 8,72 11. S vng dy trong mt lp: l2 399,5 w12 = -1 = - 1 = 81vng 2.2.46 n v 2 .d '2 Trong : l2 = l1 = 39,95 cm = 399,5 mm nv2 l s si chp trong mt vng dy, nv2 = 2 12. S lp ca dy qun: w 1030 = 13 vng n12 = 2 max = w 12 81 13. in p lm vic gia hai lp k nhau u12 = 2.w12.uv = 2.81.6,42 = 1040,04 V 14. Cn c vo u12 = 1040 V, ta chn cch in gia cc lp ca dy qun hnh ng nhiu lp, chiu dy cch in l 12=0,12 mm, s lp giy cch in l 2 lp. 15. Nh trn cp nhc im chnh ca kiu dy qun ny l to nhit km, m bo iu kin to nhit tt ta qun nn que nm. B tr dy qun nh sau: Lp th nht n lp th 11: 80.11 = 891 vng Lp th 12: 54 + 2.12 =78 vng Lp th 13: 6.12 = 72 vng --------------------------------------------------------------------------Tng cng 1030 vng 16. Chiu dy dy qun cao p Theo cng thc 13-54a a2 = d.n12 + 2.12 (n12-1) =2,46.13 + 2.0,12.(13-2) + 5= 34,8 mm=3,48 cm 17. ng knh trong ca dy qun cao p D2 = D1 + 2.a12 = 19,84 + 2.1,2 = 22,24 cm 18. ng knh ngoi ca dy qun cao p D2 = D2 + 2.a2 = 22,24 + 2.3,48 = 29,2cm 19. Khong cch gia hai tr cnh nhau C = D2 + a22 = 30,16 + 2,2 = 32,36 cm 20. Din tch b mt lm lnh ca dy qun Dy qun cao p qun trn nm, nn theo cng thc 13-59abJ2 =

12

n mn hc: thit k my bin p M2 = 1,5.t. k..(D2+D2).l2 Trong : t s tr tc dng, t=3. k l h s k n s che khut, y ta ly k = 0,88 M2 = 1,5.3.0,88..( 22,24+ 30,21).39,95.10-4 = 2,61 m2 21. Trng lng dy qun cao p D' + D" 2 GCu2 = 28.t. 2 .s 2 .w 2 max .10 -5 2 22,24 + 29,2 = 28.3. .1030.8,72.10 -5 = 194kg 2 160 161 164 165 5 14,2 12 34,8 22 40 20

3 179,2 183,2 186,2 191,2 226 S b tr qun dy v kch thc chnh ca dy qun h p v cao p

III. Tnh ton cc tham s ngn mch A. Tn hao 1. Tn hao chnh13

n mn hc: thit k my bin p - Tn hao ng trong dy qun h p Theo cng thc 13-75 PCu1 = 2,4.J2.GCu1 = 2,4.2,272.88,6 = 1096 W - Tn hao ng trong dy qun cao p PCu2 = 2,4.J2.GCu2 D '2 + D " 2 GCu2 = 28.t. .s 2 .w 2 m .10 -5 2 22,24 + 29,2 = 28.3. .982.8,72.10 -5 = 185 kg 2 PCu2 = 2,4.2,622.185 = 3048 W

2. Tn hao ph trong dy qunTn hao ph thng c ghp ni vo tn hao chnh thng qua h s tn hao kf , do vic xc nh tn hao ph chnh l xc nh h s kf - Vi dy qun h p tit din dy ch nht, theo cng thc 13-78b kf = 1+0,0952.a4.n2 b.m += .k R l1 . b chiu cao mt vng dy, b=20,5 mm . m l s thanh dn ca dy qun song song vi t thng tn, m = 18 . l chiu cao ton b dy qun, l = 39,95 cm . kR h s Ragovski, ly kR = 0,95 2,05.18 = .0,95 = 0,88 2 = 0,77 39,95 + a kch thc dy dn thng gc vi t thng tn, a = 4,1 mm + n s thanh dn ca thanh dn song song vi t thng tn, n=4. Vy: kf1 = 1+0,0952.a4.n2 = 1+0,095.0,77.(0,41)4. 42 = 1,033 -Vi dy dn cao p tit din trn, theo cng thc 13-78c kf = 1+0,0442.d4.n2 d.m += .k R l2 . d ng knh trn ca dy qun cao p, d=2,36 cm . m s thanh dn ca dy qun song song vi t trng tn, m=81 . l2 = 39,95 cm

14


0,236.81 .0,95 = 0,452 = 0,21 39,95 + n s thanh dn ca dy qun vung gc vi t trng tn, n=13 Vy: kf2 = 1+0,044.2.a4.n2=

= 1+0,095.0,21.(0,236)4. 132 = 1,0048

A)

B)

Hnh v dng xc nh tn hao ng trong cc dy qun A) Vi dy ng ch nht B) Vi dy ng trn

3. Tn hao ph trong dy dn ra

PCur = 2,4.J2.GCur +J: mt dng in ca dy qun +GCur khi lng ng dy dn ra ca dy qun GCur = lr.sr.Cu .lr: chiu di dy dn ra . khi lng ring ca ng, Cu =8900kg/m3 - Khi lng ng dy dn ra ca dy qun h p GCur1 = lr1.sr1.Cu V h p u Y nn lr1 = 7,5 l1 = 7,5.39,95 = 299,63 cm GCur1 = 299,63.159,04.8900.10-8 = 4,24 kg Vy tn hao dy dn ra h p l: PCur1 = 2,4.2,272.4,24 = 52,44 W - Khi lng ng dy dn ra ca dy qun cao p GCur2 = lr2.sr2.Cu V cao p u nn lr1 = 14.l1 = 14.39,95 = 559,3 cm GCur2 =559,3.8,72.8900.10-8 = 0,434 kg Vy tn hao dy dn ra cao p l: PCur2 = 2,4.2,622.0,434 = 7,15 W 4. Tn hao trong thng du v cc chi tit bng kim loi khc Tn hao ny rt kh xc nh chnh xc, theo cng thc gn ng 13-80: Pt = 10.k.SP l h s t l, tra bng 13.18 vi cng sut my 250 kVA, ta c k=0,012 Pt = 10.0,012.250 = 30 W

15

n mn hc: thit k my bin p 5. Tng tn hao ngn mch ton phn Pn = kf1.PCu1 + kf2.PCu2 + PCur1 + PCur2 + Pt Thay s vo ta c Pn = 1,033.1096 +1,0048.3048 +52,44 +7,15+30 = 4284,39 W So snh vi s liu u bi cho Pn = 4100, sai s Pn - Pn 4284,39 - 4100 % = .100% = .100% = 4,5% < 5%, nh vy coi Pn 4100 nh cng sut tn hao ngn mch ca my l t yu cu.

B. in p ngn mch1. Thnh phn tc dng Theo cng thc 13-81 ' Pn unr = 10.S P Pn Tn hao trong iu kin lm vic bnh thng ca dy qun cao p ch lm vic nh mc. Pn = Pn 0,05.(Pdq2.kf2) = 4284,39 0,05.(1,0048.3048) = 4131,26 W ' Pn 4131,26 = = 1,65% unr = 10.S P 10.250 2. Thnh phn phn khng 7,92.f .S' P ..a R unx = .k R u2 v Trong + f l tn s cng nghip, f =50 Hz + SP Cng sut trn mi tr ca my, SP = 83,33 kVA + h s hnh dng ca my bin p, =1,69 a + a2 1,42 + 3,48 + aR = a 12 + 1 = 1,2 + 3 3 = 2,83 + kR1 = 1-(1- e ) = -

=

a 12 + a 1 + a 2 1,2 + 1,42 + 3,48 = .l .39,95 = 0,048 16

n mn hc: thit k my bin p thay gi tr vo1 kR = 1- 0,048.(1- e 0,048 ) = 0,952 -

vy gi tr ca thnh phn phn khng 7,92.50.83,33.1,69.2,83 .0,952.10 3 unx = 2 6,42 = 3,65% 3. in p ngn mch ton phn un = u 2 + u 2 = 1,65 2 + 3,65 2 = 4 % nr nx Nh vy kt qu tnh ton trng lun vi d liu ban u cho un =4%, kt lun in p ngn mch vi cc thng s chn l tho mn yu cu bi.

III. Tnh ton lc c hc khi ngn mch1. Dng in ngn mch xc lp, theo cng thc13-85 100 100 = 22,9. = 572,5A In = I m . 4 un 2. Dng ngn mch cc i tc thi Theo cng thc 13-86 imax = 2 .kM.In

kM = (1+ e = 1,24 imax = 2 .1,24. 572,5 =1004 A 3. Lc hng knh Theo cng thc 13-87 FK = 0,628.( imax.w2)2..kR.10-6 = 0,627.(1004.982)2.1,69.0,952.10-6 =0,98 Mpa 4. ng sut nn - Trong dy qun h p FK nrHA = 2 .s1 .w 1

.u nr u nx

.1,65 ) = (1 + e 3,65 )

0,98.10 6 = = 27,26.10 6 Pa = 27,26 Mpa 6 2 .159,04.10 .36

17

n mn hc: thit k my bin p Vi dy ng : 30MPacp 18Mpa, nh vy gi tr nrHA tnh c tho mn iu kin v ng sut nn ngang trc(do lc hng knh gy nn). - Trong dy qun cao p FK nrCA = 2 .s 2 .w 20,98.10 6 = = 18,23 Mpa 2 .8,72.10 6 .982 Vi dy ng : 30MPacp 18Mpa, nh vy gi tr nrCA tnh c cng tho mn iu kin v ng sut nn ngang trc(do lc hng knh gy nn). 5. Lc chiu trc Do hai dy qun c cng chiu cao cc vng dy phn b u trn ton b chiu cao nn Ft = 0. C a Ft = FK . R 2.l 2,83 = 0,98.106. =34731 N 2.39,95 6. ng sut do lc chiu trc gy nn p dng cng thc 13-92 cho dy qun khng c nm chm gia cc bnh dy, y ta ch xt ring cho dy qun h p do n phi chu lc p in ng cc i trong dy qun, nu n tho mn th iu kin ny i vi dy cao p cung tho mn. Ft' n = " D ' + D1 . 1 .(a 1 a 11 ) 2 34731 n = =6,52 Mpa 17 + 19,84 2 2 . .10 .(1,42 0,5).10 2 mc cho php 1820 Mpa ncp nh vy n tho mn iu kin v ng sut nn dc trc.

IV. Tnh ton cui cng v h thng mch t1. Chn kt cu li thp Ta chn kt cu li thp kiu 3 pha nm tr, c 4 mi ni nghing 4 gc, l thp xen k, l thp l loi tn cn lnh 3405 c chiu dy 0,35 mm u im ca loi thp k thut in ny l sut tn hao trong li thp nh, cng sut t ho

18

n mn hc: thit k my bin p nh, t thm theo chiu cn ln nn c th lm gim tn hao st ca m.b.a.. Tr p bng bng ai thu tinh khng c tm st m. Gng p bng x p gng, Kch thc tp l thp nh hnh v, nh trn chn tit din tr c 8 bc v tit din gng c 5 bc, ta c bng tng hp kt qu tnh ton tr v gng nh sau: Th t l thp 1 2 3 4 5 6 Tr 15,36x2,24 14,08x1,575 12,48x1,19 10,08x1,225 7,52x0,84 4,48x0,595 Gng 15,36x2,24 14,08x1,575 12,48x1,19 10,08x1,225 7,52x0,84 7,52x0,84

2. Tng chiu dy cc l thp ca tit din tr (hoc gng) bT = 2.( 2,24 + 1,575 + 1,19 + 1,225 + 0,84 + 0,595 ) = 15,33 cm 3. Tng tit din cc bc trong tr Theo cng thc 13-94 SbT = 2.aT.bT =2.(15,36.2,24+14,08.1,575+1,19.12,48+10,08.1,225+7,52.0.84 4.48.0,595) = 185,53 cm2 4. Tit din c ch ca tr ST = SbT.kc kC h s p cht, y ta ly kC = 0,98 ST = 0,98.185,53 = 181,82 cm2 5. Tit din bc thang ca gng Theo cng thc 13-95 SG = kG.SbT kG = 1,02 SG = 1,02.185,53 =189,24 cm2 6. Tit din c ch ca gng SG = kC.SG

+

19

n mn hc: thit k my bin p = 0,98.189,24 = 185,46 cm2 7. Chiu dy gng, ta ly chiu dy gng bng chiu dy tr, tc l bng tng chiu dy cc l thp ca tit din tr. bG = bT = 15,33 cm 8. Tng s l thp trong tr v gng theo cng thc 13-96 b n1 = T ; T T l chiu dy mi l thp, T = 0,35 mm thay vo biu thc trn b 153,3 n1 = T = = 438 l thp T 0,35 9. Chiu cao tr st theo cng thc 13-99 lT = l + l0 + l0 l l chiu cao dy qun, l = 39,95 cm l0 l khong cch t gng trn l0 l khong cch t dy qun n gng di Ta ly l0 = l0 = 4cm lT = 39,95 + 2.4 = 47,95 cm 10. Khong cch gia hai tm tr theo cng thc 13-100 C = D2 + a22 D2 l ng knh ngoi cun cao p, D2=30,21 cm a22 l khong cch gia 2 cun cao p, a22 =2,2 cm C = 30,21+2,2 =32,41 cm Ly trn C = 32,5 cm 11. Trng lng st gc mch t chung cho gng v tr theo cng thc 13-101 G g = 2.kC. Fe.10-6.(aiT.aiG.biT) = 2.0,98.7650.10-6.(15,36.15,36.2,24+14,08.14,08.1,575+ +12,48.12,48.1,19+10,08.10,08.1,225+7,52.7,52.0,84+4,48.7,52.0,595) = 18,26 kg 12. Trng lng st trong gng Theo cng thc 13-102 GG + GG GG = - GG l trng lng phn thng nm gia hai tr bin, theo 13-102a

20

n mn hc: thit k my bin p GG = 2.(t-1).C.SG.Fe.10-6 Fe trng lng ring ca thp, Fe =7650 kg/m3 GG = 2.(3-1).32,5.185,46.7650.10-6 = 184,44 kg - GG l trng lng phn gng cc gc theo cng thc 13-102b Gg GG = 4. = 2.Gg 2 = 2.18,26 = 36,52 kg vy trng lng ca gng l GG + GG GG = = 184,44 + 36,52 = 220,96 kg 13. Trng lng st trong tr l theo cng thc 13-103 GT = GT + GT Trong - GT l trng lng phn tr nm trong chiu cao ca s mch t theo cng thc 13-103a GT = t.lT.ST.Fe.10-6 = 3.47,95.181,82.7650.10-6 = 200 kg - GT l trng lng phn tr ni vi gng theo cng thc 13-103b GT = t.(a1G.ST.Fe.10-6 Gg ) = 3.(15,36.181,82.7650.10-6- 18,26) = 9,31 kg Vy trng lng ca st trong tr l: GT = GT + GT = 200 + 9,31 =209,3 kg 14. Trng lng st ton b trong tr v gng GFe = GG + GT = 220,96 + 209,3 = 430,26 kg

V. Tnh tn hao khng ti v dng in khng ti1. Li thp lm bng tn silc m hiu 3405, dy 0,35 mm, h s t cm trong tr v gng l: - H s t cm trong tr

21

n mn hc: thit k my bin p BT =

uv 4,44.f .S T

6,42 4,44.50.181,82.10 4 = 1,59 T - H s t cm trong gng l uv BT = 4,44.f .S G 6,42 = 4,44.50.185,46.10 4 =1,56 T= 2. Theo bng V.13 vi tn 3405, dy 0,35 mm ta tra c cc sut tn hao st tng ng BT = 1,59 T pT = 1,135 W/kg, pkT = 0,0375 W/cm2 BG = 1,56 T pG = 1,074 W/kg, pkG = 0,0375 W/cm2 H s t cm trong rnh nghing l B 1,59 Bkn = T = = 1,124 T 2 2 Vi Bkn ta tra c pkn = 0,0333W/cm2 3. Cc h s xt n s gia tng tn hao trong gng v tr. Kt cu mch t chn l mch t phng ni nghing 4 gc, tr gia ni thng, li st khng t l, tn c sau khi ct, c kh bavia, t tra c cc h s: - k1 h s gia tng tn hao v cng sut t ho gng d hnh dng tit din gng nh hng n s phn b t cm trong tr v gng, v s bc thang ca tr v gng gn bng nhau (nG = 5, nT = 5) nn ly k1 = 1. - k2 l h s do tho lp gng lng dy vo tr lm cht lng l thp gim xung, vi cng sut my 250 kVA ta ly k2 = 1,01. - k3 l h s do p tr ai, h s ny ph thuc vo cng sut my, vi cng sut my 250 kVA ta ly k3 = 1,02. - k4 l h s do ct dp l tn thnh tm, vi tn c k4 =1 - k5 l h s do vic x l bavia, tn c li, c kh bavia, ly k5=1 - k6 l h s xt n cc gc ni ca mch t , tra ph lc XVIII ta c k6 = 10,45.

22


4. Tn hao khng ti P0 = k4.k5.[ pT.GT + pG.(GG - 4Gg) +

P0=1.1.[1,135.209,31+1,074.(184,44 4.18,26)+

1,135 + 1,074 .10,45.18,26 + 2 4.0,0333.257,13+1.181,82.0,06375+2.185,46.0,0615] = 655,8 W sai lch so vi thng s cho l: 655,8 610 = .100% = 7,52 % 610 4. Theo bng V.13 vi tn 3405, dy 0,35 mm ta tra c cc sut t ho tng ng BT = 1,59 T qT = 1,564 W/kg, qkT = 2,28 W/cm2 BG = 1,56 T qG = 1,447 W/kg, qkG = 2,07 W/cm2 H s t cm trong rnh nghing l B 1,59 Bkn = T = = 1,124 T 2 2 Vi Bkn ta tra c qkn = 0,142W/cm25. Cc h s xt n s gia tng cng sut t ha trong gng v tr. i vi kt cu li thp v cng ngh ch to mch t c l tn sau khi ct dp ta tra c cc h s sau: - k1 h s gia tng cng sut t ho gng do hnh dng tit din gng nh hng n s phn b t cm trong tr v gng, v s bc thang ca tr v gng gn bng nhau (nG = 5, nT = 5) nn ly k1 = 1. - k2 l h s do tho lp gng lng dy vo tr lm cht lng l thp gim xung, vi cng sut my 250 kVA ta ly k2 = 1,01. - k3 l h s do p tr ai, h s ny ph thuc vo cng sut my, vi cng sut my 250 kVA ta ly k3 = 1,02. - k4 l h s do ct dp l tn thnh tm, vi tn c k4 =1,18 - k5 l h s do vic x l bavia, tn c li, c kh bavia, ly k5=1 - k6 l h s xt n cc gc ni ca mch t , tra ph lc XVIII ta c k6 = 41,775. 6. Cng sut t ho khng ti: Theo cng thc 13-106 ta c

pK.nK.SK ].k1.k2.k3 Thay s vo biu thc:

pT + pG ' .k 6 .G g 2

+

23

n mn hc: thit k my bin p Q0 = {k4.k5.[ qT.GT + qG.(GG - 4Gg) +

qK.nK.SK }.k1k2.k3 Thay s vo biu thc trn Q0 = {1,18.1.[ 1,564.209,31 + 1,447.(184,44 4.18,26) + 1,564 + 1,447 .1,41.41,775.18,26 ]+ 4.0,142.257,13+1.2,28.181,82 2 2.2,07.185,46}.1.1,01.1,02 = 3954,54 Var7. Thnh phn tc dng ca dng in khng ti Theo cng thc 13-105 P0 ior% = 10.S P 655,8 = 0,262 % = 10.250 8. Thnh phn phn khng ca dng in khng ti Theo cng thc 13-107 Q0 iox% = 10.S P 3954,54 = 1,582 % = 10.250 9. Tr s dng in khng ti Io =

qT + qG " .k 6 .G g ] 2

+

+

i2 + i2 or ox

= 0,262 2 + 1,582 2 = 1,6 % < 2% nh vy tr s dng khng ti tho mn iu kin t ra ca bi 10. Hiu sut ca my bin p lc ti nh mc, cos = 1 theo cng thc 13-109 P0 + Pn = (1 ) S P + P0 + Pn 655,8 + 4284,39 ) = 98,06% = ( 1 250.10 3 + 655,8 + 4284,39

24


VI. Tnh ton nhit: A. Tnh ton nhit ca dy qun1. Nhit chnh trong lng dy qun vi mt ngoi ca n - Dy qun h p, tit din dy ch nht

Hnh v s dy qun h p Theo cng thc 13-110 c q. 01 = cd trong + q: Mt dng nhit trn b mt dy qun Vi dy qun ng kp khng c rnh dn du gia, theo cng thc 13-110a 107 a 2 q = 2. b .J .k f kK a' .kK l h s che kn mt ngoi ca dy qun, ly kK = 0,75 .b l chiu cao trn ca mi si chp, nh trn chn b=20mm=2 cm. .a l chiu rng ca mi si chp, a = 4,1 mm .a l chiu rng k c cch in ca mi si chp, a= 4,6mm 2 . J l mt dng h p, J =2,27 A/mm

25

n mn hc: thit k my bin p .kf l h s tn hao ph, tnh trn kf =1,033 107 4,1 q = 2. .2,27 2 .1,033 2 0,75 4,6 = 2707,45 W/m2 + c l h s dn nhit ca lp cch in, cch in gia cc lp l ba nn ta ly c = 0,25 W/m.0C + l chiu dy cch in mt pha, = 0,25 mm =0,025 cm thay s vo biu thc q. 01 = cd

2707,45.0,25.10 3 01 = =2,71 0C 0,25 - Dy qun cao p, tit din dy trn

Hnh v s dy qun cao p Kiu dy dn cao p l kiu hnh ng nhiu lp khng c rnh du ngang gia, c rnh du dc i vi ng cch in, theo cng thc 13-110c p.a 2 02 = 8. tb Trong +a2 l chiu dy ca dy qun cao p, a = a2 = 3, 48 cm +p l tn hao trong mt n v th tch ca dy qun, theo cng thc 13-112 c J 2 .d 2 .10 -8 p = 1,68 (d c + 1 ) . d, dc l ng knh trn v ng knh c bc cch in ca dy dn trn, d= 2,36 mm = 2,36.10-3m dc = 2,46 mm = 2,46.10-3 m

26

n mn hc: thit k my bin p . 1 l chiu dy tm cch in, 1 = 8 mm . J l mt dng in ca dy qun cao p, J = 2,62 A/mm2 =2,62.106 A/

m2. vy:

J 2 .d 2 p = 1,68 .10 -8 (d c + 1 ).d c (2,62.10 6 ) 2 .(2,36.10 -3 ) 2 = 1,68 .10 -8 -3 3 -3 (2,46.10 + 4.10 ).2,46.10 =4,04.104 W/m3 + tb sut dn nhit trung bnh ca dy qun khng k cch in gia cc lp. Theo cng thc 13-113 ..(d + 1 ) tb = 1 .1 + 1 .d c .1 l sut dn nhit gia cc lp dy, 1=0,25 W/ m.0C . l sut dn nhit bnh qun quy c ca dy qun c c = 0,7. trong d -d = c d 2,46 - 2,36 = =0,042 2,36 c sut dn nhit ca cc vt liu cch in vng dy,c 0 =0,17W/ m. C 0,17 = = 1,73 W/m.0C 0,7. 0,042 vy gi tr tb 1,73.0,25.(2,46.10 -3 + 4.10 3 ) tb = 1,73.4.10 -3 + 0,25.2,46.10 -3 = 0,38 W/m.0C Thay cc gi tr p, tb tnh vo biu thc 02 c p.a 2 02 = 8. tb 4,04.10 4 .(3,48.10 -2 ) 2 = 8.0,38 27

n mn hc: thit k my bin p = 16,1 0C 2. Nhit chnh ca ca mt ngoi dy qun i vi du theo cng thc 13-114, vi dy qun hnh ng khng c rnh du ngang 0d = k.q 0,6 Trong . q l mt dng nhit trn b mt dy qun . k l h s t l, y ta ly k = 0,285 - Dy qun h p: tnh c q1 = 2707,45 W/ m2 thay vo biu thc trn ta c 0d1 = 0,285. 2707,450,6 = 32,69 0C - Dy dn cao p Ta phi tnh q2 c b mt to nhit cu dy qun cao p M2 = 1,5.t. k..(D2+D2).l2 Trong : t s tr tc dng, t=3. k l h s k n s che khut, y ta ly k = 0,88 M2 = 1,5.3.0,88..( 22,24+ 30,21).39,95.10-4 = 2,61 m2 Mt khc theo cng thc P .k q= n f M vy P .k q2 = n 2 f 2 M2 P .k M2 = n 2 f 2 q2 3048.1,0048 = 1173,42 W/m2 = 2,61 Thay q2 vo biu thc tnh 0d ta c 0d2 = 0,285. 1173,420,6 = 19,8 0C 3. Nhit chnh trung bnh ca dy qun i vi du 0dtb = 0 + d - Vi dy qun h p 0dtb1 = 01 + 0d1 = 2,71 + 32,69 = 35,40C - Vi dy qun cao p 0dtb2 = 02 + 0d2

28

n mn hc: thit k my bin p = 16,1 + 19,8 = 35,90C

B. Tnh ton nhit thng du1. Chn loi thng du V cng sut trn mi tr ca my bin p l 83,33 kVA nn theo bng ta chn thng du l kiu thng c gn ng tn nhit do n c u im l h s to nhit cao, t tn nguyn liu hn so vi thng c gn ng phng. 2. Chn cc khong cch cch in t dy dn ra n vch thng, n x p gng trn c xc nh nh sau: - s1 l khong cch n vch thng cho dy dn ra cao p c Uthn2 = 20kV, theo bng XIV.6 chn s1 = 20 mm = 2cm. - s2 l khong cch n x p gng cho dy dn ra h p c Uthn1 = 5kV ta tra bng XIV.6 chn s2 = 15 mm = 1,5 cm. - s3 l khong cch t dy dn ra c bc cch in hay khng bc cch in ca dy qun h p n dy cao p vi Uthn1=5kV theo bng XIV.6 ta chn s3 = 35 mm = 2,5 cm. - s4 l khong cch t dy dn ra ca dy qun h p n vch thng, vi Uthn1=5kV theo bng XIV.7 ta chn s4= 25 mm = 2,5 cm. - d1 l ng knh dy dn ra c bc cch in ca dy qun cao p, vi Uthn2=20kV theo bng XIV.7 ta chn d2=25mm =2,5cm. - d1 l ng knh dy dn ra c bc cch in ca dy qun h p, vi Uthn1=5kV theo bng XIV.7 ta chn d1=10mm =1cm. 3. Chiu rng ti thiu ca thng du Theo hnh v s lc thng du: B = D2 + s1 + s2 + d1 + s3 + s4 = 29,2 + 2 +1,5+2,5+2,5+2+1 = 40,7 cm Ta chn B = 45 cm =0,45 m

29


Khong cch ti thiu bn trong ca my bin p

Kch thc c bn ca thng du my bin p 4. Chiu di thng du A = 2.C + B C l khong cch gia hai tr , C = 32,5 cm Vy A = 2.32,5 +45 = 110 cm Ta chn li A=110 cm = 1,1 m 5. Chiu cao rut my Ht Ht l khong cch t y thng n ht chiu cao li st, c xc nh theo cng thc sau Ht = lT +2.hG+n Trong + lT l chiu cao ca tr, nh trn xc nh c lT = 47,95 cm + hG l chiu cao ca gng, n bng b rng cc i ca tm l thp trong gng, hG = a1 = 15,36 cm + n l chiu dy tm lt gng di, chn n = 4cm Thay s vo biu thc Ht = 47,95 + 2.15,36 + 4 = 82,67 cm Chn li Ht = 84 cm = 0,84m 6. Khong cch t gng trn n np thng Hn, tra bng 13.3 vi in p ca dy qun cao p l 6,3 kV ta chn Hn = 30 cm 7. Chiu cao thng: HTh = Ht + Hn = 84 + 30 = 114 cm = 1,14 m 8. tng nhit trung bnh gia du v thng dt = 65 - 0dtb

30

n mn hc: thit k my bin p y ta s ly 0dtb ln nht l ca dy qun cao p, vy 0dtb=35,90C vy dt = 65 - 0dtb = 65 35,9 = 29,10C 9. tng nhit gi thit mt trn ca du so vi khng kh dt = 1,2.dt = 1,2.29,1 = 34,920C

Documents

Do an Mon Hoc Thiet Ke May Bien AP 0009