Trng SPKT Hng Yn N MN HCKhoa: in - in T

CHNG II

CC PHNG PHP TNH TON DY QUN MY BIN P MT PHA .

Trn thc t c rt nhiu cch tnh ton dy qun cho my bin p cm ng mt pha, ti ny chng em xin trnh by v thc hin theo phng php tnh ton dy qun da trn kch thc li thp cho trc. y l phng php hay v c nhiu u im, trc ht phng php ny ph hp vi kh nng ca chng em. V chng ta c th d dng thc hin theo phng php ny v li thp c sn xut sn, sau khi mua v chng ta s tnh ton dy qun theo kch thc sn c ca li thp.

Sau y chng em xin trnh by v phng php v cch thc tnh ton ca chng em.

2.1 o kch thc chun ca li thp loi E, I.

Hnh 2.1: Kch thc c bn ca li thp tiu chun dng E I.

K hiu v tn gi cc kch thc c bn cu li thp:

+ a : B rng tr gia ca li thp.

+ b : B dy cu li thp bin p.

+ c : B rng ca ca s li thp.

+ h : chiu cao ca s li thp.

( Ch :cc kch thc ny khi o v tnh ton u tnh theo n vi l mm hoc cm)

Cc kch thc a , c v h u c o trc tip trn mt l thp E I. Nh ti ny chng em o c s liu nh sau :

a = 4 ( cm ).

c = 2.4 ( cm ).

h = 6 ( cm ).

b = ( b dy 1 l thp)*(tng s l thp). (2.1)

b = 8 (cm )

Hnh 2.2 : Cc kch thc ngoi ca li thp.

Kch thc tng qut cu ton b l thp sau khi ghp sat c xc nh nh hnh 2.2 . Chng ta c th tnh khi lng li thp bin p ( dng tiu chun) theo h thc :

W fer = 46,8 . a2.b = 46,8 . 0.42 .0,8 = 6 (kg) (2.2)

Trong W fer l khi lng cu li thp my bin p, vi gi tr khi lng ring ca l thp l 7,8 kg/dm3 . W fer c tnh thep n v kg.

Sau khi xc nh c cc kch thc c bn ca my bin p ta tnh tit din tr gia ca li thp ch E. y chnh l tit din cho t thong mc vng xuyn qua cc b dy qun .

Gi At l tit din tr gia li thp , ta c:

At = a.b = 4. 8 = 32 (cm2) (2.3)

Trong , n v o l : At (cm2), (a) = (b) = (cm).

2.2 Xc nh gi tr Nv ( S vng dy qun to ra 1 volt sc in ng cm ng).

phn ny chng ta ch c hai thao tc c bn nh sau:

+ Chn mt t thng (hay t cm) B dng tnh ton cho li thp.

+p dng cng thc tnh c s vng dy qun to ra c 1 volt sc in ng cm ng.

Nv = (2.4)

Trong :

+Nv l s vng dy qun to ra 1 V sc in ng cm ng, n v (vng/volt).

+B : mt t thong , n v T

+At : tit din tr gia li thp , n v cm 2

Tiu chun chn B:

+ Vi l thp k thut in c b dy tiu chun t 0.5mm n 0.35mm, l thp thuc loi tole cn nng v hm lng Si t 2% n 4% th chng ta chn gi tr B t 1,0 T n 1,2 T.

+ Vi l thp k thut in c b dy tiu chun t 0.5mm n 0.35mm, l thp thuc loi tole cn lnh vi hm lng Si khong 4% th ta chn gi tr ca B t 1,4 T n 1,6 T.

ti ny chng em chn B = 1,2 T. Vy thay vo cng thc 2.4 ta c:

Nv = = 1.2 (vng/ Volt) (2.5)2.3 Xc nh s nguyn l ca my bin p, tnh ton s vng dy qun cho mi cun.

Hnh 2.3 :S nguyn l my bin p theo yu cu ca ti.

Theo l thuyt v my bin p ( c trnh by chng I) th s vng dy qun ca cc cun dy my bin p c xc nh theo sc in ng cm ng trong cc b dy s cp v th cp. Nhng theo s nguyn l cu my bin p cn thc hin th cun s cp c 8 u vo v c 3 u ra. trng hp ny ta chn mt gi tr ca in p bn s cp tnh, sau tnh tip cc gi tr cn li. ta chn : U1 = U15 = 220V, U2 = U22 =24 V. Vi U1 v U2 l in p cun s cp v th cp ta c:

N1 = Nv. U1 (2.6) N2 = 1,1 . Nv. U2 (2.7)

Trong :

+ N1 , N2 ln lt l s vng dy qun s cp v th cp.

+ S 1,1 trong cng thc (2.7) c ngha l t s chnh lch gia sc in ng ti dy qun th cp so vi in p nh mc ti th cp lc y ti.

Theo s liu tnh c cng thc (2.5) ta c Nv = 0,8 (vng/volt), ta thay vo cng thc (2.6) v (2.7) ta c:

N1 = 1,2 . 220 = 264 (vng)

N2 = 1,1 . 1,2 . 24 = 32 (vng)

Vi 10v ca cun s cp th ta cn 10 . 1,2 = 12 (vng)

Vi 12 v ca cun th cp cn 12 . 1,1 . 1,2 = 15.84 ( vng ) = 16 (vng)

Nh vy vi bn s cp c 8 mc in p vo khc nhau chnh lch nhau 10 v th c 12 vng dy ta li xut 1 u dy ra ngoi ly in.

V d :

in p U11 = 180 V tc l phi qun n 180 . 1,2 = 216 vng

in p U12 = 190 V th ta ch qun thm 12 vng l c.

Tng t nh vy vi cun th cp th ta phi qun tt c l N22 vng vi N22 = 24 . 1,1 .1,2 = 31,84 vng ( ly bng 32 vng)

ly in p ly in p 12 V th ta ly vng th 16 l c.

2.4 Chn mt dng in, c lng hiu sut, chn gi tr h s lp y tnh ton, ng knh dy qun s cp v th cp.

2.4.1 Chn mt dng in J.

Mt dng in J ph thuc vo cc thong s sau : cp cch in chu nhit ca vt liu lm dy bin p, ch lm vic my bin p lien tc , ngn hn hay lp li hay ngn hn khng lp li. kiu thng gi tn nhit cho my bin p.

Mun chn mt dng in chnh xc chng ta cn tham kho cc bn s tiu chun v mt dng in. trong ti ny chng em chn gi tr J=4.5A/mm2 tnh ton v thc hin ti.

2.4.2 Tnh chn h s lp y Kld.

a , Tnh ton gi tr din tch ca ca s li thp .

Gi Acs l din tch ca ca s li thp, ta c h thc tnh Acs nh sau:

Acs = c.h (2.8)

Trong :

(Acs ) = (mm2) ; (c) = (h) = (mm)

Vi l thp tiu chun ta c quan h gia cc kch thc c bn vi kch thc ca ca s li thp nh sau :

C =; h = ( 2.9)Nh vy chng ta c cng thc :

Acs = 0.75 .a2 = 0.75 . (402) = 1200 (mm 2) (2.10)

Gi Kld l h s lp y ca ca s li thp, ta c cng thc :

Kld = (2.11)

i vi phng php tnh ny th ta c th chn Kld trong khong t 0.36 n 0.46 l hp l. bi ny chng em chn Kld = 0.46 lm kt qu tnh ton cho cc mc sau. Sau s quay li tnh , kim tra li Kld sau.

2.4.3 c lng hiu sut .

Gi l hiu sut my bin p, theo thuyt my bin p mt pha, chng ta nh ngha hiu sut ca my bin p theo cng thc:

= (2.12)

( = (2.13)

Trong cc trng hp ti c tnh in cm, ng thi nu xem tn hao thp rt b, in khng tn t ca bin p khng cao; lc gi tr h s cng sut ti v h s cng sut pha s cp c th xem nh gn bng nhau.

Trong trng hp ny , mt cch gn ng chng ta c th vit nh sau:

= = (2.14)

Ta chn = 90% ( = 0,9)Trong :

S1 v S2 ln lt l cng sut biu kin cung cp vo pha c cp v th cp, nu p dng cng thc (2.14) chng ta c th xc nh c t s gi tr dng in qua cc dy qun s cp v th cp theo cng thc.

= . (2.15)

Vi gi tr mt dng in J = 4,5 A/mm2 chn trong cc cng on tnh ton trn, chng ta c t s tit din ca dy qun s cp v th cp t cng thc (2.15) theo l thuyt ta c:

S = = (2.16)

Gi S1 v S2 l tit din dy qun s cp v th cp, suy ra :

= = . = ()2 (2.17)

Tm li ta c quan h sau :

= (2.18)

Trong : d1 v d2 l ng knh dy trn pha s cp v th cp.

Vi dy qun s cp v th cp l dy in t tit din trn trng men, ng knh dy c tnh lp men bc v ng knh dy trn c quan h vi nhau nh sau:

dcd (mm) = d (mm) + 0.05(mm) (2.19)

T cc cng thc trn ta c cng thc tnh li h s lp y nh sau :

Kld = (2.20) N1. Scd1 + N2 . Scd2 = Kld . Acs (2.21)2.4.4 Tnh ton ng knh ca dy qun.

Theo kt qu chn la cc khu trn ta c :

= 0,9 ; Kld = 0.46

Thay vo cng thc (2.15) ta c:

= . = 0,9 . = 8.25

T l tit din dy qun ca cun th cp so vi s cp theo cng thc (2.17) ta c : = = 8.25T y ta c : 8.25 ( 8.25 (2.22)T l ng knh ca cun th cp so vi s cp theo cng thc (2.18) ta c: = = = 2.87 T cng thc (2.21) v ( 2.22 ) ta c h phng trnh :

Vi tit din trn ca dy qun nh vy ta ng knh ca dy s cp v th cp l :

d1cd = 1,128 . = 1,128 . =1.22 mm

d2cd = 1,128 . = 1.128 . = 3,406 (mm)

Trn thc t ta chn kch thc dy cho my bin p l:

d1 = 1,2 (mm) ; d2 = 3,5 (mm)

2.4.5 Kim tra li h s lp y. (Kld)

Ta c cng thc :

Kld = Lc ny ta chn c ng knh dy cun s cp v th cp nh sau:

+ Vi cun s cp : d1/ d1cd = 1,2 (mm)/ 1,204(mm)

Tit din thc ca dy s cp l : S1cd = . (1,2)2 1(mm2 )+ Vi cun th cp : d2 / d2cd = 3.4(mm)/ 3,404 (mm)

Tit din thc l : S2cd = . (3,404)2 = 8.95 (mm2)

Vy ta c :

Kld = = = 0,458 (tha mn)2.5 c lng cng sut biu kin nhn c pha th cp my bin p.

Theo yu cu ca ti th S2 = 1100 (VA)

( I2dm = = = 45.83 (A)

Theo tnh ton th :

I1 = J. . (d1)2 = 4,5 . . (1)2 = 3,53 (A)

I2 = J. . (d2)2 = 4,5 . . 3,52 = 44(A)

Cng sut biu kin cu my bin p l

S1 = U1.I1 = 220 . 3,53 = 1216 (VA)

S2 = U2.I2 = 24. 44 = 1056 (VA)

Gio Vin Hng Dn

Nguyn nh HngNhm Thc Hin : ng Vn Dng

Phm Vn Dng

Khoa Vn Dng