, Un111 %= 55 Cong suat tong S= 300KVA Tfnh tai cua moi may.
e. Sl:f' phl:I thUQC cu aM d6i VO'i R2.
I TT11B~H-t-~+++=-1---+--.!..-+~--.i-=:j I ~ ' I
r__.___.--+--1...+------1--+-......f----l.__-__ ?--~.:t=--f==-"'--~=--1
o-__,___-;:?s,J....Tl?...J_._-:;S~r;:;-3 -L-;;7S--r;,...:_,t--;:!~..L_~--.;;_" m~, fj~__::_}:___l__:s~·
0,2 0,4 0,5 0,8 t,0 1,2 f,4
Hinh 3.21 EJu'fmg c1?c tfnh M = f(s) v&i cac cJi~n tr& rotor khac nhau
N€lu rt= 0 th1 R'2 = 01 2r'2 va tT s6 R2/xn thLI'D'ng rM be do d6 Mdt di qua tri s6 Mmax v&i s khong l&n lam: sm = 0, 12 d§n 0,2. 86ng thO'i Mmm & cac dong CO' rotor day qu§n c6
di$n khang tan l&n hO'n di$n khang tan cua rotor 16ng s6c nen Mmm c6 the giam xu6ng qua gi&i h9n cho phep khi m& may, lam dong CO' khong m& may dU'Q'C, de lo9i tru dieu §y, h9n cho phep khi m& may, lam dong CO' khong m& may dU'Q'C, de lo9i tru dieu §y, can thi§t phai dU'a vao roto mot di$n tr& pht,1 rt. NhU' v$y tu bieu thCrc (3-8a), (3-7) th1 Mmax = const nhLI'ng Sm dU'Q'C tang len.
3.3.1.3. Cong thCrc Clox (Klox):
Trang truyen dong di$n vi$c xac dinh M= t(s) theo nhO,ng s6 da cho & c§m nang r§t quan tr9ng. Cac thong s6 thLI'bng dU'Q'C cho: Mdm, sdm, kM ...
N§u khong c6 cac tham s6 c§u t90 cua dong CO' R1 ,x1 ,R2,x2 ta van c6 the tfnh dU'Q'C Smax, Mmax va ve dU'Q'C d~c tfnh CO' cua may. L§y cac quan h$ (3-6) va (3-8a) chT dung d§u ( +) trLI'D'ng hqp dong CO' va bo tri s6 di$n tu ta c6:
Theo (3-7) ta c6: M Mmax 2~( R1 + ✓R~ +x~) [ R1 + Rs; r + x~ R/
0$t - -2- lam thll'a s6 chung:
8max
M M
- - = - - - - - - - - - - -
Mmax
[_s_+_sma_x_+2-R_1 smax) Smax s R2 Mmax
+ 0,2, nen 2 R1 sm r§t nh6 c6 th§ b6 qua
R2
M 2 2Mmax
--=----- ⇒ M=-----
Mmax s ·smax s smax
- - + - - - - + - -
Smax s Smax s
(3-11)
(3-11 ): la bi§u thCrc Klox d§ ve dU'ong thang CO' cua may. V&i Smax dU'Q'C tinh nhLI' sau:
smax =sam(km + ✓kfn-1)
3.3.1.4. D~c tinh CO' va v§n d~ 6n
djnh n
Ta da bi§t: M2 = Mdt - Mo n1
Do Mo << M2 nen d$c tinh CO' cua nA i=~~-~ dong ca M2 = f(n) c6 th€! coi bang n
At
Mdt = f(n) va n6 c6 d9ng nhLI' M= f(s) & h1nh 3.22
Phan tich Sl! lam vi~c 6n djnh cua n 8
dong ca:
Gia SU' dong CO' lam vi~c v&I mot nB1
moment phv tal Mc nao d6. Theo
Mc:::: f(s)
1----.,-+---,E--f---.------"'.,_
phLI'ang tr1nh can bang mo men
~ f ½-a1 Moi-1 ~ 1 M!!!!!!!!!!'!!!!!!!!!!!!!!!!!!!-!!!!!!!!
Khoa nieih - Trbong TCN KTCN Hung Voong
1m , h 3 22 . S lf am v1~c on ufn cua uQng ca 1, •A ., ,,r h , ,,,A
Page 87
d(lng CO' c6 the lam Vi$C & hai diem Ava B.
- Xet trU'&ng hQ'p may lam Vi$C & diem A: n§u vl m(lt If do nao day MCA tang McA 1 >
McA thl Ma1 < O; nA - nA 1.
T<;1i nA1 M0A1 > McA1 ⇒ Mdl > 0 ⇒ nA1 - nA nen diem A la diem lam Vi$C 6n d[nh.
0i§u ki$n lam vi$C 6n djnh: dM0 > dMc (hay dM0 < dMc J
ds ds dn dn
-Xet tm&ng hqp may lam Vi$C t<;1i diem B:
Gia SU' Mes tang d§n McB1 > M0B ⇒ Mdl < 0 ⇒ ns - ns1
T<;1i nsf Ma1 = M0s1 - McB1 < 0 - Mal am ⇒ ns1 giam n = 0 ⇒ diem B la diem lam vi$c kh6ng 6n dinh
0i§u ki$n lam vi$C kh6ng 6n d!nh:
dM0 < dMc (hay dM0 > dMc)
ds ds dn dn
3.3.2 Cac d~c tinh lam vi~c cua dc}ng CO' khong df>ng be}
Cac d~c tf nh lam Vi$c cua d(lng CO' di$n kh6ng dong bQ gom: n, M, h va coscp = f(P2) v&i U 1 = const, f 1 = const.
3.3.2.1. D~c tinh t6c dQ n = f (P2) TL.I' c6ng thCrc: TL.I' c6ng thCrc: V&i n1 -n S=-- ⇒ n1 n=n1(1-s) s = Pcu2 Pctt
Khi khong tai Pcu2
⇒ s =O,n ~ n1
::::: 0
Khi kh6ng tai Ii tLI'&ng Pcu2= 0. Khi ph1,1 tai tang Mc= Mam do hi$u suat
Hinh 3.23 Cac cloong cJ~c tfnh lam
vi{}c cua may cli{}n khong c16ng 69.
ia ,,,
0,8
0,6
(),4. O,t O,t
cua dong CO' nen s = Pcu2 ~ Pcu2 =
. Pctt P2 0
Yj
(1,5+5) %. Sf> be Crng v&i d(lng CO' - -~
cong suat l&n, sf> l&n Crng v&i dQng O,S ~O GO'
cong suat nh6 (3.+10) KW. Do d6 s rat nh6, toe dQ giam rat ft khi s giam coi quan h$ n = f(P2) la m(lt dLI'&ng thang hO'i nghieng v§ tr1,1c hoanh.
3.3.2.2. D~c tinh moment M = f (P2)
Ta da bi§t & tlnh tr<;1ng lam Vi$C 6n d[nh M = M2 + Mo khi Mc= 0 + Mam th] coi nhLI' n =
canst ( s bi§n d6i trong gi&i h9n be) nen M = f(P2) coi nhLI' mc)t dU'&ng thang (M = P2 n
9,55).
3.3.2.3. Ton hao va d~c tinh hi~u su§t cua d9ng CO' h = f (P2)
Khi may lam Vi$C c6 cac t6n hao: T6n hao dong trong stator va rotor Pcu1 va PCu2, t6n hao s§.t PFe, t6n hao CO' PCO', t6n hao ph1,1 Pf,··· 4 lo9i t6n hao d§u da c6 c6ng thCrc xac d[nh (PCu1 = m1 l1 2r1, PFe = m1 lo2rm,Pcu2 = m1 l212r'2, PcO'= PcO'- P2 - Pf) con t6n hao
pht,1 bao gom t6n hao pht,1 trong dong va sat. Cach trnh refit phLI'c tc;tp nen thLI'O'ng lay la Pt= 0,5%P1.
ThLI'O'ng thiet ke TJ max vao khoang (0,5 + 0,75) P2. HiE;lU suat cua may: