MD Mot Chieu
62
Next Néi dung Back Bµi gi¶ng m¸y ®iÖn Next Néi dung Back Bµi gi¶ng m¸y ®iÖn PhÇn 1: M¸y ®iÖn mét chiÒu PhÇn 2: M¸y biÕn ¸p
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không thể nào bỏ qua tài liệu tuyệt vời này về vấn đề máy điện một chiều. nó mang lại cho ta cảm xuc
Transcript of MD Mot Chieu
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Next Ni dungBack
Bi ging my in
NextNi dungBack
Bi ging my in
Phn 1: My in mt chiu
Phn 2: My bin p
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Phn m uBi ging my in
My in
My in tnh My in quay
My bin p My inmt chiu
My inxoay chiu
ng cmt chiu
My phtmt chiu
My inkhng
ng b
My inng b
My phtkhng
ng b
ng ckhng
ng bng cng b
My phtng b
NextNi dungBack
Bi ging my in
MF
HtiuthMBA MBA
1. Vai tr ca cc loi my in trong nn kinh t quc dn:
2. Khi nim, phn loi v phng php nghin cu my in:a, i cng v my in:- Nguyn l lm vic ca my in da trn c s ca nh lut cmng in t. S bin i nng lng trong my in c thc hinthng qua t trng trong n. to c nhng t trng mnh vtp trung, ngi ta dng vt liu st t lm mch t.
my bin p mch t l mt li thp ng yn. Cn trong ccmy in quay, mch t gm hai li thp ng trc: mt quay, mtng yn v cch nhau bng mt khe h. b, Phng php nghin cu my in:
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NextNi dungBack
Bi ging my in
3. S lc v vt liu ch to my in:Gm c vt liu tc dng, vt liu kt cu v vt liu cch in.
Vt liu tc dng: bao gm vt liu dn in v dn t dng chto dy qun v li st.Vt liu cch in: dng cch in cc b phn dn in vi ccb phn khc ca my v cch in cc l thp ca li st.Vt liu kt cu: ch to cc chi tit my v cc b phn chu lc cgii nh trc, v my, khung my.S lc c tnh ca vt liu dn t, dn in v cch in dngtrong ch to my in.a, Vt liu dn t:b, Vt liu dn in:
c, Vt liu cch in:Cp cch in Y A E B F H C Nhit (0C) 90 105 120 130 155 180 >180
Phn 1: My in mt chiu
Chng 1 : Nguyn l lm vic - kt cu c bn
Chng 2 : Dy qun My in mt chiu
Chng 3 : Cc quan h in t trong my
Chng 4 : T trng trong my in mt chiu
Chng 5 : i chiu
Chng 6 : My pht in mt chiu
Chng 7 : ng c mt chiu
Chng 8 : My in mt chiu c bit
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Bi ging my in
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Chng 1: Nguyn l lm vic- kt cu c bn
Bi ging my in
NextNi dungBack
1.1: Cu to ca my in mt chiu
1.2: Nguyn l lm vic
1-3: cc lng nh mc
1.1: Cu to ca my in mt chiu
1. Phn tnh (Stato):
NextChng IBack
a) Cc t chnh:(L b phn sinh ra t thngkch thch)
b) Cc t ph:t gia cc cc t chnh, dng ci thin i chiu.
c) Gng t (v my):
d) Cc b phn khc:Np my: Bo v an ton cho ngi v thit b.C cu chi than: a dng in t phn quay ra mch ngoi.
Phn I: my in mt chiu
Cc t chnh
Dy qun cc t ph
Dy qun cc t chnh
Cc t ph
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my in mt chiu
phn cm ng c in mt chiu
Cc tv
Bu lng
Cun dy
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phn cm ng c in mt chiu
Bu lngcc t
cun dy
v
2. Phn ng (Rto):a) Li st phn ng: Dng dn t.
+) Vi cc my cng sut va v ln ngi ta dp lthng gi dc trc. +) Vi cc my in cng sut ln cn x rnh thnggi ngang trc.b) Dy qun phn ng: L phn sinh ra scin ng v c dng in chy qua.+) Dy qun thng lm bng ng c bccch in. trnh khi quay dy qun bvng ra ming rnh thng c nm chtbng tre, g php v u dy qun thngc ai cht.+) Vi cc M cng sut nh dy qun ctit din trn, cn my c cng sut va vln dy qun c tit din hnh ch nht.
Nm
Cchinrnh
Dyqun
Li st
Rnh
L thng gi dc trc
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my in mt chiu
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phn ng ng c in mt chiu
C gp
dy qun
li thp
trc
phn ng ng c in mt chiu
C gp
cun dy
li thp
trc
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c) Vnh i chiu (Vnh gp):Dng bin i dng xoay
chiu thnh dng mt chiu.Phin gp
d) Cc b phn khc:Cnh qut: Dng lm mt.Trc my: gn li st phn ng, c gp, cnh qut v bi.
Trc lm bng thp cc bon tt.
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my in mt chiu
1.2: Nguyn l lm vic
Phn tnh: Gm 1 h thng t c 2 cc N v S.
1. Nguyn l lm vic ch my pht:
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my in mt chiu
Phn ng: Gm khung dy abcd(1phn t dy qun).
d
b
ac
e
I
eFt
FtI b
d
c
aU
+
-
n
Rt
S
N
A
B
Theo nh lut cm ng in t: tr s sc in ng trong tngthanh dn ab v cd c xc nh: e = B.l.vTrong : B l tr s cm ng t ni dy dn qut qua
l l chiu di thanh dn nm trong t trng.v l vn tc di ca thanh dn.
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tt
Khi mch ngoi c ti th ta c: U = E - IRTrong : E l sc in ng ca my pht.
IR l st p trn khung dy abcdU l in p gia 2 u cc
Khi vng dy s chu 1 lc tc dng gi l lc t: Ft = B.I.l
Tng ng ta s c m men in t: Mt = Ft.D/2.= B.I.l.D/2T hnh v ta thy ch my pht Mt ngc vi chiu quay phn ng nn n c gi l M hm.
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my in mt chiu
Sc in ng v dng xoay chiu cm ngtrong thanh dn c chnh lu thnh scin ng v dng 1 chiu nh h thngvnh gp chi than.Ta c th biu din scin ng v dng in trong thanh dn v mch ngoi nh hnh v:
N
S
F, Mtn
Nh vy: ch ng c th U > E cn ch mypht th U < E
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my in mt chiu
ch ng c Mt cng chiu vi chiuquay phn ng gi l mmen quay.Nu in p t vo ng c l U th ta c: U = E + IR
2. Nguyn l lm vic ch ng c:
N
S
F, Mt
n
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my in mt chiu
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1-3: cc lng nh mc
1. Cng sut nh mc: Pm- Ti ca M ng vi tng nhit cho php ca my theo iukin lc thit k c quy nh l cng sut nh mc ca my.- Cng sut nh mc u c tnh u ra ca my.
2. Cc i lng nh mc khc:- Cc tr s in p, dng in, tc quay, h s cng sut... ngvi Pm u l cc tr s nh mc.
Chng 2: Dy qun My in mt chiu
2.1. Nhim v, cu to, phn loi
2.2. Dy qun xp n
2.3. Dy qun xp phc tp
2.4. Dy qun sng n gin
2.5. Dy qun sng phc tp
2.6. Dy cn bng in th
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my in mt chiu
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2.1: Nhim v - cu to - phn loi1. Nhim v ca dy qun phn ng:- Sinh ra c 1 sc in ng cn thit, hay c th cho 1 dng innht nh chy qua m khng b nng qu 1 nhit nht nh sinh ra 1 mmen cn thit ng thi m bo i chiu tt, cchin tt, lm vic chc chn, an ton. Tit kim vt liu, kt cu n gin.
2. Cu to ca dy qun phn ng:- Dy qun phn ng gm nhiu phn t nivi nhau theo 1 quy lut nht nh.- Phn t dy qun l 1 bi dy gm 1 hay nhiu vng dy m 2 u ca n ni vo 2 phin gp. - Cc phn t ni vi nhau thng qua 2 phingp v lm thnh cc mch vng kn.
u ni
Cnh tc dng
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Chng 2
Nu trong 1 rnh phn ng (rnh thc) ch t 2 cnh tc dng (dy qun 2 lp) thrnh gi l rnh nguyn t. Nu trong 1 rnh thc c 2u cnh tc dng vi u = 1,2,3... th rnh thc chia thnh u rnh nguyn t.
u=1 u=2
u=3
Quan h gia rnh thc Z v rnh nguyn t Znt : Znt = u.ZQuan h gia s phn t ca dy qun S v s phin gp G: S = G.
Znt = S = G
3. Phn loi:- Theo cch thc hin dy qun:
+ Dy qun xp n v xp phc tp.+ Dy qun sng n v sng phc tp.
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+ Trong 1 s trng hp cn dng c dy qun hn hp: kt hpc dy qun xp v sng.
-
Dng xp Dng sng
- Theo kch thc cc phn t: dy qun c phn t ng u v dyqun theo cp.
4. Cc bc dy qun:- Bc dy qun th nht y1 :khong cch cnh tc dng 1 & 2 ca 1 phn t.- Bc dy qun th hai y2 : khong cch cnh tc dng th haica phn t 1 v cnh tc dng 1 ca phn t th 2.- Bc dy tng hp y : khong cch gia 2 cnh tc dng thnht ca hai phn t lin k.- Bc vnh gp yG : khong cch gia 2 thanh gp ca 1 phnt.
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my in mt chiu
Dy qun c phnt ng u
Dy qun c phnt theo cp
2.2: Dy qun xp n1. Bc cc v cc bc dy qun:a) Bc cc : L chiu di phn ng di 1 cc
D l ng knh phn ng l bc ccp l s i cc
b) Cc bc dy qun: Bc dy qun th nht y1: y1 = p2Znt
Trong : l 1 s hoc phn s y1 l 1 s nguyn.
+ Nu y1 = ta c dy qun bc .
+ Nu y1 > ta c dy qun bc di.
+ Nu y1 < ta c dy qun bc ngn.p2Znt
p2Znt
p2Znt
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1 2 3
y1y y2
my in mt chiu
= [cm..]p
D. 2
pi
= [rnh ng. t]p
Znt2
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2. S khai trin:
Khai trin dy qun xp n MMC c Znt = S = G = 16, 2p = 4.
- Bc dy qun tng hp (y) v bc vnh gp (yG ): c im ca dy qun xp n l 2 u ca 1 phn t
ni vo 2 phin gp k nhau nn yG = y = 1.
a) Tnh cc bc dy qun:y1 = = = 4 (Bc ) y2 = y1 - y = 4 -1 = 3.
y = yG = 1.p2Znt
416
b)Th t ni cc phn t:Cn c vo cc bc dy qun ta c th b tr cch ni cc phn t thc hin dy qun.
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- Bc dy qun th hai y2: Trong dy qun xp n: y1 = y2 + y y2 = y1 - y.
my in mt chiu
+y1
- Gi s ti thi im kho st phn t 1 nm trn ng trung tnhhnh hc ( l ng thng trn b mt phn ng m dc theo ncm ng t bng 0).- V tr ca cc cc t trn hnh v phi i xng nhau, khong cchgia chng phi u nhau. Chiu rng cc t bng 0,7 bc cc.V trca chi than trn phin i chiu cng phi i xng, khong cchgia cc chi than phi bng nhau. Chiu rng chi than ly bng 1 phin i chiu.- Yu cu chi than phi t v tr dng in trong phn t khi bchi than ngn mch l nh nht v sc in ng ly ra 2 u chithan l ln nht. Nh vy chi than phi t trn trung tnh hnh hcv trc chi than trng vi trc cc t. Khai trin
Lp trn 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1
5 6 7 8 9 10 11 12 13 14 15 16 1 2 3 4Lp di
c) Gin khai trin:
my in mt chiu
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SNSN
16 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
A1 +A +
A2 +B1 -
B -B2 -
n
S khai trin dy qun MMCDy qun xp n c Znt = S = G = 16, 2p = 4.
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3.Xc nh s i mch nhnh: A1
B2
A2
B1
(-)Nhn t ngoi vo dy qun phn ng cth biu th bng s sau:
- Ta thy: dy qun phn ng l 1 mch in gm 4 mch nhnhsong song hp li. (Mch nhnh song song l phn dy qun nmgia 2 chi in c cc tnh khc nhau).Nu my c 2p cc th s c 2p mch nhnh song song.
Kt lun:- Trong dy qun xp n gin th s mch nhnh song song bngs cc t hay s i mch nhnh song song bng s i cc : a = p- Nu dy qun xp tho mn 2 iu kin: chi than nm trnng trung tnh hnh hc v h thng mch t i xng th scin ng cc nhnh bng nhau v t gi tr ln nht.
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(+)
my in mt chiu
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2-3: dy qun xp phc tp1. Bc dy qun:
c im ca dy qun xp phc tp l yG = m (m = 2, 3, 4...). Thng thng ch dng m = 2. Trong nhng my cng sut tht lnmi dng m > 2.Khi m = 2 = yG:- Nu s rnh nguyn t v s phn t l chn th tac 2 dy qun xp n c lp.
- Nu s rnh nguyn t v s phn t l ta c 2 dyqun xp n nhng khng c lp m ni tip nhauthnh 1 mch kn.
Nh vy c th coi dy qun xp phc tp gm m dy qun xp n lm vic song song nh chi than. V chi than phi c b rng m ln phin gp mi c th ly in ra.
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my in mt chiu
y1y y2
1 2 3 4 5
2. S khai trin:Dy qun xp phc tp c: yG = m = 2; 2p = 4; Znt = S = G = 24.
a) Cc bc dy qun: y2 = y1 - y = 6 - 2 = 4
yG = y = 2
6424
p2Zy1 ===
b) Th t ni cc phn t:
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Khp kn+y1
Khp kn
Lp trn 1 3 5 7 9 11 13 15 17 19 21 23 1
7 9 11 13 15 17 19 21 23Lp di 1 3 5
+y1
Lp trn 2 4 6 8 10 12 14 16 18 20 22 24 2
8 10 12 14 16 18 20 22 24Lp di 2 4 6
my in mt chiu
-
Gin khai trin dy qun MMCDy qun xp phc tp yG = m = 2; 2p = 4; Znt = S = G = 24.
N SSN
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
n
A1 + B1 -
A + B -
A2 + B2 -
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 2223 24
Cc t v chi in nh dy qun xp n. Ch khc l b rng chiin 2 ln phin gp c th ly in ng thi 2 dy qun ra.
my in mt chiu
Dy qun xp phc tp do m dy qun xp n cng u chungchi than do s i mch nhnh song song ca dy qun: a = m.p.
2-4: dy qun sng n1. Bc dy qun:
y1 = .
Dy qun sng n khc vi dy qun xp n yG.p2
Znt
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Mun cho khi qun xong vng th nht u cui ca phn t th p phi k vi u u ca phn t u tin th s phin i chiu mcc phn t vt qua phi l: p.yG = G 1 yG = (G l sphin gp).
Du (+) ng vi dy qun phi. Du (-) ng vi dy qun tri.
p1G
y1 = . y2 = y - y1 = yG - y1.
y = yG =
p2Znt
p1G
my in mt chiu
y1 y2
y
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2. S khai trin:Khai trin dy qun sng n c Znt = S = G = 15; 2p = 4
a) Bc dy qun:y1 = = - = 3 (bc ngn) . y2 = y - y1 = 7 - 3 = 4.
y = yG = = = 7 (dy qun tri)
415
43
2115
p2Znt
p1G
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b) Th t ni cc phn t:
+y1
Lp trn 1 8 15 7 14 6 13 5 12 4 11 3 10
4 11 3 10 2 9 1 8 15Lp di 7 14 6
2 9
13 5
1
12
+y2
my in mt chiu
SNSN
Gin khai trin dy qun MMCDy qun sng n c Znt = S = G = 15; 2p = 4
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
B2 -A2 +
B -
B1 -
A +
A1 +
1 23 4 5 6 7 8 9 10 11 12 13 14 15
n
Quy lut ni dy ca dy qun sng n l ni tip tt c cc phnt di cc cc c cng cc tnh li ri ni vi cc phn t di cccc c cc tnh khc cho n ht.
my in mt chiu
Dy qun sng n ch c 1 i mch nhnh song song: a = 1.
-
2-5: dy qun sng phc tp1. Bc dy qun: Tng t nh vi dy qun sng n.
Ring bc vnh gp: yG pmG
=
2. S khai trin:
a) Tnh bc dy qun: y1 = = = 4 (dy qun bc ngn)
yG = 8 = y; y2 = y - y1 = 8 - 4 = 4.
p2Znt
4
2
4
18
2218
pmG
=
=
b) Th t ni cc phn t:
m = 2; 2p = 4; Znt = S = 18.
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Khp kn
Khp kn
+y1
Lp trn 1 9 17 7 15 5 13 3 11 1
5 13 3 11 1 9 17 7 15Lp di
+y1
Lp trn 2 10 18 8 16 6 14 4 12 2
6 14 4 12 2 10 18 8 16Lp di
+y2
+y2
my in mt chiu
Gin khai trin dy qun MMCDy qun sng phc tp c: m = 2; 2p = 4; Znt = S = 18
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N SSN
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
1 23 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
n
B2 -A2 +
B -
B1 -
A +
A1 +
my in mt chiu
Dy qun sng phc tp gm m dy qun sng n hp li do s i mch nhnh song song ca dy qun sng phc tp: a = m.
-
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2.6: Dy cn bng in th
1.iu kin dy qun i xng:
- Dy qun MMC tng ng nh 1 mch in gm 1 s nhnhsong song ghp li. Mi nhnh gm 1 s phn t ni tip nhau.
- Dy qun phi m bo 1 s yu cu sau:+ m bo v cm ng t: H thng mch t phi c cu to
i xng, t thng cc cc nh nhau.+ iu kin v dy qun: Tt c cc dy qun to thnh mch
nhnh phi tng ng nhau v s phn t ca cc nhnh cng phitng ng.
- iu kin bnh thng: sc in ng sinh ra trong cc mch nhnh song song bng nhau, dng in phn b u trong cc nhnh.
my in mt chiu
- Dy cn bng in th lm mt s khng i xng ca mch t trong M cn bng in th cc mch nhnh ca dyqun xp nm di cc cc t c cng cc tnh c gi l dycn bng loi 1. Bc th yt bng s phin i chiu di mii cc:
yt = a
GpG
=
- Dy cn bng lm mt s phn b khng i xng ca inp trn vnh gp gi l dy cn bng loi 2.
Bc th: yt = a
Ga
S=
2. Dy cn bng in th loi 1:
3. Dy cn bng loi 2:
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my in mt chiu
-
Chng 3:Cc quan h in t trong my
3.1: Sc in ng dy qun phn ng
3.2: M men in t - cng sut in t
3.3: Cn bng nng lng - tn hao- hiu sut
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my in mt chiu
3.1: Sc in ng dy qun phn ngSc in ng trung bnh cm ng trong 1 thanh dn c chiu
di l, chuyn ng vi vn tc v trong t trng bng: etb = Btb.l.v l bc ccD l ng knh phn ng. p l s i cc.n l tc quay phn ng(v/pht)
: t thng khe h di mi cc t (Wb)
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60np2
60nD
... =
pi
l
v =
Btb =
l
60np2 .
60np2 .. etb = .l. =
Nu gi N l tng s thanh dn ca dy qun th mi mch nhnh song song s c thanh dn ni tip nhau. Nh vy sc inng ca my: a2
N
E =
Trong : Ce = l h s ph thuc kt cu my v dy qun.
a2N
na60
pN..
a60pN
.etb = Hay E = Ce..n (V)
my in mt chiu
-
3.2. M men in t - cng sut in t1. Mmen in t:
Khi M lm vic trong dy qun phn ngs c dng in chy qua. Tc dng ca ttrng ln dy dn c dng in chy qua ssinh ra mmen in t trn trc my.
- Lc in t tc dng ln tng thanh dn: f = Btb.l.i
M = Btb. .l.N. . a2
I 2
D Thay D = v Btb = ta c:pi
p2l.
M = . .l.N. = ..I = CM. .I (Nm) l.
a2I
pi
2p2
pia2pN
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Gi N l tng s thanh dn ca dy qun v dng trong mch nhnhl: i
= I
/2a th mmen in t tc dng ln dy qun phn ng l:
Trong : CM = l h s ph thuc kt cu my.pia2pN
l t thng di mi cc t (Wb).
my in mt chiu
S
nM
B Btb
2. Cng sut in t:
Cng sut ng vi mmen in t (ly vo vi my pht v a ravi ng c) gi l cng sut in t: Pt = M.
= l tc gc phn ng.60
n2pi
pia2pN
60n2pi
a60pN
- Trong ch my pht: M ngc chiu quay vi phn ng nnng vai tr l mmen hm. My chuyn cng sut c (M.) thnh cng sut in (EI).- Trong ch ng c: M c tc dng lm quay phn ng cngchiu vi chiu quay phn ng. My chuyn cng sut in (EI) thnh cng sut c (M.)
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Pt = ..I . = .n..I = E . I
my in mt chiu
-
1. Tn hao trong MMC:
3.3: Cn bng nng lng - tn hao - hiu sut
a) Tn hao c( pc):b) Tn hao st (pFe):
c) Tn hao ng (pcu): Gm 2 phn: - Tn hao ng trn mch phn ng: pcu. = I
2RVi: R = r + rb + rf + rtx ; (r : in tr dy qun phn ng;
rf : in tr cc t ph; rb : in tr dy qun b; rtx: in tr tip xc chi than.)
- Tn hao ng trn mch kch thch:(Bao gm tn hao ng ca dy qun kch thch v ca in triu chnh trong mch kch thch): pcu.kt = Ukt Ikt
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0p
- Hai loi tn hao trn tn ti ngay c khi khng ti nn gi l tnhao khng ti: p0 = pc + pFe. Tn hao ny sinh ra M hm ngay ckhi khng ti nn gi l M khng ti: M0 =
Chng 3
my in mt chiu
d) Tn hao ph: (pf)Trong ng v trong thp u sinh ra tn hao ph. Tn hao phthng kh tnh. Ta ly pf = 1%Pm
2. Gin nng lng v hiu sut:
a) My pht in:
- Gin nng lng:
- Hiu sut:
( )11
cuFeco1
1
2
Pp1
PpppP
PP
=
++==
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Pt = P1 - (pc + pFe) = P1 - p0 = E IP2 = Pt - pcu = U.I
my in mt chiu
P1 Pt
pcupFepc
P2
-
b) ng c in:Ta c cng sut in m ng c nhn t li:
P1 = U.I = U.(I + Ikt )Vi: I = I + Ikt l dng nhn t li vo.
U l in p u cc my.Pt = P1 - (pcu. + pcu.kt)Pt = EI
Cn li l cng sut c a ra u trc: P2 = M. = Pt - (pc +pFe)
- Gin nng lng:
P1 Pt
pcu. + pcu.kt pFe pc
P2- Hiu sut:
( )11
cuFeco1
1
2
Pp1
PpppP
PP
=
++==
NextChng 3Back
my in mt chiu
4-1: t trng lc khng ti-T Trng cc t chnh-
4-2: t trng khi c ti
NextPhn IBack
Chng 4 : T trngtrong my in mt chiu
my in mt chiu
-
4-1: t trng lc khng ti
1.T trng chnh v t trng tn:
T thng chnh l t thng i qua khe h khng kh gia phn ng vcc t trong phm vi 1 bc cc.
0
T thng ca cc t c tnh nh sau: c = 0 + = 0(1+/ 0 ) = 0.t
NextChng 4Back
0Vi t = 1+ l h s tn t ca cc t chnh.
2. Sc t ng cn thit sinh ra t thng:
- Do mch t hon ton i xng v sc t ng cc cc t nhnhau nn ta ch cn tnh cho 1 i cc.
my in mt chiu
- c t thng chnh 0 ta cn cung cp cho dy qun kchthch 1 sc t ng F0 no . n gin cho vic tnh ton ta dngcch phn on mch t thnh 5 on: khe h khng kh (), rngphn ng (hr), lng phn ng (l), cc t (hc), gng t (lG).
NextChng 4Back
B
S
Khi sc t ng cn thit cho 1 i cc s tnh nh sau:F0 = I.W = H.l = 2H. + 2Hr.hr + H.l + 2Hc.hc + HG.lG
= F + Fr + F + Fc + FGTrong : h ch chiu cao, l ch chiu di. Trong mi on cng t trng c tnh: H = vi B = , S, l t thng, tit din, h s t thm ca cc on.
a) Sc t ng trn khe h F: F = 2H.* Khi phn ng nhn:
- Do khe h gia cc t v phn ng khng u: gia th khe hnh, 2 u mp cc t khe h ln: max = (1,5 2,5) nn phn b tcm nhng im thng gc vi b mt phn ng cng khc nhau.
my in mt chiu
hc hc
hrhr
l
lG
-
B
b
- n gin ta thay ng cong t cm thc tbng 1 hnh ch nht c chiu cao l B v y lb = . sao cho din tch hnh ch nht bngdin tch bao bi ng cong thc t. (b l cungtnh ton ca cc t cn l h s tnh toncung cc). Trong MMC c cc t ph th = 0,62 0,72;
MMC khng c cc t ph th = 0,7 0,8
NextBack
- Phn b t cm di 1 cc t biu din nhhnh v. T cm gia cc t c gi tr ln nhtcn 2 mp cc tr s gim dn v ng trungtnh hnh hc gia 2 cc t th bng 0.
Chng 4
my in mt chiu
Gi l l chiu di phn ng theo dc trcv lc l chiu di cc t th ta c chiu di tnhton l = . Vi l = l1- ng.bg
ng,bg l s rnh v chiu rng rnh thng gi.2
ll c+ Bl
lcl1
T cm khe h khng kh:
Sc t ng c tnh:
=
=
blSB
.
=
==
.
.....
bl2B2H2F
001
* Khi phn ng c rng: - Khi tnh ton ta phi quy i phn ngc rng v phn ng nhn bng cch tngkhe h khng kh l ' = K. vi ' cgi l tr s tnh ton ca khe h.
K l h s khe h: t1 l bc rng;br1 l chiu rng nh rng
++
=10b
10tK
1r
1
Ta c sc t ng phn ng khi c rng :
F = 2.H.' F = 2.H. K. = 2.F1. K = 2. .K' = K.
... bl0
NextChng 4Back
my in mt chiu
br1
t1
-
b) Sc t ng trn rng phn ng:
Hrtb
Hr1t1
br1
x
br2
t2
Hr2
T thng i qua 1 bc rng t1 l t = B.l.t1Xt 1 tit din ng tm vi mt phn ng,
cch nh rng 1 khong x th t thng i qua tit din gm 2 thnh phn:
xrrxt +=Chia 2 v ca (1) cho Srx (tit din rng) ta c:
(1)
rx
xr
rx
rx
rx
t
SSS
+
=
rx
rx
t BS
=
rx
rx
rx BS
=
l tr s t cm tnh ton v thc t ca rngv
NextChng 4Back
(1)
Vi phn ng c rng v rnh khi t thng i qua khe hkhng kh th phn lm 2 mch song song i vo rng v rnh phnng. Do t dn ca thp ln hn khng kh nhiu nn i b phn tthng i vo rng. t1 l bc nh rng
t2 l bc chn rng
my in mt chiu
*) ngha ca Brx:Coi ton b t thng u i qua rng phn ng (mch t cha
bo ho). Khi Btx> 1,8 Tesla th mch t bt u bo ho, t tr tngdn, ta khng th b qua t cm trn rnh. Khi thnh phn th 2 ca biu thc (1') biu din nh sau:
rxxr0rxxrrx
xr
xr
xr
rx
xr KHKBSS
SS....
=== (2)- Gi thit rng nhng mt ct hnh tr ngang rng v rnh cc cao x u l nhng mt ng tr ca t trng th khi c th xemnh Hrx = Hrx Thay vo (1'): rxxrrxrx KHBB .. '0+= (3)
Gi tr biu thc ny c th tm c t ng cong t ho B = f(H) v qua cc bc tnh c th sau:
+ V ng cong t ho ca li st phn ngKhi bit kch thc ca rng v rnh ta c: 1
klblt
SSK
crx
x
rx
xrrx ==
Vi: Kc l h s p cht li st; Stx, tx l tit din rng v bc rng cao x; l, l l chiu di thc v tnh ton ca li st.
my in mt chiu
-
B(2)
(1)
0Hrx H
rxrx0 KH ..
Brx
+ T ng cong t ho cho cc tr s ca tcm Brx ta tm c Hrx tng ng Brx theo (3).Sau v ng biu din (ng 2).
- Ngoi ra tr s t cm tnh ton ca rng cng c
th xc nh theo biu thc: crx
1
rx
trx klb
ltBS
B
=
= (4)
* Thc t khi tnh ton sc t ng rng ch cn tnh H 3 im theochiu cao rng: nh rng, chn rng v gia chiu cao rng. Khi trs tnh ton ca cng t trng trung bnh bng:
( )2rrtb1rr HH4H61H ++= (5)
Gi hr l chiu cao rng th sc t ng rng i vi 1 i cc l:Fr = 2. Hr. hr (6)
* n gin ta ch cn xc nh t cm v t trng tit din cchchn rng 1/3 lm tr s tnh ton. Khi ta c: Fr = 2. Hr1/3. hr (7)Trong my in 1 chiu t cm ni hp nht ca rng = 1,8 2,3 T.
my in mt chiu
c) Sc t ng trn lng phn ng:
T thng sau khi i qua khe h khng kh vo phn ng qua rng v rnh s phn b khng u: gn rng c t cm ln hnnhng s khc bit khng ln lm nn c th ly t cm trung bnh lng phn ng tnh ton.
T cm lng phn ng:c
00
klh2S2SB
....
=
=
=
Trong : l t thng phn ng;
l tit din lng phn ng.
20
=
cklhS .. =
Theo ng cong t ho t B H
Khi sc t ng trn lng phn ng c tnh: F = H.l
NextChng 4Back
my in mt chiu
-
d) Sc t ng trn thn cc t v gng t:
Tnh ton sc t ng cc t ta phi xt n nh hng ca tthng tn. Khi t thng cc t ln hn t thng chnh: c = 0.tvi t = 1,15 1,25 l h s tn t.
Thc t do tn ra khp cc t nn t thng cc phn trncc t v gng t cng khc nhau. Nhng n gin ho tnh ton tacoi nh trn cc t v gng t c t thng khng i. (G = 1/2c)
Ta c: v
Vi Sc v SG l tit din cc t v gng tc
cc S
B
=
G
cG S2
B
=
Sc t ng cc t v gng t: Fc = 2.Hc.hc hc: chiu cao cc tFG = HG.lG lG: chiu di trung bnh ca gng t.
NextChng 4Back
my in mt chiu
3. ng cong t ho ca MMC:
ng biu din quan h gia 0 v F0gi l ng cong t ho ca MMC.
F0
a b cmNu ko di on tuyn tnh ta c quan
h = f(F). Khi t thng t gi tr nh mcth on ab c trng cho sc t ng khe hcn on bc c trng cho sc t ng ri trncc phn st ca mch t.
ab = F F0 = ac = F + Fr + F + Fc + FGbc = Fr + F + Fc + FGt th k l h s bo ho ca mch t.
ab
ack
F
F0==
Trong cc M trit li dng vt liu v cng sut khi in
p l nh mc ta chn im lm vic l im chm bo ho (im c: im m ng cong t ho bt u cong vi k = 1,1 1,35).
NextChng 4Back
my in mt chiu
-
4-2: t trng ca my in khi c ti
1. T trng cc t chnh: N
S
T hnh v phn b t trng ca cc t chnhca my 2 cc ta thy t trng chnh nhn trccc lm trc i xng v khng thay i v trtrong khng gian.
2. T trng phn ng:
a) Chiu ca t trng phn ng:
* Khi chi than t trn trung tnh hnh hc:
Chng 4Back Next
- Trc sc t ng tng ca c dy qun sinhra lun lun trng vi trc chi than.
Trung tnhhnh hcN
S
N
n
S
my in mt chiu
* Nu dch chi than khi trung tnh hnhhc ng vi 1 on b trn phn ng:
- Phn tch sc t ng phn ng F thnh 2 thnhphn: + Sc t ng ngang trc Fq.
+ Sc t ng dc trc: Fd.
b) S phn b t trng trn b mt phn ng:
* Khi chi than trn ng trung tnh hnh hc: (*)Ta xt 1 mch vng i xng vi im gia ca 2 chi than th
1 im cch gc l x sc t ng c tnh nh sau:Fx = A.2x (A/i cc)
A = (A/cm): l ph ti ng ca phn ng.
i = l dng trong thanh dn.
.
.
DiN
pi
a2I
NextBack
Sc t ng s ln nht khi x = . Khi : F = A.2. = A..2
2
Chng 4
my in mt chiu
N
n
S
Trung tnhvt l
Trung tnhhnh hc
N
CDn
S
A B
b
b
(*)
Fd
Fq
F
-
NextChng 4Back
- Nu b qua t tr ca li thp tht tr ca mch phn ng ch cn l 2 khe h khng kh nn t cm ca phnng di mt cc t c biu din: Bx = 0.Hx = 0. = 0.
= 0. .x 2F x
2x2A.
A
Di mi bc cc trong phm vi 2b dng in sinh ra sc t ngdc trc cn trong phm vi ( - 2b) dng in sinh ra sc t ngngang trc: Fd = A.2b (A/i cc)
Fq = A. ( - 2b)
FxA/2
Bx
N S
F
* Khi chi than dch khi trung tnhhnh hc 1 khong b no :
Tm li: t trng phn ng ph thuc vo v tr ca chi than vmc ca ti.
my in mt chiu
(*)
NextChng 4Back
3. Phn ng phn ng:a) Khi chi than t trn ng trung tnh hnh hc:
mN S
F (4)
(3)(1)
(2)
m
- S phn b ca t trng do ttrng cc t chnh v t trngphn ng hp li nh sau:
(4): T trng tng khi mch tbo ho.
(1): T cm ca cc t chnh.(2): T trng phn ng.(3): T trng tng khi mch t
cha bo ho.
*)Tm li:
Khi chi than t trn trung tnh hnh hc ch c phn ng ngang trclm mo dng t trng khe h v xut hin ng trung tnh vt l
my in mt chiu
m
m
-
NextChng 4Back
b) Khi chi than dch khi trung tnh hnh hc: Phn tch sc t ng phn ng thnh 2 phn: Fq v Fd.- Thnh phn ngang trc Fq c tc dng lm mo dng t trngcc t chnh v kh t 1 t nu mch t bo ho.- Thnh phn dc trc Fd nh hng trc tip n t trng cc tchnh. N c tc dng kh t hoc tr t tu theo chiu x dch cachi than. (*)
Do yu cu ca i chiu ch cho php quay chi than theochiu quay ca phn ng trong trng hp my pht cn ng c thngc li.
my in mt chiu
4. T trng cc t ph: Sf N Nf S Sf
NextBack
F
(2)
(3)
(1)
Hb-1
Hb-2
Hb-3
(1): Sc t ng cc t chnh.
(3): Sc t ng cc t ph.(2): Sc t ng phn ng.
Hnh b-3: Phn b t cm.Hnh b-2: Sc t ng tng.
my in mt chiu
Chng 4
Tc dng ca cc t phl sinh ra 1 sc t ng trittiu t trng phn ngngang trc v to ra t trngngc chiu vi t trngphn ng khu vc ichiu.
-
NextChng 4Back
Hnh c3: Cc t ph ctc dng tr t.
Hc3
Hc2
Hc1
SN NfSfS
F
Hnh c1: Cc t phkhng nh hng ti ttrng tng.Hnh c2: Cc t ph ctc dng kh t.
my in mt chiu- Khi chi than t trn trung tnh hnh hc th cc t ph
khng nh hng n cc t chnh.
Cch ni cc t ph:
F
NfSf
N
S
5. T trng dy qun b:
(1)
(2)
F
NextChng 4Back
- Tc dng ca dy qun b lsinh ra t trng trit tiu phnng phn ng lm cho t trngkhe h v cn bn khng b modng na.
ng (1): Sc t ng caphn ng phn ng ngang.
ng (2): Sc t ng cady qun b.
ng (4): Phn b t trng tngkhi c c dy qun b v cc t ph.
Cch ni dy qun b:
ng (3): Sc t ng khi khng ti.
N NfS Sf S
(3)
my in mt chiu
(4)
-
Chng 5 : i chiu
5.1: Nguyn nhn sinh ra tia latrn vnh gp
5.2: Qu trnh i chiu
5.3: Cc phng php ci thini chiu
NextPhn IBack
my in mt chiu
5.1: Nguyn nhn gy tia la trn vnh gp
NextChng 5Back
1. Nguyn nhn v c kh:- Vnh gp khng ng tm vi trc.- S cn bng b phn quay khng tt.- B mt vnh gp khng phng do nhng phin i chiu hoc mi ca cch in gia cc phin i chiu nh ln.- Lc p chi than khng thch hp (mnh qu c th lm mn chiv vnh gp), kt chi trong hp chi, hp chi than khng c gicht hay t khng ng v tr.
2. Nguyn nhn v in:- Do sc in ng phn khng khng trit tiu ht sc in ngi chiu.- Do s phn b khng u mt dng in trn mt tip xc vquan h phi tuyn ca in tr tip xc: rtx = f(t,) vi l thng sc trng cho tc dng nhit v hin tng in phn di chi than.
my in mt chiu
-
3. Cc cp tia la in:
Bng cc cp tia la in (*)
- Ta thy tia la mnh gy hao mn nhanh chng chi than v vnhgp. Do tia la cp 2 ch cho php vi nhng ti xung ngn hn, tia la cp 3 ni chung l khng cho php.
- Ch lm vic lu di vi cp tia la 1.
NextChng 5Back
my in mt chiu
1.Mt s khi nim:
5.2: Qu trnh i chiu
a) Qu trnh i chiu:
NextChng 5Back
b) Chu k i chiu:
2i 2i 2it = 0 0 < t < Tc t = Tc
Qu trnh i chiu ca dng in khi phnt di ng trong vng trung tnh hnh hc v bchi than ni ngn mch gi l s i chiu.
i i i i i i i ii i i i
S NA2
B1
A1 S
B2
my in mt chiu
N
-
- Qu trnh i chiu ca dng in trong mi phn t tn ti trong 1 khong thi gian rt ngn.
NextChng 5Back
- Khong thi gian dng in hon thnh vic i chiu gi l chuk i chiu (Tc). l thi gian cn thit vnh gp quay i 1 gcng vi chiu rng ca chi, ngha l Tc = (1)
G
c
v
b
bc l chiu rng ca chi gp.vG l vn tc di ca vnh gp.
Gi DG l ng knh vnh gp Ta c: bG = pi. bG l bc vnh gpG : s phin gp vG = DG. pi. n = bG.G.n
GDG
Gi G l h s trng khp th: G = Thay vo (1):
Tc = = G. (2)G
c
bb
nGbb
G
c
.. nG1.
(y l chu k i chiuca dy qun xp n)
my in mt chiu
2. Cc sc in ng trong mch vng i chiu:- Sc in ng t cm eL:
dc
ccL
T
i2L
dt
diLe ==
- Sc in ng h cm eM:
===n
1n
dc
uMn
nn
1n
n
1MnM M.
T
i2e
dt
di.Mee
Mn l h s h cm gia phn t ang xt v phn t th n.in l dng in trong phn t th n
NextChng 5Back
my in mt chiu
- Sc in ng i chiu ec: sinh ra khi phn t i chiu chuynng trong t trng tng hp ti vng trung tnh:
ec = 2.W.Bc.lc.v.W: s vng ca phn t i chiu.Bc: T cm i chiu
- Sc in ng phn khng epk: epk = eL + eM m bo i chiu c tt th sc in ng phn khng
phi lun lun ngc chiu vi sc in ng i chiu.
-
3. Phng trnh i chiu:
Theo nh lut Kishop 2 vit cho mch vng phn t b:i.rpt + i1.(rtx1 + rd) - i2.(rtx2 + rd) = e (1)Trong :rpt : in tr ca phn t i chiu.rd : in tr dy ni.rtx1,2: in tr tip xc ca chi than vi phin gp 1 v 2.e : Tng cc sc in ng sinh ra trong phn t i chiu:e = eM + eL + ec = epk + ecTheo nh lut Kishop 1 vit cho cc nt 1 v 2:Nt 1: i + i - i1 = 0 (2)Nt 2: - i + i - i2 = 0 (3)
i: dng ngn mch chy trong phn t i chiu.i1, i2: dng chy qua dy ni vi cc phin i chiu 1 v 2.
NextChng 5Back
i i i i
1 2i1 i2
a b c
1 2
my in mt chiu
Cc dng i, i1, i2 c th tnh t cc phng trnh (1), (2) v (3) nu cc i lng khc bit.
i.rpt + i1.(rtx1 + rd) - i2.(rtx2 + rd) = ei + i - i1 = 0
- i + i - i2 = 0
- mc gn ng gi thit rpt 0, rd 0. Ta c: i1.rtx1 - i2.rtx2 = e i + i - i1 = 0- i + i - i2 = 0 1tx2tx
1tx2tx
1tx2tx
rr
ei.
rr
rri
+
+
+
=
V gi thit rpt 0, rd 0 nn (rtx1+ rtx2) l tng tr ca phn ti chiu khi b chi than ngn mch v dng if chnh l dng ngn mch trong phn t gy bi e
NextChng 5Back
(4)
my in mt chiu
if
-
Gi thit rng rtx1, rtx2 t l nghch vi b mt tip xc ca chiin vi phin i chiu 1 v 2. Trong qu trnh i chiu t 0 nTc vi iu kin bc = bG th cc b mt tip xc c tnh nh sau:
ST
tS
dctx =2 S
T
tTS
dc
dc1tx
=
Vi S l din tch tip xc ton phn cachi vi phin gp.
txdc
dctx
1tx1tx r
tT
Tr
S
Sr
== txdc
tx2tx
2tx rt
Tr
S
Sr ==
Thay (5), (6) vo (4) ta c:
(5)
nm
dc r
ei.
T
t21i
+
=
)1T(t
T.rr
dc
2dc
txnm
=
(6)
(7)
Vi
NextChng 5Back
K hiu rtx l in tr tip xc ton phnng vi mt tip xc ton phn S ta c:
(*)
(7): L phng trnh i chiutng qut.
my in mt chiu
SStx2
Stx1
4. Cc loi i chiu:a) i chiu ng thng:
e = 0 v dng in i chiu l
dc
i.T
t21i
=
Mt dng in 2 pha i vo v i ra bngnhau: J1 = J2 nn khng c tia la xut hin. (*)
b) i chiu ng cong:
if, rnm
0
(1)
(2)
(3)
tm e 0 nn if 0.
Dng in ph ny s cng vi dng c bnlm cho dng in i chiu tr nn phi tuynv ta c i chiu ng cong.
nmf
r
ei
=
(1): rnm(t)(2): if(t) khi e > 0(3): if(t) khi e < 0
Ta biu din if di dng sau: (e = const):
NextChng 5Back
my in mt chiu
Tc
1
t
i1t
-i
0
i22+i
i
-
* Gi s epk > ec hay e > 0 v githit rtx = const th dng i chiuc dng:
* Gi s epk < ec hay e < 0 th if idu theo ng (3). Khi dng ichiu c dng:
Ta thy 1 < 2 nn J1 < J2 => i chiuvt trc v c tia la u vo cachi.
NextChng 5Back
Ta thy 1 > 2 nn J1 > J2 => i chiumang tnh tr hon. Tia la xut hin u ra ca chi in khi phn t ri khiv tr ngn mch.
my in mt chiu
+i2
i2
0
-it
i1
Tc
1
t
i
-i
Tc
1t i1
i2
2+i
0 t
i
1.Phng php dng cc t ph:
5.3. Cc phng php ci thin i chiu
Bin php c bn ci thin i chiu trong MMC l to ra ttrng i chiu ti vng trung tnh hnh hc bng cch t nhngcc t ph gia nhng cc t chnh. Mun vy sc t ng ca cct ph Ff phi ngc chiu vi sc t ng phn ng ngang trc. V tr s ngoi vic trung ho phn ng phn ng ngang trc cn phito ra 1 t trng ph sinh ra ec lm trit tiu epk.
Mt khc : Fq v epk t l vi i do sc t ng cc t ph v ttrng i chiu cng phi bin i t l vi ph ti. Mun vy dyqun cc t ph phi ni tip vi dy qun phn ng ng thi mch t cc t ph khng bo ho. Thng thng khe h gia cc t phvi phn ng ln hn so vi cc t chnh (t 1,5 2 ln).
NextChng 5Back
my in mt chiu
-
2. Phng php x dch chi than khi trung tnh hnh hc:
my pht: Chi than ang c t trn trungtnh hnh hc. Ta xt 1 phn t dy qun:
-i ec
epk+ieq
e
NextChng 5Back
my in mt chiu
N
n
S
*) Nhn xt: Khi M lm vic ch my pht ci thin ichiu ta phi dch chi in khi trung tnh hnh hc 1 gc: = + theo chiu quay phn ng. Cn ch ng c th dch chi in i 1 gc ngc chiu quay phn ng.(: l gc gia trung tnh hnh hc v trung tnh vt l.
: l gc ng vi iu kin t trng tng bng t trng i chiu)
4. Nhng bin php khc:Chn chi than ph hp, gim sc in ng phn khng ...
NextChng 5Back
my in mt chiu
3. Dng dy qun b: Trong cc M cng sut ln v ti thay i t ngt ngi ta
thng dng dy qun b h tr cho cc cc t ph. Dy qunb c t di mt cc t chnh v s trit tiu t trng phnng di phm vi mt cc t chnh v lm cho t trng chnh hunh khng thay i. Dy qun b ni tip vi dy qun phn ngnn c th b bt c ti no.
-
Chng 6 : My pht mt chiu
6.1: Nhng khi nim c bn
6.2: c tnh c bn ca my pht mt chiu
6.3: My pht mt chiu kch t song song
NextPhn IBack
6.4: My pht mt chiu kch t ni tip
6.5: My pht mt chiu kch t hn hp
6.6: My pht mt chiu lm vic song song
my in mt chiu
NextChng 6Back
6.1: Nhng khi nim c bn
1. Phn loi: Tu theo phng php kch thch cc t chnh MFMC c phn thnh 2 loi:
a) My pht in mt chiu kch t c lp:b) My pht 1 chiu t kch:
UI
Ikt
I
Hnh a
U
I I
I
Hnh b
Trong mi trng hp cng sut kch thch chim 0,3 0,5% Pm.
+ My pht mt chiu kch thch song song: I = I + Ikt (hnh a).+ My pht mt chiu kch thch ni tip: I = Ikt = I (hnh b).+ My pht mt chiu kch thch hn hp: I = I + Iktss (hnh c).
I
IU
Ikt
Hnh c
U
I
Iktss
Iktnt
I
my in mt chiu
-
NextChng 6Back
2. Phng trnh cn bng m men:
Ta c: P1 = pc + pFe + PtChia 2 v cho : hay: Mq = M0 + Mt
Trong : Mq l mmen ca my pht in.M0 l mmen cn khng ti.Mt l mmen in t.
+
+=
dtFeco1 PPPP
Nu t M0 + Mt = MCT (mmen cn tnh) th phng trnh cn bngmmen s l: Mq = MCT
3. Phng trnh cn bng in p:
U.I = E.I - (I2rdq + UtxI)
+=
txdq
I
UrIEU
t l in tr mch phn ng
txdq
I
Urr
+= U = E - Ir
my in mt chiu
P1
pc pFe
Pt
pcu
P2
P2 = Pt - (pcu + pf)
4. Cc c tnh ca my pht 1 chiu:
C 5 dng c tnh:+ c tnh khng ti: U0 = E = f(Ikt) khi I = 0, n = const.+ c tnh ngn mch: In = f(Ikt) khi U = 0, n = const.+ c tnh ngoi: U = f(I) khi Ikt = const, n = const.+ c tnh ph ti: U = f(Ikt) khi I = const, n = const.+ c tnh iu chnh: Ikt = f(I) khi U = const, n = const.
NextChng 6Back
my in mt chiu
-
1. c tnh khng ti: U = f(Ikt) khi I = 0, n = const.
6.2: Nhng c tnh c bn camy pht mt chiu
c tnh c xc nh bng thc nghim theo s th nghim
Khi I = 0 U = E = Ce..n = Ce.. c tnh lp li dng ng cong t ho ring ca my in.
Ikt
V
A
I
A'
-IktmIktm
AU
B
0Ikt
B
my in mt chiu
NextChng 6Back
2. c tnh ngn mch: In = f(Ikt) Khi U = 0, n = const.
0
Ikt
(1)
(2)In
(1): My c kh t d.(2): My cha c kh t d.
- Do U = 0 ta c E = IR ngha lton b sc in ng sinh ra bp cho st p trn mch phn ng.
- Mt khc: dng ngn mch c hn ch bng (1,25 1,5)Im vR rt nh v vy E nh Ikt tng ng nh mch t khngbo ho. Do E t l tuyn tnh vi Ikt nn I cng t l vi Ikt ctnh c dng ng thng.
NextChng 6Back
my in mt chiu
-
Tam gic c tnh:
Dng tam gic c tnh
(1): c tnh khng ti(2): c tnh ngn mch.
ln ca AB ph thuc vo loi my, ln nht MMC khng c cc t phv dy qun b. my c cc t ph v dy qun b phn ng phn ng hu nhb trit tiu, cnh AB 0. MMC kch t hn hp, dy qun ni tip c tc dngtr t v nu sc t ng ca n ln hn AB, ngha l ngoi phn sc t ng trit
tiu nh hng ca phn ng phn ng cn sc t ng tr t th cnh AB snm v bn phi ca BC.
Gi s khi ngn mch trong phn ng c dng Im tng ng vi dng kch thchIt = OC: 1 phn OD sinh ra sc in ng khc phc in p ri trn in trphn ng Im.R = AD = BC; Phn cn li DC = AB dng khc phc phn ngphn ng lc ngn mch.
0 C D It
Enm
Inm=Im B A
U (1)
I (2)E,I
A B
U (1)I (2)
Inm=Im
Enm
0 D C It
E,I
my in mt chiu
NextChng 6Back
ABC c cnh BC t l vi dng in phn ng v cnh AB trong iu kin mch t khng bo ho t l vi phn ng phn ng (t l vi I) gi l tam gic c tnh.
3. c tnh ph ti: U = f(Ikt) khi I = const, n = const.
NextChng 6Back
Dng c tnh:(1): c tnh khng ti.(2): c tnh ph ti.
ng (2) c th xc nh khi bit ng (1) v tam gic c tnh: Gi s bit tam gic c tnh 1 ch ti no . VD ti nhmc l tam gic ABC. Ta t tam gic sao cho nh A nm trn ctnh khng ti, cc cnh AB v BC song song vi trc honh v trctung ng thi t l vi ph ti, khi tam gic dch chuyn song songvi chnh n nh C s v nn c tnh ph ti.
0
A1 B1
C1
(2)C
BA(1)U
IktIkt
Rt
I
A
V
my in mt chiu
-
4. c tnh ngoi: U = f(I) Khi Ikt = const, n = const.
Um = U0 - Um vi iu kin Ikt =Iktmgi l bin i in p nh mc:
( )%155%100U
UU%U
dm
dm0=
=
0 ImI
U
E
Um
UU0
Um
* C th dng c tnh ngoi t ctnh khng ti v tam gic c tnh:Cho OP = Ikt = constPP' = UI = 0 = E im Dt tam gic ABC c AB v BC theot l ng vi I = Im sao cho A nmtrn c tnh khng ti cn BC nmtrn ng thng ng PP' PC lin p khi I = Im ta c im D' v gc phn t th 2.
U
D'
I Im 0 P Ikt
P'A B
C
D
Im/2
my in mt chiu
Ikt
Rt
I
A
V
Ikt
0I
5. c tnh iu chnh:Ikt = f(I) Khi U = const, n = const
c tnh iu chnh cho ta bit cn phi iu chnh dng kch thch nh th no gi cho in p u ra ca my pht khng thay i khi ti thay i.
NextChng 6Back
my in mt chiu
-
1. iu kin t kch ca my:
6.3: My pht mt chiu kch t song song
m bo my t kch c cn c cc iu kin sau:- Trong my phi tn ti 1 lng t d d = (2 3)% m- Cun dy kch thch phi u ng chiu hoc my quay ng chiu sinh ra dng ikt > 0- Nu tc quay bng hng s th in tr mch kch thch phi nhhn 1 in tr ti hn no . Hoc nu in tr mch kch thch bng hng s th tc quay phi ln hn 1 tc ti hn no .
UI
Ikt
I
Nu my pht thomn 3 iu kin trn th qu trnh t kch xy ra nh sau:
NextChng 6Back
rkt2 rkt1
U = Iktrkt
Ikt0
U rthA
my in mt chiu
h mch kch thch v quay my pht n nm. Do trong my tn ti d nn trong dy qun s cm ng 1 sc in ng E v trn 2 cc my s c 1 in p U = (2 3)% Um.
Ni kn mch kch thch trong mch kch thch s c: no . Dng ny sinh ra t thng d v tng (d + d ) > d ssinh ra dng kch thch ln hn. C nh vy my s tng kch t in p u cc tng ln v my tip tc t kch cho n khi n lm vic n nh im A.Nu d ngc chiu vi d th my s khng t kch c.
2. c tnh ngoi:U = f(I) khi Ikt = const, n = const
U
Um Um(1)
(2)
I
IthImI00
(1): c tnh ngoi ca MF kch t c lp.(2): c tnh ngoi ca MF kch t song song.
NextChng 6Back
kt
'kt
r
UI =
my in mt chiu
-
UB
0 Iktm Ikt
3. c tnh khng ti:
4. c tnh iu chnh:Ikt = f(I) khi U = const, n = const(1): ca my pht kch thch song song(2): ca my pht kch thch c lp
NextChng 6Back
Ikt
0I
(1)(2)
my in mt chiu
MFMC kch thch ni tip cng thuc loi t kch.
6.4: My pht mt chiu kch t ni tip
I Ikt
I
Rt
Do dng ti I = I = Ikt nn trong my ch c 2 lng ph thuc nhau l U v I nn ta ch xy dng c tnh ngoi: U = f(I) khi n = const.
- V khi ti tng in p 2 u cc my pht thay i nhiu nn thc t t dng loi my ny.
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- Khi I = Ith mch t bo ho in p bt u gim
0
U
Ud
Ith I
my in mt chiu
-
My pht mt chiu kch thch hn hp c 2 cun dy kch thch. Tu theo cch u dy m ta c s ni thun v ni ngc.
6.5. My pht mt chiu kch t hn hpU
I
Iktss
Iktnt
I1. c tnh ngoi: U = f(I) khi n = const
0 I
U0
U(1)
(2)
(3)(4)
Khi ni thun in p c gihu nh khng i (ng (2))
Khi b tha (ng 1) in p s tng khi ti tng. iu ny c ngha quan trng trong vic truyn ti in nng i xa.
Nu ni ngc 2 dy qun kch thch (ng (4)) khi ti tng p sgim nhanh hn so vi my pht kch thch song song (ng (3))
NextBack Chng 6
my in mt chiu
U
I
Iktss
Iktnt
I
2. c tnh iu chnh:
0 I
(2)
(1)
(3)Iktng (1) : Khi ni thun 2 dy qun kch thch v b bnh thng.
(2) : Khi b tha.
(3) : Khi ni ngc 2 dy qun kch thch.
NextChng 6Back
my in mt chiu
-
1. iu kin ghp cc my pht lm vic song song:
6.6: My pht mt chiu lm vic song song
Cng cc tnh: Sc in ng ca my pht II phi bng in p U ca thanh gp.Nu ghp cc my pht kch thch hn hp lm vic song song thcn c iu kin th 3: Ni dy cn bng gia cc im m v n nhhnh b.
A A
V
F2F1
Hnh a
A A
mI n
F1 F2
Hnh b
NextBack
my in mt chiu
Chng 6
2. Phn phi v chuyn ti gia cc my pht in:
IIIII
(2)
(2')
U
(1')
(1)EII=U
I = III = II+III
Ghp my pht II lm vic song song vi my pht I.
Vic thay i EI v EII bng cch bin i dng kch t IktI v IktIIhoc bng cch thay i tc quay ca cc ng c s cp.
NextChng 6Back
my in mt chiu
Do E2=U nn my II cha tham gia pht in v ton b ti vn do my I m nhn. Lc ny c tnh ngoi ca 2 my l ng (1) v (2).
-
Chng 7 : ng c mt chiu
7.1: Nhng khi nim c bn
7.2: M my ng c in mt chiu
7.3: ng c in mt chiu kch thch song song hoc c lp
NextPhn IBack
7.4: ng c mt chiu kch thch ni tip
7.5: ng c mt chiu kch thch hn hp
my in mt chiu
1. Phn loi:
7.1: Nhng khi nim c bn
UI
Ikt
I
Hnh b
U
I I
I
Hnh c
U
I
Iktss
Iktnt
Hnh d
II
IU
IktUktHnh a
+ ng c mt chiu kch thch hn hp: I = I + Ikt (hnh d).
+ ng c mt chiu kch thch c lp: I = I (hnh a).
+ ng c mt chiu kch thch song song: I = I + Ikt (hnh b).
+ ng c mt chiu kch thch ni tip: I = I = Ikt (hnh c).
NextChng 7Back
my in mt chiu
-
2. Phng trnh cn bng p:Cng sut in a vo u ng c kch thch song song l:
P1 = U.(I + Ikt)Pt = P1- (pcu.kt + pcu.)
EI = U.(I + Ikt) - (U.Ikt + I2.R) E = U - I.R
3. Phng trnh cn bng mmen:
Pt = pFe + pc + P2
+
+=
2Fecodt PPPP
Mt = M0 + M2
t: M0 + M2 = MCT (Mmen cn tnh) Mt = MCT
Trong : M0: mmen cn khng ti.M2: mmen ph ti.
NextChng 7Back
my in mt chiu
P1 Pt
pcu. + pcu.kt pFe pc
P2
1. Yu cu khi m my:
7.2. M my ng c mt chiu
Dng m my phi c hn ch n mc nh nht trnh cho dy qun khi b chy hoc nh hng xu n i chiu.
M men m my phi c tr s cao nht c th c hon thnh qu trnh m ma ngha l t c tc quy nh trong 1 thi gian ngn nht.
2. Cc phng php m my:
- Khi m my trong mi trng hp u phi m bo c maxngha l trc khi ng ng c vo ngun in, bin tr iu chnh dng kch thch phi t v tr sao cho in tr kch thch nh nht mmen t gi tr ln nht ng vi mi gi tr ca dng phn ng.
NextChng 7Back
my in mt chiu
-
a. M my trc tip:
Ti thi im u: n = 0 E = 0 U = I.R I = Imm = I = (5 10)Im.
V dng m my ln nn phng php ny t c s dng. Ch yu dng cho ng c cng sut vi trm ot (v R tng i ln nn Imm (4 6)Im).
RU
b. M my nh bin tr:
CU
Ikt
0 T
12 3 4
5M
rmm
- Khi m my nh bin tr dng c tnh:
mmi
i
RR
EUI
+
=
Rmmi l in tr m my th i.
NextChng 7Back
my in mt chiu
- Bin tr m my c tnh sao cho: Imm = (1,4 1,7)Im i vi cc ng c cng sut ln Imm = (2 2,5)Im vi ng c cng sut nh.
Qu trnh m my c biu din nh hnh v: n
1 2 3 4 5
0 MMc
M
I
nI1
I2M1
M2
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Thng dng m my cho nhng ng c cng sut ln kt hp c vic iu chnh tc bng cch thay i in p.
c) M my bng in p thp (Umm < Um):
Phi dng 1 ngun c lp c th iu chnh in p c cung cp cho phn ng ng c. Mt ngun khc U = Um cung cp cho mch kch thch.
my in mt chiu
-
1. c tnh c:
7.3. ng c in mt chiu kch thch song song hoc c lp
T phng trnh: E = Ce..n
Thay M = CM..I ta c: (1)
=
=
e
e C
R.IU
C
En
2Me
e .C.C
R.M
C
Un
=
Vi iu kin: U = const, Ikt = const khi M (hoc I) thay i th t thng cng hu nhkhng i.
ng c ny c dng trong trng hp tc hu nh khng i khi ti thay i (my ct kim loi...)
0
nn0
Mm M(I)
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my in mt chiu
k
R.Mnn 0 =
Vi v k = Ce.CM.
=
e0
C
Un
2
(1)
2. iu kin n nh ca ng c:Xt c tnh M = f(n) ca ng c in v Mc = f(n) ca ti nh hnh v:
Ta c:
vi:
l qun tnh phn quay.
dt
djMM c
+=
g4
GDj
2
=
Trng hp a: P l im lm vic ca h thng c M = Mchay 0
dn
dM=
Nu v l do no tc tng: n = nlv + n th Mc > M ng c b ghm, tc gim dn v im P n = nlv. Ngc li: nu tc gim Mc < M ng c c gia tc v t tc lm vic.
Nh vy: iu kin lm vic n nh ca h thng l: dn
dMdndM c Mc lm tc tip tc tng mi hoc s gim tc s dn n hu qu l tc gim mi.
dn
dM
dn
dM c
dndM
dndM c>
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my in mt chiu
M
0
M McP
Hnh a
nlv
nn
Hnh b
0
MM
McP
n
nnlv
-
3. iu chnh tc : 2
Me
e .C.C
R.M
C
Un
=
a. Phng php thay i t thng:
1
n
0
n0m
n01
n02
2m
M
Bng cch thay i tr s ca bin tr trong mch kch thch. Cc ng ny c n0 > n0mv giao nhau ti 1 im trn trc honh ng vi (n = 0, I = )
R
U
b. Thay i in tr ph trn mch phn ng:
0
nn0 Rf = 0
Rf1Rf2
Rf3 M
( )k
RR.Mnn f0
+=
NextChng 7Back
my in mt chiu
c) Phng php thay i in p:
0 M
UUm n02
n01
n03n
Vic cung cp in p cho ng c c thc hin bng 1 ngun c lp bng cch ghp thnh t hp my pht - ng c.
4. c tnh lm vic:
a) c tnh tc : n = f(I) ging c tnh c:
=
.C
R.I
C
Un
e
e
b) c tnh mmen: M = f(I) khi U = Um = const M = CM..I
Do Ikt = const khi U = const = const M = f(I) l ng thng.
M
0Ic) c tnh hiu sut: = f(I)
Khi U = Um = const.max c tnh vi dng in ti I = 0,75Im. Khi tn hao khng i trong ng c(pc + pFe) bng tn hao bin i trong mch phn ng (ph thuc rdq v t l I
2 ) 0
0,75Im
I
max
NextChng 7Back
my in mt chiu
-
1. c tnh c ca ng c 1 chiu kch thch ni tip:
7.4: ng c mt chiu kch thch ni tip
V I = I = Ikt trong phm vi rng c th biu th = K.I (1)Trong : K = const khi I < 0,8Im v gim i 1 cht khi I > 0,8Im
do nh hng bo ho ca mch t.
M = CM..I. Thay phng trnh (1) vo ta c:
M
22
MC
K.M
KCM
==
MC
K.M =
2Me
e .C.C
R.M
C
Un
=
=
K.C
R
K.M.C
CU
e
e
M (2)
Nu b qua R th n t l vi hay . MU
2
2
n
CM =
Khi mch t cha bo ho c tnh c c dng hypecbol bc 2.
NextChng 7Back
my in mt chiu
0 M
n
ng c mt chiu kch thch ni tip vi c tnh c rt mm c ng dng trong nhng ni cn iu kin m my nng n v cn thay i tc trong 1 vng rng (cu trc, xe in...)
Khi n gim th M tng v ngc li. Trong trng hp mt ti (I = 0, M = 0) th n c tr s rt ln v th loi ng c ny khng cho php lm vic trong iu kin c th mt ti (ai truyn...).
2. iu chnh tc :a) iu chnh tc bng cch thay i t thng:
Vic thay i t thng trong ng c kch t ni tip c th thc hin theo 3 phng php:
NextChng 7Back
my in mt chiu
-
Mc in tr sun vo dy qun kch thch. (Hnh a)
U
RktRS
C
I
(a)
U
C
Wkt
W'kt
I
(b)
U
RS
Rkt
I
(c)
Mc in tr sun vo mch phn ng. (Hnh c)Thay i s vng dy ca dy qun kch thch. (Hnh b)
NextChng 7Back
Hai bin php u cho cng 1 kt qu: nu dng kch thch ban u l Ikt thdng sau khi iu chnh l I'kt = k.IktVi k l h s gim.
+ Trng hp a: + Trng hp b: 1RR
RkSkt
S Fd Kb > 1: b tha - ng (2 )
- Khi FCB < Fd Kb < 1: b thiu - ng (3).
ng dng:
NextChng 8Back
my in mt chiu
8.2: My pht hn mt chiuMy pht hn phi c c tnh ngoi U = f(I) c dc cao nhhnh v. My pht hn c sn xut vi:
U = 35V (U0 = 80) v I = 500A
0
U
Rt1
Rt2Rt3 I
Thc t ch to c loi my pht c bit c s nh sau:
Khi I tng t thng ca cc cc ln d gim nhiu cn t thng ca cc cc b n khng thay i (do li thp bo ho) t thng tng (d + n) gim rt nhanh khin cho UAB h thp rt nhiu nn c tnh ngoi rt dc. Ch : Khi I tng UBC cung cp cho cc dy qun kch thch vn gi khng i v n khng i.
NextChng 8Back
+ -Rt
INnNd
C
B
SdSn
A
I"I'
F1
F2
my in mt chiu
-
My pht 1 cc l loi my c bit khng vnh gp cho php t c dng in ln (n 50000A) in p thp (1 50V).
8.3: My pht mt cc
Cu to nh hnh v:Hai cc t hnh tr lng vo nhau. Thanh dn t trn hnh tr trong (Rto) (hay c thdng chnh bn thn rto thay cho thanh dn) hai u ni cht vi 2 vnh C 1 v C2.
Khi rto quay trong cc thanh dn s sinh ra sc in ng vdng in ly ra t cc chi t ln 2 vnh C1 v C2. V dng in rt ln, trnh tn hao ngi ta dng chi than bng kim loi lng (thu ngn Natri) dn dng ra ngoi.My pht 1 cc c dng cho in phn, cp in cho cc nam chm in ca thit b tng tc NextChng 8
C1C2
2
S 1
N
my in mt chiu
L my pht in dng bin i chuyn ng quay thnh tn hiu in (in p). Yu cu i vi loi my ny l phi c quan h U = f(n) l ng thng v chnh xc 0,2 0,5 %.
8.4: My pht o tc
- Khi khng ti ta c: U = E = Ce..n = Ke.n (v = const).
t
R
U
t
R
U- Khi c ti : I = U = E - .R
U = E - I.R E = U.(1 + )t
R
U
t
e
t
R
R 1
n.K
R
R 1
EU
+=
+
NextChng 8Back
my in mt chiu
n
URt =
Rt1
Rt2
Hnh a
U
n
Hnh b
Nu b qua tc dng ca phn ng phn ng v st p do tip xc gia chi than vnh gp th c tnh u ra l tuyn tnh. (h a).
gim qun tnh ca phn quay v s p mch ca t thng vin p v s tn ti ca rng trn mt phn ng ta dng rto rng.
-
Nhim v: bin i tn hiu in (in p iu khin) nhn c thnh di chuyn c hc ca trc tc ng ln cc b phn iu khin hoc iu chnh khc.
8.5: ng c tha hnh
Yu cu: tc ng nhanh, chnh xc, mmen v tc quay phi ph thuc vo in p iu khin theo quan h ng thng.Cu to: tng t nh 1 ng c kch t c lp. N c th c iu khin trn phn ng hoc trn cc t:
ng c tha hnh tc ng nhanh ngi ta ch to phn ng c qun tnh nh di dng rto rng hoc dt hnh a c mch in.
NextChng 8Back
my in mt chiu
+ Khi iu khin cc t: in p iu khin c a vo dy qun kch thch. Nh vy cng sut iu khin s nh nhng quan hn = f(Uk) khng l ng thng.
+ Khi iu khin trn phn ng in p kch thch t thng trc trn dy qun kch thch, ng c trng thi chun b tha hnh. Khi c Uk t ln dy qun phn ng lp tc ng c hot ng. Vi phng php iu khin ny: M = f(Uk) v n = f(Uk) l nhng ng thng.
