da may dien dot 2 2010 dh
DESCRIPTION
do an mon hocTRANSCRIPT
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B GIO DC O TO CNG HA X HI CH NGHA VIT NAM TRNG I HC NG c lp T do Hnh phc
PHIU CHM BI THI TUYN SINH LIN THNG
T CAO NG LN I HC S phch:.. Mn: My in.Ngnh: in Ngy thi: 28,29/08/2010
Cu Ni dung p n Thang
im im chm
1 1. Th nghim khng ti MBA (1 im) 1.1 Th nghim khng ti l xc nh h s bin p k, tn hao st t trong
li thp pFe, v cc thng s ca mba ch khng ti.
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1.2 S ni dy th nghim khng ti.
t in p U1 = U1m vo dy qun s cp, th cp h mch, cc dng c o cho ta cc s liu sau: P0 l cng sut tn hao khng ti; I0 l dng in khng ti; cn U1m v U20 l in p s cp v th cp. T ta tnh c:
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1.3 a) H s bin p k:
20
1UU
=k m
b) Dng in khng ti phn trm : %10%1100II%Idm1
00
c) Tn hao trong li thp : PFe = P0 - r1I02 P0
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1.4 d) Tng tr khng ti
+ in tr khng ti: r0 = r1 + rm = 20
0
IP
Do rm >> r1 nn gn ng ly bng: rm = r0 - r1
+ Tng tr khng ti : 0
dm10 I
UZ
+ in khng khng ti. 2020m10 rzxxx
0,25
x2 r1 r2 x1
V W A
V
S thay th mba khi khng ti v S ni dy th nghim khng ti
1U 1E
rm xm
01 II 0I
-
in khng t ha xm >> x1 nn ly gn ng bng: xm = x0
e) H s cng sut khng ti.: 0dm1
00 IU
Pcos
2. Th nghim ngn mch. (1,0) 1.5 Th nghim ngn mch l xc nh in p ngn mch phn trm
Un%, tn hao ng nh mc P m, h s cng sut cosn, in tr ngn mch rn v in khng ngn mch xn ca mch in thay th mba.
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1.6 S th nghim ngn mch: Tin hnh th nghim nh sau: Dy qun th cp ni ngn mch, dy qun
s cp ni vi ngun qua b iu chnh in p. Ta iu chnh in p vo dy qun s cp bng Un sao cho dng in trong cc dy qun bng nh mc. in p Un gi l in p ngn mch. Lc cc dng c o cho ta cc s liu sau: Un l in p ngn mch; Pn l tn hao ngn mch; I1m v I2m l dng in s cp v th cp nh mc.
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1.7 a) Tn hao ngn mch Lc th nghim ngn mch, in p ngn mch Un nh nn t thng nh,
c th b qua tn hao st t. Cng sut o c trong th nghim ngn mch Pn chnh l tn hao trn in tr hai dy qun khi mba lm vic ch nh mc. Ta c:
Pn = r1I21m + r2I22m = rnIn2 b) Tng tr, in tr v in khng ngn mch.
+ Tng tr ngn mch: Zn = m1
nIU
+ in tr ngn mch: rn = r1+ r2 = 21mIPn
+ in khng ngn mch: xn = 22 nn rZ Trong m.b.a thng r1 = r2 v x1 = x2. Vy in tr v in khng tn ca
dy qun s cp:
r1 = r2 = 2nr ; x1 = x2 = 2
nx
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1U
rn xn
nII 1
Mch in thay th m.b.a khi ngn mch v S th nghim ngn mch
A
W A
V I2m
I1m
Un
Pn B
iu chnh in p
U1
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1.8 c) H s cng sut ngn mch :
mm 1
nn IU
Pcos
d) in p ngn mch
in p ngn mch phn trm:
Un% = %100UU
%100U
IZ
1
n
1
1n
mm
m
+ in p ngn mch tc dng phn trm:
Unr% = %100UIr
1
1n m
m
+ in p ngn mch phn khng phn trm:
Unx% = %100UIx
1
1n m
m
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b. Cc iu kin lm vic song song ca my bin p: 1.9 - Cng t ni dy.
- in p nh mc s cp v th p bng nhau hoc h s MBA k bng nhau: U1I = U1II = . . .= U1n v U2I = U2II = . . . = U1n hoc kI = kII = . . . = kn.
- in p ngn mch bng nhau : UnI = UnII = . . . = Unn. Trong thc t ch c iu kin 1 phi tun th mt cch tuyt i. Cc iu kin 2, 3 c thc hin vi mt mc sai khc nht nh c qui nh trong 1 gii hn cho php.
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Cu 2 (2,0) 2.1
in tr nhnh t ha
86,44,2
28IPR 2210
10th
Tng tr nhnh t ha
67,914,2
220IUZ
10
10th
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2.2 in khng nhnh t ha 54,91RZX 2th
2thth
in tr ngn mch
04,018,43
75IPRRR 22
n1
n1'21n
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2.3 Tng tr ngn mch
195,018,434,8
IUZ
n1
n1n
0,25
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in khng ngn mch
1904,0RZX 2n2
nn
Coi '21'
21 XX;RR
02,0204,0
2RRR n'21
0952,02
1904,02
XXX n'21
2.4 H s bin p k
73,1127220
UUk
dm2
dm1 0,25
2.5 Thng s dy qun th cp cha quy i
0067,073,102,0
kRR 22
'2
2
032,073,10952,0
kXX 22
'2
2
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2.6 Hiu sut my bin p khi nh mc
987,0752885,0.9500
85,0.9500PPcos.S
cos.S
n0tdm
tdm
Khi h s ti kt = 0,7
989,075.7,02885,0.9500.7,0
85,0.9500.7,0PkPcos.S.k
cos.S.k2
n2
t0tdmt
tdmt
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2.7 in p ngn mch phn trm
%82,3100.220
4,8100.UU%U
dm1
nn
bin thin in p th cp phn trm )sin%.Ucos%.U(k%U tnXtnRt
Trong
%79,0207,0.82,3cos%.U%U nnnR
%736,3978,0.82,3sin%.U%U nnnX
Vi
978,0cos1sin
207,018,43.4,8
75I.U
Pcos
n2
n
n1n1
n1n
Thay s: 85,0cos;7,0k tt
%848,1)sin%.Ucos%.U(k%U tnXtnRt
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2.8 V35,2127.
100848,1U%.848,1U dm22
V65,12435,2127UUU 2dm22 0,25
Cu 3(1,5) 3.1 Cc yu cu khi m my:
- Mmm phi ln thch ng vi c tnh c ca ti. - Imm cng nh cng tt. - Phng php m my v cc thit b cn dng n gin, r tin v chc chn. - Tn hao cng sut trong qu trnh m my t.
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3.2 Cc phng php m my 1. M my trc tip ng c rotor lng sc: Dng in m my ln, ch dng cho cc my c cng sut nh. Nu my c cng sut ln th dng trong li in c cng sut ln. Phng php ny m my nhanh, n gin.
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3.3 2. Phng php h in p m my: Ch dng vi cc thit b yu cu moment m my nh.
a. Dng cun khng bo ha trong mch stator Theo phng php ny nh c in p ri trn cun khng nn in p trc tip t vo ng c gim i k ln. Imm gim k ln th Mmm gim k2 ln.
Phng php ch c dng trong cc trng hp m vn tr s Mmm khng c ngha quan trng.
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3.4 b. Dng bin p t ngu h U m my Theo phng php ny in p trc tip t vo ng c gim i k ln. Imm
gim k2 ln v Mmm gim k2 ln.
Khi m my bng bin p t ngu dng in trong li gim i k2 ln so vi Imm khi ni trc tip.
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3.5 c. Phng php: Y Ch s dng vi ng c c 2 cp in p 220/380 v lm vic thng
trc cp 220V. Dng in m my trong li khi ni Y nh hn nhiu khi ni 3 ln. Mmm cng gim i 3 ln. Coi phng php ny l trng hp c bit m my bng bin p t ngu c
3k BA .
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Cu 4(2,5) 4.1
Dng in nh mc
A6,2885,0.75,0.380.3
10.12.cosU3
PI3
dm
dm
0,25
4.2 Tc t trng quay
1500250.60
pf60n1 vng/pht
0,25
4.3 H s trc 0,25
-
0466,01500
14301500n
nns1
1
4.4 Mmen nh mc
Nm14,801430129550
nP9550
60/n.2PPM
dm
dm
dm
dm
dm
dmdm
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4.5 Mmen m my Mm = 1,5Mm = 120,21Nm
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4.6 Dng in m my Im = 4,7Im = 134,42A
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4.7 Mmen cn MC = 0,52Mm = 41,67Nm
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4.8 Dng my bin p t ngu m my, ta c dng in m my gim k bnh phng ln.
BA2BA,m
tt,m kII
H s bin p
64,150
42,134II
kBA,m
tt,mBA
0,25
4.9 Mmen m my khi dng my bin p t ngu
Nm69,4464,1
21,120kM
M 2BA
2tt,m
BA,m 0,25
4.10 V CBA,m MM nn ng c c th m my c khi dng my bin p t ngu m my.
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Cu 5(1,5) 5.1 Tc quay
1500250.6060
1 pfnn vng/pht 0,25
5.2 Dng in nh mc
85,319686,0.5,10.3
50000cos.3
dm
dmdm U
PI A 0,25
5.3 Cng sut biu kin ca my pht ra
53,5813986,0
50000cos
dm
dmdm
PS
KVA 0,25
5.4 Cng sut phn khng my pht ra 16,2965151,0.53,58139sin. dmdmdm SQ KVAr
0,25
5.5 Cng sut ng c s cp
78,5617989,0
500001
dmPP KW 0,25
5.6 Tng tn hao trong my 78,61795000078,561791 dmPPP KW
0,25