Tính toán bảo vệ rơle cho trạm biến áp 110KV (Đồ án tốt nghiệp)
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ISO 9001:2008
ÁP 110 KV
- 2016
ISO 9001:2008
ÁP 110 KV
H QUY
Sinh viên:
- 2016
----------------o0o-----------------
Sinh viên:
Mã sv: 1513102005
CL901
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1.
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:L
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J621.......................................................................................................... 36
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I0)...................................... 56
I. ........................................................... 57
T
áp 110 kV
4
2
3
4
.
.
0/
/ Y0
S1Nmax = 2500 MVA.
S2Nmax = 2000 MVA.
ính:
S1Nmax= 2500 MVA
SNmin = 2100 MVA
Xomax = 0,7 X1max MVA
Xomin = 0,8 X1max
SNmax = 2000 MVA
SNmin = 1600 MVA
Xomax = 0,75 X1max
X0min = 0,9 X1max
1;
S
T2
y Yo-
11-
Yo
kV
U
(C - T = 10,5 , C - H = 17 , T - H = 6 )
9x1,78 %
D1:
L = 70 Km; AC 240
Z = 0,12 + j 0,386
[ /km]
Z = 2,5 Z
[ /km]
C/UT/UH
= 115/38,5/23
D2:
L2 = 55 Km; AC 185
Z = 0,156 + j 0,394
[ /km]
Z =2Z
[ /km]
.
-
Umg
-
Ilvcb
-
BN
-
ixk
-
CMC
I
* Phía 110 kV:
- Ilvcb= Kqtsc.I
= 1,4.I
= 1,4.
=1,4.
= 0,281 kA = 281 A
- ixk =
-
1
b
max
ixk =
.1,8.4,45 = 11,33 kA
- 110 - 40/2000.
= 110 kV
I
= 2000 A
I
= 40 kA
Ildd = 40 kA
* Phía 35 kV:
- Ilvcb= Kqtsc.I
= 1,4.I
= 1,4.
= 1,4.
- ixk =
-
3
max
ixk =
.1,8.3,95 = 10,05 kA
-35-40/3200.
= 35 kV
I
= 3200 A
I
= 40 kA
Ildd = 40 kA
* Phía 22 kV:
- Ilvcb= Kqtsc.I
-
= 1,4.I
= 1,4.
= 1,4.
ixk =
-
2
max
ixk =
.1,8.4,46 = 11,81 kA
M-22-40/1200Y3.
= 22 kV
I
= 1200 A
I
= 40 kA
Ildd = 25 kA
.
-
Umg
-
Ilvcb
-
Z2
-
.k
-
) .t
BN
.k . I
ixk
:
U ,kV
T H-110M
T H-35M
T H-22M
110
35
22
5
5
5
60/1
65/1
12/4
300/1
1000/1
1500/1
nh
.
-
Umg
-
S2
HK
- 110
58
3HOM
35
06 24Y3
U , kV
110
35
24
Umax, kV
110
35
24
11000/
:100/
:100 35000/
:100/
:100 24000/
:100/
Y0/Y0/
Y0/Y0/
Y0/Y0/
600
300
300
dây
MVA
:100
2
2.1.1
nh
-
u.
-
2.1.
:
Kh
2.2.
.
2
.
I)
ch
-
-
; 50/I
)
a
c thang
51N
2
.
Hình 2.1:
1-
6-
2-
7-
3-
8-
không
9-
452
gian
phao kim
Hình 3.2:
.
/87T
Hình 2.3:
SL
N1
ISL = I = I1 - I2 = 0
2
2
ISL = I = I1 + I2
SL
=
kcb).
I > Ikcbmax
lv
h).
-Restricted_Earth_Fault)
IN ngoài
vùng
Hình 2.5:
I
0sec.
I = kat
at
IN ngoài max
= 1,3
I
= kat I
nv
kat = ( 3
I = kat
IN TG2(max)
I = kat
IN TG3(max)
Hình 2.6:
5 )I
Ilvmax
Ilvmax = 1,4 I = 1,4.
Kat
Kat = 1,1 1,2
Kmm
Kmm
Ktv
Ktv = 0,8
>/51N
0,9
Hình 2.7:
0
0
kvb
I > Ikcb.
Ikcb = (0,1 0.3).I
Nên có:
I = (0,1 0.3).I
0SC
Ikt
/49
khác
