(TIỂU LUẬN) EXERCISE NUMBER 4 – ELEARNING SUBJECT ELECTROMAGNETIC TRANSIENT PROCESS MISSION CALCULATE, SOFTWARE SIMULATION
Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (862.99 KB, 13 trang )
VIETNAM GENERAL CONFEDERATION OF LABOUR TON DUC
THANG UNIVERSITY FACULTY OF ELECTRICAL &
ELECTRONICS ENGINEERING
EXERCISE NUMBER 4 – ELEARNING
SUBJECT: ELECTROMAGNETIC TRANSIENT PROCESS
MISSION: CALCULATE, SOFTWARE SIMULATION
ADVISED BY: Dr. NGUYỄN CƠNG TRÁNG
GROUP: 06
BY:
NGUYỄN
NGỌC HỒNG VŨ, StudentID code: 41900916
NGUYỄN VĂN THẮNG, StudentID code: 41900877
NGUYỄN MINH THUẬN, StudentID code: 41900891
TRANG THANH TUẤN, StudentID code: 41900912
LÊ MINH THẮNG, StudentID code: 41900875
HO CHI MINH CITY, 2021
I. Introduce the problem and calculate
The figure:
Detailed specifications:
Generator: SrG1 = 117,5 MVA; UrG1 = 10 KV; Cosφ = 0.85; ′′= 0.1593 ,
1
′′
= 6.6 kV
System:
Transformers: SrT1 = SrT2 = 125 MVA; Uđm = 10.5/115 KV; UkT1 % = UkT2 %
= 10.5%;
T3: SrT3= 200 MVA; Uđm = 15.75/242 KV; UkT3% = 11%;
Lines: l1 = 45 km; l2 = 23km; l3= 40km; All 4 lines have X’= 0.4 Ω/km, R’ = 0.3
Ω/km, 110kV
Loads: stuck on busbar 22kV, SrL = 15 MVA, Cosφ = 0.8;
′′
.
= 176.5 MVA;
′′
Calculate short circuit:
= 0.4
=
Choose
=10KV
Ta có:
System:
Line 1:
1=
′*
1
3:
3=
′*
2
′*
1
2
′′ .
=
176,5
∗( )2=0.4*45*(
102
=0.57
10 2
) =0.15
2
110
∗( ) =0.11
Line 2:
2=
=
1=
′*
2=
2
∗(
3=
′*
3
)2=0.6*23*(
110
∗( )2=0.06
2
∗(
) =0.4*40*(
110
′ *3 ∗( )2=0.1
10 2
) =0.08
10
)2=0.13
Trans1, 2:
2 = 1 = 2=
1%
*
2
=10,5
∗
102
=0.084 100
1 100 125
Trans3:
3
=
3=
3%
*
2
=
11
∗
102
=0.055 100
3 100 200
Loads:
2
=
* Cos( )=
2
=
=
* Sin( )=
*√()2 + (
2
) − ′′
=17,1KV
Electromotive of system return to base voltage chosen:
Generator:
1=
+
1=0.14j+0.084j=0.224j
=
=10KV
2=
3=
3
3
+
12
=
13
=
23
=
+
=0.055j+0.57j=0.625j
=0.084j+5.1+4j=5.3+4.084j
4=
5=
23
+
13
+ 2=0.04+0.05j+0.625j=0.04+0.675j
3=0.02+0.03j+5.3+4.084j=5.32+4.114j
=+ +
6
4
12
=
7
5
12
1
=
=
1
√0.224
2
1
1
=
=
2
7
2
√5.7 +4.15
7
∗ 1+ 2∗ 7
=
7
1,7=
−3
1// 7=5.54*10 +0.22j
=
1,7
1,7
√(5.54∗10−3)2+0.222
1
=
6
6
=4.84∗4.54+10∗1.42=6.07
=
3
√3∗
.
′′
=
3
3
.
.
−3 2
√3∗√(6.63∗10 )
II. Simulate the problem by software.
Introduction about software MELSHORT2.
●
●
●
●
●
The Melshort provided by Mitsubishi is nearly perfect. It provides all
functions and information about short circuits.
The software can calculate short circuits at all times of breakers.
Help to choose a suitable CB (Circuit Breaker).
The software is used to design a short circuit diagram easily. In addition, the
difference between two types of CBs is clearly shown through their routing
curve diagram.
Because it is Mitsubishi’s software, the data information of each type of CB is
shown very clearly in detail along with other data to make the calculation
easier.
Simulation
1.
2. Start Melshort, software interface shown
Click the icons on the toolbar to create the diagram following the problem.
3.
Enter the parameters
○ Generator
○
System
○
Transformers (T1, T2, T3)
○
Lines
Tick in Calculation point to calculate short circuit at Line 2
4.
Click Number 2 to auto check the diagram
The software notification
Click OK then click Number 3 to calculate automatically and show the
result.
The result: Isc=22.427kA
BI. Comment, compare the results.
Compare
Actual short circuit current
Comment
●
●
Manual calculation and software simulation is not too different.
Rounding by manual calculation.
