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VIETNAM GENERAL CONFEDERATION OF LABOUR TON DUC THANG UNIVERSITY FACULTY OF ELECTRICAL & ELECTRONICS ENGINEERING EXERCISE NUMBER – 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 , ′′ = 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 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= ′* 3: 3= ′* ′* ′′ = 176,5 ∗( )2=0.4*45*( 102 =0.57 10 ) =0.15 110 ∗( ) =0.11 Line 2: 2= = 1= ′* 2= ∗( 3= ′* )2=0.6*23*( 110 ∗( )2=0.06 ∗( ) =0.4*40*( 110 ′ *3 ∗( )2=0.1 10 ) =0.08 10 )2=0.13 Trans1, 2: = = 2= 1% * =10,5 ∗ 102 =0.084 100 100 125 Trans3: = 3= 3% * = 11 ∗ 102 =0.055 100 100 200 Loads: = * Cos( )= = = * Sin( )= *√()2 + ( ) − ′′ =17,1KV Electromotive of system return to base voltage chosen: Generator: 1= + 1=0.14j+0.084j=0.224j = =10KV 2= 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 =+ + 12 = 12 = = √0.224 1 = = √5.7 +4.15 ∗ 1+ 2∗ = 1,7= −3 1// 7=5.54*10 +0.22j = 1,7 1,7 √(5.54∗10−3)2+0.222 = 6 =4.84∗4.54+10∗1.42=6.07 = √3∗ ′′ = 3 −3 √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 Start Melshort, software interface shown Click the icons on the toolbar to create the diagram following the problem Enter the parameters ○ Generator ○ System ○ Transformers (T1, T2, T3) ○ Lines Tick in Calculation point to calculate short circuit at Line Click Number to auto check the diagram The software notification Click OK then click Number 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 ... 1=0.14j+0.084j=0.224j = =10KV 2= 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 =+... = √0.2 24 1 = = √5.7 +4. 15 ∗ 1+ 2∗ = 1,7= −3 1// 7=5. 54* 10 +0.22j = 1,7 1,7 √(5. 54? ??10−3)2+0.222 = 6 =4. 84? ? ?4. 54+ 10∗1 .42 =6.07 = √3∗ ′′ = 3 −3 √3∗√(6.63∗10 ) II Simulate the problem by software. .. 0 .4 = Choose =10KV Ta có: System: Line 1: 1= ′* 3: 3= ′* ′* ′′ = 176,5 ∗( )2=0 .4* 45*( 102 =0.57 10 ) =0.15 110 ∗( ) =0.11 Line 2: 2= = 1= ′* 2= ∗( 3= ′* )2=0.6*23*( 110 ∗( )2=0.06 ∗( ) =0 .4* 40*(
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