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EEE458 MT Exam 13042020

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Dr. Ömer Cihan KΔ±vanç
EEE458
ELECTRICAL DISTRIBUTION SYSTEMS
MID-TERM EXAM
13.04.2020
1. A power plant single-line diagram is shown in Fig. 1. A breaker is connected to 10 kV bus. Calculate
the trip short-circuit current and power (πΌπ‘Ž , π‘†π‘Ž ). [πœ‡ = 0.8] (20 PTS)
π‘†π‘˜" = 3000 𝑀𝑉𝐴
110 π‘˜π‘‰
30 𝑀𝑉𝐴
π‘’π‘˜ = 12%
25 𝑀𝑉𝐴
30 𝑀𝑉𝐴
π‘₯𝑑" = 12.5%
π‘’π‘˜ = 12%
10 π‘˜π‘‰
Fig. 1 Power plant single-line diagram.
2. A 10 kV medium voltage distribution line is shown in Fig. 2. Calculate rated cable cross sections.
[𝑣 ≤ 10%, π‘₯ " = 0.4 Ω/π‘˜π‘š, k = 56 m/β„¦π‘šπ‘š2 ] (15 PTS)
10 π‘˜π‘‰
6 π‘˜π‘š
4 π‘˜π‘š
500 π‘˜π‘‰π΄
20 𝐴
cos πœ‘ = 0.8
cos πœ‘ = 0.6
Fig. 2 Medium voltage distribution line.
3. A low voltage distribution line is shown in Fig. 3. Calculate the end-line voltage value, rated cable
cross sections and total power losses. [𝑣 ≤ 3%, π‘₯ " = 0.4 Ω/π‘˜π‘š, k = 56 m/β„¦π‘šπ‘š2 ] [Rated Cable Cross
Sections: 2.5; 4; 6; 10; 16; 25; 35; 50; 70; 95 (π‘šπ‘š2 )] (15 PTS)
220
380 𝑉
50 π‘š
𝐴
30 π‘š
𝐡
20 π‘š
𝐢
M
25 𝐴
15 𝐴
Fig. 3 Low voltage distribution line.
13 π‘˜π‘‰π΄
cos πœ‘ = 0.8
4. Calculate the starting 3Ø short-circuit current value at 34.5 kV bus (B). Simulate the power system
shown in Fig. 4, using Matlab/Simulink and show the current and power flow in scopes (assuming
that there is no short-circuit anywhere). (35 PTS)
π‘†π‘˜" = 2500 𝑀𝑉𝐴
154 π‘˜π‘‰
𝑆𝐺 = 25 𝑀𝑉𝐴
2π‘₯
πœ€π‘‘" = 12.5%
𝑆𝑇𝑅 = 30 𝑀𝑉𝐴
π‘’π‘˜ = 12%
𝐴
10 π‘˜π‘‰
3 π‘₯ 31.14 π‘šπ‘š 2
3 π‘₯ 62.44 π‘šπ‘š 2
π‘₯0 = 0.4 Ω/π‘˜π‘š
𝑆𝐺 = 10 𝑀𝑉𝐴
πœ€π‘‘" = 12.5%
π‘₯0 = 0.4 Ω/π‘˜π‘š
8 π‘˜π‘š
𝜘𝐴𝐿 = 27.77 π‘š/Ωπ‘šπ‘š 2
34.5 π‘˜π‘‰
𝐡
M
𝐼𝑁 = 176 𝐴
𝐼𝐼𝐢 = 5π‘₯𝐼𝑁
Fig. 4 Distribution system single-line diagram.
5. Calculate maximum voltage drop point & value at Fig. 5. [k = 56 m/β„¦π‘šπ‘š2 , π‘žπ‘› = 25 π‘šπ‘š2 , π‘₯𝑙𝑖𝑛𝑒 =
0.1 Ω/π‘˜π‘š] Simulate the power system shown in Fig. 5, using Matlab/Simulink and show the
current and power flow in scopes. (35 PTS)
500 π‘˜π‘Š
cos πœ‘ = 0.8
630 π‘˜π‘‰π΄
400 π‘š
cos πœ‘ = 0.6
500 π‘š
800 π‘š
10 π‘˜π‘‰
600 π‘š
500 π‘š
380 π‘˜π‘Š
500 π‘˜π‘‰π΄
36 𝐴
cos πœ‘ = 0.8
Fig. 5 A distribution system.
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