# EEE458 MT Exam 13042020

```Dr. &Ouml;mer Cihan KΔ±van&ccedil;
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 , ππ ). [π = 0.8] (20 PTS)
ππ&quot; = 3000 πππ΄
110 ππ
30 πππ΄
π’π = 12%
25 πππ΄
30 πππ΄
π₯π&quot; = 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.
[π£ ≤ 10%, π₯ &quot; = 0.4 β¦/ππ, k = 56 m/β¦ππ2 ] (15 PTS)
10 ππ
6 ππ
4 ππ
500 πππ΄
20 π΄
cos π = 0.8
cos π = 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. [π£ ≤ 3%, π₯ &quot; = 0.4 β¦/ππ, k = 56 m/β¦ππ2 ] [Rated Cable Cross
Sections: 2.5; 4; 6; 10; 16; 25; 35; 50; 70; 95 (ππ2 )] (15 PTS)
220
380 π
50 π
π΄
30 π
π΅
20 π
πΆ
M
25 π΄
15 π΄
Fig. 3 Low voltage distribution line.
13 πππ΄
cos π = 0.8
4. Calculate the starting 3&Oslash; 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&quot; = 2500 πππ΄
154 ππ
ππΊ = 25 πππ΄
2π₯
ππ&quot; = 12.5%
πππ = 30 πππ΄
π’π = 12%
π΄
10 ππ
3 π₯ 31.14 ππ 2
3 π₯ 62.44 ππ 2
π₯0 = 0.4 Ω/ππ
ππΊ = 10 πππ΄
ππ&quot; = 12.5%
π₯0 = 0.4 Ω/ππ
8 ππ
ππ΄πΏ = 27.77 π/Ωππ 2
34.5 ππ
π΅
M
πΌπ = 176 π΄
πΌπΌπΆ = 5π₯πΌπ
Fig. 4 Distribution system single-line diagram.
5. Calculate maximum voltage drop point &amp; value at Fig. 5. [k = 56 m/β¦ππ2 , ππ = 25 ππ2 , π₯ππππ =
0.1 Ω/ππ] 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
cos π = 0.8
630 πππ΄
400 π
cos π = 0.6
500 π
800 π
10 ππ
600 π
500 π
380 ππ
500 πππ΄
36 π΄
cos π = 0.8
Fig. 5 A distribution system.
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