EEL303: Powe Power r Engineerin Engineering g I - Tutorial utorial 2
1. A 100 MVA, MVA, 33 kV 3 phase generator has sub-transien sub-transientt reactance of 15 %. The generator is connected to the motors through a transmission line and transformers as shown in Figure Figure 1. The motors have have rated rated inputs inputs of 30 MVA, MVA, 20 MVA MVA and 50 MVA MVA at 30 kV with with 20 % sub-tra sub-transi nsien entt reactanc reactance. e. The 3 phase phase transfo transforme rmers rs are rated at 110 MVA, MVA, 32 kV ∆ / 110 kV Y with leakage leakage reactance reactance of 8 %. The line line has a reactanc reactancee of 50 Ω. Selecting the generator rating as the base quantities in the generator circuit, determine the base quantities in other parts of the system and evaluate the corresponding p.u. T 1 line M 1 M 2 values. [X p.u = 0.06838 p.u., X p.u = 0.3886 p.u.,X p.u = 0.5509 p.u., X p.u = 0.8264 p.u., M 3 X p.u = 0.3305 p.u.]
Figure 1: Solution: Generator reactance on its own base is 0.15 p.u.
Transformer T1’s p.u. reactance on new base MVA and kV is T 1 X p.u
= 0.08 ×
100 × 110
2
32
= 0.06838 p.u.
33
Base Voltage on secondary of transformer T1 is 33 ×
110 = 113.43kV 32
Hence, the p.u. impedance of transmission line is 50 × 100 = 0.3886 p.u. 113.432 The base value of voltage for motor circuit is 113.43 ×
32 = 33kV 110
Hence, the p.u. reactance of motors are M 1 : 0.2
×
100 30
2
×
30 33
= 0.5509 p.u.
M 3 : 0.2
×
100 50
M 2 : 0.2
×
100 20
×
30
2
×
30 33
= 0.3305 p.u.
Electrical Engineering Dept - IIT Delhi
33
2
= 0.8264 p.u.
EEL303: Power Engineering I - Tutorial 2
2. The single line diagram of a power system is shown in Figure 2. The specifications are given below. G1: 80 MVA, 11 kV, X= 18 % ; T1: 20 MVA, 11/220 kV, X=12 % ; T2: 20 MVA, 220/11 kV, X=10 % ; T3: 20 MVA, 11/110 kV, X=8 % ; T4: 20 MVA, 110/11 kV, X=6 % ; M1= 40 MVA, 10.8 kV, X=15 % ; Series reactances of line 1 and line 2 are 50 and 60 Ω respectively. Take generator ratings T 1 line1 T 2 T 3 as base quantities. [X p.u = 0.48 p.u, X p.u = 0.0826 p.u., X p.u = 0.4 p.u.,X p.u = 0.32 p.u., line2 T 4 M 1 load X p.u = 0.3966 p.u., X p.u = 0.24 p.u., X p.u = 0.289 p.u., Z p.u = 3.2 + 2 .4i] (a) Draw an impedance diagram with per-unit values. (b) If the Motor is replaced by a load of 20 MVA, 11 kV 0.8 p.f lagging, calculate Z p.u of the load
Figure 2:
Solution:
(a) p.u. reactance of generator on its own base is 0.18 p.u. p.u. reactance of T1 is T 1 X pu
= 0.12 ×
80 × 20
11
2
= 0.48 p.u.
11
V base on the secondary side of T1=220 kV.
p.u. reactance of Line 1 line1 X pu =
50 × 80 = 0.0826 p.u. 2202
p.u. reactance of T2 is
11
2
= 0.1 ×
80 × 20
11
2
= 0.08 ×
80 × 20
T 2 X pu
= 0.4 p.u.
11
p.u. reactance of T3 is T 3 X pu
11
= 0.32 p.u.
Electrical Engineering Dept - IIT Delhi
EEL303: Power Engineering I - Tutorial 2
V base on the secondary side of T3=110 kV.
p.u. reactance of Line 2 line2 X pu =
60 × 80 = 0.3966 p.u. 1102
p.u. reactance of T4 is T 4 X pu
= 0.06 ×
80 × 20
11
2
= 0.24 p.u.
11
p.u. reactance of M1 is M 1 X pu
= 0.15 ×
80 × 40
10 8 .
11
2
= 0.289 p.u.
(b) 112 Z base = = 1.5125Ω 80 112 Z actual = = 4.84 + 3.63iΩ 16 − 12i Z p.u. = 3.2 + 2 .4i 3. A 100 MVA, 13.8 kV, 3 phase generator has reactance of 20 %. The generator is connected to a 3 phase transformer (T1) rated 100 MVA, 12.5 kV/110 kV with 10 % reactance. The HV side of the transformer is connected to a transmission line of reactance 100 Ω. The far end of the line is connected to a step-down transformer T2 made of 3 single phase transformers, connected in Y - ∆, each rated 30 MVA, 66 kV/10 kV with 10 % reactance. The generators supply two motors connected on the LV side of T2. The motors are rated at 25 MVA and 50 MVA both at 10 kV with 15 % reactance. Draw the reactance diagram showing all the values in p.u. Take generator rating as T 1 line T 2 M 1 base. [X p.u = 0.0820 p.u., X p.u = 0.6780 p.u., X p.u = 0.0985 p.u., X p.u = 0.5319 p.u., M 2 X p.u = 0.2659 p.u. ] Solution: Generator reactance on its base is 0.2 p.u.
p.u. reactance of T1 is T 1 X pu
= 0.1 ×
100 × 100
V base on the secondary side of T1=13.8
12 5 .
13.8
2
= 0.0820 p.u.
110 × 12 = 121.44kV .5
Electrical Engineering Dept - IIT Delhi
EEL303: Power Engineering I - Tutorial 2
100 × 100 = 0.6780 p.u. 121.442 121.44 V base on the secondary side of T2=10 × √ = 10.62kV 3 × 66 p.u. reactance of transmission line =
T 2 X pu
M 1 X pu
M 2 X pu
= 0.1 ×
100 90
2
×
= 0.15 ×
100 25
= 0.15 ×
100 50
10
= 0.0985 p.u.
10.62
10 10.62
2
×
10 10.62
2
×
= 0.5319 p.u.
= 0.2659 p.u.
4. A 3 bus system is given in Figure 3. The ratings of various components are listed below. G1: 50 MVA, 13.8 kV, X=0.15 p.u.; G2: 40 MVA, 13.2 kV, X=0.2 p.u.; G3: 30 MVA, 11 kV, X=0.25 p.u.; T1: 45 MVA, 11 kV / 110 kV, X=0.1 p.u.; T2: 25 MVA, 12.5 kV / 115 kV, X=0.15 p.u.; T3: 40 MVA, 12.5 kV / 115 kV, X=0.1 p.u. The line impedances are as shown in figure. Determine the reactance diagram based on T 1 50 MVA and 13.8 kV as base quantities in Generator 1. [X p.u = 0.0705 p.u., line12 line23 T 2 G2 T 3 X p.u = 0.1312 p.u., X p.u = 0.0656 p.u., X p.u = 0.2083 p.u., X p.u = 0.1936 p.u., X p.u = 0.0868 p.u]
Figure 3:
Solution: G1 X pu = 0.15 p.u. T 1 X pu
= 0.1 ×
50 × 45
11 13.8
2
= 0.0705 p.u.
Electrical Engineering Dept - IIT Delhi
EEL303: Power Engineering I - Tutorial 2
V base on secondary of T1 = 13.8
× 110 = 138kV 11
50 × 50 = 0.1312 p.u. 1382 25 × 50 line13 line23 X pu = X pu = = 0.0656 p.u. 1382 line12 X pu =
V base on primary of T2 = 138 T 2 X pu
G2 X pu
= 0.15 ×
50 × 25
12 5
= 0.2 ×
50 × 40
13 2
V base on primary of T3 = 138 T 3 X pu
.5 × 12 = 15kV 115 .
2
15 .
= 0.2083 p.u.
2
15
= 0.1936 p.u.
.5 × 12 = 15kV 115
= 0.1 ×
50 × 40
12 5 .
15
2
= 0.0868 p.u.
5. The primary side reactance of a 5 kVA, 200/400 V transformer is 4 Ω. Determine per-unit value of reactance referred to (a) primary side, and (b) secondary side.
[X p.u. = 0.5 p.u.]
Consider base quantities as 5 kVA and 200 V on the low voltage side of the transformer. Solution: The per-unit value of reactance referred to primary side is
4 × 0.005 = 0.5 p.u 0.22 The ohmic value of reactance referred to secondary side is 4×
400 200
2
= 16Ω
The per-unit value of reactance referred to secondary side is 16 × 0.005 = 0.5 p.u 0.42
Electrical Engineering Dept - IIT Delhi
EEL303: Power Engineering I - Tutorial 2
6. 3 single phase transformers, each of rating 11/33 kV, 100 MVA are connected in Y - ∆. If the reactance of each single phase transformer is 6 Ω as referred to LV side, calculate the p.u. reactance of the resulting 3 phase transformer. [X p.u. = 4.9589 p.u.] Solution: p.u. reactance referred to primary P rimary X p.u. =
11 ×300√ 3 × 6 = 4 9589 .
2
p.u.
The ohmic value of reactance referred to secondary side is 6×
33 11 ×
√
3
2
= 18Ω
V base on secondary is 33 kV. Secondary X p.u. =
300 332
× 18 = 4.9589 p.u.
Electrical Engineering Dept - IIT Delhi
