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- A 100 MVA, 33 kV 3 phase generator has sub-transient reactance of 15 %. The generator is connected to the motors through a transmission line and transformers as shown in Figure 1. The motors have rated inputs of 30 MVA, 20 MVA and 50 MVA at 30 kV with 20 % sub-transient reactance. The 3 phase transformers are rated at 110 MVA, 32 kV ∆ / 110 kV Y with leakage reactance of 8 %. The line has a reactance 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. values. [X p.uT^1 = 0. 06838 p.u., Xp.uline = 0. 3886 p.u.,X p.uM^1 = 0. 5509 p.u., X p.uM^2 = 0. 8264 p.u., X p.uM^3 = 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
X p.uT^1 = 0. 08 ×
×
= 0. 06838 p.u.
Base Voltage on secondary of transformer T1 is
33 ×
= 113. 43 kV
Hence, the p.u. impedance of transmission line is 50 × 100
- 432
= 0. 3886 p.u.
The base value of voltage for motor circuit is
- 43 ×
= 33kV
Hence, the p.u. reactance of motors are
M1 : 0. 2 ×
×
= 0. 5509 p.u. M2 : 0. 2 ×
×
= 0. 8264 p.u.
M3 : 0. 2 ×
×
= 0. 3305 p.u.
- 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= % ; 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 as base quantities. [X p.uT^1 = 0. 48 p.u, X p.uline 1 = 0. 0826 p.u., X p.uT^2 = 0. 4 p.u.,X p.uT^3 = 0. 32 p.u., X p.uline 2 = 0. 3966 p.u., X p.uT^4 = 0. 24 p.u., X p.uM^1 = 0. 289 p.u., Zp.uload = 3.2 + 2. 4 i]
(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 Zp.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
X puT^1 = 0. 12 ×
×
= 0. 48 p.u.
Vbase on the secondary side of T1=220 kV. p.u. reactance of Line 1
X puline 1 =
50 × 80
= 0. 0826 p.u.
p.u. reactance of T2 is
X puT^2 = 0. 1 ×
×
= 0. 4 p.u.
p.u. reactance of T3 is
X puT^3 = 0. 08 ×
×
= 0. 32 p.u.
p.u. reactance of transmission line =
100 × 100
= 0. 6780 p.u.
Vbase on the secondary side of T2=10 ×
3 × 66
= 10. 62 kV
X puT^2 = 0. 1 ×
×
= 0. 0985 p.u.
X puM^1 = 0. 15 ×
×
= 0. 5319 p.u.
X puM^2 = 0. 15 ×
×
= 0. 2659 p.u.
- 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 50 MVA and 13.8 kV as base quantities in Generator 1. [X p.uT^1 = 0. 0705 p.u., X p.uline 12 = 0. 1312 p.u., X p.uline 23 = 0. 0656 p.u., X p.uT^2 = 0. 2083 p.u., X p.uG^2 = 0. 1936 p.u., X p.uT^3 =
- 0868 p.u]
Figure 3:
Solution: X puG^1 = 0. 15 p.u.
X puT^1 = 0. 1 ×
×
= 0. 0705 p.u.
Vbase on secondary of T1 = 13. 8 ×
= 138kV
X puline 12 =
50 × 50
= 0. 1312 p.u.
X puline 13 = X puline 23 =
25 × 50
= 0. 0656 p.u.
Vbase on primary of T2 = 138 ×
= 15kV
X puT^2 = 0. 15 ×
×
= 0. 2083 p.u.
X puG^2 = 0. 2 ×
×
= 0. 1936 p.u.
Vbase on primary of T3 = 138 ×
= 15kV
X puT^3 = 0. 1 ×
×
= 0. 0868 p.u.
- 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. [Xp.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
- 22
= 0. 5 p.u
The ohmic value of reactance referred to secondary side is
4 ×
The per-unit value of reactance referred to secondary side is 16 × 0. 005
- 42
= 0. 5 p.u
