A three phase transmission line having a selfreactance of 1 Ω/km and a mutual reactance of 0.2 Ω/km. The negative and zero sequence equivalent reactances respectively are
Positive sequence reactance = X_{s} – X_{m} = 0.8 Ω/km
Negative sequence reactance = X_{s} – X_{m} = 0.8 Ω/km
Zero sequence reactance = X_{s} + 2X_{m} = 1.4 Ω/km
A balanced star connected load takes 50 A from a balanced 3phase. 4wire supply. The zero and positive sequence components (I_{ao} and I_{a1}) of the line current (I_{a}) are respectively
The given load is balanced star connected load. In balanced system only positive sequence component is present, both negative and zero sequence components are absent.
Zero sequence component, I_{ao} = 0 A
Given that, I_{a} = 50A
Positive sequence component,
I_{a1} = I_{a} = 50A
What is the per unit reactance of an alternator of the capacity is doubled and the voltage is halved?
A transmission line has an impedance of 0.3 PU on 11 KV and 200 MVA base. Find the new PU Impedance on 33 kV and 1600 MVA base
ZPU_{new} = 0.3 ×
A transmission line of inductance 0.1 H and resistance 5 ohms is suddenly shortcircuited at t = 0 at the bar end ass shown in the figure. The approximate value of the first current maximum (Maximum momentary current) is
Z = 5 + j 314 × 0.1 = 5 + j 31.4 = 31.8 ∠81°
= 3.14 sin (314 t  66) + 2.87 e^{50 t}
First current maximum of symmetrical short circuit current occurs at
57.3 × 314 t  66 = 90°
⇒ t = 0.00867 sec
First current maximum
i_{mm} = 3.14 + 2.87 e50 × 0.00867 = 5 A
Two power stations S_{1} and S_{2} are interconnected through a transmission line of per unit (p.u) reactance of 0.6. Station S_{1} has one generator of P.U. the reactance of 0.3, and station S_{2} has two generators of p.u reactance 0.6 each. For a 3 – phase symmetrical short circuit at the middle of the transmission line (point F). The change in the value of the equivalent fault reactance, with one of the generators of S_{2} removed, is _____ (in %).
Equivalent reactance with respect to fault point =(0.3 + 0.3)  (0.3 + (0.6  0.6))
=0.6  (0.3 + 0.3)
= 0.3
After removal of one of the generators of S_{2} with respect to fault point
= (0.3 + 0.3)  (0.3 + 0.6)
= (0.6)  (0.9) = 0.36
Change in the reactance =
= 20%
For a generator, the ratio of fault current for Line to Line fault and three phase fault is 0.5.
The positive sequence reactance is 0.2 μ. The negative sequence reactance is __(in p.u.)
A threephase YΔ transformer is rated 400 MVA, 220 kV/22 kV. The short circuit impedance of the transformer measured on low voltage side is 0.121 Ω and because of the low value, it is considered equal to the leakage reactance. The per unit value of reactance to represent this transformer in a system whose base on the high voltage side of the transformer is 100 MVA, 230 kV is ___ (pu)
On its own base the pu reactance of the transformer is,
On the chosen base the reactance becomes,
A 3ϕ load is connected to a 3ϕ balanced supply V_{an }= 20∠120°, V_{bn} = 20∠120°, V_{cn} = 20∠0° find the value of R
Apply KCL
I_{a} + I_{b} + I_{c} = I_{n} = 0
R = 1.154 Ω
Two 300 MVA, 22 kV, 3phase, 60 Hz alternators are connected by a transmission line having transformers at both ends as shown in the single line diagram given below
The transformer (T1) between the generator 1 and the transmission line is a ΔY connected 3phase transformer bank made up of three single phase transformers each rated at 120 MVA, 22 kV/127 kV with leakage reactance of 8%. The generator 2 is connected to the line by a 3phase, 300 MVA, 22 kV/220 kV, ΔY transformer (T2) with 10% leakage reactance. The star side of T1 and T2 are grounded. The zero sequence reactance of the transmission line is 81 Ω. Considering the MVA rating of T1 as the base, the zerosequence network of the complete system is (Assume X_{g1} and X_{g2} are the zero sequence reactance of the generators respectively)
Given that base MVA is the rating of T_{1}
Rating of single phase transformer of 3phase transformer bank = 12 MVA
Rating of 3phase transformer bank = 3 × 120 = 360 MVA
It is connected in Δ Y connection.
Primary voltage of three phase transformer bank = Primary voltage of single phase transformer = 22 kV
Secondary voltage of three phase transformer bank = √3 × secondary voltage of single phase transformer = √3 × 127kV = 220kV
X_{T1} (for single phase transformer) = j0.08
X_{T1} = (for three phase transformer)
X_{T1} = j0.08 pu
X_{T2} (old) = j0.10
= j 0.12 pu
Given that, reactance of transmission line (X_{L}) = 81Ω
As the generator reactance values are not given, let assume zero sequence reactance of generator 1 and generator 2 are X_{g1} and X_{g2} respectively.
Now, the zerosequence network becomes.
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