ELEC-E8409 HIGH VOLTAGE ENGINEERING Exercise 4 ELEC-E8409 HIGH VOLTAGE ENGINEERING

Question 1 Disruption of capacitive current occurs when an idle (no-load) cable or a storage capacitor is disconnected. Derive the maximum recovery voltage between the switch terminals. ~ C L

current voltage Capacitive load: input voltage is -90o out of phase with load current (current leading voltage). ~ C L A B

Voltage over capacitance C ~ L A B uc C Voltage over capacitance C 1.0 u [p.u.] -1.0 sin (ωt) Maximum when sin ωt = 1

Recovery voltage Voltage between terminals ~ Maximum when sin ωt = -1 uAB Recovery voltage ~ L A B Voltage between terminals Maximum when sin ωt = -1 1.0 u, i [p.u.] -1.0 Recovery voltage ωt1 i(ωt) u(ωt) In practice,

Question 2 An overhead line has characteristic impedance of 450 Ω. A 200 kV rectangular impulse 10 km in length propagates along the line. What is the energy of the impulse? How much of the energy is in the magnetic field and how much is in the electric field?

Impulse duration t = l/v (where v = c = 3∙108 m/s) 200 kV Z = 450 Ω l = 10 km Impulse duration t = l/v (where v = c = 3∙108 m/s) Impulse energy

Energy W = 3 kJ How much of the energy is in the magnetic field and how much is in the electric field? γ = capacitance per unit length λ = inductance per unit length L = inductance for length of cable (λl) Inductive Energy: Energy is distributed half into the electric field and half in the magnetic field Capacitive Energy:

Question 3 A step wave with amplitude 450 kV propagates along an overhead line to a 110/20 kV transformer. The characteristic impedance of the line is 450 Ω. The primary winding is protected by a nonlinear resistor type arrester with inception voltage Ui = 550 kV. How long does it take for the arrester to activate? How large capacitors have to be connected to the secondary winding and ground so that the capacitive overvoltage over the transformer does not exceed 75 kV (secondary test voltage)? The transformer can be described as a capacitive equivalent circuit used for impulse voltage testing. transformer C12 C10 C20 Z = 450 Ω u = 450 kV Ui =550 kV C10 =3 nF C12 =5 nF C20 =10 nF

Z u2 C12 C10 C20 Z = 450 Ω u = 450 kV Ui =550 kV C10 =3 nF C12 =5 nF transformer C12 C10 C20 Z = 450 Ω u = 450 kV Ui =550 kV C10 =3 nF C12 =5 nF C20 =10 nF Z u2

Laplace transform: where Time to activate:

How large capacitors have to be connected to the secondary winding so that the capacitive overvoltage over the transformer does not exceed 75 kV? C12 C10 C20 Cx u2 u1 C12 C10 u1 C20+Cx

Solve for Cx …

Question 4 A 110 kV transformer is protected by a surge arrester as shown in the figure. Inception voltage and residual voltage (constant) is 400 kV. Distance between the transformer and arrester is 30 m. A lightning stroke produces a wedge wave propagating along the line with steepness of 1000 kV/µs. Determine the voltage stress experienced by the transformer and the inception time of the interrupter when the transformer is represented as surge capacitance C = 0. How far can the interrupter be placed from the transformer if the transformer can withstand 550 kV? C = 0 S = 1000 kV/µs l = 30 m

Ua = S∙2τ + 2S(ti – 2τ) Voltage at point A: Inception time: C = 0 S = 1000 kV/µs l = 30 m Open line (Z = ∞) Short circuit to ground when arrester operates (Z = 0) ρ = -1 ρ = 1 A B Propagation time between AB: Voltage at point A: Arriving wave with steepness S + Positive reflected wave (2S) until arrester starts to conduct (short circuit at ti) = Ua = S∙2τ + 2S(ti – 2τ) Inception voltage of arrester Inception time:

First reflection from B arrives at A (tAB = 0.1 µs) 500 Voltage at A reaches inception voltage of arrester (residual voltage = 400 kV) 400 300 2S = 2000 kV/µs Voltage [kV] First reflection from B arrives at A (tAB = 0.1 µs) 200 Ua 100 S = 1000 kV/µs 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 Time [µs]

C = 0 S = 1000 kV/µs l = 30 m Open line (Z = ∞) Short circuit to ground when arrester operates (Z = 0) ρ = -1 ρ = 1 A B Voltage at point B: Arriving wave + Reflected positive wave (2S) until arrester operates Negative reflected wave (-2S) when arrester becomes operational = Ub = 2S(τ + ti) – 2Sτ Ub = 2S ∙ ti = 2(1000 ∙ 109)(0.3 ∙ 10-6) = 600 kV

Onset of arrester seen at B (A: ground Z = 0  ρ = -1) 800 Onset of arrester seen at B (A: ground Z = 0  ρ = -1) τ 700 600 2S = 2000 kV/µs Ub -2S Reflection from A returns to B (B: open Z = ∞  ρ = 1) Reflection at A 2S -2S Arrester becomes active at A 500 Voltage [kV] 400 300 200 100 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 Time [µs]

Solve for lmax Now Ub = Uw = 550 kV When distance l = 30 m: Ub = 600 kV > Uw = 550 kV Arrester must be brought closer to transformer lmax = maximum allowable distance to transformer τ´= propagation time of distance lmax Now Ub = Uw = 550 kV Solve for lmax