Exercise 4

1. Exercise 4 S-18.3150/S-18.3150 High Voltage Engineering

2. 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. L ~ C S-18.3150/S-18.3150 High Voltage Engineering

3. Capacitive load: input voltage is -90o out of phase with load current (current leading voltage). current voltage L A B ~ C S-18.3150/S-18.3150 High Voltage Engineering

4. L A B u [p.u.] ~ uc C 1.0 sin (ωt) Voltage over capacitance C -1.0 Maximum when sin ωt = 1 S-18.3150/S-18.3150 High Voltage Engineering

5. L A B Recovery voltage u, i [p.u.] 1.0 ~ uAB Voltage between terminals i(ωt) ωt1 Recoveryvoltage u(ωt) -1.0 Maximum when sin ωt = -1 In practice, S-18.3150/S-18.3150 High Voltage Engineering

6. 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? S-18.3150/S-18.3150 High Voltage Engineering

7. 200 kV Z = 450 Ω l = 10 km Impulse duration t = l/v (where v = c = 3∙108 m/s) Impulse energy S-18.3150/S-18.3150 High Voltage Engineering

8. 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 unitlength λ = inductance per unitlength 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: S-18.3150/S-18.3150 High Voltage Engineering

9. 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 u = 450 kV Z = 450 Ω C10 =3 nF C10 C20 Ui=550 kV C12 =5 nF C20 =10 nF S-18.3150/S-18.3150 High Voltage Engineering

10. Z transformer C12 u = 450 kV u2 Z = 450 Ω C10 =3 nF C10 C20 Ui=550 kV C12 =5 nF C20 =10 nF S-18.3150/S-18.3150 High Voltage Engineering

11. Laplace transform: where Time to activate: S-18.3150/S-18.3150 High Voltage Engineering

12. 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 u1 u2 C10 C20 Cx C12 C10 u1 C20+Cx S-18.3150/S-18.3150 High Voltage Engineering

13. Solve for Cx … S-18.3150/S-18.3150 High Voltage Engineering

14. 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? S = 1000 kV/µs l = 30 m C = 0 S-18.3150/S-18.3150 High Voltage Engineering

15. B A ρ= -1 ρ= 1 S = 1000 kV/µs Propagation time between AB: Open line (Z = ∞) l = 30 m 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) Short circuit to ground when arrester operates (Z = 0) C = 0 Inception voltage of arrester Inception time: S-18.3150/S-18.3150 High Voltage Engineering

16. 500 Voltage at A reaches inception voltage of arrester (residual voltage = 400 kV) 400 2S = 2000 kV/µs 300 Voltage [kV] First reflection from B arrives at A (tAB = 0.1 µs) 200 Ua 100 S = 1000 kV/µs 0 0 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] S-18.3150/S-18.3150 High Voltage Engineering

17. B A ρ= -1 ρ= 1 Voltage at point B: S = 1000 kV/µs Open line (Z = ∞) l = 30 m Arriving wave + Reflected positive wave (2S) until arrester operates + Negative reflected wave (-2S) when arrester becomes operational = Ub = 2S(τ + ti) – 2Sτ Short circuit to ground when arrester operates (Z = 0) C = 0 Ub = 2S ∙ ti= 2(1000 ∙ 109)(0.3 ∙ 10-6) = 600 kV S-18.3150/S-18.3150 High Voltage Engineering

18. 800 Onset of arrester seen at B (A: ground Z = 0  ρ = -1) 700 600 -2S -2S Arrester becomes active at A Reflection at A 500 Voltage [kV] 400 300 2S 2S 2S = 2000 kV/µs τ 200 Reflection from A returns to B (B: open Z = ∞  ρ = 1) 100 Ub 0 0 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] S-18.3150/S-18.3150 High Voltage Engineering

19. 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´= propagationtime of distancelmax Now Ub = Uw = 550 kV Solve for lmax S-18.3150/S-18.3150 High Voltage Engineering