1. ELEC-E8409 HIGH VOLTAGE ENGINEERING
Exercise 4
ELEC-E8409 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. ~ C L

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

4. 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

5. 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,

6. Question 2
An overhead line has characteristic impedance of 450 Ω. A 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?

7. 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

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 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:

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

10. 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

11. Laplace transform:
where
Time to activate:

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
C10
C20 Cx u2 u1
C12
C10 u1
C20+Cx

13. Solve for Cx …

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?
C = 0
S = 1000 kV/µs
l = 30 m

15. 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:

16. 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]

17. 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

18. 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]

19. 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