Download TEMPUS: SWITCHING OPERATIONS 1: Introduction In case an

TEMPUS: SWITCHING OPERATIONS
1: Introduction
In case an electrical grid supplies a sinusoidal voltage to a linear load, a sinusoidal current is
flowing (which is a steady state behavior). When a load switch or a circuit breaker opens, a
transient behavior occurs before the current becomes and remains zero. This transient will be
studied, mainly when interrupting a short circuit current and a distinction is made between
-
ohmic loads,
inductive loads,
ohmic-inductive loads.
2: An ohmic loaded grid
̂ 𝑠𝑖𝑛(𝜔𝑡) and suppose the grid
Consider a grid supplying a sinusoidal voltage 𝑢(𝑡) = 𝑈
impedance 𝑅1 is resistive. The load 𝑅2 is ohmic and 𝑅2 ≫ 𝑅1 . The circuit breaker S is closed
and the current is limited by 𝑅1 + 𝑅2 giving a current
𝑖(𝑡) =
̂
𝑢(𝑡)
𝑈
=
𝑠𝑖𝑛(𝜔𝑡) .
𝑅1 + 𝑅2
𝑅1 + 𝑅2
Figure 1: Ohmic loaded grid
In case of a short circuit between a and b in Figure 1, a short circuit current will flow which is
only limited by 𝑅1 . The large short circuit current 𝑖𝐶𝐶 (𝑡) also has the same phase as the grid
voltage 𝑢(𝑡). More precisely,
𝑖𝐶𝐶 (𝑡) =
̂
𝑢(𝑡)
𝑈
=
𝑠𝑖𝑛(𝜔𝑡) .
𝑅1
𝑅1
The normal load current 𝑖(𝑡) is flowing during hours (or more). As the short circuit occurs,
the short circuit current is flowing during for instance 0.5 seconds (notice 0.5 seconds
contains 25 periods of 20 milliseconds). The short circuit current, which is flowing during a
number of periods of the grid voltage, is visualized in Figure 2 (dotted lines). This short
circuit is detected and at an arbitrary 𝑡 = 𝑡0, the contacts of the circuit breaker will open.
After 𝑡 = 𝑡0 , due to the electric arc, the short circuit current is still flowing. As the
instantaneous value of the short circuit current passes zero, the arc extinguishes (at 𝑡 = 𝑡1 in
Figure 2). The circuit breaker avoids re-ignition of the arc implying the current remains zero.
As long as the circuit breaker is closed, there is no voltage across the closed contacts.
Between 𝑡0 and 𝑡1 , the contacts are open but the voltage across the contacts (across the arc) is
̂ 𝑠𝑖𝑛(𝜔𝑡) appears across the open contacts of
limited. From 𝑡 = 𝑡1 , the grid voltage 𝑢(𝑡) = 𝑈
the circuit breaker. Due to this voltage, no re-ignition of the arc is allowed.
Figure 2: Interrupting an ohmic short circuit current
3: An inductive loaded grid
̂ 𝑠𝑖𝑛(𝜔𝑡) and suppose the grid
Consider a grid supplying a sinusoidal voltage 𝑢(𝑡) = 𝑈
impedance 𝐿1 is inductive. The load 𝐿2 is purely inductive and 𝐿2 ≫ 𝐿1 . The circuit breaker S
is closed and the current is limited by 𝐿1 + 𝐿2 giving a steady state current
𝑖(𝑡) =
̂
𝑈
𝜋
𝑠𝑖𝑛 (𝜔𝑡 − )
𝜔(𝐿1 + 𝐿2 )
2
which lags an angle 𝜋⁄2 with respect to the voltage. Figure 3 visualizes this inductive loaded
grid. Notice in Figure 3 the parasitic capacitance 𝐶 which has no influence as long as the
circuit breaker S is closed.
Figure 3: Inductive loaded grid
In case of a short circuit between a and b in Figure 3, a short circuit current will flow which is
only limited by 𝐿1 . The large short circuit current 𝑖𝐶𝐶 (𝑡) also lags an angle 𝜋⁄2 with respect
to the grid voltage 𝑢(𝑡). More precisely,
𝑖𝐶𝐶 (𝑡) =
̂
𝑈
𝜋
𝑠𝑖𝑛 (𝜔𝑡 − ) .
𝜔𝐿1
2
The normal load current 𝑖(𝑡) is flowing during hours (or more). As the short circuit occurs,
the short circuit current is flowing during for instance 0.5 seconds (notice 0.5 seconds
contains 25 periods of 20 milliseconds). The short circuit current, which is flowing during a
number of periods of the grid voltage, is visualized in Figure 4. This short circuit is detected
and at an arbitrary 𝑡 = 𝑡0 , the contacts of the circuit breaker will open.
Figure 4: Interrupting an inductive short circuit current
After 𝑡 = 𝑡0 , due to the electric arc, the short circuit current is still flowing. As the
instantaneous value of the short circuit current passes zero, the arc extinguishes (at 𝑡 = 𝑡1 in
Figure 4). The circuit breaker avoids re-ignition of the arc implying the current remains zero.
As long as the circuit breaker is closed, there is no voltage across the closed contacts.
Between 𝑡0 and 𝑡1 , the contacts are open but the voltage across the contacts (across the arc) is
limited. From 𝑡 = 𝑡1 , a voltage appears across the open contacts of the circuit breaker. Notice
however, this voltage is not a sinusoidal voltage having a frequency of 50 Hz as it is the case
when considering an ohmic loaded grid.
At 𝑡 = 𝑡1 , the grid voltage is not zero (the grid voltage attains a maximum) but due to the
parasitic capacitor 𝐶 the voltage across the contacts does not change immediately. Notice the
high resonance frequency due to 𝐶 and 𝐿1 giving, across the contacts of the circuit breaker, a
voltage
𝑡 − 𝑡1
̂ 𝑐𝑜𝑠(𝜔(𝑡 − 𝑡1 )) − 𝑈
̂ 𝑐𝑜𝑠 (
𝑢𝑆 (𝑡) = 𝑈
).
√𝐿1 𝐶
Due to the high resonance frequency, the voltage across the circuit breaker S changes very
fast and becomes almost twice the grid peak voltage. Since this voltage across S changes very
fast and becomes almost twice the grid peak voltage, it is not obvious to avoid re-ignition of
the arc. Indeed, interrupting an inductive current is much more difficult than interrupting an
ohmic current.
3.1: Voltage across the circuit breaker
The voltage across the open circuit breaker will be calculated for all 𝑡 ≥ 𝑡1. Due to the short
circuit, the evolution of the voltage across the circuit breaker S can be modeled as visualized
in Figure 5. When opening the circuit breaker, the parasitic capacitor 𝐶 across the contacts of
the circuit breaker is important.
Define 𝑡 ′ = 𝑡 − 𝑡1 . At 𝑡 ′ = 0, 𝑢𝐶 (𝑡 ′ = 0) = 0 and 𝑖𝐿1 (𝑡 ′ = 0) = 𝑖𝐶 (𝑡 ′ = 0) = 0 and the
behavior is described by the differential equation
̂ 𝑐𝑜𝑠(𝜔𝑡′) = 𝐿1
𝑈
𝑑 𝑖𝐿1 (𝑡′)
+ 𝑢𝐶 (𝑡′)
𝑑𝑡′
and
𝑡′
𝑢𝐶 (𝑡′) = 𝑢𝐶
(𝑡 ′
𝑡′ (𝜏)
𝑖𝐿1 (𝜏)
𝑖𝐶
′
= 0) + ∫
𝑑𝜏 = 𝑢𝐶 (𝑡 = 0) + ∫
𝑑𝜏.
𝐶
𝐶
0
0
Figure 5: Modeling the open circuit breaker in case of a short circuit
By taking the Laplace transform,
̂
𝑈
𝑠
= 𝐿1 𝑠 𝐼𝐿1 (𝑠) − 𝐿1 𝑖𝐿1 (𝑡 ′ = 0) + 𝑈𝐶 (𝑠)
𝑠2 + 𝜔2
and
𝑈𝐶 (𝑠) =
1 1
(𝐼 (𝑠) + 0)
𝐶 𝑠 𝐿1
giving
̂
𝑠𝑈
1
(𝑠)
=
𝑠
𝐿
𝐼
+
𝐼 (𝑠)
1
𝐿1
𝑠2 + 𝜔2
𝑠 𝐶 𝐿1
or equivalently
𝐼𝐿1 (𝑠) =
̂
𝑠𝑈
𝑠𝐶
.
𝑠 2 + 𝜔 2 1 + 𝑠 2 𝐿1 𝐶
This implies
𝑈𝐶 (𝑠) =
̂
𝐼𝐿1 (𝑠)
𝑠𝑈
1
= 2
.
𝑠𝐶
𝑠 + 𝜔 2 1 + 𝑠 2 𝐿1 𝐶
Using partial fraction decomposition, one obtains
𝑈𝐶 (𝑠) =
̂
𝑈
𝑠
𝑠
(
−
)
1 − 𝜔 2 𝐿1 𝐶 𝑠 2 + 𝜔 2 𝑠 2 + 1
𝐿1 𝐶
and taking the inverse Laplace transform (𝜔 ≪ (1⁄𝐿1 𝐶 ))
𝑢𝐶 (𝑡′) =
̂
𝑈
𝑡′
𝑡′
̂ 𝑐𝑜𝑠(𝜔𝑡′) − 𝑈
̂ 𝑐𝑜𝑠 (
(𝑐𝑜𝑠(𝜔𝑡′)
−
𝑐𝑜𝑠
(
))
≈
𝑈
).
1 − 𝜔 2 𝐿1 𝐶
√𝐿1 𝐶
√𝐿1 𝐶
Since 𝑡 ′ = 𝑡 − 𝑡1 and 𝑢𝐶 (𝑡) = 𝑢𝑆 (𝑡) one obtains that
𝑡 − 𝑡1
̂ 𝑐𝑜𝑠(𝜔(𝑡 − 𝑡1 )) − 𝑈
̂ 𝑐𝑜𝑠 (
𝑢𝑆 (𝑡) = 𝑈
).
√𝐿1 𝐶
Notice at 𝑡 = 𝑡1 , the voltage across the circuit breaker S equals zero but a fast increase of this
̂ is reached. Due to this voltage rise, no re-ignition of
voltage occurs and almost a voltage 2𝑈
the voltage is allowed.
4: An ohmic inductive loaded current
In case the grid impedance and the load are ohmic inductive, the transient voltage across the
circuit breaker S after 𝑡1 contains
-
a 50 Hz voltage component due to the grid voltage,
a damped high frequent oscillation due to 𝐶 and 𝐿1 .
5: Interrupting the current
Until now, the zero crossing of the current is used to extinguish the arc in the circuit breaker
(or the load switch) implying it is mainly the task of the circuit breaker to avoid this arc does
not re-ignite. In other cases, the current is interrupted without the use of the zero crossings of
the current. Figure 6 visualizes an example of such a situation.
Figure 6: Interrupting the primary current of a transformer
Figure 6 visualizes the sinusoidal grid voltage source 𝑢(𝑡) and the grid impedance containing
a resistive part 𝑅 and an inductive part 𝐿. Notice the parasitic capacitances 𝐶1 and 𝐶2 . Notice
the circuit breaker S having a voltage 𝑢𝑆 (𝑡) across its contacts. The load is the primary
winding of a transformer having an unloaded secondary winding. The current consumed by
the transformer at primary side is ohmic inductive (mainly inductive).
Due to a voltage across the parasitic capacitor 𝐶1 , energy is stored in this capacitor. In case
the capacitor voltage equals 𝑢𝐶1 , the stored energy equals
𝐸(𝐶1 ) = 𝐶1
2
𝑢𝐶1
.
2
Notice the inductance 𝐿1 of the primary winding of the transformer. Due to a current 𝑖𝐿1 , the
energy stored in 𝐿1 equals
𝐸(𝐿1 ) = 𝐿1
2
𝑖𝐿1
.
2
Figure 7 visualizes the evolution of the voltage 𝑢𝑆 (𝑡) across the circuit breaker (or load
switch) and the evolution of the current 𝑖𝑆 (𝑡). Originally, the circuit breaker S is closed and
the 50 Hz current consumed by the primary winding of the transformer (by 𝐿1 ) is flowing
through S (the current through 𝐶1 is very small and can be neglected). This current lags
approximately an angle 𝜋⁄2 with respect to the applied voltage 𝑢𝑆 (𝑡) which is visualized by
dotted lines.
The circuit breaker opens which forces 𝑖𝑆 (𝑡) to zero. There is energy stored in 𝐿1 and this
energy will be transported to 𝐶1 . The voltage across 𝐶1 increases rapidly implying a large
voltage across the circuit breaker (𝐶1 is small implying a high voltage is needed to store some
energy). Due to the large voltage across the circuit breaker, an arc appears. Due to this arc,
electrical charges are drained which discharges 𝐶1 implying the voltage across S decreases.
Due to this decreased voltage, the arc extinguishes.
Figure 7: Evolution of the circuit breaker current and voltage
When the arc extinguishes, the current through 𝐿1 is not zero (although already smaller). The
energy stored in 𝐿1 will be transported to 𝐶1 . The voltage across 𝐶1 increases rapidly implying
a large voltage across the circuit breaker. Due to this second voltage peak across the circuit
breaker, a second arc appears. Also this second arc drains electrical charges from 𝐶1 implying
the voltage decreases and the arc extinguishes a second time.
The phenomenon where:
-
energy is transported from 𝐿1 to 𝐶1 ,
an arc appears and 𝐶1 is discharged,
the arc extinguishes,
repeats itself several times until the energy stored in 𝐿1 is sufficiently small. The voltage
across 𝐶1 remains sufficiently small implying no new arc appears.
Finally, an oscillatory behavior due to 𝐿1 and 𝐶1 appears. No current is flowing through the
circuit breaker anymore since no new arc appears. Due to resistances (implying losses), the
oscillation is damped. Once the high frequent oscillation has disappeared, only the 50 Hz grid
voltage appears across the circuit breaker.
6: DC voltages and DC currents
It is obvious in case of an AC current, the zero crossings of the current help to interrupt the
current. In case a DC current will be interrupted, there are no zero crossings. Interrupting a
DC current is much more difficult than interrupting an AC current (especially if the load is
inductive). Due to this reason, there are dedicated DC fuses, DC contactors and DC circuit
breakers available on the market. Due to the rise of photovoltaic installations and battery
storage installations, the use of DC fuses, DC circuit breakers and DC contactors is rising.
References
Van Dommelen D., Productie, transport en distributie van elektriciteit, Acco, Leuven, 2001.
Weedy B.M., Cory B.J., Electric Power Systems, Wiley & Sons, New York, 2001.