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Selection and application
3-8 Selecting an MCCB for capacitor circuit
3-8 Selecting an MCCB for capacitor circuit
3-8-1 Characteristics specific to capacitor
circuits
Note the following points when considering MCCBs for
capacitor circuits:
(1) Arc reignition due to recovery voltage
When a capacitor circuit shown in Fig. 3-25 is opened, it
exhibits characteristics distinctly different from inductive loads
due to the effects of residual electric charge in the capacitor.
In a single-phase circuit like that shown in Fig. 3-26, the
capacitor voltage lags 90° behind the current, and a peak
voltage exists across the capacitor terminals when the circuit
is opened. The recovery voltage appearing between the switch
contacts immediately after the circuit is opened is equal to
the difference between the capacitor residual voltage and the
supply voltage. Therefore, half a cycle after the circuit opens,
the voltage between the switch contacts rises to twice the
supply voltage or higher.
In a three-phase circuit, the recovery voltage appearing
between the contacts in the first interrupted phase could rise
as high as 2.5 times the supply voltage.
Unless the breaker contacts are fully open until half a cycle
after the capacitor circuit opens, restrike of arc will occur. If the
capacitor is discharged by damped oscillation at the oscillation
frequency according to the inductance (L) and capacitance (C)
of the circuit at re-ignition, then residual peak voltage will be
left at the terminal again if the arc is quenched (current cuts
off).
If restrike of arc is repeated, the voltage could continue to rise
to the dielectric breakdown point of the capacitor.
Hence, fast-interrupting circuit breakers with quick-make,
quick-break action are recommended for this type of circuit.
Fig. 3-25 Residual electric charge in the capacitor
Fig. 3-26 Capacitor residual voltage
Vc=Em
(2) Transient inrush current when a circuit closes
When a capacitor circuit like the one shown in Fig. 3-25
closes, the capacitor must be charged with an equivalent of
the voltage applied the instant the circuit closed. This causes
the circuit to be flooded with massive inrush current that has a
steep slope like that shown in Fig. 3-27.
If the circuit closes now with peak supply voltage present, then
the transient current at this time is expressed by the following
equation.
J = 4L/CR
2
i = (Em/LE)H
Dt
sinEt (1)
D = R/2L
E = (4L/CR
2
)/2L, E = (1/LCR
2
/4L
2
)
If
which yields
E = J/2L
Generally E > 0 (1/LC > R
2
/4L
2
) is true, and oscillating transient
current flows at the natural frequency as shown below.
Fig. 3-27 Transient current when a capacitor circuit closes
Since the natural frequency at this time is as follows: f = Z/2S,
equation (1) yields Z = E, and so
f = 1/(LCR
2
/4L
2
)/2S, f = J/4SL
Then equation (1) above yields the following equation.
i = (2Em/J)H
−tR/2L
sin (J/2L)t .....................(2)
Maximum current at this time is expressed as follows:
im = (Em/ L/C)H
R/JtanJ/R
(3)
im = (Em/ L/C)H
Dxø/E
(3-1)
The first wave peak W0 is expressed as follows:
W
0 = (2L/J)tan
−1
J/R W0 = ø/E ...........(4)
ø = tan
−1
E/D (rad)
Since the time shown in equation (4) is very short, the voltage
in equation (3) or (3)-1 is essentially V = Em.
Since H
−D·ø/E
is approximately 1, the peak transient inrush
current is derived as follows from equation (3)-1.
im | Em C/L
(5)
(Here, Em is
√
2/3
times the line voltage in a three-phase
circuit and is
√
2
times the line voltage in a single-phase
circuit.)
The preceding equations prove that transient inrush current
flowing to the capacitor is related to inductance (L), that is, it
is related to the power supply capacity and the presence or
absence of reactors connected in series with the capacitor.
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