Page 871 - Industrial Power Engineering and Applications Handbook

Making capacitor units and ratings of switching devices 25/823
a - log, 14.67
30

-
or 11.17seconds
-- 30
2.686
and discharge resistance
1
k=
R, = 11'17 (2 = 93 k12, say, 100 kC2
120 x 10-6
R If the closing period is only 15 seconds, the required discharge
k= 113
resistance will become 50 kR. The resistance loss in 100 kQ,
discharge resistance
v,"
-- i.e. 4152 or 1.72 watts
Rd 100 x 103
and 3.44 watts in a 50 kQ discharge resistance. Since the
loss is negligible the resistance may be left in the circuit
permanently.
25.7.2 The value of a series inductor
In an L and R dampening circuit the capacitor will
k= 1 discharge according to the following:

(25.7)

L VP
i.e. t = 2 x - log, - seconds (25.8)
R
.
U
k- 1 k=3 where
t =changeover time or required discharge time in
seconds
L = series inductance in the circuit in henry
R = series resistance in the circuit in Q.
ziz
vp = -. v, = 1 P.U.
43
u = residual voltage, say, 50 V or 75 V as noted earlier,
a maximum up to 10% of the rated voltage for a
quick reclosing.
R Example 25.5
k= 1 k=3 Determine the discharge device for the discharge of a three-
phase 6.6 kV, 50 Hz, 1000 kVAr, y-connected capacitor bank,
Figure 25.8 Possible arrangements to provide the discharge connected in units of 10 x 100 kVAr each, through an automatic
resistances across the capacitor terminals and values of k p.f. correction relay, having a closing cycle of 10 seconds.
Data available from the capacitor manufacturer,
C = 30 pF (for each 100 kVAr bank)
Example 25.4 Alternative 1
Consider the scheme of Example 23 4 having an automatic Consider an open delta transformer for this purpose, having
parallel switching If we assume the closing sequence cycle the following data:
to be 30 seconds, the recommended value of discharge L = 120 henry
resistance for each 20 kVAr capacitor bank having a R=90Q
capacitance of 120 pF can be determined as follows u=75v
The time of discharge with these data
t = 30 seconds
VI = 415 volts L VIJ
u = 40 volts (10% Of V,) t = 2 x - x log, - seconds
R
U
k = 1 for D configuration
30 6600 x 45
'. z= ~
120
log,
1 x log, fi 415 = 2 x - seconds
40 90 75