Basler Electric BE1-87G - Manual (Page 64)

Rt Total CT burden = R
i
+ R
r
+ R
w
(for phase faults and switching)
= 2R
i
+ 2R
r
+ R
w
(for external ground faults)
SFR Saturation factor ratio = [(Vce)
1
÷ (Vce)
2
] × [(Rt)
2
÷ (t)
1
] (if SFR is <1, use 1/SFR)
Vc CT accuracy class voltage rating
Vce Effective CT accuracy class voltage rating = Vc(N
a
÷ N)
X”
d
Sub-transient reactance
CT Quality
CTs should have an accuracy class of either C20 or better or T20 or better.
Burden Limit
The CTs should be operating below the knee point under symmetrical current conditions, that is, no ac
saturation. For a generator, ac saturation should not occur for the maximum external fault current. For a
motor, ac saturation should not occur during starting. For a shunt reactor, high current fundamental-
frequency currents will not flow unless the reactor becomes faulted.
Assuming a maximum current of 20 times the CT rating: R
t
< 0.007Vce, where V
knee
= 70% of Vce and 20
× 5 A nominal rating = 100 A so that:
R
t
= V
knee
÷Imax = 0.7Vce ÷ 100 A = 0.007Vce
Relative CT Performance
Table 5-1 lists the recommended minimum pickup settings to provide security during external faults and
switching.
Table 5-1. Recommended Pickup Settings
Vce (Lower Value)
SFR
1 1.5 2 3 4
20 0.2 0.4 0.5 0.8 1.6
50 0.2 0.4 0.4 0.5 0.8
100 0.2 0.2 0.4 0.4 0.5
200 0.1 0.1 0.2 0.4 0.5
>200 0.1 0.1 0.2 0.4 0.5
Note: Use next higher setting if CTs are a mix of C classification and T classification.
The saturation factor ratio, SFR, represents an index of the relative performance of the two sets of CTs.
This performance is a function of the relative quality of the CTs (Vce), the relative burdens (R
t
), and
neglecting the remanence. If SFR is less than 1, use 1/SFR for SFR.
Setting Example 1
Select the pickup setting for the motor application in Figure 5-7. In this application, the settings need to be
based on the probability of significant dissimilar CT saturation due to the very slow decaying dc
component of the starting current. Since the motor is not grounded, no ground current can flow during
starting of an unfaulted motor. Therefore, one-way lead burden is used to determine the total CT burden.
Each phase CT carries just the burden for the lead of that phase.
(R
t
)
1
= R
l
+ R
w
= 0.22 + 0.14 = 0.36 ; (Vce)
1
= 50 ; R
t
< 0.007(Vce)
1
= 0.35
(R
t
)
2
= 0.09 + 0.10 = 0.19 ; (Vce)
2
= 100 ; R
t
< 0.007(Vce)
2
= 0.7
Inequality (1) is met with CT #2, but not with CT #1. However, since the locked rotor current is only 4.8
times the CT rating (versus the assumption of 20 times rated for inequality (1)), the application is suitable.
SFR = (100 ÷ 50) × (0.36 ÷ 0.19) = 3.8
NOTE
Relay Style x1x-xxx-xxxxx: if R
w
is unknown, assume R
w
= 0.003 × N
a
Relay Style x2x-xxx-xxxxx: if R
w
is unknown, assume R
w
= 0.01 × N
a
5-8 BE1-87G Testing and Setting 9170800990 Rev N