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Consider the system illustrated in
ratings:
10, 000 KVA
13, 800 Volts wye/138, 000 Volts delta
Core losses = 15.0 KW at Rated Voltage
The full load current at 13, 800 volts is
10, 000 = 418 Primary Amperes.
13.8.j3
At rated voltage, the core-loss component produces an in-phase current of
13.8 ../3
Since the relay may be called on to operate at lower-than-normal voltages, the core
losses at these lower voltages should be obtained.
readily available, consider that 95% of rated voltage is a good figure and, since
the transformer core losses are roughly proportional to the square of the applied
voltage, the minimum core-loss current will be
(0.9 5) (0.95) (0.628) = 0.567 Primary Amperes.
At reduced voltage, the CCP minimum operating current goes up.
inversely proportional to the applied voltage.
case will be the nameplate value divided by 0.95, and the 0.004 ampere relay will
have a minimum pickup, at 95% voltage, of
0.004
= 0.0042 Secondary Amperes.
Based on the maximum full-load current, the smallest CT ratio that could be used is
500/5.
With this ratio, the core-loss component of current would be
Since the CCP will
satisfactory.
If the main CT•s had been selected as 800/5, then the core-loss component would have
been 0.00354 amperes and the relay would not operate.
might be possible to use suitable auxiliary cT•s to step up the current to the CCP.
However, caution should be used in the selection of these auxiliary cT•s to ensure
that no more than 5.0 amperes is supplied to the relay under full-load conditions.
Note that the above calculations are based on a balanced three-phase condition, with
equal magnitudes of core-loss current flowing in all three phases.
S AMPLE CALCULATIONS
=
0.628 Primary Amperes.
s
econ ary mperes.
d
A
pick up at 0.0042 secondary amperes, this application is
GEK-65525
Figure 1 having the
Thus, the minimum pickup in this
4
following
transformer
If this information is not
Actually, it is
If this were the case, it
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