Page 504 - Industrial Power Engineering and Applications Handbook

15/478 Industrial Power Engineering and Applications Handbook
Table 15.10 Example 15.3
Consider a protective scheme having a total VA burden of 15
and requiring an accuracy limit factor (ALF) of 20:
Circuit whose Current rating Main CT VA burden shared
current is being ratio by the main :. VA x ALF = 15 x 20 = 300
summed CTs
which is too large to design a CT adequately. In such cases
it is advisable to consider two sets of CTs, one for those
Circuit 1 1000 A 1000/1 A *25 'Oo0 relays that are set high and operate at high to very high
3400
= 7.0 VA currents (short-circuit protection relays) and the second for
all other relays that are required to operate on moderate
Circuit 2 800 A 800/1 A *25 overloads. For example, consider one set of CTs for short-
3400 circuit protection having
= 6.0 VA
Circuit 3 1600 A 1600/1 A *25 l6Oo VA = 5
3400
= 12.0 VA and ALF=20
Total load Ratio of *VA of i.e. a 5P20 CT having a product of VA x ALF of not more than
= 3400 A summation summation 150 and the other set for all the remaining protections having,
CTs = 3400/1 CTs = 25
say,
VA = 10
classes may be one of 5P, 10P and 15P. A protection CT and ALF = 5
is designated by accuracy class, followed by accuracy
limit factor, such as 5P10. The current error, phase error i.e. a 10P5 CT having a product of VA x ALF of much less
6 (Figure 15.18) and the composite error with the rated than 150.
burden in the circuit are as in Table 15.9, according to
IEC 60044-1. It should be chosen depending upon the Note
protective device and its accuracy requirement to For high set protective schemes, where to operate the
discriminate. Closer discrimination will require more protective relays, the primary fault currents are likely to be
extremely high, as in the above case. Here it is advisable to
accurate CTs. consider a higher primary current than the rated for the
protection CTs and thus indirectly reduce the ALF and the
Note product of VA x ALF. In some cases, by doing so, even one
An accuracy class beyond 1OP is generally not recommended. set of CTs may meet the protective scheme requirement.
(iv) Output and accuracy limit factors Example 15.4
Consider a system being fed through a transformer of 1500
The capabilities of a protection CT are determined by the kVA, 11/0.433 kV, having a rated LV current of 2000 A. The
primary inputs of a CT such as the primary ampere turns protection CT ratio on the LV side for the high set relay may
AT (primary current x primary number of turns), core be considered as 4000/5 A (depending upon the setting of
dimensions and the quality of laminations. All this is the relay) rather than a conventional 2000/5 A, thus reducing
roughly proportional to the product of the rated output the ALF of the previous example from 20 to 10. Now only one
(VA) and the rated accuracy limit factor of the CT. For set of 15 P10 CTs will suffice, to feed the total protective
scheme and have a VA x ALF of not more than 150.
normal use, the product of the VA burden and the accuracy
limit factor of a protection CT should not exceed 150,
otherwise it may require an unduly large and more Other considerations when selecting a
expensive CT. For example, for a 10 VA. CT, the accuracy protection CT
limit factor should not exceed 15 and vice versa. The
burden and the accuracy limit factor are thus interrelated. 1 The accuracy limit factor (ALF) will depend upon the
A decrease in burden will automatically increase its highest setting of thc protective device. For a 5 to 10
accuracy limit factor and vice versa. In a ring or bar times setting of the high set relay, the ALF will be a
primary CT, which has only one turn in the primary, the minimum of 10.
ampere turns are limited by the primary current only, thus 2 A higher ALF than necessary will serve no useful
limiting the accuracy and burden of such CTs. This is one purpose.
reason why CTs of up to 5OAare generally manufactured 3 It has been found that, except high set relays, all other
in a wound primary design, with a few turns on the relays may not require the ALF to be more than 5. In
primary side to obtain a reasonably high value of VA such cases it is worth while to use two sets of protection
burden and accuracy. For larger products than 150, it is CTs, one exclusively for high set relays, requiring a
advisable to use more than one protection CT, or use low high accuracy limit factor (ALF), and the other, with
secondary current CTs, i.e. 1 A instead of 5 A. a lower ALF, for the remaining relays. Otherwise choose
a higher primary current than rated, if possible, and
indirectly reduce the ALF as illustrated in Example
Note
For similar reasons, a measuring CT of up to 50 A primary current 15.4 and meet the requirement with just one set of
is recommended to be produced in a wound design. protection CTs.