what is the difference between 5P20 class CT , PS Class CT
and 5P10 class CT??
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Answer / amit_er1981
CT's are categorized as Protection CT, Special Protection
CT and Measuring CT. Based on this, the CT's are
classified. Here is the meaning of the CT classes:
Class 5P20:
The letter 'P' indicates it is a protection CT.
The number 5 indicates the accuracy of the CT. Most common
accuracy numbers are 5 and 10.
The number 20 (called accuracy limit factor) indicates that
the CT will sense the current with the specified accuracy
even with 20 times of its secondary current flows in the
secondary. This is required for protection CT, because the
fault current is high and the CT should be able to sense
the high fault current accurately to protect the system.
The common numbers are 10, 15, 20 and 30.
Class PS:
PS is for 'Protection Special'. This class of CT's are used
for special protection such as differential protection,
distance protection etc.
Class 1M:
The letter 'M' indicates it is a measuring CT.
The number 1 indicated the accuracy of the CT. The
measuring CT's should be more accurate than the protection
CT. The most common accuracy numbers are 0.5 and 1.
Measuring CT's are intended to work in normal and it
doesn't require working with accuracy beyond the CT
secondary current rating, so the accuracy limit factor is
not mentioned here.
Is This Answer Correct ? | 295 Yes | 18 No |
Answer / er.k.k.murty
K K. MURTY
--
5P10 & 5P20 Class CTs: They are generally used for Over
current (Instantaneous and IDMT relays) and E/F protection
of Feeders, Transformers etc.
(i) 5 signifies the % limit of composite error.
(ii) P signifies the Protection core,
(iii) 10, 20 signify the quantum of fault current as
multiple of the rated current or ALF(Accuracy Limit Factor)
up to which the CT shall have the defined percentage of
composite error i.e. ± 5%. During fault condition the fault
current is many times the rated current. Standard Accuracy
Limit factors as per IS: 2705 and IEC60044-1 are 5, 10, 15,
20 & 30.
Accuracy limit factor is inversely proportional to the rated
VA burden i.e. if the connected VA burden is lower than the
rated one, the limit of accuracy shall be maintained beyond
the declared one.
The formula for actual ALF;
ALF(ACT)= ALF(RTD) x {Sin+Sn(Rated Burden)}/{ Sin+Sa (Actual
Burden)}.
Example:- Calculate actual Accuracy limit factor if
secondary burden is reduced to 1/3rd, if CT rated as 200/5
A,5P10,15VA and Rct =0.15Ω?
ALF Rated=10(From CT data 5P10),
Sin (Burden due to internal resistance of the CT) =
(5A)squre x 0.15=25 x 0.15=3.75VA,
Sn (Rated burden) = 15VA (From CT data), Sa(Actual Burden) =
15/3= 5VA,
ALF(ACT) = ALF(RTD) x {Sin+Sn(Rated Burden)}/{ Sin+Sa(Actual
Burden)}
ALF(ACT) = 10x (3.75+15)/(3.75+5) = 10x18.75/8.75 = 21.42
It is clear from the above example that the ALF increases if
the connected actual burden is lower than the rated one. The
CT in this condition shall maintain accuracy up to fault
current of 21.42 times of rated current as against 10 times
the rated current and shall not saturate till then.
PS Class CTs: Though abbreviation PS is not elaborated
anywhere, however it may be called as “Protection Special
Class” CT core. This core is used particularly where current
balance is precisely required to be maintained. In
Differential protection, balance is the prime requirement
between secondary currents of associated CTs of either side
of the equipments. Differential and restricted E/F
Protection of Transformers and Overall Differential
Protection of Generators need such CTS.
The 5P10P or 5P20 class CTs cannot match the Characteristic
as that of PS class. The core of PS class is such that very
high current is needed for saturation of the core. Knee
point voltage of the CT is of valid importance. The
developed voltage across the relay terminals should be lower
than the Knee point Voltage of the CTs.
From: Er.K.K.Murty, B.E.(Hons)Electrical,FIE.
Retd.Chief Engineer(Testing & Commnu),M.P.State Electricity
Board.
Jabalpur.
Email ID:kkmurty05@yahoo.com
Is This Answer Correct ? | 80 Yes | 6 No |
Answer / veera
in 5p20 5 is 5% of the 20times of the primary current(i.e)
in case 100/1A CT 20*100=2000A 5% is 100A ERROR AT 2000A
fault cuurent for instantaneous trip this error is
negligible
Is This Answer Correct ? | 67 Yes | 19 No |
Answer / lim
PS the first time i know.
may be no 1 is 1% tolerence accuracy
i take the knowledge only.
Is This Answer Correct ? | 12 Yes | 68 No |
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