Interested to Buy Any Domain ? << Click Here >> for more details...
what is the CT of protection and CT of metering ? what is
class 5P10 & 5P20 and class PS ?
- 16 Answers
- 131678 Views
- JEE Main, ONGC, VRA, I also Faced
- E-Mail Answers
Answer Posted / er.k.k.murty
From: Er.K.K.Murty, B.E.(Hons)Electrical,FIE.
Retd.Chief Engineer(Testing & Commnu),M.P.State Electricity
Board.
Jabalpur(INDIA).
Email ID:kkmurty05@yahoo.com
Measuring Current transformer: This is also known as
Metering CT ,it should confirm to IS:27O5-2 or IEC-60044-1
The cross section area of the cores are so designed that
,the core saturates at about 5 times the rated current (with
low saturation current) .This would facilitate safety to the
instruments, while heavy fault current flows in the Primary
circuit.
Before going into the details of characteristic / duty to
be performed by the metering CTs, one should be aware of
some definitions / terminology relevant to the equipment.
7.1 Instrument Security Factor (IFS/FS): The ratio of
instrument limit primary Current to the rated primary current.
Note: In the event of system fault current flowing through
the primary winding of a CT, the safety of the apparatus
supplied by the CT is greatest when the value of the
instrument security factor (SF) is small.
. Rated Instrument Limit Primary Current (IPL): The Value
of the minimum primary current at which the composite error
of the measuring CT is equal to or greater than 10%,while
the secondary burden is equal to the rated burden.
Note: The composite error should be greater than 10%, in
order to protect the apparatus supplied by the CT against
the high current produced in the event of system fault.
8.0 Accuracy classes;
(a)Standard Accuracy classes: 0.1,0.2,0.5 and1.0 class-The
accuracy limits are defined from 5% of current onwards,
however the declared accuracy limits ie 0.1,0.2,0.5 & 1.0%
respectively and 5,10,30,60 minutes respectively would be at
100% to 120% of rated current at 25% to 100% of rated
burden. {Ref:IS:2705(part-2)}
(b)Special Application Accuracy classes: 0.2S and 0.5S are
used for special applications. The accuracy limits are
defined from1% of rated current, however the declared
accuracy limits ie 0.2% & 0.5%and 10&30 minutes respectively
would be at 20% to 120% rated current at 25% to 100% of
rated burden. Precision metering of High value HT consumers
and Tariff meters etc are categorized as special application
metering and need ‘S’ class CTs. {Ref:IS:2705(part-2)or
IEC:60044-1}.
Note: Almost all the Distribution Utilities follow the
convention mentioned here under;
(i) 0.2 and 0.2S are used for the voltage above 33kV level.
(ii) 0.5 and 0.5S are used up to 33kV level.
5P10 & 5P20 Class CTs: They are generally used for Over
current (Instantaneous and IDMT relays) and E/F protection
of Feeders, Transformers etc.
5 signifies the % limit of composite error
P signifies Protection core,
10, 20 signify the quantum of fault current as multiples 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%.
Standard Accuracy Limit factors as per IS: 2705 and
IEC:60044-1 are 5, 10, 15, 20 & 30.
Accuracy limit factor is inversely proportional to the rated
burden i.e. if the burden is lower than the rated one, the
limit of accuracy shall be maintained beyond the declared one.
The formulae for actual ALF;
ALFACT= ALFRTD 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)2 x
0.15=25 x 0.15=3.75VA,
Sn (Rated burden) = 15VA (From CT data),
Sa(Actual Burden) = 15÷3= 5VA,
ALFACT = ALFRTD x (Sin+Sn(Rated Burden))/( Sin+Sa(Actual
Burden))
ALFACT = 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.
| Is This Answer Correct ? | 48 Yes | 6 No |
Post New Answer View All Answers
How does zener phenomenon differ from avalanche breakdown?
critically criticize the application of ring main unit in power system
how to calculate the tarriff for an industry whose kva demand is 10 mva
what is advertised hop count
how many joint can done in HT line and what is life of the joint
a)In a 3-phase 4-wire circuit, a circuit breaker is selected to protect the circuit whose highest fault level is 6.6 MVA. Determine the required minimum breaking capacity in kA. (Note: Covert 3-phase power in terms of kA to find answer.) b)Using the below, determine the required minimum making capacity in kA as a “peak value”. Rate breaking capacity (Icn) ka: Icn<= 1.5 Factor = 1.41 1.5
what is the circutor?why we use it ?
What are the symptoms of a bad PG card of a drive (inverter) esp. of Yaskawa Ac Drives (inverters) G7 ?
Please sent me some previous year question papers of BSNL-JTO telecom and RRB in ELectrical Engineering to varshashahi@yahoo.com
Star Delta Diagram with full description
There are a trformer and an induction machine. Those two have the same supply. For which device the load current will be maximum? And why?
How to choose the vector group for the step up and step down transformers??? Tell me the which vector group is suitable for step up and step down transformers and whats the reason to choose and whats the benefits???
How should i design a 1ph & 3 ph transformer. Kindly let meknow all steps and formulaes related to design from scratch.
I have one ct of ratio 3000-2000/1 Amp its knee point voltage mentioned on name plate is 500-333/30mA-50mA at Vk/2. in secondary box there is three terminal named 1S1, 1S2, 1S3 then on which terminal obtained the knee point voltage ie 1S1-1S2 OR 1S1-1S3 & also specify the knee point voltage for the ratio 3000 from name plate.
How cmd is calculated in case of energy neters
