Guide Line to Size a Current Transformer from Siemens. This is good presentation of the CT applications and CT sizing guideline.
Guide Line to Size a Current Transformer from Siemens. This is good presentation of the CT applications and CT sizing guideline.Full description
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CT Sizing XLS fileFull description
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CT Sizing CalculationsFull description
CT
CT Sizing Sizing Calculatio n
Feeder Type :
CT use used d:
Typical Feeder/Incomer Feeder/Incomer with L ine Differential protection
500/ 50 0/1A 1A 15 15VA VA,, 5P2 5P20 0 ,Fe ,Feede ederr Diff Diff and and OC OC Protn Protn -wit -with h SIEM SIEMEN ENS S 7SD 7SD52 52
Calculation SYSTEM PARAMETERS: Norminal voltage (UN) :
33.0 kV
Norminal frequency (f N) :
50.0 Hz
Rated max. symmetrical short circuit current (I SC) :
31.5 kA
CT PARAMETERS: CT Type : Primary norminal current (I pn) :
IEC Class 5P 500 A
Secondary norminal current (I sn) :
Norminal accuracy limit factor (K nalf ) :
20
Internal resistance (R ct) :
Norminal burden (P n) :
15 VA
Internal burden (P ct) : Knee Voltage achieved (Vk'>=)
RELAY DATA : Manufacturer: SIEMENS
Type: 7SD52
Norminal current (I n): 1 A
1 A 6.00
6.0 VA 375 V
Relay burden (Prelay): 0.05 VA
CALCULATION OF TOTAL CT BURDEN : The Th e ct be beha havi viou ourr de en ends ds on th the e to tota tall co conn nnec ecte ted d bu burd rden en on se seco cond ndar ar si side de.. The total burden consists of the relay s elf-burden, additional burden of further connected relays and equipment and the wire burden: Relay burden (P relay) :
0.05 VA
Additional relay burden burden (Padd) :
VA
Wire burden : The effective wire burden will be calculated for a two wire connection of ct and relay according to length, cross-section, copper resistance and sec. norminal current: 2.50 mm 2 Ohm/ m/m m 0.00741 Oh 0.00828 Ohm/m 10 m
Cross section (a wire) : Resist Resi stan ance ce of wi wire re at 20 ºC ºC,, (R (Rw2 w20) 0) = Resistance of wire at 50 ºC, (Rw50) = Length (lwire) : 2
P wire = k wire x l wire x I sn x
Rw50
=
50 Operating Temperature ºC = Resist Resi stan ance ce Rw Rw50 50 = Rw20 (1+Tdiff) = Numerical temperature co-efficient for Cu (0.00393) 2 Eff. W ire runs in i n p.u. (kwire) :
0.17 VA
The total effective burden is given by the s um of all connected burden : Ptotal = Prelay + Padd + Pwire 0.22 VA =
CT Sizing calculation _Protection_21.12.2010_Rev0
CT Sizing Calculatio n
CT REQUIREMENTS FOR 7SD61 acc. to SIEMENS Numerical Protection Relays Catalog Kssc = K oalf K td x I scc / I pn
where
(Requirement 1)
Ipn :
CT primary norminal current
=
500 A
Iscc :
Max. outflowing short-circuit current for external faults
=
31.50 kA
=
1.2
Koalf : K td
Operating accuracy limiting factor :
K oalf (=K' SSC)
Rated transient dimensioning factor
CHECK OF CT REQUIREMENTS : - Operating accuracy limiting factor K oalf : The CT behaviour acc. to IEC-P will be described by the operating accuracy limiting factor K oalf . It depends on the nominal accuracy limiting factor K nalf , the nominal burden P n, the internal CT burden P ct and the total effective connected burden Ptotal : K oalf = K nalf P n + P ct
CT-DIMENSIONING CHECK : - Check of requirement 1 : Kssc = K oalf K td x I scc / I pn
K oalf (=K' SSC) Iscc /Ipn
= =
67.6 63.0
The value of Issc must be considered with the actual parameters of the devices that are connected. In this case the protection limit of the relay will be 20x the input and hence this must be the limit to set the Iscc/Ipn The relay settings shall be done according to this principle. This means that the value of Inst S/C that will be set will approximately lie in the range of 20 tim es Ipn and not more than this value. K td x I scc / I pn
