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CT Selection Calculation - Line

This document provides a calculation for selecting a current transformer for a line, transformer or bus. It details the system configuration, equipment ratings, fault current information, and performs checks on the CT ratio to ensure sensitivity, continuous rating and saturation limits are met.
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0% found this document useful (0 votes)
117 views7 pages

CT Selection Calculation - Line

This document provides a calculation for selecting a current transformer for a line, transformer or bus. It details the system configuration, equipment ratings, fault current information, and performs checks on the CT ratio to ensure sensitivity, continuous rating and saturation limits are met.
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Current Transformer Selection Calculation

Prepared by J. Acharya

1.0 Scope
This calculation is intended for checking proposed Current Transformer (CT) for the Line, Transformer or
Bus. One set of calculations can cover a CT or a group of CTs with identical ratio, accuracy and burden.

2.0 Power System Configuration and Equipment

Nominal system voltage (L-L) VRated  Vnom  VRated kV  400  kV


(at CT location)
6
MVA  10  V A

LineOrTransformer 
- Choose if this study is for Line or Transformer.

2.1 Maximum Load Limit

Transformer: Maximum Capacity in MVA

- For transformers, maximum limit is the highest cooled rating.


TrafoMVA  240MVA - Enter 1.5 times the maximum rating. E.g.160MVA x 1.5=240MVA

TrafoMVA
ITrafo   346.41 A
 3  Vnom 
Line: Maximum operating limit of conductor Ampere rating (e.g. rated at 85 or 100 deg C)

At conductor temp of 85 deg C and ambient temp of 40 deg C


ICond  875A When maximum loading is limited by other factors than conductor ampacity
rating, enter the corresponding maximum desinged Ampere rating of line.
NumBund  4

IConductor  ICond NumBund  3500 A

Maximum loading limis

IMAX   IConductor if LineOrTransformer = "Line"   3500 A


 
 ITrafo otherwise 

CT Check Dr. J Acharya Page 1 of 7


Bus continuous rating:

It is desirable to meet or exceed bus capacity to allow ultimate load growth. For radial and ring bus, it is
the main bus and for 1-1/2 breaker scheme, use diameter bus (not main bus).

BusRating  4000A

3.0 CT Class and Burden


4000
CT ratio selected: CTR   4000
1

4000
Full winding (max) CT ratio: CTRFull   4000
1

CT Class 
CT Class (at max ratio):
CT V_rated  CT Class V  800 V

CT continuous thermal CT RF 
Verify RF data from CT nameplate
current rating factor, RF:

CT secondary Rct  0.0012 CTR ohm  4.8 Ω


winding resistance:
CT test results are preferred.
If test report not available, assume 0.0012 Ohm/turn at 75°C

Cable length is generally short for line CTs in GIS


One-way CT cable length: LLead  50m sustation and longer for outdoor transformers.

AWG  Cable sized used later to calculate its resistance.


CT cable size:
AWG (Americal Wire Gauage) is the US standard
measure of electrical conductors size.

CT remanance factor assumed: FluxResidual  0% Residual flux can be range from zero to as high as
80%. In reality, 0-25% is assumed.

4.0 Fault current information

The fault currents are based on the fault studies using the updated short circuit model. For present fault
levels, use the scenario with maximum fault current and for future use the 20-yr outlook model or apply
25% growth factor.

Note: Always use 3I0 for the LG fault current .

4.1 Maximum Fault Current Flow through CT

Apply a close-in fault and record the fault current contribution through the CT being studied.
Conservatively, substation bus faults can be used if close-in fault results not available.

CT Check Dr. J Acharya Page 2 of 7


3PH fault: I3PH.max  51026A XtoR3PH.max  16

1LG fault: I1LG.max  43310A XtoR1LG.max  15

4.2 Minimum Fault Current through the CT that the connected relay can detect

First REMOVE the strongest source by taking the highest fault contributing line or tie-transformer
out of service and simulate faults as below.

- For line relays, apply remote line-end fault with remote terminal open and suitable fault resistance.
- For bus relays, apply a bus fault. For transformer, apply fault on opposite bus of study side.

L-L fault (smaller of ILL.min  17551A


faulted phase current):
For Line only, use LG with fault
I1LG.min  11401A resistance (5ohm-100ohm)
1LG fault (3I0 current):

4.3 Ultimate Fault Current through the CT

Ultimate (10-yr or 20-yr outlook) fault levels from the Short Circuit Study Report are preferred. If
ultimate fault levels are not available, assume 25% higher than present fault levels.

Ultimate 
Select whether ultimate fault levels are available:

If ultimate fault levels are provided, enter the values here: IUlt.3PH  63000A IUlt.1LG  63000A

I3PH.max.ult   IUlt.3PH if Ultimate = "Yes"   63000 A


 
 1.25I3PH.max otherwise 

I1LG.max.ult   IUlt.1LG if Ultimate = "Yes"   63000 A


 
 1.25I1LG.max otherwise 

For simplicity, same X/R ratio will be used for short-term and ultimate fault levels.

5.0 CT Ratio Check


5.1 CT Sensitivity Check

Imin  min ILL.min I1LG.min  11401 A

Imin
Imin.sec   2.85 A
CTR

CT Check Dr. J Acharya Page 3 of 7


A minimum of (0.1 x Inominal) is needed for a relay to detect faults. Therefore, the minimum fault
current must exceed 0.1 Amps secondary.

ResultMin  "Relay cannot detect minimum fault. Decrease CT ratio." if Imin.sec  0.1A
"CT ratio is ok" otherwise

ResultMin  "CT ratio is ok"

5.2 CT Continuous Rating Check

In order to avoid overloading the CT, the CT ratio must be selected such that its continous current
rating (including its rating factor) exceeds the maximum load.

Maximum load current (as calculated above):

IMAX  3500 A

BusRating  4000A

ICONT  max IMAX BusRating  4000 A

ResultCONT  "CT continuous rating is too low. Increase CT ratio." if CTR CTRF 1  A  ICONT
"CT ratio is ok" otherwise

ResultCONT  "CT ratio is ok"

6.0 Present CT Saturation Check


6.1 CT Ratio Check

The CT ratio shall be selected such that the ideal CT secondary fault current is less than 15 times
rated secondary current. Ultimate is 20 x rated CT secondary current.

IFault.Max  max I3PH.max I1LG.max  51026 A

IFault.Max
IRelay.Max   12.76 A
CTR

ResultCTR.Present  "Secondary fault levels are too high. Increase CT ratio." if IRelay.Max  15A
"CT ratio is ok" otherwise
ResultCTR.Present  "CT ratio is ok"

CT Check Dr. J Acharya Page 4 of 7


6.2 CT Saturation Check

Calculate effective CT voltage rating:


CTR
CT Vstd_eff   1  FluxResidual  CT V_rated  800 V
CTRFull

Calculated CT secondary circuit burden:

LLead 0.232 AWG  2.32


Rlead  e  ohm  0.164 Ω
1000ft

0.1
Rrly   ohm  0.1 Ω Example: Relay burden (0.1VA @ 1A) per SEL-421 Datasheet
2
1

6.2.1 3-Phase Fault

Calculated CT secondary burden RB.3PG  Rct  Rlead  Rrly  5.064 Ω


for a 3-phase fault:
Maximum fault current and X/R ratio I3PH.max  51026  A XtoR3PH.max  16

CT secondary voltage required to avoid saturation:

I3PH.max
Vmax_eff_3PH   1  XtoR3PH.max   RB.3PG  1098.19 V
CTR

6.2.2 Single-Phase-to-Ground Fault

Calculated CT secondary burden RB.1PG  Rct  2Rlead  Rrly  5.228 Ω


for single phase fault:

Maximum fault current and X/R ratio: I1LG.max  43310  A XtoR1LG.max  15

CT secondary voltage required to avoid saturation:

I1LG.max
Vmax_eff_1LG   1  XtoR1LG.max   RB.1PG  905.71 V
CTR

6.2.3 Result
Vmax_eff  max Vmax_eff_3PH Vmax_eff_1LG  1098.19 V

ResultSat.Present  "CT saturates. Increase CT ratio and/or class." if Vmax_eff  CT Vstd_eff


"CT ratio and class are ok" otherwise

ResultSat.Present  "CT saturates. Increase CT ratio and/or class."

CT Check Dr. J Acharya Page 5 of 7


7.0 Future ultimate CT Performance Check
7.1 CT Ratio Check
The CT ratio shall be selected such that the ideal CT secondary fault current is less than 20 times
the rated current (i.e. 20 A for 1A CT and 100A for 5A CT.

IFault.Max.Ult  max I3PH.max.ult I1LG.max.ult  63000 A

IFault.Max.Ult
IRelay.Max.Ult   15.75 A
CTR

ResultCTR.Ult  "Secondary fault levels are too high. Increase CT ratio." if IRelay.Max.Ult  20A
"CT ratio is ok" otherwise

ResultCTR.Ult  "CT ratio is ok"

7.2 CT Saturation Check

Effective CT voltage rating. CT Vstd_eff  800 V


CT secondary circuit burden: Rlead  0.164 Ω
Relay burden: Rrly  0.1 Ω

7.2.1 3-Phase Fault


Calculated CT burden for 3PH fault: RB.3PG  5.064 Ω

Maximum fault current and X/R ratio: I3PH.max.ult  63000  A XtoR3PH.max  16

CT secondary voltage required to avoid saturation:

I3PH.max.ult
Vmax_eff_3PH.ult   1  XtoR3PH.max   RB.3PG  1355.9 V
CTR
7.2.2 Single-Phase-to-Ground Fault

Calculated CT burden for LG fault: RB.1PG  5.228 Ω

Maximum fault current and X/R ratio: I1LG.max.ult  63000  A XtoR1LG.max  15

CT secondary voltage required to avoid saturation:


I1LG.max.ult
Vmax_eff_1LG.ult   1  XtoR1LG.max   RB.1PG  1317.48 V
CTR
7.2.3 Result
Vmax_eff.ult  max Vmax_eff_3PH.ult Vmax_eff_1LG.ult  1355.9 V

ResultSat.Ult  "CT saturates. Increase CT ratio and/or class." if Vmax_eff.ult  CTVstd_eff


"CT ratio and class are ok" otherwise
ResultSat.Ult  "CT saturates. Increase CT ratio and/or class."

CT Check Dr. J Acharya Page 6 of 7


8.0 Summary of Results

CT Ratio

Sensitivity Check: ResultMin  "CT ratio is ok"

Continuous Current
Rating Check: ResultCONT  "CT ratio is ok"

CT Saturation - Present Fault Levels

Ratio Check: ResultCTR.Present  "CT ratio is ok"

Saturation Check: ResultSat.Present  "CT saturates. Increase CT ratio and/or class."

CT Saturation - Ultimate Fault Levels

Ratio Check: ResultCTR.Ult  "CT ratio is ok"

Saturation Check: ResultSat.Ult  "CT saturates. Increase CT ratio and/or class."

Some digital relays have algorithms for detecting CT saturation during external faults. In these cases,
moderate CT saturation may be acceptable if the CT performs long enough for the algorithm to operate.

If above calculations indicate potential CT saturation, further investigation is recommended. The IEEE
PSRC's CT Saturation Calculator (Excel file) can be used as a tool to estimate the time the when CT
begins to saturate after fault.

CT Check Dr. J Acharya Page 7 of 7

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