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Transformer Secondary BCT:
What's Useful and Important to You




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Transformer Bushing CT Secondary Current Calculator



Given transformer data 30 /40 / 50 MVA Single Phase rating at 115 / 32 kV with a CT 1200 / 5 A at secondary side.

Solve for secondary Ampere and CT output current at 30 MVA , 40 MVA, and
50 MVA. Can you still use the same CT 1200/5 A for 40 and 50 MVA explain your answer.



Valid for 1 Φ single phase only

Secondary Ampere
=
1-Phase MVA

kV secondary voltage


Assuming for discussion purposes the CT specs is 20 times the rated nominal secondary CT output for proper operation of protection class CT. What will happen to your measurement accuracy if the secondary is 62.5 times higher than the acceptable instrument accuracy? Will your relay operate properly? Explain why it will work or why not? What is the acceptable CT saturation multiplier to make sure protection relays will operate properly?

A ,CT Secondary Rating

A, CT Primary Rating
=
A, CT secondary measured

A, XF secondary measured


When there is a short circuit the secondary voltage goes down very low. At that instant the current is approaching its maximum value. For example, at 30 MVA when the secondary voltage drop to 2 kV due to short circuit, the secondary ampere current will be 15,000. Try it.

When you download the transformer differential relay or breaker relay data you will see close to 62.5 A primary ampere reading during fault which is 62.5 A x 240 the CT ratio equals to 15,000 ampere.
CT ratio 1200/5 = 240 ; CT ratio 2000/5 = 400



3 Phase Transformer CT Output Calculator

Same Problem as above but now it is 3 phase rating:
Given transformer data 320 base MVA three phase rating at 525 / 230 kV with a CT 2000 / 5 A at primary winding side.

Using the base 320 MVA solve for primary ampere and CT output current . Assuming a voltage dip of 50 kV at the primary side. Estimate the fault current at the primary side using only the 50 kV voltage dip recorded by your Pi historian or differential relay data.

Assuming the transformer % Z ( impedance) = 6.9 % . What is the maximum available short circuit current the transformer was designed to withstand? if the voltage dip goes down to 12 kV , what is the primary short circuit current? What do you think will happen to the transformer at this fault level?



Secondary ampere per phase and secondary voltage were shown. To calculate the primary ampere use primary voltage.

Valid for 3 Φ three phase only

Secondary Ampere
=
3 Phase MVA

secondary kV * 1.732




A ,CT Secondary Rating

A, CT primary Rating
=
A, CT secondary measured

A, XF secondary measured


CT Saturation discussion
Credit to: Doug Millner P.E.


Beyond the Knee Point:
A Practical Guide to CT Saturation




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Approximately between 2.3 and 2.5 million schools globally, according to the latest available data from government and education ministry reports.


➡️ USA ~ Public 98,500 ~ Private 30,000 ~ Total 128,500 ➡️ Canada ~ Public 15,500 ~ Private 2,000 ~ Total 17,500 ➡️ Brazil ~ Public 138,000 ~ Private 40,000 ~ Total 178,000 ➡️ Vietnam ~ Public 42,000 ~ Private 8,000 ~ Total 50,000 ➡️ China ~ Public 217,200 ~ Private 152,800 ~ Total 470,000 ➡️ India ~ Public 1,022,386 ~ Private 335,844 ~ Total 1,358,230 ➡️ Japan ~ Public 30,240 ~ Private included ~ Total 30,240 ➡️ Morocco ~ Public 20,600 ~ Private 6,300 ~ Total 26,900 ➡️ Indonesia ~ Public 390,718 ~ Private included ~ Total 390,718 ➡️ Philippines ~ Public 47,831 ~ Private 13,000 ~ Total 60,831 Great Britain ~ Public 29,202 ~ Private included ~ Total 29,202 ➡️ Australia ~ Public 9,653 ~ Private included ~ Total 9,653 ➡️ Russia ~ Public 39,070 ~ Private included ~ Total 39,070 ➡️ Germany ~ Public 31,039 ~ Private included ~ Total 31,039 ➡️ Poland ~ Public 36,291 ~ Private included ~ Total 36,291 ➡️ Iran ~ Public 80,000 ~ Private included ~ Total 80,000 ➡️ France ~ Public 58,100 ~ Private included ~ Total 58,100 ➡️ Mexico ~ Public 132,505 ~ Private included ~ Total 132,505

Use an estimated range of 200 to 400 students per school if student enrollment is the only available data.


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