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Medium Voltage Application Guide

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SWITCHGEAR<br />

Exercise 3: Protection CT2 for transformer feeder circuit:<br />

Step1: Calculate transformer TXR2 nominal primary current, In (A)<br />

n<br />

The primary current for TXR2 is 105 A<br />

Step 2: Calculated maximum expected short circuit current at CT2 installation, Isc (A)<br />

Ignoring any power cable or busbar impedances:<br />

I sc<br />

I n<br />

The maximum expected short circuit current at CT2 is 2100 A<br />

Step 3: Select protection CT2 ratings<br />

Primary rated current<br />

Use a rating of 150A<br />

Secondary rated current<br />

Use a rating of 1 A<br />

Short-time withstand rating,<br />

Use a rating of 10 kA<br />

Primary circuit voltage<br />

Use a ratings of 12 kV<br />

I pr = (1.0-1.25) x I n<br />

= (1.0-1.25) x 105<br />

I sr<br />

I th ≥ I sc<br />

U p ≥ U<br />

Real output power: typically > 1 VA for digital type protection relay.<br />

Use 2.5 VA (this allows 1.5 VA for cable burden, etc)<br />

Step 4: Calculate protection class 5PX<br />

The instantaneous trip current level of protection relay OC2 is set to 10 x I n<br />

I_trip = 10 x 105<br />

= 1050 A (primary current)<br />

(Note: In most digital protection relays, the trip current levels are set with respect to the secondary current. In this<br />

case<br />

SEC<br />

The instantaneous trip current level for the CT secondary is 7 A<br />

The trip current level should fall between 100 to 50% of the accuracy limit factor (ALF).<br />

Using an ALF of 10 (5P10), the trip current level of 1050 A falls within the range of 100% to 50% ALF so a 5P10<br />

protection class CT is suitable.<br />

)<br />

)<br />

Use protection class 5P10<br />

710-12280-00A <strong>Medium</strong> <strong>Voltage</strong> <strong>Application</strong> <strong>Guide</strong> Page 117

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