Medium Voltage Application Guide
Medium Voltage Application Guide
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