CT burden calculation. How to calculate Current Transformer burden

Before starting the discussion about CT or current transformer burden calculation we should know what is a burden!!!!!!

Burden is something which irritates?!!!! Actually burden is some additional load, which you need to bear.

Just like that the current transformer also need to drive some load which is called burden of CT or current transformer. Now what are the burdens?

For knowing that we need to understand what are the elements that are connected with the current Transformer. A protection class CT or current transformer has relays connected with it. And a metering class CT has mainly ammeter and multifunction meter connected with it. All this instruments has a small amount of power demand which is imposed on the current Transformer as burden of CT.

So basically sum of power demand of the instruments connected with the CT is called burden of CT. 

Now for outdoor switchyard what happened the current Transformer is installed on field. But the relay, MFM, ammeter etc are installed a certain distance away from the CT location, inside a panel. And this distance is connected via cable. So now there is a power loss in the cable, so that loss will also be included in CT burden calculation.

This is called lead burden. However lead burden js applicable mostly for CT installed at out door. Now,

Single line diagram of metering and protection https://electricaltechnologyrishi.blogspot.com

Let us consider this picture. Here we can see a an Ammeter, a Tri vector meter and an MFM is connected in series with the current Transformer. So their VA rating will be added to the burden of CT.

Now from a typical catalogue of MFM and ammeter we have found that

Ammeter has a burden of around 0.9 VA.

MFM has burden of 0.5 VA (MFM  has two burden, one is on CT and the other one is on PT)

And TVM has a burden of say 0.6 VA.

So their total VA demand=2 VA.

Now we will calculate the lead burden. It is assumed that CTs are outdoor type, and they are connected with the meters with cable. Now we will find out the power loss in the cable.

Three line diagram of metering connection to CT https://electricaltechnologyrishi.blogspot.com

So from the above diagram we can see that each CT is connected to the meter. So for completing the operating circuit(i.e. you can see that as per the ammeter selector switch position two leads will be connected between CT and meters, so we need to consider two cable) two nos. of cable runs are required, so we need to calculate the power loss for two runs. Generally 2.5 sq. mm Cu cables are used for such connection, so we will also consider the same.

2.5 sq mm Cu cable has resistance of 9.8 Ohm/ km.

Let's say the metering panels and CTs are 100 m away so we need to calculate resistance of 200 m cable. 200 is considered for to and fro run.

So total resistance will be 9.8*200/1000=1.96 Ohm

Say CT secondary rating is 1A, so total power loss=I^2*R=1*1*1.96=1.96 VA.

So, total burden will be 2+1.96=3.96 VA

So CT selected VA should be minimum 5 VA, however considerable margins are kept to burden selection.

Say the burden is selected as 15 VA.

Now let us see the condition for protection CT.

Relay connection to the current Transformer https://electricaltechnologyrishi.blogspot.com

From the diagram we can see the connection of the Relay with the CT. Here also for calculating the lead burden our consideration will similar as that for protection.

Since the relay, meter all are housed in a same panel so lead resistance will be same for both the case.

Numerical relay has burden of around 0.2 VA. So total burden for protection CT is 1.96+0.2= 2.16 VA.

Let us select 15 VA CT. Now Selection of such higher value of CT burden will give us another advantage. We will see that.

Suppose our selected CT is 5P20 class CT, i.e. its Accuracy limit factor is 20. Now if the VA is higher the actual ALF will increase.

The corrected or actual accuracy limit factor or ALF can be found as following

ALF'=ALF*(Pi+Pn)/(Pi+Pr)

Where 

ALF'= The actual accuracy limit factor

ALF= Rated accuracy limit factor

Pi=CT internal I^2*R loss

Pn=Selected CT burden

Pr=Actual burden on CT.

Out CT is 1A secondary rated, and CT resistance shall be maximum 5 ohm (Rct<=5 ohm).

So, Pi=5VA

Pn= say 15

Pr= 2.16 VA (burden of numerical relay and cable)

And the CT is a 5P20 CT so ALF=20.

So, ALF'= 55.86

So our 5P20 CT will act like 5P56.

Its accuracy limit factor will increase. So thus the actual ALF of CT increases with higher burden selection.

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