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QUESTION: How do I apply the T60 differential protection ... · QUESTION: How do I apply the T60 differential protection element to provide efficient transformer protection? Differential

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Page 1: QUESTION: How do I apply the T60 differential protection ... · QUESTION: How do I apply the T60 differential protection element to provide efficient transformer protection? Differential

QUESTION: How do I apply the T60 differential protection element to provide efficient transformer protection?

Differential protection is the primary transformer protection, and therefore correct configuration of the T60 Percent Differential is important.

The power system, transformer, and CTs all influence the application. The T60 Percent Differential protection has trip/restrain characteristic composed by

setting a pick-up, two slopes and and associated breakpoints. The characteristic in the diagram below is defined by the following settings:

� Minimum PKP, pu � Slope 1 , %

� Break 1, pu � Break 2, pu

� Slope 2, %

Id, pu

Ir, pu

min PKP

Break 1 Break 2

Slope1

Slope 2

It is important for the user to know the meaning of the percentage settings of Slope1 and

2, and in the per unit values of the Min PKP, Break 1 and 2.

A T60 Percent Differential element per unit value refers to setting scaled to the identified UNIT CT.

The UNIT CT is the smallest ratio of CT primary rated-, to nominal winding-currents, for the transformer windings and their respective CT’s:

For example the transformer: Dy30, 100MVA, 220kV/69kV, has CT1(500:1) on the

Delta winding, and CT2(1000:1) on Wye winding.

AmpskVxMVAIDeltanom

43.262)2203/(100)( ==

AmpskVxMVAI Wyenom 73.836)693/(100)( ==

Dividing the primary current rating with nominal current:

9.143.262/500/1 )( ==Deltanom

ICT and,

195.173.836/1000/2 )( ==WyenomICT

CT2’s calculated ratio is lower and it is therefore the unit CT value is 1000 A.

Page 2: QUESTION: How do I apply the T60 differential protection ... · QUESTION: How do I apply the T60 differential protection element to provide efficient transformer protection? Differential

Now we can elaborate on the per unit settings:

For this example a Minimum Pickup setting of 0.1 pu is equal to 0.1x1000 = 100 A differential current. This setting should be larger then the transformer magnetizing

current and the steady state CT errors during no load conditions.

The Break 1 setting is based on the previously defined pu value of the full load transformer current. The information for the different MVA transformer loads, can be

obtained from the transformer’s nameplate.

The Break 2 setting is based on the saturation limit for each winding CT during external faults. The Break 2 is the minimum pu current causing CT saturation, for all CT’s.

Slope 1 and Slope 2 express the slope of the operating characteristic as a function of

differential current (Id) and restraint current (Ir)

)(2)(1 compcompiiIdvv

+=

),max( )(2)(1 compcompiiIrv

=

Slope = puIr

Id%,100*

Slope 1 is the slope setting from pickup to Break 1, and should is based on CT errors

during normal load currents and from tap changes. Slope 2 identifies the slope where the CT saturation are likely and maximum restraint is

required. A recommended setting is 98%, which gives provided stability if one CT is partly saturated during an external fault. It implies that a large differential current is

required for an differential operation .

--- end of public record ----

Author: Lubo Sevov

Date: June 25, 2001

Products: T60

Keywords: T60, transformer differential protection, setting calculations

Reviewed by: Mara Scutelnicu

Reviewed on: July 5, 2001

Approved by: Bogdan Kasztenny

Approved on: July 5, 2001