**Instrument Transformer and Power Management (P1) Course**

**Chapter (6) : Current Transformers**__6.3.3 Current Transformer Basic Formula__

The secondary output of these devices is the information used by the relays to determine the conditions existing in the plant being protected. It is necessary therefore that, the secondary output of current and voltage represent a true picture for the conditions in the primary circuit during faults as well as during normal condition.

Or, alternatively that their performance be known under extreme conditions so that any error in production in the secondary circuit can be partially or completely compensated for in the setting and characteristic of the relay.

In many applications, core saturation will almost inevitably occur during the transient phase of a heavy short circuit, The performance of the associated instrument transformers during faults is, therefore an important consideration in providing an effective relaying scheme.

The relays and their associated current transformers must be considered as a unit determining the overall performance of the protective scheme , Consequently the characteristic of the current and potential transformers at high current and low voltage, respectively must be known.

EK = 4.44 x B A F N volts = 4.44 x 1.6 A F N volts………….. (I)

Where **EK**= secondary induced vol (r.m.s value, known as the knee point voltage)

N = Number of secondary turns

F = System frequency in hertz

A = Core -cross sectional area in sq. meters

B = Flux density in tesla

The relative circuit voltage required is typically :

Where:

ZB = The connected external burden in ohms.

Zs = The C. T secondary winding impedance in ohms.

ZL = The resistance of any associated connecting leads.

In any given case, several of these quantities are known or can usually be Estimated. in order to predict the performance of the transformers.From the A.C magnetization characteristic, commonly plotted in secondary voltage versus exciting current, Es can be determined for a minimum exciting current.

-Equation (2) then indicates whether the voltage required is adequate.

**-Example :**

Assume that a bar primary type 2000 / 5A ( Cross core) current transformer having a core C.S.A. area of 20 sq.cms. is available with the secondary resistance of 0.31 ohm. the maximum current up to which the transformer must maintain its current ratio is 40,000 amperes. It's required to determine the maximum secondary burden permissible if core saturation is to be avoided.

Assume that the current transformer core will start to saturate at 1.6 tesla.

**-**

**Solution:**

From the data given N = 2000 / 5 = 400 turns

F = 50 HZ.

Secondary current ( Is) with a primary current of 40,000 A is given by:

Is = 40,000 x (5/2000) = 100 amps.

Knee point voltage EK is given form expression (I) as follows :

EK = (4.44 x 1.6 x 20 x 50 x 400) / 104= 284 volts.

Maximum burden permissible ( including C.T. secondary resistance and lead burden) is equal to 284/100 = 2.84 ohms

**.** Consequently the connected burden including that of the pilots can be as high as 2.84-0.31=2.53ohms for negligible saturation in the core. Thus it may be seen that the secondary burden and the maximum available fault current are two important criteria in determining the performance of a given current transformer

**.** A current transformer may operate satisfactorily:

a) At a high primary current where the connected secondary burden is low.

b) At a lower primary current where the secondary burden is high

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