Transcript Switchgear & Protection

Combined Selective Systems :-
Combined Selective Systems :-
Double Ended Unit Substations :-
One-Line Diagrams:Introduction
Chapter 3 (protection of industrial power system)
Lecture 3
Chapter 3 (protection of industrial power system)
Current and voltage Transformers:Current Transformer:
It is a device produces a primary current at a reduced level . A current
transformer (CT) designed for measuring purposes operates over a range of
current up to a specific rated value (circuit normal rating ) and it has a
specified errors at that value . On the other hand a protection CT is
required to operate over a range of current many times the circuit rating .
Under such condition, the flax density corresponds to advanced saturation
and the response during this and the initial transient period of short circuit
current is important .
Thus the method specifying measuring CT is not necessarily satisfactory for
those for protection , and a deep detail of knowledge of CT operation is
required to predict the performance of the protection
CT's have two important qualities:
1-CTs reduce primary current to a value which can be carried by a
small cross sectional area cable to suit protection and measuring
circuit.
2-They provide insulation berried for protection used for the high
voltage equipment.
Construction:
Fig. 3.1 Construction of a current transformer
Design:
* CTs based on the normal transformer e.m.f equation where the
average induced voltage is equal to the product of the number of
turns and the rate of change of magnetic flux (Ф)=F/R. The normal
design citation is to limit the flux to the value where saturation
commencer (known as the knee point flux) and therefore it is the
maximum value of the magnetizing current which produces this flux.
*Magnetizing current charges from zero to maximum in 1/4 cycle and
therefore the rate of change of flux is
{(Ф-0)/(1/4)}=4 Ф web/cycle
* Or at a frequency of cycle/sec
Flux=4 Фf web/sec
But Li=N Ф=Var
* Thus the induced voltage
Var= 4 ФfN
where
N=Number of turns
At knee point voltage
Vr.m.s = 1.11 Var
Vr.m.s= 1.11*4 Ф f N = 4.44 Ф f N
* In terms of flux density
Vr.m.s = 4.44 f N B*A
Example:
The flux density of electrical sheet steel is about 1.5 tesla (web/m^2) at
knee-point of CT ratio 300/1 with core area of 40*30mm^2. Determine the
knee point flux.
Solution:
B= Ф/A or Ф=B*A = 1.5*40*30*10^-6=.0018 web on f=50Hz
The knee point voltage=4.44*0.0018*300*50=120v
Short time factor
Short time factor :
CTs which are used in power system are subjected to fault currents
many times their rated current, and therefore, they must be able to
withstand the effect of this current for short time. The maximum
current which it can carry without mechanical and thermal damage is
expressed as multiple of its rated short time factor.
Burden :
The load of current transformer is called the burden and can be
expressed either as a VA load or as impedance.
Burden
For example :
A 5 VA burden on a 1A current transformer would have an impedance :
Or on a 5A CT
A ring – type CT as shown in fig ( 3.2 ) is of secondary winding
resistance (RL) , magnetizing current (Ie) and resistance (Rb) and
reactance ( Xb) burden . The primary turns must be equal the
secondary and magnetizing Ampere- turns, so, N1 I1 = N2 ( I2 + Ie)
In practice Ie very small compared with I2 and is ignored in all CT
calculation with the except of those concerned with ratio and phase
angle error. By ohm law
V2 = I2 (R2 + Rb + jxb)
The ratio error
The ratio error :
which is the difference in magnitude of I1 & I2, and Ө
which the phase angle error are apparent as shown in Fig 3.3.
Magnetizing characteristic for 100/1 A CT is shown in Fig 3.4 and it is
measured before that Ie is small compared to I2 up to and beyond the kneepoint of the characteristic.
Hence the ratio and phase angle errors will also be small.
Fig. 3.4 CT magnetizing characteristic
The ratio error
This means that the primary / secondary current relationship will be
maintained to this point.
These If I2 (R2 + Rb + jXb) = 120 V = V knee
Rb + jXb = 7.5 + j0 Ω
And if R2 = 1 Ω
Then linearity would be maintained up to a secondary current of
I2 = 120 / 8.5 = 14.1 A *
CT rating
CT rating :
Alternatively, if linearity is required up to 20 * CT rating, then the total
impedance should not exceed
R2 + Rb + jXb = 120 / (20 * 1) =6 Ω
For example:
CT of ratio 200/5, which is capable of withstanding 13000A would have
a short time factor of 65 such factor would be associated with a
period of duration of the current ( 3 seconds ). Smaller current would
be permissible for long time.
Accuracy Limit Factor
* CT used for protection should maintain its characteristic ratio up to
some multiple of its rated current. This multiple could be 10, 20, or
higher and it is known as
(Accuracy Limit Factor).
In case of measuring CTs, the small ratio error is often compensated
by modifying the ratio of the primary to secondary turns from the
nominal ratio. For example a 100/1A CT might have one primary
turn and 98 secondary turns, so the transformation ratio would be
100 to
* But it is used to supply protective relay, more burden is applied and
therefore the secondary current is reduced to 1 Amp by the
magnetizing losses.
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