The model of the Transformer for Electrical Power Systems
Analysis
Transformer Equivalent Circuit
Assumptions:
Assuming infinite conductivity and permeability
Hence no resistance and no flux leakage.
As a result of infinite conductivity, no resistance, hence
V1 = N1 dΦ/dt
V2 = N2 dΦ/dt
V3 = N3 dΦ/dt
Since all Φ are same and there is no leakage due
to infinit permeability
V1/V2 =N1/N2
V2/V3 = N2/N3
V3/V1 = N3/N1
This means that
V1 = (N1/N2)V2
V2 = (N2/N3)V3
V3 = (N3/N1)V1
Information on how the coils are wound is usually not given,
hence, polarity marks (dots) are provided
From ampere’s law
1
∫
H.dl = ienclosed
1
∫
H.dl = ienclosed
1
∫
H.dl = ienclosed
If the core center line is chosen as
the enclosed path, the magnetic field intensity is zero everywhere because of
infinite core permeability.
0 = ienclosed
0 = N1i1 + N2i2
+ N3i3
N1I1 + N2I2 + N3I3
= 0
S = V1I*1 + V2I*2
+ V3I*3
= V1I*1 + (N2/N1)V1I*2
+ (N3/N1)V1I*3
= V1/N1[N1I1 + N2I2
+ N3I3]*
=0
V2base =(N2/N1)V1base
V3base =(N3/N1)V1base
S1base = S2base = S3base =
S1Φbase
I1base = S1Φbase/V1base
I2base = S1Φbase/V2base
I3base = S1Φbase/V3base
I2base = N1/N2I1base
I3base = N1/N3I1base
V1 = (N1/N2)V2
V1/V1base= ((N1/N2)V2)/V1base
Hence
V1pu = V2pu
V1/V1base= ((N1/N3)V3)/(N1/N3)V3base
V1pu = V3pu
V1pu = V2pu =V3pu {*}
I1 + (N2/N1)I2 +
(N3/N1)I3 =0
I1 /I1base + ((N2/N1)I2)/I1base+
((N3/N1)I3)/I1base=0
I1 /I1base + ((N2/N1)I2)/(N2/N1)I2base+
((N3/N1)I3)/()N3/N1)I3base=0
I1pu + I2pu + I3pu = 0
{**}
Eqns {*} {**} can be used to construct an equivalent circuit
 |
NAME: ONOH UGOCHUKWU.K
ReplyDeleteREG NO:ESUT/2009/102228
DEPT: ELECTRICAL/ELECTRONICS ENGR
AUTO TRANSFORMER
An auto transformer differs form other transformers because the winding is electrically connected and as well coupled by a mutual flux.a variac is an auto transformer used to liquidate the voltage applied to a particular device from 0 to a certain maximum value Vmax.
1: Variacs are used in starting induction motors rating more than 5hp
2: it is used to limit high search of current
There are two practical to prove transformer parameters
A: open circuit test
B: short circuit test
OPEN CIRCUIT TEST
The high voltage winding is on open circuit while the low voltage winding is connected to a variable voltage supply.at normal frequency the primary is assumed to be in the low voltage winding...
SHORT CIRCUIT TEST
The low voltage winding is short circuited through an ammeter and the high voltage winding is connected to variable voltage supply.at normal frequency the high voltage winding is assumed to be primary winding..
EFFICIENCY
Efficiency=output/input
The loss in a transformer are limitd to the following.
1: copper loss in resistance of windings which is variable with load current
2: di-electric losses in the insulations this is the insilating materials is applicable only in the high voltage transformers
NAME:EZEGBO OKECHUKWU.O
ReplyDeleteREG: ESUT/2009/106468
DEPT: ELECTRICAL/ELECTRONICS ENGR
TRANSFORMER
A transformer is a device which consists of two or more coils placed so that they are linked by the same flux . Transformers are used to step up or step down voltage.transformers are useful in poWer system transmission and distribution
THE IDEAL TRANSFORMER
Properties:
-Permeability of the core
-The flux varies sinsodially in the core with infinit permeability of the core
-The resistance of the winding is 0
Since the winding resistance is zero,the voltage e induced in each winding by the changing flux is also a terminal voltage v of the winding
V1=e1=n1 do/dt
V2=e2=n2 do/dt
Where o is the instantanous value of the flux ni,n2 is number of turns 1 and 2