Transcript Auto-Transformer

Why Transformer Rating in kVA?
As seen, Cu loss of transformer depends on current and iron loss on voltage.
Hence, total transformer loss depends on volt-ampere (VA) and not on phase
angle between voltage and current i.e. it is independent of load power factor.
This is why rating of transformers is in kVA and not in kW.
Regulation of a Transformer Or Transformer Regulation
Regulation of a transformer is defined as the difference between the full-load and
no-load secondary terminal voltages expressed as a percentage of the full-load
voltage.
When a transformer is loaded with a constant primary voltage, the secondary
voltage decreases because of its internal resistance and leakage reactance.
Let,
0V2
= secondary terminal voltage at no-load
= E2 = E1K=KV1 because at no-load the impedance drop is negligible.
V2 = secondary terminal voltage on full load
The change in secondary terminal voltage from no-load to full load
is=0V2-V2.
This change divided by 0V2 is known as regulation ‘down’.
If this change is divided by V2 i.e. full-load secondary terminal
voltage, then it is called regulation ‘up’.
V2  V2
0
 % regulation down 
100
0V2
V V
and % regulation up  0 2 2 100
V2
Regulation is usually to be taken as regulation ‘down’.
The lesser this value, the better the transformer, because a good
transformer should be kept secondary terminal voltage as constant as
possible under all conditions of load.
Losses in Transformer
Since a transformer is a static device, there are no friction and windage losses.
Hence, the only losses occurring are:
(a) Core or Iron Loss
(b) Copper Loss
Core or Iron Loss
It includes both hysteresis loss and eddy current loss.
Because the core flux in transformer remains practically constant for all loads the
core loss is practically the same at all loads.
Hysteresis loss: Wh=Bmax1.6fV watt;
Eddy current loss: We=PBmax2f 2t2 watt
Where, V=volume of the core in m3; =Steinmetz hysteresis coefficient;
t=thikness.
These losses are minimized by using steel of high silicon content for the core and
by using very thin laminations.
Iron or core loss is found from the O.C. test.
The input of the transformer when on no-load measures the core loss.
Copper Loss
This loss is due to the ohmic resistance of the transformer
windings.
Total Cu loss=I12R1+I22R2= I12R01=I22R02.
It is clear that Cu loss is proportional to (current)2 or kVA2.
So, Cu loss at half-load is one-fourth [(1/2)2=1/4] of that at full
load.
Cu loss at one-quarter-load is one-sixteen [(1/4)2=1/16] of that at
full load.
Cu loss at five-fourths -load is twenty five by-sixteen
[(5/4)2=25/16] of that at full load.
The value of copper loss is found from the short-circuit test.
Efficiency of a Transformer
The efficiency of a transformer at a particular load and power factor is
defined as the output divided by the input- the two being measured in
the same unit (either watts or kilowatts).
output
Efficiency

Thus the equation of efficiency can be written as:
input
We can write the efficiency equation in terms of output and losses as:
output
output
Efficiency 

output  losses output  Cu Loss  IronLoss
Also, the efficiency equation in terms of input and losses can be
written as:
Efficiency,   input -losses 1 Losses
input
input
Efficiency can be computed by determining core loss from no-load
or open circuit test and Cu loss from the short-circuit test.
Condition for Maximum Efficiency
Iron loss= hysteresis loss+ eddy
Cu loss  I 2R or I 2R Wcu
1 01
2 02
current loss= Wh+We=Wi
Considering primary side, Primary input, =V1I1cos1
I 2R W
V I cos  losses V I cos   I 2R W
1 1 01 i 1 1 01 i
1
 11
 11
V I cos
V I cos
V I cos 
11
1
11
1
11
1
I 2R
W
Differentiating both sides with respect to
1
01
i
 1

V I cos  V I cos  I1, we get
11
1 11
1
W
d   R01 
i
For maximum efficiency, d  0, hence
dI
V cos  V I 2 cos 
dI
1
1
1 11
1
1
W  I 2R or W  I 2R
That means, Cu Loss  Iron Loss
i 1 01
i 2 02
The output current corresponding to maximum efficiency is: I2  Wi / R02
By proper design, it is possible to make the maximum efficiency occur at any
desired load.
All-Day Efficiency
Transformers are frequently provide service 24 hours per day,
although the load is by no means constant over this entire period.
Under such conditions, the transformer must be judges by its all day
efficiency rather than by its full-load efficiency.
All day efficiency (also known as ‘operational efficiency’ or ‘energy
efficiency’) is a ratio of energy output to energy input taken over a 24hr period.
Mathematically, the expression of all-day efficiency can be written as
follows:

 output in kWh (for 24 hours)
allday input in kWh
To find this all-day efficiency, we have to know the load cycle of the
transformer i.e. how much and how long the transformer is loaded
during 24 hours.
Example 14.13 Determine the all-day efficiency of a 10 kVA 2220/220V 60 Hz
transformer. During a day of 24 hours, it is loaded with unity power factor as under
the following condition: 2 hours at five-fourths load; 6 hours at full load; 8 hours at
half load; 4 hours at one-quarter load; 4 hours at no load. Take core loss as 153 W and
full-load copper loss as 224 W.
Solution:
The energy loss due to core loss is= 24153 = 3672 W-hr=3.672 kW-hr
Total energy loss= 3.672 + 2.548 = 6.22 kW-hr
Total energy input= 135.0 + 6.22 = 141.22 kW-hr
 output  135  0.956 or 95.6%
allday ontput 141.22

Example 32.55: Find the all-day efficiency of 500kVA distribution transformer
where copper loss at full load is 4.5kW and iron loss is 3.5kW. During a day of 24
hours, it is loaded as under
Solution: It should be noted that a load of 400 kW at 0.8 pf is equal to
800/0.8=500kVA.
Similarly, 300 kW at 0.75 pf means 300/0.75=400 kVA
and 100 kW at 0.8 pf means 100/0.8=125 kVA i.e. one fourth of the full load.
Cu loss at F.L. of 500 kVA= 4.5 kW
Cu loss at 400 kVA= 4.5(400/500)2=2.88 kW
Cu loss at 125 kVA= 4.5(125/500)2=0.281 kW
The
total
Cu
loss
in
24
hours
=(64.5)+(102.88)+(40.281)+(40)=27+28.8+1.124=56.924 kWh
The iron loss takes place throughout the day irrespective of the load on
the transformer because its primary is energized all the 24 hours.
So, iron loss in 24 hours = 243.5= 84 kWh
Total transformer loss=56.924+84=140.924 kWh
Transformer output in 24 hours= (6400)+(10300)+(4100)= 5800
kWh
output
5800
So, 


 0.976 or 97.6%
allday output  losses 5800 140.924
Auto-Transformer
The transformer with one winding only, part of this being common to
both primary and secondary, is called auto transformer.
In this transformer the primary and secondary are not electrically
isolated from each other as is the case with a 2-winding transformer.
But its theory and operation are similar to those of a two winding
transformer.
Because of one winding, it uses less copper and hence is cheaper.
Fig. 32.60 shows both step-down and step-up auto- transformer.
Fig. 32.60 (a)
Fig. 32.60 (b)
As shown in Fig. 30.60(a), AB is primary winding having N1 turns and BC is
secondary winding having N2 turns.
V N
I
2
2
Neglecting iron losses and no-load current: V  N  I1  K
1
1 2
The current in section CB is vector difference of I2 and I1.
But as the two currents are practically in phase opposition, the resultant current is (I2I1) where I2 is greater than I1.
As compared to an ordinary 2-winding transformer of same output, an autotransformer has higher efficiency but smaller size.
Moreover, its voltage regulation is also superior.
Fig. 32.60 (a)
Fig. 32.60 (b)
Saving of Copper
Volume and hence weight of Cu, is proportional to the length and area of crosssection of the conductors.
Length of conductor is proportional to the number of turns and cross section
depends on current.
Hence, weight is proportional to the product of the current and number of turns.
With reference to Fig. 32.60
Weight of Cu in section AC(N1- N2)I1;
Weight of Cu in section BC(I2- I1)N2;
So, total weight of Cu in auto-transformer (Wa) (N1- N2)I1+(I2- I1)N2;
If a two winding transformer were to perform the same duty, then
Weight of Cu in primaryN1I1;
Weight of Cu in secondaryN2I2;
Total weight of Cu (Wo) N1I1+N2I2;
( N  N )I  N (I  I )
Weight
of
Cu
in
auto
transforme
r
So, Ratio of weight 
 1 2 1 2 2 1
N I N I
Weight of Cu in ordinary- transformer
11 2 2
N I N I N I N I
2N I
2N I
1
1
2
2
2
1
2
1
2
1
21
Ratio of weight 
1
1
N I N I
N I N I
N I N I
11 2 2
11 2 2
11 2 2
N
2 2
N
I
N
2
2 1
2
K


K
;
1
Ratio of weight 1
1
1 K
N
I K
N I
2
1
1
1 2 2
N I
1 1
Weight of Cu in auto- transformer (Wa ) 
(1 K ) Weight of Cu in ordinary- transformer(Wa )
Saving  (Wo -Wa ) Wo  (1 K )Wo  KWo
Saving  K (Weight of Cu in ordinary- transformer)
Hence, saving will increase as K approaches unity.
It can be proved that power transform inductively is= input(1-K)
The rest of the power= (Kinput) is conducted directly from the source
to the load i.e. it is transferred conductively to the load.
Uses of Auto-Transformer
Auto transformers are used:
1. To give small boost to a distribution cable for the
voltage drop.
2. As auto-starter transformers to give up to 50 to 60% of
full voltage to an induction motor during starting.
3. As furnace transformers for getting a convenient supply
to suit the furnace winding from a 230 V supply.
4. As interconnecting transformer in 132 kV/330 kV
system.
5. In control equipment for 1-phase and 3-phase electrical
locomotives.
Conversion of 2-winding Transformer
into Auto-Transformer
Any two-winding transformer can be converted into an autotransformer either step-down or step-up. Fig. 32.62(a) shows such a
transformer with its polarity marking.
If we employ additive polarity between the high-voltage and lowvoltage sides, we get a step-up transformer.
If, however, we use the subtractive polarity, we get a step-down
auto-transformer.
Additive Polarity
Connections for such a polarity are shown in Fig. 32.62(b).
The circuit is re-drawn in Fig. 32.62(c) showing common terminal at
the bottom.
Because of additive polarity, V2=2400+240=2640 V and V1 is 2400 V.
As shown in Fig. 32.62(d), common current flows towards the
common terminal. The transformer acts as a step-up transformer.
Subtracting Polarity
Such a connection is shown in Fig. 32.63(a).
The circuit has been re-drawn with common polarity at top in Fig.
32.63(b) and in Fig. 32.63(c).
In this case, the transformer acts as a step-down auto-transformer.
The common current flows away from the common terminal.
Here V2= 2400-240= 2160 V.
Parallel Operation of Single-Phase Transformer
For supplying a load in excess of the rating of an existing transformer,
a second transformer may be connected in parallel with it as shown in
Fig. 32.75.
The primary windings are connected to the supply bus bars and
secondary winding are connected to the load bus-bars.
In connecting two or more than two transformers in parallel, it is
essential that their terminals of similar polarities are joined to the same
bus bars.
If this is not done, the two emfs
induced in the secondaries which are
paralleled with incorrect polarities,
will act together in the local
secondary circuit even when
supplying no load and will hence
produce the equivalent of a dead
short-circuit.
There are certain definite conditions which must be satisfied in order to
avoid any local circulating currents and to ensure that the transformers
share the common load in proportion to their kVA ratings.
The conditions are:
1. Primary windings of the transformers should be suitable for the
supply system voltage and frequency.
2. The transformer should be properly connected with regard to
polarity.
3. The voltage ratings of both primaries and secondaries should be
identical. In other words, the transformer should have the same turn
ratio i.e. transformation ratio.
4. The percentage impedances should be equal in magnitude and have
the same X/R ratio in order to avoid circulating currents and operation
at different power factors.
5. With transformers having different kVA ratings, the equivalent
impedance should be inversely proportional to the individual kVA
rating if circulating currents are to be avoided.
Three Phase Transformer
Large-scale generation of electric power is usually 3-phase at generated voltage of
11 kV or 13.2 kV or somewhat higher.
Transmission is generally accomplished at higher voltage of 110, 132, 275, 400 and
750 kV for which purpose 3-phase transformers are necessary to step up the
generated voltage to that of the transmission line.
At load centers, the transmission voltages are reduced to distribution voltages of
6.6, 4.6 and 2.3 kV.
At most of the consumers, the distribution voltages are still reduced to utilization
voltages of 440, 220 or 110 volts.
Years ago, it was a common practice to use suitably interconnected three singlephase transformers instead of a single three-phase transformer.
But these days, the latter is gaining popularity because of improvement in design
and manufacture.
As compared to a bank of single-phase transformers, the main advantages of a 3phase transformer are that it occupies less floor space for equal rating, weighs less,
cost about 15% less and further, that only one unit is to be handled and connected.
Like single-phase transformers, the three phase transformers are also of the core
type and shell type.
Three-Phase Transformer Connections
Three-phase power may be transformed by using combinations of single-phase
transformers.
The primaries may be connected in either Y (Wye) or  (Delta), and with either
type of primary connection, secondaries may likewise be connected either way.
Thus, the most common connection are (i) star/star or Y-Y, (ii) -, (iii) Y-, (iv)
-Y, (v) open delta or V-V, and (vi) Scott connection or T-T connection.
Star/Star or Y-Y
This connection is most economical for small, high-voltage
transformers because the number of turns/phase and the amount of
insulation required are minimum (as phase voltage is only 1/3 of
line voltage).
The ratio of line voltages on the primary and secondary sides is the
same as the transformation ratio of each transformer.
However, there is a phase shift of 30o between the phase voltages and
line voltages both on the primary and secondary sides.
This connection works satisfactorily only if the load is balanced.
In Fig. 33.4 is shown a bank of 3 transformers connected in Y on both the primary
and the secondary sides.
With the unbalanced load to the neutral, the neutral point
shifts thereby making the three line-to-neutral (i.e. phase)
voltages unequal.
This difficulty of shifting neutral can be obviated by
connecting the primary neutral (shown dotted in the figure)
back to the generator so that primary coil A can take its
required power from between its line and the neutral.
Another advantage of stabilizing the primary neutral by
connecting it to neutral of the generator is that it eliminates
distortion in the secondary phase voltages.
For delivering a sine wave of voltage, it is necessary to
have a sine wave of flux in the core, but on account of the
characteristic of iron, a sine wave of flux requires a third
harmonic component in the exciting current.
If the primary neutral isolated, the triple frequency current
cannot flow.
Hence, the flux in the core cannot be a sine wave and so the voltages are distorted.
But if the primary neutral earthed i.e. joined to the generator neutral, then this
provides a path for the triple-frequency currents and emfs and the difficulty is
overcome.
Delta-Delta or - Connection
This connection is economical for large, low-voltage transformers in which insulation
problem is not so urgent, because its increases the number of turns/phase.
The transformer connections and voltage triangles are shown in Fig. 33.5.
The ratio of transformation between primary and secondary line voltage is exactly the
same as that of each transformer.
Further, the secondary voltage triangle abc occupies the same relative position as the
primary voltage triangle ABC i.e. there is no angular displacement between the two.
This connection has the following advantages:
1. The third harmonic component of the magnetizing current
can flow in the -connected transformer primaries without
flowing in the line wires. Therefore the flux is sinusoidal
which results in sinusoidal voltages.
2. No difficulty is experienced from the unbalanced loading as
was the case in Y-Y connection. The three-phase voltages
remain practically constant regardless of load imbalance.
3. If one transformer becomes disabled, the system can
continue to operate in open-delta or in V-V although with
reduced available capacity.
Wye-Delta or Y- Connection
The main use of this connection is at the substation end of the
transmission line where the voltage is to be stepped down.
The primary windings is Y-connected with grounded neutral as shown
in Fig. 33.6.
The ratio between the secondary and primary
line voltage is 1/3 times the transformation
ratio of each transformer.
There is a 30o shift between the primary and
secondary line voltages which means that a Y transformer bank cannot be paralleled with
either a Y-Y or a - bank.
Also, third harmonic currents flows in the  to
provide a sinusoidal flux.
Delta-Wye or -Y Connection
This connection is generally employed where it
is necessary to step up the voltage as for
example, at the beginning of high tension
transmission system.
The connection is shown in Fig. 31.7.
In this connection, the primary and secondary
line voltages and line currents are out of phase
with each other by 30o.
Because of this 30o shift, it is impossible to
parallel such a bank with a - or Y-Y bank of
transformers even though the voltage ratios are
correctly adjusted.
The ratio of secondary to primary voltage is 3
time the transformation ratio of each
transformer.
Open-Delta or V-V Connection
If one of the transformers of a - is removed and 3-phase supply is
connected to the primaries as shown in Fig. 33.11, then three equal 3phase voltages will be available at the secondary terminals on no-load.
It is employed:
1. when the three phase load is too small to
warrant the installation of full three-phase
transformer bank.
2. when one of the transformers in a - bank is
disabled, so that service is continued although at
reduced capacity, till the faculty transformer is
repaired or a new one is sustained.
3. when it is anticipated that in future the load
will increase necessitating the closing of open
delta.
As seen from Fig. 33.12(a)
- capacity = 3VL.IL=3VL.( 3IS)=3VLIS
In Fig. 33.12(b), it is obvious that when - bank becomes V-V bank,
the secondary line current IL becomes equal to the secondary phase
current IS.
V-V capacity = 3VL.IL=3VL.IS (Since line current and phase current
are equal)
3V I
V

V
capacity
L S  1  0.577or 58%
Thus,

 -  capacity 3V I
3
L S
It means that the capacity of V-V transformer is 57.7% (or 1/3
times) of the capacity of - transformer.
The disadvantages of V-V connection are:
1. The average power factor at which the V-bank
operates is less than that of the load. This power factor
is actually 86.6 of the balanced load power factor.
2. Secondary terminal voltages tend to become
unbalanced to a great extent when the load is increased,
this happens even when the load is perfectly balanced.