Transformers
Lecture 6
Transformers utilization
As engineers, you should have seen some transformers while you driving or
walking by some substations near power plants, industrial cities or closer to your
houses!
Practical Transformer
Contrary to an ideal transformer, a practical transformer has
• Winding resistance
• Flux leakage
• Finite permeability
• Core losses
Figure 1-a
Transformer Model
• physical reasoning
• mathematic model of coupled
circuits
• Winding resistance in series
with leakage inductance
• Magnetizing inductance in
parallel with core resistance
Figure 1-b
Referred equivalent circuits
• Practical transformer is equivalent to lumped parameters
circuit and ideal transformer
Figure 2
The ideal transformer can be shifted to either side as in Figures 3 and 4and the
circuit parameters reduced to the appropriate values
E1=E’2= aE2
V’2=aV2
I’2=I2/a
X’l2=a2xl2
R’2=a2R2
Figure 3
Figure 4
Approximate Equivalent Circuits
•
•
I1R1 and I1Xl1 are small
Therefore, |E1| = |V1|
Shunt branch can be moved to
supply terminal
Figure 5-a
•
IΦ small (5% of rated current)
Shunt branch removed
Figure 5-b-c
Determination of Equivalent Circuit Parameters
• No-load test (rated voltage on one side whereas the other side is open)
Figure 6 a-b
• Short-Circuit test (rated current on one side whereas the other
side is short-circuited)
Figure 7 a-b
EXAMPLE
•
An Engineer needs to know the parameters of a 46KVA Transformer which has a
2300V/230V winding. His results are:
•
•
Open Circuit Test: 230V 11.2A 1150W
Short Circuit Test: 160V 28.0A 1150W
•
•
We must first determine which side Low or High the test was performed on.
•
In this example we see that the Open Circuit Test voltage is the same as the Low side
rated operating voltage, thus we know the test was performed on the low side and the
high side was left open.
•
Next we need to determine which side the Short Circuit Test was performed on, so
we compare the Current this time.
For the Open Circuit test we compare the tested voltage to the rated voltage of the
transformer.
•
Example (continued)
•
So now we know where the tests were performed.
•
•
•
Open Circuit Test: 230V 11.2A 1150W -H.V. Left open, and tested on Low side
Short Circuit Test: 160V 28.0A 1150W -L.V. Shorted, and tested on High side
So we know that the Open Circuit parameters are Referred to Primary.
•
So we will use the Referred to Primary parameters for the Open Circuit test.
What we know:
•
Example (continued)
•
Now remember how the Short Circuit Test proved out that the L.V. side was shorted?
•
This means our equivalent calculations are Referred to the H.V. side. So in effect what we
are calculating are our X2 and R2 values. If we want our X1 and R1 values, we must divide
by our transformation ratio.
•
To complete our Primary Referred Circuit we must find our R1 and X1 values. And given
the fact that the transformer is stepping up voltage from 230 to 2300, we can see that it is a
1:10 ratio, or N1 = 1,N2 = 10.
Example (continued)
•
To complete our Primary Referred Circuit we must find our R1 and X1 values.
And given the fact that the transformer is stepping up voltage from 230 to 2300,
we can see that it is a 1:10 ratio, or N1 = 1,N2 = 10.
•
Now of course one might think this is the end, but that is far from the truth. At
this time we know enough about the circuit to go further if we desire. From here
we can obtain the terminal voltages for simulation, a very important factor in
designing a system!
Example (continued)
•
Now of course one might think this is the end, but that is far from the truth. At this time we know enough about the
circuit to go further if we desire. From here we can obtain the terminal voltages for simulation, a very important factor in
designing a system!
•
•
Now here we must convert our resistance into a singular vector for multiplication.
We will momentarily ignore our j operator to get the magnitude of the vector
Now we can find E’1:
So you can see from the determination that 6.6 volts are due to internal losses in the transformer
itself. Knowing this magnitude, an Engineer can design a simulation with a source to the exact
specifications and losses.
.
Voltage regulation
No load V2=V1/a
Loaded V2=V1/a ± ∆V2
Voltage regulation =
Figure 8 a
Vnl
− Vfl
x 100
%
Vfl
•
Maximum voltage regulation
occur
if θL= θeq1
Figure 8 a-b
Efficiency
η=
Pout
Pout
+ Pc + P cu
V 2 I 2 cos θ 2
η=
V 2 I 2 cos θ 2 + Pc + I 22 R eq 2
Pc=Core Losses in Watts
Pcu= Copper Losses in Watts
Max efficiency occurs for:
Fixed V 2 and
θ2
Pc = I 22 R eq 2
Fixed V 2 and
I2
cos θ 2 = 1
Autotransformer
•
Another type of transformer is known as the autotransformer.
•
It consists of a single tapped primary where the center tap is common
to both primary and secondary (not isolated).
•
some of the turns on the coil are used for the primary and some are
used for the secondary.
•
an example of a variable autotransformer is known as a variac.
•
A variac is a single coil with a sweeping arm for the center common,
this allows the ratio of primary turns:secondary turns to be altered
easily.