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Why we use Per-Unit in Power Systems, Part 2c.
In this video, we will demonstrate another advantage of using Per-unit in Power systems.
Let’s start with the basics.
In order to perform power system analysis in power systems having one or more than one transformers,
which usually is the case, we have to convert a transformer into its
equivalent circuit referring to either the primary side or the secondary side of the transformer.
There are formulas to achieve this,
such as transformer turns ratio formula, voltage formula,
impedance formula etc.
Impedances on primary side should be multiplied
by the square of turns ratio to give the secondary side impedance.
We have to find the leakage reactance and magnetizing currents as well.
However, when a transformer is transformed into per unit equivalent,
the need for tedious calculations performed to
refer transformers to either the HV side or the LV side is eliminated.
The magnetizing currents are neglected
and even there is no requirement to calculate impedances on different sides individually.
Let’s see how this is made easy.
Consider a transformer drawn like the one below.
We know that to find Z2 referred to primary side, the formula is this:
Z2 is equal to the number of turns on the secondary side divided by the number of turns on the primary side squared
which is N over 1 squared times Z1 which is the impedance on the primary side.
That would give us N squared times Z1.
Next, we calculate per unit impedance of Z1:
Z1 p.u. is equal to Z1 over Z1 base
which is equal to Z1 over V1 base divided by I1 base.
And that is equal to Z1 times I1 base divided by V1 base.
And as we substitute the value of Z2 we get:
Z2 p.u. is equal to Z2 divided by Z2 base
and that is equal to Z2 divided by V2 base over I2 base.
And that is equal to Z2 times I2 base over V2 base.
And that is equal to N squared times Z1 times I2 base over V2 base.
We know that the formula for transformers that relates turns, voltages and current is equal to:
1 over N is equal to V1 base over V2 base which is equal to I2 base over I1 base.
So you see there’s an inverse there.
Substituting these values in the equation of Z2, p.u we get:
Z2 p.u is equal to N squared times Z1 times I2 base over V2 base
and that is equal to N squared times Z1 times I1 base over N times 1 over N times V1 base.
We simplify this to get simply:
Z2 p.u. is equal to Z1 times I1 base over V1 base.
Hence, we prove that the per unit impedance of primary and secondary sides are equal in magnitude.
This means that calculating per unit impedances of either side
is enough to analyze a transformer.
Let’s take an example from the book Power System Analysis by John J. Grainger (The 1994 Edition).
Example 2.5 states that a single-phase transformer is rated 110/440 V with a 2.5 kVA power rating.
The leakage reactance measured from the low voltage side is 0.06 ohms.
The question is determine the leakage reactance in per-units.
Now the solution on the low voltage side is this:
We have Zbase which is equal to 110V squared divided by 2.5 times 10 to the third VA
which is voltage squared over power.
We get 4.84 ohms.
Now the per-unit Z p.u is equal to 0.06 divided by 4.84 base and
that is equal to 0.0124 p.u.
in terms of the leakage reactance.
By doing so this two-step calculation we know that
Zpu on both sides i.e. the low voltage side and high voltage side, is equal to 0.0124 p.u.
And because of our derivation earlier in the video, we know that if we calculate it by
using the impedance referred from the high-voltage side, the calculation would be lengthy
but result is exactly the same.
This concludes that if we want to analyze a transformer,
we would simply calculate the per unit impedance
on the either side of the transformer.
In the next video, we will see the case of multiple transformers in power systems.
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