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Part 2d, Why we use Per-Units in Power Systems.
In this video, we will demonstrate another advantage of using Per-unit systems in Power System calculations.
This aspect is an extension of the previous video.
In a power system, multiple transformers result in
multiple voltage levels at different points in the system.
Per-unit calculations ease this analysis
by eliminating these voltage levels
and bringing the impedances on both sides of the transformer on a single scale.
So this is one advantage.
We demonstrated this aspect in detail in Part 2c, with a single transformer.
Here in Part 2d we will consider a power system with more than one transformer,
thus tackling multiple voltage levels.
For this, we will take an example from the book Power System Analysis by John J. Grainger.
Example 2.6 states three parts of a single phase electric system are designated A, B and C
and connected to each other through transformers, as shown in figure 2.10 below.
The transformers are rated as: A-B 10,000kVA, 13.8kV on the primary side, 138kV at the secondary side
with the leakage reactance of 10%. So this a step up transformer.
And then the transformer between region B and region C is also a 10,000kVA transformer
and it steps down the voltage from 138kV in region B to 69kV in region C.
and this transformer has a leakage reactance of 8%.
Now the question states if the base in circuit B is chosen as 10,000kVA, 138kV,
find the per unit impedance of the 300 ohm resistive load in circuit C referred to Circuit C, B and A.
and draw the impedance diagram neglecting the magnetizing current, transformer resistance and line impedance.
So now let’s solve this question.
Keep in mind that this is a single phase system so the base impedance calculations are different as compared to a three phase system.
Since we are given the resistance in circuit C,
we’ll solve for the per unit values of that resistance
at the base voltage of 69kV.
Circuit C also has a base value of power of 10,000 kVA
so the base value of impedance simply becomes as follows.
The base value of impedance in circuit C is the base value of voltage in circuit C squared divided by the base value of power in circuit C
which we know to be 69kV squared over 10,000kVA and we get 476 ohm.
Now this is the base value of impedance for circuit C.
The per unit impedance of circuit C becomes simply the actual impedance divided by the base.
So the actual impedance is 300 ohms
which is given in the problem statement
divided by 476 ohms of the base impedance which we calculated above.
And that gives us 0.63 per unit.
Now, based on Part 2c of this series,
we can stop here as we already know the per unit impedances for both transformers and the load.
The load per unit impedance was calculated as 0.63pu
and since the transformer leakage reactance was given as a percentage in the problem statement,
with the power base being the same for both transformers,
we already know they corresponds to per unit values.
So the equivalent circuit for this problem looks like this:
We have j0.1 which is the impedance of the first transformer,
and we have j0.08 pu which is the impedance of second transformer
and then we have 0.63 pu which is the resistance of the load.
So we already have the equivalent circuit for this entire problem statement.
Despite solving part of this answer,
we will go through and show the advantages of per unit systems.
Because the selection of base in various regions of the system
is determined by the turns ratio of the transformers,
and because the base kVA is the same in all regions of the system,
the per unit impedance of the load transferred to any regions of the system
will be exactly the same.
This will be illustrated in the remaining part of this tutorial.
We have calculated the per unit impedance of the circuit referred to Circuit B,
and we also know that the per unit impedance in Circuit B
would be equal to the per unit impedance in Circuit C
i.e. it will be equal to 0.63 p.u., as described in Part 2c of this series.
We first find out the base value of the impedance in Circuit B of this problem
by taking 138 kV as the base value of voltage
and the base value of power remains the same at 10,000 kVA.
So the base impedance for Circuit B is simply equals to the voltage base in Circuit B
divided by the power base for Circuit B.
The voltage base being a 138kV squared
and power base being 10,000kVA,
which gives us 1904.4 ohms.
Now if we were to calculate the actual value of impedance in circuit B
we would multiply it with the square of turns ratio of 2:1
and the actual value is 1200 ohms.
Meaning the Zb actual is simply 300 ohms which is the resistance of the load in Circuit C
times the turns ratio 2 over 1 squared and that gives us 1200 ohms.
So that is the actual resistance or the actual impedance of the 300 ohms in Circuit C reflected on Circuit B.
Lastly when we convert the actual values to per units
we find that it becomes essentially the same.
And here is how: so the per unit impedance of circuit B is equal to the actual impedance of Circuit B
divided by the base impedance of Circuit B.
so that actual impedance of 1200 ohm which we found that it was a reflection of Circuit C
and the base impedance is 1904.4 ohms
which is dependent on the voltage and the base power of circuit B
and that gives us 0.63 p.u.
So, here we have proved what we learnt in Part 2c,
which is that per unit impedances on both the HV side and LV side of the transformer will be exactly the same
just like how per unit impedance of Circuit C and per unit impedance of Circuit B are the same.
Now, we can see that Circuit A and Circuit B
are the primary and secondary sides of the first transformer, respectively.
Logic says that the per unit impedance of Circuit A will be equal to that of Circuit B.
Let’s find out if that statement is actually correct.
We have the base impedance of Circuit A to be calculated
as base voltage of Circuit A squared divided by the base power of Circuit A.
the base voltage being 13.8kV and we square that divided by 10,000kVA power
for circuit A and we get 19 ohms.
Now the actual impedance of circuit A
is basically what we calculated in terms of ohms in Circuit B to be reflected on Circuit A, right.
So this is all we are doing here is that we are reflecting the 300 ohms
from Circuit C to B and from C to A, considering the transformer turns ratio.
So, 1200 ohms time 1 over 10
which is the transformer turns ratio squared and that gives us 12 ohms.
So Za which is the per unit impedance of Circuit A
is equal to the actual impedance of Circuit A divided by the base impedance of Circuit A
which we get as 12 ohms divided by 19 ohms
and that gives us 0.63p.u!
Finally, we have the per unit impedances in all parts of the circuits
in the power system of this example.
It equals to 0.63 p.u in all parts of the circuit
which reflects that when a power system consists of
more than one transformer, the per unit impedance for every side of every transformer remains the same.
We can draw the equivalent circuit like this, which we drew before.
It's j0.1 per unit which is the first transformer’s impedance,
j0.08 per unit which is the second
and then the 0.63 per unit which is the per unit load of Circuit C.
Notice that the multiple voltage levels are eliminated
by making use of Per-units and
the circuit consists of just impedances that have been reduced to a single voltage level
that is common to the entire system.
We can also conclude that no matter how many transformers and voltage levels are added in a system,
we can find the impedance by simply doing calculation for only one side of one transformer.
See, how per unit systems can drastically reduce the mathematics involved.
In the next video, we will be dealing with an example for a per unit system in three-phase system calculations.
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