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Why we use Per-Units in Power Systems, Part 2b.
In this video, we will demonstrate the advantage of using
Per-units in Power systems and how it makes it easy to calculate
and compare different equipment.
By expressing the impedance of different electrical equipment
in per units, it is easier to compare their characteristics.
It’s worth mentioning that we’re specifically talking about comparing impedances in per-units
because the size of an electrical machine varies,
then its internal impedance, voltage drop and losses also vary.
For example, a primary circuit reactance of 0.1 ohms
would be much higher for one transformer
and on the other hand much lower for the other transformer;
and the driver for this is the Voltage, Power, and various other ratings of the two transformers.
However, if that equipment whether it be a machine, transformer, generator, capacitor bank, etc,
would have its electrical quantities converted into per unit values,
the different voltage drops and levels disappear
and the impedance can be compared over the same scale.
Now this is a huge advantage of using per-units in power systems.
Let’s consider an example from the book of
Electric Power System by B.M. Weedy (5th Edition).
In example 2.3, it says, a d.c. series machine rated at 200V, 100A
has an armature resistance of 0.1 ohms
and field resistance of 0.15 ohms.
The friction and windage loss is 1500W.
Calculate the efficiency when operating as a generator.
Let’s ignore the friction and windage losses and efficiency
and focus on calculating the % Losses instead.
We will employ two methods to do that,
one is by normal calculations
and the other is through per-units.
Now method 1, the normal power system method is that you
calculate the total resistance which is 0.1 ohms plus 0.15 ohms
which equals a total of 0.25 ohms.
And we know that power out is voltage times current.
The voltage is 200V and the current is a 100A.
Which means we get a total power out of 20,000W or 20kW.
Now the P losses, we also know is described by a common equation
called I squared R.
We get 100A squared times 0.25 ohms. 2,500W.
Now the percent losses, we also know this by a common formula
called P losses over P output times 100.
We know that the P losses were 2500W
and the P output is 20kW or 20,000W.
And we multiply that by a 100 which equals 0.125 times a 100 or 12.5%.
That are the percent losses.
Let’s solve the same example by way of per-units.
The method 2 we will use the per-unit method
and we know that to covert a power system into per units,
we require base values and actual values.
In the case of this machine we will take the ratings of the base values
and we will get the base voltage and the base current.
We will now find the per unit impedance for the machine:
So the total resistance again is 0.1 ohms plus 0.15 ohms which gives us 0.25 ohms.
And the per-unit impedance we know is simply equal to
the actual impedance divided by the base impedance
which then equals the actual impedance divided by the
base voltage over base current.
Well, we know that to be simply
equaling 0.25 ohms as the actual impedance divided by the base voltage of 200V over the base current of a 100A.
And that gives us 0.125 per units.
Now if we compare the two values between method 1 and method 2,
we will find that they are exactly the same value.
It’s exactly the same results.
This means that under the given ratings,
the per unit impedance is equal to % Losses in terms of impedances.
Thus, if we employ either of the methods in comparing more than one machines,
we can determine which machine is more efficient.
The difference is that by using per unit calculations,
we don’t have to perform extra steps
and calculate as we did in the first method.
And so this is one of the reasons why per-unit impedances or per-unit calculations
is much more desirable in power systems because they
eliminate some of the extra calculations and make the problem more simple.
Now this concludes that if we want to compare different electrical components
we could simply calculate the per-unit impedances
and find out the desirable component for our power system.
Manufacturers usually specify the impedance of a component
in percents or per units on the base of the nameplate rating.
In this video we saw how per unit systems can ease the
comparison of different electrical equipment in power systems.
In the upcoming videos we will be dealing with an
example for per unit systems for transformer calculations.
We hope, you have a continued interest in this topic and series as a student or professional.
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