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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 impedances of different electrical components 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).

*Example 2.3, pg. 62, Electric Power System by B.M. Weedy (5th Edition)*

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

** 1: Power method**

Total resistance = 0.1 + 0.15 = 0.25 ohms

P_{out} = V × I = 200V ×100A = 20,000W

P_{losses} = I^{2} × R= (100V)^{2} × 0.25Ω = 2,500W

**%Losses = P× 100 = 2,500 20,000 × 100 = 0.125 × 100 = 12.5% **

_{losses}P

_{out}

Let’s solve the same example by way of per-units.

**Method 2: Per-Unit method**

We know that to covert a power system into per units, we require base values and actual values. In the case of these machines we will take the ratings as 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:

Total resistance = 0.1 + 0.15 = 0.25 ohms

**Per unit impedance = Actual impedance Base impedance = Actual impedance Base voltage ⁄ Base current = 0.25Ω 200V ⁄ 100A =0.125 p.u **

Let’s draw a table and compare the results using the two calculation methods:

We can see that the results are exactly the same! This means that under the given ratings, the per unit impedance is equal to % Losses. Thus, if we employ either of the methods in comparing more than one machines, we can determine which machine is efficient. The difference is that by using per unit calculations, we don’t have to perform extra steps and calculations as we did in the first method.

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 systems! Manufacturers usually specify the impedance of a component in 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 a power system. In the upcoming videos we will be dealing with an example for per unit systems in transformer calculations.

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