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In the previous videos, Part 1b and Part 1c, we discussed the formulas for converting values into their per unit equivalents and described base values and then moved forward by solving a simple example. In this video, Part 1c, we will be deriving the formulae that are involved in per unit calculations of single phase and three phase systems, some of those, we have already used in the introductory example in Part 1b and it is essential to have a good understanding on these formulas because we will be using them in calculations in the upcoming videos.

First and foremost, let’s remember that if any two base quantities are given we can find other two base quantities and hence all the four quantities can be known. In electrical power system, the two base quantities that are almost always given are base power, which we denote as S base and base voltage, which we denote as V base. The calculation of other two depends on if we have a single phase system or a three phase system.

Let’s start with the simplest single phase system: We take a source with apparent power, S, and voltage, V, and connected to an impedance Z

The ratings of the source are S and V that are taken as S base and V base. The current flowing is I and the load is Z. this is neutral (point at ground element) since we are dealing with single phase.

We are given S base and V base which are the two known quantities. Let’s calculate I base and Z base which are the unknown but can be easily derived. We know the formula for power that is

**Power = Voltage × Current**

So,

**Base Power = Base Voltage × Base Current** but it can be rearranged to the following

**Base Current = Base Power Base Voltage **(equation #1)

**I S**(Equation #2)

_{base}=_{base}V

_{base}

We know the formula for Base impedance is

**Base Impedance = (Base Voltage)**(Equation #2)

^{2}Base Power

**Z V**(Equation #3)

_{base}=_{base}

^{2}S

_{base}

We can see that these all formulas were discussed in Part 1b and we know that per unit values can be calculated using given actual values and these base values.

Moving on towards three phase systems.

Let’s draw a Y connected system.

‘n’ marked is neutral so,

**V _{an} = V_{phase} or V_{∅}**

And the voltage between a and b is Line to line voltage:

**V _{ab} = V_{ll}**

Let’s recall some of the basic formulae for Y connected three phase system

**V _{ll} =√3 ×V_{∅} **(Equation #4)

**I _{l} = I_{∅}**(Equation #5)

**S _{3∅} = 3 × S_{∅} =3 × V_{∅} × I_{∅}**(Equation #6)

We will now investigate three key points associated with per unit calculations in three phase systems which will help us understand how we derive the base quantities intuitively.

Keypoint Number 1) Line to line voltage in per unit is always equal to phase voltage in per unit. To prove this let’s look at an example.

It is given that:

**V _{ll(actual) }= 100 kV**

**V _{ll(base)} = 132 kV**

So, we will calculate V_{ll(p.u)}.

**V 100 132 **=0.76 p.u

_{ll(p.u)}=These were line to line voltages.

Calculating V_{∅(p.u)} requires the actual and base values of voltage of one phase. Let’s convert the line voltage into phase voltage:

**V 100 KV √3 **

_{∅(actual)}=**V 132 KV √3 **

_{∅(base)}=So calculating V_{∅(p.u)} by:

**V 100 ⁄ √3 kV 132 ⁄ √3 kV **= 100/132=0.76 p.u

_{∅(p.u)}=Hence it is proved that Line to line voltage in per unit is equal to phase voltage in per unit.

**V _{ll(p.u)} = V_{∅(p.u)} (proven)**

Keypoint Number 2) The two base values given are almost always Sbase and Vbase. In three phase systems the Sbase is the total three phase power of the system and Vbase is the line to line voltage. These two quantities are then used to calculate Ibase and Zbase.

Keypoint Number 3) All the base impedances are taken as per phase impedances for three phase systems.

We’ll be looking deeper into Keypoint 2 and Keypoint 3 in the following derivation.

As keypoint 2 states, we are given with Sbase and Ibase in a three phase system. We will now calculate Ibase and Zbase.

Taking Equation #6 that we wrote earlier which is

**S _{3∅} = 3 × S_{∅} = 3 × V_{∅ } × I_{∅}**

Since S_{3∅} =S_{ base}in a three phase system so:

**S _{base} =3× V_{∅ } × I_{∅}**

Since we want to calculate base current. the other two quantities should be base power and base voltage. We have Sbase in the above

equation. To incorporate base voltage, we will be Substituting Equation #4 in the above equation which makes:

**S V × I **

_{base}= 3 ×_{ll}√3

_{∅}We know that V_{ll} =V_{ base } in a three phase system so:

**S V× I **

_{base}= 3 ×_{base}√3

_{∅}Which then simplifies to the following

**S V**

_{base}= √3 ×_{base}× I

_{∅}

Then we solve for the phase current which we know as:

**I S**

_{∅ }=_{base}√3 × V

_{base}

We know from eq 5 that I_ll= I_∅ which was taken as I_{base}

**I S**

_{∅ }= I_{l}= I_{base}=_{base}√3 × V

_{base}

Or we can also write, substituting V_{base} with V_{ll}

**I S**(Equation #7)

_{base }=_{3∅}√3 × V

_{ll}

Here we have the formula for Ibase.

Now let’s calculate Zbase. As keypoint 3 states that all the base impedances are in per phase so

**Z V**(Equation #8)

_{∅}=_{∅}I

_{∅}

Substituting eq 4 into eq 8 gives:

**Z V**

_{∅}=_{ll}⁄√3 I

_{∅}

We can also write this equation as:

**Z V**

_{base}=_{ll}/ √3 I

_{base}

Now let’s substitute eq 7 into the above eq:

**Z V**

_{base}=_{ll}/ √3 S

_{3∅}⁄ √3 × V

_{ll}

Rearranging this equation:

**Z V**

_{base}=_{ll}/ √3 × √3 × V

_{ll}S

_{3∅}

Root 3 root 3 cancel so it gives us:

**Z V**

_{base}=_{ll}

^{2}S

_{3∅}

or we can also write this as:

**Z V**----eq 9

_{base}=_{base}

^{2}S

_{base}

Now we went through his long and complicated derivation so we can understand exactly how these equations are derived… but at the end of the day, we are only concerned about the simplified equations which we will summarize below.

These were the formulas that we calculated for single phase and three phase systems These two formulas are same for both single phase and three phase systems provided that V base and S base are VLL and S, three phase respectively for three phase systems.

**Z V**

_{base}=_{base}

^{2}S

_{base}

In the next part of this series, we’ll be discussing the advantages of per unit systems.

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