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In the previous video, Part 1b, we discussed about the per unit formula and the base values and solved a very small part of an example. In this video, Part 1c, we will complete Part (a) and Part (b) of that same example. The example is taken from a book called Electric Machinery Fundamentals by Stephan Chapman (Edition no. 4).  Example 2.3, pg 95, Electric Machinery Fundamentals by Stephan J. Chapman (Edition 4)

We will not be solving Parts (c) and (d) since our objective is to focus on per part of this problem.. Part (a) Finding out base values for the different regions

Region 1

We solved this in our previous video, Part 1b. For Region 1, we calculated the base values for Ibase and Zbase using the given values of Vbase and Sbase. The answers were:

Ibase = 20.83 A

Zbase = 23.04 Ω

Region 2

First, let’s write down the base values that are given for Region 2. We know that the ratings of the generator both as Vbase and Sbase for region 1.. Since there is a step-up transformer between region 1 and region 2, the Vbase for Region 2 would be different however, the Sbase will remain same… in fact, the Sbase will remain the same for the entire system. So, for Region 2,

S(2,base) = 10 kVA (which will remain the same for the entire system)

Now let’s find out the base voltage by using the transformer turns ratio, we can simply find out the voltage on the secondary side of the transformer.

Vp / Vs
=
Np / Ns

Since

Vp = V(1,base) = 480 V

So:

480 V / Vs
=
1 / 10

Rearranging and solving for Vs:

Vs = 4800 V

And we know that this is the base voltage for Region 2 so:

V(2,base) = 4800 V

Calculating other base values:

I(2,base)
S(2,base) / V(2,base)
=
10 KVA / 4800 V
=2.083 A Z(2,base)
V(2,base) / I(2,base)
=
4800 V / 2.083 A
=2304 Ω

We can also calculate these values using transformer turns ratio formulas.

Region 3

The base power is same throughout the system so:

S(3,base) = 10 kVA

Since there is a step-down transformer linking Region 2 and Region 3, so we will employ the calculation process as we did in Region 2 calculation.

V p / Vs
=
N p / N s

Vp = V(2,base) = 4800 V

4800 V / Vs
=
20 / 1

Vs = 240 V

V(3,base) = 240 V

Calculating other base values:

I(3,base) =
S(3,base) / V(3,base)
=
10 KVA / 240 V
= 41.67 A
z(3,base) =
V(3,base) / I(3,base)
=
240 V / 41.67 A
= 5.76 Ω We can also calculate these values using transformer turns ratio formulas.

Now that we have completed part b of the problem by caluating the base values for every region of the power system, let’s move on to the next part.

Part (b) Converting into per unit equivalent

The per-unit equivalent circuit consists of the components of individual regions. Region 1 is the generating region, so the equivalent would be the source. Region 2 is the transmission line, so the equivalent would be the per-unit impedance of the transmission line. Region 3 consists of load, so the equivalent circuit would contain per-unit impedance of the load. We know that the generic formula to find out per-unit values is:

Per unit value=
Actual value / Base value

Region 1

We calculated Vpu in the video Part 1b that was:

V(1,p.u) = 1.0 p.u

Region 2

We will calculate the per unit impedance of the transmission line:

z(2,p.u) =
Z(2,actual) / Z(2,base)
=
20 + j60 Ω / 2304 Ω
= 0.0087 + j0.0260 p.u

Region 3

We will calculate the per unit impedance of the load connected to the power system:

z(3,p.u) =
Z(3,actual) / Z(3,base)
=
10 ∠ 30° Ω / 5.76 Ω
= 1.736 ∠ 30° p.u Now since we have calculated the per-unit values, we will draw the equivalent circuit. As you can see, the per unit equivalent circuit is more simple because we have eliminated the different voltage levels completely and thereby eliminated region 1, 2, and 3 boundaries. This simplification is one of the biggest reasons why we use per unit in power systems. In the next video we will look into the per-unit formulas for single phase and three phase systems.

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