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Introduction to Per-Unit Systems, Part 1c: Completing the Per-Unit Example.
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).
In this example, in Region 1 we have a 480 volts generator at 0 degrees and there is a step-up transformer which leads us to Region 2.
In Region 2, we have a transmission line that’s 20 ohms resistance and 60 ohms reactance.
And then there’s the step-down transformer which leads us to Region 3.
In Region 3 we have a load that’s 10 ohms at 30 degrees.
We are told in this example that the base value for the system is chosen to be 480 volts at 10 kVA at generator.
In Part (a), we are asked to find the base voltage, current, impedance and apparent power at every point in the power system.
(b) Convert the system to per unit equivalent circuit,
(c) Find the power supplied to the load in the system.
(d) Find the power lost in the transmission line.
We will not be solving Parts (c) and (d) since our objective is to focus on the per-unit part of the problem.
Now, Part (a) Find out the base values of the different regions.
We solved this in our previous video.
For Region 1, we calculated the base values for Ibase and Zbase
and using the given values of Vbase and Sbase.
The answers were simply: Ibase is equal to 20.83 Amps
and Zbase was equal to 23.04 Ohms.
This is for Region1.
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 the same.
In fact, the Sbase will remain the same for Region 1, Region 2 and Region 3.
So the Sbase for Region 2 is simply equal to 10kVA which again 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 base voltage on the secondary side of the transformer by using the following equation.
The voltage on the primary side (Vp) divided by the voltage on the secondary side (Vs) is equal to the Number of turns ratio on the primary side (Np) divided by the Number of turns ratio on the secondary side (Ns).
Since Vp is equal to V1,base which is equal to 480 volts.
We know that 480 volts divided by Vs is equal to 1 divided by 10.
Rearranging and solving for Vs, while Vs is equal to 4800 volts.
We know that this is the base voltage for Region 2 so V2, base is equal to 4800 volts.
Now, we can easily calculate the other base values.
So I2,base is equal to S2,base divided by V2,base,
which is simply equal to 10 KVA divided by 4800 volts,
which is equal to 2.083 Amps.
Z2,base is equal to V2,base divided by I2,base
which is equal to 4800 volts divided by 2.083 Amps,
which equals 2304 Ohms.
We can also calculate these values using by using the transformer turns ratio formula.
So, now that we have calculated all 4 base values for Region 2.
Let’s move on to Region 3.
The base power is the same for the entire system, which equals S3,base is equal to 10 KVA.
Again, since there is a step-down transformer linking Region 2 and Region 3,
we can employ the calculation process we did for Region 2.
Voltage on the primary side (Vp) divided by the voltage on the secondary side (Vs) is equal to the Number of primary turns (Np) divided by the Number of secondary turns (Ns).
Now, which is equal to 4800 volts divided by Vs equals 20 divided by 1.
Rearranging this and solve for Vs equals 240 volts.
So, V3,base is equal to 240 volts.
So, we can easily calculate the other base values.
I3,base is equal to S3,base divided by V3,base which is equal to 10 KVA divided by 240 volts
which equals 41.67 Amps.
And Z3,base is equal to V3,base divided by I3,base
and that equals 240 volts divided by 41.67 Amps,
that equals 5.76 ohms.
So, now that we have calculated the Vbase, Zbase, Sbase and Ibase
for Region 1, Region 2 and Region 3.
Now that we have completed PART A of the problem of the base values for every region of the power system,
let’s move on to the next part.
In Part (b) we are asked to convert into per unit equivalent system.
Now, 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.
and Region 3 consists of the load, so the equivalent circuit would contain per-unit impedance of the load.
And we know that the generic formula to find out the per-unit value is simply equals
per-unit value is equal to the actual value divided by the Base value.
And again, we calculate the base value for all of the 3 regions.
For Region 1, we calculated the Vpu in Part 1b which was given as 1 per unit.
For Region 2, we will calculate the per-unit impedance of the transmission line,
which is equal to: Z2,pu which is equal to Z2,actual divided by Z2,base.
Where the Z2,actual is equal to 20 + j60 ohms and Z2,base is equal to 2304 ohms
that is equal to 0.0087 + j0.0260 per unit
and keep in mind that this is the per unit value.
Region 3. We will calculate the per unit impedance of the load connected to the power system:
So, Z3,pu is equal to Z3,actual divided by the Z3,base,
where Z3,actual is equal to 10 ohms at 30 degrees and Z3,base is equal to 5.76 ohms.
That gives us 1.73 at the angle of 30 degrees per unit.
Since, now we have calculated the per-unit values, we will draw the equivalent circuit
and the equivalent circuit will be very very simple.
It has the voltage source of 1 per unit
and then it has a resistive component and a reactive component
at 0.0087 per unit and j0.0260 per unit
and then it has a load at 1.736 at 30 degrees per unit.
As you can see, the per-unit equivalent circuit is more simple
because we have eliminated the different voltage levels completely
and thereby eliminated the Region 1, Region 2, and Region 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 at per-unit formulas for single phase and three phase systems.
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