CHAPTER THREE a single phase per unit example. What you see before you is a simple single phase circuit that consists of a generator with some internal resistance or reactance and a load, the load is made up of 60 plus j 70 ohms or in polar coordinates 92.2 at 49 degrees in ohms. The generator is 1000 volts and zero degrees and its internal reactance is J 10 ohms. The total resistance then is the sum of the two impedances because they're in series so it's 60 plus j 80 ohms or 153 or degrees ohms. So calculating the current is simply using ohms law, which is the source voltage over top of the divided by the total impedance, which is 1000 volts at zero degrees all over 153.1. So that equals 10 amps and minus 53 degrees.

The internal or the VA or the apparent power that's delivered to the load is given by the source voltage times the current flowing in the circuit, which is 1000 at zero degrees times 10 at minus 53 degrees given us 10,000, va at minus 53 degrees. Now we're going to begin the process of normalization by converting the LM Two per unit values, starting with the selection of the base quantities to which all elements are compared to. So we're going to select 10,000 va and which is the amount of apparent power that's delivered to the load. And it is a base quantity. So it's a scalar quantity, it doesn't have any direction. So it's 10,000 va or 10 kV a.

The voltage we're going to choose is the source voltages which is one kV, which is gives us a V base of 1000 volts or one kV. Now once these two bases are picked, or set then all of the other bases are set by calculating using the VA and the voltage basis We can calculate apparent power by multiplying voltage times current. And we can calculate the current in a circuit by dividing the apparent power by the voltage. So, in calculating the AI base we can take the S base and divided by the V base which is 10,000 divided by 1000, which gives us 10 amps or one per unit of amperage. Similarly, impedance is equal to the voltage over the current and the current is given by the apparent power over the voltage. So, we can use the voltage squared all over the apparent power to calculate the impedance.

So, when we want to calculate the impedance base, we can Divide the voltage base squared by the s base. So 1000 squared divided by 10,000 is equal to 100 ohms or one per unit of impedance. Remember that base values are magnitude only. And the basic units are called base values for this circuit. Let's place the base values in the upper right hand corner so we can refer to them as we do our calculations. So we're going to begin to calculate the per unit values.

And the per unit values are calculated by taking the actual values and dividing by their base values. So if we're going to calculate the per unit value for current assets, And we're going to call it I subscript p EU. It's the actual current i all over the base current, which is 10 at minus 53 degrees all over 10 which equals to one at minus 53 degrees per unit. The impedance of the load is given by taking the actual load impedance and dividing it by the base impedance which is 92.2 at 49 degrees divided by 100 which equals point nine to two at 49 degrees per unit. Now, I'm just going to mention this we already talked about it before and that is per unit values are phasers themselves. And they have the same phase angle as the actual values themselves.

Now the next thing I'm going to do is I'm going to calculate the per unit value for the resistance component of the load and the reactance component of the load, and you use the same base values. for doing this. In other words, we will take the resistance of the load which is 60 at zero degrees and divided by the impedance base which is 100. And that gives us point six at zero degrees per unit. And likewise, I'm going to calculate the per unit value for the reactance of the load. And we can take the reactance of the load x subscript l all over Zed base which is 70 at 90 degrees divided by 100 is equal to 0.7 at 90 degrees per unit carrying on with the calculation of the per unit values, we'll now look at the power delivered to the load.

And in real terms, it's the resistance times the current squared, which is equal to 60 times 10 squared, which is equal to 6000 watts. We want to change that to the per unit value of for power, we have to divide by the base value of VA, which is 6000 over 10,000 equaling point six per unit. Another way to calculate the per unit value for power is to take the per unit value for resistance and multiply it by the per unit value of the current square. And then that is equal to point six times one squared, which is equal to 0.6 per unit which is agrees with the way we calculated just previously. If we want to calculate var load delivered. It's the reactance of the load times the current squared, which is 70 times 10 squared, or 7005 hours of inductance.

The per unit value for that is just that var load divided by the base value for VA and it's 7000 over 10,000 or 0.7 per unit of inductance. Similarly, as we did with the power we could calculate the per unit value for bars by taking the per unit value of the inductance and multiplying it by the per unit value of current squared which is point seven times one squared, which is equal to 0.7 per unit. We are now going to calculate the voltage drop across little load. And we're going to do it first in per unit values because that's what we have in front of us. And the per unit voltage drop across the load is given by the impedance of the load in per unit values times the current through it in per unit values, which is point nine to two at 49 degrees times one at minus 53 degrees, which equals point nine two to add minus four degrees per unit.

We'd now like to calculate the actual voltage that is across the load or the voltage drop across the load, and that's given by the subscript L. And all we have to do is take the printed value for the voltage and multiplied by the base value and that is equal 2.9 to two at minus four degrees. times 1000 volts, which is equal to 922 volts at minus four degrees. And that ends chapter three and our single phase per unit example