Actual Line-to-Line voltage from per unit value

Question

In this question, I am able to find the per unit phase voltages using the symmetrical components. Taking the base MVA and base voltage to be the same as that of the alternator,

\$ (I_a)_1 = \frac{E_a}{Z_1 + Z_2 + Z_0} \$ where \$ Z_1 \$, \$ Z_2 \$ and \$ Z_0 \$ are the positive negative and zero sequence components respectively.

Given that the fault occurs on phase 'a', \$ V_a = 0 \$.

Using the equations to find \$ (V_a)_1 \$, \$ (V_a)_2 \$ and \$ (V_a)_0 \$ and further finding the phase voltages, I get:

$$ V_a = 0 $$ $$ V_b = -0.2143-j0.9898 $$ and $$ V_c = -0.2143+j0.9898 $$ To find Line-to-Line voltages, \$ (V_{ab})_{pu} = V_a - V_b\$ has to be evaluated.

Thus, the actual value of the line voltage will be given by: $$ (V_{ab})_{actual} = ((V_{ab})_{pu}).(13.2) kV $$

But the solution in the book says:

I don't understand why \$ 13.2 kV \$ is divided by \$ \sqrt{3} \$, as it only yields the base phase voltage.

Is the solution wrong or am I missing something?

Answer 1

By definition the symmetrical component method uses line-neutral voltage. Symmetrical component circuits are single-phase. So, per unit voltages must be multiplied by the line-neutral voltage base to find their actual values.

In your example,

$$ V_b = (-0.2143-j0.9898)\times \frac{13.2\text{kV}}{\sqrt3}=7,718\angle-102.2° \text{V} $$ and $$ V_c = (-0.2143+j0.9898)\times \frac{13.2\text{kV}}{\sqrt3}=7,718\angle102.2° $$

Now do your subtractions to find your line-line voltages. That magic \$\sqrt3\$ business only shows up for balanced conditions when line-neutral voltage magnitudes are equal and phase angles 120° apart.

p.s. Recheck your calculation on \$Vc\$. \$Vb\$ is correct.