Chapter 12, Problem 1P

If V_ab = 400 V in a balanced Y-connected three-phase generator, find the phase voltages, assuming the phase sequence is:

1. (a) abc
2. (b) acb

To determine

Find the phase voltages for the abc sequence.

Answer to Problem 1P

For the abc sequence,

The phase voltage $\left(V_{an}\right)$ is $231\angle -30°\text{V}$.

The phase voltage $\left(V_{bn}\right)$ is $\text{231}\angle -150°\text{V}$.

The phase voltage $\left(V_{cn}\right)$ is $231\angle 90°\text{V}$ or $231\angle -270°\text{V}$.

Explanation of Solution

Given data:

For a balanced Y-connected three phase generator,

The line to line voltage (line voltage) $V_{ab}$ is $400\text{V}$.

The phase sequence is abc.

Formula used:

The line voltage in the Y-connection can be written as follows in abc sequence.

$\begin{array}{l}{V}_{ab}=\sqrt{3}{V}_p\angle 30°\text{ V}\\ V_p=\frac{V_{ab}}{\sqrt{3}\angle 30°}\text{ V}\end{array}$

Reduce the equation to find the rms value of the phase voltage.

$V_p=\frac{V_{ab}}{\sqrt{3}}\angle -30°\text{}\text{V}$                                                                                                   (1)

Here,

$V_{ab}$ is the line voltage.

In Y-connection, the phase voltages can be written as follows on positive phase sequence abc.

$V_{an}=V_p\angle 0°\text{V}$                                                                                                        (2)

$V_{bn}=V_p\angle -120°\text{V}$                                                                                                 (3)

$V_{bn}=V_p\angle +120°\text{V}=V_p\angle -240°\text{V}$ (4)

Substitute equation (1) in equations (2), (3) and (4).

The phase voltage $V_{an}$:

$\begin{array}{c}{V}_{an}=\left(\frac{V_{ab}}{\sqrt{3}}\angle -30°\text{}\right)\angle 0°\text{V}\\ =\frac{V_{ab}}{\sqrt{3}}\angle \left(-30°+0°\right)\text{V}\end{array}$

$V_{an}=\frac{V_{ab}}{\sqrt{3}}\angle -30°\text{V}$                                                                                           (5)

The phase voltage $V_{bn}$:

$\begin{array}{c}{V}_{bn}=\left(\frac{V_{ab}}{\sqrt{3}}\angle -30°\text{}\right)\angle -120°\text{V}\\ =\frac{V_{ab}}{\sqrt{3}}\angle \left(-30°-120°\right)\text{V}\end{array}$

$V_{bn}=\frac{V_{ab}}{\sqrt{3}}\angle -150°\text{V}$                                                                                          (6)

And

The phase voltage $V_{cn}$:

$\begin{array}{c}{V}_{cn}=\left(\frac{V_{ab}}{\sqrt{3}}\angle -30°\text{}\right)\angle +120°\text{V}\\ =\frac{V_{ab}}{\sqrt{3}}\angle \left(-30°+120°\right)\text{V}\end{array}$

$V_{cn}=\frac{V_{ab}}{\sqrt{3}}\angle 90°\text{V}$ (7)

Or the phase voltage $V_{cn}$ can be written as follows.

$\begin{array}{c}{V}_{cn}=\left(\frac{V_{ab}}{\sqrt{3}}\angle -30°\text{}\right)\angle -240°\text{V}\\ =\frac{V_{ab}}{\sqrt{3}}\angle \left(-30°-240°\right)\text{V}\end{array}$

$V_{cn}=\frac{V_{ab}}{\sqrt{3}}\angle -270°\text{V}$ (8)

Calculation:

Substitute $400\text{V}$ for $V_{ab}$ in equation (5) to find the phase voltage $V_{an}$.

$\begin{array}{c}{V}_{an}=\frac{400\text{}\text{V}}{\sqrt{3}}\angle -30°\text{ V}\\ =230.94\angle -30°\text{ V}\end{array}$

$V_{an}\cong 231\angle -30°\text{ V}$

Substitute $400\text{V}$ for $V_{ab}$ in equation (6) to find the phase voltage $V_{bn}$.

$\begin{array}{c}{V}_{bn}=\frac{400\text{}\text{V}}{\sqrt{3}}\angle -150°\text{V}\\ =230.94\angle -150°\text{V}\end{array}$

$V_{bn}=231\angle -150°\text{V}$

Substitute $400\text{V}$ for $V_{ab}$ in equation (7) to find the phase voltage $V_{cn}$.

$\begin{array}{c}{V}_{cn}=\frac{400\text{}\text{V}}{\sqrt{3}}\angle 90°\text{V}\\ =230.94\angle 90°\text{V}\end{array}$

$V_{cn}\cong 231\angle 90°\text{V}$

Or from equation (8) the phase voltage $V_{cn}$ can also find as follows.

$\begin{array}{c}{V}_{cn}=\frac{400\text{}\text{V}}{\sqrt{3}}\angle -270°\text{V}\\ =230.94\angle -270°\text{V}\end{array}$

$V_{cn}\cong 231\angle -270°\text{V}$

Conclusion:

Thus, for the abc sequence,

The phase voltage $\left(V_{an}\right)$ is $231\angle -30°\text{V}$.

The phase voltage $\left(V_{bn}\right)$ is $\text{231}\angle -150°\text{V}$.

The phase voltage $\left(V_{cn}\right)$ is $231\angle 90°\text{V}$ or $231\angle -270°\text{V}$.

To determine

Find the phase voltages for the acb sequence.

Answer to Problem 1P

For the acb sequence,

The phase voltage $\left(V_{an}\right)$ is $231\angle 30°\text{V}$.

The phase voltage $\left(V_{bn}\right)$ is $\text{231}\angle 150°\text{V}$.

The phase voltage $\left(V_{cn}\right)$ is $231\angle -90°\text{V}$.

Explanation of Solution

Given data:

The phase sequence is acb.

Formula used:

The line voltage in the Y-connection can be written as follows in acb sequence.

$\begin{array}{l}{V}_{ab}=\sqrt{3}{V}_p\angle -30°\text{ V}\\ V_p=\frac{V_{ab}}{\sqrt{3}\angle -30°}\text{ V}\end{array}$

Reduce the equation to find the rms value of the phase voltage.

$V_p=\frac{V_{ab}}{\sqrt{3}}\angle 30°\text{}\text{V}$                                                                                                  (9)

Here,

$V_{ab}$ is the line voltage.

In Y-connection, the phase voltages can be written as follows on positive phase sequence acb.

$V_{an}=V_p\angle 0°\text{V}$                                                                                                        (10)

$V_{cn}=V_p\angle -120°\text{V}$                                                                                                  (11)

$V_{bn}=V_p\angle +120°\text{V}=V_p\angle -240°\text{V}$ (12)

Substitute equation (9) in equations (10), (11) and (12).

The phase voltage $V_{an}$:

$\begin{array}{c}{V}_{an}=\left(\frac{V_{ab}}{\sqrt{3}}\angle 30°\text{}\right)\angle 0°\text{V}\\ =\frac{V_{ab}}{\sqrt{3}}\angle \left(30°+0°\right)\text{V}\end{array}$

$V_{an}=\frac{V_{ab}}{\sqrt{3}}\angle 30°\text{V}$                                                                                                (13)

The phase voltage $V_{cn}$:

$\begin{array}{c}{V}_{cn}=\left(\frac{V_{ab}}{\sqrt{3}}\angle 30°\text{}\right)\angle -120°\text{V}\\ =\frac{V_{ab}}{\sqrt{3}}\angle \left(30°-120°\right)\text{V}\end{array}$

$V_{cn}=\frac{V_{ab}}{\sqrt{3}}\angle -90°\text{V}$                                                                                              (14)

And

The phase voltage $V_{bn}$:

$\begin{array}{c}{V}_{bn}=\left(\frac{V_{ab}}{\sqrt{3}}\angle 30°\text{}\right)\angle +120°\text{V}\\ =\frac{V_{ab}}{\sqrt{3}}\angle \left(30°+120°\right)\text{V}\end{array}$

$V_{bn}=\frac{V_{ab}}{\sqrt{3}}\angle 150°\text{V}$ (15)

Or the phase voltage $V_{bn}$ can be written as follows.

$\begin{array}{c}{V}_{bn}=\left(\frac{V_{ab}}{\sqrt{3}}\angle 30°\text{}\right)\angle -240°\text{V}\\ =\frac{V_{ab}}{\sqrt{3}}\angle \left(30°-240°\right)\text{V}\end{array}$

$V_{bn}=\frac{V_{ab}}{\sqrt{3}}\angle -210°\text{V}$ (16)

Calculation:

Substitute $400\text{V}$ for $V_{ab}$ in equation (13) to find the phase voltage $V_{an}$.

$\begin{array}{c}{V}_{an}=\frac{400\text{}\text{V}}{\sqrt{3}}\angle 30°\text{ V}\\ =230.94\angle 30°\text{ V}\end{array}$

$V_{an}\cong 231\angle 30°\text{ V}$

Substitute $400\text{V}$ for $V_{ab}$ in equation (14) to find the phase voltage $V_{cn}$.

$\begin{array}{c}{V}_{cn}=\frac{400\text{}\text{V}}{\sqrt{3}}\angle -90°\text{V}\\ =230.94\angle -90°\text{V}\end{array}$

$V_{cn}=231\angle -90°\text{V}$

Substitute $400\text{V}$ for $V_{ab}$ in equation (15) to find the phase voltage $V_{bn}$.

$\begin{array}{c}{V}_{bn}=\frac{400\text{}\text{V}}{\sqrt{3}}\angle 150°\text{V}\\ =230.94\angle 150°\text{V}\end{array}$

$V_{bn}\cong 231\angle 150°\text{V}$

Or from equation (16) the phase voltage $V_{bn}$ can also find as follows.

$\begin{array}{c}{V}_{bn}=\frac{400\text{}\text{V}}{\sqrt{3}}\angle -210°\text{V}\\ =230.94\angle -210°\text{V}\end{array}$

$V_{bn}\cong 231\angle -210°\text{V}$

Conclusion:

Thus, for the acb sequence,

The phase voltage $\left(V_{an}\right)$ is $231\angle 30°\text{V}$.

The phase voltage $\left(V_{bn}\right)$ is $\text{231}\angle 150°\text{V}$ or $231\angle -210°\text{V}$

The phase voltage $\left(V_{cn}\right)$ is $231\angle -90°\text{V}$.