Chapter 11, Problem 1P

a)

To determine

Find the phase sequence for the given set of voltages.

Answer to Problem 1P

The phase sequence is acb sequence.

Explanation of Solution

Given data:

Consider the set of voltages.

$v_{\text{a}}=180\cos \left(\omega t+27°\right)\text{V}$        (1)

$v_{\text{b}}=180\cos \left(\omega t+147°\right)\text{V}$        (2)

$v_{\text{c}}=180\cos \left(\omega t-93°\right)\text{V}$        (3)

Calculation:

Convert the voltage in Equation (1) from cosine voltage form to phasor form.

$V_{\text{a}}=180\angle 27°\text{V}\left\{\because V_m\cos \left(\omega t+\theta \right)=V_m\angle \theta \right\}$

Convert the voltage in Equation (2) from cosine voltage form to phasor form.

$V_{\text{b}}=180\angle 147°\text{V}$

Convert the voltage in Equation (3) from cosine voltage form to phasor form.

$V_{\text{c}}=180\angle -93°\text{V}$

Subtract the phase angle of a-phase from all the phase angles of $V_{\text{a}}$, $V_{\text{b}}$, and $V_{\text{c}}$.

$\angle {V^{\prime }}_a=27°-27°$

$\angle {V^{\prime }}_a=0°$        (4)

$\angle {V^{\prime }}_b=147°-27°$

$\angle {V^{\prime }}_b=120°$        (5)

And

$\angle {V^{\prime }}_c=-93°-27°$

$\angle {V^{\prime }}_c=-120°$        (6)

From Equations (4), (5), and (6), the voltage at phase-b leads a-phase voltage by $120°$ and c-phase voltage lags a-phase voltage by $120°$. Therefore, this phase relationship is acb phase sequence or negative phase sequence.

Conclusion:

Thus, the phase sequence is acb sequence.

b)

To determine

Find the phase sequence for the given set of voltages.

Answer to Problem 1P

The phase sequence is abc sequence.

Explanation of Solution

Given data:

Consider the set of voltages.

$v_{\text{a}}=4160\cos \left(\omega t-18°\right)\text{V}$        (7)

$v_{\text{b}}=4160\cos \left(\omega t-138°\right)\text{V}$        (8)

$v_{\text{c}}=4160\cos \left(\omega t+102°\right)\text{V}$        (9)

Calculation:

Convert the voltage in Equation (7) from cosine voltage form to phasor form.

$V_a=4160\angle -18°\text{V}\left\{\because V_m\cos \left(\omega t+\theta \right)=V_m\angle \theta \right\}$

Convert the voltage in Equation (8) from cosine voltage form to phasor form.

$V_b=4160\angle -138°\text{V}$

Convert the voltage in Equation (9) from cosine voltage form to phasor form.

$V_c=4160\angle 102°\text{V}$

Subtract the phase angle of a-phase from all the phase angles of $V_{\text{a}}$, $V_{\text{b}}$, and $V_{\text{c}}$.

$\angle {V^{\prime }}_a=-18°-\left(-18°\right)$

$\angle {V^{\prime }}_a=0°$        (10)

$\angle {V^{\prime }}_b=-138°-\left(-18°\right)$

$\angle {V^{\prime }}_b=-120°$        (11)

And

$\angle {V^{\prime }}_c=102°-\left(-18°\right)$

$\angle {V^{\prime }}_c=120°$        (12)

From Equations (10), (11), and (12), the voltage at phase-b lags a-phase voltage by $120°$ and c-phase voltage leads a-phase voltage by $120°$. Therefore, this phase relationship is abc phase sequence or positive phase sequence.

Conclusion:

Thus, the phase sequence is abc sequence.