Question

Find the phase voltages, ${\mathbf{V}}_{\mathbf{an}\mathbf{}\mathbf{,}}{\mathbf{V}}_{\mathbf{bn}\mathbf{}\mathbf{,}\mathbf{}}\mathbf{and}\mathbf{}{\mathbf{V}}_{\mathbf{cn}}$whose sequence components are ${\mathbf{V}}_{\mathbf{0}}\mathbf{=}\mathbf{45}\mathbf{\angle }{\mathbf{80}}^{\mathbf{\circ }}\mathbf{,}{\mathbf{V}}_{\mathbf{1}}\mathbf{=}\mathbf{90}\mathbf{\angle }{\mathbf{0}}^{\mathbf{\circ }}\mathbf{}\mathbf{,}{\mathbf{V}}_{\mathbf{2}}\mathbf{=}\mathbf{45}\mathbf{\angle }{\mathbf{90}}^{\mathbf{\circ }}$

Short answer

The phase voltage${\mathrm{V}}_{\mathrm{an},}{\mathrm{V}}_{\mathrm{bn},}\mathrm{and}{\mathrm{V}}_{\mathrm{cn}}$are$132.46\angle {4240}^{\circ }\mathrm{V},94.6\angle 216.{39}^{\circ }\mathrm{V},\mathrm{and}99.78\angle 88.{98}^{\circ }\mathrm{V}$.

Step-by-step solution

Write the given data from the question.

A given sequence components are${\mathrm{V}}_0=45\angle {80}^{\circ },{\mathrm{V}}_1=90\angle 0^{\circ },{\mathrm{V}}_2=45\angle {90}^{\circ }$.

Determine the formula of maximum power flow mismatch.

Write the formula of ${\mathrm{V}}_{an}$.

${\mathbf{V}}_{\mathbf{an}}\mathbf{=}{\mathbf{V}}_{\mathbf{0}}\mathbf{+}{\mathbf{V}}_{\mathbf{1}}\mathbf{+}{\mathbf{V}}_{\mathbf{2}}$ ...........(1)

Here, ${\mathrm{V}}_{\mathrm{an}}$are phase voltages of an un-balanced three-phase network,, and are positive sequence voltage, negative sequence voltage and zero sequence voltage.

Write the formula of ${\mathrm{V}}_{\mathrm{bn}}$.

${\mathbf{V}}_{\mathbf{bn}}\mathbf{=}{\mathbf{V}}_{\mathbf{0}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}{\mathbf{V}}_{\mathbf{1}}\mathbf{+}{\mathbf{aV}}_{\mathbf{2}}$ ..........(2)

Here, ${\mathrm{V}}_{\mathrm{bn}}$are phase voltages of an un-balanced three-phase network,, and are positive sequence voltage, negative sequence voltage and zero sequence voltage.

Write the formula of ${\mathrm{V}}_{\mathrm{cn}}$.

${\mathbf{V}}_{\mathbf{cn}}\mathbf{=}{\mathbf{V}}_{\mathbf{0}}\mathbf{+}{\mathbf{aV}}_{\mathbf{1}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}{\mathbf{V}}_{\mathbf{2}}$ ............(3)

Here, role="math" localid="1655441352745" ${\mathrm{V}}_{\mathrm{cn}}$are phase voltages of an un-balanced three-phase network ${\mathrm{V}}_{1,}{\mathrm{V}}_{2,}\mathrm{and}{\mathrm{V}}_0$and are positive sequence voltage, negative sequence voltage and zero sequence voltage.

Determine the phase voltage, Van ,Vbn , and Vcn.

Convert the sequence voltages in polar form into rectangular form for further solving.

$$\begin{gathered} {\mathrm{V}}_0=45\angle {80}^{\circ } \\ =7.81+\mathrm{j}44.32 \end{gathered}$$ …… (4)

Here, ${\mathrm{V}}_0$is zero sequence voltage.

Write the rectangular form for the voltage${\mathrm{V}}_1$.

$$\begin{gathered} {\mathrm{V}}_1=90\angle 0^{\circ } \\ =90 \end{gathered}$$ …… (5)

Here, ${\mathrm{V}}_1$is positive sequence voltage.

Write the rectangular form for the voltage ${\mathrm{V}}_2$.

$$\begin{gathered} {\mathrm{V}}_2=45\angle {90}^{\circ } \\ =\mathrm{j}45 \end{gathered}$$ …… (6)

Here, role="math" localid="1655444239133" ${\mathrm{V}}_2$ is negative sequence voltage.

Determine the phase voltage ${\mathrm{V}}_{\mathrm{an}}$.

Substitute$7.81+\mathrm{j}44.32$ for, ${\mathrm{V}}_0=90$for ${\mathrm{V}}_1$and j45 for${\mathrm{V}}_2$ in the equation (1)

$$\begin{gathered} {\mathrm{V}}_{\mathrm{an}}=7.81+\mathrm{j}44.32+90+\mathrm{j}45 \\ =97.81+\mathrm{j}89.32 \\ =132.46\angle 42.{40}^{\circ }\mathrm{V} \end{gathered}$$

Determine the phase voltage ${\mathrm{V}}_{\mathrm{bn}}$.

Substitute$(0^{\circ }+{120}^{\circ })$ for a ,$(\angle {90}^{\circ }+{120}^{\circ })$ for ${\mathrm{a}}^2$,$45\angle {80}^{\circ }$ for,${\mathrm{V}}_0=90$ for ${\mathrm{V}}_1$and $45\angle {90}^{\circ }$for${\mathrm{V}}_2$ in the equation (2)

$$\begin{gathered} {\mathrm{V}}_{bn}=45\angle {80}^{\circ }+90(\angle 0^{\circ }+\angle {240}^{\circ })+45(\angle {90}^{\circ }+\angle {120}^{\circ }) \\ =45\angle {80}^{\circ }+90\angle {240}^{\circ }+45\angle {210}^{\circ } \\ =7.81+j44.32-45+j77.94+38.97-j22.5 \\ =-76.16-j56.12 \end{gathered}$$

As further solve as

${\mathrm{V}}_{\mathrm{bn}}=94.6\angle 216.{39}^{\circ }\mathrm{V}$

Determine the phase voltage ${\mathrm{V}}_{\mathrm{cn}}$.

Substitute role="math" localid="1655443573985" $(0^{\circ }+{120}^{\circ })$for a role="math" localid="1655443613633" $(\angle {90}^{\circ }+{120}^{\circ })$, for ${\mathrm{a}}^2$, role="math" localid="1655443660552" $45\angle {80}^{\circ }$for role="math" localid="1655443688755" ${\mathrm{V}}_0=90$, for ${\mathrm{V}}_1$and$45\angle {90}^{\circ }$ for${\mathrm{V}}_2$ in the equation (3)

role="math" localid="1655443555051" $$\begin{gathered} {\mathrm{V}}_{\mathrm{cn}}=45\angle {80}^{\circ }+90(\angle 0^{\circ }+\angle 120)+45({90}^{\circ }+\angle {240}^{\circ }) \\ =45\angle {80}^{\circ }+90\angle {120}^{\circ }+45\angle {330}^{\circ } \\ =7.81+\mathrm{j}44.32-45+\mathrm{j}77.94+38.97-\mathrm{j}22.5 \\ =1.78+\mathrm{j}99.76 \end{gathered}$$

As further solve as

${\mathrm{V}}_{\mathrm{cn}}=99.78\angle 88.{98}^{\circ }\mathrm{V}$

Therefore, the phase voltages are given as ${\mathrm{V}}_{\mathrm{an}}=132.46\angle 42.{40}^{\circ }\mathrm{V},{\mathrm{V}}_{\mathrm{bn}}=94.6\angle 216.{39}^{\circ }\mathrm{V}\mathrm{and}{\mathrm{V}}_{\mathrm{cn}}=99.78\angle 88.{98}^{\circ }\mathrm{V}$.