Question

A three-phase line with an impedance of$\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{j}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{\Omega }$feeds three balanced three-phase loads connected in parallel.

Load I: Absorbs a total of 150kW and 120Kvar :

Load 2: Delta connected with an impedance oflocalid="1655120783137" $\mathbf{(}\mathbf{150}\mathbf{-}\mathbf{j}\mathbf{48}\mathbf{)}\mathbf{\Omega }\mathbf{/}$:

Load 3: 120KVA at 0.6PF leading:

If the line-to-neutral voltage at the load end of the line is 2000 v ( rms ) , determine the magnitude of the line-to-line voltage at the source end of the line.

Short answer

The line-to-line voltage at the sending end is 3478.73V.

Step-by-step solution

Determine the concept formulas to calculate the magnitude of the line-to-line voltage at the source end of the line.

The apparent power is given by,

S = VI (1)

The complex power absorbed per phase by the load is given by,

${\mathbf{S}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{P}}_{\mathbf{1}}\mathbf{+}{\mathbf{jQ}}_{\mathbf{1}}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Here, are the real and reactive power absorbed by the load 1 respectively.

The relation between the star and delta per phase connected load.

${\mathbf{Z}}_{\mathbf{Y}}\mathbf{=}\frac{{\mathbf{Z}}_{\mathbf{∆}}}{\mathbf{3}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Here, ${\mathbf{Z}}_{\mathbf{Y}}\mathbf{}\mathbf{\&}\mathbf{}{\mathbf{Z}}_{\mathbf{∆}}$are the impedance connected in start and delta respectively.

The line to node voltage at the source end of the line is given by,

${\mathbf{V}}_{\mathbf{LM}}\mathbf{=}\mathbf{2000}\mathbf{}\mathbf{+}{\mathbf{lZ}}_{\mathbf{L}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{4}\mathbf{\right)}$

Here, ${\mathbf{Z}}_{\mathbf{L}}$is the line impedance per phase.

Line to line voltage at the sending end is given by,

${\mathbf{V}}_{\mathbf{LL}}\mathbf{=}\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{LN}}$

Calculate the magnitude of the line-to-line voltage at the source end of the line.

By using the equation (2), calculate the complex power for load 1,

$$\begin{gathered} {\mathrm{S}}_1=\frac{1}{3}(150+\mathrm{j}120) \\ {\mathrm{S}}_1=\left|\right(50+\mathrm{j}40)\mathrm{KVA} \end{gathered}$$

By using the equation (1), calculate the current $l_1$,

$$\begin{gathered} S=V{I}^{°} \\ I=\frac{S^{*}}{V} \end{gathered}$$

Substitute the values and solve as

$$\begin{gathered} l_1=\frac{{\left[\left(50+\mathrm{j}40\right)\times 1000\right]}^{*}}{2000} \\ {l}_1=\left(25+\mathrm{j}20\right)\mathrm{A} \end{gathered}$$

By using the equation (3), calculate the per phase impedance for star connection for loas 2,

$$\begin{gathered} {\mathrm{Z}}_{2\mathrm{Y}}=\frac{150-\mathrm{j}48}{3} \\ {\mathrm{Z}}_{2\mathrm{Y}}=(50+\mathrm{j}16)\mathrm{\Omega }/\mathrm{phase} \end{gathered}$$

Calculate the current $l_2$drawn from the load 2,

$$\begin{gathered} {\mathrm{l}}_2=\frac{\mathrm{V}}{{\mathrm{Z}}_{2\mathrm{Y}}} \\ {\mathrm{l}}_2=\frac{2000}{50-\mathrm{j}16} \\ {\mathrm{l}}_2=(36.29-\mathrm{j}11.61)\mathrm{A} \end{gathered}$$

The per phase power absorbed by the load 3 is given by

$$\begin{gathered} {\mathrm{S}}_3=\frac{120}{3}\angle {\cos }^{-1}\left(0.6\right) \\ {\mathrm{S}}_3=40\angle -53.13° \end{gathered}$$

Convert the complex power in cartesian form

$$\begin{gathered} {\mathrm{S}}_3=40(0.6-\mathrm{j}0.8) \\ {\mathrm{S}}_3=(24-\mathrm{j}32)\mathrm{kVA} \end{gathered}$$

The current $l_3$can be calculated as

$$\begin{gathered} l_3=\frac{S^{*}}{V} \\ {l}_3=\frac{{\left[\left(24-j32\right)\times {10}^3\right]}^{*}}{2000} \\ {l}_3=\left(12+j16\right)A \end{gathered}$$

The total current drawn by the three-load connected in the parallel.

$$\begin{gathered} l=l_1+l_2+l_3 \\ l=25+j20+36.29-j11.61+12+j16 \\ l=(73.36+j7.61)A \end{gathered}$$

By using the equation (4), calculate the line to node voltage,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{LN}}=2000+(73.36+\mathrm{j}7.61)(0.2+\mathrm{j}1) \\ {\mathrm{V}}_{\mathrm{LN}}=2008.45\angle 2.{13}^{°}\mathrm{V} \end{gathered}$$

By using the equation (5), calculate the line-to-line voltage at the sending end

$$\begin{gathered} {\mathrm{V}}_{\mathrm{LL}}=\sqrt{3}\times 2008.45 \\ {\mathrm{V}}_{\mathrm{LL}}=3478.73\mathrm{V} \end{gathered}$$

Hence the line-to-line voltage at the sending end is 3478.73.