Question

Two balanced three-phase loads that are connected in parallel are fed by a three-phase line having a series impedance of $\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{+}\mathit{j}\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{)}\mathbf{\Omega }$per phase. One of the loads absorbes $\mathbf{560}\mathbf{}\mathbf{kVA}$at 0.707 power factor lagging, and the other $\mathbf{132}\mathbf{}\mathbf{kW}$at unity power factor. The line-to-line voltage at te load end of the line is$\mathbf{2200}\sqrt{\mathbf{3}\mathbf{}}\mathit{V}$ . Compute (a) the line-to-line voltage at the source end of the line, (b) the total real and reactive power losses in the three-phase line and (c) the total three-phase real and reactive power supplied at the sending end of the line. Check that the total three-phase complex power delivered by the source equals the total three-phase complex power absorbed by the line and loads.

Short answer

(a) The source voltage is $4160\mathrm{V}$.

(b) The real and reactive power loss in line is $12\mathrm{kW}$and $81\mathrm{kVAR}$.

(c) The real and reactive power supplied by the source is$719.496\angle 41.{42}^{°}\mathrm{kVA}.$

Step-by-step solution

Determine the formulas of line current

Write the formula of phase voltage

${\mathrm{V}}_{\mathrm{ph}}=\frac{{\mathrm{V}}_{\mathrm{line}}}{\sqrt{3}}$ ……. (1)

Write the formula for power angle

$\mathrm{\theta }={\cos }^{-1}\mathrm{p}.\mathrm{f}$ ……. (2)

Write the formula of active power for three-phase

$\mathrm{p}=\mathrm{Scos\theta }$ ……. (3)

Write the formula of reactive power for three phase

$\mathrm{Q}=\mathrm{S}.\mathrm{sin\theta }$ ……. (4)

Write the formula of apparent power

$\mathrm{S}=\sqrt{{\mathrm{p}}^2+{\mathrm{Q}}^2}$ ……. (5)

Write the formla of apparent power for three-phase

$\mathrm{S}=3{\mathrm{V}}_{\mathrm{ph}}{\mathrm{l}}_{\mathrm{ph}}$ ……. (6)

Write the formula of power angle

$\mathrm{\theta }={\cos }^{-1}\frac{\mathrm{P}}{\mathrm{S}}$ …… (7)

Write the formula of line voltage

$V_L=I_{ph}{Z}_L$ …… (8)

Write the formula of source voltage

${\mathrm{V}}_{\mathrm{s}}={\mathrm{V}}_{\mathrm{L}}+{\mathrm{V}}_{\mathrm{ph}}$ …… (9)

Write the formula of apparent power loss in a line.

$S_L=3l_{ph}^2(R_L+j{X}_L)$ …… (10)

Determine the line-to-line source voltage.

$$\begin{gathered} \left(\mathrm{a}\right) \\ \mathrm{Determine}\mathrm{the}\mathrm{phase}\mathrm{voltage}. \\ \mathrm{Substitute}2200\sqrt{3}\mathrm{V}\mathrm{for}{\mathrm{V}}_{\mathrm{line}}\mathrm{in}\mathrm{equeation}\left(1\right). \\ {V}_{\mathit{p}\mathit{h}}\mathit{=}\frac{\mathit{2200}\sqrt{\mathit{3}\mathit{}}}{\sqrt{\mathit{3}}} \\ =2200\mathrm{v} \\ \mathrm{Determine}\mathrm{the}\mathrm{power}\mathrm{angle}\mathrm{at}\mathrm{first}\mathrm{load}\mathrm{using}\mathrm{equation}\left(2\right). \\ \theta =\cos \left(0.707\right) \\ ={45}^{°} \\ \mathrm{Detremine}\mathrm{the}\mathrm{active}\mathrm{power}\mathrm{at}\mathrm{first}\mathrm{source}\mathrm{using}\mathrm{equation}\left(3\right) \\ P\mathit{=}\mathit{560}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{3}}\mathit{\times }cos\mathit{}{\mathit{45}}^{\mathit{°}} \\ \mathrm{Detremine}\mathrm{the}\mathrm{reactive}\mathrm{power}\mathrm{at}\mathrm{first}\mathrm{source}\mathrm{using}\mathrm{equation}\left(4\right) \\ \mathrm{Q}=560\times {10}^3\times \sin {45}^{°} \\ =395.97\mathrm{kW} \\ \mathrm{Determine}\mathrm{the}\mathrm{total}\mathrm{apparent}\mathrm{power}\mathrm{at}\mathrm{load}\mathrm{end}\mathrm{using}\mathrm{equation}\left(5\right) \\ \mathrm{S}=\sqrt{{\left(395.97+132\right)}^2+395.{97}^2} \\ =659.59\mathrm{kVA} \\ \mathrm{Determine}\mathrm{the}\mathrm{per}\mathrm{phase}\mathrm{current}\mathrm{by}\mathrm{the}\mathrm{source}. \\ \mathrm{Substitute}2200\mathrm{V}\mathrm{for}{\mathrm{V}}_{\mathrm{ph}}\mathrm{and}659.59\mathrm{kVA}\mathrm{for}\mathrm{in}\mathrm{equation}\left(6\right). \\ 659.59\times {10}^3=3\times 2200\times l_{\mathrm{ph}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}{l}_{\mathit{p}\mathit{h}}\mathit{=}\frac{\mathit{659}\mathit{.}\mathit{59}\mathit{\times }{\mathit{10}}^{\mathit{3}}}{\mathit{3}\mathit{\times }\mathit{2200}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\mathit{00}\mathit{.}\mathit{93}\mathit{}A \\ \mathrm{Determine}\mathrm{the}\mathrm{power}\mathrm{factor}\mathrm{for}\mathrm{total}\mathrm{apparent}\mathrm{power}\mathrm{using}\mathrm{equation}\left(7\right) \\ \mathrm{\theta }={\cos }^{-1}\frac{132+3.95+97}{659.59} \\ ={\cos }^{-1}0.8 \\ =36.{86}^{°} \\ \mathrm{Determine}\mathrm{the}\mathrm{voltage}\mathrm{drop}\mathrm{on}\mathrm{the}\mathrm{line}. \\ \mathrm{Substitute}(0.4+\mathrm{j}2.7)\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{L}}\mathrm{and}99.93\angle -36.{86}^{°}\mathrm{A}\mathrm{for}{l}_{\mathit{p}\mathit{h}}\mathit{}\mathrm{in}\mathrm{equation}\left(8\right). \\ {\mathrm{V}}_{\mathrm{L}}=(0.04+j2.7\mathit{)}99.93\angle -36.{86}^{°} \\ =(0.4+j2.7)(79.954\mathit{-}j59.944) \\ =272.755\angle 44.{713}^{°}\mathrm{V} \\ \mathrm{Determine}\mathrm{the}\mathrm{phase}\mathrm{voltage}\mathrm{at}\mathrm{source}\mathrm{end}. \\ \mathrm{Substitute}2200\mathrm{V}{\mathrm{forV}}_{\mathrm{ph}}\mathrm{and}272.28\angle 44.{713}^{°}\mathrm{for}{\mathrm{V}}_{\mathrm{L}}\mathrm{in}\mathrm{equation}\left(9\right). \\ {\mathrm{V}}_{\mathrm{S}}=2200+272.28\angle 44.{13}^{°} \\ =2400\angle 4.{56}^{°}\mathrm{V} \\ \mathrm{Determine}\mathrm{the}\mathrm{line}\mathrm{to}\mathrm{line}\mathrm{voltage}\mathrm{at}\mathrm{the}\mathrm{source}\mathrm{end}. \\ {\left({\mathrm{E}}_{\mathrm{cource}}\right)}_{\mathrm{L}-\mathrm{L}}=\sqrt{3}\left(2400\right) \\ \approx 4160\mathrm{V} \\ \mathrm{Therefore},\mathrm{the}\mathrm{line}\mathrm{to}\mathrm{line}\mathrm{voltage}\mathrm{source}\mathrm{end}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{is}4160\mathrm{V}. \end{gathered}$$

Determine the power losses in the line.

(b)

Determine the power loss in line using equation (10)

$$\begin{gathered} \mathrm{S}=3\times 99.{93}^2(0.4+j2.7) \\ =12\mathrm{k}+\mathrm{j}81\mathrm{kVAR} \\ =81.88\mathrm{VA} \\ \mathrm{Therefore},\mathrm{the}\mathrm{real}\mathrm{power}\mathrm{loss}\mathrm{is}{P}_{\mathit{L}}\mathit{=}12\mathrm{kW}\mathrm{and}\mathrm{the}\mathrm{reactive}\mathrm{power}\mathrm{loss}\mathrm{is}{\mathrm{Q}}_{\mathrm{L}}=\mathrm{kVAR} \end{gathered}$$

Determine the real and reactive power supplied by the source.

(c)

Determine the real power supplied by the source.

$$\begin{gathered} P_S=P+P_L \\ =395.97\mathrm{KW}+132\mathrm{kW}+12\mathrm{kW} \\ =539.97\mathrm{kW} \\ \mathrm{Determine}\mathrm{the}\mathrm{reactive}\mathrm{power}\mathrm{supplied}\mathrm{by}\mathrm{the}\mathrm{source}. \\ {\mathrm{Q}}_{\mathrm{S}}=\mathrm{Q}+{\mathrm{Q}}_{\mathrm{L}} \\ =395.97\mathrm{kVAR}+81\mathrm{kVAR} \\ =467.97\mathrm{kVAR} \\ \mathrm{The}\mathrm{three}\mathrm{phase}\mathrm{apparent}\mathrm{power}\mathrm{is}, \\ S=\sqrt{539.{97}^2+467.{97}^2} \\ =720.46\mathrm{k}\mathrm{VA} \\ \mathrm{The}\mathrm{three}\mathrm{phase}\mathrm{sending}\mathrm{power}\mathrm{is} \\ \mathrm{S}=3\times 2400\angle 4.{56}^{°}\times 99.93\angle 36.{86}^{°} \\ =719.496\angle 41.{42}^{°}\mathrm{kVA} \end{gathered}$$

Therfeore, the power supplied by the source is same as power aborbed by the load.