Question

A balanced three-phase load is connected to a 4.16kV, three-phase, four Wire, grounded-wye dedicated distribution feeder. The load can be modeled by an impedance of ${\mathbf{Z}}_{\mathbf{L}}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{j}\mathbf{9}\mathbf{)}\mathbf{\Omega }\mathbf{/}\mathit{p}\mathit{h}\mathit{a}\mathit{s}\mathit{e}$wye-connected. The impedance of the phase conductors is $\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{j}\mathbf{1}\mathbf{)}\mathbf{\Omega }$. Determine the following by using the phase to neutral voltage as a reference and assume positive phase sequence:

(a) Line currents for phases A, B and C.

(b) Line-to-neutral voltages for all three phases at the load.

(c) Apparent, active, and reactive power dissipated per phase, and for all

three phases in the load.

(d) Active power losses per phase and for all three phases in the phase

conductors.

Short answer

(a) The line current for the phase A, B and C are $214.82\angle -63.{43}^{°},214.82\angle -183.{43}^{°}\mathrm{A}\mathrm{and}214.82\angle -303.{43}^{°}\mathrm{A}$respectively.

(b) The line to neutral voltage for phase A, B and C is , $2179.5\angle -1.01\mathrm{V},2179.5\angle -121.01\mathrm{V},2179.5\angle -241.01\mathrm{V}$respectively.

(c)The apparent, active and reactive power for the single phase is$468.2\angle 62.{42}^{°},216.77\mathrm{KW}\mathrm{and}414.99\mathrm{kVAR}$

and respectively.

The apparent, active and reactive power for three phases is $1404.6\mathrm{kVA},650.31\mathrm{kW}\mathrm{and}1244.97\mathrm{kVAR}$, and respectively.

(d) The active power loss for single phase is 13.84 kW and for three phases is 41.53kW .

Step-by-step solution

Write the all values given by the question.

Line to line voltage${\mathrm{V}}_{\mathrm{LL}}=4.16\mathrm{kV}$.

Load impedance${\mathrm{Z}}_{\mathrm{L}}=(4.7+\mathrm{j}9)\mathrm{\Omega }/\mathrm{phase}$.

The phase conductor impedance${\mathrm{Z}}_{\mathrm{load}}=(0.3+\mathrm{j}1)\mathrm{\Omega }$.

Determine the formulas to calculate the line current, line to neutral voltage for the three phase, apparent, active and reactive power for the three phases.

$$\begin{gathered} \mathbf{For}\mathbf{}\mathbf{the}\mathbf{}\mathbf{balance}\mathbf{}\mathbf{three}\mathbf{}\mathbf{phase}\mathbf{}\mathbf{system}\mathbf{,}\mathbf{}\mathbf{all}\mathbf{}\mathbf{the}\mathbf{}\mathbf{line}\mathbf{}\mathbf{current}\mathbf{}\mathbf{separated}\mathbf{}\mathbf{by}\mathbf{}\mathbf{the}\mathbf{}\mathbf{phase}\mathbf{}\mathbf{shift}\mathbf{}\mathbf{of}\mathbf{}\mathbf{}\mathbf{.}\mathbf{}\mathbf{} \\ \mathrm{The}\mathrm{line}\mathrm{to}\mathrm{neutral}\mathrm{voltage}\mathrm{for}\mathrm{the}\mathrm{phase}\mathrm{A}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\left({\mathrm{V}}_{\mathrm{AN}}\right)}_{\mathrm{Load}}={\mathrm{V}}_{\mathrm{AN}}-{\mathrm{l}}_{\mathrm{A}}{\mathrm{Z}}_{\mathrm{Load}}........\left(4\right) \\ \mathbf{For}\mathbf{}\mathbf{the}\mathbf{}\mathbf{balance}\mathbf{}\mathbf{three}\mathbf{}\mathbf{phase}\mathbf{}\mathbf{system}\mathbf{}\mathbf{line}\mathbf{}\mathbf{to}\mathbf{}\mathbf{neutral}\mathbf{}\mathbf{voltage}\mathbf{}\mathbf{separated}\mathbf{}\mathbf{by}\mathbf{}\mathbf{the}\mathbf{}\mathbf{phase}\mathbf{}\mathbf{shift}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{120}}^{\mathbf{°}}\mathbf{}. \\ \mathrm{The}\mathrm{expression}\mathrm{for}\mathrm{the}\mathrm{apparent}\mathrm{power}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mathrm{S}}_{1\varnothing }={\left({\mathrm{V}}_{\mathrm{AN}}\right)}_{\mathrm{Load}}{\mathrm{l}}_{\mathrm{A}}^{*}.......\left(5\right) \\ \mathrm{The}\mathrm{apparent}\mathrm{power}\mathrm{for}\mathrm{three}\mathrm{phases}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mathrm{S}}_{3\varnothing }=3{\mathrm{S}}_{1\varnothing }........\left(6\right) \\ \mathrm{The}\mathrm{expression}\mathrm{for}\mathrm{the}\mathrm{active}\mathrm{power}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mathrm{P}}_{1\varnothing }={\mathrm{S}}_{1\varnothing }\mathrm{c}\mathrm{o}\mathrm{s}\left(\mathrm{\theta }\right)........\left(7\right) \\ \mathrm{Here}\mathrm{\theta }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{line}\mathrm{voltage}\mathrm{and}\mathrm{current}. \\ \mathrm{The}\mathrm{expression}\mathrm{for}\mathrm{the}\mathrm{reactive}\mathrm{power}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mathrm{Q}}_{1\varnothing }={\mathrm{S}}_{1\varnothing }\sin \left(\mathrm{\theta }\right)....\left(8\right) \\ \mathrm{The}\mathrm{active}\mathrm{power}\mathrm{for}\mathrm{all}\mathrm{three}\mathrm{phase}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mathrm{P}}_{3\varnothing }=3{\mathrm{P}}_{1\varnothing }.....\left(9\right) \\ \mathrm{The}\mathrm{reactive}\mathrm{power}\mathrm{for}\mathrm{all}\mathrm{three}\mathrm{phase}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mathrm{Q}}_{3\varnothing }=3{\mathrm{Q}}_{1\varnothing }...........\left(10\right) \\ \mathrm{The}\mathrm{active}\mathrm{power}\mathrm{loss}\mathrm{for}\mathrm{single}\mathrm{phase}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\left({\mathrm{P}}_{1\varnothing }\right)}_{\mathrm{Load}}={\mathrm{l}}^2\mathrm{R}...........\left(11\right) \\ \mathrm{The}\mathrm{active}\mathrm{power}\mathrm{loss}\mathrm{for}\mathrm{three}\mathrm{phase}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mathrm{P}}_{3\varnothing }=3{\mathrm{P}}_{1\varnothing }.......\left(12\right) \end{gathered}$$

Determine the line current for the phase and.

$$\begin{gathered} \left(\mathrm{a}\right) \\ \mathrm{Calculate}\mathrm{the}\mathrm{voltage}\mathrm{between}\mathrm{phase}A\mathrm{and}\mathrm{neutral}N\mathrm{by}\mathrm{using}\mathrm{the} \\ \mathrm{equation}\left(1\right). \\ {V}_{AN}=\frac{4.16\times 1000}{\sqrt{3}}\angle 0^{°} \\ {\mathrm{V}}_{\mathrm{AN}}=2401.77\angle 0^{°}\mathrm{V} \\ \mathrm{The}\mathrm{total}\mathrm{impedance}\mathrm{would}\mathrm{be}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{Load}\mathrm{impedance} \\ \mathrm{and}\mathrm{impedance}\mathrm{of}\mathrm{phase}\mathrm{conductor}(\mathrm{by}\mathrm{using}\mathrm{equation}2). \\ Z=4.7+\mathrm{j}9+0.3+\mathrm{j}1 \\ Z=5+\mathrm{j}10 \\ \mathrm{The}\mathrm{line}\mathrm{current}\mathrm{for}\mathrm{the}\mathrm{phase}\mathrm{A}\mathrm{can}\mathrm{be}\mathrm{calculate}\mathrm{by}\mathrm{using}\mathrm{the}\mathrm{equation}\left(3\right). \\ {l}_A=\frac{2401.7\angle 0^{°}}{5+\mathrm{j}10} \\ {l}_A=214.82\angle -63.{43}^{°} \\ \mathrm{For}\mathrm{the}\mathrm{balance}\mathrm{three}\mathrm{phase}\mathrm{system}\mathrm{the}\mathrm{line}\mathrm{current}{l}_B\mathrm{and}{l}_c\mathrm{would} \\ \mathrm{be}\mathrm{shifted}\mathrm{by}{120}^{°}. \\ \mathrm{The}\mathrm{line}\mathrm{current}{l}_B\mathrm{is}, \\ {l}_B=214.82\angle {(-63.43-120)}^{°} \\ {l}_B=214.82\angle -183.{43}^{°}\mathrm{A} \\ \mathrm{The}\mathrm{line}\mathrm{current}{l}_c\mathrm{is}, \\ {l}_c=214.82\angle {(-63.43-240)}^{°} \\ {l}_C=214.82\angle -303.{43}^{°}\mathrm{A} \\ \mathrm{Hence}\mathrm{the}\mathrm{line}\mathrm{current}\mathrm{for}\mathrm{the}\mathrm{phase}\mathrm{A},\mathrm{B}\mathrm{and}\mathrm{C}\mathrm{are}214.82\angle -63.{43}^{°}, \\ 214.82\angle -183.{43}^{°}\mathrm{A},\mathrm{and}214.82\angle -303.{43}^{°}\mathrm{A}\mathrm{respectively}. \end{gathered}$$

The line to neutral voltage for the phase and A , B nad C

Calculate the apparent, active and reactive power dissipated in all three phases.

Calculate the active for los for all per phase and three phases.