Question

A balanced delta connected impedance load with $\mathbf{(}\mathbf{12}\mathbf{+}\mathbf{j}\mathbf{9}\mathbf{)}\mathbf{}\mathbf{\Omega }$ per phase is supplied by a balanced three-phase $\mathbf{60}\mathbf{}\mathbf{Hz}$, $\mathbf{208}\mathbf{}\mathbf{V}$source. (a) Calculate the line current, the total real and reactive power absorbed the load, the load power factor, and the apparent power. (b) Sketch a phasor diagram showing the line currents, the line-to-line source voltages, and the $\mathbf{∆}$ -load currents. Use ${\mathbf{V}}_{\mathbf{ab}}$as the reference.

Short answer

(a)The line current is $24.016\angle -66.{86}^0\mathrm{A}$, the real power is $6921.21\mathrm{kW}$, the reactive power is $5189.71\mathrm{kVAR}$, the apparent power is $8654\mathrm{kVA}$, and power factor is 0.8.

(b)

The phasor diagram of line currents is drawn below:

The phasor diagram of phase currents is drawn below:

The phasor diagram of line-line voltage is drawn below:

Step-by-step solution

Determine the formulas of line current

Write the formula of phase voltage in-terms of line-voltage

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

Write the formula of per-phase impedance in star connection

${\mathbf{Z}}_{\mathbf{y}}\mathbf{=}\frac{{\mathbf{Z}}_{\mathbf{∆}}}{\mathbf{3}}$ ……. (2)

Write the formula of phase current/line current for star-connection.

${\mathbf{I}}_{\mathbf{ph}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{ph}}}{\mathbf{Z}}$ ……. (3)

Write the formula of power factor

$\mathbf{p}\mathbf{.}\mathbf{f}\mathbf{=}\mathbf{cos\theta }$ ……. (4)

Write the formula of active power for three-phase

$\mathbf{P}\mathbf{=}\mathbf{3}{\mathbf{V}}_{\mathbf{ph}}{\mathbf{l}}_{\mathbf{ph}}\mathbf{cos\theta }$ ……. (5)

Write the formula of reactive power

$\mathbf{Q}\mathbf{=}\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{ph}}{\mathbf{l}}_{\mathbf{ph}}\mathbf{sin\theta }$ ……. (6)

Write the formula of apparent power

$\mathbf{S}\mathbf{=}\sqrt{{\mathbf{P}}^{\mathbf{2}}\mathbf{+}{\mathbf{Q}}^{\mathbf{2}}}$ ……. (7)

Write the formla of line current

${\mathbf{l}}_{\mathbf{ph}}\mathbf{=}\frac{{\mathbf{l}}_{\mathbf{line}}}{\sqrt{\mathbf{3}}}$ ……. (8)

Determine the line current.

(a)

Determine the phase voltage.

Substitute $208\mathrm{V}$ for ${\mathrm{V}}_{\mathrm{line}}$equation (1).

$$\begin{gathered} {\mathrm{V}}_{\mathrm{ph}}=\frac{{\mathrm{V}}_{\mathrm{line}}}{\sqrt{3}}\angle -{30}^0 \\ =\frac{208}{\sqrt{3}}\angle -{30}^0 \\ =120.08\angle -{30}^0\mathrm{V} \end{gathered}$$

Determine the per-phase impedance in star-connection.

Substitute $(12+\mathrm{j}9)\mathrm{\Omega }$ for Z equation (2).

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{y}}=\frac{{\mathrm{Z}}_{∆}}{3} \\ =\frac{(12+\mathrm{j}9)}{3} \\ =(4+\mathrm{j}3)\mathrm{\Omega } \end{gathered}$$

Determine the per-phase phase current.

Substitute $120.08\mathrm{V}$ for ${\mathrm{V}}_{\mathrm{ph}}$and $(4+\mathrm{j}3)\mathrm{\Omega }$ for ${\mathrm{Z}}_{\mathrm{y}}$ in equation (3).

$$\begin{gathered} {\mathrm{l}}_{\mathrm{ph}}=\frac{{\mathrm{V}}_{\mathrm{ph}}}{{\mathrm{Z}}_{\mathrm{y}}} \\ =\frac{120.08\angle -{30}^0}{4+\mathrm{j}3} \\ =\frac{120.08\angle -{30}^0}{5\angle 36.{86}^0} \\ =24.016\angle -66.{86}^0\mathrm{A} \end{gathered}$$

Determine the load power factor, real power, reactive power and apparent power.

Determine the load power factor.

Substitute $36.{86}^0$ for$\mathrm{\theta }$ in equation (4).

$$\begin{gathered} \mathrm{pf}=\cos \left(36.{86}^0\right) \\ =0.8 \end{gathered}$$

Determine the active power for three-phase.

Substitute $120.08\mathrm{V}$ for ${\mathrm{V}}_{\mathrm{ph}}$, $24.016\mathrm{A}$ for ${\mathrm{l}}_{\mathrm{ph}}$ and 0.8 for $\mathrm{cos\theta }$in equation (5).

role="math" localid="1655184454397" $$\begin{gathered} \mathrm{P}=3\times 120.08\times 24.016\times 0.8 \\ =6921.21\mathrm{kW} \end{gathered}$$

Determine the reactive power.

Substitute $120.08\mathrm{V}$ for ${\mathrm{V}}_{\mathrm{ph}}$,role="math" localid="1655184384783" $24.046\mathrm{A}$for ${\mathrm{l}}_{\mathrm{ph}}$ and $36.{86}^0$ for $\mathrm{\theta }$ in equation (6).

$$\begin{gathered} \mathrm{Q}=3\times 120.08\times 24.016\times \sin 36.{86}^0 \\ =5189.71\mathrm{kVAR} \end{gathered}$$

Determine the apparent power.

Substitute $6921.21\mathrm{kW}$for $\mathrm{P}$and $5189.71\mathrm{kVAR}$for Q in equation (7).

$$\begin{gathered} \mathrm{S}=\sqrt{6921.{21}^2+5189.{71}^2} \\ =8654\mathrm{kVA} \end{gathered}$$

Determine the phasor diagram of line currents, line-line source voltages and delta load currents.

(b)

The phasor diagram of line current is drawn below:

Determine the phase current in delta-connection using equation (8).

$$\begin{gathered} {\mathrm{l}}_{\mathrm{ph}}=\frac{{\mathrm{l}}_{\mathrm{line}}}{\sqrt{3}}\angle {30}^0 \\ =\frac{24.016\angle -66.{86}^0}{\sqrt{3}}\angle {30}^0 \\ =13.87\angle -36.{86}^0\mathrm{A} \end{gathered}$$

The phasor diagram of phase currents is drawn below:

The phasor diagram of line-line voltage is drawn below: