Question

Two balanced$\mathbf{Y}\mathbf{‐}$connected loads in parallel, one drawing$\mathbf{15}\mathbf{KW}$at$\mathbf{0}\mathbf{.}\mathbf{6}$power factor lagging and the other drawing$\mathbf{10}\mathbf{kvA}$at 0.8 power factor leading, are supplied by a balanced, three-phase, $\mathbf{480}\mathbf{‐}$volt source. (a) Draw the power triangle for each load • and for the combined load. (b) Determine the power factor of the combined load and State whether lagging or leading. (c) Determine the magnitude of the line current from the source (d)$\mathbf{∆}\mathbf{‐}$ connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg Of the A to make the source power factor unity? Give your answer in $\mathbf{\Omega }$.(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Short answer

(a) Therefore, the power triangle for load 1, 2 and combined load shown below

The power triangle for the load 1,

The power triangle for the load 2,

The combined power triangle

(b)The combined power factor of the loads is $0.854\mathrm{lagging}$.

(c)The magnitude of the line current from the source is$32.37\mathrm{A}$.

(d) The connected capacitors that are connected in the parallel with combined load is$49.37\mathrm{\Omega }$.

(e) The line current in each capacitor is $9.72\mathrm{A}$and line current from the source is $27.66\mathrm{A}$.

Step-by-step solution

Step 1: Draw the power triangle for each load • and for the combined load.

(a)

The reactive power absorbed by the load 1.

$\mathbf{Q}\mathbf{=}\mathbf{Ptan}\mathbf{\left(}\mathbf{cos}\left(\mathrm{pf}\right)\mathbf{\right)}$

Solve further as,

$$\begin{gathered} \mathrm{Q}=15\times {10}^3\tan \left({\cos }^{-1}0.6\right) \\ \mathrm{Q}=15\times {10}^3\times 1.33 \\ \mathrm{Q}=20\mathrm{kVAR} \end{gathered}$$

The complex power absorbed by the load 1

$$\begin{gathered} {\mathbf{S}}_{\mathbf{1}}\mathbf{=}\mathbf{P}\mathbf{+}\mathbf{jQ} \\ {S}_1=15+j20 \end{gathered}$$

Convert the above power in polar form

$$\begin{gathered} S_1=\sqrt{{15}^2+{20}^2}{\tan }^{-1}\left(\frac{20}{15}\right) \\ {S}_1=25\angle 53.13 \end{gathered}$$

The total power absorbed by the load 2,

$$\begin{gathered} S_2=10\angle -{\cos }^{-1}\left(0.8\right) \\ {S}_2=10\angle -36.86 \end{gathered}$$

The total complex power

$$\begin{gathered} \mathrm{s}={\mathrm{s}}_1+{\mathrm{s}}_2 \\ \mathrm{S}=25\angle 53.13+10\angle -36.86 \\ \mathrm{S}=26.92\angle 31.{32}^{\circ }\mathrm{kVA} \end{gathered}$$

The power triangle for the load 1,

The power triangle for the load 2,

Calculate the power factor of the combined load and State whether lagging or leading.

(b)

The combined load power factor can be calculated as

$$\begin{gathered} p.f=\cos \left(31.32\right) \\ pf=0.854\mathrm{lagging} \end{gathered}$$

Hence, the combined power factor of the loads is .

Calculate the magnitude of the line current from the source.

(c)

The line current can be calculated as

$$\begin{gathered} {\mathbf{l}}_{\mathbf{L}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{1}}}{\sqrt{\mathbf{3}{\mathbf{V}}_{\mathbf{LL}}}} \\ {\mathrm{l}}_{\mathrm{L}}=\frac{26.92\times {10}^3}{\sqrt{3}\times 480} \\ {\mathrm{l}}_{\mathrm{L}}=32.37\mathrm{A} \end{gathered}$$

Hence the magnitude of the line current from the source is$32.37\mathrm{A}$ .

Calculate the ∆‐connected capacitors that are connected in the parallel with combined load.

(d)

The total combined power in the polar form is $\left(23+\mathrm{j}14\right)\mathrm{kVAR}$.

The value of the capacitor can be calculated by the expression

$$\begin{gathered} X_C=\frac{3{\left(V_{LL}\right)}^2}{Q_C} \\ {X}_C=\frac{3\times {480}^2}{14\times {10}^3} \\ {X}_C=49.37\Omega \end{gathered}$$

Hence the capacitor value is$49.37\Omega$ .

Calculate the magnitude of the current in each capacitor and the line current from the source.

(e)

The magnitude of the current in each capacitor can be calculated as

$$\begin{gathered} {\mathrm{I}}_{\mathrm{C}}=\frac{{\mathrm{V}}_{\mathrm{LL}}}{{\mathrm{X}}_{\mathrm{C}}} \\ {\mathrm{I}}_{\mathrm{C}}=\frac{480}{49.37} \\ {\mathrm{I}}_{\mathrm{C}}=9.72\mathrm{A} \end{gathered}$$

The magnitude of the line current from the source can be calculated as

$$\begin{gathered} {\mathrm{I}}_{\mathrm{L}}=\frac{{\mathrm{P}}_{\mathrm{L}}}{\sqrt{3}{\mathrm{V}}_{\mathrm{LL}}} \\ {\mathrm{I}}_{\mathrm{L}}=\frac{23\times {10}^3}{\sqrt{3}\times 480} \\ {\mathrm{I}}_{\mathrm{L}}=27.66\mathrm{A} \end{gathered}$$

Hence the line current in each capacitor is$9.72\mathrm{A}$and line current from the source is$27.66\mathrm{A}$.