Question

shows three loads connected in parallel across a

1000-V (RMS) , 60-Hz single-phase source.

Load 1: Inductive load, 125kVA,0.28 PF lagging.

Load 2: Capacitive load, 10 kW, 40 kvar .

Load 3: Resistive load 15kW,

(a) Determine the total kW, kvar, kva, and supply power factor.

(b) In order to improve the power factor to 0.8 lagging, a capacitor of negligible resistance is connected in parallel with the above loads. Find the kvar rating of

that capacitor and the capacitance in$\mathbf{\mu F}$Comment on the magnitude of the supply current after adding the capacitor

Short answer

(a) Therefore, the KW, kvar, kVA and supply power factor are 60KW,80kvar,100kVA

and 0.6 lagging respectively.

(b) The reactive power of capacitor is 30KVAR and value of the capacitance is

$92.84\mathrm{\mu F}$and the change in the magnitude of the current is 25A .

Step-by-step solution

Calculate the total KW, KVAR, kVA and supply power factor.

(a)

Calculate the phase angle of the load 1.

$$\begin{gathered} \mathrm{\theta }={\cos }^1\left(\mathrm{pf}\right) \\ \mathrm{\theta }={\cos }^1\left(0.28\right) \\ \mathrm{\theta }=73.73\circ \end{gathered}$$

The complex power of the load 1 can be calculated as$$\begin{gathered} {\mathrm{S}}_1=125\left[\cos \left(73.73\right)+\mathrm{jsin}\left(73.73\right)\right] \\ {\mathrm{S}}_1=125\left(0.28+\mathrm{j}0.96\right) \\ {\mathrm{S}}_1=35+\mathrm{j}120\mathrm{kVA} \end{gathered}$$

The complex power of the load 2 can be calculated as

$$\begin{gathered} {\mathrm{S}}_2=\mathrm{P}-\mathrm{jQ} \\ {\mathrm{S}}_{2}10-\mathrm{j}40\mathrm{kVA} \end{gathered}$$

Since the load 3 is purely resistive, therefore the power of the load 3 is given by. ${\mathrm{S}}_{3}15\mathrm{kVA}$

The total complex power can be calculated as

$\mathrm{S}={\mathrm{S}}_1+{\mathrm{S}}_2+{\mathrm{S}}_3$

Substitute the values and solve as

$$\begin{gathered} \mathrm{S}=\left(35+\mathrm{j}120\right)+\left(10-\mathrm{j}40\right)+15 \\ \mathrm{S}=\left(60+\mathrm{j}80\right)\mathrm{kVA} \end{gathered}$$

Convert the above complex power into polar form

$\mathrm{S}=100\angle 53.13$

In the complex power (60+j80) , p = 60KW is the real power and Q = 80kvar is

the reactive power.

The supply power factor can be calculated as

$$\begin{gathered} \mathrm{\theta }=\cos \left(53.13\right) \\ \mathrm{\theta }=0.6 \end{gathered}$$

Hence, the KW, kvar, kVA and supply power factor are 60KW,80kvar, 100kVA and 0.6lagging respectively.

Calculate the kvar rating of that capacitor and the capacitance in PE Comment on the magnitude of the supply current after adding the capacitor.

(b)

The improved power factor is 0.8lagging. Calculate the new phase angle as

$$\begin{gathered} {\mathrm{\theta }}^{.}={\cos }^{-1}\left(0.8\right) \\ {\mathrm{\theta }}^{.}=36.{86}^{\circ } \end{gathered}$$

Consider the capacitor is added in the load, then the real power remains the same but reactive power is changed.

The reactive power can be calculated as

${\mathrm{Q}}_{\mathrm{c}}=\mathrm{p}\left[\tan \left(\mathrm{\theta }\right)-\tan \left(\mathrm{\theta }\right)\right]$

Substitute the values and solve as

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{c}}=60\left[\tan \left(53.13\right)-\tan \left(36.86\right)\right] \\ {\mathrm{Q}}_{\mathrm{c}}=60\times 0.5833 \\ {\mathrm{Q}}_{\mathrm{c}}=35\mathrm{KVAR} \end{gathered}$$

The value of the capacitance can be calculated as

$\mathrm{C}=\frac{{\mathrm{Q}}_{\mathrm{C}}}{2{\mathrm{\pi fV}}^2}$

Substitute the values and solve as

$$\begin{gathered} \mathrm{C}=\frac{35\times {10}^3}{2\mathrm{\pi }\times 60\times {\left(1000\right)}^2} \\ \mathrm{C}=\frac{7}{2\mathrm{\pi }\times 12\times 1000} \\ \mathrm{C}=\frac{7}{2\mathrm{\pi }\times 1000} \\ \mathrm{C}=92.85\mathrm{\mu F} \end{gathered}$$

Hence the reactive power of capacitor is 35KVAR and value of the capacitance is $92.84\mathrm{\mu F}$.

The reactive power is given by

$\mathrm{S}={\mathrm{VI}}^{\bullet }$

Derive the expression for the current

$I=\frac{{\mathrm{S}}^{\bullet }}{{\mathrm{V}}^{\bullet }}$

Substitute the values and solve as

$$\begin{gathered} I^{·}=\frac{100\times 1000\angle -53.13}{1000\angle 0^{\circ }} \\ {I}^{·}=100\angle -53.13\mathrm{A} \end{gathered}$$

After adding the capacitor, the complex power can be calculated as

$$\begin{gathered} {\mathrm{S}}^{.}=60+\mathrm{j}\left(80-35\right) \\ {\mathrm{S}}^{.}=60+\mathrm{j}45 \end{gathered}$$

After adding the capacitor, the value of the current

$$\begin{gathered} I^{·}=\frac{100\times 1000\angle -53.13}{1000\angle 0^{\circ }} \\ {I}^{·}=\frac{75\times 1000\angle -36.{86}^{\circ }}{1000\angle 0} \\ {I}^{·}=75\angle -36.{86}^{\circ }\mathrm{A} \end{gathered}$$

Change in magnitude of the current after adding the capacitor

$$\begin{gathered} I_{ch}=I-I^{.} \\ {I}_{ch}=100-75 \\ {I}_{ch}=25\mathrm{A} \end{gathered}$$

Hence, the change in the magnitude of the current is 25A .