Question

An industrial load consisting of a bank of induction motors consumes $\mathbf{50}\mathbf{KW}$at a power factor of $\mathbf{0}\mathbf{.}\mathbf{8}$lagging from a$\mathbf{220}\mathbf{-}\mathbf{V}\mathbf{,}\mathbf{60}\mathbf{-}\mathbf{Hz}\mathbf{}\mathbf{,}$single-phase source. By placing a bank of capacitors in parallel with the load, the resultant power factor is to be raised to$\mathbf{0}\mathbf{.}\mathbf{95}$ lagging. Find the net capacitance of the capacitor bank in that is required.

Short answer

Therefore, the net capacitance of the capacitor bank is $1154\mu F.$

Step-by-step solution

determine the formula to find the values of power factor, reactive power and capacitance.

The power factor is given by,

$\mathbf{pf}\mathbf{=}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{\theta }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

The reactive power consumed by inductor is given by,

$\mathit{Q}\mathbf{=}\mathbf{Ptan\theta }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

The value of capacitance is given by,
$\mathbf{C}\mathbf{=}\frac{{\mathbf{Q}}_{\mathbf{c}}}{{\mathbf{v}}^{\mathbf{2}}\mathbf{\left(}\mathbf{2}\mathbf{mf}\mathbf{\right)}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

calculate the net capacitance of the capacitor bank in that is required. 

The figure for the calculation us shown below.

By substituting the values in equation (1) calculate the power factor before adding the capacitor.

$$\begin{gathered} (p\mathit{.}f)i\mathit{=}{\cos }^{-1}(0.8) \\ {(p\mathit{.}f)}_i=36.{87}^{\circ } \end{gathered}$$

Calculate the reactive power consumed by the inductor by substitute the values in equation (2),

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{i}}=50\tan (36.{87}^{\circ }) \\ {\mathrm{Q}}_{\mathrm{i}}=37.5\mathrm{kVAR} \end{gathered}$$

By substituting the values in equation (1) calculate the power factor after adding the capacitor.

$$\begin{gathered} {(p\mathit{.}f)}_f={\cos }^{-1}(0.95) \\ {(p\mathit{.}f)}_f=18.{19}^{\circ } \end{gathered}$$

Calculate the reactive power consumed by the load

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{f}}=50\tan (18.{19}^{\circ }) \\ {\mathrm{Q}}_{\mathrm{f}}=1643\mathrm{kVAR} \end{gathered}$$

The difference in the reactive power can be calculated as

$\mathrm{Q}={\mathrm{Q}}_1-{\mathrm{Q}}_f$

Substitute the values and solve as

role="math" localid="1655110960636" $$\begin{gathered} \mathrm{Q}=37.5-16.43 \\ \mathrm{Q}=21.07\mathrm{kVAR} \end{gathered}$$

Substitute the values in equation (3) and calculate the capacitance value

$$\begin{gathered} \mathrm{C}=\frac{21.07\times {10}^3}{{220}^2\times 2\mathrm{\pi }\times 60} \\ \mathrm{C}=\frac{21.07\times {10}^3}{1.824\times {10}^7} \\ \mathrm{C}=1154\mu \mathrm{F} \end{gathered}$$

Therefore, the capacitance of the capacitor bank is$1154\mu \mathrm{F}$