Question

An industrial plant consisting primarily of induction motor loads absorbs $\mathbf{500}\mathbf{kW}$at 0.6 power factor lagging. (a) Compute the required kVA rating of a shunt capacitor to improve the power factor to 0.9 lagging. (b) Calculate the resulting power factor if a synchronous motor rated at $\mathbf{500}\mathbf{}\mathbf{hp}$with 90 % efficiency operating at rated load and at unity power factor is added to the plant instead of the capacitor. Assume constant voltage$\left(1\mathrm{hp}=0.746\mathrm{KW}\right)$

Short answer

(a) The Therefore, the kVA rating of the shunt capacitor required is$424.51\mathrm{kVA}$.

(B) The power factor after connecting synchronous motor is 0.8.

Step-by-step solution

Determine the formulas of power factor, active power and reactive power.

Write the formula of power factor in terms of active and reactive power.

$$\begin{gathered} \delta ={\cos }^{-1}(0.6) \\ =53.{13}^{°} \end{gathered}$$ ……. (1)

Write the formula of active power.

$\mathrm{P}=\mathrm{Vlcos\sigma }$ ……. (2)

Write the formula of reactive power.

role="math" localid="1655121774791" $$\begin{gathered} \mathrm{Q}=\mathrm{Vlsin\sigma } \\ =\mathrm{Vlcos\sigma }\left(\frac{\mathrm{sin\sigma }}{\cos \left(\mathrm{\sigma }\right)}\right) \\ =P\tan \sigma \end{gathered}$$ ……. (3)

Determine the product of rms voltage and current, reactive power supplied by source with or without capacitor and reactive power supplied by capacitor.

(a)

Calculate the rms voltage and current using equation (2).

Solve for the reactive power as.

$$\begin{gathered} \mathrm{Q}=\mathrm{Ptan\sigma } \\ =(500\mathrm{k})\tan 53.{13}^{°} \\ =666.67\mathrm{kvar} \end{gathered}$$

Consider the power factor angle is 0.9.

$$\begin{gathered} \sigma ={\cos }^{-1}(0.9) \\ =25.{84}^{°} \end{gathered}$$

Solve for the reactive power.

$$\begin{gathered} \mathrm{Q}=\mathrm{Ptan\sigma } \\ =(500\mathrm{k})\tan 25.{84}^{°} \\ =242.16\mathrm{kvar} \end{gathered}$$

Solve for the reactive power supplied.

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{c}}=(666.67-242.16)\mathrm{kvar} \\ =424.51\mathrm{kvar} \end{gathered}$$

Since the real power is not affected by the shunt capacitor then, the reactive power is,

$$\begin{gathered} S_{\mathrm{c}}={\mathrm{Q}}_{\mathrm{c}} \\ =424.51\mathrm{kVA} \end{gathered}$$

Therefore, the kVA rating of the shunt capacitor required is$424.51\mathrm{kVA}$

Determine the power factor after connecting synchronous motor

(b)

With 90% efficiency, the power output at synchronous motor is,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{out}}=\mathrm{n}\times {\mathrm{P}}_{\mathrm{in}} \\ =0.9\times 500\times 0.746 \\ =335.7\mathrm{kW} \end{gathered}$$

Determine the power factor using equation (1)

$$\begin{gathered} \mathrm{cos\theta }=\frac{(500\times {10}^{-3}+335.7\times {10}^3)}{\sqrt{{(500\times {10}^3+335.7\times {10}^3)}^2+(666.664\times {10}^3{)}^2}} \\ \mathrm{cos\theta }=0.8 \end{gathered}$$