Question

A three phase $\mathbf{25}\mathbf{}\mathbf{kVA}\mathbf{}\mathbf{,}\mathbf{480}\mathbf{}\mathbf{V}\mathbf{,}\mathbf{}\mathbf{60}\mathbf{}\mathbf{Hz}$alternator, operating under balanced steady-state conditions, supplies a line current of$\mathbf{20}\mathbf{}\mathbf{A}$per phase at a 0.8 lagging powerfactor and at rated voltage. Determine the power triangle for this operating condition.

Short answer

The power triangle for the given operating condition is drawn below:

Step-by-step solution

Determine the formulas of active power, reactive power, apparent power.

Write the formula of apparent power

$\mathbf{S}\mathbf{=}\sqrt{{\mathbf{P}}^{\mathbf{2}}\mathbf{+}{\mathbf{Q}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}...........\left(1\right)$

Write the formula of active power

$\mathbf{P}\mathbf{=}\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{rms}}{\mathbf{I}}_{\mathbf{rms}}\mathbf{.}\mathbf{cos\theta }\mathbf{}\mathbf{}\mathbf{}...........\left(2\right)$

Write the formula of reactive power

$\mathbf{Q}\mathbf{=}\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{rms}}{\mathbf{I}}_{\mathbf{rms}}\mathbf{.}\mathbf{sin\theta }\mathbf{}\mathbf{}\mathbf{}...........\left(3\right)$

Write the formula of power angle

$\mathbf{\theta }\mathbf{=}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{p}\mathbf{.}\mathbf{f}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}...........\left(4\right)$

Determine the power triangle.

Determine the active power.

Substitute $480\mathrm{V}$for$V_{rms}$,$20\mathrm{A}$for $l_{rms}$and 0.8 for $\cos \theta$ in equation (2).

$$\begin{gathered} P=\sqrt{3}\times 480\times 20\times 0.8 \\ =13.3\mathrm{kW} \end{gathered}$$

Determine the power angle.

Substitute 0.8 for $p.\mathrm{f}$ in equation (4).

$$\begin{gathered} \theta ={\cos }^{-1}0.8 \\ =36.86° \end{gathered}$$

Determine the reactive power.

Substitute $480\mathrm{V}$for $V_{rms}$, $20\mathrm{A}$ for $l_{rms}$ and $36.86°$for $\theta$in equation (3).

$$\begin{gathered} \mathrm{Q}=\sqrt{3}\times 480\mathrm{x}20\times \sin 36.86° \\ =9.976\mathrm{kVAR} \end{gathered}$$

Determine the apparent power.

Substitute $1.33\mathrm{kW}$for $P$and $9.976\mathrm{kVAR}$for$Q$in equation (1).

$$\begin{gathered} S=\sqrt{13.3^2+9.{976}^2} \\ =16.63\mathrm{kVA} \end{gathered}$$

Therefore, the power triangle is drawn below: