Master Balanced Three-Phase Power Systems

Demonstrative Video

Power in a Balanced System

\begin{matrix} v_{A N} & = \sqrt{2} V_{p} cos ω t v_{B N} & = \sqrt{2} V_{p} cos (ω t - 120^{\circ}) v_{C N} & = \sqrt{2} V_{p} cos (ω t + 120^{\circ}) \end{matrix}

$\begin{aligned} v_{A N}& =\sqrt{2} V_{p} \cos \omega t \\ v_{B N}&=\sqrt{2} V_{p} \cos \left(\omega t-120^{\circ}\right) \\ v_{C N}&=\sqrt{2} V_{p} \cos \left(\omega t+120^{\circ}\right) \end{aligned}$

$\begin{aligned} p&=p_{a}+p_{b}+p_{c}=v_{A N} i_{a}+v_{B N} i_{b}+v_{C N} i_{c} \\ &=2 V_{p} I_{p}\left[\cos \omega t \cos (\omega t-\theta)+\cos \left(\omega t-120^{\circ}\right) \cos \left(\omega t-\theta-120^{\circ}\right)\right. +\\ & ~~ \qquad \left.\cos \left(\omega t+120^{\circ}\right) \cos \left(\omega t-\theta+120^{\circ}\right)\right] \\ =&V_{p} I_{p}\left[3 \cos \theta+\cos (2 \omega t-\theta)+\cos \left(2 \omega t-\theta-240^{\circ}\right)+\cos \left(2 \omega t-\theta+240^{\circ}\right)\right] \\ =&V_{p} I_{p}[3 \cos \theta+\cos \alpha+\cos \alpha \cos 240^{\circ}+\sin \alpha \sin 240^{\circ}+\cos \alpha \cos 240^{\circ}\\ & \qquad \qquad -\sin \alpha \sin 240^{\circ}] \\ =&V_{p} I_{p}\left[3 \cos \theta+\cos \alpha+2\left(-\frac{1}{2}\right) \cos \alpha\right]=\boxed{3 V_{p} I_{p} \cos \theta}~~\boxed{\alpha = 2\omega t - \theta } \end{aligned}$

Note:

Important Points:

Thus the total instantaneous power in a balanced $3-\phi$ system is constant—it does not change with time as the instantaneous power of each phase does.
This result is true whether the load is Y- or $\Delta$ -connected.
This is one important reason for using a three-phase system to generate and distribute power.

$\begin{aligned} P_{p} &=V_{p} I_{p} \cos \theta \\ Q_{p} &=V_{p} I_{p} \sin \theta \\ S_{p} &=V_{p} I_{p} \\ \mathbf{S}_{p} &=P_{p}+j Q_{p}=\mathbf{V}_{p} \mathbf{I}_{p}^{*} \\ P &=P_{a}+P_{b}+P_{c} \\ &=3 P_{p} \\ &=3 V_{p} I_{p} \cos \theta=\sqrt{3} V_{L} I_{L} \cos \theta \end{aligned}$

$\begin{aligned} P_{\text {loss }} &=2 I_{L}^{2} R=2 R \frac{P_{L}^{2}}{V_{L}^{2}}\\ P_{\text {loss }}^{\prime}&=3\left(I_{L}^{\prime}\right)^{2} R^{\prime}=3 R^{\prime} \frac{P_{L}^{2}}{3 V_{L}^{2}}=R^{\prime} \frac{P_{L}^{2}}{V_{L}^{2}}\\ &\frac{P_{\text {loss }}}{P_{\text {loss }}^{\prime}}=\frac{2 R}{R^{\prime}} \\ &R=\rho \ell / \pi r^{2} \text { and } R^{\prime}=\rho \ell / \pi r^{\prime 2} \\ &\frac{P_{\text {loss }}}{P_{\text {loss }}^{\prime}}=\frac{2 r^{\prime 2}}{r^{2}} \end{aligned}$

system. system use 33% more material than If the same power loss is tolerated in both systems, then

Problem

Determine the total average power, reactive power, and complex power at the source and at the load.

$\begin{aligned} \mathbf{V}_{p} &=110 \angle 0^{\circ} \mathrm{V} \\ \mathbf{I}_{p} &=6.81 \angle-21.8^{\circ} \mathrm{A} \\ \mathbf{S}_{s} &=-3 \mathbf{V}_{p} \mathbf{I}_{p}^{*}\\ &=-3\left(110 \angle 0^{\circ}\right)\left(6.81 \angle 21.8^{\circ}\right) \\ &=-2247 \angle 21.8^{\circ}\\ &=-(2087+j 834.6) \mathrm{VA} \\ \mathbf{S}_{L} &=3\left|\mathbf{I}_{p}\right|^{2} \mathbf{Z}_{p} \\ \end{aligned}$