Balanced Three-Phase Power Systems
Demonstrative Video
Power in a Balanced System
\[\begin{aligned} v_{A N}& =\sqrt{2} V_{p} \cos \omega t \\ v_{B N}&=\sqrt{2} V_{p} \cos \left(\omega t-120^{\circ}\right) \\ v_{C N}&=\sqrt{2} V_{p} \cos \left(\omega t+120^{\circ}\right) \end{aligned}\] \[\begin{aligned} i_{a}&=\sqrt{2} I_{p} \cos (\omega t-\theta), \\ i_{b}&=\sqrt{2} I_{p} \cos \left(\omega t-\theta-120^{\circ}\right) \\ i_{c}&=\sqrt{2} I_{p} \cos \left(\omega t-\theta+120^{\circ}\right) \end{aligned}\]
\[\begin{aligned} p&=p_{a}+p_{b}+p_{c}=v_{A N} i_{a}+v_{B N} i_{b}+v_{C N} i_{c} \\ &=2 V_{p} I_{p}\left[\cos \omega t \cos (\omega t-\theta)+\cos \left(\omega t-120^{\circ}\right) \cos \left(\omega t-\theta-120^{\circ}\right)\right. +\\ & ~~ \qquad \left.\cos \left(\omega t+120^{\circ}\right) \cos \left(\omega t-\theta+120^{\circ}\right)\right] \\ =&V_{p} I_{p}\left[3 \cos \theta+\cos (2 \omega t-\theta)+\cos \left(2 \omega t-\theta-240^{\circ}\right)+\cos \left(2 \omega t-\theta+240^{\circ}\right)\right] \\ =&V_{p} I_{p}[3 \cos \theta+\cos \alpha+\cos \alpha \cos 240^{\circ}+\sin \alpha \sin 240^{\circ}+\cos \alpha \cos 240^{\circ}\\ & \qquad \qquad -\sin \alpha \sin 240^{\circ}] \\ =&V_{p} I_{p}\left[3 \cos \theta+\cos \alpha+2\left(-\frac{1}{2}\right) \cos \alpha\right]=\boxed{3 V_{p} I_{p} \cos \theta}~~\boxed{\alpha = 2\omega t - \theta } \end{aligned}\] Note: \(\cos A \cos B=\frac{1}{2}[\cos (A+B)+\cos (A-B)]\)
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.
\[\begin{aligned} P_{p} &=V_{p} I_{p} \cos \theta \\ Q_{p} &=V_{p} I_{p} \sin \theta \\ S_{p} &=V_{p} I_{p} \\ \mathbf{S}_{p} &=P_{p}+j Q_{p}=\mathbf{V}_{p} \mathbf{I}_{p}^{*} \\ P &=P_{a}+P_{b}+P_{c} \\ &=3 P_{p} \\ &=3 V_{p} I_{p} \cos \theta=\sqrt{3} V_{L} I_{L} \cos \theta \end{aligned}\] \[\begin{aligned} &\boxed{\mathbf{S} =3 \mathbf{S}_{p}=3 \mathbf{V}_{p} \mathbf{I}_{p}^{*}=3 I_{p}^{2} \mathbf{Z}_{p}=\frac{3 V_{p}^{2}}{\mathbf{Z}_{p}^{*}}}\\ &\boxed{\mathbf{S} =P+j Q=\sqrt{3} V_{L} I_{L} / \theta} \end{aligned}\]
\[\begin{aligned} P_{\text {loss }} &=2 I_{L}^{2} R=2 R \frac{P_{L}^{2}}{V_{L}^{2}}\\ P_{\text {loss }}^{\prime}&=3\left(I_{L}^{\prime}\right)^{2} R^{\prime}=3 R^{\prime} \frac{P_{L}^{2}}{3 V_{L}^{2}}=R^{\prime} \frac{P_{L}^{2}}{V_{L}^{2}}\\ &\frac{P_{\text {loss }}}{P_{\text {loss }}^{\prime}}=\frac{2 R}{R^{\prime}} \\ &R=\rho \ell / \pi r^{2} \text { and } R^{\prime}=\rho \ell / \pi r^{\prime 2} \\ &\frac{P_{\text {loss }}}{P_{\text {loss }}^{\prime}}=\frac{2 r^{\prime 2}}{r^{2}} \end{aligned}\] If the same power loss is tolerated in both systems, then \(r^2 = 2r^{\prime 2}\) \[\begin{aligned} &\frac{\text { Material for single-phase }}{\text { Material for three-phase }}\\ &=\frac{2\left(\pi r^{2} \ell\right)}{3\left(\pi r^{\prime 2} \ell\right)}=\frac{2 r^{2}}{3 r^{\prime 2}} \\ &=\frac{2}{3}(2)=1.333 \end{aligned}\] \(1\phi\) system use 33% more material than \(3\phi\) system.
Problem
Determine the total average power, reactive power, and complex power
at the source and at the load.
\[\begin{aligned} \mathbf{V}_{p} &=110 \angle 0^{\circ} \mathrm{V} \\ \mathbf{I}_{p} &=6.81 \angle-21.8^{\circ} \mathrm{A} \\ \mathbf{S}_{s} &=-3 \mathbf{V}_{p} \mathbf{I}_{p}^{*}\\ &=-3\left(110 \angle 0^{\circ}\right)\left(6.81 \angle 21.8^{\circ}\right) \\ &=-2247 \angle 21.8^{\circ}\\ &=-(2087+j 834.6) \mathrm{VA} \\ \mathbf{S}_{L} &=3\left|\mathbf{I}_{p}\right|^{2} \mathbf{Z}_{p} \\ \end{aligned}\] \[\begin{aligned} \mathbf{Z}_{p} &=10+j 8=12.81 \angle 38.66^{\circ} \\ \mathbf{I}_{p} &=\mathbf{I}_{a}=6.81 \angle-21.8^{\circ} \\ \mathbf{S}_{L} &=3(6.81)^{2} 12.81 \angle 38.66^{\circ}\\ &=1782 \angle 38.66 \\ &=(1392+j 1113) \mathrm{VA} \\ \mathbf{S}_{\ell} &=3\left|\mathbf{I}_{p}\right|^{2} \mathbf{Z}_{\ell}\\ &=3(6.81)^{2}(5-j 2)\\ &=695.6-j 278.3 \mathrm{VA} \end{aligned}\]