Unbalanced Three-Phase Systems: Analysis and Solutions
Demonstrative Video
Unbalanced Three-Phase Systems
Caused by two possible situations:
the source voltages are not equal in magnitude and/or differ in phase by angles that are unequal,
load impedances are unequal
Unbalanced \(3\phi\) systems are solved by mesh and nodal analysis
The line currents are determined by Ohm’s law \[\begin{aligned} \mathbf{I}_{a} &=\frac{\mathbf{V}_{A N}}{\mathbf{Z}_{A}}, \quad \mathbf{I}_{b}=\frac{\mathbf{V}_{B N}}{\mathbf{Z}_{B}}, \quad \mathbf{I}_{c}=\frac{\mathbf{V}_{C N}}{\mathbf{Z}_{C}} \\ \mathbf{I}_{n} &=-\left(\mathbf{I}_{a}+\mathbf{I}_{b}+\mathbf{I}_{c}\right) \end{aligned}\]
To calculate power in an unbalanced three-phase system requires that we find the power in each phase
Total power is sum of the powers in the three phases.
Problem
Find:
the line currents,
the total complex power absorbed by the load,
the total complex power absorbed by the source.
\[\begin{aligned} \text{Mesh-1} \quad (10+j 5) \mathbf{I}_{1}-10 \mathbf{I}_{2} &=120 \sqrt{3} \angle {30^{\circ}} \\ \text{Mesh-2} \quad -10 \mathbf{I}_{1}+(10-j 10) \mathbf{I}_{2} &=120 \sqrt{3} \angle-90^{\circ} \\ {\left[\begin{array}{cc} 10+j 5 & -10 \\ -10 & 10-j 10 \end{array}\right]\left[\begin{array}{l} \mathbf{I}_{1} \\ \mathbf{I}_{2} \end{array}\right] } &=\left[\begin{array}{c} 120 \sqrt{3} \angle 30^{\circ} \\ 120 \sqrt{3} \angle-90^{\circ} \end{array}\right] \\ \mathbf{I}_{1} &=56.78 \mathrm{~A} \\ \mathbf{I}_{2} &=42.75 \angle 24.9^{\circ} \mathrm{A} \end{aligned}\]
\[\begin{aligned} &\mathbf{I}_{a}=\mathbf{I}_{1}=56.78 \mathrm{~A}\\ & \mathbf{I}_{c}=-\mathbf{I}_{2}=42.75 \angle-155.1^{\circ} \mathrm{A} \\ &\mathbf{I}_{b}=\mathbf{I}_{2}-\mathbf{I}_{1}=25.46 \angle 135^{\circ} \mathrm{A} \\ &\mathbf{S}_{A}=\left|\mathbf{I}_{a}\right|^{2} \mathbf{Z}_{A}\\ &=(56.78)^{2}(j 5)=j 16,120 \mathrm{VA} \\ &\mathbf{S}_{B}=\left|\mathbf{I}_{b}\right|^{2} \mathbf{Z}_{B}\\ &=(25.46)^{2}(10)=6480 \mathrm{VA} \\ &\mathbf{S}_{C}=\left|\mathbf{I}_{c}\right|^{2} \mathbf{Z}_{C}\\ &=(42.75)^{2}(-j 10)\\ &=-j 18,276 \mathrm{VA} \\ &\mathbf{S}_{L}=\mathbf{S}_{A}+\mathbf{S}_{B}+\mathbf{S}_{C}\\ &=6480-j 2156 \mathrm{VA} \end{aligned}\] \[\begin{aligned} &\mathbf{S}_{a}=-\mathbf{V}_{a n} \mathbf{I}_{a}^{*}\\ &=-\left(120 \angle 0^{\circ}\right)(56.78)=-6813.6 \mathrm{VA}\\ &\mathbf{S}_{b}=-\mathbf{V}_{b n} \mathbf{I}_{b}^{*}\\ &=-\left(120 \angle-120^{\circ}\right)\left(25.46 \angle-135^{\circ}\right)\\ &=790-j 2951.1 \mathrm{VA}\\ &\mathbf{S}_{c}=-\mathbf{V}_{b n} \mathbf{I}_{c}^{*}\\ &=-\left(120\angle 120^{\circ}\right)\left(42.75 \angle 155.1^{\circ}\right)\\ &= - 4 5 6 . 0 3 + j 5 1 0 9 . 7 \mathrm { VA }\\ &\mathbf{S}_{s}=\mathbf{S}_{a}+\mathbf{S}_{b}+\mathbf{S}_{c}\\ &=-6480+j 2156 \mathrm{VA} \end{aligned}\]