Question 4 of 6: Three-Phase Power

Question 4 — Three-Phase Power

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A Y-connected 4.16 kV (line-to-line), 60 Hz source feeding a delta load whose each branch is $Z_\phi=50+j30\ \Omega$; $V_{an}=2401.8\ \text{V}\angle0^\circ$, positive (abc) rotation, line impedance negligible.

Find. Line-current magnitude; the three phase (L–N) and three line (L–L) voltages; the three line and three delta-phase currents; the absorbed power; the per-phase capacitor kVAR for 0.93 lagging; and the corrected line current.

Approach. Because the load is delta, each branch sees the full line voltage; find the delta phase currents $I_{ab},I_{bc},I_{ca}=V_{LL}/Z_\phi$, then the line currents $I_L=\sqrt3\,I_p$ shifted $-30^\circ$. Compute $P$ and $Q$ from $3I_p^2Z_\phi$, size the capacitors from $\Delta Q$, and re-evaluate the line current.

1. Line and phase voltages. The source phase (line-to-neutral) voltages are $V_{an}=2401.8\angle0^\circ$, $V_{bn}=2401.8\angle{-120^\circ}$, $V_{cn}=2401.8\angle120^\circ\ \text{V}$. The line (line-to-line) voltages — which appear directly across the delta branches — are $V_{ab}=4160\angle30^\circ$, $V_{bc}=4160\angle{-90^\circ}$, $V_{ca}=4160\angle150^\circ\ \text{V}$.
2. Delta phase currents. Each branch carries $$I_{ab}=\frac{V_{ab}}{Z_\phi}=\frac{4160\angle30^\circ}{58.31\angle30.96^\circ} =71.3\ \text{A}\ \angle{-0.96^\circ}$$ and by symmetry $I_{bc}=71.3\angle{-121.0^\circ}$, $I_{ca}=71.3\angle119.0^\circ\ \text{A}$.
3. Line currents. $I_a=I_{ab}-I_{ca}=\sqrt3\,I_{ab}\angle{-30^\circ}$, giving $$\boxed{|I_L|=\sqrt3(71.3)=123.6\ \text{A}}$$ with $I_a=123.6\angle{-30.96^\circ}$, $I_b=123.6\angle{-150.96^\circ}$, $I_c=123.6\angle89.0^\circ\ \text{A}$.
4. Power absorbed. Per branch $|I_p|=71.3\ \text{A}$, so $P=3|I_p|^2(50)=\boxed{763.5\ \text{kW}}$ and $Q=3|I_p|^2(30)=458.1\ \text{kVAR}$ ($S=890.4\ \text{kVA}$, pf $=0.857$ lagging).
5. Capacitor sizing to 0.93 lagging. The corrected reactive demand is $Q'=P\tan(\cos^{-1}0.93)=763.5\tan21.57^\circ=301.8\ \text{kVAR}$, so the capacitors must supply $Q_C=Q-Q'=458.1-301.8=156.3\ \text{kVAR}$ total, i.e. $$\boxed{Q_{C,\phi}=156.3/3=52.1\ \text{kVAR per phase}}$$ (across each delta branch at 4160 V this is $C=Q_{C,\phi}/(\omega V_{LL}^2)=7.99\ \mu\text{F}$).
6. Corrected line current. With $P=763.5\ \text{kW}$ at pf 0.93, $$I_L'=\frac{P}{\sqrt3\,V_{LL}\,(0.93)}=\frac{763{,}500}{\sqrt3(4160)(0.93)}=\boxed{113.9\ \text{A}}$$