Question

A balanced three-phase source is supplying 60 kVA at 0.6 lagging to two balanced Y-connected parallel loads. The distribution line connecting the source to the load has negligible impedance. Load 1 is purely resistive and absorbs 30 kW. Find the per-phase impedance of Load 2 if the line voltage is and the impedance components are in series

Answer 1

Answer 2

Load1 is pure resistive load .Load2 is unknownST=60KvA at 0.6 lagging (KvA of total load)                   ↑⏐ cos0PT=60 KvA x 0.6 =36 KwQT=60KvA x sin cos−10.6=60 KvA x 0.8=48K VAR (lagging)P1=30 kwQ1=0 KVARS1=30 KVAR (Pure resistive load)$$\begin{gathered} Loa{d}_1 is pure resistive load . \\ Loa{d}_2 is unknown \\ {S}_T=60KvA at 0.6 lagging \left(KvA of total load\right) \\ \uparrow \cos 0 \\ {P}_T=60 KvA x 0.6 =36 Kw \\ {Q}_T=60KvA x \sin {\cos }^{-1}0.6=60 KvA x 0.8=48K VAR \left(lagging\right) \\ {P}_1=30 kw \\ {Q}_1=0 KVAR \\ {S}_1=30 KVAR \left(Pure resistive load\right) \end{gathered}$$

 

To find unknown impedence 'z' :P2=PT−P1=36kw−30kw=6kwQI=QT−Q1=48 KVAR−0 KVAR=48 KVARS2=P2+JQ2=6 KW+J 48 KVAR↑⏐⏐⏐⏐⏐⏐⏐Complex power of loadPower factor of load 2=Cos tan−1 (48 KVAR(Q)6KW (P))                        cos ∅2=0.124 (lagging)Ic=line current=∣∣S2∣∣3√V2=(6x103)2+(48x103)2√3√x1203                                     =23.2156 APerphase impedence Zph=∣∣VpH∣∣∣∣IL∣∣∠cos−1(0.124)                                     =12033√x23.2156∠cos−1(0.124)                                     =(3.7097+J 29.6865)ΩPer phase impedence Zph=3.7079+J29.6865 Ω