Question
For the 3-phase network above, the three connections at theleft side (with small circles on their ends), areconnected to a 3-phase power source, and the 3-phase sourceline-to-line voltage is 460 volts at 60Hz and the phase sequenceis: a-b-c. Zdelta = 15 + 15j ohms. The 3-phase wye loadis: 250 kW at a power factor of 0.8 leading. Assume Va -n as reference (0 degrees). a) Determine Iphi , Ia - b , andIa . (Answers are: Iphi = 392.22 atangle 36.87 degrees A, Ia - b = 21.68 at angle-15 degrees A, Ia = 399.26 at angle 31.53 degreesA, but please show all steps.) b) Determine the total apparent (complex) power supplied bythe source. (Answers are: ST source = 271.16 -j166.34 kVA = 318.11 at angle -31.53 degrees kVA, but please showall steps.) Thank you.
Explanation / Answer
a)Vs= source voltage(say)=460V line to line at 60 hz(notimp.) so line to neutral voltage = 460/3=265.6V 3 phase load = 250 Kw at .8 pf leading(cos) implies per phase load= 250/3=83.3 Kw P=3VIcos=P/3=VIcos=83.3 implies I = Iphi = 83.3/265.6*.8= 392.1A at .8pf lead i.e 36.87deg Now (Ia - b)*(Zdelta)=460 V(line to line) andZdelta= 15 + 15j ohms. From here you can obtain Ia - b =460/Zdelta=....................................Ans.(itmaches the answer,check it) Next part Ia = phasor sum of Ia -b,Iphi and Ia - c Again Ia-b=Ia-c(balanced) So Ia = phasor sum of 3*Ia-b and Iphi Now do it yourself...the answer will match b)Power supplied by source = 3*Vline*Iline =3*460*Ia(from the earlier part)=Ans(Do it,the answermatches,318107.7W=318.11Kw at 31.53degabsolute)......................................ANS