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
firstly for the wye connection Vphase=460/3 volt power =250kW sine 3*Vphase*Iphase*cos=250*1000 here Iphase=I line Iline=Iphase=250*1000/(3*460*.8)=392.22A at angle 36.87(i.e. cos-1 .8) Iph=392.22 at angle of36.87deg......................(1) for delta connection Vphase=460V Iphase= 460/(15+15j)=21.68 at an angle of -60+45=-15 Iline=3*Iphase=37.55A..........................(2) Ia=Adding (1) and (2)