A balanced threephase load requires 480 kW at a leading powe

A balanced three-phase load requires 480 kW at a leading power factor of (0.8). The load is fed from a line having an impedance
of 0.005 + j0.025 /. The line voltage at the terminals of the load is 600 * (3)^0.5 V.
a)Calculate the total complex power of the loads.b)Calculate the complex power of the three-phase system.c)Calculate the power factor of the transmission line.d)Calculate the power factor at the generator.e)What is the value of the load should we should add it in parallel in order to have a unity power factor at the generator. Assume line impedance is only 0.005 .(I need clear explanation of part e)

Solution

FROM THE GIVEN DATA

ACTIVE POWER OF THE LOAD=480KW

POWER FACTOR COSq=0.8

LINE VOLTAGE VL=600*3^0.5=1039.23 VOLTS

ACTIVE POWER P=1.73*VL*IL*COSq=480*10^3

1.73*1039.23*0.8*IL=480*10^3

IL=333.33 AMPS

IMPEDANCE PER PHASE=(0.005+j0.025) ohms/phase

a) TOTAL COMPLEX POWER OF THE LOAD=1.73*VL*IL=1.73*1039.23*333.33=599.993 KVA

b) COMPLEX POWER OF THE THREE PHASE SYTEM=599.993*10^3+3(0.005+j0.025)*333.33^2=608.491 kva

c) power factor of the transmission line= R/Z=0.005/(0.005+j0.025)=0.005/0.025495=0.193

d) IL=333.33<0 AMPS

SOURCE VOLTAGE=LOAD VOLTAGE+VOLTAGE DROP ACROSS TRANSMISSION LINE

SOUCE VOLTAGE=1039.23+333.33*3*(0.005+j0.025)

source voltage=1044.52<1.37

power factor= cosine of angle between voltage and current

power factor of the generator= cos(1.37)=0.999