A three-phase transmission line has an impedance of 0.1 +j0.8 ohm per phase. The

Question

A three-phase transmission line has an impedance of 0.1 +j0.8 ohm per phase. The line feeds two balanced three-phase loads connected in parallel. The first load is absorbing a total power of 630kW at a lagging power factor of 0.6. The second load is Y-connected and has an impedance of 15.36-j4.48 ohm per phase. The line-to-neutral voltage at the load end of the line is 4kV. a) What is the line voltage at the sending end of the transmission line? b) How much power is dissipated by the transmission line? c) What is the total power supplied by the source? d) Determine the first load impedance per phase.

Explanation / Answer

load end voltage is 4kV and power of one load is 630kW at 0.6 lagging pf

so the apperant power will be= 630+j630*0.8/0.6 =630kW+j840kVar

(as reactive power=active power*sin(power factor angle/cos(power factor angle)

so impedance Z=V2/(conj(apperant power)/3) = 4000*4*3/(630-j840)=27.42+36.57j ohms

so net load impedance ZLeq=15.36-j4.48 II27.42+36.57j

ZLeq=(15.36-j*4.48)*(27.42+36.57*j )/(15.36-j*4.48+27.42+36.57*j )=13.67<00 almost

a. line voltage at sending end is

Vse=4kV(ZLeq+ZT)/ZLeq=4kV(13.67+0.1+i*0.8)/13.67=4029.261 +234.08i V( line to neutral)

b.power dissipated in transmission line is =resistance * I2 =0.1* abs( 4kV/(13.67+0.1+i*0.8))2=2892*0.1=8352.1W

c. total power supplied by source= 630kW+8.352kW=638kW

d.

load end voltage is 4kV and power of one load is 630kW at 0.6 lagging pf

so the apperant power will be= 630+j630*0.8/0.6 =630kW+j840kVar

(as reactive power=active power*sin(power factor angle/cos(power factor angle)

so impedance Z=V2/(conj(apperant power)/3) = 4000*4*3/(630-j840)=27.42+36.57j ohms

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