Characteristics of a three phase circuit

Thread Starter

labhelp

Joined Jan 28, 2010
28
In this problem i have a three phase power source supplying a balanced three phase load over a transmission line having impedance of Z=2+j20 ohms per phase. The voltage at the source end of the transmmission line is 2887 volts line to neutral. The current supplied through the transmission line is I=100 /-30° A.

I have to determine the power factor seen by the source and the line to neutral source at the load.
Also, I have to determine the power factor of the load and the real and reactive power consumed by the load.

My work:
θ=0-(-30)=30°
pf=cos(30°)= 0.866 lagging

I keep thinking that the line to neutral voltage at the load is the same as the source line to neutral but I'm not sure how to actually calculate this
 

t_n_k

Joined Mar 6, 2009
5,455
Your power factor value looks good.

You know the line impedance and the line current. Calculate the (complex) potential drop across the line impedance. Subtract this from the source voltage and you will have the load voltage. And so forth .....
 

Thread Starter

labhelp

Joined Jan 28, 2010
28
Awesome, heres my work:

Vl=100/-30*(2+j20)
=1173.2+j1632

Vln= 2887 - (1173.2+j1632)
= 1713.8-j1632
= 2366.54 /-43.6

Then I worked on the power factor for the load and got an impossible answer

Iload= (1713.8-j1632)/(2+j20) = -72.31-j92.92= 117.74 /52.11

But then I get
θ= -43.6-52.11= -95.71
which is wrong
 

t_n_k

Joined Mar 6, 2009
5,455
You have the correct load voltage.

The line and load currents will be the same.

\(

Z_{load}=\frac{V_{load}}{I_{line}}=\frac{2366.54 \angle{-43.6^o}}{100 \angle{-30^o}}=23.67 \angle{-13.6^o} \Omega

\)

The load power factor will be cos(-13.6°)=0.972 lagging
 
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