Hi all,


I have been stuck where to start on a power flow question that has two interconnected voltage sources, which is in the attachment. Also in the attachment is the equation that I have been trying to use.

Firstly i would like to know is this the correct equation I should be using to solve this? If so then i should be able to build a matrix to solve this question.

What I would like to find out is

i) the source complex power that are supplied and received (
ii) the complex power loss in the line
iii) which source receives or supplies the active and reactive power components.


The questions variables are as followed

\(Z=1+j7\)
\(Y = 0.02-j0.14\)

Polar
\(v1=120 \angle -5\)
\(v2=100 \angle 0\)
Rect
\(v1=119.54-j10.46\)
\(v2=100+j0 \)


What I have done is solved

\(i=V*Y\)

i1 = 52.1-j970 A
i2 = 114.6-j802.1 A

now I use these values and solved for

\(S=V*I\)

Which gave me 27.96*10^3-j195.722*10^3 VA

I would like to know if this is the correct way to go about solving this.

Thanks
Kayne

 

Why do you have a current i1 and current i2? There is only one current between the two sources (unless you're trying to use superposition). The current will be the difference in the voltages divided by the impedance between the voltage sources.

 

your method makes no sense. when you write equation down, you are supposed to use it. you skip that and write random 'results'. if the Z-1+j7 Ohms, then your Y is correct. but there is no mention HOW you got V and what it's value is.
also you post equation for current i and you never sow what that current is. instead of that, you show i1 and i2 (whatever that is). you never mention what they represent or how you obtained them. then you calculate S from V and I. what V and I? is I=i perhaps?

 

Both of you are correct I really was struggling to understand what going on.

So I have started again taking on your advice

Variables are

\( Z=1+j7\)
\( Y=0.02-j0.14\)

pol
\( V1=120 \angle -5\)
\( V2=100\)

rec
\( V1=119.54-j10.46\)
\( V2=100\)


Current in the circuit

\( I=\frac{V1-V2}{Z}= -1.073-j2.945 A\)

Complex power

where I` is a complex number

Sv1 = complex power in Voltage source 1
Sv2 = complex power in Voltage source 2

\( Sv1=\frac{V*I`}{2} = \frac{119.453-j10.46*-1.073-j2.945}{2}=-79.56-j170.43 kVA\)

magitudeSv1 \( = 188.1 \angle -115\)

\( Sv2=\frac{V*I`}{2} = \frac{100*-1.073-j2.945}{2}=-53.67-j147.26 kVA\)

magitudeSv2 \( = 156 \angle -110\)


So voltage source 1 is the supplying the load and voltage source 2 is receiving

Power loss in the line = I^2*R

Find Resistance

\(R= mag(Z)*cos(Z)= 1\)

\(PL=I^2*R=(-1.073-j2.945)^2*1=-7.52+j6.32 W\)

Am I on the right track now with answering this question?

Thank

 

1. I got the same answer as you for the current.

2. Why are you dividing the complex power of each source by 2?

3. You are asked for the "complex power loss" for the line. Normally I think of power loss as resistive, but nevertheless, I think you could take the magnitude of the current, square it, and multiply it times the resistance and the reactance.

For example, given the current you found, the magnitude is 3.135A. Squaring the magnitude and multiplying it times the resistance (1Ω) would give a line power loss of about 9.83 W. Do the same for the inductive reactance to find the VARs.

 

I certainly agree with mlog when questioning the division by \(2\) in the expression for \(S\); complex power in any component in an electrical circuit is \(S = V I^*\). It does not matter if the component considered is an active component (like a voltage source) or a passive component (like an impedance). As such, then, the answer for item ii) should be a complex number; if the circuit shown is a representation of a transmission line, for instance, then the impedance given corresponds to the line itself, and there is no reason to consider the loss in the line as just the real part of the total complex power. What I would say then is that \(S = I^2 Z\) because \(V = I Z\) and \(I I^* = I^2\). Also, current in the circuit shown is just one item, and that is what I call \(I\) in my post.