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I've just had and exam (Renewable Energy) and we had a problem that I'm not sure if is correct. The answers should be simple as we haven't studied tranmission lines at all.

Okay, here we go.
We have an AC signal, tranmission line with impedance and transformer at the end.
At the beginning of transmission line we have a signal with(max values):
V=240kV
I=400A
PF=0.85

First question was to calculate apparent power etc. what was straightforward, but the second question was:
There was voltage drop of 1kV on the transmission line.
The impedance of transmission line 0.1+0.1j per km.

So Vin=240kV sin(alpha)
Phase angle was arccos(0.85)=32 degrees
Iin=400A cos (alpha + 32)

How can we calculate what was length of the line?
Won't this initial phase shift somehow make it more complex?
Also, do transmission line do have inductance or they just have characteristic impedance?

Maybe I'm overthinking it and the answer should be just
V=IR -> 1000V=400A*(0.1*l) and therefore l = 1000/(400*0.1)=25km
Or maybe, as sqare root of (0.1^2+0.1^2) = 0.14
V=IR -> 1000V=400A*(0.14*l) and therefore l = 1000/(400*0.14)=17.85km

After calculating this we were asked what is be impedance on the secondary side of transformer with turn ratio = 16 (i.e. secondary voltage = 15kV).
Could you please help me with that?

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  • \$\begingroup\$ The transmission line here is a power transmission line and is modeled as a lumped element with 0.1+0.1j per km. It is not a signal transmission line with characteristic impedance and finite propagation delay. \$\endgroup\$ – rioraxe May 6 '16 at 22:31
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The answer to the second half of your question is definitely just to divide the impedance on the line side by 16^2, because the equivalent impedance ratio is the square of the transformer ratio, as described here:http://www.radioremembered.org/outimp.htm

For the first half they have effectively told you the magnitude of the answer (239kV) but not the phase angle. Using phasors: v_out=v_in-I*R

You know the phasor for the current. Graphically, R is a 45deg phasor, whose length varies with distance. Multiplying phasors adds angles, and the negative sign is a 180 deg phase shift, so you need to add 225 to the current angle to get the direction of the voltage drop phasor. (I.e. 257deg or 193 deg, I'm not sure if your current is leading or lagging, I forget the convention). Then you just need to solve for the multiplier to get a result vector of the appropriate length.

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  • \$\begingroup\$ Ah, actually, I figured out how to solve it graphically, just need to write up the math.... \$\endgroup\$ – Paul Foster May 6 '16 at 17:25
  • \$\begingroup\$ Actually, this is a bit of a pain to solve, so I'll do it when I get home from work if no one else has solved it by then. \$\endgroup\$ – Paul Foster May 6 '16 at 17:50
  • \$\begingroup\$ solve: abs(240k + x*(cos(257)+jsin(257)))=239k; sqrt((240k + xcos(257))^2+(xjsin(257))^2)=239k; According to wolfram alpha: x=(240+-239)/(jcos(13)+sin(13)); x=-IR; let y=Re(x)/Re(-I*(0.1+.1j)); y is the number of km \$\endgroup\$ – Paul Foster May 6 '16 at 18:12

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