I will first write the problem followed by the equations I'm given and then explain what troubles me.

A 3-phase transmission line of 50 Hz and nominal voltage 400kV has the following per phase parameters L=1.296 mH/km , C=0,01 μF/km. The line has no real losses meaning active power at the start of the line is equal to the active power delivered to the load. The line's length is 277.8 km and delivers power of 600 MVA to a load with an inductive power factor of 0.8 .

If the voltage at the start of the line is 400 kV find the reactive power along with the capacitance of the capacitors you need to connect to the line in order to achieve a 400 kV voltage on the load.

These are the formulas (_R means at the load, _S means at the start of the line, δ is the angle(phase difference) between Vr and Vs, S is complex power) enter image description here

β and Xe both depend on L and C

My main problem is which of the following change and what doesn't after the capacitors are added:

1)phase difference between voltages Vr and Vs

2)the current

3)the complex power delivered to the load(active and reactive specifically)

There has to be a reason I'm given the power factor before the capacitors. I can find active and reactive power but does one of them stay the same afterwards? Active power could be the same but if it were and the load still draws 600 MVA and reactive power would be the same as well and this leads to the change of Xe and phase angle δ.

I can find δ and Vr before the capacitors but it didn't help since they both change afterwards I think.

( I know I haven't included everything required to use the formulas because I don't think it's necessary for my queries. I can add them though if someone wants to give a more detailed answer)


1 Answer 1


Assuming that the load is constant power load (the power consumption is not a function of Vr).

Note that the load is higher than SIL. Hence Vr is going to be less than Vs and adding the capacitor will rise up this voltage.

Let us start from third question to first one,

3- The power delivered to load will be the same (P and Q) as per our assumption. However, the reactive power transmitted will be different as the new capacitor supplies reactive power to the load.

2- The current, you did not mention which current but I will assume it is the receiving current Ir. The current is going to decrease as the receiving voltage will definitely increases when the capacitor is connected and as we assume that the load has constant power (i.e. \$I=S^*/V^*\$).

1- Finally, the phase difference. We have \$P=\frac{Vs VR}{X} sin(\delta)\$ this leads to \$sin(\delta)=\frac{P}{\frac{Vs VR}{X}}\$ as P,X and Vs are constants you can get \$\sin(\delta) \propto \frac{1}{Vr}\$ . As Vr is going to be increased due to the additional capacitor, \$sin(\delta)\$ decreases and hence \$\delta\$ decreases also (as the stable range of operation is \$\delta < 90\$)


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