0
\$\begingroup\$

Consider the following system below, nine buses, 3 loads and 3 generators. From my understanding of reactive power, lagging power factors (per unit voltage less than 1) causes the reactive power to flow from the load to the generator which causes a voltage drop thus less than 1 per unit voltage. Inversely for over compensating for, it over excites the generator causing more voltage to be produced causing a per unit greater than 1. enter image description here So to stop reactive power flowing from the generators to the source and vice versa, the reactive power is locally generated on the same bus as the load. In the system above, I have found that if a load draws 85Mvar, and I supply 85Mvar of capacitance on the same bus, it causes a per unit greater than 1 on my generators. In the image above, with that current system, every generator has a per unit voltage greater than 1, why? Im not supplying enough capacitive Mvar for each load so the system per unit should be less than 1. Secondarily, how is the amount of capacitance calculated?

\$\endgroup\$
  • \$\begingroup\$ Look at bus 8 for example, the load is an inductive 85Mvar, I am supplying 60.4Mvar capacitance. Yet in the current configuration, this is overcompensating causing the voltage on bus 2 to be greater than its rating. \$\endgroup\$ – user160063 Feb 12 '18 at 19:52
  • \$\begingroup\$ Why do you think the voltage will be less than 1? Bus 8 has 85MVAR - 60.4VAR = 24.6 VAR going back between source and load. \$\endgroup\$ – StainlessSteelRat Feb 12 '18 at 19:59
  • \$\begingroup\$ Because that 24.6Mvar is inductive, lagging, under compensated which causes a voltage drop. \$\endgroup\$ – user160063 Feb 12 '18 at 20:13
  • \$\begingroup\$ It's been a while since I studied power flow analysis at university, and I haven't had to use it since, but you say that if you add leading Mvars equal to your lagging Mvars on the load bus you end up with over excitation on your generating buses, which means there is still an imbalance, what parameters are being used for your transmission lines? \$\endgroup\$ – JCollins Feb 12 '18 at 20:35
  • \$\begingroup\$ Could it be that the extra capacitance in the state is due to the capitance in the lines itself? Could explain the increase \$\endgroup\$ – user160063 Feb 13 '18 at 16:23
0
\$\begingroup\$

You are assuming lagging means less voltage. If leading reactive power = lagging reactive power, all power would go to load, pf=1 and unity voltage.

You need 125MW + 150MW + 150MW = 425MW of real power. You produce 182MW + 163MW + 85MW = 430MW of real power.

You need 100MVAR + 80MVAR + 85MVAR = 265MVAR of lagging reactive power. Your capacitive power correction provides 84.2MVAR + 61.2MVAR + 60.4MVAR = 205.8MVAR. You are 59.2MVAR short.

Your over-excited generator on bus 3 provides 9MVAR (an assumption on my part due to the negative), which makes you 50.2VAR short.

So generators provide 425MW and 50.2VAR, which means 427.95MVA and a pf = 0.993. 425MW goes to loads and 50.2VAR goes back and forth between loads and generators. Which is as far as my knowledge goes.

I'd guess if you provide 265MVAR - 9MVAR (bus 3 generator) = 256MVAR of capacitive power, you'd get your unity voltage.


When pf = 1, S = P and \$Q_C = Q_L\$. The voltage of the source drives the real power load. Your unity voltage. Leading reactive power of the capacitors supplies the lagging reactive power of inductors (motors).

When the pf < 1 and \$Q_C < Q_L\$, the load is inductive. Again, the capacitors supply reactive power to the loads, but there is not enough. The source must supply P and \$Q_{NET} = Q_L - Q_C\$. The source voltage, which cannot change, is providing power to drive the MW and some of the MVAR. Less of the voltage is going to the MW real power load.

Same thing would happen if \$Q_C > Q_L\$. The load is capacitive and pf < 1. The capacitors supply all of the reactive power to the loads. Now the source must supply P and \$Q_{NET} = Q_C - Q_L\$. The source voltage is providing power to drive the MW and the extra MVAR for capacitors.

This is not your case, but the same situation applies. Less of the voltage would go to the MW real power load.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy