Why the effect of a triphasic capacitor could be as twice as its rated kVar?

Please check Update 1, since if its true, then I think my issues are solved.


This is a real life case of a client of mine, which have a irrigation pumps station located in Chile. The power grid supplies their equipment through a power transformer dedicated to his consumption, where I don't know any of the transformer's plate data. The power transformer is providing in the low voltage stage a triphasic supply with neutral wire of 380 V rms among life wires with a frequency of 50 Hz (Chilean standard for triphasic low voltage), and the client consumption is registered on the high voltage side of the power transformer where just three power lines could be seen (I don't know their voltage).

About 50 meters from the power transformer, my client have two irrigation pumps connected within a cabin, each of them controlled by its own Variable Frequency Drives (VDF):

  • Pump 1: SAER model IR65-160/B 15 HP (11 kW) with a power factor \$\cos(\phi)=0.86\$, controlled by a VDF Mitsubishi Electric model FR-D740-11K
  • Pump 1: SAER model IR65-160/A 20 HP (15 kW) with a power factor \$\cos(\phi)=0.87\$, controlled by a VDF Mitsubishi Electric model FR-D740-15K

Every time the pumps are powered on by the VDF units, they took about \$\approx 30\$ seconds in achieve full load, and they work during the day at full load uninterrupted for the time required by the irrigation controllers.

The system from the Main Electric Board on forwards have a TN-S grounding system, but the original installer instead of building a grounding grid near the cabin, he connected the grounding wire to the neutral wire that is coming from the power transformer located near 50 meters away, so I think you could think in the installation of having a TN-C-S grounding scheme.

In Chile, the power factor penalization works as this:

  • If the power factor is inductive then penalization starts when \$\cos(\phi)< 0.93\$, and the penalization factor is \$\rho = 0.93-\cos(\phi)\$ (commonly used)
  • If the power factor is capacitive then penalization starts when \$\cos(\phi)< 0.95\$, and the penalization factor is \$\rho = 0.95-\cos(\phi)\$ (rare)

On my client electricity tariff segment (AT-4.3), the penalty fee due low power factor is calculated as: $$\begin{array}{r c l} \text{PF_fee} & = & \rho\, \cdot \left(C_1 + C_2 + C_3 + C_4 \right) \\ \text{PF_fee} & = & \text{fee due low power factor} \\ C_1 & = & \text{charge due kWh consumption} \\ C_2 & = & \text{charge due maximum registered power kW} \\ C_3 & = & \text{charge due maximum registered power kW on high demand hours} \\ C_3 & = & \text{charge due lecturing energy in the low voltage side} \\ \end{array}$$

Where for this client \$C_4 = 0\$ always since its power consumption is registered on the high voltage side of the power transformer.

Last year, my client electricity bills were as follows: $$\begin{array}{| c | c | c | r | r | r | c | r |} \hline \text{month} & \text{kWh} & \text{kVarh} & C_1 & C_2 & C_3 & \rho & \text{PF_fee} \\ \hline \text{jun-21} & 487.5 & 2,550 & $31,464 & $53,907 & $2,509 & 74\% & $65,031 \\ \text{jul-21} & 337.5 & 1,987.5 & $21,783 & $53,907 & $2,966 & 76\% & $59,779 \\ \text{aug-21} & 337.5 & 1,725 & $21,783 & $53,907 & $543 & 74\% & $56,412 \\ \text{sep-21} & 787.5 & 1,837.5 & $50,826 & $54,576 & $274 & 54\% & $57,065 \\ \text{oct-21} & 1,012.5 & 1,575 & $65,349 & $54,576 & $2,966 & 39\% & $47,927 \\ \text{nov-21} & 3,187.5 & 2,100 & $205,727 & $81,099 & $2,965 & 9\% & $26,081 \\ \text{dec-21} & 3,187.5 & 2,100 & $205,727 & $81,099 & $2,965 & 5\% & $17,166 \\ \text{jan-22} & 5,137.5 & 2,062.5 & $331,584 & $81,387 & $2,966 & 0\% & $0 \\ \text{feb-22} & 8,250 & 2,850 & $535,441 & $115,127 & $2,977 & 0\% & $0 \\ \text{mar-22} & 24,525 & 17,212.5 & $1,591,721 & $151,590 & $2,978 & 12\% & $209,555 \\ \text{apr-22} & 6,300 & 6,750 & $408,882 & $167,169 & $3,803 & 25\% & $144,964 \\ \text{may-22} & 450 & 2363 & $29,206 & $167,169 & $4,123 & 74\% & $148,369 \\ \hline \end{array}$$

Money is valued in Chilean pesos, if you want to express them in US$ dollars you should use the ratio \$\text{US}$\,1 = $\,950\$.

By looking at the electricity bills one can notice that on the rainy season (winter), among May and August here, the irrigation pumps were not used but the client was still charged with huge penalties due low power factor (near \$$330,000\$ in total) even when there was little energy consumption, attributable to the reactive power consumption of the power transformer.

So I started to evaluate the installation of a cheap power factor correction capacitor (bottle kind), to be permanently connected (shunt) on the main electric board previous where the VDFs are supplied (after the VDFs, since the frequency is changed, also will change the capacitance of the capacitor, so it cannot be connected between the VDF and each controlled electric motor of the pumps - it could even get damaged).

The other thing to notice is, that it were quite uneven the reactive power consumption on these winter months, even when its supposed to be just a power transformer connected 24/7. On rural areas here the power supply's quality is pretty bad, keep this idea in mind.

Main calculations

By averaging the active and reactive power consumption on the winter months I have that \$\bar{P} = 403\,\text{kWh}\$ and \$\bar{Q} = 2,156\,\text{kVarh}\$ respectively, which gives me a power factor of \$\cos(\phi) = 18\%\$ and a penalization factor of \$\rho = 75\%\$. Now, assuming an average month of \$30\$ days of 24 hours, and using the standard formula for calculating the required capacitance to achieve a power factor of \$\cos(\phi) = 0.93\$ I got that:

$$Q_c = \frac{\bar{P}}{30\cdot 24}\cdot \left(\tan\left( \cos^{-1}\left(\cos(\phi)\right)\right) - \tan\left( \cos^{-1}\left(0.93\right)\right)\right) = 2.8\,\text{kVar}$$

So I started searching for a triphasic capacitor of \$3 \, \text{kVar}\$.

After reading the VDFs installation manual, I noticed that shunt capacitor aren't recommended for power factor correction by the manufacturer, they sell some expensive inductance, or other related equipment, even some expensive capacitors for the same purpose, so I started to ask other installers if they have experience with the connection of shunt capacitors after the supply of VDFs for irrigation pumps purposes, and I found that:

  • Every VDF equipment manual asks for avoiding the shunt capacitor solutions since it could interact with the equipment harmonics and create resonances, and also generate Over-voltage alarms that make the VDFs stop running to self protect of getting burned.
  • Every installer I talked have do installed shunt capacitors previous the VDFs for power factor correction purposes without having any problem related to them.

So, with these mixed opinions, and since I wasn't requiring the capacitor for compensate the reactive power consumption of the irrigation pumps, but just for the power transformer consumption, I think in make a test field here and connect the power factor correction capacitor and see how it will behave, since it rating was the smallest available commercially here and the required current protection was about 7 amps, much lower than the currents that commonly manage the VDFs (that were the equipment I were worried the get burned - quite expensive equipment). Also the existent energy meter doesn't register harmonic levels, and the grid's electricity quality is already bad so I am expecting the equipment could manage any fluctuation introduced or potentiated by the capacitor.

So I installed finally on the Main Electric Board a bottle kind triphasic capacitor in shunt with the circuits that supply both VDFs and pumps, at the end of September, and so far everything have worked without problems (Nov 17th), both VDFs and pumps have worked with the capacitor connected and nothing wrong have happened.

The connected triphasic capacitor installed now is of \$2.5\,\text{kVar}\$ rated for \$400\,V\$ at \$50\,\text{Hz}\$ (ELECTRONICON model MKP-276.036-501700), so since at a lower voltage level it will be have even lower rating, I thought it will be a good alternative (it was the only available triphasic capacitor here with a rating lower than \$5\,\text{kVar}\$.

With this successful results, I started to evaluate the possibility of connecting additional capacitors but now for reducing the power factor of each of the irrigation pumps, by connecting them through a contactor for capacitor switching activated when the irrigation pumps are under operation.

For each pump the require capacitor rating are:

  • $$Q_1 = 11\,kW\cdot \left(\tan\left( \cos^{-1}\left(0.86\right)\right) - \tan\left( \cos^{-1}\left(0.93\right)\right)\right) = 2.18\,\text{kVar}$$
  • $$Q_2 = 15\,kW\cdot \left(\tan\left( \cos^{-1}\left(0.87\right)\right) - \tan\left( \cos^{-1}\left(0.93\right)\right)\right) = 2.57\,\text{kVar}$$

So, in principle, I have to buy two another \$2.5\,\text{kVar}\$ triphasic capacitors for keep the power factor within the range were no penalties are charged.

Unfortunately, I don't have the number of operation hours by month where each of the irrigation pumps were running, so I don't have how to figure out accurately which is the impedance values of the power transformer that supply the installations.

The big issue

A few days ago I received the electricity bill from October 2022, the first month with the capacitor connected and the irrigation pumps on operation:

$$\begin{array}{| c | c | c | c | r |} \hline \text{month} & \text{kWh} & \text{kVarh} & \rho & \text{PF_fee} \\ \hline \text{oct-22} & 2475 & 300 & 0\% & $0 \\ \hline \end{array}$$

As you can see, the installed capacitor for the power factor correction, not have only reduced the reactive power consumption of the power transformer, but also have compensated the reactive consumption of both irrigation pumps, as if it where a capacitor of at least twice its rated capacity in kVar!.

It looks good at first since it have eliminated the charges by low power factor, but the problem is that during the next winter, maybe the effective kVar compensation is going to be too high and it will get charged again but because of having a capacitive power factor out of range, and I don't know why is this over-effect happening neither have the experience to figure it out with the available information.

I were expecting with the capacitor to compensate winter months, and if it were able to operate jointly with the pumps, to keep it connected on irrigation months to get an extra income, but know it looks like it would end working exactly in the opposite way, disconnected during winter.

I have talked with other installers that have made similar capacitor installations on irrigation pumps, and they have had the same issue, having to reduce later the amount of compensated reactive power by removing some capacitors from their installed banks.

  • Is this something that always happened with power factor correction capacitors? (like linear theories underestimate the real effect by a factor of 2)
  • Is this situation something that happens only when capacitors are of low ratings?
  • Or conversely, Why is happening this situation that defies classic theory? Any clues?
  • Did I made some mistaken assumptions or miscalculated something? Hope you could explain with detail.
  • Have the same issue ever happened to you on similar installations?

Hope somebody with experience in power factor corrections could give me a clue.

Added later

This is a simplified version of the single-line diagram of the current state of the installation, with the \$2.5\ \text{kVar}\$ capacitor connected in shunt on the main electric board, before the electric boards with the VDFs that supplies each of the triphasic irrigation pumps.

simplified unilinear diagram

Update 1

I have realize that maybe the electricity bills aren't \$100\%\$ trustworthy on every months, since for \$\text{march-2022}\$ there is no way they have had all that consumption, I guess that due the pandemic some lectures were not registered on-site by the distribution company but instead estimated, and later adjusted somehow on the bills. But also, I don't think it was the case for the winter months with low consumption, and I don't know why there is that much variation on the reactive power consumed by the power transformer.

Also, after reading the answers until now, maybe I have misunderstood the analysis as follows:

If I am getting right what other users are trying to explain, it looks like even when each irrigation pump's motor have their own rated power factor \$\cos(\phi)\approx 0.85\$, since they are supplied through VDFs it looks that from the point of view of power grid supply, the pumps are behaving as they were just active power demands, so for the grid their effective power factor is \$\cos(\phi) = 0\$. The next image show what I am trying to explain here:

explanation VDF power factor hiding

If this is true it is huge, since it crashes half my analysis!, so please confirm it. If that is the case. then my issues are solved since the only reactive power consumption I see is the one coming from the power transformer, and in this scenario the capacitor is already working as designed, so there is nothing else required to do.

I hope that it is indeed the case, so please if you can elaborate if it makes sense that the only reactive power registered could be the existent in the power transformer and power lines (for this purpose, if required ignore the lectures of \$\text{march-2022}\$. If it is indeed the case, but I still going to have problems on winter months were the power transformer will work without any load, please also explain why.

PS: Thanks you for all the answers. Some have explained correctly that automatic reactive power control alternative will be a better approach, but in this case they aren't commercially plausible, since the overall power factor penalties reduction wouldn't cover the cost of those equipment at local available prices. Been really interesting to learn about them, It wouldn't be considered as solutions (neither for the problem, neither for the bounty), and I am saying this for avoid wasting your time, but being grateful about your intentions, willingness to share your knowledge, and for taking the time for your detailed answers. Hope you could understand.


2 Answers 2


... having to reduce later the amount of compensated reactive power by removing some capacitors from their installed banks.

Don't know if this can help you.
One can use variable switching as in this note and this.
You should be able to measure "cos(phi)" and control accordingly the SCR.
One can control inductors or capacitors, within needs (three phases included).
Some specs are obviously to be checked.

enter image description here

Another thing : I checked the datasheet of VFD FR-D740-11K (unless I am wrong).
This is the schematic used ...

enter image description here

The input stage is a PD3. This circuit has an "efficiency" of ~ 95%.
So, basically, no need for a capacitor bank (???).

And last, but not least, here is a simulation that can "prove" that efficiency is "lowest" (and also cos(phi) ???) when one uses a "bank" capacitor (in some "theoretical" conditions ...).

enter image description here

And for enjoying KW and KVAR ...
Here is an interactive circuit file made with microcap v12.
Double-click on the "red" 'define' to change data values manually or click and then change values with up & down arrows until "Global cos(phi)" is 1.

enter image description here


Low and medium power VFD feeds DC bus with a 3 phase full wave rectifier and a capacitor bank (and an inrush current limiting circuit). Current flows through diodes at AC voltage peak, so power factor is about 0.95 or higher, with harmonics content. The DC capacitor bank will supply reactive power to the motor and normally no additional device is necessary to improve power factor. High reactive power loads are fluorescent lighting, and direct on line motors.

The transformer with no load will have an inductive component (Lo), that probably have been compensated with the capacitor you have installed. But the power transformer and long medium voltage wiring are also serial inductive components (Lcc) that will delay the current of the rectifier, increasing reactive power if active power is high, that could explain the high reactive energy at the same months of high active energy.

So the installed capacitors balances both components with the pumps running and power factor is Ok. It is possible that when the pumps stop, the capacitors turn the power factor high capacitive, you have to test with no load.

Best solution is an automatic compensator, otherwise you will need to estimate capacitors for each VFD and switch on when VFD are running. To switch on capacitors to the grid needs to take care of inrush current. Also capacitors bank is splitted in differents filters to fit the harmonics generated from the rectifier, this is a specialized job.

For your understanding, I´ll make some rough numbers based on assumptions and random parameters.

The breaker at low voltage is rated 125 A, so power transformers should be about 85 kVA. Typical short circuit impedance should be about 5 %, that means 0.09 Ohm per phase at low voltage side. This impedance is the sum, let´s say 30 % resistive is 0.03 Ohm, and the rest (0.06 Ohm) of inductance (at 50 Hz, 0.2 mH). Transformer plate usually show no-load power losses, but no idea how inductive is this current. I will use 800 mH for no-load transformer inductance (0.33 kVA).

Motors rated power are 11 + 15 kW, these are mechanical power. Assuming pumps are loading 85% of rated power, and efficiency of motor is also 85%, we need 11 + 15 kW at the VFD output. 3x380 V rectified and filtered will be 530 V, so equivalent resistances in DC bus for these power are 25 and 21 Ohm. For faster simulation these resistance is made of 2 series, and central point connected to ground.

2.5 kVAr-400 V means 17 uF capacitors, equivalent in star is 51 uF, current at 380 V is 3.6 A, 2.4 kVAr. This capacitor bank should be connected in simulation downstream transformer impedance, but I changed upstream because waveforms are easier to see. enter image description here Here you can see phase 1 voltage, Lo and Capacitor currents. enter image description here Here you can see VFD distorted and asymmetric current. enter image description here The current peak is delayed from voltage peak about 1 mS. To know power factor, you need to calculate main and significant higher harmonics currents and phase, and power factor impact of each one. The reactive power should be close the 2.4 kVAr of the new capacitors.

If you stop the pumps, the balance will be 2.4 (capacitors) - 0.33 (Lo) = 2.07 kVAr , so maybe you need to switch off the capacitors.

Suggestion: it is possible in many VFD to read real power of output module by parameters, and also there are running-time counters (hours), than can be manually reseted. This can help in your calculations.

  • \$\begingroup\$ Thanks for answering. Let me see if I follow: what you are saying is that from the beginning, the VDFs eliminate the power factor problems from each pump, so from the point of view of the power grid supply, there are just two consumption of 11 kW and 15 kW both with power factor $FP = 0$?? If that is the case then my issues are solved, since the only reactive power consumption I see is the one coming from the power transformer, and in this scenario the capacitor is already working as designed, so there is nothing else required to do. I hope that it is indeed the case, please elaborate. \$\endgroup\$
    – Joako
    Commented Nov 20, 2022 at 0:05

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