Some electrical accounts, especially for large consumers include a charge for Power Factor issues (PF) and the charge will be seen on the account as “cents per kVAR” (kilo volt/amps/reactive). The local supply authority introduces this charge to commercial consumers to encourage the consumer to maintain PF at a tolerable value, often between .9 and 1, or penalties will apply (kVAR charges).

Power Factor correction devices are introduced at the main switch-board to maintain the PF within the desired range. The device is a stepped capacitor bank and will switch the capacitors depending on the PF at any given time. Generally an Electrical Engineer will have been involved to design and oversee the set up the PF device.

Domestic consumers are only charged for kilowatt hours (kWh) and will not need to be concerned about KVAR. But large sites may attract large kVAR charges each month, which is just a penalty for not maintaining a correct PF.

Power Factor is a figure derived from the angle that the current lags behind the voltage on a given site. The concept is a little difficult to understand for most consumers however the power triangle below will help a little.

There are three types of power to consider:

Real Power – common unit is the kilo-watt (kW). An example of a purely resistive load would be a heating element. In a resistive load the supply voltage and the load current are “in-phase” such that the peaks and troughs of the sinusoidal waveform coincide. Real Power does real work.

Reactive Power –common unit is the kilo-volt-ampere-reactive (kVAR). Loads such as motors or lamp ballasts will cause the load current to lag behind the voltage and capacitor loads will cause the current to lead the voltage.

Apparent Power – common unit is the kilo-volt-amp (kVA). The apparent power is the “vector sum “of the Real Power and the Reactive Power. The Apparent Power is the total burden placed on the local grid by the combination of the Real and Reactive Power elements.

###### Understanding the effects of introducing Solar PV and how it can affect “Power Factor” on complex Industrial/Commercial sites.

You will note if you observe the Power Triangle above, the more the Reactive Power (Q) then the longer the Apparent Power (S) line.

On a site that has large amounts of Reactive Power the burden on the local supply grid is increased. The angle θ is “opened,” thus decreasing the PF below the predetermined PF of perhaps .9

Example:

Angle θ = 22°

Power Factor = cos θ

Power Factor =cos 22°

Power Factor = .92 --- this would be acceptable and may not attract chargers.

However if this angle was “opened” due to the Reactive Power increase then:

Angle θ = 40°

Power Factor = cos θ

Power Factor = cos 40°

Power Factor = .76 --- this PF is now sub-par and would attract kVAR chargers.

This is the reason the local supply authority may require a particular site to maintain a PF of between .9 and 1 (unity) because the more the Reactive Power the more the burden of Apparent Power ( kVA ) and yet the Power ( P) may not have increased.

If this PF cannot be maintained the customer is charged a kVAr rate and this amount if shown on the electrical bill. (Penalty for sub-par Power Factor).

How will introducing solar PV affect the Power Factor?

Introducing solar PV will reduce your PF considerably and while you may see a reduction in your kWh charges your site may see a dramatic rise in charges for KVAr penalties.

The reasons and explanations for PF can be a little hard to understand, and I would recommend engaging the services of an Electrical Engineer or a Renewable Energy Engineer to help you solve the issue.

For your interest and understanding I have shown a worked example, mathematically, to prove the effects of introducing solar PV to a given site. If you are considering a Solar PV installation at your site you will need to discuss this issue with the solar company or engage the services of a company such as Living Power to ensure that PF is corrected and that additional kVAr charges will not increase when the solar PV installation is complete.

Typical Example

A certain commercial site has a large electricity bill each month that is billed for kWh and a KVAr/h rate if the PF dips below the local supply authority perimeter of .9

The commercial site owner decides to install a Solar PV array to help reduce the monthly electrical bill (kWh). A solar company is engaged and the company technician should connect a Power Analyzer to determine the overall electrical load. The electrical load is determined at 150 kVA and a PF of .92 (23°) at peak daily production. Or the instrument may show the kW load at the site to be 138 kW.

The solar company is proposing a 95 kW PV array and the installation is completed in due course.

This is how the description above of the site electrical conditions would look using the Power Triangle that was described earlier.

S (kVA) = P ( kW) +Q ( kVAR)

150 kVA = 138 kW + j58.6 kVAR(rectangular number)

Or expressed as a polar number

150 kVA at angle 23 °

Or

23°

150 kVA

As the solar begins to operate and inputs into the site 50 kW at a given time. It is important to note that Grid Tied Inverters (GTI) will only supply the kW at unity PF or 1 as they are tied to the grid frequency. Some inverters will supply kVAr’S into the AC bus, such as SMA and can help the falling PF, however in most areas the Service Rules prohibit Power Factor from becoming leading at anytime and penalties do apply.

The site electrical conditions are now:

138 kW is still the required load. The solar is providing 50 kW and remainder 88 kW is supplied form the grid. The kVAr’s are still required at 58.6 kVAr’s. The Power Triangle would look like the following.

Therefore the new Power Factor is:

PF = cos θ = cos 33.66 °

NEW POWER FACTOR = .83

The new PF is now below the required the PF of .9

As the solar input increases the PF becomes progressively sub-par.

At 80 kW of PV input the new PF would be:

Assumes required load still at 138 kW .

The required kVAr’s are still at 58.6 kVARs

Therefore:

KW required from the grid = total power required minus solar input.

58 kW = 138 kW – 80 kW solar

Therefore the new Power Factor is:

PF = cos θ = cos 45.29 °

NEW POWER FACTOR = .70

The new PF is now well below the required PF of .9 and penalties will apply.

Conclusion:

As can be seen above the introduction of a solar PV array will reduce the Power Factor and may result in penalties. The monthly bill for the kWh's may have reduced but the cost for the kVAr’s will increase.

It is important that the solar company can minimise the risk of an increase in kVAR penalties. This can be resolved easily by the introduction of a Power Factor Correction device.

Our company, Living Power, can provide advice to you or your solar contractor if required. We recommend the use of a Power Analyser BEFORE the solar project begins so that all ramifications regarding the introduction of a solar PV array can be determined and results maximised.

Complex electrical loads such as manufacturing plants, factories and process plants, require careful considerations and a sound understanding of Electrical Engineering practices to achieve good results.