enables the device to produce the desired work. The inductive circuit cyclically absorbs energy from the system (during the build-up of the magnetic field) and then re-injects that energy back into the system (during the collapse of the magnetic field). Hence the current (I) in the supply and distribution system will increase (with the additional inductive current), and so the losses in cables and transformers will rise – frequently referred to as ‘I2
R losses’ (R being the electrical
resistance). This reactive power is also known as ‘wattless’ power. It does not directly affect the useful power (watts) drawn by the appliance to produce the work, heat or light. However, the total demand on the power supply systems, measured in VA (volts x amps), and the magnitude of the flow of energy in the conductors, will increase to provide the reactive power. The total or apparent power (kVA) required by a reactive system is the sum of the ‘true’ or ‘real power’ (kW) and the reactive power (kVAR
), commonly known as ‘kilovars’.
The power factor is the ratio of the real power (kW) to the total (apparent) power (kVA). The lag (or lead) angle can be shown
using a phasor diagram (Figure 3) and simple calculations used to determine unknown values. The cosine of the angle f is the ‘displacement power factor’ (normally simplified to just ‘power factor’). So, for example, a particular 20kW
motor operates at a power factor of 0.80 (lagging). Hence, the apparent power required by the motor = 20kW/0.80 = 25kVA. If the power factor is improved to 0.95, then the corrected apparent power = 20kW/0.95 = 21kVA. In this example, the apparent power is reduced by 16% – this will not necessarily reduce energy bills, but may contribute to having smaller cables or improved tariffs.
(real power)² + (reactive power)² = (apparent power)²
Real power = 20kW f Power factor = cos f
Reactive power = 15kVAr
In commercial and industrial buildings,
a power factor of below 0.9 is considered to cause excessive running currents, and is often penalised through electricity supply tariffs by incurring a financial charge for reactive power (kVAr
) that goes above a
set proportion of the real power (kW). So, typically, a building’s power factor would be maintained at 0.95 or above by the installation of ‘power factor correction’ (PFC). The motor from the previous example
can be simplified into a resistance, R, and inductance, L, as shown in Figure 4a. When voltage is applied to the circuit, the main supply will need to provide both I2
, the ‘inductive’ current, and I3
‘resistive’ current. If, subsequently, the switch is closed,
Figure 3: Phasor representation of power (with example values taken from the uncorrected example motor described in the text)
52 CIBSE Journal June 2012
as in Figure 4b, the capacitor is instantly charged and this charge oscillates between the capacitor and inductor, creating a current through C and L. The oscillating charge provides the ‘extra current’ in I2 and the current from the main supply is reduced. This is known as static power factor correction, and is often used with individual pieces of equipment. It may not be physically practical or financially economical where there are a large number of devices requiring power factor correction. These correcting capacitors can be of
a fixed rating to match the inductive load of a large item of equipment, or be in
Figure 5: Centralised power quality correction automatically adjusts to meet the needs of the building electrical load
a bank of capacitors, mounted close to significant loads or at the building power intake switchgear. If centralised, these are likely be progressively brought into circuit by automatic controls to match inductive loads in the building. Power factor is unlikely to be a constant.
For example, motor inductance will alter with motor speed (as shown in Figure 6). This may well demand more complex solutions, as inappropriate ‘balancing’
www.cibsejournal.com I1 kW
power meter
capacitive C inductive L resistive R I2 I3
I1 = 0 kW
power meter
capacitive C inductive L resistive R
I2
I3
Figure 4a
Motor
Figure 4b Figures 4a and 4b: Adding a capacitor as a means of correcting the power factor (after1 )
Motor
, the
Apparent power = 25kVA
Main power supply
Main power supply
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