## Wednesday, January 4, 2012

### Efficiency, Power Factor, Watt, Voltampere and Alternating Current

For alternating current there are two power units, namely: watts and voltamperes (va). Sometimes this is confusing, because the watt is the same as the voltage (volt) times current (ampere). Also, efficiency and power factor (pf). The power factor is only applied to alternating current and sometimes confusing.

Efficiency is the ratio between the output power to input power

h = output power / input power

Efficiency value is ​​ranging from 0 to 1. In the real world the efficiency value is ​​typically ranged from 0.6 to 0.8, the bigger the better. Efficiency applies to direct current (dc), also applies to alternating current (ac). Power lost at the output is usually converted into heat, occurs in: coils, resistors, condensers, transformers, transistors, ICs, etc.. Power can also lost due to friction, usually occurs in: electric motors, generators, alternators, pumps, compressors, air conditioning (ac), etc..

Especially on alternating current, there is other thing that affect the power output and make it less than the input power, it is the power factor (pf):

pf = real power / apparent power = watt / voltampere

In direct current, power in watt unit is always equal to the power unit voltampere (va). Because power factor in direct current is always close to or equal to 1 (one).

In alternating current, a resistive load which is produced by resistor component, has a power factor close to 1, because the wave of voltage and the wave of current are always synchronous. So, if the equipment contains only resistors such as heater, soldering iron, incandescent lamps, etc., the power in watts can be considered equal to the power in voltamperes unit.

For inductive load, then the alternating current wave is not synchronized with the wave of voltage, current phase lags behind the voltage phaseTherefore, the voltage multiplied by amperage is not the same as watt. So power factor is lower than 1 (one).

Likewise, if the alternating current gets capacitive load, the voltage lags behind the current phase. So power factor is lower than 1 (one).

Power factor can also be calculated by the difference degree in phase of voltage and amperage:

pf = cos q

The picture below shows charts of voltage, amperage, and power, with a phase difference of 30 degrees or power factor = cos q = cos 30 = 0.87, this is a very good power factor.

The picture below shows charts of voltage, amperage, and power, with a phase difference of 90 degrees or power factor = cos q = cos 90 = 0, very low power factorWe can see power curve is going down below zero line or negative, and positive part of power chart is about the same as negative part of power chart, so effective power, or power in watt is close to zero.

This Excel file has chart to simulate voltage, current, and power for certain power factor.

One way to improve power factor is by installing parallel capacitors and located as close as possible to inductive loads (coils, motors, etc.). The greater the load the greater capacitor, therefore cost consideration will be needed. Some power loss and converted to heat and current leaks in capacitor will reduce efficiency. For electric motors, always apply load close to the load ratewhere the motor is designed to work most efficiently. Avoid equipments operate with voltage above rating.