# The Foundations and Fundamentals of Electricity: Part 2

Power is measured in “watts.”  One watt is a current flow of one ampere at one volt, a rather small unit of measure.  Therefore, to measure electrical power usage, a “kilowatt” is the term most used to express power consumption.  A kilowatt is 1000 watts.  As an example, 3000 watts is 3 kilowatts and is abbreviated as 3KW.

The meters that measure electrical usage used by utilities for billing purposes include a clock.  This adds a time element to the computation for billing customers of the utility.  This results in the “kilowatt-hour.”  A kilowatt used for one hour is called a kilowatt-hour, and is abbreviated KWH.  Electricity is sold by the KWH.  One KWH may be half a kilowatt used for two hours, two kilowatts used for a half-hour, etc.

With DC current, the current flows in the same direction.  If voltage and current are constant, it is simple to find power or wattage.  The product of volts and amps equals watts.  V x A = W.  A DC device that uses 10 amps at 120 volts consumes 1200 watts of power.

V x A = W is true only for DC circuits or in purely resistive AC circuits, such as heaters, where the voltage and amperage are in phase, that is, both rise and fall at the same time.

In AC circuits with inductive loads, such as motors, relay coils, etc., the voltage and current get out of phase, and current will lag the voltage (when the circuit is primarily inductive).  This lag difference is called the “power factor.”  Power factor is expressed as a percentage of the number of degrees of lead or lag.

To calculate wattage of an AC circuit with inductive loads, the power factor must be considered.  The equation now becomes W = V x A x PF.  (Watts = Volts x Amps x Power Factor.)

For large users, electric utilities do include power factors in their energy rates.  The utility buys generators rated in so many KVAs.  If the utility is serving so many customers with say 80% power factors, the utility is getting only 80% revenue out of a generator instead of the full 100% of which it is capable.  The 20% difference represents revenue lost on invested capital.  The electric utility could not afford many large users with poor power factors unless it can bill them for this “wattless current.”  For this reason, many large electric users invest in power factor correction devices.

The power factor is found by dividing the measured active energy utilized by the apparent energy consumed.  This is easily done using a voltmeter, ammeter, and wattmeter.  Inserting actual meter readings into the formula: V x A

Where      W        is the true power in Watts

V         is the apparent voltage in Volts

And          A         is the apparent current in Amperes

Solving the equation results in an answer giving the power factor expressed as a percentage.  As an example, Meter readings are:  120 volts, 15 amps, and 1440 watts. 120 x .15    180

An 80% power factor means the current lags the voltage by 36 degrees, or 1/10 of a cycle, or 1/360 of a second.  In a “purely resistive circuit,” watts is equal to volts times amps, so in our example, 1440 watts divided by 120 volts equals 12 amps.  Our meter reading shows 15 amps being consumed.  The difference of 3 amps is the “wattless current.”  The current does not go into producing power.  This current is continuously being transferred between the power company and the end-user. Wattless current is not billed to residential customers since the total number of inductive loads in a single residence is insignificant compared to large industrial facilities.  Commercial and industrial customers are therefore billed for the wattless current.

Power factors of 70% to 80% are not uncommon.  Because of the ever-increasing costs of producing electricity, it pays large users to invest in power factor correction devices such as power capacitors, special synchronous motors, etc.  The benefits of power factor correction are abundant:

•  Reduction of current reduces heat losses.
•  More active loads can be carried.
• Voltage drop is minimized.
• Shed peak demand.
• Releases more generating, transmission, and distribution capacity.
• The investment in corrective devices is usually recovered in two to three years.

Besides power factor being a cause of high electric bills, another cause of electrical problems for commercial industrial users is phase unbalance.  Phase unbalance is common.  A survey of electric utilities found that 40% of them allowed a generated voltage unbalance of 3%!  30% allowed an unbalance of 5%!!

Whenever phase unbalance is determined to be the Power Company’s fault, they should be notified, in writing, and prevailed upon to correct it.  Phase unbalance can quickly ruin three phase motors and cause all kinds of problems.

Often, phase unbalance is created in a building when a circuit is pulled off two legs of a three-phase circuit.  This can result in current unbalance that is dangerous to motors.  Current unbalance causes voltage unbalance and vice-versa.  Therefore, by taking voltage readings, calculations can be made to find the percentage of current unbalance.  Unbalanced currents flowing in the motor starter windings create opposing forces.  High temperatures result, and the motor “trips out,” or worse, it burns out!  If the unbalance is great enough, the motor may not be able to start.  In addition, the full load speed will be reduced.  Motor failure frequently happens.

If a motor repeatedly “trips out,” and a diagnosis of the motor shows no faults in the motor, suspect phase unbalance as a likely cause of the “trip outs.”

It is very easy to check for phase unbalance.  One of two formulas can be used to calculate the percentage of voltage unbalance and consequently, the current unbalance.  We’ll refer to them as method A and method B.

Method A

“The percentage of voltage unbalance is defined as 100 times the sum of the deviations of the three voltages from the average, divided by twice the average voltage.”

Example using method A:

Taking the three voltage readings, phase to phase, they are:

Phase A to B, 217 volts.  B to C, 221 volts, A to C 228 volts.

Add the three voltages and divide by three to get the average voltage: The sum of the deviations from average:

222 – 217 + 222 – 221 + 228 – 222 = 5 + 1 + 6 or 12

Now divide 12 by two times the average: The voltage unbalance is 2.7%

Method B

“The percentage of voltage unbalance is equal to 100 times the maximum voltage deviation from the average voltage, divided by the average voltage.”

Using the readings taken for method A example: Maximum voltage deviation from average: Identical results are obtained using either method A or method B.

The effects of the percentage of voltage unbalance can be found in many electrical publications.  The following table is a portion of one.

Table 1.

 % Voltage Unbalance % Current Rise % Temperature Rise 0.5 3.8 0.5 1 8 2 1.5 12.5 4.5 2 16.75 8 2.5 21 12.5 3 27 18

The percent temperature increase will be about two times the square of the voltage unbalance. Note that a relatively small voltage unbalance can result in a large current and temperature increase.  Finding the cause or causes of voltage unbalance can pay big dividends.

There is one more interesting property of AC voltage and current.  You may have noticed with AC current that the voltage and amperage vary within the cycle.  First they are zero, then go to maximum in one direction, back to zero, then to maximum in the other direction, and back to zero again.  Nevertheless, the effective voltage and amperage will always be less than the maximums.  These less than maximum values are the effective values and are called “Root-Mean-Squares,” abbreviated RMS.  The RMS is .707 times the maximums.

For instance, the maximum values for a 120-volt, 60-hertz, 10-amp AC circuit are 170 volts and 15 amps.  The instantaneous wattage at maximums would be 2550 watts, but the effective usable wattage is 1200.

Product Categories:
Feature this resource?:
No