Updated for 2010, this guide covers how to estimate, warranties and quantity discounts, an estimate form, design and engineering fees, service and installation hourly rates, a letter footage chart, and ...
How to minimize electrical losses and make power companies happy
Typically, a power company's goal is to sell as much electricity as it can. But power companies don't like to "make" more electricity to compensate for power that's "lost" on the way from a power station to an electrical meter, which refers to losses in the electrical distribution system.
This month's topic, saving power by reducing losses in neon circuits, requires some discussion about alternating-current (AC) theory. However, for more in-depth information, readers should consult electrical-engineering books.
"Power factor," the ratio between the actual load power and apparent load power, indicates how effectively a current is converted into useful work output.
Every wire has a resistance -- the longer and thinner the wire, the higher its resistance. And, every current that passes through a resistance creates losses. For a given wire length, resistance and losses can be reduced by increasing the wire diameter, which results in increased material costs. A power company's primary goal is to distribute power and, at the same time, incur fewer losses and minimal distribution-center costs.
Voltage, current and power
Electrical power is defined by the actual voltage at the load, multiplied by the current flowing through this load. With direct current (DC), everything is constant in time, and the measurement of current and voltage (and multiplication to determine the power drawn) is quite simple. But, in nearly all U.S. towns, AC networks are used today.
With AC, the voltage varies with time from zero through positive maximum, through zero again, through a negative maximum, back to zero again, and then repeats. In the United States, AC reverses direction 120 times every second. But what happens to the current when the voltage alternates?
With a resistance load on an alternating current source, the current rises every time the voltage rises, and vice versa. Or, simply, the current tracks the voltage.
To determine how much power is drawn, multiply the current by the voltage -- at every point on the curve -- and sum it up over a complete cycle (this is what engineers call "integrating over time"). Now, you'd expect the result to be zero, implying that positive and negative parts have the same value and would cancel each other out.
However, this isn't the case. Negative voltage multiplied by negative current renders a positive result, so the total power is positive. Consequently, the electrical meter will run. $image1
Capacitors and inductors on AC
