Wire Size and Electric Current – Joule Heating

发布日期:[13-02-07 10:15:01] 浏览人次:[]

Ever take a peek inside the toaster while you’re waiting for the toast to pop up? If so, you would have noticed a bright orange glow. That glow is produced when the toasting wires heat up, which in turn creates a nice crusty surface on your bread or waffle. It’s the same phenomenon as when the filament inside an incandescent bulb glows. The light and heat produced in both these cases are the result of the Joule, pronounced “jewel,” effect at work.

To understand Joule heating, let’s first refresh our memories as to electrical current resistance. We learned previously that wire is not a perfect conductor, and as such resistance to flow is encountered. This resistance causes power to be lost along the length of wire, in accordance with this equation:

Power Loss = I2 × R

Where I is the electric current flowing through a wire, and R is the total electrical resistance of the wire. The power loss is measured in units of Joules per second, otherwise known as watts, “watt” denoting a metric unit of power. It is named after the famed Scottish mechanical engineer, James Watt, who is responsible for inventing the modern steam engine. A Joule is a metric unit of heat energy, named after the English scientist James Prescott Joule. He was a pioneer in the field of thermodynamics, a branch of physics concerned with the relationships between different forms of energy.

Anyway, to see how the equation works, let’s look at an example. Suppose we have 12 feet of 12 AWG copper wire. We are using it to feed power to an appliance that draws 10 amperes of electric current. Going to our handy engineering reference book, we find that the 12 AWG wire has an electrical resistance of 0.001588 ohms per foot, “ohm” being a unit of electrical resistance. Plugging in the numbers, our equation for total electrical resistance becomes:

R = (0.001588 ohms per foot) × 12 feet = 0.01905 ohms

And we can now calculate power loss as follows:

Power = I2 × R = (10 amperes)2 × (0.01905 ohms) = 1.905 watts

Instead of using a 12 AWG wire, let’s use a smaller diameter wire, say, 26 AWG. Our engineering reference book says that 26 AWG wire has an electrical resistance of 0.0418 ohms per foot. So let’s see how this changes the power loss:

R = (0.0418 ohms per foot) × 12 feet = 0.5016 ohms

Power = I2 × R = (10 amperes)2 × (0.5016 ohms) = 50.16 watts

This explains why appliances like space heaters and window unit air conditioners have short, thick power cords. They draw a lot of current when they operate, and a short power cord, precisely because it is short, poses less electrical resistance than a long cord. A thicker cord also helps reduce resistance to power flow. The result is a large amount of current flowing through a superhighway of wire, the wide berth reducing both the amount of power loss and the probability of dangerous Joule heating effect from taking place.

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文章作者:Philip J.O'Keefe | 文章来源:Engineering-Expert-Witness-Blog | 责任编辑:intoner | 发送至邮箱: | 加入收藏:
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