Electrical Power and Energy
9–1 ENERGY
In Section 2–8 we discussed the conversion of electrical energy to perform work. The type of work or change accomplished involves force and motion. The following changes all require energy: mechanical movement, the production of heat or light, the production of sound, the conversion of one chemical compound into another, and the production of radio waves. The amount of energy required for these changes can be measured, although we can- not see it. How this is accomplished is the subject of this chapter.
In common usage the words work and energy have broad meanings. The physical work or energy with which we are concerned does not include such things as the work done by someone counting the cars that pass a corner, the work done in getting people to change their minds, or the energy with which one tackles an arithmetic problem. Instead, words like work and energy have precise, scientific meanings, which should be understood before we discuss the concept of power. The following sections will help clarify these ideas.
Note: Some courses of study, under constraints of time, may bypass these discussions and proceed directly to Section 9–3. However, many programs of electrical apprenticeship demand some rudimentary knowledge of physical science related to these concepts.
Mechanical Energy
The lifting of a weight illustrates the meaning of a unit of measurement of energy. One foot-pound of energy, or 1 foot-pound of work, is required to lift a 1-pound weight a distance of 1 foot. (The words work and energy can be interchanged where measurement is concerned.)
A foot-pound is defined as the energy used when a 1-pound force moves an object a distance of 1 foot. The 1-foot movement is in the same direction as the applied force.
How much work is done in lifting a 20-pound weight 5 feet vertically, as shown in Figure 9–1?
Work = Force x Distance
A 20-pound force traveling 1 foot accomplishes 20 foot-pounds of work. If the force must travel 5 feet, then 100 foot-pounds of work is accomplished.
Foot-pounds = Feet x Pounds
Potential Energy
What happens to the 100 foot-pounds of energy? It is saved up or conserved. When the 20-pound weight is lifted 5 feet, it has 100 foot-pounds of energy that it did not
have when it was on the ground. This energy is called potential energy. When the weight is permitted to fall back to earth, it delivers 100 foot-pounds of energy to the earth; see Figure 9–2.
How much work is done when a 200-pound box is dragged horizontally along the floor a distance of 6 feet? The question cannot be answered when asked in this manner. Since 200 pounds is a vertical force, it is not in the same direction as the motion. Using a spring scale, however, we find that the horizontal force is 50 pounds. The amount of work done can now be computed.
50 lb x 6 ft = 300 ft-lb
What becomes of this 300 foot-pounds of energy? The energy is converted into heat by the process of friction against the floor. The box does not gain potential energy.
Kinetic Energy
How much work is done by a ball player whose hand moves 6 feet while it is applying an average 8-pound force to a 3⁄4-pound ball?
Work = Force x Distance = 8 lb x 6 ft x 48 ft-lb
What becomes of this energy? It exists as the energy of motion of the ball and is called kinetic energy. The ball in flight has 48 foot-pounds of energy it will deliver when it strikes its target.
Let us return to the example of potential energy shown in Figure 9–2. What happens when the 20-pound weight is allowed to fall the 5-foot distance to the floor? The potential energy of the weight is 100 foot-pounds. This energy becomes the kinetic energy of motion as the weight falls through 5 feet.
The purpose of these mechanical examples is to illustrate the meaning of such terms as work, energy, potential energy, and kinetic energy. These terms are often used in electrical energy discussions, but they are easier to visualize in mechanical energy examples.
The foot-pound is the energy unit commonly used in the British system of measurements. The metric unit of energy is called a joule. Most common electrical units are based on the joule as the unit of energy. The joule is the work done when a force of 1 newton is exerted through a distance of 1 meter. (A newton is 100,000 dynes, which is about 31⁄2 ounces; a meter is about 39 inches.)
1 J = 0.738 ft-lb
1 ft-lb = 1.35 J
There are two units of measurement that are used to express heat energy. The unit in the British system for heat is called Btu (British thermal unit). A Btu, equivalent to 1,055 joules, is the amount of heat needed to raise the temperature of a pound of water 1 degree Fahrenheit.
How much heat is needed to warm 10 pounds of water from 50°F to 65°F? The temperature increase is 65° – 50° = 15°F. Therefore, 10 lb x 15°F = 150 Btu.
The metric system unit of heat is called a calorie or a gram-calorie. The calorie is the energy needed to raise the temperature of 1 gram of water 1 degree Celsius. (The calorie used in food energy calculations is a kilogram-calorie; this amount of energy will raise 1,000 grams of water 1 degree.)
1 Btu = 778 ft-lb
1 Btu = 252 cal
9–3 POWER
The word power (P), as commonly used, means a variety of things. In technical language, power means how fast work is done or how fast energy is transferred. Two useful definitions of the term power are as follows:
Power is the rate of doing work. Power is the rate of energy conversion.
It is important to understand what is meant by the statement: power is a rate. We do not buy or sell electrical power. What we buy or sell is electrical energy. Power indicates how fast the energy is used or produced. A mechanical example will illustrate the meaning of power and energy.
EXAMPLE 9–1
Given: An elevator lifts 3,500 pounds a distance of 40 feet in 25 seconds.
Find:
a. The work done in 25 seconds (foot-pounds).
b. The rate of work (foot-pounds per second).
Solution
Naturally, power can also be expressed in foot-pounds per minute. The next example will make this clear.
EXAMPLE 9–2
Given: A pump takes 20 minutes to lift 5,000 pounds of water 66 feet.
Find: The rate of doing work.
Solution
The Horsepower
When James Watt started to sell steam engines, he needed to express the capacity of his engines in terms of the horses they were to replace. He found that an average horse, working at a steady rate, could do 550 foot-pounds of work per second. This rate is the definition of 1 horsepower (hp), Figure 9–3.
Can you see that the elevator in Example 9–1 is doing work at the rate of approximately 10 horsepower? Remember, its rate of work was found to be 5,600 foot pounds per second. To convert this to horsepower, we compute
Now try to compute the horsepower rating of the pump described in Example 9–2. Your answer should be approximately 0.5 horsepower. Do you agree?
Horsepower can also be expressed in units of electrical power called watts (W).
1 hp = 746 W
Watt—The Unit of Electrical Power
Recall that the metric unit of energy is called a joule. In the metric system, power is measured in joules per second. This unit of measurement corresponds to foot-pounds per second in the British system. Because the unit joules per second is used so often, it is replaced with the single term watt. (Figure 9–4 gives a comparison of various power units with the watt.)
One watt is a rate of 1 joule per second.
Another way of explaining the term watt is to say that 1 watt of power is dissipated (in the form of heat) when 1 volt of electrical pressure forces 1 ampere of current through the resistance of 1 ohm, Figure 9–5. This relationship can be expressed by the power equation
Recall from Chapter 8 the Ohm’s law circle used as a memory device. The power equation lends itself to another such memory circle, often referred to as a PIE diagram, Figure 9–6. Cover the unknown quantity to be found and read off the equation. Thus,
Given: The label of a television that states 720 W/120 V.
Find: The current in the supply line to the TV.
Solution
A Second Equation for Power
The power equation P 5 E 3 I may be combined with Ohm’s law to yield the hybrid equation P 5 I2R. Let us see how this is derived.
Assume that we need to calculate the electrical power of a circuit for which the volt- age is unknown. From Ohm’s law we know that
P = I2R is an important formula to remember. The synonymous word power and the term I2R have found their way into the vocabulary of people in the electrical trades. Thus, we speak of I2R losses (power losses in the form of heat), I2R heating, and I2R ratings.
Mathematically, the quantities of this equation may be transposed to yield two other equations, namely:
The following examples will explain the application of these formulas in finding an unknown circuit quantity.
EXAMPLE 9–6
Given: A current of 2 amperes flowing through a 250-ohm resistance.
Find: The electrical power converted into heat by the resistor. (Remember, all resistors develop heat when a current flows through them.)
A Third Equation for Power
It is possible to develop yet a third formula by combining the power equation with Ohm’s law. Beginning with P 5 E 3 I, let us assume that the quantity I is unknown. From Ohm’s law, we substitute the equality
EXAMPLE 9–11
Given: A lamp rated at 300 watts, 120 volts.
Find: The resistance of the lamp while it is operating.
The PIRE Wheel
Up to this point we have become acquainted with 12 mathematical expressions that relate to Ohm’s law and electrical power. For your convenience, these 12 equations are summarized in the circular chart in Figure 9–7.
The four letters P, I, R, and E, shown in the inner circle, represent the unknown quantities that may need to be found. Radiating outward from each of these letters are three choices of equalities that can be used for calculating the unknown.