79. When the work in joules is known, the work in foot-pounds That is, the equivalent work done in foot-pounds is obtained by multiplying the number of joules by .7373. EXAMPLE.-Express the work done in foot-pounds in a circuit when a current of 8 amperes flows for 2 hours between potentials differing by 10 volts. force = 10 volts = = SOLUTION. Reducing the time to seconds gives 2 × 60 × 60 = 7,200 seconds t. The current 8 amperes C, and the electromotive E. Then, by formula 15, the electrical work done == 8 x 10 × 7,200 = 576,000 joules. Expressed in foot-pounds, this will be, by formula 18, F. P. = .7373 × 576,000 = 424,684.8 foot-pounds. Ans. ELECTRICAL POWER. 80. Power, or rate of doing work, is found by dividing the amount of work done by the time required to do it. In mechanics, the unit of power is called the horsepower; in electrotechnics, the unit of power is the watt. It is found by dividing the amount of electrical work done by the time required to do it. Let E the electromotive force in volts; J, the electrical work in joules; t, the time in seconds; C, the current in amperes; and W, the power in watts. By formula 15, the amount of electrical work J = CEt. Then, W = CEt = CE. (19.) The power in watts is equal to the strength of current in amperes multiplied by the electromotive force in volts. EXAMPLE.-What is the power in watts developed in a closed circuit in which a current of 12 amperes is flowing between potentials differing by 25 volts? SOLUTION. mula 19, E=25 volts and C = 12 amperes. Hence, by for W=CE=12 × 25 = 300 watts. Ans. By taking into consideration the resistance of the circuit, the equation for determining the power in watts may be expressed in two other ways: By derivation from formula 16, That is, the power in watts is equal to the strength of current in amperes squared, multiplied by the resistance in ohms. EXAMPLE. Find the power in watts in a closed circuit in which a current of 30 amperes is flowing against a resistance of 3 ohms. SOLUTION. C = 30 and R = 3. Hence, by formula 20, W = C R = 30o × 3 = 2,700 watts. Ans. By derivation from formula 17, That is, the power in watts is the quotient arising from dividing the electromotive force in volts squared by the resistance in ohms. EXAMPLE. The drop of potential in a closed circuit when a current is flowing is 20 volts and the resistance is 10 ohms; what is the power in watts expended? SOLUTION.- E 20 volts and R = = 10 ohms. Hence, by for 81. One watt equals of a horsepower; or, 1 horsepower equals 746 watts. That is, to express the rate of doing electrical work in horsepower units, find the number of watts and divide the result by 746. The horsepower may also be expressed by three other equations, by expressing the watts in terms of electromotive force, current, and resistance, as obtained from formulas 19, 20, 21, viz.: 50 amperes and electromotive force = EXAMPLE.-Given, current == 250 volts; express the power directly in horsepower units. SOLUTION.- E=250 volts; C = 50 amperes; hence, EC 250 X 50 H. P. = 746 746 = 16.756 horsepower. Ans. EXAMPLE.-Given, strength of current = 25 amperes and resistance = 14.92 ohms; express the power directly in horsepower units. SOLUTION. C = 25 amperes; R = 14.92 ohms; hence, - EXAMPLE.-Given, electromotive force = 4 ohms; express the power directly in horsepower units. = 110 volts and resistance 82. To express the power in watts when the horsepower is known, use the following formula: That is to say, the power in watts is found by multiplying the horsepower by 746. EXAMPLE.-Express the equivalent of 4.35 horsepower in watts. SOLUTION. H. P. = 4.35; by formula 23, the electrical power W = 4.35 X 746 = 3,245.1 watts. Ans. 83. The watt is too small a unit for convenient use in expressing the output of large dynamos, so the kilowatt is generally used. One kilowatt is equal to 1,000 watts or about 1 horsepower. For example, if a dynamo were rated at 75 kilowatts, it would have an output of 75,000 watts or roughly about 100 horsepower. The kilowatt-hour is a unit of work commonly used in electrical work. It is the amount of work done when 1 kilowatt is expended for 1 hour, or kilowatt for 2 hours, etc. The kilowatt-hours are, therefore, found by multiplying the average number of kilowatts by the average number of hours during which the kilowatts were expended. Since 1 kilowatt = 1,000 watts, 1 kilowatt-hour 1,000 watt-hours. Now 1 watt expended for 1 second is equal to 1 joule; hence, 1 kilowatt-hour = 1,000 × 3,600 3,600,000 joules, or 3,600,000 x .7373 =2,654,280 foot-pounds. The kilowatt-hour represents a definite amount of work, whereas the kilowatt expresses the rate at which work is done, and is, therefore, a unit of power. = HEAT AND STEAM. HEAT. NATURE OF HEAT. 1. All modern scientists and investigators agree that heat is a form of energy. It is conceived to be a motion of the molecules composing matter. All matter is composed of molecules, which, according to the generally accepted theory, are not in a state of rest, but are moving or vibrating back and forth with a greater or less velocity. It is this movement of the molecules that is generally believed to cause the sensations of warmth and cold; if the motion is slow, the body feels cold; whereas, if the motion is rapid, the body feels warm. Since a body in motion has kinetic energy and since the molecules composing matter are supposed to be in motion, each molecule possesses kinetic energy; hence, we can conceive heat to be a form of energy. 2. Temperature is a term used to indicate how hot or cold a body is; i. e., to indicate the velocity of the vibration of the molecules of a body. A body having a high temperature is said to be hot; a body having a low temperature is said to be cold. When a body, as, for example, an iron bar, receives heat from any source, its temperature rises; on the other hand, when a body loses heat, its temperature falls. COPYRIGHTED BY INTERNATIONAL TEXTBOOK COMPANY. ENTERED AT STATIONERS' HALL, LONDON 211 |