RF Cascade Workbook

Copyright

1996 -
2016

Webmaster:

Kirt
Blattenberger,

BSEE - KB3UON

RF Cafe began life in 1996 as "RF Tools" in an AOL screen name web space totaling 2 MB. Its primary purpose was to provide me with ready access to commonly needed formulas and reference material while performing my work as an RF system and circuit design engineer. The Internet was still largely an unknown entity at the time and not much was available in the form of WYSIWYG ...

All trademarks, copyrights, patents, and other rights of ownership to images and text used on the RF Cafe website are hereby acknowledged.

My Hobby Website:

AirplanesAndRockets.com

to Find What You Need.

There are 1,000s of Pages Indexed on RF Cafe !

NAVPERS 10622 - Chapter 7

Here is the "Electricity - Basic Navy Training Courses" (NAVPERS 10622) in its entirety. It should provide one of the Internet's best resources for people seeking a basic electricity course - complete with examples worked out. See copyright. See Table of Contents.

¶ U.S. GOVERNMENT PRINTING OFFICE; 1945 - 618779

**CHAPTER 7ELECTRICAL POWERFORCE**

Scientists and technicians make a point of specifically defining all the terms they use. They like their language to say exactly what they mean. This is necessary because scientists use technical terms in explaining their work. With a good working knowledge of such terms as force, power, work, emf, current, and resistance, you'll be far more savvy about your own work. Too, you'll want to be sure so that you can shoot the breeze about your job. Knowing exactly WHAT certain words MEAN helps a lot!

You often HEAR the word FORCE. But you USE force far. more) often than you hear the word. Every time you lift something, you use force. Every time you move, you have exerted force. A ship moves through the water because of force. In fact, every time anything moves or tends to move, force has been exerted. Force may be a push or a pull. FORCE then, IS THAT WHICH PRODUCES MOTION OR TENDS TO PRODUCE MOTION. Consider the force of GRAVITY. It causes bodies to move toward the earth. Suppose you put a box on a table. The box tends to move DOWNWARD because of gravity but the table exerts an UPWARD force - the two forces are balanced. However, if the box is "too heavy" for the table, the table cannot exert enough upward force to balance the pull of gravity and the table collapses.

The force exerted by a propeller is a MECHANICAL force. The explosion of hydrogen and oxygen to form water is a CHEMICAL force. The force which causes electrons to flow is an ELECTROMOTIVE force and its unit is the VOLT. There are many kinds of force, but they ALL produce or tend to produce motion.

**WORK**

WORK IS A FORCE ACTING THROUGH SPACE. Imagine that you push with. all your strength against a steel bulkhead. You probably think you've <;lone, work but technically you HAVEN'T. True, you have exerted force on the bulkhead, but since the bulkhead hasn't moved, NO WORK has been done. Now imagine that you exert the same force lifting a 200-pound shell from the deck to a shelf 4 feet high. WORK HAS BEEN DONE, because the force acted through space'. You exerted a 200pound FORCE through a DISTANCE of 4 feet.

Work = force x distance

so in this case -

Work =200 pounds x 4 feet = 800 foot-pounds (ft.-lb.)

NOTE that work and force are different. Force is exerted whenever a body is pushed or pulled BUT work is done only IF THE BODY MOVES. In electricity, the unit of work is the JOULE.

ELECTRICAL WORK = VOLTAGE x COULOMBS = JOULES

The JOULE, by itself, has relatively little use because it does not take into consideration the factor of TIME. That is, it might take 2 seconds or 2 days for 120 volts to move 1 coulomb (6.3 billion billions electrons); and in either case, you would have done 120 x 1 = 120 joules of electrical work. So time is really important - and that brings you to power.

**POWER**

POWER IS THE TIME RATE OF DOING WORK. This relation is expressed -

Power = Work/Time

You have learned that the amount of WORK done has nothing to do with the time it takes to do the work. BUT the amount of POWER depends on HOW FAST that work can be done. You know that a steam shovel has a great deal more POWER than a man. Both can do the same amount of work but the steam shovel will do it a lot FASTER. For example, say that 1,000 pounds of earth must be raised 20 feet. The work is -

Work = force x distance = 1,000 x 20 = 20,000 foot-pounds.

The steam shovel does the job in one scoop, taking 2 seconds. The man does the job in 20 minutes (or 1,200 seconds). The steam shovel has -

Power = Work/Time = 20,000/2 = 10,000 ft-lb. per second of power

The man has -

Power = Work/Time = 20,000/1,200 = 16.7 ft-lb. per second of power

The steam shovel has exerted approximately 600 times as much POWER as the man even though the WORK done by both is equal.

The mechanical unit of power - FOOT-POUNDS PER SECOND - is too small for practical use. In the early days, power was generally supplied by horses, and experiments indicated that an average horse could do 550 foot-pounds of work per second. This led to the establishment of a larger unit-the HORSEPOWER (hp).

HP = 550 FT.-LBS. PER SECOND

What was the HP of the steam shovel in the preceding example?

10,000 ft.lbs. per sec./550 ft.lbs. per second = 18.2 hp

What was the HP of the man?

16.7 ft.-lbs. per second/550 ft.-lbs. per second - 0.032 hp

Power in the electrical system is measured in WATTS.

Power = Work/Time = Volts x Coulombs/Time = Watts (w.)

Do you recognize the expression - Coulombs/Time? In Chapter 3, you learned that coulombs divided by time was the time rate of flow of current - the AMPERE. So you can substitute AMPERE for Coulomb/Time in the above equation, and the equation becomes -

Power = volts X amperes = Watts, or -

P=E x I

POWER IS AN IMPORTANT MEASURE IN ELECTRICITY. It tells you how much you can expect from a motor or generator.

Study the circuit in figure 34. It shows a motor connected to its generator; and meters are installed to read the values of current and voltage in the circuit. By multiplying the ammeter and volt-meter readings, you get the power consumed by the motor -

P = E x I = 120 x 8 = 960 watts

which means that this motor CONSUMES 960 watts of power.

By measuring the amount of mechanical work a number of electric motors did in one second, it was determined that- 746 Watts = 1 hp

Figure 34. - Power consumption of a motor.

Does the motor DELIVER 960 watts, or 960/746 = 1.29 hp, of power? No, because some of the power is lost within the motor. This loss is caused by internal heat and friction. All machines lose some power by heat and friction. If they didn't, they would be 100 percent EFFICIENT and there would be perpetual motion. EFFICIENCY is the percentage of the total input power that is actually delivered as output. Motors deliver their power at their shafts. Say that this particular motor is a one-hp job. This means that the motor DELIVERS one hp AT ITS SHAFT.

What is its efficiency?

INPUT = 960 watts

OUTPUT = 1 hp = 746 watts

EFFICIENCY = Output/Input = 746/960 = 0.777, or 77.7%

The motor is 77.7 percent efficient - in other words, it delivers 77.7 percent of the power it consumes. The balance of power - 22.3 percent is lost as heat and friction. Look at figure 35. It shows the power and the power losses as a picture. If you follow the arrows through this picture of a motor, you will find the input power is ELECTRICAL POWER. It splits up in the motor, going in two directions. The losses in the form of heat are radiated upward, and the output in the form of MECHANICAL POWER is delivered by a rotating shaft. This gives you the definition of a motor - A MACHINE WHICH CONVERTS ELECTRICAL ENERGY INTO MECHANICAL ENERGY. (Just the opposite to the action of a generator.)

Figure 35. - Delivered and lost power in a motor.

How much work is this motor capable of doing? Since 1 hp = 550 ft-lbs. per second, the motor can exert a force of 550 pounds through 1 foot of space every second. Or, 275 pounds through 2 feet of space every second. Or, 55 pounds through 10 feet of space every second. You'll notice that the force decreases as the speed increases.

To be sure you understand these terms, consider a harder example. Figure 36 shows a circuit involving a 10 hp motor, a generator, meters, and a 'l,000-foot length of double wire connecting lines.

Figure 36. - Power loss in a long line.

The generator furnishes 440 volts of emf, but the motor draws 25 amperes at only 390 volts. The difference -50 volts - is used in pushing the current through the 2,000 feet (1,000 ft. for each wire) of connecting wire.

You can calculate the resistance of this wire -

R = E/I = 50/25 = 2 Ω

which means that 50 volts of force are used in pushing 25 amperes of current through 2,000 feet of wire having 2 ohms of resistance.

How much power is consumed by the motor?

P = E x I = 390 x 25 = 9,750 w

What is the efficiency of this 10 hp motor?

Efficiency = output/input = 10 x 746/9.750 = 7.460/9,750 = 76.5%

How much power is lost in the motor by heat and friction?

Losses = input - output = 9,750 - 7,460 = 2,290 w.

What is the power consumed by the line in delivering current to the motor?

P = E x I = 50 x 25 = 1,250 w.

As a check, you know that total power furnished, minus all losses, should give the power output of the motor. In this case, 440 x 25 = 11,000 watts is the power furnished. The losses are 2,290 + 1,250 = 3,540 watts. The output then is 11,000 - 3,540 = 7,460 watts. This checks with the rated output.

Remember - whenever work must be done, power is consumed doing it. It requires work to force current through a wire - in this case, 1,250 watts of power is consumed by the wire. It requires work to overcome the friction of the motor and force current through its windings - 2,290 watts of power is consumed in doing this work. Finally, the motor is capable of furnishing 10 hp or 7,460 watts at its shaft to do work.

The power equation may be used in three forms depending on the problems to be worked -

P = EI; E = P/I; I = P/E

**EXAMPLES**

1. Determine the value of current in a 100 watt lamp on a 115 volt line.

I = P/E = 100/115 = 0.87 amp.

2. Determine the potential drop of a line which consumes 1,200 watts in carrying 60 amperes.

E = P/I = 1,200/60 = 20 v.

3. Determine the power consumed by a 440 volt motor, if it draws 22 amperes.

P = E x I = 440 x 22 = 9,680 w.

4. What is the hp of the motor in question 3?

hp = P/746 = 9,680/746 = 13 hp

**LARGE AND SMALL UNITS **

You ordinarily would measure butter by the pound and coal by the ton. Think how clumsy it would be to reverse this procedure - butter by the ton and coal by the pound. The simple units of electrical measure - volts, amperes, ohms, and watts - prove to be clumsy when very LARGE or extremely SMALL quantities are involved. A system of pre-fixes has developed for use in measuring large and small quantities of electrical units. The table below gives the common prefixes used in electricity. Each prefix can be used with any of the electrical units. For example, instead of saying "a 10,000 watt generator," it is handier to say "a 10 KILOWATT generator." Instead of writing "0.010 amperes" it is easier to write "10 MILLIAMPERES." In testing insulation you will use MEGOHMS instead of millions of ohms. In radio work, the MICRO- and MILLI- prefixes are constantly used.

MEGA ................ MILLION (1,000,000)

KILO ................. THOUSAND (1,000)

1 MILLI .............. ONE-THOUSANDTH (1/1,000)

1 MICRO ............ ONE-MILLIONTH (1/1,000,000)

**Chapter 7 Quiz**