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Electricity  Basic Navy Training Courses NAVPERS 10622 
Chapter 20

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 20 TRANSFORMERS
WHAT FOR
Transformers get their name from the kind of work they dothey TRANSFORM ELECTRICAL
POWER. Suppose you had to supply a galley stove at 220 volts, the ship's lighting
system at 110 volts, and a 6volt signalbell system. Three different generators
for three different voltages? Absolutely not! Use transformers, and then one generator
can supply all three loadseach at its proper voltage. Your generator would have
a standard output, say 110 volts. One transformer would STEP this UP to 220 volts.
Another would STEP it DOWN to 6 volts.
This sounds like something for nothingbut it isn't. TRANSFORMERS DO NOT INCREASE
OR DECREASE POWER. Remember that power is voltage multiplied by current. And when
a transformer changes voltage it also changes current. If you INCREASE the voltage,
the current is DECREASED. One goes up and the other goes down power INPUT. The VOLTAGE
and CURRENT of a circuit ARE CHANGED BY TRANSFORMERS. But the TOTAL POWER remains
the SAME.
HOW IT WORKS
The transformer is a mutual induction circuit. And that should tell you what
to expect 
1. There is an iron core.
2. There are TWO electrical circuits.
3. Energy is transformed from one circuit to the other by a field
of flux.
Look at figure 218. This is a SIMPLE transformer. Modern transformers are not
built this way, but every important transformer principle can be illustrated by
this diagram.
Figure 218.  Simple transformer.
FIRST, notice the core  it forms a closed iron path for flux. This is the path
of energy transfer between the two circuits. The efficiency of a transformer depends
on this core and it is always designed to carry the maximum flux.
SECOND, notice the two circuits  they are formed in coils around the iron of
the core. The coil, or winding, that receives energy from the source is called the
PRIMARY. In figure 218, the winding on the left is connected to the line, or source
 it is the primary. The other winding, the one connected to the load, is called
the SECONDARY. In the figure, the winding on the right supplies the load  it is
the secondary. Never let the relative number of turns on the two windings confuse
you. The source winding is always the primary and the load winding is always the
secondary. If the load and source connections of figure 218 were exchangedthe names
of the windings would also exchange.
Figure 219.  Transformer flux.
THIRD, notice the flux set up by the transformer primary. Figure 219 shows this
better than figure 218. You know that this is a mutual induction circuit. And you
know something else about ityou know that flux lines must be cut for transfer of
energy. Neither the primary nor the secondary coils can move. So the flux field
must move. And that means just one thing  ONLY A.C. and PULSATING D.C. can be used
in transformers  they produce moving fields.
Imagine that 60 cycle a.c. is being fed into the primary of figure 219. A field
blossoms out around EVERY PRIMARY TURN. The iron preserves this field and carries
it to the secondary. At the secondary EVERY TURN is cut by the field. Voltage is
induced in the secondary and power has 'been transferred from primary to secondary.
In 60 cycle a.c., you know that current changes direction 120 times each second
(two alternations to the cycle). Which means that the secondary is cut 120 times
a second. Therefore, the frequency of the secondary is EXACTLY the same as the frequency
of the primary.
HOW MUCH?
The question, "How much?" always comes up. That's because you are
using transformers to change voltage. And what's the sense of using a device unless
you know what you're going to get out of it?
Figure 220 is a typical transformer setup. Notice the voltmeter readings. They
tell the story of transformer voltage. The voltage of the primary (E_{p})
is 110 volts. The voltage of the secondary (E_{s}) is 220 volts. You'll
notice that the secondary is unloaded  an open circuit. Therefore, the secondary
current is zero. But, how much current is there in the primary? Looking at the circuit,
you see that the primary is connected directly across a 110 volt line. You'd expect
a heavy current. But you're wrong  the voltage of selfinduction (E_{si})
in the primary is very high. The tight windings and the iron core cause every turn
of the winding to be cut by almost the total of the coil's flux. The result is that
E_{si} is almost equal to E_{p}. Not only equal  but OPPOSITE.
And this nearly equal and opposite E_{si} chokes the current down to an
extremely small value. The small current that does flow is called the MAGNETIZING
CURRENT of a transformer.
Let's make figure 220 an example. The resistance of the coil is 10 ohms. And
the voltage of selfinduction is 109 volts. This gives you these values 
E_{p} = 1;1.0 v.
E_{si} = 109 v.
R = 10 Ω
I_{p} = (110  109)/10 = 1/10 amp. (by Ohm slaw)
Figure 220.  Stepup transformer.
The magnetizing current is only onetenth of an ampere.
Notice that E_{si} was subtracted from Ep. Remember that Ep and E_{si}
are OPPOSITE  they cancel each other, so one must be subtracted from the other
to get THE NET VOLTAGE ACTING ON THE CURRENT. It's like the counteremf in a dc
motor.
Now to get back to the why and wherefore of that 220 volts on the secondary.
It's like this  the magnetizing current's flux produced 109 volts of selfinduction
on the 11 turns of the primary. That's just slightly under 10 volts per turn. Call
it an even 10 volts. Is there any reason why this flux is not producing a like voltage
 10 volts per turn  on the secondary winding? No reason in the world  both coils
are on the same core, so whatever flux cuts  one, cuts the other. BUT  and it's
a big "but"  the secondary has 22 turns. And at 10 volts per turn  that's
220 volts.
This tells you that the ratio of voltages is the same as the ratio of turns.
Mathematically it says 
E_{p}/E_{s} = T_{p}/T_{s}
in which
E_{p} = voltage in primary;
E_{s} = voltage in secondary;
T_{p} = turns in primary coil;
T_{s} = turns in secondary coil.
Suppose a transformer has 600 volts and 2,400 turns on the primary. What will
be the voltage of a 400 turn secondary?
E_{p}/E_{s} = T_{p}/T_{s}
600/E_{s} = 2,400/400 = 100 volts.
Here's another way to work this problem. How many volts per turn in the primary?
600 volts divided by 2,400 turns  1/4 volt per turn. The volts per turn of the
primary and secondary are just about equal, so 400 turns multiplied by 1/4 volt
is 100 volts on the secondary. Notice that this transformer is a stepdown job.
The voltage goes from 600 volts to 100 volts.
Transformers are highly efficient. As high as 98 percent. That's why you can
ignore the difference between E_{p} and E_{si} in calculating volts
per turn.
Now put a 44ohm load on the transformer of figure 220. You have the circuit
of figure 221. The current in the secondary is 
I_{s} = E_{s}/R_{s} = 220/44 = 5 amps.
Right here is a good place for you to get this fact straight. THE SECONDARY
CURRENT IS CONTROLLED BY THE LOAD. If the load had been 22 ohms instead of 44 ohms,
the current would have been 10 amperes instead of 5 amperes.
The 5 ampere secondary current is OPPOSITE in direction to the primary current.
You can prove this by the coil hand rule. Or  reason it out this way  the current
in the secondary is produced by the voltage induced by the primary field. You know
that E_{si} is opposite to E_{p} and you know that both E_{s}
and E_{si} are produced by the same primary field. Therefore, Es must be
opposite to E_{p}. This is the same general rule you got for mutual induction
circuits.
Figure 221.  Loaded transformer.
Now, what is the EFFECT of opposite primary and secondary currents? Simply thisTHEIR
FIELDS TEND TO CANCEL EACH OTHER. The 5 ampere secondary sets up a field which destroys
some of the primary flux. Result  less E_{si} and more current in the primary.
The primary current will increase until the primary field strength balances the
secondary's field strength. Field strengths are determined by ampereturns (IT).
When the two fields are equal, their ampereturns are equal 
I_{p}T_{p} = I_{s}T_{s}
or I_{p}/Is = Ts/T_{p}
Which says that, the currents in the two windings are INVERSELY proportional
to their number of turns.
In the example 
I_{p}/I_{s} = T_{s}/T_{p}
I_{p}/5 = 22/11
and I_{p} = 10 amps.
Which means that the primary current increases to 10 amperes for a secondary
load of 5 amperes.
Suppose another transformer has a primary of 70 turns, a secondary of 350 turns,
and a 30 ampere load. What is the primary current?
I_{p}/I_{s} = T_{s}/T_{p}
I_{p}/30 = 350/70
and Ip = 150 amps.
You can prove this is correct by using the fact that primary and secondary fields
are of equal strength.
I_{p}T_{p} = I_{s}T_{s}
150 x 70 = 350 x 30
10,500 = 10,500
This brings you to an important fact about transformers. The load current in
the SECONDARY controls the current in the PRIMARY. This control is centered in the
action of the secondary flux. Every change of secondary flux changes the E_{si}
on the primary and consequently the primary current. It is an automatic control.
No load on the secondary reduces primary current almost to zero (only magnetizing
current flows). As the secondary is loaded, its current. increases and the primary
current increases with it.
INPUTOUTPUT
Since transformers are so highly efficient, their efficiency is usually considered
to be 100 percent. This assumption introduces a small error but for all practical
purposes it is close enough. Considered 100 percent efficient, the transformer's
input power and output power must be equal. Therefore 
P_{p} = P
I_{p}E_{p} = l_{s}E_{s}
Let's check this against an example. Figure 222 is a good practical problem.
And these are the things you can solve for  Es, Is, Iv, Ps, and Pv.
Figure 222.  Power in a transformer.
For E_{s} 
E_{p}/E_{s} = T_{s}/T_{p}
50/E_{s} = 250/1000
E_{s} = 200 v.
For I_{s} 
I_{s} = E_{s}/R_{s} = 200/8 = 2.5 amps.
For I_{p} 
I_{p}/I_{s} = T_{s}/T_{p}
I_{p}/2.5 = 1000/250
I_{p} = 10 amps.
For P_{s} 
P_{s} = E_{s}I_{s}
P_{s} = 200 x 2.5 = 500 w.
For P_{p} 
P_{p} = E_{p}I_{p}
P_{p} = 50 x 10 = 500 w.
Note that P_{p} = P_{s}.
SUMMARY OF THEORY
A summation of the preceding information gives you three important equations
and two important facts.
EQUATIONS 
1. E_{p}/E_{s} = T_{s}/T_{p}
2. I_{p}/I_{s} = T_{s}/T_{p}
3. I_{p}E_{p} = I_{s}E_{s}
FACTS 
1. The voltage of the secondary is always opposite in direction
to the voltage of the primary.
2. The current drawn by the primary is controlled by the secondary
load current.
MODERN CONSTRUCTION
Perhaps you are wondering why transformers are so efficient. The reason is their
lack of moving parts. When moving parts are eliminated from a device, mechanical
friction, the biggest source of loss, is gone. Losses become very small.
Modern transformers are designed with just one idea  cut down the losses to
as small a value as possible. In general, the losses that are unavoidable are divided
into two classes  IRON losses and COPPER losses.
Iron losses occur in the core. Part of these losses is due to the resistance
of the molecular magnets. They must turn over every time the a.c. reverses. In a
60cycle transformer, the molecules must shift around 120 times each second. Molecules
resist this shifting; and their desire to stand still is called HYSTERESIS. You
might say that hysteresis is really FRICTION. And you know that friction produces
heat.
The iron cores themselves act as wires. They are cut by the flux of their winding's
fields, and carry small induced currents. These induced currents are called EDDY
CURRENTS because they flow entirely within the iron core. Eddy currents produce
heat  they are really small short circuits within the core.
Thus, IRON LOSSES are made up of two factors  HYSTERESIS and EDDY CURRENTS.
Both produce heat , and both represent losses which must be subtracted from the
output power.
COPPER LOSSES occur within the windings. They are due to just one thing  the
heat generated by the current in the winding conductors. Now here is an important
fact  copper (and most other conductors) increases its resistance as it gets hotter.
This means that if the heat resulting from iron and copper losses is allowed to
accumulate, the windings will get hotter. And the hotter they get, the higher is
their resistance. Which increases the power loss due to resistance. It's a vicious
circle. Losses produce heat and heat produces even higher losses and so on.
IRON LOSSESDOWN
Hysteresis losses are reduced by using either a soft iron or a special transformer
steel containing silicon. These metals allow their molecules to shift easily  with
a minimum of friction. Eddy currents are broken up by constructing the cores out
of thin plates of iron instead of a solid piece. This LAMINATED structure breaks
up the eddy currents by insulating each plate from its neighbors. Eddy currents
still exist  but they are so small that the loss they cause can be neglected.
Figure 223.  Transformer cores.
Figure 223 shows a number of transformer cores. Notice that they are all complete
magnetic circuits, all are laminated, and all use iron generously.
COPPER LOSSES  DOWN
Reducing copper losses is a pipe. The windings are designed to be as short and
as heavy as possible. Both short turns and large wire tend to decrease resistance
and reduce heat.
Finally, the overall loss in the windings, due to accumulated heat, is cut down
by special cooling devices. Oil baths, radiators, and air blowers are all used to
keep transformers cool.
CORE SHAPES
There are three general types of core design. Each has its own special advantage.
But all put both primary and secondary windings on the same leg of iron.
Figure 224 shows all three types. A  the CORE TYPE, is best for high voltage
use. Its winding turns are short and IR drops are at a minimum. B  the SHELL type
has longer turns because the middle leg is twice the size of the outside legs. The
IR drops are larger but the flux path is shorter. Therefore, this type is best for
heavy current loads. C  the MODIFIED SHELL type is a combination of both core and
shell. The modified type has some of the advantages of both the simple types.
Figure 224A.  Core type.
Figure 224B.  Shell type.
Figure 224C.  Modified shell type.
WINDING DESIGNS
There are two winding designs which can be used on ANY of the three types of
cores. In figure 225, you'll see crosssections of both types.
Figure 225.  Types of winding design.
A  the CYLINDRICAL design consists of a primary cylinder and a secondary cylinder.
One is fitted over the top of the other. And then the whole winding assembly is
slipped over the iron core. B  the PANCAKE design separates each winding into sections
or pancakes. Then alternate pancakes of primary and secondary are slipped over the
core. In the final assembly, all pancakes of the primary are connected together
in series, and all pancakes of the secondary are. connected together in series.
The cylindrical winding is a little cheaper to construct, but it is harder to
repair. The pancake winding costs a little more but is easier to repair because
it is broken up into sections. You'll run into both designs of winding and probably
all three types of core. The best advice is to understand each, but leave them as
is.
SUMMARY
Transformers are built to do just one job  change voltage and current values.
They are designed to do it as efficiently as possible. Cores are laminated, winding
turns are shortened, and artificial cooling is used. All these make the transformer
the most troublefree and efficient device used by electricians.
Chapter 20 Quiz
(click here)

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