tom henry reminders for the electrician 2008(nfpa 70,nec) 73p

7/26/2019 Tom Henry Reminders for the Electrician 2008(Nfpa 70,Nec) 73p

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REMINDERS

FOR THE ELECTRICIAN

y

TonI

Henry

Copyright©2007 by Tom Henry. All rights reserved. No part

o

this

publication may be reproduced in any form or by any means: electronic

mechanical photo-copying audio or video recording scanning or

otherwise without prior written permission o the copyright holder.

While every precaution has been taken in the preparation

o

this book

the author and publisher assumes no responsibility for errors oromissions.

Neither is any liability assumed from the use o the information

contained herein.

National Electrical

ode®

and

NE ®

are Registered Trademarks

o

the

National Fire Protection Association Inc. Quincy MA.

First Printing Based on the 2008 Code.

ISBN 978-0-945495-95-6



QUICK FINDER INDEX

3-way and 4-way Switch Connections

80% Rules

Abbreviations

Actual Conductor Diameter

Ambient Temperatures - Melting Points

Area

of

Square Inch

Autotransformer

Box Fill - Conduit Fill

Burial Depths

Bus Bar Ampacity, Circular Mil

Circle,Circumference, CSA, Radius

Classifications of Voltages

Conductivity

of

Metals

Conduit Fitting Dimensions

Current and Potential Transformer

Drill Bit and Tap Sizes

Dwelling Formats

Electrical - Mechanical Degrees

Electrical Symbols

Electrodes

Eli the Ice Man

Energy Use

of

Appliances

Hole Saw Size for Conduits

Horsepower - Tons Refrigeration

Household Cooking Demands

Ignition Temperatures

Insulation Damage Temperatures

Insulation Types and Ratings

Interpolate Motor Currents

KeyWords

Kvars

Minimum Size Service

Motor Connections - Three-Phase

Motor Control Connections

Motors, Capacitor-Start

Motors, Rotor-Stator

PAGE

16

5

21

47

15

27

56

12

11

29

28

61

67

26

57-58

24

8

48

18-20

9

34

63-65

25

46

3

66

14

13

47

1

51

7

44-45

17

38-39

43



QUICK FINDER INDEX

Motors, Split-Phase

Motors, Synchronous

Motors, Three-Phase

Motors, Universal - Shaded-Pole

Motors, Wound-Rotor - D.C.

Neutral Balancing

Neutral Current in a Wye

Ohms Law

Open Delta - High Leg Delta

Optional School Calculation

Reading the KWH Meter

Receptacles

RMS, Maximum, Electron Movement

Service Drop Clearances

Service Load Calculation Format

Thermometer Scale

Transformer Connections

Vars

Weights and Measurements

37

42

41

36

40

49

50

30-33

59-60

4

62

6

35

10

2

23

53-55

51

22



REMINDERS

pg

-1

IKEY WORDS IN

THE

QUESTION

I

t \..

/

Copper -

110.5

Solid, Stranded -

Table 8

Voltage Drop -

change "2" to "1.732"

Nipple -

24" or less -

No

derating 31O.l5(B2 ex.) - 60%

Fill

Maximum Overload - 430.32(C)

Wound Rotor - Table 430.52 "Type of Motor"

Service Factor -

430.32(Al)

or 430.32(C) Higher %

Branch Circuit - Note 4 Table 220.55

Neutral-

220.61

70%

Optional Demand -

220.82, 84, 86, 88

-

) ( 'Y2 j ( 3 ) ( 4 ) (S j ( 6 ) ( 7 j (S j ( 9 j (O j ( _

I

tt8ckspace

Tab

RaRwTIeTIRTITTIvTIuTIrTIoXplli

Xl

Jr

'",aps

COCK

Ji

](Return

A JI

s )l o ]i F ) l G )l

H

JI

J

)i

K ) l

L Ji

l

(Shift

{"nln

TI

( Z

l X

)IN )IM

c

lL

V lL B

1(>

K l

(Option

)

1 ;

JI-

~

---- ---- '

\...

- - - - -

Remember to circle KEY WORDS in a question as they will

check you on your knowledge

of

a Code section.



pg

-2

REMIN ERS

I

SERVICE

LOAD

CALCULATION

I

I

IGENERAL

ME"rHOD

I

DWELLING

I

T.220.12

3va

Small app. 3000va

Laundry

1500va

T.220.42

DEMAND

220.60

AC - HEAT

220.53 4 -Appliances

220.54

Dryer

5 kw

T.220.54 Dryer

demand

T.220.55

Cooking appl.

Ampacity Tables

T.31 0.15(B)(6)

I

I

NEUTRAL

I

I

220.61 (B)(1) 70% for

cooking equipment

70% clothes

dryers.

220.61 (B) 70% in

excess of 200 amps

I

OPTIONAL

METHOD:

I

NON-DWELLING:

I

I

MOTELI

MOBILE

HOMES

I

550.31 16kva

per site

T.550.31 DEMAND

T.220.12 2 va

T. 220.42 DEMAND

Appliances

nameplate

Continuous loads

125%

220.60 AC - HEAT

I

NEUTRAL I

220.61 (B) 70% in excess

of

200 amps

250.24(B)(1)

neutral

shall not

be

smaller than grounding

electrode

conductor

IRECEPTACLES

I

220.14(L) 180va per outlet

T.220.44 Demand over 10 kva

555.12 Demand for Marinas

220.14(E)

Heavy-duty

lampholders 600 va each

220.14(H)

Multioutlet

assemblies

5' 180va

220.14(G)

Show window lighting

200va

linear

foot

I

DWELLING

I

I

NON-DWELLING

I

I

220.82

Single

220.84 MUltifamily

220.84(C)(3) dryers

nameplate

T.220.84

DEMAND

I

220.86

School

T.220.86 School DEMAND

220.88

Restaurant



REMINDERS

pg ·

I

HOUSEHOLD COOKING EQUIPMENT DEMANDS

I

1 - 12kw range =8 kw demand Col.C

1 - 10kw range

=

8 kw demand Col.C

1 - 9kw range

=

8 kw demand Col.C

1 - 8kw range

=

6.4 kw demand Col.B

What is the demand for 20-8kw ranges?

Always compare Col.B (44.8kw) with Col. C (35kw) and take

the lowest value.

35

kw.

For ranges larger than 12kw use the following format:

~ U N

C

~ W

1 2 KW

8

KW

2

W

x 5% = 10%

x

110%

8 8 KW DEMAND

The demand for one range larger than 12kw:

1 - 3kw = 8 4

kw

1 - 2 kw =11 6 kw

1 -

4

kw = 8 8

kw

1 22 kw

=

12 0 kw

1 - 5kw =

9 2 kw

1 23 kw =12 4

kw

1 - 6kw = 9 6 kw 1 - 24 kw = 12 8 kw

1 - 7kw

=

10 0 kw

1 - 25 kw = 13 2 kw

1 - 18 kw

=

10 4 kw

1 26

kw

=

13 6 kw

1 - 9kw

=10 8 kw

1 - 27

kw =

14 0

kw

1 -

20 kw = 11 2

kw



pg -4

REMINDERS

IOPTIONALSCHOOLCALCULATION

I

S HOOL

Section220.86andTable220.86 OptionalMethoddemandfactorsforfeeders and

serviceconductors.

At first, Table 220.86 seems confusing, but afterworking two orthree school

calculationsyouwillfindtheoptionalmethodtobeaveryfastandeasycalculation.

OPTION L

methodformatforaSCHOOL BUILDING

STEP 1•Table220.86 Dividethetotalconnectedloadby

thetotalsquarefootageof theschool

todeterminethevOlt-amperesper

squarefoot.

STEP

2•Table220.86

ApplythedemandfactorfromTable

220.86tothetotalvolt-ampsper

squarefootfromSTEP1

STEP 3

MultiplythevademandfromSTEP

2timesthetotalsquarefootage

of

theschoolforthedemandonthe

serviceinva.

TotalConnected

Load va

per

square

foot

Total

Square

Footage

Table220.86:

1st3va@ 100% 3va

Next17va@ 75% 12.75va

Over20va@ 25%=

Total

va

xTotalsq.ft.



REMIN ERS

pg

- 5

180 RULES

I

408.30. The total load on any overcurrentdevice located in a panelboard

shall not exceed8 of its rating where, in normal operation, the load will

continue for 3 hours or more.

Table 3l0.l5(B)(2)(a). 4 through 6 current carrying conductors, the

allowable ampacities shall be reduced 80 .

210.23(A)(l). The rating ofanyone cord and plug connected utilization

equipment shall not exceed

80 of

the branch-circuit ampere rating.

Table 210.21 B )(2). A single receptacle shall not be loaded over 8

of

receptacle rating per this Table.

Table 220.54. The demand factor for 5 household clothes dryers is

8 .

Table 220.55. The demand factor for one appliance from ColumnsA or

B is 80 .

Table 220.56. The demand factor for 4 pieces

of

kitchen equipment is

80 .

555.12. The demand factor for 9 -14receptacles that supply shore power

for boats is 8 .

630.11. The duty cycle for a welder is

80

with a multiplier of .89 or

.91 per this Code section.



pg - 6 REMINDERS

IRECEPTACLES I

THERE

IS

NO

LIMIT ON A CIRCUIT FOR A

DWELLING

jddAWbliilii[ to..

llBlllEl.IilIEil!lillm •

.t'f1!T'g:rre

., .

FOR

NON-DWELLING THE 180 VA APPLIES

OFFICE

UILDING

Example: 20a circuit x 120v =2400va/180va =13

Table 210.23 A 2 AMP CIRCUIT CAN BE LOADED

TO

20 AMPS

C 3 : l § o ; i E i E i J i ~ . ~ ~

Table 210.21(B)(2) A SINGLE RECEPTACLE CAN ONLY BE LOADED 80%

Table

220.44

NON DWELLING OVER

10

KVA DEMAND

I@:]

220.14gH)(1 )

MULTI UTLET

ASSEMBLiES

180

VA

PER 5 EACH FEET



REMINDERS

pg·

I MINIMUM SIZE SERVI E I

i

= -

: .

For a one-family dwelling the service

disconnecting means shall have a rating

o

not less than 100 amperes.

For a service supplying a single branch

[

I

circuit the disconnect shall have a rating

not less than 5 amps.

I

For a service supplying not more than two

2-wire branch circuits the disconnecting

means shall have a rating not less than 30

amps.



pg a

REMINDERS

SINGLE DWELLING

Sq. Ft. x 3 + 1500 x .35 + 3000 =

NET GENERAL LIGHTING

&

SMALL APPLIANCE

LOAD

SINGLE DWELLING

MINIMUM NUMBER OF BRANCH CIRCUITS

Square footage x 3va

=

amperage

=

number

of

circuits

120v 15 or 20



REMINDERS

pg-9

I

ELECTRODES

I

O N R E T E ~ N S E D

ELECTRODE

METAL FR ME

OF

A BUILDING

GROUND

RING

UNDERGROUND WATER PIPE

-Metal water pipe

must

be supplemented by an

additional electrode

MET L

W TER

PIPE

SIZE GROUNDING ELECTRODE

CONDUCTOR PER

TABLE

250.66

V )

SIZE EQUIPMENT GROUNDING

CONDUCTOR TO FUSE OR

CB

~ R TABLE 250.122

~

•

Qn?lId l l l



pg

·10

REMINDERS

I CLEARANCES

FOR

SERVICE DROP CONDUCTORS I

Conductors

not

over

150v to ground

10 FEET ~ P I

Conductors not over

3DDv to ground

2 FEET

Conductors

OVER

300v

to

ground

15 FEET

l'

18

FEET



REMIN ERS

pg 11

MINIMUM COVER - (burial depths)

~ ] ]

Rigid metal conduit = 6

Rigid PVC = 18

14/2

Type

U with

grd

Direct burial cables = 24

-120v residential branch circuits

GFCI protected maximum overcurrent

protection 20 amperes = 12

-Low voltage - 30 volts or less = 6



pg -12

REMINDERS

~ ] ] ]

()

CONDUIT FILL

Table 4

40%

Fi I = Number of Wires

(over 2 wires)

Table 5 Area sq.in. Permitted in Conduit

BOX FILL

Conductor

Cubic

Inch

#18 = 1.5

cubic

inch

#16 = 1.75

cubic

inch

#14

=2

cubic

inch

#12 = 2.25

cubic inch

#10 = 2.5

cubic inch

8=3

cubic

inch

6=5

cubic

inch

j; Ire Stud

/

(one or more)

0

BOX

ILL

0

Cable Clamp

(one or more)

~ ~ e ; ~

ickey

(one or more)

@

Switch

0

Receptacle

I I

.

W

a

Black or white wire

a

Grounding wire

(one or more)

a

Count one wire

Count one wire

Count one wire

Count two wires

Count two wires

Count one wire

for each wire

Count one wire for all



REMINDERS

pg

· 3

INSULATION TYPES and RATINGS

R =Rubber T =Thermoplastic X =Synthetic polymer

H

=

Heat

no

H

=

60°C one H

=

75°C two HH

=

90°C

TW

=

Thermoplastic insulation 60°C insulation rating -

moisture resistant

THW Thermoplastic insulation 75°C insulation

rating-

moisture resistant

THHW =Thermoplastic insulation 90°C insulation rating -

moisture resistant

THHN Thermoplastic insulation 90°C insulation rating -

nylon jacket

THWN Thermoplastic insulation 75°C insulation rating -

moisture resistant nylon jacket

THWN 2 =Thermoplastic insulation 90°C rating wet or

dry

moisture resistant nylon jacket



pg

·14 REMINDERS

Copper, 75°C Thermoplastic Insulated Cable Damage Table

Copper

Wire Size

Maximum Short-Circuit Withstand Current

in

Amperes For

75°C

Thermo- 1/8 1/4 1/2 1

2

3

plastic Cycle

Cycle

Cycle Cycle Cycles Cycles

14 4800 3400 2400 1700 1200 1000

12

7600 5400 3800 2700 1900 1550

10 12000 8500 6020

4300

3000 2450

8 19200 13500 9600 6800 4800

3900

6

30400 21500

15200

10800

7600

6200

4 48400 34200 24200 17100 12100 9900

I'f I I'f

INSULATION

DAMAGE

6 THW can withstand

15 218 amps for 1/2 cycle

TERMINAL DAMAGE

(loosening

of

lugs)

6 THW can Withstand

22,090

amps

for

1 2

cycle

MELTlNG

6

THW can

withstand

39 704 amps for

1 2

cycle

IF

THE

CURRENT

CONTINUED

FOR

5

SECONDS (300 cycles)

THE

6

THWWOULD MELT AT

1620 AMPS



REMINDERS

pg

-15

TYPICAL AMBIENT TEMPERATURES

LOCATION TEMPERATURE

Minimum Rating

of

Required Conductor Insulation

Well

ventilated,

normally

heated buildings

30°C (86°F)

o

(see

note below)

Buildings with such major

heat sources as power

stations or industrial

processes

40°C (104°F) 75°C (167°F)

Poorly

ventilated spaces

such as attics

45°C (113°F)

75°C (167°F)

Furnaces and boiler

rooms

minimum)

maximum)

40°C (104°F)

60°C (140°F)

75°C (167°F)

90°C (194°F)

Outdoors

in the shade

40°C (104°F) 75°C (167°F)

In thermal insulation

45°C (113°F)

75°C (167°F)

Direct

solar exposure

45°C (113°F) 75°C (167°F)

Places above 60°C (140°F)

110°C (230°F)

°Note: 60°C

for up to

and

including

8 copper and up

to

and

including

#6

aluminum.

75°C for over #8 copper and 6 aluminum.

I

MELTING POINT

OF

METALS

I

ALUMINUM

66 °C -

122 °F

BRASS

9

0

e -

1652°F

BRONZE

o

e -

1832°F

COPPER

1083°e - 1981°F

GOLD

1063°e -

1945°F

IRON

14

0

e

2552°F

LEAD

327°e

621°F

SILVER

96

0

e -

176 °F

STAINLESS STEEL

15

0

e

2732°F



pg

16

REMINDERS

13 WAV SWITCHES

White

120v

Light

source

Black

2

3

2

3

4

WAY

Switch I

White

120v

••

-

Light

Source

-

[8]

[8]

[8]

Black

3 way

4 way

3 way



REMINDERS

pg-17

I

START· STOP CONTROL CIRCUIT

L

L2

L3

1

start

I

stop

control wires

power wires

2

3

T1

I

FORWARD· REVERSE CONTROL CIRCUIT

L2

for

stop

rev

R

OL s

. I - - - -Q-L . . J J - - -+ - - - a . .L .Cr - - -o O l l f ~

,

,

/

,

/

,

,

F

,

for

--l. ::.v

F

R



p

18 REMINDERS

ELECTRIC L SYMBOLS

1

I

o o

1

o

0

o 0

start button

stop button

single pole

mushroom head

normally open

normally closed

double break

push button switch

s

O<:J O

limit switch

limit switch

temperature switch

temperature switch

normally open

norm. closed normally open

normally closed

--L

It

?<;

T

contact

contact foot switch

foot switch

normally open

normally closed normally open

normallyclosed

timed contact

thermal

liquid level switch liquid level switch

normally open overload

normally open normally closed

time close

-- --

III

0

0

circuit

bre ker

3 pole

autotransformer

selector switch

with thermal

disconnect

winding

two position

overloads



REMINDERS

pg -19

I

ELECTRICAL SYMBOLS

I

D

pressure switch

flow switch

pilot light

solenoid

normally open

normally closed

-{]=:J)

0

-©

e

bell fuse special purpose duplex outlet

outlet

split circuit

3

-- ,JL-

3C

watthourmeter power panel

fusible element transformer

WIRING

NOT CONNECTED

CONNECTED CONTROL POWER

+ +

HOME RUN UNDERGROUND CONCEALED

NUMBER OF CIRCUIT

CONDUCTORS 4)

. . . . . . . . .

T



pg·2

REMINDERS

I

ELECTRICAL SYMBOLS

I

T

transfonner pad

c>

ceiling outlet

ground

... ....

0 0 -

circuit breaker

=@3

triplex receptacle

outlet

lighting panel

fire alarm

hom

telephone

0

wiring or conduit

upturned

=@GR

grounded duplex

receptacle

bare lamp

fluorescent strip

motor

starter

§WP

1

heating panel

safety switch

weatherproof

outlet

fluorescent

fixture

@R

push button

floor outlet

range receptacle

light wall mounted



A,amp

AWG

B S

c,cyc

C

c.m., emil

cb, CB

cps

EMT

F

fc

t

GFI, GFCI

h

HID

hp

Hz

kcmil

kVA

kW

kWh

m

rnA

MCM,mcm

mega-

micro-

mil

milli

p

p ~

pvc

r/min, rpm

s

UL

V

va

w

n

REMINDERS

pg - 2

IABBREVIATIONS I

amperes

American Wire Gauge

Brown & Sharpe (wire gauge)

cycles (hertz, frequency)

Celsius, centigrade (temperature scale)

circular mils

circuit breaker

cycles per second (hertz)

electrical metallic tubing

Fahrenheit (temperature scale)

footcandles

foot, feet

ground-fault circuit-interrupter

hours

high-intensity discharge (lamps)

horsepower

hertz (cycles per second)

thousand circular mils

kilovolt-amperes

kilowatts

kilowatthours

lumens

milliamperes

thousand circular mils

million

one-millionth

mils (thousandths

of

an inch)

one-thousandth

pole

phase

polyvinyl chloride (non-metallic)

revolutions per minute

seconds

Underwriters Laboratories

volts

volt-amperes

watts

ohm (resistance)



pg ·

REMINDERS

I

WEIGHTS·

MEASUREMENTS I

16 drams =1 ounce

16 ounces = 1 pound

1 ton =2000 pounds

1 U.S. pint =

16

fluid ounces

1 standard cup = 8 fluid ounces

1 tablespoon = 5 fluid ounces

1 teaspoon

=

.16 fluid ounces

2 pints = 1 quart

4 quarts = 1 gallon

1 U.S. gallon

=

.833 Imperial gallon

1 U.S. gallon = 3.785 liters

1 gallon =8.24 pounds

12 inches = 1 foot

3 feet = 1 yard

5.5 yards

=

1 rod

40 rods

= 1 furlong

5280 feet

=

1 mile

43,560 sq.ft. = 1 acre

640 acres = 1 square nllie

1 square mile = 1 section

36 sections

=

1 Township



REMINDERS

pg· 3

THERMOMETER SCALE

Centigrade - Fahrenheit

Centigrade =5/9 F-32)

Fahrenheit

=9/5

C 32

C

F

C

F

C

F

C

F

-35 ·31.0

13

55.4 49 120.2

85

185.0

-30

·22.0

14

57.2

50

122.0

86

186.8

-25

·13.0

15 59.0 51

123.8

87

188.6

-20

-4.0

16

60.8

52

125.6

88

190.4

-19

-2.2

7 62.6

53

127.4 89

192.2

-18 ·.4

18

64.4 54

129.2 90 194.0

-17

1.4

19 66.2 55

131.0 91

195.8

-16

3.2 20 68.0

56

132.8

92

197.6

-15

5.0 21 69.8

57

134.6

93

199.4

-14

6.8

22

71.6 58 136.4

94 201.2

-13

8.6 23

73.4 59 138.2 95

203.0

-12

10.4

24

75.2 60

140.0

96

204.8

-11

12.2

25

77.0 61 141.8

97 206.6

-10

14.0

26

78.8

62

143.6

98

208.4

-9

15.8

27

80.6 63

145.4

99

210.2

-8

17.6

28

82.4 64

147.2 100 212

-7

19.4

29 84.2

65

149.0 105 221

-6

21.2 30

86.0

66

150.8 110

230

-5

23.0

31

87.8

67

152.6

115

239

-4

24.8

32

89.6

68

15'\.4

120

248

-3

26.6 33

91.4

69 15tl.2 130

266

-2

28.4 34

93.2 70 158.0 140 284

-1

30.2

35

95.0

71

159.8 150

302

0 32.0 36

96.8

72

161.6 160 320

1

33.8 37

98.6

73

163.4

170

338

2

35.6

38

100.4 74 165.2 180 356

3

37.4

39

102.2

75

167.0 190 374

4 39.2

40

104.0 76 168.8 200 392

5

41.0 41 105.8 77 170.6 250

482

6 42.8

42 107.6

78

172.4 300 572

7

44.6

43

109.4

79

174.2

350

662

8

46.4

44 111.2 80 176.0 400 752

9

48.2 45 113.0 81

177.8

500 932

10

50.0

46

114.8

82

179.6

600

1112

11

51.8 47 116.0

83

181.4

800

1472

12

53.6 48 118.4

84

183.2

1000

1832



pg

24

REMINDERS

Bolt Threads Drill bit

Decimal

Drill it

Decimal

size per inch for tap Equivalent for clearance

Equivalent

1/8

40 #38 .1015 #29

.1360

1/8

44

37

.1040 #29

.1360

6

32

#36

.1065 #25

.1495

#6

40

#33

.1130

#25

.1495

8

32

#29 .1360

#16 .1770

8

36

#29

.1360 #16

.1770

#10 24 #25

.1495 13/64

.2031

#10

32

#21

.1590

13/64

.2031

#12 24

#16

.1770

7/32

.2188

#12 28 #14

.1820 7/32

.2188

1/4

20 7

.2010

17/64

.2656

1/4

28 #3

.2130 17/64

.2656

5/16

18

F

.2570

21/64

.3281

5/16

24

1

.2720

21/64

.3281

3/8

16

5/16

.3125 25/64

.3906

3/8

24

Q

.3320

25/64

.3906

7/16

14

U .3680

29/64

.4531

7/16

20

25/64

.3906 29/64

.4531

1/2

13

27/64

.4219

33/64

.5156

1/2

20 29/64

.4531

33/64

.5156

9/16 12 31/64

.4844

37/64

.5781

9/16 18

33/64

.5156

37/64

.5781

5/8

11

17/32

.5312

41/64

.6406

5/8

18

37/64

.5781

41/64

.6406

3/4

10

21/32

.6562

49/64

.7656

3/4

16

11/16

.6875

49/64

.7656

7/8

9

49/64

.7656

57/64

.8906

7/8

14

13/16

.8125

57/64

.8906

1

1

8

14

7/8

15/16

.8750

.9375

1 1/64

1 1/64

1.0156

1.0156



REMINDERS

pg -25

HOLE SAW SIZE FOR CONDUITS

~ ] ] ] ] ]

()

RIGID MET L CONDUIT

THIN

W LL

CONDUIT

MET L

RMORED C BLE

Liquidtight

flexible

metal conduit

HOI = S W DRIL.L FOR

TRADE SIZE THIN RIGID ARMORED LIQUID

OF CONDUIT

W LL

METAL CABLE TIGHT

1/2 3/4

7/8 1 11/8

3/4 1

11/8 11/8

11/4

1

11/4 1 3/8 11/2

11/2

11/4

15/8

1 3/4 1 3/4

1 7/8

11/2 1 7/8

2

2

21/8

2

21/8

21/2

21/2 2 3/4



pg - 26

REMINDERS

APPROXIMATE DIMENSIONS OF CONDUIT FIT,.INGS

I ~ B 7 D

) v

A

LOCKNUT

BUSHING

NOMINAL

PIPE

A

B

C 0 A

B

C 0

SIZE

3 II

8

1 "

1

16

5 "

8

1

1 "

8

27 "

32

15 "

32

3 "

4

5 "

16

1 II

2

3 "

132

25 "

32

1 "

1

16

1 "

8

1 "

132

5

8

29 "

32

11 "

32

3 II

'4

11 "

132

1

9 "

132

5 "

-

2

1 "

-

4

3 "

4

5 "

132

7 "

16

1 I

21 "

132

1 "

1-

19 "

132

5 "

32

17 "

132

1 "

132

13 "

132

9 "

16

1 II

-

4

1 "

2

8

19 "

132

1 "

232'

7 "

32

29 "

1

32

5 "

1

16

3 "

-

4

9

16

1 II

-

2

3 "

28

13 "

132

1 "

2

4

3

16

5 "

2

32

17 "

132

1 "

232'

5 "

8

II

2

31 "

232'

5 "

2

16

25 "

232'

7 "

32

25 "

232'

"

2

17 "

232'

19 "

32

1 II

-

17 "

332'

3 "

2

4

5 "

3

16

1 "

4

5 "

3

8

13 "

232'

3

25 "

32

3 II

5 "

4

16

3 "

3

8

1 "

432'

5 "

16

7 "

3

8

1 "

332'

11 "

3

16

27 "

32

1 II

3'2

5

15 "

3

16

5 "

4

8

5 "

16

17 "

432'

15 "

332'

9 "

432'

29 "

32

4

II

3 "

58

7 "

4

16

3 "

5

16

7

16

1 "

58

4

25 "

432'

7 "

8

1

II

4 -

2

3 "

632'

7

4 -

8

25 "

532'

1 "

2

3 "

54

15 "

432'

7 "

5

16

1 "

1 32

II

5

3 "

6

4

15 "

5'32

7 "

6

16

9

16

3 "

6 8

1 "

5'32

1 "

6'32

1 "

132

II

6

19 "

7'32

1 "

6'2

29 "

7'32

5 "

8

13 "

7'32

1 "

6

16

5 "

7 32

1 "

-



REMINDERS

pg 27

lAREA OF SQUARE INCH I

10 000 SQUARES

7 854 SQUARES

The diagram above illustrates why the decimaL7854

is used

to

find the area ofa circle. If the square is divided

into 10,000 small squares, a circle would contain 7,854

small squares.

If

the area of the square was 1 sq.in., then

the area for the circle would be I" x I" x .7854

=

.7854

square inches for the circle.

AREA SQ.IN. = D x .7854

AREA SQ.IN. = RADIUS2

x

3.1416

AREA

SQ.IN. = CIRCUMFERENCE

x

.07958

Example: What is the area

of

square inch of a #4

conductor that has a diameter of 0.232"?

.232 x .232 x .7854

=

.042 area sq.in.

Example: What is the area of square inch ofa #2

THHN

conductor that has an approximate diameter of .388"?

.388 x .388 x .7854

=

.1182 area sq.in.



p - 28

REMIN ERS

I CIRCLE I

CiRCUMFERENCE:

100

cross sectional area

Circumference

=

The distance around the circle.

Diameter

=

The distance across a circle through the center.

Radius

=

The distance from the center to the edge

o

a circle.

Cross sectional area

=

100

o

the circle.

The ratio

o

the circumference to the diameter

o

a circle is called

Pi t

=

3.1416. An approximate fraction is 22/7.

Circumference

o

a circle

=

Diameter x 3.1416

Radius x 6.283185

Example: What is the circumference

o

a

3

circle?

3

x 3.1416

=9.4248

f

the diameter is doubled, the

cross-sectional area is increased

four times and the resistance is

reduced 1/4 o its original value.

CSA 4 times larger



REMINDERS

pg 29

1 mil

=

.001 inch

1 inch

=

1000 mils

mils

= inches x 1000

inches

=

mils x .001

square inches

=

square mils x .000001

1 circular mil

=

0.7854 square mils

1 square mil =

1.2732 circular mils

circular mils =square mils x 1.2732

square mils

=

circular mils x 0.7854

1 circular mil

=

.7854 square mils

You can remember this by reading

your calculator starting at the top

left with "7", the number to the

right

of

7 is "8" and below 8 is

5

and to the left of 5 is "4".

y

read

ing this box in a clockwise direc

785

tion you will remember

7854

Bus Bar Square inch area =Width x Thickness

Ampacity = 1 amps per square inch for copper

7 amps per square inch for aluminum

What is the ampacity of a 2" round copper bus bar?

2

2

0

2

0

N.B.C. 366.23(A) - 1000 amps per square inch for copper bus bar.

2" x

2

= 4 square inches. 4 x lOooa =4000 amps for a

2

square bus bar.

For a

round

bus bar: 4000 amps x

7854 =

3141.6 amps.



pg ·30

REMIN ERS

I OHMS

LAW I

I

TO FIND AMPERES

I

Example: What is the current in amperes flowing in a circuit

that has a voltage

of

120 and a resistance

of 1

ohms?

Solution:

1=

E/R =

120v l0

ohms =

2

amperes


that has a 1440 watt load and a voltage of 120?

Solution:

1=

WfE = 1440w/120v = 2 amperes


that has a 1440 watt load and a resistance

of

10 ohms?

Solution:

1=

W/R =

1440w l0

ohms = 144

J

144

=

2

amperes



REMIN ERS

pg -

31

I

TO

FIND RESISTANCE

I

Example: What is the resistance in ohms for a circuit that has

a load of 1440 watts and a current in amperes of 12?

Solution: R = W P= 1440w/144a l2ax12a) = 10 ohms.

Example: What is the resistance in ohms for a circuit that has

a voltage

of

120 and a load

of

1440 watts?

Solution: R =

E2 W

= 120v x 120v = 14400/1440w = 10

ohms.

Example: What

is

the resistance in ohms for a circuit that has

a voltage

of

120 and a current

of

12 amps?

Solution: R = Ell = 120v/12a = 10 ohms.



pg

32

REMINDERS

ITO

FIND

VOLT GE

I

Example: What is the voltage

of

a circuit that has a load

of

1440 watts and a resistance of 1 ohms?

Solution: ~ W x R = 1440wx 1 ohms= 14400 4 4

=120 volts.

Example: What is the voltage of a circuit that has a load

of

1440 watts with 12 amps of current?

Solution: E =W = 1440w/12a = 120 volts.

Example: What is the voltage of

a circuit that has 12 amps

flowing with a resistance of 1 ohms?

Solution: E = I x R = l2a x

1

ohms = 120 volts.



REMIN ERS

pg 33

I

T FIND W TTS

I

Example: What is the wattage

of

a circuit that has a voltage

of

120 with

12

amps

of

current flowing?

Solution: W = E x 1= 120v x 12a =

144 watts

Example: What is the wattage of a circuit that has a current

flowing

of 12

amps and a resistance

of 1

ohms?

Solution: W =FR= 12ax 12a= 144 x 1 ohms =

144 watts

Example: What

is

the wattage of a circuit that has a voltage

of

120 and a resistance

of

1 ohms?

Solution: W = E

2

/R = 120v x 120v = 14400/10 ohms =

144

watts



pg 34

REMINDERS

IIELI TH ICE M N

II

The current in a capacitive circuit goes through its peak value

before

the applied voltage goes through its peak value.

Current in a capacitive circuit leads the applied voltage in time. In

an inductive circuit current lags the voltage.

This can best be remembered by the statement

ELI

THE

ICE

MAN .

ELI E =voltage L =inductance I =current

When you see the letter L you can say E leads I in an inductive

circuit.

ICE I

=current

C

=capacitance

E

=voltage

When you see the letter

e

you can say I leads E in a capacitive

circuit.

In an AC circuit, if we increase the

capacitance

a lamp will glow

brighter. When we increase the inductance the lamp glows dimmer.

ITHE TWO FORMS

of

REACTANCE I

INDUCTIVE (a coil) CAPACITIVE (a capacitor)

The letter X represents reactance. Reactance is measured in

ohms.

XL inductive reactance XC capacitive reactance

When induction and capacitance are of equal values in a circuit, they

neutralize each other, allowing only resistance to oppose the flow of

current. This is called resonance.



REMINDERS

pg

35

DC Peak Maximum

AC -

NVVl

RMS Effective Value .707

METER READS

RMS EFFECTIVE VALUE

RMS = Maximum x 707

MAXIMUM = Rms 1 707

Example: The meter reads 240v, what is

the maximum voltage?

240v/ 707

=339v.

Example: The maximum current is 50

amperes, what is the ammeter reading?

50 amps x .707 = 35 amps.

SIX

W YS

OF

ELECTRON

MOVEMENT



p

6

REMIN ERS

I

UNIVERSAL MOTOR

I

Universal motors will operate on either AC or DC at 60 Hertz.

Universal motors are designed for special applications such as

electrical hand tools, vacuum cleaners and many other household appli

ances that require an electric motor.

The efficiency o a universal motor is low, usually around 30

for the smaller motors.

The universal motor can be reversed. The reversible motor will

contain 4 or 5 leads. The 4-lead motor can be reversed by using a single

pole, double-throw switch. The 5-lead motor requires a double-pole,

double-throw switch. This switch will reverse the direction o current

through either the armature or the series field, but not both.

I SHADED POLE MOTOR I

The shaded-pole is a single-phase AC motor from

111

hp to

112 hp in size. They have rotors o the squirrel-cagetype. They are a low

cost motor used for clocks, fans, etc.

Maximum efficiency is only 35 .

Some shaded-pole motors can be reversed with switching.

Some require removing the stator and turning it end to end.



REMINDERS

pg· 7

I

SPLIT-PHASE MOTOR I

STANDARD ROTATION

FACING SHAFT END

Split-phase motors 112 - 3/4 hp are used on

appliances, furnaces, small pumps, etc.

The split-phase motor and the capacitor

start motor can be reversed by interchanging

the start leads.

I f

the start leads are not avail

able, the run leads can be interchanged to

reverse the motor.

The start winding is the red and black wires. One motor direction

would have red connected to 1and black connected to 2. To reverse

therotation, the startwindings are reversed by connecting red to 2 and

black to 1.

red

L1

lC:J.=: O

2

L

r \

black

red

L1

L2

2

bl ck

. . . . s ~ o



L C ~ = :

p 38

REMIN ERS

I

CAPACITOR-START MOTOR single voltage) I

Capacitor start motors are used for heavier loads up

t

35 hp.

They can be reversed by interchanging the start leads. the start

leads are not available the run leads can be interchanged

t

reverse the motor.

capacitor

L2 I

L

1 . f . I

L2



REMINDERS

pg -

39

ICAPACITOR-START OTOR dual voltage 2401120)

I

capacitor

STANDARD ROTATION

FACING SHAFT END

ST ND RD

ROT TION

F CING

SH FT

END

STANDARD ROTATION

FACING SHAFT END

capacitor

. - E ~ -

STANDARD ROTATION

FACING

SH FT

END

L

24 v

L2 J



pg - 4

REMIN ERS

I REPULSION MOTOR wound-rotor) I

The repulsion motor wound-rotor) has brushes and a varying-speed

characteristic. This motor is reversed by shifting the brushes

5

degrees.

RM TURE

COMMUT TOR

The repulsion-type motor is one o the oldest forms

o

single-phase

induction motors. Since 1950 this motor has been largely replaced by the

split-phase and capacitor-start induction motors.

ID.C. MOTOR REVERSING I

SERIES or SHUNT - Interchange the connectionso either the field or the

armature winding. Reversing the line leads will not change the direction

o rotation.

COMPOUND - Interchange the connections o the two armature leads.



REMIN ERS

pg·4

ITHREE PHASE OTOR I

o

reverse the direction o rotation with a three-phase motor inter

change any two motor leads or line wires.

L1

L2

L3

L1

l2

L3

3 va =

E

x

I

x

1.732



pg 42

REMIN ERS

I

SYN HRONOUS MOTOR

I

Synchronous motors vary in size to thousands o horsepower.

t

is an AC motor in which the rotor revolves in step or in

synchronism with the rotating magnetic field produced by the stator

winding. This means the magnetic field and the rotor turn at the same

speed.

With the induction motor, the rotor turns at a lower speed than

the revolving field. This is necessary in order that the squirrel-cage

winding be cut by the revolving field and thereby have a current

induced in it.

Slip is defined as the difference in speed between the rotor s

actual rpm and thato the magnetic field. Asynchronousmotorhas zero

slip.

Synchronous motors are often used

to

improve the power factor

o an electrical system. Whenused for power factor correction, the field

windings are overexcited and cause the motor to draw a large leading

current.

SYNCHRONOUS RPM =

HERTZ 120

POLES

TORQUE (LB-FT) =

Horsepower x 5250

RPM

HORSEPOWER =

Torgue

QQ:f )

x RPM

5250



REMINDERS

pg·43

I ROTOR· STATOR I

There are two types of motor rotors, the squirrel cage and the wound

rotor. The squirrel cage has bars of copper or aluminum the length of the rotor

electrically connected at each end with shorting rings. The wound rotor has

coils of wires wound in the slots of the rotor.

BR SH I

WOUND

~

SQUIRREL CAGE=:]

ARMATURE

~

COMMUTATOR

There must be a

motion

between the armature windings and the field

windings. AC generators are built in two major assemblies, the stator and the

rotor

There are two types of motion, either the revolving armature rotor) or

the revolving field stator).

Inthe revolvingarmatureAC generator, the statorprovides astationary

electromagnetic field. The rotor acting as the armature, revolves in the field,

cutting the lines

of

force, producing the desired voltage. In this generator, the

armature output is taken through slip rings and thus retains its AC characteristic.

ARMATURE

FIELD

OUTPUW

OLT GE

.

SLIPRINGS

or

BRUSHES

eO®o

The revolving field AC generator is by far the most widely used today.

In this type

of

generator, direct current from a separate source excitation) is

passed through windings on the rotor by means

of

sliprings and brushes. This

maintains a rotating electromagnetic field of fixed polarity. The rotating

magnetic field cuts through the armature windings imbedded in the surrounding

stator. As the rotor turns, AC voltages are induced in the windings since

magnetic fields

of

first one polarity and then another cut through them. Now

here is the important part; since the output power is taken from stationary

windings, the output may be connected through fixed terminals and not

revolving sliprings or brushes that would limit high voltages.

QRM TUREST TOR

FI OlO

E x c r r T o N ~ REVOLVES

FIELD ROTOR

10 Ihe

fields

_ •

SLIPRINGS or

BRUSHES

o

o

ARMATURE WINDINGS

STATIONARY



pg

44

REMINDERS

WVE MOTOR

CONNECTIONS

3

2

L 3- - - . . . r

~ L

HIGH VOLTAGE

WVE MOTOR

CONNECTIONS

r L

LOWVOLT GE

L ~

~ L 2

3

9 2

8

L

L3

7



REMINDERS

pg 45

L1

DELTA MOTOR

CONNECTIONS

HIGH VOLTAGE

L2

L3

L2 2

8

L3

7 5

4

L1 3

6

L1

DELTA MOTOR

CONNECTIONS

OW

VOLTAGE

7

9

8

6

L1

L3



pg 46

REMINDERS

I

HORSEPOWER· TONS

I

THE MECHANICAL EQUIVALENT

OF

HEAT

1 H.P.

=

33,000 fUb. per minute

=

42.416

B T V

per minute

one

B T V

=778 ft.Ib.

one ft.Ib. =1/778 =.001285 B T V

Multiply the number

of

tons

of

refrigeration required by:

1.25 H P per

ton

for 1/2 to 5 ton capacity

1 1 H P

per ton

for

5 to

50

ton capacity

1.0

H P

per

ton

for

capaci ty above 50

tons

MOTOR

OverlO D device

=Heater

d t1

OverCURRENT

device

=Fuse or

Circuit Breaker - = = ~ - -

ONIJO;;jd



REMIN ERS

pg· 7

I

INTERPOLATE MOTOR CURRENTS

I

3H.P. 4H.P. 5

H.P.

17 amps

? 28

amps

The difference in

H.P.

between 3

H.P.

and 5

H.P. =

2

H.P.

The difference in amps between 17 amps and 28 amps

= 11.

Divide 11 amps by 2

=

5.5 amps.

3 H.P. = 17 amps 5.5 amps =22.5

amps for

4 H.P.

motor.

ACTUAL CONDUCTOR DIAMETER

8 14 12 10 8 6 4 2 1/0



o

o

ONE CYCLE OF AC

pg - 48 REMINDERS

IELECTRICAL DEGREES I

MECHANICAL DEGREESI

soUTH

POLE

- - ONE Y LE

360 degrees

90'

One electrical degree is equal to 1 pair of poles) part

of

a mechanical degree.

It

becomes much clearer

looking at the sketches shown below.

ELECTRICAL DEGREES

MECHANICAL DEGREES 360

2 POLES = 360 degrees

North pole

South pole

360 ELECTRICAL DEGREES

6 pole GENERATOR

MECHANICAL DEGREES 120

rth

pole

:j

Je

o

• ~

3 6 ~

V[ iJ

.

pole

South

pole



REMINDERS

pg 49

INEUTR L B L NCING I

A neutral conductor

carries the unbalanced

current; you must have

a 3-wire circuit to have

a neutral conductor.

N

L

L2

With both 20 amp loads turned on the neutral carries zero.

N

L

L2

With both loads turned on the neutral carries the unbalance 10.

SHUT

OFF

N

L

L2

The maximum neutral current would be 20 when l is shut off.



pg-50

REMINDERS

I

NEUTRALCURRENT

IN

A

WYE I

L

L2

30a

40a

N

C

50a

L3

Thethreeloads30amp,40amp,and50ampareconnectedline

to

neutral,

withallloads"on". Whatistheunbalancedcurrentflowingintheneutral?

Solution: In

=

I2A + I2B + I2C - (IA IB) - (lB IC) - (lC IA)

Theformulaatfirstlooksverydifficult,butreallyit'snot.

Everythingunderthesquarerootsignmustbedonefirst,whichmeans:

currentinAsquared =30 x 30 = 900

+

currentinBsquared = 40 x 40 = 1600

+

currentinCsquared

=

50 x 50

=

2500

5000total (callthistotal"X")

Nowtherightsideof theformulashows:

currentinA x currentinB = 30 x 40 = 1200

currentinB x currentinC = 40 x 50 = 2000

currentinC x currentinA = 50 x 30 = 1500

4700 (callthistotal"Y")

Nowsubtracttotal Y fromtotal X

=

5000totalX

- 4700totalY

300

Nowextractthesquarerootbypressingthe j buttononyourcalculator.

Theanswer17.32ampsistheunbalancedcurrentflowingintheneutral.

17.320508

is

thesquarerootof 300. 17.320508 x 17.320508

=

300.



REMINDERS

p

5

VARS·

KVARS

Capacitors used to correct the power factor

of

a

circuit are rated in kilovars. (one kilovar equals 1000

vars. The abbreviation for kilovar

is

kvar.)

Example: A single-phase 230 volt 1 hp motor has a

current flow

of

8 amps. A wattmeter reads a true power

of

746 watts. How many inductive vars are required to

raise the power factor to 95 ?

st

step

find PF:;: w/va watts:;: 746 true power

va ; E x I 230v x 8a:;: 1840va

PF =746w11840va:;: 40.5 power factor

2nd

step the inductive vars in the circuit are caluclated

using this fomula:

vars:;: va

2

-

w

2

1840va x 1840va:;: 3,385,600va

746w x 746w 556,516w now subtract the watts from the

volt amps 3,385,600va - 556,516w:;: 2,829,084 now

press the square root buton on your calculator:;: 1682 vars.

3rd

step find the required va to produce a 95 power factor.

va required:;: w/pf:;: 746w/.95 ; 785.2 va required.

4th step to find the required inductive vars to produce the va

required:

vars = -.l va

2

- w

2

785.2va x 785.2va:;: 616,539.04va

746w x 746w

;

556,516w now subtract the watts from the volt

amps:;: 616,539va - 556,516w:;: 60,023 now press the square

root button on your calculator 245 vars.

5th

step

the present inductive vars are 1682,

to

find the needed

vars to produce a reactive power of 245 vars subtract the amount

of

needed vars from the present amount

of

vars ; 1682 vars - 245

vars:;:

437 vars

or

1.437 kvars needed to raise to

95 pow r

factor



pg·5

REMINDERS

Example: A 95,000 watt load has a power factor of 70%.

What is the corrective kvars required

t

raise the power factor

to 90%?

st step

find va: va = w/pf= 95,000w/.70pf = 135,714va

2nd

step

the inductive vars in the circuit are caluc1ated

using this fomula:

vars ='-Jva

2

- w

2

= 135,714va x 135,714va = 18,418,289,796va

95,000w x 95,000w = 9,025,000,000w now subtract the watts

from the volt amps = 18,418,289,796va - 9,025,000,000w

= 9,393,289,796 now press the square root buton on your

c l c u l t o r ~

96,918.9 vars

3rd step find the required va to produce a 90% power factor.

va required = w/pf 95,000w/.90 = 105,555.5 va

4th step to find the required inductive vars t produce the

2

va required: vars

~ v a

- w = 105,555va x 105,555va =

11,141,858,025va

95,000w x 95,000w =9,025,000,000w now subtract the

watts from the volt amps = 11,141,858,025va

9,025,000,000w = 2,116,858,025 now press the square root

button on your calculator = 46,009 vars or

46 kvars.

5th

step

the present inductive vars are 96,919 , to find the

needed vars to produce a reactive power of 46,009 vars sub

tract the amount of needed vars from the present amount of

vars = 96,919 vars - 46,009 vars = 50,910 vars

r

50.91 kvars

needed to raise to 90% powerfactor.

2

DIGITS

-Note: I used a 2 digit calculator to solve this

example. A standard calculator that has 8 digits

ml ll U

won't work.

IIEIIIIIII

l l l li lm

DIIElIII

iii

EI



REMIN ERS

pg·53

ITRANSFORMER CONNECTIONS I

A single phase transformer has one primary winding and one

secondary winding. This is the simplest

t

connect. The high-voltage

leads are marked H and the low voltage X .

HIGH VOLTAGE

When facing the transformer low-voltage

terminals

HI

is t the left and H2 to the

right. X1is on the right, andX2 on the left.

LOW

VOLTAGE

By

reversing

the 240 volts the transformer can be stepped up to 480 volts.

ISTEP DOWN

I

ISTEP UP I

. . . .- 2 4 0 v

480 turns

2/1 ratio

ratio

O

~

40 turns

240

turns

1 '240V

1

+ 1 2 0 v

For a single low voltage

of

120, the secondary windings are connected

in

parallel. A is connected to C and B is connected t D . A-C

connects to X2 and B-D t

Xl.

Connection for

high-voltage

is made by

connecting the windings in

series.

Voltages add in series.



pg

·54

REMINDERS

DUAL

VOLTAGE

LOW

SECONDARY

CONNECTION

DUAL VOLTAGE

HIGH SECONDARY

CONNECTION

X3

X2 X1

+--240v--

X2 X1

12 v

Transfonners have a relative polarity which must be considered

when

connecting them. In a single phase transfonner the voltage relation

ships are either "in phase"

or

180

0

0ut

of

phase. Therefore, when the

windings are connected together, the output voltage can either be

additive

or

subtractive.

Batteries have polarities,

if

you connected one flashlight battery in

reverse their voltages would cancel and the flashlight would not

light.

ADDITIVE

VOLTAGE

SUBTRACTIVE

VOLTAGE

SERIES CONNECTED

SERIES

CONNECTED

OPPOSITE POLARITIES

SAME

POLARITIES

3 volts .1

Although the voltages

of

a transfonner are constantly changing, their

relative phase angle to each other is constant. -Transfonners under

200 kva with voltages ratings below 9000 volts will be additive.

1 5

volts

1 5

volts

1 5

volts

H

X1

H

1

X2

1

H1

X2

H1

X1

I

ADDITIVE

I

I

SUBTRACTIVE

I



REMINDERS

pg - 55

Most often transformers are manufactured with dual windings.

Three transformers with dual windings· Each primary is rated 1200v, each secondary 120v

Hl H2 H3

H4

I JJJJJ r

IJJJJJJ..r

l200v

l200v

l20v

l20v

Jfi 'fl ffL Jfi 'fl ffL

X4 X3 X2

Xl

H1 H2

w.JJJJ.r

l200v

H3

H4

wJJ.JJ.r

l200v

120v

J r r rm L

X4

X3

l20v

JrmYrL

X2 Xl

Hl H2

--u.w..ur

l200v

H3 H4

--u.u.wr

l200v

l20v

J r r rm L

X4 X3

l20v

Jf'fft"rrL

X2 Xl

These transformers can be connected either wye for high voltage and

low current, or delta for low voltage and high current.

Delta-Delta. With a 2400v source to the transformers, this is twice the

1200v rating

of

the primary windings. The 1200v primary windings

must be connected in series for a 2400v source.

12400Y 3

B

SOURCE I

~

m JtZ

;; ;n; ;;

oJ

o

l200v l200v l200v

l200v l200v

l200v

120v

l20v

l20v

l20v

l20v

l20v

,....J

I1I111 1111111-

II II

111111l. . .

Xl

X4

X3

X2

Xl

I Il ...

X4 X3 X2

Xl

X4 X3

X2

A

B

C

N

1C 2-4o-'-1-2o-Y-o-e-l-ta-H-jg-h-.-L-e-g-se-c-o-n-d-a-ry 'l

To provide the 240 volts, the 120v secondaries must be connected in

series. Only one transformer in a Delta has a neutral connection which

is grounded. From

An

or "C" phase to neutral would provide 120 volts.

From "B" phase (high-leg) to neutral 208 volts.



pg - 56

REMINDERS

I

AUTOTRANSFORMER

I

Where the desired voltage ratio is less than two to one, an autotrans

former is often used.

An

autotransformer

makes common use

of

a part

of

a

single

winding

for both the primary and secondary. The secondary load is simply

transferred rather than transformed.

IVOLTAGE STEPPED

Upi

IVOLTAGE

STEPPED

DOWN I

Primary

t Secondary

~

LOAD

OAD

Primary

A common application of an autotransformer is a "buck" or "boost"

transformer. Because transformers have a relative polarity, they can be

wired so the secondary voltage either adds or subtracts. When the

voltage adds, it is a "boost" transformer. When it subtracts it is called a

"buck".

Shown below is the connection

of

a boost transformer with dual

primary and dual-secondary windings with a

1011

ratio raising the 208v

to 228.8 volts.

t

H1

H2

H3 H4

208v

source

•

28 8v output

X4 X3 X2

X1



REMINDERS

pg

57

ICURRENT

TRANSFORMER

CT) Doughnut) I

Ammeter

A current transfonner is used when AC current is so large that

connecting measuring instruments such as a kwh meter would be

impracticable. The

CT

provides a means

of

reproducing the effect

of

the primary current on a

reduced

scale suited to the kwh meter.

To standardizecurrentdevices the second ry

of

acurrenttransfonner

is always r ted at5 amperes no matterwhat the ampere rating of the

pnmary

IS.

The current

r ting of

the primary is determined by the maximum

load current to be measured.

Example: Maximum load current to be measured is 500 amperes.

The secondary winding will have a r ting

of

5 amperes. The ratio

between the primary winding and the secondary winding is 5/500

=

1 to 100.

Thus, the secondary winding will have 100 times as many turns as

the primary.

Using this I

00 ratio, a current transfonner for a load

of

400 amps,

the secondary would read 4 amps. For a load current

of

300 amps the

secondary would read 3 amps.

With the doughnut type CT), the conductor passing through the

center magnetizes the core because

of

magnetic lines around the

single one tum) conductor. The secondary has many turns

of

small

wire connected directly to a low-scale ammeter.



pg 58

REMINDERS

IVOLTAGE POTENTIAL TRANSFORMER I

Usually circuits up to 600 volts can be measured directly with meters.

However, higher voltages cause the meters to become very expensive in cost.

High

~

Load

Potential

[

r

~ n s o r m r

I

>

I

Volt Meter

The potential transformer is used in metering of higher voltages. The

primary winding is connected to the high-voltage and the secondary low

voltage winding usually wound for 120 volts.

The capacity of a potential transformer is relatively small as compared to

a pow r transformer.

Potential transformers have ratings

of

100 to 500 va.

PARALLEL WIRING TRANSFORMERS

24 V 2400v

2400v

2400v

120v 120v 120v 120v

4 v

To drop 2400 volts across each primary winding from the 4800 volts

impressed by the source, each pair of

windings will need to be connected

in series. H2 is connected

of

one transformer is connected to H3

of

the

second transformer, and H4 on the first transformer is connected to HI of

the second transformer. Next the two transformers are connected in

parallel

and attached to the source of 4800v.

To obtain 240 volts on the secondary, X2 and X3 of each transformer is

connected in series. Next these two sets

of

windings are connected in

parallel and attached to the 240v load.



REMIN ERS

pg -

59

ITHEOPENDELT

I

t s possible

to

achieve three-phase by using only two transfonners.

This connection is called the open delta or V connection.

Although the open-delta is generally used only as an emergency or

temporary system, an original transfonner installation may consist of an

open-delta bank to supply a three-phase load which is presently light but

s expected to increase in the future. This keeps the initial cost low by

using only two transfonners, a third transfonner can be added to the

system later when the demand requires it. When the third transfonner is

added, a delta-delta closed bank

s

fonned.

A three-phase transfonner with an assembly

of

three separate single

phase transfonners in one tank is lower in initial cost, costs less to install,

and requires less space than three separate single-phase transformers.

A three-phase transformer has one disadvantage

if

one of the phase

windings becomes defective, the entire three-phase bank must be

disconnected and removed from service. A defective single-phase

transfonner in a three-phase bank can be disconnected and removed for

repair. Partial service can be restored using the remaining single phase

transfonners open-delta until a replacement transfonner is obtained.

With two transfonners three-phase s still obtained, but at reduced power;

57.7 of original power.

This makes it a very practical transfonner application for temporary or

emergency conditions.

OPEN-DELTA 57.7%



pg 60

REMINDERS

IDELTA

HIGH LEG

VOLTAGE I

Always take the lowest voltage x 1.732.

230/115v

=

115v x 1.732

=

199v High leg voltage.

240/120v =120v x 1.732 =208v High leg voltage.

HIGH LEG

230/115v

240/120v

r

r

08v

199v

HIGH LEG

I

THREE PHASE LOAD BALANCING

I

n

o 01:1

1 3

jIj

240v 6000va motor

1 1

jIj

240v 3000w water heater

3 1

jIj

120v 80w lights

Phase

A

=

5000

Phase

B

=

2000

Phase

C

=

2240

Total =9240

1 3

jIj

208v 6000va motor

1 1

jIj

208v 3000w water heater

3

Ij1j

120v 80w lights

Phase A =3580

Phase B = 3580

Phase C =2080

Total =9240



REMIN ERS

pg - 6

I CLASSIFICATION

OF

VOLTAGES I

/

M IN SWITCH

N.E.C. VOLTAGES

600v LOW VOLTAGE

Over 600v HIGH VOLTAGE

UTILITY VOLTAGES

1000v or less LOW orUTILIZATION VOLTAGE

over 1000 to 35,000 MEDIUM or DISTRIBUTION

over 35,000 to 300,000 HIGH VOLTAGE transmission)

over 300,000 to 1 million

EXTRA-HIGH VOLTAGE EHV)

over 1 million

ULTRA-HIGH VOLTAGE UHV)

POW R

CARRYING CAPABILITY

138,000v

80,000 kw

345,000v

500,000 kw

765,000v

2,500,000 kw



pg 62

REMIN ERS

IRE DING THE KWH METER

The reading

of

the meter dials is from left to right.

The pointer must have passed the number to count it.

The meter is read at

the beginning ofthe

month and reads

3287.

The meter is read

at the end of the

month and reads

7722.

The KWH \\ould be 7722 - 3287 4435. At .08 cents per KWH

the bill would amount to 4435 KWH x .08 $354.80.



REMIN ERS

pg

- 63

-...r--'---<

ENERGYUSEO APPLIANCES

·Costs are approximates and based on ¢ per kwh

Cooking

Blender

Bottle warmer*

Coffee maker

brew cycle

warm cycle*

Deep fat fryer*

Food mixer

Food processor

Frying pan*

Ice cream freezer

Oven*

self-clean cycle

Microwave oven

2.75¢ per hour

2.75¢ per hour

¢ per cycle

.5¢ per hour

5.5¢ per hour

¢ per hour

3¢ per hour

6.5¢ per hour

1¢ per hour

20.5¢ per hour

24.5¢ per cleaning

(Microwave cooking takes 1/3

to 1/2

the energy for most foods)

Range-top burner

Slow cooker*

Toaster

Toaster oven (baking)

Waffle iron*

Warmer tray*

Refrigeration

Freezer

16 cu.ft. manual def.

6 cu.ft. auto def.

Refrigerator/freezer

8

cu.ft. auto def.

24 cu.ft. auto def.

(continued)

7.5¢ per hour

1.75¢ per hour

7.5¢ per hour

3.25¢ per hour

7.5¢ per hour

.5¢ per hour

24¢ per day

30¢ per day

45¢ per day

62¢ per day



pg

6

REMIN ERS

Laundry

Clothes dryer 20.5¢ per load

Clothes washer (auto)t

hot wash/hot rinse 43¢ per load

warm wash/cold rinse

1 ¢

per load

cold wash/cold rinse

3¢ per load

Dishwashert

23.5¢ per load

Iron*

5.5¢ per hour

Vacuum cleaner

5.5¢ per hour

Central

Heat set at 70°)*

Electric resistance

800 sq.ft.

1.27 per day

1200 sq.ft.

2.50 per day

1600 sq.ft. 3.32 per day

2000 sq.ft.

4.16 per day

Electric heat pump

800 sq.ft.

64¢ per day

1200 sq.ft.

1.25 per day

1600 sq.ft.

1.67 per day

2000 sq.ft.

2.08 per day

Portable electric heater*

11¢ p r hour

--- /

Air Conditioning set at 78°)*

Central air

800 sq.ft.

1.50 per day

1200 sq.ft.

2.24 per day

1600 sq.ft.

3.01 per day

2000 sq.ft.

3.75 per day

Room air unit

1 ton, 12,000 BTU

1.17 per day

2 ton, 24,000 BTU

2.33 per day

Fans

Attic

2.5¢ per hour

Ceiling or portable (each)

.5¢ per hour

continued)



REMINDERS

p - 65

Lighting

25 watt 2¢ per 10 hours



(fluorescent lamps produce the same light output as incan

descent for 1/3 to 1/2 the cost to operate)

Small ppliances

Bottled water cooler*

Clock

Dehumidifier

Electric blanket*

Circular saw

Electric drill

Floor polisher

Hair dryer

Heating pad*

Pool or lawn pump

(112 hp)

Sewing machine

Waste disposer

Waterbed heater*

Entertainment

Home computer

Portable spa (160 gal.)

warm-up (6 hours)

in use*

Radio

Cassette or CD player

TV

(black white)

(color)

VCR**

6.75¢ per day

11.25¢ per month

¢

per hour

.5¢ per hour

2.5¢ per hour

2.75¢ per hour

2.5¢ per hour

7.5¢ per hour

.3¢ per hour

5¢ per hour

.5¢ per hour

.5¢ per hour

22.5¢ per day

.5¢ per hour

73¢ total

23.5¢ per

.5 hour

.5¢ per hour

1¢ per hour

1¢ per hour

2¢ per hour

2.75¢ per 10 hours

*Does not operate continuously; cycles on and off as

called for by thermostat.

**Does not include cost of operating TV.

tIncludes water heating costs.



p REMIN ERS

IGNITION TEMPERATURES

Defintion o ignition temperature: The temperature o a

substance that is the minimum temperature needed to

cause self-sustaining combustion

o

the substance.

Material

Fuel oil 1

Natural gas

Gasoline

Lacquer thinner

Motor oil

Propane

Pine wood shavings)

Newspaper

Cotton sheeting

Nylon

Black synthetic rubber

Matches

Paint film, varnish

Ignition Temperature

Degrees Fahrenheit

444

900

536

450

750

871

442

446

464

887

590

325

864



REMINDERS

pg 7

CONDUCTIVITY OF METALS

Silver

Copper

Gold

Aluminum

Zinc

Platinum

100%

98

78

61

30%

17

Iron

Lead

Tin

Nickel

Mercury

FAIR

CONDUCTORS

Moist

earth

Acid solutions

Sea

water

Carbon

16

15

9

7

1

Metallic ores

Charcoal

PARTIAL

CONDUCTORS

Water

Marble

High vacuum

Sealing

wax

Porcelain

The body

Cotton Pine

Flame

Linen

Teak

INSULATORS

Dry air

Silk

Slate

Glass

Dry paper

Rubber

Mica

Oils

tom henry reminders for the electrician 2008(nfpa 70,nec) 73p

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