Tom Henry Reminders for the Electrician 2008(Nfpa 70,Nec) 73p
Transcript of 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
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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
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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
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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.
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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
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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
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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.
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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.
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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 "
-
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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.
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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
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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.
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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
Example: What is the current in amperes flowing in a circuit
that has a 1440 watt load and a voltage of 120?
Solution:
1=
WfE = 1440w/120v = 2 amperes
Example: What is the current in amperes flowing in a circuit
that has a 1440 watt load and a resistance
of
10 ohms?
Solution:
1=
W/R =
1440w l0
ohms = 144
J
144
=
2
amperes
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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.
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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.
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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
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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.
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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
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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.
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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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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.
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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
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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
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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.
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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
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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.
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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
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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.
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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.
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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%
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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
7/26/2019 Tom Henry Reminders for the Electrician 2008(Nfpa 70,Nec) 73p
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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
7/26/2019 Tom Henry Reminders for the Electrician 2008(Nfpa 70,Nec) 73p
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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.
7/26/2019 Tom Henry Reminders for the Electrician 2008(Nfpa 70,Nec) 73p
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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
7/26/2019 Tom Henry Reminders for the Electrician 2008(Nfpa 70,Nec) 73p
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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)
7/26/2019 Tom Henry Reminders for the Electrician 2008(Nfpa 70,Nec) 73p
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REMINDERS
p - 65
Lighting
25 watt 2¢ per 10 hours
60 watt 5¢ per 10 hours
100 watt 8¢ 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.
7/26/2019 Tom Henry Reminders for the Electrician 2008(Nfpa 70,Nec) 73p
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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
7/26/2019 Tom Henry Reminders for the Electrician 2008(Nfpa 70,Nec) 73p
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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
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