ELECTRICAL-REFERENCES

--&+- ! .' -

REVISED 1990 EDITION

BY GEORGE V. HART

A note from the author . . . UGLY'S ELECTRICAL REFERENCES is designed to be used as a quick on-the-job reference in the electrical industry. We have tried to include the most commonly - - required information in an easy-to-read format.

Ugly's Electrical Reference is not intended to be a substitute for the National Electrical Code@.

We salute the National Fire Protection Association for I

their dedication to protecting lives and property from fire and electrical hazards through sponsorship of the National Electrical Code.

I

NATIONAL ELECTRICAL CODE' AND NEC? ARE REGISTERED TRADEMARKS OF THE NATIONAL FIRE PROTECTION ASSOCIATION, INC., O U N C Y . M A

While the au thor a n d publisher of UGLY'S I

E L E C T R I C A L R E F E R E N C E S have made efforts t o insure tha t all information in this book is clear and accurate, neither au thor nor publisher shall be held responsible for any inadvertent errors in = content; no r shall they be responsible for the interpretation o r application of material in this book. I

I

ISBN 0-9623229-1-1 ,

- UGLY'S

ELECTRICAL REFERENCES

COPYRIGHT, 1978 BY GEORGE V. HART (AUTHOR)

REVISED 1990

PRINTED IN U.S.A.

THIS BOOK MAY NOT BE REPRODUCED IN ANY FORM WITHOUT WRITTEN PERMISSION OF THE COPYRIGHT HOLDERS

GEORGE V. HART AND

SAMMIE HART

united printing arts 3509 Oak Forest Drive . Houston, Texas 77018. (713) 6884115

TABLE OF CONTENTS

TITLE - OHM'S LAW SERIES CIRCUITS PARALLEL CIRCUITS COMBINATION CIRCUITS ELECTRICAL FORMULAS TO FIND: AMPERES (I)

HORSEPOWER (HP) WATTS (P) KILO-WATTS (KW) KILO-VOLT-AMPERES (KVA) CAPACITANCE (C), AND CAPACITORS INDUCTION (L) IMPEDANCE (Z) REACTANCE (INDUCTIVE-XL, AND

CAPACITIVE-XC) RESISTOR COLOR CODE

U.S. WEIGHTS AND MEASURES METRIC SYSTEM CONVERSION TABLES METALS AND SPECIFIC RESISTANCE (K) CENTIGRADE AND FAHRENHEIT THERMOMETER

SCALES USEFUL MATH, FORMULAS THE CIRCLE FRACTIONS EQUATIONS SQUARE ROOT TRIGONOMETRY CONDUIT BENDING TAP, DRILL BIT, AND HOLE SAW TABLES MOTORS: RUNNING OVERLOAD UNITS

BRANCH CIRCUIT PROTECTIVE DEVICES DIRECT CURRENT SINGLE-PHASE (A.C.) TWO-PHASE (A.C.)

THREE-PHASE A.C. MOTORS TRANSFORMERS: CALCULATIONS

VOLTAGE DROP CALCULATIONS SINGLE-PHASE CONNECTIONS BUCK AND BOOST CONNECTIONS FULL LOAD CURRENTS THREE-PHASE CONNECTIONS TWO-PHASE CONNECTIONS TWO-PHASE AND THREE-PHASE

CONNECTIONS

PAGE - 1 - 2 3 - 4 5 - 7

8 - 12 13

14 - 19 20 - 21

2 2 23 - 24

25

- TABLE OF CONTENTS (Continued)

MISCELLANEOUS WIRING DIAGRAMS = PROPERTIES OF CONDUCTORS

ALLOWABLE AMPACITIES OF CONDUCTORS INSULATION CHARTS MAXIMUM NUMBER OF CONDUCTORS I N

CONDLJIT - - - - MAXIMUM NUMBER OF FIXTURE WIRES IN

CONDUIT TABLES. METAL BOXES

COVER REQUIREMENTS TO 600 VOLTS VOLUME REQUIRED PER CONDUCTOR CLEAR WORKING SPACE IN FRONT OF

ELECTRICAL EQUIPMENT - MINIMUM CLEARANCE OF LIVE PARTS GROUNDING

ELECTRICAL SYMBOLS HAND SIGNALS FOR CRANES AND CHERRY

PICKERS USEFUL KNOTS AMERICAN RED CROSS FIRST AID

OHM'S LAW

T H E R A T E OF T H E FLOW OF THE CURRENT I S E Q U A L TO E L E C T R O M O T I V E FORCE D I V I D E D B Y R E S I S T A N C E .

E L E C T R O M O T I V E FORCE = V O L T S = " E M V O L T S CURRENT = AMPERES = :I" AMPERES = - R E S I S T A N C E = OHMS = R" OHMS

S E R I E S C I R C U I T P A R A L L E L C I R C U I T

A S E R I E S C I R C U I T I S A C I R C U I T A P A R A L L E L C I R C U I T I S A T H A T H A S ONLY ONE P A T H THROUGH C I R C U I T T H A T H A S MORE THAN W H I C H THE ELECTRONS MAY FLOW. ONE P A T H THROUGH W H I C H THE N O T E : " T " STANDS FOR T O T A L . ELECTRONS MAY FLOW.

I E T = E l + E 2 + E 3 ET = E l = E 2 = E 3

1

NOTE: FOR A P A R A L L E L C I R C U I T H A V I N G ONLY TWO R E S I S T O R S , T H E FOLLOWING FORMULA MAY BE U S E D .

R 1 X R 2 - RT = - R 1 + R 2

-1 -

OHM'S LAW

A . WHEN VOLTS AND O m A R E KNOWN:

E V O L T S OR I = - A M P E R E S = - OHMS R

EXAMPLE: F I N D T H E C U R R E N T OF A 120 V O L T C I R C U I T W I T H A R E S I S T A N C E OF 60 OHMS.

I = - = - 12' = 2 A M P E R E S R 60

0 . WHEN WATTS A N 0 VOLTS A R E KNOWN:

P W A T T S OR I = - A M P E R E S = - V O L T S E

E X A M P L E : A 120 V O L T C I R C U I T H A S A 1440 WATT L O A D . C

D E T E R M I N E T H E C U R R E N T .

I = - = - P 1440 = 12 AMPERES E 120

C . WHEN O m A N D WATTS A R E KNOWN:

A M P E R E S = +g OR 1 =& EXAMPLE: A C I R C U I T CONSUMES 625 W A T T S THROUGH A 12.75 OHM -

R E S I S T O R . D E T E R M I N E T H E C U R R E N T .

I =E = = fi = 7 A M P E R E S 12.75

A. ONE E L E C T R I C A L HORSEPOWER = 746 W A T T S E L E C T R I C MOTORS ARE R A T E D I N HORSEPOWER.

0 . ONE K I L O W A T T = 1000 W A T T S G E N E R A T O R S A R E R A T E D I N K I L O W A T T S .

SERIES CIRCUITS

RULE 1: T H E T O T A L CURRENT I N A S E R I E S C I R C U I T I S E Q U A L TO THE CURRENT I N ANY OTHER PART OF T H E C I R C U I T .

T O T A L CURRENT = 1 ( 1 ) = 1 ( 2 ) = I ( 3 ) . AND E T C .

RULE 2 : T H E T O T A L VOLTAGE I N A S E R I E S C I R C U I T I S EQUAL TO THE SUM OF T H E V O L T A G E S ACROSS A L L P A R T S OF T H E C I R C U I T .

T O T A L VOLTAGE = E ( l ) + E ( 2 ) + E ( 3 ) , AND E T C .

RULE 3 : T H E T O T A L R E S I S T A N C E OF A S E R I E S C I R C U I T I S E Q U A L TO T H E SUM OF THE R E S I S T A N C E S O F A L L T H E P A R T S O F THE C I R C U I T .

T O T A L R E S I S T A N C E = R ( 1 ) + R ( 2 ) + R ( 3 ) , AND E T C .

FORMULAS FROM OHM'S LAW

AMPERES = R E S I S T A N C E

V O L T S R E S I S T A N C E = -

AMPERES

V O L T S = AMPERES X R E S I S T A N C E OR E = I X R

EXAMPLE: F I N D T O T A L V O L T A G E . T O T A L CURRENT. AND T O T A L R E S I S T A N C E .

E ( 2 ) = 1 0 V O L T S I ( 2 ) = 0 . 4 AMP R ( 2 ) = 2 5 OHMS

E ( l ) = 8 V O L T S E ( 3 ) = 6 V O L T S I I (1 ) = 0 . 4 AMP I ( 3 ) = 0 . 4 AMP

R ( 1 ) = 2 0 OHMS R ( 3 ) = 1 5 OHMS

E ( T ) = ? I ( T ) = ? R ( T ) = ?

C O N T I N U E D N E X T PAGE

PARALLEL CIRCUITS

-1: T H E T O T A L CURRENT I N A P A R A L L E L C I R C U I T I S E Q U A L TO THE SUM OF T H E CURRENTS I N A L L THE BRANCHES OF T H E C I R C U I T .

T O T A L CURRENT = I ( 1 ) + 1 ( 2 ) + I ( 3 ) . AND E T C .

-2: T H E T O T A L VOLTAGE ACROSS ANY BRANCH I N P A R A L L E L I S E Q U A L TO THE VOLTAGE ACROSS ANY OTHER BRANCH AND I S A L S O E Q U A L TO THE TOTAL V O L T A G E .

T O T A L V O L T A G E = E ( l ) = E ( 2 ) = E ( 3 ) . AND ETC

RULE 3: T H E T O T A L R E S I S T A N C E I N A P A R A L L E L C I R C U I T I S FOUND BY A P P L Y I N G OHM'S LAW TO THE T O T A L V A L U E S OF T H E C I R C U I T .

T O T A L R E S I S T A N C E = TOTAL VOLTAGE O R ET

RT = - T O T A L AMPERES . I T

EXAMPLE: F I N D T H E T O T A L CURRENT. T O T A L V O L T A G E . AND T O T A L R E S I S T A N C E .

I ( T ) = I ( 1 ) + I ( 2 ) + 1 ( 3 ) E ( T ) = E ( l ) = E ( 2 ) = E ( 3 ) = 2 + 1 . 5 + 1 = 1 2 0 = 1 2 0 = 1 2 0

I ( T ) = 4 . 5 AMP E ( T ) = 1 2 0 V O L T S

E ( T ) 1 2 0 V O L T S R ( T ) = - = - = 2 6 . 6 6 OHMS R E S I S T A N C E

I ( T ) 4 . 5 AMP

NOTE: I N A P A R A L L E L C I R C U I T THE T O T A L R E S I S T A N C E I S ALWAYS L E S S THAN T H E R E S I S T A N C E OF ANY BRANCH.

I F THE BRANCHES OF A P A R A L L E L C I R C U I T HAVE THE SAME R E S I S T A N C E . THEN EACH W I L L DRAW T H E SAME CURRENT.

I F T H E BRANCHES OF A P A R A L L E L C I R C U I T HAVE D I F F E R E N T R E S I S T A N C E S . T H E N EACH W I L L DRAW A D I F F E R E N T CURRENT.

I N E I T H E R S E R I E S OR P A R A L L E L C I R C U I T S . THE LARGER THE R E S I S T A N C E . T H E SMALLER THE CURRENT DRAWN.

I

PARALLEL CIRCUITS

TO D E T E R M I N E THE T O T A L R E S I S T A N C E I N A P A R A L L E L C I R C U I T WHEN THE - TOTAL CURRENT. AND TOTAL VOLTAGE ARE UNKNOWN.

1 1 - - 1

+ - + - AND E T C . TOTAL R E S I S T A N C E R ( 1 ) R ( 2 ) R ( 3 )

EXAMPLE: F I N O THE TOTAL R E S I S T A N C E .

R ( l ) = R ( 2 ) = R ( 3 ) =

6 0 OHMS 8 0 OHMS 1 2 0 OHMS I

1 4 + 3 + 2 USE LOWEST COMMON - = DENOMINATOR ( 2 4 0 ) IC

R ( T ) 2 4 0

I \= /$_ - CROSS M U L T I P L Y R ( T ) ' ' 240

9 X R ( T ) = 1 X 2 4 0 OR 9 R T = 2 4 0

D I V I D E BOTH S I D E S OF THE E Q U A T I O N B Y " 9 "

R ( T ) = 26 .66 OHMS R E S I S T A N C E

NOTE: THE T O T A L R E S I S T A N C E OF A NUMBER OF EQUAL R E S I S T O R S I N I

P A R A L L E L I S EQUAL TO THE RESISTANCE OF ONE R E S I S T O R D I V I O E D BY THE NUMBER OF RESISTORS.

T O T A L RESISTANCE = R E S I S T A N C E OF ONE R E S I S T O R

NUMBER OF R E S I S T O R S I N C I R C U I T

CONTINUED NEXT PAGE

PARALLEL CIRCUITS

FORMULA: R

R ( T ) = - N

EXAMPLE: F I N D THE T O T A L R E S I S T A N C E

THERE ARE THREE R E S I S T O R S I N P A R A L L E L . EACH HAS A V A L U E OF 1 2 0 OHMS R E S I S T A N C E . ACCORDING TO THE FORMULA. I F WE D I V I D E THE R E S I S T A N C E OF ANY ONE OF THE R E S I S T O R S BY THREE WE W I L L O B T A I N THE T O T A L R E S I S T A N C E OF THE C I R C U I T .

R R ( T ) = - OR R ( T ) =

1 2 0 N 3

T O T A L R E S I S T A N C E = 4 0 OHMS

R ( T ) = ?

NOTE: TO F I N D T H E T O T A L R E S I S T A N C E OF ONLY TWO R E S I S T O R S I N P A R A L L E L . M U L T I P L Y THE R E S I S T A N C E S . AND THEN D I V I D E THE PRODUCT B Y THE SUM OF THE R E S I S T O R S .

FORMULA: T O T A L R E S I S T A N C E = R(l) R ( 2 ) R ( 1 ) + R ( 2 )

E X A M P L E :

.,, , , -

R ( 1 ) X R ( 2 )

I R ( 1 ) = 4 0 OHMS I R ( 1 ) + R ( 2 )

4 0 X 8 0

R ( T ) = = 2 6 . 6 6 OHMS

R ( T ) = ? 1 2 0

w

COMBINATION CIRCUITS

I N C O M B I N A T I O N C I R C U I T S WE COMBINE S E R I E S C I R C U I T S W I T H P A R A L L E L C I R C U I T S . C O M B I N A T I O N C I R C U I T S MAKE I T P O S S I B L E TO O B T A I N THE I

D I F F E R E N T VOLTAGES OF S E R I E S C I R C U I T S . A N 0 D I F F E R E N T CURRENTS OF P A R A L L E L C I R C U I T S .

EXAMPLE: 1. P A R A L L E L - S E R I E S C I R C U I T :

SOLVE FOR A L L M I S S I N G V A L U E S .

E ( 3 ) = ? ' I ( 3 ) = ? ' R ( 3 ) = 1 0 OHMS

E ( 4 ) = ? I ( 4 ) = ? R ( 4 ) = 5 0 OHMS

TO SOLVE:

1. F I N D THE T O T A L R E S I S T A N C E OF EACH GROUP. BOTH GROUPS ARE S I M P L E S E R I E S C I R C U I T S , SO

I

R ( 1 ) + R ( 2 ) = R ( A ) 2 0 + 4 0 = 6 0 OHMS. T O T A L R E S I S T A N C E OF GROUP " A "

Rlo3 1 + R ( 4 ) = R ( B ) + 5 0 = 6 0 OHMS, TOTAL R E S I S T A N C E OF GROUP " 8 "

2 . RE-DRAW THE C I R C U I T . C O M B I N I N G R E S I S T O R S ( R ( 1 ) + R ( 2 ) ) AND ( R ( 3 ) + R ( 4 ) ) SO THAT EACH GROUP W I L L HAVE ONLY ONE R E S I S T O R .

- CONTINUED NEXT PAGE

COMBINATION CIRCUITS

NOTE: WE NOW HAVE A S I M P L E P A R A L L E L C I R C U I T , SO

E ( T ) = E ( A ) = E ( B ) 1 2 0 v = 120 =

WE NOW HAVE A P A R A L L E L C I R C U I T W I T H ONLY TWO R E S I S T O R S , AND THEY ARE OF EQUAL V A L U E . WE HAVE A C H O I C E OF THREE D I F F E R E N T FORMULAS THAT CAN BE USED TO SOLVE FOR THE T O T A L R E S I S T A N C E OF C I R C U I T .

( 2 ) W H E N THE R E S I S T O R S OF A P A R A L L E L C I R C U I T ARE OF EQUAL V A L U E .

R ( T ) = ! = 60 = 3 0 OHMS N 2

l\=OL OR 1 X R ( T ) = 1 X 3 0 OR R ( T ) = 3 0 OHMS - R ( T ) / '30

3 . WE NOW KNOW THE V A L U E S OF E ( T ) . R ( T ) , E ( A ) . R ( A ) . E ( B ) . R ( B ) , R ( 1 ) . R ( 2 ) , R ( 3 ) , AND R ( 4 ) . NEXT WE W I L L SOLVE FOR I ( T ) , I ( A ) , I ( B ) , I ( l ) , 1 ( 2 ) , I ( 3 ) . AND I ( 4 ) .

- 1 2 0 - E ( B ) = I ( B ) OR - - 2 I ( B ) = 2 x . R( 0 ) 6 0

I ( B ) = I ( 3 ) = I ( 4 ) OR 2 = 2 = 2 I ( 3 ) = 1 ( 4 ) =

CONTINUED NEXT PAGE

I

COMBINATION CIRCUITS

4 . WE KNOW T H A T R E S I S T O R S # I a n d # 2 OF GROUP " A " ARE I N S E R I E S . WE KNOW TOO T H A T R E S I S T O R S # 3 a n d # 4 OF GROUP " 0 " ARE I N - S E R I E S . WE HAVE D E T E R M I N E D THAT THE T O T A L R E S I S T A N C E OF GROUP " A - IS 2 A M P , AND THE T O T A L RESISTANCE O F GROUP IS 2 AMP: BY U S I N G THE S E R I E S FORMULA WE CAN S O L V E FOR THE CURRENT V A L U E OF EACH R E S I S T O R .

5 . WE WERE G I V E N THE R E S I S T A N C E VALUES OF A L L R E S I S T O R S . R ( 1 ) = 2 0 OHMS, R ( 2 ) = 4 0 OHMS. R ( 3 ) = 1 0 OHMS. AND - R ( 4 ) = 5 0 OHMS.

BY U S I N G OHM'S LAW WE CAN D E T E R M I N E THE VOLTAGE DROP ACROSS EACH R E S I S T O R .

E ( 1 ) = R ( 1 ) X I ( 1 ) E ( 3 ) = R ( 3 ) X 1 ( 3 ) = 2 0 X 2 = 1 0 X 2

E ( 1 ) = 4 0 V O L T S E ( 3 ) = 2 0 V O L T S

E ( 2 ) = R ( 2 ) X 1 ( 2 ) E ( 4 ) = R ( 4 ) X 1 ( 4 ) = 4 0 X 2 = 5 0 x 2

E ( 2 ) = 8 0 V O L T S E ( 4 ) = 1 0 0 V O L T S

EXAMPLE: 2 . S E R I E S P A R A L L E L C I R C U I T :

SOLVE FOR A L L M I S S I N G V A L U E S

E ( 2 ) = ? I ( 2 ) = ? 4 C - r R ( 2 ) = 20

E ( 1 ) = ? I ( 1 ) = ? R ( 1 ) = u

E ( 3 ) = ? E ( T ) = 1 1 0 V . 1 ( 3 ) = ? I ( T ) = ? I.r

R ( 3 ) = 30 R ( T ) = ?

GROUP " A "

- -1 0-

COMBINATION CIRCUITS

TO SOLVE: I

1. WE CAN S E E THAT R E S I S T O R S # 2 AND # 3 ARE I N P A R A L L E L . AND COMBINED THEY ARE GROUP " A " . WHEN THERE ARE ONLY TWO R E S I S T O R S . WE U S E THE FOLLOWING FORMULA.

2 . WE CAN NOW RE-DRAW OUR C I R C U I T A S A S I M P L E S E R I E S C I R C U I T

I

R ( l ) = 1 0 OHMS R ( A ) = 1 2 OHMS E ( T ) = 1 1 0 VOLTS

I I ( T ) = ? GROUP " A " R ( T ) = ?

= 3 . I N A S E R I E S C I R C U I T

R ( T ) = R ( l ) + R ( A ) OR R ( T ) = 1 0 + 1 2 OR 2 2 OHMS

BY U S I N G O H M ' S LAW

I N A S E R I E S C I R C U I T

I ( T ) = I ( 1 ) = I ( A ) OR I ( T ) = 5 AMP, I ( 1 ) = AND I ( A ) =

BY U S I N G O H M ' S LAW

E ( l ) = I ( 1 ) X R ( 1 ) = 5 X 1 0 = 5 0 V O L T S

E ( T ) - E ( 1 ) = E ( A ) OR 1 1 0 - 5 0 = 6 0 V O L T S = E ( A )

I N A P A R A L L E L C I R C U I T

E ( A ) : E ( 2 ) = E ( 3 ) OR E ( A ) = 6 0 VOLTS. E ( 2 ) = 6 0 V O L T S , AND E ( 3 ) = 6 0 VOLTS.

COMBINATION CIRCUITS

BY U S I N G O H M ' S LAW

PROBLEM: S O L V E FOR T O T A L R E S I S T A N C E RE-DRAW C I R C U I T AS MANY T I M E S AS NECESSARY CORRECT ANSWER I S 1 0 0 OHMS

G I V E N VALUES:

R - 1 = 1 5 OHMS

R - 2 = 3 5 OHMS

R - 3 = 5 0 OHMS

R - 4 = 4 0 OHMS

R - 5 = 3 0 OHMS

R - 3 R - 4 R - 5

R - 7 = 1 0 OHMS

R - 2 GROUP A

R - 1 R - 9 ,.A,. 1 Ann

R - 8 = 3 0 0 OHMS I

R - 9 = 6 0 OHMS

R - 6

R - 7

R - 8

R - T = ?

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TO FIND AMPERES

DIRECT CURRENT:

A . WHEN HORSEPOWER I S KNOWN:

AMPERES = HORSEPOWER X 7 4 6 OR I = HP X 7 4 6

V O L T S X E F F I C I E N C Y E X % E F F

WHAT CURRENT W I L L A T R A V E L - T R A I L E R T O I L E T DRAW WHEN E Q U I P P E D W I T H A 1 2 V O L T , 1 / 8 HP MOTOR, H A V I N G A 9 6 % E F F I C I E N C Y R A T I N G ?

HP X 7 4 6 7 4 6 X 1 / 8 I = - = - = -.. 9 3 ' 2 5 - 8 . 0 9 AMP E X % E F F 1 2 X 0 . 9 6 1 1 . 5 2 -

B . WHEN K I L O W A T T S ARE KNOWN:

AMPERES = K I L O W A T T S X 1 0 0 0 KW X 1 0 0 0 OR I =

V O L T S E

A 7 5 KW. 2 4 0 V O L T . D I R E C T CURRENT GENERATOR I S USED TO I

POWER A V A R I A B L E - S P E E D CONVEYOR B E L T AT A ROCK C R U S H I N G P L A N T . D E T E R M I N E THE CURRENT.

I = K W 1000 = 75 = 3 1 2 . 5 AMPERES E 2 4 0

SINGLE PHASE:

A . WHEN WATTS. VOLTS. AND POWER-FACTOR ARE KNOWN:

AMPERES = WATTS -

V O L T S X POWER-FACTOR

D E T E R M I N E THE CURRENT WHEN A C I R C U I T H A S A 1 5 0 0 WATT L O A D A POWER-FACTOR OF 8 6 % , AND OPERATES FROM A S I N G L E - P H A S E 2 3 0 V O L T SOURCE.

I = I 5 O 0 = 1500 = 7 . 5 8 AMP 2 3 0 X 0 . 8 6 1 9 7 . 8

TO FIND AMPERES

SINGLE PHASE: -- 0 . WHEN HORSEPOWER I S KNOWN:

HORSEPOWER X 7 4 6 AMPERES =

V O L T S X E F F I C I E N C Y X POWER-FACTOR

* D E T E R M I N E THE AMP-LOAD OF A S I N G L E - P H A S E . 1 / 2 H P . 1 1 5 VOLT MOTOR. THE MOTOR HAS AN E F F I C I E N C Y R A T I N G OF 9 2 % . AND A POWER-FACTOR OF 8 0 % .

I = 4 . 4 AMP

C . WHEN K I L O W A T T S ARE KNOWN:

K I L O W A T T S X 1 0 0 0 OR I = KW X 1 0 0 0 AMPERES =

V O L T S X POWER-FACTOR E X PF

A 2 3 0 VOLT S I N G L E PHASE C I R C U I T ' H A S A 1 2 KW POWER L O A D , AND OPERATES A T 8 4 % POWER-FACTOR. D E T E R M I N E THE CURRENT.

D . WHEN K I L O V O L T - A M P E R E I S KNOWN:

K I L O V O L T - A M P E R E X 1 0 0 0 KVA X 1 0 0 0 AMPERES = OR I =

V O L T S E

A 1 1 5 V O L T , 2 K V A , S I N G L E PHASE GENERATOR O P E R A T I N G A T F U L L LOAD W I L L D E L I V E R 1 7 . 4 AMPERES. ( P R O V E )

2 X 1 0 0 0 2 0 0 0 I = - = - = 1 7 . 4 AMP

1 1 5 1 1 5 - REMEMBER: BY D E F I N I T I O N AMPERES I S THE RATE OF THE FLOW OF

THE CURRENT.

TO FIND AMPERES

TWO-PHASE, FOUR WIRE:

NOTE: FOR THREE W I R E , TWO-PHASE C I R C U I T S . THE CURRENT I N THE COMMON CONDUCTOR I S 1 . 4 1 GREATER THAN I N E I T H E R OF THE OTHER TWO CONDUCTORS.

A . WHEN WATTS, VOLTS. AND POWER-FACTOR ARE KNOWN:

WATTS P AMPERES =

V O L T S X POWER-FACTOR X 2 E X P F X 2

D E T E R M I N E THE CURRENT WHEN A C I R C U I T H A S A 1 5 0 0 WATT L O A D , A POWER-FACTOR OF 8 6 % . A N 0 OPERATES FROM A TWO P H A S E . 2 3 0 - VOLT SOURCE.

P 1500 I =

- 1 5 0 0 - - E X P F X 2 2 3 0 X 0 . 8 6 X 2 3 9 5 . 6

I = 3 . 7 9 AMP I

0 . WHEN HORSEPOWER I S KNOWN:

AMPERES = HORSEPOWER X 7 4 6

V O L T S X E F F I C I E N C Y X POWER-FACTOR X 2 - OR

DETERMINE THE AMP-LOAD OF A TWO-PHASE. 1 / 2 H P . 2 3 0 VOLT -

MOTOR. THE MOTOR H A S AN E F F I C I E N C Y R A T I N G OF 9 2 % . AND A POWER-FACTOR OF 8 0 % .

HP X 7 4 6 1 / 2 X 7 4 6 I = - = -

E X % E F F X P F X 2 2 3 0 X 0 . 9 2 X 0 . 8 0 X 2

= - = 3 7 3 1.1 AMP 3 3 9 -

NOTE :

CoNsUMEo = & = POWER-FACTOR ( P F )

APPARENT POWER KVA

u

TO FIND AMPERES

T W O - P H A S E , F O U R WIRE: I

C . WHEN K I L O W A T T S ARE KNOWN:

K I L O W A T T S X 1 0 0 0 AMPERES =

V O L T S X POWER-FACTOR X 2

I = KW X 1 0 0 0

E X P F X 2

1 A 2 3 0 V O L T . TWO-PHASE C I R C U I T , H A S A 1 2 K W POWER L O A D , AND OPERATES A T 8 4 % POWER-FACTOR. D E T E R M I N E THE CURRENT.

I = - 3 1 AMP

0 . WHEN K I L O V O L T - A M P E R E I S KNOWN:

K I L O V O L T - A M P E R E X 1 0 0 0 AMPERES =

V O L T S X 2 m

OR

KVA X 1 0 0 0 I =

E X 2

1 A 2 3 0 V O L T . 4 K V A , TWO-PHASE GENERATOR O P E R A T I N G A T F U L L LOAD W I L L D E L I V E R 8 . 7 AMPERES. ( P R O V E )

4 X 1 0 0 0 I = - = 4000 = 8 . 7 AMP 2 3 0 X 2 4 6 0

TO FIND AMPERES

THREE-PHASE:

A . WHEN WATTS, VOLTS. AND POWER-FACTOR ARE KNOWN:

AMPERES = WATTS

V O L T S X POWER-FACTOR X 1 . 7 3

D E T E R M I N E T H E CURRENT WHEN A C I R C U I T HAS A 1 5 0 0 WATT LOAD. m A POWER-FACTOR OF 8 6 % . AND OPERATES FROM A T H R E E - P H A S E . 2 3 0 VOLT SOURCE.

= 4 . 4 AMP

8. WHEN HORSEPOWER I S KNOWN:

AMPERES = HORSEPOWER X 7 4 6

V O L T S X E F F I C I E N C Y X POWER-FACTOR X 1 . 7 3

OR

D E T E R M I N E THE AMP-LOAD OF A THREE-PHASE, 1 / 2 H P , 2 3 0 VOLT MOTOR. THE MOTOR HAS AN E F F I C I E N C Y R A T I N G OF 9 2 % , AND A POWER-FACTOR OF 8 0 % .

= - = 3 7 3 1 . 2 7 AMP 2 9 3

TO FIND AMPERES

m THREE-PHASE:

C . WHEN K I L O W A T T S ARE KNOWN:

K I L O W A T T S X 1 0 0 0 AMPERES =

V O L T S X POWER-FACTOR X 1 . 7 3 a

0 1

I = KW X 1 0 0 0

E X PF X 1 . 7 3

I A 2 3 0 V O L T , THREE-PHASE C I R C U I T , H A S A 1 2 KW POWER LOAD, AND OPERATES AT 8 4 % POWER-FACTOR. D E T E R M I N E THE CURRENT.

I = KW X 1 0 0 0 - 1 2 . 0 0 0 1 2 , 0 0 0 = - E X P F X 1 . 7 3 2 3 0 X 0 . 8 4 X 1 . 7 3 3 3 4 . 2 4 - I = 36AMP

0 . WHEN K I L O V O L T - A M P E R E I S KNOWN:

K I L O V O L T - A M P E R E X 1 0 0 0 = KVA X 1 0 0 0 AMPERES =

I. E X 1 . 7 3 E X 1 . 7 3

A 2 3 0 V O L T , 4 K V A , THREE PHASE GENERATOR O P E R A T I N G AT F U L L LOAD W I L L D E L I V E R 1 0 AMPERES. ( P R O V E )

4 0 0 0 I =

KVA X 1 0 0 0 = 4 X 1 0 0 0 - m E X 1 . 7 3 2 3 0 X 1 . 7 3 3 9 7 . 9

I =

NOTE: TO BETTER UNDERSTAND THE PRECEDING FORMULAS:

1. TWO-PHASE CURRENT X 2 = S I N G L E - P H A S E CURRENT. 2 . THREE-PHASE CURRENT X 1 . 7 3 = S I N G L E PHASE CURRENT. 3 . THE CURRENT I N THE COMMON CONDUCTOR OF A TWO-PHASE

( T H R E E W I R E ) C I R C U I T I S 1 4 1 % GREATER THAN E I T H E R OF THE OTHER TWO CONDUCTORS OF THAT C I R C U I T .

TO FIND HORSEPOWER

DIRECT CURRENT:

HORSEPOWER = VOLTS X AMPERES X E F F I C I E N C Y

7 4 6

A 1 2 VOLT MOTOR DRAWS A CURRENT OF 8 . 0 9 AMPERES, AND HAS AN E F F I C I E N C Y R A T I N G OF 9 6 % . DETERMINE THE HORSEPOWER.

I

HP = E X I X % E F F 1 2 X 8 . 0 9 X 0 . 9 6 9 3 . 1 9 - - 7 4 6 7 4 6 7 4 6

= 0 . 1 2 4 9 =

SINGLE-PHASE:

HP = V O L T S X AMPERES X E F F I C I E N C Y X POWER-FACTOR

7 4 6

A S I N G L E - P H A S E . 1 1 5 VOLT ( A C ) MOTOR HAS AN E F F I C I E N C Y R A T I N G - OF 9 2 % , A N 0 A POWER-FACTOR OF 8 0 % . DETERMINE THE HORSEPOWER I F THE AMP-LOAD I S 4 . 4 AMPERES.

TWO-PHASE:

HP = VOLTS X AMPERES X E F F I C I E N C Y X POWER-FACTOR X 2

7 4 6

DETERMINE THE HORSEPOWER OF A TWO-PHASE. 2 3 0 VOLT ( A C ) MOTOR. THE MOTOR H A S AN E F F I C I E N C Y R A T I N G OF 9 2 % . A POWER-FACTOR OF 8 0 % . AND AN AMP-LOAD OF 1.1 AMPERES. I

TO FIND HORSEPOWER

THREE-PHASE: .I

HP = V O L T S X AMPERES X E F F I C I E N C Y X POWER-FACTOR X 1.73

7 4 6

A T H R E E - P H A S E . 4 6 0 V O L T MOTOR DRAWS A CURRENT OF 5 2 AMPERES. THE MOTOR H A S AN E F F I C I E N C Y R A T I N G OF 9 4 % . AND A POWER FACTOR

I OF 8 0 % . D E T E R M I N E THE HORSEPOWER.

TO FIND WATTS

THE E L E C T R I C A L POWER I N ANY PART OF A C I R C U I T I S EOUAL TO T H E C U R R E N T IN T H A T P A R T MULTIPLIED B Y THE VOLTAGE A C R O S S T H A T P A R T OF THE C I R C U I T .

A WATT I S THE POWER USED WHEN ONE VOLT CAUSES ONE AMPERE TO FLOW I N A C I R C U I T .

I ONE HORSEPOWER I S THE AMOUNT OF ENERGY R E Q U I R E D TO L I F T 3 3 . 0 0 0 POUNDS. ONE FOOT. I N ONE M I N U T E . THE E L E C T R I C A L E Q U I V A L E N T OF ONE HORSEPOWER I S 7 4 5 . 6 WATTS. ONE WATT I S THE AMOUNT OF ENERGY R E Q U I R E D TO L I F T 4 4 . 2 6 POUNDS, ONE FOOT. I N ONE M I N U T E . WATTS I S POWER, AND POWER I S THE AMOUNT OF WORK DONE I N A G I V E N T I M E .

1. W H E N V O L T S A N D A M P E R E S A R E K N O W N :

A . POWER ( W A T T S ) = V O L T S X AMPERES

A 1 2 0 VOLT A - C C I R C U I T DRAWS A CURRENT OF 5 AMPERES: D E T E R M I N E THE POWER CONSUMPTION.

P = E X I = 1 2 0 X 5 = 6 0 0 WATTS

WE CAN NOW D E T E R M I N E THE R E S I S T A N C E OF T H I S C I R C U I T .

( 1 . ) POWER = R E S I S T A N C E X ( A M P E R E S ) Z

P = R X ( I ) ' OR 6 0 0 = R X 2 5

6 0 0 - = R OR R = 2 4 OHMS 2 5

( 2 . ) POWER = OR P = R E S I S T A N C E R

1 4 , 4 0 0 R X 6 0 0 = ( 1 2 0 ) ' OR R = -

6 0 0 R = 2 4 OHMS

NOTE: REFER TO FORMULAS OF THE OHM'S LAW CHART ON PAGE 1.

-22-

DIRECT CURRENT:

KILOWATTS = V O L T S X AMPERES 1 0 0 0

A 1 2 0 V O L T ( D C ) MOTOR DRAWS A CURRENT OF 4 0 AMPERES D E T E R M I N E THE K I L O W A T T S .

SINGLE-PHASE:

KILOWATTS = V O L T S X AMPERES X POWER-FACTOR

1 0 0 0

A S I N G L E - P H A S E . 1 1 5 V O L T ( A C ) MOTOR DRAWS A CURKENT OF 2 0 AMPERES. AND H A S A POWER-FACTOR R A T I N G O F 86%. D E T E R M I N E THE K I L O W A T T S .

= 1 . 9 7 8 =

TWO-PHASE:

KILOWATTS = V O L T S X AMPERES X POWER-FACTOR x 2 1 0 0 0

A TWO-PHASE, 2 3 0 V O L T ( A C ) MOTOR W I T H A POWER-FACTOR O F 9 2 % . DRAWS A CURRENT OF 5 5 AMPERES. D E T E R M I N E T H E K I L O W A T T S .

KW = E X I X P F X 2 2 3 0 X 5 5 X 0 . 9 2 X 2

1 0 0 0 1 0 0 0

TO FIND KILOWAlTS

THREE-PHASE: - K I L O W A T T S =

V O L T S X AMPERES X POWER-FACTOR X 1.73

1 0 0 0

A T H R E E - P H A S E , 4 6 0 V O L T MOTOR DRAWS A CURRENT OF 5 2 AMPERES. AND H A S A POWER-FACTOR RATED A T 8 0 % . D E T E R M I N E T H E K I L O W A T T S .

II

K I R C H H O F F ' S LAWS

F I R S T LAW ( C U R R E N T 1 - THE SUM OF THE CURRENTS A R R I V I N G AT ANY P O I N T I N A C I R C U I T MUST EQUAL THE SUM OF THE CURRENTS L E A V I N G T H A T P O I N T .

SECOND LAW ( V O L T A G E ) I

THE T O T A L VOLTAGE A P P L I E D TO ANY C L O S E 0 C I R C U I T P A T H I S ALWAYS EQUAL TO THE SUM OF THE VOLTAGE DROPS I N T H A T P A T H .

OR I

THE A L G E B R A I C SUM OF A L L THE VOLTAGES ENCOUNTERED I N ANY LOOP E Q U A L S ZERO.

TO FIND KILOVOLT-AMPERES

SINGLE-PHASE:

K I L O V O L T - A M P E R E S = V O L T S X AMPERES

1 0 0 0

A S I N G L E - P H A S E , 2 4 0 VOLT GENERATOR D E L I V E R S 4 1 . 6 6 AMPERES A T F U L L L O A D . D E T E R M I N E THE K I L O V O L T - A M P E R E S R A T I N G .

E X I 2 4 0 X 4 1 . 6 6 _ 1 0 , 0 0 0 KVA = - = - - =

1 0 0 0 1 0 0 0 1 0 0 0

T W O - P H A S E :

K I L O V O L T - A M P E R E S = A M P E R E S 1 0 0 0

A TWO-PHASE. 2 3 0 V O L T GENERATOR D E L I V E R S 5 5 AMPERES I D E T E R M I N E THE K I L O V O L T - A M P E R E S R A T I N G .

KVA = E X I X 2 - 2 3 0 X 5 5 X 2 - 2 5 , 3 0 0 - -

1 0 0 0 1 0 0 0 1 0 0 0

I = 2 5 . 3 =

THREE-PHASE:

K I L O V O L T - A M P E R E S = 1 0 0 0

A T H R E E - P H A S E , 4 6 0 VOLT GENERATOR D E L I V E R S 5 2 AMPERES D E T E R M I N E THE K I L O V O L T - A M P E R E S R A T I N G .

N O T E : KVA = APPARENT POWER = POWER BEFORE U S E D . SUCH AS THE R A T I N G OF A TRANSFORMER.

TO FIND

CAPACITANCE (C1:

a COULOMBS C = - OR C A P A C I T A N C E = - E V O L T S

C A P A C I T A N C E I S THE PROPERTY OF A C I R C U I T OR BODY T H A T P E R M I T S I T TO STORE A N E L E C T R I C A L CHARGE E Q U A L TO THE ACCUMULATED CHARGE D I V I D E D B Y T H E V O L T A G E . EXPRESSED I N FARADS. I

A . TO D E T E R M I N E T H E T O T A L C A P A C I T Y OF C A P A C I T O R S , AND / OR CONDENSERS CONNECTED I N S E R I E S .

D E T E R M I N E THE T O T A L C A P A C I T Y OF FOUR E A C H , 1 2 M I C R O F A R A D C A P A C I T O R S CONNECTED I N S E R I E S .

C ( T ) = 3 M I C R O F A R A D S

8 . TO D E T E R M I N E T H E T O T A L C A P A C I T Y OF C A P A C I T O R S . AND / OR CONDENSERS CONNECTED I N P A R A L L E L . - DETERMINE THE T O T A L C A P A C I T Y OF FOUR EACH. 1 2 M I C R O F A R A D C A P A C I T O R S CONNECTED I N P A R A L L E L .

C ( T ) = + C ( T ) = 4 8 M I C R O F A R A D S

A FARAD I S THE U N I T OF C A P A C I T A N C E OF A CONDENSER T H A T R E T A I N S ONE COULOMB OF CHARGE W I T H ONE V O L T D I F F E R E N C E OF I

P O T E N T I A L .

1 FARAD = 1 , 0 0 0 , 0 0 0 MICROFARADS

-26-

- 6-DOT COLOR CODE FOR MICA AND MOLDED PAPER CAPACITORS

D I R E C T I O N OF +- ' O " ~ T K T I 2 N D I D I G ~ \

OR C L A S S M U L T I P L I E R T O L E R A N C E L/

t---

T Y P E

J A N . M I C A

E I A . M I C A

MOLDED P A P E R

COLOR

B L A C K BROWN R E D ORANGE Y E L L O W G R E E N B L U E V I O L E T GRAY W H I T E G O L D S I L V E R B O D ' I

1ST D I G I T

0 1 2 3 4 5 6 7 8 9

Z N D D I G I T

0 1 2 3 4 5 6 7 8 9

M U L T I P L I E R

1 1 0

100 1.000

10.000 100,000

1,000,000 10,000,000

100.000.000 1.000.000.000

. 1 . O 1

l O L E R A N C E ( % )

* 1 * 2 f 3 + 4 * 5 i 6 i 7 i 6 + 9

+ l o * 2 0

C H A R A C T E R I S T I C OR C L A S S

A P P L I E S T O T E M P E R A T U R E C O E F F I C I E N T OR M E T H O D S OF T E S T I N G

- MAXIMUM PERMISSIBLE CAPACITOR KVAR FOR USE WITH

OPEN-TYPE THREE-PHASE SIXTY-CYCLE INDUCTION MOTORS

NOTE. I F C A P A C I T O R S OF A LOWER R A T I N G THAN THE V A L U E S G I V E N I N THE T A B L E ARE U S E D , T H E PERCENTAGE R E D U C T I O N I N L I N E CURRENT G I V E N I N THE T A B L E SHOULD BE REDUCED P R O P O R T I O N A L L Y . -

REPRINTED WITH PERMISSION FROM NFPA 70 1990 NATIONAL ELECTRICAL CODEv COPYRIGHT 1989 NATIONAL FIRE PROTECTION ASSOCIATION OUINCY MA 02269 THIS REPRINTED MATERIAL IS NOT THE COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIRETY

MOTOR R A T I N G

H P

1 0 1 5 2 0 2 5 3 0

4 0 5 0 6 0 7 5

1 0 0 1 2 5 1 5 0 2 0 0

1 0 1 5 2 0 2 5 3 0

4 0 5 0 6 0 7 5

1 0 0 1 2 5 1 5 0 2 0 0

3 6 0 0 RPM

MAXIMUM C A P A C I T O R

R A T I N G KVAR

3 4 5 6 7

9 1 2 1 4 1 7

2 2 2 7 3 2 . 5 4 0

R E D U C T I O N I N L I N E CURRENT

%

1 0 9 9 9 8

8 8 8 8

8 8 8 8

1 8 0 0 RPM

MAXIMUM C A P A C I T O R

R A T I N G KVAR

3 4 5 6 7

9 11 1 4 1 6

2 1 2 6 3 0 3 7 . 5

1 2 0 0 RPM

R E D U C T I O N I N L I N E CURRENT

%

11 1 0 1 0 1 0

9

9 9 8 8

8 8 8 8

M A X I M U M C A P A C I T O R

R A T I N G K V A R

3 . 5 5 6 . 5 7 . 5 9

11 1 3 1 5 1 8

2 5 3 0 3 5 4 2 . 5

9 0 0 RPM

R E D U C T I O N I N L I N E CURRENT

%

1 4 1 3 1 2 11 11

1 0 1 0 1 0 1 0

9 9 9 9

5 6 . 5 7 . 5 9

1 0

1 2 1 5 1 8 2 1

2 7 3 2 . 5 3 7 . 5 4 7 . 5

7 2 0 RPM

2 1 1 8 1 6 1 5 1 4

1 3 1 2 11 1 0

1 0 1 0 1 0 1 0

6 . 5 8 9

11 1 2

1 5 1 9 2 2 2 6

3 2 . 5 4 0 4 7 . 5 6 0

6 0 0 RPM

2 7 2 3 2 1 2 0 1 8

1 6 1 5 1 5 1 4

1 3 1 3 1 2 1 2

7 . 5 9 . 5

1 2 1 4 1 6

2 0 2 4 2 7 3 2 . 5

4 0 4 7 . 5 5 2 . 5 6 5

3 1 2 7 2 5 2 3 2 2

2 0 1 9 1 9 1 8

1 7 1 6 1 5 1 4

POWER-FACTOR CORRECTION

T A B L E V A L U E S X KW L O A O = K V A O F C A P A C I T O R S N E E D E D T O C O R R E C T FROM E X I S T I N G T O D E S I R E D POWER F A C T O R .

T Y P I C A L P R O B L E M : W I T H A L O A O OF 500 KW A T 70% POWER F A C T O R . I T I S D E S I R E D T O F I N D T H E K V A O F C A P A C I T O R S R E Q U I R E O T O C O R R E C T T H E POWER F A C T O R T O 85%.

E X I S T I N G POWER

F A C T O R %

5 0 5 2 5 4 5 5 5 6 5 8 6 0 6 2 6 4 6 5 6 6 6 8 7 0 7 2 7 4 7 5 7 6 7 8 8 0 8 2 8 4 8 5 8 6 8 8 9 0 9 2 9 4 9 5

S O L U T I O N : FROM T H E T A B L E S E L E C T T H E M U L T I P L Y I N G F A C T O R 0 .400 C O R R E S P O N D I N G T O T H E E X I S T I N G 70%. A N D T H E C O R R E C T E D 85% POWER F A C T O R . 0 .400 X 500 = 200 K V A O F C A P A C I T O R S R E Q U I R E O .

REPRINTED WITH PERMISSION FROM NFPA 70 1990. NATIONAL ELECTRICAL CODE' .COPYRIGHT 1989. NATIONAL FIRE PROTECTION ASSOCIATION. OUINCY, MA 02269 THIS REPRINTED MATERIAL IS NOT THE COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIRETY

C O R R E C T E D POWER F A C T O R

100%

1 . 7 3 2 1 . 6 4 3 1 . 5 5 8 1 . 5 1 8 1 . 4 7 9 1 . 4 0 4 1 . 3 3 3 1 . 2 6 5 1 . 2 0 1 1 . 1 6 8 1 . 1 3 9 1 . 0 7 8 1 . 0 2 0 0 . 9 6 4 0 . 9 0 9 0 . 8 8 2 0 . 8 5 5 0 . 8 0 2 0 . 7 5 0 0 . 6 9 8 0 . 6 4 6 0 . 6 2 0 0 . 5 9 4 0 . 5 4 0 0 . 4 8 5 0 . 4 2 6 0 . 3 6 3 0 . 3 2 9

95%

1 . 4 0 3 1 . 3 1 4 1 . 2 2 9 1 . 1 8 9 1 . 1 5 0 1 . 0 7 5 1 . 0 0 4 0 . 9 3 6 0 . 8 7 2 0 . 8 3 9 0 . 8 1 0 0 .749 0 . 6 9 1 0 . 6 3 5 0 . 5 8 0 0 . 5 5 3 0 . 5 2 6 0 . 4 7 3 0 . 4 2 1 0 . 3 6 9 0 . 3 1 7 0 . 2 9 1 0 . 2 6 5 0 . 2 1 1 0 . 1 5 6 0 . 0 9 7 0 . 0 3 4

90%

1 . 2 4 7 1 . 1 5 8 1 .073 1 .033 0 .994 0 . 9 1 9 0 . 8 4 8 0 .780 0 . 7 1 6 0 . 6 8 3 0 . 6 5 4 0 . 5 9 3 0 . 5 3 5 0 . 4 7 9 0 . 4 2 4 0 . 3 9 7 0 . 3 7 0 0 .317 0.265 0 .213 0 . 1 6 1 0 . 1 3 5 0 . 1 0 9 0 . 0 5 5

85%

1 . 1 1 2 1 . 0 2 3 0 . 9 3 8 0 . 8 9 8 0 . 8 5 9 0 . 7 8 4 0 . 7 1 3 0 . 6 4 5 0 . 5 8 1 0 . 5 4 8 0 . 5 1 9 0 . 4 5 8 0 . 4 0 0 0 . 3 4 4 0 . 2 8 9 0 . 2 6 2 0 . 2 3 5 0 . 1 8 2 0 . 1 3 0 0 . 0 7 8

80%

0 . 9 8 2 0 . 8 9 3 0 . 8 0 8 0 . 7 6 8 0 . 7 2 9 0 . 6 5 4 0 . 5 8 3 0 . 5 1 5 0 . 4 5 1 0 . 4 1 8 0 . 3 8 9 0 . 3 2 8 0 . 2 7 0 0 .214 0 . 1 5 9 0 . 1 3 2 0 . 1 0 5 0 . 0 5 2

75%

0 . 8 5 0 0 . 7 6 1 0 . 6 7 6 0 . 6 3 6 0 .597 0 . 5 2 2 0 . 4 5 1 0 . 3 8 3 0 . 3 1 9 0 . 2 8 6 0 . 2 5 7 0 . 1 9 6 0 . 1 3 8 0 .082 0 . 0 2 7

TO FIND

INDUCTION (L): - I N O U C T I O N I S THE PRODUCTION OF M A G N E T I Z A T I O N OF E L E C T R I F I C A - T I O N I N A BODY BY THE P R O X I M I T Y OF A M A G N E T I C F I E L D OR E L E C T R I C CHARGE. OR OF THE E L E C T R I C CURRENT I N A CONDUCTOR BY THE V A R I A T I O N OF THE MAGNETIC F I E L D I N I T S V I C I N I T Y . EXPRESSED I N HENRYS.

A . TO F I N O THE TOTAL I N D U C T I O N OF C O I L S CONNECTED I N S E R I E S . -

D E T E R M I N E THE TOTAL I N D U C T I O N OF FOUR C O I L S CONNECTEO I N S E R I E S . EACH C O I L HAS AN INDUCTANCE VALUE OF FOUR HENRYS. - L ( T ) = L i 1 ) + L j 2 ) +

+ = 16 HENRYS

B . TO F I N O THE TOTAL I N D U C T I O N OF C O I L S CONNECTEO I N P A R A L L E L .

DETERMINE THE TOTAL I N O U C T I O N OF FOUR C O I L S CONNECTED I N P A R A L L E L . EACH C O I L HAS AN INDUCTANCE VALUE OF FOUR HENRYS.

1 - OR L ( T ) X 4 = 1 X 4 OR L ( T ) = 4 L ( T ) 4

L ( T ) = 1 HENRY

AN I N O U C T I O N C O I L I S A O E V I C E . C O N S I S T I N G OF TWO CONCENTRIC C O I L S AND AN I N T E R R U P T E R . THAT CHANGES A LOW STEADY VOLTAGE I N T O A H I G H I N T E R M I T T E N T A L T E R N A T I N G I

VOLTAGE BY ELECTROMAGNETIC I N D U C T I O N . MOST OFTEN USED A S A SPARK C O I L .

TO FIND

IMPEDANCE (Z): - IMPEDANCE I S THE T O T A L O P P O S I T I O N TO AN A L T E R N A T I N G CURRENT PRESENTED B Y A C I R C U I T . EXPRESSED I N OHMS.

A . WHEN V O L T S AND AMPERES ARE KNOWN:

1 V O L T S OR Z = E IMPEDANCE = -

AMPERES I

D E T E R M I N E THE IMPEDANCE OF A 1 2 0 VOLT A-C C I R C U I T THAT DRAWS A CURRENT OF FOUR AMPERES.

3 0 OHMS

B . WHEN R E S I S T A N C E AND REACTANCE ARE KNOWN:

1

D E T E R M I N E THE IMPEDANCE OF AN A-C C I R C U I T WHEN THE

I R E S I S T A N C E I S 6 OHMS. AND THE REACTANCE I S 8 OHMS.

= 1 0 OHMS - C . WHEN R E S I S T A N C E , I N D U C T I V E REACTANCE, AND C A P A C I T I V E REACTANCE ARE KNOWN:

DETERMINE THE IMPEDANCE OF AN A-C C I R C U I T WHICH HAS A R E S I S T A N C E OF 6 OHMS. AN I N D U C T I V E REACTANCE OF 1 8 OHMS. AND A C A P A C I T I V E REACTANCE OF 1 0 OHMS.

z = 7 / R ' + ( X ( L ) - X ( C ) )'

= yS2 + ( 1 8 - 1 0 ) ' = d m = d= = 7/100 = 1 0 OHMS

TO FIND

REACTANCE (XI:

REACTANCE I N A C I R C U I T I S THE O P P O S I T I O N TO AN A L T E R N A T I N G CURRENT CAUSED BY INDUCTANCE AND C A P A C I T A N C E . EQUAL TO THE D I F F E R E N C E BETWEEN C A P A C I T I V E AND I N D U C T I V E REACTANCE. EXPRESSED I N OHMS.

A . I N D U C T I V E REACTANCE XJ.J

I N D U C T I V E REACTANCE I S THAT ELEMENT OF REACTANCE I N A C I R C U I T CAUSED BY S E L F - I N D U C T A N C E .

X ( L ) = 2 X 3 . 1 4 1 6 X FREQUENCY X I N D U C T A N C E

D E T E R M I N E THE REACTANCE OF A FOUN-HENRY C O I L ON A 6 0 CYCLE, A -C C I R C U I T .

X ( L ) = 6 . 2 8 X F X L = 6 . 2 8 X 6 0 X 4 = 1 5 0 7 OHMS

8 . C A P A C I T I V E REACTANCE

C A P A C I T I V E REACTANCE I S THAT E l C I R C U I T CAUSED B Y C A P A C I T A N C E .

REACTANCE

1 X ( C ) =

2 X 3 . 1 4 1 6 X FREQUENCY X CAPACITANCE

DETERMINE THE REACTANCE OF A FOUR MICROFARAD CONDENSER ON A 6 0 C Y C L E , A -C C I R C U I T .

= 6 6 3 OHMS 0 . 0 0 1 5 0 7 2

A HENRY I S A U N I T OF INDUCTANCE. EQUAL TO THE INDUCTANCE -

OF A C I R C U I T I N WHICH THE V A R I A T I O N OF A CURRENT A T THE RATE OF ONE AMPERE PER SECOND I N D U C E S AN ELECTROMOTIVE FORCE OF ONE V O L T .

RESISTOR COLOR CODE

r 1 1.51 O I G I T

(PERCENT)

TOLERANCE ( P E R C E N T )

2 5 %

+ 1 0 %

+ 2 0 %

COLOR

BLACK

BROWN

R E 0

ORANGE

YELLOW

GREEN

BLUE

V I O L E T GRAY

WHITE

GOLD

S I L V E R

NO COLOR

1 S T O I G I T

0

1

2

3

4

5

6

7 8

9

2ND D I G I T

0

1

2

3

4

5

6

7

8

9

M U L T I P L I E R

1

1 0

1 0 0

1 , 0 0 0

1 0 , 0 0 0

1 0 0 , 0 0 0

1 . 0 0 0 , O O O 1 0 . 0 0 0 . 0 0 0 1 0 0 . 0 0 0 . 0 0 0

1 . 0 0 0 . 0 0 0 , 0 0 0

. 1

. O 1

U.S. WEIGHTS AND MEASURES

L INEAR M E A S U R E

1 I N C H = 2 . 5 4 0 C E N T I M E T E R S 1 2 I N C H E S = 1 FOOT = 3 . 0 4 8 D E C I M E T E R S 3 F E E T = 1 YARD = 9 . 1 4 4 D E C I M E T E R S 5 . 5 YARDS = 1 ROD, P O L E , OR PERCH = 5 . 0 2 9 METERS 4 0 RODS = 1 FURLONG = 2 . 0 1 8 HECTOMETERS - 8 FURLONGS = 1 M I L E = 1 . 6 0 9 K I L O M E T E R S

M I L E M E A S U R E M E N T S

1 S T A T U T E M I L E = 5 . 2 8 0 FEET 1 SCOTS M I L E = 5 . 9 5 2 FEET 1 I R I S H M I L E = 6 . 7 2 0 FEET 1 R U S S I A N VERST = 3 . 5 0 4 FEET 1 I T A L I A N M I L E = 4 . 4 0 1 F E E T 1 S P A N I S H M I L E = 1 5 , 0 8 4 F E E T

OTHER L INEAR M E A S U R E M E N T S I

1 HAND = 4 I N C H E S 1 L I N K = 7 . 9 2 I N C H E S 1 SPAN = 9 I N C H E S 1 FATHOM = 6 F E E T 1 C H A I N = 2 2 YARDS 1 FURLONG = 1 0 C H A I N S 1 KNOT = 1 N A U T I C A L M I L E 1 CABLE = 6 0 8 F E E T

= 6 0 8 0 F E E T - SQUARE M E A S U R E

1 4 4 SQUARE I N C H E S = 1 SQUARE FOOT 9 SQUARE FEET = 1 SQUARE YARD 3 0 - 1 1 4 SQUARE YAROS = 1 SQUARE ROO

= 1 SQUARE POLE = 1 SQUARE PERCH

4 0 RODS = 1 ROOD 4 ROODS = 1 ACRE 6 4 0 ACRES = 1 SQUARE M I L E 1 SQUARE M I L E = 1 S E C T I O N 3 6 S E C T I O N S = 1 TOWNSHIP

CUBIC O R SOLID M E A S U R E

1 CU. FOOT = 1 7 2 8 CU. I N C H E S 1 CU. YARD = 2 7 CU. F E E T 1 C11. FOOT = 7 . 4 8 GALLONS - - -

1 GALLON (WATER) = 8 . 3 4 L B S . 1 GALLON ( U . S . ) = 2 3 1 CU. I N C H E S OF WATER 1 GALLON ( I M P E R I A L ) = 2 7 7 - 1 / 4 CU. I N C H E S OF WATER

- U.S. WEIGHTS AND MEASURES

- L I Q U I D M E A S U R E

1 P I N T = 4 G I L L S 1 QUART = 2 P I N T S 1 G A L L O N = 4 QUARTS 1 F I R K I N = 9 G A L L O N S ( A L E OR B E E R ) 1 BARREL = 4 2 G A L L O N S ( P E T R O L E U M OR CRUDE O I L )

DRY M E A S U R E

1 QUART = 2 P I N T S 1 PECK = 8 QUARTS _ 1 BUSHEL = 4 PECKS

W E I G H T M E A S U R E M E N T ( M A S S )

A . A V O I R D U P O I S WEIGHT: - 1 OUNCE = 16 DRAMS 1 POUND = 1 6 OUNCES 1 HUNDREDWEIGHT = 1 0 0 POUNOS 1 TON = 2 0 0 0 POUNDS

B . TROY WEIGHT:

m 1 CARAT = 3 . 1 7 G R A I N S 1 PENNYWEIGHT = 2 0 G R A I N S 1 OUNCE = 2 0 PENNYWEIGHTS 1 POUND = 1 2 OUNCES 1 LONG HUNDREO- - WEIGHT = 1 1 2 POUNDS 1 LONG TON = 2 0 LONG HUNDREDWEIGHTS

= 2 2 4 0 POUNDS

C . A P O T H E C A R I E S WEIGHT:

1 SCRUPLE = 2 0 G R A I N S = 1 . 2 9 6 GRAMS 1 DRAM = 3 SCRUPLE = 3 . 8 8 8 GRAMS 1 OUNCE = 8 DRAMS = 3 1 . 1 0 3 5 GRAMS 1 POUND = 1 2 OUNCES = 3 1 3 . 2 4 2 0 GRAMS

D . K I T C H E N W E I G H T S AND MEASURES:

1 U . S . P I N T = 1 6 F L . OUNCES 1 STANDARD CUP = 8 F L . OUNCES 1 TABLESPOON = 0 . 5 F L . OUNCES ( 1 5 CU. CMS. ) 1 TEASPOON = 0 . 1 6 F L . OUNCES ( 5 CU. CMS. )

- X r - ? - METRlC SYSTEM

PREFIXES:

I\. WEGA = 1,000,000 I 0. K I L O = 1,000 C. HECTO = 100 D. DEKA = 10

E. DECI = 0.1 F. CENT1 = 0.01 G. M I L L 1 = 0.001 H. MICRO = 0.000001

UNEAR MEASURE:

i E U N I T 'c THE METER = 39.37 INCUFQ.

CENTIM 3 10 MILLIMETERS 011 I N . * 1 DECIME 10 CENTIMETERS = 3.9370113 INS . l-wf'am ' 10 DECIMETERS = 1.0936143 YDS.

1 =,3. 1 DEKAMETER = 10 METERS = 10.936143 YDS. 1 HECTOMETER = 10 DEKAMETERS = 109.36143 YDS. 1 KILOMETER = 10 HECTOMETERS = 0.62137 MILE

I 1 MYRIAMETER = 10.000 METERS

I SQUARE MEASURE:

/ THE UNIT I S THE SQUARE METER = 1549.9969 SQ. INCHES:

I 1 SQ. CENTIMETER = 100 SQ. MILLIMETERS = 0.1550 SQ. I N . 1 SQ. DECIMETER = 100 SQ. CENTIMETERS = 15.550 SQ. INS. 1 SQ. METER = 100 SQ. DECIMETERS = 10.7639 SQ. FT. 1 SQ. DEKAMETER = 100 SQ. METERS = 119.60 SQ. YDS 1 SQ. HECTOM€TER r 100 SQ. DEKAMETERS 1 SQ. KILOMETER = 100 SQ. HECTOMETERS

1 THE UNIT I S THE "ARE" = 100 SO. METERS: I ' 1 CENTIARE = 10 MILLIARES = 10.7643 SQ. FT.

1 DECIARE = 10 CENTIARES = 11.96033 SQ. YDS. 1 ARE = 10 DECIARES = 119.6033 SQ. YDS. 1 DEKARE = 10 ARES = 0.247110 ACRES 1 HEKTARE = 10 DEKARES = 2.471098 ACRES

(HECTO-ARE) 1 SQ. KILOMETER = 100 HEKTARES = 0.38611 SQ. M I L E

THE UNIT I S THE "STERE" = 61.025.38659 CU. INS.:

1 DECISTERE = 10 CENTISTERES = 3.531662 CU. FT. 1 STERE = 10 DECISTERES = 1.307986 CU. YDS. 1 DEKASTERE = 10 STERES = 13.07986 CU. YDS.

METRIC SYSTEM

CUBIC MEASURE:

THE UNIT I S THE "METER" = 39 .37 INS.:

1 CU. CENTIMETER= 1000 CU. MILLIMETERS = 0.06125 CU. I N . 1 CU. DECIMETER = 1000 CU. CENTIMETERS = 61.1250 CU. INS. 1 CU. METER = 1000 CU. DECIMETERS = 35 .3156 CU. FT.

= 1 STERE = 1 .30797 CU. YOS.

1 CU. CENTIMETER (WATER) = 1 GRAM 1000 CU CENTIMETERS (WATER) = 1 LITER = 1 KILOGRAM 1 CU. METER ( 1 0 0 0 LITERS) = 1 METRIC TON

MEASURES OF WEIGHT:

THE UNIT r S THE GRAM = 0.035274 OUNCES:

1 MILLIGRAM = 0.015432 GRAINS 1 CENTIGRAM = 1 0 MILLIGRAMS = 0.15432 GRAINS 1 DECIGRAM = 10 CENTIGRAMS = 1 .5432 GRAINS 1 GRAM = 1 0 DECIGRAMS = 15.4323 GRAINS * 1 DEKAGRAM 1 0 GRAMS = 5.6438 DRAMS 1 1 0 DEKAGRAMS =

5 7 1 0 HECTOGRAM! MYRIAGRA = 1 0 KILOGRAMS 22.046223 POUND

I QUINTAL = 1 0 MYRIAGRAMS = 1.986412 CWT. 1 METRIC TON = 1 0 QUINTAL = 2.204.622 POUNDS I

1 GRAM = 0.56438 DRAMS 1 DRAM = 1.77186 GRAMS

= 27.3438 GRAINS 1 METRIC TON = 2.204.6223 POUNDS

MEASURE OF CAPACITY:

THE UNIT I S THE "LITER" = 1.0567 LIOUID OUARTS:

1 CENTILITER = 1 0 MILL IL ITERS = 0.338 FLUID OUNCES 1 DECILITER = 1 0 CENTILITERS = 3 . 3 8 FLUID OUNCES 1 LITER = 1 0 DECILITERS = 33 .8 FLUID OUNCES 1 DEKALITER = 1 0 LITERS = 0.284 BUSHEL 1 HECTOLITER = 1 0 DEKALITERS = 2 .84 BUSHELS 1 KILOLITER = 1 0 HECTOLITERS = 264.2 GALLONS

= MILES 9 X 8 = K 1 W K T E R S 8 5

CONVERSION TABLES

ATMOSPHERES = 33.9 X F T . OF WATER 29.92 X I N S . OF MERCURY 14.7 X L B S . PER SQ. I N .

gTJ = 252 X C A L O R I E S (GRAM) 777.5 X F T . L B S . 0.0003927 X HORSEPOWER-HOURS 1054 X J O U L E S 0.0002928 X KILOWATT-HOURS

B T U ( P E R M I N . ) = l 2 96 X F T . L B S . PER SEC. 0 02356 0 05686

X HORSEPOWER X WATTS

C A L O R I E S = 0.003968 X B T U

DYNE = GRAMS , X CM/SEC/SEC

= 9.48 X lo-" X B T U 1.0 X DYNE C E N T I M E T E R S 7 . 3 7 x loe8 X F T . L B S .

FEET OF WATER = 0.02950 X ATMOSPHERES 0.8826 X I N S . OF MERCURY 0.4335 X FT. L B S . PER SQ. I N .

FOOT POUNDS PER SECOND = 0.07717 X E T U PER M I N .

0.001818 X HORSEPOWER 0.001356 X K I L O W A T T S

FOOT CANDLE = 10.765 X LUX

HORSEPOWER = 42.44 X BTU. PER M I N . 33,000 X F T . L B S . PER M I N . 550 X F T . L B S . PER SEC.

HCIRSFPCIWFR . , - . . - - - . . - HOURS = 2547 X B T U .

1.98 X lo6 X F T . L B S 2.68 X lo6 X J O U L E S

- . . . . . - 0.7376 x FT:-LBS. 0.000278 X WATT-HOURS 1.0 X WATT-SECONDS

KILOWATT-HOURS = 3415 X B T U PER M I N . 3.6 x 1013 x E R G S

L u x = 0.929 X F T . CANDLES

CONVERSION TABLES

BTU PER. M I N . = WATTS X 0 . 0 5 6 9 2 ERGS PER. SEC. = WATTS X 1 . 0 X 10' FT. LBS. PER M I N . = WATTS X 4 4 . 2 6 HORSEPOWER = WATTS X 0 . 0 0 1 3 4

1 GRAM CALORIE = 0 . 0 0 3 9 6 4 BTU 4 . 1 8 4 JOULES

1 GRAM FORCE = 9 8 0 . 6 DYNES

1 FOOT POUND = 1 LBS. FORCE X 1 FOOT 1 . 3 5 6 JOULES

1 POUND MASS = 4 5 3 . 6 GRAMS

1 NEWTON = 1 KILOGRAM X 1 METER/SEC/SEC 1 0 0 . 0 0 0 DYNES 0 . 2 2 4 LBS. FORCE

1 SLUG = 3 2 . 2 LBS MASS 1 4 . 6 0 6 KILOGRAMS

1 KILOWATT HOUR = 3 . 6 0 0 . 0 0 0 JOULES

1 WATT = 3 . 4 1 2 BTU/HRS. 0 . 2 3 9 GRAM CALORIE/SEC.

1 BTU RAISES ONE POUND OF WATER 1' F

1 GRAM CALORIE RAISES ONE GRAM OF WATER 1 " C

1 CIRCULAR M I L = 0 . 7 8 5 4 SQ. M I L

1 SOUARE M I L = 1 . 2 7 C I R . M I L S

1 M I L = 0 . 0 0 1 I N S .

T O DETERMINE THE CIRCULAR M I L OF A CONDUCTOR

1. ROUND CONDUCTOR CM = ( DIAMETER I N M I L S )'

2. BUS BAR cM = WIDTH ( M I L S ) X THICKNESS ( M I L S )

0 . 7 8 5 4

NOTES: 1. 1 MILLIMETER 3 9 . 3 7 M I L S

2 . 1 C I R . MILLIMETER = 1 5 5 0 C I R . M I L S

3 . 1 SQ. MILLIMETER = 1 9 7 4 C I R . M I L S

METALS

E L E C . COND. % COPPER

64.9 4.42 4.9 9.32 1.50

28.0 18.0 22.7 50.1 17.8

100.00 100.00 71.2

20.6 32.5

17.6

10

10

8.35 38.7 0.9 1.80

36.1 3.0 25.0 lo-'' 17.5 28 14.4 10

106

10

2.5 3.5 3.0 13.9

METAL

ALUMINUM ANTIMONY A R S E N I C B E R Y L L I U M B I S M U T H BRASS (70-30) BRONZE (5% S N ) CADMIUM C A L C I U M COBALT COPPER

ROLLED T U B I N G

GOLD G R A P H I T E I N D I U M I R I D I U M

I R O N

M A L L E A B L E

WROUGHT

L E A D MAGNESIUM MANGANESE MERCURY MOLYBDENUM MONEL (63-37) N I C K E L PHOSPHORUS P L A T I N U M POTASSIUM S E L E N I U M S I L I C O N S I L V E R

S T E E L (CARBON)

S T A I N L E S S (18-8) ( 1 3 - C R ) (18-CR)

TANTALUM

L B S . C U . "

.0978

.2390 ,2070 .0660 .3540 .3070 ,3200 ,3120 .0560 ,3210

.3210

.3230

.6970 ,0812 ,2640 .a090

,2600

,2600

.2780

,4120 .0628 ,2600 .4930 ,3680 ,3200 .3210 ,0657 ,7750 ,0310 .I740 .0866 .3790

,2830

.2860

.2810 ,2790 .5990

SPEC. GRAV.

2.710 6.620 5.730 1.830 9.800 8.510 8.870 8.650 1.550 8.900

8.890 8,950 19.30 2.25 7.30 22.40

7.20

7.20

7.70

11.40 1.74 7.20 13.65 10.20 8.87 8.90 1.82 21.46 0.860 4.81 2.40 10.50

7.84

7.92 7.78 7.73 16.6

SYMB.

A L S B A S B E B I

CD CA CO C U

AU

I N I R

F E

P B MG MN HG MO

N I P

P T K

SE S I AG

T A

MELT

c 0

660 630

- - - -

1280 271 900 1000 321 850 1495

1083 ....

1063 3500 156 2450 1200

TO 1400 1500

TO 1600 1500

TO 1600 327 651 1245 -38.9 2620 1300 1452 44.1 1773 62.3 220 1420 960 1330

TO 1380

1500 1520 1500 2900

F "

1220 1167 - - - -

2336 520 1652 1382 610 1562 2723

1981 ....

1945 6332 311 4442 2192

TO 2552 2732

TO 2912 2732

TO 2912 621 1204 2273 -37.7 4748 2372 2646 111.4 3221 144.1 428 2588 1760 2436

TO 2516

2732 2768 2732 5414

- METALS

- SPECIFIC RESISTANCE (K)

THE S P E C I F I C R E S I S T A N C E ( K ) OF A M A T E R I A L I S THE R E S I S T A N C E

M E T A L

T E L L U R I U M T H O R I U M T I N T I T A N I U M TUNGSTEN U R A N I U M V A N A D I U M Z I N C Z I R C O N I U M

OFFERED BY A WIRE OF THIS MATERIAL WHICH IS ONE FOOT LONG WITH A - D I A M E T E R OF ONE M I L .

SYMB.

T E T H S N T I W U V

Z N Z R

NOTE: 1. THE R E S I S T A N C E OF A W I R E I S D I R E C T L Y PROPORTIONAL TO THE S P E C I F I C R E S I S T A N C E OF T H E M A T E R I A L . - 2. " K " = S P E C I F I C R E S I S T A N C E

M A T E R I A L

B R A S S

CONSTANTAN

COPPER

GERMAN S I L V E R 18%

GOLD

I R O N ( P U R E )

MAGNESIUM

M A N G A N I N

S P E C . GRAV.

6.2 11.70 7.30 4.50 19.30 18.70 5.96 7.14 6.40

" K "

43.0

295

10.8

200

14.7

60.0

276

265

MELT PO1NT

M A T E R I A L

A L U M I N U M

MONEL

N ICHROME

N I C K E L

TANTALUM

T I N

TUNGSTEN

S I L V E R

E L E C . CONO. % COPPER

1 0 ~ ~ 9.10 15.00 2.10 31.50 2.80 6.63 29.10 4.20

c 0

450 1845 232 1800 3410 1130 1710 419 1700

" K "

17.0

253

600

947

93.3

69.0

34.0

9.7

L B S . CU:

,2240 .422 ,264 .I62 ,697 .675 ,215 .258 ,231

F " '

846 3353 449 3272 - - -

2066 3110 786 3092

L.

CENTIGRADE AND FAHRENHEIT THERMOMETER SCALES

1. TEMP. C" = 5 / 9 X ( T E M P . F ' - 3 2 )

2 . TEMP. F ' = ( 9 / 5 X TEMP. C o ) + 3 2

3 . A M B I E N T TEMPERAlURE I S THE TEMPERATURE OF THE SURROUNDING COOLING MEDIUM

4 . RATED TEMPERATURE R I S E I S THE P E R M I S S I B L E R I S E I N TEMPERATURE ABOVE A M B I E N T WHEN OPERATING UNDER LOAD.

R I G H T T R I A N G L E

" C " " A "

" B " h USEFUL MATH FORMULAS

OBTUSE T R I A N G L E

SOLVE A S TWO R I G H T T R I A N G L E S

SPHERE C Y L I N D R I C A L CONE

AREA = D'X 3 . 1 4 1 6 VOLUME = AREA O F VOLUME = AREA OF VOLUME = D3 x 0 . 5 2 3 6 END X H E I G H T END X H E I G H T / 3

Q -- - f l --- " A "

"W"

E L L I P T I C A L

SOLVE THE SAME AS C Y L I N D R I C A L M = A X B X C

THE CIRCLE

D E F I N I T I O N : A CLOSED PLANE CURVE H A V I N G EVERY P O I N T AN EQUAL D I S T A N C E FROM A F I X E D P O I N T W I T H I N THE CURVE.

C IRCUMFERENCE: THE D I S T A N C E AROUND A C I R C L E . D I A M E T E R : THE D I S T A N C E ACROSS A C I R C L E THROUGH THE CENTER R A D I U S : THE D I S T A N C E FROM THE CENTER TO THE EDGE OF A

C I R C L E . ARC : A PART OF THE CIRCUMFERENCE. CHORD : A S T R A I G H T L I N E CONNECTING THE ENDS OF AN ARC. SEGMENT : AN AREA BOUNDED BY AN ARC A N 0 A CHORD. SECTOR : A PART OF C I R C L E ENCLOSE0 BY TWO R A D I I AND THE

ARC WHICH THEY CUT O F F .

CIRCUMFERENCE OF A C I R C L E = 3 . 1 4 1 6 X 2 X R A D I U S AREA OF A C I R C L E = 3 . 1 4 1 6 X R A D I U S X R A D I U S ARC LENGTH = DEGREES I N ARC X R A O I U S X 0 . 0 1 7 4 5 R A D I U S LENGTH = ONE H A L F LENGTH OF D I A M E T E R SECTOR AREA = ONE H A L F LENGTH OF ARC X R A D I U S

CHORD LENGTH = 2 1

SEGMENT AREA = SECTOR AREA M I N U S T R I A N G L E AREA.

m: 3 . 1 4 1 6 X 2 X R = 3 6 0 DEGREES

v w 0 . 0 0 8 7 2 6 6 X 2 X R OR

0 . 0 1 7 4 5 X R = 1 DEGREE

T H I S G I V E S U S THE ARC FORMULA

DEGREES X R A D I U S X 0 . 0 1 7 4 5 = DEVELOPEO LENGTH

EXAMPLE:

FOR A N I N E T Y DEGREE C O N D U I T BEND, H A V I N G A R A D I U S OF 1 7 . 2 5 " :

9 0 X 1 7 . 2 5 " X 0 . 0 1 7 4 5 = DEVELOPED LENGTH

2 7 " = DEVELOPED LENGTH

FRACTIONS

.I DEFINITIONS:

A . A F R A C T I O N I S A Q U A N T I T Y L E S S THAN A U N I T

8 . A NUMERATOR I S THE TERM OF A F R A C T I O N I N D I C A T I N G HOW MANY OF THE P A R T S OF A U N I T ARE TO B E T A K E N . I N A COMMON

a F R A C T I O N I T A P P E A R S ABOVE OR TO THE L E F T OF THE L I N E .

C . A OENOMINATOR I S THE TERM OF A F R A C T I O N I N D I C A T I N G THE NUMBER OF E Q U A L P A R T S I N T O WHICH THE U N I T I S O I V I D E U . I N A COMMON F R A C T I O N I T APPEARS BELOW OR TO THE R I G H T OF THE L I N E .

1 1 + NUMERATOR D . E X A M P L E S : ( 1 . ) - = = F R A C T I O N

2 + DENOMINATOR

( 2 . ) NUMERATOR + 1 / 2 + OENOMINATOR

TO ADD OR SUBTRACT:

T O S O L V E 112 - 2 / 3 + 3 / 4 - 5 / 6 + 7 / 1 2 = ?

A . D E T E R M I N E THE LOWEST COMMON DENOMINATOR T H A T E A C H OF THE DENOMINATORS 2 . 3 . 4 . 6 . AND 12 W I L L D I V I D E I N T O A N EVEN

.I NUMBER OF T I M E S .

THE LOWEST COMMON DENOMINATOR I S 12

B . WORK ONE F R A C T I O N A T A T I M E U S I N G THE FORMULA - T I M E S NUMERATOR OF F R A C T I O N DENOMINATOR OF F R A C T I O N

( 1 . ) 1 2 / 2 X 1 = 6 X 1 = 6 112 BECOMES 6 / 1 2

( 2 . ) 1 2 / 3 X 2 = 4 X 2 = 8 2 /3 BECOMES 8 / 1 2

( 3 . ) 1 2 / 4 X 3 = 3 X 3 = 9 3 / 4 BECOMES 9 / 1 2

( 4 . ) 1 2 / 6 X 5 = 2 X 5 = 10 5 / 6 BECOMES 10112

( 5 . ) 7 / 1 2 R E M A I N S 7 / 1 2

C O N T I N U E D N E X T PAGE

FRACTIONS

T O ADD O R S U B T R A C T (CONTINUED) :

C . WE CAN NOW CONVERT THE PROBLEM FROM I T S O R I G I N A L FORM TO I T S NEW FORM U S I N G 1 2 AS THE COMMON DENOMINATOR.

1 1 2 - 2 / 3 + 3 / 4 - 5 / 6 + 7 / 1 2 = O R I G I N A L F O R M

+ + = PRESENT FORM - 1 2

+ 2 2 - l8 = = 1 REDUCED TO LOWEST FORM

1 2 1 2 3

D . TO CONVERT F R A C T I O N S TO D E C I M A L FORM S I M P L Y D I V I D E THE I

NUMERATOR OF THE F R A C T I O N BY THE DENOMINATOR OF THE F R A C T I O N .

EXAMPLE: 1 D I V I D E D BY 3 = 0 . 3 3 = A N S .

T O MULTIPLY: - A. THE NUMERATOR OF F R A C T I O N # 1 T I M E S THE NUMERATOR OF

F R A C T I O N # 2 I S EQUAL TO THE NUMERATOR OF THE PRODUCT

8 . THE OENOMINATOR OF F R A C T I O N # 1 T I M E S THE DENOMINATOR OF F R A C T I O N # 2 I S EQUAL TO THE OENOMINATOR OF THE PRODUCT. -

C. EXAMPLE:

F R A C T I O N # 1 F R A C T I O N # 2 PRODUCT

1 NUMERATORS

1-'1 - 2 1 - = - 2-3

DENOMINATORS

TO CHANGE 1 / 3 TO D E C I M A L FORM. D I V I D E 1 B Y 3 = 0.33

FRACTIONS

TO DIVIDE: 1 -

A . THE NUMERATOR OF F R A C T I O N #1 T I M E S THE OENOMINATOR OF F R A C T I O N # 2 I S EQUAL TO THE NUMERATOR OF THE Q U O T I E N T .

8 . THE DENOMINATOR OF F R A C T I O N # I T I M E S THE NUMERATOR OF - F R A C T I O N # 2 I S EQUAL TO THE OENOMINATOR OF THE Q U O T I E N T

C . E X A M P L E :

F R A C T I O N #1 F R A C T I O N # 2 Q U O T I E N T

L- - 3

DENOMINATORS

TO CHANGE 3 / 4 TO D E C I M A L FORM. D I V I D E 3 B Y 4 = 0.75

EQUATIONS

T H E W O R D E Q U A T I O N M E A N S E Q U A L O R T H E S A M E AS.

EXAMPLE: 2 X 10 = 4 X 5 20 = 2 0

w: A . THE SAME NUMBER MAY B E ADOEO TO BOTH S I D E S OF A N E Q U A T I O N

WITHOUT CHANGING I T S V A L U E S .

E X A M P L E : ( 2 X 1 0 ) + 3 = ( 4 X 5 ) + 3 23 = 2 3

B . THE SAME NUMBER MAY B E SUBTRACTED FROM B O T H S I D E S OF AN -

E Q U A T I O N WITHOUT C H A N G I N G I T S V A L U E S .

EXAMPLE: ( 2 X 1 0 ) - 3 = ( 4 X 5 ) - 3 1 7 = 1 7

C . BOTH S I D E S OF AN E Q U A T I O N MAY BE D I V I D E D B Y THE SAME NUMBER WITHOUT CHANGING I r s V A L U E S .

2 x 1 0 4 x 5 EXAMPLE: - = -

20 2 0

0 . BOTH S I D E S OF AN E Q U A T I O N MAY B E M U L T I P L I E D BY THE SAME NUMBER WITHOUT CHANGING I T S V A L U E S .

EXAMPLE: 3 X ( 2 X 1 0 ) = 3 X ( 4 X 5 ) 60 = 6 0

E . T R A N S P O S I T I O N :

THE PROCESS OF MOVING A Q U A N T I T Y FROM ONE S I D E OF A N E Q U A T I O N TO THE OTHER S I D E OF AN E Q U A T I O N BY C H A N G I N G I T S - S I G N OF O P E R A T I O N I S TRANSPOSING.

1. A TERM MAY B E TRANSPOSED I F I T S S I G N I S C; iANGED FROM PLUS ( + ) TO M I N U S ( - ) . OR FROM M I N U S ( - ) TO P L U S ( + ) .

EXAMPLES

EQUATIONS

w: 1

E . T R A N S P O S I T I O N :

2 . A M U L T I P L I E R MAY BE REMOVED FROM ONE S I D E OF AN E Q U A T I O N BY M A K I N G I T A D I V I S O R I N THE OTHER. OR A D I V I S O R MAY BE REMOVED FROM ONE S I D E OF AN E Q U A T I O N BY M A K I N G I T A M U L T I P L I E R I N THE OTHER.

EXAMPLE: M U L T I P L I E R FROM ONE S I D E OF E Q U A T I O N BECOMES D I V I S O R I N OTHER S I D E OF THE E Q U A T I O N .

4 0 E X A M P L E : 4X = 4 0 BECOMES X = -

4

D I V I S O R FROM ONE S I D E OF M U L T I P L I E R I N OTHER S I D E

E Q U A T I O N BECOMES OF THE E Q U A T I O N .

1 EXAMPLE: = 1 0 BECOMES X = 4 X 1 0 4

w: A. A D D I T I O N :

1 1. RULE: USE THE S I G N OF THE LARGER AND ADD

EXAMPLES: + 3 - 2 + 3 3 - 2 - 9 2 -

8 . S U B T R A C T I O N :

RULE: CHANGE THE S I G N OF A S I N A D D I T I O N :

THE SUBTRAHEND AND PROCEED

EXAMPLES: + 3 - 2 + 3 - 3 - 2 + 3 + 2 <

CHANGE SUBTRAHEND AND ADO

EQUATIONS

3. (CONTINUED):

C . M U L T I P L I C A T I O N :

1. THE PRODUCT OF ANY TWO NUMBERS H A V I N G L I K E S I G N S I S P O S I T I V E . THE PRODUCT OF ANY TWO NUMBERS H A V I N G U N L I K E S I G N S I S N E G A T I V E .

EXAMPLE: ( + 3 ) X ( - 2 ) = - 6 ( - 3 ) X ( + 2 ) = - 6

D . D I V I S I O N : - 1. I F THE D I V I S O R A N 0 D I V I D E N D HAVE L I K E S I G N S , THE S I G N

OF THE Q U O T I E N T I S P O S I T I V E . I F THE D I V I S O R AND D I V I D E N D HAVE U N L I K E S I G N S . THE S I G N OF THE Q U O T I E N T I S N E G A T I V E .

EXAMPLE:

+ 6 + 6 - 6 - 6 - = - 3 - + 3 - = - 3 - + 3

- 2 + 2 + 2 - 2

SQUARE ROOT

1. GROUPING THE D I G I T S I N A NUMBER I S E S S E N T I A L I N S O L V I N G SQUARE ROOT PROBLEMS. START AT THE D E C I M A L P O I N T . AND GROUP TWO TO A GROUP TO THE L E F T . I F THERE I S A D I G I T L E F T OVER AT THE EXTREME L E F T THAT D I G I T W I L L B E C O N S I D E R E D TO BE A GROUP. START A G A I N A T THE D E C I M A L P O I N T AND GROUP TWO TO A GROUP TO THE R I G H T . I F THERE I S A D I G I T L E F T OVER A T THE EXTREME R I G H T , S I M P L Y ADD A " 0 " . WHICH W I L L NOT CHANGE THE VALUE OF THE NUMBER.

EXAMPLES: 2 3 4 . 5 6 7 GROUP AS 2 3 4 . 5 6 7 0 2 3 4 5 . 6 7 GROUP AS 2 3 4 5 . 6 7

2 . THE CONSTANT I N S O L V I N G A L L SQUARE ROOT PROBLEMS W I L L BE NUMBER "20".

3 . S O L U T I O N : 1 5 . 3 1

y m 2 0 x 1 = 2 0 1

+ 5 = 2 5 ( 1 3 4

2 0 X 1 5 3 = 3 0 6 0 - 9 0 9 + 1 = 3 0 6 1 4 7 7 0

- 3 0 6 1

( S T E P 1 . )

THE LARGEST NUMBER THAT W I L L s Q Y A R E I N T O THE F I R S T GROUP I S

1

( S T E P 2 . )

M U L T I P L Y 5 X 2 5 TO = 1 2 5 SUBTRACT 1 2 5 FROM 1 3 4 TO = 9 B R I N G DOWN T H I R D GROUP 5 6 AND OUR REMAINDER = 9 5 6

CONSTANT 2 0 X TOTAL ANSWER 1 5 I S 3 0 0 .

D I V I D E 9 5 6 RY 3 0 0 = 3 A D D - 3 0 0 ~ ~ ~ ~ - 3 i~ 3 0 3 3 0 3 W I L L D I V I D E I N T O 9 5 6 THREE T I M E S . SO 3 BECOMES THE T H I R D DIGIT.IN OUR ANSWER

( S T E P 3 . )

S u i ~ R n c i 9 0 9 - ; ~ 0 M 9 5 6 TO EQUAL 4 7 . B R I N G DOWN L A S T GROUP 7 0 TO E S T A B L I S H A REMAINDER OF 4 7 7 0 .

( S T E P 4 . ) - . CONSTANT 2 0 X TOTAL ANSWER

SUBTRACT "1" FROM "Z" * ,AND B R I N G 1 5 3 = 3 0 6 0 . DOWN SECOND GROUP " 3 4 D I V I D E 4 7 7 0 BY 3 0 6 0 = 1

ADD 3 0 6 0 AND 1 = 3 0 6 1 CONSTANT 2 0 X ANSWER 1 = 2 0 3 0 6 1 W I L L D I V I D E I N T O 4 7 7 0 ONE D I V I D E 1 3 4 BY 2 0 = 6 T I M E . 1 BECOMES THE FOURTH D I G I T ADD 2 0 AND 6 = 2 6 I N OUR ANSWER.

NOTE: 2 6 W I L L NOT D I V I D E I N T O NOTE: M U L T I P L Y 1 5 . 3 1 X 1 5 . 3 1 1 3 4 S I X T I M E S SO THEREFORE 2 5 = 2 3 4 . 3 9 6 . ADD 2 3 4 . 3 9 6 MUST BE USED. 2 5 W I L L D I V I D E AND 0 . 1 7 0 9 ( T H E REMAINDER INTO 1 3 4 F I V E T I M E S . NOW OUR OF THE P R O B L E M ) TO EQUAL ANSWER I S 5 AND WE ARE READY FOR 2 3 4 . 5 6 7 . T H I S ENABLES STEP 2 . YOU TO CHECK FOR

ACCURACY.

NATURAL TRIGONOMETRIC FUNCTIONS

1 4

1 5 1 6 1 7 1 8 1 9

2 0 2 1 2 2 2 3 2 4

2 5 2 6 2 7 2 8 2 9

3 0 3 1 3 2 3 3 3 4

3 5 3 6 3 7 3 8 3 9

4 0 4 1 4 2 4 3 4 4

4 5

ANGLE

. 2 4 1 9

, 2 5 8 8 , 2 7 5 6 . 2 9 2 4 . 3 0 9 0 , 3 2 5 6

. 3 4 2 0 , 3 5 8 4 , 3 7 4 6 , 3 9 0 7 . 4 0 6 7

. 4 2 2 6 , 4 3 8 4 . 4 5 4 0 , 4 6 9 5 . 4 8 4 8

, 5 0 0 0 , 5 1 5 0 . 5 2 9 9 , 5 4 4 6 . 5 5 9 2

. 5 7 3 6

. 5 8 7 8 , 6 0 1 8 , 6 1 5 7 . 6 2 9 3

6 4 2 8 . I3561 , 6 6 9 1 . 6 8 2 0 . 6 9 4 7

. 7 0 7 1

C O S I N E

, 9 7 0 3

9 6 5 9 . g e l 3 , 9 5 6 3

9 5 1 1 9 4 5 5

9 3 9 7 . 9 3 3 6 , 9 2 7 2

9 2 0 5 , 9 1 3 5

, 9 0 6 3 . a 9 8 8 . 8 9 1 0

8 8 7 9 , 8 7 4 6

8 6 6 0 0 5 7 2

, 8 4 8 0 . a 3 8 7 .El290

, 8 1 9 2 , 8 0 9 0 . 7 9 8 6 . 7 8 8 0 . 7 7 7 1

. 7 6 6 0 , 7 5 4 7 , 7 4 3 1 . 7 3 1 4 . 7 1 9 3

, 7 0 7 1

S I N E

. 2 4 9 3

, 2 6 7 9 2 8 6 7 3 0 5 7

. 3 2 4 9 3 4 4 3

, 3 6 4 0 3 8 3 9

, 4 0 4 0 . 4 2 4 5 . 4 4 5 2

4 6 6 3 4 8 7 7 5 0 9 5 5 3 1 7 5 5 4 3

5 7 7 4 , 6 0 0 9

6 2 4 9 6 4 9 4 6 7 4 5

7 0 0 2 7 2 6 5 7 5 3 6 7 8 1 3 8 0 9 8

8 3 9 1 8 6 9 3 9 0 0 4 9 3 2 5 9 6 5 7

1 0 0 0 0

C O T A N .

4 0 1 0 8

3 . 7 3 2 1 3 4 8 7 4 3 . 2 7 0 9 3 . 0 7 7 7 2 . 9 0 4 2

2 . 7 4 7 5 2 . 6 0 5 1 2 . 4 7 5 1 2 . 3 5 5 9 2 . 2 4 6 0

2 . 1 4 4 5 2 . 0 5 0 3 1 9 6 2 6 1 8 8 0 7 1 8 0 4 0

1 7 3 2 1 1 . 6 6 4 3 1 6 0 0 3 1 5 3 9 9 1 . 4 8 2 6

1 4 2 8 1 1 3 7 6 4 1 . 3 2 7 0 1 . 2 7 9 9 1 2 3 4 9

1 1 9 1 8 1 1 5 0 4 1 . 1 1 0 6 1 0 7 2 4 1 . 0 3 5 5

1 . 0 0 0 0

TANGENT

1 0 3 0 6

1 0 3 5 3 1 0 4 0 3 1 0 4 5 7 1 . 0 5 1 5 1 0 5 7 6

1 0 6 4 2 1 0 7 1 1 1 . 0 7 8 5 1 . 0 8 6 4 1 . 0 9 4 6

1 . 1 0 3 4 1 1 1 2 6 1 . 1 2 2 3 1 . 1 3 2 6 1 . 1 4 3 4

1 . 1 5 4 7 1 . 1 6 6 6 1 . 1 7 9 2 1 . 1 9 2 4 1 2 0 6 2

1 2 2 0 8 1 . 2 3 6 1 1 2 5 2 1 1 2 6 9 0 1 . 2 8 6 8

1 3 0 5 4 1 . 3 2 5 0 1 . 3 4 5 6 1 . 3 6 7 3 1 . 3 9 0 2

1 4 1 4 2

COSECANT

4 . 1 3 3 6

3 . 8 6 3 7 3 . 6 2 8 0 3 4 2 0 3 3 2 3 6 1 3 . 0 7 1 6

2 9 2 3 8 2 . 7 9 0 4 2 6 6 9 5 2 5 5 9 3 2 4 5 8 6

2 3 6 6 2 2 2 8 1 2 2 . 2 0 2 7 2 1 3 0 1 2 . 0 6 2 7

2 . 0 0 0 0 1 9 4 1 6 1 . 8 8 7 1 1 . 8 3 6 1 1 7 8 8 3

1 . 7 4 3 4 1 7 0 1 3 1 . 6 6 1 6 1 . 6 2 4 3 1 . 5 8 9 0

1 . 5 5 5 7 1 . 5 2 4 3 1 . 4 9 4 5 1 . 4 6 6 3 1 4 3 9 6

1 4 1 4 2

SECANT

7 6

7 5 7 4 7 3 7 2 7 1

7 0 6 9 6 8 6 7 6 6

6 5 6 4 6 3 6 2 6 1

6 0 5 9 5 8 5 7 5 6

5 5 5 4 5 3 5 2 5 1

5 0 4 9 4 8 4 7 4 6

4 5

ANGLE

TRIGONOMETRY

TRIGONOMETRY I S T H E M A T H E M A T I C S D E A L I N G W I T H THE R E L A T I O N S OF S I D E S AND A N G L E S OF T R I A N G L E S .

A T R I A N G L E I S A F I G U R E ENCLOSED BY THREE S T R A I G H T S I D E S . THE SUM OF T H E THREE A N G L E S I S 1 8 0 DEGREES. A L L T R I A N G L E S H A V E S I X P A R T S : THREE A N G L E S , A N 0 THREE S I D E S O P P O S I T E THE A N G L E S .

R I G H T T R I A N G L E S ARE T R I A N G L E S T H A T HAVE ONE ANGLE OF N I N E T Y DEGREES AND TWO ANGLES OF L E S S T H A N N I N E T Y DEGREES.

TO H E L P YOU REMEMBER T H E S I X T R I G O N O M E T R I C F U N C T I O N S : M E M O R I Z E

"OH HELL ANOTHER HOUR OF ANDY"

( O H ) O P P O S I T E S I D E

SINE U = HYPOTENUSE ( H E L L )

( A N O T H E R ) ADJACENT 'S IDE

C O S I N E o = HYPOTENUSE ( H O U R )

TANGENT 0 = A D J A C E N T S I D E ( A N D Y )

NOW U S E BACKWARDS -

"ANDY OF HOUR ANOTHER HELL OH"

( A N D Y ) A D J A C E N T S I D E

COTANGENT U = ALWAYS P L A C E THE ANGLE TO

O P P O S I T E S I D E B E S O L V E D A T T H E VERTEX

( O F ) (WHERE " X " AND " Y " CROSS)

( H O U R ) HYPOTENUSE

SECANT n = A D J A C E N T S I D E ( A N O T H E R )

( H E L L ) HYPOTENUSE

COSECANT 0 = O P P O S I T E S I D E ( O H )

NOTE: o = T H E T A = ANY ANGLE

I

BENDING OFF-SETS WITH TRIGONOMETRY

" C " " 0 " / 4 2 " -

S I D E O P P O S I T E - S I D E ADJACENT " E "

26" -

THE COSECANT OF THE ANGLE T I M E S THE O F F - S E T D E S I R E D I S EQUAL TO THE O I S T A N C E BETWEEN THE CENTERS OF THE BENDS.

EXAMPLE: TO MAKE A F I F T E E N I N C H ( 1 5 " ) O F F - S E T : U S I N G T H I R T Y ( 3 0 ) DEGREE BENDS: - 1. USE T R I G . T A B L E (PAGE 5 4 ) TO F I N D THE COSECANT OF A

T H I R T Y ( 3 0 ) OGREE ANGLE. WE F I N O I T TO B E TWO ( 2 ) . 2 . M U L T I P L Y TWO ( 2 ) T I M E S THE O F F - S E T D E S I R E D . WHICH

I S F I F T E E N ( 1 5 ) I N C H E S TO D E T E R M I N E THE D I S T A N C E BETWEEN B E N 0 " 8 " AND BEND " C " . THE ANSWER I S I

T H I R T Y ( 3 0 ) I N C H E S .

TO MARK THE C O N D U I T FOR BENDING:

1. MEASURE FROM E N 0 OF CONDUIT " A " T H I R T Y - F O U R ( 3 4 ) I N C H E S TO CENTER OF F I R S T BEND " 8 " . A N 0 MARK. -

2 . MEASURE FROM MARK " 0 " T H I R T Y ( 3 0 ) I N C H E S TO CENTER OF SECOND BEND " C " . AND MARK.

3 . MEASURE FROM MARK " C " FORTY-TWO ( 4 2 ) I N C H E S TO " 0 " . A N 0 MARK. CUT. REAM. AN0 THREAD C O N D U I T BEFORE B E N D I N G .

R O L L I N G O F F - S E T S : TO D E T E R M I N E HOW MUCH O F F - S E T I S NEEDED TO MAKE A R O L L I N G O F F - S E T :

1. MEASURE V E R T I C A L REQUIRED. USE WORK T A B L E ( A N Y SQUARE W I L L DO) AND MEASURE FROM CORNER T H I S AMOUNT AND MARK. -

2 . MEASURE H O R I Z O N T A L REQUIRED. MEASURE N I N E T Y DEGREES FROM THE V E R T I C A L L I N E MEASUREMENT ( S T A R T I N G I N SAME CORNER) AND MARK.

3 . THE D I A G O N A L O I S T A N C E BETWEEN THESE MARKS W I L L BE THE AMOUNT OF O F F - S E T REQUIREO. -

NOTE: S H R I N K I S HYPOTENUSE MTNl lS THE S I D E A D J A C E N T . -

-54-

CHICAGO-TYPE BENDERS

NINETY DEGREE BENDING

" A M TO u C - = STUB-UP " C " TO " D " = T A I L

M C M = BACK OF STUB-UP "C" = BOTTOM OF C O N D U I T

NOTE THERE ARE MANY VAR- I A T I O N S OF T H I S TYPE BENDER. BUT MOST MAN- UFACTURERS OFFER TWO S I Z E S .

THE SMALL S I Z E SHOE TAKES 1 1 2 " . 3 / 4 . AND 1" C O N D U I T .

THE LARGE S I Z E SHOE TAKES 1 - 1 1 4 " AND 1 - 1 / 2 " C O N D U I T .

TO D E T E R M I N E " T A K E - U P " AND * S H R I N K ' ' OF EACH S I Z E CONDUIT FOR A P A R T I C U L A R BENDER TO MAKE N I N E T Y DEGREE BENDS.

1. USE A S T R A I G H T P I E C E OF SCRAP C O N D U I T

2 . MEASURE EXACT LENGTH OF SCRAP C O N D U I T . " A " TO " D " . ,,"" ,. ', ,, b 1P1

O R I G I N A L MEASUREMENT

3 . PLACE C O N D U I T I N BENDER. MARK AT EDGE OF SHOE. " 0 " .

L E V E L C O N D U I T . BEND N I N E T Y , AND COUNT NUMBER OF PUMPS. AND KEEP NOTES ON EACH S I Z E C O N D U I T USED.

MAKE

5 . AFTER B E N D I N G N I N E T Y :

A . D I S T A N C E BETWEEN " 0 " AND " C " I S THE T A K E - U P

B . O R I G I N A L MEASUREMENT OF THE SCRAP P I E C E OF C O N D U I T SUBTRACTED FROM ( D I S T A N C E " A " TO " C " P L U S D I S T A N C E "C" TO " D m ) I S THE S H R I N K .

NOTE: BOTH T I M E AND ENERGY W I L L B E SAVED I F C O N D U I T CAN B E C U T , REAMED. AND THREADED BEFORE B E N D I N G .

THE SAME METHOD CAN BE USED ON H Y D R A U L I C BENDERS.

-55-

CHICAGO-TYPE BENDER

OFF-SETS

S T R A I G H I - t O G E

P O I N l " A "

U

C H I C A G O TYPE BENDER

EXAMPLE: TO BEND A 6 " OFF-SET:

1. MAKE A MARK 3" FROM C O N D U I T END P L A C E C O N l l U I T I N BENDER W I T H MARK A T O U T S I D E EDGE OF JAW. -

2 . MAKE THREE F U L L PUMPS. M A K I N G S U R t H A N D L E GOES A L L T H E WAY DOWN TO THE S T O P .

3 . REMOVE C O N D U I T FROM BENDER AND PLACE ALONG S I D E S T R A I G H T - EDGE. -

4 . MEASURE 6 " FROM S T R A I G H T - E D G E TO CENTER OF C O N D U I T . MARK P O l N T " 0 " . USE SQUARE FOR ACCURACY.

5 . MARK C t N T E R OF C O N D U I T FROM B O I H U I R E C I I O N S THROUGH BEND A S I L L U S T R A T E D BY BROKEN L I N E . WHERE L I N E S CROSS I S P O I N T " 8 " . ,

6. MEASURE FROM " A " TO " 8 " TO D E T E R M I N E D I S T A N C E FROM " D m TO " C " . MARK "C" AND P L A C E C O N D U I T I N BENDER W I T H MARK A T O U T S I D E EDGE OF J A W . AND W I T H THE K I C K P O I N T I N G DOWN. USE L E V E L TO PREVENT DOGGING C O N C U I T .

7 . REPEAT S T E P " 2 "

NOTE: 1. THERE ARE S F V E R A L METHODS OF B E N D I N G R I G I D C O N D U I T W I T H A C H I C A G O TYPE BENDER. A N 3 ANY METHOD T H A T GETS THE J O B DONE I N A M l N I M U M AMOUNT OF T I M E W I T H C R A F T S M A N S H I P I S GOOD.

2 . WHATEVER METHOD U S E D . Q U A L I T Y W I L L IMPROVE W I T H E X P E R I E N C E .

MULTI-SHOT NINETY DEGREE CONDUIT BENDING

PROBLEM: - A. TO MEASURE. THREAD. C U T . AND REAM CONOUIT BEFORE B E N D I N G B . TO ACCURATELY BEND C O N D U I T TO THE D E S I R E 0 H E I G H T OF THE

STUB-UP ( H ) , AND TO THE D E S I R E D LENGTH OF THE T A I L ( L ) .

I A. S I Z E OF C O N D U I T . 2 " 0 . SPACE BETWEEN C O N D U I T (CENTER TO C E N T E R ) . 6 " C . H E I G H T OF S T U B - U P . 36 " 0 . LENGTH OF T A I L . 48"

S O L U T I O N : u -

A . TO D E T E R M I N E R A D I U S : ( R )

CONDUIT # 1 ( I N S I D E C O N O U I T ) W I L L USE THE M I N I M U M R A D I U S UNLESS OTHERWISE S P E C I F I E D . THE M I N I M U M R A D I U S I S E I G H T T I M E S THE S I Z E OF THE C O N D U I T . P L U S O N E - H A L F THE O U T S I D E D I A M E T E R OF THC C O N D U I T . ( S E E PAGE 5 9 )

R A D I U S OF C O N D U I T # 1 = 8 X 2 " + 1 . 2 5 " = 1 7 . 2 5 "

R A D I U S OF C O N D U I T # 2 : R A D I U S # 1 + 6" = 2 3 . 2 5 "

R A D I U S OF C O N D U I T # 3 = R A D I U S # 2 + 6 " = 2 9 . 2 5 "

B . TO D E T E R M I N E DEVELOPED LENGTH: ( D L ) R A D I U S X 1 . 5 7 = D L

D L OF C O N D U I T # 1 = R X 1 . 5 7 = 1 7 . 2 5 ' ' X 1 . 5 7 = 2 7 "

DL OF CONOUIT # 2 = R X 1 .57 = 23 .25 " X 1 . 5 7 = 3 6 . 5 "

D L OF C O N D U I T # 3 = R X 1 . 5 7 = 2 9 . 2 5 " X 1 . 5 7 = 46 "

C . TO D E T E R M I N E LENGTH OF N I P P L E : ( S E E PAGE 6 1 )

LENGTH OF N I P P L E , C O N D U I T # 1 = L + H + D L - 2R = 48 " + 3 6 " + ? 7 " - 34.,!" = 7 6 . 5 " = 6 - 4 . 5

LENGTH OF N I P P L E , C O N D U I T # 2 = L + H + D L - 2 R = 54 " + 4 2 " + 3 6 . 5 " - 4 6 . 5 " = 86 " = 7 ' . 2 "

LENGTH OF N I P P L E . C O N D U I T # 3 = L t H + D L - 2: I = 60 + 48 + 46 - 5 8 . 5 "

= 9 5 . 5 - = 7 ' - 1 1 . 5 "

NOTE: 1 . FOR 9 0 DEGREE BENDS. S H R I N K = 2R - D L

2 . FOR O F F - S E T BCNDS. S H R I N K = HYPOTENUSE - S I D E

1 ADJACENT.

MULTI-SHOT NINETY DEGREE CONDUIT BENDING

LAYOUT AND B E N D I N G :

A . TO LOCATE P O I N T " 8 " . MEASURE FROM P O I N T " A " . THE LENGTH OF THE STUB-UP MINUS THE RADIUS. ON A L L THREE.CONDUIT . POINT " B " W I L L B E 1 8 . 7 5 " FROM P O I N T " A " . (PAGE 5 9 )

B . TO LOCATE P O I N T " C " . MEASURE FORM P O I N T " 0 " THE LENGTH M I N U S THE R A D I U S . ( R E F E R PAGE 6 1 ) . ON A L L ~ H R E E C O N D U I T . P O I N T " C " W I L L BE 3 0 . 7 5 " FROM P O I N T " D M . (PAGE 5 9 )

C . D I V I D E THE DEVELOPED LENGTH ( P O I N T " B " TO P O I N T " C " ) I N T O EQUAL SPACES. SPACES SHOULQ NOT BE MORE THAN 1 . 7 5 " TO PREVENT W R I N K L I N G OF THE C O N D U I T . ON C O N D U I T # 1. SEVENTEEN - SPACES OF 1 . 5 8 8 2 " EACH. WOULD G I V E U S E I G H T E E N SHOTS OF 5 DEGREES EACH. REMEMBER THERE I S ALWAYS ONE L E S S SPACE THAN SHOT. WHEN D E T E R M I N I N G THE NUMBER OF SHOTS. CHOOSE A NUMBER THAT W I L L D I V I D E I N T O N I N E T Y AN EVEN NUMBER OF T I M E S .

0 . I F AN E L A S T I C NUMBERED TAPE I S NOT A V A I L A B L E TRY THE METHOD I L L U S T R A T E D .

- A TO B = C O N D U I T # 1. DEVELOPED LENGTH = 2 7 "

C 5 T A B L E OR PLYWOOD CORNER - MEASURE FROM P O I N T " C " ( T A B L E CORNER) 1 7 I N C H E S ALONG TABLE EDGE TO P O I N T " A " AND MARK. PLACE END OF RULE AT P O I N T " A " . P O I N T " 0 " W I L L BE LOCATED WHERE 2 7 " MARK MEETS TABLE EDGE B - C . MARK ON BOARD THEN TRANSFER TO C O N D U I T .

I

MULTI-SHOT NINETY DEGREE CONDUIT BENDING

L + H + D L - 2 R = N I P P L E

1 . 5 7 X R = DL

H - R = " 8 "

L . R = " C "

" D " I 3 0 . 7 5 "

rill I

1 I I

TO LOCATE POINT " 0 "

H # 1 - RADIUS # 1 = " 8 " 36" - 1 7 . 2 5 " = "8"

1 8 . 7 5 " = " 8 "

I H # 2 - RADIUS # 2 = "8 " 4 2 " - 2 3 . 2 5 " = " 0 "

1 8 . 7 5 " = " 8 "

H # 3 - RADIUS # 3 = "8"

TO LOCATE P O I N T "C"

L # 1 - R A D I U S #1 = " C " 4 8 " - 1 7 . 2 5 " = " C "

3 0 . 7 5 " = " C "

L # 2 - RADIUS # 2 = " C " 5 4 " - 2 3 . 2 5 " = " C "

3 0 . 7 5 " = " C "

L # 3 - RADIUS # 3 = " C " 6 0 - - 2 9 . 2 5 - = - c w

1 8 . 7 5 " = " C "

POINTS "B" AND "C" ARE THE SAME DISTANCE FROM THE END ON ALL TllREE CONDUITS.

f W A W W \ \ \ \ ?.m-m

WLL N m u u

. - : : + 3 m N N .a \, \

Z Z ?.- + - 0 1 , : 1

W U N O * *

. . . . . . - - . - . . m u N m m * :- I ,\\\,, + - - ? . - m

. A a , , , , , J L L C(*_(_(NN

RUNNING OVER-LOAD UNITS

S U P P L Y S Y S T E M

E X C E P T I O N : W H E R E P R O T E C T E D BY O T H E R A P P R O V E D M E A N S

I

REPRINTED WITH PERMISSION FROM NFPA 70 1990 NATIONAL ELECTRICAL CODE' COPYRIGHT 1989 hAT OhAL F RF PROTECTIOF. ASSOCAI i l h (1, l lCV MA 02269 1115 111 I'H h l l 11 MA11 H A. S h U I 1111 COMll.tTt A h l l OFt C,A 1'0s I ON OF 1111 h l I 'A Oh 1111 H ~ l l HI h i 1 I SJRJICT \h r l CII S RI PHI SI h T l I r - ONLY BY THE STANDARD IN ITS ENTIRETY

-61 -

MOTOR BRANCH-CIRCUIT PROTECTIVE DEVICES

M A X I M U M RATING OR SETTING

REPRINTED WITH PERMISSION FROM NFPA 70 1990. NATIONAL ELECTRICAL CODE' .COPYRIGHT 1989. NATIONAL FIRE PROTECTION ASSOCIATION. OUINCY. MA 02269 THIS REPRINTED MATERIAL IS NOT THE COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIREW

-62-

TYPE OF MOTOR

S I N G L E - P H A S E , A L L T Y P E S ( N O CODE L E T T E R ) -- A L L AC S I N G L E - P H A S E AND POLYPHASE S Q U I R R E L - C A G E AND SYNCHRONOUS MOTORS W I T H F U L L - V O L T A G E , R E S I S T O R OR REACTOR S T A R T I N G ( N O CODE L E T T E R ) (CODE L E T T E R F TO V ) -- (CODE L E T T E R B TO E ) - - (CODE L E T T E R A ) - A L L AC S Q U I R R E L - C A G E AND SYNCHRONOUS MOTORS W I T H AUTOTRANSFORMER S T A R T I N G . NOT MORE THAN 3 0 AMPS ( N O CODE L E T T E R ) -- MORE THAN 3 0 AMPS ( N O CODE L E T T E R ) (CODE L E T T E R F TO V ) - ---- (CODE L E T T E R B TO E ) - - (CODE L E T T E R A ) -- H I G H - R E A C T I N G S Q U I R R E L - C A G E . NOT MORE T H A N 3 0 AMPS. ( N O CODE L E T T E R ) MORE THAN 3 0 AMPS. ( N O CODE L E T T E R ) WOUND ROTOR ( N O CODE L E T T E R ) DC ( C O N S T A N T V O L T A G E ) NO MORE THAN 5 0 HP ( N O CODE L E T T E R ) - - -- MORE THAN 5 0 HP ( N O CODE L E T T E R )

SYNCHRONOUS MOTORS OF THE 4 5 0 RPM OR L O W E R ) . T H A T S T A R T UNLOADED. DO NOT R E Q U I R E A FUSE R A T I N G OR C I R C U I T - B R E A K E R S E T T I N G I N EXCESS OF 2 0 0 PERCENT OF F U L L - L O A D CURRENT.

PERCENT

N O N T I M E D E L A Y F U S E

3 0 0

3 0 0 3 0 0 2 5 0 1 5 0

2 5 0

2 0 0 2 5 0 2 0 0 1 5 0

2 5 0

2 0 0

1 5 0

1 5 0

1 5 0

LOW-TORQUE.

OF

D U A L - ELEMENT T I M E - D E L A Y F U S E

1 7 5

1 7 5 1 7 5 1 7 5 1 5 0

1 7 5

1 7 5 1 7 5 1 7 5 1 5 0

1 7 5

1 7 5

1 5 0

1 5 0

1 5 0

LOW-SPEED

F U L L - L O A D

I N S T A N - TANEOUS T R I P BREAKER

7 0 0

7 0 0 7 0 0 1 0 0 7 0 0

7 0 0

7 0 0 7 0 0 7 0 0 7 0 0

7 0 0

7 0 0

7 0 0

2 5 0

1 7 5

TYPE

CURRENT

:yi:RSE B R E A K E R

2 5 0

2 5 0 2 5 0 2 0 0 1 5 0

2 0 0

2 0 0 2 0 0 2 0 0 1 5 0

2 5 0

2 0 0

1 5 0

1 5 0

1 5 0

( U S U A L L Y

FULL-LOAD CURRENT IN AMPERES

DIRECT-CURRENT MOTORS

HEPR hTF.1 H 111 PEHM SjUh FROM h l P A 70 Iq'ifl $.AT J\A. I.ECII4 [ A. COl)l ' COPYHll;dT 1Odq \ A I I ~ A I nl I'HuIEI T n r l n5s.11 A T or. L.I hcv MA 0276s 11, s I w R h ~ t u M A T ~ R A L .; NOT COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIRETY

DIRECT CURRENT MOTORS

I TERMINAL MARKINGS:

TERMINAL MARKINGS ARE USED TO TAG TERMINALS TO WHICH CONNECTIONS ARE TO BE MADE FROM OUTSIDE C I R C U I T S .

FACING THE END OPPOSITE THE D R I V E END) T H E STANDARD D I R E C T I O N OF SHAFT ROTATION I S COUNTER CLOCKWISE.

A - 1 AND A - 2 I N O I C A T E ARMATURE LEADS. S - 1 AN0 S - 2 I N O I C A T E S E R I E S - F I E L D LEADS. F - 1 AND F - 2 I N D I C A T E S H U N T - F I E L D LEADS. rE1 SHUNT WOUND MOTORS

TO CHANGE ROTATION. REVERSE E I T H E R ARMATURE LEADS OR SHUNT LEADS. DO REVERSE BOTH ARMATURE AND SHUNT

I F I - F~ LEADS. 1

S E R I E S WOUND MOTORS

TO CHANGE ROTATION. REVERSE E I T H E R ARMATURE LEADS OR S E R I E S LEADS. REVERSE BOTH ARMATURE AND

I $1

S E R I E S L E N S . s2 I

COMPOUND WOUND MOTORS

TO CHANGE ROTATION, REVERSE E I T H E R ARMATURE LEADS OR BOTH THE S E R I E S AND SHUNT

Fz LEADS. DO RREERSE A L L THREE SETS OF LEADS.

fidTE: STANDARD ROTATION FOR O.C. GENERATOR I S CLOCKWISE

DIRECT CURRENT MOTORS

TO REVERSE THE ROTATION OF D I R E C T CURRENT MOTORS:

I D I R E C T CURRENT MOTORS ARE REVERSED BY CHANGING THE D I R E C T I O N OF THE FLOW OF THE CURRENT THROUGH THE ARMATURE. OR F I E L D .

SHUNT - F I E L D

L1 +

A COMPOUND 0 - C MOTOR CONNECTED TO A DOUBLE-POLE. DOUBLE THROW TRANSFER SWITCH.

I TO CHANGE THE SPEED OF A D-C MOTOR:

HOLRING C O I L SHUNT - F I E L D

RESISTANCE ARMATURE

I

FULL-LOAD CURRENT IN AMPERES

SINGLE-PHASE ALTERNATING CURRENT MOTORS - - - - - -

I

T H E V O L T A G E S L I S T E D A R E R A T E D M O T O R V O L T A G E S T H E L I S T E D C U R R E N T S A R E F O R S Y S T E M V O L T A G E R A N G E S O F 1 I0 T O 120 A N D 220 T O 240

REPRINTED WITH PERMISSION FROM NFPA 70 1990. NATIONAL ELECTRICAL CODE ' . COPYRIGHT 1989, NATIONAL FIRE PROTECTION ASSOCIATION. OUINCY. M A 02269 THIS REPRINTED MATERIAL IS NOT THE COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED - ONLY BY THE STANDARD IN ITS ENTIRETY

SINGLE-PHASE USING STANDARD THREE-PHASE STARTER

-"

fGq"=J"ART me

T- m4 1

I 110 VOLT

OL I I M < I

OL

DISCONNECT SWITCH

FUSE

- AUX.

STOP T

220 VOLT

CONNECTIONS

,

I

CONNECTIONS

3 AUX. On fie 0c

I

T - 1 0 T - 3 T-2 0 7 - 4 :

I

SINGLE PHASE MOTORS

S T A R T I N G

L I N E W I N D I N G

S E C . 2 M A I N W I N D I N G

T ,

115 V O L T S JU c c w I

2 3 0 V O L T S CCW

TO REVERSE I N T E R - CHANGE 5 A N D 8

C L A S S E S OF S I N G L E PHASE MOTORS: I

1. S P L I T - P H A S E

A . C A P A C I T O R - S T A R T 0 . R E P U L S I O N - S T A R T C . R E S I S T A N C E - S T A R T 0 . S P L I T - C A P A C I T O R

2 . COMMUTATOR

A. R E P U L S I O N 0 . S E R I E S

T E R M I N A L COLOR MARKING:

T, B L U E T1 ORANGE T5 E m

T, WHITE T I YELLOW T, RED

NOTE: S P L I T - P H A S E MOTORS ARE U S U A L L Y F R A C T I O N A L HORSEPOWER. THE M A J O R I T Y OF E L E C T R I C MOTORS USED I N W A S H I N G M A C H I N E S . R E F R I G E R A T O R S . AND E T C . ARE OF T H E S P L I T - PHASE T Y P E .

TO CHANGE T H E SPEED OF A Z P L I T - P H A S E MOTOR THE NUMBER OF POLES MUST B E CHANGED.

1. A D D I T I O N OF R U N N I N G W I N D I N G 2 . TWO S T A R T I N G W I N D I N G S . AND TWO R U N N I N G W I N D I N G S 3 . CONSEQUENT POLE C O N N E C T I O N S .

1

SINGLE PHASE MOTORS

S P L I T - P H A S E : S Q U I R R E L CAGE

A . R E S I S T A N C E START:

S T A R T I N G W I N D I N G . R U N N I N G WINDING_

R E S I S T A N C E

C E N T R I F U G A L S W I T C H ( C S ) OPENS A F T E R R E A C H I N G 7 5 % OF NORMAL S P E E D .

1

T T 5 I T 8 y T T T 8

CLOCKWISE COUNTER- - CLOCKWISE L L P L L -

0 . C A P A C I T O R START:

m

S T A R T I N G W I N D I N G

N O T E : 1. A R E S I S T A N C E S T A R T MOTOR H A S A R E S I S T A N C E CONNECTED I N S E R I E S W I T H THE S T A R T I N G W I N D I N G .

2 . T H E C A P A C I T O R S T A R T MOTOR I S EMPLOYED WHERE A H I G H S T A R T I N G TORQUE I S R E Q U I R E D .

FULL-LOAD CURRENT

TWO-PHASE ALTERNATING-CURRENT MOTORS (4 WIRE)

F O R 9 0 A N D 8 0 P E R C E N T POWER F A C T O R THE A B O V E F I G U R E S S H O U L D B E M U L T I P L I E D B Y 1.1 A N D 1 . 2 5 R E S P E C T I V E L Y .

REPR hTED 1 I n PERM SSlOh FROM hFPA 70 1990 hAT OhA FLFCTR CA. COOF COPYR GhT 1989 hAT OhA. F HE PROTECT O h ASSOC AT ON OL NCY MA 02269 I d S REPR hTFD MATER A S hOT ThF rOMPLETE AND OFF CA. POS T O h OF Tnf hFPA Oh TrlE RFFFRFhCED SJRJFCT W h C d S REPRESlhTED - ONLY BY THE STANDARD IN ITS ENTIRETY

-70-

TWO-PHASE, FOUR WIRE

STANDARD THREE PHASE STARTER

* NO HEATER OR HEATER OVERLOAD RELAY NECESSARY FOR Tn

TWO-PHASE MOTORS

TWO PHASE - - - THREE W I R E

TO REVERSE T l l C D I R E C T I O N Of A TWO PHASE. THREE W I R E MOTOR INTERCHANGE THE TWO O U T S I O E MOTOR L E A D S . 1 A N 0 2 .

TWO PHASE - - - - FOUR W I R E

TO REVERSE THE D I R E C T I O N OF A TWO P H A S E , FOUR W I R E MOTOR INTERCHANGE THE L E A D S I N ONE PHASE.

- FULL LOADS CURRENT THREE-PHASE ALTERNATING CURRENT MOTORS

L i

I INDUCTION TYPE I SYNCHRONOUSTYPE SQUIRREL CAGE AND WOUND-ROTOR 'UNITY POWER FACTOR

AMPERES AMPERES I

'FOR 90 AND 80 PERCENT POWER FACTOR. THE ABOVE FIGURES SHALL BE MULTIPLIED BY 1 1 AND 1.25 RESPECTIVELY

REPRINTED WITH PERMISSION FROM NFPA 70 1990 NATIONAL ELECTRICAL CODE. COPYRIGHT 1989 NATIONAL FIRE PROTECTION ASSOCIATION OUINCY MA 02269 THIS REPRINTED MATERIAL IS NOT THE COMPLETE AN0 OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIRETY

FULL-LOAD CURRENT AND OTHER DATA

THREE PHASE A.C. MOTORS

FULL-LOAD CURRENT AND OTHER DATA

THREE PHASE A.C. MOTORS

I

NOTE:

1. W I R E S I Z E W I L L VARY D E P E N D I N G ON TYPE OF I N S U L A T I O N

2 . THE P R E C E E O I N G C A L C U L A T I O N S A P P L Y TO I N D U C T I O N T Y P E . S Q U I R R E L - C A G E . AND WOUND-ROTOR MOTORS O N L Y .

3 . THE VOLTAGES L I S T E D ARE RATEO MOTOR V O L T A G E S ; CORRESPONDING N O M I N A L SYSTEM VOLTAGES ARE 2 2 0 V TO 2 4 0 V . AND 4 4 0 V TO 4 8 0 V . - 4 . H E R T Z : PREFERRED TERMINOLOGY FOR C Y C L E S PER SECOND.

5 . FORM C O I L : C O I L MADE W I T H RECTANGULAR OR SQUARE WIRE

6 . MUSH C O I L : C O I L MADE W I T H ROUND W I R E .

S L I P : PERCENTAGE D I F F E R E N C E BETWEEN SYNCHRONOUS AND O P E R A T I N G S P E E D S .

8 . SYCHRONOUS SPEEO: MAXIMUM SPEEO FOR A . C . MOTORS OR (FREQUENCY X 1 2 0 ) / P O L E S

I 9 . F U L L LOAD SPEEO: SPEEO A T WHICH RATEO HORSEPOWER I S DEVELOPED.

10. P O L E S : NUMBER OF M A G N E T I C P O L E S S E T UP I N S I D E THE MOTOR BY THE PLACEMENT AND CONNECTION OF THE W I N D I N G S .

P

THREE PHASE A.C. MOTOR WINDINGS AND CONNECTIONS

3 . " Y " OR STAR

H I G H VOLTAGE H I G H VOLTAGE

%88 G F a * LOW VOLTAGE

NOTE: -

LOW VOLTAGE

THE MOST IMPORTANT PART OF ANY MOTOR IS THE NAME-PLATE. CHECK THE DATA GIVEN ON THE PLATE BEFORE MAKING THE CONNECTIONS. TO CHANGE ROTATION DIRECTION OF 3 PHASE MOTOR, SWAP ANY 2 T-LEADS.

THREE WIRE STOP-START STATION

WIRING DIAGRAM

CONTROL TRANSFORMER 0_3

@- 1

C I R C U I T

OVERLOAD HEATER

1 2 0 V . SCHEMATIC DIAGRAM I

GR.

Ann-,,

-

GR.

NOTE: CONTROLS AND MOTOR ARE OF DIFFERElOT VOLTAGES.

TWO THREE WIRE STOP-START STATIONS

W I R I N G DIAGRAM

SCHEMATIC DIAGRAM

START OL O L

O L

NOTE: CONTROLS AN0 MOTOR ARE OF THE SAME VOLTAGE.

HAND OFF AUTOMATIC CONTROL

0-1 0 - 2 0 - 3

W I R I N G OIAGRAM

OL

L - 1 S C H E M A T I C OIAGRAM

- -

I NOTE: CONTROLS AND MOTOR ARE OF T H E SAME VOLTAGE.

I I

START-STOP-JOG STATION

WIRING DIAGRAM

CONTROL TRANSFORMER

0-1 0-3

CIRCUIT

START

2

STOP

- GR.

OvERLOAD " I,+.' $+ HEATER

START

GR.

4 8 0 - V MOTOR

NOTE: J 0 6 AND STOP ARE MECHANICALLY INTERLOCKED. WHEN JOG I S CLOSED. STOP W I L L BE I N OPEN P O S I T I O N . BOTH JOG AND STOP ARE MOMENTARILY OPEN UPON RELEASE OF THE JOG BUTTON; THUS OPENING C I R C U I T TO THE C O I L .

- TRANSFORMER CALCULATIONS

TO B E T T E R UNDERSTAND THE F O L L O W I N G FORMULAS R E V I E W THE R U L E OF T R A N S P O S I T I O N I N E Q U A T I O N S .

A M U L T I P L I E R MAY B E REMOVED FROM ONE S I D E OF AN E Q U A T I O N B Y M A K I N G I T A D I V I S O R ON THE OTHER S I D E . OR A D I V I S O R MAY B E REMOVED FROM ONE S I D E OF A N E Q U A T I O N BY M A K I N G I T A M U L T I P L I E R ON THE OTHER S I D E . - 1. VOLTAGE AND C U R R E N T : . P R I M A R Y ( p ) AND SECONDARY ( s )

POWER ( p ) = POWER ( s ) OR E p X I p = E s X I s

- 2 . VOLTAGE AND TURNS I N C O I L :

VOLTAGE ( p ) X TURNS ( s ) = VOLTAGE ( s ) X TURNS ( p )

3 . AMPERES AND TURNS I N C O I L :

A M P E R E S ( p ) X TURNS ( p ) = AMPERES ( s ) X TURNS ( s )

VOLTAGE DROP CALCULATIONS

INDUCTANCE NEGLIGIBLE

= OROP I N C I R C U I T VOLTAGE : R E S I S T A N C E PER F T . OF CONOUCTOR (OHMS / F T . ) = CURRENT I N CONOUCTOR ( A M P E R E S ) = ONE-WAY LENGTH OF C I R C U I T ( F T . ) = CROSS S E C T I O N AREA OF CONOUCTOR ( C I R C U L A R M I L S ) = R E S I S T I V I T Y OF CONDUCTOR

A . K = 1 2 FOR C I R C U I T S LOADED TO MORE THAN 50% OF ALLOWABLE C A R R Y I N G C A P A C I T Y (COPPER CONDUCTOR)

B . K = 11 FOR C I R C U I T S LOAOEU L E S S T H A N 5 0 % OF ALLOWABLE C A R R Y I N G C A P A C I T Y (COPPER CONDUCTOR)

C. K = 1 8 FOR ALUMINUM CONDUCTORS ( 3 0 DEGREES " C " )

TWO-WIRE S I N G L E PHASE C I R C U I T S :

v = 2 K X L X I

D

THREE-WIRE S I N G L E PHASE C I R C U I T S :

V = 2 K X L X I

0

THREE-WIRE THREE PHASE C I R C U I T S :

FOUR-WIRE THREE PHASE BALANCED C I R C U I T S :

v = 2 K X L X I 1

X - D 2

N O T F : 1. FOR L I G H T I N G LOADS: VOLTAGE DROP BETWEEN ONE O U T S I D E CONOUCTOR AND N E U T R A L EQUALS O N E - H A L F OF - DROP C A L C U L A T E D BY FORMULA FOR TWO-WIRE C I R C U I T S .

2 . FOR MOTOR L E A D S : VOLTAGE OROP BETWEEN ANY TWO O U T S I D E CONDUCTORS EQUALS 0 . 8 6 6 T I M E S DROP D E T E R M I N E D BY FORMULA FOR TWO-WIRE C I R C U I T S .

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BUCK AND BOOST TRANSFORMER CONNECTONS

1 1 5 V I N P U T

1 1 0 % BOOST I

- 1 1 5 V I N P U T -

2 0 % BOOST

,115V I N P U T a

1 0 % BUCK 1 ~ OUTPUT

5% BOOST

1 1 0 % BOOST I

OUTPUT 4 k 2 3 0 V I N P U T -I

5% BUCK

OUTPUT

2 3 0 V I N P U T

OUTPUT

-84-

I N P U T

5% BOOST L:d I N P U T

1 0 % BOOST

O U T P U T

k O U T P U T 4

THREE PHASE CONNECTIONS

STAR VOLTAGE FROM " A " " 8 " . OR " C " TO GROUND = E ( G ~

VOLTAGE BETWEEN A - B . A - C . OR B-C = E ( P )

E ( P ) = E ( G ) X 1 . 7 3

E ( G ) = E ( P ) / 1 . 7 3

POWER = 3 X E ( G ) X I X COS I I

I ( P ) DELTA

I ( W ) = CURRENT OF W I N D I N G I ( P ) = CURRENT OF PHASE

D t L T A " E " = STAR " E " X 1 . 7 3 STAR " E m = D E L T A " E " / 1 . 7 3

STAR "I" = D E L T A "!" X 1 . 7 3 D E L l A " I " = STAR "I / 1 . 7 3

POWER = 3 X E ( W ) X COS O

I ( P ) I ( P ) = I ( W ) X 1 . 7 3

" C " L E Q U I V A L E N T WYE-DELTA NETWORKS

B X C a = -

K ( 2 ) A = -

K ( 1 ) a

A X C I , = -

K ( 2 ) B = -

K ( 1 )

ST

AR

-DE

LTA

Hz

THR

EE

-PH

AS

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TAN

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RD

PH

AS

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OTA

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30

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PO

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TRANSFORMER CONNECTIONS

SERIES CONNECTION OF LOW VOLTAGE WINDINGS

THREE-PHASE A D D I T I V E P O L A R I T Y

H I G H VOLTAGE - A 7 B - A As

c T T C

- b ( f I c f - L -

LOW VOLTAGE D E L T A - D E L T A

THREE-PHASE A D O I T l V E P O L A R I T Y I El

H I G H VOLTAGE - - A A A c

LOW VOLTAGE S T A R - D E L T A

NOTE: S I N G L E - P H A S E TRANSFORMERS SHOULD BE THOROUGHLY CHECKED FOR I M P E D A N C E . P O L A R I T Y . A N 0 VOLTAGE R A T I O B E F O R E I N S T A L L A T I O N . -

TRANSFORMER CONNECTIONS

SERIES CONNECTION OF LOW VOLTAGE WINDINGS

H I G H VOLTAGE A B T -

T

g r' LOW VOLTAGE

THREE PHASE A D D I T I V E P O L A R I T Y

D E L T A - S T A R

4: I

A H I G H VOLTAGE -

C

T

b ( < C I N i g -

LOW VOLTAGE I THRLE PHASE A D D I T I V E P O L A R I T Y ?Ac

STAR-STAR

NOTE: FOR A D D I T I V E P O L A R I T Y THE H - l AND THE X - 1 B U S H I N G S ARE D I A G O N A L L Y O P P O S I T E EACH OTHER.

TRANSFORMER CONNECTIONS

SERIES CONNECTION OF LOW VOLTAGE WINDINGS

H I G H VOLTAGE a A

T B - C 1 'I a c

a

& ' b i I: t C I' I'

LOW VOLTAGE

THREE PHASE S U B T R A C T I V E P O L A R I T Y

D E L T A - D E L T A

H I G H VOLTAGE

A - a

B J J a a

c J J

i b i i I C

L N I - .f LOW VOLTAGE

THREE PHASE S U B T R A C T I V E P O L A R I T Y

D E L T A - S T A R

a<:

NOTE: FOR S U B T R A C T I V E P O L A R I T Y I H E H - 1 AND THE X - 1 B U S H I N G S ARE D I R E C T L Y O P P O S I T E EACH O T H E R .

TRANSFORMER CONNECTIONS

TWO PHASE-FOUR WIRE

LOW VOLTAGE

TWO-PHASE F O U R - W I R E I S TRANSFORMED TO TWO-PHASE FOUR-WIRE OF A - D I F F E R E N T VOLTAGE W I T H NO C O N N E C T I O N BETWEEN THE TWO P H A S E S .

TWO P H A S E - - - - F O U R W I R E 1 TWO P H A S E - - - T H R E E W I R C

H I G H VOLTAGE

T 1

- - COMMON 4

rwXrrl 1 - -

LOW VOLTAGE

THE TWO PHASES ON THE LOW VOLTAGE S I D E ARE E L E C T R I C A L L Y CONNECTED. W I T H B A L A N C E D LOAD THE CURRENT I N THE COMMON W I R E I S 1 . 4 1 GREATER THAN THE CURRENT I N E I T H E R OF O U T S I D E W I R E S .

TRANSFORMER CONNECTIONS

TWO PHASE-THREE WIRE

H I G H VOLTAGE - LL - - 'T

< 6 LOW VOLTAGE

BOTH PHASES ARE E L E C T R I C A L L Y CONNECTED B Y THE COMMON. T H E COMMON I S S O M E T I M E S GROUNDED. W I T H THE B A L A N C E D L O A D THE - CURRENT I N THE COMMON I S 1 . 4 1 T I M E S T H A T I N THE O U T S I D E L E G S .

THREE PHASE-OPEN D E L T A - H I G H VOLTAGE A n - -i

T \ - w - T <

c - LOW VOLTAGE

I N T H I S OPEN D E L T A C O N N E C T I O N THE U N I T S W I L L TRANSFORM 86% OF T H E I R R A T I N G .

7

I T I S NOT NECESSARY T H A T T H E IMPEDANCE C H A R A C T E R I S T I C S B E I D E N T I C A L A S W I T H THREE U N I T BANKS.

R E G U L A T I O N OF O P E N - D E L T A BANK I S NOT AS GOOD A S A C L O S E D - D E L T A B A N K . -

- MISCELLANEOUS WIRING DIAGRAMS

NEUTRAL - LAMPS

NEUTRAL

LAMPS - 1

B A T T E R Y

(II.1 PUSH BUTTON

1

TWO 3-WAY S W I T C H E S

TWO 3-WAY S W I T C H E S

ONE 4-WAY S W I T C H

B E L L C I R C U I T

MISCELLANEOUS WIRING DIAGRAMS

NEUTRAL

115V A 'I

u r r I 1

SWITCH I

REMOTE CONTROL C I R C U I T - ONE RELAY AND ONE S W I T C H

SUPPORTS FOR RIGID METAL C O N D U I T

C O N D U I T S I Z E

112" - 3 / 4 "

1"

1 - 1 / 4 " - 1 - 1 / 2 "

2 " - 2 - 1 / 2 "

3 " AND LARGER

D I S T A N C E BETWEEN SUPPORTS

10 F E E T

12 "

14 "

16 "

20 "

I

CONDUCTOR PROPERTIES

REPRlhTFO WITH PERM.SSIOh FROM hFPA 70 1990 &AT OhA. ELECTRCA. COOEi COPVRGnT 1989 hAT DNA. F RE PROTECT D h ASSOC AT ON 01, RCV MA 02269 114 S REPR hTED MATER A. S hOT THE COMPLETE ANDOFF'CA. POS T O h Ot ThE hFPA Oh TdE REFFRFhCED SLB.lCT W d C d S RFPRESEhTED - ONLY BY THE STANDARD IN ITS ENTIRETY

COPPER

1 AMPACITIES OF SINGLE INSULATED CONDUCTOR RATED 0-2.WO VOLTS IN FREE AIR 1

COPPER

ALUMINUM OR COPPER-CLAD ALUMINUM

ALUMINUM OR COPPER-CLAD ALUMINUM

A

P OVERCURRENT PROTECTION FOR CONDUCTOR WPES MARKED ) WILL NOT EXCEED 15 AMPERES FOR SIZE 12 AWG. AND 25 AMPERES FOR &ZE 10 AWG e (-) FOR WET LOCATIONS ONLY, SEE 75% COLUMN FOR WET LOCATIONS

- INSULATION CHART

-

I

- - -

R E F E R T O P A G E 1 0 3 F O R S P E C I A L P R O V I S I O N S A N D / O R A P P L I C A T I O N S I

REPRINTED WITH PERMISSION FROM NFPA 70~1990 , NATIONAL ELECTRICAL CODE' .COPYRIGHT 1989. NATIONAL FIRE PROTECTION ASSOCIATION. OUINCY. MA 02269 THIS REPRINTED MATERIAL IS NOT THE COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIRETY

INSULATION CHART

R E F E R T O P A G E 103 F O R SPECIAL PROVISIONS AND/OR APPLICATIONS

REPRINTED WITH PERMISSION FROM NFPA 70-1990. NATIONAL ELECTRICAL CODEm .COPYRIGHT 1989. NATIONAL FIRE PROTECTION ASSOCIATION. OUINCY. MA 02269. THIS REPRINTED MATERIAL IS NOT THE COMPLETE AN0 OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIRETY.

w

INSULATION CHART

S E E A R T I C L E 339 N E C

S E E A R T I C L E 338 N.E C

R E F E R T O P A G E 103 F O R S P E C I A L P R O V I S I O N S A N D / O R A P P L I C A T I O N S

REPRINTED WITH PERMISSION FROM NFPA 70-1990. NATIONAL ELECTRICAL CODEO .COPYRIGHT 1989, NATIONAL FIRE PROTECTION ASSOCIATION, OUINCY. MA 02269. THIS REPRINTED MATERIAL IS NOT THE COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIREN.

INSULATION CHART -

WHERE ENVIROMENTAL CONDITIONS REQUIRE MAXIMUM CONDUCTOR OPERATING TEMPERATURES ABOVE 90°C

. . INSULATION AND OUTER COVERING THAT MEET THE REQUIREMENTS OF I FLAME-RETARDANT, LIMITED SMOKE AND ARE SO LISTED SHALL BE PERMIT-

TED TO BE DESIGNATED LIMITED WlTH THE SUFFIX ILS AFTER THE CODE TYPE DESIGNATION.

. . . LISTED WIRE TYPES DESIGNATED WlTH SUFFIX -2 SUCH AS RHW-2 SHALL BE PERMITTED TO BE USED AT A CONTINUOUS 90°C OPERATING TEMPERATURE WET OR DRY AMPACITIES OF THESE WIRE TYPES ARE GIVEN IN THE 90°C IN THE APPROPRIATE AMPACITY TABLE.

RlPR hTED W Tn PLHM SSlOh FROM NFPA 71)-1990 NATlOhAl FLECTR CA. COUC' COPYR GHT 1989 hAT OhAL F HC PROTECTION ASSOCIAT Oh OL NCY MA 02269 Tn S REPRlNTtll MATER AL .S NOT TtcE - COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIRETY

-1 03-

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3 4 3 4 6 5 5 8

3 4 3 3 4

- MINIMUM COVER REQUIREMENTS 0-600 VOLTS, NOMINAL - - - - - - --

COVER IS DEFINED AS THE DISTANCE BETWEEN THE TOP SURFACE OF DIRECT BURlAl CABLE. CONDUIT. OR OTHER RACEWAYS AND THE FINISHED SURFACE -

I FOR COMPLETE DETAILS REFER TO NATIONAL ELECTRICAL CODE' TABLE 300-5

WIRING METHOD

DIRECT BURIAL CABLES RIGID METAL CONDUIT INTERMEDIATE METAL CONDUIT RIGID NONMETALLIC CONDUIT (APPROVED FOR DIRECT BURIAL WITHOUT CONCRETE ENCASEMENT)

VOLUME REQUIRED PERCONDUCTOR - -- -

MINIMUM BURIAL (INCHES)

24 6 6

18

SIZE OF FREE SPACE WITHIN BOX CONDUCTOR FOR EACHCONDUCTOR

NO 18 1 5 CUBIC INCHES No 16 1 75 CUBIC INCHES No 14 2 CUBIC INCHES No 12 2 25 CUBIC INCHES NO 10 2 5 CUBIC INCHES NO 8 3 CUBIC INCHES No 6 5 CUBIC INCHES

REPRINTED WITH PERMISSION FROM NFPA 70-1990. NATIONAL ELECTRICAL CODE' .COPYRIGHT 1989 NATIONAL FIRE PROTECTION ASSOCIATION. OUINCY. MA 02269 THIS REPRINTED MATERIAL IS NOT THE COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED - ONLY BY THE STANDARD IN ITS ENTIRETY

MINIMUM DEPTH OF CLEAR WORKING SPACE IN FRONT OF

ELECTRICAL EQUIPMENT

C O N D I T I O N S : 1. EXPOSEO L I V E PARTS ON ONE S I D E AND NO L I V E OR GROUNDED P A R T S ON THE OTHER S I D E OF THE WORKING SPACE OR EXPOSEO L I V E PARTS ON BOTH S I O E S E F F E C T I V E L Y GUARDED BY S U I T A B L E WOOD OR OTHER I N S U L A T I N G M A T E R I A L S . I N S U L A T E D W I R E OR I N S U L A T E D B U S B A R S . 0 P E R A T I N G A T NOT OVER 3 0 0 V O L T S S H A L L NOT B E C O N S I D E R E D L I V E PARTS.

N O M I N A L VOLTAGE TO GROUND

6 0 1 - 2 5 0 0 2 5 0 1 - 9 0 0 0 9 0 0 1 - 2 5 . 0 0 0 2 5 . 0 0 1 - 7 5 k V A b o v e 7 5 k V

2 . EXPOSED L I V E PARTS ON ONE S I D E AND GROUNDED PARTS ON THE OTHER S I D E . CONCRETE. B R I C K , OR T I L E WALLS W I L L B E CONSIDERED A S GROUNDED SURFACES.

C O N D I T I O N S

1 2 3 FEET F E E T F E E T

3 4 5 4 5 6 5 6 9 6 8 1 0 8 1 0 1 2

3 . EXPOSED L I V E PARTS ON BOTH S I O E S OF THE WORK SPACE (NOT GUARDED A S P R O V I D E D I N C O N D I T I O N 1) W I T H THE OPERATOR BETWEEN.

REPRINTED WITH PERMISSION FROM NFPA 70 1990 NATIONAL ELECTRICAL CODE' COPYRIGHT 1989 NATIONAL FIRE PROTECTION ASSOCIATION OUINCY MA 02269 THIS REPRINTED MATERIAL IS NOT THE COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIRETY

-1 10-

- MINIMUM CLEARANCE OF LIVE PARTS

FOR S I U N I T S : ONE I N C H = 25.4 M I L L I M E T E R S . * T H E VALUES G I V E N ARE THE M I N I M U M CLEARANCE FOR R I G I D PARTS AND BARE

CONDUCTORS UNDER FAVORABLE S E R V I C E C O N O I T I O N S . THEY S H A L L BE I N C R E A S E D FOR CONDUCTOR MOVEMENT OR UNDER UNFAVORABLE S E R V I C E C O N D I T I O N S . OR - WHEREVER SPACE L I M I T A T I O N S P E R M I T . THE S E L E C T I O N OF THE A S S O C I A T E D I M P U L S E W I T H S T A N D VOLTAGE FOR A P A R T I C U L A R SYSTEM VOLTAGE I S DETERMINED BY THE C H A R A C T E R I S T I C S OF THE SURGE P R O T E C T I V E EQUIPMENT.

REPRINTED WITH PERMISSION FROM NFPA 70-1990. NATIONAL ELECTRICAL CODE' .COPYRIGHT 1989. NATIONAL FIRE PROTECTION ASSOCIATION. OUINCY. MA 02269 THIS REPRINTED MATERIAL IS NOT THE COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTEC ONLY BY THE STANDARD IN ITS ENTIRETY

N O M I N A L VOLTAGE R A T I N G .

K V

2.4-4.16 7.2

13.8 14.4 23 34.5

4 6

69

115

138

161

230

* M I N I M U M CLEARANCE OF L I V E PARTS. I M P U L S E WITHSTAND.

8 . I . L . I N C H E S

PHASE-TO-PHASE K V

INDOORS

6 0 7 5 9 5

110 125 150 200

INDOORS

4.5 5.5 7.5 9.0

10.5 12.5 18.0

PHASE-TO-GROUND

OUTDOORS

95 9 5

110 110 150 150 200

200 250

250 350

550

550 650

650 750

750 900

1050

OUTDOORS

7 7

12 12 15 15 1 8

1 8 2 1

2 1 3 1

53

5 3 6 3

6 3 7 2

7 2 8 9

105

INDOORS

3.0 4.0 5.0 6.5 7.5 9.5

13.0

OUTDOORS

6 6 7 7

10 10 13

13 17

17 25

42

42 5 0

5 0 58

58 7 1 8 3

MINIMUM SIZE EQUIPMENT GROUNDING CONDUCTORS FOR

GROUNDING RACEWAY AND EQUIPMENT

REPRINTED WITH PERMISSION FROM NFPA 70-1990. NATIONAL ELECTRICAL CODEL ,COPYRIGHT 1989. NATIONAL FIRE PROTECTION ASSOCIATION, OUINCY. MA 02269 THIS REPRINTED MATERIAL IS NOT THE COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIREN -

R A T I N G OR S E T T I N G OF A U T O M A T I C O V E R C U R R E N T

D E V I C E I N C I R C U I T A H E A D OF E Q U I P M E N T . C O N D U I T .

E T C . . N O T E X C E E D I N G ( A M P E R E S )

1 5 2 0 3 0

4 0 6 0

1 0 0

2 0 0 3 0 0 4 0 0

5 0 0 6 0 0 8 0 0

1 0 0 0 1 2 0 0 1 6 0 0

2 0 0 0 2 5 0 0 3 0 0 0

4 0 0 0 5 0 0 0 6 0 0 0

S I Z E . C O P P E R

WIRE NO,

1 4 1 2 1 0

1 0 1 0 8

6 4 3

2 1 0

2 / 0 3 / 0 4 / 0

2 5 0 k c m ~ l 3 5 0 k c m ~ l 4 0 0 k c m ~ l

5 0 0 k c m ~ l 7 0 0 k c m ~ l 8 0 0 k c m ~ l

A L U M I N U M OR C O P P E R - C L A D

ALUMINUM W I R E N O .

1 2 1 0

8

8 8 6

4 2 1

1 / 0 2 / 0 3 / 0

4 / 0 2 5 0 k c m ~ l 3 5 0 k c m ~ l

4 0 0 k c m ~ l 6 0 0 k c m ~ l 6 0 0 k c m f l

8 0 0 k c m ~ l 1 2 0 0 k c m ~ l 1 2 0 0 kern11

- GROUNDING ELECTRODE CONDUCTOR FOR AC SYSTEMS

I 'WHERE THERE ARE NO SERVICE-ENTRANCE CONDUCTORS. THE

GROUNDING ELECTRODE CONDUCTOR S I Z E S H A L L B E D E T E R M I N E D BY THE E Q U I V A L E N T S I Z E OF THE LARGEST S E R V I C E - E N T R A N C E CONDUCTOR REQUIRED FOR THE LOAD TO BE SERVED.

AII'H hIEl1 W 114 PERM SSOh FROM t4FPA 70 1990 hAT 1JhAI t. tC l l l l iAL i l lOE' COPYRIGHT lYH9 hAT ONA. I HI I'HOIICI ON ASSOC Al Cltr OIINCY MA 02269 Trl S RlPR h l l l l MA11 II A 5 hOT TriF

S I Z E OF LARGEST S E R V I C E - E N T R A N C E CONDUCTOR OR E Q U I V A L E N T AREA FOR

P A R A L L E L CONDUCTORS

I COMPLETE AND OFFICIAL POSITION OF THE NFPA ON THE REFERENCED SUBJECT WHICH IS REPRESENTED ONLY BY THE STANDARD IN ITS ENTIRETY

S I Z E OF GROUNDING ELECTRODE CONDUCTOR

COPPER

2 OR SMALLER 1 OR 0 2 / 0 OR 3 / 0 OVER 3 / 0 THRU

3 5 0 k c m l l OVER 3 5 0 k c m ~ l

THRU 6 0 0 k c m ~ l OVER 6 0 0 k c m ~ l

THRU 1 1 0 0 k c m ~ l OVER 1 1 0 0 k c m l l

COPPER

8 6 4

2

0

2 / 0 3 / 0

ALUMINUM OR COPPER-CLAD

ALUMINUM

0 OR SMALLER 2 / 0 OR 3 / 0 4 / 0 OR 2 5 0 k c m ~ l OVER 2 5 0 THRU

THRU 5 0 0 k c m l l OVER 5 0 0 kcml l

THRU 9 0 0 k c m ~ l OVER 9 0 0 k c m l l

THRU 1 7 5 0 k c m l l OVER 1 7 5 0 k c m ~ l

* A L U M I N U M OR COPPER-CLAD

ALUMINUM

6 4 2

0

3 / 0

4 / 0 2 5 0 k c m ~ l

ELECTRICAL SYMBOLS

WALL C E I L I N G S W I T C H O U T L E T S

0 O U T L E T S S I N G L E POLE S W I T C H

a @ DROP CORD s, DOUBLE P O L E S W I T C H

--@ @ FAN O U T L E T 5, THREE WAY S W I T C H

9 @ J U N C T I O N BOX s, FOUR WAY S W I T C H

@ L A N P HOLDER I s, A U T O M A T I C DOOR S W I T C H

P 0 , s LAMP HOLDER W I T H P U L L S W I T C H S E

E L E C T R O L I E R S W I T C H

-0 @ P U L L S W I T C H - sp S W I T C H AND P I L O T LAMP -0 @ L!J;O;HDISCHARGE s, KEY O P E R A T E 0 S W I T C H

-0 @ E X I T L I G H T s,, C I R C U I T B R E A K E R - 4 @ CLOCK O U T L E T WEATHER PROOF

S W C B C I R C U I T BREAKER

@ B L A N K E D O U T L E T MOMENTARY CONTACT sMc S W I T C H

:I;::; C O N V E N I E N C E REMOTE CONTROL SRc SWITCl4 el,3 S I N G L E . T R I P L E X , E T C . SWP WEATHER PROOF S W I T C H - RANGE O U T L E T FUSED S W I T C H

S W I T C H AND es CONVENIENCE O U T L E T WEATHER PROOF FUSED S W I T C H

S P E C I A L PURPOSE @ OUTLET L I G H T I N G S W I T C H

@ FLOOR O U T L E T POWER PANEL

Rpproduied From ARerlLan Standard l t I f

ELECTRICAL SYMBOLS

S I N G L E BREAK S W I T C H

MOMENTARY CONTACT S I N G L E C I R C U I T ( N . C .

MOMENTARY CONTACT MUSHROOM HEAD SW.

TWO P O S I T I O N CONTACT

FOOT S W I T C H

VACUUM S W I T C H

SW. -

P I L O T L I G H T NON-PUSH TO T E S T

P I L O T L I G H T PUSH TO T E S T

F U S E

" A " OVERLOAO THERMAL R E L A Y - L I N E C I R C U I T

" 0 " OVERLOAO CONTACT CONTROL C I R C U I T

L I Q U I D L E V E L SWITCH i"wi" A ' OVERLOAO MAGNETIC

R E L A Y - L I N E C I R C U I T

T I M E 0 SWITCH " 0 " OVERLOAO CONTACT E N E R G I Z E D CONTROL C I R C U I T

T I M E S W I T C H OE-ENERGIZED

TEMPERATURE ACTUATED SW. NOT CONNECTED

FLOW S W I T C H

L I M I T SWITCH ( N . O . )

NOT CONNECTED

CONNECTED

POWER CABLE

- CONTROL CABLE 0 AUTOMATIC HOME RUNS C A B L E

- - - - - UNDERGROUND

aTQ CONCEALED I N FLOOR 1 DOUBLE BREAK SWITCH NUMBER OF CONDUCTORS U # I N C O N D U I T ( 4 )

Reproduced f r o m A w r l ~ a n Stilndard I E E t .

ELECTRICAL SYMBOLS

3 - P O L E DRAWOUT TYPE C I R C U I T BREAKER W I T H MAGNETIC OVER-LOADS

3 - P O L E . 1-THROW FUSED SWITCH

CURRENT TRANSFORMER

P D N T E N T I A L TRANSFORMER

S O L E N O I D

C A P A C I T O R

BLOWOUT

THERMAL OVERLOAD HEATERS

2 - P O L E , 1-THROW 0 cD1' ) ) C I R C U I T BREAKER T GROUND

- " A " 1 - P O L E SW.

"8" 1 - P O L E C 8 .

" A " " 8 " I 1 1 1 k BATTERY

Reproduced fro. mer,can Standard I i t t .

-116-

Wiring Diagrams For NEMA Configurations

125V 2P.3W

GROUNOING - -

250V 2P. 3W

GROUNDING

1

Courtesy of Cooper Industr~es, Inc - Arrow Hart Wiring Gevices

- 117A -

Wiring Diagrams For NEMA Configurations

48OUAC

GROUNOING irov 2p'3w

- ' _ " I

... \ > L 1 , R L C 0 9

BOOUAC I 2P. 3W -a" -

GROUNDING ' 1 - 1

I - 11 IP

D

Courtesy of Cooper Industr~es. Inc. - Arrow Hart Wlrlng Dev~ces I

- 117B -

Wiring Diagrams For NEMA Configurations

12512501 3 P . I W

GROUNDING

4 , ,:.>OR - - V - V i 4 i : i I 4 i O I 4 . 6 #>:rnl L l : 1 3 ? . .

3 0 2501

3P. 4W GROUNDING

LI1-?OR

Cotirtesy of Cooper lndustr~es. Inc - Arrow Hart Wiring Devices

Wiring Diagrams For NEMA Configurations

Courtesy of Cooper Industries. Inc -A r row Hart Wiring Devices

- 117D -

m

Wiring Diagrams For NEMA Configurations

Courtesy of Cooper Industries. Inc. - Arrow Hart Wiring Dev~ces

HAND SIGNALS FOR

CRANES AND CHERRY PICKERS

DOG EMERGENCY E V E R Y T H I N G STOP

RETRACT BOOM

T R A V E L B O T H TRACKS

[CRAWLER CRANES O N L Y 1

I

I EXTEND S W I N G

BOOM

HAND SIGNALS Fa

CRANES AND CHERRY PICKERS

R A I S E LOAD -

MOVE SLOWLY

I

LOWER LOAD M A I N H O I S T

R A I S E BOOM AN0 LOWER LOWER BOOM AND LOAD ( F L E X F I N G E R S ) R A I S E LOAD ( F L E X

F I N G E R S )

USE WHIP L I N E

BOOM UP BOOM DOWN

USEFUL KNOTS

BOWLINE RUNNING BOWLINE BOWLINE ON THE B I G H T

CLOVE H I T C H SHEEPSHANK R O L L I N G H I T C H

S I N G L E BLACKWALL

H I T C H

CATSPAW

- .

SQUARE KNOT T IMBER H I T C H S I N G L E W I T H HALF HITCH SHEET

BEND

American Red Cross + GENERAL DIRECTIONS FOR FIRST AID: While help is being summoned, do the following: 1) Minimize further injury-move victim only if necessary for safety reasons. 2) Controi severe bleeding. 3) Maintain an open airway and give Artificial Respirat~on or CPR if necessary 4) Treat for Shock

URGENT CARE: BLEEDING

First Aid: 1) Direct Pressure and Elevation:

' Place dressing directly over the wound and elevate open woundsunless there is evidence of a fracture.

2 ) Pressure Points: If bleedlng continues after use of direct pressure and elevation. use the pressure polnts. Arm: Use the brachial artery-pushing the artery against the upper arm bone.

' Leg: Apply pressure on femoral artery, pushing it against the pelvic bone. 3) Nosebleed.

Place victim in a sitting posltlon. Apply pressure by pressing the nostrils toaether toward the middle of the nose.

POISONING Symptoms:Vomiting, heavy labored breathlng,suddenonset of painor ~llness, burns or odor around the lips or mouth, unusual behavior. First Aid: ~f ;onscious: 1 ) Give water to drink. %glass for children. 1 glass for adult 2) Call Poison Control and identify poison. If unconscious or nauseous: 1 ) Call EMS system immediately. 2) DO NOT give anything by mouth 3) Identify the poison. 4) Call Poison Control. 5) Position victim on side.

SHOCK Symploms: Cool moist skin, weak rapid pulse (over loo) , nausea, rate of breathing increased, apathetic. First Aid: 1 ) Maintain open airway, have victlm lie down. 2) Maintain normal body temperature (913.6'). If too hot, cool doWn,and iftoocold.

warm the victim, under and over.

BURNS Symptoms: Small, t h ~ n (surface) burns or large, thin burns: redness, pain, and swellina. Deep birns: blisters, deep tissue destruction, charred appearance. First Aid: Small, thin burns: 1 ) Run or pour cool water on burn. Immerse if possible. Cool until pain is reduced.

DO NOT use ice. 2) Gently pat the area dry with sterile gauze and bandage. Large, thin burns: 1 ) Cool with water immediately. DO NOT use ice. 2) Dry gently and cover with thick, dry, sterile dressing. Deep burns: 1) Cover burn with a thick, dry, sterile dressing 8 bandage. 2) Medical care is urgent. DO NOT put water directly on an open burn. 3) Heat Burn: do not remove clothing.

Chemical Burn: you must remove an infected clothing. 4) Elevate the burned areas if this does not cause pain or further injury.

ELECTRIC SHOCK Symptoms: Unconsciousness, absence of breathing 8 pulse. First Aid: 1) TURN OFF THE POWER SOURCE. 2) DO NOT approach vlctim until source of power has been turned off. 3) Drag victim clear, if necessary. 4) Administer Artificial Respiration or CPR if necessary 5) Treat for Shock. 6) Check for other injuries and seek Medical help. 7) Monitor victim till Medical help arrive.

FROSTBITE Symptoms: Flushed, wh~te, or gray skin. Pain The nose. cheeks, ears, fingers, and - toes are most likely to be affected. Pain may be felt early and then subside. Blisters may appear later. First Aid: 1) Cover the frozen part. Loosen restrictive clothing or boots. 2 ) B r ~ n g victim indoors ASAP. 3) Give thev~ct im awarm drink. (DO NOT give alcoholic beverages, tea, or coffee) 4) Immerse frozen part in warm water (102°-1050), or wrap in a sheet and warm

blankets. DO NOT rewarm if there is a possibility of refreezing. 5) Remove from water and discontinue warmlng once part becomes flushed. 6) After thaw~ng, the victlm should try to move the injured area a little. - 7) Elevate the injured area and protect from further injury. 8) DO NOT rub the frozen part.

DO NOT break the blisters. - DO NOT use extreme or dry heat to rewarm the part.

9) If f~ngers or toes are involved, when bandaging place dry, ster~ie gauze between them to keep areas separated.

HYPOTHERMIA Symptoms: Lowered body core temperature. Persistant shivering, lips may be blue. slow slurredspeech, memory lapses. Most cases occurwhen airtemperature ranges I.

from 30°-50" or water temperature is below 70° F First Aid: 1) Move victim to shelter and remove wet clothing if necessary. 2) Rewarm victim with blankets or body-to-body contact in sleeping bag. 3) If victim is conscious and able to swallow, give warm sugary liquids. 4) Keep victim warm 8 quiet. 5) DO NOT give alcoholic beverages I

6) Constantly monitor vlctim and give artificial respiration and CPR if necessary.

HEAT EXHAUSTIONIHEAT STROKE Symptoms: Heat Exhaustion: Pale clammy skin, profuse perspiration, weakness. nausea. headache. Heat Stroke: Hot dry red skin, no perspiration, rapid 8 strong pulse. High body - temDerature 1105'+). This is an Immediate life threatening emergency. . . . . ~ i r s l Aid: Heat Exhaustion: 1) Give victim cool water 11 conscious and not nauseous 2) Have victim lie down and loosen any restrictive clothing., 3) Cool body with cool water, cold packs, fans or air condltloning. Heat Stroke: 1) Get medical help as soon as possible. 2) Cool body quickly . . t h ~ s is an emergency! 3) Keep body temperature down, repeat cooling process if necessary 4) DO NOT give fluids.

ARTIFICIAL RESPIRATION IF A VICTIM APPEARS TO BE UNCONSCIOUS: Tap vct im on the shoulder and shout, A r e you okay?"

IF THERE IS NO RESPONSE:, Tilt the victim's head, chln polnting up.

Placeone hand on thevictim's jaw and theother handonthe forehead, gently l i f t~ng the jaw under the bony part near the chin, and applying the major force with the hand on the forehead This will move the tonaueawav from the back of the throat to open the airway.

-

IMMEDIATELY, LOOK, LISTEN, AND FEEL FOR BREATHING: While maintaining the head tilt, place your cheek and ear close to thevictim's mouth. Look for the chest to rise, listen for an alr exchange. and feel for the return of air on your cheek Check for 3-5 seconds.

IF THE VICTIM IS NOT BREATHING: While maintaining the head tllt position, pinch thevictlm'snose with the hand on the forehead. Open your mouth wide, take a deep breath, seal the victim's mouth wlth your own and GIVE TWO SLOW FULL breaths.

If you do not get an air exchange when you blow, it may help to retlp the head and try again.

AGAIN. LOOK. LISTEN. AND FEEL FOR BREATHING

IF THERE IS STILL NO BREATHING: Give one breath every 5 seconds for an adult; G ~ v e one breath every 4 seconds for a child; Giveone breath every 3seconds for an infant

I FIRST AID FOR CHOKING

----------------------------------------.--------..-----.--

ltv,crrrn cdrr i i ~ g h . speak, b i ~ , i l h c + DO riot !ilt~rli~rv

It ",Cl,rn c_a_"!Ol

cairgh 0 TAKE ACTION: speak - bredlhe FOR CONSCIOUS VICTIM

G1ve6-10 p - - Abdom~nal

t Thrusts m--- 7

t Repeat t

steps until e f fect~ve o r un t~ l v~ct i rn becomes unconscious. --!

TAKE ACTION: FOR UNCONSCIOUS VICTIM

t t TRY TO VENTILATE 6 10 MANUAL THRUSTS FINGER SWEEP - - - R e p e a t s teps u n t l e f f e c t ~ v e - - Cont~nue a r t ~ f ~ c ~ a l ventlation or CPR, as ~ n d ~ c a t e d C~UIIO~: ~ b d o r n ~ n a l il~rurtr may cause lnfury 0.. no! practfce on people

-1 23-

fmt aid kit

Accidents do happen. Be ready with an automobile first aid kit that makes it easy for anyone to treat almost any emergency quickly and correctly. Everythings's organized in sealed packets printed with step-by-step directions.

Your family counts on you to take care of life's emergencies-big or small-so count on the Red Cross to back you up with a complete family first aid kit for your car. van, truck, or boat. Order one right away by contacting your Red Cross chapter. Only $19.95, and satisfaction is guaranteed! In Houston. Texas and surrounding counties call (713) 526-8300.

Now you'll be ready, when it's all up to you.

+ American Red Cmm

Greater Houston Area Chapter P.O. Box 397 Houston, Texas 770014397

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a note from the publisher.. . We believe that UGLY'S ELECTRICAL

REFERENCES is the finest pocket electrical reference book available anywhere. It is our goal to continually improve this little yellow book so that it retains it's #I position in the industry for years to come.

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