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1
NEMA Junction Box Sizing,NEMA Junction Box Sizing,
Conductor Bending Radii & FiberConductor Bending Radii & Fiber
Optic CableOptic Cable
from: NEC 2002 - Article 314.28, Article 314.54 & Article 770
WSDOT
Fall / Winter 2004
Presented by: Terry Thayer
2
The clear catinkus for Junction Box Sizing and DepthThe clear catinkus for Junction Box Sizing and Depth
u The definition of length
and width
W
L1
L1
L2
Junction box
Junction box
GRS conduit grounded end bushing size table
Size diameter inches
Inches External
add this number
for end bushing
size - then
round up to
nearest quarter
end
bushing
outside
diameter
1/2 0.840 0.75 1 1/2
3/4 1.050 0.75 1 3/4
1 1.315 0.75 2
1 1/4 1.660 0.75 2 1/2
1 1/2 1.900 0.75 2 3/4
2 2.375 0.75 3 1/4
2 1/2 2.875 1.00 4
3 3.500 1.25 4 3/4
3 1/2 4.000 1.25 5 1/4
4 4.500 1.75 6 1/4
5 5.563 2.00 7 1/2
6 6.625 2.00 8 3/4
Example: A 1 1/4" GRS conduit is 1.660" + 0.75" =
2.410". Round 2.410" up to 2.5" or 2 1/2 inches.
L(1) or L(2) = (8) largest conduit in row plus the sum of all
other conduits in row. Calculate for all rows and use sum
of largest row.
W=(6) largest conduit in row plus the sum of all other
conduits in row. Calculate for all rows and use sum of
largest row. NOTE: If no conduit enters the adjacent
sides of the box the dimension is width.
adjacent side
3
Junction Box LengthJunction Box Length -- straight pullstraight pull
• If one or more of the conductors is a # 4 or larger,
• Then the length is not less than 8x the diameter of
the largest conduit if a through pull, regardless of
the number of conduits or conductors. We use this
multiplier: (8x the largest conduit plus the sum of
all the other conduit diameters in the same row) for
all wire sizes to avoid future undersized junction
box problems.
• This assumes no bending and minimal wire storage.
• (NEC 2002 - Article 314.28 for sizing of junction boxes, Article 314.54 directs us to use
methods in Article 314.28.)
44
55
66
7
Junction Box LengthsJunction Box Lengths -- Angle or U PullAngle or U Pull
or with splicesor with splices
• The distance is 6x the diameter of the largest conduit
in the row plus the sum of all the other conduit
diameters in the same row. (There are exceptions to
this in the NEC which we are not likely to encounter.)
• Each row is calculated separately and the single row
that provides the maximum distance shall be used.
• We seldom have more than 3 conduits in a side. This
minimizes conductor bending problems.
8
Exhibit 314.10
An example
showing
calculations
required by
314.28(A)(2)
for splices,
angle pulls or
U pulls.
From 2002 NEC
Handbook, page 313.
99
Water tight conduit
fitting (Meyers Hub)
10
1111
12
Junction Box DepthJunction Box Depth
• The box needs only to be as deep as is needed to install the
conduits and their fittings.
• Consider grounded end bushing size (diameter) in
relationship to the conduit.
• Always round up to the nearest even dimension. Unless
space is restricted, round up to the next (second) even
dimension. (for a 1¼ inch conduit with a 2¼ inches O.D.
grounded end bushing round up to 4 inches as a minimum
but in most cases round up again and call out 6” as the
depth)
• Junction boxes should be positioned to allow all conduits to
enter from the ends.
• This reduces the effort required by the contractor during
construction, allows for easier conductor installation and
allows us to use smaller boxes.
13
Radius of Conduit Bends - NEC 344.24 and 352.24
Conduit
Size
Conduit
Size
One shot
& full shoe
bender
One shot
& full shoe
bender
Other
Bends
Other
Bends
Metric
Designator
Trade
Size mm inch mm inch
16 1/2 101.6 4 101.6 4
21 3/4 114.3 4 1/2 127 5
27 1 146.05 5 3/4 152.4 6
35 1 1/4 184.15 7 1/4 203.2 8
41 1 1/2 209.55 8 1/4 254 10
51 2 241.3 9 1/2 304.8 12
63 2 1/2 266.7 10 1/2 381 15
78 3 330.2 13 457.2 18
91 3 1/2 381 15 533.4 21
103 4 406.4 16 609.6 24
129 5 609.6 24 762 30
155 6 762 30 914.4 36
1) Use entire
chart for GRS
conduit. (NEC
344.24)
2) For field
bending of PVC,
use the “other
bends” columns
only. (NEC
352.24)
14
Radius of Conduit Bends - NEC 353.24 (2005)
Conduit
Size
Conduit
Size
Minimum
Bending
Radius
Minimum
Bending
Radius
Metric
Designator
Trade
Size mm inch
16 1/2 250 10
21 3/4 300 12
27 1 350 14
35 1 1/4 450 18
41 1 1/2 500 20
53 2 650 26
63 2 1/2 900 36
78 3 1200 48
103 4 1500 60
1) Use this chart
for HDPE conduit
only.
15
Conductor Bending RadiiConductor Bending Radii
• There should be consideration given to the minimum
bending radii of the conductors.
• When the junction box is sized properly this will not
usually be a problem.
• This must still be checked to insure there is proper
space for conductor bend Radii.
16
Conductor Bending RadiiConductor Bending Radii
A =(6 x 4 In.) + 2 in.+2 In. +28 In. minimum
B =(6 x 4 In.) + 2 In. +2in.+ 28 In. minimum
C = 6 x 2 In. = 12 In. minimum required
between raceways enclosing the same
conductor
D = 6 x 2 In. = 12 In. minimum required
between raceways enclosing the same
conductor
E = 6 x 4 In. = 24 In. minimum required
between raceways
enclosing the same conductor
Exhibit 314.11 Raceway entries enclosing
the same conductors are required to have
a minimum separation between them. The
intent is to provide adequate space for the
conductor to make the bend.
1717
1818
1919
20
ConductorsConductors
2002 NEC 300.34 Conductor Bending Radius.
The conductor shall not be bent to a radius less than
8 times the overall diameter for nonshielded
conductors or 12 times the diameter for shielded or
lead covered conductors during or after installation.
For multicolor or multiplexed single conductor
cables having individually shielded conductors, the
minimum bending radius is 12 times the diameter of
the individually shielded or 7 times the overall
diameter, whichever is greater.
21
Conductors & CableConductors & Cable
CONDUCTORCONDUCTOR OUTSIDE DIAOUTSIDE DIA MULTIPLIERMULTIPLIER BENDING RADIUSBENDING RADIUS
# 8# 8 0.2660.266 88 2.128’’2.128’’
# 6# 6 0.3040.304 88 2.432’’2.432’’
# 4# 4 0.3520.352 88 2.816’’2.816’’
# 2# 2 0.4120.412 88 3.296”3.296”
(ind.cond.) (ind.conductor)(ind.cond.) (ind.conductor)
#2c(#14)#2c(#14) 0.326 ( 0.14)0.326 ( 0.14) 7(12)7(12) 2.282’’2.282’’
#3c(#20)#3c(#20) 0.35 ( 0.07)0.35 ( 0.07) 7(12)7(12) 2.45’’2.45’’
#4c(#18)#4c(#18) 0.41 ( 0.09)0.41 ( 0.09) 7(12)7(12) 2.87’’2.87’’
#5c(#14)#5c(#14) 0.51 ( 0.14)0.51 ( 0.14) 7(12)7(12) 3.57’’3.57’’
#7c(#14)#7c(#14) 0.55 ( 0.14)0.55 ( 0.14) 7(12)7(12) 3.85’’3.85’’
6pr(#19)6pr(#19) 0.56 ( 0.08)0.56 ( 0.08) 7(12)7(12) 3.92’’3.92’’
24 (SM) fiber24 (SM) fiber 0.470.47 2020 9.40”9.40”
48 (SM) fiber48 (SM) fiber 0.470.47 2020 9.40”9.40”
72 (SM) fiber72 (SM) fiber 0.470.47 2020 9.40”9.40”
96 (SM) fiber96 (SM) fiber 0.780.78 2020 15.60”15.60”
144 (SM) fiber144 (SM) fiber 0.780.78 2020 15.60”15.60”
22
ConductorsConductors
• There should be consideration given for the slack
required in the conductors.
• Standard specification 8-20.3(8) requires sufficient
slack wire be installed to allow any conductor to be
raised 18’’ outside of the junction box.
• Discuss this with the electrical inspectors if you
have many circuits in one junction box / conduit
run.
• If you had a through pull and 4 circuits, which does
happen, you would need space for 24’ of slack.
23
ConductorsConductors -- continuedcontinued
• There was a case with 2 conduits passing through
one junction box carrying 11 conductors; that’s
33’ of slack in an 8’’x 8’’ x 18’’ JB.
• The electrical inspectors will usually hold the
contractor to the slack called for in a standard
structure mounted (traffic barrier) junction box
(8’’x 8’’ x 18’’) because there are usually not
many conductors in the run.
• Standard concrete junction boxes have the same
slack requirements as surface mounted boxes.
24
Fiber Optic CableFiber Optic Cable
• The greater the radius of the conduit, the easier the pull.
• Use cable vaults and pull boxes wherever possible – the
conduit then enters from the side of the box and eliminates
180 degrees of sweep at each box.
• Use 36” radius conduit sweeps wherever possible. This is
the WSDOT standard for all jobs.
• When unable to use 36” radius sweep, use a minimum of 20
times the outside diameter of the cable to calculate the
smallest radius that WSDOT will allow to be used. Note:
This usually only applies inside a building or cabinet.
25
Any questions?Any questions?
26
QUIZ #1 QUESTIONSQUIZ #1 QUESTIONS
u 1) Given: one 400watt HPS luminaire @ 480VAC w/50’
pole and 16’ mast arm, #6 copper conductor, 500’ run of wire
to base of luminaire pole. (all service cabinet terminations,
junction boxes & splice distances included in these lengths)
u Find: what is the voltage drop in this circuit?
u 2) Given: ITS service cabinet @120VAC - 1750 watt load.
u Find: size branch breaker
u 3) Given: 10 Kva transformer @ 240VAC - 8500 watt ITS
service cabinet.
u Find: size transformer branch breaker in service cabinet.
u 4) Given: 1 each 2 inch dia, 1 each 1 1/2inch, 1 each 1 inch
conduit - straight pull with splices.
u Find: size junction box.
27
Quiz #2 QuestionsQuiz #2 Questions
u 1) Given: one 400watt HPS luminaire @ 240VAC w/40’
pole and 16’ mast arm, #6 copper conductor, 350’ run of wire
to base of luminaire pole. (all service cabinet terminations,
junction boxes & splice distances included in these lengths)
u Find: what is the voltage drop in this circuit?
u 2) Given: ITS service cabinet @120VAC - 1750 watt load.
u Find: size branch breaker
u 3) Given: 15 Kva transformer @ 480VAC - 10748 watt load
for 5 ITS service cabinets. (1 ramp meter, 1 Camera
Cabinet, 2 Data Stations & 1 HAR Station)
u Find: size transformer branch breaker in service cabinet.
u 4) Given: 1 each 2 1/2 inch dia, 2 each 2 inch, 1 each 1 1/2
inch conduit - straight pull with splices.
u Find: size junction box.
28
QUIZ #1QUIZ #1 QUESTION ANSWERSQUESTION ANSWERS
u 1) Given: one 400watt HPS luminaire @ 480VAC w/50’
pole and 16’ mast arm, #6 AWG copper conductor, 500’ run
of wire to base of luminaire pole. (all service cabinet
terminations, junction boxes & splice distances included in
these lengths)
u Find: what is the voltage drop in this circuit?
u Within LINE LOSS section, see illumination load chart,
slide 20, for luminaire load (amperage).
u Answer: 400watt luminaire @ 480 volts = 1.1 amp
u Vd #6 AWG =2ALR Vd= 2 x 1.1 x 500 x 0.000510
u Vd #10 AWG =2ALR Vd= 2 x 1.1 x 62 x 0.001290
u Vd #6 AWG=0.561000 volts.Vd #10 AWG=0.175956 volts
u Vd total = 0.74 volts
29
QUIZ #1QUIZ #1 QUESTION ANSWERSQUESTION ANSWERS
u2) Given: ITS service cabinet @120 VAC - 1750
watt load.
uFind: size branch breaker
uAnswer:
uUnit load=1750watts/120volts=14.58amps.
uITS breaker= 14.58 amps continuous load x 125%
factor = 18.2amps.
uBranch breaker = 20 amps
30
QUIZ #1QUIZ #1 QUESTION ANSWERSQUESTION ANSWERS
u3) Given: 10 Kva transformer @ 240 VAC - 8500
watt ITS service cabinet.
uFind: size transformer branch breaker in service
cabinet.
uAnswer:
uUnit load = 10000watts/240volts = 41.66amps.
uITS breaker = 41.66 amps load x 125% factor =
52.08 amps.
uBranch breaker = 60 amps
31
QUIZ #1QUIZ #1 QUESTION ANSWERSQUESTION ANSWERS
u 4) Given: one each 2 inch dia, one each 1 1/2inch, one each
1 inch conduit - straight pull with splices.
u Find: size junction box.
u Answer: Length = 2” x 8 + 1 1/2” + 1”
u Length = 16” + 1 1/2” + 1”. Length = 18 1/2”
u Width=2” x 6 + 1 1/2” + 1”. Width=12” +1 1/2” +1”
u Width=14 1/2”
u Depth = largest conduit is 2” in dia. Outside dia. of 2”
conduit is 2.375” or 2 3/8”. A 2” Grounded end bushing is
+ or - 3 1/4” dia.
u Box size = 18 1/2L x 14 1/2W x 4”D (use 6” for depth to
allow clearance for ground lug on j-box)
32
QUIZ #2QUIZ #2 QUESTION ANSWERSQUESTION ANSWERS
u 1) Given: one 400watt HPS luminaire @ 240VAC w/40’
pole and 16’ mast arm, #6 AWG copper conductor, 350’ run
of wire to base of luminaire pole. (all service cabinet
terminations, junction boxes & splice distances included in
these lengths)
u Find: what is the voltage drop in this circuit?
u Within LINE LOSS section, see illumination load chart,
slide 20, for luminaire load (amperage).
u Answer: 400watt luminaire @ 240 volts = 2.1 amp
u Vd #6 AWG =2ALR Vd= 2 x 2.1 x 350 x 0.000510
u Vd #10 AWG =2ALR Vd= 2 x 2.1 x 52 x 0.001290
u Vd #6 AWG=0.749700 volts.Vd #10 AWG=0.281736 volts
u Vd total = 1.03 volts
33
QUIZ #2QUIZ #2 QUESTION ANSWERSQUESTION ANSWERS
u2) Given: ITS service cabinet @120 VAC - 1750
watt load.
uFind: size branch breaker
uAnswer:
uUnit load=1750watts/120volts=14.58amps.
uITS breaker= 14.58 amps continuous load x 125%
factor = 18.2amps.
uBranch breaker = 20 amps
34
QUIZ #2QUIZ #2 QUESTION ANSWERSQUESTION ANSWERS
u3) Given: 15 Kva transformer @ 480VAC - 10748
watt load for 5 ITS service cabinets. (1 ramp meter,
1 Camera Cabinet, 2 Data Stations & 1 HAR
Station)
uFind: size transformer branch breaker in service
cabinet.
uAnswer:
uUnit load = 15000watts/480volts = 31.25amps.
uITS breaker = 31.25 amps load x 125% factor =
39.06 amps.
uBranch breaker = 40 amps
35
QUIZ #2QUIZ #2 QUESTION ANSWERSQUESTION ANSWERS
u 4) 4) Given: 1 each 2 1/2 inch dia, 2 each 2 inch, 1 each 1
1/2 inch conduit - straight pull with splices. Find: size
junction box.
u Answer: Length = 2 1/2” x 8 + (2) 2” + 1 1/2”
u Length = 20” + 4” + 1 1/2”. Length = 25 1/2”
u Width=2 1/2” x 6 +(2) 2” + 1 1/2”. Width=15” +4” +1 1/2”
u Width=20 1/2”
u Depth = largest conduit is 2 1/2” in dia. Outside dia. of
u 2 1/2” conduit is 2.875” or 2 7/8”. A 2 1/2” Grounded end
bushing is + or - 4” dia.
u Box size = 25 1/2L x 20 1/2W x 6”D (use 6” for depth to
allow clearance for ground lug on j-box)
36
Any questions?Any questions?
3710
C
D
E
3
10
C
D
E
3
Circuit A
Circuit B
Circuit D
Circuit E
Circuit C

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Pullbox sizing

  • 1. 1 NEMA Junction Box Sizing,NEMA Junction Box Sizing, Conductor Bending Radii & FiberConductor Bending Radii & Fiber Optic CableOptic Cable from: NEC 2002 - Article 314.28, Article 314.54 & Article 770 WSDOT Fall / Winter 2004 Presented by: Terry Thayer
  • 2. 2 The clear catinkus for Junction Box Sizing and DepthThe clear catinkus for Junction Box Sizing and Depth u The definition of length and width W L1 L1 L2 Junction box Junction box GRS conduit grounded end bushing size table Size diameter inches Inches External add this number for end bushing size - then round up to nearest quarter end bushing outside diameter 1/2 0.840 0.75 1 1/2 3/4 1.050 0.75 1 3/4 1 1.315 0.75 2 1 1/4 1.660 0.75 2 1/2 1 1/2 1.900 0.75 2 3/4 2 2.375 0.75 3 1/4 2 1/2 2.875 1.00 4 3 3.500 1.25 4 3/4 3 1/2 4.000 1.25 5 1/4 4 4.500 1.75 6 1/4 5 5.563 2.00 7 1/2 6 6.625 2.00 8 3/4 Example: A 1 1/4" GRS conduit is 1.660" + 0.75" = 2.410". Round 2.410" up to 2.5" or 2 1/2 inches. L(1) or L(2) = (8) largest conduit in row plus the sum of all other conduits in row. Calculate for all rows and use sum of largest row. W=(6) largest conduit in row plus the sum of all other conduits in row. Calculate for all rows and use sum of largest row. NOTE: If no conduit enters the adjacent sides of the box the dimension is width. adjacent side
  • 3. 3 Junction Box LengthJunction Box Length -- straight pullstraight pull • If one or more of the conductors is a # 4 or larger, • Then the length is not less than 8x the diameter of the largest conduit if a through pull, regardless of the number of conduits or conductors. We use this multiplier: (8x the largest conduit plus the sum of all the other conduit diameters in the same row) for all wire sizes to avoid future undersized junction box problems. • This assumes no bending and minimal wire storage. • (NEC 2002 - Article 314.28 for sizing of junction boxes, Article 314.54 directs us to use methods in Article 314.28.)
  • 4. 44
  • 5. 55
  • 6. 66
  • 7. 7 Junction Box LengthsJunction Box Lengths -- Angle or U PullAngle or U Pull or with splicesor with splices • The distance is 6x the diameter of the largest conduit in the row plus the sum of all the other conduit diameters in the same row. (There are exceptions to this in the NEC which we are not likely to encounter.) • Each row is calculated separately and the single row that provides the maximum distance shall be used. • We seldom have more than 3 conduits in a side. This minimizes conductor bending problems.
  • 8. 8 Exhibit 314.10 An example showing calculations required by 314.28(A)(2) for splices, angle pulls or U pulls. From 2002 NEC Handbook, page 313.
  • 10. 10
  • 11. 1111
  • 12. 12 Junction Box DepthJunction Box Depth • The box needs only to be as deep as is needed to install the conduits and their fittings. • Consider grounded end bushing size (diameter) in relationship to the conduit. • Always round up to the nearest even dimension. Unless space is restricted, round up to the next (second) even dimension. (for a 1¼ inch conduit with a 2¼ inches O.D. grounded end bushing round up to 4 inches as a minimum but in most cases round up again and call out 6” as the depth) • Junction boxes should be positioned to allow all conduits to enter from the ends. • This reduces the effort required by the contractor during construction, allows for easier conductor installation and allows us to use smaller boxes.
  • 13. 13 Radius of Conduit Bends - NEC 344.24 and 352.24 Conduit Size Conduit Size One shot & full shoe bender One shot & full shoe bender Other Bends Other Bends Metric Designator Trade Size mm inch mm inch 16 1/2 101.6 4 101.6 4 21 3/4 114.3 4 1/2 127 5 27 1 146.05 5 3/4 152.4 6 35 1 1/4 184.15 7 1/4 203.2 8 41 1 1/2 209.55 8 1/4 254 10 51 2 241.3 9 1/2 304.8 12 63 2 1/2 266.7 10 1/2 381 15 78 3 330.2 13 457.2 18 91 3 1/2 381 15 533.4 21 103 4 406.4 16 609.6 24 129 5 609.6 24 762 30 155 6 762 30 914.4 36 1) Use entire chart for GRS conduit. (NEC 344.24) 2) For field bending of PVC, use the “other bends” columns only. (NEC 352.24)
  • 14. 14 Radius of Conduit Bends - NEC 353.24 (2005) Conduit Size Conduit Size Minimum Bending Radius Minimum Bending Radius Metric Designator Trade Size mm inch 16 1/2 250 10 21 3/4 300 12 27 1 350 14 35 1 1/4 450 18 41 1 1/2 500 20 53 2 650 26 63 2 1/2 900 36 78 3 1200 48 103 4 1500 60 1) Use this chart for HDPE conduit only.
  • 15. 15 Conductor Bending RadiiConductor Bending Radii • There should be consideration given to the minimum bending radii of the conductors. • When the junction box is sized properly this will not usually be a problem. • This must still be checked to insure there is proper space for conductor bend Radii.
  • 16. 16 Conductor Bending RadiiConductor Bending Radii A =(6 x 4 In.) + 2 in.+2 In. +28 In. minimum B =(6 x 4 In.) + 2 In. +2in.+ 28 In. minimum C = 6 x 2 In. = 12 In. minimum required between raceways enclosing the same conductor D = 6 x 2 In. = 12 In. minimum required between raceways enclosing the same conductor E = 6 x 4 In. = 24 In. minimum required between raceways enclosing the same conductor Exhibit 314.11 Raceway entries enclosing the same conductors are required to have a minimum separation between them. The intent is to provide adequate space for the conductor to make the bend.
  • 17. 1717
  • 18. 1818
  • 19. 1919
  • 20. 20 ConductorsConductors 2002 NEC 300.34 Conductor Bending Radius. The conductor shall not be bent to a radius less than 8 times the overall diameter for nonshielded conductors or 12 times the diameter for shielded or lead covered conductors during or after installation. For multicolor or multiplexed single conductor cables having individually shielded conductors, the minimum bending radius is 12 times the diameter of the individually shielded or 7 times the overall diameter, whichever is greater.
  • 21. 21 Conductors & CableConductors & Cable CONDUCTORCONDUCTOR OUTSIDE DIAOUTSIDE DIA MULTIPLIERMULTIPLIER BENDING RADIUSBENDING RADIUS # 8# 8 0.2660.266 88 2.128’’2.128’’ # 6# 6 0.3040.304 88 2.432’’2.432’’ # 4# 4 0.3520.352 88 2.816’’2.816’’ # 2# 2 0.4120.412 88 3.296”3.296” (ind.cond.) (ind.conductor)(ind.cond.) (ind.conductor) #2c(#14)#2c(#14) 0.326 ( 0.14)0.326 ( 0.14) 7(12)7(12) 2.282’’2.282’’ #3c(#20)#3c(#20) 0.35 ( 0.07)0.35 ( 0.07) 7(12)7(12) 2.45’’2.45’’ #4c(#18)#4c(#18) 0.41 ( 0.09)0.41 ( 0.09) 7(12)7(12) 2.87’’2.87’’ #5c(#14)#5c(#14) 0.51 ( 0.14)0.51 ( 0.14) 7(12)7(12) 3.57’’3.57’’ #7c(#14)#7c(#14) 0.55 ( 0.14)0.55 ( 0.14) 7(12)7(12) 3.85’’3.85’’ 6pr(#19)6pr(#19) 0.56 ( 0.08)0.56 ( 0.08) 7(12)7(12) 3.92’’3.92’’ 24 (SM) fiber24 (SM) fiber 0.470.47 2020 9.40”9.40” 48 (SM) fiber48 (SM) fiber 0.470.47 2020 9.40”9.40” 72 (SM) fiber72 (SM) fiber 0.470.47 2020 9.40”9.40” 96 (SM) fiber96 (SM) fiber 0.780.78 2020 15.60”15.60” 144 (SM) fiber144 (SM) fiber 0.780.78 2020 15.60”15.60”
  • 22. 22 ConductorsConductors • There should be consideration given for the slack required in the conductors. • Standard specification 8-20.3(8) requires sufficient slack wire be installed to allow any conductor to be raised 18’’ outside of the junction box. • Discuss this with the electrical inspectors if you have many circuits in one junction box / conduit run. • If you had a through pull and 4 circuits, which does happen, you would need space for 24’ of slack.
  • 23. 23 ConductorsConductors -- continuedcontinued • There was a case with 2 conduits passing through one junction box carrying 11 conductors; that’s 33’ of slack in an 8’’x 8’’ x 18’’ JB. • The electrical inspectors will usually hold the contractor to the slack called for in a standard structure mounted (traffic barrier) junction box (8’’x 8’’ x 18’’) because there are usually not many conductors in the run. • Standard concrete junction boxes have the same slack requirements as surface mounted boxes.
  • 24. 24 Fiber Optic CableFiber Optic Cable • The greater the radius of the conduit, the easier the pull. • Use cable vaults and pull boxes wherever possible – the conduit then enters from the side of the box and eliminates 180 degrees of sweep at each box. • Use 36” radius conduit sweeps wherever possible. This is the WSDOT standard for all jobs. • When unable to use 36” radius sweep, use a minimum of 20 times the outside diameter of the cable to calculate the smallest radius that WSDOT will allow to be used. Note: This usually only applies inside a building or cabinet.
  • 26. 26 QUIZ #1 QUESTIONSQUIZ #1 QUESTIONS u 1) Given: one 400watt HPS luminaire @ 480VAC w/50’ pole and 16’ mast arm, #6 copper conductor, 500’ run of wire to base of luminaire pole. (all service cabinet terminations, junction boxes & splice distances included in these lengths) u Find: what is the voltage drop in this circuit? u 2) Given: ITS service cabinet @120VAC - 1750 watt load. u Find: size branch breaker u 3) Given: 10 Kva transformer @ 240VAC - 8500 watt ITS service cabinet. u Find: size transformer branch breaker in service cabinet. u 4) Given: 1 each 2 inch dia, 1 each 1 1/2inch, 1 each 1 inch conduit - straight pull with splices. u Find: size junction box.
  • 27. 27 Quiz #2 QuestionsQuiz #2 Questions u 1) Given: one 400watt HPS luminaire @ 240VAC w/40’ pole and 16’ mast arm, #6 copper conductor, 350’ run of wire to base of luminaire pole. (all service cabinet terminations, junction boxes & splice distances included in these lengths) u Find: what is the voltage drop in this circuit? u 2) Given: ITS service cabinet @120VAC - 1750 watt load. u Find: size branch breaker u 3) Given: 15 Kva transformer @ 480VAC - 10748 watt load for 5 ITS service cabinets. (1 ramp meter, 1 Camera Cabinet, 2 Data Stations & 1 HAR Station) u Find: size transformer branch breaker in service cabinet. u 4) Given: 1 each 2 1/2 inch dia, 2 each 2 inch, 1 each 1 1/2 inch conduit - straight pull with splices. u Find: size junction box.
  • 28. 28 QUIZ #1QUIZ #1 QUESTION ANSWERSQUESTION ANSWERS u 1) Given: one 400watt HPS luminaire @ 480VAC w/50’ pole and 16’ mast arm, #6 AWG copper conductor, 500’ run of wire to base of luminaire pole. (all service cabinet terminations, junction boxes & splice distances included in these lengths) u Find: what is the voltage drop in this circuit? u Within LINE LOSS section, see illumination load chart, slide 20, for luminaire load (amperage). u Answer: 400watt luminaire @ 480 volts = 1.1 amp u Vd #6 AWG =2ALR Vd= 2 x 1.1 x 500 x 0.000510 u Vd #10 AWG =2ALR Vd= 2 x 1.1 x 62 x 0.001290 u Vd #6 AWG=0.561000 volts.Vd #10 AWG=0.175956 volts u Vd total = 0.74 volts
  • 29. 29 QUIZ #1QUIZ #1 QUESTION ANSWERSQUESTION ANSWERS u2) Given: ITS service cabinet @120 VAC - 1750 watt load. uFind: size branch breaker uAnswer: uUnit load=1750watts/120volts=14.58amps. uITS breaker= 14.58 amps continuous load x 125% factor = 18.2amps. uBranch breaker = 20 amps
  • 30. 30 QUIZ #1QUIZ #1 QUESTION ANSWERSQUESTION ANSWERS u3) Given: 10 Kva transformer @ 240 VAC - 8500 watt ITS service cabinet. uFind: size transformer branch breaker in service cabinet. uAnswer: uUnit load = 10000watts/240volts = 41.66amps. uITS breaker = 41.66 amps load x 125% factor = 52.08 amps. uBranch breaker = 60 amps
  • 31. 31 QUIZ #1QUIZ #1 QUESTION ANSWERSQUESTION ANSWERS u 4) Given: one each 2 inch dia, one each 1 1/2inch, one each 1 inch conduit - straight pull with splices. u Find: size junction box. u Answer: Length = 2” x 8 + 1 1/2” + 1” u Length = 16” + 1 1/2” + 1”. Length = 18 1/2” u Width=2” x 6 + 1 1/2” + 1”. Width=12” +1 1/2” +1” u Width=14 1/2” u Depth = largest conduit is 2” in dia. Outside dia. of 2” conduit is 2.375” or 2 3/8”. A 2” Grounded end bushing is + or - 3 1/4” dia. u Box size = 18 1/2L x 14 1/2W x 4”D (use 6” for depth to allow clearance for ground lug on j-box)
  • 32. 32 QUIZ #2QUIZ #2 QUESTION ANSWERSQUESTION ANSWERS u 1) Given: one 400watt HPS luminaire @ 240VAC w/40’ pole and 16’ mast arm, #6 AWG copper conductor, 350’ run of wire to base of luminaire pole. (all service cabinet terminations, junction boxes & splice distances included in these lengths) u Find: what is the voltage drop in this circuit? u Within LINE LOSS section, see illumination load chart, slide 20, for luminaire load (amperage). u Answer: 400watt luminaire @ 240 volts = 2.1 amp u Vd #6 AWG =2ALR Vd= 2 x 2.1 x 350 x 0.000510 u Vd #10 AWG =2ALR Vd= 2 x 2.1 x 52 x 0.001290 u Vd #6 AWG=0.749700 volts.Vd #10 AWG=0.281736 volts u Vd total = 1.03 volts
  • 33. 33 QUIZ #2QUIZ #2 QUESTION ANSWERSQUESTION ANSWERS u2) Given: ITS service cabinet @120 VAC - 1750 watt load. uFind: size branch breaker uAnswer: uUnit load=1750watts/120volts=14.58amps. uITS breaker= 14.58 amps continuous load x 125% factor = 18.2amps. uBranch breaker = 20 amps
  • 34. 34 QUIZ #2QUIZ #2 QUESTION ANSWERSQUESTION ANSWERS u3) Given: 15 Kva transformer @ 480VAC - 10748 watt load for 5 ITS service cabinets. (1 ramp meter, 1 Camera Cabinet, 2 Data Stations & 1 HAR Station) uFind: size transformer branch breaker in service cabinet. uAnswer: uUnit load = 15000watts/480volts = 31.25amps. uITS breaker = 31.25 amps load x 125% factor = 39.06 amps. uBranch breaker = 40 amps
  • 35. 35 QUIZ #2QUIZ #2 QUESTION ANSWERSQUESTION ANSWERS u 4) 4) Given: 1 each 2 1/2 inch dia, 2 each 2 inch, 1 each 1 1/2 inch conduit - straight pull with splices. Find: size junction box. u Answer: Length = 2 1/2” x 8 + (2) 2” + 1 1/2” u Length = 20” + 4” + 1 1/2”. Length = 25 1/2” u Width=2 1/2” x 6 +(2) 2” + 1 1/2”. Width=15” +4” +1 1/2” u Width=20 1/2” u Depth = largest conduit is 2 1/2” in dia. Outside dia. of u 2 1/2” conduit is 2.875” or 2 7/8”. A 2 1/2” Grounded end bushing is + or - 4” dia. u Box size = 25 1/2L x 20 1/2W x 6”D (use 6” for depth to allow clearance for ground lug on j-box)