### Three Phase Circuits Ch 4 - Biosystems and Agricultural Engineering

```Branch Circuits
Chapter 2
1
Introduction
•
•
The purpose of branch circuits is to carry the current
from the service entrance panel (SEP) to the electrical
devices.
Three common types of current are used in agricultural
buildings:
– 120 volt (108 – 125)
– 240 volt (220 – 250)
– Three phase
Why are a range of voltages listed?
What happens when the voltage drops below 108 V?
What happens when the voltage goes above 124 V?
2
Service Entrance Panel (SEP)
•
•
•
•
The service entrance panel (load center) is the entry
point for the electricity into the building.
The size (amp capacity) of the load center is
determined by the number of circuits and total amp load
for the building.
Current NEC regulations require that the load center
have a master disconnect.
The entrance panel must be grounded with a NEC
approved earth connection.
What is an NEC earth connection?
3
Service Entrance Panel [SEP] “Load Center”
Service Entrance
Neutral
Master Disconnect
120/240 V Service
Service “Hot” Conductors
Metal Box
Breaker
Non-conducting base
Grounding Bar
Non-conducting Attachment
bars
Circuit Neutral &
Ground
Connections
Conducting Attachment bars
Ground
Bonding Screw
Neutral
120 V
120 V Branch Circuit
“Hot” (black)
Conductor
Earth Ground
120 V Branch Circuit
Neutral (white)
Conductor
240 V Circuit
120 V
120 V Branch Circuit Ground (bare)
Conductor
4
SEP--cont.
•
•
•
•
•
The 120/240 service is attached to
the master disconnect (breaker).
From master breaker each hot
conductor is connected to one of
the conducting breaker bars.
The 120/240 neutral conductor is
attached to the grounding bar.
A 120 volt breaker attaches by snapping onto one conducting and
one non-conducting bar in the load center.
For a 240 volt circuit two individual breakers may be used and the
levers are pined together or a combination breaker may be used.
5
Grounding
•
Branch circuits have two different types of grounds.
– System
– Equipment
System grounding is accomplished by one of the two current carrying
conductors (white).
What is another term for the system ground?
What is the insulation color of the system ground?
What is the insulation color of the equipment ground?
6
Grounding - Equipment
•
•
•
Equipment grounding is the bonding of
all non current carrying metal components
back to the SEP.
The equipment ground is designed to
provide a low resistant circuit to the earth
in the event of a short from the energized
conductor to any metallic component.
Must make a complete, low resistance
circuit from all metallic electrical devices in
the system to the earth.
Equipment ground
What hazard is created if the equipment ground is interrupted?
7
120 V Circuits
•
120 V circuits have 3 or 4 conductors:
– one energized (hot) conductor
– one neutral conductor
– one ground conductor.
Black or red
White
Bare or green
What does PVC stand for?
What other common building
8
240 V Circuits
•
240 Volt circuits have three conductors:
– Two hot
– Equipment ground
•
•
Neutral circuit is not required unless both 240 and 120
circuits are supplied by the device.
The 240 Volt electrical service to the SEP will have a
neutral so both 240 and 120 Volt branch circuits can be
used.
9
Three Circuit Types
•
•
•
General purpose branch circuits
Individual branch circuits
Motor
10
General Purpose Branch Circuits
•
•
•
•
•
Designed for temporary loads such as lights and DCOs
(Duplex Convenience Outlets) under 1500 W.
Minimum 12 AWG
Fused at 20 amps
No more than ten (10) DCOs or light fixtures per circuit.
(Fig. 2-3, pg 21)
Recommended location for DCOs (Table 1-12, pg 18).
11
Special Purpose Branch circuits
•
– Stationary motors
– Stationary appliances
•
SPOs (Special Purpose Outlets)
– Usually used for loads greater than 20 amps—240 V.
What would be an example of an SPO in an agriculture building?
12
Motor Circuits
•
Use 240 V whenever possible.
– Reduces amperage load on circuit
– Reduces stray voltage potential
•
Five (5) horsepower and larger should be 3 phase.
13
Motor Circuits—cont.
•
Branch circuits for electric motors have four (4) requirements (Fig 2-6
through 9, pg 23-26):
1
2
3
4
•
Branch circuit, short circuit protection
A disconnecting means
A controller
Summary Table 2-1, pg 26
14
Motor Circuits—Short Circuit Protection
•
Fuse or circuit breaker
– For motor circuits they must have greater capacity than full
•
devices must be able to handle temporary overload.
– Inverse time breaker
– Time delay fuses
•
Maximum size
– Inverse time breaker = 2.50 times full load current
– Time delay fuses = 1.75 times full load current.
15
Motor Circuits—Short Circuit Protection--example
•
Determine the required SCP for a 120 V circuit for a ½ horsepower,
single phase motor.
Table 2 - 3. 1/2 hp = 9.8 V
Breaker : 9.8 x 2.5 = 24.5 A
Fuse : 9.8 x 1.75 = 17.15
• Determine the required SCP for a 240 V circuit for a 1/6 horsepower,
single phase motor.

Table 2 - 3 : 1/6 hp = 2.2 V
Breaker : 2.2 x 2.5 = 5.5 A
Fuse : 2.2 x 1.75 = 3.8 A

• Smallest breaker is 10 A
• Solution—Use 10 A breaker
in the SEP and install a 4 to 6
amp fuse inline with the
motor.
16
Motor Circuits—Disconnecting Means
•
•
•
Each motor or motor circuit must have an
individual disconnecting means.
The disconnecting means must
disconnect all hot wires.
The DM must clearly indicate whether it is
on or off.
17
Motor Circuits—Disconnecting Means-cont.
•
•
Must be located within sight and within 50 feet of the controller
and the motor.
Disconnecting Means
– Stationary motors can use the circuit switch as long as correct size.
– Portable motors the plug and receptacle is acceptable.
•
The circuit switch can be a snap switch as long as the motor is 2
hp or less and its capacity is equal to or 1.25 times greater than
What is a snap switch?
18
Motor Circuits—Controller
•
•
A controller is a device used to
automatically start and stop a motor.
Only required to open enough
conductors to stop the motor.
– One wire for 120 & 240 V single phase.
•
•
Must be located within sight and 50 feet
of the motor.
Thermostats, variable speed controllers
and timers are considered to be a
controller.
Is a heater/airconditioner thermostat within sight and 50 feet of the
furnace/airconditioner?
If not, does this meet code?
19
Motor Circuits—Controller—cont.
•
•
•
•
Current rating must be greater than
or equal to motor full load rating, or
a magnetic starter must be used.
For 1/3 hp and less portable motors
the plug and receptacle can
function as the controller.
If motor is 2 hp or less, a snap
switch an serve as the controller.
If a knife switch is operated by
hand, it can serve as both the
disconnecting means and the
controller.
What is a magnetic starter?
20
Motor Circuits—Controller—Magnet Starter
21
• Motors and conductors must be
• Because motors draw more current
for starting that running, the
circuit but not allow the overload to
last long enough to damage the
motor.
When magnetic starters are used the
in the starter.
Common practice to use a heater device to trip the controller before the
conductors or motor overheats.
One hp and larger motors have specific requirements based on the design
and size of the motor.
22
•
•
For motors less than 1 hp, and manually started, the
circuit breaker or fuse can serve as the OPD.
Smaller motors may include a built in overload protection
switch.
23
•
•
Critical issue is if a manual restart or automatic restart is
used.
Manual restart is usually used unless the motor operates
a critical function such as a ventilation fan in a chicken
house.
Why?
24
Branch Circuit Conductors
25
Sizing Conductors
• Conductors are usually considered single wires.
• Cables are multiple conductors in the same sheathing.
•Conductor are sized using two systems—
 American Wire Gauge (AWG)
 circular mills (cmil).
26
Sizing Conductors—cont.
•
AWG
– Numbers run from 40 to 0000
– AWG numbers only apply to non-ferrous metals.
– The larger the number--the smaller the diameter of the wire.
•
cmils
– Circular-mils (cmils) is a unit used to describe the cross-sectional area of
wire.
– A mil = 0.001 inch
– AWG sizes greater than 0000 are sized in thousands of circular mils (kcmil)
•
AWG #8 and higher are usually multiple strands.
•
The diameter of multiple strand wire in cmils is the cmils of each strand
times the number of strands.
27
Sizing Conductors—cont.
•
The minimum size of an individual conductor is determined by
two factors.
– Ampacity
– Voltage drop
What is ampacity?
What is voltage drop?
28

Ampacity--Resistivity
•
•
•
All materials will conduct electricity.
Good conducting materials have low resistance.
The resistance of a conductor depends on the physical properties of
the material (), the length (ft) of the conductor and the cross-sectional
area of the conductor (cmils).
•
Expressed in an equation:
A = cross-sectional area in cmils = (diameter in mils)2
1mil = 0.001 in
R =
L
A
R  resistance of the conductor ( )
 = resistivity of the conductor material (
 - cmil/ft)
A = cross sectional area of conductor (cmils)
L = Length of conductor (ft)
29
Material
Example--Resistance
•
What is the resistivity of a 1/2 inch steel
rod that is 12 feet long?
– Steel = 100 ohm-cmil/foot
  cmil
 =
ft
mil = 0.50 in x
1 mil
 500 mil
0.001 in
Area = (500 mil)2 = 250,000 cmil
Silver
9.55
Copper
10.67
Gold
14.7
Aluminum
17.01
Tungsten
33.10
Platinum
66.9
Steel
100
129
Cast iron
360
Mercury
577
Electricity for Agricultural Applications, Bern
Restance 
()=
L
Resistance )
( =

Resistivity ()
A
100 x 12
250,000
 0.0048 
30
Voltage Drop
•
•
•
•
•
When electricity passes through a resistance heat is generated.
Heat is energy
The loss energy shows up as voltage drop.
All conductors have resistance = all conductors have voltage drop.
What must be avoided is excessive voltage drop.
What will cause excessive voltage drop?
What are some possible outcomes of a circuit with excessive voltage
drop?
31
Voltage Drop--Cont.
• When there is no current
flow, there is no voltage at
• A 2 % voltage drop is
considered normal.
• If the voltage drop is
more than 2% the circuit
will overheat.
32
Three Ways of Wiring Circuits
• The loads and electrical components in a circuit can be connected in
three different ways:
– Series
– Parallel
– Series-parallel (not included)
33
Series Circuit
•
•
In a series circuit the electricity has no alternative paths, all of the
electricity must pass through all of the components.
The total circuit resistance is the sum of the individual resistances.
Rt = R1 + R2 + . . . Rn
For these calculations assume no
resistance in the conductors or
connections.
Determine the total resistance for the circuit in the illustration.
Rt = 5.0 + 8.2 = 13.2 
34
Series Circuit-cont.
To the power source, a series circuit appears as one resistance.
=
•
•
In all circuits a voltage drop occurs as electricity passes through each
resistance in the circuit.
The method for calculating voltage drop in series circuits is different
than the method for parallel circuits.
35
Parallel Circuits
• In parallel circuits the electricity has alternative paths.
• The amount of current in each path is determined by the resistance of
that path. “Electricity follows the path of least resistance”
• Because there are alternative paths, the total resistance of the circuit is
not the sum of the individual resistances.
• In a parallel circuit: The inverse of the total resistance is equal to sum
of the inverse of each individual resistance.
1
1
1
1
=
+
+.. .
Rt
R1 R2
Rn

36
Parallel Circuits--cont.
An alternative equation is:
Rt =
R1 x R2
R1 + R2
When a circuit has more than two resistors, select any two and reduce
them to their equivalent resistance and then combine that resistance

with another one in the circuit until all of the resistors have been
combined.
37
Parallel Circuit Resistance
Determine the total resistance
for the circuit in the illustration.
1
1
1
1
=
+
+.. .
Rt
R1 R2
Rn
1
Rt
=
1
R1
+
1
R2
+
1
1
=
R3
2.5

1
4.0
+
1
5.2
=
21
52

13
52
+
10
52
=
44
= 0.846
52

Rt =
or

1
Rt
=
Rt =
1
R1
1
0.846
+
1
R2
1
0.842
= 1.18 
+
R3
=
1
2.5

1
4.0
+
Rt =
1
5.2
= 0.4 + 0.25 + 0.192 = 0.842
Rt =
= 1.19 
or

1
R1 x R2
R1 + R2
R1 x R2
R1 + R2
1 .54 x 5.2
1.54 + 5.2
=

2.5 x 4.0
2.5 + 4.0
8 .0
6 .74
=
10
6.5
= 1.54
 1 .19 
38
Circuits Summary
• When the source voltage, and the total resistance of the circuit is
known, amperages and voltages can be determine for any part of a
circuit.
• In a series circuit the amperage is the same at all points in the circuit,
but the voltage changes with the resistance.
• In a parallel circuit the amperage changes with the resistance, but the
voltage is the same throughout the circuit.
39
Calculating Voltage In A Series Circuit
•
•
•
What would V1 read in the illustration?
Ohm’s Law states:
E = IR
Therefore:
E
R=
•
•
•
I
At this point
there is insufficient data
because I (amp) is unknown.
Using
Ohm’s Law to solve for the
E = IR
current in the circuit:
Knowing the amount of current we
can calculate the voltage drop.

I=
E 120 V
=
 9.09 A
R 13.2 
E = IR= 9.09 A x 5.0  = 45.4 V
Note: circuit conductors behave like resistors in series.

40
Determining Voltage In A Parallel Circuit
Assuming no resistance in the conductors, the two volt meters in the
illustration will have the same value--source voltage.
41
Determining Amperage In A Series Circuit
• Determine the readings for A1 and A2 in the illustration.
• In a series circuit the electricity has no alternative paths, therefore the
amperage is the same at every point in the circuit.
• The current in the circuit is determined by dividing the voltage by
the circuit resistance.
E = IR

I=
E
12 V
=
= 1.0 A
R
1.5 + 4.2 + 6.3 
42
Determining Amperage in a Parallel Circuit
meters A1 and A2 in the circuit.
•
In a parallel circuit the amperage varies with the resistance.
•
In the illustration, A1 will measure the total circuit amperage, but A2 will
only measure the amperage flowing through the 6.3 Ohm resistor.
•
To determine circuit amperage the total resistance of the circuit must
be calculated:
Rt =
R1 x R2
R1 + R2
1.10 x 6.3
1.10 + 6.3
=
=
1.5 x 4.2
1.5 + 4.2
6.96
7.40
=
6.3
5.7
= 1.10
= 0.94 
43
Determining Amperage in a Parallel Circuit--cont.
When the total resistance is known,
the circuit current (Amp’s) can be
calculated.
Total current is:
I=
E
12 V
=
= 12.76 A
R 0.94 
A1= 12.76 A
When the circuit current (Amp’s) is known, the current for each
branch circuit can be calculated.

Branch current is:
I=
E 12 V
=
= 1.9 A
R 6.3 
A2 = 1.9
A

44
Conductor Size
The conductor size is determined by seven (7) factors.
1. the load on the circuit
2. the voltage of the circuit
3. the distance from the load to the source
4. the circuit power factor
5. the type of current (phases)
6. the ampacity of the conductor
7. the allowable voltage drop
45
Insulation
Insulation
Designation
THHN
The common
types used in
Agriculture.
90o C = 194o F
THWN
Heat resistant
thermoplastic
Temperature
o
Rating ( C)
90
Moisture and
heat resistant
thermoplastic
Non metallic
sheathe cable
75
SE
Service
entrance cable
75
USE
Underground
service
entrance cable
75
UF
Underground
feeder cable
60
Multiplex
(triples,
feeder
90
NM
60
Description and Common uses
Flame retardant, heat resistant thermoplastic insulated
individual conductor s. For use in condu it, dry and damp
locations.
Flame retardant, moisture resistant, heat resistant
thermoplastic insulated individual conductor s. For use in
condu it, dry and wet locati ons.
Two or t hree conductor s (plus bare grounding conductor ) in
a moisture resistant, flame retardant non metallic sheath.
For use in normally dry locations. Cannot be i mbedded in
pour ed concrete or used as service entrance cable. Used
in family dwellings not exceeding 3 floors above grade and
other structures. Can be used exposed or conc ealed.
Common ly three conductor s in a flame retardant, moisture
resistant covering. The neutral is braided around the two
energized conductor s. Type SE is used primarily between
an above ground point of attachment and th e service
entrance panel.
Single conductor s cabled into an unco vered assembly for
direct burial as a feeder or branch circuit or service lateral.
Covering is moisture resistant but not necessarily flame
retardant, or protected against mechanical abuse.
Two or t hree conductor s (plus bare ground) with flame
retardant, moisture resistant, fungus resistant, corrosion
resistant covering for direct bur ial as a feeder or branch
circuit. Also used for interior wiring in wet, dry or corrosive
locations. Used in livestock buildings, Cannot be exposed
to direct sunlight unless label specifies Òsunlight resistantÓ.
Cannot be used as service entrance cable Cannot be
embedded in pour ed concrete.
Two or t hree insulated aluminum conductor s wound around
a bare stranded messenger which serves as a neutral, and
support s the assembly. The messenger contains one steel
strand for strength
Electricity for Agricultural Applications, Bern
46
Environment--cont.
•
The selection of insulation is very important because the
life of the conductor is usually determined by the life of
the insulation.
•
Conductors never wear out.
•
Insulation deteriorates over time.
– Insulation reacts with oxygen, ammonia, oil, gasoline, salts,
UV and water.
47
Determining Conductor Size
•
The first step is to determine answers for five of the seven factors.
These are:
1.
2.
3.
4.
5.
•
the voltage of the circuit
the distance from the load to the source
the circuit power factor
the type of current (phases)
Once these are known, the remaining two factors are used to determine
the conductor size.
5. the ampacity of the conductor
6. the allowable voltage drop
48
Determining Conductor Size--cont.
•
– The circuit load is the amperage used by the electrical device, or the size of
over current protection device that will be used.
•
Circuit voltage
– Circuit voltage is the source voltage.
•
Distance from source
– The distance between the source and the load is not used as often as the
run.
– The run is the total amount of conductor that is used to connect the load to
the source.
•
Power factor
– The power factor for reactive loads is less than one.
– The power factor for resistance loads is equal to one.
•
The number or phases must be know.
– Three phase current can use smaller diameter wires.
49
Determining Conductor Size--cont.
•
Once values are known for the first five factors, the last
two are used to determine the minimum conductor size.
– Ampacity is the largest load that a conductor is designed to
carry regardless of length.
– Voltage drop is the amount of energy that is lost from the
electricity passing through the resistance of the conductors.
50
Ampacity
•
•
•
•
•
•
Ampacity refers to the current carrying ability of the conductor.
Ampacity is dependent on the conductor resistance, the allowable
operating temperature of the insulation and the heat dissipation ability
of the conductor.
Ampacity increases with conductor size.
Ampacity for copper is higher than the ampacity for aluminum.
Ampacity is higher for conductors which have higher temperature
ratings.
Exceeding the ampacity rating increases the heat of the insulation.
– The amount of damage that occurs is a function of the amount of overload
and the duration of the overload.
•
Ampacity ratings for conductors can be determined from tables such as
33-19. (Agricultural Mechanics)
51
Ampacity-cont.
•
•
Ampacity can be calculated, but
tables present this information.
Example: what is minimum size
of conductor with THWN
insulation in conduit, operating
on 120 volts that should be used
to carry 15 amps?
AWG 14 ??
Note: Based on ampacity, #14 is
sufficient, but according to the NEC
#12 is smallest size of wire that can
be used under any conditions using
120 V.
Pg 35 Wiring handbook
52
Sizing Conductors by Voltage Drop
•
•
Voltage drop is the result of a
current passing through a
resistance.
Example:
– What is the percent voltage drop
at the service entrance panel for
the building in the illustration?
53
Example--cont.
The first step is to determine the total
resistance of the circuit.
Copper
Size
(AWG)
Area
(cmil)
Aluminum
Number of
Strands
Weight (lb/1000
ft)
Resistance
(ohms/1000 ft)
Weight
(lb/1000 ft)
Resistance
(ohms/1000 ft)
14
4110
1
12.5
3.07
3.78
5.06
12
6530
1
19.8
1.93
6.01
3.18
10
10380
1
31.43
1.21
9.556
2.00
8
16510
1
49.98
0.764
15.20
1.26
6
26240
7
79.44
0.491
24.15
0.808
4
41740
7
126.3
0.308
38.41
0.508
3
52620
7
159.3
0.245
48.43
0.403
2
66360
7
205
0.308
62.3
0.319
In this example the
resistance for each
conductor is determined
separately.
54
Example--cont.
This circuit diagram illustrates the
resistance of the conductors.
55
Example--cont.
•
•
•
The next step is to determine
the voltage drop and the
percentage drop.
Voltage drop is:
Percent voltage drop is:
• This is an unacceptable voltage
drop.
• Picking the wire size first must not
be the best way.
R T = 0.0491   0.0491 
E = I R = 50 amp x 0.0982 = 4.91 V
The conductor size is determined by
calculating the allowable resistance for the  
desired voltage drop.
120 V - 4.91 V = 115.06 V
% VD =
120 V - 115.06 V
= 4.09 %
120 V


56
Voltage Drop Example--cont.
•
•
A voltage drop of 4.09% is excessive.
Results of excessive voltage drop.
– The heat output of a resistance heater will decrease more than 8% because
power output is proportional to the square of the voltage.
– The useable light from an incandescent lamp will drop about 10%.
•
Five (5) possible solutions:
1.
2.
3.
4.
5.
•
Use larger conductors.
Reduce the distance between the load and the source.
Use a conductor that has a lower resistance.
Use a higher voltage.
Of these 5 options, number 2 and 5 are usually the only practical
solution.
57
Voltage Drop Example--cont.
If the voltage is increased to 240 V, what will be the percent voltage drop?
RT = 0.0982 
E = IR = 50 A x 0.0982 = 4.91 V
240 V - 4.91 V = 235.09 V
%VD =
4.91 V
x 100 = 2.04 %
240 V
2.04% is an acceptable loss for this electrical service.

58
Designing For Acceptable Voltage Drop
•
•
Because all conductors have resistance, it can not be eliminated
from the circuit.
Therefore, circuits are designed for a specific voltage drop.
– Maximum of 2% in branch circuits is common standard
– NEC allows up to 3% maximum in branch circuit at farthest power outlet
– NEC allows 5% maximum drop in feeder and branch circuit combined
59
Conductor Size Equation
The equation for calculating the conductor size (cmils) for a specified
voltage drop:
Note: it is common practice to
add 10% to the length to
account for the resistance of
the connections.
A (cmil) =
xIxL
E
A = size of conductor (cmils)
 = resistiv ity of conductor
I = amperage in the circuit
L = length of indiv idual conductor
E = allowable v oltage drop

60
Conductor Size Example -1
Using the resistivity equation, determine the size of conductor that should be
used to power an grow lamp that draws 6.6 amps and is operating on single
phase and 120 volts. The grow lamp is located 75 feet from the nearest
source. Copper conductors will be directly buried and a 2% voltage drop is
acceptable.
61
Conductor Size Example-cont.
A=
IL 10.37 x 6.6 x (150 x 1.1) 11292.93
E
=
=
120 x 0.02
2.4
= 4705.39 cmil
4705.39 cmil = AWG #12

Ampacity = #12
Note: 10% has been added for connections.
Voltage Drop = #12
Copper
Size
(AWG)
Area
(cmil)
Aluminum
Number of
Strands
Weight (lb/1000
ft)
Resistance
(ohms/1000 ft)
Weight
(lb/1000 ft)
Resistance
(ohms/1000 ft)
14
4110
1
12.5
3.07
3.78
5.06
12
6530
1
19.8
1.93
6.01
3.18
10
10380
1
31.43
1.21
9.556
2.00
8
16510
1
49.98
0.764
15.20
1.26
6
26240
7
79.44
0.491
24.15
0.808
4
41740
7
126.3
0.308
38.41
0.508
3
52620
7
159.3
0.245
48.43
0.403
2
66360
7
205
0.308
62.3
0.319
62
Resistivity Equations
Because of inductance in the conductor, the equation for copper is usually
changed for design purposes to:
22 x I x
A =
E

22 = constant for copper
l = Run (distance)
E = Allowable voltage drop (V)
The resistivity equation for aluminum conductors is changed to:
A =
1.6 x 22 x I x
E
Note: in these equations the length is the one way
length,
 not the total length.
63
Conductor Size Example #2
Using the table method, determine the size of copper conductor that should
be used to provide electrical service to a livestock building that is located 65
feet from the source. The service is 120 V and the estimated load for the
building is 35 amps. UF-B cable and a 2% voltage drop will be used.
First determine minimum size based on ampacity. =
Second: determine size based on VD =
AWG #8
AWG 6
64
Conductor Size Example #3
•
•
•
Determine by calculation, using the standard equation, the size of conductor
that will be required to provide service to a 120 V, 1-1/2 hp water pump that
is located 250 feet from its source. The conductors with be copper and
directly buried. A 2% voltage drop is acceptable. The motor has a power
factor of 0.70.
The standard equation for copper requires values for amperage, voltage
and length.
The load on the circuit, amperage, must be determined first.
• 1 hp = 746 watts, but when determining conductor sizes for electric motors it
is common practice to use 1,000 watts per horsepower.
P = EI
I=
P
1500 W x 0.70
=
= 8.75 amp
E
120 V

65
Conductor size #3--cont.
For this circuit with a 8.75 amp load, the circular mills of the conductors can be
determined by:
22 x I (Amp) x (f t)
A (cmil) =
E (V)
The conductor size is:
A =
22 x I x
E
=
22 x 8.75 x (250 x 1.1)

= 22, 057.29 cmil
2.4
Size (AWG)
#6 = 26,240 cmil
#8 = 16,510 cmil.
Which size should be used?
Area (cmil)
14
4110
12
6530
10
10380
8
16510
6
26240
4
41740
3
52620
2
66360
66
Conductor Sizing Conclusion
•
•
For short distances, ampacity determines the minimum conductor size.
For long distances, voltage drop determines the minimum size.
Because long and short are relative terms, both ampacity and
voltage drop must be checked when sizing conductors.
67
Questions ?
68
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