1

Branch CircuitsChapter 2

2

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?

3

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?

4

Service Entrance Panel [SEP] “Load Center”

Master Disconnect 120/240 V Service

Service “Hot” Conductors

Metal Box

Non-conducting base

Non-conducting Attachment bars

Conducting Attachment bars

120 V Branch Circuit Ground (bare) Conductor

240 V Circuit

120 V Branch Circuit “Hot” (black) Conductor

Ground

Neutral

120 V

120 V

120 V Branch Circuit Neutral (white) Conductor

Earth Ground

Bonding Screw

Circuit Neutral & Ground Connections

Grounding Bar

Breaker

Service Entrance Neutral

5

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.

Grounding

• Branch circuits have two different types of grounds.– System– Equipment

6

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?

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.

7

Equipment ground

What hazard is created if the equipment ground is interrupted?

120 V Circuits

• 120 V circuits have 3 or 4 conductors:– one energized (hot) conductor Black or red– one neutral conductor White – one ground conductor. Bare or green

8

What other common building component is made from PVC

What does PVC stand for?

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

• Used for known specific loads– Stationary motors– Stationary appliances

• SPOs (Special Purpose Outlets)– Usually used for loads greater than 20 amps—240 V.

12

What would be an example of an SPO in an agriculture building?

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 Branch circuit, short circuit protection

2 A disconnecting means

3 A controller

4 Overload protection

• 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

load current.

• Motor starting load is higher than the running load—SCP 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.

16

€

2Table - 3. 1/2 hp = 9.8 V

Breaker: 9.8 2.5x =24.5 AFuse : 9.8 1.75x =17.15

• Determine the required SCP for a 240 V circuit for a 1/6 horsepower, single phase motor.

€

2Table - 3 : 1/6 hp =2.2 V

Breaker: 2.2 2.5x =5.5 AFuse : 2.2 1.75x =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.

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 the motor full load rating.

18

What is a snap switch?

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.

19

Is a heater/airconditioner thermostat within sight and 50 feet of the furnace/airconditioner?

If not, does this meet code?

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.

20

What is a magnetic starter?

Motor Circuits—Controller—Magnet Starter

21

Motor Circuits—Overload protection

• Motors and conductors must be protected from overloads.

• Because motors draw more current for starting that running, the overload protection device must allow temporary overload on the circuit but not allow the overload to last long enough to damage the motor.

22

When magnetic starters are used the overload protection is usually included 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.

Motor Circuits—Overload protection—cont.

• 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

Motor Circuits—Overload protection—cont.

• 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.

24

Why?

Branch Circuit Conductors

25

26

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).

27

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.

Sizing Conductors—cont.

• The minimum size of an individual conductor is determined by two factors.

– Ampacity– Voltage drop

28

What is ampacity? What is voltage drop?

29

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:

€

R =LA

R = (resistance of the conductor Ω) = (resistivity of the conductor material Ω - / )cmil ft

A = ( )cross sectional area of conductor cmils

L = ( )Length of conductor ft

A = cross-sectional area in cmils = (diameter in mils)2

1mil = 0.001 in

30

Example--Resistance

• What is the resistivity of a 1/2 inch steel rod that is 12 feet long?

– Steel = 100 ohm-cmil/foot

€

mil=0.50 in x1 mil

0.001 in=500 mil

Area = (500 )mil2 = 250,000 cmil

Electricity for Agricultural Applications, Bern

Material Resistivity ()

Silver 9.55

Copper 10.67

Gold 14.7

Aluminum 17.01

Tungsten 33.10

Platinum 66.9

Steel 100

Lead 129

Cast iron 360

Mercury 577

€

=Ω ⋅cmil

ft

€

(Restance Ω) = L

A

(ResistanceΩ) =100 12x

250, 000=0.0048 Ω

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.

31

What are some possible outcomes of a circuit with excessive voltage drop?

What will cause excessive voltage drop?

3232

Voltage Drop--Cont.

• When there is no current flow, there is no voltage at the load.

• A 2 % voltage drop is considered normal.

• If the voltage drop is more than 2% the circuit will overheat.

3333

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)

3434

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 Ω

3535

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.

3636

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.

€

1Rt

=1R1

+1

R2+ . . .

1Rn

3737

Parallel Circuits--cont.

An alternative equation is:

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.

€

Rt =R1 x R2R1 + R2

3838

Parallel Circuit Resistance

Determine the total resistance for the circuit in the illustration.

€

1Rt

=1R1

+1R2

+1R3

= 1

2.5+

14.0

+1

5.2 =

2152

+1352

+1052

= 4452

= 0.846

Rt =1

0.846=1.18 Ω

€

1Rt

=1R1

+1R2

+1R3

= 1

2.5+

14.0

+1

5.2 = 0.4 + 0.25 + 0.192 = 0.842

Rt =1

0.842=1.19 Ω

€

Rt =R1 x R2R1 + R2

=2.5 4.0x

2.5 + 4.0=

106.5

=1.54

1 .54 x5 .21.54 + 5.2

=8 .06 .74

=1 .19 Ω

€

1Rt

=1R1

+1

R2+ . . .

1Rn

€

Rt =R1 x R2R1 + R2

or

or

3939

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.

4040

Calculating Voltage In A Series Circuit

• What would V1 read in the illustration?

• Ohm’s Law states:

• Therefore:

€

R=E

I

€

E = IR

• At this point there is insufficient data because I (amp) is unknown.

• Using Ohm’s Law to solve for the current in the circuit:

• Knowing the amount of current we can calculate the voltage drop.

€

E = IR I=E

R=

120 V13.2 Ω

=9.09 A

€

E= IR=9 .09 5.0 A x Ω =45.4 V

Note: circuit conductors behave like resistors in series.

4141

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.

4242

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.

€

E = IR I=ER

=12 V

1.5 + 4.2 + 6.3 Ω= 1.0 A

• The current in the circuit is determined by dividing the voltage by the circuit resistance.

4343

Determining Amperage in a Parallel Circuit

Determine the readings

for amp 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 R2R1 + R2

=1.5 4.2x

1.5 + 4.2=

6.35.7

=1.10

1.10 6.3x

1.10 + 6.3=

6.967.40

=0.94 Ω

4444

Determining Amperage in a Parallel Circuit--cont.

When the total resistance is known, the circuit current (Amp’s) can be calculated.

Branch current is:

€

I =E

R=

12 V6.3 Ω

=1.9 A A2 = 1.9 A

A1= 12.76 A

€

I =E

R=

12 V0.94 Ω

=12.76 ATotal current is:

When the circuit current (Amp’s) is known, the current for each branch circuit can be calculated.

45

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

The type of insulation is determined by the environment.

46

Insulation

The common types used in Agriculture.

Insulation Designation

Trade Name

Temp era ture Ra ting (oC)

Des cription and Common uses

THHN Hea t resis tant the rmop las tic

90 Flame re ta rdant, hea t resis tan t the rmop las tic insulated ind ividu al conductor s . For use in condu it, dry and damp loc a tion s .

THWN Moisture and hea t resis tant the rmop las tic

75 Flame re ta rdant, mo isture resis tant, heat resis tant the rmop las tic insulated ind ividu a l conductor s . For use in condu it, dry and we t location s .

NM Non m e tallic shea the cable

60 Two or t hree conductor s (plus ba re ground ing conductor ) in a mo isture resis tant, flame re ta rdant non me ta llic shea th. For u se in no rmally dry loc a tion s. Cannot be imb edd ed in pour ed concr e te or use d as se rvice entranc e cable. Use d in family dwelling s not excee ding 3 floor s abo ve grade and oth e r st ructur es . Can be use d expo se d or conc eale d.

SE Se rvice entranc e cable

75 Common ly three conductor s in a flame re ta rdant, mois ture resis tant co ve ring . The neutral is braided a round the two ene rgize d conductor s . Type SE is use d primarily be twee n an abo ve ground po int of a tta chm ent and th e se rvice entranc e panel.

USE Und e rground se rvice entranc e cable

75 S ing le conductor s cabled into an unco ve red asse mb ly for direct bu rial as a fee de r or branch circu it or se rvice la te ra l. Co ve ring is mo is ture resis tant bu t not necessa rily flame re ta rdant, or prot ected against mech anical abu se .

UF Und e rground fee de r cable

60 Two or t hree conductor s (plus ba re ground ) with flame re ta rdant, mo is ture resis tant, fungu s resis tant, co rrosion resis tant co ve ring for direct bur ial as a fee der or branch circu it. Also use d for interior wiring in we t, dry or corro sive loc a tion s . Use d in lives tock bu ilding s , Cannot be expo se d to d irect sun light un less la bel specifies “sun light r esis tant”. Cannot be use d as se rvice entranc e cable Cannot be emb edd ed in pour ed concr e te .

Multiplex (triples , qu adrup lex)

Ove rhea d fee de r

90 Two or t hree insulated a luminum conductor s wound a round a ba re s trand ed messe ng e r which se rves as a neutral, and support s the asse mb ly. The messe ng e r cont ains on e s teel s trand for s trength

Electricity for Agricultural Applications, Bern

90o C = 194o F

47

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.

48

Determining Conductor Size

• The first step is to determine answers for five of the seven factors. These are: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)

• 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

49

• Circuit load– 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.

Determining Conductor Size--cont.

50

• 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.

Determining Conductor Size--cont.

51

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)

52

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

53

• 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?

Sizing Conductors by Voltage Drop

54

Example--cont.

The first step is to determine the total resistance of the circuit.

Copper Aluminum

Size (AWG)

Area (cmil)

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.

55

Example--cont.

This circuit diagram illustrates the resistance of the conductors.

56

Example--cont.

• The next step is to determine the voltage drop and the percentage drop.

• Voltage drop is:

• Percent voltage drop is:

€

RT = 0.0491 Ω + 0.0491 Ω

€

E = I R = 50 0.0982 amp x Ω = 4.91 V

€

120 V - 4.91 V = 115.06 V

€

% VD =120 V - 115.06 V

120 V= 4.09 %

• This is an unacceptable voltage drop.

• Picking the wire size first must not be the best way.

The conductor size is determined by calculating the allowable resistance for the desired voltage drop.

57

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. Decrease the load.

2. Use larger conductors.

3. Reduce the distance between the load and the source.

4. Use a conductor that has a lower resistance.

5. Use a higher voltage.

• Of these 5 options, number 2 and 5 are usually the only practical solution.

58

Voltage Drop Example--cont.

If the voltage is increased to 240 V, what will be the percent voltage drop?

€

RT = 0.0982 ΩE = IR = 50 0.0982 A x Ω = 4.91 V240 V - 4.91 V = 235.09 V

%VD =4.91 V240 V

100x = 2.04 %

2.04% is an acceptable loss for this electrical service.

59

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

Conductor Size Equation

The equation for calculating the conductor size (cmils) for a specified voltage drop:

60

€

( )A cmil = x I x L

E

A = ( )size of conductor cmils

= resistivity of conductor

I = amperage in the circuit

L = length of individual conductor

E = allowable voltage drop

Note: it is common practice to add 10% to the length to account for the resistance of the connections.

61

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.

62

Conductor Size Example-cont.

€

A =ρIL

E=

10.37 x 6.6 x (150 x 1.1)

120 x 0.02=

11292.93

2.4= 4705.39 cmil

4705.39 cmil = AWG #12

Ampacity = #12

Voltage Drop = #12 Note: 10% has been added for connections.

Copper Aluminum

Size (AWG)

Area (cmil)

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

63

Resistivity Equations

Because of inductance in the conductor, the equation for copper is usually changed for design purposes to:

€

A =22 x I xl

E

The resistivity equation for aluminum conductors is changed to:

€

A =1.6 22 x x I x l

E

Note: in these equations the length is the one way length, not the total length.

22 = constant for copperI = Circuit load (amp)l = Run (distance)E = Allowable voltage drop (V)

64

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

65

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=PE

=1500 0.70W x

120 V= 8.75 amp

66

Conductor size #3--cont.

For this circuit with a 8.75 amp load, the circular mills of the conductors can be determined by:

The conductor size is:

€

A =22 x I xl

E=

22 8.75 (250 1.1)x x x2.4

= 22, 057.29 cmil

#6 = 26,240 cmil

#8 = 16,510 cmil.

€

( )A cmil =22 ( ) x I Amp xl ( )ft

( )E V

Size (AWG) Area (cmil)

14 4110

12 6530

10 10380

8 16510

6 26240

4 41740

3 52620

2 66360

Which size should be used?

67

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.

68

Questions ?