Topic 5 three phase, four wire systems topic b5

Solve problems in single and threephase low voltage circuits

Topic 5: Three-Phase, Four-Wire
Systems


Three-Phase, Four-Wire Systems








A three-phase, four-wire system consists of three ACTIVE
conductors plus a NEUTRAL conductor (note: the earth
conductor does not normally carry current therefore is excluded).
This type of system must be a STAR-connected system, due to
the existence of the neutral conductor, and is most often
connected to unbalanced loads.
Unbalanced loads produce an out-of-balance current at the starpoint (the phases have unequal current flows in them, thus they
cannot take away all of the current that is supplied by the other
phases).
The NEUTRAL conductor’s primary purpose is to carry this outof-balance current away from the star-point.

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Nathan Condie

2

Determining the Neutral Current

ZA

ZB
12Ω

5Ω
Pƒ=1

For the circuit diagram drawn,
determine the following:

ZC

1) Current flowing INTO the
star point IABC

8Ω

Pƒ=1

Pƒ=1

2) Neutral current IN
3) Phase angle ØN

L1

L2

L3

N

3Ø, 400 V
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Nathan Condie

3


Determining the Neutral Current


Process
 Step 1: Calculate the current flowing in each phase using Ohm’s
Law.
 Step 2: Convert the power factor for each load into a phase
angle.
 Step 3: Draw a scaled phasor diagram (based on V as the
A
reference phasor) of Currents at their appropriate phase angle to
their respective phase voltage.
 Step 4: Use phasor addition to find the resultant current (I )
ABC
 Step 5: Draw this current 1800 out-of-phase and label I .

N
 Step 6: Measure and label the phase angle (Ø ) compared to the
N
reference.

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Nathan Condie

4


Determining the Neutral Current
VC
IC in-phase
IABC
VA (Reference)
IA in-phase
IN
IB in-phase

IAB

VB
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Neutral Current: Answers

Answers:
ZA

ZB
12Ω

5Ω
Pƒ=1

L1

1) IABC = 23.4A

ZC
8Ω

Pƒ=1

2) Neutral current IN = 23.4A

Pƒ=1

L2

L3

3) Phase angle ØN = 1590 Lag

VA
N

3Ø, 400 V
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Nathan Condie

6


Neutral Current Exercises


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Complete the exercises on three-phase,
four-wire systems (calculating neutral
current)

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7


Neutral Current Exercise


Exercise 2

ZA

ZB

ZC

10Ω

10Ω

10Ω

Pƒ=
0.8lag

Pƒ=
0.8lag

Pƒ=
0.8lag

L1

For the circuit diagram drawn,
determine the following:

L2

L3

1) IA, IB, IC
2) Ø1, Ø2, Ø3,
3) Draw a scaled phasor
diagram to determine IABC
N

4) Neutral current IN
5) Phase angle ØN

3Ø, 400 V
20/02/17

Nathan Condie

8


Answer – balanced load
VC
IC 370 lag

VA (Reference)

IA 370 lag

IB 37 lag
0

IAB
VB

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Neutral Current Exercise


ANSWER 2
Answers:
1) IA=IB=IC=23.1A
ZA

ZB

ZC

10Ω

10Ω

10Ω

Pƒ=
0.8lag

Pƒ=
0.8lag

Pƒ=
0.8lag

2) ØA =ØB =ØC=36.90E Lag
3) IABC = 0A (BALANCED
LOAD)
4) Neutral current IN = 0.0A

L1

L2

L3

N

5) Phase angle ØN = 00 INPHASE

3Ø, 400 V
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Nathan Condie

10


Neutral Current Exercise


Exercise 3

ZA

ZB

ZC

20Ω

20Ω

20Ω

Pƒ=
0.8lag

Pƒ=
0.6lag

Pƒ=
0.4lag

L1

For the circuit diagram drawn,
determine the following:

L2

L3

1) IA, IB, IC
2) Ø1, Ø2, Ø3,
3) Draw a scaled phasor
diagram to determine IABC
N

4) Neutral current IN
5) Phase angle ØN

3Ø, 415 V
20/02/17

Nathan Condie

11


Neutral Current Exercise


ANSWER 3
Answers:
1) IA=IB=IC=12A
ZA

ZB

ZC

20Ω

20Ω

20Ω

2) ØA=36.90 lag VA, ØB =53.10
lag VB, ØC=66.40 lag Vc

Pƒ=
0.8lag

Pƒ=
0.6lag

Pƒ=
0.4lag

3) IABC = 4.9A
4) Neutral current IN = 4.9A

L1

L2

L3

N

5) Phase angle ØN = 1680 lag
reference

3Ø, 415 V
20/02/17

Nathan Condie

12


Neutral Current Exercise


Exercise 4

ZA

ZB

ZC

10Ω

15Ω

20Ω

Pƒ= 1

Pƒ=
0.8lag

Pƒ=
0.6lag

L1

For the circuit diagram drawn,
determine the following:

L2

L3

1) IA, IB, IC
2) Ø1, Ø2, Ø3,
3) Draw a scaled phasor
diagram to determine IABC
N

4) Neutral current IN
5) Phase angle ØN

3Ø, 400 V
20/02/17

Nathan Condie

13

Neutral Current Exercise


ANSWER 4
Answers:
1) IA=IB=IC=12A
ZA

ZB

ZC

10Ω

15Ω

20Ω

2) ØA=36.9 lag VA, ØB =53.1 lag
VB, ØC=66.40E lag Vc

Pƒ= 1

Pƒ=
0.8lag

Pƒ=
0.6lag

3) IABC = 4.9A
4) Neutral current IN = 4.9A

L1

L2

L3

N

5) Phase angle ØN = 1680 lag
reference

3Ø, 400 V
20/02/17

Nathan Condie

14


Neutral Current Exercise


Exercise 5

ZA

ZB
30Ω

40Ω

ZC

1) IA, IB, IC

50Ω

Pƒ=
Pƒ=
0.5lag 0.8lead

L1

For the circuit diagram drawn,
determine the following:

2) Ø1, Ø2, Ø3,

Pƒ=
0.9lag

L2

L3

3) Draw a scaled phasor

diagram to determine IABC
N

4) Neutral current IN
5) Phase angle ØN

3Ø, 11kV
20/02/17

Nathan Condie

15


Neutral Current Exercise


ANSWER 5
Answers:
ZA

ZB
30Ω

40Ω

ZC
50Ω

Pƒ=

Pƒ=
0.5lag 0.8lead

L1

2) ØA=600 lag VA, ØB =36.90
lead VB, ØC=25.80 lag Vc

Pƒ=
0.9lag

L2
3Ø, 11kV

20/02/17

1) IA=158.8A; IB=211.7A;
IC=127A

L3

3) IABC = 241A
N

4) Neutral current IN = 241A
5) Phase angle ØN = 1130 lead
reference

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16


Functions of a Neutral
Conductor


Functions of a Neutral Conductor








To carry the OUT-OF-BALANCE current away from
the star-point.
To maintain the phase voltages equal in value.
To allow the connection of single-phase loads.
To carry fault currents in a distribution system that
uses the MEN system.
To carry third harmonic currents produced in circuits
with certain types of loads.

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Neutral Conductor Size Specifications:
AS3000
New Specifications AS3000:2007 Clause 3.5.2
 Multiphase Circuits
 The current-carrying capacity of the neutral conductor
of the consumer’s mains, sub-mains, and final subcircuits shall be not less than the CCC of the largest
associated active conductor.

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Effects of a Broken or High
Impedance Neutral
A.
B.

Effects on the load
Effects on the MEN system


Effects of a Broken or High Impedance
Neutral

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Effects of a Broken or High Impedance
Neutral


Effects on Three-Phase Load


The out-of-balance current MUST be reabsorbed
by the star system





Higher currents will flow in the phases.
The phase voltages will become irregular




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causes system instability.

some phase voltages higher, some lower.

The star-point will “FLOAT” above zero potential.

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Effects of a Broken or High Impedance
Neutral

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Effects of a Broken or High Impedance
Neutral


Effects on MEN System








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The out-of-balance current will attempt to travel via the
MEN link through to the earth electrode, and back to the
supply transformer.
Since the earth return path is normally poor, insufficient
current will flow thus causing the star-point to float above
zero potential.
This floating voltage will be impressed onto the earthing
system, causing it to become live.
This represents a significant risk of electric shock. Note:
The circuit protection is unlikely to recognize this situation
as a fault, thus will remain energized.

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Effects of a Broken or High Impedance
Neutral

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