Download Analyse interconnected systems

Contents
Introduction
3
Interconnected systems
4
Fault conditions
14
Blown supply fuse (star load)
14
Blown supply fuse (delta load)
21
Reversed phase winding (star load)
27
Reversed phase winding in delta
29
Open delta connection
31
Summary
34
Answers
40
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Introduction
The power source used to supply loads, whether single-phase, two-phase or
three-phase, is typically a star connected system. This system allows for any
of the three types of loads to be connected.
For single-phase and two-phase loads, the relationship of line to phase
values is as described for a star system. For three-phase star connected
loads, the relationship of line to phase values is the same for the load as for
the supply. However, for three-phase delta connected loads, there is a
complication, as the load phase values will not equal the supply phase
values. This is addressed in the first part of this section.
Later in the section we will look at various fault conditions that may arise.
At the end of this section you should be able to:

draw the typical combinations of three phase interconnected systems
using star-connection and delta-connection

show the relationship between line and phase voltages and line and
phase currents in the typical interconnected systems using starconnections and delta-connections

give example of loads in typical power systems

state the limitations and uses of open delta connections

state the effect of a reversed phase winding of a delta connected
transformer
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Interconnected systems
This term applies to systems which have one type of arrangement, star or
delta, for power delivery, and the other type at the load. Figure 1 shows star
supply–delta load. Figure 2 shows delta supply–star load.
Figure 1: Star supply-delta load
Figure 2: Delta supply-star load
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As you can see from Figures 1 and 2, there is no connection point for a
neutral. The first impression is that the loads connected in this type of
system must be balanced, and usually they are, but there are some instances
where the load is unbalanced.
Typically, the star system is used for the supply, with the three-phase loads
being in either delta or star. If the supply is in star, any type of load may be
connected. If the phase load is unbalanced, the supply must be in star. The
types of loads that can be connected to a star supply were discussed in
Section 3. They are:

single-phase loads where the supply is one active and a neutral (eg
appliances such as toasters)

single-phase loads where the supply is two actives (eg welders)

two-phase loads, where the supply to the load is two actives and a
neutral (eg cooking ranges)

balanced three-phase loads, where the supply is three actives (eg
three-phase motors)

unbalanced three-phase loads, where the supply is three actives and a
neutral (eg three-phase and one-phase load combinations).
Some of these loads are shown in Figure 3.
Figure 3: Connection of loads to three-phase four-wire supply
To determine the phase and line values of current and voltage for either the
load or the supply, we use the methods already discussed in this module.
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Example 1
A three-phase star connected transformer develops 240 V in each phase
winding and supplies a three-phase delta connected motor. The impedance
of each stator windings is 24 . Determine the:
(a)
line potential of the supply
(b)
phase potential at the load
(c)
phase currents in the load
(d)
line currents from the supply
(e)
phase currents in the supply.
Solution
Step 1
Sketch the circuit (Figure 4).
Figure 4: Circuit for example 1
Step 2 Determine the line potential of the supply. Remember that in star,
VLine  3 Vphase
As the supply is always balanced,
VAB  VBC  VCA  415 V
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Step 3 Determine the phase potential at the load. Remember that in
delta, VLine  V phase and that as the voltages supplied to the load are
balanced, the phase voltages must also be balanced.
VAB  VA  415 V
VBC  VB  415 V
VCA  VC  415 V
Step 4 Determine the phase currents using Ohm’s law.
I phase 
V phase
Z phase
415
24
 17.3 A

As each phase has the same impedance,
I A  I B  IC  17.3 A
Step 5 Determine the line currents from the supply. Remember that in a
balanced delta load I line  3 I phase , and that as the phase currents in
the load are equal, the line currents supplied to the load are also
equal.
I A  3 I AB
 3 17.3
 30 A
I B  3 I BC
 3 17.3
 30 A
I C  3 I CA
 3 17.3
 30 A
Step 6 Determine the phase currents in the supply. Remember that in star
I line  I phase , and that as the line currents supplied to the load are
equal, the phase currents in the supply are also equal.
I A phase  I A line  30 A
I B phase  I B line  30 A
I C phase  I C line  30 A
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If the delta connected load is unbalanced the problem is still solved in the
same way, the only difference being that the phase currents will not be equal
and the line currents will not be equal.
Example 2
In this example we will consider an extremely unbalanced load where each
phase has a different impedance and a widely different phase angle. You
will not be examined with this type of problem, but it will help you to
understand what the effect of an unbalanced load is. Work through the
example step by step. The end result provides a good reason why threephase loads should be as balanced as possible.
A three-phase star connected transformer develops 240 V, 50 Hz in each
phase winding and supplies a three-phase delta connected load. The load is
made up of a 20  resistor, a 25  impedance with a power factor of 0.8
lagging, and a 160 µF capacitor. Determine the:
(a)
line potential of the supply
(b)
phase potential at the load
(c)
phase currents in the load
(d)
line currents from the supply
(e)
phase currents in the supply.
Solution
Step 1
Sketch the circuit (Figure 5).
Step 2 Determine the line potential of the supply.
Recall that in star, VLine  3 Vphase
 VAB  3 VA
 3  240
 415 V
As the supply is always balanced,
VAB  VBC  VCA  415 V
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Figure 5: Circuit for example 2
Step 3 Determine the phase potential at the load. Remember that in delta,
VLine  V phase , and that as the voltages supplied to the load are
balanced, the phase voltages must also be balanced.
VAB  VA  415 V
VBC  VB  415 V
VCA  VC  415 V
Step 4 Determine the phase impedances.
Z A  RA2  X A2
 202  02
 20 
Z B  25 
ZC  RC 2  X C 2
Now:
XC 
1
2 fC
1
2    50 160 10 –6
 19.9 

 Z  02  19.92
C
 19.9 
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Step 5
Determine the phase angle of the current using the
equation I  V – Z :
  VBC  ZBC
 00
 0
IBC  VBC – ZBC
 –120–36.8
 –156.8
ICA  VCA –ZCA
 120–(–90)
 210
Step 6
Determine the phase currents in the load using Ohm’s law.
I phase 
V phase
Z phase
As each phase has the same voltage,
I AB 
VAB
Z Ab
415
20
 20.75 A
V
 BC
Z BC

I BC
415
25
 16.6 A
V
 CA
Z CA

I CA
415
19.9
 20.8 A

Step 7 Determine the line currents from the supply. Remember that in an
unbalanced delta load, we have
I A  I AB – I CA
I B  I BC – I AB
I C  I CA – I BC
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and that the calculations must be done using a phasor diagram.
Select a suitable scale—we will use 5 A10 mm. The scaled values are
20.75 10
5
 41.5 mm at 0
16.6 10

5
 33.2 mm at –156.8
20.8 10

5
 41.6 mm at 210
I AB 
I BC
I CA
Step 8 Draw the phasor and add the phase currents to determine the line
current. This is shown in Figure 6.
Figure 6: Phasor diagram for example 2
Step 9 Measure the line currents.
I A  80 mm
I B  73 mm
I C  9.5 mm
Convert the scaled values of line current to real current values in amperes:
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80  5
10
 40 A
73  5
IB 
10
 36.5 A
9.5  5
IC 
10
 4.7 A
IA 
Step 10 Determine the phase currents in the supply. Remember that in star,
I Line  I phase , and that as the line currents supplied to the load are
equal, the phase currents in the supply are also equal.
I A phase  I A line  40 A
I B phase  I B line  36.5 A
I C phase  I C line  4.7 A
The results show two obvious problems with connecting this load to the
supply.
12

There are two heavily loaded phases in the supply, while the third
phase has very little load. This gives rise to problems at the point of
supply.

The line current in line A and line B (the two most heavily loaded
lines) exceeds the nominal amount of 3 times the phase current.
This can cause extra heating of the supply cables and will increase
voltage drop in those two lines, so that the line terminal voltage at
the load could drop at one or two points. This may cause the load to
malfunction.
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Student exercise 1
A three-phase star connected transformer develops 415 V between the line terminals
and supplies a three-phase delta connected motor. The impedance of the stator
windings is 15 . Determine the:
(a) phase potential of the supply
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
(b) phase potential at the load
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
(c) phase currents in the load
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
(d) line currents from the supply
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
(e) phase currents in the supply.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
Check your answers with those given at the end of the section.
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Fault conditions
There are a number of fault conditions that can arise within a three-phase
system. They can be loosely classified as follows:

blown supply fuse

reversed phase winding in star

reversed phase winding in delta

open delta condition.
Blown supply fuse (star load)
When looking at this problem, we must begin by determining what type of
load—star or delta—is connected to the supply system. They react
differently when the supply loses one of the three phases. How the load
reacts also depends on whether the supply is three-phase three wire or threephase four wire. We will consider how to deal with each case, beginning
with star loads.
If a three-phase three wire supply loses one line to a star connected load, the
load will appear to the supply as a single phase load, with one phase of the
load having no power supplied to it. This is illustrated in Figure 7, where
line A is open circuit to the load.
Figure 7: Supply fault to star load (three-phase three-wire)
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If we change the diagram in Figure 7 so that only the phases supplied with
voltage are shown, we can see that the result of an open circuit in the line, or
a blown supply fuse, is that two phases of the load are in series with each
other. This is shown in Figure 8.
Figure 8: Simplified diagram of fault condition
In the circuit in Figure 8 the B phase and the C phase in the load are
connected in series with each other, and there is line voltage applied to this
combination. The line voltage will be divided proportionally across the two
load impedances, and both load phases will have the same current flowing in
them, regardless of whether they are balanced or unbalanced.
The result is that phase voltages across the loads may be different,
depending on whether the load is balanced or unbalanced. In this text, we
will use balanced loads.
Example 3
Three 24  heating elements are connected in star configuration to a threephase three wire 415 V, 50 Hz supply. If the fuse in line B blows, determine
the:
(a)
voltage across each phase of the load
(b)
phase currents in the load
(c)
line currents supplied to the load.
Solution
Step 1
Draw the circuit as in Figure 9, circling or highlighting the open
circuit fuse.
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Figure 9: Circuit diagram for example 3
Step 2
Redraw the circuit as in Figure 10 to show the phases still supplied
with voltage.
Figure 10: Redrawn circuit for example 3
Step 3 Determine the total impedance of the load. For a series circuit use
the equation Zt  Z1  Z 2 .
Zt  Z A  ZC
As the load is resistive, Z equals R.
Z A  RA  24 
Z C  RC  24 
Zt  24  24
 48 
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Step 4
Determine the phase voltages using the voltage divider equation:
VA 
Vt Z A
Zt
415  24
48
 207.5 V

VC 
Vt Z C
Zt
415  24
48
 207.5 V

VB = 0 V, as there is no current path.
Step 5 Determine the phase currents using Ohm’s law.
I phase 
IA 
V phase
Z phase
VA
ZA
207.5
24
 8.6 A

As the two phases are in series, the current in the A phase is the same as the
current in the C phase.
I A  IC  8.6 A
Step 6 Determine the line currents. In a series circuit there is only one
current:
I line  I phase  8.6 A
If the supply to the load is a three-phase four-wire star system and one line
becomes open circuit, the load will not be classed as a single phase load.
There will be a potential from each line terminal of the load still supplied
with power to the neutral point in the load. This is shown in Figure 11.
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Figure 11: Fault in supply to star load (three-phase four-wire)
The two phases of the load still connected to the supply will each react
individually as if there was no problem.
If we look at the circuit of Figure 11, we can see that the phase voltages
across the two phases supplied with power are unchanged by the loss of one
phase. Since the phase voltages in those phases are unchanged, the current
flowing in them is also unchanged. However, the current in the neutral
conductor will now be the phasor sum of two currents, not three. This means
that the neutral current will acquires a substantial value.
Example 4
Three 24  heating elements are connected in star configuration to a threephase four wire 415 V, 50 Hz supply. If the fuse in line C blows, determine
the:
(a)
voltage across each phase of the load
(b)
phase currents in the load
(c)
line currents supplied to the load
(d)
neutral current.
Solution
Step 1
18
Draw the circuit as in Figure 12 and circle or highlight the open
circuit fuse.
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Figure 12: Circuit for example 4
Step 2
Determine the phase voltages using the equation
Vline  3 V phase
VAB
3
 240
VA 
VB  VA  240 V
Step 3 Determine the phase currents using Ohm’s law.
I phase 
IA 
V phase
Z phase
VA
ZA
240
24
 10 A

IB 
VC
ZC
240
24
 10 A

IC  0 A
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Step 4
Determine the line currents. As the load is star connected, the line
currents must equal the phase currents.
I line  I phase  10 A
Step 5
Determine the neutral current. Remember that this must be done
using a phasor diagram.
I N  I A  I B  IC
Because the impedance of each phase is purely resistive, there is no phase
separation between the phase currents and their corresponding phase
voltages. Therefore:
 A  0
B  –120
C  120
The currents may therefore be written as:
I A  10 A at 0
I B  10 A at –120
I C  0 A at 120
Convert the current values to lengths, using a scale of 2 A10 mm.
10 10
2
 50 mm at 0
IA 
10 10
2
 50 mm at –120
IB 
0 10
2
 0 mm at 120
IC 
Draw the phasor diagram as in Figure 13, and add the phase quantities to
determine the neutral current value.
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Figure 13: Phasor circuit drawn for example 4
Measure the length of the neutral phasor:
I N  50 mm
and convert it to a real value.
50  2
10
 10 A
IN 
As you can see, the phase values and the line values of current and voltage
remain the same, but the neutral current increases from 0 A (because it was
a balanced load initially) to 10 A.
Blown supply fuse (delta load)
As there is no connection point for a neutral conductor in a delta load, the
only type of supply available to it is three-phase three-wire. If the supply
loses one line to a delta connected load, the load, like the star load just
discussed, will appear to the supply as a single phase load. This is shown in
Figure 14, where line A is open circuit to the load.
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Figure 14: Fault in supply to delta load
If we draw the circuit in Figure 14 so that only the lines supplying voltage to
the load are shown, as in Figure 15(a), we can see that the result of an open
circuit in the line or a blown supply fuse is that two phases of the load are
placed in series with each other, and that this series combination is in
parallel with the third phase. This is shown more clearly in Figure 15(b).
Figure 15: Simplified diagram of delta load fault condition
As you can see, if the fuse in line A blows, the A phase and the C phase in
the load become connected in series with each other and this combination is
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connected in parallel with the B phase. The circuit has line voltage applied
to the combination. The result of this is that the load phase voltages for the
A phase and the C phase are equal, and that the load phase current in the A
phase is the same as that in the C phase.
Remember a series circuit is also called a voltage divider circuit, because the
line voltage will be divided proportionally across the load impedances. As a
delta load (a three-phase motor for example) is typically balanced,
VA  VC  0.5 VB .
For the B phase, the phase voltage is equal to the line voltage, and the phase
current is the normal load current. The line current will decrease from
3I phase to 1.5 I phase in two lines and drop to zero in the line with the blown
fuse.
Example 5
Three 24  heating elements are connected in delta to a three-phase three
wire 415 V, 50 Hz supply. If the fuse in line B blows, determine the:
(a)
voltage across each phase of the load
(b)
phase currents in the load
(c)
line currents supplied to the load.
Solution
Step 1 Draw the circuit as in Figure 16, and circle or highlight the open
circuit fuse.
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Figure 16: Circuit for example 5
Step 2 Redraw the circuit as in Figure 17 to show the situation more
clearly.
Figure 17: Redrawn circuit for example 5
Step 3 Determine the impedances of the load. For the series circuit use the
equation Zt  Z1  Z 2
As the load is a heating load, Z is equal to R, since the inductive
reactance (XL) is 0 .
Z A  RA
Z C  RC
Z AC  RA  RC
 24  24
 48 
Z B  24 
Step 4 Determine the phase voltages for the series phases of the load using
the voltage divider equation:
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ZA 
Vt  Z A
Zt
415  24
48
 207.5 V

VC 
Vt  Z C
Zt
415  24
48
 207.5 V

VB  415 V
Determine the phase currents using Ohm’s law.
Step 5
I phase 
IA 
V phase
Z phase
VA
ZA
207.5
24
 8.6 A

As the two phases are in series, the current in the C phase of the load is
the same as the current in the A phase of the load.
I C  I A  8.6 A
IB 
VB
ZB
415
24
 17.3 A

Step 6
Determine the line currents. As the load now appears to be a single
phase load, and in a parallel circuit the total current is the sum of
the branch currents, the line currents must equal the sum of the
branch (phase) currents:
I line  I A  I B
 8.6  17.3
 25.9 A
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Student exercise 2
1
Three 41.5  heating elements are connected in delta configuration to a three-phase
three-wire 415 V, 50 Hz supply. If the fuse in line C blows, determine the:
(a) voltage across each phase of the load
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
(b) phase currents in the load
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
(c) line currents supplied to the load.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
2
Three 41.5  heating elements are connected in star configuration to a three-phase
four wire 415 V, 50 Hz supply. If the fuse in line B blows, determine the:
(a) voltage across each phase of the load
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
(b) phase currents in the load
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
(c) line currents supplied to the load
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_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
(d) neutral current.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
Check your answers with those given at the end of the section.
Reversed phase winding (star load)
We know that the voltages developed in a three phase star winding are
displaced by 120°. However, this will only occur if the correct method of
connection is used. That method is to connect either all the starts or all the
finishes of the windings together to form the star point. This is shown in
Figure 18.
Figure 18: Correct star connections
This connection allows each of the phase voltages to be displaced by 120°,
as shown in Figure 19.
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Figure 19: Phase voltages for correct connection
If, however, one of the phases is reversed in an alternator or supply
transformer, the resultant phasor causes the line voltages to change in both
magnitude and phase displacement. This is illustrated in Figure 20, which
shows the C phase as the reversed phase, and in Figure 21, which shows the
phasor addition to determine the line voltages.
Figure 20: Phase voltages with C phase winding reversed
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Figure 21: Line voltages with reversed winding
As you can see from Figure 21, only one line potential is correct (VBC), with
the other two line potentials (VAB and VCA) lower than normal. This can
prevent the load from functioning as intended. A polarity check must be
performed on the windings before connection. This is covered in the
resource module Transformers.
Reversed phase winding in delta
The voltages developed in a three phase delta winding are also displaced by
120°, provided again that the correct method of connection is used. The
method is to connect the start of one winding to the finish of the previous
winding to form a closed loop called a delta connection. This is shown in
Figure 22.
Figure 22: Correct delta connection
This connection allows each of the phase voltages to be displaced by 120° as
shown in Figure 23. The phasor sum of these voltages is 0 V.
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Figure 23: Phase voltages for correct connection
If, however, one of the phases is reversed in a supply transformer, the phase
voltages, when added on the phasor, will total twice the normal phase
voltage value. This is illustrated in Figure 24, which shows the reversed
phase, and Figure 25, which shows the phasor addition to determine the total
voltage.
Figure 24: One winding reversed
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Figure 25: Phase voltages with reversed winding
The large resultant voltage only exists within the delta connection, and as
the windings are of low impedance, this will create a very large circulating
current in the transformer. The result will be that the windings will burn out.
To find out whether the connection has been made correctly, a simple but
effective method is to break the delta loop and place a voltmeter across the
break. If the voltmeter reads zero, the connection is correct. If the voltmeter
reads twice Vphase, then one phase in the delta connection is reversed. This is
illustrated in Figure 26.
Figure 26: Checking for correct connections
Open delta connection
In the open delta connection, one phase of a delta connected load is open
circuited, either accidentally or deliberately.
This is shown in Figure 27. It should not be mistaken for a blown line fuse.
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Figure 27: Open delta load
A three phase supply to a load can be obtained using two single phase
transformers. The circuit for the open delta using two single phase
transformers is shown in Figure 28. In this circuit, the two primaries are
connected in series. Three terminals are thus created, one at the unconnected
end of each primary and one at the junction of the primaries. The load is
connected in a similar fashion to series connected secondaries.
Figure 28: Open delta supply
This type of connection has the disadvantage of not producing a symmetrical
supply, and so it does not have the same power capabilities as a star or delta
system. The available power from the open delta system is 57.7% of what
would be available from a supply connected in delta with three windings in
the supply transformer. For this reason, the system is rarely used.
More often an open delta connection is produced as the result of an accident;
for example, where one winding of a motor burns out. If the load becomes
open delta connected, the supply (three phase three wire) is unaffected. The
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EGG202A: 13 Analyse interconnected systems
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line potential is also unaffected, but the value of the line currents now equals
that of the phase currents, and the normal relationship of I line  3 I phase ,
does not apply.
As one phase of the load is not functioning, the load output is reduced to
57.7% of the rated delta connected output. The result of this is that the
machine is placed into an overload condition with normal full load. The
machine therefore heats up, with the possibility that further damage will
occur.
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Summary

Interconnected systems are systems with either a star supply and a
delta load or a delta supply and a star load.

The most common supply system is star connected.

The types of loads that can be connected to a star connected system
are:
–
–
–
–
–
single-phase loads where the supply is one active and a neutral
single-phase loads where the supply is two actives
two-phase loads where the supply is two actives and a neutral
balanced three-phase loads where the supply is three actives
unbalanced three-phase loads where the supply is three actives
and a neutral

In star, VLine  3 Vphase and I Line  I phase .

In delta, VLine  V phase and I Line  3 I phase .

To determine the current drawn from the supply, the load current
value(s) must first be determined, using Ohm’s law.

There are a number of fault conditions:
–
–
–
–
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blown supply fuse
reversed phase winding in star
reversed phase winding in delta
open delta condition

For three-wire star connected supplies, if the load is star connected
and a line fuse blows, the two remaining phases of the load revert to
a single phase load supplied with line potential. The voltage across
each load component will be one half of line voltage.

For four-wire star connected supplies, if the load is star connected
and a line fuse blows, the load will become a two phase load. The
voltage across each phase of the load will be normal for two phases
and zero for the third.

If the load is delta connected and a line fuse blows, the load will be
referred to as a single phase load, with two original phases of the
load in series and this series arrangement in parallel with the third
original phase of the load. If the load is balanced, the voltage across
EGG202A: 13 Analyse interconnected systems
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one original phase of the load will be line voltage, while the other
two original phases will have one half line voltage.

A reversed phase winding in star causes the line potential for two
lines to decrease, while the third line value is unaltered. This will
cause the load to operate at a decreased level or rate.

A reversed phase winding in delta causes the total voltage to alter
from zero volts to twice line potential. This will result in a very large
circulating current in the supply transformer, creating heat and
eventually resulting in its destruction.

A voltmeter can be used to determine if a supply transformer has a
reversed winding.

Open delta connections may be created deliberately but are typically
the result of a fault in the load. The result of an open delta
connection is that the load output is reduced to 57.7% of the rated
output.

In open delta, the line current equals the phase current.
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Check your progress
In questions 1 to 6, place the letter matching your answer in the brackets provided.
1
The resultant voltage around the closed loop of a delta connected transformer with one
phase winding reversed is:
(a) zero
(b) half the phase voltage
(c) equal to the phase voltage
(d) twice the phase voltage.
2
( )
The angle between line voltages obtained from a star connected transformer with one
phase reversed will be:
(a) 0 degrees
(b) 30 degrees
(c) 120 degrees
(d) unequal.
3
( )
The magnitude of line voltages of a star connected alternator with one winding
reversed will be:
(a) equal
(b) unequal
(c) 1.73 times phase voltage
(d) the phasor sum of the phase voltages.
4
( )
If a delta connected load has one supply line open circuited then there will be:
(a) zero current in all phases
(b) zero current in one phase
(c) equal current in all phases
(d) equal current in two phases.
5
( )
The available power from an open delta connected transformer compared to a delta
connected transformer is:
(a) 33.3%
(b) 50%
(c) 57.7%
(d) 66.6%.
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( )
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6
The line current of an open delta system is:
(a) equal to the phase current
(b) 1.73 times the phase current
(c) twice the phase current
(d) half the phase current.
7
( )
A three-phase star connected transformer develops 240V in each phase winding and
supplies a three-phase delta connected motor. The motor stator winding impedances
are each 20 Ω. Determine the line current taken by the motor when it is started.
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8
If the motor in question 7 is reconnected in star determine the new line current taken
by the motor at start.
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9
A delta connected heater is supplied from a three-phase, three-wire 415V 50Hz
system. Each element has a resistance of 24 Ω. If the fuse in line C blows, determine
the current in each of the heating elements.
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10 Three 24 ohm heating elements are connected in star to a 415V three phase supply.
Assuming the fuse in line C blows then determine the load phase voltages and line
currents for a:
(a) 4-wire supply system
(b) 3-wire supply system
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11 A heater on low heat has three elements connected in star to a 415 volt 3-phase supply
and takes a current of 10 A per phase. Calculate the line current when the elements are
connected in delta for high heat.
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12 Draw the phasor diagram (tip to tail) of the phase voltages of a delta connected supply
with one winding reversed.
13 Draw the phasor diagram of the line voltages of a star connected system with one
winding reversed.
Answers to Check your progress are at the end of the section.
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Answers
Student exercise 1
(a)
VA = 240 V, VB = 240 V, VC = 240 V
(b)
VAB = 415 V, VBC = 415 V, VCA = 415 V
(c)
IAB = 27.7 A, IBC = 27.7 A, ICA = 48 A
(d)
IA = 48 A, IB = 48 A, IC = 48 A
(e)
IA = 48 A, IB = 48 A, IC = 48 A
Student exercise 2
1 (a)
VAB = 415 V, VBC = 207.5 V, VCA = 207.5 V
(b)
IAB = 10 A, IBC = 5 A, ICA = 5 A
(c)
IA = 15 A, IB = 15 A, IC = 0 A
2 (a)
VA = 240 V, VB = 0 V, VC = 240 V
(b)
IA =5.8 A, IB = 0 A, IC = 5.8 A
(c)
IA =5.8 A, IB = 0 A, IC = 5.8 A
(d)
IN = 5.8 A
Check your progress
1 (d)
2 (d)
3 (b)
4 (d)
5 (c)
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EGG202A: 13 Analyse interconnected systems
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6 (a)
7 35.9 A
8 12 A
9 IA and IC both = 8.64 A, IB = 17.29 A
10 (a)
VA = 240V, VB = 240V, VC = 0 V
IA = 10 A, IB = 10 A, IC = 0 A
(b)
VA = 207.5 V, VB = 207.5 V, VC = 0 V
IA = 8.64 A, IB = 8.64 A, IC = 0 A
11
IL = 30 A
12
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