Three phase power Lecture by cardiff University engineering

a

1. Spring Semester 2023/24
EN1215/EN1216
Power Engineering
Dr Stephen Robson
Room E/3.26
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2. Aims of this section
This section is the course content for Week 16 - week 2 of the
power theme. It aims to introduce the concept of three phase power.
Objectives of this section
By the end of this section, you should have a good understanding of
three phase power. You should understand the basic engineering
and economic reasons for the dominance of three phase systems in
modern power systems. You should understand the difference
between∆and star connected systems. You should be able to
simplify a balanced three phase system into three single phase
systems.
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3. Nomenclature
Voltage is a measurement between two points. In a single
phase system, there is only one logical place to measure
voltage. However, three-phase systems introduce some
ambiguity because there are different ways to measure
voltage.
We will use double subscript notation
1
to describe the
voltage between two points. For exampleVabis the voltage
between pointsaandb. Ifaandbare designated phases,
this represents the line to line or simply the line voltage,
VL. Similarly, if the voltage is between one of the phases
and neutral (n), for exampleVan, this is called the phase
voltage, orVΦ
Note that the above also applies for current. We will look
at∆andYsystems, where we will define line and phases
voltages and currents with care. n
n
Line Voltage
Phase Voltages
Line Voltage
Line Voltage
1
For more info on double subscript notation, see Grainger and Stephenson Section 1.3 3 / 30

4. Why do we use Three-Phase Power?
▶The purpose of a power system is to deliver power from generator to customer as cost
effectively, efficiently and reliably as possible.
▶Each conductor has acost, and each system configuration has acapacity
▶A three-phase system can deliver 3 times as much power as a single-phase system but
with the addition of just 1 more conductor.
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5. Above: a three-phase double circuit line on the National Grid transmission system.
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6. Cardiff University 11 kV three-phase test facility at Llanrumney playing fields (wood pole
line with distribution transformer).
6 / 30

7. Basic Principles of Three Phase PowerVaVcVbVaVbVc
~
Va=V60
~
Vb=V6120
~
Vc=V6240

7 / 30

8. Representation of Three-phase voltages
Vab=
√
3Van∠30
◦
Vbc=
√
3Vbn∠30
◦
Vca=
√
3Vcn∠30
◦Van
a
b
c
V
bn
Vbc
Vca
Vab
Vcn
8 / 30

9. Worked Example 1.22
In a balanced three-phase circuit the voltageVabis 173.2∠0
◦
V. Determine all the voltages
and currents in a Y-connected load havingZL=10∠20
◦
Ω.
Vab=173.2∠0
◦
Vbc=173.2∠240
◦
Vca=173.2∠120
◦
Van=100∠−30
◦
Vbn=100∠210
◦
Vcn=100∠90
◦ n
n
2Example 1.2 in ’Power System Analysis’ by Grainger and Stephenson - see Simulink model PE161.slx
9 / 30

10. Worked Example 1.2 (continued)3
In a balanced three-phase circuit the voltageVabis 173.2∠0
◦
V. Determine all the voltages
and currents in a Y-connected load havingZL=10∠20
◦
Ω.
To calculate the current, we can see that the voltage across
each load isVan=100∠−30
◦
,Vbn=100∠210
◦
and
Vcn=100∠90
◦
. Use Ohms Law:
Ian=
Van
Z
=
100∠−30
◦
10∠20
◦
=10∠−50
◦
A Ibn=
Vbn
Z
=
100∠210
◦
10∠20
◦
=10∠190
◦
A Icn=
Vcn
Z
=
100∠90
◦
10∠20
◦
=10∠70
◦
A n
n
3Example 1.2 in ’Power System Analysis’ by Grainger and Stephenson - see Simulink model PE161.slx
10 / 30

11. Worked Example 1.3
Use the current calculations of worked example 1.2 to prove that it has has a neutral current
of zero.
The impedances of the load in worked example 1.2 are the same (and there is no line
impedance), so the system is balanced. The current into the neutral is the sum of the three
currents, e.g.Ian+Ibn+Icn
Ian+Ibn+Icn=10∠−50
◦
+10∠190
◦
+10∠70
◦
It is easier to convert to rectangular form to do addition:
Ian+Ibn+Icn= (6.43−j7.66) + (−9.85−j1.74) + (3.42+j9.4) =0+j0A
11 / 30

12. Worked Example 1.44
The terminal voltage of a Y-connected load consisting of three equal impedances of
20∠30
◦
Ωis 4.4 kV line to line. The impedance of each of the three lines connecting the load
to a bus at a substation isZL=1.4∠75
◦
Ω. Find the line-to-line voltage at the substation bus. n
n
per-phase equivalent
4.4 kV
2.54 kV
?
?
4Example 1.3 in ’Power System Analysis’ by Grainger and Stephenson - see Simulink model PE162.slx
12 / 30

13. Worked Example 1.4 (continued)
The terminal voltage of a Y-connected load consisting of three equal impedances of
20∠30
◦
Ωis 4.4 kV line to line. The impedance of each of the three lines connecting the load
to a bus at a substation isZL=1.4∠75
◦
Ω. Find the line-to-line voltage at the substation bus.
The image in the previous slide shows how we can convert the three-phase system into a
per-phase equivalent (as long as the system is balanced). This greatly simplifies the
calculations. Our aim is to find the line-to-line voltage at the bus (represented as generators).
However, please bear in mind that the per-phase equivalent shows phase voltages, so we need
to convert the line-to-line voltage of 4.4 kV to its phase equivalent:
Van=
4.4kV
√
3
=2.54kV
With this, we can calculate the current through the load based on Ohms law:
Ian=
2.54∠0
◦
kV
20∠30
◦
=127∠−30
◦
A
Where the above assumes that the phase voltage is the reference phasor with angle 0
◦
.
13 / 30

14. Worked Example 1.4 (continued)
The terminal voltage of a Y-connected load consisting of three equal impedances of
20∠30
◦
Ωis 4.4 kV line to line. The impedance of each of the three lines connecting the load
to a bus at a substation isZL=1.4∠75
◦
Ω. Find the line-to-line voltage at the substation bus.
Now we have the current, we can work backwards. Remember we are trying to figure out the
voltage at the bus. We know that the current through the load is the very same current as
passed through the line impedance. The voltage across the load will differ from the voltage
at the bus by the voltage drop across the line impedance, e.g:
Van+IanZL=2540∠0
◦
+ (127∠−30
◦
)·(1.4∠75
◦
) =2670∠2.7
◦
V
This gives us the phase voltage - but the question wanted to the line-to-line voltage. We can
easily convert:
√
3·2.67=4.62kV
14 / 30

15. Star Versus Delta ConfigurationsVaVcVbIa(t)Ib(t)Ic(t) VbcVacVabIa(t)Ib(t)Ic(t)
▶“Wye”, “Y” or “Star” connection
▶VL=
√
3VΦ
▶IL=IΦ
▶P=
√
3VLILcos(θ)W
▶Q=
√
3VLILsin(θ)VAr
▶“Delta”, “∆” connection
▶VL=VΦ
▶IL=
√
3IΦ
▶P=
√
3VLILcos(θ)W
▶Q=
√
3VLILsin(θ)VAr
15 / 30

16. Delta (∆) to Star Conversion
▶To simplify the analysis of a three-phase system into three
independent single phase systems spaced 120
◦
apart, both the
generator and load must be star configured.
▶If they are∆configured, it is possible to convert to star.
▶The relationship betweenZ∆andZYis:ZY=
Z∆
3
for balanced
systems5ZZZIcˆtIaˆtIbˆtZ)Z)Z)IcˆtIaˆtIbˆt
5See Grainger and Stephenson Table 1.2 for conversion factors for unbalanced systems 16 / 30

17. Delta (∆) to Star Conversion: Example444Ic(t)Ia(t)Ib(t)1:331:331:33Ic(t)Ia(t)Ib(t)
17 / 30

18. Practical Considerations
Most domestic customers are supplied with a single phase load. Such a single phase
connection can be ‘tapped’ from a three-phase system. It is challenging to ensure that a
distribution system remains balanced when various loads are being tapped off.
There are various trade-offs associated with the use of a star or delta connection. A star
connection can be earthed at its neutral point. This simplifies the detection of earth faults. It
also makes it easy to distribute to lots of single-phase loads because you can think of it as
three separate single phase systems.
∆connections are usually preferred for motors. Each phase sees the full line voltage. Note
that a motor connected in a star configuration will see a lower phase voltage than the
line-to-line voltage being supplied.
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19. Power in Three-phase systems
Phase A pan(t) =Van(t)Ian(t) =VΦIΦcos(θ)−VΦIΦcos(2ωt+θ)
Phase Bpbn(t) =Vbn(t)Ibn(t) =VΦIΦcos(θ)−VΦIΦcos(2ωt+θ−240)
Phase Cpcn(t) =Vcn(t)Icn(t) =VΦIΦcos(θ)−VΦIΦcos(2ωt+θ−120)
Total P(t) =3VΦIΦcos(θ)
For “Y” connections:
VΦ=
VL
√
3
p(t) =
√
3VLILcos(θ)
19 / 30

20. Worked Example 1.56
A balanced, star-connected 3-phase source having a line voltage of 415 V supplies a
balanced,∆-connected 3-phase load having an impedance per phase Z=3+j4, through a
three-phase line of impedance per phase 0.1+j0.5. Find:
a) The equivalent star load.
b) The line currents
c) The phase voltages at the load terminals.
d) The currents in each phase of the delta load.
e) The real, reactive and apparent power absorbed by the load.
f) The real, reactive and apparent power supplied by the source3+j43+j43+j4Va0.1+j0.5iaVc0.1+j0.5icVb0.1+j0.5ib
6see Simulink model PE163.slx
20 / 30

21. Worked Example 1.5 (continued)
A balanced, star-connected 3-phase source having a line voltage of 415 V supplies a
balanced,∆-connected 3-phase load having an impedance per phase Z=3+j4, through a
three-phase line of impedance per phase 0.1+j0.5. Find:
a) The equivalent star load.3+j43+j43+j4Ic(t)Ia(t)Ib(t)1+j1.331+j1.331+j1.33Ic(t)Ia(t)Ib(t)
21 / 30

22. Worked Example 1.5 (continued)
A balanced, star-connected 3-phase source having a line voltage of 415 V supplies a
balanced,∆-connected 3-phase load having an impedance per phase Z=3+j4, through a
three-phase line of impedance per phase 0.1+j0.5. Find:
a) The equivalent star load.
b) The line currents
The line currents are found by using the single phase equivalent of the wye generator, wye
load system. In the Figure below, we see that this contains the per phase impedance in series
with the equivalent wye impedance. The line voltage of 415 V is converted to phase voltage
by divding by
√
3.415
p
3
60

1 +j1:330:1 +j0:5ia
22 / 30

23. Worked Example 1.5 (continued)
A balanced, star-connected 3-phase source having a line voltage of 415 V supplies a
balanced,∆-connected 3-phase load having an impedance per phase Z=3+j4, through a
three-phase line of impedance per phase 0.1+j0.5. Find:
a) The equivalent star load.
b) The line currents
ia=
Va
Zline+Zphase
=
415
√
3
∠0
◦
(0.1+j0.5) + (1+j1.33)
=
240∠0
◦
2.13∠59
◦
A
ia=112.6∠−59
◦
A
The other 2 phases are 120
◦
shifted in phase, hence:
ib=112.6∠(−59
◦
−120
◦
) =112.6∠−179
◦
A
ic=112.6∠(−59
◦
−240
◦
) =112.6∠−299
◦
A
23 / 30

24. Worked Example 1.5 (continued)
A balanced, star-connected 3-phase source having a line voltage of 415 V supplies a
balanced,∆-connected 3-phase load having an impedance per phase Z=3+j4, through a
three-phase line of impedance per phase 0.1+j0.5. Find:
a) The equivalent star load.
b) The line currents
c) The phase voltages at the load terminals.
To find the phase voltages at the load, we can again use the single phase equivalent circuit to
find the voltage across the load on phase a, then generalise for the other 2 phases. Using
circuit analysis:
V
load(a)=240∠0
◦
−iaZline
=240∠0
◦
−(112.6∠−59
◦
)(0.51∠78.7
◦
)
=240∠0
◦
−57.43∠(−59
◦
+78.7
◦
) =240∠0
◦
−57.43∠19.7
◦
=185.9−j19.35
V
load(a)=186.9∠−5.9
◦
24 / 30

25. Worked Example 1.5 (continued)
A balanced, star-connected 3-phase source having a line voltage of 415 V supplies a
balanced,∆-connected 3-phase load having an impedance per phase Z=3+j4, through a
three-phase line of impedance per phase 0.1+j0.5. Find:
a) The equivalent star load.
b) The line currents
c) The phase voltages at the load terminals.
V
load(a)is the load across an equivalentwye connectedload. But the actual load is
Delta connected. To convert back to∆, we must consider that the voltage across the
terminals will be
√
3 larger and 30
◦
phase shifted because the voltage across a∆configured
load is actually the line voltage. Hence:
Vab=
√
3V
load(a)∠30
◦
=
√
3·186.9∠(−5.9
◦
+30
◦
)
Vab=323.7∠24.1
◦
Vbc=323.7∠(24.1
◦
−120
◦
) =323.7∠−95.4
◦
Vca=323.7∠(24.1
◦
−240
◦
) =323.7∠−215.9
◦
25 / 30

26. Worked Example 1.5 (continued)
A balanced, star-connected 3-phase source having a line voltage of 415 V supplies a
balanced,∆-connected 3-phase load having an impedance per phase Z=3+j4, through a
three-phase line of impedance per phase 0.1+j0.5. Find:
a) The equivalent star load.
b) The line currents
c) The phase voltages at the load terminals.
d) The currents in each phase of the delta load.
To calculate the current in the load, use Ohm’s law:
iab=
Vab
Zab
=
323.7∠24.1
◦
3+j4
=65∠−29
◦
ibc=
Vbc
Zbc
=
323.7∠−95.4
◦
3+j4
=65∠−149
◦
ica=
Vca
Zca
=
323.7∠−215.9
◦
3+j4
=65∠−269
◦
26 / 30

27. Worked Example 1.5 (continued)
A balanced, star-connected 3-phase source having a line voltage of 415 V supplies a
balanced,∆-connected 3-phase load having an impedance per phase Z=3+j4, through a
three-phase line of impedance per phase 0.1+j0.5. Find:
a) The equivalent star load.
b) The line currents
c) The phase voltages at the load terminals.
d) The currents in each phase of the delta load.
e) The real, reactive and apparent power absorbed by the load.
The power factor angle of the load (the phase difference between voltage and current through
the load) is the angle ofZphase, (53.1
◦
). Then:
Pload=3VIcosθ=3·186.9·112.6·cos(53.1
◦
) =37.9kW
Qload=3VIsinθ=3·186.9·112.6·sin(53.1
◦
) =50.5kVar
Sload=
q
P
2
load
+Q
2
load
=63.14kVA
27 / 30

28. Worked Example 1.5 (continued)
A balanced, star-connected 3-phase source having a line voltage of 415 V supplies a
balanced,∆-connected 3-phase load having an impedance per phase Z=3+j4, through a
three-phase line of impedance per phase 0.1+j0.5. Find:
a) The equivalent star load.
b) The line currents
c) The phase voltages at the load terminals.
d) The currents in each phase of the delta load.
e) The real, reactive and apparent power absorbed by the load.
f) The real, reactive and apparent power supplied by the source
FindPsource,Qsource,Ssourcesupplied by the generator.
S=VI
∗
=3·240∠0
◦
·112.6∠59
◦
=81.07∠59
◦
=41.75+j69.5kVA
Therefore:
Psource=41.75kV
Qsource=69.5kVars
28 / 30

29. Worked Example 1.6
480 V, star-connected 3-phase generator delivers a current of 35 A at 0.95 power factor.
Calculate the real, reactive and apparent power delivered by the generator
If the voltage is given, but it is not specified whether it is line voltage or phase voltage, the
default assumption is that it isLine Voltage. SoVL=480V,IL=35A,cosθ=0.95 and
θ=18.19
◦
P=
√
3VLILcosθ
P=
√
3·480·35·0.95=27.64W
Q=
√
3VLILsinθ
Q=
√
3·480·35·0.31=9.08kVar W
S=
√
3VLIL=29.09kVA
29 / 30

30. Synchronous GeneratorRotating MachineLinesCircuit Breaker 30 / 30