EET 103/4 ELECTRICAL TECHNOLOGY – Polyphase Systems
-----------------------------------------------------------------------------------------------------------Example 23.1
The phase sequence of the Y-connected generator in Fig. 23.13 is ABC.
a.
Find the phase angles θ2 and θ3.
b.
Find the magnitude of the line voltages.
c.
Find the line currents.
Fig. 23.13
Example 23.1 – Solutions
a.
For an ABC phase sequence
θ2 = -120o
b.
c.
and
θ3 = +120o
E L  3E  1.73120V   208V
Therefore,
E AB  E BC  ECA  208V
Vø = Eø.
Therefore,
Van = EAN
IøL =
Vbn = EBN
Vcn = ECN
Van 120V0
120V0


 24A - 53.13
Z an 3  j 4 553.13
V
120V  120
Ibn = bn 
 24A - 173.13
Z bn
553.13
V
120V  120
Icn = cn 
 24A66.87
Z cn
553.13
Ian =
And since IL = IøL,
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IAa = Ian = 24A - 53.13
IBb = Ibn = 24A - 173.13
ICc = Icn = 24A66.87
EET 103/4 ELECTRICAL TECHNOLOGY – Polyphase Systems
-----------------------------------------------------------------------------------------------------------Example 23.2
For the three-phase system in Fig. 23.15:
a.
b.
c.
Find the phase angles θ2 and θ3.
Find the current in each phase of the load.
Find the magnitude of the line currents.
Fig. 23.15
Example 23.2 – Solutions
a.
b.
For an ABC sequence,
θ2 = -120o
and
Vø = EL.
Therefore,
Vab = EAB
θ3 = +120o
Vca = ECA
Vbc = EBC
The phase currents are
Vab 150V0
150V0


 15A - 53.13
Z ab 6  j8 1053.13
V
150V  120
Ibc = bc 
 15A - 173.13
Z bc
1053.13
V
150V  120
Ica = ca 
 15A66.87
Z ca
1053.13
Iab =
c.
I L  3I   1.7315A   25.95A
Therefore,
IAa = IBb = ICc = 25.95 A
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EET 103/4 ELECTRICAL TECHNOLOGY – Polyphase Systems
-----------------------------------------------------------------------------------------------------------Example 23.3
For the Δ-Δ system shown in Fig. 23.20:
a.
b.
c.
Find the phase angles θ2 and θ3 for the specified phase sequence.
Find the current in each phase of the load.
Find the magnitude of the line currents.
Fig. 23.20
Example 23.3 – Solutions
a.
b.
For an ACB sequence,
θ2 = +120o
and
Vø = EL
Therefore,
Vab = EAB
θ3 = -120o
Vca = ECA
Vbc = EBC
The phase currents are
Z  Z ab  Z bc  Z ca

Iab =
Z R ZC
505  90  25  90  3.54  45

Z R  ZC
5  j 5
7.071  45
Vab
120V0

 33.9A45
Z ab 3.54  45
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EET 103/4 ELECTRICAL TECHNOLOGY – Polyphase Systems
------------------------------------------------------------------------------------------------------------
Vbc 120V  120

 33.9A165
Z bc 3.54 - 45
V
120V  120
Ica = ca 
 33.9A - 75
Z ca 3.54 - 45
Ibc =
c.
I L  3I   1.7334A   58.82A
Therefore,
IAa = IBb = ICc = 58.82 A
Example 23.4
For the Δ-Y system shown in Fig. 23.21:
a.
b.
Find the voltage across each phase of the load.
Find the magnitude of the line voltages.
Fig. 23.21
Example 23.4 – Solutions
a.
IøL = IL
Therefore
Ian = IAa = 2A0
Ibn = IBb = 2A  120
Icn = ICc = 2A  120
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EET 103/4 ELECTRICAL TECHNOLOGY – Polyphase Systems
-----------------------------------------------------------------------------------------------------------The phase voltages are
Van = IanZan = 2A010  53.13  20V - 53.13
Vbn = IbnZbn = 2A  12010  53.13  20V - 173.13
Vcn = IcnZcn = 2A  12010  53.13  20V66.87
b.
E L  3V  1.7320V   34.6V
Therefore,
EBA = ECB = EAC = 34.6 V
Example 23.5
For the Y-connected load in Fig. 23.23:
a.
b.
c.
d.
Find the average power to each phase and the total load.
Determine the reactive power to each phase and the total reactive power.
Find the apparent power to each phase and the total apparent power.
Find the power factor of the load.
Fig. 23.23
EET103/4 Sem 1 2008/2009
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EET 103/4 ELECTRICAL TECHNOLOGY – Polyphase Systems
-----------------------------------------------------------------------------------------------------------Example 23.5 – Solutions
a. The average power is
P  V I  cos IV  100V20Acos 53.13  20000.6  1200W
OR
P  I  R  20A  3  4003  1200W
OR
P 
2
2
VR
60V 2  1200W

R
3
2
PT  3P  31200 W   3600W
OR
PT  3E L I L cos  IV  1.732173.2V 20A 0.6  3600W
b. The reactive power is
Q  V I  sin  IV  100V20Asin 53.13  20000.8  1600VAR
OR
Q  I  X   20A  4  4004  1600VAR
2
2
QT  3Q  31600VAR   4800VAR
OR
QT  3E L I L sin  IV  1.732173.2V 20A 0.8  4800VAR
c. The apparent power is
S   V I   100V 20A   2000 VA
S T  3S  32000VA   6000 VA
OR
S T  3E L I L  1.732 173.2V 20A   6000VA
d. The power factor is
Fp 
PT
3600W

 0.6 lagging
ST 6000VA
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EET 103/4 ELECTRICAL TECHNOLOGY – Polyphase Systems
-----------------------------------------------------------------------------------------------------------Example 23.6
For the Δ-Y connected load in Fig. 23.25:
a. Find the total average, reactive and apparent power.
b. Find the power factor of the load.
Fig 23.25
Example 23.6 – Solutions
a. Consider the Δ and Y separately.
For the Δ:
Z   6  j8  10  - 53.13
E
200 V
I  L 
 20 A
Z
10 
PT  3I 2 R  320 6  7200 W
2
QT  3I 2 X   320 8  9600 VAR C 
2
S T  3V I   320020  12,000 VA
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EET 103/4 ELECTRICAL TECHNOLOGY – Polyphase Systems
-----------------------------------------------------------------------------------------------------------For the Y:
ZY  4  j3  5 36.87
EL
200 V
3
3 116 V
I 


 23.12 A
ZY
5
5
PTY  3I 2 R  323.12 4  6414.41 W
2
QTY  3I 2 X   323.12 3  4810.81 VAR L 
2
S TY  3V I   311623.12  8045.76 VA
For the total load:
PT  PT  PTY  7200 W  6414.41 W  13,614.41 W
QT  QT  QTY  9600 VAR( C)  4810.81 VAR( L)  4789.19 VAR( C)
S T  PT2  S T2  (13,614.41) 2  (4789.19) 2  14,432.2 VA
b. The power factor is
Fp 
PT
13,614.41W

 0.943 leading
ST 14,432.20VA
Example 23.7
Each transmission line of the three-wire, three-phase system in Fig. 23.36 has an
impedance of 15 Ω + j 20 Ω. The system delivers a total power of 160 kW at 12,000 V to
a balanced three-phase load with a lagging power factor of 0.86.
a. Determine the magnitude of the line voltage EAB of the generator.
b. Find the power factor of the total load applied to the generator.
c. What is the efficiency of the system?
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EET 103/4 ELECTRICAL TECHNOLOGY – Polyphase Systems
------------------------------------------------------------------------------------------------------------
Fig. 23.26
Example 23.7 – Solutions
a. V (load ) 
VL
3

12,000 V
 6936.42 V
1.73
PT (load )  3V I  cos 
And
I 
PT
160,000

 8.94 A
3V cos  36936.420.86
Since   cos 1 0.86  30.68 , assigning V  V 0 , a lagging power factor results
in
I   8.94 A - 30.68
For each phase, the system will appear as shown in Fig. 23.27, where
Fig. 23.27
The loading on each phase of the system in Fig. 23.26.
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EET 103/4 ELECTRICAL TECHNOLOGY – Polyphase Systems
-----------------------------------------------------------------------------------------------------------E AN  I  Z line  V
 8.94  30.682553.13  6936.420
 223.522.45  6936.420
 206.56  j85.35  6936.42
 7142.98  j85.35
 7143.5 V0.68
Then
b.
E AB  3Eg  1.737143.5  12,358.26 V
PT  Pload  Plines
 160 kW  3I L  Rline
2
 160 kW  3 8.94  15
2
 160,000 W  3596.55 W
 163,596.55 W
PT  3VL I L cos  T
PT
163,596.55
cos  T 

3VL I L 1.7312,358.268.94
And
FP  cos  T  0.856
c.

 0.86 of the load
Po
Po
160 kW


 0.978
Pi Po  Plosses 160 kW  3596.55 W
 97.8%
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